A Level Mathematics for AQA Student Book 2 (Year 2) with Cambridge Elevate Edition (2 Years) (AS/A Level Mathematics for AQA) [Student ed.] 1316644693, 9781316644256, 9781316644690, 9781316644621, 9781316644638

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Brighter Thinking

A Level Mathematics for AQA Student Book 2 (Year 2) Stephen Ward, Paul Fannon, Vesna Kadelburg and Ben Woolley

Contents Introduction How to use this resource Working with the large data set 1 Proof and mathematical communication 1: A reminder of methods of proof 2: Proof by contradiction 3: Criticising proofs 2 Functions 1: Mappings and functions 2: Domain and range 3: Composite functions 4: Inverse functions 3 Further transformations of graphs 1: Combined transformations 2: The modulus function 3: Modulus equations and inequalities 4 Sequences and series 1: General sequences 2: General series and sigma notation 3: Arithmetic sequences 4: Arithmetic series 5: Geometric sequences 6: Geometric series 7: Infinite geometric series 8: Mixed arithmetic and geometric questions 5 Rational functions and partial fractions 1: An extension of the factor theorem 2: Simplifying rational expressions 3: Partial fractions with distinct factors 4: Partial fractions with a repeated factor 6 General binomial expansion 1: The general binomial theorem 2: Binomial expansions of compound expressions Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 7 Radian measure 1: Introducing radian measure 2: Inverse trigonometric functions and solving trigonometric equations 3: Modelling with trigonometric functions 4: Arcs and sectors 5: Triangles and circles 6: Small angle approximations 8 Further trigonometry 1: Compound angle identities 2: Double angle identities 3: Functions of the form 4: Reciprocal trigonometric functions 9 Calculus of exponential and trigonometric functions 1: Differentiation 2: Integration

10 Further differentiation 1: The chain rule 2: The product rule 3: Quotient rule 4: Implicit differentiation 5: Differentiating inverse functions 11 Further integration techniques 1: Reversing standard derivatives 2: Integration by substitution 3: Integration by parts 4: Using trigonometric identities in integration 5: Integrating rational functions 12 Further applications of calculus 1: Properties of curves 2: Parametric equations 3: Connected rates of change 4: More complicated areas 13 Differential equations 1: Introduction to differential equations 2: Separable differential equations 3: Modelling with differential equations 14 Numerical solution of equations 1: Locating roots of a function 2: The Newton–Raphson method 3: Limitations of the Newton–Raphson method 4: Fixed-point iteration 5: Limitations of fixed-point iteration and alternative rearrangements 15 Numerical integration 1: Integration as the limit of a sum 2: The trapezium rule 16 Applications of vectors 1: Describing motion in two dimensions 2: Constant acceleration equations 3: Calculus with vectors 4: Vectors in three dimensions 5: Solving geometrical problems Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 17 Projectiles 1: Modelling projectile motion 2: The trajectory of a projectile 18 Forces in context 1: Resolving forces 2: Coefficient of friction 3: Motion on a slope 19 Moments 1: The turning effect of a force 2: Equilibrium Focus on … Proof 3 Focus on … Problem solving 3 Focus on … Modelling 3 Cross-topic review exercise 3

20 Conditional probability 1: Set notation and Venn diagrams 2: Two-way tables 3: Tree diagrams 21 The normal distribution 1: Introduction to normal probabilities 2: Inverse normal distribution 3: Finding unknown or 4: Modelling with the normal distribution 22 Further hypothesis testing 1: Distribution of the sample mean 2: Hypothesis tests for a mean 3: Hypothesis tests for correlation coefficients Focus on … Proof 4 Focus on … Problem solving 4 Focus on … Modelling 4 Cross-topic review exercise 4 Practice papers Formulae Answers Gateway to A Level revision sheets Gateway to A Level revision sheets answers Worked solutions for book chapters 1 Proof and mathematical communication 2 Functions 3 Further transformations of graphs 4 Sequences and series 5 Rational functions and partial fractions 6 General binomial expansion 7 Radian measure 8 Further trigonometry 9 Calculus of exponential and trigonometric functions 10 Further differentiation 11 Further integration techniques 12 Further applications of calculus 13 Differential equations 14 Numerical solution of equations 15 Numerical integration 16 Applications of vectors 17 Projectiles 18 Forces in context 19 Moments 20 Conditional probability 21 The normal distribution 22 Further hypothesis testing Worked solutions for cross topic reviews Cross topic review exercise 1 Cross topic review exercise 2 Cross topic review exercise 3 Cross topic review exercise 4 Worksheet answers Support sheets Extension sheets Working with the large data set

Working with the large data set answers Acknowledgements Copyright

Introduction You have probably been told that mathematics is very useful, yet it can often seem like a lot of techniques that just have to be learnt to answer examination questions. You are now getting to the point where you will start to see where some of these techniques can be applied in solving real problems. However as well as seeing how maths can be useful we hope that anyone working through this book will realise that it can also be incredibly frustrating, surprising and ultimately beautiful. The book is woven around three key themes from the new curriculum: Proof Maths is valued because it trains you to think logically and communicate precisely. At a high level maths is far less concerned about answers and more about the clear communication of ideas. It is not about being neat – although that might help! It is about creating a coherent argument which other people can easily follow but find difficult to refute. Have you ever tried looking at your own work? If you cannot follow it yourself it is unlikely anybody else will be able to understand it. In maths we communicate using a variety of means – feel free to use combinations of diagrams, words and algebra to aid your argument. And once you have attempted a proof, try presenting it to your peers. Look critically (but positively) at some other people’s attempts. It is only through having your own attempts evaluated and trying to find flaws in other proofs that you will develop sophisticated mathematical thinking. This is why we have included lots of common errors in our Work it out boxes – just in case your friends don’t make any mistakes! Problem solving Maths is valued because it trains you to look at situations in unusual, creative ways, to persevere and to evaluate solutions along the way. We have been heavily influenced by a great mathematician and maths educator George Polya who believed that students were not just born with problem-solving skills – they were developed by seeing problems being solved and reflecting on their solutions before trying similar problems. You may not realise it but good mathematicians spend most of their time being stuck. You need to spend some time on problems you can’t do, trying out different possibilities. If after a while you have not cracked it then look at the solution and try a similar problem. Don’t be disheartened if you cannot get it immediately – in fact, the longer you spend puzzling over a problem the more you will learn from the solution. You may never need to integrate a rational function in the future, but we firmly believe that the problem solving skills you will develop by trying it can be applied to many other situations. Modelling Maths is valued because it helps us solve real-world problems. However, maths describes ideal situations and the real world is messy! Modelling is about deciding on the important features needed to describe the essence of a situation and turning that into a mathematical form, then using it to make predictions, compare to reality and possibly improve the model. In many situations the technical maths is actually the easy part – especially with modern technology. Deciding which features of reality to include or ignore and anticipating the consequences of these decisions is the hard part. Yet it is amazing how some fairly drastic assumptions – such as pretending a car is a single point or that people’s votes are independent – can result in models which are surprisingly accurate. More than anything else this resource is about making links – links between the different chapters, the topics covered and the themes above, links to other subjects and links to the real world. We hope that you will grow to see maths as one great complex but beautiful web of interlinking ideas. Maths is about so much more than examinations, but we hope that if you take on board these ideas (and do plenty of practice!) you will find maths examinations a much more approachable and possibly even enjoyable experience. However always remember that the results of what you write down in a few hours by yourself in silence under exam conditions is not the only measure you should consider when judging

your mathematical ability – it is only one variable in a much more complicated mathematical model!

How to use this resource Throughout this resource you will notice particular features that are designed to aid your learning. This section provides a brief overview of these features.

In this chapter you will learn how to: review proof by deduction, proof by exhaustion and disproof by counterexample use a new method of proof called proof by contradiction criticise proofs.

Learning objectives A short summary of the content that you will learn in each chapter.

Before you start… Student Book 1, Chapter 1

You should be able to use logical connectors.

1 Insert a and b:

or

in the places marked

a b Student Book 1, Chapter 1

You should be able disprove a statement by counterexample.

2 Disprove the statement: ‘Apart from there are no other integers that can be written as both and . ’

Before you start Points you should know from your previous learning and questions to check that you’re ready to start the chapter.

WORKED EXAMPLE The left-hand side shows you how to set out your working. The right-hand side explains the more difficult steps and helps you understand why a particular method was chosen.

PROOF Step-by-step walkthroughs of standard proofs and methods of proof.

WORK IT OUT Can you identify the correct solution and find the mistakes in the two incorrect solutions?

Key point A summary of the most important methods, facts and formulae.

Common error Specific mistakes that are often made. These typically appear next to the point in the Worked example where the error could occur.

Tip Useful guidance, including ways of calculating or checking answers and using technology.

Each chapter ends with a Checklist of learning and understanding and a Mixed practice exercise, which includes past paper questions marked with the icon . In between chapters, you will find extra sections that bring together topics in a more synoptic way. FOCUS ON… Unique sections relating to the preceding chapters that develop your skills in proof, problemsolving and modelling.

CROSS-TOPIC REVIEW EXERCISE Questions covering topics from across the preceding chapters, testing your ability to apply what you have learned.

You will find practice paper questions towards the end of the resource, as well as a glossary of key terms (picked out in colour within the chapters), answers and full worked solutions. Maths is all about making links, which is why throughout this resource you will find signposts emphasising connections between different topics, applications and suggestions for further research.

Rewind Reminders of where to find useful information from earlier in your study.

Fast forward Links to topics that you may cover in greater detail later in your study.

Focus on… Links to problem-solving, modelling or proof exercises that relate to the topic currently being studied.

Did you know? Interesting or historical information and links with other subjects to improve your awareness about how mathematics contributes to society.

Worksheet A support sheet for each chapter contains further worked examples and exercises on the most common question types. Extension sheets provide further challenge for the most ambitious.

Gateway to A Level GCSE transition material which provides a summary of facts and methods you need to know before you start a new topic, with worked examples and practice questions.

Colour coding of exercises The questions in the exercises are designed to provide careful progression, ranging from basic fluency to practice questions. They are uniquely colour-coded, as shown here.

1

A sequence is defined by .

2

Show that

3

Show that

4

Prove by induction that

5

Prove by induction that

6

Prove that

7

Use the principle of mathematical induction to show that

. Use the principle of mathematical induction to prove that

. 8

Prove that

9

Prove that

Black – practice questions which come in several parts, each with subparts i and ii. You only need attempt subpart i at first; subpart ii is essentially the same question, which you can use for further

practice if you got part i wrong, for homework, or when you revisit the exercise during revision. Green – practice questions at a basic level. Blue – practice questions at an intermediate level. Red – practice questions at an advanced level. Yellow – designed to encourage reflection and discussion.



Working with the large data set As part of your course you will work with a large data set. At the time of this Student Book’s publication (2017), this data set is on the purchasing of different types of food in different parts of the country in different years – this context may change in future years, but the techniques for working with large data sets will stay the same. This is an opportunity to explore statistics in real life. As well as using the ideas from Chapters 20 and 22 you will use these data to look at four key themes.

Practical difficulties with data The real world is messy. Often there are difficulties with being overwhelmed by too much data, or there are errors, missing items or ambiguous labels. For example, how do you deal with the fact that milk drinks and milk substitutes are combined together in some years if you want to compare regions over time? When grouping data for a histogram, how big a difference does where you choose to put the class boundaries make?

Using technology Modern statistics is heavily based on using technology. You will use spreadsheets and graphing packages, looking at the common tools which help simplify calculations and present data effectively. One important technique you can employ with modern technology is simulation. You will look more closely at hypothesis testing by using the data set to simulate the effect of sampling on making inferences about the population.

Thinking critically about statistics Why might you want to use a pie chart rather than a histogram? Whenever you calculate statistics or represent data sets graphically, some information is lost and some is highlighted. In modern statistics it is important to ask critical questions about how evidence provided by statistics is used to support arguments. One important part of this is the idea of validating statistics. For example, what does it mean when it says that there has been a 100% decrease in the amount of sterilised milk purchased? You will look at ways to interrogate the data to try to understand it more.

Statistical problem solving Technology can often do calculations for you. The art of modern statistics is deciding what calculations to do on what data. You rarely have exactly the data you want, so you have to make indirect inferences from the data you do have. For example, you probably will not see a newspaper headline saying ‘the correlation coefficient between amount of bread purchased and amount of confectionary purchased is – 0.52’, but you might see one saying ‘Filling up on carbs reduces snacking!’. Deciding on an appropriate statistical technique to determine whether bread purchases influence confectionary purchases and then interpreting results is a skill which is hard to test but very valuable in real world statistics. There are lots of decisions to make. Should you use the mass of butter bought? Or the mass compared to 2001? Or the mass as a percentage of all fats? The answer to your main question depends on such decisions. You will explore all of these themes with examples and questions in the Elevate Section. You need to get used to working with the variables and contexts presented in the large data set.

1 Proof and mathematical communication In this chapter you will learn how to: review proof by deduction, proof by exhaustion and disproof by counterexample use a new method of proof called proof by contradiction criticise proofs.

Before you start… Student Book 1, Chapter 1

You should be able to use logical connectors.

1 Insert or in the places marked a and b:

a b Student Book 1, Chapter 1

You should be able disprove a statement by counterexample.

2 Disprove the statement: ‘Apart from there are no other integers that can be written as both and . ’

Student Book 1, Chapter 1

You should be able to prove a statement by deduction.

3 Prove that the sum of any two odd numbers is always even.

Student Book 1, Chapter 1

You should be able to prove a statement by exhaustion.

4 Use proof by exhaustion to prove that is a prime number.

Developing proof One of the purposes of this chapter is to provide revision of the material from Student Book 1. It draws on all chapters from that book but, in particular, it builds on the fundamental ideas of proof from Chapter 1. This chapter introduces a new and very powerful method of proof that mathematicians often rely on: proof by contradiction.

Section 1: A reminder of methods of proof In Student Book 1, Chapter 1, you met proof by deduction, proof by exhaustion and disproof by counterexample. The questions in Exercise 1A show how these methods can be used in topics from throughout Student Book 1. EXERCISE 1A 1

Use a counterexample to show that this statement is not true:

2

The velocity of a particle after time is given by

. Prove that the particle never returns to

its original position. 3

a Prove from first principles that if

then

b Use a counterexample to show that if

. then it is not necessarily true that

.

4

Prove that

.

5

Prove by exhaustion that any square number is either a multiple of or one more than a multiple of .

6

A set of data has mean , mode and median . Consider this statement: Prove this statement or use a counterexample to disprove it.

7

The diagram shows a triangle

where

lie on the circle with centre .

is the midpoint of is the angle

.

a Use the cosine rule to prove that b Show that

.

c Hence prove that

8

.

In quadrilateral

is the origin and

is the midpoint of . a Show that b Hence prove that

.

is the midpoint of

. is a parallelogram.

are the position vectors of points is the midpoint of

and .

and is the midpoint of

c If 9

is a rectangle, what can be said about the quadrilateral

a Use a counterexample to disprove the statement b Prove that if c If

10

does it mean that

a Show that b Hence prove that

11



, then

Prove algebraically that if take is one.

? .

. ?

where is a constant to be determined. is prime if and only if

.

then the sum of the probabilities of the different values can

Section 2: Proof by contradiction Proof by contradiction starts from the opposite of the statement you are trying to prove, and shows that this results in an impossible conclusion. WORKED EXAMPLE 1.1

Use proof by contradiction to prove that there are an infinite number of prime numbers. Assume that there is a largest prime number, .

Proof by contradiction always starts by assuming the opposite of what you want to prove.

Construct another number that is the product of all the prime numbers up to and

Now set about trying to find a larger prime than .

including . Consider



. This is one greater than a

number divisible by all the primes up to and including , so it cannot be divisible by any of the primes up to and including .

Therefore is either itself prime, or is divisible by primes larger than . Either way, a prime larger than has been discovered which contradicts the premise that there is a largest prime number. Therefore there are an infinite number of prime numbers.

Here the contradiction to the original assumption (that there is a largest prime) occurs.

Did you know? A variation on the proof in Worked example 1.1 can be found in Euclid’s masterpiece The Elements, a textbook written in around but still in use in many schools in the first half of the 20th century!

WORKED EXAMPLE 1.2

WORKED EXAMPLE 1.2

Using the fact that if

is even then so is , prove that

is irrational.

Start by assuming the opposite of what you want to prove. Assume that

A number is rational if it can be written as a fraction – and this can always be cancelled down until the numerator and denominator share no common factors.

where and are integers

with no common factors. Squaring both sides:

     (1) This means that is even so must also be even. Therefore

, so

Substituting into

.

:

Using the fact given.

This means that is even, so must be even.

Using the given fact again.

This has shown that both and are even, so they share a factor of . This contradicts the

Here the contradiction (to the fact that and share no common factors) arises.

original assertion, so it must be incorrect. Therefore

cannot be written as .

EXERCISE 1B

EXERCISE 1B 1

Prove that if

2

Prove that

3

Prove that there is an infinite number of even numbers.

4

Prove that the sum of a rational and irrational number is irrational.

5

Prove that if

6

Prove that

7

Prove that

is even then is also even. is irrational.

is even, with

integers, then at least one of them is even.

is irrational. is irrational.

Worksheet See Support Sheet 1 for an example of the same type as Q7 and for further practice questions on proof by contradiction. 8

Suppose that is a composite integer. Prove that has a prime factor less than or equal to

9

Prove that if any

10

Prove that

11

a Show that if

.

dates are chosen, some three must be within the same month. is never if and are whole numbers. is a solution to the equation

then

.

b Explain why there is no solution to this equation when is odd and is odd. c Prove that there are no rational solutions to 12



Prove that if a triangle has three sides, triangle.

and , such that

. , then it is a right-angled

Section 3: Criticising proofs In Student Book 1, Chapter 1, you were introduced to the notation used in logic: means that statements and are equivalent. means if is true then so is . means if is true then so is . When looking at proof (including solving equations, which is a type of proof!) you have probably looked out for errors in areas such as arithmetic or algebra. You now also need to look for errors in logic. WORKED EXAMPLE 1.3

Yas was solving the equation

. Find the errors in her working.

In line the symbol should be : if is negative line is correct but line is not possible.

This is an error in logic.

In line the symbol should be then

: if

This is an error in logic.

In line the square root of

should be

Because some of the implications go only one way, the final solution might not work in the original equation. It should be checked.

.

This is an arithmetic error. This is an error in logic.

EXERCISE 1C 1

Lambert was asked to solve the equation

Here is his working.

Line 1:   Line 2: 



Line 3: 



Line 4: 



Line 5: 



or

a By checking his solutions, find the correct solution. b In which line of working is his mistake? 2

Craig was asked to solve the equation

a Show that

. Here is his solution.

is also a solution to the original equation.

b What logical symbol should Craig have used in the second line? 3

Freja was asked to solve the equation

. Here is her working.

Line 1:   Line 2: 

 

Line 3: 

 

Line 4: 

 

Line 5: 

 

Line 6: 

 

Line 7: 

 

a By checking her solutions find the correct solution. b In which line of working is her mistake? 4

Jamie was asked to solve

. Here is her working.

Line 1:   Line 2: 

 

Line 3: 

 

Line 4: 

 

Line 5: 

 

Line 6: 

 

Line 7: 

 

In which line of working did Jamie make a mistake? 5

Andrew was asked to prove the statement: ‘The function is his working.

has a minimum at

’. Here

Line 1:   Line 2:   Line 3:  Line 4:   Line 5:    Line 6:      Line 7:  So it is a minimum. Describe the errors in this proof. 6

Ann was asked to answer the question: ‘ If

is a factor of

is a factor of

. Find the remaining factor.’

then:

    Comparing coefficients of Therefore the remaining factor is

.

Brian claims that Ann’s solution isn’t complete. Explain why he is correct, and give a full solution. 7

Find the error in this proof that

is irrational.

Line 1:  Assume that

where and are integers with no common factors.

Line 2:  Squaring both sides gives Line 3:  So Line 4:  This means that is even so must also be even. Line 5:  You can then write that

, so

Line 6:  Substituting this into equation Line 7:  So

.

gives

.

Line 8:  This means that is even, so must be even. Line 9:  But you have shown that both and are even, so they share a factor of . Line 10: This contradicts the original assertion, so

cannot be written as .

Checklist of learning and understanding You should be able to apply counterexamples, proof by exhaustion and proof by deduction to material from Student Book 1. Proof by contradiction is a method of proof that works by showing that assuming the opposite of the required statement leads to an impossible situation. When criticising proofs, look out for flaws in logic as well as mistakes in algebra or arithmetic.



Mixed practice 1 1

Which symbol should be used to replace ‘?’ in this working?

A B C D 2

Prove that

3

Prove that

4

Prove that there is an infinite number of square numbers.

5

Find the error(s) in this working to solve

is always even. is irrational.

for

.

Line 1:  

6

Line 2: 

 

Line 3: 

 

Line 4: 

 

Line 5: 

 

Fermat says that if is prime then

is prime.

Which value of provides a counterexample to this statement? A B C D 7 8

Prove that

is a parallelogram with at the origin and . is the midpoint of and is the point on Prove that

9 10

is irrational. and

is a straight line.

Prove that if and are whole numbers a Prove that if then

are the position vectors of points such that is .

is either odd or a multiple of .

is a polynomial of finite order, with integer coefficients, and is an integer,

is an integer.

b Use a counterexample to show that this statement is not always correct. If is a polynomial, where integer coefficients. 11

Consider this working to solve Line 1: 

 

Line 2: 

 

Line 3: 

 

is an integer whenever is an integer, then

must have

Line 4: 

 

Line 5: 

 

Line 6: 

 

a In which lines are there mistakes? b Rewrite the solution correctly, making appropriate use of logical connectors. 12

This proof is trying to demonstrate that there are an arbitrary number of consecutive composite (non-prime) numbers. Line 1: Consider for Line 2: 

is divisible by

Line 3: So the numbers

are not prime.

Line 4: Therefore, this is a list of consecutive composite numbers. Which is the first line to contain an error? 13

Prove that if

14

a By considering a right-angled triangle, prove that if is an acute angle then

and are integers such that

b Hence prove that 15

Prove that for values of between

, then either or is even.

is a rational number. and

.

Worksheet See Extension Sheet 1 to complete the details of a couple of famous proofs.



2 Functions In this chapter you will learn how to: distinguish between mappings and functions determine whether a function is one-to-one or many-to-one find the domain and range of a function find composite functions find the inverse of a function.

Before you start… GCSE

You should be able to interpret function notation.

1 Given that

, evaluate:

a b

Student Book 1, Chapter 1

You should be able to use interval notation to write inequalities.

2 Use interval notation to write these inequalities. a

and

b

or

3 Express

Student Book 1,

You should be able to complete

Chapter 3

the square.

in the form

Student Book 1, Chapter 3

You should be able to solve quadratic inequalities.

4 Solve the inequality

Student Book 1, Chapter 7

You should be able to rearrange exponential and log expressions.

5 Make the subject of each equation.

. Hence state the coordinates of the turning point of .

a b

Student Book 1, Chapter 13

You should be able to establish where a function is increasing/decreasing.

6 Find the range of -values for which is an increasing function.

Why study functions? Doubling, adding five, finding the largest prime factor – these are all instructions that can be applied to numbers to produce a numerical result. This idea comes up a lot in mathematics. The formal study of it leads to the concept of a function. You will use functions whenever you need to express how one quantity changes with another, whether it is how the strength of the gravitational force varies with distance, or how the amount of paint needed depends on the area of the wall.



Section 1: Mappings and functions A mapping is any rule that assigns to each input value ( ) one or more output values ( ). For example:

factor of . In the first of these, each -value maps to a single -value. But in the second and the third this is not always the case. In the second example, when could be or . In the third example, when could be

or .

Mappings such as

, where there is only one -value for each -value, are called functions.

The easiest way to identify if a mapping is a function is to look at its graph, and apply the vertical line test.

: a function

: not a function

a factor of : not a function

Key point 2.1 A mapping is a function if every -value maps to a single -value. Vertical line test: if a mapping is a function, any vertical line will cross its graph at most once.

The most common way to describe a function is by using function notation, for example . An alternative way of writing this is . For a specific value of the function, you would write , or

.

is the image of .

Gateway to A Level See Gateway to A Level section F for a reminder about basic techniques with functions.

Once you have decided that a mapping is a function, you can ask whether each output comes from just one input. To check this, apply the horizontal line test.

Key point 2.2 A function is: one-to-one if every -value corresponds to only one -value. many-to-one if there are some -values that come from more than one -value. Horizontal line test: if a function is one-to-one, any horizontal line will cross the graph at most once.

A mapping in which a single input corresponds to more than one output (so is not a function) is called

one-to-many.

WORKED EXAMPLE 2.1

Which of these graphs represent functions? Classify those that are functions as one-to-one or many-to-one. a

b

c

a

Draw several vertical lines and see how many times they cross the graph.

Any vertical line meets the graph at most once, therefore it is a function. Draw several horizontal lines and see how many times they cross the graph.

Some horizontal lines meet the graph at more than one point, therefore it is a many-to-one function. b



Draw several vertical lines and see how many times they cross the graph.

Some vertical lines meet the graph more than once, therefore it is not a function. Draw several vertical lines and see how many times they cross the graph. Be careful: a vertical line through an open circle does not count as an intersection.

c

Any vertical line meets the graph at most once, therefore it is a function.

Tip Remember that an open circle on a graph means that point is not a part of the graph, and a solid circle means that it is.

Draw several horizontal lines and see how many times they cross the graph.

Any horizontal line meets the



graph at most once, therefore it is a one-to-one function.

When a mapping is given by an equation linking and you may not know how to draw its graph. In that case, use the equation and ask whether there are any values of for which you can find more than one value of . WORKED EXAMPLE 2.2

Determine whether each mapping is a function. a b a For

could be or

So this is not a function. b

.

Is there more than one value of for each value of ? Think of an example.

There is only one value of for each value of . So this is a one-to-one function.

Is there more than one value of for each value of ? Find the value of for a given value of . Finding the cube root does not introduce other possible values (unlike finding the square root, which does).

EXERCISE 2A 1

Classify each graph as a mapping or a function. If you decide a graph is a function, state if it is one-to-one or many-to-one. a

b

c

d

e

f

2

Determine whether each mapping is a function ( denotes the input and the output). a i ii b i ii c i ii

3

Classify each function as one-to-one or many-to-one. a i ii b i ii c i ii

4

Determine whether each mapping is a function ( denotes the input and the output). a i ii b i ii c i ii

5

Which of these statements is correct? A The square root of is . B The square root of is

.

Would it be possible to define the ‘square root’ function so that the square root of is



?

Section 2: Domain and range As well as telling you what to do with the input, for a function to be completely defined it needs to tell you what type of input is allowed to go into the function.

Key point 2.3 The set of allowed input values is called the domain of the function.

WORKED EXAMPLE 2.3

Sketch the graph of

over each domain.

a b a

Sketch the graph where all real numbers are allowed for .

Discard the part of the graph outside the required domain. Since the endpoint is not included, label it with an open circle.

Use the same original graph as in a, but this time it only exists at whole numbers, so label them with solid circles.

b

The largest possible domain of a function will be all real numbers unless it contains a mathematical operation that can’t accept certain types of number. The three most common restrictions on the domain are: you cannot divide by zero you cannot find the square root of a negative number you cannot take the logarithm of a non-positive number.

