zotero-db/storage/HKEX6F4Z/.zotero-ft-cache

Mathematics Higher Level
for the IB Diploma
Paul Fannon, Vesna Kadelburg, Ben Woolley and Stephen Ward
Not for printing, sharing or distribution.

cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Mexico City Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK www.cambridge.org Information on this title: www.cambridge.org/9781107661738 © Cambridge University Press 2012 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 2012 Printed and bound in Great Britain by the MPG Books Group A catalogue record for this publication is available from the British Library ISBN 978-1-107-66173-8 Paperback with CD-ROM for Windows® and Mac® Cover image: Craig Jewell/Shutterstock 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. notice to teachers Worksheets andcopies of them remain in the copyright of Cambridge University Press and such copies may not be distributed or used in any way outside the purchasing institution.
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Contents

Introduction

vii

Algebra

(Topic 1: Algebra)

1 Counting principles

1 1D Counting selections

11

1A The product principle and the addition principle
1B Counting arrangements
1C Algebra of factorials

1E Exclusion principle

15

1 1F Counting ordered selections

18

6 1G Keeping objects together or separated

21

9

Algebra, functions and equations

(Topics 1: Algebra & 2: Functions & equations)

2 Exponents and logarithms

28

2A Laws of exponents

28

2B Exponential functions

33

2C The value e

39

2D Introducing logarithms

40

2E Laws of logarithms

44

2F Graphs of logarithms

49

2G Solving exponential equations

50

3 Polynomials

58

3A Working with polynomials

58

3B Remainder and factor theorems

63

3C Sketching polynomial functions

68

3D The quadratic formula and the discriminant 76

4 Algebraic structures

87

4A Solving equations by factorising

87

4B Solving equations by substitution

89

4C Features of graphs

91

4D Using a graphical calculator to solve equations 94

4E Solving simultaneous equations

by substitution

97

4F Systems of linear equations

101

4G Solving inequalities

110

4H Working with identities

113

5 The theory of functions
5A Relations, functions and graphs 5B Function notation 5C Domain and range 5D Composite functions 5E Inverse functions 5F Rational functions
6 Transformations of graphs
6A Translations 6B Stretches 6C Reflections 6D Modulus transformations 6E Consecutive transformations 6F Reciprocal transformations 6G Symmetries of graphs and functions
7 Sequences and series
7A General sequences 7B General series and sigma notation 7C Arithmetic sequences 7D Arithmetic series 7E Geometric sequences

118
118 123 125 131 133 141
154
155 157 161 163 167 174 178
190
190 193 195 199 201

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Contents iii

7F Geometric series 7G Infinite geometric series 7H Mixed questions

205 8 Binomial expansion
207 8A Introducing the binomial theorem 211 8B Applying the binomial theorem
8C Products of binomial expansions 8D Binomial expansions as approximations

217
217 219 223 226

Geometry

(Topics 3: Circular functions & trigonometry & 4: Vectors)

9 Circular measure and trigonometric functions
9A Radian measure 9B Definitions and graphs of sine and cosine
functions 9C Definition and graph of the tangent
function 9D Exact values of trigonometric functions 9E Transformations of trigonometric graphs 9F Modelling using trigonometric functions 9G Inverse trigonometric functions
10 Trigonometric equations and identities
10A Introducing trigonometric equations 10B Harder trigonometric equations 10C Trigonometric identities 10D Using identities to solve equations
11 Geometry of triangles and circles
11A Right-angled triangles 11B The sine rule 11C The cosine rule 11D Area of a triangle 11E Trigonometry in three dimensions 11F Length of an arc 11G Area of a sector 11H Triangles and circles

232
232
239

12 Further trigonometry
12A Double angle identities 12B Compound angle identities 12C Functions of the form a sin x + b cos x 12D Reciprocal trigonometric functions

248 13 Vectors
251 13A Positions and displacements
254 13B Vector algebra
261 13C Distances
264 13D Angles

273
273 281

13E Properties of the scalar product 13F Areas 13G Properties of the vector product
14 Lines and planes in space

291 14A Vector equation of a line 297 14B Solving problems with lines
14C Other forms of equation of a line

304
305 307 312 318

14D Equation of a plane
14E Angles and intersections between lines and planes
14F Intersection of three planes
14G Strategies for solving problems with lines and planes

321

326

330

333

346
347 354 359 364
375
375 384 389 393 397 402 404
413
413 420 432 439
449 457
463

iv Contents

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Algebra

(Topic 1: Algebra)

15 Complex numbers
15A Definition and basic arithmetic of i 15B Geometric interpretation 15C Properties of complex conjugates 15D Complex solutions to polynomial
equations

476 15E Sums and products of roots of polynomials 498

476 15F Operations in polar form

504

482 15G Complex exponents

510

487 15H Roots of complex numbers

512

15I Using complex numbers to derive

493

trigonometric identities

517

Calculus

(Topic 6: Calculus)

16 Basic differentiation and its applications
16A Sketching derivatives 16B Differentiation from first principles 16C Rules of differentiation 16D Interpreting derivatives and second
derivatives 16E Trigonometric functions 16F The exponential and natural logarithm
functions 16G Tangents and normals 16H Stationary points 16I General points of inflexion 16J Optimisation
17 Basic integration and its applications
17A Reversing differentiation 17B Constant of integration 17C Rules of integration 17D Integrating x–1 and ex 17E Integrating trigonometric functions 17F Finding the equation of a curve 17G Definite integration 17H Geometrical significance of definite
integration 17I The area between a curve and the y-axis 17J The area between two curves

527
527 535 538
541 547
549 550 554 560 562
569
569 571 572 575 576 577 579
581 588 591

18 Further differentiation methods 599

18A Differentiating composite functions

using the chain rule

599

18B Differentiating products using the

product rule

604

18C Differentiating quotients using the

quotient rule

607

18D Implicit differentiation

611

18E Differentiating inverse trigonometric

functions

616

19 Further integration methods
19A Reversing standard derivatives
19B Integration by substitution
19C Using trigonometric identities in integration
19D Integration using inverse trigonometric functions
19E Other strategies for integrating quotients
19F Integration by parts

622
622 626
634
641
644 649

20 Further applications of calculus
20A Related rates of change 20B Kinematics 20C Volumes of revolution 20D Optimisation with constraints

659
659 663 668 674

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Contents

v

Probability and statistics

(Topic 5: Statistics & probability)

21 Summarising data
21A Some important concepts in statistics 21B Measures of spread 21C Frequency tables and grouped data

689
689 691 695

22 Probability

705

22A Introduction to probability

705

22B Combined events and Venn diagrams

711

22C Tree diagrams and finding the intersection 715

22D Independent events

720

22E Counting principles in probability

722

22F Conditional probability

724

22G Further Venn diagrams

727

22H Bayes’ theorem

732

23 Discrete probability distributions 743

23A Random variables

743

23B Expectation, median and variance of a discrete

random variable

747

23C The binomial distribution

751

23D The Poisson distribution

758

24 Continuous distributions

769

24A Continuous random variables

769

24B Expectation and variance of continuous random

variables

773

24C The normal distribution

777

24D Inverse normal distribution

783

Algebra

(Topic 1: Algebra)

25 Mathematical induction
25A The principle of mathematical induction 25B Induction and series 25C Induction and sequences 25D Induction and differentiation 25E Induction and divisibility 25F Induction and inequalities

791
791 793 795 799 801 803

26 Questions crossing all chapters

809

Examination tips

820

Answers

821

Index

911

Acknowledgements

917

Terms and conditions of use for the CD-ROM

918

vi Contents

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Introduction

Structure of the book
The book is split roughly into four blocks: chapters 2 to 8 cover algebra and functions; chapters 9 to 14 cover geometry; chapters 16 to 20 cover calculus; and chapters 21 to 24 cover probability and statistics. Chapters 1, 15 and 25; on counting principles, complex numbers and induction (respectively), bring together several areas of the course and chapter 26 includes questions that mix different parts of the course – a favourite trick in International Baccalaureate® (IB) examinations.
You do not have to work through the book in the order presented, but (given how much the International Baccalaureate® likes to mix up topics) you will find that several questions refer to material in previous chapters. In these cases a ‘rewind panel’ will tell you that the material has been covered previously so that you can either remind yourself or decide to move on.
We have tried to include in the book only the material that will be examinable. There are many proofs and ideas which are useful and interesting, and these are on the CD-ROM if you would like to explore them.
Each chapter starts with a list of learning objectives to give you an idea about what the chapter contains. There is also an introductory problem that illustrates what you will be able to do after you have completed the chapter. Some introductory problems relate to ‘real life’ situations, while others are purely mathematical. You should not expect to be able to solve the problem, but you may want to think about possible strategies and what sort of new facts and methods would help you. The solution to the introductory problem is provided at the end of the chapter, after the summary of the chapter contents.
Key point boxes
The most important ideas and formulae are emphasised in the ‘KEY POINT’ boxes. When the formulae are given in the Formula booklet, there will be an icon: ; if this icon is not present, then the formulae are not in the Formula booklet and you may need to learn them or at least know how to derive them.
Worked examples
Each worked example is split into two columns. On the right is what you should write down. Sometimes the example might include more detail then you strictly need, but it is designed to give you an idea of what is required to score full method marks in examinations. However, mathematics is about much more than examinations and remembering methods. So, on the left of the worked examples are notes that describe the thought processes and suggest which route you should use to tackle the question. We hope that these will help you with any exercise questions that differ from the worked examples. It is very deliberate that some of the questions require you to do more than repeat the methods in the worked examples. Mathematics is about thinking!

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Introduction vii

Signposts
There are several boxes that appear throughout the book.
Theory of knowledge issues
Every lesson is a Theory of knowledge lesson, but sometimes the links may not be obvious. Mathematics is frequently used as an example of certainty and truth, but this is often not the case. In these boxes we will try to highlight some of the weaknesses and ambiguities in mathematics as well as showing how mathematics links to other areas of knowledge.

From another perspective
The International Baccalaureate® encourages looking at things in different ways. As well as highlighting some international differences between mathematicians these boxes also look at other perspectives on the mathematics we are covering: historical, pragmatic and cultural.

Research explorer
As part of your course, you will be asked to write a report on a mathematical topic of your choice. It is sometimes difficult to know which topics are suitable as a basis for such reports, and so we have tried to show where a topic can act as a jumping-off point for further work. This can also give you ideas for an Extended essay. There is a lot of great mathematics out there!

exam hint

Exam hint
Although we would encourage you to think of mathematics as more than just learning in order to pass an examination, there are some common errors it is useful for you to be aware of. If there is a common pitfall we will try to highlight it in these boxes. We also point out where graphical calculators can be used effectively to simplify a question or speed up your work, often referring to the relevant calculator skills sheet on the CD-ROM.

Fast forward / rewind

Mathematics is all about making links. You might be interested to see how something you have just learned will be used elsewhere in the course, or you may need to go back and remind yourself of a previous topic. These boxes indicate connections with other sections of the book to help you find your way around.

How to use the questions

The colour-coding

The questions are colour-coded to distinguish between the levels.
Black questions are drill questions. They help you practise the methods described in the book, but they are usually not structured like the questions in the examination. This does not mean they are easy, some of them are quite tough.

viii Introduction

© Cambridge University Press 2012

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Each differently numbered drill question tests a different skill. Lettered subparts of a question are of increasing difficulty. Within each lettered part there may be multiple roman-numeral parts ((i), (ii),...) , all of which are of a similar difficulty. Unless you want to do lots of practice we would recommend that you only do one roman-numeral part and then check your answer. If you have made a mistake then you may want to think about what went wrong before you try any more. Otherwise move on to the next lettered part.
Green questions are examination-style questions which should be accessible to students on the path to getting a grade 3 or 4.
Blue questions are harder examination-style questions. If you are aiming for a grade 5 or 6 you should be able to make significant progress through most of these.
Red questions are at the very top end of difficulty in the examinations. If you can do these then you are likely to be on course for a grade 7.
Gold questions are a type that are not set in the examination, but are designed to provoke thinking and discussion in order to help you to a better understanding of a particular concept.
At the end of each chapter you will see longer questions typical of the second section of International Baccalaureate® examinations. These follow the same colour-coding scheme.
Of course, these are just guidelines. If you are aiming for a grade 6, do not be surprised if you find a green question you cannot do. People are never equally good at all areas of the syllabus. Equally, if you can do all the red questions that does not guarantee you will get a grade 7; after all, in the examination you have to deal with time pressure and examination stress!
These questions are graded relative to our experience of the final examination, so when you first start the course you will find all the questions relatively hard, but by the end of the course they should seem more straightforward. Do not get intimidated!
Calculator versus non-calculator questions
In the final examination there will be one paper in which calculators are not allowed. Some questions specifically need a calculator but most could appear in either the calculator or the noncalculator paper.
Some questions are particularly common in the non-calculator paper and you must be able to know how to deal with these. They are highlighted by the non-calculator symbol.
Some questions can be done in a clever way using a calculator or cannot be realistically done without using a calculator. These questions are marked with a calculator symbol.
In the final examination you will not get this indication, so you must make sure you learn which types of questions have an easy calculator method. The calculator skills sheets on the CD-ROM can help with this.
For the questions that do not have either calculator icon, you should mix up practising with and without a calculator. Be careful not to become too reliant on your calculator. Half of the core examination is done without one!

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Introduction ix

On the CD-ROM
On the CD-ROM there are various materials that you might find useful.
Prior learning
The International Baccalaureate® syllabus lists what candidates should know before taking the examination. The topics on the list are not all explicitly covered in the syllabus, but their knowledge may be required to answer examination questions. Don’t worry, you do not need to know all this before starting the course, and we have indicated in the rewind boxes where a particular concept or skill is required. On the CD-ROM you will find a self-assessment test to check your knowledge, and some worksheets to help you learn any skills that you might be missing or need to revise.
Option chapters
Each of the four options is covered by several chapters on the CD-ROM.
Coursebook support
Supporting worksheets include: • calculator skills sheets that give instructions for making optimal use of some
of the recomended graphical calculators • extension worksheets that go in difficulty beyond what is required at
International Baccalaureate® • fill-in proof sheets to allow you to re-create proofs that are not required in the
examination • self-discovery sheets to encourage you to investigate new results for yourself • supplementary sheets exploring some applications, international and
historical perspectives of the mathematics covered in the syllabus.
e-version
A flat pdf of the whole coursebook (for days when you don’t want to carry the paperback!). We hope you find Higher Level Mathematics for the IB diploma an interesting and enriching course. You might also find it quite challenging, but do not get intimidated, frequently topics only make sense after lots of revision and practice. Persevere and you will succeed.
The author team.

x Introduction

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1Counting principles
Introductory problem If a computer can print a line containing all 26 letters of the alphabet in 0.01 seconds, estimate how long it would take to print all possible permutations of the alphabet.
Counting is one of the first things we learn in mathematics and at first it seems very simple. If you were asked to count how many people there are in your school, this would not be too tricky. If you were asked how many chess matches would need to be played if everyone were to play everyone else, this would be a little more complicated. If you were asked how many different football teams could be chosen, you might find that the numbers become far too large to count without coming up with some clever tricks. This chapter aims to help develop strategies for counting in such difficult situations.
1A The product principle and the addition principle
Counting very small groups is easy. So, we need to break down more complicated problems into counting small groups. But how do we then combine these together to come up with an answer to the overall problem? The answer lies in using the product principle and the addition principle, which can be illustrated using the following menu.

In this chapter you will learn:
• how to break down complicated questions into parts that are easier to count, and then combine them together
• how to count the number of ways to arrange a set of objects
• the algebraic properties of a useful new tool called the factorial function
• in how many ways you can choose objects from a group
• strategies for applying these tools to harder problems.
Counting sometimes gets extremely difficult. Are there more whole numbers or odd numbers; fractions or decimals? Have a look at the work of Georg Cantor, and the result may surprise you!

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1 Counting principles 1

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The word analysis literally means ‘breaking up’. When a problem is analysed it is broken down into simpler parts. One of the purposes of studying mathematics is to develop an analytical mind, which is considered very useful in many different disciplines.

MENU

Mains
Pizza Hamburger Paella
Desserts
Ice-cream Fruit salad

n(A) means the size of the set of options of A. See Prior Learning Section G on the CD-ROM.

Anna would like to order a main course and a dessert. She can choose one of three main courses and one of two desserts. How many different choices could she make? Bob would like to order either a main course or a dessert. He can choose one of the three main courses or one of the two desserts; how many different orders can he make?
We can use the notation n(A) to represent the number of ways of making a choice about A.
The product principle tells us that when we want to select one option from A and one option from B we multiply the individual possibilities together.
KEY POINT 1.1
The Product principle (AND rule)
The number of ways of in which both choice A and choice B can be made is the product of the number of options for A and the number of options for B. n(A AND B) = n(A) × n(B)

2 Topic 1: Algebra

The addition principle tells us that when we wish to select one option from A or one option from B we add the individual possibilities together.
The addition principle has one essential restriction. You can only use it if there is no overlap between the choices for A and the choices for B. For example, you cannot apply the addition principle to counting the number of ways of getting an odd
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number or a prime number on a die. If there is no overlap between the choices for A and for B, the two events are mutually exclusive.
KEY POINT 1.2
The Addition Principle (OR rule) The number of ways of in which either choice A or choice B can be made is the sum of the number of options for A and the number of options for B. If A and B are mutually exclusive then n(A OR B) = n(A) + n(B)

The hardest part of applying either the addition or product principle is breaking the problem down and deciding which principle to use. You must make sure that you have included all of the cases and checked that they are mutually exclusive. It is often useful to rewrite questions to emphasise what is required, ‘AND’ or ‘OR’.

Worked example 1.1

An examination has ten questions in section A and four questions in section B. How many different ways are there to choose questions if you must:
(a) choose one question from each section? (b) choose a question from either section A or section B?

