Cambridge International AS & A Level Mathematics: Revision Guide for Pure Mathematics 3 (Preview)

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EXAM PREP

Cambridge International AS & A Level

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MATHEMATICS Yong Yau

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Revision Guide for Pure Mathematics 3

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with Lee Lip Seong, Amy Khoo

Copyright Š 2020 by Sunway University Sdn Bhd Published by Sunway University Press An imprint of Sunway University Sdn Bhd No. 5, Jalan Universiti Sunway City 47500 Selangor Darul Ehsan Malaysia press.sunway.edu.my

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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.

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Past question papers contained in this publication are reproduced by permission of Cambridge Assessment International Education.

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All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, now known or hereafter invented, without permission in writing from the publisher.

ISBN 978-967-5492-07-5

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Perpustakaan Negara Malaysia

Cataloguing-in-Publication Data

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Yong, Yau Cambridge International AS & A-Level MATHEMATICS : Revision Guide for Pure Mathematics 3 / Yong Yau, Lee Lip Seong, Amy Khoo.

ISBN 978-967-5492-07-5 1. Mathematics--Problems, exercises, etc.

I. Lee, Lip Seong II. Khoo, Amy. III. Title.

510.76

Edited by Sarah Loh, Lim Bing Hai Designed by Rachel Goh Typeset by Helen Wong Printed by Nets Printwork Sdn Bhd, Selangor

Cover image: Ana Prego/Shutterstock.com Image used under licence from Shutterstock.com

Foreword According to Albert Einstein, “Pure Mathematics is, in its way, the poetry of logical ideas.”

This book aims to support students preparing for the Pure Mathematics 3 component of the Cambridge International AS and A Level Mathematics (9709) examination. It is a culmination of over 40 years of teaching experience by award-winning lecturer Yong Yau and I am confident you will benefit greatly from this book. Under Yong Yau’s leadership, the A Level Mathematics department at Sunway College, Malaysia, has, since 2008, consistently produced high-

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performing students with many winning Cambridge Top in the World and Top in Malaysia awards for Mathematics.

Yong Yau and his team are passionate educators who believe in the importance of developing students’ confidence and building a solid foundation in the subject through employing

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effective pedagogy. Our students’ results are a reflection of the time and effort put in by these lecturers. I believe the insights provided in this book will be invaluable in helping you understand the subject in preparation for the Cambridge International AS and A Level

Elizabeth Lee (Dr) Chief Executive Officer

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examination.

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Sunway Education Group

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Contents Contents Cambridge International AS & A Level Mathematics 9709 Syllabus

vii

How to use this book

viii

Chapter 1 The Modulus Function

1

Definition of modulus function

1 1

Solving equations involving modulus functions

4

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Sketching graphs of modulus functions Solving inequalities involving modulus functions

Chapter 2 Polynomials and Polynomial Equations Equality and operations

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Definition of polynomials

17 17 17

Remainder theorem and factor theorem

21

Factorisation of polynomials

23

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Polynomial equations and inequalities

Chapter 3 The Binomial Series

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Binomial theorem

27

37 37 40

Using partial fractions in binomial expansion

46

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Expressing rational expressions in partial fractions

51

Exponential functions

51

Logarithmic functions

52

Graphs of exponential and logarithmic functions

53

Solving exponential and logarithmic equations

55

Transforming a non-linear relationship into linear form

62

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Chapter 4 Exponential and Logarithmic Functions

Chapter 5 Trigonometry

iv 4

9

69

Graphs of secant, cosecant and cotangent functions

69

Fundamental identities

72

Harmonic form

77

Use of harmonic form

80

Contents

Chapter 6 Differentiation

91

Derivative of logarithmic function

91

Derivative of exponential function

93

Derivative of trigonometric function

93

Techniques of differentiation

94

Derivative of tan x

109

Implicit and parametric differentiations

112

Applications of differentiation

Chapter 7 Integration

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Standard formulae and rules

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−1

117

131 131 132

Integrating exponential functions

141

Integrating trigonometric functions

144

Integration by parts

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Integrating rational functions

