This series has been developed specifically for the Cambridge International AS & A Level Mathematics (9709) syllabus

*976*
*319*
*42MB*

*English*
*Pages 370*
*Year 2018*

*Table of contents : CoverTitle pageCopyrightContentsSeries introductionHow to use this bookAcknowledgements1 Algebra2 Logarithmic and exponential functions3 TrigonometryCross-topic review exercise 1 4 Differentiation 5 Integration 6 Numerical solutions of equationsCross-topic review exercise 27 Further algebra8 Further calculusCross-topic review exercise 39 Vectors10 Differential equations11 Complex numbersCross-topic review exercise 4Pure Mathematics 2 Practice exam-style paperPure Mathematics 3 Practice exam-style paperAnswersGlossaryIndex*

Cambridge International AS & A Level Mathematics:

Pure Mathematics 2 & 3 Coursebook

Sue Pemberton Julianne Hughes Series Editor: Julian Gilbey

Cambridge International AS & A Level Mathematics:

Pure Mathematics 2 & 3 Coursebook

University Printing House, Cambridge CB2 8BS, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 79 Anson Road, #06–04/06, Singapore 079906 Cambridge University Press is part of the University of Cambridge. It furthers the University’s mission by disseminating knowledge in the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org Information on this title: www.cambridge.org/9781108407199 © Cambridge University Press 2018 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 2018 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Printed in the United Kingdom by Latimer Trend A catalogue record for this publication is available from the British Library ISBN 978-1-108-40719-9 Paperback Cambridge University Press has no responsibility for the persistence or accuracy of URLs for external or third-party internet websites referred to in this publication, and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Information regarding prices, travel timetables, and other factual information given in this work is correct at the time of first printing but Cambridge University Press does not guarantee the accuracy of such information thereafter. ® IGCSE is a registered trademark Past exam paper questions throughout are reproduced by permission of Cambridge Assessment International Education. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. The questions, example answers, marks awarded and/or comments that appear in this book were written by the author(s). In examination, the way marks would be awarded to answers like these may be different. NOTICE TO TEACHERS IN THE UK It is illegal to reproduce any part of this work in material form (including photocopying and electronic storage) except under the following circumstances: (i) where you are abiding by a licence granted to your school or institution by the Copyright Licensing Agency; (ii) where no such licence exists, or where you wish to exceed the terms of a licence, and you have gained the written permission of Cambridge University Press; (iii) where you are allowed to reproduce without permission under the provisions of Chapter 3 of the Copyright, Designs and Patents Act 1988, which covers, for example, the reproduction of short passages within certain types of educational anthology and reproduction for the purposes of setting examination questions.

Contents

Contents Series introduction How to use this book

vi viii

Acknowledgements

x

1 Algebra

1

1.1 The modulus function

2

1.2 Graphs of y = f( x ) where f( x ) is linear

7

1.3 Solving modulus inequalities

8

1.4 Division of polynomials

11

1.5 The factor theorem

14

1.6 The remainder theorem

18

End-of-chapter review exercise 1

22

2 Logarithmic and exponential functions

25

2.1 Logarithms to base 10

26

2.2 Logarithms to base a

30

2.3 The laws of logarithms

33

2.4 Solving logarithmic equations

35

2.5 Solving exponential equations

38

2.6 Solving exponential inequalities

40

2.7 Natural logarithms

42

2.8 Transforming a relationship to linear form

44

End-of-chapter review exercise 2

50

3 Trigonometry

52

3.1 The cosecant, secant and cotangent ratios

53

3.2 Compound angle formulae

58

3.3 Double angle formulae

61

3.4 Further trigonometric identities

66

3.5 Expressing a sin θ + b cos θ in the form R sin(θ ± α ) or R cos(θ ± α )

68

End-of-chapter review exercise 3

74

Cross-topic review exercise 1

76

iii

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

4 Differentiation

80

4.1 The product rule

81

4.2 The quotient rule

84

4.3 Derivatives of exponential functions

87

4.4 Derivatives of natural logarithmic functions

91

4.5 Derivatives of trigonometric functions

95

4.6 Implicit differentiation

99

4.7 Parametric differentiation

103

End-of-chapter review exercise 4

108

5 Integration

111

5.1 Integration of exponential functions 1 5.2 Integration of ax + b 5.3 Integration of sin( ax + b ), cos( ax + b ) and sec 2 ( ax + b )

112

5.4 Further integration of trigonometric functions

121

P2 5.5 The trapezium rule

131

6 Numerical solutions of equations

P3

133

6.1 Finding a starting point

135

6.2 Improving your solution

140

6.3 Using iterative processes to solve problems involving other areas of mathematics

149

End-of-chapter review exercise 6

155

Cross-topic review exercise 2

158

7 Further algebra

165

7.1 Improper algebraic fractions

166

7.2 Partial fractions

168

7.3 Binomial expansion of (1 + x ) for values of n that are not positive integers

174

7.4 Binomial expansion of ( a + x ) n for values of n that are not positive integers

177

7.5 Partial fractions and binomial expansions

179

End-of-chapter review exercise 7

182

n

P3

118 126

End-of-chapter review exercise 5 iv

115

8 Further calculus 8.1 Derivative of tan −1 x 1 8.2 Integration of 2 x + a2 k f ′( x ) 8.3 Integration of f( x )

184 186 187 189

Contents

8.4 Integration by substitution

191

8.5 The use of partial fractions in integration

194

8.6 Integration by parts

197

8.7 Further integration

201

End-of-chapter review exercise 8

203

P3

Cross-topic review exercise 3

205

P3

9 Vectors

209

P3

P3

P3

9.1 Displacement or translation vectors

211

9.2 Position vectors

220

9.3 The scalar product

225

9.4 The vector equation of a line

231

9.5 Intersection of two lines

236

End-of-chapter review exercise 9

240

10 Differential equations

243

10.1 The technique of separating the variables

245

10.2 Forming a differential equation from a problem

252

End-of-chapter review exercise 10

262

11 Complex numbers

265

11.1 Imaginary numbers

267

11.2 Complex numbers

269

11.3 The complex plane

273

11.4 Solving equations

283

11.5 Loci

288

End-of-chapter review exercise 11

297

Cross-topic review exercise 4

300

Pure Mathematics 2 Practice exam-style paper

307

Pure Mathematics 3 Practice exam-style paper

308

Answers

310

Glossary

353

Index

355

v

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Series introduction Cambridge International AS & A Level Mathematics can be a life-changing course. On the one hand, it is a facilitating subject: there are many university courses that either require an A Level or equivalent qualification in mathematics or prefer applicants who have it. On the other hand, it will help you to learn to think more precisely and logically, while also encouraging creativity. Doing mathematics can be like doing art: just as an artist needs to master her tools (use of the paintbrush, for example) and understand theoretical ideas (perspective, colour wheels and so on), so does a mathematician (using tools such as algebra and calculus, which you will learn about in this course). But this is only the technical side: the joy in art comes through creativity, when the artist uses her tools to express ideas in novel ways. Mathematics is very similar: the tools are needed, but the deep joy in the subject comes through solving problems. You might wonder what a mathematical ‘problem’ is. This is a very good question, and many people have offered different answers. You might like to write down your own thoughts on this question, and reflect on how they change as you progress through this course. One possible idea is that a mathematical problem is a mathematical question that you do not immediately know how to answer. (If you do know how to answer it immediately, then we might call it an ‘exercise’ instead.) Such a problem will take time to answer: you may have to try different approaches, using different tools or ideas, on your own or with others, until you finally discover a way into it. This may take minutes, hours, days or weeks to achieve, and your sense of achievement may well grow with the effort it has taken. vi

In addition to the mathematical tools that you will learn in this course, the problem-solving skills that you will develop will also help you throughout life, whatever you end up doing. It is very common to be faced with problems, be it in science, engineering, mathematics, accountancy, law or beyond, and having the confidence to systematically work your way through them will be very useful. This series of Cambridge International AS & A Level Mathematics coursebooks, written for the Cambridge Assessment International Education syllabus for examination from 2020, will support you both to learn the mathematics required for these examinations and to develop your mathematical problem-solving skills. The new examinations may well include more unfamiliar questions than in the past, and having these skills will allow you to approach such questions with curiosity and confidence. In addition to problem solving, there are two other key concepts that Cambridge Assessment International Education have introduced in this syllabus: namely communication and mathematical modelling. These appear in various forms throughout the coursebooks. Communication in speech, writing and drawing lies at the heart of what it is to be human, and this is no less true in mathematics. While there is a temptation to think of mathematics as only existing in a dry, written form in textbooks, nothing could be further from the truth: mathematical communication comes in many forms, and discussing mathematical ideas with colleagues is a major part of every mathematician’s working life. As you study this course, you will work on many problems. Exploring them or struggling with them together with a classmate will help you both to develop your understanding and thinking, as well as improving your (mathematical) communication skills. And being able to convince someone that your reasoning is correct, initially verbally and then in writing, forms the heart of the mathematical skill of ‘proof’.

Series introduction

Mathematical modelling is where mathematics meets the ‘real world’. There are many situations where people need to make predictions or to understand what is happening in the world, and mathematics frequently provides tools to assist with this. Mathematicians will look at the real world situation and attempt to capture the key aspects of it in the form of equations, thereby building a model of reality. They will use this model to make predictions, and where possible test these against reality. If necessary, they will then attempt to improve the model in order to make better predictions. Examples include weather prediction and climate change modelling, forensic science (to understand what happened at an accident or crime scene), modelling population change in the human, animal and plant kingdoms, modelling aircraft and ship behaviour, modelling financial markets and many others. In this course, we will be developing tools which are vital for modelling many of these situations. To support you in your learning, these coursebooks have a variety of new features, for example: ■ Explore activities: These activities are designed to offer problems for classroom use. They require thought and deliberation: some introduce a new idea, others will extend your thinking, while others can support consolidation. The activities are often best approached by working in small groups and then sharing your ideas with each other and the class, as they are not generally routine in nature. This is one of the ways in which you can develop problemsolving skills and confidence in handling unfamiliar questions. ■ Questions labelled as P , M or PS : These are questions with a particular emphasis on ‘Proof’, ‘Modelling’ or ‘Problem solving’. They are designed to support you in preparing for the new style of examination. They may or may not be harder than other questions in the exercise. ■ The language of the explanatory sections makes much more use of the words ‘we’, ‘us’ and ‘our’ than in previous coursebooks. This language invites and encourages you to be an active participant rather than an observer, simply following instructions (‘you do this, then you do that’). It is also the way that professional mathematicians usually write about mathematics. The new examinations may well present you with unfamiliar questions, and if you are used to being active in your mathematics, you will stand a better chance of being able to successfully handle such challenges. At various points in the books, there are also web links to relevant Underground Mathematics resources, which can be found on the free undergroundmathematics.org website. Underground Mathematics has the aim of producing engaging, rich materials for all students of Cambridge International AS & A Level Mathematics and similar qualifications. These high-quality resources have the potential to simultaneously develop your mathematical thinking skills and your fluency in techniques, so we do encourage you to make good use of them. We wish you every success as you embark on this course. Julian Gilbey London, 2018

Past exam paper questions throughout are reproduced by permission of Cambridge Assessment International Education. Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. The questions, example answers, marks awarded and/or comments that appear in this book were written by the author(s). In examination, the way marks would be awarded to answers like these may be different.

vii

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

How to use this book Throughout this book you will notice particular features that are designed to help your learning. This section provides a brief overview of these features. This book covers both Pure Mathematics 2 and Pure Mathematics 3. One topic (5.5 The trapezium rule) is only covered in Pure Mathematics 2 and this section is marked with the icon P2 . Chapters 7–11 are only covered in Pure Mathematics 3 and these are marked with the icon P3 . The icons appear in the Contents list and in the relevant sections of the book.

In this chapter you will learn how to:

■ ■ ■ ■

PREREQUISITE KNOWLEDGE

formulate a simple statement involving a rate of change as a differential equation ﬁnd, by integration, a general form of solution for a ﬁrst order differential equation in which the variables are separable use an initial condition to ﬁnd a particular solution interpret the solution of a differential equation in the context of a problem being modelled by the equation.

Where it comes from

What you should be able to do

Check your skills

IGCSE® / O Level Mathematics

Perform long division on numbers and find the remainder where necessary.

1 Using long division, calculate:

c 4283 ÷ 32 IGCSE / O Level Mathematics

Learning objectives indicate the important concepts within each chapter and help you to v uv u navigate through the coursebook. KEY POINT 8.4

I t r

∫

ti

∫

Key point boxes contain a l AS & A L v l M t m ti s: Pure M t summary of the most important methods, facts and formulae.

Sketch straight-line graphs.

m ti s 2 & 3

WORKED EXAMPLE 1.3 Solve x + 3 + x + 5 = 10. Answer x + 3 + x + 5 = 10

e double angle formulae. W 2 ≡ − 2

Key terms are important terms in the topic that you are learning. They are highlighted in orange bold. The glossary contains clear definitions of these key terms. EXPLORE 6.1

2 Sketch the graph of y = 2 x − 5.

Prerequisite knowledge exercises identify prior learning that you need to have covered before starting the chapter. Try the questions to identify any areas that you need to review before continuing with the chapter.

The formula for integration by parts: dv du u dx = uv − v dx dx dx

viii

a 4998 ÷ 14 b 10 287 ÷ 27

u

Subtract x + 5 from both sides.

x + 3 = 10 − x + 5

Split the equation into two parts.

x + 3 = 10 − x + 5

(1)

x + 3 = x + 5 − 10

(2)

Using equation (1):

x

Worked examples provide step-by-step approaches to answering questions. The left side shows a fully worked solution, while the right side contains a commentary explaining each step in the working.

Anil is trying to find the roots of f( x u) = xtan x = 0 between 0 and 2π radians. He states:

v

f(3.1) . 0 and f(3.2) . 0. There is no change of sign so there is no root of tan x = 0 between 3.1 and 3.2. Mika states: y = tan x and it meets the x-axis where x = π radians so Anil has made a calculation error. Discuss Anil and Mika’s statements.

Explore boxes contain enrichment activities for extension work. These activities promote group-work and peerto-peer discussion, and are intended to deepen your understanding of a concept. (Answers to the Explore questions are provided in the Teacher’s Resource.) °

°

TIP Solving trigonometric inequalities is not in the Cambridge syllabus. You should, however, have the skills necessary to tackle these more challenging questions.

Tip boxes contain helpful guidance about calculating or checking your answers.

How Series to use introduction this book

FAST FORWARD

REWIND

You need to have studied the skills in Chapter 8 before attempting questions 13 and 14.

You might need to look back at Chapter 3 before attempting question 11.

Rewind and Fast forward boxes direct you to related learning. Rewind boxes refer to earlier learning, in case you need to revise a topic. Fast forward boxes refer to topics that you will cover at a later stage, in case you would like to extend your study.

Throughout each chapter there are multiple exercises containing practice questions. The questions are coded: PS

These questions focus on problem-solving.

P

These questions focus on proofs.

M

These questions focus on modelling.

DID YOU KNOW?

You should not use a calculator for these questions.

The relationship between the Fibonacci numbers and the golden ratio produces the golden spiral.

These questions are taken from past examination papers.

This can be seen in many natural settings, one of which is the spiralling pattern of the petals in a rose flower. From a tight centre, petals flow outward in a spiral, growing wider and larger as they do so. Nature has made the best use of an incredible structure. This perfect arrangement of petals is why the small, compact rose bud grows into such a beautiful flower.

Did you know? boxes contain interesting facts showing how Mathematics relates to the wider world.

Checklist of learning and understanding

●

∫

∫ 1 ∫ e ∫ dx = a e + c 1 1 ∫ ax +∫ b dx = a ln ax + b

●

∫

∫

Integration formulae ●

●

∫e

x

ax + b

dx = e x + c

1 dx = ln x + c x 1 1 cos x dx = sin x + c

∫

1 sin x dx = − cos x + c

ax + b

cos(ax + b ) dx =

The End-of-chapter review contains exam-style questions covering all topics in the chapter. You can use this to check your understanding of the topics you have covered. The number of marks gives an indication of how long you should be spending on the question. You should spend more time on questions with higher mark allocations; questions with only one or two marks should not need you to spend time doing complicated calculations or writing long explanations.

+c

1 sin(ax + b ) + c a

1 sin(ax + b ) dx = − cos(ax + b ) + c a

∫ ∫

END-OF-CHAPTER REVIEW EXERCISE 7 1

Expand (1 − 2 x )−4 in ascending powers of x, up to and including the term in x 3, simplifying the coefficients. [4]

2

Expand

3

Expand (2 − x )(1 + 2 x ) 2 in ascending powers of x, up to and including the term in x 2, simplifying the coefficients.

31−

(6x ) in ascending powers of x up to and including the term in x 3, simplifying the coefficients. [4] Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2011

At the end of each chapter there is a Checklist of learning and understanding. The checklist contains a summary of the concepts that were covered in the chapter. You can use this to quickly check that you have covered the main topics.

−

n

3

4

Expand

5

Expand

1 + 3x in ascending powers of x up to and including the term in x 2, simplifying the coefficients. [4] (1 + 2 x ) Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2013 1 in ascending powers of x, up to and including the term in x 3 , simplifying the coefficients. 1 + 3x 10

x

E

Extension material goes beyond the syllabus. It is highlighted by a red line to the left of the text.

x CROSS-TOPIC REVIEW EXERCISE 4 2 P3

This exercise is for Pure Mathematics 3 students only. 1

Relative to the origin O, the position vectors of the points A and B are given by −5 OA = 0 3

Web link boxes contain links to useful resources on the internet.

1 and OB = 7 . 2

a

Find a vector equation of the line AB.

b

The line AB is perpendicular to the line L with vector equation: m 4 r = 2 +µ 3 −3 9

[3]

C

. m

WEB LINK Try the Equation or identity?(II) resource on the Underground Mathematics website

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2016

Cross-topic review exercises appear after several chapters, and cover topics from across the preceding chapters. x

x p

z

q

D

[4]

ix

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Acknowledgements The authors and publishers acknowledge the following sources of copyright material and are grateful for the permissions granted. While every effort has been made, it has not always been possible to identify the sources of all the material used, or to trace all copyright holders. If any omissions are brought to our notice, we will be happy to include the appropriate acknowledgements on reprinting. Past examination questions throughout are reproduced by permission of Cambridge Assessment International Education. Thanks to the following for permission to reproduce images: Cover image Mint Images – Frans Lanting/Getty Images Inside (in order of appearance) sakkmesterke/Getty Images, Don Johnston/Getty Images, Photo by marianna armata/Getty Images, DEA PICTURE LIBRARY/Getty Images, malcolm park/Getty Images, agsandrew/ Shutterstock, Stephan Snyder/Shutterstock, alengo/Getty Images, Moritz Haisch/EyeEm/Getty Images, Rosemary Calvert/Getty Images, bluecrayola/Shutterstock, PinkCat/Shutterstock, Maurice Rivera/EyeEm/Getty Images, PASIEKA/Getty Images, Alfred Pasieka/Science Photo Library/Getty Images

x

1

Chapter 1 Algebra In this chapter you will learn how to:

■ ■ ■

understand the meaning of x , sketch the graph of y = ax + b and use relations such as a = b ⇔ a 2 = b2 and x − a , b ⇔ a − b , x , a + b in the course of solving equations and inequalities divide a polynomial, of degree not exceeding 4, by a linear or quadratic polynomial, and identify the quotient and remainder (which may be zero) use the factor theorem and the remainder theorem.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE® / O Level Mathematics

Perform long division on numbers and find the remainder where necessary.

1 Using long division, calculate: a 4998 ÷ 14 b 10 287 ÷ 27 c 4283 ÷ 32

IGCSE / O Level Mathematics

Sketch straight-line graphs.

2 Sketch the graph of y = 2 x − 5.

Why do we need to study algebra?

−b ± b2 − 4ac , for solving quadratic 2a equations. Versions of this formula were known and used by the Babylonians nearly 4000 years ago. You have previously learnt the general formula, x =

It took until the 16th century for mathematicians to discover formulae for solving cubic equations, ax 3 + bx 2 + cx + d = 0 , and quartic equations, ax 4 + bx 3 + cx 2 + dx + e = 0. (These formulae are very complicated and there is insufficient space to include them here but you might wish to research them on the internet.) 2

Mathematicians then spent many years trying to discover a general formula for solving quintic equations, ax5 + bx 4 + cx 3 + dx 2 + ex + f = 0. Eventually, in 1824, a mathematician managed to prove that no general formula for this exists. In this chapter you will develop skills for factorising and solving cubic and quartic equations. These types of equations have various applications in the real world. For example, cubic equations are used in thermodynamics and fluid mechanics to model the pressure/volume/temperature behaviour of gases and fluids. You will also learn about a new type of function, called the modulus function.

1.1 The modulus function The modulus of a number is the magnitude of the number without a sign attached. The modulus of 3 is written 3 . 3 = 3 and −3 = 3. It is important to note that the modulus of any number (positive or negative) is always a positive number. The modulus of a number is also called the absolute value. The modulus of x, written as x , is defined as: x if x ù 0 x = − x if x , 0

WEB LINK Explore the Polynomials and rational functions station on the Underground Mathematics website.

Chapter 1: Algebra

EXPLORE 1.1 You are given these eight statements: 1

a+b = a + b

2

a−b = a − b

3

ab = a × b

4

a = a ÷ b , if b ≠ 0 b

5

a

6

a

7

a+b ø a + b

8

a−b ø a−c + c−b

2

= a2

n

= a n , where n is a positive integer

You must decide whether a statement is Always true •

Sometimes true

Never true

If you think that a statement is either always true or never true, you must give a clear explanation to justify your answer. 3

•

If you think that a statement is sometimes true you must give an example of when it is true and an example of when it is not true.

The statement x = k , where k ù 0 , means that x = k or x = − k . This property is used to solve equations that involve modulus functions. If you are solving equations of the form ax + b = k , you can solve the equations using ax + b = k

and ax + b = − k

If you are solving harder equations of the form ax + b = cx + d , you can solve the equations using ax + b = cx + d

and ax + b = −( cx + d )

When solving these more complicated equations you must always check that your answers satisfy the original equation.

FAST FORWARD You will see why it is necessary to check your answers when you learn how to sketch graphs of modulus functions in Section 1.2.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 1.1 Solve: 2x − 1 = 3

a

b

x − 4 = 2x + 1

Answer 2x − 1 = 3

a

2 x − 1 = 3 or 2 x − 1 = −3 2 x = −2 2x = 4 x = −1 x = 2 CHECK: 2 × 2 − 1 = 3 ✓ and 2 × ( −1) − 1 = 3 ✓ The solution is: x = −1 or x = 2 x − 4 = 2x + 1

b

x − 4 = 2 x + 1 or x − 4 = −(2 x + 1) x = −5 3x = 3 x =1 x =1 CHECK: −5 − 4 = 2 × ( −5) + 1 ⨯ and 1 − 4 = 2 × 1 + 1 ✓

4

The solution is: x = 1

EXPLORE 1.2 Using a

2

= a 2 you can write a 2 − b2 as a

2

2

− b .

Using the difference of two squares you can now write: a 2 − b2 =

(a

− b

)( a

+ b

)

Using the previous statement, explain how these three important results can be obtained: (The symbol ⇔ means ‘is equivalent to’.) •

a = b ⇔ a 2 = b2

•

a . b ⇔ a 2 . b2

•

a , b ⇔ a 2 , b2 , if b ≠ 0 .

Worked example 1.2 shows you how to solve equations of the form cx + d = ex + f . KEY POINT 1.1 To solve equations of the form cx + d = ex + f 2

a = b ⇔a =b

2

we can use the rule:

Chapter 1: Algebra

WORKED EXAMPLE 1.2 Solve the equation 3x + 4 = x + 5 . Answer Method 1 3x + 4 = x + 5 3x + 4 = x + 5 or 3x + 4 = −( x + 5) 2x = 1 4x = −9 or 9 1 x= x = − 2 4 CHECK: 3 ×

1 1 9 9 +4 = +5 ✓, 3×− +4 = − +5 ✓ 2 2 4 4

Solution is: x =

1 9 or x = − 2 4

Method 2 3x + 4 = x + 5 (3x + 4) = ( x + 5) 2

Use a = b ⇔ a 2 = b2. 2

9x + 24x + 16 = x 2 + 10 x + 25 2

8x 2 + 14x − 9 = 0

Factorise.

(2 x − 1)(4x + 9) = 0 9 1 x= or x = − 2 4

WORKED EXAMPLE 1.3 Solve x + 3 + x + 5 = 10. Answer x + 3 + x + 5 = 10

Subtract x + 5 from both sides.

x + 3 = 10 − x + 5

Split the equation into two parts.

x + 3 = 10 − x + 5

(1)

x + 3 = x + 5 − 10

(2)

Using equation (1): −7 + x + x + 5 = 0 x + 5 = 7 − x x + 5 = 7 − x or x + 5 = −(7 − x ) 2 x = 2 or 0 = −12 x =1

Split this equation into two parts. 0 = −12 is false.

5

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Using equation (2): x + 5 = x + 13

Split this equation into two parts.

x + 5 = x + 13 or x + 5 = −( x + 13) 0=8

or

0 = 8 is false.

2 x = −18 x = −9

The solution is x = 1 or x = −9.

EXERCISE 1A 1 Solve. a

4x − 3 = 7

x +2 =3 3 2 Solve. 2x + 1 =5 a x−2 d

d

3x − 5 = x + 1

3x − 2 =4 5

b

1 − 2x = 5

c

e

x + 2 2x − =2 3 5

f 2 x + 7 = 3x

b

3x − 1 =1 x+5

c 2−

e

x+ x+4 =8

f

x+2 =5 x−3

8 − 1 − 2x = x

3 Solve. 6

a

2x + 1 = x

b

3 − 2 x = 3x

c 2x − 5 = 1 − x

d

3x + 5 = 1 + 2 x

e

x −5 = 2 x +1

f

3 2x − 1 =

1 x−3 2

4 Solve. a

x2 − 2 = 7

b

5 − x2 = 3 − x

c

x2 + 2x = x + 2

d

x2 − 3 = 2x + 1

e

2 x 2 − 5x = 4 − x

f

x2 − 7x + 6 = 6 − x

5 Solve the simultaneous equations. a

x + 2y = 8

b

x+2 + y=6 6 Solve the equation 5 x − 1

3x + y = 0 y = 2x2 − 5

2

+ 9 x − 1 − 2 = 0.

7 a Solve the equation x 2 − 5 x + 6 = 0 . b Use graphing software to draw the graph of y = x 2 − 5 x + 6 . c

Name the equation of the line of symmetry of the curve.

8 Solve the equation 2 x + 1 + 2 x − 1 = 3. PS

9 Solve the equation 3x − 2 y − 11 + 2 31 − 8x + 5 y = 0 .

Chapter 1: Algebra

y

1.2 Graphs of y = f(x) where f(x) is linear

y = |x|

Consider drawing the graph of y = x . First draw the line y = x .

x

O

Then reflect, in the x-axis, the part of the line that is below the x-axis.

WORKED EXAMPLE 1.4 1 x − 1 , showing the points where the graph meets the axes. 2 1 x − 1 in an alternative form. Use your graph to express 2 Sketch the graph of y =

Answer 1 First sketch the graph of y = x − 1. 2 1 The line has gradient and a y-intercept of −1. 2 You then reflect in the x-axis the part of the line that is below the x-axis. 1 x − 1 can be written as: The graph shows that 2 1 x − 1 if x ù 2 2 1 x −1 = 2 − 1 x − 1 if x , 2 2

y y=

1 x –1 2

1

O

2

x

−1

7

In Worked example 1.1, we found that there were two roots, x = −1 and x = 2, to the equation 2 x − 1 = 3.

y

These can also be found graphically by finding the x-coordinates of the points of intersection of the graphs of y = 2 x − 1 and y = 3 as shown.

3

y = |2x – 1|

4 y=3

2 1 –2

Also in Worked example 1.1, we found that there was only one root, x = 1, to the equation x − 4 = 2 x + 1. This root can be found graphically by finding the x-coordinates of the points of intersection of the graphs of y = x − 4 and y = 2 x + 1 as shown.

y

–1 O –1

1

2

3

y = 2x + 1

6 5

y = |x – 4|

4 3 2 1 –1 O –1

x

1

2

3

4

5

6

7

8

9

x

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 1B 1 Sketch the graphs of each of the following functions showing the coordinates of the points where the graph meets the axes. Express each function in an alternative form. 1 a y= x+2 b y = 3−x c y = 5− x 2 2 a Complete the table of values for y = x − 3 + 2 . x y

0 5

1

2 3

3

4

5

6

b Draw the graph of y = x − 3 + 2 for 0 ø x ø 6 . c

Describe the transformation that maps the graph of y = x onto the graph of y = x − 3 + 2.

3 Describe fully the transformation (or combination of transformations) that maps the graph of y = x onto each of these functions. a

y = x +1 + 2

b

y = x−5 −2

c

y=2− x

d

y = 2x − 3

e

y = 1− x + 2

f

y =5−2 x

4 Sketch the graphs of each of the functions in question 3. For each graph, state the coordinates of the vertex. 5 f( x ) = 5 − 2 x + 3 for 2 ø x ø 8 Find the range of function f . 8

6 a Sketch the graph of y = 2 x − 2 + 1 for −2 , x , 6, showing the coordinates of the vertex and the y-intercept. b On the same diagram, sketch the graph of y = x + 2. c

Use your graph to solve the equation 2 x − 2 + 1 = x + 2 .

7 a Sketch the graph of y = x − 2 for −3 , x , 6 , showing the coordinates of the vertex and the y-intercept. b On the same diagram, sketch the graph of y = 1 − 2 x . c

Use your graph to solve the equation x − 2 = 1 − 2 x .

8 a Sketch the graph of y = x + 1 + x − 1 . b Use your graph to solve the equation x + 1 + x − 1 = 4.

1.3 Solving modulus inequalities Two useful properties that can be used when solving modulus inequalities are: a ø b ⇔ −b ø a ø b

and

a ù b ⇔ a ø − b or a ù b

The following examples illustrate the different methods that can be used when solving modulus inequalities.

Chapter 1: Algebra

WORKED EXAMPLE 1.5 Solve 2 x − 5 , 3. Answer Method 1: Using algebra 2x − 5 , 3 −3 , 2 x − 5 , 3 2 , 2 x , 8 1 , x , 4 Method 2: Using a graph

y

The graphs of y = 2 x − 5 and y = 3 intersect at the points A and B.

4

2 x − 5 if x ù 2 1 2 2x − 5 = −(2 x − 5) if x , 2 1 2 At A, the line y = −(2 x − 5) intersects the line y = 3. −(2 x − 5) −2 x + 5 2x x

y = |2x – 5|

5

3

A

y=3

B

2 1

=3 =3 =2 =1

O –1

1

2

3

4

5

x

9

At B, the line y = 2 x − 5 intersects the line y = 3. 2x − 5 = 3 2x = 8 x=4 To solve the inequality 2 x − 5 , 3 you must find where the graph of the function y = 2 x − 5 is below the graph of y = 3. Hence, 1 , x , 4.

WORKED EXAMPLE 1.6 Solve the inequality 2 x − 1 ù 3 − x . Answer Method 1: Using algebra 2x − 1 ù 3 − x

Use a ù b ⇔ a 2 ù b2.

(2 x − 1)2 ù (3 − x )2 4x 2 − 4x + 1 ù 9 − 6x + x 2 3x 2 + 2 x − 8 ù 0 (3x − 4)( x + 2) ù 0

Factorise.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

y = 3x2 + 2x – 8

4 Critical values are −2 and . 3 4 Hence, x ø −2 or x ù . 3

+

+ −2

4 3

−

Method 2: Using a graph The graphs of y = 2 x − 1 and y = 3 − x intersect at the points A and B. 2 x − 1 if x ù 1 2 2x − 1 = 1 −(2 x − 1) if x , 2 if x ù 3 x − 3 3−x = x−3 = −( x − 3) if x , 3 At A, the line y = −(2 x − 1) intersects the line y = −( x − 3).

y

2x − 1 = x − 3 x = −2 At B, the line y = 2 x − 1 intersects the line y = −( x − 3).

7 6

2 x − 1 = −( x − 3) 3x = 4 10

y = |2x – 1|

8

y = |3 – x| 5

A

4 x = 3 To solve the inequality 2 x − 1 ù 3 − x you must find where

4 3 2

the graph of the function y = 2 x − 1 is above the graph

1

of y = 3 − x . 4 Hence, x ø − 2 or x ù . 3

–3

–2

–1 O

B

1

EXERCISE 1C 1

y 8 y = |x – 1|

7 6 5 4 3 2 y = 2|x – 4|

1 –1 O

1

2

3

4

5

6

7

8

9 x

The graphs of y = x − 1 and y = 2 x − 4 are shown on the grid. Write down the solution to the inequality x − 1 . 2 x − 4 . 2 a On the same axes, sketch the graphs of y = 2 x − 1 and y = 4 − x − 1 . b Solve the inequality 2 x − 1 . 4 − x − 1 .

2

3

4

5

6

7

8

x

Chapter 1: Algebra

3 Solve. (You may use either an algebraic method or a graphical method.) a

2x + 1 ø 3

b

2−x ,4

c

3x − 2 ù 7

c

x − 7 − 2x ø 4

4 Solve. (You may use either an algebraic method or a graphical method.) a

2x − 5 ø x − 2

b

3 + x . 4 − 2x

5 Solve. (You may use either an algebraic method or a graphical method.) a

x + 4 , 2x

b

2x − 5 . 3 − x

c

3x − 2 ø 1 − 3x

d

x −1 .2 x − 4

e

3 3 − x ù 2x − 1

f

3x − 5 , 2 2 − x

6 Solve 2 x + 1 + 2 x − 1 . 5.

1.4 Division of polynomials A polynomial is an expression of the form an x n + an −1x n −1 + an − 2 x n − 2 + … + a2 x 2 + a1x1 + a0 where: ● ● ● ● ●

x is a variable n is a non-negative integer the coefficients an , an −1, an − 2 , … , a2 , a1, a0 are constants an is called the leading coefficient and an ≠ 0 a0 is called the constant term.

The highest power of x in the polynomial is called the degree of the polynomial. For example, 5x 3 − 6x 2 + 2 x − 6 is a polynomial of degree 3. In this section you will learn how to use long division to divide a polynomial by another polynomial. Alternative methods are possible, this is shown in Worked example 1.10. DID YOU KNOW? Polynomial functions can be used to predict the mass of an animal from its length.

WORKED EXAMPLE 1.7 Divide x 3 + 2 x 2 − 11x + 6 by x − 2 . Answer Step 1: x2 x − 2 x + 2 x 2 − 11x + 6

)

3

x3 − 2x2 2

4x − 11x

Divide the first term of the polynomial by x: x3 ÷ x = x2 . Multiply ( x − 2) by x 2 : x 2 ( x − 2) = x 3 − 2 x 2 . Subtract: ( x 3 + 2 x 2 ) − ( x 3 − 2 x 2 ) = 4x 2 . Bring down the −11x from the next column.

11

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Step 2: Repeat the process x 2 + 4x x − 2 x + 2 x 2 − 11x + 6

)

Divide 4x 2 by x: 4x 2 ÷ x = 4x.

3

Multiply ( x − 2) by 4x: 4x( x − 2) = 4x 2 − 8x .

x3 − 2x2 4x 2 − 11x

Subtract: (4x 2 − 11x ) − (4x 2 − 8x ) = −3x .

4x 2 − 8x

Bring down the 6 from the next column.

− 3x + 6 Step 3: Repeat the process x 2 + 4x − 3 x − 2 x + 2 x 2 − 11x + 6

)

3

x3 − 2x2 4x 2 − 11x

Divide −3x by x: −3x ÷ x = −3.

4x 2 − 8x

Multiply ( x − 2) by −3: −3( x − 2) = −3x + 6.

− 3x + 6 − 3x + 6

Subtract: ( −3x + 6) − ( −3x + 6) = 0.

0 Hence, ( x 3 + 2 x 2 − 11x + 6) ÷ ( x − 2) = x 2 + 4x − 3. 12

Each part in the calculation in Worked example 1.7 has a special name: ( x 3 + 2 x 2 − 11x + 6) ÷ ( x − 2) = ( x 2 + 4x − 3) dividend

divisor

quotient

There was no remainder in this calculation so we say that x − 2 is a factor of x 3 + 2 x 2 − 11x + 6. WORKED EXAMPLE 1.8 Find the remainder when 2 x 3 − x + 52 is divided by x + 3. Answer There is no x 2 term in 2 x 3 − x + 52 so we write it as 2 x 3 + 0 x 2 − x + 52 . Step 1: 2x2 x + 3 2 x 3 + 0 x 2 − x + 52

)

2 x 3 + 6x 2 − 6x 2 − x

Divide the first term of the polynomial by x: 2 x 3 ÷ x = 2 x 2 . Multiply ( x + 3) by 2 x 2 : 2 x 2 ( x + 3) = 2 x 3 + 6x 2 . Subtract: (2 x 3 + 0 x 2 ) − (2 x 3 + 6x 2 ) = −6x 2 . Bring down the − x from the next column.

Chapter 1: Algebra

Step 2: Repeat the process 2 x 2 − 6x x + 3 2 x + 0 x 2 − x + 52

)

3

Divide −6x 2 by x: −6x 2 ÷ x = −6x .

2 x 3 + 6x 2

Multiply ( x + 3) by −6x : −6x( x + 3) = −6x 2 − 18x .

− 6x 2 −

x

Subtract: ( −6x 2 − x ) − ( −6x 2 − 18x ) = 17 x .

− 6x 2 − 18x

Bring down the 52 from the next column.

17 x + 52 Step 3: Repeat the process 2 x 2 − 6x + 17 x + 3 2 x + 0 x 2 − x + 52

)

3

2 x 3 + 6x 2 − 6x 2 −

x Divide 17x by x: 17 x ÷ x = 17.

− 6x 2 − 18x

Multiply ( x + 3) by 17: 17( x + 3) = 17 x + 51.

17 x + 52 17 x + 51

Subtract: (17 x + 52) − (17 x + 51) = 1 .

1 The remainder is 1. The calculation in Worked example 1.8 can be written as:

13

(2 x 3 − x + 52) = ( x + 3) × (2 x 2 − 6x + 17) + 1 dividend

divisor

quotient

remainder

KEY POINT 1.2 The division algorithm for polynomials is: dividend = divisor × quotient + remainder

EXERCISE 1D 1 Simplify each of the following. a

( x 3 + 5x 2 + 5x − 2) ÷ ( x + 2)

b

( x 3 − 8x 2 + 22 x − 21) ÷ ( x − 3)

c

(3x 3 − 7 x 2 + 6x − 2) ÷ ( x − 1)

d

(2 x 3 − 3x 2 + 8x + 5) ÷ (2 x + 1)

e

(5x 3 − 13x 2 + 10 x − 8) ÷ (2 − x )

f

(6x 4 + 13x − 19) ÷ (1 − x )

2 Find the quotient and remainder for each of the following. a

( x 3 + 2 x 2 + 5x − 4) ÷ ( x + 1)

b

(6x 3 + 7 x 2 − 6) ÷ ( x − 2)

c

(8x 3 + x 2 − 2 x + 1) ÷ (2 x − 1)

d

(2 x 3 − 9x 2 − 9) ÷ (3 − x )

e

( x 3 + 5x 2 − 3x − 1) ÷ ( x 2 + 2 x + 1)

f

(5x 4 − 2 x 2 − 13x + 8) ÷ ( x 2 + 1)

3 Use algebraic division to show these two important factorisations. a

x 3 − y3 = ( x − y )( x 2 + xy + y2 )

b

x 3 + y3 = ( x + y )( x 2 − xy + y2 )

DID YOU KNOW? The number 1729 is the smallest number that can be expressed as the sum of two cubes in two different ways.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

4 a Use algebraic division to show that x − 2 is a factor of 2 x 3 + 7 x 2 − 43x + 42 . b Hence, factorise 2 x 3 + 7 x 2 − 43x + 42 completely. 5 a Use algebraic division to show that 2 x + 1 is a factor of 12 x 3 + 16x 2 − 3x − 4. b Hence, solve the equation 12 x 3 + 16x 2 − 3x − 4 = 0 . 6 a Use algebraic division to show that x + 1 is a factor of x 3 − x 2 + 2 x + 4 . b Hence, show that there is only one real root for the equation x 3 − x 2 + 2 x + 4 = 0 . 7 The general formula for solving the quadratic equation ax 2 + bx + c = 0 is x =

−b ± b2 − 4ac . 2a

There is a general formula for solving a cubic equation. Find out more about this formula and use it to solve 2 x 3 − 5x 2 − 28x + 15 = 0 .

1.5 The factor theorem In Worked example 1.7 you found that x 3 + 2 x 2 − 11x + 6 is exactly divisible by x − 2 . ( x 3 + 2 x 2 − 11x + 6) ÷ ( x − 2) = x 2 + 4x − 3 This can also be written as: x 3 + 2 x 2 − 11x + 6 = ( x − 2)( x 2 + 4x − 3) If a polynomial P(x ) divides exactly by a linear factor x − c to give the polynomial Q(x ), then 14

P(x ) = ( x − c ) Q(x ) Substituting x = c into this formula gives P(c ) = 0 . This is known as the factor theorem. KEY POINT 1.3 If for a polynomial P(x ) , P(c ) = 0 then x − c is a factor of P(x ) .

For example, when x = 2, x 3 + 2 x 2 − 11x + 6 = (2)3 + 2(2)2 − 11(2) + 6 = 8 + 8 − 22 + 6 = 0 Therefore ( x − 2) is a factor of x 3 + 2 x 2 − 11x + 6. The factor theorem can be extended to: KEY POINT 1.4 b If for a polynomial P(x ) , P = 0 then ax − b is a factor of P(x ) . a

2 , 3 3 2 2 2 2 8 8 3x 3 − 2 x 2 − 3x + 2 = 3 − 2 − 3 + 2 = − − 2 + 2 = 0 3 3 3 9 9

For example, when x =

Hence 3x − 2 is a factor of 3x 3 − 2 x 2 − 3x + 2.

Chapter 1: Algebra

WORKED EXAMPLE 1.9 Use the factor theorem to show that x − 3 is a factor of x 3 + 2 x 2 − 16x + 3 . Answer Let f( x ) = x 3 + 2 x 2 − 16x + 3

f(3) = (3)3 + 2(3)2 − 16(3) + 3 = 27 + 18 − 48 + 3 =0 Hence x − 3 is a factor of x 3 + 2 x 2 − 16x + 3 .

If f(3) = 0 , then ( x − 3) is a factor.

EXPLORE 1.3

A x 3 + x 2 − 3x + 1

B 2 x 3 + 7 x 2 + x − 10

C 3x 3 + 7 x 2 − 4

D 6x 3 + x 2 − 17 x + 10

E 6x 4 + x 3 − 8x 2 − x + 2

F 3x 3 + x 2 − 10 x − 8

Discuss with your classmates which of the expressions have: 1

x − 1 as a factor

2

x + 2 as a factor

15

3 2 x − 1 as a factor.

WORKED EXAMPLE 1.10 a Given that 3x 2 + 5x − 2 is a factor of 3x 3 + 17 x 2 + ax + b , find the value of a and the value of b. b Hence, factorise the polynomial completely. Answer a Let f( x ) = 3x 3 + 17 x 2 + ax + b . Factorising 3x 2 + 5x − 2 gives (3x − 1)( x + 2). Hence, 3x − 1 and x + 2 are also factors of f( x ). 1 Using the factor theorem, f = 0 and f( −2) = 0. 3 3

2

1 1 1 1 f = 0 gives 3 + 17 + a + b = 0 3 3 3 3 1 17 a + + +b = 0 9 9 3 a + 3b = −6

(1)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

f( −2) = 0 gives 3( −2)3 + 17( −2)2 + a ( −2) + b = 0 −24 + 68 − 2 a + b = 0 2 a − b = 44

(2)

(1) + 3 × (2) gives 7 a = 126 a = 18 Substituting in (2) gives b = −8. Hence, a = 18 and b = −8. b Method 1: Using long division x+4 3x 2 + 5x − 2 3x 3 + 17 x 2 + 18x − 8

)

3x 3 + 5x 2 − 2 x 12 x 2 + 20 x − 8 12 x 2 + 20 x − 8 0 Hence, 3x 3 + 17 x 2 + 18x − 8 = (3x 2 + 5x − 2)( x + 4) = (3x − 1)( x + 2)( x + 4) 16

Method 2: Equating coefficients If 3x 2 + 5x − 2 is a factor, 3x 3 + 17 x 2 + 18x − 8 can be written as 3x 3 + 17 x 2 + 18x − 8 = (3x 2 + 5x − 2)( cx + d )

coefficient of x 3 is 3, so c = 1 since 3 × 1 = 3

constant term is −8, so d = 4 since −2 × 4 = −8

Hence, 3x 3 + 17 x 2 + 18x − 8 = (3x − 1)( x + 2)( x + 4)

WORKED EXAMPLE 1.11 Solve x 3 + 4x 2 − 19x + 14 = 0. Answer Let f( x ) = x 3 + 4x 2 − 19x + 14. The positive and negative factors of 14 are ±1, ± 2, ±7 and ±14. f(1) = (1)3 + 4(1)2 − 19(1) + 14 = 0, so x − 1 is a factor of f( x ). The other factors of f( x ) can be found by any of the following methods.

Chapter 1: Algebra

Method 1: Trial and error f(2) = (2)3 + 4(2)2 − 19(2) + 14 = 0, so x − 2 is a factor of f( x ) f( −7) = ( −7)3 + 4( −7)2 − 19( −7) + 14 = 0, so x + 7 is a factor of f( x ). Hence, f( x ) = ( x − 1)( x − 2)( x + 7). Method 2: Long division x 2 + 5x − 14 x − 1 x 3 + 4x 2 − 19x + 14

)

x 3 − x 2 5x 2 − 19x 5x 2 − 5x

− 14x + 14 − 14x + 14 0

Hence, f( x ) = ( x − 1)( x 2 + 5x − 14) = ( x − 1)( x − 2)( x + 7) Method 3: Equating coefficients Since x − 1 is a factor, x 3 + 4x 2 − 19x + 14 can be written as: 17

x 3 + 4x 2 − 19x + 14 = ( x − 1)( ax 2 + bx + c )

coefficient of x 3 is 1, so a = 1 since 1 × 1 = 1

constant term is 14, so c = −14 since −1 × −14 = 14

x 3 + 4x 2 − 19x + 14 = ( x − 1)( x 2 + bx − 14) Equating the coefficients of x 2 gives 4 = −1 + b, so b = 5 . Hence, f( x ) = ( x − 1)( x 2 + 5x − 14) = ( x − 1)( x − 2)( x + 7) The solutions of the equation x 3 + 4x 2 − 19x + 14 = 0 are: x = 1, x = 2 and x = −7.

EXERCISE 1E 1 Use the factor theorem to show that x + 1 is a factor of x 4 − 3x 3 − 4x 2 + 5x + 5. 2 Use the factor theorem to show that 2 x − 5 is a factor of 2 x 3 − 7 x 2 + 9x − 10. PS

3 Given that x + 4 is a factor of x 3 + ax 2 − 29x + 12 , find the value of a.

PS

4 Given that x − 3 is a factor of x 3 + ax 2 + bx − 30, express a in terms of b.

PS

5 Given that 2 x 2 + x − 1 is a factor of 2 x 3 − x 2 + ax + b , find the value of a and the value of b.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PS

6 Given that x − 3 and 2 x + 1 are factors of 2 x 3 + px 2 + (2q − 1)x + q: a Find the value of p and the value of q. b Explain why x + 2 is also a factor of the expression.

PS

7 It is given that x + a is a factor of x 3 + 4x 2 + 7 ax + 4a . Find the possible values of a.

PS

8 a It is given that x + 1 is a common factor of x 3 + px + q and x 3 + (1 − p )x 2 + 19x − 2q. Find the value of p and the value of q. b When p and q have these values, factorise x 3 + px + q and x 3 + (1 − p )x 2 + 19x − 2q completely.

PS

9 Given that x − 1 and x + 2 are factors of x 4 − x 3 + px 2 − 11x + q : a Find the value of p and the value of q. b When p and q have these values, factorise x 4 − x 3 + px 2 − 11x + q completely. 10 Solve. a

x 3 − 5x 2 − 4x + 20 = 0

b

x 3 + 5x 2 − 17 x − 21 = 0

c

2 x 3 − 5x 2 − 13x + 30 = 0

d

3x 3 + 17 x 2 + 18x − 8 = 0

e

x 4 + 2 x 3 − 7 x 2 − 8x + 12 = 0

f

2 x 4 − 11x 3 + 12 x 2 + x − 4 = 0

P

11 You are given that the equation x 3 + ax 2 + bx + c = 0 has three real roots and that these roots are consecutive terms in an arithmetic progression. Show that 2 a 3 + 27c = 9ab.

PS

12 Find the set of values for k for which the equation 3x 4 + 4x 3 − 12 x 2 + k = 0 has four real roots.

18

1.6 The remainder theorem Consider f( x ) = 2 x 3 − 5x 2 − 2 x + 12. Substituting x = 2 in the polynomial gives f(2) = 2(2)3 − 5(2)2 − 2(2) + 12 = 4 . When 2 x 3 − 5x 2 − 2 x + 12 is divided by x − 2 , there is a remainder: 2 x 2 − x − 4 x − 2 2 x 3 − 5x 2 − 2 x + 12

)

2 x 3 − 4x 2

− x2 − 2x − x2 + 2x − 4x + 12 − 4x + 8 4

The remainder is 4. This is the same value as f(2). f( x ) = 2 x 3 − 5x 2 − 2 x + 12 can be written as: f( x ) = ( x − 2)(2 x 2 − x − 4) + 4 In general: If a polynomial P(x ) is divided by x − c to give the polynomial Q( x ) and a remainder R, then: P( x ) = ( x − c )Q( x ) + R

Chapter 1: Algebra

Substituting x = c into this formula gives P( c ) = R . This leads to the remainder theorem. KEY POINT 1.5 The remainder theorem: If a polynomial P( x ) is divided by x − c, the remainder is P( c ).

The remainder theorem can be extended to: KEY POINT 1.6 b If a polynomial P( x ) is divided by ax − b , the remainder is P . a

WORKED EXAMPLE 1.12 Use the factor theorem to find the remainder when 4x 3 − 2 x 2 + 5x − 8 is divided by x + 1. Answer Let f( x ) = 4x 3 − 2 x 2 + 5x − 8

19

Remainder = f( −1) = 4( −1)3 − 2( −1)2 + 5( −1) − 8 = −4 − 2 − 5 − 8 = −19

WORKED EXAMPLE 1.13 f( x ) = 6x 3 + ax 2 + bx − 4 When f( x ) is divided by x − 1, the remainder is 3. When f( x ) is divided by 3x + 2 , the remainder is −2. Find the value of a and the value of b. Answer f( x ) = 6x 3 + ax 2 + bx − 4 When f( x ) is divided by x − 1 the remainder is 3, hence f(1) = 3 . 6(1)3 + a (1)2 + b(1) − 4 = 3 a+b=1

(1)

2 When f( x ) is divided by 3x + 2 the remainder is −2, hence f − = −2. 3

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3

2

2 2 2 6 − + a − + b − − 4 = −2 3 3 3 2 a − 3b = 17

(2)

2 × (1) − (2) gives 5b = −15 b = −3 Substituting b = −3 in equation (1) gives a = 4. a = 4 and b = −3 .

EXERCISE 1F 1 Find the remainder when: a 6x 3 + 3x 2 − 5x + 2 is divided by x − 1 b x 3 + x 2 − 11x + 12 is divided by x + 4 c

x 4 + 2 x 3 − 5x 2 − 2 x + 8 is divided by x + 1

d 4x 3 − x 2 − 18x + 1 is divided by 2 x − 1 2 a When x 3 − 3x 2 + ax − 7 is divided by x + 2 , the remainder is −37. Find the value of a. b When 9x 3 + bx − 5 is divided by 3x + 2 , the remainder is −13. Find the value of b. 20

3

f( x ) = x 3 + ax 2 + bx − 5 f( x ) has a factor of x − 1 and leaves a remainder of −6 when divided by x + 1. Find the value of a and the value of b.

4 The polynomial 3x 3 + ax 2 + bx + 8 , where a and b are constants, is denoted by f( x ) . It is given that x + 2 is a factor of f( x ), and that when f( x ) is divided by x − 1 the remainder is 15. Find the value of a and the value of b. 5 The function f( x ) = ax 3 + 7 x 2 + bx − 8, where a and b are constants, is such that 2 x + 1 is a factor. The remainder when f( x ) is divided by x + 1 is 7. a Find the value of a and the value of b. b Factorise f( x ) completely. 6 The polynomial x 3 + 6x 2 + px − 3 leaves a remainder of R when divided by x + 1 and a remainder of −10R when divided by x − 3. a Find the value of p. b Hence find the remainder when the expression is divided by x − 2 . 7 The polynomial x 3 + ax 2 + bx + 2, where a and b are constants, is denoted by f( x ) . It is given that x − 2 is a factor of f( x ), and that when f( x ) is divided by x + 1 the remainder is 21. a Find the value of a and the value of b. b Solve the equation f( x ) = 0, giving the roots in exact form.

Chapter 1: Algebra

8 The polynomial 2 x 3 + ax 2 + bx + c is denoted by f( x ). The roots of f( x ) = 0 are −1, 2 and k. When f( x ) is divided by x − 1, the remainder is 6. a Find the value of k. b Find the remainder when f( x ) is divided by x + 2 . PS

9

P( x ) = 2 x + 4x 2 + 6x 3 + … + 100 x50 Find the remainder when P( x ) is divided by ( x − 1).

PS

10 P( x ) = 3( x + 1)( x + 2)( x + 3) + a ( x + 1)( x + 2) + b( x + 1) + c It is given that when P( x ) is divided by each of x + 1, x + 2 and x + 3 the remainders are −2, 2 and 10, respectively. Find the value of a, the value of b and the value of c.

Checklist of learning and understanding The modulus function ●

The modulus function x is defined by:

x if x ù 0

x =

− x if x , 0 The division algorithm for polynomials ●

P( x )

= D( x ) ×

Q( x )

+

21

R( x )

dividend = divisor × quotient + remainder The factor theorem ●

If for a polynomial P( x ) , P( c ) = 0 then x − c is a factor of P( x ).

●

If for a polynomial P( x ), P

b = 0 then ax − b is a factor of P( x ). a

The remainder theorem If a polynomial P( x ) is divided by x − c, the remainder is P( c ). b . ● If a polynomial P( x ) is divided by ax − b, the remainder is P a ●

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 1 1

Solve the equation 2 x − 3 = 5x + 1 .

[3]

2

Solve the inequality 5x − 3 ù 7.

[3]

3

Solve the inequality 2 x − 3 , 2 − x .

[3]

4

Solve the equation x 2 − 14 = 11.

[4]

5

The polynomial ax 3 − 13x 2 − 41x − 2 a , where a is a constant, is denoted by p( x ).

6

7 8

22

9

a

Given that ( x − 4) is a factor of p( x ) , find the value of a.

[2]

b

When a has this value, factorise p( x ) completely.

[3]

The polynomial 6x 3 − 23x 2 − 38x + 15 is denoted by f( x ). a

Show that ( x − 5) is a factor of f( x ) and hence factorise f( x ) completely.

b

Write down the roots of f x

( ) = 0.

The polynomial x 3 − 5x 2 + ax + b is denoted by f( x ). It is given that ( x + 2) is a factor of f( x ) and that when f( x ) is divided by ( x − 1) the remainder is −6. Find the value of a and the value of b.

[4] [1] [5]

The polynomial x 3 − 5x 2 + 7 x − 3 is denoted by p( x ). a

Find the quotient and remainder when p( x ) is divided by ( x 2 − 2 x − 1).

[4]

b

Use the factor theorem to show that ( x − 3) is a factor of p( x ).

[2]

The polynomial 4x 4 + 4x 3 − 7 x 2 − 4x + 8 is denoted by p( x ) . a

Find the quotient and remainder when p( x ) is divided by ( x 2 − 1).

[3]

b

Hence solve the equation 4x 4 + 4x 3 − 7 x 2 − 4x + 3 = 0.

[3]

10 The polynomial x 4 − 48x 2 − 21x − 2 is denoted by f( x ). a

Find the value of the constant k for which f( x ) = ( x 2 + kx + 2)( x 2 − kx − 1) .

[3]

b

Hence solve the equation f( x ) = 0. Give your answers in exact form.

[3]

11 The polynomial 2 x 4 + 3x 3 − 12 x 2 − 7 x + a is denoted by p( x ) . a

Given that (2 x − 1) is a factor of p( x ), find the value of a.

[2]

b

When a has this value, verify that ( x + 3) is also a factor of p( x ) and hence factorise p( x ) completely.

[4]

3

2

12 The polynomial 3x + ax − 36x + 20 is denoted by p( x ) . a

Given that ( x − 2) is a factor of p( x ) , find the value of a.

[2]

b

When a has this value, solve the equation p( x ) = 0 .

[4]

3

2

13 The polynomial 2 x + 5x − 7 x + 11 is denoted by f( x ). a

Find the remainder when f( x ) is divided by ( x − 2).

[2]

b

Find the quotient and remainder when f( x ) is divided by ( x 2 − 4x + 2) .

[4]

3

2

14 The polynomial ax + bx − x + 12 is denoted by p( x ) . a

Given that ( x − 3) and ( x + 1) are factors of p( x ) , find the value of a and the value of b.

[4]

b

When a and b take these values, find the other linear factor of p( x ) .

[2]

Chapter 1: Algebra

15 The polynomial 6x 3 + x 2 + ax − 10 , where a is a constant, is denoted by P( x ) . It is given that when P( x ) is divided by ( x + 2) the remainder is −12. a

Find the value of a and hence verify that (2 x + 1) is a factor of P( x ) .

[3]

b

When a has this value, solve the equation P( x ) = 0.

[4]

16 The polynomial 2 x 3 + ax 2 + bx + 6 is denoted by p( x ) . a

Given that ( x + 2) and ( x − 3) are factors of p( x ) , find the value of a and the value of b.

[4]

b

When a and b take these values, factorise p( x ) completely.

[3]

17 The polynomials P( x ) and Q( x ) are defined as: P( x ) = x 3 + ax 2 + b and Q( x ) = x 3 + bx 2 + a. It is given that ( x − 2) is a factor of P( x ) and that when Q( x ) is divided by ( x + 1) the remainder is −15. a

Find the value of a and the value of b.

[5]

b

When a and b take these values, find the least possible value of P( x ) − Q(x ) as x varies.

[2]

18 The polynomial 5x 3 − 13x 2 + 17 x − 7 is denoted by p( x ) . a

Find the quotient when p( x ) is divided by ( x − 1), and show that the remainder is 2.

[4]

b

Hence show that the polynomial 5x 3 − 13x 2 + 17 x − 9 has exactly one real root.

[3]

19 The polynomial 4x 3 + kx 2 − 65x + 18 is denoted by f( x ). a

Given that ( x + 2) is a factor of f( x ), find the value of k.

[2]

b

When k has this value, solve the equation f( x ) = 0.

[4]

c

Write down the roots of f( x 2 ) = 0 .

[1]

20 The polynomial 2 x 3 − 5x 2 + ax + b , where a and b are constants, is denoted by f( x ). It is given that when f( x ) is divided by ( x + 2) the remainder is 8 and that when f( x ) is divided by ( x − 1) the remainder is 50. a

Find the value of a and the value of b.

[5]

b

When a and b have these values, find the quotient and remainder when f( x ) is divided by x 2 − x + 2.

[3]

21 The polynomial 2 x 3 − 9x 2 + ax + b , where a and b are constants, is denoted by f( x ). It is given that ( x + 2) is a factor of f( x ), and that when f( x ) is divided by ( x + 1) the remainder is 30. a

Find the value of a and the value of b.

[5]

b

When a and b have these values, solve the equation f( x ) = 0.

[4]

22 The polynomial x 3 + 3x 2 + 4x + 2 is denoted by f( x ). i

Find the quotient and remainder when f( x ) is divided by x 2 + x − 1.

[4]

ii

Use the factor theorem to show that ( x + 1) is a factor of f( x ).

[2]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 June 2010

23

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

23 The polynomial 4x 3 + ax 2 + 9x + 9, where a is a constant, is denoted by p( x ). It is given that when p( x ) is divided by (2 x − 1) the remainder is 10. i

Find the value of a and hence verify that ( x − 3) is a factor of p( x ) .

[3]

ii

When a has this value, solve the equation p( x ) = 0 .

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2011 24 The polynomial ax 3 − 5x 2 + bx + 9, where a and b are constants, is denoted by p( x ) . It is given that (2 x + 3) is a factor of p( x ) , and that when p( x ) is divided by ( x + 1) the remainder is 8. i

Find the values of a and b.

[5]

ii

When a and b have these values, factorise p( x ) completely.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 June 2013

24

25

Chapter 2 Logarithmic and exponential functions In this chapter you will learn how to:

■ ■ ■ ■

understand the relationship between logarithms and indices, and use the laws of logarithms (excluding change of base) understand the definition and properties of e x and ln x, including their relationship as inverse functions and their graphs use logarithms to solve equations and inequalities in which the unknown appears in indices use logarithms to transform a given relationship to linear form, and hence determine unknown constants by considering the gradient and/or intercept.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE / O Level Mathematics

Use and interpret positive, negative, fractional and zero indices.

1 Evaluate. a 5−2 2

b 83 c 70 2 Solve 32 x = 2 . IGCSE / O Level Mathematics

Use the rules of indices.

3 Simplify. a 3x −4 × b

2 21 x ÷ 2 x −2 5

2 x5 c 3

What is a logarithm? 26

3

FAST FORWARD

At IGCSE / O Level we learnt how to solve some simple exponential equations. For example, 32 x = 2 can be solved by first writing 32 as a power of 2. In this chapter we will learn how to solve more difficult exponential equations such as a x = b where there is no obvious way of writing the numbers a and b in index form with the same base number. In order to be able to do this we will first learn about a new type of function called the logarithmic function. A logarithmic function is the inverse of an exponential function. At IGCSE / O Level we learnt about exponential growth and decay and its applications to numerous real-life problems such as investments and population sizes. Logarithmic functions also have real-life applications such as the amount of energy released during an earthquake (the Richter scale). One of the most fascinating occurrences of logarithms in nature is that of the logarithmic spiral that can be observed in a nautilus shell. We also see logarithmic spiral shapes in spiral galaxies, and in plants such as sunflowers.

2.1 Logarithms to base 10 Consider the exponential function f( x ) = 10 x . To solve the equation 10 x = 6, we could make an estimate by noting that 10 0 = 1 and 101 = 10 , hence 0 , x , 1. We could then try values of x between 0 and 1 to find a more accurate value for x. The graph of y = 10 x could also be used to give a more accurate value for x when 10 x = 6. y 10 8 6 4

y = 10x

2 –0.5

2 21 x 3

O

0.5

0.78

1 x

You will encounter logarithmic functions again when studying calculus in Chapters 4, 5 and 8.

WEB LINK Explore the Exponentials and logarithms station on the Underground Mathematics website.

Chapter 2: Logarithmic and exponential functions

From the graph, x ≈ 0.78 . There is, however, a function that gives the value of x directly: If 10 x = 6 then x = log10 6.

TIP

log10 6 is read as ‘log 6 to base 10’.

log10 6 can also be written as log 6 or lg 6.

log is short for logarithm. On a calculator, for logs to the base 10, we use the log or lg key. So if 10 x = 6 then

x = log10 6 x = 0.778 to 3 significant figures.

KEY POINT 2.1 The rule for base 10 is: If y = 10 x then x = log10 y

y

y = 10x

This rule can be described in words as: y=x

log10 y is the power that 10 must be raised to in order to obtain y. For example, log10 100 = 2 since 100 = 10 2.

1

y = 10 x and y = log10 x are inverse functions.

O

x

We say that y = 10 ⇔ x = log10 y. This is read as: y = 10 x is equivalent to saying that x = log10 y. y

EXPLORE 2.1 1 Discuss with your classmates why each of these two statements is true. 10 log

log10 10 x = x for x ∈ ℝ

10

x

= x for x . 0

2 a Discuss with your classmates why each of these two statements is true. log10 10 = 1

log10 1 = 0

b Discuss what the missing numbers should be in this table of values, giving reasons for your answers. log10 0.01

log10 0.1

log10 1

log10 10

log10 100

log10 1000

…..

…..

0

1

…..

…..

y = log10 x 1

x

27

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 2.1 a Convert 10 x = 58 to logarithmic form. b Solve 10 x = 58 giving your answer correct to 3 significant figures. Answer a Method 1 10 x = 58 Step 1: Identify the base and index

base = 10 , index = x

Step 2: Start to write in log form

x = log10 ?

Step 3: Complete the log form

x = log10 58

TIP In log form the index is always on its own and the base goes at the base of the logarithm.

10 x = 58 ⇔ x = log10 58 Method 2 10 x = 58 log10 10 x = log10 58 x = log10 58

Take logs to base 10 of both sides. Use log10 10 x = x.

28

b

10 x = 58 x = log10 58 x ≈ 1.76

WORKED EXAMPLE 2.2 a Convert log10 x = 3.5 to exponential form. b Solve log10 x = 3.5 giving your answer correct to 3 significant figures. Answer a Method 1 log10 x = 3.5 Step 1: Identify the base and index

base = 10, index = 3.5 TIP

Step 2: Start to write in exponential form

103.5 = ?

In log form the index is always on its own.

Chapter 2: Logarithmic and exponential functions

x = 103.5

Step 3: Complete the exponential form log10 x = 3.5 ⇔ x = 103.5 Method 2 log10 x = 3.5 10 log

10

x

Write each side as an exponent of 10.

= 103.5

Use 10 log

x = 103.5 b

10

x

= x.

log10 x = 3.5 x = 103.5 x ≈ 3160

WORKED EXAMPLE 2.3 Find the value of: a

log10 10 000

b log10 0.0001

c log10 1000 10.

Answer log10 10 000 = log10 10 4 =4

Write 10 000 as a power of 10, 10 000 = 10 4.

b

log10 0.0001 = log10 10 −4 = −4

Write 0.0001 as a power of 10, 0.0001 = 10 −4 .

c

log10 1000 10 = log10 103.5 = 3.5

Write 1000 10 as a power of 10.

a

29

1000 10 = 103 × 10 0.5 = 103.5.

EXERCISE 2A 1 Convert from exponential form to logarithmic form. a

102 = 100

b

10 x = 200

c

10 x = 0.05

2 Solve each of these equations, giving your answers correct to 3 significant figures. a

10 x = 52

b

10 x = 250

c

10 x = 0.48

c

log10 x = −0.6

3 Convert from logarithmic form to exponential form. a

log10 10 000 = 4

b

log10 x = 1.2

4 Solve each of these equations, giving your answers correct to 3 significant figures. a

log10 x = 1.88

b

log10 x = 2.76

c

log10 x = −1.4

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

5 Without using a calculator, find the value of: a

b log10 0.0001

log10 100

(

d log10

3 10

)

e

(

log10

100 3 10

)

(

)

c

log10 10 10

f

100 log10 1000

6 Given that the function f is defined as f : x ֏ 10 x − 3 for x ∈ ℝ, find an expression for f −1( x ) .

WEB LINK Try the See the power resource on the Underground Mathematics website.

7 Given that p and q are positive and that 4(log10 p )2 + 2(log10 q )2 = 9, find the greatest possible value of p.

PS

2.2 Logarithms to base a In the last section you learnt about logarithms to the base of 10. The same principles can be applied to define logarithms in other bases. KEY POINT 2.2 If y = a x then x = log a y

30

log a a = 1

log a 1 = 0

log a a x = x

a log a x = x

The conditions for log a x to be defined are: ● ●

a . 0 and a ≠ 1 x.0

We say that y = a x ⇔ x = log a y. WORKED EXAMPLE 2.4 Convert 53 = 125 to logarithmic form. Answer Method 1 53 = 125 Step 1: Identify the base and index

base = 5, index = 3

Step 2: Start to write in log form

3 = log5 ?

Step 3: Complete the log form

3 = log5 125

3

5 = 125 ⇔ 3 = log5 125 Method 2 53 = 125 log5 53 = log5 125 3 = log5 125

Take logs to base 5 of both sides. Use log5 5x = x .

TIP In log form the index is always on its own and the base goes at the base of the logarithm.

Chapter 2: Logarithmic and exponential functions

WORKED EXAMPLE 2.5 Convert log3 x = 2.5 to exponential form. Answer Method 1 log3 x = 2.5 Step 1: Identify the base and index

base = 3, index = 2.5

Step 2: Start to write in exponential form

32.5 = ?

Step 3: Complete the exponential form

32.5 = x

TIP In log form the index is always on its own.

log3 x = 2.5 ⇔ 32.5 = x Method 2 log3 x = 2.5 3log

3

x

= 32.5

Write each side as an exponent of 3. Use 3log

3

x

=x

x = 3

2.5

WORKED EXAMPLE 2.6 Find the value of: a log2 16

31

1 b log3 9

Answer a

log2 16 = log2 2 4 = 4

b

log3

1 = log3 3−2 9 = −2

Write 16 as a power of 2, 16 = 2 4 .

Write

1 1 as a power of 3, = 3−2. 9 9

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 2.7 3x . Simplify log x x Answer

( )

3x −2 = log x x 3 log x x =−

3

x as a power of x. x

Write

2 3

EXERCISE 2B 1 Convert from exponential form to logarithmic form. a

52 = 25

b

2 4 = 16

c

e

8x = 15

f

xy = 6

g

1 243 ab = c

3−5 =

1 1024 =7

d

2 −10 =

h

x5y

log16 4 =

2 Convert from logarithmic form to exponential form. a

log2 8 = 3

b

log3 81 = 4

c

log8 1 = 0

d

e

log8 2 =

1 3

f

log2 y = 4

g

log a 1 = 0

h

1 2 log x 5 = y

log2 x = 3

b

log3 x = 2

c

log 4 x = 0

d

log x 49 = 2

32

3 Solve. a

4 Solve. a

log3 ( x + 5) = 2

log2 (3x − 1) = 5

b

c

log y (7 − 2 x ) = 0

5 Find the value of: a

log3 27

b log6 36

e

1 log 4 16

f

(

log5 5 5

)

c

log 4 2

d log2 0.125

g

3 log3 3

37 h log7 7

c

log x x 2 x

g

x log x 3 x

2

6 Simplify. a

log x x 3

e

1 log x 3 x

b log x 3 x 2

f

log x

(

x

5

)

(

)

d log x

1 x4

x2 h log x x

7 Given that the function f is defined as f : x ֏ 1 + log2 ( x − 3) for x ∈ ℝ, x > 3, find an expression for f −1( x ). PS

8 Solve. a

log2 (log3 x ) = −1

b

log 4 23− 5 x = x 2

3

Chapter 2: Logarithmic and exponential functions

PS

9

log3 9

log2 8

log3 20

log3 4

log2 2

log 4 3

log2 3

Without using a calculator, arrange these logarithms in ascending order. Add some more logarithms of your own in the correct position in your ordered list.

WEB LINK Try the Logarithm lattice resource on the Underground Mathematics website.

EXPLORE 2.2 1 Prya and Hamish are asked what they think log10 12 + log10 4 simplifies to. log10 12 + log10 4 = log10 (12 + 4) = log10 16

Prya says that:

Hamish says that: log10 12 + log10 4 = log10 (12 × 4) = log10 48 Use a calculator to see who is correct and check that their rule works for other pairs of numbers. 2 Can you predict what log10 12 − log10 4 simplifies to? Use a calculator to check that you have predicted correctly and check for other pairs of numbers.

2.3 The laws of logarithms If x and y are both positive and a . 0 and a ≠ 1, then the following laws of logarithms can be used. KEY POINT 2.3 Multiplication law

Division law

Power law

log a (xy ) = log a x + log a y

x log a = log a x − log a y y

log a ( x ) m = m log a x

Proofs: Multiplication law

Division law

Power law

log a ( xy )

x log a y

log a ( x ) m

= log a ( a log = log a ( a

a

x

× a log

log a x + log a y

= log a x + log a y

)

a

y

)

a = log a log a

log a x

= log a ( a log

a

a

y

x − log a y

= log a x − log a y 1 Using the power law, log a = log a x −1 x = − log a x

(

= log a ( a log = log a ( a

x m

)

m log a x

= m log a x )

a

)

)

33

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Another useful rule to remember: KEY POINT 2.4 1 log a = − log a x x

WORKED EXAMPLE 2.8 Use the laws of logarithms to simplify these expressions. a log2 3 + log2 5

b

log3 8 − log3 4

c 2 log5 2 + log5 3

b

log3 8 – log3 4

c

Answer a

log2 3 + log2 5 = log2 (3 × 5)

8 = log3 4 = log 3 2

= log 2 15

2 log5 2 + log5 3 = log5 22 + log5 3 = log5 (4 × 3) = log5 12

WORKED EXAMPLE 2.9 34

Given that log 4 p = x and log 4 q = y, express in terms of x and/or y a

log 4 p5 − log 4 q 2

64 c log 4 . p

b log 4 p + 5 log 4 3 q

Answer

a

log 4 p5 − log 4 q 2 = 5 log 4 p − 2 log 4 q = 5x − 2 y

b

log 4 p + 5 log 4 3 q

c

1 5 log 4 p + log 4 q 2 3 1 5 = x+ y 2 3 =

64 log 4 p = log 4 43 − log 4 p = 3−x

EXERCISE 2C 1 Write as a single logarithm. a

log2 7 + log2 11

d 2 log3 8 − 5 log3 2

b log6 20 − log6 4

c

3 log5 2 − log5 4

e 1 + 2 log2 3

f

2 − log 4 2

c

log2 60 − log2 5

2 Write as a single logarithm, then simplify your answer. a

log2 40 − log2 5

b log6 20 + log6 5

Chapter 2: Logarithmic and exponential functions

3 Simplify. a 3 log5 2 +

1 log5 36 − log5 12 2

b

1 1 log3 8 + log3 18 − 1 2 2 log5 8 . log5 0.25

4 Express 8 and 0.25 as powers of 2 and hence simplify 5 Simplify. log2 27 a log2 3

b

log3 128 log3 16

c

log3 25 log3 0.04

d

log5 1000 log5 0.01

6 Given that log8 y = log8 ( x − 2) − 2 log8 x, express y in terms of x. 7 Given that log3 ( z − 1) − log3 z = 1 + 3 log3 y, express z in terms of y. 8 Given that y = log5 x, find in terms ofy: a

b log5 (25x )

x

c

x x log5 125

d log x 125

9 Given that x = log 4 p and y = log 4 q, express in terms of x and/or y: a

log 4 (64 p )

2 b log 4 q

c

pq

d log 4 p2 − log 4 4 q

(

)

10 Given that log a x = 7 and log a y = 4 , find the value of: a PS

x log a y

x b log a 2 y

c

(

log a x 2 y

)

y d log a 3 . x

35

11 x

−log2 5

y

log3 2 − log2 5

2 log2 5 − 3 log3 2

z

The number in the rectangle on the side of the triangle is the sum of the numbers at the adjacent vertices. For example, x + y = − log2 5. Find the value of x, the value of y and the value of z.

2.4 Solving logarithmic equations We have already learnt how to solve simple logarithmic equations. In this section we will learn how to solve more complicated equations. Since log a x only exists for x . 0 and for a . 0 and a ≠ 1, it is essential when solving equations involving logs that all roots are checked in the original equation.

WEB LINK Try the Summing to one and the Factorial fragments resources on the Underground Mathematics website.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 2.10 Solve: a

2 log8 ( x + 2) = log8 (2 x + 19)

b

4 log x 2 − log x 4 = 2

Answer a

2 log8 ( x + 2) = log8 (2 x + 19)

Use the power law.

log8 ( x + 2)2 = log8 (2 x + 19)

Use equality of logarithms.

( x + 2)2 = 2 x + 19

Expand brackets.

x 2 + 4x + 4 = 2 x + 19 x 2 + 2 x − 15 = 0 ( x − 3)( x + 5) = 0 x = 3 or x = −5 Check when x = 3: 2 log8 ( x + 2) = 2 log8 5 = log8 25 is defined log8 (2 x + 19) = log8 25 is defined. So x = 3 is a solution, since both sides of the equation are defined and equivalent in value. Check when x = −5: 2 log8 ( x + 2) = 2 log8 ( −3) is not defined. 36

So x = −5 is not a solution of the original equation. Hence, the solution is x = 3 b

4 log x 2 − log x 4 = 2 log x 2 4 − log x 22 = 2 log x 2

4−2

Use the power law. Use the division law.