Fast forward In Sections 3 and 4 of this chapter, you’ll see that you don’t always use the largest possible domain when forming composite or inverse functions.

WORKED EXAMPLE 2.4

Find the largest possible domain of Need So the domain is

. You can only take the log of a positive number.

.

Sometimes you need to apply more than one restriction. WORKED EXAMPLE 2.5

What is the largest possible domain of

? Write your answer using interval

notation. Need       

Look for any of the three potential problems listed. There will be division by zero when

Need         Domain:

There will be the square root of a negative number when .

.

and

Both these restrictions are needed. This describes two intervals: from from to infinity, excluding .

to and

Tip Remember that

, for example, is just an alternative notation for

.

Once you have fixed the domain of a function, you can ask what values can come out of the function.

Key point 2.4 The set of all possible outputs of a function is called the range.

Tip Be aware that the range will depend upon the domain.

The easiest way of finding the range is to sketch the graph. WORKED EXAMPLE 2.6

Find the range of

if the domain is:

a b

.

a

Sketch the graph

Range:

Use the graph to state which -values can occur. Sketch the graph

for the given domain.

for the given domain.

b

Use the graph to state which -values can occur.

Range:

Tip Graph-plotting software or graphical calculators can be useful when investigating the domain and range of functions.

Tip Although it is the values of the graph of that you are considering when finding the range, always make sure you give the range in terms of rather than , for example, , not .

EXERCISE 2B 1

State the largest possible domain and range of each function. a b c d

Rewind Exponential functions were covered in Student Book 1, Chapter 8. 2

Find the largest possible domain of each function. a i ii b i

ii c i ii d i ii e i ii f

i ii

3

Find the range of each function. a i ii b i ii c i ii d i ii

4

Find the largest possible domain and range of each function. a i ii b i ii c i ii d i ii

Rewind These questions require knowledge of completing the square and the solution of quadratic inequalities, covered in Student Book 1, Chapter 3. 5

6

a Write

in the form

.

b Hence state the range of the function

.

Find the domain and range of the function

.

7 8 9

The function is given by

. Find the domain of the function.

Find the largest possible domain of the function a Sketch the graph of

.

.

b Hence find the largest possible domain of the function 10 11

.

Find the largest possible domain of the function

.

Find the largest set of values of such that the function given by values.

12

a State the domain of the function i ii

.

b Write an expression for



.

if:

takes real

Section 3: Composite functions After applying a function to a number, you can apply another function to the result. The resulting function is called a composite function.

Key point 2.5 If you apply the function to and then the function to the result, you write: or

It can be useful to refer to

or

as the inner function and

as the outer function.

Common error If you have the composite function

, make sure you apply first and then . Don’t work

left to right! Be careful:

and

are not the same function.

WORKED EXAMPLE 2.7

If

and

find:

a b c

. Evaluate

a

and then apply to the result.

Note that you do not need to work out the general expression for . b

Replace in

with the expression for

.

c

Replace in

with the expression for

.

WORK IT OUT 2.1 Two functions are defined by

and

. Find

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

You can also compose a function with itself.

Key point 2.6 can also be written as

.

WORKED EXAMPLE 2.8

Given that

find and simplify expression for

.

means do

, then apply f to the result.

Multiply top and bottom of the fraction by to simplify the denominator.

For the composite function to exist, the range of must lie entirely within the domain of otherwise you would be trying to put values into which it cannot take.

,

Tip Whichever notation is being used, remember the correct order: the function nearest to acts first!

WORKED EXAMPLE 2.9

The functions and are defined by

and

a Explain why the composite function

is not defined.

b Find the largest possible domain for which a Need

.

is defined. In this case, state the range of Check whether the output from is within the domain of .

.

But, for example, the domain of .

, and this is not in

It is enough to find one counterexample to show that is not true for all . You need

b For largest possible domain,

or

.

This is a quadratic inequality; the solution consists of two separate intervals.

On this domain, the range is

When is defined, it can take any nonnegative real value.

.

A more complex problem is to recover one of the original functions when you are given a composite function. The best way to do this is to use a substitution. WORKED EXAMPLE 2.10

If

find

. Substitute

the inner function

Rearrange to get Replace all s.

Write the answer in terms of .

EXERCISE 2C 1

Given that a i ii

and

find:

.

b i .

ii 2

Given that

and

find:

a i ii b i .

ii 3

a Given that

and

find:

i .

ii

b Given that

and

, find:

i .

ii 4

Given that

and

find:

a

5

b

.

Find

, given these conditions.

a i ii b i ii c i ii d i ii 6

Given that

and

7

Given that

and

8

The function is defined by

, solve the equation , solve the equation . Find an expression for

. . in terms of in each case.

a b 9

Functions and are defined by undefined for .

and

a Find the value of and the value of . b Find the range of 10

.

and a Explain why

exists but

does not.

b Find the largest possible domain for so that

is defined.

. The composite function

is

11



Let and be two functions. Given that

and

, find

.

Section 4: Inverse functions Functions transform an input into an output, but sometimes you can reverse this process to allow you to say which input produced a given output. When this is possible, you do it by finding the inverse function, usually labelled

.

For example, if then is a number that, when put into , produces the output words, you are looking for a number such that . Hence .

. In other

Common error Make sure you don’t get confused about this notation. With numbers, the superscript ‘ denotes reciprocal, for example, With functions,

’

.

denotes the inverse function of .

Finding the inverse function To find the inverse function you must rearrange the formula to find the input ( ) in terms of the output ( ).

Key point 2.7 To find the inverse function

given an expression for

:

1 start with 2 rearrange to get (the input) in terms of (the output) 3 give

by replacing the s with s.

WORKED EXAMPLE 2.11

Find the inverse function of

. Start with

.

Rearrange to make the subject. Express both sides as powers of e to remove ln. Write the resulting function in terms of .

WORK IT OUT 2.2 Find the inverse function of

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

The relationship between and Once you know how to find inverse functions there are a couple of very important facts you need to know about them. When you are finding the inverse function you switch the inputs and the outputs, so on the graph you switch the - and -axes.

Key point 2.8 The graph of

is a reflection of the graph of

in the line

When you apply a function and then its inverse you get back to where you started.

.

Key point 2.9 , for all .

WORKED EXAMPLE 2.12

The graph of

is shown. Sketch the graphs of:

a b

a

The graph of line of

b

Simplify

is a reflection in the .

to

.

The fact that the graphs of and are reflections of each other gives you a very useful technique to solve some equations that would otherwise involve complicated (or impossible) algebra. WORKED EXAMPLE 2.13

The diagram shows the graph of the function

a On the same axes, sketch the graph of b Solve the equation

.

.

.

a

The graph of is the reflection of the graph of in the line .

b

Finding the equation for the inverse function would involve solving a complicated cubic equation. Luckily, you can see from the graph that the graphs of and intersect on the line . This means that solving the equation is equivalent to solving the equation .

The reflection in the line coordinates.

swaps the domain and the range of a function, because it swaps and

Key point 2.10 The domain of The range of

is the same as the range of

.

is the same as the domain of

.

This means that, when asked to find the domain and range, you can work either with f or with whichever one is easier.

,

WORKED EXAMPLE 2.14

The function is defined by

for

a Find an expression for

.

and state its domain and range.

b Find the range of . Set

a

.

Make the subject.

Replace with . The domain of

The range of

is

is

b The range of is

EXERCISE 2D 1

Find a i ii b i ii c i ii d i ii

if:

.

In the expression for cannot be zero.

, the denominator

.

The range of is the same as the domain of , which was given in the question.

.

The range of f is the same as the domain of , which you found in part a.

e i ii f

i ii

g i ii h i ii 2

Sketch the inverse of each function. a

b

c

d

3

For each function, find the expression for range for both and .

and hence state the largest possible domain and

a i ii b i ii c i ii d i ii 4

This table gives selected values of the one-to-one function

.

 

a Evaluate b Evaluate

Worksheet See Support Sheet 2 for a further example of finding inverse functions and their domains, and for more practice questions.

5

The function is defined by Evaluate

6

.

.

Given that

:

a find the inverse function b state the domain and range of

.

7

Given functions

8

Let and be to functions such that Let .

and

, find the function

.

is defined, and suppose that both

and

both exist.

Prove that 9

.

The diagram shows the graph of graph.

. The lines

a On the same axes, sketch the graph of b State the domain and range of c Solve the equation 10

and

are the asymptotes of the

.

. .

The functions and are defined by

.

a Calculate b Show that

11

The function is defined for

12

Let a Find b Let

13

by

Find an expression for

.

, for . . Find

, giving your answer in the form

The function is defined by

for

. The graph of

diagram. a On the same set of axes, sketch the graph of b Solve the equation

.

When does the inverse function exist?

, where

.

. is shown in the

All functions have inverse mappings, but these inverse mappings are not necessarily themselves functions. Since an inverse function is a reflection in the line , for the result to pass the vertical line test, the original function must pass the horizontal line test. But, as you saw in Section 1, this means it must be a one-to-one function.

Fast forward This idea will be applied in Chapter 7 when inverses of trigonometric functions are defined.

Key point 2.11 Only one-to-one functions have inverse functions.

This leads to one of the most important uses of domains. By restricting the domain you can turn any function into a one-to-one function, which allows you to find its inverse function. WORKED EXAMPLE 2.15 a Find the largest value of such that the function b Find

is one-to-one.

for this value of and state its range.

a

Sketch the graph of … but eliminate the points towards the right of the graph that cause the horizontal line test to fail.

Decide which section remains. b

Follow the standard procedure for finding inverse functions.

Take the negative root, as

Since

Write The range of

WORK IT OUT 2.3

is

.

.

.

The range of

is the domain of .

Find the inverse function of

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3 It doesn’t exist.

It should be clear from the horizontal line test that a function is one-to-one if f either increases or decreases throughout its domain. As soon as there is a turning point, the function is no longer one-to-one (and therefore has no inverse). WORKED EXAMPLE 2.16

, for all real . Prove that

has an inverse function. f must either be increasing or decreasing so find to determine which.

for all

is one-to-one for all .

So has an inverse function.

Complete the square. Note that this is a common way of showing that a function is non-negative. (In this case it actually factorises.) Although , the gradient is never negative and so is one-to-one.

EXERCISE 2E 1

Find the value of that gives the largest possible domain such that the inverse function exists. For this domain, find the inverse function. a b c

d 2

For each function shown in the diagrams, determine the largest possible domain of the given form for which the inverse function exists. For that domain, sketch the inverse function. a domain:

b domain:

c domain:

3

Given that

for

:

a find b explain why

does not exist.

The domain of is changed to

so that

now exists.

c Find the largest possible value of . 4

A function is self-inverse if

for all in the domain.

Find the value of the constant so that 5

A function is defined by

is a self-inverse function. where is a constant.

Find the set of values of for which

exists for all .

Checklist of learning and understanding A mapping is a function if every -value maps to a single -value. The output value corresponding to the input is called the image of . A function is: one-to-one if every -value corresponds to only one -value many-to-one if there are some -values that come from more than one -value. The vertical and horizontal line tests can be applied to graphs of mappings. Vertical line test: if a mapping is a function, any vertical line will cross its graph at most once. Horizontal line test: if a function is one-to-one, any horizontal line will cross the graph at most once. The set of allowed input values of a function is called the domain. The set of all possible outputs of a function is called the range. The composite function formed by applying to and then to the result is written as or or . The inverse,

, of a function is such that

To find the inverse function

, for all .

given an expression for

1

start with

2

rearrange to get (the input) in terms of (the output)

3

give

:

by replacing the s with s.

Only one-to-one functions have inverse functions. The graph of The domain of The range of



is a reflection of the graph of

in the line

is the same as the range of

.

is the same as the domain of

.

.

Mixed practice 2 1 Which one of these statements about the inverse function,

is true?

A B C doesn’t exist

D 2

Find the inverse of each function. a b

3

The diagram shows three graphs. is part of the graph of

.

is part of the graph of

.

is the reflection of graph in line . Write down: a the equation of in the form b the coordinates of the point where cuts the -axis.

4

Let

5

If

, and

. Solve the equation and

.

:

a evaluate b find an simplify an expression for c state the geometric relationship between the graphs of

and

d i find an expression for ii find the range of iii find the domain of e solve the equation 6

.

The functions and are defined with their respective domains by:

, for , for real values of a State the range of . b i Find

.

ii Solve the equation c The inverse of is i Find

.

.

.

ii Solve the equation

. [© AQA 2012]

7

The functions and are defined with their respective domains by of and

for all real values

for real values of

a Explain why does not have an inverse. b The inverse of is

. Find

c State the range of

.

.

d Solve the equation

. [© AQA 2011]

8

The function is given by a Write

, for

in the form

.

.

b Find the inverse function c State the domain of

. .

and

9

a Find and simplify: i ii the range of iii iv v

.

b Explain why

does not exist.

c i Find the form of

.

ii State the geometric relationship between the graphs of iii State the domain of iv State the range of 10

and

.

. .

The functions and are defined over the domain of all real numbers by and . a Write b Hence sketch the graph of of the turning point.

in the form

.

, labelling all axis intercepts and the coordinates

c State the range of

and

.

d Hence or otherwise find the range of 11

The curve with equation a The inverse of is i Find

.

, where

, is shown in the sketch.

.

.

ii State the range of

.

, indicating the value of the -coordinate of the iii Sketch the curve with equation point where the curve intersects the -axis. b The function is defined by i Find

, for all real values of .

, giving your answer in the form

ii Write down an expression for .

, where

and are integers.

, and hence find the exact solution of the equation

[AQA 2013]

12 a Write

in the form

.

b Hence or otherwise find the range of

.

c By using the largest possible domain of the form 13

a Show that if

then

. By replacing all instances of with satisfies.

c By solving these two identities simultaneously, express 14

The functions is defined for

and are given by except for the interval

in terms of .

and

. The function

.

a Calculate the values of and of . b Find the range of 15

.

An odd function is any function a Show that

.

.

b A function satisfies the identity find another identity that

, find the inverse function

that satisfies

.

is an odd function.

b What type of symmetry must the graph of any odd function have?

c Given any function

, show that

An even function is any function which satisfies d Show that

is an odd function. .

is an even function.

e What type of symmetry must the graph of any even function have? f

Given any function

show that

is an even function.

g Hence or otherwise show that any function can be written as the sum of an even function and an odd function.

Worksheet See Extension Sheet 2 for a selection of more challenging problems.



3 Further transformations of graphs In this chapter you will learn how to: draw a graph after two (or more) transformations find the equation of a graph after a combination of transformations sketch graphs of functions involving the modulus (absolute value) use modulus graphs to solve equations and inequalities.

Before you start… Student Book 1, Chapter 5

You should be able to recognise a graph transformation from the

1 The graph of diagram.

is shown in the

equation.

Sketch the graph of: a b Student Book 1, Chapter 5

You should be able to change the equation of a graph to achieve a given transformation.

2 A graph has equation . Find the equation of the graph after: a a translation of units in the positive -direction, followed by b a horizontal stretch with scale factor .

Student Book 1, Chapter 1

You should be able to use interval notation to express solutions of inequalities.

3 Solve each inequality and write the solution using interval notation. a b

and

Combining transformations In this chapter the transformations you met in Student Book 1, Chapter 5, will be combined to produce a sequence of transformations of the original graph. The modulus function will also be introduced; this is used in many contexts where a quantity needs to be positive, for example, the total distance travelled must be the sum of positive quantities, even though the particle may change direction. You have also seen this idea used to find the area between a curve

and the -axis when the enclosed region is below the axis.

Section 1: Combined transformations In this section you will look at what happens when two transformations are applied to a graph. An important question to consider is, does the order in which the two transformations are done affect the outcome? To investigate this question, first consider transformations of a single point.

The point

is translated two units up and then reflected in the -axis. The new point is

The point

is reflected in the -axis first and then translated two units up. The new point is

The point

is translated two units up and then reflected in the -axis. The new point is

.

.

.

The point

is reflected in the -axis first and then translated two units up. The new point is

Key point 3.1 When two vertical transformations or two horizontal transformations are combined, changing the order affects the outcome. When one vertical and one horizontal transformation are combined, the outcome does not depend on the order.

Combining one vertical and one horizontal transformation WORKED EXAMPLE 3.1

The diagram shows the graph of the function

.

On separate diagrams, draw the graph of: a b

Both parts involve one horizontal and one vertical transformation. Since the order doesn’t matter, deal with the transformation in brackets first. a

: translate units to the left

is replaced by , so the graph is translated to the left.

.

b

: vertical stretch with scale factor

Multiplying the whole expression by results in all -values doubling.

: horizontal stretch with scale factor

is replaced by , so the graph is stretched horizontally.

: reflection in the -axis

Making the -coordinate negative results in a reflection in the -axis.

Combining two vertical transformations To transform the graph of into the graph of first multiply by and then add on . The flow chart shows the order of operations and the corresponding transformations.

Notice that this follows the normal order of operations: multiplication is done before addition. WORKED EXAMPLE 3.2

A graph has equation

. Find the equation of the graph after these transformations.

a A vertical stretch with scale factor followed by a translation with vector b A translation with vector a Vertical stretch:

followed by a vertical stretch with scale factor . Vertical stretch: multiply the function by .

becomes

Vertical translation: .

becomes

The final equation is

b Vertical translation: .

Vertical translation: add to the whole expression. You could perform the two transformations in one go: The new equation is .

becomes

Vertical stretch:

.

This time add to the expression first.

Then multiply the whole expression by .

becomes

. The final equation is

.

Again, you can write this in one go:

.

Combining two horizontal transformations To transform the graph of into the graph of you first replace with and then replace all occurrences of by . The flow chart shows the order of operations and the corresponding transformations.

Notice that the transformations are in the ‘wrong’ order: the addition is done before the multiplication. WORKED EXAMPLE 3.3

The diagram shows the graph of

. Sketch the graph of

.

This is a combination of two horizontal transformations, so deal with the addition first.

is replaced by

horizontal translation

Replace with Change

is replaced by

horizontal stretch, scale factor

Replace with Change

. to

.

to



This graph shows the final answer.

If you want to perform a horizontal stretch before a translation you need to use brackets correctly in the equation. WORKED EXAMPLE 3.4

The graph of is transformed by a horizontal stretch with scale factor and then translated units to the right. Find the equation of the resulting graph. A horizontal stretch with scale factor is achieved by replacing by .

Replace by : is changed to Replace by

.

:

is changed to The final equation is

. .

A horizontal translation by units is achieved by replacing by .

Note that if the translation had been performed before the stretch, the resulting equation would have been .

Make sure that you know the correct order of resolving transformations.

Key point 3.2 Vertical transformations follow the normal order of operations as applied to arithmetic. Horizontal transformations are done in the opposite order.

WORK IT OUT 3.1 Describe the sequence of two transformations that transform the graph of of . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Add to ; this is a horizontal translation units to the left. Replace by

; this is a horizontal stretch with scale factor .

Solution 2 Replace by

; this is a horizontal stretch with scale factor .

Add to ; this is a horizontal translation units to the left. Solution 3 Add to ; this is a horizontal translation units to the left. Replace by ; this is a horizontal stretch with scale factor .

EXERCISE 3A 1

Here are the graphs of

and

.

to the graph

Sketch the graphs of: a i ii b i ii c i ii d i ii e i ii 2

Given that

, express each of these functions as

of transformations mapping

and hence describe the sequence

to the given function.

a i ii b i ii 3

Given that

, give the function

that represents the graph of

a i translation

, followed by a vertical stretch of scale factor

ii translation

, followed by a vertical stretch of scale factor

b i vertical stretch of scale factor , followed by a translation ii vertical stretch of scale factor , followed by a translation

after:

c i reflection through the horizontal axis, followed by a translation ii reflection through the horizontal axis, followed by a translation d i reflection through the horizontal axis, followed by a vertical stretch of scale factor , followed by a translation ii reflection through the horizontal axis followed by a translation

, followed by a vertical

stretch, scale factor . 4

Given that

, express each of these functions as

transformation mapping

and hence describe the

to the given function.

a i ii b i ii 5

Given that a i translation ii translation

, give the function

that represents the graph of

after:

, followed by a horizontal stretch of scale factor , followed by a horizontal stretch of scale factor

b i horizontal stretch of scale factor , followed by a translation ii horizontal stretch of scale factor , followed by a translation c i translation

, followed by a reflection in the -axis

ii reflection through the vertical axis, followed by a translation 6

Find the resulting equation after the graph of transformations.

is transformed through each sequence of

a A vertical translation units up, then a vertical stretch with scale factor relative to the -axis. b A vertical stretch with scale factor , followed by a vertical translation units up. c A horizontal stretch with scale factor relative to the -axis, then horizontal translation units to the left. d A horizontal translation units to the left, followed by a horizontal stretch with scale factor . 7

a The graph of

is transformed through a horizontal stretch with scale factor , followed by a

vertical stretch with scale factor . Find the equation of the resulting graph. Can you explain why this is the case? b The graph of is transformed through a horizontal translation units to the left, followed by a vertical stretch with scale factor . The equation of the resulting graph is again . Find the value of . c The graph of

is translated units up. What transformation (other than a translation

units down) will return the graph to its original position? Use technology to sketch the graphs and see why this is the case.

8

The diagram shows the graph of

.

On separate axes sketch the graphs of: a b

9

Sketch each graph. a b c In each case, indicate clearly the positions of the vertical asymptote and the -intercept.

10

The graph of

is translated units to the right and then reflected in the -axis. Find the

equation of the resulting graph, in the form 11

The graph of function

.

is transformed by this sequence:

translation by reflection in horizontal stretch with scale factor . The resultant function is

.

Find the values of and . 12

The graph of function

is transformed by this sequence:

reflection through translation by horizontal stretch with scale factor . The resultant function is Find the values of 13

Given that transforming

.

and . , give in simplest terms the formula for

by this sequence of transformations:

vertical stretch, scale factor

, which is obtained by

translation by horizontal stretch, scale factor . 14

a Describe a sequence of two transformations that transforms the graph of

to the graph

of b Sketch the graph of c Sketch the graph of intercepts.



marking clearly the positions of any asymptotes and -

Section 2: The modulus function The modulus (or absolute value) of a number is an operation that leaves positive numbers alone but makes negative numbers positive. You use to denote the modulus of number ; for example, and .

Key point 3.3 The modulus function can be defined as:

The graph of

is shown. It is useful to call the red branch the reflected branch (as it is the

reflection of the graph The domain of

in the -axis) and the blue branch the unreflected branch.

is all real numbers, whilst the range is all positive numbers and zero.

You can combine this with the rules for transforming graphs to sketch some other functions involving the modulus.

Did you know? You can think of the modulus function as giving the distance of a number from the zero on the number line. A similar idea was applied to vectors in Student Book 1. But the distance between two objects is not always easy to define; for example, what length gives the ‘distance’ between two points on the surface of the Earth? This question leads to the idea of a metric. You might want to find out about the Minkowski Metric for finding the distance between two points in space–time in the Theory of Relativity.

WORKED EXAMPLE 3.5

Sketch the graphs of: a b In each case find the -intercept of the graph.

a

-intercept:

has been replaced by , so the graph of is shifted three units to the right.

b

-intercept:

has been replaced by , so the graph of is shifted two units to the left. Subtracting then shifts the graph units down.

You can also think about applying the modulus function to the graph of the function inside the modulus symbol. Any parts of the original graph that are below the -axis will have their -values changed from negative to positive so those parts of the graph will be reflected in the -axis.

Key point 3.4 To sketch a graph of , start with the graph of parts that are below the -axis.

and reflect in the -axis any

Tip In practice it is usually easier to sketch modulus graphs by reflecting parts of the original graph that are below the -axis rather than thinking of transforming

.

WORKED EXAMPLE 3.6

Sketch the graph of

, indicating the intercepts with the coordinate axes. Start by drawing the graph of This crosses the -axis at at .

. and the -axis

The part of the graph to the right of the intercept has negative -values; taking the modulus will make those positive so this part of the graph needs to be reflected in the axis.

Using modulus notation in inequalities The modulus function is often used as an alternative way of writing some inequalities; for example, you can write concisely as .

Fast forward You will use inequalities of this form with sequences (Chapter 4) and the binomial expansion (Chapter 6).

You can extend this notation to write other inequalities that represent a single interval. For example, means , which can be rearranged into .

The number is in the middle of the interval units away from each end. So the inequality can be read as: ‘the distance between and is less than .

Rewind You have already used the modulus of a vector to represent the distance between two points in Student Book 1, Chapter 15.

Similarly, the inequality or

represents the region outside the interval

, i.e. the region

Key point 3.5 The modulus inequality: is equivalent to is equivalent to

or

Fast forward If you study Further Mathematics, you will extend the modulus notation to measure distances between points in the complex plane.

WORKED EXAMPLE 3.7 a Write the inequality: i

in the form

ii

in the form

b Write the interval

or

.

using an inequality of the form is between

a i

. and .

Rearrange to leave on its own.

ii

is greater than or less than

or or

.

Rearrange again.

or is the number in the middle of the interval.

b

is the distance from the middle of the interval to one end. So

.

The end-points of the interval are included.

EXERCISE 3B 1

Sketch each graph, showing the axes intercepts. a i ii b i ii c i ii d i

 

ii 2

Write the equation of each graph in the form a i

ii

b i

ii

.

c i

ii

3

Write these inequalities in the form

.

a i ii b i ii 4

Write these inequalities in the form

or

.

a i ii b i ii 5

Write these statements in the form a i ii b i ii

6

Write these statements in the form a i ii b i ii

.

7

Sketch the graph of: a b showing the coordinates of any axes intercepts.



8

Sketch the graph of

9

Sketch the graph of axes.

10

Sketch the graph of

, labelling any intercepts with the coordinate axes. . Give the coordinates of the points where the graph crosses the

.

Section 3: Modulus equations and inequalities You can use graphs to solve equations and inequalities involving the modulus function. You need to use the graph to decide whether the intersection is on the reflected or the unreflected part of the graph. If it is on the unreflected part you can rewrite the equation without the modulus sign in. If it is on the reflected part you need to replace the modulus sign by a minus sign.

Key point 3.6 When solving an equation involving a modulus function, sketch the graph first.

WORKED EXAMPLE 3.8

Solve the equation Sketch the graphs of

and

.

There are two intersection points: : the blue graph intersects the reflected part of the red graph : the blue line intersects the unreflected part of the red graph. Write a separate equation for each.

For the reflected part, make the equation negative.

For the unreflected part just remove the modulus sign.

So the solution is

or

You can also solve the intersection of two modulus graphs. WORKED EXAMPLE 3.9

Solve the equation

.

Sketch the graphs of

and

There are two intersections: : the unreflected blue line and the reflected red line : the unreflected blue line and the unreflected red line.

You need the reflected part of the red graph, so make the equation negative.

You need the unreflected parts for both graph, so remove the modulus signs.

The solution is



or .

WORK IT OUT 3.2 Solve the equation

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution:

Solution 2

Solution:

or

Solution 3

Solution:

or

To solve inequalities, sketch the graphs and find their intersections. Then decide on which parts of the graph the inequality is satisfied.

Rewind This is the same method that you use to solve quadratic inequalities – see Student Book 1, Chapter 3.

WORKED EXAMPLE 3.10

Solve Sketch the graphs of and , highlighting where the inequality is satisfied.