Integration by substitution

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Chapter 8 Differential Equations

152 158

167 167

Rate of change

173

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General and particular solution

Chapter 9 Numerical Solutions of Equations

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The graphical method and change of sign Iterative method

183 183 188

Chapter 10 Vectors

203

Definition of vectors

203

Vectors in two-dimensional space

208

Vectors in three-dimensional space

212

Scalar product of two vectors

216

The vector equation of a straight line

228

Parallel, intersecting or skew lines

231

The perpendicular distance from a point to a line

235

The angle between two lines

240

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Contents

Chapter 11 Complex Numbers

247

Fundamental processes

247

The conjugate of complex numbers

250

Argand diagram, modulus and argument

256

Square roots of complex numbers

263

Solving polynomial equations with complex roots

271

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Definition of complex numbers

Solving quadratic equations with non-real coefficients

274

Geometrical effects

277

Locus problems

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Inequalities of a complex number

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Answers

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247

293 301

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Cambridge International AS & A Level Mathematics (9709) Syllabus Assessment for this syllabus consists of six examination papers: Description

Paper 1: Pure Mathematics 1

• Compulsory for AS Level and A Level • 60% of the AS Level; 30% of the A Level

Paper 2: Pure Mathematics 2

• Offered only as part of AS Level • 40% of the AS Level

Paper 3: Pure Mathematics 3

• Compulsory for A Level • 30% of the A Level

Paper 4: Mechanics

• Offered only as part of AS Level or A Level • 40% of the AS Level; 20% of the A Level

Paper 5: Probability & Statistics 1

• Compulsory for A Level • 40% of the AS Level; 20% of the A Level • Offered only as part of A Level • 20% of the A Level

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Paper 6: Probability & Statistics 2

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Component

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These six papers can be combined in specific ways to obtain either the AS Level Mathematics qualification (by taking only two papers) or the A Level Mathematics qualification (by taking a total of four papers). The following are permitted combinations you can opt for:

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AS Level

A Level

Option 1* • Paper 1 • Paper 2

or

Option 2 • Paper 1 • Paper 4

Option 1 • Paper 1 • Paper 4 • Paper 3 • Paper 5 or Option 2 • Paper 1 • Paper 5 • Paper 3 • Paper 6†

or

Option 3 • Paper 1 • Paper 5

* Choosing Option 1 of the AS Level route leads to an AS Level only qualification and will not allow you to progress to a full A Level qualification. †

Paper 6 depends on content knowledge from Paper 5 and as such, there is no option to combine Paper 4 and Paper 6.

Information on this page is adapted from the 2020–2022 Syllabus of the Cambridge International AS & A Level Mathematics 9709 available on the Cambridge Assessment International Education website.

vii 7

How to use this book This book is tailored for students undertaking the Cambridge International AS & A Level Mathematics (9709) examination. Filled with tips and step-by-step guidance, this book is a comprehensive revision guide covering the 2020–2022 syllabus for Pure Mathematics 3 (Paper 3).

Key features of the book: CHAPTER 1 The Modulus Function

hint!

Hint!

This box offers tips and advice to aid your learning

For this type of inequality, it

Example 12 Solve the inequality | 8 – 3x | < 2.

is better to use Method 1 to avoid dividing by a negative number.

Solution Method 1 (Using Rules 1 and 2)

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| 8 – 3x | = | 3x – 8 |

| 8 – 3x | < 2 ⇒ | 3 x – 8 | < 2 –2 < 3x – 8 < 2 6 < 3x < 10

coMMon errors

Common Errors This is an alert on common errors committed by students in the examination

2<x<

The following is incorrect: –10 < –3x < –6 10 <x<2 3

Students tend to forget

10 3

Method 2 (Using Rule 2) | 8 – 3x | < 2 ⇒ –2 < 8 – 3x < 2 –10 < –3x < –6

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reversing the inequality signs

10 >x>2 3

after multiplying or dividing by a negative value.

or

2<x<

10 3

Example 13 Solve the inequality | x + 2 | < | 5 – 2x |.