=2

2

log x 2 = 2 log x 4 = 2

Convert to exponential form.

2

x =4 x = ±2 Since logarithms only exist for positive bases, x = −2 is not a solution. Check when x = 2: 4 log2 2 − log2 4 = 4 − 2 log2 2 = 4−2 =2 So x = 2 satisfies the original equation. Hence, the solution is x = 2

Chapter 2: Logarithmic and exponential functions

EXERCISE 2D 1 Solve. a

log5 x + log5 3 = log5 30

b

log3 4x − log3 2 = log3 7

c

log2 (2 − x ) + log2 9 = 2 log2 8

d

log10 ( x − 4) = 2 log10 5 + log10 2

2 Solve. a

log10 (2 x + 9) − log10 5 = 1

b

log3 2 x − log3 ( x − 1) = 2

c

log2 ( 5 − 2 x ) = 3 + log2 ( x + 1 )

d

log5 (2 x − 3) + 2 log5 2 = 1 + log5 (3 − 2 x )

3 Solve. a

log2 x + log2 ( x − 1) = log2 20

b

log5 ( x − 2) + log5 ( x − 5) = log5 4x

c

2 log3 x − log3 ( x − 2) = 2

d

3 + 2 log2 x = log2 (3 − 10 x )

4 Solve. a

log x 40 − log x 5 = 1

b

log x 36 + log x 4 = 2

c

log x 25 − 2 log x 3 = 2

d

2 log x 32 = 3 + 2 log x 4

5 Solve.

PS

a

(log2 x )2 − 8 log2 x + 15 = 0

b

(log10 x )2 − log10 ( x 2 ) = 3

c

(log5 x )2 + log5 ( x 3 ) = 10

d

4(log2 x )2 − 2 log2 ( x 2 ) = 3

b

4x = 2 y

6 Solve the simultaneous equations. a

xy = 81

3 log10 y = log10 x + log10 2

log x y = 3 c

log 4( x − y ) = 2 log 4 x log 4( x + y + 9) = 0

PS

d

log10 x = 2 log10 y log10 (2 x − 17 y ) = 2

x4 7 Given that log10 ( x 2 y ) = 4 and log10 3 = 18 , find the value of log10 x and of log10 y. y

37

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2.5 Solving exponential equations At IGCSE / O Level you learnt how to solve exponential equations whose terms could be converted to the same base. In this section you will learn how to solve exponential equations whose terms cannot be converted to the same base. WORKED EXAMPLE 2.11 Solve, giving your answers correct to 3 significant figures. a

5 2 x +1 = 7

b

32 x = 4x + 5

Answer 52 x +1 = 7

a

log 52 x +1 = log 7 (2 x + 1) log 5 = log 7 2 x log 5 + log 5 = log 7 x=

Take logs to base 10 of both sides. Use the power rule. Expand the brackets. Rearrange to find x.

log 7 − log 5 2 log 5

x = 0.105 to 3 significant figures. 32 x = 4x + 5

b

log 32 x = log 4x + 5

38

Take logs to base 10 of both sides. Use the power rule.

2 x log 3 = ( x + 5) log 4

Expand the brackets.

2 x log 3 = x log 4 + 5 log 4

Collect x terms on one side.

x(2 log 3 − log 4) = 5 log 4

Divide both sides by (2 log 3 − log 4).

5 log 4 x= 2 log 3 − log 4 x = 8.55 to 3 significant figures.

WORKED EXAMPLE 2.12 Solve the equation 2(32x ) + 7(3x ) = 15, giving your answers correct to 3 significant figures. Answer 2(32 x ) + 7(3x ) − 15 = 0 2 y2 + 7 y − 15 = 0

Use the substitution y = 3x. Factorise.

(2 y − 3)( y + 5) = 0 y = 1.5 or y = −5 When y = 1.5 3x = 1.5 x log 3 = log1.5 x=

log1.5 = 0.369 … log 3

Take logs of both sides.

Chapter 2: Logarithmic and exponential functions

When y = −5 There are no solutions to this equation since 3x is always positive.

3x = −5 Hence, the solution is x = 0.369 to 3 significant figures.

EXERCISE 2E 1 Solve, giving your answers correct to 3 significant figures. a

5x = 18

b

2 x = 35

c

32x = 8

d

2 x +1 = 25

e

32 x − 5 = 20

f

3x = 2 x +1

g

5x + 3 = 7 4 − 3x

h

41− 3x = 3x − 2

i

2 x = 2(5x )

j

4x = 7(3x )

k

2 x +1 = 3(5x )

l

5(2 x − 3 ) = 4(3x + 4 )

2 a Show that 2 x +1 + 6(2 x −1 ) = 12 can be written as 2(2 x ) + 3(2 x ) = 12. b Hence solve the equation 2 x +1 + 6(2 x −1 ) = 12, giving your answer correct to 3 significant figures. 3 Solve, giving your answers correct to 3 significant figures. a

2 x + 2 + 2 x = 22

b

3x +1 = 3x −1 + 32

c

2 x +1 + 5(2 x −1 ) = 2 4

d

5 x + 53 = 5 x + 2

e

4 x −1 = 4 x − 4 3

f

3x +1 − 2(3x − 2 ) = 5

4 Use the substitution y = 2 x to solve the equation 22x + 32 = 12(2 x ). 5 Solve, giving your answers correct to 3 significant figures. a

52x − 5(5x ) + 6 = 0

b

22x + 5 = 6 × 2 x

c

32x = 6 × 3x + 7

d

42x + 27 = 12(4x )

6 Use the substitution u = 2 x to solve the equation 22 x − 5(2 x +1 ) + 24 = 0. 7 Solve, giving your answers correct to 3 significant figures. a

22 x − 2 x +1 − 35 = 0

b

32 x − 3x +1 = 10

c

2(5x +1 ) − 52 x = 16

d

42 x +1 = 17(4x ) − 15

8 Solve, giving your answers correct to 3 significant figures. a

4x + 2 x − 12 = 0

b

4x + 15 = 4(2 x +1 )

d

3(16x ) − 10(4x ) + 3 = 0

2(9x ) = 3x +1 + 27 x correct to 3 significant figures. 9 For each of the following equations find the value of y b 7 x = (2.7) y c 42 x = 35 y a 3x = 7 y c

10 Solve the equation x 2.5 = 20 x1.25, giving your answers in exact form. 11 Solve, giving your answers correct to 3 significant figures. a

4x − 8 = 2

b

2(3x ) − 4 = 8

c

3 2x − 5 = 2x

d

3x + 4 = 3x − 9

e

5(2 x ) + 3 = 5(2 x ) − 10

f

2 x + 2 + 1 = 2 x + 12

g

32 x = 6 3 x

h

4

( ) + 16

x

( ) + 14

=5 2

x

39

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PS

12 Given that 2 4 x +1 × 32 − x = 8x × 35 − 2 x , find the value of: a 6

PS

x

WEB LINK

b x

Try the To log or not to log? resource on the Underground Mathematics website.

13 Solve the equation 8(8x −1 − 1) = 7(4x − 2 x +1 ).

2.6 Solving exponential inequalities To solve inequalities such as 2 x . 3 we take the logarithm of both sides of the inequality. We can take logarithms to any base but care must be taken with the inequality symbol when doing so. If the base of the logarithm is greater than 1 then the inequality symbol remains the same. If the base of the logarithm is between 0 and 1 then the inequality symbol must be reversed. This is because y = log a x is an increasing function if a . 1 and it is a decreasing function when 0 , a , 1. ● ●

Hence, when taking logarithms to base 10 of both sides of an inequality the inequality symbol is unchanged. It is also important to remember that if we divide both sides of an inequality by a negative number then the inequality symbol must be reversed. WORKED EXAMPLE 2.13 40

Solve the inequality 0.6x , 0.7, giving your answer in terms of base 10 logarithms. Answer 0.6x , 0.7 log 0.6x , log 0.7 x log 0.6 , log 0.7 x.

log 0.7 log 0.6

x . 0.698 (correct to 3 significant figures)

Take logs to base 10 of both sides, inequality symbol is unchanged. Use the power rule for logs. Divide both sides by log10 0.6. log 0.6 is negative so the inequality symbol is reversed.

Chapter 2: Logarithmic and exponential functions

WORKED EXAMPLE 2.14 Solve the inequality 4 × 32 x −1 . 5, giving your answer in terms of base 10 logarithms. Answer 4 × 32 x −1 . 5 5 32 x −1 . 4

Divide both sides by 4. Take logs to base 10 of both sides and use the power rule.

5 4 5 ( 2 log 3 ) x − log 3 . log 4 5 log + log 3 4 x. 2 log 3 (2 x − 1) log 3 . log

Expand brackets. Rearrange.

Simplify.

15 4 x. log 9 log

EXERCISE 2F 1 Solve the inequalities, giving your answer in terms of base 10 logarithms. x 2 a 2x , 5 b 5x ù 7 c .3 d 0.8x , 0.3 3 2 Solve the inequalities, giving your answer in terms of base 10 logarithms. 3− x 5 c 2 × 52 x +1 ø 3 d 7 × .4 a 85 − x , 10 b 32 x + 5 . 20 6 2

3 Solve 5x . 2 x , giving your answer in terms of logarithms. P PS

15 log 2 × 2 ù 5 is x ù . 4 Prove that the solution to the inequality 3 9 log 8 5 In this question you are not allowed to use a calculator. 2 x −1

1− 3x

You are given that log10 4 = 0.60206 correct to 5 decimal places and that 10 0.206 , 2 . a Find the number of digits in the number 4100. b Find the first digit in the number 4100.

41

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2.7 Natural logarithms

FAST FORWARD

There is another type of logarithm to a special base called e. The number e is an irrational number and e ≈ 2.718. The function y = e x is called the natural exponential function. Logarithms to the base of e are called natural logarithms. ln x is used to represent log e x. KEY POINT 2.5 If y = e x then x = ln y

y

y = ex y=x

1 O

y = ln x 1

x

The curve y = ln x is the reflection of y = e x in the line y = x. 42

y = ln x and y = e x are inverse functions. All the rules of logarithms that we have learnt so far also apply for natural logarithms. We say that y = e x ⇔ x = log e y. DID YOU KNOW? The Scottish mathematician John Napier (1550 −1617) is credited with the discovery of logarithms. His original studies involved logarithms to 1 the base of . e

DID YOU KNOW? The number e is also known as Euler’s number. It is named after Leonhard Euler (1707 −1783), who devised the following formula for calculating the value of e: 1 1 1 1 1 e = 1+ + + + + + … 1 1× 2 1× 2 × 3 1× 2 × 3 × 4 1× 2 × 3 × 4 × 5

The number e is a very important number in Mathematics as it has very special properties. You will learn about these special properties in Chapters 4 and 5.

Chapter 2: Logarithmic and exponential functions

EXERCISE 2G 1 Use a calculator to evaluate correct to 3 significant figures: a

e3

b e2.7

c

e 0.8

d e −2

c

ln 0.9

d ln 0.15

c

5e ln 6

d e

c

e3 ln x = 64

d

e − ln x = 3

c

e x +1 = 8

d

e2 x − 3 = 16

c

e 2 x −1 = 6

d

e2

c

5 × e2 x + 3 , 1

2 Use a calculator to evaluate correct to 3 significant figures: a

ln 3

b ln1.4

3 Without using a calculator, find the value of: 1

a

e ln 2

b e 2

ln 9

− ln

1 2

4 Solve. a

e ln x = 5

b

ln e x = 15

5 Solve, giving your answers correct to 3 significant figures. a

e x = 18

b

e2x = 25

6 Solve, giving your answers in terms of natural logarithms. a

e x = 13

b

e3x = 7

1

x+3

=4

7 Solve, giving your answers in terms of natural logarithms. a

e x . 10

b e5 x − 2 ø 35

43

8 Solve, giving your answers correct to 3 significant figures. a

ln x = 5

b

ln x = −4

c

ln( x − 2) = 6

d

ln(2 x + 1) = −2

9 Solve, giving your answers correct to 3 significant figures. a

ln(3 − x 2 ) = 2 ln x

b

2 ln( x + 2) − ln x = ln(2 x − 1)

c

2 ln( x + 1 ) = ln ( 2 x + 3 )

d

ln(2 x + 1) = 2 ln x + ln 5

e

ln( x + 2) − ln x = 1

f

ln(2 + x 2 ) = 1 + 2 ln x

b

ln( y + 2) − ln y = 1 + 2 ln x

e2x − 5e x + 6 = 0

10 Express y in terms of x for each of these equations. a

2 ln( y + 1) − ln y = ln( x + y )

11 Solve, giving your answers in exact form. a

e2x + 2e x − 15 = 0

b

c

6e2x − 13e x − 5 = 0

d e x − 21e − x = 4

12 Given that the function f is defined as f : x ֏ 5e x + 2 for x ∈ R, find an expression for f −1( x ). PS

13 Solve 2 ln(3 − e2x ) = 1, giving your answer correct to 3 significant figures.

PS

14 Solve the simultaneous equations, giving your answers in exact form. a

2 ln x + ln y = 1 + ln 5 ln10 x − ln y = 2 + ln 2

PS

15 Solve ln(2 x + 1) ø ln( x + 4).

b

e3x + 4 y = 2e2 x − y e2 x + y = 8e x + 6 y

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2.8 Transforming a relationship to linear form When we collect experimental data for two variables we often wish to find a mathematical relationship connecting the variables. When the data plots on a graph lie on a straight line the relationship can be easily found using the equation y = mx + c, where m is the gradient and c is the y-intercept. It is more common, however, for the data plots to lie on a curve. Logarithms can be used to convert some curves into straight lines. This is the case for relationships of the form y = kx n and y = k ( a x ) where k, a and n are constants. Logarithms to any base can be used but it is more usual to use those that are most commonly available on calculators, which are natural logarithms or logarithms to the base 10. WORKED EXAMPLE 2.15 Convert y = aebx , where a and b are constants, into the form Y = mX + c. Answer y = aebx Taking natural logarithms of both sides gives: ln y = ln( aebx ) ln y = ln a + ln ebx ln y = ln a + bx

44

ln y = bx + ln a Now compare ln y = bx + ln a with Y = mX + c: ln y = b x + ln a Y = m X + c The non-linear equation y = ae

bx

becomes the linear equation:

Y = mX + c, where Y = ln y, X = x, m = b and c = ln a

It is important to note: ●

●

The variables X and Y in Y = mX + c must contain only the original variables x and y. (They must not contain the unknown constants a and b.) The constants m and c must contain only the original unknown constants a and b. (They must not contain the variables x and y.) WORKED EXAMPLE 2.16

(

)

The variables x and y satisfy the equation y = a e − bx , where a and b are constants. The graph of ln y against x 2 is a straight line passing through the points (0.55, 0.84) and (1.72, 0.26) as shown in the diagram. 2

ln y (0.55, 0.84) (1.72, 0.26)

Find the value of a and the value of b correct to 2 decimal places.

O

x2

Chapter 2: Logarithmic and exponential functions

Answer y = a × e − bx

2

Taking natural logarithms of both sides gives: 2

ln y = ln( a × e − bx ) ln y = ln a + ln e − bx

2

ln y = ln a − bx 2 ln y = −bx 2 + ln a Now compare ln y = −b x 2 + ln a with Y = mX + c: ln y = − b x 2 + ln a Y = m X + c Gradient = m =

0.26 − 0.84 = −0.4957 … 1.72 − 0.55

DID YOU KNOW?

∴ b = 0.50 to 2 decimal places

When one variable on a graph is a log but the other variable is not a log, it is called a semilog graph. When both variables on a graph are logs, it is called a log–log graph.

Using Y = mX + c, X = 0.55, Y = 0.84 and m = −0.4957 … 0.84 c ∴ ln a a Hence a = 3.04 and b = 0.50

= −0.4957 … × 0.55 + c = 1.1126 … = 1.1126 … = e1.1126… to 2 decimal places.

WORKED EXAMPLE 2.17 x

5

10

20

40

80

y

2593

1596

983

605

372

The table shows experimental values of the variables x and y. a By plotting a suitable straight-line graph, show that x and y are related by the equation y = k × x n , where k and n are constants. b Use your graph to estimate the value of k and the value of n. Answer a

y = k × xn

Take natural logarithms of both sides. n

ln y = ln( k × x ) ln y = ln k + ln x n ln y = n ln x + ln k

Use the multiplication law. Use the power law.

45

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Now compare ln y = n ln x + ln k with Y = mX + c: ln y = n ln x + ln k Y = m X + c Hence the graph of ln y against ln x needs to be drawn where: ● gradient = n ● intercept on vertical axis = ln k Table of values is ln x

1.61

2.30

3.00

3.69

4.38

ln y

7.86

7.38

6.89

6.41

5.92

ln y 10 8 6 4 2

46 O

ln x 1

2

3

4

5

The points form an approximate straight line, so x and y are related by the equation y = k × x n . 5.92 − 7.86 ≈ −0.70 to 2 significant figures. 4.38 − 1.61 ln k = intercept on vertical axis

b n = gradient ≈

ln k ≈ 9 k ≈ e9 k ≈ 8100 to 2 significant figures.

DID YOU KNOW? The distributions of a diverse range of natural and man-made phenomena approximately follow the power law y = ax b . These include the size of craters on the moon, avalanches, species extinction, body mass, income and populations of cities.

Chapter 2: Logarithmic and exponential functions

EXERCISE 2H 1 Given that a and b are constants, use logarithms to change each of these non-linear equations into the form Y = mX + c. State what the variables X and Y and the constants m and c represent. (Note: there might be more than one method to do this.) a

y = e ax + b

e

a = ex

2

+ by

b

y = 10 ax − b

c

y = a( x −b )

d

y = a(b x )

f

x a yb = 8

g

xa y = b

h

y = a (e − bx )

2 The variables x and y satisfy the equation y = a × x n, where a and n are constants. The graph of ln y against ln x is a straight line passing through the points (0.31, 4.02) and (1.83, 3.22) as shown in the diagram. Find the value of a and the value of n correct to 2 significant figures.

ln y

(0.31, 4.02) (1.83, 3.22)

ln x O

3 The variables x and y satisfy the equation y = k × e n ( x − 2), where k and n are constants. The graph of ln y against x is a straight line passing through the points (1, 1.84) and (7, 4.33) as shown in the diagram. Find the value of k and the value of n correct to 2 significant figures.

ln y

(7, 4.33)

(1, 1.84)

O

x

4 Variables x and y are related so that, when log10 y is plotted on the vertical axis and x is plotted on the horizontal axis, a straight-line graph passing through the points (2, 5) and (6, 11) is obtained. a Express log10 y in terms of x. b Express y in terms of x, giving your answer in the form y = a × 10bx . 5 Variables x and y are related so that, when ln y is plotted on the vertical axis and ln x is plotted on the horizontal axis, a straight-line graph passing through the points (2, 4) and (5, 13) is obtained. a Express ln y in terms of x. b Express y in terms of x. 6 The variables x and y satisfy the equation 52 y = 32 x +1 . By taking natural logarithms, show that the graph of ln y against ln x is a straight line, and find the exact value of the gradient of this line and state the coordinates of the point at which the line cuts the y-axis. 7 The mass, m grams, of a radioactive substance is given by the formula m = m0 e − kt , where t is the time in days after the mass was first recorded and m0 and k are constants. The table below shows experimental values of t and m. t

10

20

30

40

50

m

40.9

33.5

27.4

22.5

18.4

a Draw the graph of ln m against t. b Use your graph to estimate the value of m0 and k. c

The half-life of a radioactive substance is the time it takes to decay to half of its original mass. Find the half-life of this radioactive substance.

47

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

8 The temperature, T °C , of a hot drink, t minutes after it is made, can be modelled by the equation T = 25 + ke − nt , where k and n are constants. The table below shows experimental values of t and T . t

2

4

6

8

10

T

63.3

57.7

52.8

48.7

45.2

a Convert the equation to a form suitable for drawing a straight-line graph. b Draw the straight-line graph and use it to estimate the value of k and n. c

Estimate: i

the initial temperature of the drink

ii

the time taken for the temperature to reach 28 °C

iii the room temperature.

48

Chapter 2: Logarithmic and exponential functions

Checklist of learning and understanding The rules of logarithms ●

If y = a x then x = log a y.

●

log a a = 1, log a 1 = 0, log a a x = x , a log a x = x

●

Product rule: log a (xy ) = log a x + log a y

●

x Division rule: log a = log a x − log a y y

●

Power rule:

●

1 Special case of power rule: log a = − log a x x

log a (x ) m = m log a x

Natural logarithms

y

●

Logarithms to the base of e are called natural logarithms.

●

e ≈ 2.718

●

ln x is used to represent log e x.

1

●

If y = e x then x = ln y.

O

●

All the rules of logarithms apply for natural logarithms.

y = ex y=x y = ln x 1

x

Transforming a relationship to linear form ●

Logarithms can be used to convert relationships of the form y = kx n and y = k ( a x ) , where k , a and n are constants, into straight lines of the form Y = mX + c , where X and Y are functions of x and of y.

49

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 2 1

Solve the inequality 2 x . 7 , giving your answer in terms of logarithms.

[2]

2

Given that ln p = 2 ln q − ln(3 + q ) and that q . 0 , express p in terms of q not involving logarithms.

[3]

3

Solve the inequality 3 × 23x + 2 , 8, giving your answer in terms of logarithms.

[4]

4

Use logarithms to solve the equation 5 x + 3 = 7 x −1 , giving the answer correct to 3 significant figures.

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 November 2015 5

Solve the equation 6(4x ) − 11(2 x ) + 4 = 0, giving your answers for x in terms of logarithms where appropriate. [5]

6

Solve the equation ln(5x + 4) = 2 ln x + ln 6 .

7

[5]

ln y (6, 10.2)

(0, 2.0) ln x

O

50

The variables x and y satisfy the equation y = Kx m , where K and m are constants. The graph of ln y against ln x is a straight line passing through the points (0, 2.0) and (6, 10.2), as shown in the diagram. Find the values of K and m, correct to 2 decimal places. [5] Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2011 8

i

Given that y = 2 x , show that the equation 2 x + 3(2 − x ) = 4 can be written in the form y2 − 4 y + 3 = 0.

ii

[3]

Hence solve the equation 2 x + 3(2 − x ) = 4, giving the values of x correct to 3 significant figures where appropriate.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2010 x [3] 9 Given that (1.2)x = 6 y, use logarithms to find the value of correct to 3 significant figures. y 10 The polynomial f( x ) is defined by f( x ) = 12 x 3 + 25x 2 − 4x − 12. i ii

Show that f( −2) = 0 and factorise f( x ) completely. y

y

y

[4] y

Given that 12 × 27 + 25 × 9 − 4 × 3 − 12 = 0, state the value of 3 and hence find y correct to 3 significant figures.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2011

Chapter 2: Logarithmic and exponential functions

11 Solve the equation 4 − 2 x = 10, giving your answer correct to 3 significant figures.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2012 12 Use logarithms to solve the equation e x = 3x − 2 , giving your answer correct to 3 decimal places.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q1 November 2014 13 Using the substitution u = 3x , solve the equation 3x + 32 x = 33x giving your answer correct to 3 significant figures.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2015 14 The variables x and y satisfy the equation 5 y = 32 x − 4. a By taking natural logarithms, show that the graph of y against x is a straight line and find the exact value of the gradient of this line. [3] b This line intersects the x-axis at P and the y-axis at Q. Find the exact coordinates of the midpoint of PQ. 15

[3]

ln y

(3.1, 2.1)

(2.3, 1.7) O

x

The variables x and y satisfy the equation y = K ( b x ) , where K and b are constants. The graph of ln y against x is a straight line passing through the points (2.3, 1.7) and (3.1, 2.1), as shown in the diagram. Find the values of K and b, correct to 2 decimal places. [6]

51

52

Chapter 3 Trigonometry In this chapter you will learn how to:

■ ■

understand the relationship of the secant, cosecant and cotangent functions to cosine, sine and tangent, and use properties and graphs of all six trigonometric functions for angles of any magnitude use trigonometrical identities for the simplification and exact evaluation of expressions and in the course of solving equations, and select an identity or identities appropriate to the context, showing familiarity in particular with the use of: sec 2 θ = 1 + tan2 θ and cosec 2 θ = 1 + cot 2 θ the expansion of sin( A ± B ), cos( A ± B ) and tan( A ± B ) the formulae for sin 2 A, cos 2 A and tan 2 A the expression of a sin θ + b cos θ in the forms R sin(θ ± α ) and R cos(θ ± α ).

• • • •

Chapter 3: Trigonometry

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

Pure Mathematics 1 Coursebook, Chapter 5

Sketch graphs of the sine, cosine and tangent functions.

1 Sketch the graphs of these functions for 0° ø x ø 360°. a b c

y = 2 sin x + 1 y = tan( x − 90°) y = cos(2 x + 90°)

Pure Mathematics 1 Coursebook, Chapter 5

Find exact values of the sine, cosine and tangent of 30° , 45° and 60° and related angles.

2 Find the exact values of:

Pure Mathematics 1 Coursebook, Chapter 5

Solve trigonometric equations using sin θ the identities tan θ ≡ and cos θ sin2 θ + cos2 θ ≡ 1.

3 Solve for 0° ø x ø 360° .

a sin120° b cos 225° c tan150° a 5 sin x = 3 cos x b 2 cos2 x − sin x = 1

Why do we study trigonometry?

REWIND

In addition to the three main trigonometrical functions (sine, cosine and tangent) there are three more commonly used trigonometrical functions (cosecant, secant and cotangent) that we will learn about. These are sometimes referred to as the reciprocal trigonometrical functions. We will also learn about the compound angle formulae, which form a basis for numerous important mathematical techniques. These techniques include solving equations, deriving further trigonometric identities, the addition of different sine and cosine functions and deriving rules for differentiating and integrating trigonometric functions, which we will cover later in this book. As you know, scientists and engineers represent oscillations and waves using trigonometric functions. These functions also have further uses in navigation, engineering and physics.

3.1 The cosecant, secant and cotangent ratios There are a total of six trigonometric ratios. You have already learnt about the ratios sine, cosine and tangent. In this section you will learn about the other three ratios, which are cosecant (cosec), secant (sec) and cotangent (cot). KEY POINT 3.1 The three reciprocal trigonometric ratios are defined as: cosec θ =

1 sin θ

sec θ =

1 cos θ

Consider the right-angled triangle: y x y sin θ = cos θ = tan θ = r r x r r x cosec θ = sec θ = cot θ = y x y

cot θ =

1 cos θ = tan θ sin θ

r y θ x

This chapter builds on the work we covered in the Pure Mathematics 1 Coursebook, Chapter 5.

WEB LINK Explore the Trigonometry: Compound angles station on the Underground Mathematics website.

53

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

The following identities can be found from this triangle: x 2 + y2 = r 2

Divide both sides by x 2 .

2

y r 1+ = x x

2

Use

r y = sec θ . = tan θ and x x

1 + tan2 θ ≡ sec 2 θ x 2 + y2 = r 2 2

Divide both sides by y2 .

x r y + 1 = y

2

Use

x r = cot θ and = cosec θ . y y

cot 2 θ + 1 ≡ cosec 2 θ We will need these important identities to solve trigonometric equations later in this section. KEY POINT 3.2 1 + tan2 θ ≡ sec 2 θ cot 2 θ + 1 ≡ cosec 2 θ

The graphs of y = sin θ and y = cosec θ y y = cosec θ

54 1

y

1

y = sin θ

O

θ 90

180

270

O 90

360

–1

–1

The function y = cosec θ is the reciprocal of the function y = sin θ . The function y = sin θ is zero when θ = 0°, 180°, 360°, … , this means that y = cosec θ 1 is not defined at each of these points because is not defined. Hence vertical asymptotes 0 need to be drawn at each of these points.

180

270

360 θ

Chapter 3: Trigonometry

The graphs of y = cos θ and y = sec θ We can find the graph of y = sec θ in a similar way from the graph of y = cos θ . y y = sec θ y 1

1

y = cos θ

O 180

90

270

O

360 θ

180

90

–1

270

360 θ

–1

The graphs of y = tan θ and y = cot θ We can find the graph of y = cot θ from the graph of y = tan θ . y

y y = tan θ

y = cot θ

O 90

180

270

55

O

360 θ

180

90

WORKED EXAMPLE 3.1 Find the exact value of cosec 240° . Answer 1 sin 240° 1 = –sin 60° 1 = 3 − 2 2 = − 3

cosec 240° =

y

A

S 240° 60° O

T

x

C

270

360 θ

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 3.2 Solve sec 2 x − tan x − 3 = 0 for 0° ø x ø 360°. Answer sec 2 x − tan x − 3 = 0

Use 1 + tan2 x ≡ sec 2 x.

tan2 x − tan x − 2 = 0

Factorise.

(tan x − 2)(tan x + 1) = 0 tan x = 2 or tan x = −1

y 3

y = tan x y=2

2

tan x = 2 ⇒ x = 63.4° (calculator) or x = 63.4° + 180° = 243.4°

1

tan x = −1 ⇒ x = − 45° (not in required range) or x = − 45° + 180° = 135° or x = 135° + 180° = 315°

–45 –90

63.4 90

O

180

270

x

360

y = –1

–1 –2

The values of x are 63.4°, 135°, 243.4°, 315°.

–3

EXERCISE 3A 56

1 Find the exact values of: a sec 60° e

cosec135°

2 Find the exact values of: π a cosec 6 5π e sec 6

b cosec 45° f

cot 330°

b cot f

π 3

cosec

c

g sec 150°

d sec 300° h cot( −30°).

π 4 4π cot 3

2π 3 π h sec − . 6

c

cosec x = −3

d

3sec x − 4 = 0

c

cot x = 2

d

2 cot x + 5 = 0

c

cot 2 x = 1

d

2 sec 2 x = 3

b

sec(2 x + 60°) = −1.5 for 0° ø x ø 180°

d

2 cosec(2 x − 1) = 3

c 7π 4

cot120°

g

sec

d cot

3 Solve each equation for 0° ø x ø 360°. a sec x = 3

b

cot x = 0.8

4 Solve each equation for 0 ø x ø 2π . a cosec x = 2

b

sec x = −1

5 Solve each equation for 0° ø x ø 180°. a cosec 2 x = 1.2

b

sec 2 x = 4

6 Solve each equation for the given domains. a c

cosec( x − 30°) = 2 for 0° ø x ø 360° π cot x + = 2 for 0 ø x ø 2 π 4

for − π ø x ø π

Chapter 3: Trigonometry

7 Solve each equation for −180° ø x ø 180°. cosec 2 x = 4

b

sec 2 x = 9

c

d sec x = cos x

e

cosec x = sec x

f

a

1 x = 4 2 2 tan x = 3 cosec x

9 cot 2

8 Solve each equation for 0° ø θ ø 360° . a 2 tan2 θ − 1 = sec θ c

2 cot 2 θ − cosec θ = 13

e tan2 θ + 3sec θ = 0

b

3 cosec 2 θ = 13 − cot θ

d

cosec θ + cot θ = 2 sin θ

f

3 sec 2 θ = cosec θ

9 Solve each equation for 0° ø θ ø 180°. a sec θ = 3 cos θ − tan θ c

sec 4 θ + 2 = 6 tan2 θ

b

2 sec 2 2θ = 3tan 2θ + 1

d

2 cot 2 2θ + 7 cosec 2θ = 2

b

3 cot 2 θ + 5 cosec θ + 1 = 0

10 Solve each equation for 0 ø θ ø 2 π . a

tan2 θ + 3sec θ + 3 = 0

11 a Sketch each of the following functions for the interval 0 ø x ø 2π . i

y = 1 + sec x

ii

y = cot 2 x

π iii y = 2 cosec x − 2

1 π x vi y = sec 2 x + 2 4 b Write down the equation of each of the asymptotes for your graph for part a vi. iv y = 1 − sec x

P

P

y = 1 + cosec

12 Prove each of these identities. a

sin x + cos x cot x ≡ cosec x

b

cosec x − sin x ≡ cos x cot x

c

sec x cosec x − cot x ≡ tan x

d

(1 + sec x )(cosec x − cot x ) ≡ tan x

b

sec 2 x + sec x tan x ≡

13 Prove each of these identities. 1 ≡ sin x cos x a tan x + cot x c e g

PS

v

1 − cos2 x ≡ 1 − sin2 x sec 2 x − 1 sin x ≡ cosec x 1 − cos2 x 1 1 + ≡ 2 cosec 2 x 1 + cos x 1 − cos x

d f h

1 1 − sin x

1 + tan2 x ≡ sec x cosec x tan x 1 + sin x 2 ≡ ( tan x + sec x ) 1 − sin x cos x cos x + ≡ 2 sec x 1 + sin x 1 − sin x

14 Solve each equation for 0° ø θ ø 180°. a

6 sec 3 θ − 5 sec 2 θ − 8 sec θ + 3 = 0

b

2 cot 3 θ + 3 cosec 2 θ − 8 cot θ = 0

57

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3.2 Compound angle formulae EXPLORE 3.1 1 Sunita says that: sin( A + B ) = sin A + sin B Use a calculator with A = 20° and B = 50° to prove that Sunita is wrong. 2 The diagram shows ∆PQR .

R A B

RX is perpendicular to PQ. ∠A and ∠B are acute angles.

q

h

p

QR = p, PR = q and RX = h. a Write down an expression for h in terms of: i q and A

P

X

Q

p and B 1 b Using the formula area of triangle = ab sin C , we can write: 2 1 Area of ∆PQR = pq sin( A + B ). 2 ii

Find an expression in terms of p, q, A and B for the area of: i c

ii

∆PRX

∆QRX

Use your expressions for the areas of triangles PQR, PRX and QRX to prove that: sin( A + B ) ≡ sin A cos B + cos A sin B

58

3 In the identity sin( A + B ) ≡ sin A cos B + cos A sin B , replace B by − B to show that: sin( A − B ) ≡ sin A cos B − cos A sin B 4 In the identity sin( A − B ) ≡ sin A cos B − cos A sin B , replace A by (90° − A) to show that: cos(A + B ) ≡ cos A cos B − sin A sin B 5 In the identity cos( A + B ) ≡ cos A cos B − sin A sin B , replace B by − B to show that: cos(A − B ) ≡ cos A cos B + sin A sin B 6 By writing tan( A + B ) as tan( A + B ) ≡

sin( A + B ) sin( A − B ) and tan( A − B ) as , show that: cos( A + B ) cos( A − B )

tan A − tan B tan A + tan B and tan( A − B ) ≡ 1+ tan A tan B 1− tan A tan B

KEY POINT 3.3 Summarising, the six compound angle formulae are: sin( A + B ) ≡ sin A cos B + cos A sin B sin( A − B ) ≡ sin A cos B − cos A sin B cos( A + B ) ≡ cos A cos B − sin A sin B cos( A − B ) ≡ cos A cos B + sin A sin B tan A + tan B tan( A + B ) ≡ 1 − tan A tan B tan A − tan B tan( A − B ) ≡ 1 + tan A tan B

Chapter 3: Trigonometry

WORKED EXAMPLE 3.3 Find the exact value of: a sin105°

b

cos 42° cos12° + sin 42° sin12°

Answer a

sin105° = sin(60° + 45°) = sin 60° cos 45° + cos 60° sin 45° 3 1 1 1 = + 2 2 2 2 =

b

3 +1 2 2

cos 42° cos12° + sin 42° sin12°

Use cos ( A − B ) ≡ cos A cos B + sin A sin B.

= cos(42° − 12°) = cos 30° =

3 2

WORKED EXAMPLE 3.4 Given that sin A = a sin( A + B )

4 5 and cos B = , where A is obtuse and B is acute, find the value of: 5 13 b cos( A − B ) c tan( A − B )

Answer y

y S 5 4 –3 T

a

B 12

O

O

x

4 3 4 , cos A = − , tan A = − 5 5 3

sin( A + B ) = sin A cos B + cos A sin B 4 5 3 12 = + − 5 13 5 13 16 =− 65

b

13

A

C

sin A =

A

S

A

cos( A − B ) = cos A cos B + sin A sin B 3 5 4 12 = − + 5 13 5 13 33 = 65

T

sin B =

x

5

C

12 5 12 , tan B = , cos B = 5 13 13

59

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

c

tan( A − B ) =

tan A − tan B 1 + tan A tan B

− 4 − 12 3 5 = 4 12 1+ − 3 5 =

56 33

WORKED EXAMPLE 3.5 Solve the equation sin(60° − x ) = 2 sin x for 0° , x , 360° . Answer sin(60° − x ) = 2 sin x

Expand sin(60° − x ). Use sin 60° =

sin 60° cos x − cos 60° sin x = 2 sin x 3 1 cos x − sin x = 2 sin x 2 2

3 1 and cos 60° = . 2 2

Multiply both sides by 2.

3 cos x − sin x = 4 sin x

Add sin x to both sides.

3 cos x = 5 sin x

Rearrange to find tan x .

60

3 = tan x 5 x = 19.1° or x = 180° + 19.1° The values of x are 19.1° and 199.1° .

EXERCISE 3B 1 Expand and simplify cos( x + 30°). 2 Without using a calculator, find the exact value of each of the following: a sin 20° cos 70° + cos 20° sin 70°

b sin172° cos 37° − cos172° sin 37°

c

cos 25° cos 35° − sin 25° sin 35°

d cos 99° cos 69° + sin 99° sin 69°

e

tan 25° + tan 20° 1 − tan 25° tan 20°

f

tan 82° − tan 52° 1 + tan 82° tan 52°

b tan 75°

c

cos105°

19 π 12

g

cot

3 Without using a calculator, find the exact value of: a sin 75° e

sin

π 12

4 Given that cos x =

f

cos

7π 12

d tan( −15°) h sin

7π 12

4 and that 0° , x , 90° , without using a calculator find the exact value of cos( x − 60°). 5

Chapter 3: Trigonometry

4 5 63 , where A and B are acute, show that sin( A + B ) = and sin B = . 5 13 65 12 3 and sin B = , where A is obtuse and B is acute, find the value of: 6 Given that sin A = 13 5 b cos( A − B ) c tan( A + B ) a sin( A + B )

5 Given that sin A =

7 Given that cos A = − a sin( A + B )

4 8 and sin B = − and that A and B lie in the same quadrant, find the value of: 5 17 b cos( A + B ) c tan( A − B )

8 Given that tan A = t and that tan( A − B ) = 2, find tan B in terms of t. 9 If cos( A − B ) = 3 cos( A + B ), find the exact value of tan A tan B. 10 a Given that 8 + cosec 2 θ = 6 cot θ , find the value of tan θ . b Hence find the exact value of tan(θ + 45°). 11 a Given that x is acute and that 2 sec 2 x + 7 tan x = 17, find the exact value of tan x . b Hence find the exact value of tan(225° − x ). 12 a Show that the equation sin( x + 30°) = 5 cos( x − 60°) can be written in the form cot x = − 3 . b Hence solve the equation sin( x + 30°) = 5 cos( x − 60°) for −180° , x , 180°. 13 Solve each equation for 0° ø x ø 360°. a

cos( x + 30°) = 2 sin x

b

cos( x − 60° ) = 3 cos x

c

sin(30° − x ) = 4 sin x

d

cos( x + 30°) = 2 sin( x + 60°)

14 Solve each equation for 0° ø x ø 180°. a

2 tan(60° − x ) = tan x

b

2 tan(45° − x ) = 3tan x

c

sin( x + 60°) = 2 cos( x + 45°)

d

sin( x − 45°) = 2 cos( x + 60°)

e

tan( x + 45°) = 6 tan x

f

cos( x + 225°) = 2 sin( x − 60°)

15 Solve the equation tan( x − 45°) + cot x = 2 for 0° ø x ø 180°. P

16 Use the expansions of cos(5x + x ) and cos(5x − x ) to prove that cos 6x + cos 4x ≡ 2 cos 5x cos x.

PS

17 Given that sin x + sin y = p and cos x + cos y = q, find an expression for cos( x − y ), in terms of p and q.

3.3 Double angle formulae We can use the compound angle formulae to derive three new identities. Consider

sin( A + B ) ≡ sin A cos B + cos A sin B

If B = A,

sin( A + A) ≡ sin A cos A + cos A sin A ∴ sin 2 A ≡ 2 sin A cos A

Consider

cos( A + B ) ≡ cos A cos B − sin A sin B

If B = A,

cos( A + A) ≡ cos A cos A − sin A sin A ∴ cos 2 A ≡ cos2 A − sin2 A

61

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

tan A + tan B 1 − tan A tan B tan A + tan A tan( A + A) ≡ 1 − tan A tan A 2 tan A ∴ tan 2 A ≡ 1 − tan2 A tan( A + B ) ≡

Consider If B = A,

These three formulae are known as the double angle formulae. We can use the identity sin2 A + cos2 A ≡ 1 to write the identity cos 2 A ≡ cos2 A − sin2 A in two alternative forms: Using sin2 A ≡ 1 − cos2 A gives: cos 2 A ≡ 2 cos2 A − 1 Using cos2 A ≡ 1 − sin2 A gives: cos 2 A ≡ 1 − 2 sin2 A Summarising, the double angle formulae are: KEY POINT 3.4 sin 2 A ≡ 2 sin A cos A

cos 2 A ≡ cos2 A − sin2 A

tan 2 A ≡

2 tan A 1 − tan2 A

≡ 2 cos2 A − 1 ≡ 1 − 2 sin2 A

62

WORKED EXAMPLE 3.6

Given that sin x = − a sin 2 x

3 and that 180° , x , 270°, find the exact value of: 5 x b cos 2 x c tan 2

Answer y A

S

–4 x

O

–3

5

From the diagram: 3 3 4 sin x = − , cos x = − and tan x = 5 5 4

C

T

a

sin 2 x = 2 sin x cos x 3 4 24 = 2 − − = 5 5 25

b

cos 2 x = cos2 x − sin2 x 2

2

4 3 7 = − − − = 5 5 25

Chapter 3: Trigonometry

x 2 c tan x = 2 x 1 − tan 2 x 2 tan 3 2 = 4 1 − tan2 x 2 x x 2 x 3tan + 8 tan − 3 = 0 2 2 x x 3tan − 1 tan + 3 = 0 2 2 x 1 x tan = or tan = −3 2 3 2 2 tan

x

If 180° , x , 270° , then 90° ,

Use 2 A = x in the double angle formula.

Rearrange.

Factorise. Solve.

x , 135°, which means 2

x is in the second quadrant. 2 x ∴ tan = −3 2 that

WORKED EXAMPLE 3.7 63

Solve the equation sin 2 x = sin x for 0° ø x ø 360°. Answer sin 2 x = sin x 2 sin x cos x = sin x 2 sin x cos x − sin x = 0

Use sin 2 x = 2 sin x cos x. Rearrange. Factorise.

sin x(2 cos x − 1) = 0 sin x = 0 or cos x =

1 2

sin x = 0 ⇒ x = 0°, 180°, 360° cos x =

1 ⇒ x = 60° 2 or x = 360° − 60° = 300°

The values of x are 0°, 60°, 180°, 300°, 360°.

y 1 y=

0.5 O

60 90

180

270

–0.5 y = cos x –1

When solving equations involving cos 2 x it is important that we choose the most suitable expansion of cos 2 x for solving that particular equation.

360 x

1 2

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 3.8 Solve the equation cos 2 x + 3sin x = 2 for 0 ø x ø 2 π. Answer cos 2 x + 3sin x = 2

Use cos 2 x = 1 − 2 sin2 x .

1 − 2 sin2 x + 3sin x = 2

Rearrange.

2 sin2 x − 3sin x + 1 = 0

Factorise.

(2 sin x − 1)(sin x − 1) = 0 sin x = sin x =

y

O

π 5π or x = π − = 6 6

–0.5

x π 6

π 2

π

π π 5π , , . 6 2 6

64

EXERCISE 3C 1 Express each of the following as a single trigonometric ratio. 2 tan17° a 2 sin 28° cos 28° b 2 cos2 34° − 1 c 1 − tan2 17° 4 2 Given that tan x = , where 0° , x , 90°, find the exact value of: 3 a sin 2 x b cos 2 x c tan 2 x 3 Given that cos 2 x = − a sin 2 x 4 Given that cos x = − a sin 2 x

d tan 3x

527 and that 0° , x , 90°, find the exact value of: 625 b tan 2 x c cos x d tan x

3 and that 90° , x , 180°, find the exact value of: 5 1 b sin 4x c tan 2 x d tan x 2

5 Given that 3 cos 2 x + 17 sin x = 8, find the exact value of sin x. 6 Solve each equation for 0° ø θ ø 360°. a

2 sin 2θ = cos θ

3π 2

2π y = sin x

–1

π 2

The values of x are

y=

0.5

1 π ⇒ x = 2 6

sin x = 1 ⇒ x =

y=1

1

1 or sin x = 1 2

b

2 cos 2θ + 3 = 4 cos θ

c

2 cos 2θ + 1 = sin θ

1 2

Chapter 3: Trigonometry

7 Solve each equation for 0° ø θ ø 180°. a

2 sin 2θ tan θ = 1

b

3 cos 2θ + cos θ = 2

c

2 cosec 2θ + 2 tan θ = 3sec θ

d

2 tan 2θ = 3 cot θ

e

tan 2θ = 4 cot θ

f

cot 2θ + cot θ = 3

8 Express cos2 2 x in terms of cos 4x. 9 a Use the expansions of cos(2 x + x ) to show that cos 3x ≡ 4 cos3 x − 3 cos x. b Express sin 3x in terms of sin x. 10 Solve tan 2θ + 2 tan θ = 3 cot θ for 0° ø θ ø 180°. 11 a Prove that tan θ + cot θ ≡

2 . sin 2θ

b Hence find the exact value of tan

π π + cot . 12 12

12 a Prove that 2 cosec 2θ tan θ ≡ sec 2 θ . b Hence solve the equation cosec 2θ tan θ = 2 for −π , θ , π. 13 a Prove that cos 4x + 4 cos 2 x ≡ 8 cos 4 x − 3. b Hence solve the equation 2 cos 4x + 8 cos 2 x = 3 for −π ø x ø π. 14 a Show that θ = 18° is a root of the equation sin 3θ = cos 2θ .

65

b Express sin 3θ and cos 2θ in terms of sin θ . c

Hence show that sin18° is a root of the equation 4x 3 − 2 x 2 − 3x + 1 = 0.

d Hence find the exact value of sin18°. PS

15 Find the range of values of θ between 0 and 2 π for which cos 2θ . cos θ .

PS

16 Solve the inequality cos 2θ − 3sin θ − 2 ù 0 for 0° ø θ ø 360°.

PS

17

A b

B

3x

x a

C

Prove that a = 4b cos 2 x cos x. PS

18 a Use the expansions of cos(2 x + x ) and cos(2 x − x ) , to prove that: cos 3x + cos x ≡ 2 cos 2 x cos x b Solve cos 3x + cos 2 x + cos x . 0 for 0° , x , 360°.

PS

19 Solve the inequality cos 4θ + 3 cos 2θ + 1 , 0 for 0° ø θ ø 360°.

TIP Solving trigonometric inequalities is not in the Cambridge syllabus. You should, however, have the skills necessary to tackle these more challenging questions.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3.4 Further trigonometric identities This section builds on our previous work by using the compound angle formulae and the double angle formulae to prove trigonometric identities. WORKED EXAMPLE 3.9

Prove the identity tan 3x ≡

3tan x − tan3 x . 1 − 3tan2 x

Answer LHS ≡ tan(2 x + x ) ≡

≡

Use the compound angle formula for tan.

tan 2 x + tan x 1 − tan 2 x tan x 2 tan x + tan x 1 − tan2 x 2 tan x 1 − tan x 2 1 − tan x

≡

2 tan x + tan x(1 − tan2 x ) 1 − tan2 x − 2 tan2 x

≡

3tan x − tan3 x 1 − 3tan2 x

66

Use the double angle formula. Multiply numerator and denominator by (1 − tan2 x ).

Expand the brackets and simplify.

≡ RHS

WORKED EXAMPLE 3.10

Prove the identity

2 sin( x − y ) ≡ cot y − cot x. cos( x − y ) − cos( x + y )

Answer 2 sin( x − y ) cos( x − y ) − cos( x + y ) 2 sin x cos y − 2 cos x sin y ≡ (cos x cos y + sin x sin y ) − (cos x cos y − sin x sin y )

LHS ≡

≡

2 sin x cos y − 2 cos x sin y 2 sin x sin y

2 sin x cos y 2 cos x sin y − 2 sin x sin y 2 sin x sin y cos y cos x ≡ − sin y sin x ≡ cot y − cot x ≡ RHS ≡

Use compound angle formulae. Simplify denominator. Separate fractions. Simplify fractions.

Chapter 3: Trigonometry

EXPLORE 3.2

Odd one out 1 − cos 2x sin 2x sin 2x + sin x cos 2x + cos x + 1

cosec 2x − cot 2x 1 cosec 2x + cot 2x

2

sin 2x 1 + cos 2x

sec2 x sin 2x cot x − 2 cot 2x

Find the trigonometric expression that does not match the other six expressions. Create as many expressions of your own to match the ‘odd one out’. (Your expressions must contain at least two different trigonometric ratios.) Compare your answers with your classmates.

EXERCISE 3D 67

P

1 Prove each of these identities. a

tan A + cot A ≡ 2 cosec 2 A

b 1 − tan2 A ≡ cos 2 A sec 2A

c

tan 2 A − tan A ≡ tan A sec 2 A 1 cot 2 A + tan A ≡ cosec A sec A 2 sec 2A cosec 2 A ≡ 4 cosec 2 2 A

d (cos A + 3sin A)2 ≡ 5 − 4 cos 2 A + 3sin 2 A

e g P

2 Prove each of these identities. sin A cos A 2 cos( A − B ) a + ≡ cos B sin B sin 2 B

f

cosec 2 A + cot 2 A ≡ cot A

h sin( A + B ) sin( A − B ) ≡ sin2 A − sin2 B

b

cos A + sin A ≡ sec 2 A + tan 2 A cos A − sin A

c

1 − tan2 A ≡ cos 2 A 1 + tan2 A

d

cos 2 A + sin 2 A − 1 ≡ tan A cos 2 A − sin 2 A + 1

e

sin 3A + sin A ≡ cos A 2 sin 2 A

f

cos 3A − sin 3A ≡ cos A + sin A 1 − 2 sin 2 A

g

cos 2 A + 9 cos A + 5 ≡ 2 cos A + 1 4 + cos A

h

cos3 A − sin3 A 2 + sin 2 A ≡ cos A − sin A 2

P

3 Use the fact that 4 A = 2 × 2 A to show that: sin 4 A ≡ 8 cos3 A − 4 cos A b cos 4 A + 4 cos 2 A ≡ 8 cos 4 A − 3 a sin A

P

4 Prove the identity 8 sin2 x cos2 x ≡ 1 − cos 4x.

P

1 5 Prove the identity (2 sin A + cos A) ≡ (4 sin 2 A − 3 cos 2 A + 5). 2

P

6 Use the expansions of cos(3x − x ) and cos(3x + x ) to prove the identity:

2

cos 2 x − cos 4x ≡ 2 sin 3x sin x

WEB LINK Try the Equation or identity?(II) resource on the Underground Mathematics website.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3.5 Expressing a sin θ + b cos θ in the form R sin(θ θ ± α ) or R cos(θθ ± α ) In this section you will learn how to solve equations of the form a sin θ + b cos θ = c. EXPLORE 3.3 Use graphing software to confirm that the graph of y = 3sin θ + 4 cos θ for −90° ø θ ø 360° is: y 5

−90

y = 3 sin θ + 4 cos θ θ

O

180

90

270

360

−5

1 What are the amplitude and period of this graph? 2 Discuss with your classmates how this graph can be obtained by transforming the graph of: y = sin θ

a

b

y = cos θ

3 Use your knowledge of transformations of functions to write the function y = 3sin θ + 4 cos θ in the form: y = R sin(θ + α )

a

b R sin(θ + β )

(You will need to use your graph to find the approximate values of α and β .)

68

The formula for replacing a sin θ + b cos θ by R sin(θ + α ) is derived as follows. Let a sin θ + b cos θ = R sin(θ + α ) where R . 0 and 0° , α , 90°. R sin(θ + α ) = R(sin θ cos α + cos θ sin α ) ∴ a sin θ + b cos θ = R sin θ cos α + R cos θ sin α Equating coefficients of sin θ :

R cos α = a

R sin α = b Equating coefficients of cos θ : sin α b b = ⇒ tan α = (2) ÷ (1): cos α a a Squaring equations (1) and (2) and then adding gives:

(1) (2)

a 2 + b2 = R 2 cos2 α + R 2 sin2 α = R 2 (cos2 α + sin2 α )

Use cos2 α + sin2 α = 1.

= R2 Hence, R =

a 2 + b2

∴ a sin θ + b cos θ can be written as R sin(θ + α ) where R =

a 2 + b2 and tan α =

b a

Chapter 3: Trigonometry

EXPLORE 3.4 Using a similar approach, show that: 1

a sin θ − b cos θ ≡ R sin(θ − α ) where R =

2

a cos θ + b sin θ ≡ R cos(θ − α ) where R =

3

a cos θ − b sin θ ≡ R cos(θ + α ) where R =

b a b a 2 + b2 and tan α = a b a 2 + b2 and tan α = a

a 2 + b2 and tan α =

Summarising the results from Explore 3.4: KEY POINT 3.5 a sin θ ± b cos θ ≡ R sin(θ ± α ) and a cos θ ± b sin θ ≡ R cos(θ ∓ α ) where R =

a 2 + b 2 , tan α =

b and 0° , α , 90°. a

WORKED EXAMPLE 3.11 a Express 2 sin θ − 3 cos θ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°. Give the value of α correct to 2 decimal places. b Hence solve the equation 2 sin θ − 3 cos θ = 2 for 0° , θ , 360°. Answer a

2 sin θ − 3 cos θ = R sin(θ − α ) ∴ 2 sin θ − 3 cos θ = R sin θ cos α − R cos θ sin α Equating coefficients of sin θ : R cos α = 2

(1)

Equating coefficients of cos θ : R sin α = 3

(2)

3 (2) ÷ (1): tan α = ⇒ α = 56.31° 2 Squaring equations (1) and (2) and then adding gives: 32 + 22 = R 2 ⇒ R = 13 ∴ 2 sin θ − 3 cos θ = 13 sin(θ − 56.31°) b

2 sin θ − 3 cos θ = 2 13 sin(θ − 56.31°) = 2 2 13 θ − 56.31° = 23.09°, 180° − 23.09° = 23.09°, 156.91° θ = 79.4°, 213.2°

sin(θ − 56.31°) =

69

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

We can find the maximum and minimum values of the expression a sin θ + b cos θ by writing the expression in the form R sin(θ + α ). From the Pure Mathematics Coursebook 1, Chapter 5, Section 5.4, we know that −1 ø sin(θ + α ) ø 1, hence −R ø R sin(θ + α ) ø R . This can be summarised as: KEY POINT 3.6 The maximum value of a sin θ + b cos θ is

a 2 + b 2 and occurs when sin(θ + α ) = 1.

The minimum value of a sin θ + b cos θ is − a 2 + b 2 and occurs when sin(θ + α ) = −1.

WORKED EXAMPLE 3.12 Express 2 cos θ − sin θ in the form R cos(θ + α ), where R . 0 and 0° , α , 90°. Hence, find the maximum and minimum values of the expression 2 cos θ − sin θ and the values of θ in the interval 0° , θ , 360° for which these occur. Answer 2 cos θ − sin θ = R cos(θ + α ) ∴ 2 cos θ − sin θ = R cos θ cos α − R sin θ sin α 70

Equating coefficients of cos θ : R cos α = 2

(1)

Equating coefficients of sin θ : R sin α = 1 1 ⇒ α = 26.57° (2) ÷ (1): tan α = 2 Squaring equations (1) and (2) and then adding gives:

(2)

22 + 12 = R 2 ⇒ R = 5 ∴ 2 cos θ − sin θ = 5 cos ( θ + 26.57° ) 22 + ( −1)2 and occurs when cos (θ + 26.57°) = 1 θ + 26.57° = 360° θ = 333.4° Minimum value is − 22 + ( −1)2 and occurs when cos(θ + 26.57°) = −1 θ + 26.57° = 180° θ = 153.4°

Maximum value is

∴ Max value =

5 when θ = 333.4°, min value = − 5 when θ = 153.4°

DID YOU KNOW? The French mathematician Joseph Fourier (1768−1830) found a way to represent any wave-like function as the sum (possibly infinite) of simple sine waves. The sum of sine functions is called a Fourier series. These series can be applied to vibration problems. Fourier is also credited with discovering the greenhouse effect.

Chapter 3: Trigonometry

EXERCISE 3E 1 a Express 15 sin θ − 8 cos θ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°. Give the value of α correct to 2 decimal places. b Hence solve the equation 15 sin θ − 8 cos θ = 10 for 0° , θ , 360°. 2 a Express 2 cos θ − 3sin θ in the form R cos(θ + α ), where R . 0 and 0° , α , 90°. Give the exact value of R and the value of α correct to 2 decimal places. b Hence solve the equation 2 cos θ − 3sin θ = 1.3 for 0° , θ , 360°. 3 a Express 15 sin θ − 8 cos θ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°. Give the value of α correct to 2 decimal places. b Hence solve the equation 15 sin θ − 8 cos θ = 3 for 0° , θ , 360°. c

Find the greatest value of 30 sin θ − 16 cos θ as θ varies.

4 a Express 4 sin θ − 6 cos θ in the form R sin(θ − α ), where R . 0 and 0° , α , 90°. Give the exact value of R and the value of α correct to 2 decimal places. b Hence solve the equation 4 sin θ − 6 cos θ = 3 for 0° , θ , 180°. c

Find the greatest and least values of (4 sin θ − 6 cos θ )2 − 3 as θ varies.

5 a Express 3sin θ + 4 cos θ in the form R sin(θ + α ), where R . 0 and 0° , α , 90°. Give the value of α correct to 2 decimal places. b Hence solve the equation 3sin θ + 4 cos θ = 2 for 0° , θ , 360°. c

Find the least value of 3sin θ + 4 cos θ + 3 as θ varies.

6 a Express cos θ + 3 sin θ in the form R cos (θ − α ), where R . 0 and 0 , α , Give the exact values of R and α . π 1 1 ≡ sec 2 θ − . b Hence prove that 2 ( cos θ + 3 sin θ ) 4 3

π . 2

7 a Express 8 sin 2θ + 4 cos 2θ in the form R sin(2θ + α ), where R . 0 and 0° , α , 90°. Give the exact value of R and the value of α correct to 2 decimal places. b Hence solve the equation 8 sin 2θ + 4 cos 2θ = 3 for 0° , θ , 360°. 10 as θ varies. c Find the least value of (8 sin 2θ + 4 cos 2θ )2 8 a Express cos θ − 2 sin θ in the form R cos(θ + α ) , where R . 0 and 0° , α , 90° . Give the exact value of R and the value of α correct to 2 decimal places. b Hence solve the equation cos θ − 2 sin θ = −1 for 0° , θ , 360°. 1 c Find the least value of 2 as θ varies. ( 2 cos θ − 2 sin θ )

71

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

9 a Express cos θ − sin θ in the form R cos(θ + α ), where R . 0 and 0 , α , Give the exact values of R and α . 1 b Show that one solution of the equation cos θ − sin θ = 2 interval 0 < θ < 2 π. c

6 is θ =

π . 2

19 π and find the other solution in the 12

State the set of values of k for which the equation cos θ − sin θ = k has any solutions.

10 a Express sin θ − 3 cos θ in the form R sin(θ − α ), where R > 0 and 0° , α , 90°. Give the exact value of R and the value of α correct to 2 decimal places. b Hence solve the equation sin θ − 3 cos θ = −2 for 0° , θ , 360°. c

Find the greatest possible value of 1 + sin 2θ − 3 cos 2θ as θ varies and determine the smallest positive value of θ for which this greatest value occurs.

11 a Express

5 cos θ + 2 sin θ in the form R cos(θ − α ), where R . 0 and 0° , α , 90°.

Give the value of α correct to 2 decimal places. b Find the smallest positive angle θ that satisfies the equation c

Find the smallest positive angle θ that satisfies the equation

5 cos θ + 2 sin θ = 3. 1 1 5 cos θ + 2 sin θ = −1. 2 2

12 a Given that 3sec θ + 4 cosec θ = 2 cosec 2θ , show that 3sin θ + 4 cos θ = 1. b Express 3sin θ + 4 cos θ in the form R sin(θ + α ) , where R . 0 and 0° , α , 90°. Give the value of α correct to 2 decimal places.

72

c PS

Hence solve the equation 3sec θ + 4 cosec θ = 2 cosec 2θ for 0° , θ , 360°.

13 a Express sin(θ + 30°) + cos θ in the form R sin(θ + α ), where R . 0 and 0° , α , 90°. Give the exact values of R and α . b Hence solve the equation sin(θ + 30°) + cos θ = 1 for 0° , θ , 360°.

PS

14 a Find the maximum and minimum values of 7 sin2 θ + 9 cos2 θ + 4 sin θ cos θ + 2. b Hence, or otherwise, solve the equation 7 sin2 θ + 9 cos2 θ + 4 sin θ cos θ = 10 for 0° , θ , 360°.

Chapter 3: Trigonometry

Checklist of learning and understanding Cosecant, secant and cotangent cosec θ =

1 sin θ

sec θ =

y

1 cos θ

cot θ =

1 tan θ

y

y y = cosec θ

y = sec θ y = cot θ

1 O −1

1 θ 180

90

270

360

O −1

θ 90

180

270

360

O

θ 90

180

270

360

Trigonometric identities ●

1 + tan2 x ≡ sec 2 x

●

1 + cot 2 x ≡ cosec 2 x

Compound angle formula ●

sin( A + B ) ≡ sin A cos B + cos A sin B

●

sin( A − B ) ≡ sin A cos B − cos A sin B

●

cos( A + B ) ≡ cos A cos B − sin A sin B

●

cos( A − B ) ≡ cos A cos B + sin A sin B

●

●

tan A + tan B 1 − tan A tan B tan A − tan B tan( A − B ) ≡ 1 + tan A tan B tan( A + B ) ≡

Double angle formulae ●

sin 2 A ≡ 2 sin A cos A

●

cos 2 A ≡ cos2 A − sin2 A ≡ 1 − 2 sin2 A ≡ 2 cos2 A − 1

●

tan 2 A ≡

2 tan A 1 − tan2 A

θ ± α ) or R cos (θθ ± α ) Expressing a sin θ + b cos θ in the form R sin(θ ●

a sin θ ± b cos θ = R sin(θ ± α )

●

a cos θ ± b sin θ = R cos(θ ∓ α ) where R =

a 2 + b 2 and tan α =

b a

73

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 3 1

Sketch the graph of y = 3sec(2 x − 90°) for 0° , x , 180°.

[3]

2

By expressing the equation cosec θ = 3sin θ + cot θ in terms of cos θ only, solve the equation for 0° , θ , 180°.

[5]

4

Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2016 1 [5] Given that cos A = , where 270° , A , 360°, find the exact value of sin 2 A. 4 2 [6] Solve the equation 2 tan x + sec x = 1 for 0° ø x ø 360°.

5

Solve the equation 2 cot 2 x + 5 cosec x = 10 for 0° , x , 360°.

6

a

Prove that sin( x + 60°) + cos( x + 30°) ≡ 3 cos x .

[3]

b

3 Hence solve the equation sin( x + 60°) + cos( x + 30°) = for 0° , x , 360°. 2

[3]

a

Prove that sin(60° − x ) + cos(30° − x ) ≡ 3 cos x .

[3]

b

2 Hence solve the equation sin(60° − x ) + cos(30° − x ) = sec x for 0° , x , 360°. 5

[3]

i

Show that the equation tan( x + 45°) = 6 tan x can be written in the form 6 tan2 x − 5 tan x + 1 = 0.

[3]

ii

Hence solve the equation tan( x + 45°) = 6 tan x , for 0° , x , 180°.

[3]

3

7

8

[6]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2010 9

a

Prove that tan( x + 45°) − tan(45° − x ) ≡ 2 tan 2 x.

[4]

b

Hence solve the equation tan( x + 45°) − tan(45° − x ) = 6 for 0° , x , 180°.

[3]

Express 3 cos θ + sin θ in the form R cos(θ − α ), where R . 0 and 0° , α , 90°, giving the exact value of R and the value of α correct to 2 decimal places.

[3]

Hence solve the equation 3 cos 2 x + sin 2 x = 2, giving all solutions in the interval 0° ø x ø 360° .

[5]

74

10 i ii

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2013 11 a b

Prove that cos(60° − x ) + cos(300° − x ) ≡ cos x .

[3]

Hence i find the exact value of cos15° + cos 255°

[2]

1 ii solve the equation cos(60° − x ) + cos(300° − x ) = cosec x for 0° , x , 180°. 4 2 sin 2θ − 3 cos 2θ + 3 ≡ 4 cos θ + 6 sin θ . 12 a Prove the identity sin θ π b Express 4 cos θ + 6 sin θ in the form R cos(θ − α ), where R .0 and 0 , α , . 2 Give the exact value of R and the value of α correct to 2 decimal places.

[3] [3]

[4]

2

c 13

2 sin 2θ − 3 cos 2θ + 3 Write down the greatest value of . sin θ

[1]

i Express 4 sin θ − 6 cos θ in the form R sin(θ − α ) , where R . 0 and 0° , α , 90° . Give the exact value of R and the value of α correct to 2 decimal places. ii Solve the equation 4 sin θ − 6 cos θ = 3 for 0° ø θ ø 360°.

[3] [4]

iii Find the greatest and least possible values of (4 sin θ − 6 cos θ ) + 8 as θ varies. 2

[2]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 June 2011

Chapter 3: Trigonometry

i By first expanding sin(2θ + θ ), show that sin 3θ = 3sin θ − 4 sin3 θ . [4] 2 sin θ 1 , the equation x 3 − x + 3 = 0 can be written in ii Show that, after making the substitution x = 6 3 3 [1] the form sin 3θ = . 4 1 iii Hence solve the equation x 3 − x + 3 = 0, giving your answers correct to 3 significant figures. [4] 6 Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2014 1 ≡ sec x. [3] 15 a Prove the identity sin( x + 30°) + cos( x + 60°) 2 = 7 − tan2 x for 0° , x , 360°. b Hence solve the equation [6] sin( x + 30°) + cos( x + 60°) 14

16 a b

Prove the identity cosec 4 x − cot 4 x ≡ cosec 2 x + cot 2 x.

[3]

Hence solve the equation cosec x − cot x = 16 − cot x for 0° , x , 180°.

[6]

4

4

17 The curves C1 and C2 have equations y = 1 + 4 cos 2 x and y = 2 cos2 x − 4 sin 2 x . a b

c

Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation 3 cos 2 x + 4 sin 2 x = 0.

[3]

Express 3 cos 2 x + 4 sin 2 x in the form R sin(2 x + α ), where R . 0 and 0° , α , 90° . Give the exact value of R and the value of α correct to 2 decimal places.

[3]

Hence find all the roots of the equation 3 cos 2 x + 4 sin 2 x = 0 for 0° , x , 180°.

[3] 75

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

CROSS-TOPIC REVIEW EXERCISE 1 1

Find the set of values of x satisfying the inequality 3 x − 1 , 2 x + 1 .

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q1 November 2012 2

Solve the equation 2 3x − 1 = 3x , giving your answers correct to 3 significant figures.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2013 3

i

Solve the equation 3x + 4 = 3x − 11 .

[3]

ii

Hence, using logarithms, solve the equation 3 × 2 y + 4 = 3 × 2 y − 11 , giving the answer correct to 3 significant figures.

[2]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 June 2015 4

i

Solve the equation 2 x − 1 = 3 x .

[3]

ii

Hence solve the equation 2 5x − 1 = 3 5x , giving your answer correct to 3 significant figures.

[2]

Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2016 ln y

5

76

(5, 4.49)

(0, 2.14)

O

x

( )

The variables x and y satisfy the equation y = A b x , where A and b are constants. The graph of ln y against x is a straight line passing through the points (0, 2.14) and (5, 4.49), as shown in the diagram. Find the values of A and b, correct to 1 decimal place.

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2012 6

i ii

It is given that x satisfies the equation 32x = 5(3x ) + 14. Find the value of 3x and, using logarithms, find the value of x correct to 3 significant figures. 2x

Hence state the values of x satisfying the equation 3

= 5(3

x

) + 14.

[4] [1]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 November 2016 7

The variables x and y satisfy the equation x n y = C , where n and C are constants. When x = 1.10 , y = 5.20 , and when x = 3.20, y = 1.05. i

Find the values of n and C .

[5]

ii

Explain why the graph of ln y against ln x is a straight line.

[1]

Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2010 8

Given that 3e x + 8e − x = 14, find the possible values of e x and hence solve the equation 3e x + 8e − x = 14 correct to 3 significant figures.

[6]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2016

Cross-topic review exercise 1

9

The angles θ and φ lie between 0° and 180°, and are such that tan(θ − φ ) = 3 and tan θ + tan φ = 1. Find the possible values of θ and φ .

[6]

Cambridge International A Level Mathematics 9709 Paper 31 Q3 November 2015 10

i

Solve the equation 4x − 1 = x − 3 .

[3]

ii

Hence solve the equation 4 y +1 − 1 = 4 y − 3 correct to 3 significant figures.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2013 11

The polynomial 4x 3 + ax 2 + 9x + 9 , where a is a constant, is denoted by p( x ) . It is given that when p( x ) is divided by (2 x − 1) the remainder is 10. i

Find the value of a and hence verify that ( x − 3) is a factor of p( x ) .

[3]

ii

When a has this value, solve the equation p( x ) = 0 .

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2011 12

The polynomial 2 x 3 − 4x 2 + ax + b, where a and b are constants, is denoted by p( x ) . It is given that when p( x ) is divided by ( x + 1) the remainder is 4, and that when p( x ) is divided by ( x − 3) the remainder is 12. i

Find the values of a and b.

ii

When a and b have these values, find the quotient and remainder when p( x ) is divided by x 2 − 2 .

(

[5]

)

77

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2012 13

i ii

Express cos x + 3 sin x in the form R cos ( x − α ), where R . 0 and 0° , α , 90° , giving the exact value of R and the value of α correct to 2 decimal places.

[3]

Hence solve the equation cos 2θ + 3 sin 2θ = 2, for 0° , θ , 90°.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2011 14

It is given that 2 ln(4x − 5) + ln( x + 1) = 3 ln 3. i

Show that 16x 3 − 24x 2 − 15x − 2 = 0.

ii

By first using the factor theorem, factorise 16x − 24x − 15x − 2 completely.

[3] 3

2

iii Hence solve the equation 2 ln(4x − 5) + ln( x + 1) = 3 ln 3.

[4] [1]

Cambridge International A Level Mathematics 9709 Paper 31 Q6 June 2014 15

The polynomial 8x 3 + ax 2 + bx − 1, where a and b are constants, is denoted by p( x ) . It is given that ( x + 1) is a factor of p( x ) and that when p( x ) is divided by ( 2 x + 1 ) the remainder is 1. i

Find the values of a and b.

[5]

ii

When a and b have these values, factorise p( x ) completely.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2015

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

16

The polynomial 3x 3 + 2 x 2 + ax + b, where a and b are constants, is denoted by p( x ) . It is given that ( x − 1) is a factor of p( x ) , and that when p( x ) is divided by ( x − 2) the remainder is 10. i

Find the values of a and b.

[5]

ii

When a and b have these values, solve the equation p( x ) = 0 .

[4]

Cambridge International AS and A Level Mathematics 9709 Paper 21 Q7 November 2010 17

i

ii

The polynomial x 3 + ax 2 + bx + 8, where a and b are constants, is denoted by p( x ) . It is given that when p( x ) is divided by ( x − 3) the remainder is 14, and that when p( x ) is divided by ( x + 2) the remainder is 24. Find the values of a and b.

[5]

When a and b have these values, find the quotient when p( x ) is divided by x 2 + 2 x − 8 and hence solve the equation p( x ) = 0.

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2013 18

i ii

Given that ( x + 2) and ( x + 3) are factors of 5x 3 + ax 2 + b, find the values of the constants a and b.

[4]

When a and b have these values, factorise 5x 3 + ax 2 + b completely, and hence solve the equation 53 y +1 + a × 52 y + b = 0, giving any answers correct to 3 significant figures.

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 November 2014 78

19

i

Find the quotient and remainder when x 4 + x 3 + 3x 2 + 12 x + 6 is divided by ( x 2 − x + 4) .

[4]

ii

It is given that, when x 4 + x 3 + 3x 2 + px + q is divided by ( x 2 − x + 4) , the remainder is zero. Find the values of the constants p and q.

[2]

iii When p and q have these values, show that there is exactly one real value of x satisfying the equation x 4 + x 3 + 3x 2 + px + q = 0 and state what that value is.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2015 20

The angle α lies between 0° and 90° and is such that 2 tan2 α + sec 2 α = 5 − 4 tan α . i

Show that 3tan2 α + 4 tan α − 4 = 0 and hence find the exact value of tan α .

[4]

ii

It is given that the angle β is such that cot(α + β ) = 6. Without using a calculator, find the exact value of cot β .

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2014 21

i

Express 5 sin 2θ + 2 cos 2θ in the form R sin(2θ + α ), where R . 0 and 0° , α , 90°, giving the exact value of R and the value of α correct to 2 decimal places.

Hence

[3]

solve the equation 5 sin 2θ + 2 cos 2θ = 4, giving all solutions in the interval 0° ø θ ø 360°, 1 as θ varies. iii determine the least value of (10 sin 2θ + 4 cos 2θ )2 ii

[5] [2]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2013

Cross-topic review exercise 1

22

The polynomial p( x ) is defined by p( x ) = ax 3 + 3x 2 + bx + 12, where a and b are constants. It is given that ( x + 3) is a factor of p( x ). It is also given that the remainder is 18 when p( x ) is divided by ( x + 2). i

Find the values of a and b.

ii

When a and b have these values,

[5]

a show that the equation p( x ) = 0 has exactly one real root,

[4]

b solve the equation p(sec y ) = 0 for −180° , y , 180°.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2016

79

80

Chapter 4 Differentiation In this chapter you will learn how to:

■ ■ ■

differentiate products and quotients use the derivatives of e x , ln x, sin x, cos x, tan x, together with constant multiples, sums, differences and composites find and use the first derivative of a function, which is defined parametrically or implicitly.

Chapter 4: Differentiation

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

Pure Mathematics 1 Coursebook, Chapter 7

Differentiate x n together with constant multiples, sums and differences.

1 Differentiate with respect to x. 3 a y = 5x 3 − 2 + 2 x x b

Pure Mathematics 1 Coursebook, Chapter 7

Differentiate composite functions using the chain rule.

y=

x 8 − 4x 5 + x 2 2x3

2 Differentiate with respect to x. a (3x − 5)4 b

4 1 − 2x

Pure Mathematics 1 Coursebook, Chapter 7

Find tangents and normals to curves.

3 Find the equation of the normal to the curve y = x 3 − 5x 2 + 2 x − 1 at the point (1, −3) .

Pure Mathematics 1 Coursebook, Chapter 8

Find stationary points on curves and determine their nature.

4 Find the stationary points on the curve y = x 3 − 3x 2 + 2 and determine their nature. 81

Why do we study differentiation? In this chapter we will learn how to differentiate the product and the quotient of two simple functions. We will also learn how to find and use the derivatives of exponential, logarithmic and trigonometric functions. All of the functions that we have differentiated so far have been of the form y = f( x ) . In this chapter we will also learn how to differentiate functions that cannot be written in the form y = f( x ). For example, the function y2 + 2 xy = 4 . Lastly, we will learn how to find and use the derivative of a function where the variables x and y are given as a function of a third variable. Although these topics might seem like they are just for Pure Mathematics problems, differentiation is used for calculations to do with quantum mechanics and field theories in Physics.

4.1 The product rule The function y = ( x + 1)4 (3x − 2)3 can be considered as the product of two separate functions: y = uv where u = ( x + 1)4 and v = (3x − 2)3 To differentiate the product of two functions we can use the product rule.

REWIND This chapter builds on the work from Pure Mathematics 1 Coursebook, Chapters 7 and 8 where we learnt how to find and use the derivative of x n .