Find the intersection points. is on the reflected part of the red line.

is on the unreflected part of the red line. Describe the highlighted region in terms of . Remember that you can use either inequality notation or interval notation to do this.

EXERCISE 3C 1

Solve each equation. a i

ii b i ii c i ii d i ii e i ii f

i ii

2

Solve each inequality. Write your answer using the stated notation. a Use interval notation. i ii b Use inequality notation. i ii c Use interval notation. i ii d Use inequality notation. i ii e Use interval notation. i ii f

Use inequality notation. i ii

3

a Solve the equation b Solve the inequality

4

a Solve the equation b Solve the inequality

. Write your answer using interval notation.

Worksheet See Support Sheet 3 for a further example of modulus inequalities and for more practice questions. 5

a Sketch the graph of

.

b Hence solve the inequality 6

Solve the inequality

7

Given that

8

Solve the equation

. .

, find in terms of the solution of the inequality , where

.

.

Checklist of learning and understanding Two (or more) transformations of graphs can be combined: translations, stretches and reflections, both horizontal and vertical. The order in which transformations occur may affect the outcome. One horizontal and one vertical transformation can be done in either order. Changing the order of two horizontal or two vertical transformation affects the outcome. For a function of the form

the stretch is performed before the translation.

For a function of the form

the translation is performed before the stretch.

The modulus function can be used to reflect the part of the graph below the -axis so that the whole graph is above it. To solve equations and inequalities involving the modulus function always use graphs. You need to decide whether the solutions are on the reflected or unreflected part of the graph.



Mixed practice 3 1

The graph of

is shown.

Sketch on separate diagrams the graphs of: a b 2

Solve the equation Choose from these options. A

or

B

or

C D 3

The graph of

is transformed by applying a translation with vector

followed by

a vertical stretch with scale factor . Find the equation of the resulting graph in the form . 4

Find two transformations whose composition transforms the graph of of .

5

a On the same set of axes sketch the graphs of b Hence solve the inequality

6

and

to the graph

.

.

The diagram shows the graphs of

and

.

a Find the -coordinates of the points of intersection of the graphs of b Hence, or otherwise, solve the inequality

.

and

.

[© AQA 2013] 7

Describe a sequence of two transformations that map the graph of

onto the graph of

. Choose from these options. a Horizontal stretch, scale factor followed by translation left by . b Horizontal stretch, scale factor followed by translation right by . c Translation left by followed by horizontal stretch, scale factor d Translation right by followed by horizontal stretch, scale factor . 8

Solve the inequality

9

a Describe two transformations that transform the graph of .

. to the graph of

b Describe two transformations that transform the graph of .

to the graph of

c Hence describe a sequence of transformations that transforms the graph of to the graph of . 10

a State the sequence of three transformations that transforms the graph of graph of . Hence sketch the graph of . b Solve the equation

.

c Write down the solution of the inequality 11

to the

.

a Describe a transformation that transforms the graph of

to the graph of

.

b Sketch on the same diagram the graphs of: i ii Mark clearly any asymptotes and -intercepts on your sketches. c The graph of the function produce the graph of

has been translated and then reflected in the -axis to .

i State the translation vector. ii If 12

find constants

such that

a Describe a sequence of geometrical transformations that maps the graph of

. onto

the graph of

.

b Find the exact values of the coordinates of the points where the graph of crosses the coordinate axes. [© AQA 2008] 13

Solve the equation

14

Sketch the graph of

. .

Worksheet See Extension sheet 3 for more challenging questions on modulus graphs and equations.



4 Sequences and series In this chapter you will learn how to: determine the behaviour of some sequences use a sigma notation for series work with sequences with a constant difference between terms work with finite series with a constant difference between terms work with sequences with a constant ratio between terms work with finite and infinite series with a constant ratio between terms apply sequences to real-life problems.

Before you start… GCSE

You should be able to find the formula for the term of a linear sequence.

1 Find the formula for the each sequence.

term of

a b

GCSE

You should be able to use term-to-term rules to generate sequences.

2 Find the second and third terms of the sequence defined by:

GCSE

You should be able to solve linear simultaneous

3 Solve the simultaneous equations:

equations. Student Book 1, Chapter 3

You should be able to solve quadratic equations and inequalities.

4 Find the smallest positive integer that satisfies the inequality .

Student Book 1, Chapter 7

You should be able to solve exponential equations and inequalities.

5 Find the smallest integer value of such that .

Chapter 3

You should be able to use modulus notation.

6 List all integers that satisfy

.

Modelling with sequences When you drop a ball, it will bounce a little lower each time it hits the ground. The heights the ball reaches after each bounce form a sequence. Although the idea of a sequence may just seem to be about abstract number patterns, it has a remarkable number of applications in the real world – from calculating mortgages to estimating harvests on farms.



Section 1: General sequences You can describe sequence such as

in two different ways:

by a term-to-term rule: by a position-to-term rule (also called the formula for the

term):

Gateway to A Level For a reminder of how these two types of rules work, see Gateway to A Level section X. There is also further practice at the beginning of Exercise 4A.

Did you know? Term-to-term rules can involve more than one previous term. You may already know the example of the famous Fibonacci sequence, with . This is based on a model Leonardo Fibonacci made for the breeding of rabbits, and has found many applications from describing the pattern made by the scales on pine cones to a proof of the infinity of prime numbers. There is also a beautiful link to the golden ratio: .

Tip is the standard notation for the

term of a sequence. So in the sequence etc.

Increasing, decreasing and periodic sequences Sequences can behave in many different ways but you need to be aware of three possibilities.

Key point 4.1 An increasing sequence is one where each term is larger than the previous one: for all . A decreasing sequence is one where each term is smaller than the previous one: for all . A periodic sequence in one where the terms start repeating after a while: some number (the period of the sequence).

WORKED EXAMPLE 4.1

for

Find the first five terms of each sequence, and describe the behaviour of the sequence. a b c a

Substitute each term into the formula to get the next term.

The sequence is increasing.

The terms are clearly getting larger.

b

Substitute

The sequence is decreasing.

The terms are halving each time so are getting smaller.

c

Substitute each term into the formula to get the next term.

The sequence is periodic, with period .

The sequence starts to repeat after two terms:

to calculate the terms.

In some cases, you might need to provide a proof that a sequence is increasing or decreasing, rather than just listing a few terms and making a conclusion.

Focus on … Focus on … Problem solving 1 looks further at the technique of using small cases to establish patterns in sequences.

Key point 4.2 To prove that a sequence is increasing or decreasing, find

.

If

then

and the sequence is increasing,

If

then

and the sequence is decreasing.

WORKED EXAMPLE 4.2

The

term of a sequence is defined by

.

Prove that the sequence is decreasing. You need to show that . Create a common denominator and combine the fractions.

for all is decreasing.

Since negative.

and

, the fraction is



The limit of a sequence In Worked example 4.1, the sequence in part b decreases but the terms remain positive although they get closer and closer to zero. The sequence converges to zero. By contrast, the sequence in part a diverges – the terms increase without a limit.

Fast forward You will see in Chapter 14 how you can use convergent sequences to solve some equations.

You can use your calculator to investigate the long-term behaviour of sequences, by generating a large number of terms until you can see what is going on. But you also need to be able to show that a convergent sequence has a particular limit, by solving an equation. If a sequence is converging, then, in the limit as (and equal to the limit).

, consecutive terms

and

will be the same

Tip Use the table function on your calculator if the sequence is given by the or the ANS button if you have a term-to-term rule.

term formula,

Key point 4.3 To find the limit, , of a convergent sequence defined by a term-to-term rule, set and solve for .

WORKED EXAMPLE 4.3

A convergent sequence is defined by

.

By setting up and solving an equation, find the limit, , as Replace

Let

. and

by .

Solve the equation for .

The limit as

is

.



EXERCISE 4A 1

Write out the first five terms of each sequence, defined by a term-to-term rule. a i ii b i ii c i ii d i ii

2

Write out the first five terms of each sequence, defined by an

term formula.

a i ii b i ii c i ii d i ii 3

Using your calculator, determine the behaviour of each sequence. a i

ii b i ii 4

A sequence is defined by a Write down

and

.

b Show that 5

.

for a constant . Work out the value of .

A sequence is defined by

with

.

a Find the fifth term of the sequence. b Describe the long-term behaviour of the sequence. 6

A sequence is defined by

, where k is a constant.

The limit of the sequence as

is

.

Find the value of . 7

a A sequence is defined by Show that b Given that

8

. and that the sequence is increasing, find the value of .

A sequence is defined by a Find

and

b State the value of 9

where is a constant.

. .

.

A sequence is defined by a If

.

is chosen so that the sequence converges, find the limit.

b Determine, explaining your reasoning fully, the range of value of converges. 10

The

term of a sequence is defined by

for which the sequence

.

Prove that the sequence is increasing.

Worksheet See Extension sheet 4 for some more challenging questions on the long term behaviour of sequences and series.



Section 2: General series and sigma notation If interest is paid on money in a bank account each year, these interest payments form a sequence. While it may be useful to know how much interest is paid in each year, you might be even more interested to know how much will be accumulated altogether. This is one of many examples of a situation where you might want to sum a sequence. The sum of a sequence up to a certain point is called a series. The symbol is often used to denote the sum of the first terms of a sequence; that is:

Instead of seeing form, using sigma notation.

, you will often see exactly the same expression in a shorter

Key point 4.4 Sigma notation is a shorthand way to describe a series:

where the Greek capital sigma means ‘add up’. is a placeholder – it shows what changes with each new term. is the first value taken by – where counting starts. is the last value taken by – where counting ends.

Tip Don’t be intimidated by this notation. If you are in doubt, try writing out the first few terms longhand.

There is nothing special about the letter here; you could use any letter but and are the most usual. You may also see the or being missed out above and below the sigma. WORKED EXAMPLE 4.4

 Find the value of

. Put the starting value, , into the expression to be summed, . The end value has not been reached, so put in . The end value has still not been reached, so put in . The end value has been reached, so stop and evaluate.

WORKED EXAMPLE 4.5

Write the series

in sigma notation.

General term

Describe the general term of the series in the variable .

Starts at

Note the first value of .

Ends at

Note the final value of . Summarise in sigma notation.

EXERCISE 4B

EXERCISE 4B 1

Evaluate each expression. a i

ii

b i

ii

c i

ii 2

Write each expression in sigma notation. Be aware that there is more than one correct answer. a i ii   b i ii c i ii

3

A sequence is defined by Find

4

.

.

A sequence is defined by

where is a constant.

a Show that

.

b Given that

, find the value of .

5

Find the exact value of

6

a Find the exact value of

, giving your answer in the form

b Hence find the exact value of



.

.

.

Section 3: Arithmetic sequences Arithmetic sequences have a common difference between consecutive terms: to get from one term to the next you add the common difference (which may be negative). For example: Add to each term. Subtract from each term. In general, if the first term is and the common difference is , then term

term

This shows how to form the

term

term

term

term of an arithmetic sequence.

Key point 4.5 The

term of an arithmetic sequence with first term and common difference is:

Common error The formula for the not .

term involves

and not

. So, for example,

and

You should already be familiar with these sequences and know how to find a formula for the although you might have used a slightly different method in the past.

term,

Gateway to A Level For practice at identifying arithmetic sequences and finding their th term, see Gateway to A Level Section X.

WORKED EXAMPLE 4.6

An arithmetic progression has first term and common difference . Find the term with the value

. You need to find when Use Solve the equation.

So

is the

term.



with

. and

.

Tip Arithmetic progression is just another way of saying arithmetic sequence.

You will often need to set up simultaneous equations to find and .

Worksheet See Support Sheet 4 for a further example of using simultaneous equations to find a and .

WORKED EXAMPLE 4.7

The fifth term of an arithmetic sequence is and the eighth term is

.

a Find the first term, , and common difference, . b Hence find the

term.

a

Use

with

You know that

.

.

Repeat for the eighth term. :

Solve simultaneously.

Now form the term formula. Note that you could tidy up by expanding the brackets and simplifying: . However, there is no need as you are not asked to state the formula at all. b

Substitute

.

WORKED EXAMPLE 4.8

The first three terms of an arithmetic sequence are

.

Find the possible values of . The difference between terms must be constant . Rearrange and solve the quadratic equation.

EXERCISE 4C

EXERCISE 4C 1

Find the general formula for the arithmetic sequence, given these conditions. a i first term , common difference ii first term

, common difference

b i first term

, common difference

ii first term

, common difference

c i first term , second term ii first term , second term d i first term , second term ii first term

, second term

e i third term , eighth term ii fifth term , eighth term 2

How many terms are there in each sequence? a i ii b i first term , common difference , last term ii first term , ninth term

3

, last term

An arithmetic sequence has and

as its first two terms, respectively.

a Write down, in terms of , an expression for the

term,

.

b Find the number of terms of the sequence that are less than 4

The

term of an arithmetic sequence is

Find the value of the 5

term is

.

term.

The eighth term of an arithmetic sequence is Which term has the value

6

and the

.

and the

term is

.

?

The heights of the rungs above ground in a ladder forms an arithmetic sequence. The third rung is above the ground and the rung is above the ground. Given that the top rung is above the ground, how many rungs does the ladder have?

7

The first four terms of an arithmetic sequence are constants. Find and .

8

A book starts at page and is numbered on every page. a Show that the page numbers for the first

pages contain

b If the total number of digits used for page numbers is book?



, and

, where and are

digits.

, how many pages are there in the

Section 4: Arithmetic series When you add up terms of an arithmetic sequence you get an arithmetic series. Just as it was useful to have a formula for the term, so it is useful to have a formula that will find the sum of the first terms. There are two different versions of this.

Key point 4.6 For an arithmetic series with first term and common difference :

or

where is the last term of the series. These will be given in your formula book.

Focus on … The first formula is proved in Focus on … Proof 1.

The second formula follows immediately from the first, since the last term, , is just the arithmetic sequence:

term of the

Tip You will need the first formula when you know or want , and the second when you know or want .

WORKED EXAMPLE 4.9

Find the sum of the first difference .

terms of an arithmetic progression with first term and common

Use

with

You must be able to work backwards too and find how many terms there are in the series, given the sum of the series. Remember that the number of terms can only be a positive integer. WORKED EXAMPLE 4.10

An arithmetic sequence has first term and common difference .

. The sum of the first terms is

Find the value of . Use

with

Simplify and then solve the equation.

must be positive.

Sometimes you have to interpret the question carefully to see that it is about an arithmetic series. WORKED EXAMPLE 4.11

Find the sum of all the multiples of between

This is an arithmetic series with

and

and

.

Write out the first few terms to see what is happening.

. You need to know how many terms there are in this series, so set in the formula and solve for .

Now use

, with .

EXERCISE 4D 1

Find the sum of each arithmetic sequence. a i ii b i ii c i ii d i ii

2

An arithmetic sequence has first term and common difference . How many terms are required to get a sum of: a i

ii b i ?

ii 3

For the arithmetic series

4

The sum of the first terms of a series is given by

find the value of for which

.

, where

a Find the first three terms of the series. b Find an expression for the 5

term of the series.

The second term of an arithmetic sequence is . The sum of the first four terms is

.

Find the first term, , and the common difference, , of the sequence. 6

The fourth term of an arithmetic sequence is

. The sum of the first

terms is

.

Find the first term, , and the common difference, , of the sequence. 7

The sum of the first three terms of an arithmetic sequence is is

, and the sum of the first

terms

.

Find the first term, , and common difference, . 8

Prove that the sum of the first odd numbers is

9

Find the largest possible value of the sum of the arithmetic sequence

. .

10

Find the least number of terms for which the sum of the series .

11

The sum of the first terms of an arithmetic sequence is

12

A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence. The angle of the largest sector is twice the angle of the smallest sector.

is greater than

. Find the

term

.

Find the size of the angle of the smallest sector. 13

a Find the sum of all multiples of between and

.

b Hence find the sum of all integers between and 14

The ratio of the fifth term to the

term of a sequence in an arithmetic progression is

. If each

term of this sequence is positive, and the product of the first term and the third term is

, find the

sum of the first 15

terms of this sequence.

a Find an expression for the sum of the first

b Hence find the exact value of for which 16



that are not divisible by .

terms of the series

.

Find the sum of all three-digit numbers that are multiples of

but not

.

Section 5: Geometric sequences A geometric sequence has a common ratio between consecutive terms: to get from one term to the next you multiply by the common ratio. For example:

In general, if the first term is and the common ratio is , then: term

term

term

From this you can see how to form the

term

term

term of a geometric sequence.

Key point 4.7 The

term of a geometric sequence with first term and common ratio is:

Common error The formula for the .

term involves

and not

. So, for example,

and not

WORKED EXAMPLE 4.12

For the geometric sequence a find a formula for the b hence find the a

b

term

term of the sequence. If the common ratio is not obvious, divide the second term by the first term (or by , etc.) to find it. Now use

.

Substitute in

.

As with arithmetic sequences, you will often need to set up simultaneous equations.

Worksheet See Support Sheet 4 for a further example of using simultaneous equations to find and .

WORKED EXAMPLE 4.13

The seventh term of a geometric sequence is

. The ninth term is

.

Find the possible values of the common ratio. Use

with

But you know that

. .

Repeat for the ninth term. Solve simultaneously. Note that dividing the equations is a quick way of eliminating and solving for .

Tip Notice that the question asked for values rather than value, so you should expect to get at least two answers.

WORKED EXAMPLE 4.14

Three consecutive terms of a geometric sequence with common ratio

are

a Find the possible values of . b For each value of , find the common ration of the sequence. a

The ratio between terms must be constant . Rearrange and solve the quadratic.

b When

,

When

,

Find the ratio of the first two terms for each value of in turn.

When you are asked which term satisfies a particular condition, you normally need to use logarithms.

Rewind See Student Book 1, Chapter 7, if you need a reminder of how to solve exponential equations.

WORKED EXAMPLE 4.15

A geometric sequence has first term What is the first term that is less than

and common ratio . ? Express the condition for the inequality. Use

with

and

term as an

The

unknown is in the power so solve using logarithms.

is negative so when you divide by it, you must remember to change the inequality sign.

is an integer, so you need the first integer greater than .

EXERCISE 4E 1

Find an expression for the

term of each geometric sequence.

a i ii b i ii c i ii d i ii e i ii 2

How many terms are there in each geometric sequence? a i

ii

 

b i ii c i ii 3

The second term of a geometric sequence is and the fifth term is a Find a formula for the b Hence find the

4

.

term of the sequence.

term.

The third term of a geometric progression is

and the fifth term is

.

a Find the possible values of the first term and the common ratio. b Hence find the two possible values of the eighth term. 5

The first three terms of a geometric sequence are

.

a Find the possible values of . b In each case, find the tenth term of the sequence. 6

The third term of a geometric sequence is Which term has the value

7

and the sixth term is

?

A geometric sequence has first term and common ratio a Show that

.

.

.

.

b Hence find the value of . 8

A sequence is defined by

.

What is the first term of the sequence to exceed million? 9



What is the first term of the sequence

to be less than

?

10

The difference between the fourth and the third terms of a geometric sequence equals one quarter of the second term. Find the common ratio.

11

The three terms are in arithmetic progression. The three terms progression. Find the values of and , given that .

12

The sum of the first terms of an arithmetic sequence is given by the formula . Three terms of this sequence, and , are consecutive terms in a geometric sequence. Find .

are in geometric

Section 6: Geometric series Just as for an arithmetic series, a geometric series is the sum of the terms in a geometric sequence.

Tip In general, use the first of these formulae when the common ratio is less than one and the second when the common ratio is greater than . This avoids having to work with negative numbers.

Key point 4.8 The formula for the sum

of the first terms in a geometric sequence is:

or equivalently:

The first formula will be given in your formula book.

Focus on … This formula is proved in Focus on … Proof 1.

WORKED EXAMPLE 4.16

Find the exact value of the sum of the first six terms of the geometric sequence with first term and common difference . and formula as

. Use the first sum

.

Again, you will need to use logarithms if you are asked to find the term number. WORKED EXAMPLE 4.17

How many terms are needed for the sum of the geometric series

to exceed

? You need , but you know and . Use the second sum formula and express the condition as an inequality. The unknown is in the power, so use logarithms to solve the inequality.

But is a whole number, so needed.

terms are

EXERCISE 4F 1

Find the sum of each geometric series. (There may be more than one possible answer!) a i ii b i ii c i first term , common ratio ii first term

, common ratio

d i third term

, fifth term

ii ninth term 2

, last term , last term

terms.

, thirteenth term

, last term

.

Find the possible values of the common ratio if: a i first term is

, sum of the first two terms is

ii first term is , sum of the first two terms is

3

b i first term is

, sum of the first three terms is

ii first term is

, sum of the first three terms is

The

term,

.

, of a geometric sequence is given by

.

a Find the common ratio . b Hence, or otherwise, find an expression for 4

A series is defined by

, the sum of the first terms of this sequence.

.

a State the values of and . b Hence find the value of to three significant figures.

5

The first term of a geometric sequence is and the sum of the first three terms is common ratio.

6

The sum of the first three terms of a geometric sequence is is

7

. Find the

and the sum of the first four terms

. Find the first term and the common ratio.

The sum of the first four terms of a geometric sequence is and the sum of the first six terms is .

, the sum of the first five terms is

,

a Find the common ratio. b Find the sum of the first two terms. 8

The fifth term of a geometric sequence is

and the sixth term is

.

a Find the common ratio and first term.

9

b Find the smallest value of so that the sum of the first terms exceeds

.

The sum of the first three terms of a geometric sequence with common ratio sum of the first six terms is .

is

a Show that

.

b Hence find the sum of the first 10

terms.

a Show that in a geometric sequence with common ratio , the ratio of the sum of the first terms to the sum of the next is . b The sum of the seventh term and four times the fifth term equals the eighth term. Find the ratio of the sum of the first 10 terms to the sum of the next 10 terms.



and the

Section 7: Infinite geometric series If you keep adding together terms of any arithmetic sequence the sum increases (or decreases if is negative) without limit. This can happen when you add the terms of a geometric sequence too, but there is also a possibility that the sum will converge. Looking at the formula for the sum of a geometric sequence,

you can see that the only part that is affected by making bigger is . When you raise most numbers to a large power the result gets bigger and bigger. The exception is when is a number between and ; in this case, gets smaller as increases – in fact it tends to .

Key point 4.9 As increases the sum of a geometric series converges to

This is called the sum to infinity of the series. This will be given in your formula book.

Tip The condition that

is just as important as the formula itself.

WORKED EXAMPLE 4.18

The sum to infinity of a geometric sequence is and the second term is

. Find the common

ratio. Use

From

with

  

Use

:

Solve simultaneously.

with



Substituting into



:



But since the sum to infinity exists,



Check that the series actually converges for the values of found.

Remember that some questions may focus on the condition for the sequence to converge as well as the value that it converges to. WORKED EXAMPLE 4.19

The geometric series

converges. Find the range of possible values of

. Identify . Since the series converges:

Use the fact that the series converges. Solve the inequality.

EXERCISE 4G 1

Find the value of each infinite geometric series, or state that it is divergent. a i ii b i ii c i ii d i ii e i ii

2

Find the values of that allow each geometric series to converge. a i ii b i ii c i ii d i ii e i

.

ii f

i ii

g i ii h i ii i

i ii

3

Find the sum to infinity of the geometric sequence

4

The first and fourth terms of a geometric series are

and

respectively.

Find the sum to infinity of the series. 5

The fifth term of a geometric sequence is

and the seventh term is .

Find the two possible values of the sum to infinity of the series. 6

A geometric series has first term and common ratio , with

and

. The second term is

and the eighth term is . a Find and . b Find the exact value of the sum to infinity of the series, giving your answer in the form , where and are constants to be stated. 7

A geometric sequence has all positive terms. The sum of the first two terms is infinity is . Find the value of:

and the sum to

a the common ratio b the first term. 8

The sum to infinity of a geometric series is terms are positive.

. The sum of the first four terms is

and all the

Find the difference between the sum to infinity and the sum of the first eight terms. 9

Consider the infinite geometric series

.

a For what values of does the series converge? b Find the sum of the series if

.

10

The sum of an infinite geometric sequence is the first term.

11

An infinite geometric series is given by

, and the sum of the first three terms is

. Find

.

a Find the values of for which the series has a finite sum. b When .

, find the minimum number of terms needed to give a sum that is greater than

12 Evaluate: a b 13

The common ratio of the terms in a geometric series is

.

a State the set of values of for which the sum to infinity of the series exists. b If the first term of the series is

, find the value of for which the sum to infinity is

14 a Show that

.

b Hence find the exact value of . 15



Show that for a convergent geometric series with first term .

and sum to infinity

.

Section 8: Mixed arithmetic and geometric questions Be very careful when dealing with sequences and series questions. It is vital that you: identify whether it is a geometric or an arithmetic sequence identify whether it is asking for a term in the sequence or the sum of terms in the sequence translate the information given in the question into equations.

Tip With questions like that in Worked example 4.20, think carefully about whether the amount you are calculating is for the beginning or the end of a year.

WORKED EXAMPLE 4.20

A savings account pays annual compound interest, added at the end of each year. If £ is paid into the account at the start of the first year, how much will there be in the account at the start of the seventh year?

The balance of the account at the beginning of:

This is a geometric sequence with .

Each year the balance of the account is increased by the same percentage, so this gives a geometric sequence. It is a good idea to write out the first few terms in questions like this to make sure you know what is happening.

and

The balance at the start of the seventh year is . £

WORK IT OUT 4.1 A car loses value at the rate of per year. It was bought for £ In which year will its value fall below £ ?

at the beginning of 2017.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

EXERCISE 4H 1

Philippa invests £

at

compound interest for years.

a How much interest does she get paid in the sixth year? b How much does she get back after years? 2

Lars starts a job on an annual salary of £ a How much will his

year’s salary be?

and is promised an annual increase of £

.

b After how many complete years will he have earned a total of £ million? 3

A sum of £

is invested at a compound interest rate of

per annum.

a Write down an expression for the value of the investment after full years. b What will be the value of the investment at the end of years? c The value of the investment will exceed £

after full years.

i Write an inequality to represent this information. ii Calculate the minimum value of . 4

Each row of seats in a theatre has 200 more seats than the row in front of it. There are the front row and the designer wants the capacity to be at least .

seats in

a How many rows are required? b Assuming the rows are equally spread, what percentage of people are seated in the front half of the theatre? 5

A sum of £

is invested.

a If the interest is compounded annually at a rate of investment after years.

per year, find the total value of the

b If the interest is compounded monthly at a rate of

per month, find the minimum number of

months for the value of the investment to exceed . 6

A marathon is a

mile race. In a training regime for a marathon a runner runs mile on his first

day of training and each day increases his distance by of a mile. a After how many days has he run for a total of b On which day does he first run over 7

miles?

miles?

Aaron and Blake each open a savings account. Aaron deposits £ increases his deposits by £ each month. Blake deposits £

in the first month and then

in the first month and then increases his deposits by

each month.

No interest is paid on balances. Show that the first time Blake will have more money in his account than Aaron is after 8

A ball is dropped from m in the air. Each time it bounces up to a height of height.

months.

of its previous

a How high does it bounce on the fourth bounce? b How far has it travelled when it hits the ground for the ninth time? c Give one reason why this model is unlikely to be accurate after 20 bounces. 9

Samantha puts £ end of each year

into a bank account at the beginning of each year, starting in 2010. At the interest is added to the account.

a Show that at the beginning of 2012 there is

in the account.

b Find an expression for the amount in the account at the beginning of year . c When Samantha has a total of at least £

in her account at the beginning of a year she will

start looking for a house. In which year will this happen?