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Solution

Example 5 4 in partial fractions. (x – 3)(x + 1)

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Express

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CHAPTER 3 The Binomial Series

Solution

Since the degree of the numerator is less than the degree of the

Method 1 (Using graphical method) Sketch the graphs of y = | x + 2 | and y = | 5 – 2x |. y=|x+2| ⇒ y=

(x + 2), if x ≥ –2, –(x + 2), if x < –2.

(5 – 2x), if x ≤

y = | 5 – 2x | ⇒ y =

5 , 2 5 .

2

–(5 – 2x), if x >

10

denominator, we can use Rule 2.

Corresponding to the linear factors (x – 3) and (x + 1) in the denominator, there are partial fractions of the form

A and B , x–3 x+1

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where A and B are constants to be determined.

Thus

A-Level Maths-Chap 1.indd 10

A B 4 ≡ + x–3 x+1 (x – 3)(x + 1)

To find the values of A and B: l Method 1

There are three methods in which you can find the values of the constants A and B.

Worked Examples This book contains step-bystep worked examples that demonstrate how a particular problem can be solved

A B 4 ≡ + x–3 x+1 (x – 3)(x + 1) ≡

A(x + 1) + B(x – 3) (x – 3)(x + 1)

4 ≡ A(x + 1) + B(x – 3) l Equating the constant terms:

4 = A – 3B ---- (1)

Equating the coefficients of x:

0 = A + B ---- (2)

Note that this result can be obtained by multiplying both sides by the denominator of the rational function.

Solving the equations (1) and (2), (1) – (2): 4 = –4B ⇒ B = –1

Alternative Methods In many of the worked examples, you can learn different methods to finding the solution to a problem

When B = –1, from (2) ⇒ A = 1 ∴

4 1 1 ≡ – (x – 3)(x + 1) x – 3 x + 1

Method 2 4 ≡ A(x + 1) + B(x – 3) ---- (*) For this method, you can substitute the value of x to make each factor to zero: x + 1 = 0 ⇒ x = –1 x–3=0 ⇒ x=3

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A-Level Maths-Chap 3.indd 41

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How to use this book

CHAPTER 10 Vectors

Position Vector

Concise Explanations Each chapter has brief but comprehensive explanations on key concepts and topics

To specify the position of a point p, we introduce a point of reference which → is the origin O. The directed line segment Op is called the position vector of → p. Similarly, the position vector of another point q is Oq . p

O q

Vectors in Two-Dimensional Space Let i and j be unit vectors in the positive directions of the x-axis and y-axis

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respectively. Let p be a point with coordinates (x, y). y p (x, y)

y

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j

Extra Notes Numerous side notes can be found throughout this book to boost your understanding

The vector xi + yi can also be x written as . y

i

x

N

x

→ → ON has the same direction as i and ON = x. Thus, ON = xi. → → Np has the same direction as j and Np = y. Thus, Np = yj.

By the triangle law of addition, → → → Op = ON + Np

()

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O

l = xi + yi In general, if a point p has coordinates (x, y) then the position vector of → p is Op = xi + yi. From the figure, → |Op| = Op = √ ON2 + Np2 = √ x2 + y2

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208

CHAPTER 2 Polynomials and Polynomial Equations

Exercise

A-Level Maths-Chap 10.indd 208

Given that (2x + 1) is a factor of f(x), where f(x) = 6x3 + cx2 + 3. Find the

2.

The cubic polynomial x3 + ax2 – 3x – 2, where a is a constant, is denoted

6/10/19 8:19 PM

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1.

value of c and factorise f(x) completely.

Exercises Every chapter contains exam-style exercises that allow you to practise what you have learned

by f(x). Given that (x + 1) is a factor of f(x), (a) find the value of a,

(b) factorise f(x) completely.