WEB LINK Explore the Calculus of trigonometry and logarithms and the Chain rule and integration by substitution stations on the Underground Mathematics website.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

KEY POINT 4.1 The product rule is: dv du d +v ( uv ) = u dx dx dx

Some people find it easier to remember this rule as: ‘(first function × derivative of second function) + (second function × derivative of first function)’ So for y = ( x + 1)4 (3x − 2)3, dy d d = ( x + 1)4 (3x − 2)3 + (3x − 2)3 ( x + 1)4 dx dx dx differentiate second + differentiate first first second = ( x + 1)4 × 9(3x − 2)2 + (3x − 2)3 × 4( x + 1)3 = ( x + 1)3 (3x − 2)2 [ 9( x + 1) + 4(3x − 2) ] = ( x + 1)3 (3x − 2)2 (21x + 1)

The product rule from first principles Consider the function y = uv where u and v are functions of x. A small increase δx in x leads to corresponding small increases δy, δu and δv iny, u and v. y + δy = ( u + δu) ( v + δv) 82

Expand brackets and replace y with uv.

uv + δy = uv + uδv + vδu + δuδv δy = uδv + vδu + δuδv

Subtract uv from both sides. Divide both sides by δx.

δy δv δu δv ---------------(1) =u +v + δu δx δx δx δx

As δx → 0, then so do δy, δu and δv and dy δu du dv δy δv → , → and → . δx δx dx δx dx dx Equation (1) becomes: dy dv du dv =u +v +0 dx dx dx dx dv du dy =u +v ∴ dx dx dx

WORKED EXAMPLE 4.1 dy when y = (2 x − 1) 4x + 5 . dx Answer

Find

1

y = (2 x − 1)(4x + 5) 2 1

1 dy d d = (2 x − 1) [(4x + 5) 2 ] + (4x + 5) 2 (2 x − 1) dx x d d x second differentiate first differentiate second + first

Chapter 4: Differentiation

first

differentiate second

+

1 = (2 x − 1) (4x + + 2 1 − 5) 2 (4)

second

(

differentiate first

)

4x + 5 (2)

Chain rule.

2(2 x − 1) = + 2 4x + 5 4x + 5 2(2 x − 1) + 2(4x + 5) = 4x + 5 12 x + 8 = 4x + 5

Write as a single fraction. Simplify the numerator.

WORKED EXAMPLE 4.2 Find the x-coordinate of the points on the curve y = (2 x − 3)2 ( x + 5)3 where the gradient is 0. Answer y = (2 x − 3)2 ( x + 5)3 dy d d [( x + 5)3 ] + ( x + 5)3 [(2 x − 3)2 ] = (2 x − 3)2 dx d d x x second first differentiate second + differentiate first = (2 x − 3)2 [3( x + 5)2 (1)] +( x + 5)3 [2(2 x − 3)1(2)] Chain rule

83

Chain rule

= 3(2 x − 3)2 ( x + 5)2 + 4(2 x − 3)( x + 5)3

Factorise.

= (2 x − 3)( x + 5) [3(2 x − 3) + 4( x + 5)]

Simplify.

2

= (2 x − 3)( x + 5) (10 x + 11) 2

dy = 0 when (2x − 3)( x + 5)2 (10x + 11) = 0 dx x+5 = 0 2x − 3 = 0 10 x + 11 = 0 3 x= x = −5 x = −1.1 2

EXERCISE 4A 1 Use the product rule to differentiate each of the following with respect to x: a

x( x − 2)5

b 5x(2 x + 1)3

c

d ( x − 1) x + 5

e

2x − 1

f

g ( x − 3)2 ( x + 2)5

h (2 x − 1)5 (3x + 4)4

i

x

3

2 Find the gradient of the curve y = x 2 x + 4 at the point ( −3, 9) .

x x+2 x ( x 2 + 2)3 (2 x − 5)(3x 2 + 1)2

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3 Find the equation of the tangent to the curve y = (2 − x )3 ( x + 1)4 at the point where x = 1. 4 Find the gradient of the tangent to the curve y = ( x + 2)( x − 1)3 at the point where the curve meets the y-axis. 5 Find the x-coordinate of the points on the curve y = (3 − x )3 ( x + 1)2 where the gradient is zero. 6 Find the x-coordinate of the point on the curve y = ( x + 2) 1 − 2 x where the gradient is zero. PS

7 a Sketch the curve y = ( x − 1)2 (5 − 2 x ) + 3. b The curve y = ( x − 1)2 (5 − 2 x ) + 3 has stationary points at A and B. The straight line through A and B cuts the axes at P and Q. Find the area of the triangle POQ.

4.2 The quotient rule We can differentiate the function y =

x2 − 5 by writing the function in the form 2x + 1

y = ( x 2 − 5)(2 x + 1)−1 and then by applying the product rule. x2 − 5 as the division (quotient) of two separate Alternatively, we can consider y = 2x + 1 functions: u y= where u = x 2 − 5 and v = 2 x + 1. v To differentiate the quotient of two functions we can use the quotient rule. 84

KEY POINT 4.2 The quotient rule is: d u = dx v

v

du dv − u dx dx v2

Some people find it easier to remember this rule as (denominator × derivative of numerator) − (numerator × derivative of denominator) (denominator)2

The quotient rule from first principles u where u and v are functions of x. v A small increase δx in x leads to corresponding small increases δy, δu and δv in y, u and v. Consider the function y =

Chapter 4: Differentiation

y + δy = δy = δy = δy = δy =

δy = δx

u + δu Subtract y from both sides. v + δv u u + δu Replace y with . −y v v + δv u + δu u − Combine fractions. v + δv v v ( u + δu) − u( v + δv ) Expand brackets and simplify. v ( v + δv ) vδu − uδv Divide both sides by δx. v 2 + vδv δu δv v − u δx δx ---------------------(1) v 2 + vδv

As δx → 0 , then so do δy, δu and δv and

dy δu dv du δy δv → → → , . and dx δx dx dx δx δx

Equation (1) becomes d u = dx v

v

du dv − u x d dx v2

WORKED EXAMPLE 4.3

Find the derivative of y =

x2 − 5 . 2x + 1

85

Answer y=

x2 − 5 2x + 1 denominator

dy = dx

(2 x + 1)

differentiate numerator

×

numerator −

d ( x 2 − 5) − ( x 2 − 5) × dx (2 x + 1)2 denominator squared

=

2

(2 x + 1)(2 x ) − ( x − 5)(2) (2 x + 1)2

=

4x 2 + 2 x − 2 x 2 + 10 (2 x + 1)2

=

2( x 2 + x + 5) (2 x + 1)2

differentiate denominator

d (2 x + 1) dx

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 4.4

Find the derivative of y =

( x + 2)3 . x −1

Answer y=

( x + 2)3 x −1 denominator

differentiate numerator

d x −1 × [( x + 2)3 ] dx

dy = dx

numerator −

− ( x + 2)

2 x − 1 ) (

3

differentiate denominator

×

d x − 1 dx

denominator squared

( =

1 1 − x − 1 [3( x + 2)2 (1)] − ( x + 2)3 ( x − 1) 2 (1) 2 x −1

)

3( x + 2)2 x − 1 − =

86

=

=

( x + 2)3 2 x − 1

x −1 6( x + 2)2 ( x − 1) − ( x + 2)3 2( x − 1) x − 1

Multiply numerator and denominator by 2 x − 1. Factorise the numerator.

( x + 2)2 [6( x − 1) − ( x + 2)] 3

2( x − 1) 2 =

( x + 2)2 (5x − 8) 3

2( x − 1) 2

EXPLORE 4.1 x2 − 5 and In Worked examples 4.3 and 4.4 we differentiated the functions y = 2x + 1 ( x + 2)3 y= using the quotient rule. Now write each of these functions as a product, x −1 such as y = ( x 2 − 5)(2 x + 1)−1, and then use the product rule to differentiate them. Do you obtain the same answers? Discuss with your classmates which method you prefer for functions such as these: the quotient rule or the product rule.

Chapter 4: Differentiation

EXERCISE 4B 1 Use the quotient rule to differentiate each of the following with respect to x: 2x + 3 3x − 5 x2 − 3 b c a x−4 2−x 2x − 1 e

1 − 2x2 ( x + 4)2

f

5x 4 ( x 2 − 1)2

2 Find the gradient of the curve y =

g

3x 2 − 7 x x2 + 2x + 5

d

3x + 1 2 − 5x

h

( x + 4)2 ( x 2 + 1)3

x−5 1 at the point 2, − . x+4 2

3 Find the coordinates of the points on the curve y =

( x − 1 )2

where the tangent is parallel to the x-axis. 2x + 5 1 − 2x at which the gradient is 1. 4 Find the coordinates of the points on the curve y = x−5 x−4 at the point where the curve crosses the y-axis. 5 Find the equation of the tangent of the curve y = 2x + 1 6 Differentiate with respect to x: x b a 5x − 1

x −1 2x + 3

c

3 − x2 x2 − 1

x +1 where the gradient is 0. x −1 x2 + 1 at the point ( −1, 2). 8 Find the equation of the normal to the curve y = x+2

d

5( x − 1)3 x+2

7 Find the x-coordinate of the point on the curve y =

PS

9 The line 2 x − 2 y = 5 intersects the curve 2 x 2 y − x 2 − 26 y − 35 = 0 at three points. a Find the x-coordinates of the points of intersection. b Find the gradient of the curve at each of the points of intersection.

4.3 Derivatives of exponential functions The derivative of e x In Chapter 2, we learnt about the natural exponential function f( x ) = e x . This function has a very special property. If we use graphing software to draw the graph of f( x ) = e x together with the graph of its gradient function we find that the two curves are identical. KEY POINT 4.3 Hence, we have the rule: d (e x ) = e x dx

87

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An explanation of how this rule can be obtained is as follows. Consider the function f( x ) = e x and two points whose x-coordinates are x and x + δx where δx is a small increase in x. y

f(x) = ex

f(x + dx) f(x)

O

x

x + dx

x

f(x + δx ) − f( x ) dy e x +δx − e x e x (eδx − 1) = lim = lim = lim δx → 0 δx → 0 δx δx dx δx → 0 (x + δx ) − x Now consider δx e

eδx − 1 for small values of δx. δx 0.1

0.01

0.001

0.0001

−1 δx

1.051709

1.005 017

1.000 500

e

Consider the function y = e f( x ). y = eu

u = f( x )

where

dy = eu du

du = f ′( x ) dx dy dy du = × dx du dx = e u × f ′( x )

Using the chain rule:

= f ′( x ) × e f( x )

d f( x ) [e ] = f ′( x ) × e f( x ) dx KEY POINT 4.4 In particular, d ax + b ] = a e ax + b [e dx

1.000 050

−1 → 1. δx

δx

The derivative of e f( x )

Let

lim

δx → 0

δx

From the table, we can see that as δx → 0, dy ∴ = ex dx 88

TIP f( x + δx) − f( x ) ( x + δx) − x

means the limit of f( x + δx) − f( x ) as ( x + δx ) − x δx → 0 .

Chapter 4: Differentiation

WORKED EXAMPLE 4.5 Differentiate with respect to x. b xe −5 x

a e2 x

c

e 3x x2

Answer a

d 2x 2x 2x (e ) = 2 × e = 2e dx ւ ց

differentiate index original function

b

d d d ( xe −5 x ) = x × (e −5 x ) + e −5 x × (x) dx dx dx = x × ( −5e −5 x ) + e −5 x × (1)

Product rule.

= −5xe −5 x + e −5 x = e −5 x (1 − 5x )

c

d e 3x dx x 2

d 3x d ( x2 ) (e ) − e3x × x2 × d x d x = x4

Quotient rule.

x 2 × 3e3x − e3x × 2 x x4 3x e (3x − 2) = x3

=

89

WORKED EXAMPLE 4.6 The equation of a curve is y = e2 x − 9e x + 7 x . Find the exact value of the x-coordinate at each of the stationary points and determine the nature of each stationary point. Answer y = e2 x − 9e x + 7 x

Differentiate.

dy = 2e2 x − 9e x + 7 dx d2 y = 4e2 x − 9e x dx 2 Stationary points occur when

dy = 0. dx

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2e2 x − 9e x + 7 = 0

Factorise.

(2e x − 7)(e x − 1) = 0 (2e x − 7) = 0 or (e x − 1) = 0 7 or e x = 1 2 7 x = ln or x = 0 2 d2 y When x = 0 , is negative ⇒ maximum point. dx 2 d2 y 7 When x = ln , is positive ⇒ minimum point. 2 dx 2 ex =

EXERCISE 4C 1 Differentiate with respect to x. a

e5 x

d 3e −5 x

90

2

−3

g

ex

j

2(e3x − 1)

b e −4 x

c

2e6 x

x 4e 2

f

e2 x − 7

i

5 ex −

l

5(e x − 2 x )

e

h 2 x + 3e k

x

3e2 x + e −2 x 2

1 e2 x

2

2 a Sketch the graph of the function y = 1 − e2 − x . b Find the equation of the normal to the curve y = 1 − e2 − x at the point where y = 0. M

3 The mass, m grams, of a radioactive substance remaining t years after a given time, is given by the formula m = 300e −0.00012t . Find the rate at which the mass is decreasing when t = 2000 . 4 Differentiate with respect to x. a

xe x

b x 2 e 3x

d 2 x e x

e

e6x x

ex − 1 ex + 2

h

xe3x +

g

5 Find the gradient of the curve y =

e6x 2

c

5xe −2 x

f

e −2 x x

i

x2ex − x ex + 2

8 at the point where x = 0. 5 + e2 x

6 Find the exact coordinates of the stationary point on the curve y = xe x . 7 The curve y = 2e2 x + e − x cuts the y-axis at the point P. Find the equation of the tangent to the curve at the point P and state the coordinates of the point where this tangent cuts the x-axis. 8 Find the exact coordinates of the stationary point on the curve y = (x − 4)e x and determine its nature. 9 Find the exact coordinates of the stationary point on the curve y =

e2 x and determine its nature. x2

Chapter 4: Differentiation

10 The equation of a curve is y = x 2 e − x . a Find the x-coordinates of the stationary points of the curve and determine the nature of these stationary points. b Show that the equation of the normal to the curve at the point where x = 1 is e2 x + ey = 1 + e2 . 11 Find the exact value of the x-coordinates of the points on the curve y = x 2 e −2 x at which 12 Find the coordinates of the stationary point on the curve y = d (2 x ) = 2 x ln 2. dx

P

13 By writing 2 as e ln 2 , prove that

PS

14 The equation of a curve is y = x(3x ) .

e 2 x −1 . x

d2 y = 0. dx 2

Find the exact value of the gradient of the tangent to the curve at the point where x = 1.

4.4 Derivatives of natural logarithmic functions EXPLORE 4.2 y 4 3

y = ln x

2 1 –1 O –1

91 1

2

3

4

5

6

7

8

9 x

–2 –3 –4

Here is a sketch of the graph of y = ln x. Can you sketch a graph of its derivative (gradient function)? Discuss your sketch with your classmates. Now sketch the graphs using graphing software. Can you suggest a formula for the derivative of ln x ? (It might be helpful to look at the coordinates of some of the points on the graph of the derivative.)

The derivative of ln x In Chapter 2, you learnt that if y = ln x , then e y = x. dy In Section 4.3, you learnt that if y = e x , then = ex . dx Using these two results we can find the rule for differentiating ln x . y = ln x e y = x dx e y = dy dy 1 = dx x

Differentiate both sides with respect to y. Rearrange and replace e y by x.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

KEY POINT 4.5 1 d (ln x ) = dx x

The derivative of ln f(x ) Consider the function y = ln f( x ). Let y = ln u

where

dy 1 = du u

u = f( x ) du = f ′( x ) dx

Using the chain rule:

dy du dy = × dx du dx 1 = × f ′( x ) u f ′( x ) = f( x )

d f ′( x ) [ln f( x )] = dx f( x ) KEY POINT 4.6 92

In particular, a d [ln( ax + b )] = dx ax + b

WORKED EXAMPLE 4.7 Differentiate with respect to x. a ln 2 x

b ln(4x − 3)

c

ln 5 − x

Answer a

b

d 2 (ln 2 x ) = dx 2x 1 = x d 4 [ln(4x − 3)] = dx 4x − 3

‘Inside’ differentiated. ‘Inside’

Chapter 4: Differentiation

c Method 1: 1

− 1 (5 − x ) 2 ( −1) d ln 5 − x = 2 dx 5−x 1 =− 2(5 − x )

‘Inside’ differentiated. ‘Inside’

Method 2: using the rules of logarithms before differentiating. 1 d d ln 5 − x = ln(5 − x ) 2 dx dx d 1 = ln(5 − x ) dx 2 1 d [ln(5 − x )] = × 2 dx 1 −1 = × 2 5−x 1 =− 2(5 − x )

Use ln a m = m ln a

‘Inside’ differentiated. ‘Inside’

WORKED EXAMPLE 4.8 93

Differentiate with respect to x. a 2 x 4 ln 5x

b

ln 2 x x3

Answer a

b

d d d (2 x 4 ln 5x ) = 2 x 4 × (ln 5x ) + ln 5x × (2 x 4 ) dx dx dx 5 = 2 x 4 × + ln 5x × 8x 3 5x = 2 x 3 + 8x 3 ln 5x d d 3 3 d ln 2 x x × dx ( ln 2 x ) − ln 2 x × dx ( x ) = dx x 3 ( x 3 )2 2 x 3 × − ln 2 x × 3x 2 2 x = x6 2 2 x − 3x ln 2 x = x6 1 − 3 ln 2 x = x4

Product rule.

Quotient rule.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 4D 1 Differentiate with respect to x. a

PS

ln 3x

b ln 7 x

c

ln(2 x + 1)

d 5 + ln( x 2 + 1)

e

ln(2 x − 1)2

f

ln x − 3

g

ln( x + 3)5

h

2 3x + ln x

i

2 5x + ln 1 − 2x

j

ln(ln x )

k

ln 2 − x

l

ln(5x + ln x )

(

)2

2 The answers to question 1 parts a and b are the same. Why is this the case? How many different ways can you justify this? 3 Differentiate with respect to x. a

x ln x

b 2 x 3 ln x

c

x ln(2 x + 1)

d 3x ln 2 x

e

x ln(ln x )

f

ln 5x x

2 ln x

h

ln(3x − 2) x

i

ln(2 x + 1) 4x − 1

g

4 a Sketch the graph of the function y = ln(2 x − 3). b Find the gradient of the curve y = ln(2 x − 3) at the point where x = 5. 94

5 Find the gradient of the curve y = e2 x − 5 ln(2 x + 1) at the point where x = 0. 6 A curve has equation y = x 2 ln 5x. dy d2 y and at the point where x = 2. Find the value of dx dx 2 7 The equation of a curve is y = x 2 ln x. Find the exact coordinates of the stationary point on this curve and determine whether it is a maximum or a minimum point. ln x . Find the exact coordinates of the stationary point on this curve and x determine whether it is a maximum or a minimum point.

8 The equation of a curve is y =

9 Find the equation of the tangent to the curve y = ln(5x − 4) at the point where x = 1. 10 Use the laws of logarithms to help differentiate these expressions with respect to x. 1 a ln 5x − 1 b ln c ln [ x( x + 1)5 ] 3x + 2 1 − 3x 2x + 3 x( x − 2) d ln e ln f ln x −1 x2 x + 4 g

3−x ln ( x + 4) ( x − 1)

h

8 ln 2 ( x + 1) ( x − 2)

dy , in terms of x, for each of the following. dx e y = 2x2 − 1 b e y = 3x 3 + 2 x

i

( x + 2) (2 x − 1) ln x( x + 5)

c

e y = ( x + 1)( x − 5)

11 Find a

TIP Take the natural logarithm of both sides of the equation before differentiating.

Chapter 4: Differentiation

12 A curve has equation x =

PS

Find the value of

1 y (2 x − 3) e + 4 . 5

dy when x = 1. dx

4.5 Derivatives of trigonometric functions EXPLORE 4.3 Graphing software has been used to draw the graphs of y = sin x and y = cos x for 0 ø x ø 2 π together with their gradient (derived) functions. y

y

O

π 2

π

O

2π x

3π 2

y = cos x

1

gradient function

1

y = sin x

gradient function π 2

π

3π 2

2π x

–1

–1

Discuss with your classmates why, for the function y = sin x , you would expect the graph of the gradient function to have this shape. Do the same for the graph of y = cos x . Can you suggest how to complete the following formulae? d d (sin x ) = (cos x ) = dx dx

The derivative of sin x Consider the function f( x ) = sin x, where x is measured in radians, and two points whose x-coordinates are x and x + δx where δx is a small increase in x. y f(x + δx)

f(x) = sin x

f(x)

O

x

x + δx

x

f( x + δx ) − f( x ) dy = lim dx δx → 0 ( x + δx ) − x = lim

sin( x + δx ) − sin x δx

= lim

sin x cos δx + cos x sin δx − sin x δx

δx → 0

δx → 0

= cos x

Expand sin( x + δx ) As δx → 0, cos δx → 1 and sin δx → δx.

95

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

KEY POINT 4.7 d (sin x ) = cos x dx Similarly, we can show that d (cos x ) = − sin x dx

We can find the derivative of tan x using these two results together with the quotient rule. d d sin x (tan x ) = dx dx cos x =

cos x ×

Use the quotient rule.

d d (sin x ) − sin x × (cos x ) dx dx 2 (cos x )

=

cos x × cos x − sin x × ( −sin x ) cos2 x

=

cos2 x + sin2 x cos2 x

Use cos2 x + sin2 x = 1 .

=

1 cos2 x

Use

1 = sec x. cos x

= sec 2 x

96

KEY POINT 4.8 d (tan x ) = sec 2 x dx

WORKED EXAMPLE 4.9 Differentiate with respect to x. a

2 sin x

b

x cos x

c

tan x x2

Answer a

b

d d (2 sin x ) = 2 (sin x ) dx dx = 2 cos x d d d ( x cos x ) = x × (cos x ) + cos x × (x) dx dx dx = − x sin x + cos x

Product rule.

d

(3 + 2 sin x )5

Chapter 4: Differentiation

d d ( x2 ) x2 × (tan x ) − tan x × d tan x x x d d = dx x 2 x4

c

Quotient rule.

x 2 × sec 2 x − tan x × 2 x x4 x sec 2 x − 2 tan x = x3 =

d

d [(3 + 2 sin x )5 ] = 5(3 + 2 sin x )4 × 2 cos x dx = 10 cos x(3 + 2 sin x )4

Chain rule.

Derivatives of sin( ax + b ), cos( ax + b ) and tan( ax + b ) Consider the function y = sin( ax + b ) where x is measured in radians. Let y = sin u

where

dy = cos u du Using the chain rule:

u = ax + b du =a dx

dy dy du = × dx du dx = cos u × a

97

= a cos( ax + b ) TIP

KEY POINT 4.9

It is important to remember that, in calculus, all angles are measured in radians unless a question tells you otherwise.

d [sin( ax + b )] = a cos( ax + b ) dx

Similarly, it can be shown that: d [cos( ax + b )] = − a sin( ax + b ) dx d [tan( ax + b )] = a sec 2 ( ax + b ) dx

WORKED EXAMPLE 4.10 Differentiate with respect to x. a

2 sin 3x

b

4x tan 2 x

Answer a

d d (2 sin 3x ) = 2 (sin 3x ) dx dx = 2 × cos 3x × (3) = 6 cos 3x

c

π cos 2 x − 4 x2

d

(3 − 2 cos 5x )4

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

b

d d d (4x tan 2 x ) = 4x × (tan 2 x ) + tan 2 x × (4x ) dx dx dx = 4x × sec 2 2 x (2) + tan 2 x × (4)

Product rule.

= 8x sec 2 2 x + 4 tan 2 x

c

d π cos 2 x − x 2 × x d d 4 = 2 x dx

π π d 2 cos 2 x − 4 − cos 2 x − 4 × dx [ x ] ( x 2 )2

Quotient rule.

π π −2 x 2 sin 2 x − − 2 x cos 2 x − 4 4 = x4 π π −2 x sin 2 x − − 2 cos 2 x − 4 4 = 3 x d

4 3 d 3 − 2 cos 5x ) = 4 ( 3 − 2 cos 5x ) × 10 sin 5x ( dx = 40 sin 5x(3 − 2 cos 5x )3

Chain rule.

EXERCISE 4E 98

1 Differentiate with respect to x. a

2 + sin x

b 2 sin x + 3 cos x

d

3sin 2x

e

g

tan( 3x + 2 )

π h sin 2 x + 3

4 tan 5x

c

2 cos x − tan x

f

2 cos 3x − sin 2 x

i

π 2 cos 3x − 6

c

sin2 x − 2 cos x

f

π 3 cos 4 x + 2 tan2 2 x − 4

2 Differentiate with respect to x. a

sin3 x

b

d

( 3 − cos x )4

e

5 cos2 3x π 2 sin3 2 x + 6

3 Differentiate with respect to x. a e i

x sin x 5 cos 3x sin x 3x − 1

b f j

5x cos 3x x cos x 1 sin3 2 x

c g k

x 2 tan x tan x x 3x sin 2 x

d x cos3 2 x h l

sin x 2 + cos x sin x + cos x sin x − cos x

4 Differentiate with respect to x. c

etan 3x

d e( sin x − cos x )

e x sin 2 x

g

e x (2 cos x − sin x )

h x 3e cos x

x ln(sin x )

k

cos 2 x e2 x+1

l

a

esin x

b e cos 2 x

e

e x cos x

f

i

ln(cos x )

j

5 Find the gradient of the curve y = 3 sin 2 x − 5 tan x at the point where x = 0.

x sin 2 x e2 x

Chapter 4: Differentiation

π 6 Find the exact value of the gradient of the curve y = 2 sin 3x − 4 cos x at the point , −2 . 3

P

P

7 Given that y = sin2 x for 0 ø x ø π , find the exact values of the x-coordinates of the points on the curve 3 . where the gradient is 2 5 8 Prove that the gradient of the curve y = is always positive. 2 − tan x d 1 , find (sec x ). 9 a By writing sec x as x d cos x d 1 , find b By writing cosec x as (cosec x ). dx sin x d cos x , find (cot x ). c By writing cot x as dx sin x π π 10 Prove that the normal to the curve y = x sin x at the point P , intersects the x-axis at the point ( π, 0). 2 2 11 The equation of a curve is y = 5 sin 3x − 2 cos x . Find the equation of the tangent to the curve at the point π , −1 . Give the answer in the form y = mx + c , where the values of m and c are correct to 3 significant figures. 3 12 A curve has equation y = 3 cos 2 x + 4 sin 2 x + 1 for 0 ø x ø π . Find the x-coordinates of the stationary points of the curve, giving your answer correct to 3 significant figures. π . Find the exact value of the x-coordinate of the stationary 2 point of the curve and determine the nature of this stationary point.

13 A curve has equation y = e x cos x for 0 ø x ø

14 A curve has equation y = point of this curve.

sin 2 x π for 0 ø x ø . Find the exact value for the x-coordinate of the stationary 2x e 2

π e 3x for 0 , x , . Find the exact value for the x-coordinate of the stationary sin 3x 2 point of this curve and determine the nature of this point.

15 A curve has equation y =

PS

16 A curve has equation y = sin 2 x − x for 0 ø x ø 2π . Find the x-coordinates of the stationary points of the curve, and determine the nature of these stationary points.

PS

17 A curve has equation y = tan x cos 2 x for 0 ø x ,

π . Find the x-coordinate of the stationary point on the 2 curve giving your answer correct to 3 significant figures.

4.6 Implicit differentiation All the functions that we have differentiated so far have been of the form y = f( x ) . These are called explicit functions as y is given explicitly in terms of x. However, many functions cannot be expressed in this form, for example x 3 + 5xy + y3 = 16 . When a function is given as an equation connecting x and y, where y is not the subject, it is called an implicit function. dv du dy = dy × du and the product rule d ( uv ) = u +v The chain rule are used dx du dx dx dx dx extensively to differentiate implicit functions. This is illustrated in the following Worked examples.

99

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WORKED EXAMPLE 4.11 Differentiate each of these expressions with respect to x. b 4x 2 y5

a y3

c x 3 + 5xy

Answer a

b

c

d d dy ( y3 ) = ( y3 ) × dx dy dx dy = 3 y2 dx d d d (4x 2 y5 ) = 4x 2 ( y5 ) + y5 (4x 2 ) dx dx dx d = 4x 2 ( y5 ) + 8xy5 dx dy = 20 x 2 y 4 + 8xy5 dx d d d ( x 3 + 5xy ) = ( x3 ) + (5xy ) dx dx dx d d ( y) + y (5x ) = 3x 2 + 5x dx dx dy = 3x 2 + 5x + 5y dx

Chain rule.

Product rule. Chain rule:

dy d ( y5 ) = 5 y 4 . dx dx

Use the product rule for

d (5xy ). dx

100

WORKED EXAMPLE 4.12 Find

dy for the curve x 3 + y3 = 4xy . Hence find the gradient of the curve at the point (2, 2). dx

Answer x 3 + y3 = 4xy d d ( x3 ) + ( y3 ) dx dx d dy 3x 2 + ( y3 ) × dy dx dy 3x 2 + 3 y2 dx dy 2 dy − 4x 3y dx dx dy (3 y2 − 4x ) dx dy dx

d (4xy ) dx d d ( y) + y (4x ) = 4x dx dx dy = 4x + 4y dx

Differentiate with respect to x.

=

Rearrange terms.

= 4 y − 3x 2

Factorise.

= 4 y − 3x 2

Rearrange.

=

4 y − 3x 2 3 y 2 − 4x

When x = 2 and y = 2,

Chapter 4: Differentiation

dy 4(2) − 3(2)2 = = −1 dx 3(2)2 − 4(2) Gradient of the curve at the point (2, 2) is −1.

WORKED EXAMPLE 4.13 The equation of a curve is 3x 2 + 2 xy + y2 = 6. Find the coordinates of the two stationary points on the curve. Answer 3x 2 + 2 xy + y2 = 6 d d d (3x 2 ) + (2 xy ) + y2 dx dx dx d d d 6x + 2 x ( y) + y y2 ( 2x ) + dx dx dx dy dy 6x + 2 x + 2y + 2y dx dx dy dy x +y dx dx dy (x + y) dx dy dx Stationary points occur when

Differentiate with respect to x.

( ) = ddx (6)

Use product rule for

( )=0

Use chain rule for

Divide by 2 and rearrange.

= − y − 3x

Factorise.

= − y − 3x

Rearrange.

=

− y − 3x x+y

101

dy = 0. dx − y − 3x = 0 y = −3x

6x 2 = 6 x = ±1 The stationary points are ( −1, 3) and (1, −3).

EXERCISE 4F 1 Differentiate each expression with respect to x. y5

d 2 + sin y

b x 3 + 2 y2

c

5x 2 + ln y

6x 2 y 3

f

y2 + xy

e

d ( y2 ). dx

=0

Substituting y = −3x into the equation of the curve gives: 3x 2 + 2 x( −3x ) + ( −3x )2 = 6

a

d (2 xy ). dx

g

x 3 − 7 xy + y3

h x sin y + y cos x

i

x 3 ln y

j

x cos 2 y

k 5 y + e x sin y

l

2 xe cos y

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

a

dy for each of these functions. dx x 3 + 2 xy + y3 = 10

b

x 2 y + y 2 = 5x

c

2 x 2 + 5xy + y2 = 8

d

x ln y = 2 x + 5

e

2e x y + e2 x y3 = 10

f

ln( xy ) = 4 − y2

g

xy3 = 2 ln y

h

ln x − 2 ln y + 5xy = 3

2 Find

3 Find the gradient of the curve x 2 + 3xy − 5 y + y3 = 22 at the point (1, 3). 4 Find the gradient of the curve 2 x 3 − 4xy + y3 = 16 at the point where the curve crosses the x-axis. 5 A curve has equation 2 x 2 + 3 y2 − 2 x + 4 y = 4 . Find the equation of the tangent to the curve at the point (1, −2) . 6 The equation of a curve is 4x 2 y + 8 ln x + 2 ln y = 4. Find the equation of the normal to the curve at the point (1, 1). 7 The equation of a curve is 5x 2 + 2 xy + 2 y2 = 45. a Given that there are two points on the curve where the tangent is parallel to the x-axis, show by differentiation that, at these points, y = −5x. b Hence find the coordinates of the two points. 8 The equation of a curve is y2 − 4xy − x 2 = 20. a Find the coordinates of the two points on the curve where x = 4.

102

b Show that at one of these points the tangent to the curve is parallel to the x-axis. c

Find the equation of the tangent to the curve at the other point.

9 The equation of a curve is y3 − 12 xy + 16 = 0. a Show that the curve has no stationary points. b Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. 10 Find the gradient of the curve 5e x y2 + 2e x y = 88 at the point (0, 4). 11 The equation of a curve is x 2 − 4x + 6 y + 2 y2 = 12 . Find the coordinates of the two points on the curve at 4 which the gradient is . 3 12 The equation of a curve is 2 x + y ln x = 4 y . Find the equation of the tangent to the curve at the point with 1 coordinates 1, . 2 PS

PS

13 The equation of a curve is y = x x . Find the exact value of the x-coordinate of the stationary point on this curve. d2 y determine the nature of each of 14 Find the stationary points on the curve x 2 − xy + y2 = 48. By finding dx 2 these stationary points.

Chapter 4: Differentiation

4.7 Parametric differentiation Sometimes variables x and y are given as a function of a third variable t. For example, x = 1 + 4 sin t and y = 2 cos t. The variable t is called a parameter and the two equations are called the parametric equations of the curve. We can find the curve given by the parametric equations x = 1 + 4 sin t, y = 2 cos t for 0 ø t ø 2π by finding the values of x and y for particular values of t. t

0

1 π 4

1 π 2

3 π 4

π

5 π 4

3 π 2

7 π 4

2π

x

1

3.83

5

3.83

1

−1.83

−3

−1.83

1

y

2

1.41

0

−1.41

−2

−1.41

0

1.41

2

Plotting the coordinates on a grid gives the following graph. y 3 2 1 –4

–3

–2

–1 O –1

1

2

3

4

5

6 x

–2 –3

EXPLORE 4.4 1 Use tables of values for t, x and y to sketch each of the following curves. Use graphing software to check your answers. a x = t 3,

y = t2

y = t3 − t 1 1 x=t+ , y=t− t t

b x = t2 , c

2 Use graphing software to draw the curves given by these parametric equations. a x = sin t ,

y = sin 2t

b x = sin 2t , y = sin 3t c

x = sin 4t, y = sin 3t

3 The curves in question 2 are called Lissajous curves. Use the internet to find out more about these curves. (A Lissajous curve is shown at the start of this chapter.) When a curve is given in parametric form, in terms of the parameter t, we can use the dy chain rule to find in terms of t. dx

103

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 4.14 The parametric equations of a curve are x = t 2 , y = 5 − 2t. dy in terms of the parameter t. a Find dx b Find the equation of the tangent to the curve at the point where t = 2. Answer

y= dy = dx = =

104

dx = 2t dt dy = −2 5 − 2t ⇒ dt dy dt × dt dx 1 −2 × 2t 1 − t

x = t2

a

⇒

Chain rule.

1 b When t = 2, gradient = − , x = 4 and y = 1. 2 Using y − y1 = m( x − x1 ) 1 y − 1 = − ( x − 4) 2 1 y = − x + 3 2

WORKED EXAMPLE 4.15 π 3π The parametric equations of a curve are x = 1 + 2 sin2 θ , y = 4 tan θ for , θ , . 2 2 dy a Find in terms of the parameter θ . dx b Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. Answer a

dx = 4 sin θ cos θ dθ dy 4 ⇒ = 4 sec 2 θ = dθ cos2 θ

x = 1 + 2 sin2 θ ⇒ y = 4 tan θ

dy dy dθ = × dx dθ dx 1 4 = 2 × 4 sin θ cos θ cos θ 1 = sin θ cos3 θ

Chain rule.

Chapter 4: Differentiation

b Tangent is parallel to the y-axis when: sin θ cos3 θ = 0 sin θ = 0 or cos θ = 0 sin θ = 0 ⇒ θ = π ⇒ (1, 0) There are no solutions to cos θ = 0 in the range

π 3π . ,θ , 2 2

∴ The point where the tangent is parallel to the y-axis is (1, 0).

EXERCISE 4G dy 1 For each of the following parametric equations, find in terms of the given dx parameter. a

x = 2t 3, y = t 2 − 5

b

x = 2 + sin 2θ , y = 4θ + 2 cos 2θ

c

x = 2θ − sin 2θ , y = 2 − cos 2θ

d

x = 3tan θ , y = 2 sin 2θ

e

x = 1 + tan θ , y = cos θ

f

x = cos 2θ − cos θ , y = sin2 θ

g

x = 1 + 2 sin2 θ , y = 4 tan θ

h

x = 2 + e − t , y = et − e − t

i

x = e 2t , y = t 3 et + 1 5 x = ln(1 − t ), y = t

j

x = 2 ln(t + 3) , y = 4et

l

x = 1 + t , y = 2 ln t

k

2 The parametric equations of a curve are x = 3t, y = t 3 + 4t 2 − 3t. Find the two values of t for which the curve has gradient 0. 3 The parametric equations of a curve are x = 2 sin θ , y = 1 − 3 cos 2θ . Find the π exact gradient of the curve at the point where θ = . 3 4 4 The parametric equations of a curve are x = 2 + ln(t − 1), y = t + for t . 1. t Find the coordinates of the only point on the curve at which the gradient is equal to 0. 5 The parametric equations of a curve are x = e2t , y = 1 + 2tet . Find the equation of the normal to the curve at the point where t = 0. 6 The parametric equations of a curve are x = e2t , y = t 2 e −t − 1. a Show that

dy t (2 − t ) = . dx 2e3t

b Show that the tangent to the curve at the point (1, −1) is parallel to the x-axis and find the exact coordinates of the other point on the curve at which the tangent is parallel to the x-axis.

105

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

7 A curve is defined by the parametric equations x = tan θ , y = 2 sin 2θ , for π 0 ,θ , . 2 dy = 4 cos2 θ (2 cos2 θ − 1). dx b Hence, find the coordinates of the stationary point. a Show that

8 The parametric equations of a curve are x = t + 4 ln t, y = t +

9 , for t . 0. t

dy t2 − 9 = 2 . dx t + 4t b The curve has one stationary point. Find the y-coordinate of this point and determine whether it is a maximum or a minimum point. a Show that

9 The parametric equations of a curve are x = 1 + 2 sin2 θ , y = 1 + 2 tan θ . Find π the equation of the normal to the curve at the point where θ = . 4 10 The parametric equations of a curve are x = 2 sin θ + cos 2θ , y = 1 + cos 2θ , for π 0 øθ ø . 2 2 sin θ dy . a Show that = dx 2 sin θ − 1 b Find the coordinates of the point on the curve where the tangent is parallel to the x-axis. 3 3 c Show that the tangent to the curve at the point , is parallel to the 2 2 y-axis.

106

PS

11 The parametric equations of a curve are x = ln(tan t ), y = 2 sin 2t, for π 0,t , . 2 dy = sin 4t . dx b Hence show that at the point where x = 0 the tangent is parallel to the x-axis.

a Show that

WEB LINK Try the Parametric points resource on the Underground Mathematics website.

Chapter 4: Differentiation

Checklist of learning and understanding Product rule ●

dv du d +v ( uv ) = u dx dx dx

Quotient rule

●

d u = dx v

v

du dv − u dx dx v2

Exponential functions ●

d (e x ) = e x dx

d e ax + b = ae ax + b dx

d f( x ) [e ] = f ′( x ) × e f( x ) dx

Logarithmic functions ●

1 d (ln x ) = dx x

a d [ln( ax + b )] = dx ax + b

f ′( x ) d [ln(f( x ))] = f( x ) dx

Trigonometric functions ●

d (sin x ) = cos x dx

d [sin( ax + b )] = a cos( ax + b ) dx

●

d (cos x ) = − sin x dx

d [cos( ax + b )] = − a sin( ax + b ) dx

●

d (tan x ) = sec 2 x dx

d [tan( ax + b )] = a sec 2 ( ax + b ) dx 107

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 4 1

The parametric equations of a curve are

i ii

2

3

9 x = 1 + ln(t − 2), y = t + , for t . 2. t 2 dy (t − 9)(t − 2) . = Show that dx t2 Find the coordinates of the only point on the curve at which the gradient is equal to 0.

[3] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2010 dy when x = 4 in each of the following cases: Find the value of dx [4] i y = x ln( x − 3), x −1 . [3] ii y = x +1 Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2011 The parametric equations of a curve are x = e3t , y = t 2 et + 3. dy t (t + 2) i Show that = . dx 3e2t ii Show that the tangent to the curve at the point (1, 3) is parallel to the x-axis.

[2]

iii Find the exact coordinates of the other point on the curve at which the tangent is parallel to the x-axis.

[2]

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2011 108

4

Find the gradient of each of the following curves at the point for which x = 0. i ii

y = 3sin x + tan 2 x 6 y= 1 + e2 x

[3] [3] Cambridge International AS & A Level Mathematics 9709 Paper 21 Q2 June 2014

5

The equation of a curve is 2 x 2 + 3xy + y2 = 3. i ii

Find the equation of the tangent to the curve at the point (2, −1), giving your answer in the form ax + by + c = 0, where a, b and c are integers.

[6]

Show that the curve has no stationary points.

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2014 6

The equation of a curve is y3 + 4xy = 16 . dy 4y =− 2 . i Show that dx 3 y + 4x ii Show that the curve has no stationary points.

[2]

iii Find the coordinates of the point on the curve where the tangent is parallel to the y-axis.

[4]

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2015 7

The equation of a curve is y = 6 sin x − 2 cos 2 x. 1 Find the equation of the tangent to the curve at the point π, 2 . Give the answer in the 6 form y = mx + c , where the values of m and c are correct to 3 significant figures.

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2015

Chapter 4: Differentiation

8

y A

D

B

O

x

C

The parametric equations of a curve are x = 6 sin2 t, y = 2 sin 2t + 3 cos 2t, for 0 ø t , π. The curve crosses the x-axis at the points B and D and the stationary points are A and C , as shown in the diagram. dy 2 i Show that = cot 2t − 1. [5] dx 3 [3] ii Find the values of t at A and C , giving each answer correct to 3 decimal places. iii Find the value of the gradient of the curve at B.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 November 2015 9

The equation of a curve is 3x 2 + 4xy + y2 = 24. Find the equation of the normal to the curve at the point (1, 3), giving your answer in the form ax + by + c = 0 where a, b and c are integers.

[8]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2016 10

y

P –1

The diagram shows the curve y = i

ii

By first differentiating

O

1 x

1 − x . 1+ x

1− x dy , obtain an expression for in terms of x. Hence show that the gradient of 1+ x dx

the normal to the curve at the point ( x, y ) is (1 + x ) (1 − x 2 ) .

[5]

The gradient of the normal to the curve has its maximum value at the point P shown in the diagram. Find, by differentiation, the x-coordinate of P.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2010

109

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

11 The parametric equations of a curve are x = 3(1 + sin2 t ), y = 2 cos3 t. Find

dy in terms of t, simplifying your answer as far as possible. [5] dx Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2011

12 The equation of a curve is ln( xy ) − y3 = 1. dy y = . i Show that dx x(3 y3 − 1) ii Find the coordinates of the point where the tangent to the curve is parallel to the y-axis, giving each coordinate correct to 3 significant figures.

[4] [4]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2012 13 The curve with equation 6e2 x + ke y + e2 y = c , where k and c are constants, passes through the point P with coordinates (ln 3, ln 2). i

Show that 58 + 2 k = c .

[2]

ii

Given also that the gradient of the curve at P is −6, find the values of k and c.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2011

110

111

Chapter 5 Integration In this chapter you will learn how to:

P2

■ ■ ■

1 extend the idea of ‘reverse differentiation’ to include the integration of e ax + b , , ax + b 2 sin( ax + b ) , cos( ax + b ) and sec ( ax + b ) use trigonometrical relationships in carrying out integration understand and use the trapezium rule to estimate the value of a definite integral.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

Chapter 4

Differentiate standard functions.

1 Find

Chapter 4

Use standard trigonometric relationships to prove trigonometric identities.

a

dy when: dx y = 3sin 2 x − 5 cos x

b

y = e5 x − 2

c

y = ln(2 x + 1)

2 Prove that cos 4x + 4 cos 2 x ≡ 8 cos 4 x − 3.

Why do we study integration?

REWIND

In this chapter we will learn how to integrate exponential, logarithmic and trigonometrical functions and how to apply integration to solve problems. It is important that you have a good grasp of the work covered in Chapter 3 on trigonometry, as you will be expected to know and use the trigonometrical relationships when solving integration problems.

∫ is used when we are not able to find the value of ∫

b

P2 Lastly, we will learn how to find an estimate for 112

f( x ) d x using a numerical method. This

This chapter builds on the work we did in Pure Mathematics 1 Coursebook, Chapter 9, where we learnt how to integrate ( ax + b ) n , where n ≠ −1.

a b

f( x ) d x using an algebraic method.

a

WEB LINK

5.1 Integration of exponential functions Since integration is the reverse process of differentiation, the rules for integrating exponential functions are: KEY POINT 5.1

∫e

x

∫e

dx = e x + c

ax + b

dx =

1 ax + b e +c a

REWIND In Chapter 4, we learnt the following rules for differentiating exponential functions:

WORKED EXAMPLE 5.1 Find: a

∫e

3x

dx

b

∫e

b

∫e

−2 x

dx

−2 x

dx =

c

∫e

c

∫e

5x − 4

dx

5x − 4

dx =

d (e x ) = e x dx d (e ax + b ) = ae ax + b dx

Answer a

∫e

3x

dx =

1 3x e +c 3

Explore the Calculus of trigonometry and logarithms station on the Underground Mathematics website.

1 −2 x +c e −2 1 = − e −2 x + c 2

1 5x − 4 e +c 5

Chapter 5: Integration

WORKED EXAMPLE 5.2

Evaluate

∫

2

9e3x −1 d x.

1

Answer

∫

2

1

2

9 9e3x −1 d x = e3x −1 3 1

Substitute limits.

= (3e5 ) − (3e2 ) = 3e2 (e3 − 1)

WORKED EXAMPLE 5.3 y

y = x2 + e–2x

O

1

x

2

Find the area of the shaded region. 113

Answer Area =

∫

2

( x 2 + e −2 x ) d x

1 2

x 3 1 −2 x = − e 2 3 1

Substitute limits.

8 1 1 1 = − e −4 − − e −2 3 2 3 2 7 1 1 = + e −2 − e −4 3 2 2 ≈ 2.39 units2

Simplify.

EXERCISE 5A 1 Find: a d g

∫ e dx ∫ 4e dx ∫ e dx 2x

1 x 2

3 x −1

b e h

∫e ∫ 2e ∫ 6e

−4 x

dx

−x

dx

2 − 3x

dx

c f i

∫ 6e ∫e ∫ 2e

3x

dx

2x + 4

dx

8x − 3

dx

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2 Find: a d

∫e

−x

∫

4 + e2 x dx e2 x

(1 + e x ) dx

∫ 5e (2 + e x

b

) dx

8 − ex dx 2e2 x

∫

e

3x

c f

∫ (e

2x

− 1)2 dx

(e x − 2)2 dx e2 x

∫

3 Evaluate: a

∫

2

∫

3

∫

1

e 3x d x

0

d

e 4x d x

c

0

e1+ 2 x d x

∫

e

(e x + e2 x )2 d x

∫

h

1

0

0

4 A curve is such that

∫

ln 2

∫

1

∫

2

5e −2 x d x

0

1

0

0

g

1 2

∫

b

8

e

2 x −1 d x

f

(e x + 1)2 d x

0

2

2e x − 5 d x ex

i

0

(e x + 1)2 dx e2 x

dy = 6e2 x + 2e − x . Given that the curve passes through the point (0, 2), find the equation dx

of the curve. 5 A curve is such that

d2 y dy −2 x . Given that = −8 when x = 0 and that the curve passes through the 2 = 20e dx dx

5 point 1, 2 , find the equation of the curve. e 114

6 Find the exact area of the region bounded by the curve y = 1 + e2 x − 5 , the x-axis and the lines x = 1 and x = 3. 1x

7

y = 2e 2 – 2x + 3

y

O

3

x

Find the exact area of the shaded region. 8 a b

Show that

d ( xe x − e x ) = xe x . dx

y y = xex

O

3

x

Use your result from part a to evaluate the area of the shaded region. 9 a

Find

∫

a

(4e −2 x + 5e − x ) d x, where a is a positive constant.

0

b Hence find the value of

∫

∞

0

(4e −2 x + 5e − x ) d x .

Chapter 5: Integration

PS

y

10

M

x

O

The diagram shows the curve y = 2e x + 8e − x − 7 and its minimum point M. Find the area of the shaded region. P PS

y

11

y = ex

3 2

ln 2

O

ln 3

x

The diagram shows the graph of y = e x. The points (ln 2, 2) and (ln 3, 3) lie on the curve. a Find the value of

∫

ln 3

e x d x.

b Hence show that

2

ln 2

5.2 Integration of

∫

3

1 ax + b

Since integration is the reverse process of differentiation, the rules for integration are:

REWIND In Chapter 4, we learnt the following rules for differentiating logarithmic functions: d 1 (ln x ) = , x . 0 x dx

KEY POINT 5.2 1

∫

∫ x dx = ln x + c, x . 0

1 1 dx = ln( ax + b ) + c, ax + b . 0 a ax + b

WORKED EXAMPLE 5.4 Find each of these integrals and state the values of x for which the integral is valid. 2 4 6 dx a b dx c dx x 2x + 1 2 − 3x

∫

∫

∫

Answer a

2

1

∫ x dx = 2 ∫ x dx = 2 ln x + c, x . 0

b

4

27 . ln y dy = ln 4e

1

∫ 2x + 1 dx = 4∫ 2x + 1 dx 1 = 4 ln(2 x + 1) + c 2 = 2 ln(2 x + 1) + c, x . −

Valid for 2 x + 1 . 0 . 1 2

+

d [ ln( ax + b ) ] = ax dx a = , ax + b . 0 ax + b It is important to remember that ln x is only defined for x . 0.

115

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

6

c

1

∫ 2 − 3x dx = 6∫ 2 − 3x dx 1 = 6 ln(2 − 3x ) + c −3 = −2 ln(2 − 3x ) + c, x ,

Valid for 2 − 3x . 0. 2 3

EXPLORE 5.1

∫x

−1

dx.

She tries to use the formula

∫x

Fiona is asked to find

n

1 x n +1 + c to obtain her answer. n +1

dx =

∫ (2x + 3) dx. 1 ( ax + b ) She tries to use the formula ( ax + b ) dx = ∫ a ( n + 1) −1

Fiona is also asked to find

n

n +1

+ c to obtain her

answer.

Discuss with your classmates why Fiona’s methods do not work.

116

EXPLORE 5.2

Rafiu and Fausat are asked to find Rafiu writes:

∫ 2(3x1+ 1) dx = ∫ =

Fausat writes:

1

∫ 2(3x + 1) dx. 1 dx 6x + 2

1 ln(6x + 2) + c 6

∫ 2(3x1+ 1) dx = 21 ∫

1 dx 3x + 1

1 1 = ln(3x + 1) + c 2 3 1 = ln(3x + 1) + c 6 Decide who is correct and discuss the reasons for your decision with your classmates. 3

−2

1 d x. x

y 2

1 we can see that the shaded areas x that represent the two integrals are equal in magnitude. However, one of the areas is −2 3 1 1 dx = − below the x-axis, which suggests that d x. −3 x 2 x

1

Consider the integrals

∫

2

1 d x and x

∫

−3

From the symmetry properties of the graph of y =

∫

∫

–3

–2

–1 O –1 –2

1

y=

1 x

2

3 x

Chapter 5: Integration

Evaluating

∫

3

∫

3

1 d x gives: x

2

3 1 3 d x = ln x 2 = ln 3 − ln 2 = ln 2 x

2

∫

This implies that

−2

1 3 2 d x = − ln = ln . 2 3 x

−3

WEB LINK

If we try using integration to find the value of

∫

−2

−3

∫

−2

−3

A more in-depth explanation for this formula can be found in the Two for one resource on the Underground Mathematics website.

1 d x we obtain: x

−2 1 2 d x = ln x −3 = ln( −2) − ln( −3) = ln 3 x

There is, however, a problem with this calculation in that ln x is only defined for x . 0 so ln( −2) and ln( −3) do not actually exist. 1 dx = ln x + c . x In conclusion, the integration formulae are written as:

Hence for x , 0, we say that

∫

TIP

KEY POINT 5.3

∫

1 dx = ln x + c x

∫

It is normal practice to only include the modulus sign when finding definite integrals.

1 1 dx = ln ax + b + c ax + b a

WORKED EXAMPLE 5.5

Find the value of

∫

3

2

6 d x. 2 − 3x

Answer

∫

3

2

3

6 6 dx = ln 2 − 3x 2 − 3x −3 2 = ( −2 ln 7) − ( −2 ln 4) = −2 ln 7 + 2 ln 4 4 = 2 ln 7

Substitute limits.

Simplify.

EXERCISE 5B 1 Find: a d

6

∫ x dx 6 ∫ 2x − 5 dx

2 Evaluate: 5 1 dx a 0 x +2 d

∫ ∫

4

2

2 dx 4x + 1

b e

b

1

∫ 2x dx 5 ∫ 2 − 3x dx ∫ ∫

4

1

e

2

0

c f

1 dx 2x + 1

c

3 dx 2x − 7

f

1

∫ 3x + 1 dx 3 ∫ 2(5x − 1) dx 4

3 dx −1 2 x + 5

∫ ∫

3

2

4 dx 3 − 2x

117

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3 Evaluate: a

∫

10

∫

3

4

b

1

c

2 + 5 dx 3x − 2

3 − 4 dx x 2x − 1

3

2x − 1 + 4 dx 2x + 3 −1

∫

4 a Given that

4x A ≡2+ , find the value of the constant A. 2x − 1 2x − 1

∫

b Hence show that

5

1

4x d x = 8 + ln 9. 2x − 1

5 a Find the quotient and remainder when 6x 2 + 5x is divided by 2 x − 5.

∫

b Hence show that

1

0

6 A curve is such that

6x 2 + 5x 23 5 − 25 ln . dx = 3 2x − 5 2

dy 3 = 2x + . dx x+e

Given that the curve passes through the point (e, e2 ), find the equation of the curve. 118

PS

7

y

O

1

k

x

The diagram shows part of the curve y =

2 . Given that the shaded region has area 4, find the value of k. x+3

dy 2 = 3 − . The tangents to the curve at the points dx x P and Q intersect at the point R. Find the coordinates of R.

8 The points P (1, −2) and Q(2, k ) lie on the curve for which

PS

5.3 Integration of sin( ax + b ), cos( ax + b ) and sec 2 ( ax + b ) In Chapter 4, we learnt how to differentiate some trigonometric functions: d (sin x ) = cos x dx d d (cos x ) = − sin x dx d (tan x ) = sec 2 x dx

d [sin( ax + b )] = a cos( ax + b ) dx dx d [cos( ax + b )] = − a sin( ax + b ) dx dx d [tan( ax + b )] = a sec 2 ( ax + b ) dx

Chapter 5: Integration

TIP

KEY POINT 5.4

∫

Since integration is the reverse process of differentiation, the rules for integrating are: 1

∫ cos x dx = sin x + c

∫ cos(∫ax + b) dx = a sin(ax + b) + c

∫ sin x dx = − cos x + c

+ b ) dx = − cos(ax + b ) + c ∫ sin(ax a ∫

∫ sec

∫ sec (ax + b) dx = a tan(ax + b) + c

2

It is important to remember that the formulae for differentiating and integrating these trigonometric functions only apply when x is measured in radians.

1

x dx = tan x + c

1

2

WORKED EXAMPLE 5.6 Find: a

∫ sin 2x dx

b

∫ cos 3x dx

b

∫ cos 3x dx = 3 sin 3x + c

c

∫ 4 sec

c

∫ 4 sec

2

x dx 2

Answer a

1

∫ sin 2x dx = − 2 cos 2x + c

1

2

x dx = 4 2

x dx 2 x = 4 × 2 tan + c 2

∫ sec

= 8 tan

WORKED EXAMPLE 5.7

Find the exact value of

∫

π 4

(3 − 2 sin 2 x ) d x.

0

Answer

∫

π 4

0

π

4 2 (3 − 2 sin 2 x ) d x = 3x + cos 2 x 2 0

3π π = + cos − ( 0 + cos 0 ) 4 2 3π = + 0 − (0 + 1) 4 3π = −1 4

Substitute limits.

2

x +c 2

119

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 5C 1 Find: x

a

∫ sin 3x dx

b

∫ cos 4x dx

c

∫ sin 2 dx

d

∫ 3sin 2x dx

e

∫ 5 cos 3x dx

f

∫ sec

g

∫ 2 cos(1 − 5x ) dx

h

∫ 3sin(2x + 1) dx

i

∫ 2 sec (5x − 2) dx

2

2 x dx 2

2 Evaluate: a

∫

1 π 6

b

cos 4x d x

0

0

d

∫

1 π 2

(sin 2 x + cos x ) d x

e

0

3 a

∫

2 π 3

Find

∫

1 π 6

1 sin x d x 2

c

(cos 2 x + sin x ) d x

f

∫

1 π 6

sec 2 2 x d x

0

0

1 π 4 1 − π 4

∫

(5 − 2 sec 2 x ) d x

d ( x sin x + cos x ). dx

b Hence find the exact value of

∫

1 π 3

x cos x d x.

0

120

4 A curve is such that

dy = 1 − 3sin 2 x. dx

π Given that the curve passes through the point , 0 , find the equation of the curve. 4 5 A curve is such that Given that

d2 y = −12 sin 2 x − 2 cos x. dx 2

dy π = 4 when x = 0 and that the curve passes through the point , −3 , find the equation of 2 dx

the curve. π π dy = 4 sin 2 x − . 6 The point , 5 lies on the curve for which 2 dx 2 a Find the equation of the curve. b Find the equation of the normal to the curve at the point where x = PS

7

π . 3

y

O

x

The diagram shows part of the curve y = 1 + 3 sin 2 x + cos 2 x. Find the exact value of the area of the shaded region.

Chapter 5: Integration

PS

8

y

M

π

O

x

The diagram shows part of the curve y = 3sin 2 x + 6 sin x and its maximum point M. Find the exact area of the shaded region. P PS

9

y √3 2

y = sin x

1 2

π 6

O

π 3

x

π π 1 3 lie on the curve. The diagram shows part of the graph of y = sin x. The points , and , 6 2 3 2 a Find the exact value of

b Hence show that P PS

10

∫

3 2 1 2

∫

π 3 π 6

sin x d x.

(sin −1 y ) d y =

(

)

π 2 3 −1 − 12

3 −1 . 2

y y = cos x

√2 2 1 2

O

π 4

π 3

x

π π 1 2 and , lie on the curve. The diagram shows part of the graph of y = cos x. The points , 3 2 4 2 a Find the exact value of

b Hence show that

∫

2 2 1 2

∫

π 3 π 4

cos x d x.

(cos −1 y ) d y =

π (3 2 − 4) + 24

3− 2 . 2

5.4 Further integration of trigonometric functions Sometimes, when it is not immediately obvious how to integrate a trigonometric function, we can use trigonometric identities to rewrite the function in a form that we can then integrate.

121

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

To integrate sin2 x and cos2 x we must use the double angle identities: cos 2 x ≡ 1 − 2 sin2 x and cos 2 x ≡ 2 cos2 x − 1 Rearranging these two identities gives: 1 1 sin2 x ≡ (1 − cos 2 x ) and cos2 x ≡ (1 + cos 2 x ) 2 2 WORKED EXAMPLE 5.8 Find: a

∫ sin

2

x dx

2

x dx =

b

∫ 3 cos 2x dx 2

Answer a

b

∫ sin

1 2 1 = 2 1 = 2

∫ (1 − cos 2x ) dx

3

3 2 3 = 2

= 122

1 (1 − cos 2 x ). 2

x − 1 sin 2 x + c 2 1 x − sin 2 x + c 4

∫ 3 cos 2x dx = 2 ∫ (1 + cos 4x ) dx 2

Use sin2 x ≡

Use cos2 2 x ≡

1 (1 + cos 4x ). 2

x + 1 sin 4x + c 4 3 x + sin 4x + c 8

WORKED EXAMPLE 5.9

Find

∫ sin

4

x dx .

Answer sin 4 x = (sin2 x )2

Use sin2 x ≡ 2

∫

1 = (1 − cos 2 x ) 2 1 = (1 − 2 cos 2 x + cos2 2 x ) 4 1 1 1 = 1 − 2 cos 2 x + + cos 4x 4 2 2 3 1 1 = − cos 2 x + cos 4x 8 2 8 3 1 1 sin 4 x dx = − cos 2 x + cos 4x dx 8 2 8

∫

=

1 3 1 sin 4x + c x − sin 2 x + 4 32 8

1 (1 − cos 2 x ). 2

Use cos2 2 x ≡

1 (1 + cos 4x ). 2

Chapter 5: Integration

To integrate tan2 x we use the identity 1 + tan2 x ≡ sec 2 x. Rearranging this identity gives tan2 x ≡ sec 2 x − 1. WORKED EXAMPLE 5.10

Find the exact value of

∫

π 4

5 tan2 x d x.

0

Answer

∫

π 4

5 tan x d x = 2

0

∫

π 4

(5 sec 2 x − 5) d x

Use tan2 x ≡ sec 2 x − 1.

0 π

= 5 tan x − 5x 4 0

Substitute limits.

5π − (0 − 0) = 5 − 4 5π = 5 − 4

Worked example 5.11 uses the principle that integration is the reverse process of differentiation. WORKED EXAMPLE 5.11 123

a Show that if y = sec x then

dy = tan x sec x. dx

b Hence find the exact value of

∫

1 π 4

(cos 2 x + 5 tan x sec x ) d x.

0

Answer y =

a

1 cos x

Use the quotient rule.

(cos x ) (0) − (1) ( − sin x ) dy = dx cos2 x sin x = cos2 x = tan x sec x b

∫

1 π 4

(cos 2 x + 5 tan x sec x ) d x

0 1 4 1 = sin 2 x + 5 sec x 2 0

1 = + 5 2 − (0 + 5) 2 9 =5 2− 2

Use

∫ tan x sec x dx = sec x.

π

Substitute limits.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 5.12 a Prove that 2 cosec 2 x tan x ≡ sec 2 x. b Hence find the exact value of

1 π 3 1 − π 3

∫

(5 + cosec 2 x tan x ) d x.

Answer a

2 cosec 2 x tan x ≡

sin x and cos x sin 2 x = 2 sin x cos x.

2 tan x sin 2 x

Use tan x =

2 sin x (2 sin x cos x ) cos x 1 ≡ cos2 x ≡

Simplify.

≡ sec 2 x b

1 π 3 1 − π 3

∫

(5 + cosec 2 x tan x ) d x =

1 π 3 1 − π 3

∫

5 + 1 sec 2 x d x 2 1

Use part a.

π

3 1 = 5x + tan x − 1 π 2

Simplify.

3

5π 3 5π 3 = + − − − 2 3 2 3

124

=

10 π + 3 3

WORKED EXAMPLE 5.13 a Use the expansions of cos(3x − x ) and cos(3x + x ) to show that: cos 2 x − cos 4x ≡ 2 sin 3x sin x b Hence show that

∫

1 π 4

0

1 sin 3x sin x d x = . 4

Answer a

cos 2 x − cos 4x ≡ cos(3x − x ) − cos(3x + x ) ≡ (cos 3x cos x + sin 3x sin x ) − (cos 3x cos x − sin 3x sin x ) ≡ 2 sin 3x sin x

b

∫

1 π 4

sin 3x sin x d x =

0

1 2

∫

1 π 4

(cos 2 x − cos 4x ) d x 1

4 1 1 1 = sin 2 x − sin 4x 2 2 4 0

1 1 − 0 − (0 − 0) 2 2 1 = 4

=

Use part a.

0 π

Simplify.

Chapter 5: Integration

EXERCISE 5D 1 Find: a

∫ 3 cos

2

x dx

b

∫ 4 cos

x dx 2

c

∫

sin2 3x dx

d

∫ 2 tan

2

x dx

e

∫ 6 tan (3x ) dx

f

∫

cos 4 x dx

2

2

2 Find the value of: a

d

∫ ∫

1 π 3

2

sin x d x

b

0 1 π 8

cos2 2 x d x

e

0

∫

1 π 3 1 π 6

∫

1 π 6

2

3tan x d x

c

∫

1 π 3

cos2 x d x

0

4 sin2 x d x

f

0

∫

1 π 6

tan2 2 x d x

0

3 Find the value of: a

d

4 a

∫ ∫

1 π 6 0 1 π 3 0

1 dx cos2 x − cos2 x

b

(cos x − 3sin x )2 d x

e

5 a

6 a

∫

1 π 4 1 π 6

cosec 2 2 x d x =

Show that tan x + cot x ≡

b Hence show that

∫

0 1 π 6 0

(cos2 x + sin 2 x ) d x

1 π 2 1 π 3

∫

1 + cos 4 x dx cos2 x

3 . 6

2 . sin 2 x

8 d x = 1. tan x + cot x

Use the expansions of sin(5x + 2 x ) and sin(5x − 2 x ) to show that: sin 7 x + sin 3x ≡ 2 sin 5x cos 2 x

b Hence show that

7 a

1 π 3 1 π 6

∫

11 − 3 3 . 42

Show that sin 3x ≡ 3sin x − 4 sin3 x.

b Hence show that

∫

1 π 3

0

8 a

2 sin 5x cos 2 x d x =

2 sin3 x d x =

5 . 12

Show that cos 3x ≡ 4 cos3 x − 3 cos x.

b Hence show that

∫

1 π 6

0

(4 cos3 x + 2 cos x ) d x =

c

∫

1 π 4

(2 sin x + cos x )2 d x

0

cos x dy , show that if y = cot x then = − cosec 2 x. d x sin x

By differentiating

b Hence, show that

∫

1 π 6

17 . 6

f

∫

1 π 6

0

5 dx 1 + cos 4x 125

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PS

9 The diagram shows part of the curve y = cos x + 2 sin x. Find the exact volume of the solid formed when the shaded region is rotated through 360 ° about the x-axis.

y

O

PS

10 a

b Show that if y = sec x then c

Hence show that

∫

1 π 6

0

P2

1π 2

1 Show that sec x + sec x tan x ≡ . 1 − sin x 2

x

dy = sec x tan x. dx

4 d x = 2(1 + 3 ). 1 − sin 2 x

5.5 The trapezium rule We already know that the area, A, under the curve y = f( x ) between x = a and x = b can be found by evaluating

∫

y

b

y = f(x)

f( x ) d x. Sometimes we might not be able to find the value of

a

∫

b

A

f( x ) d x by the integration methods that we have covered so far or maybe the function

a

cannot be integrated algebraically. In these situations an approximate answer can be found using a numerical method called the trapezium rule. 126

This numerical method involves splitting the area under the curve y = f( x ) between x = a and x = b into equal width strips. y

y = f(x) y0

y1 h

O

y3

y2 h

y4

h

h

y5 h

a

y6 h b

x

The area of each trapezium-shaped strip can be found using the formula: 1 area = (sum of parallel sides) × width 2 The sum of the areas of all the strips gives an approximate value for the area under the curve and hence an approximate value for

∫

b

f( x ) d x.

a

b−a and the 6 length of the vertical edges (ordinates) of the strips are y0 , y1, y2 , y3 , y4 , y5 and y6 . For the previous diagram, there are 6 strips, each of width h, where h =

The sum of the areas of the trapezium-shaped strips is: h h h h h h ( y0 + y1 ) + ( y1 + y2 ) + ( y2 + y3 ) + ( y3 + y4 ) + ( y4 + y5 ) + ( y5 + y6 ) 2 2 2 2 2 2 h = [ y0 + y6 + 2( y1 + y2 + y3 + y4 + y5 )] 2

O

a

b

x

Chapter 5: Integration

When there are n strips and the ordinates are y0 , y1, y2 , y3 , … , yn −1, yn then the sum of the areas of the strips is: h h h h h ( y0 + y1 ) + ( y1 + y2 ) + ( y2 + y3 ) + … + ( yn − 2 + yn −1 ) + ( yn −1 + yn ) 2 2 2 2 2 Hence we can find an approximate value for

∫

b

f ( x ) d x using: a

KEY POINT 5.5

TIP • The width of the strip is the interval along the x-axis.

b−a h [ y0 + yn + 2( y1 + y2 + y3 + … + yn −1 )] where h = . n 2 a An easy way to remember this rule in terms of the ordinates is:

∫

b

f (x ) dx ≈

∫

b

• The number of ordinates is one more than the number of strips.

f ( x ) d x ≈ half width of strip × (first + last + twice the sum of all the others).

a

• You should not use a numerical method when an algebraic method is available to you, unless you are specifically asked to do so.

WORKED EXAMPLE 5.14 The diagram shows the curve y = x − 2 ln x.

y

Use the trapezium rule with 2 intervals to estimate the shaded area, giving your answer correct to 2 decimal places. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value.

127 O

0.5

3.5

Answer a = 0.5, b = 3.5 and h = 1.5 y

O

0.5

2

3.5

x

0.5

2

3.5

y

1.8863

0.6137

0.9945

y0

y1

y2

x

h [ y0 + y2 + 2 y1 ] 2 1.5 ≈ [1.8863 + 0.9945 + 2 × 0.6137] 2 ≈ 3.08115 …

Area ≈

≈ 3.08 (to 2 decimal places) It can be seen from the diagram that this is an over-estimate since the top edges of the strips all lie above the curve.

x

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 5.15 y

The diagram shows the curve y = 2 4x − x 2 . Use the trapezium rule with 4 intervals to estimate the value of

∫

4

2 4x − x 2 d x giving your answer correct

O

0

4x

to 2 decimal places. State, with a reason, whether the trapezium rule gives an under-estimate or an overestimate of the true value. Answer a = 0, b = 4 and h = 1

y

x

0

1

2

3

4

y

0

2 3

4

2 3

0

y0

y1

y2

y3

y4

∫

4

h 2 1 ≈ 2 ≈4

2 4x − x 2 d x ≈

0

O

1

2

3

4x

[ y0 + y4 + 2( y1 + y2 + y3 )]

(

)

0 + 0 + 2 2 3 + 4 + 2 3 3+4

≈ 10.93 (to 2 decimal places)

128

It can be seen from the diagram that this is an under-estimate since the top edges of the strips all lie below the curve. Worked example 5.14 involves a ‘concave’ curve and the diagram shows that the trapezium rule gives an over-estimate for the area. Worked example 5.15 involves a ‘convex’ curve and the diagram shows that the trapezium rule gives an under-estimate for the area. If a curve is partially convex and partially concave over the required interval it is not so easy to predict whether the trapezium rule will give an under-estimate or an over-estimate for the area. The more strips that are used, the more accurate the estimate will be. EXERCISE 5E 1 Use the trapezium rule with 2 intervals to estimate the value of each of these definite integrals. Give your answers correct to 2 decimal places. a

∫

4

∫

7

x2 − 2 dx

b

1

e x − 12 d x

c

0

2

d

∫

2

2 dx ln( x + 5)

e

∫

1 π 3

0

∫

4

∫

12

0

1 dx 2 + tan x

f

2

2 dx 3 + ex log10 x d x

DID YOU KNOW? If you double the number of strips in a trapezium rule approximation, the error reduces to approximately one quarter the previous error.

Chapter 5: Integration

2

y

O

x

The diagram shows part of the curve y = x 2 e − x . Use the trapezium rule with 4 intervals to estimate the value of

∫

5

x 2 e − x d x , giving your answer correct to

1

2 decimal places. 3 a

Use the trapezium rule with 6 intervals to estimate the value of

∫

7 π 4 1 π 4

1 cosec x d x, giving your answer correct to 2 decimal places. 2

1 b Use a sketch of the graph of y = cosec x for 0 ø x ø 2 π, to explain 2 whether the trapezium rule gives an under-estimate or an over-estimate of the true value of

4

∫

7 π 4 1 π 4

1 cosec x d x. 2

y

129

1 π 2

O

x

The diagram shows the curve y = e x cos x for 0 ø x ø rule with 3 intervals to estimate the value of

∫

1 π 2

1 π. Use the trapezium 2

e x cos x d x, giving your answer

0

correct to 2 decimal places. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of

∫

1 π 2

e x cos x d x.

0

5

y

O

1

2

3

x

ex . Use the trapezium rule with 4 The diagram shows part of the curve y = 2x intervals to estimate the area of the shaded region, giving your answer correct to 2 decimal places. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the shaded area.

WEB LINK Try the When does the trapezium rule give the exact answer here? resource on the Underground Mathematics website.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Checklist of learning and understanding

∫

Integration formulae

P2

1

●

∫e

●

∫

∫ e ∫ dx = a e + c 1 1 ∫ ax +∫ b dx = a ln ax + b

●

∫

∫ cos(ax + b) dx = a sin(ax + b) + c

●

∫ sin x dx = − cos x + c

∫ sin(∫ax + b) dx = − a cos(ax + b) + c

●

∫ sec

∫ sec (ax + b) dx = a tan(ax + b) + c

x

ax + b

dx = e x + c

1 dx = ln x + c x 1 1 cos x dx = sin x + c

2

ax + b

+c

1

1

x dx = tan x + c

2

1

The trapezium rule ●

The trapezium rule can be used to find an approximate value for

∫

b

f( x ) d x. If the region under

a

the curve is divided into n strips each of width h, then: b

∫ f(x ) dx ≈ 2 [ y h

0

a

130

+ yn + 2( y1 + y2 + y3 + … + yn −1 )] where h =

b−a . n

Chapter 5: Integration

END-OF-CHAPTER REVIEW EXERCISE 5

∫

1 π 3 1 π 6

4 3 π − . 3 6

(1 + 2 tan2 x ) d x =

1

Show that

2

Find the exact value of

∫

−2

−4

[4]

4 − 2 x d x , giving your answer in the form a + ln b , where a, b and c 1 − 2x

c are integers. P2

[4]

3

y

R

O

1

x

The diagram shows part of the curve y = 2 − x 2 ln( x + 1) . The shaded region R is bounded by the curve and by the lines x = 0, x = 1 and y = 0 . Use the trapezium rule with 4 intervals to estimate the area of R giving your answer correct to 2 decimal places. State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of R. 4

∫

Find the exact value of

1 π 6

[4]

(cos 3x − sin 2 x ) d x.

[5]

9 2 23 e − 12e + . 2 2

[5]

0

5

∫

Show that

1

0

6

a

Find

∫

k

(3e x − 2)2 d x =

131

(5e −2 x + 2e −3x ) d x , where k is a positive constant.

[4]

0

b

Hence find the exact value of

∫

∞

(5e −2 x + 2e −3x ) d x.

[1]

0

7

Show that

∫

4

3 3 21 d x = ln . 5x + 1 5 11

2

8

[5]

a

Find the value of the constant A such that

b

Hence show that

8x A ≡ 4+ . 2x + 5 2x + 5

3

11 8x d x = 8 − 10 ln . 7 2x + 5 3 Show that 12 sin2 x cos2 x ≡ (1 − cos 4x ) . 2

∫

[2] [5]

1

9

i ii

Hence show that

∫

1 π 3 12 sin2 1 π 4

x cos2 x d x =

π 3 3 . + 16 8

[3] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 June 2013 10 a b

Find

∫ 4e (3 + e x

Show that

1 π 4 1 − π 4

∫

2x

) dx.

[3]

(3 + 2 tan2 θ ) dθ =

1 (8 + π ) . 2

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2011

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

11 i ii

Show that (2 sin x + cos x )2 can be written in the form Hence find the exact value of

∫

1 π 4 0

5 3 + 2 sin 2 x − cos 2 x. 2 2

(2 sin x + cos x )2 d x .

[5] [4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2012 12 i ii

By differentiating

cos x dy , show that if y = cot x then = − cosec 2 x. sin x dx

[3]

By expressing cot 2 x in terms of cosec 2 x and using the result of part i, show that

∫

1 π 2 1 π 4

cot 2 x d x = 1 −

1 π. 4

[4]

iii Express cos 2 x in terms of sin2 x and hence show that Hence, using the result of part i, find

∫

1 1 can be expressed as cosec 2 x. 1 − cos 2 x 2

1 dx. 1 − cos 2 x

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 June 2010 13 i ii

132

∫

Find

1 + cos 4 2 x dx. cos2 2 x

[3]

Without using a calculator, find the exact value of

∫

14

4

2 + 6 d x, giving your answer in the form 3x − 2

b

ln( ae ), where a and b are integers.

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2016 14

y Q

π

O

x

1 1 The diagram shows the curve y = x sin x, for 0 ø x ø π . The point Q π, π lies on the curve. 2 2 i

Show that the normal to the curve at Q passes through the point ( π, 0).

ii

Find

d (sin x − x cos x ). dx

iii Hence evaluate

∫

1 π 2

[5] [2]

x sin x d x.