Describe the problem accurately
‘AND’ means we should apply the product principle: n (A) × n (B)

(a) Choose one question from A AND one from B
Number of ways = 10 × 4 = 40

(10 ways) (4 ways)

Describe the problem accurately
‘OR’ means we should apply the addition principle: n (A) + n (B)

(b) Choose one question from A OR one from B
Number of ways = 10 + 4 = 14

(10 ways) (4 ways)

In the example above we cannot answer a question twice so there are no repeated objects; however, this is not always the case.

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1 Counting principles 3

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Worked example 1.2

In a class there are awards for best mathematician, best sportsman and nicest person. Students can receive more than one award. In how many ways can the awards be distributed if there are twelve people in the class?

Describe the problem accurately

Choose one of 12 people for the best mathematician AND one of the 12 for best sportsmen AND one of the 12 for nicest person.

(12 ways) (12 ways) (12 ways)

Apply the product principle

12 × 12 × 12 = 1728

Exercise 1A

This leads us to a general idea. KEY POINT 1.3
The number of ways of selecting something r times from n objects is nr.
1. If there are 10 ways of doing A, 3 ways of doing B and 19 ways of doing C, how many ways are there of doing (a) (i) both A and B? (ii) both B and C? (b) (i) either A or B? (ii) either A or C?
2. If there are 4 ways of doing A, 7 ways of doing B and 5 ways of doing C, how many ways are there of doing (a) all of A, B and C? (b) exactly one of A, B or C?
3. How many different paths are there (a) from A to C? (b) from C to E? (c) from A to E?

A
4 Topic 1: Algebra

B

C

D

E

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4. Jamil is planting out his garden and needs one new rose bush

and some dahlias.There are 12 types of rose and 4 varieties of

dahlia in his local nursery. How many possible selections does

he have to choose from?

[3 marks]

5. A lunchtime menu at a restaurant offers 5 starters, 6 main courses and 3 desserts. How many different choices of meal can you make if you would like

(a) a starter, a main course and a dessert?

(b) a main course and either a starter or a dessert?

(c) any two different courses?

[6 marks]

6. Five men and three women would like to represent their club

in a tennis tournament. In how many ways can one mixed

doubles pair be chosen?

[3 marks]

7. A mathematics team consists of one student from each of years 7, 8, 9 and 10. There are 58 students in year 7, 68 in year 8, 61 in year 9 and 65 in year 10.
(a) How many ways are there of picking the team?
Year 10 is split into three classes: 10A (21 students), 10B (23 students) and 10C (21 students).
(b) If students from 10B cannot participate in the competition, how many ways are there of picking the team? [4 marks]

8. Student passwords consist of three letters chosen from A to Z, followed by four digits chosen from 1–9. Repeated characters are allowed. How many possible passwords are there? [4 marks]

9. A beetle walks along the edges from the base to the tip of an

D

octahedral sculpture, visiting exactly two of the middle vertices

(A, B, C or D). How many possible routes are there? [6 marks] A

C B

10. Professor Square has 15 different ties (seven blue, three red and five green), four waistcoats (red, black, blue and brown) and 12 different shirts (three each of red, pink, white and blue). He always wears a shirt, a tie and a waistcoat.

(a) How many different outfits can he make?

Professor Square never wears any outfit that combines red with pink.

(b) How many different outfits can he make with this

limitation?

[6 marks]

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1 Counting principles 5

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11. How many different three digit numbers can be formed using the digits 1,2,3,5,7

(a) once only?

(b) if digits can be repeated?

[6 marks]

12. In how many ways can (a) four toys be put into three boxes? (b) three toys be put into five boxes?

[6 marks]

If you look in other textbooks you may see permutations referred to in other ways. This is an example of a word that can take slightly different meanings in different countries. The definition here is the one used in the International Baccalaureate® (IB).
n! occurs in many other mathematical situations. You will see it in the Poisson distribution (see Section 23D), and if you study option 9 on calculus, you will see that it is also important in a method for approximating functions called Taylor series.

1B Counting arrangements
The word ‘ARTS’ and the word ‘STAR’ both contain the same letters, but arranged in a different order. They are both arrangements (also known as permutations) of the letters R, A, T and S. We can count the number of different permutations.
There are four possibilities for the first letter, then for each choice of the first letter there are three options for the second letter (because one of the letters has already been used). This leaves two options for the third letter and then the final letter is fixed. Using the ‘AND rule’ the number of possible permutations is 4 × 3 × 2 × 1 = 24.
The number of permutations of n different objects is equal to the product of all positive integers less than or equal to n. This expression is abbreviated to n! (pronounced ‘n factorial’).
KEY POINT 1.4
The number of ways of arranging n objects is n! n! = n(n − 1)(n − 2). . . × 2 × 1

Worked example 1.3

A test has 12 questions. How many different arrangements of the questions are possible?

Describe the problem accurately

Permute (arrange) 12 items Number of permutations = 12!
= 479 001 600

6 Topic 1: Algebra

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In examination questions you might have to combine the idea of permutations with the product and addition principles.

Worked example 1.4

A seven-digit number is formed by using each of the digits 1–7 exactly once. How many such numbers are even?

Describe the problem accurately: even numbers end in 2, 4 or 6

Pick the final digit to be even

(3 ways)

Only 6 digits left to arrange

AND then permute the remaining 6 digits

(6! ways)

Apply the product principle

3 × 720 = 2160 possible even numbers

This example shows a very common situation where there is a constraint, in this case we have to end with an even digit. It can be more efficient to fix each part of the constraint separately, instead of searching all the possibilities for the ones which are allowed.

exam hint FvqyafemyauocCcrsoicaouythcaoutlsokecslrltalcerwiuhyaoiurtl.oaaigfll3lsuaAlltestolhtgudlooturevshe.rnpeekottSsrnitutmykyehoogioeelewlhu6sr!,
CD-ROM.

Worked example 1.5 How many permutations of the word SQUARE start with three vowels?

Describe the problem accurately Apply the product principle

Permute the three vowels at the beginning (3! ways) AND Permute the three consonants at the end (3! ways)
Number of ways = 3! × 3! = 36

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1 Counting principles 7

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Exercise 1B
8 Topic 1: Algebra

1. Evaluate: (a) (i) 5! (b) (i) 2 × 4! (c) (i) 6! – 5!

(ii) 6! (ii) 3 × 5! (ii) 6! – 4 × 5!

2. Evaluate: (a) (i) 8! (b) (i) 9 × 5! (c) (i) 12! – 10!

(ii) 11! (ii) 9! × 5 (ii) 9! – 7!

3. Find the number of ways of arranging:

(a) 6 objects

(b) 8 objects

(c) 26 objects.

4. (a) How many ways are there of arranging seven textbooks on a shelf?

(b) In how many of those arrangements is the single biggest

textbook not at either end?

[5 marks]

5. (a) How many five-digit numbers can be formed by using each of the digits 1–5 exactly once?
(b) How many of those numbers are divisible by 5? [5 marks]

6. A class of 16 pupils and their teacher are queuing outside a cinema.

(a) How many different arrangements are there?

(b) How many different arrangements are there if the teacher

has to stand at the front?

[5 marks]

7. A group of nine pupils (five boys and four girls) are lining up for a photograph, with all the girls in the front row and all the boys at the back. How many different arrangements are there? [5 marks]

8. (a) How many six-digit numbers can be made by using each of the digits 1–6 exactly once?

(b) How many of those numbers are smaller than 300 000?

[5 marks]

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9. A class of 30 pupils are lining up in three rows of ten for a class

photograph.

How many different arrangements are possible?

[6 marks]

10. A baby has nine different toy animals. Five of them are red

and four of them are blue. She arranges them in a line so that

the colours are arranged symmetrically. How many different

arrangements are possible?

[7 marks]

1C Algebra of factorials
To solve more complicated counting problems we often need to simplify expressions involving factorials. This is done using the formula for factorials, which you saw in Key point 1.4:
n! n(n 1)(n − 2)…× 2 ×1

Worked example 1.6

(a) Evaluate 9! ÷ 6!
n!
(b) Simplify (n − 3)!

(c) Write 10 × 11 × 12 as a ratio of two factorials.

Write in full and look for common factors in denominator and numerator

(a) 9 8 × 7 6 × 5… 2 × 1 6 × 5…× 2 1
= 9 × 8 × 7 = 504

Write in full and look for common factors in denominator and numerator

(b) n × (n − 1) × (n − 2) × (n − 3)…× 2 × 1 (n 3) × (n − 4)…× 2 × 1
= n(n − 1)(n − 2)

Reverse the ideas from (b)

(c)

10

×

11

×

12

=

1

×

2…× 9 10 1 × 2…×

9

11

12

= 12! 9!

You can usually solve questions involving sums or differences of factorials by looking for common factors of the terms. It is important to understand the link between one factorial and the next:
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1 Counting principles 9

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KEY POINT 1.5 (n + 1)! = (n + 1) × n!

Worked example 1.7

Simplify: (a) 9! − 7!

(b) ( )! ! + ( )! = 72 × 7!

Link 9! and 7!
Take out the common factor of 7!
Link (n + 1)!, n! and (n − 1)!
Take out the common factor of (n − 1)!

(a) ! 9 8! = 9 × 8 × 7!
9 7! = 72 × 7! 7
= (72 − 1) × 7! = 71 × 7!
(b) n! n (n )! (n + 1)! (n + 1) × n! = (n + 1) × n × (n − 1)! (n + 1)! − n! + (n 1)! = n(n + 1)(n − 1)! − n(n − 1)! + (n − 1)! = (n − 1)!(n2 + n − n + 1) = (n − 1)!(n2 + 1)

0! is defined to be 1. Why might this be?
1 ! is π 22 To explore why you need to look at the Gamma function.

Being able to simplify expressions involving n! is useful because n! becomes very large very quickly. Often factorials cannot be evaluated even using a calculator. For example, a standard scientific calculator can only calculate up to 69! = 1.71 × 1098 (to 3 significant figures).
exam hint
Remember that you will lose one mark per paper in the IB if you give any answers to less than three significant figures (3SF) (unless the answer is exact of course!). See Prior learning Section B on the CD-ROM if you need a reminder about significant figures.

Exercise 1C
10 Topic 1: Algebra

1. Fully simplify the following fractions:

(a) (i) 7! 6
(b) (i) 8! 6!

(ii) 12! 11!
(ii) 11! 8!

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(c) (i) 100! 98! n!
(d) (i) (n −1)!

51! (ii)
49! ( )! (ii)
a !

(e) (i) a! (a )!

(ii) (b )! (b )!

(f) (i) ( )! ( )!

(ii) ( )! ( )!

2. Write these as a ratio of 2 factorials:

(a) (i) 7 × 8 × 9

(ii) 6 × 5 × 4 × 3

(b) (i) 1013 × 1014 × 1015 × 1016

(ii) 307 × 308 × 309

(c) (i) (n + 2)(n + 3)(n + 4)(n + 5)

(ii) n(n – 1)(n + 1)(n + 2)

3. Explain why 6! – 5! = 5 × 5!

Simplify in a similar fashion:

(a) (i) 10! – 9! (b) (i) 14! – 3 × 13!

(ii) 12! – 10! (ii) 16! + 5 × 15!

(c) (i) 11! + 11 × 9!

(ii) 12! – 22 × 10!

(d) (i) n! – (n – 1)!

(ii) (n + 2)! – n!

4. Solve n! = 20 where n is a positive integer. (n )!

5. Solve ( )! = 990. ( )!

6. Solve n! (n − 1)! = 16(n )! for n ∈N.

[5 marks] [6 marks] [6 marks]

1D Counting selections
Suppose that three pupils are to be selected from a class of 11 to attend a meeting with the Head Teacher. How many different groups of three can be chosen?
In this example we need to choose three pupils out of 11, but they are not to be arranged in any specified order. The order is not important; the selection of Ali, Bill and then Cathy is the same as the selection of Bill, Cathy and then Ali. This sort of selection is called a combination. In general, the formula for

the number of ways of choosing r objects out of n is given the

symbol

⎛ ⎝⎜

n⎞ r ⎠⎟

,

or

nCr

(pronounced

‘n

C

r’

or

‘n

choose

r’).

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1 Counting principles 11

eTywhxoehauaFmtthoierthmifsiounrulamsteubdloaof,okbrl.eutt

tells not

KEY POINT 1.6

The number of ways of choosing r objects out of n is

⎛ n⎞ ⎝r⎠

=

n!
r!(n −

r )!

⎛ n⎞ Does learning the terminology of n! and ⎝⎜ r ⎠⎟ give you new mathematical knowledge?
Does this mathematical language clarify or muddle the principles behind it? Is it necessary to have ‘expert language’ in all subjects? Does giving something a label help you think about it?

Worked example 1.8
A group of 12 friends want to form a team for a five-a-side football tournament. (a) In how many different ways can a team of five be chosen? (b) Rob and Amir are the only goalkeepers and they cannot play in any other position. If the team
of five has to contain exactly one goalkeeper, how many different ways can the team be chosen?

Describe the problem accurately Describe the problem accurately

(a) Choosing 5 players from 12

Number

of

ways

=

⎛ 12⎞ ⎝⎜ 5⎠⎟

=

792

(b) Pick a goalkeeper, then fill in the rest of the team

Split into cases to satisfy the condition
With Amir picked there are four remaining slots, but Rob cannot be picked so ten players remain Apply the addition principle and
product principle

Amir is in goal AND Choose four other players OR Rob is in goal. AND Choose four other players

(1 way) (? ways)
(1 way) (? ways)

Number of ways with Amir in goal

=

⎛ ⎝⎜

10⎞ 4 ⎠⎟

Number

of

ways

with

Rob

in

goal

=

⎛ 10⎞ ⎝⎜ 4⎠⎟

Total

number

of

ways

=

1

×

⎛ ⎝⎜

10⎞ 4⎠⎟

+

1

×

⎛ ⎝⎜

10⎞ 4 ⎠⎟

= 210 + 210 = 420

12 Topic 1: Algebra

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exam hint

You

can

calculate

⎛ n⎞ ⎝⎜ r ⎠⎟

on

your

calculator

using

a

button like this; see Calculator skills sheet 3 on the

CD-ROM.

We will use the idea of combinations to help us expand brackets in chapter 8 and to find probabilities in Section 23C.

Exercise 1D

1. Evaluate:

⎛ 7⎞

(a) (i)

⎝⎜ 2⎠⎟

(b) (i)

3

×

⎛ ⎝⎜

6⎞ 3⎠⎟

(c)

(i)

⎛ 5⎞ ⎝⎜ 0⎠⎟

×

⎛ ⎝⎜

9⎞ 5⎠⎟

(d)

(i)

⎛ 5⎞ ⎝⎜ 2⎠⎟

+

⎛ 9⎞ ⎝⎜ 5⎠⎟

⎛12⎞

(ii)

⎝⎜ 5 ⎠⎟

(ii)

10

×

⎛ ⎝⎜

6⎞ 5⎠⎟

(ii)

⎛10⎞ ⎝⎜ 8 ⎠⎟

×

⎛ 3⎞ ⎝⎜ 1⎠⎟

(ii)

⎛ 6⎞ ⎝⎜ 0⎠⎟

+

⎛ 7⎞ ⎝⎜ 3⎠⎟

2. (a) (i) In how many ways can six objects be selected from eight?
(ii) In how many ways can five objects be selected from nine?
(b) (i) In how many ways can either three objects be selected from ten, or seven objects be selected from 12?
(ii) In how many ways can either two objects be selected from five, or three objects be selected from four?
(c) (i) In how many ways can five objects be selected from seven and three different objects be selected from eight?
(ii) In how many ways can six objects be selected from eight and three different objects be selected from seven?

3. Solve:

(a) (i)

⎛ n⎞ ⎝⎜ 2⎠⎟

= 91

(b) (i)

⎛ n⎞ ⎝⎜ 3⎠⎟

= 1330

(ii)

⎛ n⎞ ⎝⎜ 2⎠⎟

=

351

(ii)

⎛ n⎞ ⎝⎜ 3⎠⎟

=

680

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1 Counting principles 13

14 Topic 1: Algebra

4. An exam paper consists of 15 questions. Students can select

any nine questions to answer. How many different selections

can be made?

[4 marks]

5. Suppose you are revising for seven subjects, but you can only complete three in one evening.

(a) In how many ways can you select three subjects to do on Monday evening?

(b) If you have to revise mathematics on Tuesday, in how

many ways can you select the subjects to do on Tuesday

evening?

[5 marks]

6. In the ‘Pick’n’Mix’ lottery, players select seven numbers out of 39. How many different selections are possible? [4 marks]

7. There are 16 boys and 12 girls in a class. Three boys and two

girls are needed to take part in the school play. In how many

different ways can they be selected?

[5 marks]

8. A soccer team consists of one goalkeeper, four defenders, four

midfielders and two forwards. A manager has three goalkeepers,

eight defenders, six midfielders and five forwards in the squad.

In how many ways can she pick the team?

[5 marks]

9. A school is planning some trips over the summer. There are 12 places on the Greece trip, ten places on the China trip and ten places on the Disneyland trip. There are 140 pupils in the school who are all happy to go on any of the three trips.

In how many ways can the spaces be allocated?

[5 marks]

10. A committee of three boys and three girls is to be selected from a class of 14 boys and 17 girls. In how many ways can the committee be selected if:

(a) Ana has to be on the committee?

(b) the girls must include either Roberta or Priya,

but not both?

[6 marks]

11. Raspal’s sweet shop stocks seven different types of 2¢ sweets and five different types of 5¢ sweets. If you want no more than one of each sweet, how many different selections of sweets can be made when spending:

(a) exactly 6¢?

(b) exactly 7¢?

(c) exactly 10¢?

(d) at most 5¢?

[6 marks]

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12. An English examination has two sections. Section A has five questions and section B has four questions. Four questions must be answered in total.

(a) How many different ways are there of selecting four questions to answer if there are no restrictions?