The cubic polynomial x3 + ax2 + bx – 8, where a and b are constants, has

4.

The polynomial 2x3 + ax2 + bx – 4, where a and b are constants, is

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3.

factors (x + 1) and (x + 2). Find the values of a and b.

denoted by p(x). The result of differentiating p(x) with respect to x is denoted by p’(x). It is given that (x + 2) is a factor of p(x) and of p’(x). (a) Find the values of a and b. (b) When a and b have these values, factorise p(x) completely.

Cambridge International AS & A Level Mathematics 9709 Paper 32 Q5 November 2009

5.

The polynomial x4 + 3x2 + a, where a is a constant, is denoted by p(x).

Past Year Questions Assess yourself with questions from actual past examination papers

It is given that x2 + x + 2 is a factor of p(x). Find the value of a and the other quadratic factor of p(x).

Cambridge International AS & A Level Mathematics 9709 Paper 3 Q2 November 2007

6.

Show that both (x – √3) and (x + √3) are factors of x4 + x3 – x2 – 3x – 6.

Hence, write down one quadratic factor of x4 + x3 – x2 – 3x – 6, and find a second quadratic factor of this polynomial.

7.

It is given that f(x) = x4 – 3x3 + ax2 + 15x + 50, where a is a constant, and that (x + 2) is a factor of f(x). (a) Find the value of a. (b) Show that f(5) = 0 and factorise f(x) completely into exact linear factors.

8.

List of Acronyms Used LHS Left hand side RHS Right hand side

The cubic polynomial 2x3 + ax2 – x – 2 has a factor (x – 2). Find the value of the constant a. Show that, for this value of a, the equation 2x3 + ax2 – x – 2 = 0 has only one real root.

35

A-Level Maths-Chap 2.indd 35

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CHAPTER 1 The Modulus Function

1 The Modulus Function Definition of Modulus Function The modulus of any number, positive or negative, is the number without a sign attached. In other words, the modulus of any number is the absolute

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value of the number such as | 3 | = 3 and | –4 | = 4.

The modulus of a function f(x), denoted by | f(x) |, is defined by f(x), if f(x) ≥ 0, and –f(x), if f(x) < 0.

≡

|x–1| ≡

x, if x ≥ 0 –x, if x < 0 (x – 1), if x ≥ 1 –(x – 1), if x < 1 1 2 1 –(2x + 1), if x < – 2

| 2x + 1 | ≡

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(2x + 1), if x ≥ –

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|x|

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For example:

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Sketching Graphs of Modulus Functions Sketch the graph of y = | f(x) |, where f(x) = ax + b.

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By definition,

y = | ax + b | ⇒ y =

b , a b –(ax + b), if x < – . a

(ax + b), if x ≥ –

The graph of y = | ax + b | consists of two line segments: y=

b a b –(ax + b), for x < – a (ax + b), for x ≥ –

These two line segments form a V-shape graph with its vertex at the point where x = –

b . a

1

CHAPTER 1 The Modulus Function

Example 1 Sketch the graph of y = | x |.

Solution y=|x| ⇒ y=

x, if x ≥ 0 –x, if x < 0 y

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y=|x|

x

V

0

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Example 2

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Sketch the graph of y = | 2x + 1 |.

Solution

1 2 1 –(2x + 1), if x < – 2

(2x + 1), if x ≥ – y = | 2x + 1 | ⇒ y =

y y = |2x + 1|

1

−1 2 2

0

x

CHAPTER 1 The Modulus Function

Example 3 Sketch the graph of y = | x – 2 | + 1.

Solution (x – 2) + 1, if x ≥ 2

y=|x–2|+1 ⇒ y=

–(x – 2) + 1, if x < 2 x – 1, if x ≥ 2

⇒ y=

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–x + 3, if x < 2 y

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y = |x – 2| + 1

1

2

x

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0

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3

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Example 4

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Sketch the graph of y = 2 – | 2x – 1 |.