[3]

0

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 November 2010 15 i

ii

Using the expansion of cos(3x − x ) and cos(3x + x ), prove that Hence show that

∫

1 π 3 1 π 6

sin 3x sin x d x =

1 3. 8

1 (cos 2 x − cos 4x ) ≡ sin 3x sin x. 2

[3]

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2010

133

Chapter 6 Numerical solutions of equations In this chapter you will learn how to:

■ ■ ■

locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change understand the idea of, and use the notation for, a sequence of approximations that converges to a root of an equation understand how a given simple iterative formula of the form xn +1 = F( xn ) relates to the equation being solved, and use a given iteration, or an iteration based on a given rearrangement of an equation, to determine a root to a prescribed degree of accuracy.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE / O Level Mathematics

Substitute numbers for letters in complicated formulae and algebraic expressions, including those involving logarithmic functions, exponential functions and major and minor trigonometric functions.

1 Evaluate each of the following expressions when x = 1.5. 2 1 a x+ 2 3 x

IGCSE / O Level Mathematics

Rearrange complicated formulae and equations.

b 100 sin x − 99x, (x in radians) c

e−x

d

ln x + 5

2 Rearrange each formula so that x is the subject. a b c

Pure Mathematics 1 Coursebook, Chapters 1 and 2

134

Understand the relationship between a graph and its associated algebraic equation, and use the relationship between points of intersection of graphs and solutions of equations.

y = 4x 2 + 13, given x > 0 y − 3x + 7 w y 1 + 3= 3 5 x 0=

3 Show that the x-coordinates of the solutions to the simultaneous equations: y = x 2 − 6x + 10 y=x also satisfy the equation x 2 − 7 x + 10 = 0.

Pure Mathematics 1 Coursebook, Chapter 1

Know that the values of x that make both sides of an equation equal are called the roots of the equation.

4 Find the roots of the equation x 2 − 7 x + 10 = 0.

Also, the roots of f( x ) = 0 are the x-intercepts on the graph of y = f( x ). IGCSE / O Level Mathematics Pure Mathematics 1 Coursebook, Chapter 5

Chapters 2 and 3

Sketch graphs of quadratic or cubic functions or the trigonometric functions sine, cosine, tangent.

Sketch graphs of exponential or logarithmic functions or the trigonometric functions cosecant, secant, cotangent.

5 Sketch the graph, y = f( x ), of each function. a

f( x ) = x 3 + 1

b

f( x ) = cos 2 x − 3

c

f( x ) = 4x 2 − 16x

6 Sketch the graph, y = g( x ), of each function. ex a g( x ) = 2 b g( x ) = ln( x − 1) c

g( x ) = cosec x

Chapter 6: Numerical solutions of equations

Why solve equations numerically? You are used to solving equations using direct, algebraic methods such as factorising or using the quadratic formula. You might be surprised to learn that not all equations can be solved using a direct, algebraic process. Numerical methods are ways of calculating approximate solutions to equations. They are extremely powerful problem-solving tools and many are available. They are widely used in engineering, computing, finance and many other applications. A well-programmed computer is quickly able to solve a complicated equation using a numerical method. So why do we need to understand how the method works? Knowing how a numerical method works helps us to: ● ● ● ●

understand when the results are likely to be reasonable understand how to use available software correctly select an appropriate method when choices are available write our own programs when we need to do so, if we have the programming skills!

To be a powerful problem-solving tool, the method must be used correctly. This might mean the method needs to work quickly or work to obtain a particular level of accuracy. When used incorrectly, numerical methods might be very slow or even fail to work, as we will see later in this chapter. In this chapter we will focus on one numerical method, the numerical solution of an equation using an iterative formula.

135

6.1 Finding a starting point As we will see throughout this chapter, a good starting point is very important. We need a method to find an approximate solution to an equation first. We might find this using, for example, a graphical approach or the change of sign approach. For suitable functions, each of these approaches will produce two values: one above and one below the solution that we are finding. WORKED EXAMPLE 6.1 By sketching graphs of y = x 3 and y = 4 − x, show that the equation x 3 + x − 4 = 0 has a root α between 1 and 2. Answer y y = x3 y = 4–x

4

O

TIP

1

2

4

x

Make sure your sketch shows all the key features of each graph, such as the coordinates of turning points and the coordinates of any intercepts.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

There is one point of intersection and its x-coordinate is between 1 and 2. At this point x 3 = 4 − x . This means that the equation x 3 = 4 − x has one root between 1 and 2. Rearranging this equation gives x 3 + x − 4 = 0. Hence x 3 + x − 4 = 0 has a root, α , between 1 and 2.

WORKED EXAMPLE 6.2 Show, by calculation, that the equation f( x ) = x5 + x − 1 = 0 has a root α between 0 and 1. Answer Find the value of α such that f(α ) = 0.

Why? Think about it and check the explanation that follows.

f(0) = 05 + 0 − 1 = −1

As α is between 0 and 1, then f(0) might have the opposite sign to f(1).

f(1) = 15 + 1 − 1 = 1

136

This conclusion is important as it completes the argument.

The change of sign indicates the presence of a root, so 0 , α , 1.

To see why the change of sign method works in this case, look at the graph of y = f( x ) . y f(1) is positive

O

1

x

There are no breaks f(0) is negative in the graph: it is continuous.

Since it starts below the x-axis at x = 0 and finishes above the x-axis at x = 1, it must cross the x-axis somewhere in between, say at x = α , where 0 , α , 1. So α is a root of the equation, that is, f(α ) = 0.

Chapter 6: Numerical solutions of equations

WORKED EXAMPLE 6.3 a By sketching a suitable pair of graphs, show that the equation cos x = 2 x − 1 (where x is in radians) has 1 only one root for 0 ø x ø π . 2 b Verify by calculation that this root lies between x = 0.8 and x = 0.9. Answer a Draw the graphs of y = cos x and y = 2 x − 1.

Why are these graphs suitable?

y y = 2x – 1 1 y = cos x O

0.5

π 2

x

–1

The graphs intersect once only and so the equation 1 cos x = 2 x − 1 has only one root for 0 ø x ø π . 2 b cos x = 2 x − 1 so cos x − 2 x + 1 = 0 Let f( x ) = cos x − 2 x + 1 and so f( x ) = 0 then f(0.8) = cos 0.8 − 2(0.8) + 1 = 0.0967… f(0.9) = cos 0.9 − 2(0.9) + 1 = −0.1783… Change of sign indicates the presence of a root.

EXERCISE 6A 1 a On the same axes, sketch the graphs of y = 1 + x and y = x 2. b Using your answer to part a, explain why the equation x 2 − 1 + x = 0 has two roots. c

Show, by calculation, that the smaller of these two roots lies between x = −1 and x = 0 .

2 a By sketching graphs of y = x 3 + 5x 2 and y = 5 − 2 x, determine the number of real roots of the equation x 3 + 5x 2 + 2 x − 5 = 0. b Verify by calculation that the largest root of x 3 + 5x 2 + 2 x − 5 = 0 lies between x = 0 and x = 2 . 3 The curve y = x 3 + 5x − 1 cuts the x-axis at the point (α , 0). a By sketching the graph of y = x 3 and one other suitable graph, deduce that this is the only point where the curve y = x 3 + 5x − 1 cuts the x-axis. b Using calculations, show that 0.1 , α , 0.5.

137

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

4 a On the same axes, sketch the graphs of y = ln( x + 1) and y = 3x − 4. b Hence deduce the number of roots of the equation ln( x + 1) − 3x + 4 = 0 . 5 a By sketching graphs of y = e x and y = x + 6, determine the number of roots of the equation e x − x − 6 = 0. b Show, by calculation, that e x − x − 6 = 0 has a root that lies between x = 2.0 and x = 2.1. 6 a Show, by calculation, that ( x + 2)e5 x = 1 has a root between x = 0 and x = −0.2. b Show, by sketching the graph of y = e5 x and one other suitable graph, that this is the only root of this equation. 7 a Show by calculation that the equation cos−1 2 x = 1 − x has a root α , where 0.4 ø α ø 0.5. b The diagram shows the graph of y = cos−1 2 x where −0.5 ø x ø 0.5. y π 3π 4 π 2 π 4

138

–1

–0.5

O

0.5

1

x

By drawing a suitable graph on a copy of the diagram, show that α is the only root of cos −1 2 x = 1 − x for −0.5 ø x ø 0.5. sin x π 8 The curve with equation y = where x , − intersects the line y = 1 at 2 2x + 3 the point (α , 1). π radians and one a By sketching the graph of y = sin x for −2 π , x , − 2 other suitable graph, deduce that this is the only point where the curve sin x y= intersects the line y = 1. 2x + 3 b Using calculations, show that −2 , α , − 1.9. PS

9 a By sketching a suitable pair of graphs, show that the equation x 3 + 4x = 7 x + 4 has only one root for 0 ø x ø 5 . b Verify by calculation that this root lies between x = 2 and x = 3.

PS

10 a Show graphically that the equation 2 x = x + 4 has exactly two roots. b Show, by calculation, that the larger of the two roots is between 2.7 and 2.8.

PS

11 a By sketching two suitable graphs on the same diagram, show that the π equation cot x = x 2 has one root between 0 and radians. 2 b Show, by calculation, that this root, α , lies between 0.8 and 1.

REWIND You might need to look back at Chapter 3 before attempting Question 11.

Chapter 6: Numerical solutions of equations

PS

12 a By sketching a suitable pair of graphs, deduce the number of roots of the equation x = tan 2 x for 0 , x , 2 π. b Verify, by calculation, that one of these roots, α , lies between 2.1 and 2.2.

PS

13 a Show graphically that the equation cosec x = sin x has exactly two roots for 0 , x , 2 π. b Use an algebraic method to find the value of the larger root correct to 3 significant figures.

PS

14 f( x ) = 20 x 3 + 8x 2 − 7 x − 3 a The equation 20 x 3 + 8x 2 − 7 x − 3 = 0 has exactly two roots. Without factorising the cubic equation, show, by calculation, that one of these roots is between 0.5 and 1. b Show, by sketching the graph of y = f( x ) for −1 ø x ø 1, that the smaller root is between −1 and 0. c

PS

Explain why it is not necessary to use a numerical method to find the two solutions of this equation.

15 A guitar tuning peg is in the shape of a cylinder with a hemisphere at one end.

r mm

The cylinder is 20 mm long and the whole peg is made from 800 mm 3 of plastic. The base radius of the cylinder is r mm . a Show that πr 3 + 30 πr 2 − 1200 = 0. b Show that the value of r is between 3 mm and 4 mm.

EXPLORE 6.1 Anil is trying to find the roots of f( x ) = tan x = 0 between 0 and 2π radians. He states: f(3.1) . 0 and f(3.2) . 0. There is no change of sign so there is no root of tan x = 0 between 3.1 and 3.2. Mika states: y = tan x and it meets the x-axis where x = π radians so Anil has made a calculation error. Discuss Anil and Mika’s statements.

DID YOU KNOW? Interestingly, cubic and quartic equations can be solved algebraically. This has been known since the 16th century and was shown by the Italian mathematicians Niccolò Fontana (Tartaglia) and Lodovico Ferrari. Quintic equations cannot be solved algebraically in general. This was shown by the Norwegian mathematician Niels Abel and the French mathematician Évariste Galois in the early 19th century.

20 mm

139

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

6.2 Improving your solution Once we have an approximate location for a solution, we can improve it. One common way to do this is to use an iterative process, finding values that are closer and closer to α , the root we are looking to find. For example, we have shown, by calculation, that the equation f( x ) = x5 + x − 1 = 0 has a root α between 0 and 1. This is a reasonable starting point but probably not accurate enough if we wanted a solution to this equation in the real world. What if we needed to know the value correct to 1 significant figure? How could we find it? Let us consider a rearrangement of f( x ) = x5 + x − 1 = 0. The root we are trying to find is a value of x, so rearranging x5 + x − 1 = 0 into the form x = F( x ) seems reasonable (in other words, x = some function of x). One possible rearrangement of this form is x = 5 1 − x . As we already know that 0 , α , 1, a good starting point is x = 0.5. Now we can test our value to see how accurate it is. How? ●

Work out the value of 51−

●

0.5 =

0.5 = 0.870 55...

F( x ) = 5 1 − x

Are x and F( x ) the same correct to 1 significant figure?

0.870 55 …

0.5 ●

1 − x for the starting value.

Compare it with x. x

140

5

5

No

Are the values the same to the required number of significant figures or decimal places? Yes – then use this value as the estimate for the root. No – then find a more accurate value.

Look at the graphs of y = x and y = 5 1 − x on the same diagram. y 1

The graphs show that the values x and F( x ) are not very close together as the vertical distance between the line and curve is quite large.

0.5

y=x

y=51–x O

1

x

REWIND Remember from Section 6.1 that α is just the special name given to the solution that you are looking for.

Chapter 6: Numerical solutions of equations

How do we improve the accuracy of the estimate for the root? We can see in the following diagram that 0.870 55... is a better approximation for the root α than 0.5. Is 0.870 55... close enough? y

1

The vertical distance between the line and curve is now smaller, so the values are converging.

0.5

y=x

y=51–x 1

O

Testing x = 0.870 55... gives x

F( x ) = 5 1 − x

51−

x

141

x = 5 1 − 0.870 55... = 0.664 38...

Are x and F( x ) the same correct to 1 significant figure?

0.5

0.870 55...

No

0.870 55...

0.664 38...

No

We can see that 0.664 38... is a better approximation of the root α than 0.870 55... because the values are closer together. Is 0.664 38... close enough? When we continue testing and comparing values, we find: x

F( x ) = 5 1 − x

Are x and F( x ) the same correct to 1 significant figure?

0.664 38...

0.80383...

No

0.80383...

0.72197...

No

0.72197...

0.77413...

No

0.77413...

0.742 62...

No

0.742 62...

0.762 27...

No

0.762 27...

0.750 26...

Yes

Each test is called an iteration. Each iteration gives a value of x that is closer to the actual value of the root than the previous value of x. It seems as if the value of α is 0.8 correct to 1 significant figure.

TIP Using a table is helpful, but still very repetitive and a lot of work!

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

This (enlarged) diagram shows what is happening with each new value. y

1

start

0.5

y=x

142

y = 51 – x

O

1

x

Each value that comes from F( x ) is used as the next value for x. The spiral is often called the cobweb pattern.

Summary When trying to solve f( x ) = 0, rearrange the equation to the form x = F( x ). When α is the value of x such that x = F( x ) then α is also a solution of f( x ) = 0.

Subscript notation and formulae In order to avoid unnecessary repetition and to make sure the process is recorded accurately, the method outlined is usually described using subscript notation. The first approximation of the root is called x1 (sometimes it is called x0). The next approximation of the root is called x2 , and so on. A relationship such as x = F( x ) where F( x ) = 5 1 − x is the basis for an iterative formula: xn +1 = F( xn ) = 5 1 − xn

Chapter 6: Numerical solutions of equations

Looking at the values we already have we can see how much easier this is to record: x1 = 0.5

xn +1 = F( xn ) = 5 1 − xn

Are they the same correct to 1 significant figure?

x2 = 0.870 55...

No

x3 = 0.664 38...

No

x4 = 0.80383...

No

x5 = 0.72197...

No

x6 = 0.77413...

No

x7 = 0.742 62...

No

x8 = 0.762 27...

No

x9 = 0.750 26...

Yes

In this format we test the accuracy of the values by comparing each value of xn in the sequence with the previous value. TIP Does your calculator have an Ans key? Using the Ans key reduces the number of key strokes needed to use an iterative process. • Type the value for x1 into your calculator and press = . e.g. 0.5 =

143

• Type the formula into your calculator using Ans instead of xn . e.g.

51−

Ans

Now each time you press = the calculator gives you the next value of xn generated (so x2 , x3 , x4 , …). Try this now and check that you generate the values that are given in the table.

WORKED EXAMPLE 6.4 The equation x 2 + x − 3 = 0 has a negative root, α . a Show that this equation can be rearranged as x =

3 − 1. x

3 − 1 with a starting value of x1 = −2.5 xn to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places, where appropriate.

b Use the iterative formula xn +1 =

Answer a

x2 + x − 3 = 0

Divide through by x.

2

x x 3 0 + − = x x x x 3 x +1− = 0 x x=

3 −1 x

Isolate x. ✓

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

b

x1 = −2.5

xn +1 =

3 −1 xn

Are these the same correct to 2 decimal places?

x2 =

3 − 1 = −2.2 −2.5

No

x3 =

3 − 1 = −2.3636 −2.2

No

x4 = −2.2692

No

x5 = −2.3220

No

x6 = −2.2920

No

x7 = −2.3089

No

x8 = −2.2993

No

x9 = −2.3047

Yes

The root seems to be α = −2.30 to 2 decimal places. We can prove this, using a change of sign test: When α = −2.30 then −2.305 , α , − 2.295 and so f( −2.305) = ( −2.305)2 + ( −2.305) − 3 = 0.00802... 2

144

f( −2.295) = ( −2.295) + ( −2.295) − 3 = −0.0279... Change of sign indicates presence of root, therefore α = −2.30, to 2 decimal places.

TIP Any value between −2.305 and −2.295 must round to −2.30 to 2 decimal places. The change of sign test is an important test. If the root passes the change of sign test it must be correct.

WORKED EXAMPLE 6.5 The equation e x − 1 = 2x has a root α = 0. a Show by calculation that this equation also has a root, β , such that 1 , β , 2 . b Show that this equation can be rearranged as x = ln(2 x + 1) . c Use an iteration process based on the equation in part b, with a suitable starting value, to find β correct to 3 significant figures. Give the result of each step of the process to 5 significant figures. Answer a Let f( x ) = e x − 1 − 2 x = 0 and look for a change of sign. f(1) = e − 1 − 2 = −0.2817… f(2) = e2 − 1 − 2(2) = 2.38905… Change of sign indicates presence of root.

Chapter 6: Numerical solutions of equations

b

ex − 1 = 2x

Isolate e. Take logs to base e.

ex = 2x + 1 x = ln(2 x + 1)

✓

c A suitable iterative formula is xn +1 = ln(2 xn + 1) with a starting value of x1 = 1.5. (Any value between 1 and 2 would be an acceptable starting value.) x1 = 1.5

xn +1 = = ln(2 xn + 1)

Are these the same correct to 3 significant figures?

x2 = ln(2 × 1.5 + 1) = 1.38629… = 1.3863

No

x3 = ln(2 × 1.38629… + 1) = 1.32776… = 1.3278

No

x4 = 1.29623… = 1.2962

No

x5 = 1.27884… = 1.2788

No

x6 = 1.26910… = 1.2691

No

x7 = 1.26362… = 1.2636

No

x8 = 1.26051… = 1.2605

Yes

The root seems to be β = 1.26 correct to 3 significant figures. We can prove this, using a change of sign test: When β = 1.26 then 1.255 , β , 1.265 and so: f(1.255) = e1.255 − 1 − 2(1.255) = −0.0021616… f(1.265) = e1.265 − 1 − 2(1.265) = 0.01309…

Change of sign indicates presence of root, therefore β = 1.26 correct to 3 significant figures. You might notice that the pattern of convergence is different in this example. This is clear when you look at the graphs of y = x and y = ln(2 x + 1) on the same grid. The type of convergence shown in the graph on the right is often called a staircase pattern. y 2

1

y = ln(2x + 1)

y=x O

1

2 x

145

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 6B 1 The equation x 3 + 5x − 7 = 0 has a root α between 1 and 1.2. 7 − xn 3 can be used to find the value of α . The iterative formula xn +1 = 5 a Using a starting value of x1 = 1.1, find and write down, correct to 4 decimal places, the value of each of x2 , x3 , x4 , x5 , x6. b Write the value of x6 correct to 2 decimal places and prove that this is the value of α correct to 2 decimal places. 2 a Show that the equation ln( x + 1) + 2 x − 4 = 0 has a root α between x = 1 and x = 2 . b Use the iterative formula xn +1 = correct to 3 decimal places.

4 − ln( xn + 1) with an initial value of x1 = 1.5 to find the value of α 2

3 The equation ( x − 0.5) e3x = 1 has a root α . a Show, by sketching the graph of y = e3x and one other suitable graph, that α is the only root of this equation. b Use the iterative formula xn +1 = e −3x + 0.5 with x1 = 0.5 to find the value of α correct to 2 decimal places. Give the value of each of your iterations to 4 decimal places. n

4 a By sketching a suitable pair of graphs, show that the equation x 3 + 10 x = x + 5 has only one root that lies between 0 and 1. 146

5 − xn 3 , with a suitable value for x1, to find the value of this root correct 9 to 4 decimal places. Give the result of each iteration to 6 decimal places.

b Use the iterative formula xn +1 =

5 The equation cos−1 3x = 1 − x has a root α . π π and . 15 12 1 b Show that the given equation can be rearranged into the form x = cos (1 − x ). 3 1 c Using the iterative formula xn+1 = cos (1 − xn ) with a suitable starting value, x1, find the value of α 3 correct to 3 decimal places. Give the result of each iteration to 5 decimal places. a Show, by calculation, that α is between

6 The terms of a sequence with first term x1 = 1 are defined by the iterative formula: x n +1 =

5 − 2 xn − xn 3 5

The terms converge to the value α . a Use this formula to find the value of α correct to 2 decimal places. Give the value of each term you calculate to 4 decimal places. b The value α is the root of an equation of the form ax 3 + bx 2 + cx + d = 0 , where a, b, c and d are integers. Find this equation. 7 The equation x 4 − 1 − x = 0 has a root, α , between x = 1 and x = 2 . a Show that α also satisfies the equation x = 4 1 + x . b Write down an iterative formula based on the equation in part a. c

Use your iterative formula, with a starting value of x1 = 1.5, to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

Chapter 6: Numerical solutions of equations

8 The equation cosec x = x 2 has a root, α , between 1 and 2. 1 1 The equation can be rearranged either as x = sin−1 2 or x = . x sin x a Write down two possible iterative formulae, one based on each given rearrangement. Use the starting value 1.5. b Show that one of the formulae from part a fails to converge. c

Show that the other formula from part a converges to α and find the value of α correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

9 The terms of a sequence, defined by the iterative formula xn +1 = ln( xn 2 + 4), converge to the value α . The first term of the sequence is 2. a Find the value of α correct to 2 decimal places. Give each term of the sequence you find to 4 decimal places. b The value α is a root of an equation of the form x 2 = f( x ). Find this equation. sin ( x − 1) + 1 = 0 has a root, α , between x = 1 and x = 1.4. 2x − 3 3 − sin ( x − 1) a Show that α also satisfies the equation x = . 2 b Using an iterative formula based on the equation from part a, with a suitable starting value, find the value of α correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

10 The equation

x

PS

3 11 The graphs of y = − 1 and y = x intersect at the points O(0, 0) and A. 2 a Sketch these graphs on the same diagram. b Using logarithms, find a suitable iterative formula that can be used to find the coordinates of the point A. c

PS

Calculate the length of the line OA, giving your answer correct to 2 significant figures. Give the value of any iterations you calculate to a suitable number of significant figures.

12 The diagram shows a container in the shape of a cone with a cylinder on top. The height of the cylinder is 3 times its base radius, r. The volume of the container must be 5500 cm 3. The base of the cone has a radius of r cm. a Write down an expression for the height of the cone in terms of r. 3r

b Show that 2 πr 3 + 11πr 2 − 5500 = 0 . c

33 cm

The equation in part b has a root, α .

r

5500 − 11πr 2 . 2π d Using an iterative formula based on the equation in part c, and a starting value of 8, find the value of α correct to 3 decimal places. Give the value of each of your iterations to an appropriate degree of accuracy. Show that α is also a root of the equation r =

3

e Explain what information is given by your answer to part d, in the context of the question. PS

13 The equation x 3 − 7 x 2 + 1 = 0 has two positive roots, α and β , which are such that α lies between 0 and 1 and β lies between 6 and 7. By deriving two suitable iterative formulae from the given equation, carry out suitable iterations to find the value of α and of β , giving each correct to 2 decimal places. Give the value of each of your iterations to 4 decimal places.

147

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PS

14 a Show graphically that the equation cot x = sin x has a root, α , which is such that 0 , α , b Show that the equation in part a can be rearranged as x = sin−1 cos x . c

π . 2

Using an iterative formula based on the equation in part b, with an initial value of 0.9, find the value of α correct to 2 decimal places. Give the value of each iteration to an appropriate number of decimal places.

d The equation given in part a can be rearranged in other ways. Find an example of a formula, based on a different rearrangement of the equation given in part a, that also converges to the value of α , with an initial value of 0.9, in fewer iterations than the formula used in part c.

EXPLORE 6.2 You are going to explore why using rearrangements of equations into the form x = F( x ) sometimes fails. Possible causes of failure to consider are the starting value and the formula itself. Discuss why formulae based on: ●

x = 51− x

●

x = ln(2 x + 1) (the staircase pattern)

(the cobweb pattern)

each converged to a root. Look at the equation e x = 3x + 1 and some related graphs. 148

y y y = 3x + 1

y = ex

O

y=

ex – 1 3

y=x

x

O

x

ex − 1 . 3 Does this give a convergent iterative formula? Explain your answer.

Discuss the rearrangement x =

Look at the equations in Exercise 6B. Using one of the equations in questions 1 to 5 or 7 to 14, draw the graph of y = x and the graph of a function F( x ) , where x = F( x ) is a rearrangement of the equation given and for which x = F( x ) will fail to converge. Explain any similarities or differences between your example and the rearrangement x =

ex − 1 . 3

Chapter 6: Numerical solutions of equations

6.3 Using iterative processes to solve problems involving other areas of mathematics WORKED EXAMPLE 6.6 2

2

1 1 − cos x 1 + cos x + = cot 2 x + . a Show that 2 2 sin x 2 sin x 2 2 1 − cos x 1 + cos x + b Hence, given that α is a root of the equation 2 sin x = x, show that α is also a root of 2 sin x π 2 the equation x = tan−1 for 0 , x , . 2 2x − 1 π 2 for 0 , x , . Verify by calculation c It is given that α is the only root of the equation x = tan−1 2 2x − 1 that the value of α lies between 0.9 and 1.0. d Using an iterative formula based on the equation in part c, find the value of α correct to 3 significant figures. Answer 2

a

1 − cos x 1 + cos x 2 sin x + 2 sin x

Square.

=

1 + 2 cos x + cos2 x 1 − 2 cos x + cos2 x + 4 sin2 x 4 sin2 x

Collect terms.

=

2 + 2 cos2 x 4 sin2 x

Simplify.

=

1 + cos2 x 2 sin2 x

Write as separate fractions.

=

cos2 x 1 + 2 sin2 x 2 sin2 x

Simplify.

cosec 2 x cot 2 x + 2 2 2 1 + cot x cot 2 x = + 2 2 1 = cot 2 x + 2 =

b

2

cot 2 x +

1 =x 2

Write in terms of cotx. Collect terms.

Rearrange.

cot 2 x = x −

1 2

Square root. 1 2

Left-hand side: write in terms of tan x. Right-hand side: write as a single fraction.

cot x =

x−

1 = tan x

2x − 1 2

Take the reciprocal of each side.

tan x =

2 2x − 1

Make x the subject.

x = tan−1

2 2x − 1

149

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

c

2 and so f( x ) = 0 then: 2x − 1

Let f( x ) = x − tan −1

f(0.9) = 0.9 − tan −1 f(1) = 1 − tan −1

2 = −0.10685… 2(0.9) − 1

2 = 0.44683… 2(1) − 1

Change of sign indicates the presence of a root. x1 = 0.95

150

2 2 xn − 1

xn +1 = tan −1

Are they the same correct to 3 significant figures?

x2 = tan−1

2 = 0.979 92… 2(0.95) − 1

No

x3 = tan−1

2 = 0.964 94… 2(0.97992…) − 1

No

x4 = 0.972 34…

No

x5 = 0.968 66…

No

x6 = 0.970 48…

No

x7 = 0.969 58…

Yes

α seems to be 0.970 correct to 3 significant figures. We can prove this, using a change of sign test: When α = 0.970 then 0.9695 , α , 0.9705 and so f(0.9695) = 0.9695 − tan−1

2 = −0.000571... 2(0.9695) − 1

f(0.9705) = 0.9705 − tan−1

2 = 0.000924... 2(0.9705) − 1

Change of sign indicates presence of root. Therefore α = 0.970 correct to 3 significant figures.

WORKED EXAMPLE 6.7 The diagram shows the design of a company logo, ABC .

A

AB is an arc of a circle with centre C. The area of the unshaded segment is to be the same as the area of the shaded triangle. Angle ACB is θ radians.

C

θ

a Show that θ = 2 sin θ . b Showing all your working, use an iterative formula based on the equation in part a, with an initial value of 1.85, to find θ correct to 3 significant figures. c Hence find the length of AB, given that AC is 8 cm.

B

Chapter 6: Numerical solutions of equations

Answer a

1 2 1 r sin θ = r 2 (θ − sin θ ) 2 2

Divide through by

sin θ = θ − sin θ

Add sin θ .

θ = 2 sin θ

b

θ 1 = 1.85

θ n + 1 = 2 sin θ n

1 2 r . 2

✓

Are they the same correct to 3 significant figures?

θ 2 = 2 sin1.85 = 1.922 55…

No

θ 3 = 2 sin1.922 55… = 1.877 53…

No

θ 4 = 1.906 64…

No

θ5 = 1.888 26…

No

θ 6 = 1.900 05…

No

θ 7 = 1.892 56…

No

θ8 = 1.897 35…

No

θ 9 = 1.894 30…

No

θ10 = 1.896 25…

No

θ11 = 1.895 00…

Yes

θ seems to be 1.90 correct to 3 significant figures. We can prove this, using a change of sign test with f(θ ) = θ − 2 sin θ = 0: When θ = 1.90 then 1.895 < θ < 1.905 and so f(1.895) = 1.895 − 2 sin1.895 = −0.000809... f(1.905) = 1.905 − 2 sin1.905 = 0.01565... Change of sign indicates presence of root, therefore θ = 1.90 correct to 3 significant figures.

151

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

c Using trigonometry: A

8 cm 0.95 C

B

0.5 AB 8 16 sin 0.95 = AB sin 0.95 =

Multiply by 16.

AB = 13.014..... = 13 cm (correct to 2 significant figures)

EXERCISE 6C e2 x + x has a stationary point with x-coordinate lying between 1 and 2. x3 3 x a Show that the x-coordinate of this stationary point satisfies the equation x = + 2 x . 2 e b Using a suitable iterative process based on the equation given in part a, find the value of the x-coordinate of this stationary point correct to 4 decimal places. Give each iteration to 6 decimal places.

1 The curve with equation y = 152

2 The parametric equations of a curve are x = t 2 + 6, y = t 4 − t 3 − 5t. The curve has a stationary point for a value of t that lies between 1 and 2. a Show that the value of t at this stationary point satisfies the equation t =

3

3t 2 + 5 4

b Use an iterative process based on the equation in part a to find the value of t correct to 3 decimal places. Show the result of each iteration to 6 decimal places. c

Hence find the coordinates of the stationary point, giving each coordinate correct to 1 significant figure.

3 In the diagram, triangle ABC is right-angled and angle BAC is θ radians. The point O is the mid point of AC and OC = r. Angle BOC is 2θ radians and BOC is a sector of the circle with centre O. The area of triangle ABC is 2 times the area of the shaded segment.

B

a Show that θ satisfies the equation sin 2θ = θ . π b This equation has one root in the interval 0 , θ , . 2 Use the iterative formula θ n+1 = sin 2θ n to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

2θ

θ A

O

r

C

Chapter 6: Numerical solutions of equations

4 The diagram shows the curve y = x 2 cos 4x for π 0 ø x ø . The point P is a maximum point. 8 a Show that the x-coordinate of P satisfies the equation 4x 2 tan 4x = 2 x.

y

P

b Show also that the x-coordinate of P satisfies 1 1 the equation x = tan−1 . 2x 4 c Using an iterative formula based on the equation in part b with initial value x1 = 0.3 find the x− - coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. P3

π 8

O

x

d Use integration by parts twice to find the exact π area enclosed between the curve and the x-axis from 0 to . 8

EXPLORE 6.3

The golden ratio φ

1

φ–1

153 1

1

1

When a rectangle with sides in the ratio 1: φ (phi) can be cut into a square of side 1 and a smaller rectangle, also with sides in the ratio 1: φ , the original rectangle is called a golden rectangle and the ratio 1: φ is called the golden ratio. φ −1 1 = . This means that φ 1 By using a suitable iterative formula based on this ratio, with a starting value of 1, find the value of φ correct to 5 significant figures. Explore the link between the golden ratio and the sequence of Fibonacci numbers: 1, 1, 2, 3, 5, 8, ..., ...

DID YOU KNOW? The relationship between the Fibonacci numbers and the golden ratio produces the golden spiral. This can be seen in many natural settings, one of which is the spiralling pattern of the petals in a rose flower. From a tight centre, petals flow outward in a spiral, growing wider and larger as they do so. Nature has made the best use of an incredible structure. This perfect arrangement of petals is why the small, compact rose bud grows into such a beautiful flower.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Checklist of learning and understanding ●

The process of finding a sequence of values that become closer and closer to a point of intersection of y = x and y = F( x ) is called an iterative process. Each time you generate a value of x you carry out an iteration.

●

When the point of intersection is at x = α and your iterations are values that are getting closer and closer to α , you are converging to α .

●

Start with x1

work out F( x1 ).

Is it accurate enough?

Yes → Stop, No →

●

Let x2 = F( x1 )

work out F( x2 ).

Is it accurate enough?

Yes → Stop, No →

●

Let x3 = F( x2 )

work out F( x3 ).

Is it accurate enough?

Yes → Stop, No →

●

Let xn+1 = F( xn )

●

The relationship xn +1 = F( xn ) is called an iterative formula.

●

The sequence of values of x given by this formula are called x1, x2 , x3 , ..., xn.

●

When the value found is accurate enough, then it is usually given a particular name such as α .

●

The success and speed of the iterative process depends on the function chosen for F( xn ) and the value chosen for x1.

…

●

154

work out F( xn+1). Is it accurate enough?

Yes → Stop.

Chapter 6: Numerical solutions of equations:

END-OF-CHAPTER REVIEW EXERCISE 6 1

6 The terms of the sequence generated by the iterative formula xn +1 = 7 converge to α . a

x + 1 with initial value x = 1.5, 1 n x 3 n

Use this formula to find α correct to 2 decimal places. Give the result of each iteration to an appropriate number of decimal places.

State an equation satisfied by α , and hence find the exact value of α . x The equation x 3 + − 8 = 0 has one real root, denoted by α . 2 a Find, by calculation, the pair of consecutive integers between which α lies.

b 2

b c 3

a b c d

4

Show that, if a sequence of values given by the iterative formula xn +1 = it converges to α .

8 1 − converges, then xn 2

[3] [2]

[2]

[2]

Use this iterative formula to determine α correct to 1 decimal place. Give the result of each iteration to 4 decimal places.

[3]

By sketching a suitable pair of graphs, show that the equation e2 x +1 = 14 − x 3 has exactly one real root.

[3]

Show by calculation that this root lies between 0.5 and 1.

[2]

ln(14 − x 3 ) − 1 Show that this root also satisfies the equation x = . 2 Use an iteration process based on the equation in part c, with a suitable starting value, to find the root correct to 4 decimal places. Give the result of each step of the process to 6 decimal places.

The sequence of values given by the iterative formula xn+1 = value x1 = 1, converges to α .

[1]

[3]

2

xn (1 + sec xn ) − tan xn with initial sec 2 xn − 1

Use this formula to find α correct to 2 decimal places, showing the result of each iteration to 4 decimal places. π b Show that α also satisfies the equation tan x − 2 x = 0 where 1 ø x , . 2 ln x 2 − 6 = x − 5 has a root α , such that p , α , q, where p and q are consecutive The equation 2 integers.

a

5

6

[3] [2]

a

Show that α also satisfies the equation x = ln x + 2.

[1]

b

Find the value of p and the value of q.

[2]

c

Use the iterative formula based on the equation in part a with starting value x = p, to find the value of α correct to 3 decimal places.

[3]

d

Hence find an approximate value of y that satisfies the equation 5 y − 2 = y ln 5.

[2]

a

By sketching a suitable pair of graphs, show that the equation 7 − x 6 = x 2 − 1 has exactly two real roots, α and β , where α is a positive constant.

[3]

b c

d

8 − x. x Write down an iterative formula based on the equation in part a. Use this formula, with a suitable starting value, to find the value of α correct to 3 significant figures. Give the result of each iteration to 5 significant figures. Show that α satisfies the equation x =

5

Hence write down the value of β correct to 3 significant figures.

[1]

[3] [1]

155

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

7

a b

8

1 + 2e6 x dx = 0.6 , where the constant a > 0. 0 3x + 5 1 5 Show that a = ln 2.8 + ln . 3a + 5 6 Use an iterative formula based on the equation in part a, with a starting value of a = 0.2, to find the value of a correct to 3 decimal places. Give the result of each iteration to an appropriate number of decimal places.

It is given that

∫

a

[4]

[3]

A curve has parametric equations 1 x = , y = t +1 t The point P on the curve has parameter p. It is given that the gradient of the curve at P is −0.4. 1

a b

9

156

2 5 ( p + 1) 4 . 5 Use an iterative process based on the equation in part a to find the value of p correct to 3 decimal places. Use a starting value of 1 and show the result of each iteration to 5 decimal places. Show that p =

In the diagram, A, B and C are points on the circumference of the circle with centre O and radius r. The shaded region, ABC , is a sector of the circle with centre C. Angle OCA is equal to θ radians. 3 The area of the shaded region is equal to of the area of the circle. 8 3π . a Show that θ = cos−1 32θ b

Verify by calculation that θ lies between 0.8 and 1.2 radians.

c

Use the iterative formula θ n+1 = cos−1

[3] [3]

A

θ r

O

C

[3] B

[2]

3π to find the value of θ correct to 3 decimal places. 32θ n Give the result of each iteration to 5 decimal places.

10 The functions f and g are defined, for 0 ø x ø π , by f( x ) = e x − 2 and g( x ) = 5 − cos x.

[3]

y

The diagram shows the graph of y = f( x ) and the graph of y = g( x ).

y = 5 – cos x

The gradients of the curves are equal both when x = p and when x = q. a

Given that p , q , verify by calculation that p is 0.16 correct to 2 decimal places.

y = ex–2

b

Show that q satisfies the equation q = 2 + ln(sin q ) .

c

Given also that 1.5 , q , 2.5 , use the iterative formula qn+1 = 2 + ln(sin qn ) to calculate q correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

O

11 The equation x = cos x + sin x has a root α that lies between 1 and 1.4. π a Show that α is also a root of the equation x = 2 cos x − . 4 π b Using the iterative formula xn+1 = 2 cos xn − , find the value of α , giving your answer 4 correct to 2 decimal places.

[4] x

[1] [3]

[3]

[3]

Chapter 6: Numerical solutions of equations:

P3

12 a

b

c

d

π π π π By sketching each of the graphs of y = sec x and y = − x + x for − , x ø on the 2 4 2 2 π π same diagram, show that the equation sec x = − x + x has exactly two real roots in 2 4 π π the interval − , x ø . 2 2

[3]

π π Show that the equation sec x = − x + x can be written in the form 2 4 2 2 πx + π − 8 sec x . x= 8 π π π π The two real roots of the equation sec x = − x + x in the interval − , x ø are denoted 2 4 2 2 α and β .

[1]

Verify by calculation that the smaller root, α , is −0.21 correct to 2 decimal places.

[2]

Using an iterative formula based on the equation given in part b, with an initial value of 1, find the value of β correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

[3]

FAST FORWARD You need to have studied the skills in Chapter 8 before attempting questions 13 and 14.

P3

13

157

y

1

y = cos3 x

R

O

α

π 6

π 3

x

The diagram shows part of the curve y = cos3 x, where x is in radians. The shaded region between the curve, the axes and the line x = α is denoted by R. The area of R is equal to 0.3. α 0.9 a Using the substitution u = sin x, find cos3 x d x . Hence show that sin α = . 3 − sin2 α 0

∫

b

P3

0.9 Use the iterative formula α n+1 = sin−1 with α1 = 0.2, to find the value of α correct 2 3 − sin α n to 3 significant figures. Give the result of each iteration to 5 significant figures.

14 It is given that a b

∫

a

1

[3]

ln x d x = 5, where a is a constant greater than 1.

4+a . ln a Use an iterative formula based on the equation in part a to find the value of a correct to 3 decimal places. Use an initial value of 5 and give the result of each iteration to 5 decimal places. Show that a =

[6]

[5] [3]

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

CROSS-TOPIC REVIEW EXERCISE 2 1

1+ x 1 The equation of a curve is y = for x . − . Show that the gradient of the curve is 1 + 2x 2 always negative.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q1 November 2013 2

y

O

1π 4

x

1π 2

1 1 tan 2 x for 0 ø x ø π . Find the x-coordinates 2 2 of the points on this part of the curve at which the gradient is 4. The diagram shows the part of the curve y =

158

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q3 November 2011 3

2 The parametric equations of a curve are x = ln(1 − 2t ), y = , for t , 0. t dy 1 − 2t i Show that = 2 . dx t ii Find the exact coordinates of the only point on the curve at which the gradient is 3.

[3] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2012 4

The equation of a curve is x 2 y + y2 = 6x. dy 6 − 2 xy i Show that = . dx x 2 + 2 y ii Find the equation of the tangent to the curve at the point with coordinates (1, 2), giving your answer in the form ax + by + c = 0.

[4] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2010 P2 5

Use the trapezium rule with two intervals to estimate the value of your answer correct to 2 decimal places. (e x − 2)2 ii Find dx . e2 x

i

∫

∫

1

0

1 d x , giving 6 + 2e x

[3] [4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2012

Cross-topic review exercise 2

6

i ii

1 By sketching a suitable pair of graphs, show that the equation ln x = 4 − x has exactly 2 one real root, α .

[2]

Verify by calculation that 4.5 , α , 5.0.

[2]

iii Use an iterative formula xn +1 = 8 − 2 ln xn to find α correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2015 7

1 1 , y = tan3 t, where 0 ø t , π. The parametric equations of a curve are x = 2 cos3 t dy = sin t. i Show that dx ii Hence show that the equation of the tangent to the curve at the point with parameter t is y = x sin t − tan t.

[4]

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2014 8

∫

a

It is given that the positive constant a is such that (4e2 x + 5) d x = 100 . −a 1 i Show that a = ln(50 + e −2 a − 5a ) . 2 1 ii Use the iterative formula an +1 = ln(50 + e −2 a − 5an ) to find α correct to 3 decimal places. 2 Give the result of each iteration to 5 decimal places.

[4]

n

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2016 9

1 The equation of a curve is y = e −2 x tan x, for 0 ø x , π. 2 dy and show that it can be written in the form e −2 x ( a + b tan x )2, i Obtain an expression for dx where a and b are constants. ii

Explain why the gradient of the curve is never negative.

iii Find the value of x for which the gradient is least.

[5] [1] [1]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2015 10

i

Prove that tan θ + cot θ ≡

ii

Hence

2 . sin 2θ

1 1 a find the exact value of tan π + cot π, 8 8

b evaluate

∫

1 π 2

0

6 dθ . tan θ + cot θ

[3]

[2]

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2014 11

The equation of a curve is xy( x − 6 y ) = 9a 3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point.

[7]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2016

159

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

12

i

6 intersects the line y = x + 1 at the point P. x2 Verify by calculation that the x-coordinate of P lies between 1.4 and 1.6.

ii

Show that the x-coordinate of P satisfies the equation x =

The curve with equation y =

6 . x + 1

6 x + 1 , with initial value x1 = 1.5, to determine n the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

[2] [2]

iii Use the iterative formula xn +1 =

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2010 13

The parametric equations of a curve are x = e2t, y = 4tet. i ii

dy 2(t + 1) = . dx et Find the equation of the normal to the curve at the point where t = 0 .

Show that

[4] [4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2013 14

For each of the following curves, find the exact gradient at the point indicated: i ii

160

1 y = 3 cos 2 x − 5 sin x at π, −1 , 6 3 3 x + 6xy + y = 21 at (1, 2).

[3] [5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q4 November 2014 15

i

By differentiating

ii

Show that

1 dy , show that if y = sec x then = sec x tan x. cos x dx

1 ≡ sec x + tan x. sec x − tan x

iii Deduce that

[1]

1 ≡ 2 sec 2 x − 1 + 2 sec x tan x. (sec x − tan x )2

iv Hence show that

∫

1 π 4

0

[2]

(

)

1 1 8 2−π . 2 dx = 4 (sec x − tan x )

[2]

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2012 16

1 A curve is defined by the parametric equations x = 2 tan θ , y = 3 sin 2θ , for 0 ø θ , π. 2 dy = 6 cos 4 θ − 3 cos2 θ . i Show that dx ii Find the coordinates of the stationary point. 3 3 . iii Find the gradient of the curve at the point 2 3, 2

[4] [3] [2]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2016

Cross-topic review exercise 2

17

y

M x

O 1

x

The diagram shows the curve y = 4e 2 − 6x + 3 and its minimum point M. i ii

Show that the x-coordinate of M can be written in the form ln a, where the value of a is to be stated.

[5]

Find the exact value of the area of the region enclosed by the curve and the lines x = 0 , x = 2 and y = 0 .

[4]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q5 June 2012 161

18

The equation of a curve is x 3 − 3x 2 y + y3 = 3. x 2 − 2 xy dy = 2 . i Show that dx x − y2 ii

[4]

Find the coordinates of the points on the curve where the tangent is parallel to the x-axis.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2016 19

1 , y = (t + 2 ) (2t + 1)2 The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1. A curve has parametric equations x = 1

i ii

1 Show that p = ( p + 2) 6 − . 2 Use an iterative process based on the equation in part i to find the value of p correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.

[6] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2012 20

i

Find a b

P2

ii

e2 x + 6 dx , e2 x

∫ ∫ 3 cos

2

[3]

x dx.

[3]

Use the trapezium rule with 2 intervals to estimate the value of your answer correct to 2 decimal places.

∫

2

1

6 d x , giving ln( x + 2)

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 November 2013

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

21

i

By sketching a suitable pair of graphs, show that the equation e2 x = 14 − x 2 has exactly two real roots.

Show by calculation that the positive root lies between 1.2 and 1.3. 1 iii Show that this root also satisfies the equation x = ln(14 − x 2 ). 2 iv Use an iteration process based on the equation in part iii, with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places. ii

[3] [2] [1]

[3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q7 June 2011 22

By sketching each of the graphs y = cosec x and y = x( π − x ) for 0 , x , π, show that the equation cosec x = x( π − x ) has exactly two real roots in the interval for 0 , x , π. 1 + x 2 sin x ii Show that the equation cosec x = x( π − x ) can be written in the form x = . π sin x iii The two real roots of the equation cosec x = x( π − x ) in the interval 0 , x , π are denoted by α and β , where α , β . 1 + xn 2 sin xn to find α correct to 2 decimal places. a Use the iterative formula xn +1 = π sin xn Give the result of each iteration to 4 decimal places.

i

b Deduce the value of β correct to 2 decimal places.

[3] [2]

[3] [1]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2014

162

23

y O x

P

The diagram shows part of the curve with parametric equations x = 2 ln(t + 2), y = t 3 + 2t + 3. i

Find the gradient of the curve at the origin.

[5]

ii

At the point P on the curve, the value of the parameter is p. It is given that the gradient 1 of the curve at P is . 2 1 a Show that p = − 2. 2 3p + 2 b By first using an iterative formula based on the equation in part a, determine the coordinates of the point P. Give the result of each iteration to 5 decimal places and each coordinate of P correct to 2 decimal places.

[1]

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2015

Cross-topic review exercise 2

24

i

Prove that 2 cosec 2θ tan θ ≡ sec 2 θ .

ii

Hence

[3]

a solve the equation 2 cosec 2θ tan θ = 5 for 0 , x , π. 1

b find the exact value of

∫

6

[3]

π

2 cosec 4x tan 2 x d x.

[4]

0

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2015 25

i

Prove the identity cos 4θ + 4 cos 2θ ≡ 8 cos 4 θ − 3.

ii

Hence a solve the equation cos 4θ + 4 cos 2θ = 1 for −

b find the exact value of

∫

1 π 4

[4]

1 1 π ø θ ø π. 2 2

[3]

cos 4 θ dθ .

[3]

0

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2011 26

i

ii

By first expanding cos(2 x + x ) , show that cos 3x = 4 cos3 x − 3 cos x.

Hence show that

∫

1 π 6

[5] 163

(2 cos3 x − cos x ) d x =

0

5 . 12

[5]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 November 2011 y

27

T1 T3

x

O T4 T2

The diagram shows the curve y = shown.

1 − x 10e 2

sin 4x for x ù 0. The stationary points are labeled T1, T2 , T3 , … as

i

Find the x-coordinates of T1 and T2 , giving each x-coordinate correct to 3 decimal places.

[6]

ii

It is given that the x-coordinate of Tn is greater than 25. Find the least possible value of n.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2014

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

28

The equation of a curve is y = the gradient of the curve is

1 . 2

3x 2 . At the point on the curve with positive x-coordinate p, x2 + 4

48 p − 16 p2 + 8 .

i

Show that p =

ii

Show by calculation that 2 , p , 3.

iii Use an iterative formula based on the equation in part i to find the value of p correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

[5] [2] [3]

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q6 June 2016 29

i

By differentiating

1 dy , show that if y = sec θ then = tan θ sec θ . cos θ dθ

[3]

ii

Hence show that

d2 y = a sec 3 θ + b sec θ , giving the values of a and b. dθ 2

[4]

iii Find the exact value of

∫

1 π 4

(1 + tan2 θ − 3 sec θ tan θ ) dθ .

[5]

0

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q8 November 2012 164

165

Chapter 7 Further algebra P3 This chapter is for Pure Mathematics 3 students only.

In this chapter you will learn how to:

■ ■

recall an appropriate form for expressing rational functions in partial fractions, and carry out the decomposition, in cases where the denominator is no more complicated than: ( ax + b )( cx + d )( ex + f ) ( ax + b )( cx + d )2 ( ax + b )( cx 2 + d ) use the expansion of (1 + x ) n , where n is a rational number and x , 1.

• • •

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

Pure Mathematics 1 Coursebook, Chapter 6

Equate coefficients of polynomials.

1 Find the value of A, B and C for: a

Ax 2 − 3x + C ≡ 6x 2 + Bx − 9

b (2 − A)x 2 + 5x + 2C ≡ 3x 2 − 3Bx + 8 Pure Mathematics 1 Coursebook, Chapter 6

Expand ( a + b ) n where n is a positive integer.

2 Find the first 3 terms, in ascending powers of x, in the expansion of: a

(1 + 2 x )7

b (3 − 2 x )5 Chapter 1

Divide polynomials.

3 Find the quotient and remainder when x 2 − 8x + 4 is divided by x − 3.

Why do we study algebra? At IGCSE / O Level we learnt how to add and subtract algebraic fractions. In this chapter we will learn how to do the ‘reverse process’. This reverse process is often referred to as splitting a fraction into its partial fractions. In Mathematics it is often easier to deal with two or more simple fractions than it is to deal with one complicated fraction. 166

In the Pure Mathematics 1 Coursebook, Chapter 6, you learnt how to find the binomial expansion of ( a + b ) n for positive integer values of n. After working through this chapter you will be able to expand expressions of the form (1 + x ) n for values of n that are not positive integers (providing x , 1). Combining your partial fraction and binomial expansion skills will enable you to obtain 2x − 1 series expansions of complicated expressions such as . 2 x 2 + 3x − 20

7.1 Improper algebraic fractions A numerical improper fraction is defined as a fraction where the numerator ù the 11 denominator. For example, is an improper fraction. This fraction can be expressed as 5 1 2 + , which is the sum of a positive integer and a proper fraction. 5 So how do we define an algebraic improper fraction? KEY POINT 7.1 P( x ) , where P( x ) and Q( x ) are polynomials in x, is said to be an algebraic Q( x ) improper fraction if the degree of P( x ) ù the degree of Q( x ) . The algebraic fraction

x 3 − 3x 2 + 7 is an improper algebraic fraction because the x−2 degree of the numerator (3) is greater than the degree of the denominator (1). For example, the fraction

FAST FORWARD In Chapter 8 you will be shown another use for partial fractions: how to integrate rational expressions 2x − 1 such as 2 x 2 + 3x − 20 by first splitting the expression into partial fractions.

WEB LINK Explore the Polynomials and rational functions station on the Underground Mathematics website.

Chapter 7: Further algebra

We can use long division to write the fraction a proper algebraic fraction.

x 3 − 3x 2 + 7 as the sum of a polynomial and x−2

x2 − x − 2 x − 2 x − 3x 2 + 0 x + 7

)

3

x3 − 2x2 − x2 + 0x − x2 + 2x − 2x + 7 − 2x + 4 3 ∴

3

2

3 x − 3x + 7 = x2 − x − 2 + x−2 x−2

EXPLORE 7.1 1 Discuss with your classmates which of the following are improper algebraic fractions. 1 2x + 1

x 2 − 4x 5 − x2

3x x−5

6x 3 − 2 x + 1 2x 2 − 1

2x − 3 ( x + 2)( x − 1)

x 3 + 2x2 − 7 − ( x + 2)( x + 1)

4x 5 − 1 3x 7 + 2

2x 4 − 8 2 x − 2x − 1

167

2 Write each improper fraction in question 1 as the sum of a polynomial and a proper fraction.

EXERCISE 7A 1 Express each of the following improper fractions as the sum of a polynomial and a proper fraction. a

8x 2x − 5

b

6x + 1 3x + 2

c

4x 3 − 3 2x + 1

d

x 3 + 4x 2 + 3x − 1 x2 + 2x + 3

e

7 x 3 + 2 x 2 − 5x + 1 x2 − 5

f

x 4 + 2x2 − 5 x2 + 1

2 Given that

x3 + x2 − 7 D ≡ Ax 2 + Bx + C + , find the values of A, B, C and D. x−3 x−3

3 Given that

x 4 + 5x 2 − 1 E ≡ Ax 3 + Bx 2 + Cx + D + , find the values of A, B, C , D and E. x +1 x +1

4 Given that

2 x 4 + 3x 3 + 4x 2 + 5x + 6 Cx + D ≡ Ax + B + 3 , find the values of A, B, C and D. x3 + 2x x + 2x

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

7.2 Partial fractions At IGCSE / O Level, we learnt how to add and subtract algebraic fractions. The following table shows three different cases for addition of algebraic fractions. Case 1 5 4 + x+3 x−2 5( x − 2) + 4( x + 3) = ( x + 3)( x − 2) 9x + 12 = ( x + 3)( x − 2) Denominator in answer has: two distinct linear factors.

Case 2

Case 3

3 2 + x + 1 ( x + 1)2 3( x + 1) + 2 = ( x + 1)2 3x + 5 = ( x + 1)2

2 3 + x − 1 x2 + 5 2( x 2 + 5) + 3( x − 1) = ( x − 1)( x 2 + 5)

Denominator in answer has: one linear repeated factor.

Denominator in answer has: one linear factor and one quadratic factor.

=

2 x 2 + 3x + 7 ( x − 1)( x 2 + 5)

In this section we will learn how to do the reverse of this process. We will learn how to split (or decompose) a single algebraic fraction into two or more partial fractions.

Case 1: Proper fraction where the denominator has distinct linear factors If the denominator of a proper algebraic fraction has two distinct linear factors, the fraction can be split into partial fractions. KEY POINT 7.2 168

px + q A B ≡ + ( ax + b )( cx + d ) ax + b cx + d

This rule can be extended for 3 or more linear factors. KEY POINT 7.3 px + q

( ax + b ) ( cx + d ) ( ex + f )

≡

A B C + + ax + b cx + d ex + f

WORKED EXAMPLE 7.1

Express

2 x − 13 in partial fractions. (2 x + 1)( x − 3)

Answer A B 2 x − 13 ≡ + (2 x + 1)( x − 3) 2 x + 1 x − 3 2 x − 13 ≡ A( x − 3) + B (2 x + 1)

Multiply throughout by (2 x + 1)( x − 3). (1)

Chapter 7: Further algebra

Method 1: Choose appropriate values for x. Let x = 3 in equation (1): 2(3) − 13 = A(3 − 3) + B (6 + 1) −7 = 7 B B = −1 Let x = −

1 in equation (1): 2

1 1 2 − − 13 = A − − 3 + B ( −1 + 1) 2 2 7 −14 = − A 2 A=4 2 x − 13 4 1 ≡ − ∴ (2 x + 1)( x − 3) 2 x + 1 x − 3 Method 2: Equating coefficients. 2 x − 13 ≡ A( x − 3) + B (2 x + 1)

Expand brackets and collect like terms.

2 x − 13 ≡ ( A + 2 B )x − 3A + B

Equate the coefficients of x and the constants.

(2) A + 2B = 2 (3) −3A + B = −13 A = 4, B = −1 2 x − 13 4 1 ≡ − ∴ (2 x + 1)( x − 3) 2 x + 1 x − 3

169

Solve equations (2) and (3) simultaneously.

Case 2: Proper fraction with repeated linear factor in the denominator If the denominator has a repeated linear factor, the fraction can be split into partial fractions using the rule: KEY POINT 7.4 px + q A B ≡ + ax + b ( ax + b )2 ( ax + b )2

WORKED EXAMPLE 7.2

Express

2 x − 11 in partial fractions. ( x − 4)2

Answer 2 x − 11 A B ≡ + x − 4 ( x − 4)2 ( x − 4)2 2 x − 11 ≡ A( x − 4) + B

Multiply throughout by ( x − 4)2 . (1)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Let x = 4 in equation (1): 2(4) − 11 = A(4 − 4) + B B = −3 Let x = 0 in equation (1): 2(0) − 11 = A(0 − 4) − 3 A = 2 2 x − 11 2 3 ≡ − ∴ x − 4 ( x − 4)2 ( x − 4)2

WORKED EXAMPLE 7.3

Express

1 − 9x − 8x 2 in partial fractions. ( x − 2)(2 x + 3)2

Answer

1 − 9x − 8x 2 A B C ≡ + + x − 2 2 x + 3 (2 x + 3)2 ( x − 2)(2 x + 3)2

Multiply throughout by ( x − 2)(2 x + 3)2 .

1 − 9x − 8x 2 ≡ A(2 x + 3)2 + B ( x − 2)(2 x + 3) + C ( x − 2)

(1)

Let x = 2 in equation (1): 170

1 − 9(2) − 8(2)2 = A(4 + 3)2 + B (2 − 2)(4 + 3) + C (2 − 2) −49 = 49 A A = −1 Let x = −

3 in equation (1): 2 2

3 3 3 3 1 − 9 − − 8 − = A( −3 + 3)2 + B − − 2 ( −3 + 3) + C − − 2 2 2 2 2 7 7 − = − C 2 2 C =1 Let x = 0 in equation (1): 1 = 9 A − 6B − 2C B = −2 ∴

1 − 9x − 8x 2 1 2 1 ≡− − + x − 2 2 x + 3 (2 x + 3)2 ( x − 2)(2 x + 3)2

(2)

Substitute A = −1 and C = 1 in equation (2).

Chapter 7: Further algebra

Case 3: Proper fraction with a quadratic factor in the denominator that cannot be factorised In this syllabus you will only be asked questions where the quadratic factor is of the form cx 2 + d. If the denominator of a proper algebraic fraction has a quadratic factor of the form cx 2 + d that cannot be factorised, the fraction can be split into partial fractions using the following rule. KEY POINT 7.5 px + q A Bx + C ≡ + ( ax + b )( cx 2 + d ) ax + b cx 2 + d

WORKED EXAMPLE 7.4

Express

8 + 2x − x2 in partial fractions. ( x − 1)( x 2 + 2)

Answer 8 + 2x − x2 A Bx + C ≡ + ( x − 1)( x 2 + 2) x − 1 x 2 + 2 8 + 2 x − x 2 ≡ A( x 2 + 2) + Bx( x − 1) + C ( x − 1)

Multiply throughout by ( x − 1)( x 2 + 2). (1)

Let x = 1 in equation (1): 8 + 2(1) − (1)2 = A(12 + 2) + B (1)(1 − 1) + C (1 − 1) 9 = 3A A=3 Let x = 0 in equation (1): 8 = 6−C C = −2 Let x = −1 in equation (1): 5 = 9 + 2B + 4 B = −4 ∴

−4x − 2 4x + 2 8 + 2x − x2 3 3 ≡ + 2 ≡ − x − 1 x2 + 2 ( x − 1)( x 2 + 2) x − 1 x +2

Case 4: Improper fractions If an algebraic fraction is improper we must first express it as the sum of a polynomial and a proper fraction and then split the proper fraction into partial fractions.

171

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 7.5

Express

2 x 2 + 5x − 11 in partial fractions. x2 + 2x − 3

Answer The fraction is improper. Hence it must first be written as the sum of a polynomial and a proper fraction. 2 x 2 + 2 x − 3 2 x 2 + 5x − 11

)

2 x 2 + 4x − 6 x−5 2 x + 5x − 11 x−5 ≡2+ 2 x2 + 2x − 3 x + 2x − 3 2

Split the proper fraction

x−5 into partial fractions: x2 + 2x − 3

x−5 A B ≡ + ( x + 3)( x − 1) x + 3 x − 1 x − 5 ≡ A( x − 1) + B ( x + 3)

Multiply throughout by ( x + 3 )( x − 1 ). (1)

Let x = 1 in equation (1): −4 = 4B B = −1 172

Let x = −3 in equation (1): −8 = −4 A A=2 x−5 2 1 ≡ − ( x + 3)( x − 1) x + 3 x − 1 ∴

2 1 2 x 2 + 5x − 11 ≡2+ − 2 x +3 x −1 x + 2x − 3

Application of partial fractions Partial fractions can be used to help evaluate the sum of some sequences. WORKED EXAMPLE 7.6 1 1 1 can be written in partial fractions as − , find the sum of the first x( x + 1) x x +1 1 1 1 1 n terms of the series + + + + ……… . Hence, state the sum to infinity. 1× 2 2 × 3 3 × 4 4 × 5

Given that the algebraic fraction

Answer Sn =

1 1 1 1 1 1 + + + + … + + 1× 2 2 × 3 3 × 4 4 × 5 ( n − 1) n n( n + 1)

1 1 1 1 1 1 1 1 1 1 1 1 = − + − + − + − + … + − + − 1 2 2 3 3 4 4 5 n − 1 n n n + 1 1 = 1− n +1

Chapter 7: Further algebra

As n → ∞,

1 →0 n +1

∴ S∞ = 1

DID YOU KNOW? The series in Worked example 7.6 is called a telescoping series, because its partial sums only have a fixed number of terms after cancellation.

EXERCISE 7B 1 Express the following proper fractions as partial fractions. a

6x − 2 ( x − 2)( x + 3)

b

7 x + 12 2 x( x − 4)

c

15x + 13 ( x − 1)(3x + 1)

d

x −1 (3x − 5)( x − 3)

e

6x 2 + 5x − 2 x( x − 1)(2 x + 1)

f

11x + 12 (2 x + 3)( x + 2)( x − 3)

2 Express the following proper fractions as partial fractions. a

2x ( x + 2)2

b

11x 2 + 14x + 5 (2 x + 1)( x + 1)2

c

x2 − 2 x( x − 1)2

d

36x 2 + 2 x − 4 (2 x − 3)(2 x + 1)2

e

3 ( x + 2)( x − 2)2

f

3x + 4 ( x + 2)( x − 1)2

3 Express the following proper fractions as partial fractions. a

2 x 2 − 3x + 2 x( x 2 + 1)

b

3x 2 + 4x + 17 (2 x + 1)( x 2 + 5)

c

2 x 2 − 6x − 9 (3x + 5)(2 x 2 + 1)

d −

6x 2 − 21x + 50 (3x − 5)(2 x 2 + 5)

4 Express the following improper fractions as partial fractions. a

2 x 2 + 3x + 4 ( x − 1)( x + 2)

b

x2 + 3 x2 − 4

5 Find the values of A, B, C and D such that

c

22 − 17 x + 21x 2 − 4x 3 ( x − 4)( x 2 + 1)

d

4x 3 + x 2 − 16x + 7 2 x 3 − 4x 2 + 2 x

4x 3 − 9x 2 + 11x − 4 B C D ≡ A+ + 2 + . x x 2x − 1 x 2 (2 x − 1)

6 a Factorise 2 x 3 − 3x 2 − 3x + 2 completely. b Hence express

x 2 − 13x − 5 in partial fractions. 2 x − 3x 2 − 3x + 2 3

7 a Factorise 2 x 3 − 11x 2 + 12 x + 9 completely. b Hence express

24 − x in partial fractions. 2 x − 11x 2 + 12 x + 9 3

8 The polynomial p( x ) = 2 x 3 + 5x 2 + ax + b is exactly divisible by 2 x + 1 and leaves a remainder of 9 when divided by x + 2. a Find the value of a and the value of b. b Factorise p( x ) completely. c

Express

120 in partial fractions. p( x )

173

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PS

2 in partial fractions. x( x + 2) b Find an expression for the sum of the first n terms of the series: 2 2 2 2 2 + … + + + + 1× 3 2 × 4 3 × 5 4 × 6 5 × 7

9 a Express

c PS

Find the sum to infinity of this series.

1 1 1 1 + + + + … is a telescoping series. If it is a 1× 2 × 3 2 × 3 × 4 3 × 4 × 5 4 × 5 × 6 telescoping series, find an expression for the sum of the first n terms and the sum to infinity.

10 Investigate whether the series

EXPLORE 7.2 There is another method that can be used to decompose a proper algebraic fraction where the denominator contains only distinct linear factors. This alternative method is called Heaviside’s cover-up method. Find out more about this method and use it to 4x 2 + 15x + 29 in partial fractions. express ( x − 2)( x + 1)( x + 3)

7.3 Binomial expansion of (1 + x ) n for values of n that are not positive integers 174

REWIND In the Pure Mathematics 1 Coursebook, Chapter 6, we learnt that the binomial expansion of (1 + x ) n, where n is a positive integer is: n n n n n (1 + x ) n = + x + x 2 + x 3 + … + x n 0 3 1 2 n and that this terminating series can also be written in the form: (1 + x ) n = 1 + nx +

n( n − 1) 2 n( n − 1)( n − 2) 3 x + x + … + x n. 2! 3!

The general binomial theorem can be used for any rational value of n . It states that: KEY POINT 7.6 (1 + x ) n = 1 + nx +

n( n − 1) 2 n( n − 1)( n − 2) 3 x + x + … where n is rational and x , 1 2! 3!

There are two very important differences when n is not a positive integer. These are: ● ●

the series is infinite the expansion is only valid for x , 1 .

The proof of the general binomial theorem is beyond the scope of this course. Consider, however, the following two examples where n is not a positive integer n = −3 and n = 1 . 2

Chapter 7: Further algebra

(1 + x ) −3 = 1 + ( −3)x +

( −3)( −4) 2 ( −3)( −4)( −5) 3 ( −3)( −4)( −5)( −6) 4 x + x + x +… 2! 3! 4!

= 1 − 3x + 6x 2 − 10 x 3 + 15x 4 + … (1 +

1 x)2

1 − 1 − 3 − 5 1 − 1 − 3 1 − 1 2 2 2 2 2 2 3 2 2 2 2 4 1 x+ x + x + x +… = 1+ 2 2! 3! 4! 1 1 3 5 4 1 x − x + … = 1 + x − x2 + 2 8 16 128

We can see that in each case the sequence of coefficients forms an infinite sequence. When x , 1, the binomial series converges and has a sum to infinity (in a similar way to a geometric series converging for r , 1, where r is the common ratio). WORKED EXAMPLE 7.7

DID YOU KNOW?

Expand (1 − 3x )−2 in ascending powers of x up to and including the term in x 3, and state the range of values of x for which the expansion is valid. Answer Use n = −2 and replace x by −3x in the binomial formula: (1 − 3x ) −2 = 1 + ( −2)( −3x ) +

( −2)( −3) ( −2)( −3)( −4) ( −3x )2 + ( −3x )3 + … 2! 3!

= 1 + 6x + 27 x 2 + 108x 3 + … Expansion is valid for −3x , 1 1 ∴ x , . 3

WORKED EXAMPLE 7.8 Find the first four terms in the expansion of (1 + 2 x 2 )−4 and state the range of values of x for which the expansion is valid. Answer Use n = −4 and replace x by 2 x 2 in the binomial formula: ( −4)( −5) ( −4)( −5)( −6) (1 + 2 x 2 ) −4 = 1 + ( −4)(2 x 2 ) + (2 x 2 )2 + (2 x 2 )3 + … 2! 3! = 1 − 8x 2 + 40 x 4 − 160 x 6 + … Expansion is valid for 2 x 2 , 1 ∴ x ,

1 2

Isaac Newton is credited with the generalised binomial theorem for any rational power. A poet once wrote that ‘Newton’s Binomial is as beautiful as the Venus de Milo. The truth is that few people notice it.’

175

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 7.9 Find the coefficient of x 3 in the expansion of

x−5 . 1 + 2x

Answer x−5 = ( x − 5)(1 + 2 x )−1 1 + 2x Use n = −1 and replace x by 2 x in the binomial formula: ( −1)( −2) ( −1)( −2)( −3) (1 + 2 x ) −1 = 1 + ( −1)(2 x ) + (2 x )2 + (2 x )3 + … 2! 3! = 1 − 2 x + 4x 2 − 8x 3 + … ( x − 5)(1 + 2 x )−1 = ( x − 5)(1 − 2 x + 4x 2 − 8x 3 + …) = x(1 − 2 x + 4x 2 − 8x 3 ) − 5(1 − 2 x + 4x 2 − 8x 3 + …) Term in x 3 is x(4x 2 ) + ( −5)( −8x 3 ) = 44x 3. ∴ The coefficient of x 3 is 44.

EXERCISE 7C 1 Expand the following, in ascending powers of x, up to and including the term in x 3. State the range of values of x for which each expansion is valid.

176

a

(1 + x )−2

d

1+ x 2

g

2 (1 − 4x )2

b

(1 + 3x )−1

c

(1 − 2 x )−4

e

1 + 2x

f

3

h

1 − 3x (1 + 2 x )3

i

−3

1 − 3x 1+ x 1 − 2x

2 Find the first 3 terms, in ascending powers of x, in the expansion of each of the following expressions. State the range of values of x for which each expansion is valid. a

(1 + x 2 )−3

b

3

1 − 2x2

3 Find the first 5 terms, in ascending powers of x, in the expansion of

c

(

1 − 4x 2

)

3

2 + 3x . 1 − 5x 2

4 Can the binomial expansion be used to expand each of these expressions? a

(3x − 1)−2

b

2x − 1

Give reasons for your answers. 5 Expand (2 x − 1)−3 in ascending powers of x, up to and including the term in x 3. 1

6 Show that the expansion of (1 + x ) 2 is not valid when x = 3. 3

PS

7 Given that (1 − 3x )−4 − (1 + 2 x ) 2 ≈ 9x + kx 2 for small values of x, find the value of the constant k.

PS

8 Given that

b a + ≈ −3 + 12 x for small values of x, find the value of a and the value of b. 1 − x 1 + 2x

Chapter 7: Further algebra

PS

9 When (1 + ax )−3, where a is a positive constant, is expanded the coefficients of x and x 2 are equal. a Find the value of a. b When a has this value, obtain the first 5 terms in the expansion. 2

PS

10 When (3 − 2 x )(1 + ax ) 3 is expanded the coefficient of x 2 is −15. Find the two possible values of a.

PS

11 The first 3 terms in the expansion of (1 + ax ) n are 1 − 24x + 384x 2. a Find the value of a and the value of n. b Hence find the term in x 3.

7.4 Binomial expansion of ( a + x ) n for values of n that are not positive integers To expand ( a + x ) n , where n is not a positive integer, we take a outside of the brackets: x ( a + x )n = a n 1 + a

n

Hence the expression is: KEY POINT 7.7 2 3 x x n( n − 1) x n( n − 1)( n − 2) x + ⋯ for + ( a + x ) n = a n 1 + n + ,1 a a a 2! 3! a

WORKED EXAMPLE 7.10 Expand (2 + x )−3 in ascending powers of x, up to and including the term in x 3 and state the range of values of x for which the expansion is valid. Answer x (2 + x )−3 = (2)−3 1 + 2

−3

2 3 x ( −3)( −4) x ( −3)( −4)( −5) x x = (2)−3 1 + ( −3) + ,1 + + … for 2 2! 2 3! 2 2

1 8

3 3 2 5 3 1 − 2 x + 2 x − 4 x + … 1 3 3 2 5 3 = − x+ x − x + … valid for x , 2 8 16 16 32 =

177

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 7.11 Find the coefficient of x 2 in the expansion of

1 − 3x . 4−x

Answer 1

− 1 − 3x = (1 − 3x )(4 − x ) 2 4−x

= (1 − 3x )(4)

−

1 2

1− x 4

Factor the 4 out of ( 4 − x ) −

1 2

Simplify ( 4 )

−

1 2

−

1 2.

1 to . 2

− 1 − 3 2 1 1 x 2 2 x − + … = (1 − 3x ) 1 + − − + 4 2 4 2 2! 1 1 3 2 x + … = (1 − 3x ) 1 + x + 2 8 128 1 1 3 2 3 1 3 2 x + … − x 1 + x + x + … = 1 + x + 2 8 128 8 128 2 Coefficient of x 2 is

1 3 3 1 45 × − × =− 2 128 2 8 256

178

EXERCISE 7D 1 Expand the following, in ascending powers of x, up to and including the term in x 3 . State the range of values of x for which each expansion is valid. a

(2 + x )−2

b

d

3

3x + 8

e

(5 − 2 x )−1 4 (3 − x )3

c f

9−x 1 + 2x (2 x − 5)3

2 Find the first 3 terms, in ascending powers of x, in the expansion of each of the following expressions. State the range of values of x for which each expansion is valid. a

(2 + x 2 )−2

b

3

8 − 3x 2

c

(

3 − x2

)

5

3 Expand (1 − 2 x ) 3 8 − 2 x in ascending powers of x, up to and including the term in x 3 . 4 a Expand (2 − x )−1 and (1 + 3x )−2 giving the first 3 terms in each expansion. b Use your answers to part a to find the first 3 terms in the expansion of (2 − x )−1(1 + 3x )−2, stating the range of values of x for which this expansion is valid. 1 5 + x + bx 2 . 4 4 Find the value of a, the value of b and the term in x 3 .

PS

5 The first 3 terms in the expansion of ( a − 5x )−2 are

PS

6 In the expansion of 3 + ax where a ≠ 0 , the coefficient of the term in x 2 is 3 times the coefficient of the term in x 3. Find the value of a.

Chapter 7: Further algebra

−1

PS

2 2 7 a Find the first 4 terms in the expansion of 1 + , where , 1. x x 2 b Show that 1 + x

−1

−1

≡

x x x ≡ 1+ . x+2 2 2 −1

x x x 1 + , where , 1. 2 2 2 d Explain why your expansions are different.

c

Find the first 4 terms in the expansion of

EXPLORE 7.3 1 , discuss why f( x ) can be written in each of (1 + x )(1 + 2 x ) the following ways:

1 Given that f( x ) =

(1 + x)−1 (1 + 2x)−1 f(x)

[1 + (3x + 2x2)]−1 2(1 + 2x)−1 − (1 + x)−1

2 Use the binomial theorem to expand each of these three expressions in ascending powers of x, up to the term in x 3. 179

3 Discuss with your classmates which method you would prefer if you were asked for: a a linear approximation for f( x ) b a quadratic approximation for f( x ) c

a cubic approximation for f( x ).