(b) How many different ways are there of selecting four

questions if there must be at least one question answered

from each section?

[6 marks]

13. 15 points lie around a circle. Each point is connected to every

other point by a straight line. How many lines are formed in

this way?

[4 marks]

14. Ten points are drawn on a sheet of paper so that no three lie in a straight line. By connecting up the points: (a) how many different triangles can be drawn? (b) how many different quadrilaterals can be drawn? [7 marks]

15. At a party, everyone shakes hands with everyone else.

If 276 handshakes are exchanged, how many people are

there at the party?

[6 marks]

16. A group of 45 students are to be seated in three rows of 15 for a school photograph. Within each row, students must sit in alphabetical order according to name, but there is no restriction determining the row in which a student must sit.

How many different seating arrangements are possible,

assuming no students have identical names?

[6 marks]

1E Exclusion principle
The exclusion principle is a way to count what you are interested in by first counting what you are not interested in. This is often needed for counting where a certain property is prohibited (not allowed). KEY POINT 1.7
The Exclusion principle
Count what you are not interested in and subtract it from the total.

Imagine that a five-digit code is formed using each of the digits 1−5 exactly once. If we wanted to count how many such codes do not end in ‘25’ we could work out all of the possible options that do not end in ‘25’:
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1 Counting principles 15

Worked example 1.9a How many five-digit codes formed by using each of the digits 1–5 exactly once do not end in ‘25’?

Describe the problem accurately
Use the product and addition principles

Pick the final digit from {1, 2, 3, 4} AND arrange the remaining four digits OR Pick the final digit as 5 AND pick the penultimate digit from {1, 3, 4} AND arrange the remaining three digits

(4 ways) (4! ways)
(1 way) (3 ways) (3! ways)

(4 × 4!) + (1 × 3 × 3!) = 114

Worked example 1.9b

This method works, but it is long and complicated. An easier way is to use the exclusion principle.

How many five-digit codes formed by using each of the digits 1–5 exactly once do not end in ‘25’?

Describe the problem accurately
Use the product and exclusion principles

Count permutations of five digit codes then EXCLUDE cases where
the last two digits are ‘25’ AND arrange the remaining three digits

(5! ways) (1 way)
(3! ways)

5! – 1 × 3! = 114

Many questions ask us to use the exclusion principle in a situation containing an ‘at least’ or ‘at most’ restriction.

Worked example 1.10

Theo has eight different jigsaws and five different toy bears. He chooses four things to play with. In how many ways can he make a selection with at least two bears?

Describe the problem accurately

Count combinations of four toys from thirteen

⎛ 13⎞ ⎝⎜ 4 ⎠⎟

ways

16 Topic 1: Algebra

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continued . . .
Apply the product and addition principles

Then EXCLUDE Combinations with no bears

⎛ 5⎞ ⎝⎜ 0⎠⎟

ways

AND Four jigsaws

⎛ 8⎞ ⎝⎜ 4⎠⎟

ways

OR

Combinations with one bear

⎛ 5⎞ ⎝⎜ 1 ⎠⎟

ways

AND Three jigsaws

⎛ 8⎞ ⎝⎜ 3⎠⎟

ways

Number

of

ways

=

⎛ 13⎞ ⎝⎜ 4⎠⎟

−

⎛ ⎝⎜

⎛ ⎝⎜

5⎞ 0⎠⎟

⎛ 8⎞ ⎝⎜ 4⎠⎟

+

⎛ ⎝⎜

5⎞ 1 ⎠⎟

⎛ 8⎞ ⎞ ⎝⎜ 3⎠⎟ ⎠⎟

= 714 – (1 × 70 + 5 × 56) = 365

exam hint
This example highlights the ambiguity of the English language. Can you see why we exclude (no bears and four jigsaws) OR (one bear and three jigsaws) rather than (no bears and four jigsaws) AND (one bear and three jigsaws)?
Be careful when you read questions. For example, the opposite of a word beginning in a consonant OR ending in a consonant is not a word beginning in a vowel OR ending in a vowel. For example, the word ‘LOVE’ fits both descriptions. If you are interested there is formal work on this, called De Morgan’s laws, in the Sets, Relations and Groups option (option 8).

Exercise 1E

1. How many of the numbers between 101 and 800 inclusive are

not divisible by 5?

[4 marks]

2. How many permutations of the letters of the word JUMPER do

not start with a J?

[5 marks]

3. A bag contains 12 different chocolates, four different mints and six different toffees. Three sweets are chosen. How many ways are there of choosing
(a) all not chocolates? (b) not all chocolates? [6 marks]

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1 Counting principles 17

4. How many permutations of the letters of KITCHEN

(a) do not begin with KI? (b) do not have K and I in the first two letters?

[6 marks]

5. A committee of five people is to be selected from a class of 12

boys and nine girls. How many such committees include at

least one girl?

[6 marks]

6. In a word game there are 26 letter tiles, each with a different

letter. How many ways are there of choosing seven tiles so that

at least two are vowels?

[6 marks]

7. A committee of six is to be selected from a group of ten men and 12 women. In how many ways can the committee be chosen if it has to contain at least two men and one woman? [6 marks]

8. Seven numbers are chosen from the integers 1−19 inclusive. How many have

(a) at most two even numbers? (b) at least two even numbers?

[7 marks]

9. How many permutations of the letters DANIEL do not begin

with D or do not end with L?

[6 marks]

Worked example 1.11

1F Counting ordered selections
Sometimes we want to choose a number of objects from a bigger group but the order in which they are chosen is important. For example, finding the possibilities for the first three finishers in a race or forming numbers from a fixed group of digits. The strategy for dealing with these situations is first to choose from the larger group and then permute the chosen objects.

A class of 28 pupils has to select a committee consisting of a class representative, a treasurer, a secretary and a football captain. Each post needs to be a different person. In how many different ways can the four posts be filled?

Select four people and then allocate them to different jobs. Note that this is the
same as selecting four people and then permuting them

Choose 4 from 28
AND Permute those 4

⎛ ⎝⎜

⎞ 4⎠⎟

ways

(4! ways)

Apply the product principle

Number of ways

=

⎛ ⎝⎜

28⎞ 4⎠⎟

× 4!

= 491 400

18 Topic 1: Algebra

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Suppose that the whole set has size n and we are selecting and

ordering a subset of size r. The number of ways of doing this

is given the symbol nPr and is described as ‘the number of permutations of r objects out of n΄. We can use the formula

for

⎛ n⎞ ⎝⎜ r ⎠⎟

to

deduce

a

formula

for

n Pr :

there

are

⎛ n⎞ ⎝⎜ r ⎠⎟

ways of

choosing the objects in the subset, and these can be arranged in

r! ways. Therefore, nPr

⎛ n⎞ ⎝r⎠

r! or:

KEY POINT 1.8

n Pr

=

n!
(n r)!

This can be written, using factorial algebra, to:
nPr n(n 1)(n − 2)…(n − r + 1)
Written in this form you can see that the formula for nPr is an application of the ‘AND rule’. For example, suppose that there are five people in a race and we want to count the number of possibilities for the first three positions. We can reason that if there are five options for the winner then, for each winner, there are four options for the second place and three options for the third place. This gives 5 4 × 3 = 5P3.

Exercise 1F

1. Evaluate:

(a) (i) 6P1 (b) (i) 8P2 (c) (i) 10P3

(ii) 5P1 (ii) 11P2 (ii) 12P3

2. Find the number of permutations of: (i) 4 objects out of 10 (ii) 6 objects out of 7.

3. Find the number of ways of selecting when order matters: (i) 3 objects out of 5 (ii) 2 objects out of 15.

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1 Counting principles 19

20 Topic 1: Algebra

4. Solve the following equations:

(a) (i) nP2 = 42

(b) (i) nP3

n P2

(ii) nP2 = 90

(ii) nP3

n P2

5. In a ‘Magic Sequence’ lottery draw there are 39 balls numbered

1–39. Seven balls are drawn at random. The result is a

sequence of seven numbers which must be matched in the

correct order to win the grand prize. How many possible

sequences can be made?

[4 marks]

6. A teacher needs to select four pupils from a class of 24 to

receive four different prizes. How many possible ways are there

to award the prizes?

[4 marks]

7. How many three-digit numbers can be formed from digits 1–9

if no digit can be repeated?

[4 marks]

8. Eight athletes are running a race. In how many different ways

can the first three places be filled?

[4 marks]

9. An identification number consists of two letters followed by

four digits chosen from 0−9. No digit or letter may appear

more than once. How many different identification numbers

can be made?

[6 marks]

10. Prove that Pn−1 nPn.

[3 marks]

11. Three letters are chosen from the word PICTURE and

arranged in order. How many of the possible permutations

contain at least one vowel?

[6 marks]

12. Eight runners compete in a race. In how many different ways

can the three medals be awarded if James wins either a

gold or a silver?

[6 marks]

13. A class of 18 needs to select a committee consisting of a

President, a Secretary and a Treasurer. How many different

ways are there to form the committee if Rajid does not want to

be President?

[6 marks]

14. Solve the equation 2nP3 6Pn , justifying that you have found

all solutions.

[6 marks]

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1G Keeping objects together or separated
How many letter arrangements of the word ‘SQUARE’ have the Q and U next to each other? How many have the vowels all separated?
When a problem has constraints like this we need some clever tricks to deal with them.
The first type of problem we will look at is where objects are forced to stay together. The trick is to imagine the letters in the word SQUARE as being on tiles. If the Q and the U need to be together, we are really dealing with five tiles, one containing QU:
S QU A R E
We must also remember that the condition is satisfied if Q and U in are in the other order.
S UQ A R E

Is mathematics about finding answers or having strategies and ideas? Has the balance between the two changed as you have progressed through learning mathematics?

Worked example 1.12 How many permutations of the word ‘SQUARE’ have the Q and the U next to each other?

Describe the problem accurately Apply the product principle

Find the number of permutations of the five ‘tiles’ (S, QU, A, R, E) AND Find the number of permutations of the letters QU on the ‘double tile’

(5! ways) (2! ways)

Number of ways = 5! × 2! = 240

KEY POINT 1.9
If a group of items have to be kept together, treat the items as one object. Remember that there may be permutations of the items within this group too.

Another sort of constraint is where objects have to be kept apart. This is not the opposite of objects being kept together unless we are separating only two objects. For example, the opposite of three objects staying together includes having two of them together and the third one apart. So when dealing with keeping objects apart we need to focus on the gaps that the critical objects can fit into.
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1 Counting principles 21

Consider the question of how many permutations of the word SQUARE have none of the vowels together. We first permute all of the consonants. One such permutation is

S

Q

R

possible positions for vowels
There are four gaps in which we can put the vowels. We only have three vowels and so only need to choose three of the gaps, and then decide in what order to insert the vowels.

Worked example 1.13 How many permutations of the word ‘SQUARE’ have none of the vowels together?

Describe the problem accurately
Apply the product principle

Permute 3 consonants AND
Choose 3 out of the four gaps
AND Permute the three vowels to put into the gaps

3! ways

⎛⎞ ⎝⎜ 3⎠⎟

ways

3! ways

Total number of ways 3! × 4 × 3! = 144

KEY POINT 1.10
If k objects have to be kept apart, permute all the other objects and then count the permutations of k ‘gaps’.

Exercise 1G

1. In how many ways can 14 people be arranged in a line if

Joshua and Jolene have to stand together?

[6 marks]

2. Students from three different classes are standing in the lunch queue. There are six students from 10A, four from 10B and four from 10C. In how many ways can the queue be arranged if students from the same class have to stand together? [6 marks]

22 Topic 1: Algebra

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3. In how many ways can six Biology books and three Physics

books be arranged on the shelf if the three Physics books are

always next to each other?

[6 marks]

4. In a photo there are three families (six Greens, four Browns

and seven Grays) arranged in a row. The Browns have had an

argument so no Brown will stand next to another Brown. How

many different permutations are permitted?

[6 marks]

5. In a cinema there are 15 seats in a row. In how many ways can seven friends be seated in the same row if

(a) there are no restrictions?

(b) they all want to sit together?

[6 marks]

6. Five women and four men stand in a line.
(a) In how many arrangements will all four men stand next to each other?
(b) In how many arrangements will all the men stand next to each other and all the women stand next to each other?

(c) In how many arrangements will all the men be apart?

(d) In how many arrangements will all the men be apart and

all the women be apart?

[6 marks]

Summary
• The product principle (‘AND rule’) states the total number of possible outcomes when there are two choices (A and B) to be made is the product of the number of outcomes for each choice:
n(A AND B) n(A) n(B)
• The addition principle (‘OR rule’) states that the number of outcomes when we are interested in either the first choice A OR the second choice B being made is the sum of the number of outcomes for each choice. A and B must be mutually exclusive.
n(AOR B) n(A) n(B)
• The number of permutations (specific arrangements) of n different objects is n!; n! = n (n − 1) (n − 2) . . . × 2 × 1
• The link between one factorial and the next can be shown algebraically as: (n + 1)! (n + 1) × n!

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1 Counting principles 23

• If we are selecting r out of n objects, and the order in which we do so does not matter, we ⎛ n⎞
describe this as a combination and there are ⎝⎜ r ⎠⎟ or nCr ways to do this. This has the formula

⎛ n⎞ ⎝⎜ r ⎠⎟

=

(n

n!
r )!r !

• There are nPr ways of selecting r out of n objects when the order in which we select does matter.

n Pr

=

(n

n!
r )!

• The exclusion principle says that to find the number of permutations which do not have a certain property, find the number of permutations that do have the property and subtract it from the number of all possible permutations.
• If a group of items have to be kept together, treat them as one object. Remember that there may be permutations within this group too.
• If objects have to be kept apart, permute all the other objects and then count the permutations of the ‘gaps’.

Introductory problem revisited
If a computer can print a line containing all 26 letters of the alphabet in 0.01 seconds, estimate how long it would take to print all possible permutations of the alphabet.
There are 26! ways to permute all the letters of the alphabet. A calculator tells us that this is approximately 4.03 × 1026 ways. At 0.01 seconds per line, this is 4.03 × 1024 seconds; 4.67 × 1019 days or 1.28 × 1017 years. The age of the universe is estimated at around 14 billion years, so this is approximately 10 million times longer than the universe has existed – it is hard to understand such large numbers!

According to the Guinness World RecordsTM (Book of Records) the largest number which has ever been found to have a use comes from the field of counting. It is called Graham’s Number and it is so large that it is not possible to write it out in normal notation even using all the paper in the world, so a new system of writing numbers had to be invented to describe it.

24 Topic 1: Algebra

© Cambridge University Press 2012 Not for printing, sharing or distribution.

Mixed examination practice 1

Short questions

1. Seven athletes take part in the 100 m final of the Olympic games. In how many

ways can three medals be awarded?

[4 marks]

2. In how many ways can five different letters be put into five different envelopes? [5 marks]

3. In how many ways can ten cartoon characters stand in a queue if Mickey, Bugs Bunny and Jerry must occupy the first three places in some order? [5 marks]

4. How many three digit numbers contain no zeros?

[6 marks]

5. A committee of four children is chosen from eight children. The two oldest

children cannot both be chosen. Find the number of ways the committee may

be chosen.

[6 marks]

(© IB Organization 2003)
6. Solve the equation (n + 1)! 30(n −1)! for n ∈N.

(Remember: N is the set of natural (whole non-negative) numbers.) [5 marks]

7. How many permutations of the word ‘CAROUSEL’ start and end in a

consonant?

[5 marks]

8.

Solve

the equation

⎛ n⎞ ⎝⎜ 2⎠⎟

= 105.

[6 marks]

9. A group of 15 students contains seven boys and eight girls. In how many ways can a committee of five be selected if it must contain at least one boy? [6 marks]

10. Abigail, Bahar, Chris, Dasha, Eustace and Franz are sitting next to each other

in six seats in a cinema. Bahar and Eustace cannot sit next to each other. In

how many different ways can they permute themselves?

[6 marks]

11. A committee of five is to be selected from a group of 12 children. The two

youngest cannot both be on the committee. In how many ways can the

committee be selected?

[6 marks]

12. A car registration number consists of three different letters followed by five

digits chosen from 1−9 (the digits can be repeated). How many different

registration numbers can be made?

[6 marks]

13. A van has eight seats, two at the front, a row of three in the middle and a row

of three at the back. If only 5 out of a group of 8 people can drive, in how

many different ways can they be arranged in the car?

[6 marks]

14. Ten people are to travel in one car (taking four people) and one van (taking

six people). Only two of the people can drive. In how many ways can they

be allocated to the two vehicles? (The permutation of the passengers in each

vehicle is not important.)

[7 marks]

© Cambridge University Press 2012 Not for printing, sharing or distribution.

1 Counting principles 25

Long questions
1. Five girls, Anya, Beth, Carol, Dasha and Elena, stand in a line. How many possible permutations are there in which (a) Anya is at one end of the line? (b) Anya is not at either end? (c) Anya is at the left of the line or Elena is on the right, or both? [9 marks]

2. In how many ways can five different sweets be split amongst two people if

(a) each person must have at least one sweet?

(b) one person can take all of the sweets?

(c) one of the sweets is split into two to be shared, and each person

gets two of the remaining sweets?

[9 marks]

3. In a doctor’s waiting room, there are 14 seats in a row. Eight people are waiting to be seen.

(a) In how many ways can they be seated?

(b) Three of the people are all in the same family and they want to sit together. How many ways can this happen?

(c) The family no longer have to sit together, but there is someone with a very

bad cough who must sit at least one seat away from anyone else. How many

ways can this happen?

[8 marks]

4. (a) Explain why the number of ways of arranging the letters RRDD, given that

all

the

R’s

and

all

the

D’s

are

indistinguishable

is

⎛ ⎝⎜

4⎞ 2⎠⎟

.

(b) How many ways are there of arranging n R’s and n D’s?