Solution

2 – (2x – 1), if x ≥

y = 2 – | 2x – 1 | ⇒ y =

1 2

2 – [–(2x – 1)], if x <

1 2

1 2 1 1 + 2x, if x < 2

3 – 2x, if x ≥ ⇒ y=

3

CHAPTER 1 The Modulus Function

y 2

1

1 2

0

1 2

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–

3 2

x

y = 2 –|2x – 1|

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Solving Equations Involving Modulus Functions Equations involving modulus functions can be solved using graphical

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methods or applying the following important rules: Rule 1: | a – b | = | b – a | hint!

Rule 2: | f(x) | ≤ a Rule 3: | f(x) | ≥ a

|b–a|=

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⇒ –a ≤ f(x) ≤ a

|a–b|=

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Rule 1: | a – b | = | b – a |

Proof:

=

⇒ f(x) ≥ a or f(x) ≤ –a

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Rule 4: | f(x) |2 = [f(x)]2

=

(a – b), if a ≥ b –(a – b), if a < b (b – a), if b ≥ a –(b – a), if b < a –(a – b), if a ≤ b (a – b), if a > b (a – b), if a > b –(a – b), if a ≤ b

=|a–b| Rule 2: | f(x) | ≤ a, where a is a positive constant implying that –a ≤ f(x) ≤ a.

Proof:

Recall that | f(x) | =

f(x), if f(x) ≥ 0, –f(x), if f(x) < 0.

Hence, | f(x) | ≤ a ⇒ f(x) ≤ a, –f(x) ≤ a ⇒ f(x) ≤ a, f(x) ≥ –a ⇒ –a ≤ f(x) ≤ a Rule 3: | f(x) | ≥ a, where a is a positive constant implying that f(x) ≥ a or f(x) ≤ –a.

Proof: 4

| f(x) | ≥ a ⇒ f(x) ≥ a, –f(x) ≥ a ⇒ f(x) ≥ a, f(x) ≤ –a

CHAPTER 1 The Modulus Function

Rule 4: | f(x) |2 = [f(x)]2 Proof:

f(x), if f(x) ≥ 0,

Recall that | f(x) | = | f(x) |2 =

–f(x), if f(x) < 0.

[f(x)]2, if f(x) ≥ 0 [–f(x)]2, if f(x) < 0

= [f(x)]2 Example 5

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hint!

Solve the equation | 2x – 3 | = 5.

Solution | 2x – 3 | = 5 ⇒ | 2x – 3 |2 = 52 ⇒ (2x – 3)2 = 25 2x – 3 = ±5 2x = 8

Method 1.

or

2x = 3 – 5 2x = –2

x = –1

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x=4

constant, it is easier to use

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2x = 3 + 5

| f(x) | = a, where a is a positive

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Method 1 (Using Rule 4)

To solve equations of the form

Method 2 (Using the definition of modulus) or 2x – 3 = –5

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| 2x – 3 | = 5 ⇒ 2x – 3 = 5

2x = –2

x=4

x = –1

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2x = 8

Example 6

Solve the equation | 0.2x – 0.4 | = 1.

Solution | 0.2x – 0.4 | = 1 ⇒ 0.2x – 0.4 = 1 0.2x = 1.4 x=7

or

0.2x – 0.4 = –1 0.2x = –0.6 x = –3

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CHAPTER 1 The Modulus Function

Example 7 Sketch the graph of y = | 2x – 3 |. Hence, determine the values of x for which | 2x – 3 | = x.

Solution 3 .

2 3 –(2x – 3), if x < .

2

| 2x – 3 | =

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(2x – 3), if x ≥

The values of x which satisfy the equation | 2x – 3 | = x are the x-coordinates of the points of intersection between the graphs of

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y = | 2x – 3 | and y = x.