7.5 Partial fractions and binomial expansions One of the most common reasons for splitting a fraction into partial fractions is so that binomial expansions can be applied. WORKED EXAMPLE 7.12 10 x + 1 , express f( x ) in partial fractions and hence obtain the expansion of f( x ) in (1 − 2 x )(1 + x ) ascending powers of x, up to and including the term in x 3. State the range of values of x for which the expansion is valid. Given that f( x ) =

Answer 10 x + 1 A B ≡ + (1 − 2 x )(1 + x ) 1 − 2 x 1 + x 10 x + 1 ≡ A(1 + x ) + B (1 − 2 x )

Multiply throughout by (1 − 2 x )(1 + x ). (1)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Let x = −1 in equation (1): −9 = 3B B = −3 Let x =

1 in equation (1): 2

3 A 2 A=4 6=

∴

10 x + 1 4 3 ≡ − ≡ 4(1 − 2 x )−1 − 3(1 + x )−1 (1 − 2 x )(1 + x ) 1 − 2 x 1 + x

Expanding (1 − 2 x )−1 gives: ( −1)( −2) ( −1)( −2)( −3) ( −2 x )2 + ( −2 x )3 + … 2! 3! 1 1 = 1 + 2 x + 4x 2 + 8x 3 + … valid for −2 x , 1 i.e. − , x , 2 2

(1 − 2 x ) −1 = 1 + ( −1)( −2 x ) +

Expanding (1 + x )−1 gives: ( −1)( −2) 2 ( −1)( −2)( −3) 3 x + x 2! 3! = 1 − x + x2 − x3 + … valid for x , 1 i.e. − 1 , x , 1

(1 + x )−1 = 1 + ( −1)x + 180

∴

10 x + 1 = 4(1 + 2 x + 4x 2 + 8x 3 + …) − 3(1 − x + x 2 − x 3 + …) (1 − 2 x )(1 + x ) = 1 + 11x + 13x 2 + 35x 3 + …

The expansion is valid for the range of values of x satisfying both −

−1

Hence, the expansion is valid for −

−

1 2

0

1 2

1 1 , x , and −1 , x , 1 2 2 1

1 1 ,x, . 2 2

EXERCISE 7E 1 a Express

7x − 1 in partial fractions. (1 − x )(1 + 2 x )

b Hence obtain the expansion of

2 a Express

7x − 1 in ascending powers of x, up to and including the term in x 3. (1 − x )(1 + 2 x )

5x 2 + x in partial fractions. (1 − x )2 (1 − 3x )

b Hence obtain the expansion of

5x 2 + x in ascending powers of x, up to and including the term in x 3. (1 − x )2 (1 − 3x )

Chapter 7: Further algebra

3 a Express

7 x 2 + 4x + 4 in partial fractions. (1 − x )(2 x 2 + 1)

b Hence obtain the expansion of

4 a Express

7 x 2 + 4x + 4 in ascending powers of x, up to and including the term in x 3 . (1 − x )(2 x 2 + 1)

19 − 7 x − 6x 2 in partial fractions. (2 x + 1)(2 − 3x )

b Hence obtain the coefficient of x 2 in the expansion of

5 a Express

21 in partial fractions. ( x − 4)( x + 3)

b Hence obtain the expansion of

6 a Express

19 − 7 x − 6x 2 . (2 x + 1)(2 − 3x )

24 in ascending powers of x, up to and including the term in x 2. ( x − 4)( x + 3)

6x 2 − 24x + 15 in partial fractions. ( x + 1)( x − 2)2

b Hence obtain the expansion of

6x 2 − 24x + 15 in ascending powers of x, up to and including the term in x 2. ( x + 1)( x − 2)2

181

Checklist of learning and understanding Partial fractions Case 1: If the denominator of a proper algebraic fraction has two distinct linear factors, the px + q A B fraction can be split into partial fractions using the rule ( ax + b )( cx + d ) ≡ ax + b + cx + d ● Case 2: If the denominator of a proper algebraic fraction has a repeated linear factor, the px + q A B ≡ + fraction can be split into partial fractions using the rule ax + b ( ax + b )2 ( ax + b )2 ● Case 3: If the denominator of a proper algebraic fraction has a quadratic factor of the form cx 2 + d that cannot be factorised, the fraction can be split into partial fractions using the rule px + q A Bx + C ≡ + ( ax + b )( cx 2 + d ) ax + b cx 2 + d

●

●

Case 4: If an algebraic fraction is improper we must first express it as the sum of a polynomial and a proper fraction and then split the proper fraction into partial fractions.

Binomial expansion of (1 + x ) n for values of n that are not positive integers ●

(1 + x ) n = 1 + nx +

n( n − 1) 2 n( n − 1)( n − 2) 3 x + x + … for x , 1. 2! 3!

Binomial expansion of (a + x ) n for values of n that are not positive integers ●

2 3 x x n( n − 1) x n( n − 1)( n − 2) x + … for + ( a + x ) n = a n 1 + n + , 1. a a a 2! 3! a

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 7 1

Expand (1 − 2 x )−4 in ascending powers of x, up to and including the term in x 3, simplifying the coefficients. [4]

2

Expand

31−

(6x ) in ascending powers of x up to and including the term in x 3, simplifying the coefficients. [4] Cambridge International A Level Mathematics 9709 Paper 31 Q1 June 2011

3

3 − 2x ) 2

Expand (2 − x )(1 + in ascending powers of x, up to and including the term in x 2, simplifying the coefficients.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q2 November 2016

182

1 + 3x in ascending powers of x up to and including the term in x 2, simplifying the coefficients. [4] (1 + 2 x ) Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2013

4

Expand

5

Expand

6

Expand

10 in ascending powers of x, up to and including the term in x 2, simplifying the coefficients. [4] (2 − x )2

7

Expand

1 in ascending powers of x, up to and including the term in x 2, simplifying the coefficients. [4] 4 − 5x

8

Express

C A B 12 in the form + + . x x2 2x − 3 x 2 (2 x − 3)

[4]

9

Express

8x 2 + 4x + 21 Bx + C A in the form + 2 . x+2 ( x + 2)( x 2 + 5) x +5

[4]

10 Express

1 in ascending powers of x, up to and including the term in x 3 , simplifying the coefficients. 1 + 3x

7 x 2 − 3x + 2 in partial fractions. [5] x( x 2 + 1) Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2013

11 Given that 12 Express

[4]

9x 3 − 11x 2 + 8x − 4 B C D ≡ A+ + 2 + , find the values of A, B, C and D. 2 x x 3x − 2 x (3x − 2)

4x 2 − 5x + 3 in partial fractions. ( x + 2)(2 x − 1)

[5] [5]

2

13 Show that for small values of x 2 , (1 − 2 x 2 )−2 − (1 + 6x 2 ) 3 ≈ kx 4 , where the value of the constant k is to be determined.

[6]

Cambridge International A Level Mathematics 9709 Paper 31 Q3 June 2015

i

4x + 12 . ( x + 1)( x − 3)2 Express f( x ) in partial fractions.

[5]

ii

Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 2 .

[5]

14 Let f( x ) =

2

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2016

Chapter 7: Further algebra

i

2x2 − 7x − 1 . ( x − 2)( x 2 + 3) Express f( x ) in partial fractions.

[5]

ii

Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 2 .

[5]

15 Let f( x ) =

Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2013

i

3x . (1 + x )(1 + 2 x 2 ) Express f( x ) in partial fractions.

[5]

ii

Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 3.

[5]

16 Let f( x ) =

Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2010

183

184

Chapter 8 Further calculus P3 This chapter is for Pure Mathematics 3 students only.

In this chapter you will learn how to:

■■ ■ ■ ■ ■

use the derivative of tan −1 x 1 extend the ideas of ‘reverse differentiation’ to include the integration of 2 x + a2 kf ′ ( x ) , and integrate such functions recognise an integrand of the form f( x ) use a given substitution to simplify and evaluate either a definite or an indefinite integral integrate rational functions by means of decomposition into partial fractions recognise when an integrand can usefully be regarded as a product, and use integration by parts.

Chapter 8: Further calculus

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

Chapter 4

Differentiate e x , ln x, sin x, cos x and tan x.

1 Differentiate with respect to x. a sin 3x b ex c

2

+1

ln(5x − 3)

d tan 2 x − 5 cos x Chapter 5

1 , sin( ax + b ), ax + b cos( ax + b ) and sec 2 ( ax + b ) and use trigonometrical relationships in carrying out integration. Integrate e ax + b ,

2 Find each of these integrals. a b c d

Chapter 7

Express rational functions as partial fractions.

∫ e dx ∫ sin 2x dx 5 ∫ 3x − 2 dx ∫ sec (3x ) dx 5 x +1

2

2

3 Express in partial fractions. 2x a ( x − 1)( x + 3) 4x 2 − 6 b x( x 2 + 2) 5 c x( x − 1)2 3x 2 − 3x − 3 d x2 − x − 2

Why do we study calculus? This chapter will further extend your knowledge and skills in differentiation and integration. In particular, we will learn about integration by substitution and integration by parts, which can be considered as the ‘reverse process’ of the chain rule and the product rule for differentiation that we learnt about earlier in this course. These new rules will enable you to integrate many more functions than you could previously. The skills that we learn in this chapter will be very important when studying Chapter 10. Chapter 10 is about solving differential equations, many of which model real-life situations. Integration by parts plays a major role in engineering and its application is often found in problems including electric circuits, heat transfer, vibrations, structures, fluid mechanics, air pollution, electromagnetics and digital signal processing.

WEB LINK Explore the Chain rule and integration by substitution and the Product rule and integration by parts stations on the Underground Mathematics website.

185

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

8.1 Derivative of tan −1 x The function y = tan −1 x can be written as: tan y = x

Differentiate both sides with respect to x.

dy sec 2 y =1 dx dy 1 = dx sec 2 y

Use sec 2 y = 1 + tan2 y.

dy 1 = dx 1 + tan2 y

Use tan y = x .

dy 1 = dx 1 + x 2 KEY POINT 8.1 1 d (tan −1 x ) = 2 dx x +1

WORKED EXAMPLE 8.1 Differentiate with respect to x. 186

a

tan −1 3x

b

tan −1 x

c

x tan −1 x−2

Answer a

b

d 1 × 3 (tan −1 3x ) = dx (3x )2 + 1 3 = 9x 2 + 1

(

)

d tan −1 x = dx

1

(

x

)2 + 1

1

×

1 1 × x +1 2 x 1 = 2 x ( x + 1) =

Chain rule.

1 −2 x 2

Chain rule.

Chapter 8: Further calculus

d dx

c

x −1 tan x − 2 =

1

( x x− 2 )

2

+1

×

d x dx x − 2

( x − 2)2 ( x − 2)(1) − ( x )(1) × ( x − 2)2 x + ( x − 2)2 −2 = 2 x 2 − 4x + 4 −1 = 2 x − 2x + 2

=

2

Chain rule and quotient rule.

Simplify.

EXERCISE 8A 1 Differentiate with respect to x. x 3

a

tan −1 2 x

b

tan −1 5x

c

tan −1

d

tan −1( x − 1)

e

tan −1 x 2

f

2x tan −1 x + 1

2 Differentiate with respect to x. tan −1 2 x c e x tan −1 x x x 3 Find the equation of the tangent to the curve y = tan −1 at the point where x = 2. 2 a

x tan −1 x

b

−1

4 Show that the tangent to the curve y = tan x at the point where x = −1 is perpendicular to the normal to the curve at the point where x = 1. Find the x-coordinate of the point where this tangent and normal intersect.

PS

8.2 Integration of

x2

1 + a2

Since integration is the reverse process of differentiation we can say that: KEY POINT 8.2

∫

1 dx = tan −1 x + c x2 + 1

This can be extended to: 1 1 x dx = tan −1 + c a a x2 + a2

∫

WORKED EXAMPLE 8.2

Find

∫

1 dx. x 2 + 25

Answer

∫

1 1 x dx = tan −1 + c 5 5 x 2 + 25

REWIND In Section 8.1 we learnt that 1 d (tan −1 x ) = 2 . dx x +1

187

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 8.3

∫ 2x

Find

1 +3

2

dx.

Answer

∫ 2x

1 2

+3

dx =

1 2

∫

=

1 × 2

=

1 2

1 x2 +

3 2

dx

2 tan −1 3 2 tan −1 3

x 23 + c 2 x +c 3

WORKED EXAMPLE 8.4

Find the exact value of

0

Answer

∫

2

0

∫

2

1 d x. x +4 2

2

1 1 x d x = tan −1 2 2 0 x +4 2

1 1 = tan −1 1 − tan −1 0 2 2 1 π 1 = × − × 0 2 4 2 π = 8

188

EXERCISE 8B 1 Find the following integrals. 1 dx a x2 + 9

∫

d

∫ 9x

2

1 dx + 16

1

b

∫ 16 + x

e

∫

1 dx 4x 2 + 1

dx

c

∫

1 dx 4x 2 + 3

f

∫ 2 + 3x

2

1

2

dx

2 Find the exact value of each of these integrals. a

∫

3

0

1 dx 2 x +9

b

3 The diagram shows part of the curve y =

∫

1 2

0

2

2 4x + 1 2

c

dx

∫

2

− 2

1 dx 3x + 2 2

y

. Find the exact value

x2 + 1 of the volume of the solid formed when the shaded region is rotated completely about the x-axis.

–1

O

1

x

Chapter 8: Further calculus

8.3 Integration of

k f ′(x) f(x)

Since integration is the reverse process of differentiation and ln x only exists for x . 0, we can say that: KEY POINT 8.3

∫

REWIND

f ′( x ) d x = ln f( x ) + c f( x )

This can be extended to: k f ′( x ) dx = k ln f( x ) + c f( x )

In Chapter 4, we learnt that d [ ln f( x ) ] = ff(′(xx)) . dx

∫

WORKED EXAMPLE 8.5

Find

∫x

2x dx. +3

2

Answer If f( x ) = x 2 + 3 then f ′( x ) = 2 x . ∴

∫

2x dx = ln x 2 + 3 + c x +3 2

WORKED EXAMPLE 8.6

Find

∫ tan 3x dx.

Answer

∫ tan 3x dx = ∫

sin 3x dx cos 3x

If f( x ) = cos 3x then f ′( x ) = −3sin 3x. 1

∫ tan 3x dx = − 3 ∫ =−

−3sin 3x dx cos 3x

1 ln cos 3x + c 3

189

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 8.7

Find

3x 2 + 5x dx. 3 + 5x 2

∫ 2x

Answer If f( x ) = 2 x 3 + 5x 2 then f ′( x ) = 6x 2 + 10 x. 3x 2 + 5x 1 6x 2 + 10 x x = d dx 2 x 3 + 5x 2 2 2 x 3 + 5x 2 1 = ln 2 x 3 + 5x 2 + c 2

∫

∫

WORKED EXAMPLE 8.8

Find

1

∫

0

2e x dx 1 + ex

Answer If f( x ) = 1 + e x then f ′( x ) = e x.

∫

1

0

2e x dx = 2 1 + ex

190

∫

1

0

ex dx 1 + ex 1

= 2 ln 1 + e x 0 = 2 ln(1 + e) − 2 ln 2 = 2 ln

Substitute limits. Simplify the logarithms.

1+ e 2

EXERCISE 8C 1 Find the following integrals. a d

∫

4x − 10 dx x 2 − 5x + 1

f

∫

sec 2 x dx 1 + tan x

∫

ln 2

c

∫

1

∫

6x 2 dx x3 − 1

b

∫ 1 + sin x dx

c

∫

cot x dx

e

x dx 2 − x2 3x 2 dx x3 + 2

cos x

∫

2 Find the exact value of each of these integrals. a d

∫ ∫

1π 4 1π 6 2

1

tan x d x

b

∫

2

0

x +1 dx x + 2x − 1 2

3 Show that 4 Show that

∫ ∫

e

∫

1π 4 1π 6

ex + e− x dx = ln(1 − e2 x ) − x + c. ex − e− x 1π 4 1π 3

sin x cos x 1 d x = − ln 2. 1 − 2 cos 2 x 8

cot x d x

0

f

0

e2 x dx 1 + e2 x

2x + 3 dx ( x + 1)( x + 2)

Chapter 8: Further calculus

PS

y

5

p

O

x

The diagram shows part of the curve y =

4x . Find the exact value of p for which the shaded region has x2 + 1

an area of 4.

8.4 Integration by substitution

REWIND

Integration by substitution can be considered as the reverse process of differentiation by the chain rule. This method is used when a simple substitution can be applied that will transform a difficult integral into an easier integral. The integral must be completely rewritten in terms of the new variable.

In the Pure Mathematics 1 Coursebook Chapter 7, we learnt the chain rule for differentiation: dy dy du = × dx du dx

WORKED EXAMPLE 8.9

Use the substitution u = 2 x + 1 to find

∫

191

4x 2 x + 1 dx.

Answer u = 2 x + 1 ⇒ x = u = 2 x + 1 ⇒

∫ 4x

u −1 2

du du = 2 ⇒ dx = dx 2

2 x + 1 dx =

u − 1 2

∫4

u

du 2

Rewrite the integral in terms of u and simplify.

1 3 2 2 u − u du

=

∫

=

2 2 2 2 u − u +c 5 3

=

2 2 (2 x + 1) 2 − (2 x + 1) 2 + c 5 3

5

Integrate with respect to u.

3

5

Rewrite in terms of x. 3

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 8.10

Use the substitution x = 2 tan u to find

∫

8 dx. x +4 2

Answer x = 2 tan u ⇒ x 2 = 4 tan2 u dx x = 2 tan u ⇒ = 2 sec 2 u ⇒ dx = 2 sec 2 u du du 8 8 2 sec 2 u du dx = 2 2 x +4 4 tan u + 4

∫

∫

Rewrite the integral in terms of u and simplify.

∫

16 sec 2 u du 4 tan2 u + 4

Use tan2 u + 1 ≡ sec 2 u.

=

∫

16 sec 2 u du 4 sec 2 u

Simplify.

=

∫ 4 du

=

Integrate with respect to u.

= 4u + c

Rewrite in terms of x.

x = 4 tan −1 + c 2

192

WORKED EXAMPLE 8.11

Use the substitution u = sin x to find

∫ sin 2x cos x dx. 2

Answer u = sin x ⇒ sin2 2 x = (2 sin x cos x )2 = 4 sin2 x cos2 x = 4u2 (1 − u )2 u = sin x ⇒

du du = cos x ⇒ dx = dx cos x du

∫ sin 2x cos x dx = ∫ sin 2x cos x cos x = sin 2 x du ∫ = (4u − 4u ) du ∫ 2

2

Replace dx by

du and simplify. cos x

2

2

4

4 3 4 5 u − u +c 3 5 4 4 = sin3 x − sin5 x + c 3 5 =

Replace sin2 2 x by 4u2 − 4u 4. Integrate with respect to u. Rewrite in terms of x.

TIP When using substitution in definite integrals you must also remember to convert the limits to limits for u.

Chapter 8: Further calculus

WORKED EXAMPLE 8.12

Use the substitution u = x + 1 to find

∫

3

0

x d x. x +1

Answer u = x + 1 ⇒ x = u − 1 du u = x +1⇒ = 1 ⇒ dx = du dx Find the new limits for u: x = 3 ⇒ u = 3+1= 4 x = 0⇒ u = 0+1=1 x=3 u=4 x u −1 dx = du x u + 1 x=0 u =1

∫

∫

=

∫

4

1

Write the integral in terms of u and simplify.

1 1 − 2 2 u − u du

Integrate with respect to u.

4

1 2 3 = u 2 − 2u 2 1 3

Evaluate the function at the limits.

16 2 = − 4 − − 2 3 3 8 = 3

193

EXERCISE 8D 1 Given that I =

x

∫

x −3 2

1 2

dx, use the substitution u = x 2 − 3 to show that I =

∫

1 du. u

Hence find I . 2 Find these integrals, using the given substitution. x a dx, u = x + 2 ( x + 2)4

∫

c

∫ cos x sin x dx, u = sin x

e

∫ 5x + 1 dx, u = 5x + 1

5

5x

b

∫x

1 − 2 x 2 dx, u = 1 − 2 x 2

d

∫e

e x + 2 dx, u = e x + 2

f

∫x

x

3x − 1 dx, u = 3x − 1

1

x2 d x, use the substitution x = sin θ to show that I = 0 1 − x2 Hence find the exact value of I .

3 Given that I =

∫

∫

1π 2

sin2 θ dθ .

0

4 Find the exact value of each of these integrals. a

∫

1π 2

cos3 x d x, u = sin x

b

0

c

∫

2

1

∫

4

0

3 d x, u = x 2− x

(

)

x

d

∫

0

x d x, u = 3 − x 3− x π

x2 x2 2 x cos d x , u = cos 4 4

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

e

g

∫ ∫

1 3 1 6

π

π

1

0

2

sec 2 x tan3 x d x , u = tan x

f

x2 , x = 2 cos θ 4 − x2

h

5

∫ ( x + 4 ) dx, x = 2 tan θ ∫ x(2 − x ) dx, x = 2 sin θ −2

2

2

2

5 Use the substitution u = x − 2 to find the exact value of

∫

2

1

2

0

4 d x. 1 + ( x − 2)2

6 The diagram shows part of the curve y = 2 sin2 x cos3 x. Using the substitution u = sin x, find the exact area of the shaded region.

y

π x 2

O

ln x . Using the substitution x u = ln x, find the exact value of the volume of the solid formed when the shaded region is rotated completely about the x-axis.

y

7 The diagram shows part of the curve y =

O cos x

194

sin x . The curve 8 The diagram shows part of the curve y = 3 e has rotational symmetry of order 2 about the origin. Using the substitution u = cos x, find the exact value of the total area bounded by the curve and the x-axis between x = −π and x = π.

−π

EXPLORE 8.1 2 dx. x −1 Try using some of the integration techniques that you have learnt so far to find this integral and discuss what happens with your classmates.

∫

2

2 dx, the denominator x 2 − 1 can be factorised to ( x − 1)( x + 1). x2 − 1 2 Hence the fraction 2 can be split into partial fractions. x −1

In the integral

∫

A B 2 ≡ + x −1 x −1 x +1 2 ≡ A( x + 1) + B ( x − 1) 2

Let x = 1 in equation (1): 2 = 2A A=1

Multiply throughout by ( x − 1)( x + 1). (1)

x

y

8.5 The use of partial fractions in integration

Consider

e

O

π x

Chapter 8: Further calculus

Let x = −1 in equation (1): 2 = −2 B B = −1 2 1 1 ∴ 2 ≡ − x −1 x −1 x +1 1 1 2 − dx = Hence dx 2 x − 1 x + 1 x −1 = ln x − 1 − ln x + 1 + c

∫

∫

= ln

x −1 +c x +1

WORKED EXAMPLE 8.13

Find

3−x

∫ (x + 1)(x

2

dx.

+ 3)

Answer First split into partial fractions. 3−x A Bx + C ≡ + ( x + 1)( x 2 + 3) x + 1 x 2 + 3

Multiply throughout by ( x + 1)( x 2 + 3) .

3 − x ≡ A( x 2 + 3) + Bx( x + 1) + C ( x + 1) Let x = −1

4 = 4A

⇒ A=1

Let x = 0

3 = 3A + C

⇒ C = 0

Let x = 1

2 = 4 A + 2 B + 2C

⇒ B = −1

∴

3−x

∫ (x + 1)(x

2

+ 3)

1

x

195

∫ x + 1 − x + 3 dx 1 1 2x = ∫ x + 1 dx − 2 ∫ x + 3 dx

dx =

2

2

= ln x + 1 −

1 ln x 2 + 3 + c 2

WORKED EXAMPLE 8.14

Find

9

∫ (x + 2)(x − 1)

2

dx.

Answer First split into partial fractions. 9 A B C ≡ + + x + 2 x − 1 ( x − 1)2 ( x + 2)( x − 1)2 9 ≡ A( x − 1)2 + B ( x + 2)( x − 1) + C ( x + 2)

Multiply throughout by ( x + 2)( x − 1)2 .

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Let x = 1 Let x = −2 Let x = 0

∴

9 = 3C 9 = 9A 9 = A − 2 B + 2C 9

∫ (x + 2)(x − 1)

dx =

2

=

⇒ C = 3 ⇒ A = 1 ⇒ B = −1 1

1

1

dx

3

∫ x + 2 − x − 1 + (x − 1)

2

1

∫ x + 2 − x − 1 + 3(x − 1)

−2

dx

= ln x + 2 − ln x − 1 − 3( x − 1)−1 + c = ln

3 x+2 − +c x −1 x −1

EXERCISE 8E 1 Find each of these integrals. 1 a dx x(2 − x )

∫

d

196

x2 + 2x − 5 dx ( x − 3)( x 2 + 1)

∫

2 Evaluate each of these integrals. 2 5x + 13 dx a x ( + 2)( x + 3) 1

∫

d g

3x + x dx 2 −1 ( x − 2)( x + 1)

∫

1

∫

3

2

b e

b

∫ ∫

2

∫

1

∫

1

0

2

1 − x − 2x2 dx x 2 (1 − x )

2x − 5

∫ (x + 2)(x − 1) dx

e

0

h

0

c

6x 2 − 19x + 20 dx ( x + 1)( x − 2)2

f

4x + 5 dx (2 x + 1)( x + 2)

c

x +x+2 dx ( x + 1)(1 + x 2 )

f

1 − 2x dx (2 x + 1)( x + 1)2

i

2(4x − 9)

∫ (2x + 1)(x − 5) dx 2 x 2 + 4x − 21 dx ( x − 1)( x + 4)

∫ ∫

2

∫

0

∫

2

∫

2

1

2

−2

1

4 − 3x dx (2 x − 1)( x + 2) 8 dx (2 − x )(4 + x 2 ) 8x 2 − 3x − 2 dx 4x 3 − 3x + 1

3 By first dividing the numerator by the denominator, evaluate each of these integrals. a

∫

3

2

x2 + 2x + 3 dx ( x − 1)( x + 2)

4 Show that

∫

3 2

2 2

5 Show that

∫

1 2

π

0

6 Show that

∫

ln 2

0

b

∫

3

1

x+3 2 d x = ln 3. x2 − 2

cos x 1 d x = ln 2. 9 − sin2 x 6 1 e2 x 27 d x = ln . 2 20 (1 + e )(2e x + 1) x

2 x 2 + 5x + 1 dx ( x + 1)( x + 2)

c

1

4x 2 + 2 x − 5 dx ( x + 1)(2 x − 1)

Chapter 8: Further calculus

8.6 Integration by parts Consider

∫xe

2x

REWIND

dx . The function to be integrated is clearly the product of two simpler 2x

functions x and e . If we try using a substitution to simplify the integral we will find that it does not help. To integrate this function we require a new technique. Integrating both sides of the equation for the product rule for differentiation with respect to x and using the fact that integration is the reverse process of differentiation we can say that: uv =

dv

In Chapter 4, we learnt the product rule for differentiation: dv du d +v . ( uv ) = u dx dx dx

du

∫ u dx dx + ∫ v dx dx

Rearranging gives the formula for integration by parts: KEY POINT 8.4 dv

du

∫ u dx dx = uv − ∫ v dx dx The following Worked examples show how to use this formula. WORKED EXAMPLE 8.15

Find

∫xe

2x

dx .

197

Answer u=x

⇒

dv = e2 x ⇒ dx

du =1 dx v=

1 2x e 2

dv

du

∫ dx dx = uv − ∫ v dx dx: 1 1 ∫ x e dx = 2 xe − ∫ 2 e

Substitute into u

2x

2x

=

2x

dx

1 1 x e2 x − e2 x + c 2 4

dv carefully. dx du dv x ln x dx. If we let u = x and = ln x , then finding is dx dx

It is important to choose our values for u and Consider trying to find

∫

easy but we do not know how to integrate ln x to find v. Hence we must let u = ln x and dv = x. This is demonstrated in Worked example 8.16. dx

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 8.16

Find

∫ x ln x dx .

Answer du 1 = dx x 1 2 v= x 2

u = ln x ⇒ dv =x ⇒ dx Substitute into

dv

du

∫ u dx dx = uv − ∫ v dx dx: 1 1 1 ∫ x ln x dx = 2 x ln x − ∫ 2 x ⋅ x dx 1 1 = x ln x − ∫ 2 x dx 2 2

2

2

=

1 2 1 x ln x − x 2 + c 2 4

WORKED EXAMPLE 8.17

198

Find

∫ ln x dx.

Answer First write

∫ ln x dx as ∫ 1 ⋅ ln x dx.

u = ln x ⇒

du 1 = dx x

dv =1 ⇒ dx

v=x

Substitute into

dv

du

∫ u dx dx = uv − ∫ v dx dx: 1 ∫ ln x dx = x ln x − ∫ x ⋅ x dx = x ln x − 1 dx ∫ = x ln x − x + c

Chapter 8: Further calculus

Sometimes we might need to use integration by parts more than once to integrate a function. WORKED EXAMPLE 8.18

Find

∫x

2

sin x d x .

Answer u = x2 dv = sin x dx

du = 2x dx

⇒

⇒ v = − cos x dv

du

∫ u dx dx = uv − ∫ v dx dx ∫ x sin x dx = −x cos x − ∫ − 2x cos x dx = − x cos x + 2 x cos x dx ∫

Substitute into

2

2

2

Use integration by parts a second time to find u = 2x dv = cos x dx

⇒

(1)

∫ 2x cos x dx

du =2 dx

⇒ v = sin x

Substitute into

dv

199

du

∫ u dx dx = uv − ∫ v dx dx

∫ 2x cos x dx = 2x sin x − ∫ 2 sin x dx = 2 x sin x + 2 cos x + c Substituting in (1) gives:

∫x

2

sin x dx = − x 2 cos x + 2 x sin x + 2 cos x + c

WORKED EXAMPLE 8.19

Given that I =

∫

ln x 1 dx, use integration by parts to show that I = (ln x )2 + c. x 2

Answer u = ln x 1 dv = dx x

⇒

1 du = dx x

⇒ v = ln x

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Substitute into

dv

du

∫ u dx dx = uv − ∫ v dx dx ln x I = (ln x ) − ∫ x dx 2

Rearrange.

2 I = (ln x )2 + k 1 k I = (ln x )2 + 2 2 1 ∴ I = (ln x )2 + c 2

Divide both sides by 2.

WORKED EXAMPLE 8.20

Find the exact value of

∫

1π 2

4x sin 2 x d x.

0

Answer u = 4x

⇒

du =4 dx

dv 1 = sin 2 x ⇒ v = − cos 2 x dx 2 200

dv

du

∫ u dx dx = uv − ∫ v dx dx ∫ 4x sin 2x dx = −2x cos 2x + ∫ 2 cos 2x dx

Substitute into

= −2 x cos 2 x + sin 2 x + c ∴

∫

1π 2

0

1π

4x sin 2 x d x = −2 x cos 2 x + sin 2 x 02

= ( −π cos π + sin π ) − (0 + sin 0) =π

EXERCISE 8F 1 Find each of these integrals. a

∫ 3xe

d

∫ x sin 2x dx

x

dx

b

∫ x cos x dx

c

∫ x ln 2x dx

e

∫x

ln x dx

f

∫

∫

2

x ln 2 x d x

c

∫

1

b

∫

1

∫

0

3

ln x dx x

2 Evaluate each of these integrals. a

∫

1π 6

x cos 3x d x

1

0

d

∫

3

1

ln x d x

e

0

xe2 x d x

0

(2 − x )e −3x d x

f

−1 5

(1 − 3xe −5 x ) d x

Chapter 8: Further calculus

3 Use integration by parts twice to find the exact value of each of these integrals. a

∫

2

∫

e

2

∫

b

(ln x ) d x

x 2 cos 2 x d x

c

0

1

d

1π 4

x(ln x )2 d x

∫

e

0

−∞

0

∫

π

∫

π

x 2 sin x d x

0

x2ex dx

f

e x sin x d x

0

4 Find the exact value of the area for each of these shaded regions. a

b

y

y

y = x2 ln x

y = x sin x

O

c

e

y y=

e

O

π

O

d

ln x x

x

x

y y = x2 e–x

O

4

x

201

y

5

O

x

The diagram shows the curve y = e − x x + 2 . Find the exact value of the volume of the solid formed when the shaded region is rotated completely about the x-axis.

8.7 Further integration Some integrals can be found using different integration techniques. The activities in Explore 8.2 look at why our answers to an indefinite integral might ‘appear’ to be different when using different integration techniques when they are actually equivalent. EXPLORE 8.2

1 Nadia and Richard are asked to find

∫ x(x − 2) dx. 7

Nadia uses the substitution u = x − 2 and her answer is:

1 1 ( x − 2)9 + ( x − 2)8 + c 9 4

Richard uses integration by parts and his answer is:

1 1 x( x − 2)8 − ( x − 2)9 + c 8 72

Use each of their methods to show that they are both correct and then use algebra to explain why their answers are equivalent.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2 a

i

4x + 1 dx using: 2 + x −1 the substitution u = 2 x 2 + x − 1

ii

partial fractions.

∫ 2x

Find

4x + 1 dx. 2 + x −1 Use algebra to show that your two answers to part a are equivalent.

∫ 2x

b Suggest another method for finding c

Explore 8.3 is designed to make you think about the different integration techniques that might be available to you for a particular integral. This activity expects you to be able to recognise the appropriate substitution to use when integrating by substitution. EXPLORE 8.3 Discuss with your classmates which integration methods can be used to find each of these integrals.

∫ 202

2x + 3 dx x + 3x − 10 2

∫

3x 2 dx x3 + 1

∫ ln 5x dx

∫x

2 x + 1 dx

∫x

2

2x + 3 dx + 3x − 10

∫ cos

5

x sin x d x

∫ x e

x2

3

∫ (2x + 1)

5

dx

∫ x sin 4x dx

Find each integral.

Checklist of learning and understanding Integrating ●

∫

k f ′( x ) f (x)

kf ′( x ) dx = k ln f( x ) + c f( x )

Integration by substitution ●

Substitutions can sometimes be used to simplify the form of a function so that its integral can be easily recognised. When using a substitution, we must ensure that the integral is completely rewritten in terms of the new variable before integrating.

Integrating rational functions ●

Some rational functions can be split into partial fractions that can then be integrated.

Integration by parts ●

dv

du

∫ u dx dx = uv − ∫ v dx dx

∫

2x dx x+5

∫

3x − 2 dx

dx

∫

sin x dx 3 + cos x

Chapter 8: Further calculus

END-OF-CHAPTER REVIEW EXERCISE 8

1

1 2

∫

Find the exact value of

xe −2 x d x.

[5]

0

Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2016 2

∫

Find the exact value of

4

1

ln x d x. [5] x Cambridge International A Level Mathematics 9709 Paper 31 Q3 November 2013

Use the substitution u = 1 + 3tan x to find the exact value of

4

i Prove that cot θ + tan θ ≡ 2 cosec 2θ . ii

5

1π 4

(1 + 3 tan x ) [5] dx cos2 x 0 Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2014

3

Hence show that

Let I =

∫

1

0

i

∫

1π 3 1π 6

∫

[3]

1 ln 3. [4] 2 Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2013

cosec 2θ dθ =

x2 d x . (4 − x 2 )

Using the substitution x = 2 sin θ , show that I =

∫

1π 6

4 sin2 θ dθ .

[3]

0

ii

Hence find the exact value of I .

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2010 6

The integral I is defined by I =

∫

2

4t 3 ln(t 2 + 1) dt.

0

i

Use the substitution x = t + 1 to show that I = 2

∫

5

(2 x − 2) ln x d x.

[3]

1

ii

Hence find the exact value of I .

[5] Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2011

7

2 in partial fractions. ( x + 1)( x + 3)

i

Express

ii

2 1 1 1 1 Using your answer to part i, show that . ≡ + + 2 − x + 1 x + 3 ( x + 3)2 ( x + 1) ( x + 1)( x + 3)

[2] 2

iii Hence show that

∫

1

0

7 4 3 dx = − ln . 12 2 ( x + 1)2 ( x + 3)2

[2] [5]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2010 8

a Find

∫ (4 + tan 2x ) dx. 2

b Find the exact value of

∫

[3] 1π 2 1π 4

sin ( x + 61 π ) d x. sin x

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2015

203

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

9

i ii

Prove the identity tan 2θ − tan θ ≡ tan θ sec 2θ . Hence show that

∫

1π 6

0

[4]

1 3 [4] ln . 2 2 Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2016

tan θ sec 2θ dθ =

10

y M

O

x

2 1

The diagram shows part of the curve y = (2 x − x 2 )e 2

x

and its maximum point M .

i

Find the exact x-coordinate of M .

[4]

ii

Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2016 2 11 By first expressing 4x + 5x + 3 in partial fractions, show that 2 x 2 + 5x + 2

∫

4

0

4x 2 + 5x + 3 d x = 8 − ln 9. 2 x 2 + 5x + 2

[10]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2012 204

y

12

2

O

x

M

The diagram shows the curve y = x 3 ln x and its minimum point M . i Find the exact coordinates of M. ii

[5]

Find the exact area of the shaded region bounded by the curve, the x-axis and the line x = 2.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2010 y

13

M

R 1

O

p

x

2

x The diagram shows the curve y = for x ù 0, and its maximum point M . The shaded region R is 1 + x3 enclosed by the curve, the x-axis and the lines x = 1 and x = p. i ii

Find the exact value of the x-coordinate of M . Calculate the value of p for which the area of R is equal to 1. Give your answer correct to 3 significant figures.

[4] [6]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2015

Cross-topic review exercise 3

CROSS-TOPIC REVIEW EXERCISE 3 P3 This exercise is for Pure Mathematics 3 students only. 1

Expand (1 − 3x ) −5 in ascending powers of x, up to and including the term in x 3.

2

Evaluate

1

1

0

3x + 1

∫

2

d x.

[3] [3]

3

Expand

32 in ascending powers of x, up to and including the term in x 3. ( x + 2)3

[4]

4

Expand

1 in ascending powers of x, up to and including the term in x 2 . 4 − 2x

[4]

5

Evaluate

∫

1 π 2

x 2 sin x d x .

[5]

0

6

Expand

7

Show that

1 + 2x in ascending powers of x, up to and including the term in x 3. 1− x

∫

1

(x + 2)e −2 x d x =

0

8

i

ii

5 7 . − 4 4e2

[5]

[5]

1 in ascending powers of x, up to and including the term in x 2 , simplifying ( 1 − 4x ) the coefficients. 1 + 2x . Hence find the coefficient of x 2 in the expansion of ( 4 − 16x ) Expand

[3] [2]

Cambridge International A Level Mathematics 9709 Paper 31 Q2 June 2012 9

When (1 + ax ) −2, where a is a positive constant, is expanded in ascending powers of x, the coefficients of x and x 3 are equal. i

Find the exact value of a.

[4]

ii

When a has this value, obtain the expansion up to and including the term in x 2 , simplifying the coefficients.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2012 10

y

O

α

x

1 1 The diagram shows the curve y = 8 sin x − tan x for 0 ø x , π. The x-coordinate of the maximum 2 2 point is α and the shaded region is enclosed by the curve and the lines x = α and y = 0 . 2 [3] i Show that α = π. 3 ii Find the exact value of the area of the shaded region. [4] Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2012

205

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

11

∫

Show that

5

0

1

12

Let I =

∫

2

0

a b 13

4 5 − 3x d x = 4 ln . 3 ( x + 1)(3x + 1)

[7]

4x 2 d x. 1 − x2

Using the substitution x = sin θ , show that I = 3 π Hence show that I = − . 3 2 a

∫

It is given that

∫

π 6

4 sin2 θ dθ .

[3]

0

[4]

x ln x d x = 30 , where a . 1.

1

119 . 2 ln a − 1

a

Show that a =

b

Use the iterative formula an +1 =

[5]

119 to determine the value of a correct to 2 ln an − 1 2 decimal places. Give the result of each iteration correct to 4 decimal places.

[3]

y

14

M

206 O 2 2−x

The diagram shows the curve y = x e i ii

x

and its maximum point M.

Show that the x-coordinate of M is 2. Find the exact value of

∫

2

x 2 e2 − x d x .

[3] [6]

0

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2015 15

Let f( x ) =

12 + 8x − x 2 . (2 − x )(4 + x 2 )

i

Express f( x ) in the form

ii

Show that

∫

1

0

Bx + C A + . 2−x 4 + x2

25 f( x ) d x = ln . 2

[4] [5]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2011 16

It is given that

∫

a

1

i ii

ln(2 x ) d x = 1, where a . 1.

1 ln 2 x exp 1 + , where exp( x ) denotes e . a 2 1 ln 2 to determine the value of a correct to Use the iterative formula an +1 = exp 1 + an 2 2 decimal places. Give the result of each iteration to 4 decimal places. Show that a =

[6]

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2014

Cross-topic review exercise 3

17

4 + 12 x + x 2 in partial fractions. (3 − x )(1 + 2 x )2

i

Express

ii

Hence obtain the expansion of

[5]

4 + 12 x + x 2 in ascending powers of x, up to and (3 − x )(1 + 2 x )2

including the term in x 2 .

[5] Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2014

18

Let f( x ) =

x 2 − 8x + 9 . (1 − x )(2 − x )2

i

Express f( x ) in partial fractions.

[5]

ii

Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 2 .

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2014 19

20

Let f( x ) =

5x − 2 . ( x − 1)(2 x 2 − 1)

a

Express f( x ) in partial fractions.

[5]

b

Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 3.

[5]

a

Given that

b

Hence show that

B C D x3 − 2 ≡ A+ + 2 + , find the values of the constants A, B, C and D. x x 2x − 1 x (2 x − 1) 2

∫

2

1

21

3 9 1 x3 − 2 d x = − 2 ln + ln 3 . 2 4 4 x (2 x − 1)

[5] [5]

2

y

O

e

M

x

The diagram shows the curve y = x 2 ln x and its minimum point M. i

Find the exact values of the coordinates of M.

[5]

ii

Find the exact value of the area of the shaded region bounded by the curve, the x-axis and the line x = e.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2011 22

i

Show that

∫

4

4x ln x d x = 56 ln 2 − 12.

[5]

2 1

ii

Use the substitution u = sin 4x to find the exact value of

∫

24

π

cos3 4x d x.

[5]

0

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2013

207

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

23

i

Express 4 cos θ + 3 sin θ in the form R cos(θ − α ), where R . 0 and 0 , α , the value of α correct to 4 decimal places.

ii

1 π. Give 2 [3]

Hence a solve the equation 4 cos θ + 3 sin θ = 2 for 0 , θ , 2 π, b find

50

∫ (4 cos θ + 3 sin θ ) dθ . 2

[4] [3]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 June 2013

208

209

Chapter 9 Vectors P3 This chapter is for Pure Mathematics 3 students only.

In this chapter you will learn how to:

■ ■ ■ ■ ■ ■ ■ ■

use standard notations for vectors, in two dimensions and three dimensions add and subtract vectors, multiply a vector by a scalar and interpret these operations geometrically calculate the magnitude of a vector and find and use unit vectors use displacement vectors and position vectors find the vector equation of a line find whether two lines are parallel, intersect or are skew find the common point of two intersecting lines find and use the scalar product of two vectors.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE / O Level Mathematics

Recall and apply:

1 The diagram shows a cuboid with CD = 5cm, DH = 6 cm and EH = 3 cm.

● ●

Pythagoras’ theorem simple trigonometry

B

to solve problems in two and three dimensions.

5 cm

C

F

G A

E 3 cm

D

6 cm

H

Find the angle between AG and AH. IGCSE / O Level Mathematics Pure Mathematics 1 Coursebook, Chapter 3

Show familiarity with the following areas of coordinate geometry in two dimensions: ●

●

210 ●

IGCSE / O Level Mathematics

Use and apply the general equation of a straight line in gradient/intercept form. Calculate the coordinates of the midpoint of a line. Find and interpret the gradients of parallel and perpendicular lines.

Divide a quantity in a given ratio.

2 Three points A, B and C are such that A(1, 7) , B (7, 3) and C (0, −6) . a Find the equation of the perpendicular bisector of the line AB. Give your answer in the form y = mx + c. b Write down the equation of the line parallel to AB that passes through the point C .

3 The diagram shows a cuboid with CD = 5 cm, DH = 6 cm and EH = 3 cm.

Recall and apply properties of prisms, other solids and plane shapes, such as trapezia and parallelograms.

B

5 cm

C

F

G A

E 3 cm

D

6 cm

H

The point M lies on DH such that DM : MH = 1:2. The point N is the midpoint of EF . a Write down the length of MH . b Find the exact length of NH . IGCSE / O Level Mathematics

Translate points in 2-dimensional space using column vectors.

4 The point P has coordinates (3, 9). The vector −1 PQ = . 3 Write down the coordinates of the point Q.

Chapter 9: Vectors

Why are vectors important? Vectors are used to model many situations in the world around us. They are used by engineers to ensure that buildings are designed safely. They are used in navigation to ensure that planes and boats, for example, arrive at their destinations in differing winds or water currents. There are countless examples of the use of vectors to model the movement of objects in mathematics and physics. Displacement, velocity and acceleration are all examples of vector quantities. Vectors are essential to describing the position and location of objects. A boat lost at sea can be found if its bearing and distance from each of two fixed points is known. These are just a few examples. Our world is 3-dimensional, as far as we know, and in this chapter we will be studying vectors in two and three dimensions. As the mathematics of vectors can be applied to all vectors, the study of vector geometry and algebra in general is important. It is helpful if you have a good understanding of simple coordinate geometry before you start working with vectors because you will be able to make comparisons between the coordinate geometry that you have previously studied and the vector geometry introduced in this chapter. Recall that a vector has both magnitude (size) and direction. You should have some knowledge of how to work with vectors in 2-dimensional space from your previous studies. You will be reminded of this here, before being introduced to vectors in 3-dimensional space.

WEB LINK Explore the Vector geometry station on the Underground Mathematics website.

9.1 Displacement or translation vectors A movement through a certain distance and in a given direction is called a displacement or a translation. The magnitude or length of a displacement represents the distance moved.

Examples of displacement or translation vectors ●

●

Sami walks 100 m in a straight line on a bearing of 045° from place A to place B. Jan is also standing at place A. Jan walks directly to Sami. Jan therefore also walks for 100 m on a bearing of 045° from place A. Jan follows the same displacement vector as Sami. The diagram shows two arrows. y 3 Q

M

2 1

–2

–1

O

1

2

3

x

–1 –2 L –3

P

211

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

The arrow from L to M is the same length and has been drawn in the same direction as the arrow from P to Q. This means that the points L and P have been displaced or translated by the same vector to M and Q, respectively. The arrow LM and the arrow PQ represent the same displacement.

Notation using upper case

The order of the letters is important as ML would represent a displacement from M to L.

TIP

A displacement from L to M is written as LM.

In some books LM is used for LM .

The arrow above the upper case letters always goes from left to right. Reverse the letters to show the opposite direction. TIP

In the previous diagram, LM = PQ . Even though the arrows are drawn on different parts of the page, the displacement shown by each is the same.

Components of a displacement in two dimensions Any displacement can be represented by the displacement in the x-direction followed by the displacement in the y-direction. These are called the components of the displacement. 3 In the following example, the displacement vector AB = . 1

When you write your vectors using the correct notation, you are less likely to make errors.

y

3 is the x-component and 1 is the y-component of the displacement. As 3 and 1 are both positive, the movement is to the right and up, respectively.

3

212

B

2 1

A –3

–2

–1 O

1

2

x

Components of a displacement in three dimensions In a 3-dimensional system of coordinates, a point P is located using three coordinates, ( x, y, z ). The axis for the third coordinate, the z-coordinate, intersects with the x- and y-axes at the origin and the three axes are perpendicular to each other. The axes may be rotated, as shown. y

z z P(x, y, z)

y

P(x, y, z)

P(x, y, z)

y

x x

z

A displacement vector in a 3-dimensional system needs to have three components. For example, the point P (1, 3, 7) translates to the point Q(5, 3, 8) by the displacement 4 vector PQ = 0 . 1

x

Chapter 9: Vectors

Notation using lower case Sometimes lower case letters are used for vectors. In text, these are in bold type. For example, a or b. When written with a pen or pencil, they should be underlined with a bar or a short, wavy line instead. For example, a or b. Sometimes the bar or wavy line is written over the top of the ɶ lower case letter. This diagram shows three vectors, a, b and c .

a

b

c

The vectors a and b have the same direction but not the same magnitude or length. Vectors a and b are parallel but not equal. In fact, a is half the length of b. When the length of a is doubled, the vectors are equal. In other words b = 2a . When a is scaled so that it is the same length as b, then the vectors are equal.

a b

a

3 of the length of vector b, but it has opposite direction. 4 3 In other words c = − b. 4 A quantity (usually a number) used to scale a vector is called a scalar. So in ‘2a’ the vector a is being multiplied by the scalar 2.

Vector c is

Parallel vectors

213

TIP This may also be written, for example, as 4c = −3b. Can you write an expression for b in terms of c ?

Two vectors are therefore parallel provided that one is a scalar multiple of the other. For example, when a = λ b, where λ is some scalar, then a and b are parallel.

Combining displacements (adding and subtracting) 4 Starting at point A, make the displacement AB = followed by the displacement 2 3 BC = . 7

C

What is the result?

7 The overall displacement shown by AC, is 9 resultant vector.

●

●

. This is usually called the

The x-component of the resultant vector is easily found by adding the x-components of AB and BC. The y-component of the resultant vector is easily found by adding the y-components of AB and BC.

B

A

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

This operation is called vector addition. KEY POINT 9.1 The formula for vector addition:

AB + BC = AC

0 , as there is no overall displacement. In other words AC + CA =

The displacement AC followed by the displacement CA must return you to point A. 0

7 −7 0 −7 = , so CA = + −9 −9 0 9

.

As −7 and −9 are both negative, the movement is to the left and down, respectively. 7 7 Alternatively, we may write − 9 9

0 0 = . In other words AC − AC = 0 0

.

It is reasonable, therefore, to state that CA = − AC . KEY POINT 9.2

214

CA = − AC is always true.

Taking the negative of a vector changes the direction of the vector and does not change its length.

Vectors AC and CA have opposite direction. As a − b is the same as a + − b, vector subtraction is defined as the addition of the negative of a vector. The displacement a − b is therefore the displacement a followed by the displacement −b. EXPLORE 9.1 a, b and c are displacement vectors such that c = a + b. Using diagrams to help you, show that a = c − b and that a + b = b + a.

KEY POINT 9.3 a + b = b + a means that the addition of vectors is commutative.

Chapter 9: Vectors

WORKED EXAMPLE 9.1 Q

R

M

a P

b

S

PQRS is a quadrilateral. M is the midpoint of RS. 1 PQ = a, PS = b and PM = (2a + 5b ). 4 a Find an expression in terms of a and b for SM. Give your answer in its simplest form.

b Find an expression for QR. c Describe quadrilateral PQRS, justifying your answer. Answer a

b

c

SM = − PS + PM or PM − PS 1 SM = (2a + 5b ) − b 4 1 5 1 1 SM = a + b − b = a + b 2 4 2 4 QR = QP + PS + SR QR = − PQ + PS + 2SM 1 1 QR = −a + b + 2 a + b 2 4 3 QR = b 2 3 QR = PS so the lines QR and PS are parallel and SM = SP + PM

Substitute SP = − PS . Substitute given vectors in a and b. Expand and simplify.

215

Use SR = 2SM and QP = − PQ. Substitute known vectors in a and b. Expand and simplify.

2 PQRS is a trapezium.

Magnitude of a vector The magnitude of a vector is the length of the line segment representing the vector. This is found using the components of the displacement and Pythagoras’ theorem. Notation: The magnitude of a vector p is written as p . This is also known as the modulus of the vector.

TIP A diagram can often be useful in vector questions.

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WORKED EXAMPLE 9.2 4 3 a Find the magnitude of each of the vectors a = , b = − and a – b. 3 4 −1 5 b p = 0 and q = 8 . Find the magnitude of the vector p + q. −3 8 Answer a

Draw a diagram showing the x- and y-components. a

3

4

|a| =

42 + 32 = 25 = 5

Apply Pythagoras’ theorem. Again, draw a diagram to show the x- and y-components.

b 4

216

3

b =

( −3)2 + 42 =

25 = 5

Apply Pythagoras’ theorem. The diagram has been given to help you to see the components of the vector a − b.

3

a

–b –4

4 3

7 = a−b = −1

b

72 + ( −1)2 =

5 −1 4 p+q = 0 + 8 = 8 8 −3 5 4 8 = 5

and

42 + 82 + 52 = 105

49 + 1 =

50 = 5 2

Apply Pythagoras’ theorem. (Notice that a − b is not the same as a − b .)

It is not a good idea to try to draw a diagram here.

Apply Pythagoras’ theorem in 3 dimensions.

Chapter 9: Vectors

Unit vectors A unit vector is any vector of magnitude or length 1. Notation: aˆ (read as a hat) is the unit vector in the direction of a. To find a unit displacement vector in a particular direction: ●

find the magnitude of the displacement then divide the components of the vector by the magnitude.

p1 This means that, for a vectorp p = p2 , the corresponding unit vector is pˆ = p3

●

p1 p p2 . p p3 p

WORKED EXAMPLE 9.3 Find the unit vector in the direction

5 . 2

Answer The diagram shows that, geometrically, we are finding lengths of similar triangles when finding components of unit vectors.

3 2 1

2 3

√5 3

√5

5 = 2

(

5 ) + 22 = 2

Unit vector in the direction

9 =3

Apply Pythagoras’ theorem to find the magnitude of the given vector.

5 5 3 . is 2 2 3

Divide each component by the magnitude.

WORKED EXAMPLE 9.4 3 Points A, B and C are such that AB = 4 1

11 and BC = −2 . −1

Find the unit displacement vector in the direction AC.

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Answer B

AC is the resultant vector formed from AB + BC .

C

A

AC = AB + BC 3 11 14 AC = 4 + −2 = 2 1 −1 0

AC = 142 + 22 + 02 =

Substitute column vectors and sum components.

200 = 10 2

Now find the magnitude by applying Pythagoras’ theorem.

Unit displacement vector in the direction of AC is

Divide each component by the magnitude of AC .

14 1 2 . 10 2 0

EXERCISE 9A 218

1 The diagram shows the vectors AB and BC on 1 centimetre squared paper.

A

C

Using 1cm as 1 unit of displacement:

a write AB and BC as column vectors b write down the column vector AC .

2

The diagram shows the vectors DE and DF on 1 centimetre squared paper.

B

F

E

D

Using 1cm as 1 unit of displacement:

show that DF − DE = EF .

a find the column vector EF b

Chapter 9: Vectors

3

Q

R P

Justify the statement QR = PR − PQ. P

4 In triangle ABC , X and Y are the midpoints of AB and AC, respectively.

A

AX = a and AY = b. X

a Use a vector method to prove that BC is parallel to XY .

XY = k BC

b

Y

B

C

Write down the value of the constant k. PS

5 The diagram shows a cuboid, ABCDEFGH. N

F

G

z y

E

H D

C M

A

219

B

x

O

M is the midpoint of BC and the point N is on FG such that FN : NG is 1:3. 12 Given that AG = 4 , find the displacement vector: 2

6

AM

ABCDEFGH is a regular octagon. AB = p, AH = q, HG = r and HC = s.

a

b AN.

A

p

B

q H

C

a Using the vectors p, q, r and s, write down expressions for: i

BC

ii

EB.

b Explain why s = kp and find the exact value of k.

s r D

G

F

P

E

7 Use a vector method to show that, when the diagonals of a quadrilateral bisect one another, then the opposite sides are parallel and equal in length.

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9.2 Position vectors A position vector is simply a means of locating a point in space relative to an origin, usually O. This means a position vector is fixed in space. Position vectors are sometimes called location vectors. Up to this point, the displacement vectors we have been working with have been free vectors. This means they are not fixed to any particular starting point or origin. Many free vectors can represent the same displacement. Position vectors are different because they start at a given point or origin. This fixed starting point means there is only one arrow that represents each vector.

REWIND Look back at vectors LM and PQ at the start of this chapter.

Since we can apply the same mathematics to all vectors, the results derived for displacement vectors (free vectors) also apply to position vectors. The fixed starting point is often called O, as in the origin of a system of Cartesian coordinates. 1 So a point with coordinates A(1, 2, 3) can be represented by the position vector OA = 2 . 3

Position vector of any point in a plane Consider this case in 2-dimensional space. P

220 a O

b

Given that a and b are position vectors relative to the origin O and that P is any other point in the same plane as a and b, then the position vector of P can be written using a and b.

This is because OP is some scalar multiple of a plus some scalar multiple of b:

OP = ma + nb, where m and n are scalars. All the other vectors in this plane can be written using combinations of the vectors a and b.

Cartesian components When the vectors a and b are perpendicular to each other and ● unit vectors ●

the plane looks like the usual set of x- and y-axes (called the Oxy plane) we use for Cartesian coordinates. In this case, the unit vector in the x-direction is called i and the unit vector in the y-direction is called j.

Chapter 9: Vectors

y P(x, y)

j

x

i

O

In this case, OP is some scalar multiple of i plus some scalar multiple of j: OP = xi + yj, where x and y are scalars.

4 The point P with coordinates (4, 5) has position vector 5 4 i + 5 j.

, which can also be written as

This can be extended to three dimensions. The unit vector in the z-direction is called k. z Q(x, y, z)

y k O

221

j i

x

In this case, OQ is some scalar multiple of i plus some scalar multiple of j plus some scalar multiple of k:

OQ = xi + yj + zk, where x, y and z are scalars. 4 The point Q with coordinates (4, 2, 3) has position vector 2 , which can also be written as 4 i + 2 j + 3k. 3 WORKED EXAMPLE 9.5 The position vectors of the points A, B and C are given by a, b and c , respectively, where 2 0 a = 5 , b = 4 −3 6

−4 and c = 2 24

.

a Write a, b and c in terms of i, j and k. b Giving your answers in terms of i, j and k, find the vectors representing: i

AB

ii AC

c Describe the relationship between the points A, B and C , justifying your answer.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Answer a

a = 2 i + 5 j − 3k, b = 4 j + 6 k and c = −4 i + 2 j + 24 k.

b

i

B

A simple sketch should help you to calculate the answer correctly.

A O

AB = AO + OB = OB − OA Substitute the Cartesian form. AB = b − a = 4 j + 6 k − (2 i + 5 j − 3k ) Expand and simplify. AB = −2 i − j + 9 k Substitute the Cartesian form. ii AC = AO + OC = OC − OA AC = c − a = −4 i + 2 j + 24 k − (2 i + 5 j − 3k ) Expand and simplify. AC = −6 i − 3 j + 27 k AC = 3AB and so these vectors are parallel. Since the vectors also have point A in common, then A, B and

c

C must be collinear (so they lie on the same straight line).

WORKED EXAMPLE 9.6 222

The diagram shows a cube OABCDEFG with sides of length 4 cm. Unit vectors i, j and k are parallel to OC , OA and OD, respectively. The point M lies on FB such that FM = 1 cm, N is the midpoint of OC and P is the centre of the square face DEFG .

Express each of the vectors NP and MP in terms of i, j and k. E

F P G

D

k O

A

j i

M

C

Answer Using the displacement of P from N and M :

NP = 2 j + 4 k F

E P G

D

k O

j

A

i

M

B P′ N

C

B

N

The cube is of side 4 cm and P ′ is the centre of the square AOCB. NP is parallel to the y-direction and so the first part of the vector NP is 2j. P is directly above this point P ′ and so the second component P ′P is parallel to the z-direction and is +4k.

Chapter 9: Vectors

MP = −2 i − 2 j + k Starting with the x-direction, move 2i left. This is the opposite to the positive direction indicated by the red arrow, so the first component is −2i .

F

E P G

D

M

In the y-direction, again we are moving in the negative direction, so the second component is −2j. A k O

i

In the z-direction, we must move a distance equal to MF . This is 1 unit and so the third component is +k.

B

j

C

N

Or we can use position vectors:

NP = OP − ON

OP is 2 units in x-direction, 2 units in y-direction and 4 units in the z-direction and ON is 2 units in x-direction.

NP = OP − ON = 2 i + 2 j + 4 k − 2 i NP = 2 j + 4 k MP = OP − OM

Subtract vectors.

OP = 2 i + 2 j + 4 k and ON = 2 i

Simplify.

OM is 4 units in x-direction, 4 units in y-direction and 3 units in z-direction.

OP = 2 i + 2 j + 4 k and OM = 4 i + 4 j + 3k MP = OP − OM = 2 i + 2 j + 4 k − (4 i + 4 j + 3k ) MP = −2 i − 2 j + k

223

Subtract vectors. Expand and simplify.

EXERCISE 9B

1 Relative to an origin O, the position vectors of the points A and B are given by OA = 5 i + 3 j and OB = 5 i + λ j + µ k .

a In the case where λ = 0 and µ = 7, find the unit vector in the direction of AB. b Given that triangle OAB is right-angled with area 5 34 and µ = 10, find the value of λ . 2 The diagram shows a cuboid OABCDEFG with a horizontal base OABC . The cuboid has a length OA of 8 cm, a width AB of 4 cm and a height BF of d cm. The point M is the midpoint of CE and the point N is the point on GF such that GN = 6 cm. The unit vectors i, j and k are parallel to OA, OC and OD, respectively. It is given that OM = 4 i + 2 j + k.

G

6 cm

N

F d cm

D

E C

B

M

k j

4 cm

a Write down the height of the cuboid.

b Find ON. PS

c

Find the unit vector in the direction MN.

O

i

8 cm

A

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

3 The position vectors of the points A, B, C and D are given by a, b, c and d, respectively, where 5 −6 13 2 a = 2 , b = 5 , c = −3 and d = −6 4 0 0 4

a i

.

Find AD , BC , AB and DC .

ii

Deduce that ABCD is a parallelogram.

b Find the coordinates of i

the point M , the midpoint of the line AB 1 the point P such that BP = BD . 3

ii

4 Relative to an origin O, the position vector of A is i − 2 j + 5 k and the position vector of B is 3i + 4 j + k.

a Find the magnitude of AB. b Use Pythagoras’ theorem to show that OAB is a right-angled triangle. c PS

Find the exact area of triangle OAB.

5 The position vectors of the points A, B and C are given by a = 5 i + j + 2 k, b = 7 i + 4 j − k and c = i + j + qk. Given that the length of AB is equal to the length of AC, find the exact possible values of the constant q.

224

6 The diagram shows a triangular prism, OABCDE . The uniform cross-section of the prism, OAE, is a right-angled triangle with base 24 cm and height 7 cm. The length, AB, of the prism is 20 cm. The unit vectors i, j and k are parallel to OA, OC and OE, respectively. The point N divides the length of the line DB in the ratio 2:3.

D E

k C

j

B 20 cm

O

a Find the magnitude of DB.

N 7 cm

i

24 cm

A

b Find ON.

Given that the magnitude of OP is equal to the magnitude of OQ, find the value of the constant a.

PS

7 Three points O, P and Q are such that OP = 2 i + 5 j + ak and OQ = i + (1 + a ) j − 3k.

PS

8 Relative to an origin O, the position vectors of the points P and Q are given by OP =

−6 k −2 8(1 + k )

2 k + 13 and OQ = −8 . −32 k

Given that OPQ is a straight line: a find the value of the constant k

b write each of OP and OQ in the form xi + yj + zk c

find the magnitude of the vector PQ.

Chapter 9: Vectors

PS

9 An ant has an in-built vector navigation system! It is able to go on complex journeys to find food and then return directly home (to its starting point) by the shortest route. An ant leaves home and its journey is represented by the following list of displacements: 1 −3 6 −1 i + 3j + 6k 10i 4i − k 0 −5 The ant finds food at this point. What single displacement takes the ant home? One unit of displacement is 10 cm. What is the shortest distance home from the point where the ant finds food?

EXPLORE 9.2 A system of objects with the properties: • can be represented by a column of numbers • can be combined using addition • can be multiplied by a number to give another object in the same system is called a vector space. Do quadratic expressions of the form ax 2 + bx + c , where a, b and c are real constants, form a vector space? Explain your answer.

9.3 The scalar product We have seen how to add and subtract vectors and also how to multiply a vector by a scalar. It is also possible to multiply two vectors. It is possible to find two different products by multiplying a pair of vectors. The first product gives an answer that is a vector. This is called the vector product or cross product (for example, the cross product of the vectors a and b is a × b). The cross product is not required for Pure Mathematics 3. The second product gives an answer that is a scalar. This is called the scalar product or dot product (the dot product of the vectors a and b, for example, is a ⋅ b).

Introductory ideas The angle between two vectors is defined to be the angle made ● ●

either when the direction of each vector is away from a point or when the direction of each vector is towards a point.

θ

Directions both away from a point.

θ

Directions both towards a point.

As vectors have magnitude and direction we can think of the scalar product as being multiplication in relation to a particular direction.

225

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A useful way to understand this is to think about the work done by a person when they pull an object along a flat surface using a rope: work done by the person = force needed to pull the object × distance the object moves

Direction of pull a

a θ

x

Displacement of object b

The diagram shows vector a representing the pull and vector b representing the displacement of the object. The direction of the pull is not in the same as the direction of the displacement. This means that only some of the force the person uses to pull the object makes it move. In fact, only the x-component of the pull is in the direction of the displacement. a is the magnitude of the pull. Using the angle made between the two directions, we can see that the magnitude of the x-component of the pull is x = a cos θ . b is the magnitude of the displacement, which is a measure of the distance moved. Applying this information to the formula, we can see that the work done by the person =

226

(a

cos θ

)

b.

Finding the scalar (number) that is work done by the person is one application of the scalar product. KEY POINT 9.4 In general, the geometric definition of the scalar product of two vectors is: a⋅b = a

b cos θ

This formula allows us to calculate the value that represents the impact of one vector on the other, taking account of direction. WORKED EXAMPLE 9.7 8 p = −6 −10

4 , q = −3 and the angle, θ , between the two vectors is 45°. Find the value of the scalar product, p ⋅ q. 0

Answer p ⋅ q = p q cos θ p =

82 + ( −6)2 + ( −10)2 = 10 2 and q =

42 + ( −3)2 + 02 = 5

2 = 50 p ⋅ q = ( 10 2 ) (5) cos 45° = ( 10 2 ) (5) 2

Chapter 9: Vectors

Special case 1: Parallel vectors

a

When two parallel vectors, a and b, have the same direction (not opposite directions), θ = 0° or 0 radians.

b

a ⋅ b = a b cos 0

Therefore

a⋅b = a b This also means that, for example: a ⋅ a = a a or a

2

a

Special case 2: Perpendicular vectors When two vectors, a and b, are perpendicular, θ = 90° or

π radians. 2

b

a ⋅ b = a b cos 90° a⋅b = 0 KEY POINT 9.5 This is very important. When two vectors are perpendicular, their scalar product is zero. It is also true that when the scalar product is zero, the two vectors are perpendicular.

EXPLORE 9.3 Show that the scalar product has the following properties: a ⋅b = b ⋅a a ⋅ mb = ma ⋅ b = m( a ⋅ b ), where m is a scalar a ⋅ (b + c ) = a ⋅ b + a ⋅ c (a + b ) ⋅ ( c + d ) = a ⋅ c + a ⋅ d + b ⋅ c + b ⋅ d

Scalar product for component form When we write a = a1i + a2 j + a3 k and b = b1i + b2 j + b3 k the properties of the scalar product give an algebraic method of finding the same value that we found using a b cos θ . a ⋅ b = ( a1i + a2 j + a3 k )( b1i + b2 j + b3 k ) = a1i(b1i + b2 j + b3 k ) + a2 j(b1i + b2 j + b3 k ) + a3 k( b1i + b2 j + b3 k ) = a1b1i ⋅ i + a1b2 i ⋅ j + a1b3 i ⋅ k + a2b1j ⋅ i + a2b2 j ⋅ j + a2b3 j ⋅ k + a3b1k ⋅ i + a3b2 k ⋅ j + a3b3 k ⋅ k However, as i is parallel to i, j is parallel to j and k is parallel to k then: i⋅i = j⋅ j = k ⋅k = 1 Also, as i, j and k are perpendicular to each other, scalar products such as i ⋅ j, j ⋅ k, etc. are all 0. a ⋅ b = a1b1i ⋅ i + a1b2 i ⋅ j + a1b3 i ⋅ k + a2b1j ⋅ i + a2b2 j ⋅ j + a2b3 j ⋅ k + a3b1k ⋅ i + a3b2 k ⋅ j + a3b3 k ⋅ k

227

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KEY POINT 9.6 In general, the algebraic definition of the scalar product of two vectors is: a ⋅ b = a1b1 + a2b2 + a3b3

a1 b1 When written in column form, a = a2 and b = b2 a3 b3

, so it is perfectly valid to

use the components of column vectors in the algebraic form of the scalar product a ⋅ b = a1b1 + a2b2 + a3b3. We now have a way to find the dot product when we do not know the angle between the two vectors. Sometimes the two versions of the scalar product are written together like this: a ⋅ b = a1b1 + a2b2 + a3b3 = a b cos θ This can be useful as it allows us to find the angle between the two vectors when we know the components of the vectors. WORKED EXAMPLE 9.8 228

p = 8 i − 6 j − 10 k and q = 4 i − 3 j Show that the angle, θ , between the two vectors is 45°. Answer p ⋅ q = p1q1 + p2q2 + p3q3

Find the value of the scalar product p ⋅ q.

p ⋅ q = 8(4) + ( −6)( −3) + ( −10)(0) = 32 + 18 = 50

Find the magnitude of p and of q using Pythagoras’ theorem.

p = 82 + ( −6)2 + ( −10)2 = 10 2 q =

42 + ( −3)2 + 02 = 5

50 = ( 10 2 ) (5) cos θ 50 = cos θ 50 2 1 = 45° θ = cos −1 2

Use p ⋅ q = p q cos θ . Make cos θ the subject. Solve for θ .

Chapter 9: Vectors

WORKED EXAMPLE 9.9 Relative to the origin O, the position vectors of the points A and B are given by 5 OA = −2 4

−1 and OB = 3 . −1

TIP Find angle OBA. Answer A

OB and AB are the two directions towards the point B, so these are the vectors we need to use.

A sketch will help you to make sure you choose a correct pair of directions for your scalar product.

O B

Find OB ⋅ AB.

First of all, find the column vector AB .

−6 −1 5 AB = OB − OA = 3 − −2 = 5

−1 Using the given vector OB = 3 , −1 find the value of OB ⋅ AB.

−1

4

−5

229

OB ⋅ AB = ( −1)( −6) + (3)(5) + ( −1)( −5) = 26

Find the magnitude of OB and of AB using Pythagoras’ theorem.

AB

−6 = 5 = −5

( −6)2 + 52 + ( −5)2 =

OB

−1 = 3 = −1

( −1)2 + 32 + ( −1)2 = 11

26 = 86 11 cosOBA 26 = cosOBA 86 11 Angle OBA = 32.3°

86

Use OB ⋅ AB = OB AB cos OBA

Make cosOBA the subject. Solve for angle OBA.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 9C 1 Find which of the following pairs of position vectors are perpendicular to one another. For any position vectors that are not perpendicular, find the acute angle between them. a

2

7 2 a= 8 , b = −1 −2 3

b

5 −3 c = 12 , d = −1 13 3

4 −4 e = −9 , f = −2 1 −2

c

−5 Relative to the origin O, the position vectors of the points A and B are given by OA = 0

3

1 and OB = 7 . 2

Find angle BOA. 3 Given that the vectors 6 i − 2 j + 5 k and ai + 4 j − 2 k are perpendicular, find the value of the constant a. 4 Relative to the origin O, the position vectors of the points P and Q are given by 5k OP = −3 7k + 9

k , OQ = k + 2 −1

where k is a constant.

a Find the values of k for which OP and OQ are perpendicular to one another. b Given that k = 2, find the angle, θ , between OP and OQ.

5 The diagram shows a cuboid OABCDEFG in which OC = 6 cm, OA = 4 cm and OD = 3 cm. Unit vectors i, j and k are parallel to OC , OA and OD respectively. The point M lies on FB such that FM = 1cm, N is the midpoint of OC and P is the centre of the rectangular face DEFG.

230

F

E P D

G A

Using a scalar product, find angle NPM.

k

B

j

O

i

D

C

a Find each of the vectors OM and NG.

N

9 The diagram shows a triangular prism, OABCDE . The uniform cross-section of the prism, OAE, is a rightangled triangle with base 77 cm and height 36 cm. The length, AB, of the prism is 60 cm. The unit vectors i, j and k are parallel to OA, OC and OE, respectively. The point N divides the length of the line DB in the ratio 4:1 and M is the midpoint of DE .

Use a scalar product to find angle MAN .

B

j

k

b Find the angle between the directions OM and NG.

F

E

8 The diagram shows a cube OABCDEFG with side length 4 units. The unit vectors i, j and k are parallel to OA, OC and OD, respectively. The point M is the midpoint of GF and AN : NB is 1:3.

PS

M

G

7 Explain why a ⋅ b ⋅ c has no meaning.

C

N

6 The position vector a is such that a = 4 i − 8 j + k. Find a ⋅ j and hence find the angle that a makes with the positive y-axis.

M

O

A

i D

M E

36 cm N

k C

j O

i

77 cm

B A

60 cm

Chapter 9: Vectors

PS

V

10 The diagram shows a pyramid VOABC . The base of the pyramid, OABC , is a rectangle. The unit vectors i, j and k are parallel to OA, OC and OV , respectively. The position vectors of the points A, B, C and V are given by

N

OA = 6 i, OC = 3 j and OV = 9 k. The points M and N are the midpoints of AB and BV , respectively.

k

a Find the angle between the directions of AN and OC. b

Find the vector MN.

c

The point P lies on OV and is such that PNM is a right angle. Find the position vector of the point P.

C

j

B M

O

A

i

9.4 The vector equation of a line We already know that, in 2-dimensional space, a line can be represented by an equation of the form y = mx + c or ax + by = c . In 3-dimensional space, we need a new approach because introducing the third dimension gives an equation of the form ax + by + cz = d , which is actually the Cartesian equation of a plane. One way around this problem is to write the equation of the line using vectors. The equation of a line in vector form is valid in both 2- and 3-dimensional space. So what information do we need to be able to do this? When a line passes through a fixed given point and continues in a known direction then it is uniquely positioned in space.

231

Therefore, to form the vector equation of a straight line, we need:

●

b1 a point on the line, A( a1, a2 , a3 ), and a direction vector, b = b2 , that is parallel to the line. b3

A (a1, a2, a3)

R (x, y, z)

b1 b = b2 b3

O (0, 0, 0)

R is any point on the line and so OR represents the position vector of any point on the line.

To reach R, starting at O, take the path OR = OA + AR.

OA is a vector we know. To move from A to R, we need to travel in the direction of b. Multiplying b by different scalars changes the position of R on the line, so simply gives us different points on the line. Using the parameter t to represent the variable scalar

OR = OA + tb

TIP Sometimes Greek letters such as λ and µ are used as parameters.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

KEY POINT 9.7 Using lower case notation for vectors, the vector equation of the line is:

r = a + tb ●

Or two points on the line, A( a1, a2 , a3 ) and C ( c1, c2 , c3 ) . One can be used as the fixed given point and the direction can be found from the displacement between the two points. C (c1, c2, c3)

A (a1, a2, a3)

R (x, y, z)

O (0, 0, 0)

The situation is almost the same. In this case, however, we do not know the direction of the line. We can find the direction using the position vectors of the points A and C . Using the parameter t as before,

OR = OA + tAC .

TIP Any known point can be used for the position vector, so you could also use OR = OC + tAC

KEY POINT 9.8 232

Using lower case notation for vectors, the vector equation of a line is: r = a + t( c − a )

In each case, the structure is: general point on line = position of known point + some way along a known direction

Parametric form R( x, y, z ) represents any point on the line.

x The position vector of this point can be written as r = y or r = xi + yj + zk. z In the equation r = a + tb , using the position vectors in component form: b1 x a1 y = a + t b or xi + yj + zk = a i + a j + a k + t ( b i + b j + b k ) 1 2 3 1 2 3 2 2 z a3 b 3 Simplifying the right-hand side of this: x a1 + tb1 y = a + tb 2 2 z a3 + tb3

xi + yj + zk = a1i + tb1i + a2 j + tb2 j + a3 k + tb3 k or xi + yj + zk = ( a + tb ) i + ( a + tb ) j + ( a + tb ) k 1 1 2 2 3 3

Comparing the components: x = a1 + tb1 y = a2 + tb2 z = a3 + tb3 These three equations are called the parametric equations of the line.

Chapter 9: Vectors

WORKED EXAMPLE 9.10 A line passes through the point with position vector 3i – 2 j + 4 k and is parallel to the vector i – j + k. Find: a the vector equation of the line b the parametric equations of the line. Answer Using t as the parameter: a General point = r

General point on line =

Known point = 3i – 2j + 4k

position of known point

Known direction = i – j + k

+ some way along a known direction.

3 1 r = −2 + t −1 4 1

b

3 1 r = −2 + t −1 or r = 3i – 2 j + 4 k + t ( i − j + k ) 4 1 x 3+t y = −2 − t or z 4 + t xi + yj + zk = (3 + t ) i + (–2 − t ) j + (4 + t ) k x = 3+t

y = −2 − t

Write r as a column vector and collect together the x-, y- and z-components of the right-hand side.