(c) A miner is digging a tunnel on a four by four grid. He starts in the top left box and wants to get to the gold in the bottom right box. He can only tunnel directly right or directly down one box at a time. How many different routes can he take?

(d) What will be the general formula for the number of routes when digging on

an n by m grid?

[10 marks]

26 Topic 1: Algebra

© Cambridge University Press 2012 Not for printing, sharing or distribution.

5. 12 people need to be split up into teams for a quiz.

(a) Show that the number of ways of splitting them into two groups of the

same

size

is

1 2

⎛12⎞ ⎝⎜ 6⎠⎟

.

(b) How many ways are there of splitting them into two groups of any size (but there must be at least one person in each group)?

(c) How many ways are there of splitting them into three groups of

four people?

[9 marks]

6. (a) How many different ways are there to select a group of three from a class of 31 people?

(b) In another class there are 1540 ways of selecting a group of three people. How many people are there in the class?

(c) In another class the teacher noted that the number of ways to select a group

of size three is 100 times larger than the number of people in the class. How

many people are in the class?

[9 marks]

© Cambridge University Press 2012 Not for printing, sharing or distribution.

1 Counting principles 27

In this chapter you will learn:
• some rules for dealing with exponents
• about a function where the unknown is in the exponent
• about the value e and some of its properties
• how to undo exponential functions using an operation called a logarithm
• the rules of logarithms
• about graphs of logarithms
• to use logarithms to find exact solutions to simple exponential equations.

2Exponents and logarithms
Introductory problem A radioactive substance has a half-life of 72 years. A 1 kg block of the substance is found to have a radioactivity of 25 million Becquerel (Bq). How long, to the nearest 10 years, would it take for the radioactivity to have fallen below 10 000 Bq?
Many mathematical models (biological, physical and financial in particular) involve the concept of continuous growth or decay where the rate of growth/decay of the population is linked to the size of that population. You may have met similar situations already, for example when a bank account earns compound interest: the increase in amount each year depends upon how much is in the account. Any similar situation is governed by an exponential function, which you will learn about in this chapter.
We shall also look at how we can find out how long an exponential process has been occurring, using a function called a logarithm.

See Prior learning Section C on the CDROM which looks at exponents.

2A Laws of exponents
The exponent of a number shows you how many times the number is to be multiplied by itself. You will have already met some of the rules for dealing with exponents before, and in this section we shall revisit and extend these rules. (In other courses, the exponent might have been called the ‘index’ or ‘power’.)

28 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

A number written in exponent form is one which explicitly looks like:
an n is referred to as the exponent or power a is referred to as the base
an is pronounced ‘a to the exponent n’ or, more simply, ‘a to the n’. To investigate the rules of exponents let us consider an example:

Worked example 2.1

Simplify. (a) a3 a4

(b) a3 a4

(c) (a4 )3

(d) a4 a3

Mathematics is often considered a subject without ambiguity.
However, the value of 00 is undetermined; it depends upon how you get there!

Use the idea from part (a)

(a) a3 a4 = (a × a × a × a) × (a × a × a) = a7

(b) a3

a4 = a × a × a = 1 = a−1 a×a×a ×a a

( ) (c) a4 3 = a4 a4 a4 = a12

(d) a4 a3 = a3(a + 1)

The example above suggests some rules of exponents. KEY POINT 2.1
am an = am + n
KEY POINT 2.2 am an = am n

It is questionable whether in part (d) we have actually simplified the expression. Sometimes the way mathematicians choose to simplify expressions is governed by how it looks as well as how it is used.

KEY POINT 2.3 ( )n am × n

1
We can use Key point 2.3 to justify the interpretation of an
( ) as the nth root of a, since a1 n a1×n a. This is exactly the

property we require of the nth root of a. So, we get the rule:

( ) am

m
an .

exam hint
Tgbthihiinrofveeaoeysbwtenkooyrlueritoihtnuteus.leedtMihcetsiaeraaea2nsfkcr6oeetui(ryos2mNsoen3uuusO)rt2l,heacTeeaam.nngd. if yrweowauyrsiseteewiit(l2la3bs)e22i6ym.opBuoocrthatannt!

© Cambridge University Press 2012

2 Exponents and logarithms 29

Not for printing, sharing or distribution.

Worked example 2.2
2
Evaluate 643.

Use am /n

⎛ ⎝

a

1 n

⎞ ⎠⎟

m

to split the

calculation into two steps

2
64 3

=

⎛ 1⎞2 ⎝⎜ 64 3 ⎠⎟

= (4)2

2
So, 64 3 = 16

You must take care when expressions with different bases are to be combined by multiplication or division, for example 23 × 42. The rules such as ‘multiplication means add the exponents together’ are only true when the bases are the same. You cannot use this rule to simplify 23 × 42.
There is however, a rule that works when the bases are different but the exponents are the same.
Consider the following example: 32 52 = 3 3 5 5
= 3×5×3×5 = 15 ×15 = 152
This suggests the following rules:
KEY POINT 2.4
a bn = (ab)n

KEY POINT 2.5

a

bn

=

⎛ ⎝

a⎞ b⎠

n

30 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

Exercise 2A

1. Simplify the following, leaving your answer in exponent form.

(a) (i) 64 63

(ii) 53 55

(b) (i) a3 a5

(ii) x6 x3

(c) (i) 711 7−14

(ii) 57 5 2

(d) (i) x4 x 2

(ii) x8 x 3

(e) (i) g −3 × g −9

(ii) k k−6

2. Simplify the following, leaving your answer in exponent form.

(a) (i) 64 63

(ii) 53 55

(b) (i) a3 a5

(ii) x6 x3

(c) (i) 57 5 2

(ii) 711 7−4

(d) (i) x4 x 2

(ii) x8 x 3

(e) (i) 2 5 2 7

(ii) 3 6 38

(f) (i) g −3 ÷ g −9

(ii) k k6

3. Express the following in the form required.

(a) (i) (23 )4 as 2n

(ii) (32 )7 as 3n

(b) (i) (5−1)4 as 5n (c) (i) ( ) 11−2 −1 as 11n
(ii) ( ) 13−3 −5 as 13n (d) (i) 4 (25 )3 as 2n
(ii) 3 5 (9 )1 −4 as 3n

(ii) (7−3 )2 as 7n

(e) (i) (42 )3 312 as 6n (ii) (63 )2 (22 )3 as 3n

4. Simplify the following, leaving your answer in exponent form with a prime number as the base.

(a) (i) 45

(ii) 97

(b) (i) 83

(ii) 165

(c) (i) 42 83

(ii) 95 272

(d) (i) (e) (i) (f) (i)

4 3 85

⎛ 1⎞3

⎝ 4⎠

⎛ ⎝

1 8

⎞ ⎠

2

÷

⎛ ⎝

1 4

⎞ ⎠

4

(ii) 37 9 2

(ii)

⎛ 1⎞3 ⎝ 9⎠

(ii)

97

×

⎛ ⎝

1⎞ 3⎠

4

© Cambridge University Press 2012

2 Exponents and logarithms 31

Not for printing, sharing or distribution.

In Section 2G you will see that there is an easier way to solve equations like this when you have a calculator and can use logarithms.

5. Write the following without brackets or negative exponents:

(a) (i) (2x2 )3

(ii) (3x4 )2

(b) (i) 2(x2 )3

(ii) 3(x4 )2

(3a3 )4
(c) (i) 9a2
(d) (i) (2x)−1

( 4 x )4 (ii) 8(2x)4

(ii)

⎛ 3 ⎞ −2 ⎝⎜ y ⎠⎟

(e) (i) 2x−1

(f) (i)

5

÷

⎛ ⎝⎜

3 xy

2

⎞ ⎠⎟

2

(ii) 3 y −2

(ii)

⎛ ⎝

ab 2

⎞ ⎠

3

÷

⎛ ⎝

a b

⎞ ⎠

2

(g) (i)

⎛ ⎝⎜

2 q

⎞ ⎠⎟

2

÷

⎛ ⎝

p 2

⎞ ⎠

− 3

(ii)

⎛ ⎝

6 x

⎞ ⎠

4

÷

⎛ ⎝

2

×

32 x

⎞ ⎠

−3

6. Simplify the following:
1
(a) (i) (x6 )2

(b) (i) ( ) 4x10 0 5

(c) (i)

⎛

27

x

9

⎞

−

1 3

⎝⎜ 64 ⎠⎟

4
(ii) (x9 )3

( ) (ii)

8x12

− 1 3

(ii)

⎛ x 4 ⎞ −1 5 ⎝⎜ y8 ⎠⎟

7. Solve for x, giving your answer as a rational value:

(a) (i) 8 32 (b) (i) 1 = 7
49x

(ii) 25x = 1 125
(ii) 1 = 8 16x

(c) (i) 2 3x = 162

(ii) 3 5x = 0 12

(d) (i) 2 5x 1 = 250

(ii) 5 3x+2 = 14

(e) (i) 16 + 2 = 2x+1

(ii) 100x+5 = 103x−1

(f) (i) 6x+1 = 162 × 2x

(ii) 41 5 2 × 16x−1

8. Any simple computer program is able to sort n input values in

k n1 5 microseconds. Observations show that it sorts a million

values in half a second. Find the value of k.

[3 marks]

9. A square-ended cuboid has volume xy2, where x and y are

lengths. A cuboid for which x = 2y has volume 128 cm3.

Find x.

[3 marks]

32 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

10. The volume and surface area of a family of regular solid shapes are related by the formula V kA1 5, where V is given in cm3 and A in cm2.

(a) For one such shape, A = 81 and V = 243. Find k.

(b) Hence determine the surface area of a shape
with volume 64 cm3. 3

[4 marks]

11. Prove that 2350 is larger than 5150.

[5 marks]

12. Given that there is more than one solution value of x

to the equation 4 b 8x, find all possible values

of a and b.

[5 marks]

2B Exponential functions
In most of the equations that you have met so far, the unknown appears as the base, for example x3 = 27.
In an exponential function, the unknown appears in the exponent, leading to a fundamentally different type of function.
The general form of a simple exponential function is f (x) = ax.
We will only consider situations when a is positive, because otherwise some exponents cannot be easily defined (for example, we cannot square root a negative number).
Here is the graph of y = 2x:

y

9

8

7

6

5

4

3

2

y = 2x

1

−3 −2 −1

x

1

2

3

1
What is ( 1)2?
( )1
What about (− ) 4?
2
What about (−1)4?
Not all mathematics is unambiguous!

© Cambridge University Press 2012

2 Exponents and logarithms 33

Not for printing, sharing or distribution.

y

=

(

1 2

)x

y y = 5x

y = 2x

y

=

(

3 2

)x

x

You may observe that the blue line is a reflection of the black line. You will see why this is the case in chapter 6.

For very large positive values of x, the y-value approaches infinity. For very large negative values of x, the y-value approaches (but never reaches) zero.
A line that a function gets increasingly close to but never reaches is called an asymptote. In this case we would say that the x-axis is an asymptote to the graph.
If we look at the graphs of other exponential functions with different bases we can begin to make some generalisations.
For all the graphs y = ax:
• The y-intercept is always (0, 1) because a0 = 1. • The graph of the function lies entirely above the x-axis since
ax > 0 for all values of x. • The x-axis is an asymptote. • If a > 1, then as x increases, so does f(x). This is called a
positive exponential. • If 0 < a < 1, then as x increases, f(x) decreases. This is called a
negative exponential.
Many mathematical models use the characteristics of exponential functions: as time (t) increases by a fixed value, the value we are interested in (N ) will change by a fixed factor, called the growth factor. Exponential functions can therefore be used to model many physical, financial and biological forms of exponential growth (positive exponential models) and exponential decay (negative exponential models).
To model more complex situations we may need to add more constants to our exponential equation.
⎛t⎞
We can use a function of the form N Ba⎝ k⎠.
We can interpret the constants in the following way:
• When t = 0, N = B so B is the initial value of N. • When t = k, N = Ba so k is the time taken for N to increase by
a factor of a. • If k = 1 then a is the growth factor. • As long as k is positive when a > 1 the function models
exponential growth. • When 0 1 the function models exponential decay. • When modelling exponential decay there might sometimes
be a background level. This means the asymptote is not N = 0.
We can change the asymptote to N = c by adding on a constant.

34 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

⎛t⎞
The new function is then N Ba⎝ k⎠ + c.
B represents how much N starts above the background level, so the initial value is B + c.
KEY POINT 2.6
⎛t⎞
If N Ba⎝ k⎠ + c:
• The background level (and the asymptote) is c. • The initial value is B + c. • k is the time taken for the difference between N and the
background level to increase by a factor of a. • If a > 1 this models exponential growth. • If 0 1 this models exponential decay.
In many applications we are given certain information and need to find the constants in the model.

exam hint SsChohAengomDtneaeothslrihwtaea-ispafootRygCeartrsppauoowOgyhtsyk2aorghmuytenmhoplMa,hrmcnosoyxavrputykconwehpiyartltimoeataototeotlttbouulshoctthaiuceosrefithyeheofuetsreqncoscilwstlmgaladhouawsdyoakrteanolpioftacoriwrotktloeruexihglpu.ioshcrl.nfietuhatnteagnmTelsty.rodhnse,asritst

Worked example 2.3
A population of bacteria in a culture medium doubles in size every 15 minutes.
(a) Write down a model for N, the number of bacteria in terms of time, t, in hours. (b) If there are 1000 bacterial cells at 08:00, how many cells are there at
(i) 08:15? (ii) 09:24?

There is a constant increasing factor so use an exponential growth model
Every time t increases by 0.25, N doubles
All details are relative to a start time 08:00, so set t = 0 at that time
Remember to convert minutes to hours

(a) Let N be the number of cells at time t hours
⎛t⎞
N Ba⎝ k⎠
Doubles every quarter hour a = 2, k = 0.25 ∴ N B × 24t
(b) When t 0, N = 1000 = B N = 1000 × 24t
(i) When t = 0.25, N = 2000 cells (ii) When t = 1.4, N = 48 503 cells

© Cambridge University Press 2012

2 Exponents and logarithms 35

Not for printing, sharing or distribution.

You may also be given a model and have to interpret it.

Unfortunately, it might not be given in exactly the same form as

⎛

t⎞

in Key point 2.6 ⎝⎜ N Bak ⎠⎟ , so you will have to use the rules of

exponents to rewrite it in the correct form.

Worked example 2.4

The temperature in degrees (θ ) of a cup of coffee a time t minutes after it was made is modelled using the function:
θ = 70 × 3−βt + 22, β > 0
(a) Show that θ – 22 follows an exponential decay. (b) What was the initial temperature of the coffee? (c) If the coffee is left for a very long time, what temperature does the model predict it will reach? (d) Find, in terms of β, how long it takes for the temperature difference between the coffee and
the room to fall by a factor of 9. (e) Coffee made at the same time is put into a thermos flask which is much better insulated
than the cup. State with a reason whether β would increase or decrease.

Use rules of exponents to rewrite the exponent in the
⎛t⎞
form N Ba⎝⎜ k⎠⎟

(a) θ − 22 = 70 × 3−βt
( ) = 70 × 3−1 βt

t

=

70

×

⎛ ⎝

1⎞ 3⎠

β

−1

which is exponential decay since the base is between 0

and 1 and β > 0

Initial means t = 0

(b) When t = 0, θ = 70 × 30 + 22 = 92°

The asymptote is given by the background level
Use algebra to rewrite the
t
exponent in the form N Bak

(c) 22 degrees

t

(d)

θ

−

22

=

70

⎛ ⎝

1 3

⎞ ⎠

β−1

So the time to fall by a factor of 3 is 1 β

To fall by a factor of 9 is a fall by a factor of 3 followed
by another fall by a factor of 3. Therefore the time to fall by a factor of 9 is 2 × 1 minutes.
β

36 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

continued . . .

(e) The time taken to fall by any given factor will be
longer for a better insulated container so 1 must be larger so β must be smaller. β

When modelling exponential growth or decay, you may be given a percentage increase or decrease. This needs to be converted into a growth factor to be used in the exponential model.
Worked example 2.5
A car costs $17 500. It then loses value at a rate of 18% each year. (a) Write a model for the value of the car (V) after n years in the form V kan. (b) Hence or otherwise find the value of the car after 20 years.

Find the growth factor Use initial value information

(a) The growth factor is 1 − 18 = 0.82 100
When n = 0, V k = 17500 V = 17500 × 0.82n

Substitute for n

(b) After 20 years, V = $330.61

Exercise 2B
1. Using your calculator, sketch the following functions for −5 ≤ ≤ 5 and 0 ≤ y ≤ 10.

Show all the axis intercepts and state the equation of the horizontal asymptote.

(a) (i) y = 1 5x

(ii) y = 3x

(b) (i) y = 2 × 3x

(c) (i)

y

=

⎛ ⎝

1⎞ 2 ⎠

x

(ii) y = 6 ×1.4x

(ii)

y

=

⎛ ⎝

2⎞ 3 ⎠

x

(d) (i) y = 5 + 2x

(ii) y = 8 + 3x

(e) (i) y = 6 − 2x

(ii) y = 1 − 5x

2. An algal population on the surface of a pond grows by 10%

every day, and the area it covers can be modelled by the

equation y k ×1.1t, where t is measured in days. At 09:00 on

Tuesday it covered 10 m2. What area will it cover by 09:00 on

Friday?

[4 marks]

© Cambridge University Press 2012

2 Exponents and logarithms 37

Not for printing, sharing or distribution.

3. The air temperature T oC around a light bulb is given by equation

T

=

A

+

B

×

2 −

x k

where x is the distance from the surface of the light bulb in millimetres. The background temperature in the room is a constant 25 oC, and the temperature on the surface of the light bulb is 125 oC.

(a) You find that the air temperature 3 mm from the surface of the bulb is only 75 oC. Find the integer values of A, B and k.

(b) Determine the air temperature 2 cm from the surface of the bulb.