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y

The gradient of y = x is less than the gradient of y = 2x – 3. Thus, the line

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A 0

At point A, y = x and y = –(2x – 3). x = –(2x – 3) =1 At point B, y = x and y = 2x – 3. x = 2x – 3 =3

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B

3

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y = x will intersect the line segment y = 2x – 3.

y = |2x – 3| y=x

x 3 2

CHAPTER 1 The Modulus Function

Example 8 Sketch the graphs of 2y = x + 2 and 2y = | x – 3 | on the same diagram, then calculate the coordinates of the point of intersection of the two graphs.

Solution –(x – 3), if x < 3. y

A 1 0

The line 2y = x + 2 is parallel to the line segment 2y = (x – 3), for x > 3.

x

3

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–2

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2y = |x – 3|

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3 2

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2y = x + 2

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(x – 3), if x ≥ 3,

2y = | x – 3 | ⇔ 2y =

At point A, 2y = x + 2 and 2y = –(x – 3), x < 3.

x=

1 2

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x + 2 = –(x – 3)

1 , 2 1 2y = + 2 2 5 y= 4

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When x =

( 12 , 54 ).

Hence, the point of intersection of the two graphs is A

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CHAPTER 1 The Modulus Function

Example 9 Solve the equation | 2x3 – 9 | = 7, showing all your working.

Solution | 2x3 – 9 | = 7 ⇒ 2x3 – 9 = 7 or

2x3 – 9 = –7

2x3 = 16

x=1

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x=2

2.

Solve the equation | 2x + 1 | = 7.

3.

Solve the equation | x3 – 14 | = 13, showing all your working.

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Solve the equation | 2x – 8 | = 5.

R 5.

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x3 = 1

1.

4.

8

x3 = 8

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Exercise 1.1

2x3 = 2

Cambridge International AS & A Level Mathematics 9709 Paper 21 Q1 June 2012

Solve the equation | 2x + 3 | = | x + 8 |.

Cambridge International AS & A Level Mathematics 9709 Paper 23 Q2(i) November 2015

Solve the equation | x + a | = | 2x – 5a |, giving x in terms of the positive constant a.

Cambridge International AS & A Level Mathematics 9709 Paper 23 Q1 June 2017

CHAPTER 1 The Modulus Function

Solving Inequalities Involving Modulus Functions Inequalities involving modulus functions can be solved using graphical methods or the aforementioned important rules.

Example 10

hint!

Find the set of values of x for which | 2x – 7 | < 3.

For this type of inequality, it is easier to use Method 1.

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Solution Method 1 (Using Rule 2) | 2x – 7 | < 3 ⇒ –3 < 2x – 7 < 3 2<x<5 Method 2 (Using Rule 4) | 2x – 7 | < 3 ⇒

(2x – 7)2 < 32

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4x2 – 28x + 49 < 9

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4 < 2x < 10

4x2 – 28x + 40 < 0

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x2 – 7x + 10 < 0

(x – 2)(x – 5) < 0

2

5

x

R

2<x<5 l

Example 11

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Find the set of values of x for which | 2x – 8 | > 5.

Solution

| 2x – 8 | > 5 l ⇒ 2x – 8 > 5 2x > 13 13 x> 2

or

Using rule 3

2x – 8 < 5 2x < 3 x<

3 2

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CHAPTER 1 The Modulus Function

hint!

Example 12 Solve the inequality | 8 – 3x | < 2.

For this type of inequality, it is better to use Method 1 to avoid dividing by a negative

Solution

number.

Method 1 (Using Rules 1 and 2) | 8 – 3x | = | 3x – 8 | | 8 – 3x | < 2 ⇒ | 3 x – 8 | < 2

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–2 < 3x – 8 < 2 6 < 3x < 10

coMMon errors

2<x<

The following is incorrect: –10 < –3x < –6

10 3

Method 2 (Using Rule 2)

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10 <x<2 3

| 8 – 3x | < 2 ⇒ –2 < 8 – 3x < 2

Students tend to forget

–10 < –3x < –6

reversing the inequality signs

10 > x > 2 or 3

after multiplying or dividing

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by a negative value.