Compare components.

z = 4+t

WORKED EXAMPLE 9.11 The position vectors of the points A and B are given by a = 3i + 2 j − 7 k and b = i − 4 j + 8 k, respectively.

a Write down the vector AB. b Find the vector equation of the line AB. Answer a

AB = b − a AB = i − 4 j + 8 k − (3i + 2 j − 7 k ) AB = −2 i − 6 j + 15 k

Substitute for a and b. Expand and simplify.

233

lines is the angle between the two direction vectors of the Cambridge International AS & Alines. Level Mathematics: Pure Mathematics 2 & 3

b Using a = 3i + 4 j − 7 k as the known point, t as the parameter and AB from part a as the known direction:

General point = r

General point on line =

Known point = 3i + 4 j − 7 k

position of known point

Known direction = −2 i − 6 j + 15 k

+ some way along a known direction.

r = 3i + 2 j − 7 k + tAB

REWIND Recall, you could also use b = i − 4 j + 8 k as the known point.

r = 3i + 2 j − 7 k + t ( −2 i − 6 j + 15 k )

KEY POINT 9.9 The angle between two lines is the angle between the two direction vectors of the lines.

WORKED EXAMPLE 9.12 The vector equation of the line L1 is r = 3i + j − k + λ ( i − 8 j + 5 k ). a The line L2 is parallel to the line L1 and passes through the point A(0, 2, 5) . Write down the vector equation of the line L2 . 234

b Find the angle between the line L1 and the line r = 4 i + 3 j − 2 k + µ (5 i − j + 2 k ) . Answer a Using µ as the parameter, General point is r

General point on line =

Known point is 2 j + 5 k

position of known point

The direction of the line L1 is i − 8 j + 5 k

+ some way along a known direction point.

The line L2 is therefore r = 2 j + 5k + µ ( i − 8 j + 5k ) b Find the angle between d1 = i − 8 j + 5 k and d 2 = 5 i − j + 2 k. d1 ⋅ d 2 = (1)(5) + ( −8)( −1) + 5(2) = 23

Find the scalar product d1 ⋅ d 2 Now find the magnitude of d1 and d 2 using Pythagoras’ theorem

d1 = 12 + ( −8)2 + (5)2 = 3 10 d 2 = 52 + ( −1)2 + 22 =

30

23 = 3 10 30 cos θ cos θ =

23 3 10 30

23 cos −1 = 63.727 … ° = 63.7° 3 10 30

Use d1 ⋅ d 2 = d1 d 2 cos θ . Make cos θ the subject. Solve for θ .

Chapter 9: Vectors

EXERCISE 9D 1 Write down the vector equation for the line through the point A that is parallel to vector b when: a

A(0, −1, 5) and b = 2i + 6j – k

b A( 0, 0, 0 ) and b = 7i – j – k c

A(7, 2, −3) and b = 3i – 4k.

2 Write each of your answers to question 1 in parametric form. 3 Show that the line through the points with position vectors 9i + 2j – 5k and i + 7j + k is parallel to the line r = λ (16 i − 10 j − 12 k ). 4 Find the parametric equations of each of the lines. a

r = 2 i + 13 j + k + t ( i + j − k )

b

r = 10 j + t (2 i + 5 j )

c

r = i − 3 j + t (2 i + 3 j + 4 k )

5 a Find the parametric equations of the line L through the points A(0, 4, −2) and B (1, 1, 6). b Hence find the coordinates of the point where the line L crosses the Oxy plane. PS

6 The line L1 has vector equation r = 2 i – 3 j + k + λ (6 i + j + 3k ) and the line L2 has parametric equations x = µ + 4, y = µ − 7, z = 3µ . 235

a Show that L1 and L2 are not parallel. b Find the angle between the line L1 and the line L2 . PS

7 The vector equation of the line L1 is given by x −3 4 y = 0 + t −1 . z 8 −3 a Find the vector equation of the line L2 that is parallel to L1 and which passes through the point A(5, −3, 2) . b Show that A(5, −3, 2) is the foot of the perpendicular from the point B (4, −7, 2) to the line L2 .

PS

8 The points A, B and C are position vectors relative to the origin O, given by 1 3 0 OA = 1 , OB = −1 and OC = 2 . 5

2

3

a Find a vector equation for the line L passing through A and B. b The line through C , perpendicular to L, meets L at the point N. Find the exact coordinates of N .

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

9.5 Intersection of two lines In 2-dimensional space, two lines are either parallel or they must intersect at a point. In 3-dimensional space it might be that: ● ● ●

they are parallel or they are not parallel and intersect or they are not parallel and do not intersect.

Lines that are not parallel and do not intersect are called skew. Parallel lines have the same direction vectors and so if a pair of lines is parallel, this can easily be seen by examining the vector equations of the lines.

Non-parallel intersecting lines In 2-dimensional space, to find the point of intersection of a pair of non-parallel lines using their Cartesian equations ( ax + by = c ), we solve the equations of the lines simultaneously. In other words, we put the two equations equal to each other. The process is the same in 3-dimensional space and using the vector equations of the lines. WORKED EXAMPLE 9.13 Given that the lines with vector equations r = i − 2 j − 4k + λ (−i − j + k ) 236

and r = 2 i − 2 j + k + µ ( i + 2 j − 7 k ) intersect at the point A, find the coordinates of A. Answer The lines are equal at the point of intersection, so write the vector equation in parametric form so that the components can be equated. r x y z

= i − 2 j − 4k + λ (−i − j + k ) = 1− λ = −2 − λ = −4 + λ

r x y z

= 2i − 2 j + k + µ( i + 2 j − 7k) = 2+µ = −2 + 2µ = 1 − 7µ

Now equate the components. 1− λ = 2 + µ

(1)

−2 − λ = −2 + 2 µ

(2)

−4 + λ = 1 − 7 µ (3) Solve these equations to find the value of λ and of µ: From (1): Substitute into (3):

λ = −1 − µ −4 − 1 − µ = 1 − 7 µ µ =1 6µ = 6

Chapter 9: Vectors

When µ = 1: λ =

µ = 1: λ = −1 − 1 = −2

Check:

λ = = −2 x = 1 − ( −2) = 3 y = −2 − ( −2) = 0 z = −4 − 2 = −6

µ =1

TIP

x = 2 +1 = 3 y = −2 + 2 = 0 z = 1 − 7 = −6

It is good to check that the value of λ and of µ that we have found give us the same point.

The lines intersect at the point A(3, 0, −6).

Skew lines Skew lines have no point of intersection. When two lines are skew, they are not parallel and they do not intersect. WORKED EXAMPLE 9.14 Show that the lines with vector equations r = 4 i + 7 j + 5 k + λ ( −2 i + j + 3k ) r = i + j + 10 k + µ ( −6 i + j + 4 k ) 237

do not intersect. Answer As −2 i + j + 3k is not a scalar multiple of −6 i + j + 4 k the lines are not parallel. Writing the equations in parametric form: r x y z

= 4 i + 7 j + 5 k + λ ( −2 i + j + 3k ) = 4 − 2λ = 7+λ = 5 + 3λ

r x y z

= i + j + 10 k + µ ( −6 i + j + 4 k ) = 1 − 6µ = 1+ µ = 10 + 4µ

Now equate the components. 4 − 2 λ = 1 − 6µ 7 + λ = 1+ µ 5 + 3λ = 10 + 4µ

(1) (2) (3)

Attempt to solve these equations to find the value of λ and of µ that should give a consistent point: From (2): Substitute into (3): When µ = −23:

λ = µ − 6 5 + 3( µ − 6) = 10 + 4µ µ = −23 λ = −23 − 6 = −29

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Check:

λ = = −29

µ = = −23

x = 4 − 2( −29) = 62 y = 7 − 29 = −22

x = 1 − 6( −23) = 139

z = 5 + 3( −29) = −82

z = 10 + 4( −23) = −82

y = 1 + ( −23) = −22

The values of λ and µ do not give a consistent point, so there is no point of intersection and the lines are skew.

TIP

EXERCISE 9E 1 Find whether the following pairs of lines are parallel, intersecting or skew. State the position vector of the common point of any lines that intersect. a

238

PS

x = 3λ + 4

x = µ +1

y = 5λ + 1

y= µ+4

z = 4λ − 3

z = 2µ + 5

b

r = i − 5 j + 4 k + λ (3i − 4 j + 9 k )

r = i − 5 j + 4 k + λ ( −6 i + 8 j − 18 k )

c

r = 7 i + 2 j + 6 k + λ (i + 3 j + 9 k )

r = 4 j + k + µ (8 i + j + 14 k )

d

r = 2 i − 3j + 4k + λ (6i + 4 j −2 k)

r = 4i − 9 j + 2 k + µ (−i − 8 j − k )

2 Given that the lines with vector equations r = 2 i + 9 j + k + λ ( i – 4 j + 5k ) and r = 11i + 9 j + pk + µ ( − i − 2 j + 16 k ) intersect at the point P, find the value of p and the position vector of the point P. 3 Three lines, L1, L2 and L3 , have vector equations r = 16 i − 4 j − 6 k + λ ( −12 i + 4 j + 3k ) r = 16 i + 28 j + 15 k + µ (8 i + 8 j + 5 k ) r = i + 9 j + 3k + ν (4 i − 12 j − 8 k ) The 3 points of intersection of these lines form an acute-angled triangle. For this triangle, find: a the position vector of each of the three vertices b the size of each of the interior angles c

the length of each side.

4 The line L1 passes through the points (3, 7, 9) and ( −1, 3, 4) . a Find a vector equation of the line L1. b The line L2 has vector equation r = i + 2 j + k + µ (3 j + 2 k ). Show that L1 and L2 do not intersect.

• Always carry out the check with both λ and µ as it is easy to make a mistake. • Re-read the question to check whether you were asked to find the position vector or the coordinates of the point of intersection and give your answer in the correct form.

Chapter 9: Vectors

5 The point A has coordinates (1, 0, 5) and the point B has coordinates ( −1, 2, 9). a Find the vector AB. b Write down a vector equation of the line AB. c

Find the acute angle between the line AB and the line L with vector equation r = i + 3 j + 4 k + µ ( − i − 2 j + 3k ) .

d Find the point of intersection of the line AB and the line L.

Checklist of learning and understanding

●

For position vectors OA = a and OB = b, AB = b − a.

●

The magnitude (length or size) is worked out using Pythagoras’ theorem: When a = a1i + a2 j + a3 k , a =

a12 + a2 2 + a32 .

●

Two vectors are parallel if one is a scalar multiple of the other.

●

Problem solving with angles will involve the use of the scalar product. When a = a1i + a2 j + a3 k and b = b1 i + b2 j + b3 k a ⋅ b = a1b1 + a2 b2 + a3b3 = a

b cos θ 239

●

Vector equation of a line: r = a + tb

●

To find the point of intersection of two intersecting lines, solve the parametric equations of the lines simultaneously.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 9 1

Relative to the origin O, the position vectors of points A, B and C are given by 3m 2 1 OA = 3 , OB = and OC = 1

7

2

4 −m − m( m + 1)

.

a

It is given that AB = OC . Find the angle OAB.

[4]

b

Find the vector equation of the line AC.

[3]

With respect to the origin O, the position vectors of the points A and B are given by −2 OA = 0

a b

1 and OB = −1 .

6

4

Find whether or not the vectors OA and OB are perpendicular. i

[2]

Write down the vector AB.

[1]

ii Find a vector equation of the line AB. c

[2]

The vector equation of the line L is given by r = 7 i − 12 j + 7 k + µ ( −3i + 10 j − 5k ).

240

Show that the lines AB and L intersect and find the position vector of the point of intersection. 3

H E N

G

F

k C

M

j D

[5]

O

B

A

i

The diagram shows a prism, ABCDEFGH, with a parallelogram-shaped uniform cross-section.

The point E is such that OE is the height of the parallelogram. The point M is such that OM is parallel to DC and N is the midpoint of DE . The side OD has a length of 5 units. The unit vectors i, j and k are parallel to OA, OM and OE, respectively.

The position vectors of the points A, E , C and M are given by OA = 9 i, OM = 15 j and OE = 12 k.

a

Express the vectors AH and NH in terms of i, j and k.

[2]

b

Use a vector method to find angle AHN .

[5]

c

Write down a vector equation of the line AH.

[2]

Chapter 9: Vectors

4

The position vectors of A, B and C relative to an origin O are given by 8 2 6 OA = 3 , OB = n and OC = 9 −1 0 2

where n is a constant.

5

a

Find the value of n for which AB = CB .

[4]

b

In this case, use a scalar product to find angle ABC .

[3]

a

Given the vectors 8 i − 2 j + 5 k and i + 2 j + pk are perpendicular, find the value of the constant p.

[3]

b

The line L1 passes through the point ( −3, 1, 5) and is parallel to the vector 7 i − j − k. i Write down a vector equation of the line L1.

[2]

ii The line L2 has vector equation r = i − 2 j + 2 k + µ ( i + 8 j − 3k ). Show that L1 and L2 do not intersect. 6

[4]

The origin O and the points A, B and C are such that OABC is a rectangle. With respect to O, the position vectors of the points A and B are −4 i + pj − 6 k and −10 i − 2 j − 10 k. a

Find the value of the positive constant p.

[3]

b

Find a vector equation of the line AC.

[3]

c

Show that the line AC and the line, L, with vector equation r = 3i + 7 j + k + µ ( −4 i − 4 j − 3k )

d 7

intersect and find the position vector of the point of intersection.

[5]

Find the acute angle between the lines AC and L.

[4]

Relative to the origin O, the points A, B, C and D have position vectors

OA = 4 i + 2 j − k, OB = 2 i − 2 j + 5 k, OC = 2 j + 7 k, OD = −6 i + 22 j + 9 k.

8

a

Use a scalar product to show that angle ABC is a right angle.

[3]

b

Show that AD = kBC , where k is a constant, and explain what this means.

[2]

c

The point E is the midpoint of the line AD. Find a vector equation of the line EC .

[4]

With respect to the origin O, the points A, B, C , D have position vectors given by

OA = −3i + j + 8 k, OB = −10 i + 2 j + 15 k, OC = 5 i − 2 j − 2 k, OD = i + 6 j + 10 k. a

Calculate the acute angle between the lines AB and CD.

[4]

b

Prove that the lines AB and CD intersect.

[5]

c

The point E has position vector 5 i + 3 j + 4 k. Show that the perpendicular distance from E to the line CD is equal to 5 .

[5]

241

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

9

PQRS is a rhombus. The coordinates of the vertices P, Q and S are (9, 2, 4), ( −0.5, 6, 6.5) and (4.5, −4, −3.5) , respectively.

a

Find the vectors PQ and PS.

[2]

b

Find the coordinates of R.

[3]

c

Show that PQRS is in fact a square and find the length of the side of the square.

[3]

d

The point T is the centre of the square. Find the coordinates of T .

[3]

e

The point V has coordinates (5, 17.5, −13.5). i Find a vector equation of the line VT .

[3]

ii Verify that T is the foot of the perpendicular from V to PR.

[2]

iii Describe the solid VPQRS .

[1]

10 The points P and Q have coordinates (0, 19, −1) and ( −6, 26, −11) , respectively. The line L has vector equation r = 3i + 9 j + 2 k + λ (3i − 10 j + 3k ).

242

a

Show that the point P lies on the line L.

[1]

b

Find the magnitude of PQ.

[2]

c

Find the obtuse angle between PQ and L.

[3]

d

Calculate the perpendicular distance from Q to the line L.

[5]

11 The points A and B have coordinates (7, 1, 6) and (10, 5, 1), respectively. a

Write down a vector equation of the line AB.

b

The point P lies on the line AB. The point Q has coordinates (0, −5, 7).

c

[3]

Given that PQ is perpendicular to AB, find a vector equation of the line PQ.

[6]

Find the shortest distance from Q to the line AB.

[1]

12 The line L1 has vector equation r = 3i + 2 j + 5 k + λ (4 i + 2 j + 3k). The points A(3, p, 5) and B ( q, 0, 2), where p and q are constants, lie on the line L1. a

Find the value of p and the value of q.

[2]

The line L2 has vector equation r = 3 j + k + µ (7 i + j + 7 k ). b

Show that L1 and L2 intersect and find the position vector of the point of intersection.

[5]

c

Find the acute angle between L1 and L2 .

[4]

243

Chapter 10 Differential equations P3 This chapter is for Pure Mathematics 3 students only.

In this chapter you will learn how to:

■ ■ ■ ■

formulate a simple statement involving a rate of change as a differential equation find, by integration, a general form of solution for a first order differential equation in which the variables are separable use an initial condition to find a particular solution interpret the solution of a differential equation in the context of a problem being modelled by the equation.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE / O Level Mathematics

Express direct and inverse variation in algebraic terms and use this form of expression to find unknown quantities.

1 The weight, w newtons, of a sphere is directly proportional to the cube of its radius, r cm. When the radius of the sphere is 60 cm, its weight is 4320 newtons. Find an equation connecting w and r.

IGCSE / O Level Mathematics

Understand and apply ● laws of indices ● laws of logarithms

Chapter 5

Integrate ● standard functions ● using substitution ● using parts

2

x + ln 3 4 Find and simplify an expression for y in terms of x. ln y − ln(20 − y ) =

3 Find: a b c

Chapter 8

Decompose a compound fraction with denominator ● ( + b )( )( cx + d )( )( ex + f ) ● ( + b )( )( cx + d )2 ● ( + b )( )( cx 2 + d ) into its partial fractions and integrate the result.

Chapter 1

Sketch graphs of standard functions e.g. involving ln f( x ) and e f( x ) (including understanding end behaviour as x → ± ∞).

244

1

∫ 3 − 1 dx sin x dx ∫ cos x ∫ xe dx. −

4 Find

2

dx. ∫ (x − 2)( 2 )(3 )( 3x + 1)

2

5 a Sketch the graph of y = e x . b

12e x → k as x → ∞. 3 + ex Find the value of the constant k.

What is a differential equation? dy d2 y or as one of its terms is called a dx dx 2 dy differential equation. When the equation has the first derivative as one of its terms, dx dy it is called a fi rst order differential equation. For example, the equation = 3x 2 is an dx alternative way of representing the relationship y = x 3 + C , where C is a constant. An equation that has a derivative such as

To solve a differential equation, we integrate to find the direct relationship between x and y, such as y = x 3 + C . The direct equation y = x 3 + C is called the general solution dy of the differential equation = 3x 2 . To find a particular solution, we must have further dx information that will enable us to find the value of C . The solution of an algebraic equation is a number or set of numbers. The solution of a differential equation is a function.

Chapter 10: Differential equations

In this chapter, we will learn the initial skills needed to master the solution of one type of first order differential equations. These simple skills will be the foundation for further study of this subject. You might wonder about the usefulness of such seemingly abstract equations in the real world. In fact, differential equations are excellent mathematical models for many real-life situations and are essential to solve many problems. First order differential equations can model rates of change, in other words, the rate at which one quantity is increasing or decreasing with respect to another quantity. There are many examples in the physical world that can be modelled using such relationships. A few examples are the rate of increase of the number of bacteria in a culture, simple kinematics problems, the rate of decrease of the value of a car and the rate at which a hot liquid cools.

10.1 The technique of separating the variables dy Solving a differential equation such as = 6( x + 1)2 can be done using an dx anti-differentiation approach. We know, from the Pure Mathematics 1 Coursebook, Chapter 9, that if we integrate both sides with respect to x, we have: d dx = 6( x + 1)2 dx

∫

∫

6( x + 1)3 +C 3 y = 2( x + 1)3 + C y=

245

Notice that the first line on the left-hand side is

∫

d

dx =

∫ 6(x + 1) dx and from 2

dy anti-differentiation we can see that, if we integrate with respect to x we obtain an dx expression for y . This means that

dy

∫ dx d

= ... . is equivalent to

∫ dy = ..., which is y = …

This result is generally true. KEY POINT 10.1 dy Integrating an expression of the form f( y ) with respect to x is equivalent to integrating f( y ) dx with respect to y. dy Therefore f( y ) dx is equivalent to f( y ) d y. y dx

∫

∫

dy dy = 6( x + 1)2 is easy to solve because is equal to a dx dx function that we know how to integrate. This is not always the case.

The differential equation

dy = xy when y . 0 ? dx As it is written, we cannot integrate both sides with respect to x and find a solution. How do we solve the differential equation

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Separating the variables in the expression xy means that we are then able to integrate. dy = xy dx 1 dy =x y dx

Separate the variables x and y. Form integrals of both sides with respect to x.

dy dx ∫ x dx ∫ 1y ddy x ∫ y dy ∫ x dx ln y + A =

x2 +B 2

ln y =

x2 +C 2

x2

y=e2

The modulus of y is not required as we know that y . 0. The constants of integration can be combined as one term and we only need to write a constant on one side.

+C x

2

y = De 2

246

Integrate each side.

where D = eC

This is the general solution of the differential equation as we do not know the value of the arbitrary constant C .

When we have further information, we can find the particular solution of the differential solution given by that information. dy = xy given that y = 2 when x = 0 . For example, solve dx Using this information: ln y = y=

x2 02 + C and so ln 2 = + C , C = ln 2 2 2 x2 + ln 2 e2

This is the particular solution of the differential equation.

x2

y = e 2 × e ln 2

Note that you often need to write your answer as an expression for y in terms of x. This commonly means using laws of indices or logarithms.

x2

y = 2e 2

By separating the variables we are rearranging the differential equation to the form dyy f( y ) = g( x ). dx

f( y )

∫ f(

d

dy = g( x ) dx dx

∫ f( y) dy

∫ gg((x ) dx ∫ g(x ) dx

(Note: Other variables may be used such as v and t.)

Chapter 10: Differential equations

So, in the case where the derivative is

dy : dx

●

form an expression in x only on one side of the equation

●

form an expression in y ×

dyy on the other side of the equation. dx

This gives us a form of the equation that can be integrated on both sides, one side with respect to x and the other side with respect to y. WORKED EXAMPLE 10.1

The variables x and y satisfy the differential equation

dy 1 + x = . dx 2y

It is given that y = 3 when x = 0. Solve the differential equation and obtain an expression for y in terms of x. Answer Even though the question does not ask for it directly, you are expected to use the initial condition that you have been given to find a particular solution. dy 1 + x = dx 2y

2y

∫

Separate the variables x andy. 2 . It was not Here we have moved the 2y essential to move the 2, we could have moved the y only. This would be just as valid as it would also give expressions that can be integrated.

dy = 1+ x dx d

Form integrals of both sides with respect to x.

∫ ( 1 + x ) dx ∫ 2 y dy = ∫ ( 1 + x ) dx dx =

x2 +C 2 02 9= 0+ + C, 2 x2 y2 = x + +9 2 y2 = x +

Integrate each side. y = 3 when x = 0 .

C =9

1

2 x2 y= x+ +9 2

You can check this by differentiating.

247

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 10.2 Find the general solution of the differential equation ( x 2 + 11) y3 + x 2

dy = 0. dx

Answer ( x 2 + 11)) y3 + x 2

dy =0 dx

Separate the variables x and y.

dy = −( x 2 11) y3 dx 1 dy ( 2 11) =− ** 3 y dx x2

x2

∫ ∫y −

−3

d

−3 −

dy = =

dx = =

∫ (−−x

∫ (−−x

−2 −

Form integrals of both sides with respect to x. Here we are assuming that y ≠ 0 .

−2

− 1) dx

− 1) dx

Integrate each side.

1 1 = − x +C 2 y2 x

Is this the complete solution?

248

Earlier in the solution (at line **) we assumed that y ≠ 0. However, when y = 0, the original differential equation dyy 0 becomes = 2 = 0 . Since y = 0 is the x-axis, it is true that its gradient is 0 at all points and so y = 0 is also dx x a solution of the differential equation and must be included for the general solution to be complete. 1 1 The complete general solution is, therefore, − 2 = − x + C or y = 0. x 2y dy In Worked example 10.2, we saw that, when rearranging to the form f( y ) = g( x ) , there dx 1 are cases where f( y ) is a function such as, for example, 3 . In these cases, a value (or y values) of y must be excluded for the separation of the variables to take place. As a final step, when finding the general solution in these cases, we must check if the value (or values) we have excluded when separating the variables gives (or give) a valid solution of the original differential equation. When they do, we should include them. WORKED EXAMPLE 10.3 dx x((300 300 − x ) = and x = 50 when t = 0. Find the particular solution of the dt 600 differential equation, giving your answer as an expression for x in terms of t.

It is given that, for 0 , x , 300 ,

Answer dx x((300 300 − x ) = dt 600

Separate the variables. The derivative is with respect to t and so the expression in x needs to be moved to the left-hand side.

Chapter 10: Differential equations

∫

1 dx 1 = x((300 300 − x ) dt 600 1 d 1 t x( x) 1 1 x x( x)

Form integrals of both sides with respect to t. Note that it is simpler to leave the 600 on the right-hand side here.

∫ ∫ dt ∫ dt

dt

∫

Write the expression on the left in terms of its partial fractions.

1 1 1 1 x 300 x x 1 1 t +C ( ln x − ln( − x ) = 600 300 1 ln x − ln(300 ln( − x) = t + C 2 x 1 ln = t +C 300 − x 2 50 C = ln , C = ln l 0.2 300 − 50 x 1 ln = t + ln 0.2 300 − x 2

∫

Integrate each side. The modulus signs are not required as we know that 0 , x , 300 .

x = 50 when t = 0 .

1

t + ln 0.2 x = e2 300 − x 1

t x = e 2 × e ln 0.2 300 − x 1

t

x = 0. 0.2e 2e 2 ((300 300 − x ) 1

1

t

x + 0. 0.2e 2e 2 x = 60 60e 2 1

t

1

x(1 + 0.2e 2 ) = 60e 2

t t 1

x=

60e 2

t 1

1 + 0.2e 0 2

t

EXERCISE 10A 1 Find the general solution of each of the following differential equations. dy ds dA = 2 πr a = x3 − 6 b c s2 = cos(t + 5) dx dt dr dyy dV dy y +1 =1 e f x + sec x sin y =2 d V −1 = d x dt dx x 2 Find the particular solution for each of the following differential equations, using the values given. a

cos2 x

b

xy

dy = y given y = 5 when x = 0 dx

dy x3 − x = given y = 4 when x = 3 d 1− y

dy = e x − y and y = ln 2 when x = 0 , obtain an expression for y in 3 Given that dx terms of x.

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Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

4 The gradient of a curve at the point ( x, y ) is 5 y3x. x a Write down the differential equation satisfied by y. b Given that the curve passes through the point (1, 1), find y2 as a function of x. 5 a Express

1 A B in the form + . (2 − x )( x + 1) 2 − x x +1

TIP

b The variables x and t satisfy the differential equation dx = (2 − = − x)(x dt

1).

It is given that t = 0 when x = 0.

PS

i

Solve this differential equation and obtain an expression for x in terms of t.

ii

State what happens to the value of x when t becomes large.

6 The variables w and t satisfy the differential equation dw = 0.001(100 − w )2 . dt Given that when t = 0 w = 0 , show that the solution to the differential equation 1000 can be written in the form w = 100 − . t + 10 dx 7 Solve the differential equation 4 = (3x 2 − x ) cos t given that x = 0.4 when dt t = 0. Give your answer as an expression for x in terms of t.

250

8 Find the general solution of the differential equation xy

dy x 2 − 1 = y +1 . dx e

dy 1 + y2 = dx 4 + x 2 and passes through the point (0, 1). Give your answer in the form y = f( x ).

9 Find the equation of the curve that satisfies the equation

dy + 2 y = 1. 1 dx

10 Find the general solution of the equation x 3

dx = 4x cos2 t given that x = 1 when t = 0. dt Give your answer as an expression for x in terms of t.

11 Solve the differential equation

12 a Use the substitution u = t 2 to find b Solve the differential equation PS

∫ te

t2

dt . 2

dx t ( e t + 55) = given that x = 1 when t = 0. dt x2

13 At a point with coordinates ( x, y ) the gradient of a particular curve is directly proportional to xy2. At a particular point P with coordinates (1, 3) it is known that the gradient of this curve is 6. Find and simplify an expression for y in terms of x. 14 The variables x and t satisfy the differential equation dx (3x + 2 x 3 ) = k (3x 2 + x 4 ) dt for x . 0 where k is a constant. When t = 0 , x = 1 and when t = 0.5, x = 2 . Solve the differential equation, finding the exact value of k , and show that 72t =

3x 2 + x 4 . 4

xɺ is an alternative dx . notation for dt

Chapter 10: Differential equations

15 a Use the substitution u

ln x to find

1

∫ x lnx dx.

b Given that x . 0, y . 0 , find the general solution of the differential equation dy y ln x = . dx x M

16 A type of medicine that reacts to strong light can be used to treat some skin problems. It is important that, after treatment with this type of medicine, the person stays out of bright light until the amount of medicine remaining in the skin is at a safe level. Jonty is treated with a medicine of this type. After the treatment, 65 mg remains in the treated area of skin. The rate at which the medicine disappears from the treated area of Jonty’s skin dm = −0.05 m, where m is the amount of is modelled by the differential equation dt medicine, in milligrams, in the skin t hours after the treatment. Jonty is safe to go out in bright light when no more than 6 mg of the medicine remains in his skin.

TIP Notice that the rate at which the medicine disappears is with respect to time but the words ‘with respect to time’ are usually not included.

How long must Jonty wait before he goes out in bright light?

EXPLORE 10.1

251

Newton’s law of cooling or heating Many differential equations arise from modelling natural laws and relationships. One example is Newton’s law of cooling or heating: The rate of change of temperature of an object is proportional to the difference in temperature of the object and its surroundings. We define

T as the temperature of the object t

as time

T0 as the temperature of the object when t = 0 s

as the constant temperature of the surroundings

k

as the constant of proportionality.

Discuss the following. ● ● ●

●

Why it is necessary to define the variables and constants. dT Why the differential equation = k (T − ) , k . 0 is satisfied by T . dt Other possible correct ways to represent Newton’s law of cooling or heating as a differential equation with this set of defined variables and constants. Why the general solution of the given form of the differential equation may be written as: T − s = Ae kt where A

eC .

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

●

● ●

How many arbitrary constants there are in the general solution of any first order differential equation. Why the modulus in the general solution is required. How the general solution can be used to represent the temperature of an object cooling down or warming up.

10.2 Forming a differential equation from a problem The key to forming a differential equation from a problem is the successful interpretation of the language used in the question. For example, when you are given a problem such as: The rate of increase of the number of bacteria in a dish is k times the number of bacteria in the dish. The number of bacteria present in the dish at time t is x… Replacing the words the number of bacteria in a dish by x the problem becomes: The rate of increase of x is k times x.

252

This rate of increase of x is with respect to time, t, and is much easier to translate into a dx differential equation: = kx . dt WORKED EXAMPLE 10.4 Velocity is defined to be the rate of change of displacement with respect to time. The velocity of a particle at time t seconds is inversely proportional to its displacement, s metres, from an origin, O. a Form a differential equation to represent this data. b Initially, the particle has a displacement of 2 metres from O. After 1 second, the particle has a displacement of 5 metres from O. Solve the differential equation giving your answer in the form t = f( s ) . c Find t when the displacement of the particle from O is 10 metres. Answer a The velocity of a particle at time t seconds is inversely proportional to its displacement, s metres, from the origin, O. v=

ds ds 1 ds k and so ∝ , which means that = dt dt s dt s

At the moment, you do not have enough information to find the constant of proportionality, k

Chapter 10: Differential equations

b

ds k = dt s ds s =k dt

Separate the variables s and t. Form integrals of both sides with respect to t.

ds

∫ s t dt ∫ k dt ∫ s ds ∫ k dt

Integrate each side.

s2 = kt + C 2 s2 C = 2, = kt + 2 2 52 = k + 2, k = 10.5 2 s2 21 = t+2 2 2 s2 − 4 t= 21

s = 2 when t = 0 . s = 5 when t = 1.

Write in the form t = f ( s ) .

c Find t when s = 10 t=

100 − 4 = 4.5714... = 4.57 seconds correct to 3 significant figures. 21

WORKED EXAMPLE 10.5 The population density, P, of a town can be modelled as a continuous variable. In the year when records began: •

the population density, P, of the town was increasing at the rate of 0.5 people per hectare per year

•

the population density of the town was 2 people per hectare.

It is suggested that, t years after records began, the rate of increase of the population density is inversely proportional to the square of the population density. a Form a differential equation to model this information. b Solve the differential equation, giving your answer in the form P = f(t ) . c The table shows the actual population density of the town, every 5 years for the first 20 years after records began. t

5

10

15

20

P

3.5

4

5.5

7

Comment on how good a fit the model is to the actual data in the longer term.

253

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Answer a The rate of increase of P is inversely proportional to P 2 : dP 1 ∝ 2 dt P When t = 0 , dP k = 2 dt P dP 2 = 2 dt P b

dP k = 2 dt P dP = 0.5 and P = 2 , therefore: dt 0.5 =

k ,k =2 4

dP 2 = 2 dt P P2

Separate the variables P and t.

dP =2 dt

Form integrals of both sides with respect to t.

dP

∫ P t dt ∫ 2 dt ∫ P dP 2 ∫ dt 254

P3 3 8 3 P3 3 P3

Integrate each side. t = 0, P = 2

= 2t + C =C 8 3 = 6t + 8

Write in the form P = f(t ) .

= 2t +

P=

3

6t + 8

c c.................................................. t

5

10

15

20

P

3.5

4

5.5

7

Model prediction (to 2 significant figures)

3.4

4.1

4.6

5.0

In the longer term, the model is not a particularly good fit as the values it generates when time increases are under-estimates of the known data.

Chapter 10: Differential equations

WORKED EXAMPLE 10.6 The value, $V , of a motorbike as it gets older can be modelled as a continuous variable. A model is suggested to estimate the rate of decrease of the value of the motorbike. This model is that the value V when the motorbike is t months old decreases at a rate proportional to V . a Write down a differential equation that is satisfied by V . b The motorbike had a value of $10 000 when it was new. Solve the differential equation, showing that V = 10 000e− kt , where k is a positive constant. c After 3 years the motorbike has lost half of its value when new. Find the value of the motorbike when it was 2 years old. Answer a The rate of decrease of V is proportional to V . dV dV ∝ V and so = − kV , where k is a positive constant. dt dt b

∫

dV = − kV dt 1 dV = −k V dt 1 d dt = − k dt

∫ 1 ∫ dV = ∫ − k dt ln V = − kt kt + C

This describes exponential decay. The rate is decreasing and so the constant of proportionality is negative.

Separate the variables V and t. Form integrals of both sides with respect to t.

Integrate each side. t = 0 , V = 10 000

ln 10 000 = C

For this particular solution, V is positive and so the modulus is not required.

ln V = −k kt + ln 10 000

Write in the form V = f(t ) .

V = e − kt + ln 10 000 V = = e − kt × × e ln 10 000 V = 10 000e − kt c V = 10 000e− kt

t = 3, V = 5000

5000 = 10 000e 5000 = e −3 k 10 000

−3k

ln 0.55 = −3k k=

− ln 0.5 3

V = 10 000e

−2

− ln 0.5 3

V = $6299.605 … . = $6300 correct to 3 significant figures

255

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 10B M

1 Write the following as differential equations. Do not solve the equations. a For height h and time t: The rate of decrease of the height of water in a tank is proportional to the square of the height. b For number of infected cells n and time t: The rate of increase of the number of infected cells in a body is proportional to the number of infected cells in the body at that time. c

For velocity v and time t: The rate of decrease of the velocity of a particle moving in a fluid is proportional to the product of v and v + 1.

d For volume V and time t: The rate at which oil is leaking from an engine is proportional to the volume of oil in the engine at that time. e For cost C and time t: The rate at which the cost of an item increases is proportional to the cube of the cost at that time. f 256

For speed v and time t: The rate of change of the speed of a skydiver just after he jumps out of a t

plane is inversely proportional to e 2 . M

2 A large tank is full of water. The tank starts to leak. t seconds after the tank starts to leak the depth of water in the tank is x metres. The rate of decrease of the depth of water is proportional to the square root of the depth of water. a Write down a differential equation relating x, t and a constant of proportionality. b At the instant when water starts to leak from the tank, the depth of water is 6.25 metres. Given that 120 seconds after the tank starts to leak, the depth of water is 4 metres, solve the differential equation, expressing t in terms of x. c

M

Find the time it takes for the depth of the water to halve.

3 The cost of designing an aircraft, $A per kilogram, at time t years after 1950 can be modelled as a continuous variable. The rate of increase of A is directly proportional to A. a Write down a differential equation that is satisfied by A. b In 1950, the cost of designing an aircraft was $12 per kilogram. Show that A = 12e kt , where k is a positive constant. c

After 25 years, the cost of designing an aircraft was $48 per kilogram. Find the cost of designing an aircraft after 60 years.

Chapter 10: Differential equations

M

4 A liquid is heated so that its temperature is x °C after t seconds. It is given that the rate of increase of x is proportional to (100 − x ). The initial temperature of the liquid is 25 °C. a Form a differential equation relating x, t and a constant of proportionality, k, to model this information. b Solve the differential equation and obtain an expression for x in terms of t and k. c

After 180 seconds the temperature of the liquid is 85 °C. Find the temperature of the liquid after 195 seconds.

d The model predicts that x cannot exceed a certain temperature. Write down this maximum temperature. M

5 The diagram shows an inverted cone filled with liquid paint.

10 cm

An artist cuts a small hole in the bottom of the cone and the liquid paint drips out at a rate of 16 cm 3 per second. At time t seconds after the hole is cut, the paint in the cone is an inverted cone of depth h cm. dV 4 = πh2 . a Show that dh 9 dh . b Hence find an expression for dt c Solve the differential equation in part b, giving t in terms of h.

15 cm

d Find the length of time it takes for the depth of the paint to fall to 7.5 cm. M

6 The size of a population, P, at time t minutes is to be modelled as a continuous variable such that the rate of increase of P is directly proportional to P.

257

a Write down a differential equation that is satisfied by P. b The initial size of the population is P0 . Show that P = P0 e kt , where k is a positive constant. c

M

The size of the population is 1.5P0 after 2 minutes. Find when the population will be 3P0 .

7 A ball in the shape of a sphere is being filled with air. After t seconds, the radius of the ball is r cm. The rate of increase of the radius is inversely proportional to the square root of its radius. It is known that when t = 4 the radius is increasing at the rate of 1.4 cm s−1 and r = 7.84 . a Form a differential equation relating r and t to model this information. b Solve the differential equation and obtain an expression for r in terms of t. c

M

How much air was in the ball at the start?

8 Maria makes a large dish of curry on Monday, ready for her family to eat on Tuesday. She needs to put the dish of curry in the refrigerator but must let it cool to room temperature, 18 °C, first. The temperature of the curry is 94 °C when it has finished cooking. At 6 pm, Maria places the hot dish in a sink full of cold water and keeps the water at a constant temperature of 7 °C by running the cold tap and stirring the dish from time to time. After 10 minutes, the temperature of the curry is 54 °C.

TIP You must define your own variables in this question as they have not been defined for you.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

It is given that the rate at which the curry cools down is proportional to the difference between the temperature of the curry and the temperature of the water in the sink. At what time can Maria put her dish of curry in the refrigerator? M

9 The half-life of a radioactive isotope is the amount of time it takes for half of the isotope in a sample to decay to its stable form. Carbon-14 is a radioactive isotope that has a half-life of 5700 years. It is given that the rate of decrease of the mass, m, of the carbon-14 in a sample is proportional to its mass. A sample of carbon-14 has initial mass m0. What fraction of the original amount of carbon-14 would be present in this sample after 2500 years?

M

10 Anya carried out an experiment and discovered that the rate of growth of her hair was constant. At the start of her experiment, her hair was 20 cm long. After 20 weeks, her hair was 26 cm long. Form and solve a differential equation to find a direct relationship between time, t, and the length of Anya’s hair, L.

M

11 A bottle of water is taken out of a refrigerator. The temperature of the water C. The bottle of water is taken outside to drink. The air in the bottle is 4 °C temperature outside is constant at 24 °C. It is given that the rate at which the water in the bottle warms up is proportional to the difference in the air temperature outside and the temperature of the water in the bottle.

258

After 2 minutes the temperature of the water in the bottle is 11°C. a How long does it take for the water to warm 20 °C ? b According to the model, what temperature will the water in the bottle eventually reach if the air temperature remains constant and the water is not drunk? M

12 The number of customers, n, of a food shop t months after it opens for the first time can be modelled as a continuous variable. It is suggested that the number of customers is increasing at a rate that is proportional to the square root of n. a Form and solve a differential equation relating n and t to model this information. b Initially, n = 0, and after 6 months the food shop has 3600 customers. Find how many complete months it takes for the number of customers to reach 6800. c

M

The food shop has a capacity of serving 14 000 customers per month. Show that the model predicts the shop will have reached its capacity sometime in the 12th month.

13 Doubling time is the length of time it takes for a quantity to double in size or value. The number of bacteria in a liquid culture can be modelled as a continuous variable and grows at a rate proportional to the number of bacteria present. Initially, there are 5000 bacteria in the culture. After 3.5 hours there are a million bacteria. a What is the doubling time of the bacteria in the culture? b What assumption does the model make about the growth of the bacteria and how realistic is this assumption?

Chapter 10: Differential equations

EXPLORE 10.2

Modelling growth Many of the models we have been using have assumed exponential growth. This type of growth is unlimited as the rate of growth of a population is proportional to the size of the population at that time. This means that

dP = kP. dt

Is an exponential growth model appropriate for modelling the population of the planet Earth in the long term? It is thought that the maximum capacity of the Earth is about 10 billion people. In 1987, the population of the Earth was about 5 billion people and the growth rate of the population was about 1.84%. Some people think that the problem of the Earth’s increasing population can be solved by sending some people to live on other planets, such as Mars. In 2031, the first people to live on Mars will launch into space. The environment on Mars is different to Earth and many challenges must be overcome. This will all take time. When a population is modelled as growing exponentially at a constant rate of r % per 70 year, it takes about years for the population to double. r ● Use the data given so far to investigate whether going to live on Mars is a good solution to the population problem. ● Comment on any assumptions you have made in your investigation. There is statistical evidence that the population of the Earth is not growing at a constant rate each year. In 2017, the growth rate had decreased to 1.11% per year. The population needs resources to survive and so the growth rate has decreased as resources have decreased. A different model, logistic growth, can be used when resources are limited. This model takes into account the changing growth rates of the population.

259

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Here are sketch graphs of exponential and logistic growth models: Exponential growth

Population size

Maximum possible population

O

Time

Logistic growth

Maximum possible population Population size

260

O

Time

For example, for a population of size P, dP rP(( k − P ) = dt k where the constant r is the maximum natural growth rate (birth rate – death rate) and the constant k is the maximum capacity of the limited environment. ●

Show that as time increases this model suggests that P tends to k.

●

Comment on the limitations of this model.

Chapter 10: Differential equations

Checklist of learning and understanding General procedure to solve a differential equation using the technique of separating the variables dyy = g( x ). dx

●

Separate the variables f( y )

●

Integrate both sides

●

Solve to find a general solution.

●

Substitute any given conditions to find a particular solution.

●

Check your answer by differentiating.

dy

∫ f( y x dx ∫ g(x ) dx ∫ f( y) dy ∫ g(x ) dx.

Other points to remember ●

Take care with modulus signs in expressions such as ln f( x ) .

●

When finding a general solution, check to see if any values excluded when the variables are separated are possible solutions.

261

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

END-OF-CHAPTER REVIEW EXERCISE 10 1

The variables x and y satisfy the differential equation dyy = x2e y+2x dx and it is given that y = 0 when x = 0. Solve the differential equation and obtain an expression for y in terms of x.

2

[9]

The number of birds, x, in a particular area of land is recorded every year for t years. x is to be modelled as a continuous variable. The rate of change of the number of birds over time is modelled by dx x( 2500 x) x = . dt 5000 It is given that x = 500 when t = 0.

3

a Find an expression for x in terms of t.

[9]

b How many birds does the model suggest there will be in the long term?

[1]

Given that y = 2 when x = 0, solve the differential equation dy = 1 + y4 dx obtaining an expression for y 4 in terms of x. y3

4 262

5

[6]

The gradient of a curve is such that, at the point ( x, y ) , the gradient of the curve is proportional to x y . At the point (3, 4) the gradient of this curve is −5. a Form and solve a differential equation to find the equation of this curve.

[7]

3 4) . b Find the gradient of the curve at the point ( −3,

[1]

a

Given that x , 5 , find

50

dx. ∫ (5 − x )(10 )( − x )

[3]

b A chemical reaction takes place between two substances A and B. When this happens, a third substance, C , is produced. After t hours there are 5 − x grams of A, there are 10 − x grams of B and there are x grams of C present. The rate of increase of x is proportional to the product of 5 − x and 10 − x. Initially, x = 0 and the rate of increase of x is 1 gram per hour. i

Show that x and t satisfy the differential equation 50

ii

Solve this equation, giving x in terms of t.

dx = (5 − x )( )(10 10 − x ) . dt

[6]

iii According to the model, approximately how many grams of chemical C are produced by the reaction? 6

[1]

[1]

The variables x and y are related by the differential equation dyy xe x = . dx 5 y 4 It is given that y = 4 when x = 0 . Find a particular solution of the differential equation and hence find the value of y when x = 3.5 . [8]

7

a

Use the substitution u =

b

Given that

y 4 − 1 to find

dyy (2 x + 1) y 4 − 1 = , y= dx xy3

∫

y3 dy. y4 − 1

3 when x = 1, find a relationship between x and y.

[1] [6]

Chapter 10: Differential equations

8

9

The height, h metres, of a cherry tree is recorded every year for t years after it is planted. It is thought that the height of the tree is increasing at a rate proportional to 8 − h. When the tree is planted it is 0.5 metres tall and after 5 years it is 2 metres tall. a Form and solve a differential equation to model this information. Give your answer in the form h = f(t ) .

[6]

b According to the model, what will the height of the cherry tree be when it is fully grown?

[1]

The variables x and y satisfy the differential equation dyy = e3( x + y ) and y = 0 when x = 0 . dx Solve the differential equation and obtain an expression for y in terms of x.

[6]

10 At the start of a reaction, there are x grams of chemical X present. At time t seconds after the start, the rate of decrease of x is proportional to xt. a Using k as constant of proportionality, where k . 0, form a differential equation to model this reaction rate.

[1]

b Solve the differential equation obtaining a relation between x, t and k.

[5]

c Initially, there are 150 grams of chemical X present and after 10 seconds this has decreased to 120 grams. Find the time after the start of the reaction when the amount of chemical X has decreased to 1 gram.

[3]

11 a b c

Using partial fractions find

∫ P(5

263

1 P P)

dP .

[4]

dP Given that P = 3 when t = 0, solve the differential equation = P (5 − P ), obtaining an expression dt for P in terms of t.

[4]

Describe what happens to P when t → ∞.

[1]

12 The population of a country was 50 million in 2000 and 60 million in 2010. The rate of increase of the population is modelled by dP kP (100 − P ) = dt 100 Use the model to predict the population of the country in 2025. 13 a

Use the substitution u = x 2 to find Given that

2

d dx.

[3]

dyy x sin x = , y = −1 when x = 0, find a relationship between x and y. dx y

[5]

2

b

∫ x sisinn x

[7]

14 Solve the differential equation dyy tan x = 3y dx e given that x = 0 when y = 0. Give your answer in the form y

f( x). x

[6]

15 There are 2000 mice in a field. At time t hours, x mice are infected with a disease. The rate of increase of the number of mice infected is proportional to the product of the number of mice infected and the number of mice not infected. Initially 500 mice are infected and the disease is spreading at a rate of 50 mice per hour. a Form and solve a differential equation to model this data. Give your answer in the form t = f( x ).

[7]

b Find how long it takes for 1900 mice to be infected.

[1]

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

16 The variables x and y are related by the differential equation dyy x = 1 − y2 . dx When x = 2 , y = 0. Solve the differential equation, obtaining an expression for y in terms of x.

[8]

Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2012 17 The variables x and θ satisfy the differential equation dx = ( + 2 ) sin2 2θ , dθ and it is given that x = 0 when θ = 0. Solve the differential equation and calculate the value of x when θ =

264

1 π giving your answer correct to 3 significant figures. π, [9] 4 Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2015

265

Chapter 11 Complex numbers P3 This chapter is for Pure Mathematics 3 students only.

In this chapter you will learn how to:

■ ■ ■ ■ ■ ■ ■ ■

understand the idea of a complex number, recall the meaning of the terms real part, imaginary part, modulus, argument and conjugate and use the fact that two complex numbers are equal if and only if both real and imaginary parts are equal carry out operations of addition, subtraction, multiplication and division of two complex numbers expressed in Cartesian form x + iy use the result that, for a polynomial equation with real coefficients, any non-real roots occur in conjugate pairs represent complex numbers geometrically by means of an Argand diagram carry out operations of multiplication and division of two complex numbers expressed in polar form r(cos θ + isin θ ) = re iθ find the two square roots of a complex number understand in simple terms the geometrical effects of conjugating a complex number and of adding, subtracting, multiplying and dividing two complex numbers illustrate simple equations and inequalities involving complex numbers by means of loci in an Argand diagram.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PREREQUISITE KNOWLEDGE Where it comes from

What you should be able to do

Check your skills

IGCSE / O Level Mathematics

Add, subtract and multiply pairs of binomial expressions.

1 Simplify each of the following. a

( a + bx ) + (2 a − 3bx )

b ( a + bx )(2 a − 3bx ) Assumed prior knowledge (not in the IGCSE / O Level course)

Manipulate expressions involving surds and calculate with surds.

2 Simplify.

b c IGCSE / O Level Mathematics

Draw simple loci.

●

it is always the same distance away from a fixed point is a circle it is always equidistant from two fixed points is a perpendicular bisector.

266

Pure Mathematics 1 Coursebook, Chapter 5

Understand the convention for positive and negative angles and use trigonometry to find the size of an angle that is not acute.

Chapter 9

●

●

Find the magnitude and direction of a vector. Know and use the relationship AB = b − a when the position vector of A is a and the position vector of B is b.

(2 −

5 4− 3 5+2 3

) (2 +

5

)

3 The diagram shows a triangle ABC . A

Understand, for example, that the locus of a point which moves so that ●

32 − 3 8

a

B

C

Copy the diagram and shade the region that is less than 4 cm from A and nearer to B than to C . 4 a Solve tan x = −1 for −π , x , π radians. 3 b Find π − sin −1 . 4 5 The point A has position vector a = 3i + 4 j and the point B has position vector b = 5 i − 7 j. Find: a the magnitude of a b the angle made between a and the positive x-axis

c BA

What are complex numbers? To understand the structure of a complex number we must first introduce a new set of numbers, the set of imaginary numbers. The number system we use today has developed over time as new sets of numbers have been needed to model new situations. Early mathematicians used natural numbers for counting. Later, zero and negative numbers were accepted as valid numbers and the set of integers was established. In Pythagoras’ time, mathematicians understood that rational numbers existed as they were the values in between integers. Pythagorean mathematicians thought that all numbers could be written as rational numbers. These same mathematicians were shocked by the discovery of irrational numbers such as 2 and

Chapter 11: Complex numbers

they did not trust them as valid quantities at the time. Even as late as the 17th and 18th centuries, some famous mathematicians still doubted that negative numbers were of any use and thought that subtracting something from nothing was an impossible operation! We will start with some simple facts. ● ● ● ●

In the set of natural or counting numbers (ℕ) there are no negatives and no zero. In the set of integers (ℤ) there are no part or fractional values. In the set of rational numbers (ℚ) there are no irrational quantities such as π. In the set of real numbers (ℝ) there are no numbers that when squared result in a negative value.

Until now, the numbers with which we have been working have all been real numbers. That is, they belong to the set of real numbers, which includes the rational and irrational numbers. Real numbers can be represented on a 1-dimensional diagram called a number line. In the 17th century, the set of imaginary numbers (I) was accepted as a valid number set. René Descartes described the collection of number sets that had been accepted as valid before that point as the set of real numbers. The new set, which he did not like very much, he described as imaginary. He intended this to be an insult as he thought they were not of great importance. Many other famous mathematicians agreed with him. They were, however, quite wrong about that. This name has often confused students who are new to this area of study.

So what is an imaginary number and how is it related to a complex number? In the new set of imaginary numbers there are numbers that when squared result in a negative value. KEY POINT 11.1 The set of complex numbers combines the real numbers and the imaginary numbers into a 2-dimensional rather than a 1-dimensional number system. Descartes described the numbers in the set of complex numbers as the sum of a real part and an imaginary part. Complex numbers are represented as points on a plane rather than points on a number line because of their 2 -dimensional nature.

Clearly, accepting new sets of numbers as being valid is something mathematicians have struggled with over time. You might not feel comfortable with this new idea. The set of real, imaginary and complex numbers does have application in the real world. Without imaginary and, therefore, complex numbers there would be no mobile telephones, for example, as they are used to find audio signals. Complex numbers are used to generate fractals, which are used in the study of weather systems and earthquakes. Fractals are also used to make realistic computer-generated images for movies.

11.1 Imaginary numbers In the set of imaginary numbers we introduce a new number i and we define i2 = −1. This means that i =

−1.

Note that: ● ●

Engineers call this quantity j as i is used in engineering for a different quantity. Just as with some other important numbers, i is described as a letter rather than a digit.

267

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Consider the equation x 2 + 1 = 0 . How many solutions does this equation have in the set of real numbers? x2 + 1 = 0

x 2 = −1

x = ± −1

No solutions in the set of real numbers.

How many solutions does this equation have in the set of imaginary numbers? x2 + 1 = 0

x 2 = −1

x = ± −1 = ± i

Two solutions in the set of imaginary numbers.

We can use laws of surds to extend the use of i to find other imaginary numbers. WORKED EXAMPLE 11.1 Write the following numbers in their simplest form. a

−9

b

−5

TIP c

−18

Answer a

When solving x 2 + 1 = 0 we needed ± in the solution. Why do we not need it here?

−9 = 9 × −1 = 9 × −1 = 3i

b

−5 = =

268

c

−18 =

5 × −1

When the multiplier is a simple surd, it is better to write the i before the surd to avoid any confusion with 5i or we can write ( 5 ) i.

5 × −1 = i 5 2 × 9 × −1

=

2 × 9 × −1

=

2 ×3×i = 3 2 i

(

)

The imaginary number i is called the unit imaginary number as it has a coefficient of 1. All other imaginary numbers are multiples of i. Imaginary numbers are therefore of the form bi or iy , where b is a real number or y is a real number. We calculate with imaginary numbers in the same way that we calculate with real numbers. TIP Many calculators are able to calculate in the set of complex numbers. You must practise without a calculator but can use one to check your answer. You need to demonstrate your ability to work with these numbers without a calculator. It is therefore important to practise writing down your method steps. Working without a calculator will also help to improve your understanding and recall of this subject (and do not forget that not all calculators are able to calculate with complex numbers).

Simplest form for an imaginary number means using simple surd form and i.

Chapter 11: Complex numbers

WORKED EXAMPLE 11.2 Without using a calculator, find: a

(3i2 ) + (3i)2

b

−8i + ( − 4i)3

c

9i + 16i 4i

d i −6 .

Answer a

(3i2 ) + (3i)2 = 3i2 + 9i2 = 12i2 = −12

b

−8i + ( − 4i)3 = −8i + 64i = 56i 9i+16i = 4i

c d

i −6 =

25i = 4i

25 5 = 4 2

1 1 1 = = −1 3 6 = i ( −1)3 ( i2 )

EXERCISE 11A Do not use a calculator in this exercise. 1 Write the following numbers in their simplest form. 36 − a −144 b 81

c

−90

d

−16 + −81 269

2 Simplify. a

−5i2 + 3i8

b

9i − ( i 2

)3

c

100i 4 − 16i2 4

b

4x 2 + 7 = 0

c

12 x 2 + 3 = 0

d

−5 6i2

3 Solve. a

x2 +

64 =0 25

EXPLORE 11.1 Without using a calculator, find i200.

11.2 Complex numbers A complex number is defined as a number of the form x + iy , where x and y are real values and i is the unit imaginary number. Note: Other general cases are common too, such as a + bi, where a and b are real. A complex number is often denoted by the letter z so z = x + iy : ● ●

the real number x is called the real part of z, Re z the real number y is called the imaginary part of z, Im z.

When x = 0, then z = iy and z is completely imaginary. When y = 0, then z = x and z is completely real.

TIP Sometimes iy is considered to be Im z.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

The relationship between the set of complex numbers (ℂ) and other sets of numbers is often represented using nested rings. 0

Equal complex numbers Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. So when z1 = x + iy and z2 = a + bi then z1 = z2 means that x = a and y = b.

TIP

Complex conjugate The complex conjugate of z = x + iy is defined as z* = x − iy.

Sometimes z * is written as z .

WORKED EXAMPLE 11.3 Solve the equation 5 z 2 + 14 z + 13 = 0. Answer Using the quadratic formula a = 5, b = 14, c = 13.

270

z =

−14 ± 142 − 4(5)(13) 10

z =

−14 ± −64 10

z =

−14 ± 64 × −1 10

z =

−14 ± 8i 10

z=−

7 4i 7 4i + or − − 5 5 5 5

Note that the two roots are complex conjugates of each other.

WORKED EXAMPLE 11.4 (2 x + y ) + i( y − 5) = 0 Find the value of x and the value of y. Answer Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. (2 x + y ) + i( y − 5) = 0 + 0i Therefore 2x + y = 0

(1)

and y−5= 0

(2)

From (2): y = 5 Substituting into (1): 2 x + 5 = 0, x = −

5 2

Chapter 11: Complex numbers

Calculating with complex numbers The processes we need to use are the same as those we use when manipulating expressions involving surds. For example, when z1 = 2 + 3i and z2 = 8 − i we have:

Addition

z1 + z2 = 2 + 3i + 8 − i

Collect like terms.

Subtraction

z1 + z2 = 10 + 2i z1 − z2 = 2 + 3i − (8 − i)

Collect like terms.

z1 − z2 = −6 + 4i Multiplication

z1z2 = (2 + 3i)(8 − i)

Expand the brackets and simplify.

z1z2 = 16 − 2i + 24i − 3i z1z2 = 16 + 22i + 3 z1z2 = 19 + 22i z1z1* = (2 + 3i)(2 − 3i) z1z1* = 4 − 6i + 6i − 9i

2

2

i2 = −1 It is important to show all your working out clearly. Expand the brackets and simplify. i2 = −1

z1z1* = 4 + 9 z1z1* = 13 Division

z2 8−i = z1 2 + 3i

Multiply the numerator and denominator by the complex conjugate of z1.

z2 (8 − i)(2 − 3i) = z1 (2 + 3i)(2 − 3i)

Expand the brackets and simplify.

z2 16 − 24i − 2i + 3i2 = z1 13

i2 = −1

z2 16 − 24i − 2i − 3 = z1 13

We have already found 13 so we did not show full working. You should usually show all your working to demonstrate you understand the method you are using.

z2 13 − 26i = = 1 − 2i z1 13 KEY POINT 11.2

When we multiply a complex number by its conjugate the result is always a real number as, if we let z = x + iy, and so z* = x − iy, then zz* = ( x + iy )( x − iy ) = x 2 + y 2 . Since x and y are real, x 2 + y 2 is also real.

271

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 11B Do not use a calculator in this exercise. 1

z1 = 5 − 3i and z2 = 1 + 2i

Write each of the following in the form x + iy , where x and y are real values. z1 b z1* − z2 c z1z2 d a z1 + z2* z2 2 Solve these equations. a

z 2 + 2 z + 13 = 0

b

z 2 + 4z + 5 = 0

c

2z2 − 2z + 5 = 0

d

z 2 − 6 z + 15 = 0

e

3z 2 + 8 z + 10 = 0 f

2 z 2 + 5z + 4 = 0

3 Find the value of x and the value of y. a

( x + 2 y ) + i(3x − y ) = 1 + 10i

c

( x − y ) + (2 x − y )i = −1

b

( x + y − 4) + 2 xi = (5 − y )i

4 Write each of the following in the form x + iy , where x and y are real values. a

(1 + 3i)(2 − i)

b

(4 − 5i)(4 + 5i)

c

(7 − 3i)2

d

( 3 − i )4

e

17 − i 3+i

f

−1 + 11i 6 − 5i

g

13(1 + i) 2 + 3i

h

(3 − 2i)2 5+i

272

1 2 6 + i is a solution of the equation 5 z 2 − 2 z + 5 = 0. 5 5 b Write down the other solution of the equation.

5 a Show that z =

PS

PS

6 Find the quadratic equations that have the following roots. a

α = −7i

c

α = 2 − 3i

β = 7i β = 2 + 3i

b

α = 1 + 5i

d

α =−

TIP

β = 1 − 5i

5 31 − i 2 2

β =−

5 31 + i 2 2

7 Find the complex number z satisfying the equation z + 4 = 3iz*. Give your answer in the form x + iy , where x and y are real.

PS

8 Find the complex number u, given that u(3 − 5i) = 13 + i.

PS

9 z = 5 + i 3 is a root of a quadratic equation. Find this quadratic equation.

M

10 In an electrical circuit, the voltage (volts), current (amperes) and impedance (ohms) are related by the equation voltage = current × impedance The voltage in a particular circuit is 240 V and the impedance is 48 + 36i ohms. Find the current.

( z − α )( z − β ) = 0 z 2 − (α + β ) z + αβ = 0

Chapter 11: Complex numbers

11.3 The complex plane

DID YOU KNOW?

As complex numbers have two dimensions, one real and one imaginary, a diagrammatical representation must have two dimensions. This means it has to be a plane and not a line. Jean-Robert Argand (1768 −1822) is credited for the diagram used to do this.

Argand diagrams The complex number x + iy can be represented by the Cartesian coordinates ( x, y ). The horizontal axis is called the real axis and the vertical axis is called the imaginary axis. For example, we can represent the numbers z1 = 2 + 3i and z2 = 5 − 3i by the points A(2, 3) and B (5, − 3) on an Argand diagram.

Real numbers can be ordered along a number line. Complex numbers cannot be ordered as they are represented by points in a plane rather than on a line.

Im(z) 4 A

3

We can also represent the numbers z1 = 2 + 3i and 2 z2 = 5 − 3i by the position vectors OA = and 3 5 . OB = −3

TIP

2 1

–1 O –1

1

2

3

4

5

Re(z)

–2

Im(z)

–3

B

4

When drawing an Argand diagram, make sure you choose equal scales for the real and imaginary axes. This will give you a good picture and not stretch any shape you draw.

–4

A 3 2 1

273 –1 O –1

1

2

3

4

5

Re(z)

–2 –3

B

–4

Modulus-argument form When a complex number is written as x + iy where x and y are real values, we say it is in Cartesian form. There are other ways of writing complex numbers. The vector representation of a complex number on an Argand diagram helps us to understand a different representation called the modulus-argument form. Before we do this, we need to define the modulus and the argument of a complex number.

Im(z)

The modulus of a complex number x + iy is the magnitude x of the position vector . y Therefore, the modulus of x + iy is defined to be Notation: The modulus of z = x + iy is z . Also note that the plural of modulus is moduli.

2

y

2

x +y . θ O

x

Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

x The argument of a complex number x + iy is the direction of the position vector . y More precisely, it is the angle made between the positive real axis and the position vector. The principal argument, θ , of a complex number is an angle such that −π , θ ø π. Sometimes it might be more convenient to give θ as an angle such that 0 ø θ , 2 π. Usually, the argument is given in radians. Notation: The argument of z = x + iy is arg z. We can use trigonometry to find the argument. For example, when θ is acute, tan θ =

y . x

A diagram must be drawn first to check the position of the complex number. WORKED EXAMPLE 11.5 Find the modulus and argument of each of the following complex numbers. a

5 + 12i

b

−3 + 4i

c

12 − 5i

d

− 4 − 3i

Answer A simple sketch of the Argand diagram is always necessary when finding the argument. a

5 + 12i =

52 + 122 = 169 = 13

From the definition of the modulus.

Im(z)

The diagram shows that the angle is in the first quadrant and so the angle is acute and positive.

274

12

θ O

5

Re(z)

12 = 1.1760... arg(5 + 12i) = tan −1 5 = 1.18 radians correct to 3 significant figures b

−3 + 4i =

( −3)2 + 42 =

25 = 5

From the definition of the modulus. The diagram shows that the angle is in the second quadrant and is obtuse and positive. This is calculated by subtracting the value of the acute angle α from π. Remember that the principal argument is −π , θ ø π.

Im(z)

4 θ

α 3

θ = π−α

O

From the definition of the argument.

Re(z)

Chapter 11: Complex numbers

c

4 arg( −3 + 4i) = π − tan −1 = 2.214... 3 = 2.21 radians correct to 3 significant figures

From the definition of the argument.

12 − 5i = 122 + ( −5)2 = 169 = 13

From the definition of the modulus.

Im(z)

The diagram shows that the angle is in the fourth quadrant and is acute and negative. Again, recall that the principal argument is −π , θ ø π.

12 O

Re(z)

α

5

θ = −α 5 = − 0.3947... arg(12 − 5i) = − tan −1 12

From the definition of the argument.

= − 0.395 radians correct to 3 significant figures d

− 4 − 3i =

( −4)2 + ( −3)2 =

From the definition of the modulus.

25 = 5

Im(z)

The diagram shows the angle is in the third quadrant and is obtuse and negative. Again recall that the principal argument is − π , θ ø π.

4 α

O θ

275

Re(z)

3

θ = −π + α 3 arg( − 4 − 3i) = −π + tan −1 = −2.498... 4 = −2.50 radians correct to 3 significant figures

From the definition of the argument.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Using trigonometry, we can now write a complex number in modulus-argument form. x x cos θ = and sin θ = . r r Therefore, x = r cos θ and y = r sin θ .

Im(z)

r

Substituting these expressions into the Cartesian form of a complex number, we have z = x + iy

y

z = r cos θ + i( r sin θ )

θ

z = r(cos θ + isin θ )

O

x

where r is z and θ is arg z. When a complex number is written as z = r(cos θ + isin θ ) where r is z and θ is arg z, we say it is in modulus-argument form. WORKED EXAMPLE 11.6 u = 6 − 3i and w = −7 + 5i Write each of u and w in modulus-argument form. Answer u = 6 − 3i r= u =

62 + ( −3)2

=

45 = 3 5

276

From the definition of the modulus.

Im(z)

The diagram shows the angle to be acute and negative. 6

O

α

Re(z)

3

θ = −α 3 arg(6 − 3i) = − tan −1 6 = − 0.4636...

From the definition of the argument.

= − 0.464 radians correct to 3 significant figures u = 3 5(cos( − 0.464) + i sin( − 0.464)) w = −7 + 5i r= w = =

( −7)2 + 52 74

From the definition of the modulus.

Re(z)

Chapter 11: Complex numbers

Im(z)

The diagram shows the angle to be obtuse and positive.

5 θ

α 7

O

Re(z)

θ = π−α 5 arg( −7 + 5i) = π − tan −1 7 = 2.5213...

From the definition of the argument.

= 2.52 radians correct to 3 significant figures w=

74(cos(2.52) + i sin(2.52))

Exponential form The mathematician Leonhard Euler (1707–1783) discovered the relationship cos θ + isin θ = e iθ . This is the basis for another representation, called the exponential form. z = r(cos θ + isin θ ) z = re iθ

277

where r is z and θ is arg z. The coordinates ( r, θ ), where r is z and θ is arg z, are called the polar coordinates. When a complex number is written as z = re iθ , where r is z and θ is arg z, we say it is in exponential form. The modulus-argument and exponential forms of a complex number are polar forms. A polar form uses polar coordinates; a Cartesian form uses Cartesian coordinates. WORKED EXAMPLE 11.7 iπ

z1 = 1 + i

z2 = 5e 3

π π z3 = 2 cos + isin 12 12

a Write z1 in

b Write z2 in

c Write z3 in

•

modulus-argument form

•

modulus-argument form

•

exponential form

•

exponential form.

•

Cartesian form.

•

Cartesian form.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Answer a

Im(z)

The diagram shows the angle to be acute and positive. 1

θ O

1

z1 = 12 + 12 =

Re(z)

From the definition of the modulus.

2

arg(1 + i) = tan −1( 1 ) =

π 4

From the definition of the argument.

Modulus-argument form: π π 2 cos + isin 4 4

z1 =

Exponential form: iπ

z1 = 278

b

2e 4

π 3 Modulus-argument form: z2 = 5 θ =

π π z2 = 5 cos + isin 3 3

Comparing the modulus-argument form with x + iy .

Im(z)

y 5 π 3 x

O

x = 5 cos x=

5 2

π 3

y=

y = 5 sin 5 3 2

Cartesian form: z2 =

5 3 5 + i 2 2

Re(z)

π 3

Evaluate. Write in Cartesian form, x + iy.

Chapter 11: Complex numbers

π 12

z3 = 2 θ =

c

iπ

Comparing the modulus-argument form with x + iy .

Exponential form: z3 = 2e 12 Im(z)

2

y

π 12 x

O

x = 2 cos x=

Re(z)

π π y = 2 sin 12 12

6+ 2 y= 2

Cartesian form: z3 =

Evaluate.

6− 2 2 6+ 2 +i 2

Write in Cartesian form, x + iy. 6− 2 2

WORKED EXAMPLE 11.8

279

u = 5e2i and w = 10e −3i a Find the modulus and argument of: w ii i uw u w b Show u, w, uw and on a single Argand diagram. u Answer a i

uw = 5 × e2i × 10 × e −3i uw = 50 × e

Using laws of indices.

TIP

2i − 3i

The exponential form of a complex number is very useful when multiplying or dividing two complex numbers.

uw = 50e − i ii

w 10 × e −3i = u 5 × e2i w = 2 × e −3i − 2i u w = 2e −5i u −5 is less than −π and so this argument is not a principal argument. The equivalent principal argument, in this example, is θ = 2 π − 5 w = 2e(2π−5)i u

Using laws of indices. Im(z)

θ α O

Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

u = 5(cos 2 + i sin 2) = −2.08 + 4.55i

b

w = 10(cos( −3) + i sin( −3)) = −9.90 − 1.41i uw = 50(cos( −1) + i sin( −1)) = 27.0 − 42.1i w = 2(cos(2π − 5) + i sin(2π − 5)) = 0.567 + 1.92i u Im(z)

TIP

A ≈ ( −2, 4.5) B ≈ ( −10, 1.5)

It is not essential to find the Cartesian form, but you might find it easier to plot in the correct quadrant if you do.

C ≈ (27, 42) D ≈ (0.5, 2)

C

A

D

B O

Re(z)

Using laws of indices and the exponential form of a complex number, we can derive some very important and useful results.

280

Let the complex number z1 be such that z1 = r1 and arg z1 = θ1 and let the complex number z2 be such that z2 = r2 and arg z2 = θ2 then: z r e iθ r z1 = r1e iθ , z2 = r2 e iθ , z1z2 = r1e iθ × r2 e iθ = r1r2 e i(θ +θ ) and 1 = 1 iθ = 1 e i(θ z2 r2 r2 e Therefore, the complex number z1z2 is such that: 1

1

2

1

2

1

2

1

−θ 2 )

2

z1z2 = r1r2 = z1 z2 and arg ( z1z2 ) = θ1 + θ 2 = arg z1 + arg z2, z and the complex number 1 is such that: z2 z1 z1 r = 1 = z2 r2 z2

z and arg 1 = θ1 − θ 2 = arg z1 − arg z2. z2

These results greatly simplify the multiplication and division of complex numbers. We can see that, when in exponential form, ● ●

to multiply two complex numbers we multiply the moduli and add the arguments to divide two complex numbers we divide the moduli and subtract the arguments. WORKED EXAMPLE 11.9 The complex numbers u and w are given by u = 9 + 12i and w = −5 − 3i. Find the modulus and argument of: a uw

b

u w

Answer u =

92 + 122 =

w =

( −5)2 + ( −3)2 =

225 = 15 34

By definition for the moduli of u and w.

Chapter 11: Complex numbers

Im(z)

From the diagram, the angle for u is acute and positive.

12

θ O

9

Re(z)

12 = 0.9272... arg u = tan −1 9 = 0.927 radians correct to 3 significant figures From the diagram, the angle for w is obtuse and negative.

Im(z)

5 α

O θ

Re(z)

3

θ = −π + α 3 arg w = −π + tan −1 = −2.601 5 = − 2.60 radians correct to 3 significant figures a

uw = u w = 15 34

arg( uw ) = arg u + arg w = 0.9272… + ( −2.601…)

281

Using z1z2 = r1r2 = z1 z2 Using arg( z1z2 ) = θ1 + θ 2 = arg z1 + arg z2

= −1.67 radians correct to 3 significant figures b

u u 15 34 15 = = = w w 34 34 u arg = arg u − arg w = 0.9272… − ( −2.601…) w = 3.528 radians However, 3.528… is more than π and so this argument is not a principal argument. The equivalent principal argument, in this example, is

Using

z z1 r = 1 = 1 z2 r2 z2

z Using arg 1 = θ1 − θ 2 = arg z1 − arg z2 z2 Im (z)

θ = −(2 π − 3.528…) = −2.754… = −2.75 radians correct to 3 significant figures

α O θ

Re (z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 11C 1

u = 5 − 2i a On an Argand diagram, show the points A, B and C representing the complex numbers u, u* and − u, respectively. b The points ABCD form a rectangle. Write down the complex number represented by the point D.

2

z1 = 1 + 5i and z2 = −7 − i a On an Argand diagram, show the points P and Q representing the complex numbers z1 and z2 , respectively. b Write down the complex number represented by the point R, the midpoint of PQ.

3 Find the modulus and argument of each of the following complex numbers. a

−12 + 5i

b 5i

d

60 − 11i

e

g

5 + 2i

h

4

c

8 + 15i

−9 − 40i

f

−1 − i 3

−24 − 7i

i

k(1 − i), where k . 0

Im(z) 4 A 3 2 1 –3 –2 –1 O –1

282

B 1

2

3

Re(z)

–2 –3

C

–4

a Write each of the complex numbers shown in the Argand diagram in modulus-argument form. b Show that ABC is a right-angled, isosceles triangle. 5 Write each of these complex numbers in Cartesian form. a

π π 3 cos + i sin 3 3

b

3π 3π + i sin 5 cos 8 8 iπ

− e iπ d 3e 4 2 6 Given that z(4 − 9i) = 8 + 3i, find the value of z 2 and the value of arg z 2 .

c

3 − 7i π π 7 w = 5 cos + isin and z = 6 6 5 − 2i a Write w in exponential form. b Write z in exponential form. z . w π 8 The complex number a + bi is denoted by z. Given that z = 5 and arg z = , find the value of 6 a and the value of b. c

PS

P

Find and simplify an expression for

9 Given that z = r(cos θ + isin θ ), find and simplify an expression for z a zz * b z* 10 z = r(cos θ + i sin θ ) a Show that z 2 = r 2 (cos 2θ + i sin 2θ ).

b Show that z −

1 = i(2 r sin θ ). z

Chapter 11: Complex numbers

EXPLORE 11.2 Using Argand diagrams and position vectors, investigate the transformations of the plane that are: ●

●

the geometrical relationship between a complex number and its conjugate the geometrical effects of combining two complex numbers by addition, subtraction, multiplication and division.

11.4 Solving equations We have already solved quadratic equations with complex roots. From the Pure Mathematics 1 Coursebook, Chapter 1, we know that when solving quadratic equations for solutions that are real:

b 2 − 4 ac

Nature of roots

Roots

.0

2 real and distinct roots

α and β

=0

2 real and equal roots (repeated roots)

α and β , α = β

,0

0 real roots 283

We can now update this for solving quadratic equations for solutions that are complex:

b 2 − 4 ac

Nature of roots

Roots

.0

2 real and distinct roots

α and β

=0

2 real and equal roots (repeated roots)

α and β , α = β

,0

2 complex roots of the form x ± iy with y ≠ 0 (a complex conjugate pair). This means that one root is the conjugate of the other root.

α and α *

Using the set of complex numbers to find solutions, a quadratic equation always has two roots.

Higher order polynomial equations We solved cubic and quartic equations for real roots in Chapter 1. We will now consider solving these equations for complex roots. The Fundamental Theorem of Algebra states that a polynomial equation of degree n has n complex roots.