Checklist of learning and understanding A sequence can be defined by a formula for the

term or by a term-to-term rule.

A sequence is: increasing if each term is larger than the previous one: decreasing if each term is smaller than the previous one: periodic if the terms start repeating after a while: period of the sequence).

for some number (the

A series is a sum of terms in a sequence and it can be described in a shorthand way using sigma notation.

Arithmetic sequences have a constant difference, , between consecutive terms. If you know the first term, , the

term is:

If you know the first term and the common difference, the sum of all terms in the sequence is:

If you know the first and last term

, the sum of all terms in the sequence is:

Geometric sequences have a constant ratio, , between consecutive terms. If you know the first term, , the

term is:

The sum of the first terms is:

or equivalently:

If



, the series converges and the sum to infinity is given by:

Mixed practice 4 1

The

term of a geometric sequence is

.

The sum to infinity is . Find the value of . Choose from these options. A B C D 2

The third term of an arithmetic sequence is and the sum of the first

terms is

. Find

the first term and the common difference. 3

The fourth term of an arithmetic sequence is the first nine terms.

4

The third term of a geometric sequence is

and the ninth term is

. Find the sum of

and the sixth term is .

a Find the first term and the common ratio. b Find the sum to infinity. 5

The fifth term of an arithmetic sequence is three times the second term. Find the ratio:

6

The

term of a geometric sequence is

a Find the value of

and show that

, where .

b Write down the common ratio of the geometric sequence. c i Show that the sum of the first an integer. ii Hence find the value of

terms of the geometric sequence is

, where is

. [© AQA 2007]

7

The sum of the first three terms of a geometric sequence is Find the common ratio, . Choose from these options. A B C D

8

Find the exact value of

.

. The sum to infinity is

.

Choose from these options. A B C D 9 10

Which is the first term of the sequence

less than

?

Ben builds a pyramid out of toy bricks. The top row contains one brick, the second row contains three bricks and each row after that contains two more bricks than the previous row. a How many bricks are there in the b If a total of

row?

bricks are used how many rows are there?

c In Ben’s largest ever pyramid he noticed that the total number of bricks was four more than four times the number of bricks in the last row. What is the total number of bricks? 11

Kenny is offered two investment plans, each requiring an initial investment of £ Plan A offers a fixed return of £ Plan B offers a return of

.

per year.

each year, reinvested in the plan.

Over what period of time is plan A better than plan B? 12

The first three terms of a geometric sequence are

.

a Find the two possible values of . b Given that it exists, find the sum to infinity of the series. 13

Evaluate

14

Find the sum of all the integers between

15

A sequence is defined by a If

The

that are divisible by .

.

, find the range of possible values of the constant .

b Given that 16

and

, find the limit of the sequence as tends to infinity.

term of a sequence is

. The sequence is defined by

where and are constants. The first two terms of the sequence are given by The limit of a Show that

as tends to infinity is

and

.

.

.

b Find the value of

. [© AQA 2013]

17

An arithmetic series has first term and common difference . The sum of the first five terms of the series is a Show that b Given also that the

.

. term of the series is

, find the value of .

c

The

term of the series is

. Given that

and

, find the value of

.

[© AQA 2014] 18

Prove that the sequence defined by

is increasing.

19

A geometric sequence and an arithmetic sequence both start with a first term of . The third term of the arithmetic sequence is the same as the second term of the geometric sequence. The fourth term of the arithmetic sequence is the same as the third term of the geometric sequence. Find the possible values of the common difference of the arithmetic sequence.

20

The first, second and fourth terms of a geometric sequence form consecutive terms of an arithmetic sequence. Given that the sum to infinity of the geometric exists, find the exact value of the common ratio.

21

Find an expression for the sum of the first

giving your answer in the form 22

terms of the series

, where

.

A student writes ‘ ’ on the first line of a page, then the next two integers ‘ ’ on the second line of the page then the next three integers ‘ ’ on the third line. He continues this pattern. a How many integers are there on the b What is the last integer on the

line?

c What is the first integer on the

line?

line?

d Show that the sum of all the integers on the

line is

.

e The sum of all the integers on the last line of the page is page? 23

Selma has a mortgage of £ pays £ .

. At the end of each year

. How many lines are on the

interest is added, then Selma

a Show that at the end of the third year the amount owing is: £

b Find an expression for how much is owing at the end of the c After how many years will the mortgage be paid off?

year.

5 Rational functions and partial fractions In this chapter you will learn how to: manipulate rational functions, including by using polynomial division with remainders use the factor theorem to find factors of the form decompose rational functions into a sum of algebraic fractions when the denominator contains distinct linear factors decompose rational functions into a sum of algebraic fractions when the denominator contains repeated linear factors.

Before you start… GCSE

You should be able to add algebraic fractions.

1 Simplify into one fraction

Student Book 1, Chapter 3

You should be able to factorise quadratic

2 Factorise

.

expressions. Student Book 1, Chapter 4

You should be able to carry out polynomial division.

3 Simplify

Student Book 1, Chapter 4

You should be able to use the factor theorem.

4 Find a linear factor of

.

What are rational functions? A rational function is a fraction in which both the denominator and numerator are polynomials. These tend to have graphs with asymptotes, which makes them applicable to a wide range of real-world situations from economics to medicine – for example, the function

is often used to model the

amount of anaesthetic in a patient after time . This chapter will focus on applying algebraic techniques to change rational functions into forms that will allow more advanced manipulations later in the course. To do this, you will review a very important theorem about polynomials – the factor theorem.

Focus on … Focus on … Modelling 1 compares models using rational functions to those using exponential functions.



Section 1: An extension of the factor theorem The version of the factor theorem that you met in Student Book 1 can be extended slightly to include factors where the coefficient of is not .

Key point 5.1 If

then

is a factor of

.

Rewind In Student Book 1, Chapter 4, the factor theorem stated that if factor of .

then

is a

WORKED EXAMPLE 5.1 a Show that

is a factor of

.

b Hence solve

.

a

By the factor theorem, if is a factor of

Therefore

is a factor of

, then

.

. Use polynomial division to find the remaining factor.

b

So, Factorise the quadratic.

Solve as normal.

Tip If you do not like long division you can write multiply out and compare coefficients.

as

, then

Rewind You met polynomial division in Student Book 1, Chapter 4.

WORK IT OUT 5.1 If

, state a linear factor of the polynomial

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 By the factor theorem,

is a factor of

.

Solution 2 is a factor of

.

Solution 3 By the factor theorem,

is a factor of

.

The factor theorem also works in reverse, so if you know that

is a factor then

.

WORKED EXAMPLE 5.2

If

is a factor of

, find the value of . can be written as the factor theorem says that . Solve the resulting equation.

EXERCISE 5A

so

EXERCISE 5A 1

Show that each expression has the given factor and hence factorise completely. a i

has factor

ii

has factor

b i

has factor has factor

ii 2

Show that

3

is a factor of

is a factor of

. Hence factorise it completely. .

Find the value of . 4

a

is a factor of

.

Find the value of . b Hence factorise 5

a Show that

. is a factor of

.

b Hence factorise 6

Show that

7

Prove that if

8

Given that

9

a

is a factor of

and

is a factor of is a factor of are factors of

. Hence solve then

.

.

, find an expression for in terms of . .

Find the values of and . b Hence solve 10



Show that

. is a factor of

and hence solve

.

Section 2: Simplifying rational expressions You need to be able to simplify and manipulate rational expressions, using your knowledge of algebra – in particular, always look to factorise expressions and simplify. WORKED EXAMPLE 5.3

Simplify

. Always try to factorise first. Look out for the difference of two squares! Divide the top and bottom of the first fraction by the factor of . To divide, flip the second fraction and multiply. There is a factor of in the denominator of the first fraction and the denominator of the second fraction. Cancel these when the fractions are being multiplied.

You can use polynomial division to simplify rational functions in which the degree of the numerator is at least the degree of the denominator. This turns the rational function into the sum of a polynomial and a simpler rational function called the remainder term.

Tip In Worked example 5.3 you used an identity sign to emphasise that the expression is just being simplified rather than solving it, but you will often see an equals sign used instead, and this is an acceptable alternative.

Key point 5.2 If

is a polynomial then

is also a polynomial and is called the quotient. is the remainder.

WORKED EXAMPLE 5.4

Write

in the form

. Use polynomial division.

So

The remainder found in polynomial division forms the numerator of the rational expression.

WORK IT OUT 5.2 Simplify

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3 Cancelling

from top and bottom:

EXERCISE 5B 1

Simplify each expression. a i ii b i ii c i

ii d i ii e i ii f

i ii

g i ii 2

Simplify each expression. a i ii b i ii c i ii d i ii e i ii

3

Simplify each expression. a i ii b i ii c i ii d i

ii 4

Write each expression in the form a i ii b i ii

5

Simplify

6

Simplify

7

a Simplify

.  . .

b Hence or otherwise solve

.

8

Solve

9

Find the quotient and remainder when

10

Find the quotient and remainder when

11

Write

12 13

Show that

.

in the form

is divided by is divided by

.

.

is a factor of

a Show that

.

and hence simplify

is a factor of

b Hence simplify

.

. .

14

The remainder when

15

When the polynomial is divided by is divisible by .

16

a A car travels at speed for then at speed expression for the average speed during this journey.

is divided by

is . Find the value of and the quotient.

the quotient is the same as the remainder. Prove that

for

. Find and simplify an

b If the average speed equals the arithmetic mean of the two speeds, show that either the speeds in the two sections of the journey are equal or the speed in the second section is twice that of the first section.



Section 3: Partial fractions with distinct factors You already know how to write a sum of two algebraic fractions as a single fraction, for example:

Gateway to A Level For a reminder of adding algebraic fractions, see Gateway to A Level Section Y.

However there are situations where you need to reverse this process. The method for doing this involves partial fractions. To do this, you need to know the form the partial fractions will take.

Fast forward You will see that you need to do this with binomial expansions in Chapter 6 and integration in Chapter 11. You will also use it to help sum some series if you study Further Mathematics.

Key point 5.3 decomposes into decomposes into

Once you know the form you are aiming for, write it as an identity and multiply both sides by the denominator of the original fraction. You can then either compare coefficients or substitute in convenient values to find the values of and .

Rewind Comparing coefficients was discussed in Student Book 1, Chapter 1.

WORKED EXAMPLE 5.5

Write

in partial fractions. Write the expression in the correct form (given in Key point 5.3). Multiply both sides by fractions.

When

:

to remove

As this is an identity, you can substitute in any value of . Choosing makes the coefficient of zero, so you can find .

When

Similarly, choosing can find .

:

eliminates so you

State the final answer.

So

You can only write in terms of partial fractions using this method if the degree of the numerator is less than the degree of the denominator. If this is not the case, you need to use algebraic manipulation to turn it into the required form. WORKED EXAMPLE 5.6

a Show that b Hence express

can be written in the form

.

in partial fractions. Make a link between the numerator and denominator.

a

You can then split the expression into two fractions.

First work with the fraction from part a. If you want to use partial fractions you must factorise the denominator.

b

Now use the form given in Key point 5.3. Multiply by the denominator to remove fractions. When

Substitute in

When

Substitute in

When

, to eliminate and .

, to eliminate and .

Substitute in



, to eliminate and .

Therefore Remember that the original expression had an as well.

So

EXERCISE 5C 1

Write each expression in terms of partial fractions. a i ii b i ii c i ii d i ii

2

Write each expression in terms of partial fractions. a i ii b i ii c i ii

3

Given that

4

Express

5

Simplify

.

6

Decompose

into partial fractions.

7

Write

8

a Show that b Hence write

find the values of and . in terms of partial fractions.

in partial fractions. . in terms of partial fractions.

9

Expand in partial fractions

10

Expand in partial fractions

11

a Show that

where is a constant. where is a constant.

can be written in the form

determined. b Hence write 12



in terms of partial fractions.

What happens if you try to write

in the form

where is a constant to be

Section 4: Partial fractions with a repeated factor If there is a repeated factor in the denominator you need to use an alternative expression for the partial fractions.

Key point 5.4 decomposes into

WORKED EXAMPLE 5.7

Express

in partial fractions. Write the expression in the correct form (given in Key point 5.4). Multiply both sides by to eliminate fractions.

When

:

Choose

so that the coefficients

of and are zero.

When

When

:

:

Choose so that the coefficients of and are zero.

There is no value of that can make the coefficients of and zero, so you have to choose another simple alternative. often makes the arithmetic simple. Substitute the values of and found above.

So

EXERCISE 5D

Write out the final answer.

EXERCISE 5D 1

Write each expression in terms of partial fractions. a i ii b i ii c i ii d i ii

2

Split

3

Express

4

Write

5

a Use the factor theorem to show that

into partial fractions. in partial fractions. in terms of partial fractions.

b Hence factorise c Write

is a factor of

.

. in partial fractions.

Worksheet See Support Sheet 5 for a further example of partial fractions with a repeated factor and for more practice questions. 6

a Use the factor theorem to show that b Hence express

7

a Write b Hence write

8

Write

.

in partial fractions. in the form

.

in partial fractions.

a Prove that if a function can be written as b Write

9

is a factor of

in the form in terms of partial fractions.

it can also be written as .

.

Checklist of learning and understanding The factor theorem can be used to check for factors of polynomials: if is a factor of

then

.

A rational function is a fraction in which both the denominator and numerator are polynomials. In arithmetic they follow all the same rules as normal fractions. If the degree of the numerator is equal to or greater than the degree of the denominator you can use polynomial division to simplify the function. . If the rational function has a numerator with degree less than the denominator it can be decomposed into partial fractions. decomposes into

.

decomposes into decomposes into



. .

Mixed practice 5 1

What is the expression

equal to?

Choose from these options. A B C D 2

a Simplify

.

b Hence find the prime factors of 3

a Show that

.

is a factor of

b Factorise

.

c Solve

.

d Simplify

.

4

Write

in partial fractions.

5

Express

in partial fractions.

6

Write

7

Write

8

Decompose

9

Express

.

in terms of partial fractions. in partial fractions. into partial fractions. in the form

. [© AQA 2014]

10

The polynomial

is defined by

a Find the remainder when b i Show that ii Express

is divided by

. .

. as a product of three linear factors.

iii Simplify . [© AQA 2008] 11

What is the expression Choose from these options. A

equal to?

B C D 12

a Express

in the form

b Express

, where and are integers.

in the form

c Hence express

, where and are integers.

as the sum of a linear function and partial fractions. [© AQA 2012 (adapted)]

13

a Simplify

.

b Hence write 14

in terms of partial fractions

a Use the factor theorem to show that b Hence factorise

15

.

d Write

in partial fractions.

a Show that

is a factor of

.

.

c Write

in partial fractions.

a Write

in the form

b Hence write 17

.

.

c Solve

b Solve

16

is a factor of

.

in partial fractions.

a

is divisible by

and by

. Find the values of and

. b Fully factorise c Solve 18

.

.

a Write

in terms of partial fractions.

b Hence write 19

in terms of partial fractions.

a Use polynomial division to simplify b Hence or otherwise write

20 21

Write

.

in partial fractions.

in terms of partial fractions. can be written as

Find the value of

.

and .

22

The remainder when

23

The remainder when a polynomial

is divided by

is

is divided by

. Write in terms of . is the same as the quotient. Find

the value of

.

Prove that if

is linear then

24

25

.

If two resistors with resistance and are connected in parallel the combined system has resistance . These are related by the equation:

a Find and simplify an expression for b Hence prove that

in terms of

and

.

.

Worksheet See Extension sheet 5 for more questions on continued fractions.



6 General binomial expansion In this chapter you will learn how to: expand where is any rational power decide when a binomial expansion will converge use partial fractions to write expressions in the form required for the binomial expansion use binomial expansions to approximate functions.

Before you start… GCSE

You should be able to simplify expressions with exponents.

1 Simplify

Student Book 1, Chapter 9

You should be able to use binomial expansions for positive integers.

2 Expand

Chapter 3

You should be able to write inequalities, using the modulus

3 Write

.

.

in the form

function. Chapter 5

You should be able to write an expression in partial fractions.

4 Write

in the form .

Extending the binomial theorem In Student Book 1 the binomial expansion was simply a quick way to expand brackets. In this chapter it is extended to allow you to approximate many other functions with polynomials.

Section 1: The general binomial theorem In Student Book 1 you met the binomial expansion when raising a sum of two terms to a positive integer power. For example:   You also saw that the first three binomial coefficients could be written as:

Continuing this pattern of the expressions for the coefficients, and using them in the same binomial expansion formula as before, gives a formula that also works for any rational number.

Tip ! is the symbol for factorial. 2!   means 3!   means In your formula book you will see this written as

and

etc.

Key point 6.1 If  for any rational value of . This will be given in your formula book.

Notice that this if is not a positive whole number this is an infinitely long polynomial.

Tip The condition that is important. Although you can always create the polynomial on the right-hand side, the series will only converge when , so the expansion is only valid if .

WORKED EXAMPLE 6.1

Find the first four terms in ascending powers of in the binomial expansion of

for

. Rewrite the fraction in the form Use Key point 6.1 with Be careful with powers of

.

. :

.

The formula in Key point 6.1 only works if the first number in the brackets is . If it is something else then you have to factorise the expression to turn the first number into . The range over which the expansion converges might also then change.

Rewind The idea of a sequence converging was covered in Chapter 4, Section 1.

WORKED EXAMPLE 6.2 a Find the first three terms, in ascending powers of , in the binomial expansion of

.

b Find the values of for which this expansion is valid. a

Rewrite the square root in the form . Take out a factor of inside the bracket to leave . Make sure the factor comes out as Now use Key point 6.1 with

.

.

Be careful with powers of :

.

Finally, multiply out the bracket. b

For

, use

.

Multiply through by . You could also write

.

Binomial expansions such as these produce infinite polynomials so you might wonder how they can be useful. If is small then very large powers of are extremely small, so they can be neglected. You can therefore use the first few terms of the binomial expansions to approximate the original expression.

Did you know? You might like to try finding

on your calculator and seeing how close this is to the

binomial approximation. Your calculator uses a method very similar to the binomial approximation to work out square roots.

WORKED EXAMPLE 6.3 a Find the first two terms, in ascending powers of , of the binomial expansion of b Use this polynomial to approximate

to three decimal places.

.

Rewrite the square root in the form

a

.

Take out a factor of inside the bracket to leave

Make sure the factor comes out as

Now use Key point 6.1 with

.

.

.

Finally, multiply out the bracket. b Need such that

.

Find the value of that will give

Substitute

.

into

EXERCISE 6A 1

Find the first three terms of the binomial expansion of each expression, stating the range of convergence. a i ii b i ii c i ii d i ii

2

Find the expansion of

in ascending powers of up to and including the term in

Worksheet See Support Sheet 6 for a further example of an expansion when the constant term isn’t 1, and for more practice questions.

3

Find the expansion of

in ascending powers of up to and including the term in

4

a Find the first three terms in the expansion of b Find the values of for which this expansion is valid.

.

.

.

5

a Use the binomial expansion to show that

 .

b State the range of values for which this expansion converges. c Deduce the first three terms of the binomial expansion of

.

d Use the first three terms of the expansion to find an approximation for places. 6

a Find the first four terms of

to four decimal

in ascending powers of .

b State the range of values for which this expansion is valid. c Hence approximate: i ii

to five decimal places to four decimal places.

7

The cubic term in the expansion of

8

Given that the expansion of

is

9

Given that the expansion of

is

10

is

. Find the value of . find the value of . find the value of .

Assume that By squaring both sides and comparing coefficients find the values of Does this agree with the binomial expansion of



?

and

.

Section 2: Binomial expansions of compound expressions Binomial expansions can be part of a larger expression. WORKED EXAMPLE 6.4

Find the quadratic term in the binomial expansion of

if

. Use the laws of indices to create expressions of the form . Expand both brackets: the first bracket and

in in

the second.

Simplify each series before continuing. Quadratic term: There are three ways to get an term.

Sometimes you have to use partial fractions to write an expression in a form where the binomial expansion is possible.

Rewind Partial fractions were covered in Chapter 5, Sections 3 and 4.

WORKED EXAMPLE 6.5

a Express

in the form

.

b Hence find the first two terms in the expansion of

.

c State the range of values for which this expansion converges. a

Apply the standard method for finding partial fractions. Substituting in

:

Substituting in

:

Therefore:

Find the binomial expansion of each term separately.

b



So:

Combine the two expansions.

c The first expansion converges if

so

.

Consider the range of convergence of each term separately.

The second expansion converges if so

.

Therefore both will converge if

.

Think about which range provides the limiting factor:

WORK IT OUT 6.1 Find the first three non-zero terms in ascending powers of in the binomial expansion of . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution A Set

. Then you need the expansion of

Substituting back for :

.

Solution B

Solution C

EXERCISE 6B

EXERCISE 6B 1

Find the first three terms of the expansion of

in ascending powers of .

2

Find the first three terms of the expansion of over which the expansion converges.

in ascending powers of , stating the range

3

Find the first three terms of the expansion of

4

Find the first three terms of the expansion of

5

a Express

. .

in the form

.

b Hence find the first three terms in the binomial expansion of

.

c Write down the set of values for which this expansion is valid. 6

a Express

in partial fractions.

b Find the first three terms in ascending powers of the binomial expansion of

.

c Find the values of for which this expansion converges. 7

Find the first three terms of the binomial expansion for

.

8

a Find the first two terms in ascending powers of the expansion of

.

b Over what range of values is the expansion valid? c By substituting in 9

10

find an approximation to

to two decimal places.

The first three non-zero terms of the expansion of . Is it possible to find a binomial expansion for

are

. Find the value of

?

Checklist of learning and understanding If for any rational value of . You can use the first few terms of a binomial expansion to approximate the expression for small values of . You can find a series expansion for rational functions by first decomposing them into partial fractions.



Mixed practice 6 1

Which expression has a binomial expansion valid for

?

Choose from these options. A B C D 2

Find the first four terms in ascending order in the expansion of

, stating the range of

-values for which the expansion converges. 3

Find the first three non-zero terms in ascending order in the expansion of

, stating the

range of -values for which this is valid. 4

a Express

in the form

.

b Hence find the first four terms in the expansion of

.

5

Find the first three non-zero terms in ascending order in the expansion of

6

Find the quadratic term in the expansion of

.

.

Choose from these options. A B C D 7

Which expression has a binomial expansion with no linear term? Choose from these options. A B C D

8

a Find the first three terms of the binomial expansion of b Hence find an approximation for

9

a Write

.

, to two decimal places, showing your reasoning.

in partial fractions.

b Hence find the first three terms of the expansion of c Write down the set of -values for which this expansion is valid.

.

10

Find the first three terms in the binomial expansion of

, stating the range of values

for which it is valid. 11

The first three terms in the binomial expansion of

are

. Find

the values of and . 12

a i Find the binomial expansion of

up to and including the term in

ii Hence find the binomial expansion of b Hence show that constants

.

up to and including the term in

.

for small values of , stating the values of the

and . [© AQA 2010]

13

a Find the binomial expansion of

up to and including the term in

.

b Find the binomial expansion of

up to and including the term in

.

c Use your answer from part b to find an estimate for integers.

in the form , where and are

[© AQA 2012] 14

a

Find the first three non-zero terms of the binomial expansion of

b By setting

, find an approximation for

15

Given that the expansion of

16

Find the first three terms of the expansion of

17

a

to five decimal places.

is

, find the value of . .

can be written in the form Find the values of

b Show that

.

and and state the set of values of for which this converges.

can be written in the form

can be written in the form

c

Find the values of

.

.

.

and and state the set of values of for which this converges.

d Use an appropriate expansion to approximate

to four decimal places, showing your

reasoning. e Use an appropriate expansion to approximate

to five significant figures, showing your

reasoning. 18

In special relativity the energy of an object with mass and speed is given by:

where

is the speed of light.

a Find the first three non-zero terms of the binomial expansion, in increasing powers of , stating the range over which it is valid. Let

be the expansion containing two terms and

terms. b By what percentage is

bigger than

if is:

be the expansion containing three

i

of the speed of light

ii

of the speed of light?

c Prove that

.

Worksheet See Extension sheet 6 for a look at Babylonian multiplication.



FOCUS ON … PROOF 1

PROOF 1

Proof of the sum of arithmetic series You are going to demonstrate that, for an arithmetic series with first term and common difference , the sum of the first terms is:

Copy the proof and fill in the gaps. Write out the first couple of terms of the series and the last couple of terms. Writing these terms in the opposite order has no effect on the outcome. Adding the two expressions gives identical terms. Collect like terms. Dividing by 2 gives the final result.

Did you know? Carl Friedrich Gauss (1777–1855) was amongst the most eminent mathematicians of the 19th century. His many contributions to mathematics included great strides in number theory, statistics and physics. He was a child prodigy and there is a famous legend about a lesson where his teacher was hoping to keep him quiet by asking him to add together all of the numbers from to  . The teacher was somewhat disappointed when he replied with the correct answer within seconds. It is believed that he applied a procedure similar to the one used in this proof.

PROOF 2

Proof of the sum of geometric series You are going to demonstrate that, for a geometric series with first term and common ratio the sum of the first terms is:

Copy the proof and fill in the gaps. Write out the first few terms and the last few terms of the sum.

Subtracting removes all the terms in common between the two series. Factorise both sides. Divide by

QUESTIONS 1

This proof doesn’t work when when

2

Does this proof work when

3

This proof appeals to the very important mathematical idea of self-similarity – looking to get similar structures in two different ways so that things cancel out. Use this idea to evaluate these expressions. a b



. At which stage does it break down? What is the formula for

? ? Can you find a simplified version of the formula in this case?

FOCUS ON … PROBLEM SOLVING 1

Trying small cases In this section you will look at solving problems about sequences, but the ideas apply in other contexts too. In addition to thinking about trying small cases to spot patterns in sequences, you should also remember several other problem-solving ideas. In particular: introduce letters to represent unknowns look for things which stay the same persevere. WORKED EXAMPLE 1

The terms of a sequence of positive integers

Find Let Then

follow the rule:

. and

.

First thoughts are that is a very large number and you obviously can’t work out all the terms on the way up to it. Also you are not told what and are, so perhaps it doesn’t matter. You’ll need to give them names to work with them though. Keep working out some more cases and hope you see something useful! It’s worth simplifying these expressions to help anything useful stand out.

Persevere….

The algebra gets messy, but it works out quite nicely in the end!

This is interesting – you’ve got back to followed by .

Since and and the sequence only depends on the previous two terms, it will continue to repeat every five terms.

Importantly, So

QUESTIONS

1

The sequence of integers

satisfies

and

where is a positive

constant. Given that

, find the value of .

2

Find a formula for the sum of the first odd numbers.

3

An expression for the sum of the first terms of an arithmetic sequence is given by Find an expression for the

4

term of the sequence.

Rebecca can walk up stairs one at a time or two at a time. For example, to go up stairs she might go or . Her house has stairs with



.

steps. How many different ways can she go up the stairs?