2<x<

10 3

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Example 13

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Solve the inequality | x + 2 | < | 5 – 2x |.

Solution Method 1 (Using graphical method) Sketch the graphs of y = | x + 2 | and y = | 5 – 2x |. y=|x+2| ⇒ y=

(x + 2), if x ≥ –2, –(x + 2), if x < –2.

5 , 2 5 –(5 – 2x), if x > .

2

(5 – 2x), if x ≤ y = | 5 – 2x | ⇒ y =

10

CHAPTER 1 The Modulus Function

y

B

y = |5 – 2 x| 5

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A y = |x + 2|

0

2

1

5 2

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2

7

x

V

Notice that the graphs intersect at the points A and B. At point A: y = x + 2 and y = 5 – 2x x=1

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x + 2 = 5 – 2x

At point B: y = x + 2 and y = –(5 – 2x) x=7

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x + 2 = 2x – 5

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The solutions for | x + 2 | < | 5 – 2x | are the portions of the graph of hint!

y = | x + 2 | that lie below the graph of y = | 5 – 2x |. ∴x<1

or x > 7

Method 2 should be used if the sketch of the graphs is not

Method 2 (Using Rule 4)

required.

| x + 2 | < | 5 – 2x | ⇒

(x + 2)2 < (5 – 2x)2 x2 + 4x + 4 < 25 – 20x + 4x2 3x2 – 24x + 21 > 0 x2 – 8x + 7 > 0 (x – 1)(x – 7) > 0 x<1

or

x>7l

1

x

7

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CHAPTER 1 The Modulus Function

Example 14 Solve the inequality | 2x – 1 | < 3x.

Solution Method 1 (Using Rule 4) | 2x – 1 | < 3x can only be true if 3x ≥ 0 or x ≥ 0 since | 2x – 1 | is always If x ≥ 0, (2x – 1)2 < (3x)2 4x2 – 4x + 1 < 9x2 5x2 + 4x – 1 > 0

x < –1

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(5x – 1)(x + 1) > 0

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positive or zero.

or

x>

1 5

1 5

l Therefore, the solution is x >

x

1 .

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Method 2 (Using Rule 2)

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–1

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However, x ≥ 0 so x < –1 is rejected.

| 2x – 1 | < 3x can only be true if x ≥ 0.

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If x ≥ 0,

| 2x – 1 | < 3x ⇒ –3x < 2x – 1 < 3x –3x < 2x – 1

or

2x – 1 < 3x

1 < 5x 1 x> 5 1 Since x ≥ 0, therefore x > .

5

x > –1

Method 3 (Using graphical method) Sketch the graphs of y = | 2x – 1 | and y = 3x.

1 , 2 1 –(2x – 1), if x < .

2

(2x – 1), if x ≥ y = | 2x – 1 | ⇒ y =

12

CHAPTER 1 The Modulus Function

y

y = 3x

y = | 2x – 1 |

1

A

The gradient of y = 3x is greater than the gradient of the line

l

segment y = 2x – 1. Thus, the line segment y = 2x – 1.

x

1 5

1 2

IE

0

W

y = 3x will not intersect the line

At the point of intersection, A, of the graphs of y = 3x and y = –(2x – 1), 3x = –(2x – 1) 1 5

V

x=

| 2x – 1 | < 3x are the portions of the graph of y = | 2x – 1 | that lie below 1 .

5

R

E

the graph of y = 3x and hence, x >

Example 15

P

Sketch the graph of y = | 2x – 3 |. Determine the values of x for which | 2x – 3 | = 3. Hence, or otherwise, find the set of values of x such that | 2x – 3 | > x.

Solution 3 , 2 3 –(2x – 3), if x < .