(c) Sketch a graph of air temperature against distance. [10 marks]

4. A tree branch is observed to bend as the fruit growing on it increases in size. By estimating the mass of the developing fruit, and plotting the data over time, a student finds that the height in metres of the branch tip above the ground closely follows the graph of:

h = 2 − 0.2 ×1.60 2m

where m is the estimated mass, in kilograms, of fruit on the branch.

(a) Plot a graph of h against m.

(b) What height above ground level is a branch without fruit?

(c) The total mass of fruit on the branch at harvest was 7.5 kg. Find the height of the branch immediately prior to harvest.

(d) The student wishes to estimate what mass of fruit would

cause the branch tip to touch the ground. Why might his

model not be suitable to assess this?

[10 marks]

5. (a) Sketch the graph of y = 1+ 161−x2.

Label clearly the horizontal asymptote and maximum value.

(b) Find all values of x for which y = 3.

[6 marks]

6. A bowl of soup is served at a temperature of 55 oC in a room

with a constant air temperature of 20 oC. Every 5 minutes,

the temperature difference between the soup and the room

air decreases by 30%. Assuming the room air temperature is

constant, at what temperature will the soup be seven

minutes after serving?

[7 marks]

38 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

7. The speed (V metres per second) of a parachutist t seconds after jumping from an aeroplane is modelled by the expression:

V = 40(1 − 3−0 1t )

(a) Find his initial speed. (b) What speed does the model predict that he will
eventually reach?

[6 marks]

2C The value e

In this section we introduce a mathematical constant, e, which will be used extensively in the rest of this chapter and in many other chapters throughout the course.

Consider the following different situations, each of which is typical of early population growth of a cell culture.

(a) There is a 100% increase every 100 seconds. (b) There is a 50% increase every 50 seconds. (c) There is a 25% increase every 25 seconds.

Although these may at first appear to be equivalent statements, they are subtly different because of the compounding nature of percentage increases.

If we begin with a population of size P, then after 100 seconds, we have
(a) P ( ) P

(b) P

⎛ ⎝

1⎞ 2 2⎠

. 5P

(c) P

⎛ ⎝

1⎞4 4⎠

.P

To

generalise

the

situation,

if

we

considered

an

increase

of

1 n

%

which occurred n times every 100 seconds, the population after

100 seconds

would

be

given

by

P × ⎛⎝1+

1 n

⎞ ⎠

n
.

It may seem from the above that as n increases, the overall

increase over 100 seconds will keep on getting larger, and this is

indeed the case, but not without limit.

© Cambridge University Press 2012

2 Exponents and logarithms 39

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π and e have many similar properties. Both are irrational, meaning that they cannot be written as a ratio of two whole numbers and both are transcendental, meaning that they cannot be written as the solution to a polynomial equation. The proof of these facts is intricate but beautiful.
In chapter 16 you will see that e plays a major role in studying rates of change.

In fact, as can be seen by taking greater and greater values of n,

the resultant increase factor

⎛⎝1

+

1 n

⎞ ⎠

n

tends towards a value of

approximately 2.71828182849 which, much like π, is such an

important part of mathematical studies that it has been given its

own letter, e.

KEY POINT 2.7

e = 2 71828182849…

Although e has many important properties in mathematics, it is still just a number, and so all the standard rules of arithmetic and exponents apply. You are not likely to find an exam question just on e, but the number e is likely to appear in questions on many other topics.
exam hint
In questions involving the number e you may be asked to either give an exact answer (such as e2) or to use your calculator, in which case you should usually round the answer to 3 significant figures.

Exercise 2C

1. Find the values of the following to 3 significant figures:

(a) (i) e + 1 (b) (i) 3e (c) (i) e2 (d) (i) 5e0 5

(ii) e − 4
(ii) e 2
(ii) e−3
(ii) 3 e7

2. Evaluate 6 (

). What do you notice about this result?

See Supplementary sheet 2 ‘Logarithmic scales and log-log graphs‘ on the CDROM if you would like to discover more about logarithms for yourself.

2D Introducing logarithms
In this section we shall look at a new operation which reverses the effect of exponentiating (that is, raising a number to an exponent).
If you are asked to solve x2 3, x ≥ 0
you can either find a decimal approximation (using a calculator or trial and improvement) or you can use the square root symbol: x = 3.

40 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012

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This restates the problem as ‘x is the positive value which when squared gives 3’. Similarly, if asked to solve:
10x = 50 you could use trial and improvement to seek a decimal value:
101 = 10 102 = 100 So x is between 1 and 2: 101 5 = 31.6 101 6 = 39.8 101 7 = 50.1
So the answer is around 1.7. However, just as with squares and square roots there is also a function to answer the question ‘What is the number, which when put as the exponent of 10, gives this value?’ The function is called a base-10 logarithm, written log10. So in the above example: 10x = 50 so x = log10 50.
This means that y = 10x may be re-expressed as x l ( y).
In fact, the base involved need not be 10, but could be any positive value other than 1.
KEY POINT 2.8
b ax ⇔ x ga b
It is worth noting that the two most common bases have abbreviations for their logarithms. Since we use a decimal system of counting, base 10 is the default base for a logarithm, so that log10 x is generally written more simply as just log x, called the common logarithm. Also e, encountered in Section 2C, is considered the ‘natural’ base, and its counterpart the natural logarithm is denoted by ln x.
KEY POINT 2.9 log10 x is often written as log x. loge x is often written as ln x.

The symbol ⇔ has a very specific meaning (implies and is implied by) in mathematical logic. It means that if the left-hand side is true then so is the right-hand side, and also if the righthand side is true then so is the left-hand side.
This is the form written in the Formula booklet; you only need to remember that it means you can switch between them.
See Key point 2.18.

© Cambridge University Press 2012

2 Exponents and logarithms 41

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Since taking a logarithm reverses the process of exponentiation, it follows that: KEY POINT 2.10
log (ax ) x
KEY POINT 2.11 aloga x x

These are sometimes referred to as the cancellation principles.

This sort of ‘cancellation’, similar to stating that (for positive x)

( ) n xn

n
x = n x , is frequently useful when simplifying

logarithm expressions, but you can only apply it when the base of the logarithm and the base of the exponential match.

Worked example 2.6
Evaluate: (a) log5 625 (b) log8 16
Express the argument of the logarithm in exponent form with the same base
Apply the cancellation principle loga (ax) = x
16 is not a power of 8, but they are both powers of 2
We need to convert 24 to an exponent of 8 = 23 Rewrite 4 as 3 × 4 and use amn = (am)n 3
Apply the cancellation principle loga (ax) = x

(a) log5 625 = log5 54

= 4 (b) log8 (16) = log8 (24)

=

log8

⎛ ⎝⎜

3× 4

⎞ ⎠⎟

=

log8

⎛ 4⎞ ⎝⎜ 83 ⎠⎟

=4 3

Actually, you can find the logarithm of a negative number, but the answer turns out not to be real; it is a complex number, which is a new type of number you will meet in chapter 15.

Whenever you raise a positive number to a power, positive or negative, the answer is always positive. Therefore we currently have no answer to a question such as 10x = –3. This means: KEY POINT 2.12
The logarithm of a negative number or zero has no real value.

42 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012

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Exercise 2D

1. Evaluate the following:

(a) (i) (b) (i) (c) (i) (d) (i)
(e) (i)

log3 27

log5 5

log12 1

log

3

1 3

log4 2

(f) (i) log8 8

(g) (i) log8 4

(h) (i) log25 125

(i) (i) log4 2 2 (j) (i) log 0.2

(ii) log4 16

(ii) log3 3

(ii) log15 1

(ii)

log

4

1 64

(ii) log27 3

(ii) log2 2

(ii) log81 27

(ii) log16 32

(ii) log9 81 3

(ii) log 0.5

2. Use a calculator to evaluate each of the following, giving your

answer correct to 3 significant figures:

(a) (i) log 50

(ii)

log

⎛ ⎝

1⎞ 4⎠

(b) (i) ln 0 1

(ii) ln 10

3. Simplify the following expressions:

(a) (i) 7 l g 2 log x

(ii)

(b) (i) (log ) (log y + 3) (ii)

(c) (i) log logb

(ii)

log logb

2 l g 3log x
(log x + 2)2 (log a)2 −1
log a −1

4. Make x the subject of the following:

(a) (i) log3 x y (b) (i) loga x 1+ y (c) (i) logx 3y 3

(ii) log4 x 2 y (ii) loga x y2 (ii) logx y = 2

5. Find the value of x in each of the following:

(a) (i) (b) (i) (c) (i)

log2 x = 32 log5 25 = 5x logx 36 = 2

(ii) log2 x = 4

(ii) log49 7 2x

(ii)

log x

10

=

1 2

exam hint Ololonggmx oibssutdtcteoannloc,utaeladntdowrlinsthxa by a ln button. eRibwsxeemjaautyemsretmaaahsbteenaidrunnmtytthhbaveeat sr‘ralisoamogbelcxea.’n

© Cambridge University Press 2012

2 Exponents and logarithms 43

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6. Solve the equation log (9 1) = 3.

[4 marks]

7.

Solve the equation log8

1−

x

=

1 .

3

[4 marks]

8. Find the exact solution to the equation ln(3 1) = 2.
[5 marks]

9. Find all values of x which satisfy (log3 x)2 = 4.
10. Solve the simultaneous equations: log x log5 y = 6 log x log5 y = 2
11. Solve the equation 3(1+ log ) = 6 + log x.

[5 marks]
[6 marks] [5 marks]

12. Solve the equation logx 4 9.

[4 marks]

13. The Richter scale is a way of measuring the strength of earthquakes. An increase of one unit on the Richter scale corresponds to an increase by a factor of 10 in the strength of the earthquake. What would be the Richter level of an earthquake which is twice as strong as a level 5.2 earthquake? [5 marks]

2E Laws of logarithms
Just as there are rules which hold when working with exponents, so there are corresponding rules which apply to logarithms. These are derived from the laws of exponents and can be found on the Fill-in proof sheet 18. ‘Differentiating logarithmic functions graphically’ on the CD-ROM.
• The logarithm of a product is the sum of the logarithms.

KEY POINT 2.13

log

loga x + loga y for x, y > 0

44 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

• The logarithm of a quotient is the difference of the logarithms.

KEY POINT 2.14

log x y

loga x

loga y for x, y > 0

• The logarithm of a reciprocal is the negative of the logarithm.

KEY POINT 2.15

log 1 x

loga x for x > 0

• The logarithm of a exponent is the multiple of the logarithm.

KEY POINT 2.16

log

loga x for x > 0

exam hint

These not in

formulae are the Formula

booklet.

exam hint IwtdOmseiolhisoniimastgwiehttmpaexiyvltkprihoefleoyolruolrygoilgtscoagycaxgtna.oon(tr+mxnitttroomhl+yotmkogytnosn)y.oiwnotro

• The logarithm of 1 is always 0, irrespective of the base. KEY POINT 2.17
loga 1 0

The rules of logarithms can be used to manipulate expressions and solve equations involving logarithms.

Worked example 2.7

If x l g10 a and y

b,

express

log10

100a2 b

in terms of x, y and integers.

Use laws of logs to isolate log10 a and log10 b First, use log x = log x − log y y ...then log xy = log x + log y…

log 100a2 b

log10 (

2 ) − log10 b

= log

+ log10 a2 − log10 b

… then log xp = p log x…

= log

+ log10 a − log10 b

… then calculate log10100 … then write in terms of x and y

= 2 + 2log10 − log10 b = 2 + 2x − y

© Cambridge University Press 2012

2 Exponents and logarithms 45

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Worked example 2.8
Solve the equation log log2 (x + 4) 5.

Rewrite one side as a single logarithm using log x + log y = log xy
Undo the logarithm by exponentiating both sides with base 2 (to balance, you must also exponentiate 5 by base 2) ...
and use the cancellation principle
Use standard methods for quadratic equations
Check your solution in the original equation

log log2 (x + 4) 5
⇔ l g2 ( ( + )) = 5
⇔ 2l ( ( + )) = 25
⇔ 2 + 4x = 32
⇔ 2 + 4x − 32 = 0
⇔ ( + )( − ) = 0
⇒ = −8 o x = 4
When x = −8 : LHS : log (−8) log2( 4) are not real so this solution does not work When x = 4: LHS : log 4 log2 8 = 2 3 = 5 = RHS

exam hint
Checking your solutions is more than just looking for an arithmetic error – as you can see from the example above, false solutions can occur through correct algebraic manipulation.

You will notice that although we have discussed logarithms for a general base, a, your calculator may only have buttons for the common logarithm and the natural logarithm (log x and ln x).
To use a calculator to evaluate, for example, log5 20 you can use the change of base rule of logarithms.

KEY POINT 2.18

Change of base rule for logarithms:

logb

a

=

logc logc

a b

46 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

So, we can calculate log5 20 using the logarithm to the base 10:

log5 20

=

log 20 log 5

= 1.86

(3

SF)

The change of base rule is useful for more than just evaluating logarithms.

For an insight into what mathematics was like before calculators, have a look at Supplementary sheet 3, ‘A history of logarithms‘ on the CD-ROM.

Worked example 2.9 Solve the equation log3 x + log9 x = 2.
We want to have logarithms involving just one base, so use the change of base rule to turn logs with base 9 into logs with base 3
Collecting the logs together
Exponentiate both sides with base 3

log9

x

=

log3 log3

x 9

=

log3 2

x

Therefore: log log9 x = 2

⇔

log3

x

+

log3 2

x

=

2

⇔

3l 2

g3

=2

⇔ l g3

=4 3

4
⇔ = 3 = 4.33 (3SF)

Equations combining logarithms and quadratics can be found in Section 4B.

Exercise 2E

1. Given b > 0, simplify each of the following:

(a) (i) logb b4 (b) (i) log b b3

(ii) logb b

(ii) log

logb2 b

2. If x l g a, y b and z l g c, express the following in terms of x, y and z:

(a) (i) log b7

(ii) log a b

(b) (i)

log

⎛ ⎝

ab2 c

⎞ ⎠

(ii)

log

⎛ ⎝

a2 bc3

⎞ ⎠

© Cambridge University Press 2012

2 Exponents and logarithms 47

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(c) (i)

log

⎛ ⎝

100 bc5

⎞ ⎠

(d) (i) log a3 l g ab2

(e) (i) loga a b

(f) (i) logab (ba )

(ii) log( ) log(2c2 )

(ii) log( b)

(ii)

logb

⎛ ⎝

a bc

⎞ ⎠

(ii) logab ac2

(5ac)

3. Solve the following for x:

(a) (i)

log3

⎛ ⎝

2 2

+ −

x⎞ x ⎠

=

3

(ii) log2 (7

(b) (i) log log3 (x − 6) 1 (ii) log

4) = 5

log8

⎛ ⎝

1⎞ x⎠

=

1

(c) (i) log
(d) (i) log
(e) (i) log (
(ii) 2 log(

log4 x

(ii) log

log8 x = 2

(ii) log

) log3 ( 1) = 2
2) log x = 0

log2 x = 4 log32 x = 0.5

(f) (i) log( )

log x

(ii) log(3 6) log 3 1

4. Find the exact solution to the equation 2 l l 9 = 3,

giving your answer in the form AeB where A and B

are rational numbers.

[5 marks]

5. If a = ln2 and b = ln5, find in terms of a and b:

(a) ln50

(b) ln0.16

[6 marks]

6. Solve log logx 2.

[5 marks]

7. Prove that if a by = (ab)xy where a, b > 1 then x + y = 1

or x = y = 0.

[5 marks]

8. Evaluate log 1 log 2 + log 3 log 4 …+ log 8 log 9 .

2345

9 10

[4 marks]

9. Given that log logb a, and that a,b ≠ 1 and a ≠ b,

find b in terms of a.

[5 marks]

48 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

2F Graphs of logarithms
It is important also to know the graph of the logarithm function, and the various properties of logarithms which we can deduce from it.
Below are the graphs of y log x, y lo x and y ln x:

y

5

4

3

y = log2(x)

2

y = ln(x)

1

y = log(x)

0

x

5

10

15

20

−1

−2

−3

−4

−5

These three curves all have a similar shape.

The

change

of

base

rule

(Key

point

2.18)

states

that

logb

a

=

logc logb

a .
a

So,

log2

x

=

log log

x 2

and

ln x = log x log e

and all other logarithm

functions are simply multiples of the common logarithm

function y lo x.

The graph shows the following important facts about the logarithm function (Key point 2.19).

KEY POINT 2.19
If y = loga x then:
• The graph of y against x crosses the x-axis at (1, 0), because loga 1 = 0 (for any positive value of a).
• Log x is negative for 0 < x < 1 and positive for x > 1. • The graph of the function lies entirely to the right of the
y-axis, since the logarithm of a negative value does not produce a real solution. • The logarithm graph increases throughout: as x tends to infinity so does y. • The y-axis is an asymptote to the curve.

In chapter 6 you will see how this type of change causes a vertical stretch of the graph.
exam hint Vadateorhiekaserlncatnotneieianeocgmcslvaawact.tpehlsuraTonyieahltcharmhoschtihtmyasowupaomnrriitrrcosatdtipdoztttwegeetihoosrfirlahpnyotanetyttlofpaadfsyraihlyoot .mu
You may observe that a logarithm graph is the reflection of an exponential graph. You will see why this is the case in chapter 5.

© Cambridge University Press 2012

2 Exponents and logarithms 49

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Exercise 2F

It is unlikely you will find exam questions testing just this topic, but you may be required to sketch a graph involving a logarithm as a part of another question.

1. Sketch the following graphs, labelling clearly the vertical asymptote and all axis intercepts.

(a) (i) y log(x2)

(ii) y l (x3)

(b) (i) y l 4x

(ii) y l 2x

(c) (i) y l (x − 2) (ii) y l (x +1) 2. Why are the graphs of y = log (x2) and y = 2log (x) different?