TIP The degree of a quadratic equation is 2, the degree of a cubic equation is 3… the degree of a polynomial equation is the same as its largest power.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

WORKED EXAMPLE 11.10 a Show that z = 4 is a root of the equation z 3 − 3z 2 − 3z − 4 = 0. b Find all complex solutions of z 3 − 3z 2 − 3z − 4 = 0. Answer a Let f( z ) = z 3 − 3z 2 − 3z − 4 then f(4) = 43 − 3(42 ) − 3(4) − 4 = 64 − 48 − 12 − 4 = 64 − 64 = 0 ✓ b As z = 4 is a root then z − 4 is a factor, therefore z 3 − 3z 2 − 3z − 4 = ( z − 4)( z 2 + az + 1) z 3 − 3z 2 − 3z − 4 = z 3 + −

az 2 + 2

4z −

Multiply out and compare. z 4az − 4

z 3 + ( a − 4) z 2 + (1 − 4a ) z − 4 z 3 − 3z 2 − 3z − 4 = z 3 + ( a − 4) z 2 + (1 − 4a ) z − 4

284

−3 = a − 4, therefore a = 1

This can also be done by synthetic or long division. Compare the coefficients of z 2 . Check, using the coefficients of z.

−3 = 1 − 4(1) ✓ ( z − 4)( z 2 + z + 1) = 0 z =4

Given.

z2 + z + 1 = 0 −1 ± 12 − 4(1)(1) 2 −1 ± −3 z = 2 −1 ± i 3 z = 2 z =

KEY POINT 11.3 When a polynomial equation with real coefficients has complex roots, they occur in complex conjugate pairs.

A cubic equation has three roots (by the Fundamental Theorem of Algebra). A cubic equation can always be written as the product of a linear factor and a quadratic factor. This means that a cubic equation must always have at least one real root.

Chapter 11: Complex numbers

Cubic equations either have ● ●

TIP

three real roots or one real root and two complex roots of the form x ± iy with y ≠ 0 (a complex conjugate pair).

A quartic equation has four roots (by the Fundamental Theorem of Algebra).

To understand this, think about the shape of the graph of a cubic equation.

Quartic equations either have ● ● ●

four real roots or two real and two complex roots of the form x ± iy with y ≠ 0 (a complex conjugate pair) or four complex roots of the form x ± iy with y ≠ 0 (two complex conjugate pairs).

A quartic equation can never have one real and three complex roots. Make sure you understand why this is true. WORKED EXAMPLE 11.11 It is given that z − 2i is a factor of the equation z 4 + 13z 2 + 36 = 0. Solve this equation. Answer As z − 2i is a factor z + 2i must also be a factor

Complex roots of the form x ± iy with y ≠ 0 occur in pairs.

( z − 2i)( z + 2i) = z 2 + 2iz − 2iz − 4i 2

285

2

= z +4 z 4 + 13z 2 + 36 = ( z 2 + 4)( z 2 + kz + 9)

Multiply out and compare.

z 4 + 13z 2 + 36 = z 4 + kz 3 + 9 z 2 + 4 z 2 + 4 kz + 36 z 4 + kz 3 + 13z 2 + 4 kz + 36

This can also be done by synthetic or long division.

z 4 + 13z 2 + 36 = z 4 + kz 3 + 13z 2 + 4 kz + 36

Compare the coefficients of z 3.

k=0

Check, using the coefficients of z.

0 = 4(0) ✓ z 4 + 13z 2 + 36 = ( z 2 + 4)( z 2 + 9) = 0 z = ±2i

From factors given.

2

z + 9 = 0 so z = ± −9 z = ±3i z = ±2i, z = ±3i

4 complex roots of the form x ± iy with y ≠ 0.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Finding the square roots of a complex number We can find the square roots of a complex number. WORKED EXAMPLE 11.12

(

)

Find the square roots of −2 − 2 3 i and show the position of these roots on an Argand diagram. Answer Find x + iy, where x and y are real, such that x + iy = ± −2 − (2 3 )i . TIP

x + iy = ± −2 − (2 3 )i

Square both sides.

( x + iy )2 = −2 − (2 3 )i

Multiply out the left-hand side.

x 2 + 2 xyi + i2 y2 = −2 − (2 3 )i

Collect real and imaginary parts.

( x 2 − y2 ) + (2 xy )i = −2 − (2 3 )i

Equate real and imaginary parts.

x 2 − y2 = −2

(1)

2 xy = 2 3

(2)

Solve simultaneously.

286

From (2): y =

3 3 and so x 2 − 2 = −2 x x x 4 − 3 = −2 x 2 u2 + 2u − 3 = 0

Multiply by x 2 . Rearrange and let u = x 2 . Solve.

( u + 3)( u − 1) = 0 u = −3, u = 1 x 2 = −3, x 2 = 1 x = ±1 y=

1 O –2 –1 –1 –2

1

2 Re(z)

Substitute into y =

3 . x

3 =± 3 ±1

Therefore, the square roots of −2 − (2 3 )i are 1 + i 3 and −1 − i 3 . Im(z) 2

x is real, so disregard x 2 = −3 .

Notice that the modulus of the two square roots is the same and the difference between the arguments of the square roots is π radians.

When we want to work in exponential form, we square re iθ to obtain r 2 e i(2θ ). We then equate this to the complex number whose square roots we are finding.

Chapter 11: Complex numbers

The cube roots of one (unity) By the Fundamental Theorem of Algebra, z 3 − 1 = 0 has three roots. We know that one of these roots is 1, since z = 3 1 = 1 is the real solution we are familiar with. To find the two complex roots of the form x ± iy with y ≠ 0, we can use this method: As z = 1 is a root then z − 1 is a factor and so z 3 − 1 = ( z − 1)( z 2 + kz + 1)

Multiply out and compare.

z 3 − 1 = z 3 + kz 2 + z − z 2 − kz − 1 3

2

z + ( k − 1) z + (1 − k ) z − 1 3

z − 1 = z 3 + ( k − 1) z 2 + (1 − k ) z − 1 k − 1 = 0, k = 1 0 = 1−1 ✓ 3

2

z − 1 = ( z − 1)( z + z + 1)

This can also be done by synthetic or long division. Compare the coefficients of z 2 . Check, using the coefficients of z. Solve.

−1 ± 12 − 4(1)(1) −1 ± −3 ,z= 2 2 −1 + i 3 −1 − i 3 z = 1, z = ,z= 2 2

z = 1, z =

287

These roots can be very useful when solving related equations. WORKED EXAMPLE 11.13 Solve z 3 = 8 . Answer z =

3

8 = 23 1

z = 2 × 1, z = 2 ×

As z is 2 × 3 1 replace 3 1 by each of the cube roots of 1 we have already found. −1 + i 3 −1 − i 3 ,z = 2× 2 2

z = 2, z = −1 + i 3, z = −1 − i 3

Simplify.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

EXERCISE 11D 1 a

Given that z1 = − i is a root of the equation z 3 + z 2 + z + k = 0, find the value of the constant k.

b Write down the other complex root of the equation and find the third root, stating whether it is real or complex. 2 Solve: a ( z − 5)3 = 8

b

(2 z + 3)3 =

1 . 64

3 Find the roots of the equation 2 z 2 + z + 3 = 0 , giving your answer in modulus-argument form. 4 It is given that z − 3 is a factor of z 3 − 3z 2 + 25 z − 75. Solve z 3 − 3z 2 + 25 z − 75 = 0 . 5

( x + iy )2 = 55 + 48i Find the value of x and the value of y, when x and y are real and positive.

6 It is given that 2 z + 1 is a factor of 2 z 3 − 11z 2 + 14 z + 10 . Solve 2 z 3 − 11z 2 + 14 z + 10 = 0 . 7 Given that z = 3i is a root, solve z 4 − 2 z 3 + 14 z 2 − 18 z + 45 = 0 . 8 Find the square roots of:

288

a

24 − 10i

b

7+ 6 2 i

c

5 −i 6 4

d

7 − 24i

e

−4 + 2 5 i

(

f

1 2i e 2

(

)

)

π

9 z 3 − 12 z 2 + pz + q = 0 where p and q are real constants. a Given that z = 2 + i is a root of this equation, write down another root of the equation. b Find the value of p and the value of q. c PS

Represent the roots of this equation on an Argand diagram.

10 f( z ) = z 4 − z 3 + 20 z 2 − 16 z + 64 = ( z 2 + a )( z 2 + bz + 4) Solve the equation f( z ) = 0.

11.5 Loci A locus is a path traced out by a point as it moves following a particular rule. The rule is expressed as an inequality or an equation. In order to draw the correct locus on an Argand diagram, we need to interpret the inequality or equation we have been given.

Chapter 11: Complex numbers

Circles with centre (0, 0) ●

z = r,

z , r,

z ør

where r is a constant.

In this case, the length of the position vector representing the complex number, z, is the constant value r. The argument of z can vary. A point that moves so that is it always the same distance from a fixed point is a circle. The fixed point is the centre of the circle. The fixed distance is the radius of the circle. Recall that a position vector is simply a means of locating a point in space relative to an origin, usually O. For the complex number z = x + iy , where x and y are real, the locus of the point P( x, y ) that satisfies: ● ●

●

z = r is the circumference of a circle with centre (0, 0) and radius r z , r is all points within a circle with centre (0, 0) and radius r, but the circumference is not included z ø r is all points within a circle with centre (0, 0) and radius r or the circumference of the circle.

WORKED EXAMPLE 11.14 On separate Argand diagrams, sketch these loci. z =5

z ,3

c

z = 5 is the circumference of a circle, centre (0, 0), radius 5.

Im(z)

a

b

z ø4

289

Answer a

6

It is only the circle, not the points within the circle.

5

The circle is drawn as a solid line as it is included.

3

4

2 1 –6

–5 –4

–3 –2

–1 O –1 –2 –3 –4 –5 –6

1

2

3

4

5

6 Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

b

z , 3 is a circle, centre (0, 0), radius 3.

Im(z) 4

It is only the points within the circle. 3

The circumference of the circle is drawn as a dotted line as it is not included.

2 1

–4

–3

–2

–1

O

1

2

3

4Re(z)

–1 –2 –3 –4

c

Im(z) 5

z ø 4 is a circle, centre (0, 0), radius 4. It is all points within the circle and on the circumference.

4

The circumference of the circle is drawn as a solid line as it is included.

2

3

1

290 –5

–4

–3

–2

–1 O –1

1

2

3

4

5 Re(z)

–2 –3 –4 –5

Circles with centre (a, b) ●

z − z1 = r, z − z1 , r, z − z1 ø r where r is a constant and z1 is the fixed point a + bi.

Im(z) P(x, y)

We need to understand the vector representation of the complex number z − z1. Let P ( x, y ) be the moving point that represents z = x + iy on an Argand diagram. Let Q( a, b ) be the fixed point that represents z1 = a + bi on the Argand diagram.

Q(a, b)

z = OP and z1 = OQ

QP = z − z

Therefore, QP = − z1 + z 1

O

Re(z)

Chapter 11: Complex numbers

We now interpret z − z1 = r as:

the length of the vector QP is r. For the complex number z = x + iy , where x and y are real, and the fixed point z1 = a + bi , where a and b are real, the locus of the point P ( x, y ) that satisfies ● ●

●

z − z1 = r is the circumference of a circle with centre ( a, b ) and radius r z − z1 , r is all points within a circle with centre ( a, b ) and radius r, but the circumference is not included z − z1 ø r is all points within a circle with centre ( a, b ) and radius r or the circumference of the circle.

WORKED EXAMPLE 11.15 On separate Argand diagrams, sketch these loci. z − (2 + 4i) = 3

a

b

z − 3i , 4

z + 5 ø5

c

Answer a

Im(z)

z1 = 2 + 4i z − (2 + 4i) = 3 is the circumference of a circle, centre (2, 4), radius 3.

7

It is only the circle, not the points within the circle.

5

6

291

4

The circle is drawn as a solid line as it is included.

3 2 1 –2

b

–1 O –1

z1 = 3i

1

2

3

4

5

6 Re(z)

Im(z) 7

z − 3i , 4 is a circle, centre (0, 3), radius 4.

6

It is only the points within the circle.

5

The circumference of the circle is drawn as a dotted line as it is not included.

4 3 2 1

–4

–3

–2

–1

O –1

1

2

3

4 Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

c

Take care with signs.

z − z1 = z + 5 = z − ( −5) z1 = −5

Im(z) 4 3

z + 5 ø 2 is a circle, centre ( −5, 0), radius 2.

2

It is all points within the circle and on the circumference. The circumference of the circle is drawn as a solid line as it is included.

1 –7 –6 –5 –4 –3 –2 –1 O –1

1

2 Re(z)

–2 –3 –4

Half-lines and part-lines ●

Im(z)

arg( z − z1 ) = α , where α is a fixed angle and z1 is the fixed point a + bi.

P(x, y)

With z − z1 defined as previously

z = OP, z1 = OQ and QP = z − z1

α

292 Q(a, b)

O

Re(z)

We now define arg( z − z1 ) as the angle made between the direction parallel to the real axis and the direction of the vector QP .

Points below Q are not part of the locus as the angle made between these points and the relevant horizontal line is π − α , not α . ●

arg( z1 − z ) = β , where β is a fixed angle and z1 is the fixed point a + bi.

Im(z)

With z1 − z = −( z − z1 ),

z = OP, z1 = OQ and PQ = z1 − z

P(x, y) β

Q(a, b) O

We now define arg( z1 − z ) as the angle made between the direction parallel to the real axis and the direction of the vector PQ.

Points above P are not part of the locus as the angle made between these points and the relevant horizontal line is π − β , not β .

Re(z)

Chapter 11: Complex numbers

WORKED EXAMPLE 11.16 On a single Argand diagram, sketch the loci z = 4 and arg( z + 2 + 3i) =

π . 4

Show that there is only one complex number, z , that satisfies both loci. Label this point as P on your diagram. Answer z =4

Circle, centre (0, 0), radius 4 π 4 z − z1 = z + 2 + 3i

arg( z + 2 + 3i) =

= z − ( −2 − 3i) z1 = −2 − 3i Half-line from ( −2, −3) at an angle of

π radians. 4 Im(z) 6 5 4

293

3 P

2 1 –5

–4

–3

–2

–1 O –1

1

2

3

4

5

6

7

8 Re(z)

–2 π 4 –3 –4 –5 –6

Perpendicular bisectors ●

z − z1 = z − z2 , z − z1 , z − z2 , and z2 is the fixed point c + di.

z − z1 ø z − z2

where z1 is the fixed point a + bi

Let P ( x, y ) be the moving point that represents z = x + iy on an Argand diagram, Q( a, b ) be the fixed point that represents z1 = a + bi on the Argand diagram, R( c, d ) be the fixed point that represents z2 = c + di on the Argand diagram, then

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

●

●

●

when z − z1 = z − z2 , the point P ( x, y ) moves so that it is always the same distance from Q as it is from R. The locus is, therefore, the perpendicular bisector of the line QR. z − z1 , z − z2 is all points such that P is nearer to Q than to R. The perpendicular bisector marks the boundary of this region but is not included. z − z1 ø z − z2 is all points such that P is nearer to Q than to R or equidistant from Q and R. The perpendicular bisector marks the boundary of this region and is included.

WORKED EXAMPLE 11.17 On an Argand diagram, sketch the locus z − 4 + i = z + 5i . Find the Cartesian equation of this locus. Answer Write as z − z1 and find z1

z − 4 + i = z − (4 − i) z1 = 4 − i z + 5i = z − ( −5i)

Write as z − z2 and find z2 .

z2 = −5i The locus is the perpendicular bisector of (4, −1) and ( 0, −5 ).

Midpoint of QP, (4, −1) and (0, −5), 294

is

4 + 0 −1 + ( −5) , = (2, −3) 2 2

Gradient of QP = 1 and so the gradient of perpendicular bisector = −1. Cartesian equation is y + 3 = −1( x − 2) y = −x − 1 Im(z) 3 2 1 –1 O –1 –2 –3 –4 –5 Q

1

2

3

4 P

5

6

7 Re(z)

Chapter 11: Complex numbers

EXERCISE 11E 1 Describe, in words, the locus represented by each of the following. π a arg( z − 2 + 3i) = b z − 10 , z − 6i 12 5π z+6−i =7 d arg( z ) = c 12 2 On different Argand diagrams, sketch the following loci. Shade any regions where needed. π b arg( z + 3i) = − a z − (2 + i) = z − (4 + 3i) 3 z + 4 − 5i ø 2 d z − 2i ø z − 4 c PS

3 On a single Argand diagram, sketch the loci z = 4 and z + 2 = z − 4 . Hence determine complex numbers that satisfy both loci, giving your answers in Cartesian form. 4 On an Argand diagram, sketch the locus z = z + 8i . Find the Cartesian equation of this locus. 5 Sketch the locus z − (3 − 6i) = 3 on an Argand diagram. Write down the Cartesian equation of this locus. 5π and z − 5i = 4 on an Argand diagram. Determine whether or not there 6 is a complex number, z, that satisfies both loci.

PS

6 Sketch the loci arg( z + 4 + 2i) =

PS

7 For complex numbers z satisfying z − 8 − 16i = 2 5 , find the least possible value of z and the greatest possible value of z . 295

PS

PS

8 By sketching the appropriate loci on an Argand diagram, find the value of z that satisfies z = 13 and π arg( z − 12) = . 2 π 9 On a single Argand diagram, sketch the loci z + 3 − 2i = 4 and arg( z + 1) = . Hence determine the value 4 of z that satisfies both loci, giving your answer in Cartesian form. 10 The complex number z is represented by the point P ( x, y ) on an Argand diagram. It is given that the locus of P is z − 5 − 5i = 5 . a Write down the Cartesian equation of the locus of P. b Sketch an Argand diagram to show the locus of P. c

Write down the greatest possible value of arg z and the least possible value of arg z.

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Checklist of learning and understanding i2 = −1 Cartesian form: x + iy where x and y are real values.

Arithmetic operations on z1 = a + bi and z 2 = c + d i : ●

Addition:

z1 + z2 = ( a + c ) + ( b + d )i

●

Subtraction:

z1 − z2 = ( a − c ) + ( b − d )i

●

Multiplication:

z1z2 = ( ac − bd ) + ( ad + bc )i

●

Division:

z1 ( ac + bd ) + ( bc − ad )i = z2 c2 − d 2

●

Modulus:

z =

●

Argument:

x 2 + y2

Found using a diagram with tan θ =

y x

z1z2 = r1r2 = z1 z2 and arg( z1z2 ) = θ1 + θ 2 = arg z1 + arg z2, z1 z1 r = 1 = z2 r2 z2

z and arg 1 = θ1 − θ 2 = arg z1 − arg z2 z2

Polar forms

296

●

Modulus-argument form: r(cos θ + i sin θ )

●

Exponential form:

re iθ

Roots of equations occur in complex conjugate pairs ●

Quadratic equations have

2 real

or

2 complex roots of the form x ± iy with y ≠ 0.

●

Cubic equations have

3 real

or

1 real and 2 complex roots of the form x ± iy with y ≠ 0.

●

Quartic equations have

4 real

or

2 real and 2 complex roots of the form x ± iy with y ≠ 0,

or

4 complex roots of the form x ± iy with y ≠ 0.

The cube roots of one are: z = 1, z =

−1 + i 3 −1 − i 3 ,z= 2 2

Chapter 11: Complex numbers

END-OF-CHAPTER REVIEW EXERCISE 11 1

a b c

2

a

b

3

[2]

On a sketch of an Argand diagram, shade the region whose points represent complex numbers satisfying the inequality z + 1 − 5i ø 2.

[3]

The complex number z is defined as z = k − 6i, where k is a real value. z Find and simplify expressions, in terms of k, for zz * and , giving your answers in the z* form x + iy where x and y are real. 5π 5π and w = 2e iπ. + i sin The complex numbers u and w are defined as u = 4 cos 12 12 u Find and simplify expressions for uw and , giving your answers in the form re iθ , w where r . 0 and −π , θ ø π.

b

Represent w and w* by points P and Q on an Argand diagram with origin O and describe the polygon OPQ. u Given also that u = −3 − i, write the complex number in the form r( cos θ + i sin θ ), w where r . 0 and −π , θ ø π.

a

Find the real values x and y such that z* = (2 x + 1) + (4x + y )i.

b

On an Argand diagram, show the points A, B and C representing the complex numbers z, z * and −z. What type of triangle is ABC ? z* Without using a calculator, express −z i in the form x + iy where x and y are real ii in the form r (cos θ + i sin θ ), where r . 0 and −π , θ ø π.

[5]

[2] [4]

[2] [3]

[3] [3]

z 2 + (4 3 ) z + 13 = 0 a

Find the roots of this equation, giving your answers in the form x + iy where x and y are real.

[2]

b

OB representing On an Argand diagram with origin O, show the position vectors OA and the roots of the equation. Describe the geometrical relationship between OA and OB.

[2]

Find the modulus and argument of each root.

[3]

c 6

[3]

z = 2 − 5i

c

5

[3]

The complex number w = 1 + 2i. a

4

5 − 2i in the form x + iy where x and y are real numbers. 1 + 3i Solve w2 − 2w + 26 = 0.

Express

z = 4 3 − 4i a b

Find the exact values of the modulus and argument of z. z π π + i sin , write in the form re iθ , where r . 0 and −π , θ ø π. Given that w = 2 2 cos 12 12 w

[3] [3]

297

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

7

a b

8

a

Find the complex number w satisfying the equation w * −2 − 2i = 3iw. Give your answer in the form x + iy where x and y are real. π i On a single Argand diagram, sketch the loci z − 3 − 3i = 2 and arg( z − 3 − 3i) = . 3 ii Hence determine the value of z that satisfies both loci, giving your answer in the form x + iy , where x and y are real.

i

Show that z − 3 is a factor of 2 z 3 − 4 z 2 − 5 z − 3. 3

9

a b

10 a

2

11 a

[2]

Solve (3z + 1)3 = −27.

[3]

It is given that w = 1 is a root of the equation f( w ) = 2w 4 + 5w3 − 2w2 + w − 6 = 0 . Show that w + 3 is a factor of 2w 4 + 5w3 − 2w2 + w − 6.

i

On an Argand diagram, sketch the locus z + 1 − i 3 = 1 .

[4] [2] [1]

iii Find the maximum value of arg z.

[2]

i

3 7 + i is a root of the equation z 2 + pz + q = 0 , where p and q 2 2 are real constants, find the value of p and the value of q.

Given that z1 = −

i

Find the roots of the equation z 3 + 1 = 0 .

[3] [1] [3] [2]

5−i

2 5 z = − i. 3 z* 3 z z b Find the value of . and arg z* z * z and its conjugate. c Find and simplify a quadratic equation with integer coefficients that has roots z* k − 4i where k is an integer. 13 The complex number z is defined by z = 2k − i 7 a The imaginary part of z is Im z = . Find the value of k. 5 b Find the argument of z. a

[1]

ii Write down the minimum value of arg z.

ii On an Argand diagram, show the points A, B and C representing the roots of the equation. What type of triangle is ABC ? 12 z =

[1]

Given that z1 = 5 − 3i and z1z2 = 21 + i , find z2 , giving your answer in the form x + iy , where x and y are real.

ii Find z1 . b

[5]

[4]

ii Solve the equation. b

[2]

ii Solve 2 z − 4 z − 5 z − 3 = 0.

i

298

[4]

( x + iy )2 = 7 − (6 2 )i Given that x and y are real numbers, find the values of x and the values of y.

b

[5]

Show that

[3] [2] [2]

[4] [3]

Chapter 11: Complex numbers

14 i

ii

Without using a calculator, solve the equation 3w + 2iw* = 17 + 8i, where w* denotes the complex conjugate of w. Give your answer in the form a + bi. In an Argand diagram, the loci 1 π and z − 3 = z − 3i 6 intersect at the point P. Express the complex number represented by P in the form re iθ , giving the exact value of θ and the value of r correct to 3 significant figures.

[4]

arg( z − 2i) =

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2013 15 Throughout this question the use of a calculator is not permitted. i

The complex numbers u and v satisfy the equations u + 2v = 2i and iu + v = 3.

ii

Solve the equations for u and v, giving both answers in the form x + iy, where x and y are real.

[5]

On an Argand diagram, sketch the locus representing complex numbers z satisfying z + i = 1 3 and the locus representing complex numbers w satisfying arg(w − 2) = π. Find the least value 4 of z − w for points on these loci.

[5]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 November 2013 299

16 Throughout this question the use of a calculator is not permitted. The complex numbers w and z satisfy the relation w=

z+i . iz + 2

i

Given that z = 1 + i, find w, giving your answer in the form x + iy, where x and y are real.

[4]

ii

Given instead that w = z and the real part of z is negative, find z, giving your answer in the form x + iy, where x and y are real.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 November 2014

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

CROSS-TOPIC REVIEW EXERCISE 4 2 P3

This exercise is for Pure Mathematics 3 students only. 1

Relative to the origin O, the position vectors of the points A and B are given by −5 OA = 0

3

1 and OB = 7 . 2

a

Find a vector equation of the line AB.

b

The line AB is perpendicular to the line L with vector equation: m 4 r = 2 +µ 3 −3 9 i

.

Find the value of m.

[3]

ii Show that the line AB and the line L do not intersect. 2

[4]

Given that y = 0 when x = 1, solve the differential equation dy = y2 + 4, dx

xy

obtaining an expression for y2 in terms of x.

300

[3]

[6]

Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2010 3

The point P ( p, q, −1) lies on the line L with vector equation r = i − j + 2 k + λ ( i + 8 j + k ). a b

Find the value of each of the constants p and q.

[2]

Find the unit vector in the direction OQ.

[2]

The position vector of Q, relative to the origin O, is OQ = −10 i + j − 5 k. i

ii Find angle POQ and hence find the exact area of triangle POQ. 4

[5]

The complex number z is given by z= i ii

( 3 ) + i.

Find the modulus and argument of z.

[2]

The complex conjugate of z is denoted by z *. Showing your working, express in the form x + iy , where x and y are real, iz * b . [4] a 2 z + z *, z iii On a sketch of an Argand diagram with origin O, show the points A and B representing 1 [3] the complex numbers z and i z * respectively. Prove that angle AOB = π . 6 Cambridge International A Level Mathematics 9709 Paper 31 Q6 November 2010

Cross-topic review exercise 4

5

The variables x and θ are related by the differential equation sin 2θ

dx = ( x + 1) cos 2θ , dθ

1 1 where 0 , θ , π . When θ = π, x = 0. Solve the differential equation, obtaining an 2 12 expression for x in terms of θ , and simplifying your answer as far as possible.

[7]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 November 2011 6

PQRS is a parallelogram. The vertices, P, Q and R have position vectors, relative to an origin O, 3 1 2 OP = 4 , OQ = 2 and OR = −1 . 5 0 5 a Find OS.

[2]

b

Find the lengths of the sides of the parallelogram.

[3]

c

Find the interior angles of the parallelogram.

[3]

(1 + 2i)2 . 2+i Without using a calculator and showing your working, express u in the form x + iy, where x and y are real.

[4]

Sketch an Argand diagram showing the locus of the complex number z such that z − u = u .

[3]

7

The complex number u is defined by u = i ii

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2012 8

The complex number 2 + 2i is denoted by u. i

Find the modulus and argument of u.

[2]

ii

Sketch an Argand diagram showing the points representing the complex numbers 1, i and u. Shade the region whose points represent the complex numbers z which satisfy both the inequalities | z − 1 | ø | z − i | and | z − u | ø 1.

[4]

iii Using your diagram, calculate the value of | z | for the point in this region for which arg z is least.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2010 9

A certain substance is formed in a chemical reaction. The mass of substance formed t seconds after the start of the reaction is x grams. At any time the rate of formation of the substance dx is proportional to (20 − x ). When t = 0, x = 0 and = 1. dt i

Show that x and t satisfy the differential equation

ii

dx = 0.05(20 − x ) . dt Find, in any form, the solution of this differential equation.

[2] [5]

iii Find x when t = 10, giving your answer correct to 1 decimal place.

[2]

iv State what happens to the value of x as t becomes very large.

[1]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2010

301

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

10

The line L1 has vector equation r = −2 j + k + λ ( i + 3k ). The line L2 passes through the point P ( −2, 1, −1) and is parallel to L1.

11

a

Write down a vector equation of the line L2 .

[2]

b

Find the shortest distance from P to the line L1.

[5]

In a certain country the government charges tax on each litre of petrol sold to motorists. The revenue per year is R million dollars when the rate of tax is x dollars per litre. The variation of R with x is modelled by the differential equation dR 1 = R − 0.57 , dx x where R and x are taken to be continuous variables. When x = 0.5, R = 16.8. i

Solve the differential equation and obtain an expression for R in terms of x.

[6]

ii

This model predicts that R cannot exceed a certain amount. Find this maximum value of R.

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 November 2014 12 302

i

9 3 + 9i . Find, showing all your working, 3−i an expression for z in the form re iθ , where r . 0 and −π , θ ø π,

[5]

ii

the two square roots of z, giving your answers in the form re iθ , where r . 0 and −π , θ ø π.

[3]

The complex number z is defined by z =

Cambridge International A Level Mathematics 9709 Paper 31 Q5 June 2014 13

The variables x and y are related by the differential equation dy 6 ye 3 x = . dx 2 + e3x Given that y = 36 when x = 0, find an expression for y in terms of x.

[6]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2014 14

a

Show that the straight line L with vector equation 1 r= 2+λ −10

b 15

8 −1 3

intersects with the line through the points A and B with coordinates (0, 2, 7) and (7, 1, 27), respectively and find the position vector of this point of intersection.

[5]

Find the acute angle between these two lines.

[4]

dy = y(1 − 2 x 2 ) , and it is given dx that y = 2 when x = 1. Solve the differential equation and obtain an expression for y in terms of x in a form not involving logarithms. The variables x and y satisfy the differential equation x

[6]

Cambridge International A Level Mathematics 9709 Paper 31 Q4 June 2016

Cross-topic review exercise 4

16

6 − 3i . 1 + 2i 1 i Showing all your working, find the modulus of u and show that the argument of u is − π. 2 1 ii For complex numbers z satisfying arg( z − u ) = π, find the least possible value of z . 4 iii For complex numbers z satisfying z − (1 + i)u = 1, find the greatest possible value of z . The complex number u is defined by u =

[4] [3] [3]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2011 17

With respect to the origin O, the position vectors of the points A, B, C and D are given by 3 2 OA = 1 , OB = m , OC = 5

18

4

1 2 −5 and OD = −4 −2 m m + 11

.

a

In the case where ABC is a right angle, find the possible values of the constant m.

[3]

b

In the case where D is the midpoint of the line BC, find the value of the constant m.

[2]

c

In the case where m = −5, find whether the lines AB and CD intersect.

[5]

The complex number w is defined by w =

22 + 4i . (2 − i)2

Without using a calculator, show that w = 2 + 4i. 1 3 It is given that p is a real number such that π ø arg(w + p ) ø π . Find the set of 4 4 possible values of p.

[3]

iii The complex conjugate of w is denoted by w *. The complex numbers w and w * are represented in an Argand diagram by the points S and T respectively. Find, in the form z − a = k, the equation of the circle passing through S, T and the origin.

[3]

i ii

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q8 June 2015 19

Given that y = 1 when x = 0, solve the differential equation dy = 4x(3 y2 + 10 y + 3), dx obtaining an expression for y in terms of x.

[9]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2015 20

60º

h

C

A tank containing water is in the form of a cone with vertex C . The axis is vertical and the semi-vertical angle is 60° , as shown in the diagram. At time t = 0, the tank is full and the depth of water is H . At this instant, a tap at C is opened and water begins to flow out. The volume of water in the tank decreases at a rate proportional to h where h is the depth of water at time t. The tank becomes empty when t = 60.

303

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

i

Show that h and t satisfy a differential equation of the form 3

− dh = − Ah 2 , dt

where A is a positive constant.

[4]

ii

[6] Solve the differential equation given in part i and obtain an expression for t in terms of h and H . 1 iii Find the time at which the depth reaches H . 2 1 [The volume V of a cone of vertical height h and base radius r is given by V = πr 2 h. ] [1] 3 Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2013 21

a b

22

Find, in the form r = a + λ b, a vector equation of the line AB where the points have coordinates A(2, 5, 7) and B (9, −1, −2).

[3]

Find the obtuse angle between the line AB and a line in the direction of i + 3 j + 2 k.

[4]

The complex number 3 − i is denoted by u. Its complex conjugate is denoted by u * . i

On an Argand diagram with origin O, show the points A, B and C representing the complex numbers u, u * and u * − u respectively. What type of quadrilateral is OABC ? u* in the form x + iy , ii Showing your working and without using a calculator, express u where x and y are real. u* , prove that iii By considering the argument of u 3 1 tan −1 = 2 tan −1 . 4 3

304

[4] [3]

[3]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2015 23

The complex number 1 + ( 2 )i is denoted by u. The polynomial x 4 + x 2 + 2 x + 6 is denoted by p( x ). i ii

Showing your working, verify that u is a root of the equation p( x ) = 0, and write down a second complex root of the equation. Find the other two roots of the equation p( x ) = 0.

[4] [6]

Cambridge International A Level Mathematics 9709 Paper 31 Q9 November 2012

Cross-topic review exercise 4

24

G

F

N h cm D

E C

k O

B M

j

4 cm i

A

10 cm

The diagram shows a cuboid OABCDEFG with a horizontal base OABC. The cuboid has a length OA of 10 cm, a width AB of 4 cm and a height BF of h cm. The point M is the midpoint of CB and the point N is the point on DG such that ON = j + 3k. The unit vectors i, j and k are parallel to OA, OC and OD, respectively.

a

25

Write down the value of h.

[1]

b

Find the unit vector in the direction ON .

[2]

c

Find angle NME .

[4]

d

Find, in the form r = a + λ b , a vector equation of the line MF .

[3]

The variables x and y are related by the differential equation dy 6xe3x . = dx y2

305

It is given that y = 2 when x = 0. Solve the differential equation and hence find the value of y when x = 0.5, giving your answer correct to 2 decimal places.

[8]

Cambridge International A Level Mathematics 9709 Paper 31 Q7 June 2012 26

i ii

Showing all your working and without the use of a calculator, find the square roots of the complex number 7 − (6 2 )i. Give your answers in the form x + iy, where x and y are real and exact. a On an Argand diagram, sketch the loci of points representing complex numbers w 3 and z such that w − 1 − 2i = 1 and arg( z − 1) = π. 4 b Calculate the least value of w − z for points on these loci.

[5] [4] [2]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2016 27

−2 The line L1 has vector equation r = 3 + λ 0

1 −1 . 2

−4 −1 The line L2 has vector equation r = 5 + µ 0 . 1 m a

In the case where m = 2, show that L1 and L2 do not intersect.

[4]

b

Find the value of m in the case where L1 and L2 intersect.

[2]

c

For your value of m from part b, find the acute angle between L1 and L2 .

[4]

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

28

The number of birds of a certain species in a forested region is recorded over several years. At time t years, the number of birds is N, where N is treated as a continuous variable. The variation in the number of birds is modelled by dN N (1800 − N ) = . dt 3600 It is given that N = 300 when t = 0. i

Find an expression for N in terms of t.

[9]

ii

According to the model, how many birds will there be after a long time?

[1]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2011 29

i ii

Showing your working, find the square roots of the complex number 1 − (2 6 )i . Give your answers in the form x + iy, where x and y are exact.

[5]

On a sketch of an Argand diagram, shade the region whose points represent the complex numbers z which satisfy the inequality z − 3i ø 2. Find the greatest value of arg z for points in this region.

[4]

Cambridge International A Level Mathematics 9709 Paper 31 Q10 November 2011 30 306

Liquid is flowing into a small tank which has a leak. Initially the tank is empty and t minutes later, the volume of liquid in the tank is V cm3. The liquid is flowing into the tank at a constant rate of 80 cm3 per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to kV cm3 per minute where k is a positive constant. i

ii

Write down a differential equation describing this situation and solve it to show that 1 V = (80 − 80e − kt ). k 4 − 4e −15 k . It is observed that V = 500 when t = 15, so that k satisfies the equation k = 25

[7]

Use an iterative formula, based on this equation, to find the value of k correct to 2 significant figures. Use an initial value of k = 0.1 and show the result of each iteration to 4 significant figures. [3] iii Determine how much liquid there is in the tank 20 minutes after the liquid started flowing, and state what happens to the volume of liquid in the tank after a long time. [2] Cambridge International A Level Mathematics 9709 Paper 31 Q10 June 2013

Pure Mathematics 2 Practice exam-style paper

PURE MATHEMATICS 2 PRACTICE EXAM-STYLE PAPER Time allowed is 1 hour 15 minutes (50 marks) cos x d show that (cot x ) = − cosec 2 x. sin x dx

1

By writing cot x as

2

Solve the inequality 3x − 1 ù 2 x . 2x

−3

x +1

[4]

= 10 giving the value of x correct to 3 significant figures.

3

Solve the equation 3

4

The polynomial x 3 + 8x 2 + px − 25 leaves a remainder of R when divided by x − 1 and a remainder of −R when divided by x + 2.

5

[3]

[5]

a Find the value of p.

[4]

b Hence, find the remainder when the polynomial is divided by x + 3.

[2]

The sequence of values given by the formula xn +1 =

8 xn 2 , 3sec xn

with initial value x1 = 1, converges to α .

6

7

8

a Use this formula to calculate α correct to 2 decimal places, showing the result of each iteration to 4 decimal places.

[3]

b State an equation satisfied by α and hence find the value of α correct to 7 decimal places.

[3]

The parametric equations of a curve are x = ln(2t + 1), y = t − e2t . dy in terms of t. a Find an expression for dx b Find the equation of the normal to the curve at the point where t = 0.

[4]

307

[4]

a Express 8 sin θ + 6 cos θ in the form R sin( θ + α ) , where R . 0 and 0° , α , 90°, giving the value of α correct to 2 decimal places.

[3]

b Hence solve the equation 8 sin θ + 6 cos θ = 7 giving all solutions in the interval 0° , θ , 360°.

[4]

c Write down the greatest value of 8 sin θ + 6 cos θ + 3 as θ varies.

[1]

a Show that cos 3x ≡ 4 cos3 x − 3 cos x.

[5]

b Hence show that

∫

π 2

0

cos3 x d x =

2 . 3

[5]

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

PURE MATHEMATICS 3 PRACTICE EXAM-STYLE PAPER Time allowed is 1 hour 50 minutes (75 marks) 1 2

Use logarithms to solve the equation e2 x = 2 x + 5 , giving your answers correct to 3 decimal places. 5

25 ( ln 25 − 1 ). 4 The polynomial 2 x 3 − 9x 2 + ax + b, where a and b are constants, is denoted by f( x ). It is given that x − 4 is a factor of f( x ) , and that when f( x ) is divided by ( x − 1) the remainder is −12 . Find the value of a and the value of b. Show that

∫

x ln x d x =

[3] [5]

0

3

4

Solve the equation 2 sin( x − 60°) = 3 cos x for −180° ø x ø 180°.

[5] [5]

y

5

P

O

x

Q

The diagram shows the curve y2 = 4x 2 − x 4 + 5. The curve is symmetrical about both axes. The point P is one of the curve’s maximum points and the point Q is one of the curve’s minimum points. Find the exact distance between the points P and Q.

308

6

f( x ) =

7 x − 5x + 27 ( x − 2) x 2 + 5

(

)

a Express f( x ) in partial fractions.

[5]

b Hence obtain the expansion of f( x ) in ascending powers of x, up to and including the term in x 2 .

[5]

7

y

π 2

O

The diagram shows the curve y = x sin 2 x for 0 ø x ø

x

π . 2

dy dy d2 y and show that x 2 − 2x + 2 1 + 2 x 2 y = 0. dx dx dx 2 b Find the area of the region enclosed by this part of the curve and the x-axis.

(

a Find

8

[7]

2

a

)

[5] [5]

The complex numbers u and w are such that: 3u − iw = 15 u + w = 5 + 10i. Without using a calculator, solve these equations to find u and w, giving each answer in the form x + iy, where x and y are real. −

[5]

π i 4

b The complex number z is defined by z = 2 e . On an Argand diagram, show the points A, B and C representing the complex numbers z, z * and z 2 , respectively and find the length of the longest side of triangle ABC .

[4]

Pure Mathematics 3 Practice exam-style paper

9

a Given the vectors 5 i + 7 j + pk and i − 2 j + pk are perpendicular, find the possible values of the constant p.

[3]

b The line L1 passes through the point (5, 0, 2) and is parallel to the vector i − 6 j + k. i

Write down a vector equation of the line L1.

ii

The line L2 has vector equation:

[2]

r = 4 j + 3k + µ( 3i + 9 j + k ) Show that L1 and L2 do not intersect.

[4]

10 A liquid is heated so that its temperature, x °C , at time t seconds satisfies the differential equation dx = α (100 − x ) dt where α is a positive constant. The temperature at t = 0 is 25 °C. 75 = α t. a Show that ln 100 − x

[6]

b It is given that x = 500α when t = 2 . Show that α = 0.2 − 0.15e −2α .

[2]

c Use an iterative formula based on the equation in part b to find the value of α correct to 2 significant figures. Use a starting value of 0.1 and show the result of each iteration to 4 significant figures.

[3]

d Find the temperature of the liquid when t = 30 .

[1] 309

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Answers

8

x=±

1 Algebra

9

x = 7, y = 5

Prerequisite knowledge

Exercise 1B

1

a 357

1

b 381

c 133 remainder 27 2

2

3

310

4

5 2 d −15, 3

b −2, 3

c −6,

e −20, 40

f

9 11 , 7 3 d 1, 3

b −1, 3

c

e 2

f

1 3 6 d −4, − 5

b −3,

a −1,

a

a −1, −

3 5

x − 3 if x ù 3 y= 3 − x if x , 3 c ∨-shaped graph, vertex = (10, 0), y-intercept = 5

22 3

7

1 x − 5 if x ù 10 2 y= 5 − 1 x if x , 10 2

7 23 , 4 6 −7, 3

c 2, 4

e −7, 1

f

0,

12 13

2

1 1 −1 − 33 ) , ( 2 2

(

d

3 − 1, 1 + 5

e 1, 2, 1 − 3, 1 + 3 f

7

3

0, 2, 6

8 16 a x = 0, y = 4 or x = − , y = 3 3 5 15 b x = −1, y = 3 or x = − , y = 2 2 4 6 x = or x = 5 5 a x = ±2, x = ±3 b

x

0

1

2

3

4

5

6

y

5

4

3

2

3

4

5

b ∨-shaped graph, vertex = (3, 2), y-intercept = 5 3 c Translation 2

33 − 1 )

c −2, ±1

6

a

a ±3 b −1, 2,

5

a ∨-shaped graph, vertex = ( −2, 0), y-intercept = 2 x + 2 if x ù − 2 y= −( x + 2) if x , − 2 b ∨-shaped graph, vertex = (3, 0) , y-intercept = 3

5 Straight line, gradient 2, crossing axes at , 0 2 and (0, −5).

Exercise 1A 1

3 4

y 6

4

5 −1 a Translation b Translation 2 −2 0 c Reflection in x-axis, translation 2 d Stretch, stretch factor 2, with y = 0 invariant, 0 translation . −3 −2 . e Reflection in x-axis, translation 1 f Stretch, stretch factor 2, with y = 0 invariant, 0 reflection in x-axis, translation 5 a ∨-shaped graph, vertex = ( −1, 2) b ∨-shaped graph, vertex = (5, −2) c ∧-shaped graph, vertex = (0, 2) d ∨-shaped graph, vertex = (0, −3) e ∧-shaped graph, vertex = ( −2, 1)

–6

c x=0

–4

–2

O

2

4

6 x

f 5

∧-shaped graph, vertex = (0, 5)

3 ø f( x ) ø 14

Answers

6

a, b

y

3

8

6

4

(0, 5) 4

5

2 (2, 1) –2 –1 O

1

2

3

4

5

6 x

c x = 1 or x = 5 6 7

a, b

x,−

y 4

2

(0, 2)

1

1 (2, 0) –3 –2 –1 O

1

2

3

4

5

6

7

1 3

b x , 2 or x .

2

5 5 or x . 4 4

b x 2 − 5x + 7

c 3x 2 − 4x + 2

d x2 − 2x + 5

−6x 3 − 6x 2 − 6x − 19

a Quotient = x 2 + x + 4, remainder = −8 b Quotient = 6x 2 + 19x + 38, remainder = 70 5 1 5 c Quotient = 4x 2 + x + , remainder = 2 4 4 d Quotient = −2 x 2 + 3x + 9, remainder = −36

6 5 4 3

e Quotient = x + 3 , remainder = −10 x − 4

2

f

1 –3 –2 –1 O

1

2

3 x

b x=±2

Quotient = 5x 2 − 7 , remainder = 15 − 13x

3

a Proof

4

a Proof

b Proof

b ( x − 2)(2 x − 3)( x + 7)

Exercise 1C 1

3,x,7

2

a

8 3

d 3,x,7 9 f 1 , x , 5

a x 2 + 3x − 1

f y

y 4

y = |2x – 1|

3

y = 4 – |x – 1|

a Proof

b ±

6

a Proof

b Proof

7

−3,

1 , 5 2

2

Exercise 1E

1

1

Proof

2

Proof

3

a = −4 1− b a= 3

–4 –3 –2 –1 O –1

1

2 or x . 2 3

2

3

4

5 x

4 1 , − 3 2

5

5

b x,−

b x.

e −5x 2 + 3x − 4

8 x

c x = ±1 a

b −2 , x , 6

Exercise 1D

3

8

a −2 ø x ø 1 5 c x ø − or x ù 3 3 7 øxø3 a 3 c x ù1 4 or x . 4 a x,− 3 1 c x ù 2 e x ø 2 or x ù 8

4

311

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

5

a = −2, b = 1

6

a

7

−4, 0, 1

8

a

p = −1, q = −6

b Proof

4

±5, ± 3

5

a a=6 b (3x + 1)(2 x + 3)( x − 4)

6

p = −7, q = −6

7

1 , ±5 3 a = −10, b = 8

8

a Quotient = x − 3, remainder = 2 x − 6

b ( x + 1)( x + 2)( x − 3) and ( x + 1)( x + 3)( x + 4) 9

a

p = −19, q = 30

b ( x − 1)( x + 2)( x − 5)( x + 3) 10

a ±2, 5 5 c − , 2, 3 2 e −3, ±2, 1

11

Proof

12

0, k ,5

b −7, − 1, 3 1 d −4, −2, 3 1 f − , 1, 4 2

312

c 4

b ±

b Proof 9

Exercise 1F b 8

a (3x − 1)(2 x + 3)( x − 5)

d −

31 4

a Quotient = 4x 2 + 4x − 3, remainder = 5

3 b − , −1, 2 10 a k = 7 1 b ( −7 − 2 1 − 3 , (7 + 2 11 a a = 6

1

a 6

2

a a=5

3

a = 2, b = 2

12

a a=7

4

a = 4, b = 0

13

a 33

5

a a = 6, b = −14

b 57

a

7

9

a a = 5, b = −15 −7 + 53 −7 − 53 b 2, , 2 2 5 b −36 a k= 2 2550

10

a = 2, b = −4, c = −2.

2 3

4 2 , 3 7 4 x ø − or x ù 2 5 5 1, x , 3 −

53 )

14

a a = 3, b = −10

15

a a = −21

16

a a = −3, b = −11

b −5,

2 , 2 3

b 3x − 4 5 1 b − , − , 2 3 2

b (2 x − 1)( x + 2)( x − 3) 17

a a = 2, b = −16

18

a Quotient = 5x 2 − 8x + 9

b −18

b Proof 19

a k = −29 1 c ±3, ± 2

20

a a = 3, b = 50

End-of-chapter review exercise 1 1

1 1 −7 + 41 ) , ( 7 − 53 ) , ( 2 2

b Quotient = 2 x + 13 , remainder = 41x − 15

6

8

41 ) ,

b (2 x − 1)( x + 3)( x − 2)( x + 1)

b b=8

b (3x − 4)(2 x + 1)( x + 2) p = 14

1 ,1 2

b −2,

1 , 9 4

b Quotient = 2 x − 3, remainder = 56 − 4x 21

a a = −11, b = 30

22

i

x + 2, 3x + 4

b −2,

3 , 5 2

ii Proof

+

Answers

23

i

−16

24

i

a = 2, b = − 6

ii −

1 3 , ,3 2 2

e −2 6

ii (2 x + 3)( x − 1)( x − 3)

a 3 e −6

2 Logarithmic and exponential functions Prerequisite knowledge 1 2 3

a

1 25

b 4

1 5 7 − a 2x 2

b

1 5

15

c

8x 27

a 2 = log10 100

b x = log10 200

c x = log10 0.05 2

a 1.72

8

a

b x = 10

4

a 75.9

b 575

7

10 10

9 2

1 2 log 4 3, log2 2, log 3 4, log 2 3, log 3 9, log 3 20, log 2 8 b −3,

3

f

2

a 3

b 2 log6 10

c 2 + log2 3

3

a log5 4 or 2 log5 2

6 c 1.5

7

0.5

8

f

9

b log 3 4 or 2 log3 2 3 23 , 2 −2 , − 2 7 a 3 b 4 x−2 y= x2 1 z= 1 − 3 y3 a 5y 3 c y−3 2 a 3+x c 4x + y

a 2 = log5 25

b 4 = log2 16

1 243 e x = log8 15

d −10 = log 2

g b = log a c

h 5 y = log x 7

c −5 = log3

a 2 =8

f

4

=4

e

g a0 = 1

1 83

10 1 1024

y = log x 6

b 3 = 81

3

d

h

e log 2 18

Exercise 2B

1 16 2

d −4

d log 3 2

5

−0.6

6

2

5 2 2 g 3 c

c log5 2

1.2

a 2 b −4 1 d e 21 3 3 −1 f ( x ) = log10 ( x + 3)

1

4 3

h −

b log 6 5

c 0.0398 5

1 2

a log 2 77

1

4

a 10 000 = 10 c x = 10

4

f −1( x ) = 3 + 2 x −1

b 2.40

c −0.319 3

g −

Exercise 2C 5 x2

Exercise 2A 1

7

9

c 1

3 2 1 b 3 5 f 2 f

=2

a 8

b 9

c 1

4

a 4

b 11

5

a 3

b 2

c 3 1 c 2

11

24 = y

d 7

d −3

d −

3 2 313

b 2+y 3 d y 1 b −y 2 1 d 2x − y − 1 2 9 b − 2 5 d 3

x = 2 log3 2 − 2 log2 5, y = log 2 5 − 2 log 3 2, z = log 2 5 − log 3 2

c 8 =1

h xy = 5

3

c 16

0

f

a 3

c −1

log 4 8

Exercise 2D 1

a 10

2

a

3

a 5

41 2

b 3.5 b

9 7

b 10

46 9 3 c − 10 c −

c 3, 6

d 54 3 2 1 d 4 d

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

4

a 8

5

a 8, 32 c

6

b 12

5 3

d 4

b 0.1, 1000

1 , 25 3125

1

1 , y = 1 2 25 625 d x= , y = 2 4

a x = 3, y = 27

Exercise 2F

2 2

d 2 2,

2

log10 x = 3, log10 y = −2

a 1.80

b 5.13

c 0.946

d 3.64

e 3.86

f 1.71

g 0.397

h 0.682

i

−0.756

k −0.443

l

−15.4

j

314

6.76

2

a Proof

b 1.26

3

a −0.322

b 1.11

c 1.83

d 1.03

e 3.21

f

0.535

4

2, 3

5

a 0.431, 0.683

b 0, 2.32

c 0, 1.77

d 0.792, 1.58

2, 2.58

7

a 2.81

b 1.46

c 0.431, 1.29

d 0.792, 0.161

a 1.58

b −0.792, 0.792

c 1.58, 2.32

d 1.37

8

9

a 1.77

b 0.510

c 1.98

10

0, 20 0.8

11

a 1.29, 1.66

b 1.63

c 1.91, 2.91

d 0.834

e −0.515

f 1.87

g ±1.89

h ±2.81

12

a 13.5

b 1.45

13

0, 1, 2

log 3 2 log 3

b xù

log 7 log 5

d x.

log 0.3 log 0.8

1 log 8

1 log 20 − 5 2 log 3

3 1 log 2 c xø − 1 2 log 5 4 log 7 d x.3− 5 log 6 3

x , 0 or x .

4

Proof

5

a 61

log 2 log 5 b 1

Exercise 2G 1

6

log 5 log 2

a x.5 − b x.

Exercise 2E 1

a x, c x,

b x=

c x = 4, y = −12 7

c

2

a 20.1

b 14.9

c 2.23

d 0.135

a 1.10

b 0.336

c −0.105

d −1.90

3

a 2

b 3

c 30

4

a 5

b 15

c 4

5

a 2.89

b 1.61

c 1.08

6

a ln13 1 (1 + ln 6) c 2

7

a x . ln10 1 (2 + ln 35) 5 1 c x , − (3 + ln5) 2 b xø

1 ln 7 3 d 2[ln(4) − 3] b

d 2 1 d 3 d 2.89

Answers

8

9

10

11

a 148

b 0.0183

c 405

d −0.432 b 5.70

c 1.41

d 0.690

e 1.16

f 1.08

13

0.151

15

5

b ln 2, ln 3

Gradient =

7

a

ln m 4 3

−1

2 1

5 e 1 b x = 2 ln 2, y = − ln 2 5 a x = e, y =

−

O

30

40

50 t

a ln(T − 25) = − nt + ln k 315

b k = 45, n = 0.08 c i

=

ii

m = a, c = −b

70 °C 34 minutes

iii 25 °C

c ln y = −b ln x + ln a, Y = ln y, X = ln x,

=

End-of-chapter review exercise 2

m = −b, c = ln a d ln y = x ln b + ln a, Y = ln y, X = x,

l

1

x.

log 7 log 2

2

p=

q2 3+q

m = ln b, c = ln a e x = −by + ln a, Y = x , X = y, m = −b, 2

=

−b, c = ln a

a ln 8 f ln y = − ln x + , Y = ln y, X = ln x, m = b b ln8 a = n , m = − , c = b b g ln x = − y ln a + ln b, Y = ln x, X = y, = l = , m = − ln a, c = ln b h ln y = −bx + ln a, Y = ln y, X = x, = m = −b, c = ln a

20

c 35 days

b log y = ax − b, Y = log y, X = x, m = a,

2

10

b m0 = 50 , k = 0.02

1 ,xø3 2

a ln y = ax + b, Y = ln y, X = x, m = a, c = b

m

ln 3 ln 3 , y-intercept = 0, ln5 2 ln5

d ln 7

8

=

x3 e2

6

Exercise 2H 1

x

a ln y = 3 ln x − 2 b y=

a ln 3 5 c ln 2

3 x+2 2

b y = 100 × 10 2

1 a y= x−2 2 b y= 2 ex − 1

12

a log10 y =

3

a 1.22

x−2 f ( x ) = ln 5

14

4

=

7

3 log 2 x,− log 8 20.1 2 log 2 − log 3 − 1, log 2 4 3 K = 7.39, m = 1.37

,3 4 5 6

2

a = 66, n = −0.53

8

i

3

k = 9.5, n = 0.42

9

9.83

Proof

ii 0, 1.58

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

10

i

ii 3 y =

2 , y = −0.369 3

11

3.81

12

22.281

13

0.438 2 ln 3 a ln5 2 ln 3 b 1, − ln5 K = 1.73, b = 1.65

14

15

Exercise 3A

( x + 2)(4x + 3)(3x − 2)

1

3

4

y 3

5

2 1 O –1

316

b

c −

d 2

e

2

f

− 3

c

2

f

− 2

6 90

180

270

360 x

2 3

a 2

g

Prerequisite knowledge a

2

d −

3 Trigonometry 1

b

g − 2

1 3

1 3

h − 3 1 b 3 2 e − 3 2 h 3

a 70.5°, 289.5°

b 51.3°, 231.3°

c 199.5°, 340.5°

d 41.4°, 318.6°

π 5π , 6 6 c 0.464, 3.61 a

b π d 2.76, 5.90

a 28.2°, 61.8°

b 37.8°, 142.2°

c 22.5°, 112.5°

d 24.1°, 155.9°

a 60°, 180°

b 35.9°, 84.1°

c 2.82, 5.96 d −2.28, −1.44, 0.865, 1.71

y

7

a −150°, − 30°, 30°, 150° b −109.5°, − 70.5°, 70.5°, 109.5°

O 90

180

270

c −112.6°, 112.6°

360 x

d −180°, 0°, 180° e −135°, 45° f

c

y 1

8

−60°, 60°

a 48.2°, 180°, 311.8° b 31.0°, 153.4°, 211.0°, 333.4°

O 90

180

270

c 19.5°, 160.5°, 203.6°, 336.4°

360 x

d 60°, 180°, 300°

–1

2 3

3 2 a 31.0°, 211.0° a

e 107.6°, 252.4° b −

2 2

c −

3 3

f 9

27.2°, 152.8°

a 41.8°, 138.2° b 13.3°, 22.5°, 103.3°, 112.5°

b 30°, 150°, 270°

c 45°, 60°, 120°, 135° d 97.2°, 172.8° 10

1 3

a 2

a

2π 4π , π, 3 3

b

7 π 11π , 6 6

Answers

11

vi

y

a i

y

6

3 4 2 2 1 O –2

π – 2

π

3π – 2

2π x

O –1

π – 2

π

2π x

3π – 2

–4 –2 –6 –3

ii

y

π 5π 9π 13π , x= , x= , x= 8 8 8 8 a Proof b Proof b x=

12

c Proof O

π – 2

π

3π – 2

2π x

13

14 y

iii

1

4 2

–2

a Proof

b Proof

c Proof

d Proof

e Proof

f

g Proof

h Proof

a 48.2°, 180°

2 π – 2

π

3π – 2

2π x

a 1 d

3

–6 y 6 4 2 O –2

π – 2

π

3π – 2

2π x

2

O

π – 2

π

3π – 2

2π x

6+ 2 4

b 2+ 3

c

g −2 + 3

h

33 65 77 a 85

c f

e

a

2 2

e 1

d −2 + 3

6 7

4

3 2

5

–6 y 6

a

317

b

4+3 3 10 Proof

4

–4

v

b 45°, 63.4°, 161.6°

3 1 cos x − sin x 2 2

–4

iv

Proof

Exercise 3B

6

O

d Proof

6− 2 4 6+ 2 4

16 65 36 b 85 b

1 2 3 3

f

2− 6 4 6− 2 4

33 56 13 c 84 c −

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

b 2

12

t−2 2t + 1 1 2 1 a 3 3 a 2 a Proof

13

a 19.1°, 199.1°

b 70.9°, 250.9°

c 5.9°, 185.9°

8 9 10 11

14

8 9

b sin 3x ≡ 3sin x − 4 sin3 x 10

35.3°, 60°, 120°, 144.7°

11

a Proof

b 4

12

a Proof

b −

d 150°, 330°

13

a Proof

a 38.4°, 111.6°

b 18.4°, 116.6°

14

a Proof

c 16.0°

d 35.0°

e 18.4°, 26.6°

f

15

22.5°, 112.5°

16

Proof

17

p2 + q 2 − 2 2

b −0.2 b −30°, 150°

1 2

3

4

5 6

a sin56° b cos 68° 24 a b 25 24 c − d 7 336 a b 625 7 c d 25 24 a − b 25 24 c d 7 1 3 a 14.5°, 90°, 165.5°, 270°

c Proof d

b 60°, 300° c 48.6°, 131.4°, 270° 7

c tan 34°

5 −1 4

16

2π 4π ,θ , 3 3 210° ø θ ø 330°

17

Proof

18

a Proof

15

7 25 44 − 117 336 − 527 24 7 336 625

2π π π , − , , 3 3 3 5π π π b − , − , , 6 6 6

b 0° , x , 45° or 120° , x , 135°

−

19

45° , θ , 135° or 225° , θ , 315°

Exercise 3D a Proof

b Proof

c Proof

d Proof

e Proof

f

g Proof

h Proof

a Proof

b Proof

c Proof

d Proof

e Proof

f

g Proof

h Proof

3

a Proof

b Proof

4

Proof

5

Proof

6

Proof

1

2

2

Proof

Proof

a 30°, 150° b 33.6°, 180°

Exercise 3E

c 30°, 150°

1

a 17 sin(θ − 28.07°)

d 33.2°, 90°, 146.8°

2

a

3

a 17 sin(θ − 28.07°)

e 39.2°, 90°, 140.8° f

24.9°, 98.8°

2π 3 5π 6

b sin 3θ ≡ 3sin θ − 4 sin3 θ , cos 2θ ≡ 1 − 2 sin2 θ

74.1°

Exercise 3C 318

1 (1 + cos 4x ) 2 a Proof

13 cos(θ + 56.31°)

c 34

b 64.1°, 172° b 12.6°, 234.8° b 38.2°, 197.9°

Answers

4 5

a 2 13 sin( θ − 56.31 °) c 49, −3

b 80.9°

a 5 sin(θ + 53.13°)

b 103.3°, 330.4°

5

41.8°, 138.2°, 194.5°, 345.5°

6

a Proof

7

a Proof

c −2

b 61.3°, 118.7°, 241.3°, 298.7°

6

π a 2 cos θ − 3

7

a 4 5 sin(2θ + 26.57°)

8

b 66.9°, 176.5°, 246.9°, 356.5° 1 c 8 a 3 cos(θ + 54.74°) b 70.5°, 180°

9

b Proof

π 2 cos θ + 4

a

b

23π 12

i

9

a Proof

10

i

a Proof 2 b i 2 ii 15°, 75°

12

a Proof c 52

13

a 3 cos(θ − 41.81°)

319

a Proof 3 sin(θ + 60°)

13

a

14

a 10 + 5, 10 − 5

b 5 sin(θ + 53.13°)

14

b 84.7°, 335.3°

15

b 18.4°, 45°, 198.4°, 225°

a Proof a Proof

b 21.8°, 161.6°

17

a Proof

b 5 sin(2 x + 36.87°)

c 71.6°, 161.6°

Cross-topic review exercise 1

y

3

5

2 ,x,4 5 0.631, − 0.369 7 i x= 6 2 i −2, 5 A = 8.5, b = 1.6

6

i

1.77

7

i

n = 1.50, C = 6.00

3

O

4

ii Proof

16

2

3

Proof

b 60°, 104.5°, 255.5°, 300°

1

2

i

iii 0.322, 0.799, − 1.12

End-of-chapter review exercise 3

–3

R = 2 13, α = 56.31°

iii 60, 8

c 115.3°, 318.4°

1

i

ii 80.9°, 211.7°

b 41.8°

c 302.6° 12

R = 10, α = 18.43°

11

b 32.3°, 290.8°

c 1 + 10, 80.8° 11

b 35.8°, 125.8°

b 2 13 cos(θ − 56.31°)

10 sin(θ − 71.57°)

a

ii 18.4°, 26.6°

ii 34.6°, 163.8°, 214.6°, 343.8°

c − 2økø 2 10

Proof

8

1 6

c

b 30°, 330°

45

90

135

131.8° 15 − 8 0°, 131.8°, 228.2°, 360°

180 x

4

ii 0.222 ii −0.569

ii ±1.77

ii n ln x + ln y = ln C is linear in ln y and ln x. 8

−0.405, 1.39

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

9

θ = 135°, φ = 63.4° or θ = 53.1°, φ = 161.6°

10

i

11

i

2 4 − , 3 5 a = −16

12

i

a = −4, b = 6

ii −0.161 ii

3 1 , − ,3 2 2

ii quotient = 2 x − 4, remainder = −2 13

i

b

R = 10, α = 71.57°

i

x − 5 y = 16

4

(0, 2) maximum, (2, −2) minimum

Exercise 4A 1

c

3x + 4 2 x+2

e

x 2 (7 x − 3) 2x − 1

f

(13x 2 + 2)( x 2 + 2)2 2 x

iii 2

16

17 320

18

i

a = 6, b = −3

a = −17, b = 12 4 ii x = −3, x = , x = 1 3 i a = 1, b = −10

h 2(2 x − 1)4 (3x + 4)3 (27 x + 14)

2

−1.5

ii quotient = x − 1, x = 1, x = 2, x = −4

3

16x + y = 32

4

5

i

i

i

a = 19, b = −36

i

5

quotient = x 2 + 2 x + 1, remainder = 5x + 2

ii p = 7, q = 4

6

iii x = −1

7

2 3

ii −

i

21

i

22

ii 13.1°, 55.1°, 193.1°, 235.1° 1 iii 116 i a = 2, b = −5

3 −1, , 3 5 1 − 3 y

a

8

29 sin(2θ + 21.80°)

b 109.5°

4 Differentiation Prerequisite knowledge 6 1 a 15x 2 + 3 + x x b

2(3x 2 + 1)(15x 2 − 30 x + 1)

9 20

(1, 3)

5 4 1 x − 4x − 2 2 2x

a 12(3x − 5)3

y = (x – 1)2(5 – 2x) + 3 (2, 4)

O

x

b 2

ii a Proof

2

3x + 9 2 x+5

g ( x − 3)( x + 2)4 (7 x − 11)

20

1

d

ii ( x + 1)(4x + 1)(2 x − 1)

ii ( x + 2)( x + 3)(5x − 6), 0.113 19

b 5(2 x + 1)2 (8x + 1)

a (6x − 2)( x − 2)4

Proof

ii ( x − 2)(4x + 1)2

15

3

( 1 − 2x ) 2

3

ii 61.2°, 10.4° 14

4

Exercise 4B 1

11 ( x − 4)2

b

1 (2 − x )2

2( x 2 − x + 3) (2 x − 1)2

d

11 (2 − 5x )2

2(8x + 1) ( x + 4)3

f

−

a − c

e −

20 x 3 ( x 2 − 1)3

Answers

g

13x 2 + 30 x − 35 ( x 2 + 2 x + 5)2

h −

4

2( x + 4)(2 x 2 + 12 x − 1) ( x 2 + 1)4

a xe x + e x

b 3x 2 e 3x + 2 xe3x

c e −2 x (5 − 10 x )

d

e x (2 x + 1) x

f

−

3

( −6, −7), (1, 0)

e6 x (6x − 1) x2 3e x g x (e + 2)2

4

(2, 1), (8, −5)

i

5

y = 9x − 4

6

a

2

e

1 4

−5x − 1 2 x (5x − 1)2

c −

x+4

b

(2 x +

x( x + 1)

5( x − 1)2 (5x + 13)

2

d

3

3

8

3y = x + 7

9

a −3, 1, 5 1 5 b 9, − , − 3 3

6

−1, − 1 e

7

y = 3x + 3, ( −1, 0)

8

(3, − e3 ) minimum

9

(1, e2 ) minimum

10

a x = 0 minimum, x = 2 maximum b Proof

11

1 1 x = 1− , x = 1+ 2 2

b −4e −4 x

12

1 , 2 2

c 12e6 x

d −15e −5 x

13

Proof

14

3 + 3 ln 3

x 2e 2

g 2 xe x i

2

−3

5 ex + 2e −2 x 2

a

f

2e2 x− 7

h

3e x 2+ 2 x

j

6e3x

a

2

10 xe x − 10

l y=1 2

O

y = 1 – e2–x 1 – e2

b y = −x + 2 0.0283 grams per year

x

1 x

2 2x + 1 4 e 2x − 1 5 g x+3 2 i 5− 2x − 1 1 k x ( x − 2) c

y

321

Exercise 4D 1

k 3e2 x − e −2 x

3

4 9

a 5e5 x

e

2

2 x 2 e x + 5xe x + 2 xe2 x − e x − 2 (e x + 2)2

−

Exercise 4C 1

h 3xe3x + 3e6 x + e3x

5

3

2( x + 2) 2

( x 2 − 1) 2 7

3 3) 2

e −2 x (4x + 1) 2x x

b

1 x

2x x +1 1 f 2( x − 3) 1 h 3− x 1 j x ln x 1 + 5x l x(5x + ln x ) d

2

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2

1 1 1 − − x − 3 x + 4 x −1 2 1 h − − x +1 x − 2 1 2 1 1 i + − − x + 2 2x − 1 x x + 5

Possible justification: ln 3x = ln 3 + ln x and ln 7 x = ln 7 + ln x d d 1 (ln 3x ) = (ln 3 + ln x ) = 0 + and dx dx x

g

d d 1 (ln 7 x ) = (ln 7 + ln x ) = 0 + dx dx x 3

a 1 + ln x c e

2x + ln(2 x + 1) 2x + 1 1 + ln(ln x ) ln x

g −

4

b 2 x 2 (1 + 3 ln x ) d 3(1 + ln 2 x ) f

1 − ln5x x2

2 x(ln x )2

i

2(4x − 1) − 4(2 x + 1)ln(2 x + 1) (2 x + 1)(4x − 1)2 x=

2

c 12

1

3x − (3x − 2)ln(3x − 2) x 2 (3x − 2)

y

a

3 2

a cos x

b 2 cos x − 3sin x

c −2 sin x − sec 2 x

d 6 cos 2x

f

2 O 1

2

3

4

5 x

π −6 sin 3x − 6

π h 2 cos 2 x + 3

a 3sin2 x cos x

b −15 sin 6x

c 2 sin x(1 + cos x )

d 4 sin x (3 − cos x )3

–1

π π e 12 sin2 2 x + cos 2 x + 6 6

–2

f

2 7

3

π π −12 cos3x sin x + 8 tan 2 x − sec 2 2 x − 4 4

a x cos x + sin x b 5(cos 3x − 3x sin 3x )

5

−8

6

2 + 4 ln10, 3 + 2 ln10

c x 2 sec 2 x + 2 x tan x d cos2 2 x(cos 2 x − 6x sin 2 x )

1 1 e , − e , minimum 2

e 15 tan 3x sec 3x

f

x sec x − tan x x2

h

2

8

e, 1 , maximum e

g

9

y = 5x − 5

i

10

a

e

−2(3sin 3x + 2 cos 2 x )

g 3sec 2 (3x + 2)

y = 1n(2x – 3)

1

c

9x 2 + 2 3x 3 + 2 x

−5

i

7

b

e 20 sec 2 5x

322

b

4x 2x2 − 1 2x − 4 ( x + 1)( x − 5)

Exercise 4E

h

a

11

5 2(5x − 1) 1 5 + x x +1 3 2 − 3x − 1 x

3 3x + 2 2 1 d − 2x + 3 x − 1 1 1 1 f + − x x−2 x+4

j

b −

(3x − 1) cos x − 3sin x (3x − 1)2 −6 cot 2 x cosec 3 2 x

sec x( x tan x + 1) 1 + 2 cos x

( 2 + cos x )2

k 3(1 − 2 x cot 2 x ) cosec 2 x l 4

2 sin 2 x − 1

a cos x esin x

b −2 sin 2 xe cos 2 x

Answers

c 3sec 2 3 xetan 3x

dy dy dy + 12 xy3 f 2y +x +y dx dx dx dy dy g 3x 2 − 7 x − 7 y + 3 y2 dx dx dy dy + sin y − y sin x + cos x h x cos y dx dx x 3 dy i + 3x 2 ln y y dx dy j −2 x sin 2 y + cos 2 y dx dy dy + e x cos y + e x sin y k 5 dx dx dy l −2 x sin y e cos y + 2e cos y dx e 18x 2 y2

d (cos x + sin x )e(sin x − cos x ) e (cos x − sin x )e x f

(2 cos 2 x + sin 2 x )e x

g e x (cos x − 3sin x ) h x 2 (3 − x sin x )e cos x i

−tan x

j

x cot x + ln(sin x ) 2(sin 2 x + cos 2 x ) e 2 x +1 (1 − 2 x ) sin 2 x + 2 x cos 2 x e2 x

k − l 5

1

6

2 3−6

7

π π , 6 3

8

Proof

9

a tan x sec x

10

Proof

11

y = −13.3x + 12.9

12

0.464, 2.03

13

π x = , maximum 4

14

x=

15

π x= , minimum 12

17

b −cot x cosec x

π 8

π 5π 7π maximum, x = minimum, x = 6 6 6 11π minimum maximum, x = 6 0.452

Exercise 4F 1

2 y(e x y2 + 1) 3e x y 2 + 2 y4 g 2 − 3xy3 11 − 25

3

dy b 3x 2 + 4 y dx dy d cos y dx

b

5 − 2 xy x2 + 2 y

d

y(2 − ln y ) x

y 2 xy2 + x y(5xy + 1) h x(2 − 5xy ) f

−

323

4

3

5

y=

1 9 x− 4 4

6

y=

3 5 x+ 8 8

7

a Proof

b ( −1, 5), (1, −5)

8

a (4, 18), (4, −2)

b Proof

y = 4x + 2

c

x=

dy a 5 y4 dx 1 dy c 10 x + y dx

3x 2 + 2 y 2x + 3 y2 4x + 5 y c − 5x + 2 y a −

e −

c −cosec 2 x

16

2

9

a Proof

b (1, 2)

10

−

11

(6, −3), ( −2, 0)

12

5x − 8 y = 1

13

e −1

14

( −4, −8) minimum, (4, 8) maximum

44 21

Exercise 4G 1

a

1 3t

b

2 − 2 sin 2θ cos 2θ

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

sin 2θ 1 − cos 2θ sin θ e − sec 2 θ 2 sec 2 θ g sin 2θ 1 2 −t t e (t + 3) i 2 5(1 − t ) k t2

d

c

f

4 cos 2θ 3sec 2 θ sin 2θ sin θ − 2 sin 2θ

h − e 2t − 1 j l

2et ( t + 3)

−3,

3

3 3

4

(2, 4)

5

x+y=2

ii (5.47, 0.693)

13

i

Proof

ii k = 5, c = 68

5 Integration Prerequisite knowledge

2

a 6 cos 2 x + 5 sin x

1

7

a Proof

b (1, 2)

8

b 6, minimum

10

a Proof 1 y=− x+4 2 a Proof

11

a Proof

Proof

1

i

Proof

2

i

4

3

i

Proof −6

iii (e , 4e

b (1, 2)

c Proof

a x − e−x + c c x − e2 x +

3

e g

5

ii −3

5

i

5x + 4 y − 6 = 0

ii Proof

6

i

Proof

7

y = 8.66x − 2.53

8

Proof 13 iii − 9 5x − 9 y + 22 = 0 1 i − 3 1 − x (x + 1) 2 dy = − cos t dx

c

+ 3)

i

i

a

h iii ( −3, −2)

ii Proof

ii 0.294, 1.865

e 4x +c 4

1 −2 x x e (e − 4) + c 2

e

1 −4 x e +c 4

c 2e3x + c

e −2e−x + c

h −2e2 − 3x + c

b Proof

−2

x

1 2x+4 e +c 2

f

ii (1, 6) 2 ii 25 ii Proof

b −

d 8e 2 + c

2

4

11

1 2x e +c 2

a

End-of-chapter review exercise 4

10

2 2x + 1

1

a Proof

9

b 5e5 x− 2

Exercise 5A

6

9

Proof

c

4 b e 4 , 2 − 1 e

324

i

1

4 t

1 3

2

12

1 3x −1 e +c 3 1 8x − 3 i e +c 4 5 x 3x b e (e + 8) + c 4

g

d x − 2e −2 x + c f

x − 2e−2 x + 4e−x + c

1 6 1 2 (e − 1) b (e − 1) 3 4 15 1 d e(e6 − 1) 8 2 1 4 4e − f ( −3 + 4e + e2 ) 2 e 1 ( −17 + 6e2 + 8e3 + 3e 4 ) 12 19 25 9 1 2 − − + 2e2 i − − 2 2e2 2 2e 4 e2

4

y = 3e2 x − 2e − x + 1

5

y = 5e −2 x + 2 x − 2

6

1 1 4 − 3 + e 2 e

7

4e 2 − 4

8

a Proof

b 1 + 2e3

9

a 7 − 2e −2 a − 5e − a

b 7

3

ii x =

1 2

Answers

10

6 − 7 ln 2

11

a 1

6 b Proof

Exercise 5B 1

a 6 ln x + c c e

2

a c e

3

a c

1 ln(3x + 1) + c 3 5 − ln(2 − 3x ) + c 3 7 ln 2 3 13 ln 2 3 3 7 − ln 2 3 5 14 12 + ln 3 5 4 + ln81

1 ln x + c 2 d 3 ln(2 x − 5) + c b

3 ln(5x − 1) + c 10 1 b ln 3 2 1 17 d ln 2 9 f

f

−ln 9

b ln

7 8 9 10

5

a Quotient = 3x + 10 , remainder = 50

1

k = 4e − 3

8

(2 ln 2, 2 ln 2 − 3)

b Proof

e 2 tan 3x − 6x + c

2

c

Exercise 5C 1

a c e g i

2

a c

3 8 3 2

e 1−

d 2

3

a x cos x

4

y=x+

5

1 b sin 4x + c 4 3 d − cos 2x + c 2 1 f tan 2x + c 2 3 h − cos(2 x + 1) + c 2

b 1

3 4

3 1 1 x + sin 2 x + sin 4x + c 8 4 32 π π 3 b 2 3− − a 6 8 6

f

2

1 − cos 3x + c 3 x −2 cos + c 2 5 sin 3x + c 3 2 − sin(1 − 5x ) + c 5 2 tan(5x − 2) + c 5

3 3 x + sin 2 x + c 2 4 b 2 x + 2 sin x + c a

1 1 x− sin 6x + c 2 12 d 2 tan x − 2 x + c

b Proof

7

b Proof

c

a A=2

y = x 2 + 3 ln( x + e) − 3 ln 2e

b Proof

Exercise 5D

27 25

4

6

π a y = 5 − 2 cos 2 x − 2 π b x + 2 y = 10 − 2 3 + 3 π + 3 2 21 4 1 a ( 3 − 1) 2 1 a ( 3− 2) 2

5π −4 2 1 b ( 3π − 3 ) 6 f

3 π cos 2 x − 4 2 y = 3sin 2 x + 2 cos x − 2 x + π − 3

e 3

3 π + 8 6 1 ( 2π − 3 3 ) 6

d f

1 ( 2π − 5 3 ) 24 1 c (2 + 5 π ) 8

1 ( 6 + 2π + 3 3 ) 24 5π 9 d − − 3 3 4 b

a

4

1 f ( 2 π + 11 3 ) 24 a Proof b Proof

5

a Proof

b Proof

6

a Proof

b Proof

7

a Proof

b Proof

8

a Proof

b Proof

9

5π π2+ 4

10

a Proof

e

b Proof

2+π 16 1 (3 3 − π) 6

c Proof

5 3 4

325

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Exercise 5E 1

a 5.22

b 17.09

c 0.92

d 5.61

e 0.40

f 7.68

14

i

Proof

ii x sin x

15

i

Proof

ii Proof

2

1.55

6 Numerical solutions of equations

3

a 6.76

Prerequisite knowledge

b Over-estimate since the top edges of the strips all lie above the curve

1

y

2

35 27 c 0.223130

O

x 2π

π

4

1.77, under-estimate since top edges of the strips all lie below the curve

5

4.07, over-estimate since top edges of the strips all lie above the curve

a x= 5

3

Proof

4

x = 2, x = 5

5

a

7 8

a A = −20

b Proof

9

i

Proof

ii Proof

10

a 12e x +

11

i

12

i

3 4 5 6

13

b x=

y

1 x

O

Proof 81 12 + ln 25 1.81, under-estimate since top edges of the strips all lie below the curve 1 12 Proof 1 (19 − 15e −2 k − 4e −3k ) a 6 19 b 6 Proof

2

d 5.405 465

y − 13 4 1+ y 27

End-of-chapter review exercise 5 1

b −48.7505 …

a

c x=

326

iii 1

4 3x e +c 3

Proof

y

O

2π

–2 –4

y

c

–2

O

b Proof 5 1 π+ 8 4 ii Proof ii

Proof 1 iii − cot x + c 2 1 1 1 i tan 2 x + x + sin 4x + c 2 2 8

ii ln(16e20 )

b

–16

2

4

6

x

x

y+7 3

Answers

6

2

y

a

a

y 20 y = x3 + 5x2

10

–5

O

x

y = 5 – 2x

0.5 x

O

3 points of intersection, so 3 roots. y

b

b Let f( x ) = x 3 + 5x 2 + 2 x − 5 = 0 then f(0) = 0 3 + 5(0)2 + 2(0) − 5 = −5 and

O

1

f(2) = 23 + 5(2)2 + 2(2) − 5 = 27. Change of sign indicates presence of root.

x

2

3

a

y y = x3

c

y

1 –1 O

x

1

1 O –1

2π

y = 1 – 5x

x

One point of intersection so one solution only of x 3 + 5x − 1 = 0. b Let f( x ) = x 3 + 5x − 1 then

Exercise 6A 1

f(0.1) = 0.13 + 5(0.1) − 1 = −0.499 and

a

y

f(0.5) = 0.53 + 5(0.5) − 1 = 1.625. Change of sign indicates presence of root.

y = x²

4

a

y

y = 3x – 4 y = ln(x + 1)

–1 O 1

y = √1 +x

–1 O

x

4 3

x

–4

b 2 points of intersection so 2 roots. c Let f( x ) = x 2 − 1 + x = 0 then f( −1) = ( −1)2 − 1 + ( −1) = 1 and f(0) = (0) − 1 + 0 = −1. Change of sign indicates presence of root. 2

b ln( x + 1) = 3x − 4 in 2 places and so ln( x + 1) − 3x + 4 = 0 has 2 roots.