FOCUS ON … MODELLING 1

Modelling with rational functions You should already be familiar with exponential models, where the rate of change of a quantity is proportional to the quantity itself. In Student Book 1 they were used to model situations such as chemical reactions, population growth and cooling of a liquid. Here are some graphs resulting from exponential models.

But exponential equations are not the only ones that result in graphs of this shape. Here are the graphs of some rational functions.

Notice that for a simple rational function, the rate of change is proportional to the square of its value. For example, if

then

.

In this section you will meet an example of a model using a rational function and then learn a technique that allows you to decide whether an exponential or a rational model is a better fit for given data. QUESTIONS 1

In biochemistry, the Michaelis–Menten model is used for the rate of reactions involving enzymes. It relates the reaction rate, , to the concentration of the substrate, :

where and are constants that depend on the substances involved in the reaction. a Given that

and

(in suitable units):

i find the rate of reaction when the concentration is ii find the concentration for which the rate of reaction is iii sketch the graph of against . b What does represent? In Student Book 1, Chapter 8, you learnt a method for determining the parameters in an exponential

model by using logarithms to turn it into an equation of a straight line. If

, then

You can use a similar idea for a rational function of the form of the Michaelis–Menten model. If then

Hence the graph of against is a straight line with gradient and vertical axis intercept .

2

The data in the table can be modelled by an equation of the form

.

Draw the graph of against . Hence determine the values of the constants and . The technique of transforming a graph into a straight line can also be used to decide whether a rational or an exponential function is a better model for a set of data. 3

Each graph shows a set of experimental data. Plot

against and against .

Hence decide whether an exponential or a rational function is a better model for the data.

.

a

b

Tip Consider



.

CROSS-TOPIC REVIEW EXERCISE 1 1

Prove that if is rational and

2

The sum of the first terms of an arithmetic sequence is given by

is irrational, then must be irrational. .

Find the common difference of the sequence. 3

Find an expression for the mean value of the first terms of a geometric series with first term and common ratio .

4

The function is defined by

.

The function is defined by

, where and are constants.

Given that of . 5

, find the possible values of and the corresponding values

The function is defined by . a Find the largest possible domain of f and the corresponding range. b i State why the inverse function exists. ii Find the inverse function, stating its domain. c Find the set of values of for which

6

If

7

a Sketch the graph of

sketch

for

. .

b Solve the equation

.

c Solve the inequality 8 The polynomial

.

.

defined by

.

a i Find ii Show that

is a factor of

.

b Simplify

giving your answer in a fully factorised form. [© AQA 2010] 9

a Find the range of the function

.

b Prove that there is no real value of such that terms of an arithmetic sequence. 10

The sequence

is defined by

a Find the exact value of

b Find the exact value of 11

and

are consecutive

. .

.

Find the range of values of for which the series

converges.

12

Functions g and h are defined by

and

.

a Find constants and such that

.

b Using transformations, or otherwise, sketch the graph of

. Label the axis intercepts

and state the range of . c Find the domain and range of 13

.

The functions and are defined by Find the value of such that

14

and

respectively.

has equal roots.

Daniel has a new kitten. In the first week the kitten eats the amount of food it needs increases by each week. a Show that in the fourth week the kitten will need Daniel bought a

of cat food. As the kitten grows,

of cat food.

bag of cat food and wants to estimate how many full weeks the food will

last. b Show that the number of weeks, , satisfies the inequality

.

c Find the number of whole weeks the bag of food can be expected to last. d Is this number likely to be an overestimate or an underestimate? Explain your answer. 15

It is given that a Express

. in the form

, where and are integers.

b i Find the first three terms of the binomial expansion of where and are rational numbers.

in the form

,

ii State why the binomial expansion cannot be expected to give a good approximation to at . [© AQA 2013] 16 Find the exact value of . 17

This table shows the values and gradient of

at various points.

a

Is

a one-to-one function?

b Evaluate c The graph

is formed by translating the graph of

by a vector

and then

reflecting it in the -axis. Find 18

a Find the coordinates of the image of reflection in the -axis. b

is a one-to-one function. The graph of Find the equation of the resulting graph.

after a reflection in the line

is rotated

anticlockwise.

followed by a

19

a If the polynomial

is divided by

the remainder is a linear function.

Explain why this statement can be written as

where

is a

polynomial. b Find an expression for

in terms of

c Hence show that the remainder when

and

is divided by

d State the condition which must be satisfied if

. is

is to be a factor of

. .

7 Radian measure In this chapter you will: learn about different units for measuring angles, called radians learn how to calculate certain special values of trigonometric functions in radians learn how to use trigonometric functions in modelling real-life situations learn how to solve geometric problems involving circles learn that trigonometric functions can be approximated by polynomials revise solving trigonometric equations.

Before you start… 1 What is the exact value of

Student Book 1,

You should be able to define

Chapter 10

trigonometric functions beyond acute angles, including exact values.

Student Book 1, Chapter 10

You should be able to solve trigonometric equations.

2 Solve

Student Book 1, Chapter 11

You should be able to use the sine and cosine rules.

3 Find the smallest angle in a triangle with sides and .

Chapter 3

You should be able to identify transformations of graphs.

4 The graph of is translated in the positive direction and stretched vertically with the scale factor of . Find the equation of the new graph.

Chapter 6

You should be able to use the binomial expansion for negative and fractional powers.

5 Find the first three non-zero terms in the expansion of

for

?

.

, in ascending powers

of .

What are radians? Measuring angles is related to measuring lengths around the perimeter of the circle. This observation leads to the introduction of a new unit for measuring angles, the radian, which is a more useful unit of measurement than degrees in advanced mathematics.

Section 1: Introducing radian measure The measure of for a full turn may seem a little arbitrary – there are many other ways of measuring sizes of angles. In advanced mathematics, the most useful unit for measuring angles is the radian. This measure relates the size of the angle to the distance moved by a point around a circle. Consider a circle with centre and radius (this is called the unit circle), and two points, and , on its circumference. As the line rotates into position , point moves a distance equal to the length of the arc . The measure of the angle in radians is simply this arc length. If point makes a full rotation around the circle, it will cover the distance equal to the length of the circumference of the circle. As the radius of the circle is , the length of the circumference is . Hence a full turn measures

radians.

You can then deduce the sizes of other common angles in radians; for example, a right angle is one quarter of a full turn, so it measures

radians. Although sizes of common angles measured in

radians are often expressed as fractions of , you can also use decimal approximations. Thus a right angle measures approximately radians. WORKED EXAMPLE 7.1 a Convert

to radians.

b Convert

radians to degrees.

a

What fraction of a full turn is Calculate the same fraction of

? .

This is the exact answer. You can also find the decimal equivalent, to three significant figures. b

What fraction of a full turn is Calculate the same fraction of

radians? .

Key point 7.1 radians To convert from degrees to radians, divide by

and multiply by .

To convert from radians to degrees, divide by and multiply by

.

Did you know? There are many different measures of angle. One historical attempt was gradians, which split a right angle into units.

WORKED EXAMPLE 7.2

Mark on the unit circle the points corresponding to these angles, measured in radians. A B C D A: radians is one half of a full turn. B: is one quarter of the full turn and the minus sign represents clockwise rotation. C:

, so point represents a full turn followed by another quarter of a turn.

D:

, so point represents two full turns followed by another th of a turn.

You can still find the values of trigonometric functions in radians, just as you could in degrees, but you’ll need to make sure you change your calculator to radian mode. However, as well as getting values of trigonometric functions from your calculator you also need to recognise a few special angles for which you can find exact values. The method relies on properties of special right-angled triangles.

Tip Make sure you can change your calculator between degree and radian modes.

WORKED EXAMPLE 7.3

Find the exact values of

and

.

If a right-angled triangle has one angle of then the third angle is , so this is half of an equilateral triangle. You can choose any length for the side of the equilateral triangle. Let

; then

Use Pythagoras’ theorem to find

.

.

You can now use the definitions of tan in right-angled triangles.

and

The results for other values follow similarly.

Key point 7.2 Radians Degrees

not defined

Gateway to A Level See Gateway to A Level Section Z for revision of working with exact values in degrees.

You can find corresponding values for angles greater than , and derive other properties, by using the symmetries of the trigonometric graphs. You therefore need to recognise -intercepts and turning points of these graphs, in radians.

Key point 7.3

WORKED EXAMPLE 7.4

Given that

, find the values of:

a b

. Label and

a

so

.

on the horizontal axis.

The graph has the same height at the two points. Label and

on the horizontal axis.

b

WORK IT OUT 7.1 Given that find the value of . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2 From the graph,

Solution 3

EXERCISE 7A 1

Draw a unit circle for each part and mark the points corresponding to each angle. a i ii b i ii c i ii d i ii

2

Express each angle in radians, giving your answers in terms of . a i ii b i ii c i ii d i ii

3

Express each angle in radians, correct to three decimal places. a i ii b i ii c i ii d i ii

4

Express each angle in degrees. a i ii b i ii c i ii d i ii

5

6

Sketch the graph of: a i

for

ii

for

b i

for

ii

for

Given that a

. find, without using a calculator, the value of:

b c d 7

Given that

find, without using a calculator, the value of:

a b c d 8

Given that

find, without using a calculator, the value of:

a b c d 9

Without using a calculator, find the exact values of: a i ii b i ii c i ii

10

Without using a calculator, evaluate each expression, simplifying as far as possible. a b c

11

Show that

12

Show that

13

Simplify

14

Simplify the expression

. . . .



Section 2: Inverse trigonometric functions and solving trigonometric equations When solving trigonometric equations in Student Book 1, you used

to solve equations of the form . Based on the work in Chapter 2 you can see that this is an example of an inverse trigonometric function.

Since the original function needs to be one-to-one for the inverse function to exist, it is clear that you will need to consider the sine graph on a restricted domain. From the graph you can see that a suitable domain is

. (There are other options but this is

chosen by convention.) You can now draw the graph of the inverse function as the reflection of the graph of the original function in the line . From the graph you can identify the domain and range of the inverse function.

Key point 7.4 The inverse function of Its domain is

is

and its range is

. .

Rewind You met inverse functions in Chapter 2, Section 4.

You can carry out a similar analysis for cosine and tangent functions to identify the domains on which their inverse functions are defined.

Key point 7.5 The inverse function of Its domain is

is

and its range is

. .

Tip You will also see

and

used for the inverse sine, cosine and tangent

functions.

Key point 7.6 The inverse function of Its domain is and its range is

is

. .

You should remember what happens when you compose a function with its inverse: Applying this to inverse trigonometric functions, you get:

.

You need to be a little more careful when composing the other way round: for example, only when is in the restricted domain used to define the arcsin function, i.e.

.

WORK IT OUT 7.2 Evaluate

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

WORKED EXAMPLE 7.5

Solve the equation Evaluate the arcsin terms on the RHS.

Rearrange. Taking sin of both sides undoes the arcsin.

As you learnt in Student Book 1, Chapter 10, trigonometric equations can have more than one solution.

Key point 7.7 Equation First solution,

Second solution,

Further solutions

Add or subtract multiples of

.

Add or subtract multiples of

.

Add or subtract multiples of .

WORKED EXAMPLE 7.6

Find the values of between

and

for which

. Sketch the graph. Note there are three solutions.

Three solutions. Use inverse cos to find the first solution. Use the symmetry of the graph to find the other solutions.

WORK IT OUT 7.3 Solve

for

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

So

Solution 2

Adding on would take us outside of the required region so this is the only solution. Solution 3

are the only solutions in the required range.

EXERCISE 7B 1

Use your calculator to evaluate these ratios, in radians, correct to three significant figures. a i ii b i ii

2

Without using a calculator, find the exact value in radians. a i ii b i ii c i ii

3

Without using a calculator, find the exact value. a i ii b i

ii c i ii d i ii 4

Without using a calculator, solve each equation. a b c

5

Without using a calculator, find the values of between and

for which:

a i ii b i ii c i ii d i ii 6

Solve these equations in the given interval, giving your answers to three significant figures, where appropriate. Do not use graphs on your calculator. a i

for

ii

for

b i ii

for for

c i

or

ii

for

d i

for

ii

for

7

Find the exact values of

8

Sketch the graph of

9

Find the exact solutions to

for which for

. .

for

.

10

a If

show that

.

b Find all solutions to

for

.

Worksheet See Support Sheet 7 for a further example of solving equations in radians and for more practice questions. 11

Find the exact solutions of the equation

12

Leo says that

13

Find the exact solutions of the equation

14

a Prove by a counter example that b Express

.

for all . Prove by counter example that this statement is false.

in terms of

c Solve the equation



for

for .

. .

.

Section 3: Modelling with trigonometric functions Trigonometric functions can be used to model real-life situations that show periodic behaviour, for example, the height of the tide, or the motion of a point on a Ferris wheel. To do this, you need to consider trigonometric functions with different periods and amplitudes. Using your knowledge of combining transformations from Chapter 3, you can adjust both the amplitude and period of and .

Key point 7.8 The functions

and

have amplitude and period

.

Rewind You know, from Student Book 1, Chapter 10, that amplitude .

and

have period

and

WORKED EXAMPLE 7.7 a Sketch the graph of

for

.

b Write down the amplitude and the period of the function. a Vertical stretch with scale factor Horizontal stretch with scale factor

Start with the graph of and consider what transformations to apply to it.

b Amplitude = 4

As well as vertical and horizontal stretches, you can also apply translations to graphs. They will leave the period and the amplitude unchanged, but will change the positions of maximum and minimum points and the axes intercepts. WORKED EXAMPLE 7.8

a Sketch the graph of

for

.

b Find the coordinates of the maximum and the minimum points on the graph. a

The equation is of the form

.

This represents a translation of units to the right and units up.

The maximum and minimum points of

b Maximum point:

and

are

. You need to apply the

same translation to these points. So the maximum point is

.

Minimum point:

So the minimum point is

You can use your knowledge of transformations of graphs to find an equation of a function, given its graph. WORKED EXAMPLE 7.9

The graph shown has the equation Find the values of

.

and .

is the amplitude, which is half the difference between the minimum and maximum values. is related to the period, which is the distance between the two consecutive maximum points. The formula is

.

represents the vertical translation of the graph. It is the value halfway between the minimum and the maximum values.

You will now use these ideas to create mathematical models of periodic motion, such as motion around a circle, oscillation of a particle attached to the end of a spring, water waves or heights of tides. In practice, you would collect experimental data to sketch a graph and then use your knowledge of trigonometric functions to find its equation. You can then use the equation to do further calculations. WORKED EXAMPLE 7.10

The height of water in a harbour is

at high tide, and

shows how the height of water changes with time over

at low tide, hours later. The graph hours.

a Find the equation for height (in metres) in terms of time (in hours) in the form b Find the first two times after the high tide when the height of water is

.

.

is the ‘central value’, half-way between minimum and maximum values.

a

is the amplitude, which is half the distance between the minimum and maximum values.

The period is

.

So b

Form an equation with

.

Rearrange the equation into the form

.

The high tide is when , so you want the first two answers with . Find two possible values of . Solve for .

The height of the water will be hours and hours after the high tide.

WORKED EXAMPLE 7.11

A point moves with constant speed around a circle of radius

, starting from the positive

and taking seconds to complete one full rotation. Let be the height of the point above the . Find an equation for in terms of time. Draw a diagram; you can see that , where is the angle between the radius and the . You now need to find how depends on time.

From the diagram: Let be the time, measured in seconds.

You have information about the values of at the start and seconds later.

So

As the point moves with constant speed, is directly proportional to so .

.

Hence

.

EXERCISE 7C 1

State the amplitude and the period of each function, where is in radians. a b c d

2

Sketch each graph, giving coordinates of maximum and minimum points. a i ii b

iii

for

c i ii

for

d i ii 3

for

The depth of water in a harbour varies during the day and is given by the equation , where is measured in metres and in hours after midnight. a Find the depth of the water at low and high tide. b At what times does high tide occur?

4

A small ball is attached to one end of an elastic spring, and the other end of the spring is fixed to the ceiling. The ball is pulled down and released, and starts to oscillate vertically. The graph shows how the length of the spring, , varies with time. The equation of the graph is . a Write down the amplitude of the oscillation. b How long does it take for the ball to perform five complete oscillations?

5

This graph has equation

6

This graph has equation

for

. Find the values of and .

for

. Find the values of and .

7

a On the same set of axes sketch the graphs of

and

b Hence state the number of solutions of the equation

for

.

for

c Write down the number of solutions of the equation

.

for

.

8

The graph shows the height of water below the level of a walkway as a function of time. The equation of the graph is of the form . Find the values of and .

9

a Sketch the graph of

for

.

b Find the coordinates of the maximum and minimum points on the graph. c Write down the coordinates of the maximum and minimum points on the graph of for 10

.

A point moves around a vertical circle of radius

, as shown in the diagram. It takes

to

complete one revolution. a The height of the point above the

is given by

, where is time measured in

seconds. Find the values of and . b Find the time during the first revolution when the point is

below the

.

11

A ball is attached at its top and bottom to elastic strings, with each string attached at the far end and under tension. When the ball is pulled down and released, it starts moving up and down, so that the height of the ball above the ground is given by the equation

, where is

measured in centimetres and is time in seconds. a Find the least and greatest height of the ball above ground. b Find the time required to complete one full oscillation. c Find the first time after the ball is released when it reaches the greatest height. 12

A Ferris wheel has radius and the centre of the wheel is minutes to complete a full rotation.

above ground. The wheel takes

Seats are attached to the circumference of the wheel. Let radians be the angle the radius connecting a seat to the centre of the wheel makes with the downward vertical, and let be the height of the seat above ground. a Find an expression for in terms of . b Initially the seat is at the lowest point on the wheel. Assuming that the wheel rotates at constant speed, find an expression for in terms of , where is the time measured in minutes. c Write down an expression for in terms of . For how long is the seat more than ground?



above

Section 4: Arcs and sectors The diagram shows a circle with centre and radius , and points and on its circumference. The part of the circumference between points and is called an arc of the circle. You can see that there are in fact two such parts; the shorter one is called the minor arc, and the longer one the major arc. The minor arc subtends an angle at the centre of the circle.

The ratio of the length of the arc to the circumference of the whole circle is the same as the ratio of angle to the angle measuring a full turn. If is the length of the arc, and angle is measured in radians, this means that

. Rearranging this equation gives the formula for the arc length.

Key point 7.9 The length of an arc is

where is the radius of the circle and is the angle subtended at the centre, measured in radians.

WORKED EXAMPLE 7.12

Arc of a circle with radius 5 cm subtends an angle of the diagram. a Find the length of the minor arc

.

b Find the length of the major arc

.

a

radians at the centre, as shown on

Use the formula for the length of an arc.

b

The angle subtended by the major arc is equal to a full turn minus the smaller angle. A full turn is radians.

A sector is a part of a circle bounded by two radii and an arc. As with arcs, you can distinguish between a minor sector and a major sector. The ratio of the area of the sector to the area of the whole circle is the same as the ratio of angle to the angle measuring a full turn. If is the area of the sector, and angle is measured in radians, this means that

. Rearranging this equation gives the formula for the area of the sector.

Key point 7.10 The area of a sector of a circle is

where is the radius of the circle and is the angle subtended at the centre, measured in radians.

Common error You can only use the formulae for arc length and area of sector in Key points 7.9 and 7.10 if the angle is in radians.

WORKED EXAMPLE 7.13

A sector of a circle has perimeter sector. radians

and angle at the centre

. Find the area of the

To use the formula for the area of a sector, the angle needs to be in radians. You are given the perimeter, and you know the formula for it. Use this to find .

Now use

.

EXERCISE 7D 1

Calculate the length of the minor arc subtending an angle of radians at the centre of the circle of radius . a b

2

Points and lie on the circumference of a circle with centre and radius radians. Calculate the length of the major arc .

. Angle

is

a b 3

Points

and lie on the circumference of a circle with centre and radius

, and

.

Calculate the area of the minor sector a b 4

Points and lie on the circumference of a circle with centre and radius angle is radians. Calculate the area of the major sector .

. The size of the

a b 5

Calculate the length of the minor arc

6

In the diagram, the radius of the circle is Calculate the size of the angle a in radians b in degrees.

:

in the diagram.

and the length of the minor arc

is

.

7

Points

and lie on the circumference of a circle with centre and radius

major arc 8

is

. Calculate the size of the smaller angle

. The length of the

.

Points and lie on the circumference of the circle with centre . The length of the minor arc is

and

radians. Find the radius of the circle.

9

A circle has centre and radius . Points and lie on the circumference of the circle so that the area of the minor sector is . Calculate the size of the angle .

10

Points and lie on the circumference of a circle with radius . The area of the major sector is . Find the size of the smaller angle in degrees.

11

In the diagram, the length of the major arc

12

A sector of a circle with angle

13

The perimeter of the sector shown in the diagram is

14

The diagram shows an equilateral triangle with side , and three arcs of circles with centres at the vertices of the triangle. Calculate the perimeter of the figure.

is

radians has area

. Find the radius of the circle.

. Find the radius of the circle. . Find its area.

15

Calculate the perimeter of the figure shown in the diagram.

16

A sector of a circle with angle

17

The diagram shows a triangle and the segment of a circle. Find the exact perimeter of the figure.

18

A sector of a circle has perimeter

has area

. Find the radius of the circle.

and angle at the centre

radians. Find the radius

of the circle. 19

Find the area of the shaded region.

20

A sector of a circle has perimeter circle.

21

Points and lie on the circumference of a circle with centre and radius between the areas of the major sector and the minor sector is

and area

. Find the possible values of the radius of the

. The difference . Find the size of the

angle 22

.

A cone is made by rolling a piece of paper shown in the diagram.

       If the cone is to have height



and base diameter

, find the size of the angle marked .

Section 5: Triangles and circles In this section you will look at two other important parts of circles: chords and segments. You will need to combine the results about arcs and sectors with the formulae for lengths and angles in triangles.

WORKED EXAMPLE 7.14

The diagram shows a sector of a circle of radius Find:

and the angle at the centre is

radians.

a the perimeter b the area of the shaded region.

The perimeter is made up of the arc chord .

a

The formula for the length of the arc is Using the cosine rule:

b

and the

.

The chord is the third side of the triangle . As two sides and the angle between them are known, you can use the cosine rule. Remember that the angle is in radians.

If you subtract the area of triangle left with the area of the segment. Area of a sector = Area of a triangle =

. .

, you are

Worked example 7.15 shows how to solve more complex geometry problems, by splitting up the figure into basic shapes such as triangles and sectors.

Rewind See Student Book 1, Chapter 11, if you need a reminder of the cosine rule and the formula for the area of the triangle.

WORKED EXAMPLE 7.15

The diagram shows two equal circles of radius

such that the centre of one circle is on the

circumference of the other. a Find the exact size of angle

in radians.

b Calculate the exact area of the shaded region.

a

The only thing you know is the radius of the circle, so draw all the lengths that are equal to the radius.

The lengths are all equal to the radius of the circle. Therefore triangle is equilateral. b

The shaded area is made up of two segments, each with angle at the centre

radians.

You can find the area of one segment by using the formula. Remember to use the exact value of

.

The shaded area consists of two segments.

EXERCISE 7E 1

Find the length of the chord

.

a i

ii

b i

ii

2

Find the perimeters of the minor segments from question 1.

3

Find the areas of minor segments from question 1.

4

A circle has centre and radius . Chord subtends angle at the centre of the circle. Given that the area of the minor segment is , show that .

5

Two circles, with centres and , intersect at and . The radii of the circles are and .

a Show that

.

b Find the size of angle

.

c Find the area of the shaded region.

Worksheet See Extension Sheet 7 for some more challenging questions of this type.



and

,

Section 6: Small angle approximations The diagram shows the graphs of and very close to each other. This means that

. As you can see, near the origin, the two graphs are for close to zero.

PROOF 3

By considering the diagram, prove that, for small values of

If the angle of the sector is radians then its area is

.

The area of the triangle is

.

For small values of , the sector and the triangle have approximately equal areas. So write down the expression for each.

.

When is small, the sector and the triangle have nearly the same area. Hence,

For small values of , the graph of looks like a (negative) parabola. You can find its equations by looking at the sector and the triangle again.

PROOF 4

Prove that, for small values of

. appears in the cosine rule, so compare the length of the arc and the straight line .

The length of the arc

is .

The length of the line is When is small:

For small , the arc and the line have approximately equal lengths.

It is also possible to derive a similar approximation for approximations, are summarised in Key point 7.11.

. All three results, called the small angle

Key point 7.11 For small , measured in radians:

These will be given in your formula book.

You can use these results to find approximate values of trigonometric functions.

Rewind This is similar to using the binomial expansion to find approximate values of powers, as you learnt to do in Chapter 6.

WORKED EXAMPLE 7.16 a Use a small angle approximation to estimate the value of

.

b Find the percentage error in your estimate. a

Using

, as

is close to .

Use a calculator to find the actual value.

b

Fast forward The ‘exact’ value of you get from the calculator in Worked example 7.16 is in fact also an approximation. It is obtained from a generalisation of small angle approximations, called Maclaurin series, which you will study if you take the Further Mathematics course.

Just as with the binomial expansion, you can replace by another function, such as multiply two expansions together.

or

. You can also

WORKED EXAMPLE 7.17

Assuming that is sufficiently small that terms in

and higher can be ignored, find an

approximate expression for: a b a

Replace by

in

.

b

Use the result from a and

.

Expand the brackets, but only keep terms up to .

You need to be a little more careful when dividing two approximate expressions – you may need to turn the quotient into a product before using the binomial expansion.

Rewind See Chapter 6 for a reminder of the binomial expansion with negative powers.

WORKED EXAMPLE 7.18

Find an approximate expression for

given that is small enough to neglect the terms in

and above. Replace and approximations.

by their small angle

To complete the expansion, turn division into multiplication.

Now do the binomial expansion. Remember that the bracket needs to be in the form .

Expand the bracket and ignore any terms in or higher.

EXERCISE 7F 1

Find the approximate value of each expression. a i ii b i ii c i ii

2

Find the small angle approximation for each expression. a i ii b i ii c i ii

3

Assuming is sufficiently small so that terms in and above can be ignored, find an approximate expression for each expression. a i ii b i ii c i ii

4

a Find a small angle approximation for b Hence find an approximate value of

5

a Given that is sufficiently small, write down an approximate expression for

in the form

. b Use your expression with 6

to find an approximate value of .

a Find an approximate expression for and higher can be ignored.

when is sufficiently small so that the terms in

b Find the percentage error when this approximation is used to estimate the value of: i .

ii 7

Given that is close to zero, find an approximate expression for

, ignoring powers of

higher than . 8

Find an approximate expression for

9

Let be a small angle, measured in degrees. Use small angle approximations to find an approximate expression for and .

10

a Find an approximate expression for including .

when is close to zero.

for small values of , including terms up and

b Hence find an approximate value of 11

Given that is close to zero so that terms in expression for

12

. and higher can be ignored find an approximate

.

Find an approximate expression for

when is sufficiently small to ignore the terms in

or higher. 13

a For each of the three small angle approximations, find the largest value of for which the approximation error is less than . b i Show that using the small angle approximation for gives the correct small angle approximation for .

and the identity

ii Why is it not appropriate to use in front of the square root?

Checklist of learning and understanding The radian is defined in terms of the distance travelled around the circle, so that a full turn radians. To convert from degrees to radians, divide by

and multiply by .