2G Solving exponential equations
One of the main uses of logarithms is to solve equations with the unknown in the exponent. By taking logarithms, the unknown becomes a factor, which is easier to deal with.

Worked example 2.10

Solve the equation 3 2x = 5x−1 , giving your answer in the form log p where p and q are

rational numbers.

log q

Since the unknown is in the exponent, taking logarithms is a good idea
Use the rules of logarithms to simplify the expression
Expand the brackets and get all of the x’s on one side and everything else on the other side Factorise and divide to find x
Use the rules of logarithms to write it in the correct form

log(3 2x ) log(5x−1)

log log2x = log(5x−1)
log log2 = (x − 1)log5

log log2 = xlog5 log5 log3 log5 = xlog5 log2

x = log3 + log5 log5 − log2

x = log15

log ⎛⎝

5 2

⎞ ⎠

50 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

exam hint A common mistake is to say that log(3 × 2x) is log3 × log2x Make sure you learn the rules of logarithms carefully.
You can use similar ideas to solve inequalities with the unknown in the exponent:

Logarithmic scales make it easy to compare values which are very different. Some of these scales are explored in Supplementary sheet 2 on the CD-ROM.

Worked example 2.11

The number of bacteria in a culture medium is given by N = 1000 × 24t, where t is the number of hours elapsed since 08:00. At what time will the population first exceed one million?

Simplify the equation where possible Take logarithms of each side

1000 × 24t 1 000 000 ⇔ 24t ≥ 1000
⇔ log( ) ≥ log1000

Use the rule loga x p = p loga x
Note that log 2 > 0, so the inequality remains in the same orientation when we divide

4 l g2 3 ⇔ ≥ 3 = 2.49(3SF)
4 log 2

To answer the question, convert 2.49 to hours and minutes, then add it to 08:00

10:29

You know that you can always apply the same operation to both sides of an equation, but in this worked example we took logarithms of both sides of an inequality. You might like to investigate for which operations this is valid.

exam hint
Whenever dividing an inequality by a logarithm it is important to remember to check if it is positive or negative.

There is another type of exponential equation which you will meet – a disguised quadratic equation. These are explored in Section 4B.

© Cambridge University Press 2012

2 Exponents and logarithms 51

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Exercise 2G

1. Solve for x, giving your answer correct to 3 significant figures.

(a) (i) 3 4x = 90

(ii) 1000 × 1 02 10 000

(b) (i) 6 73 1 = 1.2

(ii) 5 22 5 = 94

(c) (i) 32 4x 1

(ii) 5 61 x

(d) (i) 3 23 = 7 33x−2 (ii) 4 8x−1 = 3 52x+1

2. In a yeast culture, cell numbers are given by N = 100e1 03t, where t is measured in hours after the cells are introduced to the culture.

(a) What is the initial number of cells?

[1 mark]

(b) How many cells will be present after 6 hours?

[1 mark]

(c) How long will it take for the population to exceed one

thousand?

[2 marks]

3. A rumour spreads exponentially through a school. When school begins (at 9 a.m.) 18 people know it. By 10 a.m. 42 people know it.

(a) How many people know it at 10.30?

[3 marks]

(b) There are 1200 people in the school. According to the

exponential model, at what time will everyone know the

rumour?

[2 marks]

4. In an experimental laboratory, a scientist sets up a positive

feedback loop for a fission reaction and extracts heat to control

the experiment and produce power. When the reaction is

established, and while sufficient fuel is present, the power he can

siphon off is given by P = 32(

− ), where P is measured in

units of energy per second and t in seconds.

(a) How much energy is being produced after 2 minutes? [1 mark]

(b) The equipment reaches a dangerous temperature when

P exceeds 7 × 105. For how long can the experiment

safely be run?

[2 marks]

5. The weight of a block of salt W in a salt solution after t seconds is given by:

W ke−0 01t

(a) Sketch the graph of W against t.

[2 marks]

(b) How long will it take to reach 25% of its original weight?

[2 marks]

52 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

6. Solve the equation 5 4x 1 = 1 , giving your answer in the 32 x
form x = ln p , where p and q are rational numbers. [5 marks] ln q
7. Solve the equation 1 = 3 495−x , giving your answer in the 7x
form a + log7 b where a, b ∈Z.

8. A cup of tea is poured at 98 oC. After two minutes it has reached

94 oC. The difference between the temperature of the tea and the

room temperature (22 oC) falls exponentially. Find the time it

takes for the tea to cool to 78 oC.

[5 marks]

9. (a) Show that the equation 3x = 3 − x has only one solution.

[2 marks]

(b) Find the solution, giving your answer to 3 significant

figures.

[4 marks]

Summary

• In this chapter, we revisited the rules for exponents.

am an = am+n

am an = am n
a bn = (ab)n

a

bn

=

⎛ ⎝

a⎞ b⎠

n

( )n am×n

( ) m
a

n a m = n am

• Exponential equations can be used to model growth and decay of some simple real-life systems, taking as a general form the function:
⎛t⎞
N Ba⎝ k⎠ + c
• loga b asks the question: ‘What exponent do I have to raise a to in order to get b?’ b ax ⇔ x ga b
• e is the mathematical constant, (Euler’s number): e = 2.71828182849… • log10 x (common logarithm) is often written as log x.

© Cambridge University Press 2012

2 Exponents and logarithms 53

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• loge x (natural logarithm) is often written as ln x. • Logarithms undo the effect of exponentiating and vice versa:
( ) loga ax = ⇔ x = aloga x
• Logarithms obey these rules, which are not covered in the Formula booklet.

log log ⎛ x ⎞
⎝ y⎠

loga x + loga y loga x loga y

log

⎛ 1⎞ ⎝ x⎠

loga x

log

loga x

loga 1 0

• Taking the logarithm of a negative number or zero does not give a real value.

•

There is also the change of base formula (not in the Formula and a related rule for exponents (is in the Formula booklet):

booklet): ax = ex lna

logb

a

=

logc logc

a b

• The following properties of the logarithm function can be deduced from its graph, if y = loga x then: – The graph of y against x crosses the x-axis at (1,0)

– Log x is negative for 0 < x < 1 and positive for x > 1 – The graph lies to the right of the y-axis; the y-axis is an asymptote to the curve

– The logarithm graph increases throughout; as x tends to infinity so does y.

• Logarithms are used to solve many exponential equations.

Introductory problem revisited
A radioactive substance has a half-life of 72 years. A 1 kg block of the substance is found to have a radioactivity of 25 million Becquerel (Bq). How long, to the nearest 10 years, would it take for the radioactivity to have fallen below 10 000 Bq?

Establish the exponential equation
Initial condition gives B Every time t increases by 72,
R falls by 50%

Let R be the radioactivity after t years
t
R Bak
When t = 0, R = 25 × 106 = B
a = 0.5,k = 72
t
C = 25 × 106 × 0 572

54 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

continued...
The unknown is in the exponent so take logarithms
Now rearrange to find t Note that log(0.5) < 0, so the inequality is
reversed when dividing through

Require R ≤ 104
t
25 × 106 × 0 572 ≤ 104
t
⇔ 0 5 ≤ 0.0004

⇔

⎛ log⎝⎜

t
0.5 72

⎞ ⎠⎟

≤

log(0.0004)

t log(0.5) ≤ log(0.0004)
72

⇔

≥

72log(0.0004)
log(0.5)

=

812

7

It takes around 810 years for the radioactivity to fall below 10 000 Bq.

© Cambridge University Press 2012

2 Exponents and logarithms 55

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Mixed examination practice 2

Short questions
( ) 1. Solve log5 x2 + = 2 .
2. If a log x, b y and c a, b, c and integers:

l g z (all logs base 10) find in terms of

[4 marks]

x2 y (a) log
z

(b) log 0.1x

3. Solve the simultaneous equations:

ln x + ln y2 = 8

(c)

log100

⎛ ⎝

y z

⎞ ⎠

[6 marks]

ln x2 + ln y = 6

[6 marks]

4. If y ln x − ln(x + 2) + l ( x ), express x in terms of y.

[6 marks]

5. Find the exact value of x satisfying the equation

23 2 32x 3 = 36x−1

giving your answer in simplified form ln p , where p q ∈Z. ln q

[5 marks]

6. Given loga b c and logb a c −1 for some value c, where 0 < a < b,

express a in terms of b.

[6 marks]

7.

Solve p

the

equation

9

log5

x

=

25

logx

5,

expressing

your

answers

in

the

form

5q, where p q ∈Z.

[6 marks]

(© IB Organization 2006)

8. Find the exact solution to the equation ln x = 4 logx e .

[5 marks]

Long questions

1. The speed of a parachutist (V) in metres per second, t seconds after jumping is modelled by the expression:

V = 42(1 − −0 2t )

(a) Sketch a graph of V against t. (b) What is the initial speed? (c) What is the maximum speed that the parachutist could reach?

When the parachutist reaches 22 ms-1 he opens the parachute. (d) How long is he falling before he opens his parachute?

[9 marks]

56 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

2. Scientists think that the global population of tigers is falling exponentially. Estimates suggest that in 1970 there were 37 000 tigers but by 1980 the number had dropped to 22 000.

(a) Form a model of the form T kan connecting the number of tigers (T) with the number of years after 1970 (n).

(b) What does the model predict that the population will be in 2020?

(c) When the population reaches 1000 the tiger population will be described as ‘near extinction’. In which year will this happen?

In the year 2000 a worldwide ban on the sale of tiger products was implemented, and it is believed that by 2010 the population of tigers had recovered to 10 000.

(d) If the recovery has been exponential find a model of the form T kam connecting the number of tigers (T) with the number
of years after 2000 (m).

(e) If in each year since 2000 the rate of growth has been the same, find

the percentage increase each year.

[12 marks]

3. (a) If ln y 2 ln x + ln3 find y in terms of x.
(b) If the graph of ln y against ln x is a straight line with gradient 4 and y-intercept 6, find the relationship between x and y.
(c) If the graph of ln y against x is a straight line with gradient 3 and it passes through the point (1, 2), express y in terms of x.
(d) If the graph of ey against x2 is a straight line through the origin with gradient 4, find the gradient of the graph of y against ln x. [10 marks]

© Cambridge University Press 2012

2 Exponents and logarithms 57

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In this chapter you will learn:
• the formal definition of a polynomial
• operations with polynomials
• a trick for factorising polynomials and finding remainders
• how to sketch the graphs of polynomials
• how to identify the number of solutions of a quadratic equation.

3Polynomials
Introductory problem Without using your calculator, solve the cubic equation
x3 − 13x2 + 47x − 35 = 0
You may think that your calculator can ‘magically’ do any calculation, such as sin45 or e−2 or ln2. However, like any computer, it is built from a device called a logic circuit which can only really do addition. If you repeatedly do addition you get multiplication and if you repeatedly do multiplication you raise to a power. Any expression which only uses these operations is called a polynomial. What is particularly surprising is that all the other operations your calculator does can be approximated very accurately using these polynomials.

exam hint Tpschooaemllyloeenrdtodimmietesriasodl afeislgasroee.

3A Working with polynomials
The polynomial functions of x make up a family of functions, each of which can be written as a sum of non-negative integer powers of x. Polynomial functions are classified according to the highest power of x occurring in the function, called the order of the polynomial.

General form of the polynomial Order a0
ax + b 1 ax2 + bx + c 2 ax3 + bx2 + cx + d 3 ax4 + bx3 + cx2 + dx + e 4

Classification Constant polynomial Linear polynomial Quadratic polynomial Cubic polynomial Quartic polynomial

Example

y = 5

y = x + 7

y = −3x2 + 4x − 1

y = 2x3 + 7x

y

=

x4

−

x3

+

2x

+

1 2

58 Topic 2: Functions and equations

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The letters a, b, c, … are called the coefficients of the powers of x, and the coefficient of the highest power of x in the function (a in the table above) is called the lead coefficient and the term containing it is the leading order term.
Coefficients can take any value, with the restriction that the lead coefficient cannot equal zero; a polynomial of order n which had a lead coefficient 0 could be more simply written as a polynomial of order (n − 1). The sign of the lead coefficient dictates whether the polynomial is a positive polynomial (a > 0) or a negative polynomial (a < 0).
Adding and subtracting two polynomials is straightforward – it is just collecting like terms. For example:
(x4 3x2 1) − (2x3 x2 2) = x4 x3 + 4x2 − 3.
Multiplying is a little more difficult. Below is one suggested way of setting out polynomial multiplication to ensure that you include all of the terms.

The Greeks knew how to solve quadratic equations, and general cubics and quartics were first solved in 14th Century Italy. For over three hundred years nobody was able to find a general solution to the quintic equation, until in 1821 Niels Abel used a branch of mathematics called group theory to prove that there could never be a ‘quintic formula’.

Worked example 3.1
Expand (x3 3x2 2)(x2 5x 4).

Multiply each term in the 1st bracket by the whole of the
2nd bracket

(x3 + 3x2 − 2)(x2 − 5x + 4) = x3 (x2 − 5x + 4) + 3x2 (x2 − 5x + 4) − 2 (x2 − 5x + 4) = x5 − 5x4 + 4x3
+ 3x4 − 15x3 + 12x2 − 2x2 + 10x − 8
= x5 − 2x4 − 11x3 + 10x2 + 10x − 8

It is important to be able to decide when two polynomials are ‘equal’. x2 and 3 2 may take the same value when x = 1 o 2 but when x = 3 they do not. However, 3 2 and 4x (x 2) are always equal, no matter what value of x you choose.
This leads to what seems like a very obvious definition of equality of polynomials:
KEY POINT 3.1
Two polynomials being equal means that they have the same order and all of their coefficients are equal.

You will find that this type of equality is called an identity and is looked at in more detail in Section 4H.

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3 Polynomials 59

Ideas similar to comparing coefficients will be applied to vectors (chapter 13) and complex numbers (chapter 15).

However, this definition can be used to solve some quite tricky problems. We tend to use it in questions which start from telling us that two polynomials are equal; we can then compare coefficients.

Worked example 3.2
If (ax)2 + bx + 3(ax2 + x) x − 2x2 find the values of a and b.

Rearrange each side to make the coefficients clear
Compare coefficients and solve the resulting equations

LHS: a2x2 + bx + 3ax2 + 3x = a2 x2 + 3ax2 + bx + 3x = x2(a2 + 3a) + x(b + 3)

x2: (a2 + 3a) = −2

⇔ a2 + 3a + 2 = 0

⇔ (a + 1)(a + 2) = 0

⇔ a = −1 or a = −2

x:

b + 3 = 4

⇔ b = 1

A very important use of this technique is factorising a polynomial if one factor is already known. To do this we need to know what the remaining factor looks like. For example, if a cubic has one linear factor, the other factor must be quadratic.

Worked example 3.3 ( ) is a factor of x4 3x2 2x − 6. Find the remaining factor.

The remaining factor must be a cubic. We can write the original function as a product of (x − 1) and a general cubic
Multiplying out the right hand side and grouping like terms allows us to
compare coefficients

x4 3x2 2x 6 (x 1)(ax3 + x2 + x + )
= ax4 + bx3 + cx2 + dx −ax3 − bx2 − cx − d
= ax4 + x3 (b − a) + x2 (c − b) + x(d − c) − d

60 Topic 2: Functions and equations

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continued . . . Remember that the coefficient of x3 in the original expression is zero!
Answer the question

Compare x4: a = 1
Compare x3: b − a = 0 b = 1
Compare x2: c − b = 3 c = 4
Compare x: d − c = 2 d = 6
The remaining factor is x3 x2 + 4x 6

‘Finding the remaining factor’ is another way of asking you to divide. We can conclude from this example that:
x4 3x2 2x − 6 = x3 x2 + 4x 6 x −1

Exercise 3A

1. Decide whether each of the following expressions are polynomials. For those that are, give the order and the lead coefficient.

(a) x3 3x2 + 2x (c) 5x2 x 3 (e) 4e 3e2x

(b) 1 3x − x5 (d) 9x4 − 5
x
(f) x4 5x2 3 x

(g) 4x5 3x3 + 2x7 4

(h) 1

2. Expand the brackets for the following expressions:

(a) (i) ( )( 2

)

(ii) ( )( 2

)

(b) (i) (

)( 3

2

)

(ii) (

)( 3 2 )

eNeccxxhooaaetnimcmcsktepahulnienstiiwnttnehgertemtchpoeruelvdious that d = 6.

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3 Polynomials 61

(c) (i) (b (ii) ( 2
(d) (i) ( (ii) (

b )(b b )

)( 2

)

2 ( 4 2x3 + 1)

3 (x3 x − 1)

3. Find the remaining factor if (a) (i) x3 3x2 11x 2 has a factor of x − 2 (ii) x3 4x2 3x − 18 has a factor of x + 3 (b) (i) 6x3 13x2 2x − 6 has a factor of 2 3 (ii) 25x3 5x2 10 2 has a factor of 5 1 (c) (i) x4 5x3 9x2 2x 21 has a factor of x − 3 (ii) x4 5x3 5x2 + 3x 28 has a factor of x − 4 (d) (i) x4 3x3 12x2 15x 35 has a factor of x2 3x 7 (ii) x3 2x2 3x − 6 has a factor of x2 + 3

4. Given that the result of each division is a polynomial, simplify each expression.

(a) (i) x4 x3 − 2x2 3x − 6 (ii) x4 3x3 2x2 + 2x 4

x −2

x+2

(b) (i) 2x4 27x2 36 x2 + 12

(ii) 2x3 3x2 − 27 2x2 3x + 9

5. Find the unknown constants a and b in these identities.

(a) (i) ax2 + bx = 4x2 6x

(ii) ax2 + 4 3x b

(b) (i) ax2 + bx = 4x bx2 − ax (ii) ax2 + 2bx 6x = 0

(c) (i) (ax + 1)2 + 3bx = 2ax2 − 2x + 1

(ii) (x + a)2 + b x2 + 4x + 9

(d) (i) ax2 + bx − 2ax 2x x2

(ii) ax2 − 3bx2 bx 4x = x2 + 7x

(e) (i) (ax)2 − (bx)2 + bx = 2x

(ii) (ax + b)2 x2 − 20x + 25

6. In what circumstances might you want to expand brackets? In what circumstances is the factorised form better?

7. (a) Is it always true that the sum of a polynomial of order n and a polynomial of order n − 1 has order n?
(b) Is it always true that the sum of a polynomial of order n and a polynomial of order n has order n?