2

(2x – 3), if x ≥ y = | 2x – 3 | ⇒ y =

When | 2x – 3 | = 3, from the graph, –(2x – 3) = 3

or 2x – 3 = 3

x=0

x=3 13

CHAPTER 1 The Modulus Function

y

y = | 2x – 3 | y=x B

3

A

The gradient of y = x is less than

l

y = 2x – 3. Thus, the line y = x will

0

intersect both the line segments of y = | 2x – 3 |.

W

the gradient of the line segment

1

3 2

x 3

At point A, y = x and y = –(2x – 3) =1

hint!

At point B, y = x and y = 2x – 3

Since the answer can be

x = 2x – 3

V

obtained entirely from the

=3

sketch (diagram), it is not necessary to use Method 1

| 2x – 1 | > x are the portions of the graph of y = | 2x – 3 | that lie above

or Method 2 as shown in

E

the graph of y = x and hence, x < 1 or x > 3.

R

Example 16.

Compare this question to that of

P

Example 15.

IE

x = –(2x – 3)

Example 16

l Solve the inequality | 3x – 2 | > x.

Solution Method 1 | 3x – 2 | > x is valid for all negative values of x.

Squaring both sides give, (3x – 2)2 > x2 9x2 – 12x + 4 > x2 8x2 – 12x + 4 > 0 2x2 – 3x + 1 > 0

1 2

14

1

x

(2x – 1)(x – 1) > 0 l x<

1 2

or

x>1

CHAPTER 1 The Modulus Function

Method 2 (Using Rule 3) | 3x – 2 | > x is valid for all negative values of x. | 3x – 2 | > x ⇒ (3x – 2) > x

or

(3x – 2) < – x

x>1

x<

1 2

Method 3 (Using graphical method) Sketch the graphs of y = | 3x – 2 | and y = x.

2 , 3 2 –(3x – 2), if x < .

3

y = | 3x – 2 | ⇒ y =

W

(3x – 2), if x ≥

y

IE

y = |3x – 2|

2

V

y=x

l

B

1 2

2 3

the gradient of the line segment y = 3x – 2. Thus, the line y = x will intersect the graph of y = | 3x – 2 | at two different points.

x

1

R

0

E

A

The gradient of y = x is less than

At point A, y = –(3x – 2) and y = x x = –(3x – 2) 1 2

P =

At point B, y = 3x – 2 and y = x x = 3x – 2 =1 | 3x – 2 | > x are the portions of the graph of y = | 3x – 2 | that lie above the graph of y = x and hence, x <

1 or x > 1.

2

15

CHAPTER 1 The Modulus Function

Exercise 1.2 Find the set of values of x for which | 3x – 7 | < 4.

2.

Solve the inequality | 2x – 9 | > 3.

3.

Find the set of values of x satisfying the inequality | 7 – 3x | < 3.

4.

Find the set of values of x for which | x + 3 | > 2 | x – 3 |.

5.

Solve the inequality | x | < 4 | x – 3 |.

6.

Solve the inequality | 2x – 1 | > | x |.

7.

Solve the inequality | x + 1 | < 4 | x – 1 |.

8.

Sketch the graph of y = | x + 2 | and hence, solve the inequality

IE

W

1.

| x + 2 | > 2x + 1.

9.

Solve the equation 4| x | = | x – 1 |. On the same diagram, sketch the

V

graphs of y = 4| x | and y = | x – 1 |. Hence, or otherwise, solve the inequality 4| x | > | x – 1 |.

Sketch the graph of y = | 2x – 5 |. Determine the values of x for which

E

10.

| 2x – 5 | = x. Hence, or otherwise, find the set of values of x such that

R

|2x – 5 | > x.

P

11.

16

Sketch, on a diagram, the graphs of y = | x + 2 | and x + 2y = 6.

Hence, or otherwise, solve the inequality | x + 2 | <

1 (6 – x).

2

12.

Find the set of values of x for which x > | 3x – 8 |.

13.

Find the set of values of x for which | 3x – 2 | < 1 – 4x.