To convert from radians to degrees, divide by and multiply by

.

For some real numbers the three functions have exact values, which you should learn. Radians Degrees sin cos tan

undefined

The functions

and

have:

amplitude central value minimum value

and maximum value

period Properties of inverse trigonometric functions: Inverse function

Domain

Range

Trigonometric equations can be solved in radians. Equation First solution,

Second solution,

Further solutions Add or subtract multiples of . Add or subtract multiples of . Add or subtract multiples of .

If is the radius of the circle and is the angle subtended at the centre, measured in radians, then the length of an arc is For small , measured in radians:



and the area of a sector is

.

Mixed practice 7 1

What is in radians? Choose from these options. a b c d

2

The height of a wave (in metres) at a distance metres from a buoy is modelled by the function

.

a State the amplitude of the wave. b Find the distance between consecutive peaks of the wave. 3

Solve the equation

4

Use small angle approximations to estimate the value of

5

In the diagram,

for

.

is a rectangle with sides

are circular arcs, and a Write down the size of

and

. .

is a straight line.

and

radians. .

b Find the area of the whole shape. c Find the perimeter of the whole shape.

6

A sector has perimeter

7

The diagram shows a circle with centre and radius radians at the centre of the circle.

and radius

Find: a the area of the shaded region b the perimeter of the shaded region.

. Find its area. . The chord

subtends angle

8

The diagram shows a sector

of a circle with centre .

The radius of the circle is 6 cm and the angle a Find the area of the sector

.

b i Find the length of the arc

.

.

ii Hence show that the perimeter of the sector an integer.

where is

[© AQA 2011] 9

How many solutions for are there to the equation

?

Choose from these options. A B C D Infinitely many 10

The diagram shows the graph of the function

11

The shape of a small bridge can be modelled by the equation

. Find the values of and .

, where is the

height of the bridge above water, and is the distance from one river bank, both measured in metres. a Find the width of the river. b A barge has height above the water level. Find the maximum possible width of the barge so it can pass under the bridge. c Another barge has width can pass under the bridge? 12

. What is the maximum possible height of the barge so it

A runner is jogging around a level circular track. His distance north of the centre of the track, in metres, is given by where is measured in seconds.

a How long does is take the runner to complete one lap? b What is the length of the track? c At what speed is the runner jogging? 13

Let

.

a State the period of the function. b Find the coordinates of the points where the graph of . c Hence sketch the graph of, maximum and minimum points. 14

, for

Find an approximate expression for

for

crosses the

, showing the coordinates of the

, assuming is small enough to ignore terms in

and higher. 15

Two circles have equal radius and intersect at points and . The centres of the circles are and , and . a Explain why

is also

b Find the length

in terms of .

c Find the area of the sector

.

.

d Find the area of the overlap of the two circles. 16

In the diagram, is the centre of the circle and If

17

, and the circle has a radius of

a Sketch the graph of

is the tangent to the circle at .

, find the area of the shaded region.

, where is in radians. State the coordinates of the end

points of the graph. b Sketch the graph of points of the graph.

, where is in radians. State the coordinates of the end

[© AQA 2013] 18

Find the exact values of

19

Find the exact solutions to the equation

20

a Given that

satisfying the equation

. for

, show that

.

.

b Hence solve the equation your values of in radians to three significant figures.

in the interval

, giving

[© AQA 2013] 21

a Given that is small enough so that the terms in approximate expression for . b Hence find an estimate for

.

and higher can be neglected, find an

22

Two circular cogs are connected by a chain as shown in diagram A. The radii of the cogs are and and the distance between their centres is . Diagram shows the quadrilateral

. Line

is drawn parallel to

.

A

B

a Write down the size of b Explain why c Hence find the length

in radians, giving a reason for your answer.

. .

d Find the size of the angle marked , giving your answer in radians correct to four significant figures. e Calculate the length of the chain 23

a Write down an expression for b Show that c Hence solve the equation



. .

. for

.

8 Further trigonometry In this chapter you will learn how to: work with trigonometric functions of sums and differences of two angles, for example, work with trigonometric functions of double angles, for example, work with sums of trigonometric functions, for example, work with reciprocal trigonometric functions, for example,

.

Before you start… Student Book 1, Chapter 10

You should be able to use the identities and .

1 Given that is an acute angle with , find the exact value of: a b

Chapter 7; Student

You should know and be able to

Book 1, Chapter 10

use graphs of trigonometric functions, in degrees and radians.

Chapter 7; Student Book 1, Chapter 10

You should be able to solve trigonometric equations in degrees and radians.

2 State the coordinates of the minimum point on the graph of for

.

3 Solve each equation. a b

for , for

Combining trigonometric functions Periodic phenomena, such as water waves, sound waves or motion around a circle, can be modelled by sine and cosine functions. There are many situations in which several functions need to be combined together. For example, the interference of two waves can be modelled by adding the functions that describe their shapes. Some fairground rides, such as the waltzer, have groups of seats arranged in a circle that rotates on top of a larger rotating platform. In this chapter you will learn how to simplify expressions involving sums and products of trigonometric functions.

Section 1: Compound angle identities Use your calculator, working in radians, to find: and and

.

There seems to be no obvious connection between the values of, for example,

Tip You will also see these identities referred to as compound angle formulae.

In fact, though, there are formulae, called compound angle identities, to express the functions of the individual angles.

in terms of

Focus on… See Focus on... Proof 1 for the proofs of these results.

Key point 8.1

These will be given in your formula book.

In practice, you only need to prove the identities for to prove those for

and

Then you can use them

Common error Notice the signs in the cosine identities: in the identity for the sum the minus sign is used, and in the identity for the difference the plus sign.

PROOF 5

Use the identity

to prove that .

Replace by

as required.

in the first identity:

The only difference between the two expressions on the left-hand side is the sign of , so replace by . Now use the symmetries of the sin and cos graphs: and .

One of the simplest applications is to calculate exact values of trigonometric functions.

Rewind You met exact values of trigonometric functions in Chapter 7, Section 1.

WORKED EXAMPLE 8.1

Find the exact values of: a b

.

a

You only know the exact values of the sin of a few angles: . You can write as a sum of two of those.

b

Look for a way to make angles.

from the special

WORK IT OUT 8.1 Expand

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

Solution 3

You can use the sine and cosine compound angle identities to derive new ones.

Key point 8.2

These will be given in your formula book.

WORKED EXAMPLE 8.2

Prove that

. Express tan in terms of sin and cos. Use the identities for sin and cos. You want to express this in terms of . Looking at the top of the fraction, if you divide by you will get in the first term, and if you divide by you will get in the second term. So divide top and bottom by .

Convert back to tan. as required.

EXERCISE 8A 1

Without using a calculator, express each angle in the form and . a b c d

, giving exact values of

2

Without using a calculator, find the exact value of each expression. a b c d

3

Without using a calculator, find the exact value of

for the angles shown in the diagrams.

a

b

4

a Show that

.

b Simplify 5

a Express

. in terms of

b Given that

.

, find two possible values of

c Hence solve the equation 6

for

.

.

Write each of these expressions as a single trigonometric function, and hence find its maximum value and the smallest positive value of for which it occurs. a b

7 8

Assuming is small enough so that the terms in expression for

in the form

and higher can be ignored, find an approximate .

a Show that

.

b Hence solve the equation

for

.

Rewind Small angle approximations for sin and cos were covered in Chapter 7, Section 6. 9

a Show that

.

b Hence solve the equation



for

.

Section 2: Double angle identities If you set in the compound angle identity for you get an identity for the same with cos and tan giving a set of double angle identities.



. You can do

Key point 8.3

A useful application of these identities is finding exact values of half-angles.

Tip You can convert between the three different forms of the identity by using , but they are used so often that it is best to just remember all three.

WORKED EXAMPLE 8.3

Using the exact value of

, show that

. You know that

, so use a double

angle formula to relate this to

.



You have to choose between the positive and negative square root here. so take the positive square root.

Recognising the form of double angle identities can be useful in solving trigonometric equations. WORKED EXAMPLE 8.4

Solve the equation

for

. Write the left-hand side in terms of only one trig function.

Rearrange so that the left-hand side is , which is equal to . Follow the standard procedure and look at the graph to find that that there are four solutions in the given domain.

If an equation contains both only .

and

, you can use identities to turn it into an equation involving

WORKED EXAMPLE 8.5

Find exact solutions of the equation

for

.

As before, write the equation in terms of only one trig function. (Note that and are not the same function!). The left-hand side involves a double angle, so choose the identity involving just . This is a quadratic in

When

When

:

.

Solve each equation separately.

: Remember to list all the solutions at the end.

Although they are called the double angle identities, you can also use them with higher multiples.

Fast forward In Chapter 11 you will use double angle identities to integrate some trigonometric functions.

WORKED EXAMPLE 8.6

Find an expression for

in terms of:

a b

.

a

, so use one of the double angle identities. Since you want an expression involving only cos it has to be .

b From part a:

Use the same double angle formula again to replace in the answer in part a.

You can use a combination of double angle and compound angle identities to derive ‘triple-angle identities’ . WORKED EXAMPLE 8.7 a Show that

.

b Solve the equation a

for

. If you write you can use the compound angle identity. You need an expression involving only single angles so use the double angle identities. Since the answer only involves sin, use . You want only sin in the expression so replace with .

b

You need an equation in only one trig function, so use the result in .

This is a cubic in so rearrange to make the right-hand side equal to zero and then factorise.

When

:

Solve these two equations in the usual way.

When

:

The double angle formulae allow you to simplify some complicated expressions involving functions and inverse functions.

Rewind You met inverse trigonometric functions in Chapter 7, Section 2.

WORKED EXAMPLE 8.8

Let

. Find an expression for:

a b in terms of . a

This is of the form , where . So use the double angle identity involving only (as you don’t want to introduce sin).

b

This is of the form with , so use the sin double angle identity this time.

You need to convert expression involving , with

into an . Use .

When taking the square root, you have to choose between the positive and negative roots. Substituting into

:

EXERCISE 8B

1

a i Given that

, find the exact value of

ii Given that

, find the exact value of

b i Given that

and

. .

, find the exact value of

.

2

ii Given that

and

, find the exact value of

.

c i Given that

and

, find the exact value of

.

ii Given that

and

, find the exact value of

.

Without using a calculator, find the exact value of: a i ii b i .

ii 3

Simplify each expression, using a double angle identity. a b c d

4

Use double angle identities to solve each equation. a

for

b

for

c

for

d 5

for

Prove each identity. a b c d

6

Find all the values of

7

a Use the identity

that satisfy the equation to show that

b Show that Show that

.

9

a Given that

, find the possible values of

b Solve the equation

11

a Express

in terms

b Express

in terms of

Express a b

.

.

.

8

10

.

in terms of:

for . .

.

, giving your answer in terms of .

12

a Show that: i .

ii b Express 13

Given that

14

Show that:

in terms of

.

and

, express

in terms of and .

a b 15

a Show that b Hence show that



. . for

.

Section 3: Functions of the form In this section you will look at a useful method for dealing with sums of trigonometric functions. Suppose you are trying to solve the equation . You know that you need to try to write everything in terms of one trigonometric function so start by considering the identities met so far. You cannot just replace with a function of (or with a function of ) as the only identity you have linking these is and there is neither nor in the equation. So instead look for an identity involving both and . From the list of compound angle identities there are several options try one of the two that has a .

Tip Notice that



would work here as well as it also has a

.

Now, compare this with the equation:

You can see that the left-hand side of the equation would be the same as the compound angle expression if you could find so that and . This is not possible because so adjust the original identity by multiplying by a constant, :

Now you have: and which constitutes a pair of simultaneous equations in two unknowns ( and ). To find you can use

:

           

To find eliminate by dividing one equation by the other:

So, you have seen that you can write equation as:

as

which allows you to rewrite the

This is now a type of equation that you know how to solve. You can use the same procedure with a cos compound angle identity, and you can apply it to differences as well as sums of trigonometric functions.

Key point 8.4 To write

in the form

or

:

1 Expand the brackets using a compound angle identity. 2 Equate coefficients of 3 To get , use

and

to get equations for

and

.

.

4 To get , use

.

Tip You will usually be told which identity to use. However, if not, choose one that gives the same sign as the expression you have. In Worked example 8.9, work as it gives a sign.

would also

WORKED EXAMPLE 8.9 a Write

in the form

b Hence solve

.

for You are told to use , so first expand this using the compound angle identity.

a

Comparing this with

:

Compare coefficients of

and

Find by using

Find by using

.

.

.



b

Use the answer from part a.

   Since    

Now follow the standard procedure. Sketch the cos graph on the domain of .

  

WORKED EXAMPLE 8.10 a Write

in the form

where

b Hence find the maximum value of occurs.

.

and the smallest positive value of for which it

a

Follow the standard procedure. So, start by expanding .  Comparing this with

:

Compare coefficients of

 

Find by using

 

Find by using

and

.

.

.

  b Maximum value

 This value occurs when

EXERCISE 8C

Since sin has a maximum value of 1, the maximum value of this function must be . Solve value of .

for the smallest positive

EXERCISE 8C 1

Express in the form

where

and

where

and

where

and

where

and

:

i 2

. ii Express in the form

:

i .

ii 3

Express in the form

:

i .

ii 4

Express in the form

:

i .

ii 5

a Express

in the form

, where

and

.

b Hence give details of two successive transformations that transform the graph of the graph of . 6

a Express

in the form

, where

b Hence find the range of the function 7

a Express

in the form

and

a Express

in the form

.

. , where

and

b Hence find the smallest positive value of for which 8

into

, where

. .

and

.

b Hence find the coordinates of the minimum and maximum points on the graph of for .

Worksheet See Support Sheet 8 for a further example of solving equations of the form and for more practice questions. 9 10



Find, to 3 significant figures, all values of in the interval [0, 2π] for which Solve the equation

for

.

.

Section 4: Reciprocal trigonometric functions Since

you can do all the calculations you need with just sine and cosine. However, having

the notation for

can simplify many expressions.

Similarly, since expressions of the form

and

often occur, it seems sensible to

have notations for these as well. Using the graphs of sine, cosine and tangent and what you know about reciprocal transformations, you can draw the graphs of and .

Key point 8.5 The reciprocal trigonometric functions are: secant: cosecant: cotangent:

Common error It is easy to think that

has domain corresponding to

It has range

. Be careful it’s the other way round!

for integer (it has vertical asymptotes at every multiple of ).

It has range

has domain corresponding to

and

.

for integer (it has vertical asymptotes at every multiple of ). .

has domain corresponding to

for integer (it has vertical asymptotes at every multiple of ).

It has range .

Perhaps the most common usage of these functions is in the identities that can be deduced from the familiar by dividing through by and respectively (see Key point 8.6).

Key point 8.6

Tip The inverse trigonometric functions follow the same conventions as the normal trigonometric functions so .

WORKED EXAMPLE 8.11

Solve the equation

for

.

so

.

As always you want only one trig function in the equation. Use to achieve this. You do not know how to find inverse sec, so express in terms of .

When

: no solutions.

When

Finally solve each equation separately.

:

WORKED EXAMPLE 8.12

Show that You have a formula for introducing cos.

, so start by

Use the version that only contains cos as only sec is wanted in the answer: . You need from

in the answer so change back

using

.

Finally, simplify by multiplying top and bottom of the fraction by .

EXERCISE 8D 1

Giving your answers to four significant figures, find the value of: a i ii b i ii c i ii

2

.

Find the exact value of: a i ii

b i ii c i ii d i ii

.

3

Find the values of

4

Solve these equations for

and

.

, giving your answers to s.f.

a i ii b i ii c i ii d i ii 5

.

Find the exact solution of each equation for

.

a i ii b i ii c i ii d i ii 6

a i Given that

and

, find the exact value of

.

ii Given that

and

, find the exact value of

.

b i Given that

and

, find the exact value of

.

ii Given that

and

, find the exact value of

.

c i Given that ii Given that

and

, find the exact value of

and

, find the exact value of

d i Given that

, find the possible values of

.

ii Given that

, find the possible values of

.

7

Prove that

8

Solve the equation

9

Find, to three significant figures, the values of in the interval

10

Show that

11

Prove that

12

Prove that

13

a Given that

for

. for which

.

. . . , show that

a Show that

a Prove that

. for

.

.

b Hence solve the equation 15

.

.

b Hence solve the equation 14

.

for

.

.

b Hence solve the equation

for

.

16

Given that is small enough to neglect the terms in and above, find an approximate expression for sec .

17

Solve the equation

18

Find the inverse function of

for

.

in terms of the arccosine function.

Checklist of learning and understanding Compound angle identities (be careful to get the signs right!):

Double angle identities:

One particular application is to write :

in the form

or

Expand the brackets, using a compound angle identity. Equate coefficients of To get , use

and

to get equations for

and

.

To get , use

.

Reciprocal trigonometric functions are defined by:

. Two identities for reciprocal trigonometric functions can be derived from : .



.

Mixed practice 8 1

What is the maximum value of

?

Choose from these options. a b c d 2

a Use the identity for

to prove that

b Find the exact solutions of the equation 3

a Write

in the form

for

.

.

b Hence find the exact value of 4

.

for which

.

The circle shown in the diagram has centre and radius . a Write down the lengths of

and

in terms of and .

b Write down an expression for the area of the triangle

.

c Write down an expression for the area of the triangle

.

d Hence find the ratio of the two areas in the form

5

a Use the identity for b Write down the value of c Hence find the exact value of

6

to show that

where

.

.

. .

A water wave has the profile shown in the graph, where represents the height of the wave, in metres, and is the horizontal distance, also in metres.

a Given that the equation of the wave can be written as and . b

, find the values of

A second wave has the profile given by the equation

. Write down the

amplitude and the period of the second wave. When the two waves combine a new wave is formed, with the profile given by c Write the equation for in the form

, where

and

. .

d State the amplitude and the period of the combined wave. e Find the smallest positive value of for which the height of the combined wave is zero. f 7

Find the first two positive values of for which the height of the combined wave is

a Express in the form value of to the nearest .

, where

b Hence find the values of in the interval

and

.

, giving your

for which

giving your values of to the nearest degree. [© AQA 2013] 8

Simplify

.

Choose from these options. A B C D 9

a Use the identity for

to show that

b Hence solve the equation 10

a Write

as a product of a linear and a quadratic factor.

b Show that

11

.

c Write down the exact value of

.

d Hence find the exact value of

and

.

By forming and solving a quadratic equation, solve the equation in the interval , giving the values of in radians to three significant figures. [© AQA 2013]

12

a Show that the equation

can be written in the form

b Hence, or otherwise, solve the equation

giving all values of in radians to two decimal places in the interval

.

[© AQA 2012] 13

a Express

in the form

.

b The function is defined by

.

Using your answer to part a, find: i the maximum value of

, giving your answer in the form

where

ii the smallest value of for which this maximum occurs, giving your answer exactly, in terms of . 14

a Write down an expression for b Show that

. .

c Hence find an expression for 15

a i Expand

.

.

ii Hence show that

.

iii State the value of for which the equality holds. A picture of height

hangs on the wall so that the lower end of the picture is

floor. An observer sits on the floor at a distance angle , as shown in the diagram.

b Show that

above the

from the wall, and sees the picture at an

.

c Use the answer to part a to show that

.

d Hence find the value of that gives the largest possible value of

.

e State the distance the observer should stand from the wall in order to see the picture at the largest possible angle, and find the value of that angle.

Worksheet See Extension Sheet 8 for a selection of more challenging problems.

9 Calculus of exponential and trigonometric functions In this chapter you will learn how to: differentiate integrate

and and

review applications of differentiation to find tangents, normals and stationary points review applications of integration to find the equation of a curve and areas.

Before you start… Student Book 1, Chapter 7

You should be able to use rules of indices and logarithms.

1 Write

Student Book 1,

You should be able to

2

Chapter 12

differentiate

Student Book 1, Chapter 14

You should be able to integrate for .

3

Chapters 7, 8

You should be able to use compound angle formulae

4 Use small angle approximations to find

.

in the form

Differentiate

Find the exact value of

the approximate value of

.

.

.

and small angle approximations.

Extending differentiation and integration You have already seen many situations that can be modelled using trigonometric, exponential and logarithm functions. For example, in Student Book 1, Chapter 8, you studied exponential models for population growth, and in Chapter 7 of this book you saw how to use sine and cosine functions to model periodic motion, such as motion around a circle, the oscillation of a particle attached to the end of a spring, or the heights of tides. Often, you need to find the rate of change of quantities in such models; for example, at what rate is the population increasing after three years? Finding rates of change involves differentiation. Integration reverses this process – given the rate of change, you can find the equation for the original quantity. So far you have learnt how to differentiate and integrate functions of the form . In this section rules of differentiation and integration will be extended to include a wider variety of functions.

Section 1: Differentiation You already know, from Student Book 1, that the rate of growth of an exponential function is proportional to the value of the function. In particular, for the rate of growth equals the value. Key point 9.1 gives this in terms of differentiation.

Rewind In fact you already know, from Student Book 1, Chapter 8, that if

then

. You

will see how this arises in the next chapter.

Key point 9.1 If

then

The derivative of the natural logarithm function follows from the fact that of .

is the inverse function

Rewind You learnt about inverse functions and their graphs in Chapter 2, Section 4.

Key point 9.2 If

then

PROOF 6

Let point on the graph of have coordinate . Let point be the point on the graph of which is the reflection of in the line . The reflection swaps and coordinates, so has -coordinate .

The gradient of

at is .

From Chapter 2, since they are inverse ,

functions, you know how the graphs of and

are related. So you can use the

gradient of the first graph to find the gradient of the second graph.

You know that the gradient of equals the -value.

at

The gradient of the reflected line:

To see what happens to the gradient on reflection in , consider reflecting a triangle. This shows that, if a line with gradient is reflected in line is

The gradient of the graph

at is .

the gradient of the reflected

.



This is divided by the -coordinate of , so the gradient of

is .

The rules for differentiating sums and constant multiples of expressions still apply. Sometimes you will need to simplify an expression before differentiating. WORKED EXAMPLE 9.1

Differentiate

. You can only differentiate and , so you need to use rules of indices and logarithms to simplify the expression. is multiplied by a constant

.

is a

constant, so its derivative is . The derivative of

is

.

To differentiate and you need to remember differentiation from first principles, which you met in Student Book 1, Chapter 12. You also need to use compound angle formulae and small angle approximations.

Rewind You can recap compound angle formulae in Chapter 8, Section 1, and small angle approximations in Chapter 7, Section 6.

WORKED EXAMPLE 9.2

Use differentiation from first principles to prove that the derivative of measured in radians.

is

, where is

Use

.

Use small angle approximations:

Let tend to .

Differentiating

follows the same method. You can also prove the result for

in this way.

Fast forward You can also prove the result for

using the quotient rule, which you will meet in

Chapter 10.

Key point 9.3 If

then

If

then

If

then

Worked examples 9.3 and 9.4 should help you remember two applications of differentiation: finding equations of tangents and normals and finding stationary points.

Tip These formulae only apply if is in radians.

WORKED EXAMPLE 9.3

Find the equations of the tangent and the normal to the graph of the function the point . Give your answer in the form

, where

are integers.

You need the gradient, which is When

,

at

.

To find the equation of a straight line you also need coordinates of one point. The tangent passes through the point on the graph where Its -coordinate is . Tangent:

Put all the information into the equation of a line.

Normal:

The gradient of the normal is

and it

passes through the same point.

WORKED EXAMPLE 9.4

Find the coordinates of the stationary point on the graph of

and determine its

nature.

For stationary points,

First find

.

Rewrite

as

: and multiply through by

is only defined for

.

Find the -value. The stationary point is

Use laws of logs to rewrite the -coordinate. To determine the nature of the stationary point, evaluate the second derivative at .

The stationary point is a minimum.

EXERCISE 9A 1

Differentiate each function. a i ii b i

.

ii c i ii d i ii e i ii f

i ii

2

Simplify each function and then differentiate it. a i ii b i ii c i ii d i ii e i ii f

i ii

3

Find the exact value of the gradient of the graph of

at the point

4

Find the exact value of the gradient of the graph of

5

Find the value of where the gradient of

6

Find the value of where the gradient of

7

Find the rate of change of

8

Find the rate of change of

9

Find the equation of the tangent to the curve

that is parallel to

10

Find the equation of the tangent to the curve

that is parallel to

11

Given

12

Find the equations of the tangent and the normal to the graph of

when is

.

. is .

at the point

.

at the point

, find the values of for which

Give all the coefficients in an exact form.

.

. . . . at

.

13

Given that

14

Find and classify the stationary points on the curve

15

Show that the function

16

Find the range of the function

17

Find and classify the stationary points of:

, solve the equation

. in the interval

has a stationary point with –coordinate

. .

.

a b 18

.

The volume of water, in millions of litres, in a tidal lake is controlled by a dam. It can be modelled by , where is the time, in days, after the dam is first opened. a What is the smallest volume of the lake? b A hydroelectric plant produces an amount of electricity proportional to the rate of flow of water. Find the time, in the first days, when the plant is producing maximum electricity.



Section 2: Integration All the differentiation results from Section 1 can be reversed to integrate several new functions.

Key point 9.4

Note the modulus sign in

. This is because the in

cannot be negative, but the in

can. Taking the modulus means that you can find the area between the -axis and the negative part of the curve

. (You can see that this is the same as the corresponding region on the positive -axis.)

You can combine these facts with rules of integration you already know. Sometimes you will need to rewrite an expression in a different form before you can integrate it. WORKED EXAMPLE 9.5

Find

. You don’t know how to integrate a quotient, but you can split the fraction and integrate each term separately.

WORK IT OUT 9.1 Decide which statement is correct. Identify the errors made in the incorrect statements.

Statement 1

Statement 2

Statement 3

Statement 4

Remember that integration is the reverse of differentiation, so if you know the derivative of a function you can use integration to find the function itself. WORKED EXAMPLE 9.6

A curve passes through the point

and its gradient is given by

. Find the

equation of the curve. Integrate

to get .

Don’t forget the constant of integration. Using

:

Use given values of and to find the constant of integration.

Write the full equation.

You can also use integration to find the area between the curve and the -axis. This involves evaluating definite integrals. It is always a good idea to draw a diagram to make sure you are finding the correct area.

Rewind You know from Student Book 1, Chapter 14, that when a curve is below the x-axis the integral is negative.

WORKED EXAMPLE 9.7

Find the exact area enclosed between the -axis, the curve

and the lines

and

.

Sketch the graph and identify the required area.

Integrate and write in square brackets.

Evaluate the integrated expression at the upper and lower limits and subtract the value at the lower limit from the value at the upper limit.

WORK IT OUT 9.2 Three students are considering the problem, ‘Evaluate

’.

Discuss their solutions. What is the value of the area shaded in the second solution? Solution 1

Solution 2 This is the area between the graph and the -axis:

You need to calculate the two parts separately and add them up:

But And

is not defined, so there is no answer.

Solution 3 You are trying to integrate between domain of . So there is no answer.

EXERCISE 9B 1

Find each integral. a i ii b i ii c i ii d i ii e i ii f

i ii

2

Find each integral. a i ii b i ii

and , but this includes

, which is not in the

c i ii d

i ii

e i ii f

i ii

3

Find the exact value of each definite integral. a i ii b i ii c i ii

d i ii e i ii f

i ii

g i ii 4

Find the exact value of the area enclosed by the curve . Give your answer in the form

5

Find the area enclosed by the curve

, the -axis and the lines

. , the -axis and the line

.

and

6

Find the exact value of

7

Find the exact value of

8

a Evaluate

. .