62 Topic 2: Functions and equations

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3B Remainder and factor theorems
We saw in the last section that we can factorise polynomials by comparing coefficients. For example, if we know that ( ) is one factor of x3 x2 x 2, we can write
x3 x2 x 2 (x + 2)(ax2 + bx + c) and compare coefficients
to find that the other factor is ( 2 ).
If we try to factorise x3 2x2 x 5 using ( ) as one factor, we find that it is not possible; ( ) is not a factor of x3 2x2 x 5. However, using the factorisation of x3 x2 x 2 we can write:
x3 2x2 x 5 (x + 2)(x2 1) + 3

The number 3 is the remainder – it is what is left over when we try to write x3 2x2 x 5 as a multiple of ( ). In the last section we saw that factorising is related to division. In this case, we could say that:

x3 2x2 x 5 = (x2 + 1) i h
x+2

id 3

This is similar to the concept of a remainder when dividing numbers: for example, 25 = 3 7 + 4, so we would say that 4 is the remainder when 25 is divided by 7.

We can find the reminder by including it as another unknown coefficient. For example, to find the reminder when x3 2x2 x 5 is divided by ( ), we could write
x3 2x2 x 5 (x + 2)(ax2 + bx + c) R

then expand and compare coefficients. This is not a quick task. Luckily there is a shortcut which can help us find the reminder without finding all the other coefficients. If we substitute in a value of x that makes the first bracket equal to zero, in this case x = −2, into the above equation, it becomes
3 (0)( 2 + b + ) + R
so R = 3. This means that R is the value we get when we substitute x = −2 into the polynomial expression on the left. Fillin proof sheet 6 on the CD-ROM, ‘Remainder theorem’, shows you that the same reasoning can be applied when dividing any polynomial by a linear factor. This leads us to the Remainder theorem.

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3 Polynomials 63

exam hint

Notice

that

x

=

b a

is the value makes ax −

which b = 0.

KEY POINT 3.2
The remainder theorem
The remainder when a polynomial expression is divided by ( b) is the value of the expression when x = b.
a

Worked example 3.4

Find the remainder when x3 + 2x 7 is divided by x + 2.

Use the remainder theorem by rewriting the divisor in the form (ax − b)…

( + 2) (x − (−2))

… then substitute the value of x (obtained from

x = b ) into the expression when x = b

a

a

When x = −2 (− )3 + 2 × (− ) + 7 = −5
So the remainder is −5

If the remainder is zero then (ax − b) is a factor. This is summarised by the factor theorem.

KEY POINT 3.3

The factor theorem

If the value of a polynomial expression is zero when x = b,

then ( b) is a factor of the expression.

a

Worked example 3.5

Show that 2x − 3 is a factor of 2x3 13x2 19 6.

To use the factor theorem we need to substitute in x = 3 2

When

x

=

3 2:

2

×

⎛ ⎝

3⎞ 2⎠

3

−

13

⎛ ⎝

3⎞ 2⎠

2

+

19

×

⎛ ⎝

3⎞ 2⎠

−

6

= 27 − 117 + 57 − 6 = 0 442

64 Topic 2: Functions and equations

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continued . . .

Therefore, by the factor theorem, ( 2x3 13 2 19 6.

) is a factor of

We can also use the factor theorem to identify a factor of an expression, by trying several different numbers. Once one factor has been found, then comparing coefficients can be used to find the remaining factors. This is the recommended method for factorising cubic expressions on the non-calculator paper.

Worked example 3.6 Fully factorise x3 3x2 33x 35.

When factorising a cubic with no obvious factors we must put in some numbers and hope that we can apply the factor theorem
We can rewrite the expression as
(x +1) × general quadratic and compare
coefficients

When x = 1 the expression is −64 When x = 2 the expression is −81 When x = −1 the expression is 0 Therefore x + 1 is a factor.
x3 3x2 33x 35
= (x + 1)( x2 + x + ) = ax3 + ( + ) x2 + ( + ) x + c
a , b = 2, c = −35
x3 3x2 33x 35 (x 1)(x2 + 2x 35)

The remaining quadratic also factorises

= ( + )( + )(x − 5)

exam hint If the expression is going to factorise easily then you only need to try numbers which are factors of the constant term.
A very common type of question asks you to find unknown coefficients in an expression if factors or remainders are given.

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3 Polynomials 65

Worked example 3.7
x3 x2 + ax + b has a factor of (x − 1) and leaves a remainder of 17 when divided by (x − 2). Find the constants a and b.

Apply factor theorem

when x = 1 :

1 + 4 + a + b = 0

⇔ a + b = −5

(1)

Apply remainder theorem

when x = 2 :

8 + 16 + 2a + b = 17

⇔ 2a + b = −7

(2)

Two equations with two unknowns can be solved simultaneously

(2) − (1)

a = −2 b = −3

Exercise 3B

1. Use the remainder theorem to find the remainder when: (a) (i) x2 + 3x 5 is divided by x + 1 (ii) x2 + x − 4 is divided by x + 2 (b) (i) x3 6x2 4x + 8 is divided by x − 3 (ii) x3 7x2 11x is divided by x − 1 (c) (i) 6x4 7x3 − 5x2 5x + 10 is divided by 2 3 (ii) 12x4 10x3 + 11x2 − 5 is divided by 3 1 (d) (i) x3 is divided by x + 2 (ii) 3x4 is divided by x − 1

2. Decide whether each of the following expressions are factors of 2x3 73 3x + 2.

(a) x

(b) x − 1

(c) x + 1
(e) x + 2 (g) x + 1
2 (i) 2x 1

(d) x − 2 (f) x − 1
2 (h) 2x 1
(j) 3 1

66 Topic 2: Functions and equations

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3. Fully factorise the following expressions: (a) (i) x3 x2 x 2 (ii) x3 x2 − 4x 4 (b) (i) x3 7x2 16x 12 (ii) x3 6x2 12x 8 (c) (i) x3 3x2 12x 10 (ii) x3 x2 2x − 15 (d) (i) 6x3 11x2 6x − 1 (ii) 12x3 13x2 − 37x − 30

4. 6 3 2 b 8 has a factor (x + 2) and leaves a remainder of −3 when divided by (x − 1). Find a and b.

[5 marks]

5. x3 x2 + ax + b has a factor of (x − 2) and leaves a

remainder of 15 when divided by (x − 3).

Find a and b.

[5 marks]

6. The polynomial x kx k has a factor (x − k). Find the

possible values of k.

[5 marks]

7. The polynomial x (k + 1)x 3 has a factor
(x − k + 1). Find k.

[6 marks]

8. x3 ax bx + 168 has factors (x − 7) and (x − 3). (a) Find a and b. (b) Find the remaining factor of the expression.

[6 marks]

9. x3 ax2 + 9x + b has a factor of (x − 11) and leaves a remainder of −52 when divided by (x + 2).

(a) Find a and b.

(b) Find the remainder when x3 ax2 + 9x + b is divided

by (x − 2).

[6 marks]

10. f (x) = x3 + ax2 + 3x + b.

The remainder when f (x) is divided by (x + 1) is 6. Find

the remainder when f (x) is divided by (x − 1).

[5 marks]

11. The polynomial x2 5x 6 is a factor of 2x3 15x2 + ax + b. Find the values of a and b.

[6 marks]

f(x) is just a name given to the expression. You will learn more about this notation in chapter 5.

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3 Polynomials 67

The word quadratic indicates that the term with the highest power in the equation is x2. It comes from the Latin quadratus, meaning square.
x ∈R means that x can be any number. See Prior learning Section G on the CD-ROM for the meaning of other similar statements.

3C Sketching polynomial functions

The simplest polynomial with a curved graph is a quadratic.

Let us look at two examples of quadratic functions:

y1 = 2x2 + 2x – 4 and y2 = –x2 + 4x – 3 (x ∈R)
You can use your calculator to plot the two graphs. See Calculator sheet 2 on the CD-ROM to see how to sketch graphs on your GDC.

y

y

y1
x

y2
x

Vertex

Line of Symmetry
n y = xn

2

y

y =x2

These two graphs have a similar shape, called a parabola. A parabola has a single turning point (also called the vertex) and a vertical line of symmetry. The most obvious difference is that y1 has a minimum point, whereas y2 has a maximum point. This is because of the different signs of the x2 term.
The graphs of other polynomial functions are also smooth curves. You need to know the shapes to expect for these graphs.

Positive degree n polynomial

Negative degree n polynomial

Number of

Number

times graph can of

cross or touch turning

the x-axis

points

0, 1 or 2

1

x

68 Topic 2: Functions and equations

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3

y

y =x3

x

4

y

y =x4

x

1, 2, or 3

0 or 2

0, 1, 2 or 3 1 or 3

The constant term in the polynomial expression gives the position of the y-intercept of the graph (where the curve crosses the y-axis). This is because all other terms contain x, so when x = 0 the only non-zero term is the constant term. For example, 2x3 3x + 5 5 when x = 0.

Worked example 3.8

Match each equation to the corresponding graph, explaining your reasons. (a) y 3x2 − 4x − 1 (b) y 2x2 − 4x (c) y x2 − 4x + 2

y

y

y

A

B

x

x

C x

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3 Polynomials 69

continued . . . Graph B is the only positive quadratic
We can distinguish between the other two graphs based on their y-intercept

Graph B shows a positive quadratic, so graph B corresponds to equation (a).
Graph A has a positive y-intercept, so graph A corresponds to equation (c). Graph C corresponds to equation (b).

Factorising is covered in Prior learning Section N.
y

To sketch the graph of a polynomial it is also useful to know its x-intercepts. These are the points where the graph crosses the x-axis, so at those points y = 0. For this reason, they are called zeros of the polynomial. For example, the quadratic polynomial x2 5x 6 has zeros at x = 2 and x = 3. They are the roots (solutions) of the equation x2 5x 6 = 0 and can be found
from the factorised form of the polynomial: (x 2)(x − 3).
KEY POINT 3.4
The polynomial a(x − p )(x − p )(x − p3 )… has zeros
p p2, p3,…
If some of the zeros are the same, we say that the polynomial has
a repeated root. For example, the equation (x 1)2 (x + 3) has a
repeated (double) root x = 1 and a single root x = −3. Repeated roots tell us how the graph meets the x -axis.

y

y

−2

2

x

−3

x

−2

1

x

y = (x − 2)(x + 2)

y = −(x + 3)(x + 3)

y = (x − 1)3(x + 2)

If a polynomial has a factor If a polynomial has a double (x − a) then the curve passes factor (x − a)2 then the curve straight through the x-axis at a. touches the x-axis at a.

If a polynomial has a triple factor (x − a)3 then the curve passes through the x-axis at a, flattening as it does so.

70 Topic 2: Functions and equations

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KEY POINT 3.5
The stages of sketching graphs of polynomials are:
• classify the order of the polynomial and whether it is positive or negative to deduce the basic shape
• set x = 0 to find the y-intercept • write in factorised form • find x-intercepts • decide on how the curve meets the x-axis at each
intercept • connect all this information to make a smooth curve.

Worked example 3.9
Sketch the graph of y (1 x)(x − 2)2.
Classify the basic shape Find y-intercept Find x-intercepts
Decide on shape of curve at x-intercepts
Sketch the curve exam hint Ahohtaahpriascnnveoktvlteoedeieecnatrtsoactocarschrlle,pylarbpepsdillemectuaros.toc,bpaehsIxeomscthilcrmamalnuteuasopsaudrtntaeast.ebtttxeleyis

This is a negative cubic When x = 0 y = 1 × (−2)2 = 4 When y = 0, x = 1 or x = 2 At x = 1 curve passes through the axis At x = 2 curve just touches the axis
y
4
12
x

Sometimes we need to deduce possible equations from a given curve.

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3 Polynomials 71

We will examine the Fundamental theorem of algebra in more detail in Section 15D.

The first thing we should ask is what order polynomial we need to use. There is a result called the Fundamental Theorem of Algebra, which states that a polynomial of order n can have at most n real roots. So, for example, if the given curve has three x -intercepts, we know that the corresponding polynomial must have degree at least 3. We can then use those intercepts to write down the factors of the polynomial.
KEY POINT 3.6
To find the equation of a polynomial from its graph:
• use the shape and position of the x-intercepts to write down the factors of the polynomial
• use any other point to find the lead coefficient.

Worked example 3.10 Find a possible equation for this graph.

y
(1, 24)

−1

2

3x

Describe x-intercepts
Convert this to a factorised form Use the fact that when x = 1, y = 24

Single root at x = 2 and x = 3 Double root at x = −1
∴ y = k (x + 1)2 (x − 2) (x − 3)
24 = k × (2)2 × (−1) × (−2) ⇔ 24 = 8k
⇔ k = 3 So the equation is y = 3 (x + 1)2 (x − 2) (x − 3)

72 Topic 2: Functions and equations

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Exercise 3C

1. Sketch the following graphs, labelling all axis intercepts.

(a) (i) y 2(x − 2)(x − 3)

(ii) y 7(x − 5)(x + 1)

(b) (i) y 4(5 x)(x 3)(x − 3)

(ii) y 2(x −1)(2 − x)(x − 3)

(c) (i) y (x − 4)2
(d) (i) y = x (x2 + 4) (e) (i) y (1 x)2 (1 + x)

(ii) y (x − 2)2 (ii) y = (x + 1)(x2 3x + 7) (ii) y (2 x)(3 − x)2

2. Sketch the following graphs, labelling all axis intercepts.
(a) (i) y x (x 1)(x − 2)(2x − 3) (ii) y (x + 2)(x + 3)(x − 2)(x − 3)
(b) (i) y 4(x − 3)(x 2)(x + 1)(x + 3)
(ii) y 5x (x + 2)(x − 3)(x − 4) (c) (i) y (x − 3)2 (x 2)(x − 4)
(ii) y x2 (x 1)(x + 2)
(d) (i) y 2(x + 1)3 (x − 3)
(ii) y x3 (x − 4) (e) (i) y (x2 + 3x + 12)(x + 1)(3x −1)
(ii) y = (x + 2)2 (x2 + 4)

3. Match equations and corresponding graphs.

(i) A: y x2 − 3x + 6

B: y 2x2 − 3x + 3

C: y x2 − 3x + 6

y 1
6
x

y 2
6
x

y 3
x
−3

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3 Polynomials 73

(ii) A: y
1

x2 + 2x − 3
y
3

x

B: y x2 + 2x + 3
y 2
3
x

C: y = x2 + 2x + 3
y 3
x
−3

4. The diagrams show graphs of quadratic functions of the form y = ax bx + c. Write down the value of c, and then find the values of a and b.

(a) (i)

y

(ii) y

6

x

−2

1

3

1

x

(b) (i)

y

1

x

−1

4

−10
(ii) y
1

x

−1

1

5. Find lowest order polynomial equation for each of these graphs.

(a) (i)

y

(ii)

y

16

−2

1

4x

−2

3

x

(1, −36)

74 Topic 2: Functions and equations

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(b) (i)

y

(ii)

y

−2

1

x

(−1, 10)

8

1

x

−4 −2

(c) (i)

y

(ii)

y

x

−1

2

−4

(d) (i)

y

(ii)

(1, 12)

x

−3 −2

1

2

−1

x

2

−2

y
48

−1

x

2 34

(e) (i)

y

(ii)

(2, 12)

3

x

1

y
(1, 15)

x

−

1 2

2

(f) (i)

y

(ii) y

18

x

−1

3

x
13
−9

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3 Polynomials 75

6. (a) Show that (x − 2) is a factor of f (x) = 2x3
(b) Factorise f (x). (c) Sketch the graph y = f (x).

5x2 + x + 2.

7. Sketch the graph of y 2(x + 2)2 (3 − x), labelling clearly

any axes intercepts.

[5 marks]

8. The two graphs below each have equations of the form y = px3 + qx2 + rx + s. Find the values of p, q, r and s for each graph.

(a)

y

(b)

y

36 (2, 4)

3

x

−2

3

x

0

[10 marks]

9. (a) Factorise fully x4 q4 where q is a positive constant.

(b) Hence or otherwise sketch the graph y x4 − q4, labelling

any points where the graph meets an axis.

[5 marks]

10. (a) Sketch the graph of y (x − p)2(x − q) where p < q.

(b) How many solutions does the equation ( p)2( q) = k

have when k > 0?

[5 marks]

exam hint Dtqrf3tooyauo(s.grinano7nmfir’ddlgtavnuirsfepudaltpaoyrhst)eieeocifnooxnuaadrat–chKacetcetuoreotqaresrysowueliocsaapeiulltsosuhtoliekaantoeiie.notgnodtnrs
SsthheeeeeCCtsDa4l-cRuaOlnaMdto.6r on

3D The quadratic formula and the discriminant

It is not always possible to find zeros of a polynomial using

factorising. Try factorising the two quadratic expressions

f x) = x2 3x − 3 and g(x) = x2 3x + 3. It appears that

neither of the expressions can be factorised, but sketching the

graphs reveals that y1 has two zeros, while y2 has no zeros.

y

y

y = g(x)

x

x

y = f (x)

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In the case where the polynomial is a quadratic we have another option – using the quadratic formula to find the zeros (roots): KEY POINT 3.7
The zeros of ax2 + bx + c are given by the quadratic formula
x = −b b2 − 4ac 2a

You can find the proof of this formula in Fill-in proof 3 ‘Proving quadratic formula’.
The solution to quadratic equations was known to the Greeks, although they did everything without algebra, using geometry instead. They would have described the quadratic equation x2 + 3x 10 as ‘the area of a square plus three times its length measures ten units’.