327

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

5

a

One point of intersection for −0.5 ø x ø 0.5 and so only one root in this domain.

y y = ex y=x+6

8

y

a

6 y = sin x 1

–2π

–6

O

–π

x

x

O

y = 2x + 3

Graphs intersect at 2 points, so 2 roots.

π One point of intersection for −2π , x , − 2 sin x and so only one root of 1 = on this 2x + 3 domain. Also, should x be less than −2 π or π should x be greater than − , the line and 2 curve will not intersect again and so this is the sin x only point of intersection of y = and 2x + 3 y = 1.

b Let f( x ) = e x − x − 6 then f(2.0) = e2 − 2 − 6 = −0.610 … and f(2.1) = e2.1 − 2.1 − 6 = 0.0661 … Change of sign indicates presence of root. 6

a Let f( x ) = ( x + 2)e5 x − 1 = 0 then f(0) = (0 + 2)e 0 − 1 = 1 and f( −0.2) = ( −0.2 + 2)e −1 − 1 = −0.337 … Change of sign indicates presence of root. y

b

sin x − 1 = 0 then 2x + 3 sin( −2) f( −2) = − 1 = −0.09070 … and −4 + 3 sin( −1.9) f( −1.9) = − 1 = 0.1828 … Change of −3.8 + 3 sign indicates presence of root.

b Let f( x ) =

y = e5x

328 1

y=

1

–2

x

O

x+2

9

y

a

y = x3 + 4x

Graphs intersect at 1 point, so 1 root only. 7

a Let f( x ) = cos −1 2 x − 1 + x = 0 then f(0.4) = cos −1(0.8) − 1 + 0.4 = 0.0435 … and

4

−1

f(0.5) = cos 1 − 1 + 0.5 = −0.5. Change of sign indicates presence of root.

O y = 7x + 4

5 x

One point of intersection for 0 ø x ø 5 and so only one root.

y

b

–5

π

b Let f( x ) = x 3 − 3x − 4 then f(2) = 23 − 3(2) − 4 = −2 and

3π/4

f(3) = 33 − 3(3) − 4 = 14 π/2

Change of sign indicates presence of root.

π/4 y=1–x –1

–0.5

O

0.5

1

x

Answers

10

y

a

13

a

y y = cosec x

y=x+4

4 y=2

y = sin x

x

1 –4

2π

x

x

O

Two points of intersection, so two roots.

2 points of intersection for 0 , x , 2 π and so 2 roots.

b Let f( x ) = 2 − x − 4 = 0 then x

f(2.7) = 22.7 − 2.7 − 4 = −0.2019 … and

b Let f( x ) = cosec x − sin x

f(2.8) = 2 − 2.8 − 4 = 0.1644 … Change of sign indicates presence of root.

3π 3π 3π f = cosec − sin =0 2 2 2 3π = 4.71 correct to 3 significant figures. 2 a f( x ) = 20 x 3 + 8x 2 − 7 x − 3 and so

2.8

11

π

O

y

a

y = cot x

14

f(0.5) = 20(0.5)3 + 8(0.5)2 − 7(0.5) − 3 = −2 and f(1) = 20 + 8 − 7 − 3 = 18. Change of sign indicates presence of root.

y = x2

π 2

O

y

b

x

y = (5x – 3)(2x + 1)2

One point of intersection so one root. b Let f( x ) = cot x − x 2 then

O

f(0.8) = cot 0.8 − 0.82 = 0.3312 … and f(1) = cot1 − 12 = −0.3579 … Change of sign indicates presence of root. 12

a

–3

c Proof

y y=x

15

π

2π x

3 points of intersection for 0 , x , 2π and so 3 roots. b Let f( x ) = x − tan 2 x = 0 then f(2.1) = 2.1 − tan 4.2 = 0.3222… and f(2.2) = 2.2 − tan 4.4 = −0.8963… Change of sign indicates presence of root.

4 3 πr + πr 2 (20) a 800 = 3 2 b f( r ) = πr 3 + 30 πr 2 − 1200 = 0 and f(3) = −266.946 … and f(4) = 509.026 …

y = tan 2x

O

x

Change of sign indicates presence of root. Or a suitable pair of graphs drawn.

Exercise 6B 1

a 1.1338, 1.1085, 1.1276, 1.1133, 1.1240 b 1.12, f(1.115) = (1.115)3 + 5(1.115) − 7 = −0.0388 … , f(1.125) = (1.125)3 + 5(1.125) − 7 = 0.0488 … Change of sign indicates presence of root.

329

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2

a

f(1) = ln(2) + 2 − 4 = −1.306 …

11

y 6

a

f(2) = ln(3) + 6 − 4 = 3.0986 … Change of

3

sign indicates presence of root. b 1.535

4

a Proof

2

–4 –3 –2 –1 O –1 –2

b 0.5382

a Proof

b Proof

π → 0.234 48, 0.240 34, 0.24169, 0.242 00, 0 15 00, 0.242 07 → 0.242 π or → 0.24656, 0.24311, 0.242 33, 0.24215, 0 12 15, 0.24210 → 0.242 or 0.2 → 0.232 24, 0.23982, 0.24157, 0 .24157, 0.24197, 0.242 06, 0.242 09 → 0.242

b x n +1 =

a Proof

41+

b e.g. xn =

1 a xn +1 = sin −1 2 or xn +1 = xn

12

6

7 x

(8.040 36, 8.040 38), … → 8.040 e The radius of the cone that would give a container of the required volume 13

14

a Proof

b 1 → 1.5, 1.2603, 1.3713, 1.3186, 1.3434, 1.3317, 1 , 1.3372, 1.3346, 1.3358, 1.3352, … → 1.34 or 1.4 → 1.3053, 1.3497, 1.3287, 1.3386, 1 .3386, 1.3339, 1.3361, 1.3351, 1.3356, … → 1.34 or 1.2 → 1.4007, 1.3050, 1.3499, 1.3286, 1 .3286, 1.3386, 1.3339, 1.3361, 1.3351, 1.3356, … → 1.34

7 xn 2 − 1 → β = 6.98

a Proof

b Proof

0.9082 … , 0.9015 … , 0.9069 … , 0.9026 … , 0.9061 … , 0.9032 … , 0.9055 … , 0.9037 … , 0.9052 … , 0.9040 … , 0.9049 … , → 0.90 1 d e.g. xn +1 = tan −1 → 0.9063 … , sin xn 0.9038 … , 0.9048 … , 0.9044 … , 0.9045 … , 0.9045 … , → 0.90

9

b x =e −4

3

c

24, 1.068 22) → 1.068

x

xn 3 + 1 → α = 0.39, 7

e.g xn +1 = xn +1 =

1 → 1.00125, 1.089 70, 1.06210, 1 sin x 10, 1.070 04, 1.067 69, 1.068 38, (1.06818, 1.068 24, 1

a 2.0794, 2.1192, 2.1390, 2.1489, 2.1538, 2.1563, 2.1563, 2.1575, 2.1581 → 2.16

b Proof

d 8 → 8.058 69, 8.032 00, 8.04419, 8.038 63, 8.04117, 8.040 01, 8.040 54, 8.040 30, 8.040 41,

c x=

10

5

ln( xn + 1) ln(1.5)

a 33 − 3r → c Proof

1 sin xn

1 b x = sin −1 2 → 0.4605 …, which leads to x inverse sine of a value . 1.

2

4

→ c A(3.94, 3.94) → OA = 5.6

xn

c 1.2574, 1.2258, 1.2214, 1.2208, 1.2208… → 1.22 8

3

–5

b x 3 + 5x 2 + 2 x − 5 = 0 7

2

–4

→

a 0.6325, 0.8345, 0.7416, 0.7885, 0.7658, 0 .7658, 0.7771, 0.7716, 0.7743, (0.7730, …) → 0.77

330

1

–3 y=x

c

6

A

1

a Proof

5

()

3

b 0.7231, 0.6142, 0.6584, 0.6387, 0.6472, 0 , 0.6435, 0.6451, 0.6444, 0.6447 so 0.64 4

3x y = 2– – 1

5

Exercise 6C 1

a Proof 3 x + n with x1 = 1.5 → 1.574 681, 1 2 e2 x … 1.568122, 1 , 1.567 522, 1.568184, , 1.568128 → 1.5681 …

b e.g. xn +1 =

n

…

Answers

2

a Proof b

e.g. 1.5 → 1.432164, 1.407 497, 1.398 602, 1.395 404, 1.394 256, 1.393844, → 1.393696, 1.393643 → 1.394

a

7

a Proof

8 a Proof

b Proof

c 0.3 → 0.257 594, 0.273768, → 0.267 462, ( 462, (0.269 900, 0.268 954, 0.269 321, 0

1

→ 9

2

4

3

→ …→

→ 7 x = 6x + 6 → x = 6 → α = 4 6 → a 1, 2 b Proof 10

+

a Proof

c e.g. 2 → 1.9049, 1.9431, 1.9290, 1.9344, (1.9324, 1.9332, 1.9329, 1.9330, 1.9329, 1.9329) → 1.93 11

c 3 → 3.098 61, 3.130 95, 3.14134, 3.144 65, 3 65, 3.145 70, 3.146 04, 3.14614, 3.14618, … → 3.146

a Proof b e.g. 1.2 → 1.294 39 … , 1.234 93 … , 1.27370 … , 1.248 93 … , 1.264 98 … , 1.254 67 … ,

…

1.26133 … , 1.257 04 … , (1.25981 … , 1.258 02 … , 1.25917 … , 1.258 43 … ,

b 3 and 4

d x = 5 y = 3.146, y = log5 3.146 = 0.71

−

= 0 → f(0.155) = a e.g. f( x ) = e x − 2 − sin x →

= sin x (take logs to base e) → b e.g. e x − 2 … x − 2 = ln(sin x ) → x = 2 + ln(sin x ) → q q = 2 + ln(sin q ) when x = q.

) = e2(0.5)+1 − 14 + (0.5)3 = −6.4859 … → f(1) = 2 = e2(1)+1 − 14 + (1)3 = 7.08553 … change of sign indicates presence of root. c Proof ln(14 − xn 3 ) − 1 with 2 x1 = 0.75 → 0.804 230, 0.800 597, 0.800 858, 0 858, 0.800 839, 0.800 840, … → 0.8008

5

−1

e 0.155 − 2 − sin 0.155 = 0.00364 … f(0.165) = e 0.165 − 2 − sin 0.165 = −0.00463 … Change of sign indicates presence of a root.

a Proof

a 1 → 1.1825, 1.1692, 1.1662, 1.1657, 1 , 1.1656, … → 1.17 b Proof

−

331

d e.g. xn +1 =

4

3π = 0 → f(0.8) = 0 32θ

c 1 → 0.99715, 0.996 22, 0.995 92, 0.99582, 0 82, 0.995 79, 0.995 78, 0.995 78, … → 0.996

4

b f( x ) = e2 x+1 − 14 + x 3 = 0 → f(0.5) = e2

a Proof

) = 0.8 − cos −1

=

c e.g. 1.5 → 2.1985, 1.7717, 2.0039, 1.8688, 1 1.9445, (1.9011, 1.9256, 1.9117, 1.9196) → 1.9 3

→

3π = −0.11882 … , f(1.2) = 1 32(0.8) … 3π ) = 1.2 − cos −1 = 0.1475 … Change of 32(1.2) sign indicates presence of root.

−1

x4 + 1 6 1 x + 3 → 7x = 6 → x 3 7 x

4

a Proof

b f(θ ) = θ − cos −1

a 1.5 → 1.5397, 1.5546, 1.5606, 1.5632, 1.5643, 1 , 1.5647, 1.5649, 1.5650, 1.5650, … → 1.57 … b x=

…

b 1 → 1.06366, 1.072 02, 1.07311, 1.07324, 1 24, 1.07327 → 1.073 →

321, 0.269179) → 0.27 d 0.007 30 correct to 3 significant figures

End-of-chapter review exercise 6

ln(3x + 5) e6 x + = 0.6 a 3 3 0 b 0.2 → 0.164 717 … , 0.165896 … , 0 → … , 0.165857 … , (0.165858 …), → 0.166

π → 1, 0.9093, 0.9695, 0.9330, 0.9567, 0 4 .9567, 0.9419, 0.9514, 0.9454, 0.9493 → 0.95 b e.g. x1 =

4

b Proof

c e.g. 1.5 → 1.3083, 1.3689, 1.3495, 1.3557, 1 .3557, 1.3537, 1.3543, (1.3541), … → 1.35 d −1.35

c (8, −6) 3

a Proof

6

12

1.258 91 … , …) → 1.26 … a Proof b Proof

q

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

π π c e.g. f( x ) = sec x − − x + x 2 4

9

= −0.00928... Change of sign indicates presence of root. d 1 → 0.4102, 0.6822, 0.6936, 0.6913, (0.6918, 0.6917, 0.6918, …) → 0.69 13

a Proof b 0.2 → 0.30889, 0.314 70, 0.315 09, 09, 0.315 11 → 0.315

14

a

a 1

[ x ln x − x ]

=5

b 5 → 5.592 01, 5.572 41, 5.572 39, 5 , 5.572 39 → 5.572 332

→

→

Cross-topic review exercise 2 1

Proof

2

π π , 6 3

3

i

Proof

ii (ln 3, −2)

4

i

Proof

ii 2 x − 5 y + 8 = 0

5

a 0.11

dy = e −2 x (sec 2 x − 2 tan x ) = e −2 x (1 − tan x )2 dx

ii e −2 x . 0 and (1 − tan x )2 ù 0

ππ π f( −0.215) = sec( −0.215) − + 0.215 − 0.215 2 4 = 0.00495... π π f( −0.205) = sec( −0.205) − + 0.205 − 0.205 2 4

i

1 π 4 Proof

iii x = 10

i

ii a

i

3

11

( −3a, − a )

12

i

Proof

13

i

Proof

14

i

−

15

i

Proof

ii Proof

iii Proof

iv Proof

ii Proof

11 3 2

iii 1.54 1 1 ii y = − x + 2 2 ii −

5 6

iii −

ii (2, 3)

3 8

16

i

Proof

17

i

a=9

ii 8e − 14

18

i

Proof

ii ( −2, − 1), (0, 1.44)

19

i

Proof

ii 0.678 −2 x

20

x − 3e +c 3 sin 2 x 3x b + +c 4 2 ii 4.84

21

i

i

a

y

b x + 4e − x − 2e −2 x + c 6

b

2 2

x

O

y

ii Proof 22

i

iii Proof

iv 1.26

y y = cosec x

x

O

y = x(π – x)

ii Proof

iii 4.84 O

7

i

Proof

8

i

Proof

ii 1.854

π x

ii Proof ii Proof iii a

0.66

b

2.48

Answers

23

24

5 2 ii a i

b i

27

i i

1

1.11, 2.03 1 2

3

Proof

ii a 26

b ( −5.15, − 7.97)

Proof

Proof

ii a

25

Exercise 7B

i

−0.572, 0.572

Proof

b

3 1 π+ 32 4

4 2 + b x+3 x−2 7 6 c − x − 1 3x + 1 1 1 d − 2( x − 3) 2(3x − 5) 2 3 4 e + − x x − 1 2x + 1 2 2 1 f − + 2x + 3 x + 2 x − 3 a

ii Proof 2

0.362 or 20.7°, 1.147 or 65.7°

ii 33

a b

28

i

Proof

29

i

Proof

ii Proof

iii 2.728 c

ii a = 2, b = −1

iii 4 − 3 2

d

7 Further algebra

e

Prerequisite knowledge

f

1

a A = 6, B = −3, C = −9 b A = −1, B = −

2

3

5 ,C =4 3

b

a 1 + 14x + 84x 2 b 243 − 810 x + 1080 x 2

3

c

Quotient = x − 5, remainder = −11

d

Exercise 7A 1

a 4+

20 2x − 5

4 b 2−

7 1 − 2 2(2 x + 1) 4x + 7 d x+2− 2 x + 2x + 3 30 x + 11 e 7x + 2 + 2 f x −5 c

a

3 3x + 2

a b

2x2 − x +

c d x2 + 1 −

2

A = 1, B = 4, C = 12, D = 29

3

A = 1, B = −1, C = 6, D = −6, E = 5

4

A = 2, B = 3, C = −1, D = 6

6 x2 + 1

2 4 − x + 2 ( x + 2)2 3 4 2 + − 2 x + 1 x + 1 ( x + 1)2 2 3 1 − + − x x − 1 ( x − 1)2 5 4 1 + − 2 x − 3 2 x + 1 (2 x + 1)2 3 3 3 − + 16( x + 2) 16( x − 2) 4( x − 2)2 2 2 7 − + + 9( x + 2) 9( x − 1) 3( x − 1)2 2 3 − x x2 + 1 3 2 + 2x + 1 x2 + 5 1 2 − 3x + 5 2 x 2 + 1 7 3 − 2 2 x + 5 3x − 5 3 2 2+ − x −1 x + 2 7 7 1+ − 4( x − 2) 4( x + 2) 2 3x − 1 −4 + + 2 x − 4 x +1 1 2 7 2+ + − 2 x x − 1 ( x − 1)2

5

A = 2, B = −3, C = 4, D = −1

6

a (2 x − 1)( x − 2)( x + 1) b

7

5 3 1 − + 2x − 1 x − 2 x + 1

a (2 x + 1)( x − 3)2 b

5 3 − x − 4 2x

2 1 3 − + 2 x + 1 x − 3 ( x − 3)2

333

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

8

a a = −4, b = −3

8

a = 2, b = −5

9

a a=−

b (2 x + 1)( x + 3)( x − 1)

9

10

6 32 10 c − + x + 3 2x + 1 x − 1 1 1 a − x x+2 3n 2 + 5 3 1 1 b or − − 2(n + 1)( n + 2) 2 ( n + 1) n + 2 3 c 2 1 1 1 1 Telescoping series, Sn = − , + 2 2 n + 1 n + 2 1 S∞ = 4

b 1+

a = −9, a = 5

11

a a = 8, n = −3

1

1 3

c 1 + 8x + 40 x 2 + 160 x 3 , x ,

1 2

3 3 5 x + x 2 − x 3 , x , 2 2 2 4 1 1 2 1 3 e 1 + x − x + x , x , 2 2 2 1 5 f 1 − x − x 2 − x 3 , x , 3 3

d 1− 334

2

1 g 2 + 16x + 96x 2 + 512 x 3 , x , 4 1 2 3 h 1 − 9x + 42 x − 152 x , x , 2 5 2 1 3 i 1 + 2 x + x + 4x , x , 2 2 2

a 1 − 3x 2 + 6x 4 , x , 1 1 2 2 4 4 x − x , x , 3 9 2 1 2 4 c 1 − 6x + 6x , x , 2 15 3 75 4 2 2 + 3x + 5x + x + x 2 4

1 1 3 2 1 3 x − x , x ,2 − x+ 4 4 16 8 1 2 4 2 8 3 5 b x+ x + x , x , + 5 25 125 625 2 1 1 2 1 c 3− x− x − x3 , x , 9 6 216 3888 1 1 2 5 3 8 d 2+ x− x + x , x , 4 32 768 3 4 4 8 2 40 3 e + x+ x + x , x ,3 27 27 81 729 1 16 84 2 64 3 5 f − − x− x − x , x , 125 625 3125 3125 2 a

1 1 2 3 4 − x + x , x , 2 4 4 16 1 1 4 8 x , x , b 2 − x2 − 4 32 3 15 3 2 15 3 4 x + x , x , 3 c 9 3− 2 8 a

3

2−

25 23 2 67 3 x+ x + x 6 72 2592

4

a

1 1 1 + x + x 2 , 1 − 6x + 27 x 2 2 4 8

b

1 11 97 2 1 − x+ x , x , 2 4 8 3

b 1−

3 4

5

a = 2, b =

a Yes, (3x − 1)−2 = ( −1)−2 (1 − 3x )−2 = (1 − 3x )−2

6

a = −2

b No, 2 x − 1 = real number.

7

−1 1 − 2 x and

5

−1 − 6x − 24x 2 − 80 x 3

6

Proof

7

k=

177 2

−1 is not a

b −5120 x 3

Exercise 7D

a 1 − 2 x + 3x 2 − 4x 3 , x , 1 b 1 − 3x + 9x 2 − 27 x 3 , x ,

3 3 5 15 4 x + x2 + x3 + x 2 2 4 16

10

Exercise 7C 1

1 2

75 125 3 , x 16 8

2 4 8 + 2 − 3 b Proof x x x x x2 x3 x 4 c − + − 2 4 8 16 2 x , 1 gives x , − 2 or x . 2 , d , 1 gives x 2 −2 , x , 2 . The two ranges do not overlap. a 1−

Answers

Exercise 7E 1

2

3

4

5

6

2 3 − 1 − x 1 + 2x b −1 + 8x − 10 x 2 + 26x 3 a

2−

13

k = 16

14

i

2 1 3 + − 1 − 3x 1 − x (1 − x )2 b x + 10 x 2 + 43x 3 a

5 3x − 1 + 1 − x 1 + 2x2 b 4 + 8x + 7 x 2 − x 3 a

6 5 + 1 + 2 x 2 − 3x 237 b 8 a 1+

3 3 − x−4 x+3 7 7 91 2 b − + x− x 4 48 576 a

5 1 3 + − a x + 1 x − 2 ( x − 2)2 15 69 2 b − 6x + x 4 16

ii 15

16

40 3 x 3

1 − 2 x − 4x 2 −

3

2 − 7 x + 18x 2

4

1 + 2x −

5

3 27 2 135 3 1− x + x − x 2 8 16

6

5 5 15 2 + x+ x 2 2 8

7

1 5 75 2 + x+ x 2 16 256

8

A=−

9

A = 5, B = 3, C = −2

10

2 5x − 3 + x x2 + 1

11

A = 3, B = −1, C = 2, D = −2

3 2 x 2

i

1 3 12 + + x + 1 x − 3 ( x − 3)2 4 4 4 − x + x2 3 9 3 1 3x − 1 − + 2 x−2 x +3 1 5 17 2 + x+ x 6 4 72 1 2x + 1 − + 1 + x 1 + 2x2

ii 3x − 3x 2 − 3x 3

8 Further calculus Prerequisite knowledge 1

2

1 + 8x + 40 x 2 + 160 x 3

2

i ii

End-of-chapter review exercise 7 1

23 9 − 5(2 x − 1) 5( x + 2)

12

3

2

+1

a 3 cos 3x 5 c 5x − 3

b 2 xe x

1 5 x +1 e +c 5 5 c ln(3x − 2) + c 3

x 1 − sin 4x + c 2 8 1 d tan 3x + c 3

a

a c

3 1 + 2( x + 3) 2( x − 1) 5 5 5 − + x x − 1 ( x − 1)2

d 2 sec 2 2 x + 5 sin x b

7x 3 − x +2 x 1 1 + d 3− x +1 x − 2

b

2

Exercise 8A 1

a

2

2

4x + 1 3 c x2 + 9 2x e x4 + 1 x a + tan −1 x x2 + 1 2 x − (4x 2 + 1) tan −1 2 x b x 2 (4x 2 + 1) 1 c ex 2 + tan −1 x x +1

3

x − 4y = 2 − π

4

3+ π 5

8 16 , B = −4, C = 3 3

2

5 25x 2 + 1 1 d x2 − 2x + 2 2 f 5x 2 + 2 x + 1 b

335

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Exercise 8B 1

x 1 a tan −1 3 3

1 3x d tan −1 4 12 6 6 f x tan −1 6 2

1 tan −1(2 x ) 2 2 3x 3 tan −1 6 3

c e

π 12

2

a

3

2π 2

b

π 4

c

6 π 9

Exercise 8C 1

b ln(1 + sin x ) + c

c 2 ln( x 2 − 5x + 1) + c

d ln(sin x ) + c

3

1 ln(2 − x 2 ) + c 2 1 3 a ln 2 2 1 5 c ln 2 2 1 ln 2 e 2 Proof

4

Proof

5

p=

336

5

π

6

4 15

7

π 3

8

6e −

1

2

b ln5 d

1 7 ln 2 2

f

ln 3

f

3

sin6 x d +c 6 1 1 e x + − ln(5x + 1) + c 5 5 c

5

π 4

4

a c

1 − (1 − 2 x 2 ) 2 + c 6 3

2 x (e + 2) 2 + c 3

2 +c 2−x 2 x − 3 ln( x − 1) + ln( x + 4) + c

4

100 27 9 3 c ln 16 π e + ln 2 4 1 + ln 4 g 6 9 1 i ln − 2 3 16 a 1 + ln 5 Proof

5

Proof

6

Proof

3

b ln10

a ln

7 ( π − 4 ln 3) 10 1 f ( π + ln 4) 4 9 3 h 2 ln − 4 2 d

b 4 − ln

Exercise 8F

3

2 2 (3x − 1) 2 + (3x − 1) 2 + c 45 27

1

a 3xe x − 3e x + c x sin x + cos x + c 1 2 1 c x ln 2 x − x 2 + c 2 4 1 1 d sin 2 x − x cos 2 x + c 2 4 b

2 3 3 ln( 3 + 2 2 ) 2

1 x +c ln 2 2−x

e 5 ln( x + 1) + ln(2 − x ) +

2

b

a

d ln( x − 3) + 2 tan −1 x + c

e2 − 1

3x + 2 a − +c 6( x + 2)3

3

6 e

c 2 ln(2 x 2 − 9x − 5)

x2 − 3 + c

f

h

5 ( π + 2) 32 π 2

b 3 ln( x + 2) − ln(1 − x ) + c

ln(1 + tan x ) + c

f

Exercise 8D 1

f

Exercise 8E

a 2 ln( x 3 − 1) + c

e − 2

20 9 1 g ( 2π − 3 3 ) 6 e

x 1 b tan −1 4 4

b

18 ln 3 − 16

d

2 2

12 5

c

2 − ln 2

Answers

e f 2

3

1 4 1 4 x ln x − x +c 16 4 2 x ln x − 4 x + c

1 a (π − 2) 18 1 3 b ln16 − ln 2 − 2 4 1 2 c (e + 1) 4 5 2 − 3 e 9 9e a 2(ln 2 − 1)

2

c π −4 2

e 2 4

5

1 (1 + 2e3 ) 9

c

1 2

i

13

i

−1 1 3 e , − 3e 3 x= 2

ii 4 ln 2 −

1

ii p = 3.40

1 + 15x + 135x 2 + 945x 3

d ln 27 − 2

3 π 9

8 f 25 1 b ( π2 − 8) 32

2 3

4 − 6x + 6x 2 − 5x 3

1 2 d e 4

4

1 1 3 2 + x+ x 2 8 64

5

π−2

6

1+

7

Proof

8

i

1 + 2 x + 6x 2

9

i

a=

10

i

Proof

11

Proof

1 (1 + e π ) 2

b π d

26 2− 4 e

1 π(e 4 − 5) 4

End-of-chapter review exercise 8

3 3 15 3 x + x2 + x 2 8 16

ii 5

2 2

ii 1 − 2 x + ii 8 + 2 ln

1

1 1 −1 − e 4 2

12

a Proof

b Proof

2

4(ln 4 − 1)

13

a Proof

b 6.56

14

i

3

14 9

15

i

4

i

16

i

17

i

Proof

ii Proof

5

i

Proof

3 1 ii π− 3 2

6

i

Proof

ii 15 ln5 − 4

7

8

1 1 i − x +1 x + 3 iii Proof

ii Proof

ii 18

Proof

i

10

i

11

Proof

5 −1

i ii

1 tan 2 x + c 2 1 1 1 b π 3 − ln 8 2 2 a 3x +

9

15 16

Cross-topic review exercise 3

f

a

12

19

a b

ii Proof

20

ii 24 − 8e 21

ii 2e2 − 10

Proof 3 4x + 2 − x 4 + x2 Proof

ii Proof ii 1.94

1 3 1 + − 3 − x 2(1 + 2 x ) 2(1 + 2 x )2 4 8 1 2 − x+ x 3 9 27 2 1 3 − + 1 − x 2 − x (2 − x )2 9 5 39 2 + x+ x 4 2 16 3 6x + 1 − 2 x − 1 2x − 1 −2 + 3x − x 2 + 9x 3

1 15 , B = 4, C = 2, D = − 2 2 b Proof a A=

i

1 1 e , − 2e

1 2

ii

1 (2e3 + 1) 9

3 2 x 2

337

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

22

i

Proof

23

i

5 cos(θ − 0.6435)

ii

Exercise 9B

11 96

1.80, 5.77

ii a

a

2

a d =2

2 tan(θ − 0.6435) + c

b

c

9 Vectors Prerequisite knowledge 1 2 3 4

36.7° (correct to 1 decimal place) 3 2 b y = − x−6 a y = x −1 2 3 61 b a 4 cm 2 (2, 12)

1 ( −3 j + 7 k ) 58

1

3

b λ =3

b ON = 6 i + 4 j + 2 k

2 2 1 i+ j+ k 3 3 3

a i

AD

−6 2 = d − a = −6 − 2 0 0

−8 = −8 0

= 128 = 8 2,

AB

13 2 = b−a = 5 − 2 4 0

11 = 3 4

= 112 + 32 + 42 = 146,

Exercise 9A

5 5 a AB = BC = 2 −3

1

−8 5 13 BC = c − b = −3 − 5 = −8 4 4 0

= 8 2,

10 b AC = −1

338

5 −6 DC = c − d = −3 − −6 4 0

2

3 4

−7 a EF = 3

3 10 = −7 b DF − DE = − 5 2 3 QR = PR − PQ = PR + QP = QP + PR QED a XY = b − a and BC = 2 b − 2a = 2( b − a ) BC is a scalar multiple of XY , therefore BC is parallel to XY . 1 2 12 2 0

5 6

a i

q+s−p

7

Proof

= 112 + 32 + 42 = 146 ii Opposite sides are parallel and equal in length.

b i

11 7.5 2 1 1 OM = OA + AB = 2 + 3 = 3.5 2 4 2 0

2

1 BD 3 −6 13 13 1 5 + −6 − 5 3 0 4 4

3 b 4 2 ii

p − 2q − r − s

b For example, angle AHC = 45° (Interior angle of a regular octagon = 135° and angle GHC = 90° ) and the exterior angle (at A) is 45° and so the line segments AB and HC are parallel. k = 1+ 2

11 = 3 4

so M (7.5, 3.5, 2)

b k= a

ii OP = OB + =

−19 13 20 1 1 = 5 + −11 = 4 3 3 −4 4 8 20 4 8 so P , , 3 3 3

Answers

4

5 6

a

2 i + 6 j − 4k =

b

OA OB

4 + 36 + 16 = 56 = 2 14

= 1 + 4 + 25 =

1 = 88.7° correct to BOA = cos −1 34 54 1 decimal place.

30,

= 9 + 16 + 1 = 26 and 30 + 26 = 56 1 c × 30 × 26 = 195 2 16 + ( q − 2)2 = 22 and so q = 2 ± 6.

3

6a + ( −2)(4) + (5)( −2) = 0 a = 3

4

a 5 k 2 − 3( k + 2) − (7 k + 9) = 0 5 k 2 − 10 k − 15 = 0 k 2 − 2 k − 3 = 0 ( k + 1)( k − 3) = 0 → k = −1 or k = 3

a 25 cm

b ON = 9.6 i + 20 j + 4.2 k 7 8

b

4 + 25 + a = 1 + (1 + a ) + ( −3) so a = 9. 2

a

2

2

1 OP = λOQ and using the y-component, λ = .

2

5

9

2

2

−15 = 97.4° 638 21

NP = 2 j + 3k and MP = −3i − 2 j + k and so NP ⋅ MP = 2( −2) + 3(1) = −1, NP = 13, MP = 14 −1 = 94.2509 … = 94.3° NPM = cos −1 13 14

c PQ = 9 i − 6 j + 12 k and 0 Home is the null displacement 0 . Total 0 13 vector sum is 8 so to get home the 0 −13 displacement is −8 . The distance home is 0 10 169 + 64 = 153 cm , correct to the nearest cm.

2

→ θ = cos −1

92 + ( −6)2 + 122 = 3 29

2

2

=

23

OP ⋅ OQ = 10(2) + ( −3)(4) + (23)( −1) = −15

Hence, −6 k =

PQ

2 4 −1

OP = 10 + ( −3) + 23 = 638, OQ = 2 + 4 + ( −1) = 21

4

1 1 (2 k + 13), k = − and 4 2 1 1 checking 8(1 + k ) = ( −32 k ) gives k = − . 4 2 3 b OP = −2 = 3i − 2 j + 4k and 4 12 OQ = −8 = 12 i − 8 j + 16 .k 16

10 OP = −3 , OQ =

6

a ⋅ j = (4)(0) + ( −8)(1) + (1)(0) = −8 a =

42 + ( −8)2 + 12 = 81 = 9 , j = 1

−8 = 152.733 … = 152.7° correct to θ = cos −1 9 1 decimal place. 7

a ⋅ b is a scalar and the dot product is a product of two vectors.

8

a OM = 2 i + 4 j + 4 k, NG = −4i + 3 j + 4 k b

OM ⋅ NG = 2( −4) + 4(3) + 4(4) = 20, 20 OM = 6, NG = 41, cos = 58.6° −1

Exercise 9C 1

correct to 1 decimal place.

a a ⋅ b = 0, a ⊥ b

9

12 = 81.4° b θ = cos −1 13 2 19 c e ⋅ f = 0, e ⊥ f 2

OA = 34 , OB

OB ⋅ OA = ( −5)(1) + (0)(7) + (3)(2) = 1 = 54

6 41

AM = −77 i + 30 j + 36 k and DB = 77 + 36 = 85 so 1 1 AN = 60 j + BD = 60 j + ( −77 i + 36 k ) 2

2

5 5 77 36 i + 60 j + k =− 5 5 77 36 = 3245 + 30(60) + 36 AM ⋅ AN = −77 − 5 5 AM = 25 13

339

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

AN

=

OA + tAB =

1 0 t 4 + t −3 = 4 − 3t −2 8 −2 + 8t x=t y = 4 − 3t z = −2 + 8t

3889

3245 = 54.7° correct to MAN = cos −1 25 13 3889 1 decimal place. 10

a

−3 AN = 1.5 , AN

=

3 14 2

4.5 4.5 = 74.5° correct to cos −1 3 × 1.5 14 1 decimal place.

b

c

PN ⋅ MN = 0, PN =

−3 0 4.5

1 4 1 13 1 3 13 , 0 , →x = , y = 4− = → 4 4 4 4 4

b Oxy plane → z = 0, −2 + 8t = 0, t =

6

MN =

3 1.5 , 4.5 − p

3( −3) + 4.5(4.5 − p ) = 0, p = 2.5, OP =

0 0 2.5

1

a r = ( µ + 4) i + ( µ − 7) j + (3µ ) k → r = 4 i − 7 j + µ ( i + j + 3k ) → direction is i + j + 3k which is not a scalar multiple of 6 i + j + 3k, so the lines are not parallel. 16 = 44.7° correct to 1 decimal b cos −1 11 46 place.

7

Exercise 9D

x 5 4 a y = −3 + t −1 −3 z 2

a r = − j + 5 k + λ (2 i + 6 j − k ) b r = λ (7 i − j − k )

340

b

BA ⋅ d

c r = 7 i + 2 j − 3k + λ (3i − 4 k ) 2

a x = 2λ

b x = 7λ

y = −1 + 6λ

y = −λ

z =5−λ

z = −λ

c x = 7 + 3λ y=2 z = −3 − 4λ 3

4

c x = 1 + 2t y = −3 + 3t z = 4t 5

1 a AB = −3 so 8

L2

1 4 = 4 ⋅ −1 0 −3 = 1(4) + (4)( −1) + (0)( −3) = 0

8

a

−3 −3 0 AB = −2 so OB + tAB = −1 + t −2 −3

b x = 2t y = 10 + 5t z=0

2

–3t + 3 b ON = –2t + 1 –3t + 5

Direction of line through 9 i + 2 j – 5 k and i + 7 j + k is, for example, 8 i – 5 j – 6 k. The direction of this line is a scalar multiple of 16 i − 10 j − 12 k and so the lines are parallel. a x = 2+t y = 13 + t z = 1− t

and so

−3

–3t + 2 , and then CN = –2t − 1 –3t + 2

Since CN is perpendicular to L,

( −3t + 2 ) ( −3 ) + ( −2t − 1 ) ( −2 ) + ( −3t + 2 ) ( −3 ) = 0 22t − 10 = 0, so t = 3 5 ON = 1 + 5

11

5 , 11 18 −3 1 1 −2 = 11 −3 30

Answers

Exercise 9E 1

19 = 62.3923 … ° = 62.4° b cos −1 41 41 correct to 1 decimal place

a Skew b Parallel c Intersecting (8, 5, 15)

5

d Intersecting (5, − 1, 3) 2

p = −80, P (5, − 3, 16)

3

a 4 i − 3k, − 8 i + 4 j, 12 j + 5 k

4

a 6 b r = −4i + 6j − 6 k + λ ( −2i − 14 j + 2 k )

b 55.8°, 72.3°, 51.9°

c

c 13, 4 14, 3 17

d 50.6°

a r = 3i + 7 j + 9 k + λ (4 i + 4 j + 5 k ) b Proof

5

b i ii 6

4 5 r = −3i + j + 5 k + λ (7 i − j − k ) Proof

a 8 − 4 + 5p = 0 → p = −

a

7

a

−5 i − j − 5 k

−2 2 AB ⋅ CB = −4 ⋅ −4

AB = −2 i + 2 j + 4 k

6 −2

= ( −2)(2) + ( −4)( −4) + (6)( −2) = 0

b e.g. r = i + 5 k + λ ( −2 i + 2 j + 4 k ) 10 = 56.938 … ° = 56.9° correct c cos 2 6 14 to 1 decimal place −1

b

d (0, 1, 7)

−2 −10 AD = 20 and BC = 4 , AD = 5BC .

2 10 The lines AD and BC are parallel.

−1 c OE = 12 r = − i + 12 j + 4 k + λ ( i − 10 j + 3k ) 4 a 36.3°correct to 1 decimal place.

End-of-chapter review exercise 9 1

40 = 47.2466 … ° = 47.2° a cos −1 2 14 62

8

b Point of intersection is (4, 0, 1)

correct to 1 decimal place b 2

c Foot of perpendicular is N(3, 2, 4) and

r = 2 i + 3 j + 7k + λ (2 i − 5 j − 13k )

EN

a Not perpendicular as

3 i AB = −1

OA ⋅ OB = ( −2)(1) + (0)( −1) + (6)(4) = 22 ≠ 0 b

−2

ii

r = −2 i + 6 k + λ (3i − j − 2 k )

c (4, –2, 2) 3

a b

AH = −9 i + 15 j + 12 k, NH = 2.5 i + 15 j + 6 k 274.5 = 37.6695 … ° = 37.7° 1069 × 15 2 2 correct to 1 decimal place

a

−4.5 −9.5 PQ = PS = −6 4 2.5

a n=7

−7.5

b R( −5, 0, − 1) c Proof and side length = 7.5 2 d T (2, 1, 1.5) e i e.g. r = v + λ ( t − v )

cos −1

c e.g. r = 9 i + λ ( −9 i + 15 j + 12 k ) 4

9

= −2 i − j = 2 2 + 12 = 5

ii iii

5 −3 r = 17.5 + λ −16.5 −13.5 15 Proof Right, squared-based pyramid

341

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

10

a When λ = −1 the position vector given is P. 185

b

d

s3 = sin(t + 5) + C 3 ln V = 2t + c or V = 0

e

ln y + 1 = ln x + C or y = −1

f

− cos y = (1 − x ) sin x − cos x + C

a

y = 5etan x

b

y2 2 y 2 x3 54 − = −x− 2 5 3 5

c

c 143.0° correct to 1 decimal place d Foot of perpendicular is ( −3, 29, − 4) so perpendicular distance is 67. 11

a r = 7 i + j + 6 k + λ ( 3i + 4 j − 5 k )

2

5

b P (4, − 3, 11) and r = −5 j + 7 k + µ (4 i + 2 j + 4 k )

PQ

c 12

=6

a p = 2, q = −1 b 7 i + 4 j + 8k 51 = 17.8584 … ° = 17.9° c cos −1 29 99 correct to 1 decimal place.

3

y = ln(e x + 1)

4

a b

5

a

10 Differential equations w = 0.02 r 3 x

342

2

3

y=

60e 4

6

Proof

7

x=

8

b

− ln cos x + c

9

c

− xe − x − e − x + c

10

4

2 ( ln x − 2 − ln 3x + 1 ) + c 7

5

a

y

11 12

ii

x→2

2 1

6 − e4

x 3e 4

1+ 1 ln 3x − 1 + c 3

a

2(e3t − 1) e3t + 2

x=

b i

Prerequisite knowledge 1

dy = 5 y3x dx 1 y2 = 6 − 5x 2 1 1 + 3(2 − x ) 3( x + 1)

sin t

ye y + 1 − e y + 1 =

x2 − ln x + C 2

π x 1 y = tan tan −1 + 2 2 4 1 ln 1 − 2 y = 2 + B x t+

1

sin 2t

x=e 2 1 t e +C a 2 x3 1 t 5t 2 1 b = e + − 3 2 2 6 3 y= 2 − x2 2

2

13 1 O

b

k = ln 7 ,

15

a

k = 12

b 16

Exercise 10A 1

x

14

x4 − 6x + C 4

a

y=

b

A = πr + C

1 ln(3x 2 + x 4 ) = t ln 7 + ln 2 2

ln ln x + C y = B ln x where B = eC

48 hours (to the nearest hour)

Exercise 10B

2

1

a

dh = − kh2 where k > 0 dt

Answers

b

dn = kn where k > 0 dt

9

c

dv = − kv ( v + 1) where k > 0 dt

10

d

dV = − kV where k > 0 dt

e

dC = kC 3 where k > 0 dt

f

2

a

k dv = t where k > 0 dt e2 dx = − k x where k > 0 dt

11

3

a

12

c 4

13

c

86.9 °C

1

a

0.458 hours correct to 3 significant figures

5 1 1 x y = − ln − e2 x + e2 x − x 2 e2 x 4 4 2 2

2

a

y 4 = 17e 4 x − 1

4

a

dy = kx y ; 24 y = 93 − 5x 2 dx

b 5 5

dP = kP where k is a positive constant dt

a 10(ln 10 − x − ln 5 − x ) + C b i Proof t

ii x =

c

5.42 minutes correct to 3 significant figures

a

dr 3.92 = dt r

b

r = ( 5.88t − 1.568 ) 3

t

3

b Proof

10(e 10 − 1) t

2e 10 − 1 iii 5 grams 6

y5 = xe x − e x + 1025; y = 4.06 correct to 3 significant figures

7

a

1 2

y4 − 1 + C

b

1 2

y 4 − 1 = 2 x + ln x + 2 − 2

2

c V = 10.3 cm 3 correct to 3 significant figures 8

x=

b As t → ∞, x → 2500

dh dV dh −36 = × = dt dt dV πh2 π h3 1125 − c t= 36 3 d 85.9 seconds correct to 3 significant figures

7

343

2500e 2 4 + e2

a Proof

a

t = 11 ⇒ n = 12100 and t = 12 ⇒ n = 14 400

− kt

b

6

c

t

d 100 °C 5

dn = k n where k is a positive constant; dt 2 n = kt + C

End-of-chapter review exercise 10

dx a = k (100 − x ) where k is a positive dt constant x = 100 − 75e

a

b Unlimited growth, unrealistic

$334 correct to 3 significant figures

b

a 7.47 minutes correct to 3 significant figures

b 9

dA = kA where k is a positive constant dt

b Proof

dL = k where k is a positive constant; dt L = 0.3t + 20

b 24 °C

b t = 600 − 240 x c 176 seconds correct to 3 significant figures

37 or 74% 50

Approximately

At about 6.34 pm

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

8

a

dh = k (8 − h ) where k > 0; dt

3

A

1 ln 0.8 t

h = 8 − 7.5e 5

b As t → ∞, h → 8 9

y=−

10

a b c

11

1 ln(2 − e3x ) 3

dx = − kxt dt kt 2 ln x = − +C 2 47.4 seconds correct to 3 significant figures

1 ( ln P − ln 5 − P 5 15e5t b P= 2 + 3e5t c As t → ∞, P → 5 a

a −

a 5 53.1° correct to 1 decimal place

a

−

b

y2 1 = − cos x 2 + 1 2 2

1 cos x 2 + c 2

c −2 i + 11j

Exercise 11A 1

ln(1 − 3 ln cos x ) 3

2

c (3 10 )i

2 i 3 d 13i

a 8

b (9 + 2 2 )i

a 12i

c 15

a

dx 15 3x = k (2000 − x )x, t = ln dt 2 2000 − x

3

b 30.3 hours correct to 3 significant figures 16 17

y=

x2 − 4 x2 + 4

θ 1 − sin 4θ + ln 2 , x = 1.09 correct 2 8 to 3 significant figures ln x + 2 =

11 Complex numbers

29

8 i 5 1 c i 2 a

1

a 3a − 2bx

b 2 a 2 − abx − 3b 2 x 2

2

a −2 2

b −1

d

5 6

b i

7 2

2

a 6 − 5i

b 4+i

c 11 + 7i

d −

a −1 ± ( 2 3 )i

b −2 ± i

1 3 ± i 2 2 4 14 e − ± i 3 3

c

1

b

Exercise 11B

Prerequisite knowledge

c 2− 3

π 3π , 4 4

b 0.927 radians correct to 3 significant figures or

13

y=

C

b 2.29 radians correct to 3 significant figures 5

73.4 million correct to 3 significant figures

14

4

)+C

12

344

B

3

a x = 3, y = −1

1 13 − i 5 5

d 3±i 6 f

−

5 7 ± i 4 4

b x = 1, y = 3

c x = 1, y = 2 4

a 5 + 5i

b 41

Answers

c 40 − 42i

d 28 − 96i

e 5 − 2i

f

−1 + i

g 5−i

h

1 5 − i 2 2 1 2 6 − i 5 5

a Proof

b z=

6

a z + 49 = 0

b z − 2 z + 26 = 0

c z 2 − 4 z + 13 = 0

d z 2 + 5 z + 14 = 0

2

h (25, − 2.86)

4

2 k, − π 4 10 ( cos(1.89) + i sin(1.89) )

a

10 ( cos(0.322) + i sin(0.322) )

10 ( cos ( −1.25) + i sin( −1.25) ) b AC is a straight line, midpoint O as z1 = z3 .

7 8

1 + 2i

9

z 2 − 10 z + 28 = 0

10

3.2 − 2.4i amps

π 1 Angle AOB = π − tan −1 (3) − tan −1 = . 3 2 Triangles AOB and AOC are isosceles, since z1 = z2 and z2 = z3 . QED. 5

Exercise 11C All angles are given in radians correct to 3 significant figures where rounded. a

g (3, 0.730)

2

1 3 x= , y= 2 2

1

f

i

5

6

3 3 3 + i 2 2 1 c − 2 73 z2 = , arg z 2 = 3.02 97

a

7

B

2

a 5e 6

–3

–2

–1 O –1

1

2

3

4

b ( − u )* = −5 − 2i a

2e

−

iπ 4

5 πi 12

c 8

a=

5 3 5 , b= 2 2

9

a r2

b cos 2θ + i sin 2θ

10

a Proof

b Proof

Im(z) 5

Exercise 11D

P

All angles are given in radians correct to 3 significant figures where rounded.

4 3

1 2

2 –7

–6

Q

–5

–4

–3

–2

–1

O

1 Re(z)

π b 5, 2

c (17, 1.08)

d (61, − 0.181)

+

11 −25 + i 3 −25 − i 3 , z= , z= 8 16 16

6 (cos1.78 + i sin1.78), 2 6 ), (cos( −1.78) + i sin( −1.78)) 2 4 z = 3, z = 5i, z = −5i

3

b −3 + 2i

a z = 7, z = 4 + i 3, z = 4 − i 3 b z=−

–1

a (13, 2.75)

a 1 b z2 = i, z3 = −1 and real

1

3

3 2 3 2 − i 2 2

2e 5

5 Re(z) A

–2

2

d

b −

1 –4

b 1.91 + 4.62i

iπ

Im(z) C

–5

2π 2, − 3

e (41, −1.79)

−

345

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

5

x = 8, y = 3

6

z = −0.5, z = 3 + i, z = 3 − i

7

z = 3i, z = −3i, z = 1 + 2i, z = 1 − 2i

8

a −5 + i, 5 − i

b

Im(z)

O

b −3 − i 2, 3 + i 2

(0, −3)

– −π 3

3 3 2 −i , − 2+i c 2 2 d 4 − 3i, − 4 + 3i e 1 + i 5, − 1 − i 5 f 9

π i e4 ,

2 2

2 2

c

Im(z)

−3 π i e 4

7 6

a z =2−i c

Re(z)

b

p = 37, q = −40

5 4

Im(z)

3

2

2 1

B 1

C O

1

2

–1

3

4

5

6

7

8

9 Re(z)

–7 –6 –5 –4 –3 –2

A

–1 O –1

2

3 Re(z)

–2

–2

346

z

1

d 15 1 10 a = 16, b = −1, z = 4i, z = −4i, z = + i, z 2 2 1 15 i, z = − i 2 2

Im(z) 5 4 3 2

Exercise 11E 1

1

π a A half-line from (2, − 3) at an angle of 12 radians.

–1 O –1

a

4

Im(z) 5

3

5

6

7 Re(z)

1±i√15

4 3 2

Im(z)

1

7 6

–5 –4 –3 –2

5 4

–1 O –1 –2

3 –3

2

–4

1 –1 O –1

3

–3

5π d A half-line from (0, 0) at an angle of 12 radians. 2

2

–2

b The region to the right of the perpendicular bisector of the points (0, 6) and (10, 0) c A circle, centre ( −6, 1), radius 7.

1

1

2

3

4

5

6

7 Re(z)

–5

1

2

3

4

5 Re(z)

Answers

Im(z)

4

z = 12 + 5i

Im(z)

8

y = –4

15 10

O

Re(z)

5 –15 –10 –5 O –5

(0, –4)

10 15 Re(z)

5

–10 –15 (0, –8)

5 –3

–2

Im(z)

9

Im(z)

(x –

–1 O –1

1

2

3

4

3)2 5

+ (y +

6)2

z = 1 + 2i

=9

7 Re(z)

6

π – 4

–2

(–1, 0) O

Re(z)

–3 –4 –5

10 –6

a ( x − 5)2 + ( y − 5)2 = 25 b

–7

Im(z)

–8 –9

347 –10

6

No

Im(z)

5π 6

O O

Re(z)

Re(z)

(–4, –2)

c Least arg z = 0 , greatest arg z = (0, –5)

π 2

End-of-chapter review exercise 11 All angles are given in radians correct to 3 significant figures where rounded.

7

1

Im(z) least = 6√5 greatest = 10√5

20

a −0.1 − 1.7i b w = 1 − 5i, w = 1 + 5i c

Im(z) 7 6

10

5 4 3

O

10 Re(z)

2 1 –4

–3

–2

–1 O

1

2

3 Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

2

a zz* = k 2 + 36, b uw = 8e

3

a

( k 2 − 36) − 12 ki z = z* k 2 + 36

( − 712π ) i, u = 2e( − 712π ) i

6

π a 8, − 6

7

a w = −1 + i

w

( − π4 ) i

Im(z)

b i

Isosceles triangle

Im(z)

b 2 2e

P

2 1

–1

O

π– 3

1 Re(z)

–1

(3, 3)

–2 Q

b 4

3 3 2 cos π + i sin π 4 4

u = w

a x=

O

ii z = 4 + ( 3 + 3 ) i

1 , y=3 2

b C

Im(z) 5

8

Right-angled

B

3

9

1 1

2

3 Re(z)

b z=− 10

–3 –4

3 3 4 1 1 , z= − i, z = + i 3 6 2 6 2

f ( −3) = 0

a i

ii z = 1, z = −3, z = −

A

–5

i, z = −

z 21 20 i − 29 29 ii cos ( −0.761 ) + i sin ( −0.761 )

c i

1 15 − i 4 4

b i

(–1, √3)

Im(z) A

1 O

–2√3

B

O

ii Min. arg z =

–1

Reflection in the real axis z1 = z2 = 13 , arg z1 = 2.86, arg z2 = −2.86

Re(z)

π 2 π π π 5π iii Max. arg z = + + = 2 6 6 6

Re(z)

11

c

15 1 + i, z 4 4

Im(z)

a z1 = −2 3 + i, z2 = −2 3 − i b

1 i 1 i + , z3 = − − 2 2 2 2

a z2 = 3 + 2i

–2

5

2 or x = 3, y = − 2

ii z1 = 3, z2 = −

2

–3 –2 –1 O –1

a x = −3, y = b i Proof

4

348

Re(z)

a i ii

z 2 + 3z + 4 = 0 z1 = 2

Answers

1+ i 3 1− i 3 , z= 2 2

z = −1, z =

b i ii

16

i

3 1 + i 2 2

ii −

3 1 + i 2 2

Im(z)

Cross-topic review exercise 4 B

1

1 C O

–1

1

2

Re(z)

−5 a r= 0+λ 3 m = −2

b i

ii Proof

A

–1

2

y2 = 4( x 2 − 1)

3

a

p = −2, q = −25

Equilateral 12

14 ( −10 i + j − 5 k ) 42

b i

a Proof b

z z = −0.841 = 1, arg z* z*

ii Angle POQ = 90°; 63 5 4

c 3z 2 − 4 z + 3 = 0 13

a k = −1

14

i

7 − 2i

ii

1 πi 6.69e 4

3 i + 2 2

b iii

Im(z)

2 1

y=x

× O –1

4 P

y= 1 x+2 √3

2 1 O –1

3

6 Re(z)

i

7

1

5 Re(z)

i ii

3π 4 1

4

33,

26

c 65.8°, 114.2° correct to 1 decimal place

π 4

O

3

2 sin 2θ − 1

4 a OS = −3 0 b

2

–1

2

6

Im(z)

–2

1

x=

u = −2 − 2i, v = 1 + 2i

ii

B(1, √3) × ×A(√3, 1)

5

–2

15

349

Im(z) 3

6 5

π or 30° or 0.524 radians 6

z = 2 , arg z =

i

ii a 3 3 + i

b arg z = 0.862

3

6 7 −1

2

−

2 11 + i 5 5 Im(z) 4

3 Re(z)

–1 –2

π 3 2 Least z − w = 3sin − 1 = −1 4 2

3 2× 1 –3 –2 –1 O

1

2 Re(z)

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

8

i

u = 8 or 2 2, arg u =

ii

Im(z) 3

π or 45° 4

π 3 2 Least z = 3sin = 4 2 Im(z) Re(z)

O

2 1 ×

3 × 1

–1 O –1

2

3

iii z = 8 − 1 =

4

5 Re(z)

7

iii Circle, centre 3 − 3i, radius 1.

Im(z) 3

Im(z)

2 √8

–1 O –1

i

–1 O –1

1

1

9

1

2

3

4

Proof

a r = −2 i + j − k + µ ( i + 3k )

11

ln R = ln x − 0.57 x + 3.80 R = xe 1 e −1+ 3.80 = 28.850... ii R = 0.57

2

3

–3

4 Re(z)

×

–4

−0.57 x + 3.80

17

a m = −1, m = −4 b m = −3

π i 9e 3

c AB and CD do not intersect. π i 3e 6

or re iθ =

π i − πi 3e 6

=

5 − πi 3e 6

13

y = 4(2 + e3x )2

14

a −7 3 −13

15

ln y = ln x − x 2 + 1 + ln 2 and y = 2 xe − x

16

i

18

i

Proof

ii −6 ø p ø 2 b 50.1°

u =3

Im(z) 4

2

3π – 4

+1

. –4

ii Half-line from (0, − 3) Im(z) 2 1

–3

1

–2

265 5

b

i

–2

32 + ( −3)2 + 1 = 3 2 + 1

Im(z)

10

–1 O –1

4 Re(z)

×

Greatest z = –1 O –1

ii re iθ =

3

–4

iv t becomes very large, x approaches 20

i

2

–3

5 Re(z)

iii 7.9

12

1

–2

|z|

ii − ln(20 − x ) = 0.05t − ln 20

350

–π 4

2

1 π – 4

3

4

5

6 Re(z)

O

π – 4 4 Re(z)

Answers

iii w* = 2 − 4i

26

Im(z) 6

The roots are: −3 + i 2 and 3 − i 2.

ii

w − (1 + 2i) = 1: circle, centre (1, 2) radius 1 3 π : half-line, above and to the left 4 of the point (1, 0) arg( z − 1) =

×

4

i

2 Im(z) 3 –4

× O

–2

2

4

6

8

10 Re(z) 2

–2

1

×

–4 –6

–2

y=

20

i

3e −1 16 x 3−e

b Least w − z =

2

27

Proof 5

5

1 2 2 2 h =− H 2t + H 2 t = 5 150 5

5 60 H 2

−

5 60 h 2

a Proof c 73.2°

5

H2

a r = 2 i + 5 j + 7 k + λ (7 i − 6 j − 9 k )

1

28

29

N =

× O

1

2

–1

3

i

1800e 2 5+

t

351

1 t e2

The roots are: − 3 + i 2 and − 3 + i 2 .

ii Circle, centre (0, 3), radius 2. Greatest value of arg z = 2.3005... = 2.30 correct to 3 significant figures.

B ×

1

–1

i

ii 1800

Im(z) 2 ×C

i

2 −1

b m = −4

b 127.0° 22

3 Re(z)

–1

iii t = 49.3933... = 49.4 21

2

16 x 2

5

ii

3– π 4 × 1

O

–1

z−5 = 5 19

×

4 Re(z)

Im(z) 6

×A

4

OABC is a parallelogram.

×

23

u* ii = 0.8 + 0.6i iii Proof u i Second root is 1 − ( 2 )i .

2

–4

ii Other two roots are −1 ± i . 24

a h=3 b

10 ( j + 3k ) 10

c 94.9° d r = 5 i + 4 j + λ (5 i − 3k ) 25

y3 2 10 = 2 xe3x − e3x + , y = 2.437... = 2.44 3 3 3 correct to 3 significant figures

30

i

–2

O

2

4 Re(z)

dV = 80 − kV dt

4 − 4e −15 k ; 0.14 25 iii V = 540 cm 3 correct to 2 significant figures, V 80 = 570 cm3 approaches the value given by k correct to 2 significant figures. ii kn +1 =

n

Cambridge International AS & A Level Mathematics: Pure Mathematics 2 & 3

Pure Mathematics 2 Practice exam-style paper 1

Proof

2

xø

3

1.46

4

a

5

1 or x ù 1 5

b 71

a 1 → 1.2003, 1.1794, 1.1895, 1.1849, 1.1871, 1 .1871, 1.1861, (1.1865, 1.863), ... → 1.19 b Answer x =

6

a

a = 7, b = −12

4

−101.9°, 78.1°

5

2 11

6

a

5 2x − 1 + x − 2 x2 + 5

27 17 x 117 x 2 − − 10 20 200 dy a = sin 2 x + 2 x cos 2 x dx →

b −

p = −17

cos x =

3

8x 2 8 obtains sec x = 3 3 sec x

b 8

π 4

a u = 2 + i, w = 3 + 9i b A(1, − 1), B (1, 1), C (0, − 2); BC = 10

3 → 1.186 399 6 8

1 (2t + 1)(1 − 2e2t ) 2

7

b y = 2x − 1

9

a

p = ±3

b i

r = 5i + 2 k + λ ( i − 6 j + k )

ii Proof 7

a 10 sin(θ + 36.87)°

b 7.56°, 98.7°

c 13 352

8

b Proof →

Pure Mathematics 3 Practice exam-style paper 2.652

2

Proof

a Proof b Proof

a Proof

1

10

c 0.1 → 0.077 19, 0.071 46, 0.069 98, 0.069 59, 0 , 0.069 49, 0.069 46 → 0.069 d 90.5 °C

Glossary A Absolute value: the non-negative value of a number without regard to its sign P( x ) Algebraic improper fraction: the algebraic fraction , Q( x ) where P( x ) and Q( x ) are polynomials in x, is said to be an improper fraction if the degree of P( x ) ù the degree of Q( x ) Arbitrary constant: a constant that may be assumed to be any value Argument of a complex number: the argument of a complex number x + iy is the direction of the position x vector y

Direction of a vector: this is represented by the arrow on the vector and can be, for example, the angle of the path that takes you from one point to another along the length of the vector Dividend: the quantity being divided by another quantity Division algorithm for polynomials: dividend = divisor × quotient + remainder Divisor: the quantity by which another quantity is to be divided Double angle formulae: sin 2 A ≡ 2 sin A cos A cos 2 A ≡ cos2 A − sin2 A tan 2 A ≡

E C Complex number: a number that can be written in the form x + iy where x and y are real (this form of a complex number is called the Cartesian form) Compound angle formulae: sin( A + B ) ≡ sin A cos B + cos A sin B sin( A − B ) ≡ sin A cos B − cos A sin B cos( A + B ) ≡ cos A cos B − sin A sin B cos( A − B ) ≡ cos A cos B + sin A sin B tan A + tan B tan( A + B ) ≡ 1 − tan A tan B tan A − tan B tan( A − B ) ≡ 1 + tan A tan B Converging: the sequence of values generated by an iterative formula is convergent or said to be converging to a root if the values in the sequence approach the actual value of the root 1 Cosecant: cosec θ = sin θ Cotangent: cot θ =

1 cos θ = tan θ sin θ

D Decompose: split a single algebraic fraction into two or more partial fractions Degree: the highest power of x in the polynomial is called the degree of the polynomial. For example, 5x 3 − 6x 2 + 2 x − 6 is a polynomial of degree 3.

2 tan A 1 − tan2 A

Explicit functions: functions of the form y = f( x ) are called explicit functions as y is given explicitly in terms of x

F Factor: a portion of a quantity that, when multiplied by other factors, gives the entire quantity Factor theorem: if for a polynomial P( x ), P( c ) = 0 then x − c is a factor of P( x )

G General binomial theorem: n( n − 1) 2 n( n − 1)( n − 2) 3 (1 + x ) n = 1 + nx + x + x + … 2! 3! where n is rational and x , 1 General solution: a solution for a differential equation that works for any value of the constant C

I Imaginary number: any multiple of the unit imaginary number i where i2 is defined to be −1 Implicit function: when a function is given as an equation connecting x and y, where y is not the subject Integration by parts:

dv

du

∫ u dx dx = uv − ∫ v dx dx

Integration by substitution: can be considered as the reverse process of differentiation by the chain rule Iteration: an iteration is one trial in a process that is repeated such as when using an iterative formula (see iterative process)

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Iterative formula: this type of formula is one that is used repeatedly. The output from each stage is used as the input for the next stage. It is commonly used in numerical methods.

Particular solution: a solution for a differential equation that works for a specific value of C Polynomial: an expression of the form an x n + an −1x n −1 + an − 2 x n − 2 + … + a2 x 2 + a1x1 + a0

Iterative process: a process for finding a particular result by repeating trials of operations. Each repeat trial is called an iteration and each iteration should produce a value closer to the result being found.

Position vector: a means of locating a point in space relative to an origin

L Laws of logarithms: Multiplication law

Division law

Power law

log a ( xy ) =

x log a = y

log a ( x ) m =

log a x + log a y

log a x − log a y

m log a x

Locus: a locus is a path traced out by a point as it moves following a particular rule. The rule is expressed as an inequality or an equation. Logarithm: the power to which a base needs to be raised to produce a given value

M 354

Magnitude of a vector: the length or size of the vector Modulus: the magnitude of the number without a sign attached. The modulus of a number is also called the absolute value. Modulus of a complex number: the modulus of the complex number x + iy is the magnitude of the position x vector y

N Natural exponential function: the function y = e x Natural logarithms: logarithms to the base of e. ln x is used to represent log e x

P Parallel vectors: two vectors are parallel when one is a scalar multiple of the other Parameter: sometimes variables x and y are given as a function of a third variable t. The variable t is called a parameter and the two equations are called the parametric equations of the curve. Partial fraction: each of two or more fractions into which a more complex fraction can be decomposed as a sum

Principal argument: the principal argument, θ , of a complex number is an angle such that − π , θ ø π. Sometimes it might be more convenient to give θ as an angle such that 0 ø θ , 2π . (Usually the argument is given in radians.) d dv du ( uv ) = u +v Product rule: dx dx dx

Q Quotient: a result obtained by dividing one quantity by another du dv v − u d u d d x x Quotient rule: = dx v v2

R Real number: the set of all rational and irrational numbers Remainder: the amount left over after a division Remainder theorem: if a polynomial P( x ) is divided by x − c, the remainder is P( c ) Resultant vector: the combination of two or more single vectors

S Scalar: a quantity (often a number) that is used to scale a vector 1 Secant: sec θ = cos θ

T Trapezium rule: this numerical method involves splitting the area under the curve y = f( x ) between x = a and x = b into equal width strips

U Unit vector: a vector with magnitude or length one

V Vector addition: the process of finding the sum of two or more vectors Vector subtraction: the addition of the negative of a vector

Index

Index absolute value see modulus function area under a curve trapezium rule 126–8 see also integration Argand diagrams 273 argument of a complex number 274 binomial expansion 181 of ( a + x ) n where n is not a positive integer 177–8 of (1 + x ) n where n is not a positive integer 174–6 and partial fractions 179–80 calculus see differential equations; differentiation; integration chain rule 88 change of sign method, numerical solutions 136 circles, loci 289–92 cobweb pattern, iterative processes 142 complex conjugate pairs 284 complex conjugates 270 complex numbers ( z ) 267, 269–70 argument of 274 calculating with 271, 280–1 Cartesian form 273 cube roots of one 287 equal numbers 270 loci 288–94 modulus of 273 polar forms 277 roots of polynomials 283–5 square roots of 286 complex plane Argand diagrams 273 exponential form 277–81 modulus-argument form 273–7 components of a displacement 212 compound angle formulae 58–60 cosecant (cosec) 53–6 graph of 54 cosine (cos) compound angle formulae 58–60 derivative of cos x 96 derivative of cos(ax + b ) 97 double angle formula 61, 62–4 graph of 55 integration 118–19, 122 cotangent (cot) 53–6 graph of 55 cube roots of one (unity) 287 cubic equations 284–5

degree of a polynomial 11 differential equations 244–5 forming an equation from a problem 252–5 general solution 246 modelling growth 259–60 Newton’s law of heating or cooling 251–2 particular solution 246 separating the variables 245–9 differentiation chain rule 88 derivative of e f( x ) 88–90 derivative of e x 87–8 derivative of ln f( x ) 92–3 derivative of ln x 91–2 derivative of tan −1 x 186–7 derivatives of trigonometric functions 95–8 implicit 99–101 parametric 103–5 product rule 81–3 quotient rule 84–6 displacement (translation) vectors 211–12 addition and subtraction 213–15 components of 212 magnitude of 215–16 notation 212, 213 unit vectors 217–18 dividend 12 division law, logarithms 33 divisor 12 dot product see scalar product e (Euler’s number) 42 derivative of e f( x ) 88–90 derivative of e x 87–8 integration of exponential functions 112–13 equations exponential 38–9 of the form a sin θ + b cos θ = c 68–70 involving the modulus function 3–6 logarithmic 35–6 parametric 103, 232–4 quadratic 283 see also differential equations; numerical solutions of equations Euler, Leonhard 42, 277 explicit functions 99 exponential equations 38–9 exponential form of the complex plane 277–81 exponential growth 259–60 exponential inequalities 40–1

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factor theorem 14–17 factors 12 first order differential equations 244 Fourier, Joseph 70 fractals 267 general binomial theorem 174 general solution of a differential equation 244, 246 golden ratio (1:φ ) 153 golden spiral 153 graph sketching 135–7 graphs of the modulus function 7 of reciprocal trigonometric functions 54–6 transforming a relationship to linear form 44–6 growth models 259–60

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i 267–9 imaginary numbers 266–9 implicit differentiation 99–101 implicit functions 99 improper algebraic fractions 166–7 partial fractions 171–2 indices, relationship to logarithms in base 10 26–9 in other bases 30–2 inequalities exponential 40–1 involving the modulus function 8–10 integration 1 115–17 of ax + b 1 of 115–17 x 1 187–8 of 2 x + a2 of exponential functions 112–13 k f ′( x ) 189–90 f( x ) solving differential equations 245–9 trapezium rule approximation 126–8 of trigonometric functions 118–19, 121–4 use of partial fractions 194–6 using different techniques 201–2 integration by parts 197–200 integration by substitution 191–3 intersecting lines, vector equations 236–7 iterations 141 iterative formulae 142 iterative processes 140–1 cobweb pattern 142 staircase pattern 145 of

subscript notation 142–3 trigonometric equations 149–52 using a calculator 143 worked examples 143–5 loci 288 circles with centre (0, 0) 289–90 circles with centre ( a, b ) 290–2 half-lines and part-lines 292–3 perpendicular bisectors 293–4 logarithmic equations 35–6 logarithms (logs) 26 to base 10 26–9 to base a 30–2 laws of 33–4 natural 42 derivative of ln f( x ) 92–3 derivative of ln x 91–2 solving exponential equations 38–9 solving exponential inequalities 40–1 transforming a relationship to linear form 44–6 logistic growth models 259–60 magnitude (modulus) of a vector 215–16 modulus function (absolute value) 2–3 graphs of y = f( x ) 7 solving equations 3–6 solving inequalities 8–10 modulus-argument form of the complex plane 273–7 multiplication law, logarithms 33 Napier, John 42 natural exponential function 42 natural logarithms 42 derivative of ln f( x ) 92–3 derivative of ln x 91–2 Newton, Isaac 175 law of heating or cooling 251–2 numerical solutions of equations 135 iterative processes 140–5 starting points 135–7 parallel lines, vector equations 236 parallel vectors 213 scalar product 227 parametric differentiation 103–5 parametric equations 103 of a line 232–4 partial fractions 181 application of 172–3 and binomial expansions 179–80 Heaviside’s cover-up method 174

Index

improper 171–2 with a quadratic factor in the denominator that cannot be factorised 171 with repeated linear factor in the denominator 169–70 use in integration 194–6 where the denominator has distinct linear factors 168–9 particular solution of a differential equation 244, 246 perpendicular bisectors 293–4 perpendicular vectors, scalar product 227 polar coordinates 277 polar forms of a complex number 277 polynomials 11 complex roots 283–5 division algorithm 13 division of 11–13 factor theorem 14–17 remainder theorem 18–20 population growth models 259–60 position vectors 220 Cartesian components 220–3 power law 33, 46 principal argument of a complex number 274 problem solving, forming differential equations 252–5 product rule 81–3 quadratic equations, complex roots 283 quartic equations 285 quotient 12 quotient rule 84–6 reciprocal trigonometric ratios 53–6 remainder theorem 18–20 resultant vectors 213–15 roots of polynomials 283–5 scalar product (dot product) 225–6 for component form 227–9 parallel vectors 227 perpendicular vectors 227 secant (sec) 53–6 graph of 55 integration 118–19, 123–4 separating the variables, differential equations 245–9 sine (sin) compound angle formulae 58–60 derivative of sin x 95–6 derivative of sin(ax + b ) 97 double angle formula 61, 62–4 graph of 54 integration 118–19, 122, 124

skew lines, vector equations 237–8 square roots of a complex number 286 staircase pattern, iterative processes 145 subscript notation, iterative processes 142–3 substitution, integration by 191–3 tangent (tan) compound angle formulae 58–60 derivative of tan x 96 derivative of tan −1 x 186–7 derivative of tan( ax + b ) 97 double angle formula 62–3 graph of 55 integration 118–19, 123 telescoping series 172–3 translation vectors see displacement vectors trapezium rule 126–8 trigonometry compound angle formulae 58–60 derivatives of trigonometric functions 95–8 double angle formulae 61–4 equations of the form a sin θ + b cos θ = c 68–70 integration of trigonometric functions 118–19, 121–4 proving identities 66 reciprocal functions 53–6 using iterative processes 149–52 unit vectors 217–18 vector equation of a line 231–2 intersection of two lines 236–7 parametric form 232–4 skew lines 237–8 vector spaces 225 vectors addition and subtraction 213–15 angle between 225 displacement (translation) vectors 211–18 magnitude of 215–16 multiplication by a scalar 213 notation 212, 213 parallel 213, 227 perpendicular 227 position vectors 220–3 scalar product 225–9 unit vectors 217–18 uses of 211

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