.

b State the value of the area between the graph of

, the -axis and the lines

and

. 9

Find the equation of the curve given that

and the curve passes through the point

. 10

Find

.

11

The derivative of the function

is

.

a Find an expression for all possible functions b If the curve

.

passes through the point

find the equation of the curve.

Worksheet See Support Sheet 9 for a further example of finding the equation of a curve and for more practice questions. 12

The diagram shows part of the curve

.

Find the area of the shaded region, enclosed between the curve, the -axis and the lines

and

.

Fast forward Areas of regions that are partly above and partly below the x-axis are discussed in more detail in Chapter 12, Section 4.

13

The diagram shows a part of the curve with equation a Show that the curve crosses the -axis at

and

. .

b Find the exact value of the area of the shaded region. Give your answer in the form where and are rational numbers.

,

14

Show that the value of the integral

is independent of .

15

The gradient of the normal to a curve at any point is equal to the coordinate at that point. If the curve passes through the point find the equation of the curve in the form where is a rational function.

Checklist of learning and understanding The derivatives of basic trigonometric, exponential and logarithm functions are:

The results for

and

can be derived using differentiation from first principles.

The integrals of basic trigonometric, exponential and logarithm functions are:

The rules of differentiation and integration can be combined to find: rates of change equations of tangents and normals stationary points the equation of a curve with a given gradient the area between a curve and the -axis.



Mixed practice 9 1 Find . Choose from these options. A B C D 2

Find the equation of the tangent to the curve

3

If

4

i

Find the exact value of the gradient of the graph of

ii

Find the exact value of the gradient of the graph

and

at the point where

.

find

5

Find the exact value of the integral

6

The diagram shows the curve with equation

at the point when

.

.

, which crosses the -axis at and

.

Find the exact value of the shaded area. 7

Find

when

. [© AQA 2013]

8

A curve has equation i

Find

ii

Find

.

. . [© AQA 2006]

9

Find the x-coordinate of the stationary point on the curve Choose from these options. A B

, for

.

C D 10

Find the indefinite integral

.

11

Find and classify the stationary points on the curve

in the interval

. 12

Find and classify the stationary points on the curve

for

.

Give only the -coordinates, and leave your answers in terms of . 13

Find the equation of the normal to the curve Give your answer in the form

14

at the point where

and are integers.

The population of bacteria, in thousands, at a time , in hours, is modelled by

a i Find the initial population of bacteria. ii At what time does the number of bacteria reach b i Find

million?

.

ii Find the time at which the bacteria are growing at a rate of million per hour. c i Find

and explain the physical significance of this quantity.

ii Find the minimum number of bacteria, justifying that it is a minimum. 15

The diagram shows the graphs of

and

.

Find the exact area of the shaded region.

16

Find

.

Worksheet See Extension Sheet 9 for a selection of more challenging problems.

10 Further differentiation In this chapter you will learn how to: use the chain rule to differentiate composite functions differentiate products and quotients of functions work with implicit functions and their derivatives differentiate inverse functions.

Before you start… Chapter 9; Student Book 1, Chapter 12

You should be able to differentiate the functions .

1 Differentiate these expressions. a b c d

Chapter 9; Student Book 1, Chapter 13

You should be able to use differentiation to find the equations of tangents, normals and stationary points.

2 A curve has equation

.

a Find the equations of the tangent and the normal at the point where . b Find the coordinates of the stationary point and show that it is a minimum point.

Student Book 1, Chapter 10

You should know basic trigonometric identities.

3 Simplify these expressions. a b

Chapter 8

You should know the definitions of reciprocal trigonometric functions.

4 Write these in terms of

and

a b

Chapter 5

You should be able to simplify expressions involving fractions and surds.

5 Simplify each expression.

a

b

GCSE; Student Book 1,

You should be able to change the subject of a formula.

6 Make the subject of each formula. a

.

b

Chapter 7

Differentiating more complex functions In chapter 9 you learnt how to differentiate a variety of functions. You can also differentiate sums, differences and constant multiples of those functions, for example, . The next question to ask is: How can you differentiate products and quotients of functions, for example, or

?

You know from Student Book 1, Chapter 12, that you cannot just differentiate the separate components of a product (or quotient) and then multiply (or divide) the results, so you need a new rule for these situations. Before looking at products and quotients you will learn how to differentiate composite functions, such as or . You will also learn how to differentiate equations such as subject, and apply this method to differentiating inverse functions.

without making the

Section 1: The chain rule To differentiate

you could expand the brackets and differentiate term by term. But

often this is either too difficult or not possible; for example,

cannot be ‘expanded’ and

would lead to a very long expression. What about functions such as or You can differentiate no rules, so far, telling you what to do when is replaced by or .

and

, but you have

These functions may seem quite different but they do have something in common – they are all composite functions.  

where

      

where

       

where

To differentiate a composite function, think about how a change in affects . If changes by causes to change by you can see that:

which in turn causes to change by

this

. If you multiply the two rates of change

This leads to the rule for differentiating composite functions.

Key point 10.1 The chain rule states that if

where

, so that

, then

WORKED EXAMPLE 10.1

Differentiate

.

where

This is a composite function.

Use the chain rule.

Write the answer in terms of . There is no need to expand the brackets.

WORKED EXAMPLE 10.2

Differentiate where

. This is a composite function. Use the chain rule.

Write the answer in terms of and rearrange.

Rewind For a reminder of how to differentiate

and

see Chapter 9, Section 1.

WORKED EXAMPLE 10.3

Differentiate

.

where

This is a composite function. Use the chain rule.

Write the answer in terms of and rearrange.

Tip Once you get used to using the chain rule, you won’t need to write the ‘inner function’ each time. In particular, since the derivative of is just , you can now immediately recognise that:

Sometimes you need to apply the chain rule more than once. WORKED EXAMPLE 10.4

Differentiate

. Remember that

where

and

means

.

This is a composite of three functions. Use the chain rule with three derivatives.

Write everything in terms of and simplify.

You can use the chain rule to differentiate reciprocal trigonometric functions. WORKED EXAMPLE 10.5

WORKED EXAMPLE 10.5

Show that Express where

in terms of

, as you know how to differentiate that.

This is a composite function… …so apply the chain rule.

The answer contains

, which is

.

The proofs for the other two reciprocal trigonometric functions follow the same pattern, giving the results shown in Key point 10.2.

Key point 10.2

These will be given in your formula book.

EXERCISE 10A 1

Use the chain rule to differentiate each expression with respect to . a i ii b i ii c i ii d i ii e i ii f

i

ii g i ii h i ii i

i ii

j

i ii

k i ii l

i ii

m i ii n i ii 2

Use the chain rule to differentiate each expression with respect to . a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii h i ii 3

Differentiate each expression, using the chain rule twice. a i ii

b i ii c i ii d i ii 4

Find the equation of the tangent to the graph of

5

Find the equation of the normal to the curve

6

For what values of does the function

7

Find the coordinates of the stationary points on the curve with equation

8

Find the coordinates of the stationary points on the graph of

9

at the point where

.

at the point where

.

have a gradient of ? . .

A particle moves in a straight line with displacement at time t from O given by Find, at time

.

seconds:

a the velocity of the particle b the acceleration of the particle.

Rewind You met kinematics in Student Book 1, Chapter 16.

10 11 12

Find the exact coordinates of stationary points on the curve

for

Find the exact coordinates of the stationary point on the curve

.

A population of bacteria increases exponentially so that the number of bacteria, , after minutes is given by

.

Find the size of the population at the moment when its rate of increase is 13

.

Given that

bacteria per minute.

:

a find b solve the equation 14

for

A non-uniform chain hangs from two posts. Its height above the ground satisfies the equation . The left post is positioned at

, and the right post is positioned at

.

a State, with reasons, which post is taller. b Show that the minimum height occurs when

.

c Find the exact value of the minimum height of the chain. 15

a Solve the equation

for

, giving your answers in terms of .

b Find the coordinates of the stationary points of the curve your answers correct to three significant figures.

for

, giving

c Hence sketch the curve

for

.

Did you know? Many people think that a chain fixed at both ends will hang as a parabola, but it can be proved that it hangs in the shape of the curve in question 14, called a catenary. The proof requires techniques from a mathematical area called differential geometry.



Section 2: The product rule There is also a rule for differentiating products, such as

or

Key point 10.3 The product rule states that if

then

WORKED EXAMPLE 10.6

Differentiate: a b

.

a Let

and

Then

It doesn’t matter which function you call and which .

.

and Applying the product rule.

So

b Let

and

Then

is a product of two functions, so use the product rule.

.

and

So

WORK IT OUT 10.1 Differentiate

.

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

Solution 2

Solution 3

When differentiating a more complicated product, you may need to use the chain rule as well as the product rule. When the function involves powers, the two terms in the product rule often have a common factor. WORKED EXAMPLE 10.7

Differentiate Let

and

and factorise your answer. This is a product, so use the product rule.

is a composite function, so use the chain rule.

Now apply the product rule.

You are asked to factorise the answer, so instead of expanding the brackets look for common factors.

EXERCISE 10B

EXERCISE 10B 1

Differentiate each function. Use the product rule. a i ii b i ii c i ii d i ii

2

Find

and fully factorise your answer.

a i ii b i ii 3

Differentiate

4

Given that

5

Find the -coordinates of the stationary points on the curve

6

Find the exact values of the -coordinates of the stationary points on the curve

7

Find the derivative of

8

a Given that b Hence find

9

, giving your answer in the form , find

in the form

Given that

11

a Write

.

find

.

.

show that in the form

.

where and are constants to be found.

. .

c Find the exact coordinates of the stationary points of the curve and

.

with respect to .

b Hence or otherwise find

a If

is a polynomial.

.

Find the exact coordinates of the minimum point of the curve

10

12

where

.

are positive integers find the -coordinate of the stationary point of the curve in the domain .

b Sketch the graph in the case when

and

.

c By considering the graph or otherwise, determine a condition involving and/or to determine when this stationary point is a maximum.



Section 3: Quotient rule To differentiate a quotient such as

you can express it as:

and use a combination of the chain rule and the product rule. However, there is a shortcut that allows you to differentiate quotients directly.

Key point 10.4 The quotient rule states that if

then

This will be given in your formula book.

WORKED EXAMPLE 10.8

Differentiate

. Use the quotient rule and simplify as far as possible. and

Use the quotient rule, making sure to get and the right way round.

.

Use the chain rule to differentiate .

Notice that a factor of

In Chapter 9, it was stated that the derivative of is use the quotient rule, together with the derivatives of

can be cancelled.

Worked example 10.9 shows how you can and , to prove this result.

WORKED EXAMPLE 10.9

Prove that

. You know how to differentiate . Use the quotient rule.

and

, so use

Notice that

WORK IT OUT 10.2 Differentiate

.

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

So:

Solution 2

So:

Solution 3

The quotient rule, just like the product rule, often leads to a long expression. If you are required to simplify it, you may need to work with fractions and roots, as in Worked example 10.10. WORKED EXAMPLE 10.10

Differentiate

giving your answer in the form

where

.

Use the quotient rule.

As you want a square root in the answer, turn the fractional powers back into roots. Remove fractions within fractions by multiplying top and bottom by . Notice that

.

Tip You only need to use the quotient rule when differentiating a quotient of two functions. If a quotient is made up of a constant and a function you can do it more simply. For example, is just

, which you can differentiate by using the chain rule.

Worksheet See Support Sheet 10 for a further example of combining the chain rule with the product or quotient rule and some more practice questions.

EXERCISE 10C

EXERCISE 10C 1

Differentiate each equation. Use the quotient rule. a i ii b i ii c i ii d i ii

2

Find the equation of the normal to the curve in the form

at the point where

, giving your answer

, where and are exact.

3

Find the coordinates of the stationary points on the graph of

4

The graph of

5

Find the exact coordinates of the stationary point on the curve

6

Find the range of values of for which the function

7

Given that

has gradient at the point

show that

.

and

. Find the value of . and determine its nature. is increasing.

, stating clearly the value of the constants

. 8

Show that if the curve

has a maximum stationary point at

has a minimum stationary point at



as long as

.

then the curve

and

Section 4: Implicit differentiation The functions you have differentiated so far have always been of the form . From time to time you will come across functions that are written differently. For example, the equation of the circle shown in the diagram is . Such functions are said to be implicit whereas those in the form are said to be explicit. The gradient of the tangent at any point on the circle is still given by

. Rather than attempting to

rearrange the equation to make it explicit, you can just differentiate term by term with respect to :

You need to take care when you come to rule:

as it is not a function of here you will need the chain

This will generally be the case when differentiating terms involving .

Key point 10.5 When differentiating implicitly, you need to use:

WORKED EXAMPLE 10.11

Find an expression for

for the circle

.

Differentiate each term. Use the chain rule on the term involving .

Now rearrange the equation to find

.

Notice that the expression for

will often be in terms of both and

Sometimes you may need to use the product rule as well as the chain rule in this process of implicit differentiation. WORKED EXAMPLE 10.12

Find an expression for

if Differentiate term by term, using the chain rule on all terms involving . is a product, so use the product rule together with the chain rule on all terms involving . Group the terms involving

.

If you are only interested in the gradient at a particular point, or are given the gradient and need to find the -and -coordinates, you can substitute the given value into the differentiated equation without rearranging it. WORKED EXAMPLE 10.13

For the curve with equation

:

a find the gradient at the point b find the coordinates of the point where the gradient is . Differentiate each term with respect to . The term will require the product rule. Use the chain rule on terms involving .

Group terms involving a When

and

b When

:

:

.

Substitute in the numbers before rearranging.

Put in the given value of the gradient.

This is a second equation relating and . You can solve it simultaneously with the original equation. Substitute into

Usin

, the coordinates are

Remember to find both - and - coordinates.

and

You can also use the method in part b to find stationary points. WORKED EXAMPLE 10.14

Find the coordinates of the stationary points on the curve

.

Differentiate each term with respect to but be aware that the term will require the product rule. Use the chain rule on all terms involving .

For stationary points,

You know the value of

When

You have found a relationship between and at the stationary points but to find the points substitute back into the original function.

:

.

is a stationary point. When

:

is a stationary point.

One application of implicit differentiation is to differentiate exponential functions with base other than .

Key point 10.6

PROOF 7

Let Then

As this is an exponential expression, take ln of both sides.

Now use implicit differentiation. Remember that

is a constant.

Rewind This confirms that the gradient of any exponential function is proportional to the -value, as you learnt in Student Book 1, Chapter 8.

EXERCISE 10D 1

Find the gradient of each curve at the given point. a i

at at

ii b i

at at

ii c i

at at

ii

2

d i

at

ii

at

Find

in terms of and .

a i ii b i ii c i ii d i ii 3

Find the coordinates of stationary points on the curves given by each implicit equation. a b

4

Find the exact value of the gradient at the given point. a i

at

ii

at

b i ii

when when

c i

when

ii

when

d i

when

ii

when

5

Find the gradient of the curve

6

A curve has equation a Show that

at the point

.

. .

b Find the equation of the tangent to the curve at the point 7

A curve has equation

. Point has coordinates

.

a Show that point lies on the curve. b Find the equation of the normal to the curve at . 8

Find the gradient of the curve with equation

9

Find the equation of the tangent to the curve with equation

10

Find the equation of the tangent to the curve with equation

11

A curve has implicit equation

12

Find the coordinates of the stationary point on the curve given by

13

The line is tangent to the curve which has the equation

. Find an expression for

at the point at the point at the point in terms of and . . when

and

.

a Find the equation of . b Show that meets again at the point with an -coordinate which satisfies the equation . c Find the coordinates of the point .



.

Section 5: Differentiating inverse functions Another application of the chain rule is to differentiate inverse functions. If derivative of is

and the derivative of

is

In Chapter 9, the result for the derivative of is the inverse function of .

then

. But the chain rule says that:

was just stated. You can prove it by using the fact that

Key point 10.7 The derivative of the inverse function is:

WORKED EXAMPLE 10.15

Use the derivative of to prove that the derivative of Let Then

is .

Rewrite in terms of (the inverse function of ln).

. .

Now differentiate with respect to . Use the inverse function rule. But

, so

The answer needs to be in terms of .

Fast forward If you study Further Mathematics you will use this same method to differentiate inverse trigonometric functions.

EXERCISE 10E

. The

EXERCISE 10E 1

If you prefer, you can differentiate inverse functions by using implicit differentiation. For example, given that : a express in terms of b use implicit differentiation to find

2

If

3

If

, find

; write your answer in terms of .

in terms of .

, find the exact value of

at the point where

, giving your answer in the form

, where is an integer. 4

Given that

:

a by considering

, prove that

has an inverse function

b find the gradient of the graph of 5

6

a Given that

at the point where

write down an expression for

b Hence prove that

.

A function is defined by

.

.

.

a Prove that f is an increasing function. b Show that the point c 7

Find the exact value of the gradient of the graph of

Given that a find

. at the point

:

in terms of

b hence find 8

lies on the graph of

Given that

in terms of . is an increasing function, prove that

is also an increasing function.

.

Checklist of learning and understanding The chain rule is used to differentiate composite functions. If

where

then:

The product rule is used to differentiate two functions multiplied together. If

then:

The quotient rule is used to differentiate one function divided by another. If then:

The derivatives of the reciprocal trigonometric functions are:

The derivative of an exponential function is:

To differentiate functions given implicitly: differentiate each term use the chain rule for any term containing

Differentiation of inverse functions:





Mixed practice 10 ;

1

Find the value of the constant Choose from these options. A B C D 2 Find

at the point where

. Choose from these options.

A B C D 3

a Find

when:

i ii b Hence find

for

, giving your answer in the form

, where

and

are integers to be found. 4

Find the exact value of the gradient of the curve with equation

5

A curve has equation

when

.

.

a Show that the point

lies on the curve.

b Find the gradient of the curve at . 6

The curve has equation

. The tangent to the curve at the point

the coordinate axes at points and . Find the area of the triangle 7

Find the exact gradient of the curve with equation

8

A curve has equation a Find

crosses

.

at the point where

.

.

.

b i Find an equation of the tangent to the curve .

at the point on the curve where

ii This tangent intersects the -axis at the point . Find the exact value of the coordinate of the point . [© AQA 2012] 9 10

The graph of

has a stationary point when

. Find the value of .

Find the exact coordinates of the stationary point on the curve with equation

.

11

a Find the value of so that

.

b Hence show that the derivative of

is

.

Rewind Compare the derivation in Question 10 with the one using implicit differentiation shown in Proof 7.

12

A particle moves in a straight line with velocity given by

, where

. Find, as an exact value, the minimum value of the velocity during the motion, fully justifying your answer. 13

At time seconds, the displacement of a particle from is

.

a Find the value of t when the particle is instantaneously at rest. b Find the particle’s maximum speed for 14

A curve has equation

, fully justifying your answer.

.

a Write down the equation of the vertical asymptote of the curve. b Use differentiation to find the coordinates of the stationary points on the curve. c Determine the nature of the stationary points. d Sketch the graph of 15

.

a Prove that the derivative of

is

.

b Find the equation of the tangent to the graph of 16

A function is defined by a Find

and hence prove that

b Find the gradient of 17

for

at the point where

.

.

has an inverse function.

at the point

.

A curve is defined by the equation

.

Find the coordinates of the two stationary points of this curve. [© AQA 2012] 18

A curve is given by the implicit equation

.

a Find the coordinates of the stationary points on the curve. b Show that, at the stationary points,

.

c Hence determine the nature of the stationary points. 19

a Let

. By first expressing in terms of , prove that

b Find the equation of the normal to the curve 20

a Given that

, find

.

at the point where

. Give your answer in the form

where and

are integers. The diagram shows a part of the graph of the curve

.

.

b Find the coordinates of the stationary point on the curve. c Find the shaded area enclosed by the curve, the -axis and the lines

and

Worksheet See Extension Sheet 10 for questions on some properties of e that use differentiation.



.

11 Further integration techniques In this chapter you will learn how to integrate using: known derivatives the chain rule in reverse a change of variable (substitution) the product rule in reverse (integration by parts) trigonometric identities the separation of a fraction into two fractions.

Before you start… Chapter 9

You should be able to differentiate and integrate polynomial, exponential and trigonometric functions.

1 Find: a .

b 2 Given that

, find:

a b Chapter 10

You should be able to use the chain rule for differentiation.

.

3 Differentiate: a b

Chapter 8

You should be able to use double angle formulae.

4 Given that possible values of

Chapter 2, Chapter 5

You should be able to split an expression into partial fractions.

5 Write

find the

in partial

fractions.

Integrating more complex functions Having extended the range of functions you can differentiate, you now need to do the same for integration. In some cases you will be able to use the results from Chapter 10 directly, but in many others you will require new techniques. In this chapter you will look at each of these in turn and then you will face the challenge of selecting the appropriate technique from the not inconsiderable list of options you have built up.

Section 1: Reversing standard derivatives You already know how to integrate many functions, by reversing the corresponding differentiation results.

In Chapter 10 you differentiated too, and add these results to your list.

and

. You can now reverse these standard derivatives,

Thinking about reversing the chain rule for differentiation allows you to go one step further and deal with integrals such as

.

You know that the answer must include you differentiate

:

Since you do not want the in front of

as sin is the integral of

, but this is not the final answer. If

. , divide by (or multiply by ).

WORKED EXAMPLE 11.1

Find

. integrates to

but, differentiating back,

the chain rule will also give a factor of (the derivative of multiplying by .

WORKED EXAMPLE 11.2

Find

.

). Remove this by

integrates to

but, differentiating back,

the chain rule will also give a factor of (the derivative of

). Remove this by

multiplying by .

You may notice a pattern here: you always end up integrating the function and then dividing by the coefficient of . This is indeed a general rule when the ‘inside’ function is of the form .

Key point 11.1

where

is the integral of

Tip Note that this rule only applies when the ‘inside’ function is of the form

.

WORKED EXAMPLE 11.3

Find

. Integrate

to

and divide by the

coefficient of .

WORK IT OUT 11.1 Find

. Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution A

Solution B

Solution C

WORKED EXAMPLE 11.4

Find

.

Integrate to coefficient of .

EXERCISE 11A 1

Find each indefinite integral. a i ii b i ii c i ii d i ii e i

ii 2

Work out each integral. a i ii b i ii c i ii d i

and divide by the

ii 3

Find each integral. a i ii b i ii c i ii d i ii

4

Integrate each expression. a b c d e f

5

Find the exact value of

6

Find the exact area enclosed by the graph of

7

Find the area enclosed by the -axis and the curve with equation

8

Given that is



. , the -axis and the lines

and the area between the -axis, the lines , find the value of correct to three significant figures.

and .

and the graph of

.

Section 2: Integration by substitution The shortcut for reversing the chain rule (Key point 11.1) works only when the derivative of the ‘inside’ function is a constant. This is because a constant factor can ‘move through the integral sign’, for example:

Fast forward Another method for integrating products is integration by parts, which is the reverse of the product rule. You will meet this in Section 3.

This cannot be done with a variable:

is not the same as

. So you need a different

rule for integrating a product of two functions. In some cases this can be achieved by extending the principle of reversing the chain rule, leading to the method of integration by substitution.

Reversing the chain rule When using the chain rule to differentiate a composite function, you differentiate the outer function and multiply this by the derivative of the inner function; for example:

Rewind See Chapter 10, Section 1, for a reminder of the chain rule for differentiation.

You can think of this as using a substitution Look now at

. Since

. Thus the integral becomes You know how to integrate instead of

, and then

.

is a composite function it can be written as

, where

.

, but you need to integrate with respect to , so you should have

. Although du and dx are not the same thing, they are related because

Tip When making a substitution of this type, only replace the inner function with . Any other instances of will cancel out when changing to .

You can then rearrange this to make the subject so that it can be replaced in the integral:

Substituting all of this into the integral gives:

.

It follows from the chain rule that

so:

Notice that having found an answer in terms of u, you need to put it back in terms of . In practice, this method can be shorted by appreciating that gives

can be split up just like a fraction, so

.

This is illustrated in Worked examples 11.5 and 11.6. WORKED EXAMPLE 11.5

Find each integral.  a b a Let

.

Think of

as

function is

Then

; therefore the inner

.

Make the substitution.

Write the answer in terms of . b Let Then

.

is a composite function with inner function

.

Make the substitution.

Write the answer in terms of .

When limits are given, you must ensure they are changed too. In this case there is no need to change

back to the original variable at the end. WORKED EXAMPLE 11.6

Evaluate

, giving your answer in the form

Let

.

The ‘inner’ function is

.

Then Limits:

Write the limits in terms of . Make the substitution. Simplify:

This particular case of substitution, in which the top of the fraction is the derivative of the bottom, is definitely worth remembering:

Key point 11.2

You can use this result to integrate some trigonometric functions.

Tip You do not have to make a substitution if you can see that the expression is of the form , i.e. with both a function and its derivative present. You can just go straight to the answer, using the reverse chain rule:

WORKED EXAMPLE 11.7

Show that

. If

then

.

So you can write the integral in the form .

Use the result of Key point 11.2. The answer is in terms of sec , so remember that

EXERCISE 11B

and use

.

EXERCISE 11B 1

Use either a suitable substitution or the reverse chain rule to find each integral. a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii h i ii i

i ii

2

Find

.

3

Find the exact value of

4

Show that

5

Find

6

Find

, showing all your working. where is an integer to be found.

.

General substitution

.

In all the examples so far, after the substitution, the part of the integral that was still in terms of cancelled with a similar term coming from

. For example, in Worked example 11.5 part b,

. This will always happen when one part of the expression to be integrated is an exact multiple of the derivative of the inner function. In some cases the remaining -terms will not cancel and you will have to express in terms of . The full method of substitution will then be as summarised in Key point 11.3.

Key point 11.3 Integration by substitution: 1 Select a substitution (if not already given). 2 Differentiate the substitution and write 3 Replace

in terms of

.

by the expression in 2, and replace any obvious occurrences of .

4 Change the limits from to . 5 Simplify as far as possible. 6 If any terms with remain, write them in terms of . 7 Do the new integral in terms of . 8 Write the answer in terms of .

Tip If you are not told which substitution to use, look for a composite function and take ‘inner’ function.

WORKED EXAMPLE 11.8

Find

, using the sustitution

. Differentiate the substitution and write in terms of (step 2).

Replace parts that you have expressions for, and simplify if possible (steps 3 to 5).

There is still an remaining, so replace it by using

(step 6).

Now everything is in terms of so you can integrate (step 7). Remember that .

Write the answer in terms of , using (step 8).

A substitution can be given as in terms of

, rather than in terms .

WORKED EXAMPLE 11.9

Use the substitution

(with

) to evaluate

.

Differentiate the substitution. Note: now you are using

.

Find limits for . This involves solving an equation. Replace that

by .

and by

. Use the fact

Simplify if possible. Now everything is in terms of so you can integrate. Remember the modulus sign with the ln.