Worked example 3.11

Use the quadratic formula to find the zeros of x2 5x 3.

It is not obvious how to factorise the quadratic expression, so use the quadratic formula

x = 5 ± (− )2 − 4 × 1 × (− )
2 = 5 ± 37
2

The roots are

5 37 = 5.54 and 5 37 = −0.541 (3SF)

2

2

exam hint
In International Baccalaureate® exams you should either give exact answers (such as 5 37 ) or round your
2 answers to 3 significant figures, unless you are told otherwise. See Prior learning Section B on the CD-ROM for rules of rounding.

Let us examine what happens if we try to apply the quadratic formula to find the zeros of x2 3x 3:

x = 3 ± (−3)2 − 4 × 1 × 3 = 3 ± −3

2

2

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3 Polynomials 77

In chapter 15 you will meet imaginary numbers, a new type of number which makes it possible to find zeros of functions like this.
y

As the square root of a negative number is not a real number, it follows that the expression has no real zeros.
Looking more closely at the quadratic formula, we see that it can be separated into two parts:

x = −b ± b ac

2a

2a

The line of symmetry of the parabola lies halfway between the two roots:

KEY POINT 3.8

The line of symmetry of y = ax bx + c is x=− b 2a

The second part of the formula involves a root expression:
b ac.
The expression b ac is called the discriminant of the quadratic (often symbolised by the Greek letter Δ) and
b ac is the distance of the zeros from the line of 2 a
symmetry x = − b . 2a
The square root of a negative number is not a real number, so if the discriminant is negative, there can be no real zeros and the graph will not cross the x-axis. If the discriminant is zero, the graph is tangent to the x-axis at a point which lies on the line of symmetry (it touches the x-axis rather than crossing it).
The graphs below are examples of the three possible situations:

y

y

x

=

−

b 2a

(2, 2)

y = x2 − 4x + 6 Δ = −8

x

x =2

(2, 0)

y = x2 − 4x + 4 Δ=0 x

√

√

−b− Δ

Δ

2a

2a

0

√

√

Δ

−b+ Δ

2a

2a

x

y = x2 − 4x

Δ = 16

x= 2

(2, −4)
x =2

78 Topic 2: Functions and equations

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KEY POINT 3.9

For a quadratic expression ax2 + bx + c, the discriminant is:

Δ = b − ac .

• If Δ < 0 • If Δ = 0 • If Δ > 0

the expression has no real zeros the expression has one (repeated) zero the expression has two distinct real zeros

There is also a formula for solving cubic equations called ‘Cardano’s Formula’. It too has a discriminant which can be used to decide how many solutions there will be:
b c2 − 4ac3 4b d − 27a d 2 +18abcd This gets a lot more complicated than the quadratic version!

Worked example 3.12
Find the exact values of k for which the quadratic equation kx2 − (k )x 3 = 0 has a
repeated root.

Repeated root means that Δ = b − 4ac = 0
We can form an equation in k using a = k and b = −(k + 2)

b2 − 4 ac = 0
(k )2 (k)(3) = 0

This is a quadratic equation

k2 k 4 − 12k = 0 k2 8k 4 = 0

It doesn’t appear to factorise, so use the quadratic formula to find k

k = 8 ± 82 − 4 × 4 2
= 8 ± 48 2
(± )
= 2
=4±2 3

When Δ < 0, the graph does not intersect the x-axis, so it is either entirely above or entirely below it. The two cases are distinguished by the value of a.
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Questions involving discriminants often lead to quadratic inequalities, which are covered in Prior learning Section Z.
3 Polynomials 79

KEY POINT 3.10 For a quadratic function with Δ < 0: if a > 0 then y > 0 for all x if a < 0 then y < 0 for all x

Worked example 3.13
Let y 3x2 + kx − 12. Find the values of k for which y < 0 for all x.
y is a negative quadratic. y < 0 means that the graph is entirely below the x-axis. This will happen when y = 0 has
no real roots
This is a quadratic inequality. A sketch of the graph will help

No real roots, ∴ Δ < 0 k2 − 4( 3)(−12) < 0 k2 < 144
⇒ k2 − 144 < 0
y

k

−12

12

∴ −12 < k < 12

Exercise 3D

1. Evaluate the discriminant of these quadratic expressions.

(a) (i) x2 + 4x 5

(ii) x2 6x 8

(b) (i) 2 2

6

(ii) 3 2

10

(c) (i) 3x2 6x + 3

(ii) 9 2 6x + 1

(d) (i) 12 − x − x2

(ii) −x2 − 3x + 10

2. State the number of zeros for each expression in Question 1.

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3. Find the exact solutions of these equations.

(a) (i) x2 3x 1 = 0 (ii) x2 x − 1 0

(b) (i) 3 2

2= 0

(ii) 2x2 6x + 1 0

(c) (i) 4 x 3x2 = 0 (ii) 1 x 2x2 = 0

(d) (i) x2 3 4x

(ii) 3 x 2x2

4. Find the values of k for which: (a) (i) the equation x2 − x + k = has two distinct real roots (ii) the equation 3x2 − 5x + k = 0 has two distinct real roots
(b) (i) the equation x2 2x + (2k 1) = 0 has equal roots (ii) the equation 2x2 3x − (3k 1) = 0 has equal roots
(c) (i) the equation −x2 + 3x + (k −1) = 0 has real roots (ii) the equation −2x2 + 3x − (2k + 1) = 0 has real roots
(d) (i) the equation 3k 2 3x + 2 0 has no real solutions (ii) the equation kx2 + 5x 3 = 0 has no real solutions
(e) (i) the quadratic expression (k )x2 + 3x + 1 has a repeated zero
(ii) the quadratic expression −4x2 + 5x + (2k − 5) has a repeated zero
(f) (i) the graph of y x2 − 4x ( k + 1) is tangent to the x-axis
(ii) the graph of y kx2 + x − 4 is tangent to the x-axis (g) (i) the expression −3x2 + 5k has no real zeros
(ii) the expression 2k 2 3 has no real zeros

Some of the many applications of quadratic equations are explored in Supplementary sheet 1 ‘The many applications of quadratic equations’.

5. Find the values of parameter m for which the quadratic

equation mx2 − 4x 2m = 0 has equal roots.

[5 marks]

6. Find the exact values of k such that the equation
−3x2 + (2k + 1)x − 4k = 0 has a repeated root.

[6 marks]

7. Find the range of values of the parameter c such that

2x2 3x + (2c 1) ≥ 0 for all x.

[6 marks]

8. Find the set of values of k for which the equation

x 2kx 6k = 0 has no real solutions.

[6 marks]

9. Find the range of value of k for which the quadratic
equation kx2 − (k )x 1 = 0 has no real roots. [6 marks]

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3 Polynomials 81

10. Find the range of values of m for which the equation

mx2 + mx − 2 0 has one or two real roots.

[6 marks]

11. Find the possible values of m such that mx2 + 3x 4 < 0

for all x.

[6 marks]

12. The positive difference between the zeros of the quadratic expression x kx + 3 is 69. Find the possible values of k.
[5 marks]

Summary
• Polynomials are expressions involving only addition, multiplication and raising to a power.
• If two polynomials are equal we can compare coefficients. This can be used to divide two polynomials.
• The remainder theorem says that the remainder when a polynomial is divided by ax − b is the value of the polynomial expression when x = b. a
• The factor theorem says that if a polynomial expression is zero when x = b then ax − b is a a
factor of the expression.
• The graphs of polynomial are best sketched using their factorised form. Look for repeated factors and find where the graph crosses the axes.
• You can use the shape and position of the x-intercepts to find the equation of a polynomial graph.
• The number of solutions of a quadratic equation depends on the value of the discriminant, Δ = b − ac: – If Δ > 0 there are two distinct solutions. – If Δ = 0 there is one solution. – If Δ < 0 there is no solution.

82 Topic 2: Functions and equations

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Introductory problem revisited
Without using your calculator, solve the cubic equation x3 x2 + 47x − 35 = 0
If we put x = 1 into this expression we get 0, so we can conclude that x − 1 is a factor. The remaining quadratic factor can be found by comparing coefficients; it is x2 − 12x + 35, which factorises to give (x − 5)(x − 7). The equation therefore becomes (x −1)(x − 5)(x − 7) = 0 which has solutions x = 1, 5, or 7.

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3 Polynomials 83

Mixed examination practice 3

Short questions

1. A quadratic graph passes through the points (k, 0) and (k + 4, 0). Find in

terms of k the x-coordinates of the turning point.

[4 marks]

y
2. The diagram shows the graph of y = ax2 + bx + c

Complete the table to show whether each

x

expression is positive, negative or zero.

expression positive negative zero a c b2 – 4ac b

[6 marks] (© IB Organization 2000)

3. The diagram shows the graph with equation y = ax Find the values of a, b, c, d and e. y

bx3 + cx

dx + e. [6 marks]

27

13

x

−3

4. The remainder when (ax + b)3 is divided by (x − 2) is 8 and the remainder when it is divided by (x + 3) is −27. Find the values of a and b.

[5 marks]

5. (a) Show that (x − 2) is a factor of f (x) = x3
(b) Factorise f (x). (c) Sketch the graph of y = f (x).

4x2 + x + 6.

[7 marks]

84 Topic 2: Functions and equations

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6. The remainder when (ax + b)4 is divided by (x − 2) is 16 and the remainder

when it is divided by (x + 1) is 81. Find the possible

values of a and b.

[6 marks]

7. Sketch the graph of y (x − a)2 (x − b)(x − c) where b a < c. [5 marks]

8. Find the exact values of k for which the equation 2kx2 (k 1)x + 1 0 has

equal roots.

[5 marks]

9. Find the set of values of k for which the equation 2 2 k 6 = 0 has no

real roots.

[6 marks]

10. Find the range of values of k for which the quadratic function
x ( k 1)x + 5 has at least one real zero.

[6 marks]

11. The polynomial x2 4x 3 is a factor of the polynomial x3 ax2 + 27x + b.

Find the values of a and b.

[6 marks]

12. Let α and β denote the roots of the quadratic equation x kx (k − 1) 0. (a) Express α and β in terms of the real parameter k. (b) Given that α β 2= 17, find the possible values of k.

[7 marks]

13. Let q(x) = kx2 + (k )x − 2. Show that the equation q(x) = 0 has real

roots for all values of k.

[7 marks]

14. Find the range of values of k such that for all x, − ≤ x2.

[7 marks]

Long questions

1. (a) Find the coordinates of the point where the curve y = x2 + bx − a crosses the y-axis, giving your answer in terms of a and/or b.

(b) State the equation of the axis of symmetry of x bx − a, giving your answer

in terms of a and/or b.
(c) Show that the remainder when x positive.

bx − a is divided by x − a is always b

(d) The remainder when x bx − a is divided by x a is −9. Find the possible values that b can take. [14 marks]

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3 Polynomials 85

2. (a) Show that for all values of p, ( ) is a factor of
f (x) = x3 + ( p − 2)x2 + (5 2p)x −10.
(b) By factorising f(x), or otherwise, find the exact values of p for which the equation
x3 ( p − 2)x2 (5 2p)x 10 0
has exactly two real roots.
(c) For the smaller of the two values of p found above, sketch the graph of y = f(x). [10 marks]

3. (a) On the graph of y = x2 + 4x + 5 prove that there is no value of x for which

y = 0.

x+2

(b) Find the equation of the vertical asymptote of the graph.

(c) Rearrange the equation to find x in terms of y. (d) Hence show that y cannot take values between −2 and 2.

(e) Sketch the graph of y = x2 + 4x + 5. x+2

[18 marks]

4. Let f (x) = x4 + x3 + x2 + x + 1.
(a) Evaluate f ( ).
(b) Show that (x ) f (x) x5 −1.
(c) Sketch y x5 − 1.

(d) Hence show that f x) has no real roots.

[10 marks]

86 Topic 2: Functions and equations

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4Algebraic structures
Introductory problem What information can be extracted from the equation ax + b = 3x + 2? What information can be extracted from the identity ax + b = 3x + 2, for all x?
Equations are the building blocks of mathematics. There are many different types; some have no solutions, some have many solutions. Some have solutions which cannot be expressed in terms of any function you have met.
Graphs are an alternative way of expressing a relationship between two variables. If you understand the connection between graphs and equations, and can switch between the two representations, you will have a wider variety of tools to solve
mathematical problems. The International Baccalaureate® places
great emphasis on using graphical calculators to analyse graphs.
Inequalities share many of the same properties as equations. However, a few important differences mean you need different techniques to solve some inequalities.
In much of mathematics we do not distinguish between equations and identities, although they are fundamentally different. In this chapter we shall explore some of these differences and look at the different techniques we can apply.
4A Solving equations by factorising
We start by looking at some common algebraic methods for solving equations. You have already used factorising to sketch polynomials, and in this chapter we will apply the same principles in a wider context.

In this chapter you will learn:
• some standard algebraic strategies for solving equations
• how to sketch graphs, and some limitations of graphical calculators
• how to use a graphical calculator to solve equations
• how to solve systems of up to three linear simultaneous equations
• algebraic and graphical strategies for solving inequalities
• how to prove and work with identities.

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4 Algebraic structures 87

This is true for real numbers, but if you know that the numbers you are looking for are integers, then knowing the product of two numbers is far more useful.

If two numbers multiply together to give 5, what can you say

about those two numbers?

They could be 1 and 5, 10 and 1 , π and 5 ; there are actually

2

π

an infinite number of possibilities. So if you know that two

numbers multiply together to give 5 it does not help to find

what those numbers are. However, if two numbers multiply

together to give zero, this is only possible if one of them is zero.

If we can factorise an expression which is equal to zero then this is a very powerful tool for solving equations.

Worked example 4.1
( ) Solve the equation e ln(x) (2x −1) = 0.

Set the three factors equal to 0 then solve

Either ex = 0

(1)

each separately

or ln(x) − 1 = 0

(2)

or 2x 1 = 0

(3)

From (1) From (2) From (3)
List the solutions

ex = 0 has no solution
ln(x) = 1
⇔ =e
2x 1 ⇔ =1
2 x = e or 1
2

exam hint
If you divide by a function instead of looking at factors then it is possible that you might lose solutions. For example, consider the equation x3 = x. If you divide both sides by x you get x2 = 1 so x = ±1. However, if you put x = 0 into the original equation you will see that this is also a solution. The correct method for solving this equation is by factorisation.

x3 x

⇒

x3 − x = 0

⇒

x (x2 −1) 0

⇒ x (x −1)(x 1) = 0

⇒ x = 0 x 1 or x = −1

88 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.

Exercise 4A

1. Solve the following equations:

(a) (i) 3( 3)3 = 0

(ii) −4(x + 1)5 = 0

(b) (i) 7(2x 1)(5x + 3)2 0 (ii) 5(3 − x)2 (2x 6)2 = 0

( )( ) (c) (i) (l g3 )( ) 0 (ii) x

x− =0

(d) (i) x (x2 − 3) 7(x2 − 3) (ii) 5 ( 2 5x 4) 6(x2 − 5 4)

(e) (i) 6 4 × 3 0

(ii) 2 5x − 7 10 0

2. Solve the following equations:

(a) (i) x3

x2 11x

(ii) x3 x2 − 17x = 15

(b) (i) x3 5x2 7x − 2 0

(ii) x3 6x2 7x − 2 0

3. (a) Find the roots of these equations:

(i) x3 6x2 11x 6 (ii) x3 2x2 6 = 5x

(b) Find the zeros of these polynomials:

(i) x3 x2 − x − 1 (ii) x3 3x2 10x 24

You saw how to use the factor theorem to factorise polynomials in Section 3B.

4. Solve (3 )1 x2 −4 = 1.

[4 marks]

5. Solve the equation x x = 4x.

4B Solving equations by substitution
There are certain types of equation which you should know how to solve; in this section we shall focus particularly on quadratic equations as we have a formula to solve them.
We shall also see how a substitution can turn some complicatedlooking equations into quadratic equations.

Worked example 4.2

Solve the equation x4

x 2.

A substitution y = x2 turns this into a quadratic equation since x4 = y2

If y = x2 the equation becomes y2 − 4 = 3y

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4 Algebraic structures 89

continued . . . This is a standard quadratic equation
Use the substitution to find x

y2 − 3y − 4 = 0 ⇒ (y + 1) (y − 4) = 0 ⇒ y = −1 or y = 4
Therefore x2 = 4 or x2 = −1 (reject) ⇒ x2 = 4
x = 2 or x = −2

Other substitutions may not be so obvious.
In particular, it is quite common to be given an exponential equation which needs a substitution. Look out for an ax and an
a2x or an (a2 )x term, both of which can be rewritten as (ax )2.

Worked example 4.3
Solve the equation 2(4x + 2) 9 × 2x.
A substitution y = 2x turns this into a quadratic equation since 4x = 22x = y2
This is a standard quadratic equation
Use the substitution to find x

If y = 2x the equation becomes 2(y2 + 2) = 9y
2y2 − 9y + 4 = 0 ⇒ (2y − 1) (y − 4) = 0
y = 1 or y = 4 2
Therefore 2x = 1 or 2 4 2
⇒ = −1 or x = 2

Exercise 4B

1. Solve the following equations, giving your answers in an exact form.

(a) (i) a4 a2 + 21 = 0 (ii) x4 7x2 12 0

(b) (i) 2x6 7x3 = 15 (c) (i) x2 − 4 = 2
x2

(ii) a6 7a3 8 (ii) x2 + 36 = 12
x2

90 Topic 1: Algebra, and Topic 2: Functions and equations © Cambridge University Press 2012 Not for printing, sharing or distribution.