EXERCISE 11C 1

Find each integral. Use the given substitution. a i     ii b i ii

2

Use the given substitution to find each integral. a i ii b i ii

3

Find each integral. Use an appropriate substitution. a i

ii b i ii c i ii 4

Use the given substitution to evaluate each definite integral. a i ii b i ii c i ii

5

Use the substitution

6

a Show that

to find

is a factor of

. .

b Find



7

Use the substitution

8

Show that

9

By using the substitution

to find

. where

, find the exact value of

and are integers to be found. .

Section 3: Integration by parts In Section 2, you saw cases in which products of functions can be integrated by using the reverse chain rule or a substitution. But you still cannot work out integrals such as

or

.

In order to integrate these, return to the product rule for differentiation:

Integrating with respect to you get:

Tip This working shows that integration by parts can be thought of as the ‘reverse product rule’. In practice, though, while the idea of integrating using the ‘reverse chain rule’ is used, the ‘reverse product rule’ is not used; the parts formula is used instead.

Key point 11.4 The integration by parts formula is:

This will be given in your formula book.

WORKED EXAMPLE 11.10

Find

. This is a product to which the reverse chain rule cannot be applied, so try integration by parts. Apply the formula.

In Worked example 11.10, taking to be worked well because formula the resulting integral was just

is a constant, so after applying the

. You could only do this because

integrate, so you could find . In some examples this is not possible.

WORKED EXAMPLE 11.11

is easy to

WORKED EXAMPLE 11.11

Find

. This is a product, so use integration by parts. You cannot take

because you don’t know

how to integrate

, so try choosing

them the other way round. Apply the formula. Always simplify before integrating.

The strategy for choosing which function is and which is

is summarised in Key point 11.5.

Key point 11.5 When using integration by parts for

, choose

in all cases except when

You can even use this strategy to integrate

by itself.

WORKED EXAMPLE 11.12

Use integration by parts to find

. The trick is to write as a product of and so that you can use integration by parts. As suggested in Key point 12.5, let

Apply the formula.

EXERCISE 11D

.

EXERCISE 11D 1

Find each integral. Use integration by parts. a i ii b i ii c i ii d i ii

2

Use integration by parts to evaluate

3

Find the exact value of

. .

Repeated integration by parts It may be necessary to use integration by parts more than once. As long as the integrals are becoming simpler each time, you are on the right track!

WORKED EXAMPLE 11.13

Find the exact value of This is a product to which you cannot apply the reverse chain rule, so try integration by parts. Choose to be the polynomial. Apply the formula. Put in the limits on the

part straight away.

You have to integrate a product again, so use integration by parts for the second time. Choose to be the polynomial again. (Notice that if you were to change and choose you would end up back where you started!) So,

Apply the formula again and use the limits.

Therefore,

Put both integrals together, making sure to keep track of negative signs by using brackets appropriately.

Common error When using integration by parts twice, remember to apply the negative sign in front of the integral to all of the second application of the parts formula.

EXERCISE 11E

EXERCISE 11E 1

Use integration by parts twice to find each integral. a b c d

2

Use integration by parts to find each integral. a b

3

Evaluate each interval exactly. a b

4

When using the integration by parts formula, you start with

and find . Why not include a

constant of integration when you do this? Try a few examples adding happens. 5

Find

6

Evaluate

.

7

Evaluate

.

8

a Show that

to and see what

.

.

b Hence find 9 10

Use the substitution Let

to find the exact value of

.

a Use integration by parts twice to show that b Hence find



.

.

.

Section 4: Using trigonometric identities in integration The most common use of trigonometric identities in integration is integrating the squares of trigonometric functions. WORKED EXAMPLE 11.14

Find

. You can integrate .

In Section 2 you integrated expressions such as derivative

cancels with the

by using the substitution

. But if you try to integrate just

not work. You could try rewriting it as

, so use

; the

the same substitution does

, but you don’t know how to do this either.

The trick is to notice that appears in one of the versions of the double angle formulae for , and you know how to integrate . This leads to a method for integrating both and

, which you need to remember.

Key point 11.6 To integrate

use

To integrate

use

.

Rewind Double angle identities were covered in Chapter 8, Section 2.

WORKED EXAMPLE 11.15

Find

. You can find an alternative expression for by using a double angle identity.

Remember to divide by the coefficient of when integrating

.

Sometimes trigonometric identities turn out to be useful in integrals that do not appear to involve

trigonometry. WORKED EXAMPLE 11.16

Use the substitution

to find

.

Make the substitution. Simplify each part separately first. Use Differentiate the substitution.

Limits:

Find the limits for .

So:

Simplify fully before attempting to integrate.

Now use the method from Key point 11.6:

Remember to divide by when integrating .

In integrals of this type, you may need to use trigonometric identities again to write the answer in terms of at the end.

Tip The substitution in Worked example 11.16 is useful in other integrals involving expressions like .

WORKED EXAMPLE 11.17

a Show that b Use the substitution

. to find

a

There is no obvious identity linking the functions on the left to the ones on the right. So write everything in terms of sin and cos.

b

Differentiate the substitution and express in terms of .

.

Replace those parts that you already have expressions for. There are no instances of remaining, so you can integrate. Notice that

Using the result from part a, you now have a standard derivative (Section 11.1). To express the answer in terms of you need to link to .

WORK IT OUT 11.2 Three students integrate

in three different ways.

Which is the correct solution? Identify the errors made in the incorrect solutions.

Amara uses reverse chain rule with

Ben uses reverse chain rule with

:

:

Carlos uses a double angle formula:

EXERCISE 11F 1

Simplify to get a standard integral, and then integrate. a b c d e

2

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate. a b c d e

3

Find each integral. a i ii b i ii

4

Find the exact value of each of the following. a i

ii

b i ii 5

Use the given substitution to find each of the following. a i ii

b i ii

c i ii 6

Find

7

a Show that

.

b Hence find

8

. .

Given that

for some positive , find the least value of that satisfies the

equation. 9

a Use the formula for

to show that

.

b Hence find 10

Use the substitution

11

Use the substitution

12

a Show that

to find to find .

b Hence find the exact value of 13

.

.

The diagram shows a part of the circle with centre at the origin and radius . a Write the equation of the curve in the form b Use integration to prove that the shaded area is

. .



Section 5: Integrating rational functions You can already use Key point 11.4 to work out integrals such as

where the derivative of the

denominator is a multiple of the numerator. You can do this with a substitution or directly with the reverse chain rule. If the numerator or the denominator is changed just a little, this may not work any more. For example, using

in

gives

, so the s do not cancel.

An alternative method is to split the fraction into two. You can do this either by splitting the numerator, or by using partial fractions. You will look at partial fractions first.

Rewind See Chapter 5, Sections 3 and 4, for a reminder of partial fractions.

WORKED EXAMPLE 11.18

Use partial fractions to find

in the form

.



Comparing coefficients of



Use

.

Remember that you can also use partial fractions with a non-linear denominator. WORKED EXAMPLE 11.19

Find the exact value of

. There are three fractions.

and

.

Remember the modulus signs when integrating .

To integrate

write it as

.

Before starting the complicated partial fraction process, it is worth checking whether you can simplify the fraction. WORKED EXAMPLE 11.20

Find

. It is generally a good idea to check whether polynomials factorise.

You now have a standard integral, just remember to divide by the coefficient of .

You will now look at improper fractions. You can integrate these by splitting them into a polynomial plus a proper fraction, or by splitting a numerator so that one part cancels with the denominator.

Rewind See Chapter 5, Section 2, for a reminder of division of improper rational functions.

WORKED EXAMPLE 11.21

Find: a b a

b

. Polynomial division gives quotient remainder .

and

In this case you can perform polynomial division informally by splitting the numerator in a convenient way.

EXERCISE 11G 1

State whether each integral can be done by using a substitution (or reversing the chain rule). For those that can, carry out the integration. a b c d e f g

2

By first simplifying, find each integral. a b c d

3

Find each integral by splitting into partial fractions. a i ii b i ii c i ii d i ii e i

ii 4

Find each integral by first performing polynomial division (or splitting the numerator). a i ii b i ii

5

Find the remaining integrals from question .

6

a Write

as a sum of partial fractions.

b Hence find 7

giving your answer in the form

Find the exact value of

.

.

Worksheet See Support Sheet 11 for further examples of integrating rational functions and for more practice questions.

8

Find the exact value of

9

a Split

into partial fractions.

b Given that 10

Find the exact value of rational numbers.

.

, find the value of . . Give your answer in the form

, where and are

Checklist of learning and understanding When using integration by substitution remember to: differentiate the substitution and express

in terms of

simplify the resulting expression find the limits for . Two special cases of integration by substitution should be remembered: where

is the integral of

Integration by parts can be used for some products:

In integrals of the form

, take

unless

.

You may need to do integration by parts more than once. Some integrals can be simplified by using a trigonometric identity. The particularly useful ones to remember are: for

for

and

use a rearrangement of the double angle formula:

use the substitution

.

Some rational fractions can be integrated by splitting them into partial fractions. Look out for improper fractions where you may be able to split the numerator



Mixed practice 11 1

Find

.

Choose from these options. A B C D 2

Find the exact value of

.

3

Use integration by parts to find

4

Given that

5

Find the exact value of

6

Find each integral.

.

calculate, to three significant figures, the value of . .

a b 7

By using integration by parts, find

. [© AQA 2012]

8

a Given that

, find

b Hence find

.

, giving your answer in the form

, where and are

rational numbers. [© AQA 2012] 9

Find an expression in terms of and for

.

Choose from these options. A B C D 10

Find

11

a Simplify

.

b Hence find 12

Use the substitution

. . to find

.

13

a Write

in partial fractions.

b Hence find, in the form 14

Find

15

Using the substitution

16

a Given that

, the exact value of

.

. , or otherwise, find

.

can be written in the form

, where and are

integers, find the values of and . b Hence or otherwise, find

. [© AQA 2006]

17

Use the substitution

to find the exact value of

. [© AQA 2014]

18

A particle moves in a straight line with acceleration given by

When

is at rest.

Find the distance travels in the first 3 seconds of its motion. 19

Find

.

20

Given that

21

a Use a suitable substitution to find

, find the exact value of . .

b The ellipse shown in the diagram has equation

.

Use integration to prove that the area enclosed by the ellipse is 22

Let a Find

and

.

.

b By using a substitution

find

.

.

c Hence find

.

Worksheet See Extension Sheet 11 for a selection of more challenging problems.

12 Further applications of calculus In this chapter you will learn how to: use the second derivative to determine the shape of a curve use a parameter to describe curves calculate rates of change of related quantities find the area between two curves, or between a curve and the -axis.

Before you start… Chapter 10

You should be able to find the first and second derivatives of various

1 Differentiate each function. a

functions including using the chain, product and quotient rules. Chapter 8

You should be able to use trigonometric identities to simplify

b 2 Solve the equation .

for

expressions and solve equations. Student Book 1, Chapter 14

Chapter 11

You should be able to find the area between a curve and the axis.

3 Find the area between the -axis and the graph of , between

You should be able to integrate

4 Integrate these expressions.

and

various functions, use substitution and integration by parts.

.

a b c

More uses of differentiation and integration In this chapter you will use some of the ideas you have already met: from differentiation to establish further properties of curves and to look at rate of change calculations that involve more than one variable from integration to calculate more complex areas between curves. You will be introduced to a new way of describing the equation of a curve, which has applications in mechanics and engineering, for example, in circular motion and projectile motion.

Worksheet See Extension Sheet 12 for some uses of differentiation in economics.



Section 1: Properties of curves You already know that the first derivative tells you whether a function is increasing or decreasing. But an increasing function can have two different shapes:

Similarly, a decreasing function can do so in two ways:

The curves and both curve upwards; they are called convex curves. The curves and , which both curve downwards, are called concave curves. The two convex curves have something in common: Their gradients are increasing (curve has negative gradient, which becomes less negative as you move to the right). So the rate of change of gradient is positive. But this rate of change is measured by the second derivative, so

for curves and . Similarly for curves and , the second derivative is negative.

Key point 12.1 A curve that curves upwards is called convex and has A curve that curves downwards is called concave and has

. .

WORKED EXAMPLE 12.1

Find the values of corresponding to the convex sections of the curve Find the second derivative.

.

The curve is convex when

.

This is a quadratic inequality, so factorise and sketch the graph.



or

When a curve changes from convex to concave (or vice versa) the second derivative changes sign. So there is a point where

. This is called a point of inflection.

          The diagrams show that a point of inflection can be horizontal (so it is a stationary point) or nonhorizontal. Notice that if you try to draw a tangent at a point of inflection, it will cross the curve.

Key point 12.2 For a point of inflection:

WORKED EXAMPLE 12.2

The curve

has two points of inflection for

. Find their coordinates.

Find the second derivative.

The points of inflection satisfy

The points of inflection are . WORKED EXAMPLE 12.3

and

.

Remember to find the -coordinates as well.

WORKED EXAMPLE 12.3

The curve has two points of inflection. Find their coordinates and determine whether they are stationary or non-stationary. Hence sketch the curve. Rather than using the product rule, expand the brackets before differentiating.

Points of inflection have When

and

. So

Stationary points have

. .

is a stationary point of inflection. When

and

. So

is a non-stationary point of inflection. To sketch the curve, you also need the -intercepts and the stationary points.

-intercepts: Stationary points:

When

, so this is a

maximum point.

You already know that

is a point of inflection.

Use the second derivative to find out about the other stationary point. The curve has a point of inflection at the origin and a maximum point. The second point of inflection is between those two, where the curve changes from convex to concave.

Be careful not to assume that when a stationary point has

, the point is always a point of

inflection. The point could be a local maximum, a local minimum or a point of inflection. To establish which it is, consider the gradient on either side of the stationary point.

Common error At a point of inflection

, but just because

follow that it is a point of inflection.

WORKED EXAMPLE 12.4

at a particular point it does not

a Find the coordinates of the stationary point of

.

b Determine the nature of this stationary point. Stationary points have

a For stationary points

:

.



Find the -coordinate. Therefore, the stationary point is

.

b Find the nature of this point.

Use

to determine its nature.

When So, examine:

, no conclusion can be made.

Check the gradient on either side instead. The gradient goes from negative to positive, so this is a minimum point.

is a minimum. WORK IT OUT 12.1 Show that the function nature.

has a stationary point at the origin, and determine its

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

Solution 2

Solution 3

So this is a point of inflection.

Test the derivative on either side:

Test the second derivative on either side:

The curve goes from increasing

The curve changes from concave to convex, so this is a point of inflection.

to increasing, so this is a point of inflection.

EXERCISE 12A 1

Describe each section of the given curve. Use one or more of the words ‘increasing’, ‘decreasing’, ’convex’, and ‘concave’. a

i

to

ii

to

iii

to

i

to

ii

to

iii

to

b

2

Mark these points on each curve. , local maximum point   , stationary point of inflection   , non-stationary point of inflection a

b

3

Mark one or more points on the curve and

that satisfy the given conditions.

and and and and

4

Find the coordinates of the point of inflection on the curve

5

The curve

6

Show that all points of inflection on the curve

7

Find the set of values of corresponding to the convex sections of the curve

8

Find the coordinates of the points of inflection on the curve

.

has two points of inflection. Find their coordinates. lie on the -axis. .

for

. Justify

carefully that these points are points of inflection. 9

Show that the graph of

has only one stationary point for

, and that this is a

point of inflection. 10

Find the -coordinates of the stationary points on the graph of

and determine their

nature. 11

The curve

12

The curve

13

Find the set of values of corresponding to the concave section of the curve

14

The graph shows

has a stationary point of inflection. Show that is concave for

. Find the value of .

.

On a copy of the diagram: a mark points corresponding to a local minimum of

as

b mark points corresponding to a local maximum of

as

c mark points corresponding to a point of inflection of



.

as .

.

Section 2: Parametric equations You should by now be very familiar with the idea of using equations to represent graphs and using differentiation to find their gradients. You also know that some curves cannot be represented by equations of the form circle with radius centred at the origin has an implicit equation

. For example, a

.

Rewind For a reminder of implicit functions and how to differentiate them, see Chapter 10, Section 4.

But there is another way to describe points on this circle, using the angle from the horizontal. Let be a point on the circle and let be the angle

makes with the -axis.

Then the coordinates of are

.

Both and have been expressed in terms of a third variable, . These are called parametric equations and is called the parameter. You can check that these coordinates satisfy the original equation of the circle:

The last equation, involving just and , is called a Cartesian equation. Each value of the parameter corresponds to a single point on the curve. Worked example 12.5 illustrates how to use parametric equations. WORKED EXAMPLE 12.5

A curve has parametric equations

.

a By filling in the table, mark the points corresponding to the given parameter values. Hence sketch the curve. Point

 

 

 

 

 

 

 

 

 

 

b Find the parameter value(s) at the points: i ii a

. Point

Use the equations to find and for each value of .      

  Sketch the curve.

Substitute

b i

There are two possible values of , so you need to check which one gives the correct value of .

When When So the value of is

into the equation for and solve.

. Substitute

ii

When When So the possible values of are or .

into the equation for and solve.

Although each parameter value gives a single point, there can be several parameter values giving the same point. Notice that

also when

, but

there.

You can sometimes convert parametric equations into a Cartesian equation by eliminating the parameter. The two most common ways of doing this are either by substitution, or by using a trigonometric identity.

Common error When you are given - and -coordinates and want to find at that point, don’t assume that the first value of you get is correct – check it works in both the and equation.

WORKED EXAMPLE 12.6

Find the Cartesian equation of each curve: a b

.

a

Make the subject of the -equation and substitute it into the equation.

b

There is a trigonometric identity relating tan and sec, so write terms of .

The Cartesian equation does not need to be in the form equation containing just and is fine.

in

; any

In the first example of parametric equations of a circle, the parameter represented the angle between the radius and the -axis. In other contexts the parameter can represent distance along the curve, or time taken to travel to that point; sometimes the parameter has no obvious meaning, but in all cases it determines where you are on the curve.

Fast forward Kinematics in two dimensions and projectiles are covered in more detail in Chapter 17.

One important application of parametric equations is in studying motion in two dimensions. WORKED EXAMPLE 12.7

A ball is thrown upwards at an angle, from the point coordinates of the ball are given by

. The -axis represents ground level. The , where represents the time in

seconds. a Find the coordinates of the ball seconds after projection. b Find the two times when the ball is

above ground.

c Show that the path of the ball is a parabola, and find its Cartesian equation, giving the coefficients to significant figures. a When

Substitute in

:

.

and The coordinates are

. The height is measured by the -coordinate.

b and The two times are

and

.

Rearrange the first equation to make the subject and substitute into the second.

c

As the equation is quadratic, the path of the ball is a parabola.

EXERCISE 12B 1

By creating a table of values sketch the curves given by these parametric equations. a i ii

2

for for

b i

for

ii

for

c i

for

ii

for

Find the parameter value at the given point on each curve. a i

point point

ii b i ii

3

point point

c i

point

ii

point

Find a Cartesian equation for each parametric curve. a i ii b i

 

ii c i

 

ii d i

 

ii e i

 

ii 4

A ball is projected from point on the horizontal ground and moves in the vertical plane. The ball follows a parabolic path with parametric equations , where is the time after the projection. The units on the and -axes are metres, and time is measured in seconds. a Find the height of the ball above ground seconds after projection. b Find the distance from of the point where the ball hits the ground.

5

A straight line has parametric equations

.

a Find the -intercept of the line. b Find the distance between the points on the line with represent? c Find the Cartesian equation of the line in the form

and

. What does the parameter

.

Differentiating parametric equations When a curve is given parametrically you can find its gradient,

, by using the chain rule to take

account of the parameter .

Key point 12.3

Tip To differentiate when there are three (or more) variables involved, always use the chain rule. You will see this in the next section as well with connected rates of change.

WORKED EXAMPLE 12.8

A curve has parametric equations a at the point where b at the point a

. Find the gradient of the curve:

  .

Find

and

.

Use the chain rule. Remember that





b To find :

The formula for the gradient is in terms of , so you need to find the value of first.

at this point.

You need the value of that satisfies both equations. Use the equation from part a.

WORK IT OUT 12.2 Find the equation of the tangent to the curve with parametric equations point

.

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

Equation of the tangent:

Solution 2

When But When Tangent:

Solution 3

When

at the

Then Tangent:

Curves given by parametric equations can have both horizontal and vertical tangents. You can find these by looking at the values of

and

.

Key point 12.4 At a point where the tangent is parallel to the -axis,

.

At a point where the tangent is parallel to the -axis,

.

WORKED EXAMPLE 12.9

A curve has parametric equations

for

.

a Find the equation of the tangent parallel to the -axis. b Show that the curve has no tangents parallel to the -axis. If a tangent is parallel to the -axis then

a

. This tangent is vertical, so it is of the form . The equation of the tangent is

. If a tangent is parallel to the -axis then

b

. But

for all .

Hence

, so there are no horizontal

The exponential function is always positive.

tangents.

You an use parametric equations to describe some interesting curves and to prove properties of their tangents and normals. In many of these proofs you need to work with the general value of the parameter.

Did you know? The curve in Worked example 12.10 is called an astroid. A ladder sliding down a wall is always tangent to this curve.

WORKED EXAMPLE 12.10

A curve has parametric equations

, for

. is a point of the curve.

a Find, in terms of , the equation of the tangent to the curve at . The tangent at a point meets the -axis at and the -axis at . b Prove that the length of

a

does not depend on the position of on the curve.

First find

and

. Treat

as

and use the chain rule, and likewise for

Then use the chain rule to find

Equation of the tangent:

b At

.

The equation of the tangent is . Set

and

to find axis intercepts.

Divide both sides by

.

.

Factorise and use

.



At

The distance between is:

and

Now you have the coordinates of and so you can find the distance between them.

This length does not depend on , so it is the same for every position of .

EXERCISE 12C

1

Find an expression for

in terms of the parameter ( or ) for these parametric curves.

a i ii b i ii c i ii 2

Find the gradient of each curve at the given point. (You need to find the parameter value first.) a i ii b i ii

3

point point point point

c i

point

ii

point

A curve has parametric equations

.

Find the equation of the normal to the curve at the point where 4

A curve has parametric equations

.

Find the equation of the tangent to the curve at the point where 5

The tangent to the curve with parametric equations coordinate axes at the points

.

. at the point

and . Find the exact area of triangle

crosses the

.

Worksheet See Support Sheet 12 for a further example of finding tangents and normals with parametric equations and for more practice questions.

a Find the equation of the normal to the curve with equation

6

at the point

.

b Find the coordinates of the point where the normal crosses the curve again. 7

A parabola has parametric equations

.

The normal to the parabola at the point where

crosses the parabola again at the point .

Find the coordinates of . 8

Prove that the curve with parametric equations

has no tangents parallel

to the -axis. Let be the point on the curve

9

with coordinates

.

The tangent to the curve at meets the -axis at point and the -axis at point . Prove that . 10

Point a Let

lies on the parabola with parametric equations

be the point where the normal to the parabola at crosses the -axis. Find, in terms of

and , the coordinates of b

.

.

is the perpendicular from to the -axis. Prove that the distance

.

Integrating parametric equations When the equation of a curve is given in a parametric form, you can still use integration to find the area between the curve and the -axis.

Key point 12.5 The area under the curve

, between points

and

, is given by:

where and are the parameter values at the points on the curve where

and

.

Rewind This is very much like the idea used in integration by substitution, in Chapter 11, Section 2, to change from to .

WORKED EXAMPLE 12.11

The diagram shows a part of the curve with parametric equations have -coordinates and .

. Points and

a Find the values of the parameter at and . b Find the area bounded by the curve, the -axis and the -axis.

a At At

Notice that there are two possible values of that give only one of them is consistent with .

, but

Use the formula for the parametric integration, with the parameter values at and .

b

EXERCISE 12D 1

Find the area of the shaded region for each curve, with the given parametric equations. a i

ii

b i

ii

c i

ii

2

a Find the coordinates of the points where the curve with parametric equations crosses the -axis. b Find the area enclosed by the curve and the -axis.

3

A curve is defined by the parametric equations a Find the exact values of at the points where

. and

.

b Find the area bounded by the curve, the -axis and the line

4

.

The graph shows the curve with parametric equations:

a Show that the shaded area is given by:

b Hence find the exact value of .

. 5

Point

lies on the curve with parametric equations

a Find the parameter value at . b The diagram shows a part of the curve and the normal at . Find the area of the shaded region. c Find the Cartesian equation of the curve.



.

Section 3: Connected rates of change So far you have only looked at functions of one variable; for example, you can use differentiation to find how quickly the velocity of an object changes with time. In this section, you will use the chain rule to solve problems where there are several related variables. Consider, for example, an inflating balloon. You can control the amount (V) of gas in the balloon, but you may want to know how fast the radius is increasing. These are two different rates of change, but intuitively they are linked – the faster the gas fills the balloon the faster the radius will increase.

Rewind You met this idea of writing rates of change in terms of derivatives in Student Book 1, Chapter 12.

The two derivatives need to be linked:

and

. You can do this by using the chain rule and the

geometric context.

Key point 12.6 To link variables and via their rates of change, use the chain rule:

Tip In these practical applications, the rate of change is nearly always with respect to time, . For example the rate of change of volume, , is

.

WORKED EXAMPLE 12.12

A spherical balloon is being inflated with air at a rate of radius increasing when the radius is ?

per minute. At what rate is the

When solving problems in context you should always start by defining the variables.

per minute

The balloon being inflated at a rate of

per minute

means you know the rate of change of volume, Since the balloon is spherical:

.

Use the geometric context and differentiate to find

.

so You want the rate of change of the radius,

. Relate this to

the two derivatives you already have using the chain rule.

Substitute into the chain rule. Remember that

.

You want the rate of change of the radius when the radius is . So the radius is increasing at about per minute.

Sometimes you need to combine the chain rule with the product or the quotient rule. WORKED EXAMPLE 12.13

The length of a rectangle is increasing at a constant rate of , and the width is decreasing at a constant rate of . Find the rate of change of the area of the rectangle at the instant when the length is

and the width is

.

Start by defining the variables.

Write the given information in terms of derivatives. The rate of change of is . As the width is decreasing, the rate of change will be negative. Use the product rule to differentiate . You want

When

and

.

Use the given values of and .

The area is decreasing at the rate of .

WORKED EXAMPLE 12.14

As a conical ice stalactite melts, the rate of decrease of height, , is decrease of the radius of the base, , is . At what rate is the volume of the stalactite decreasing when the height is radius is ?

and the rate of

and the base

  Write the given information in terms of derivatives. Remember that decrease means a negative derivative. Use the geometric context to relate the variables. You want

so differentiate both sides with

respect to . This requires the product rule and the chain rule. Substitute in the given values.

The volume is decreasing at

.

EXERCISE 12E

1

In each case, find an expression for

in terms of .

a i ii b i ii 2

a i Given that ii Given that b i If

and

ii If

and

and and

3

, find , find

when

when when

and

. .

. .

, find the possible values of .

, find the value of for which

a i Given that

, find

when

ii Given that

, find

when

b i Given that

when

, find

c i Given that ii Given that

, find

, find

when

and and and and

. . . .

ii Given that

, find

c i Given that

and that

ii Given that

and that

when

. , find , find

. .

4

A circular stain is spreading so that the radius is increasing at the constant rate of the rate of increase of the area when the radius is .

5

The area of a square is increasing at the constant rate of side of the square when the length of the side is .

6

A spherical ball is inflated so that its radius increases at the rate of of change of the volume when the radius is .

7

A rectangle has length and width . Both the length and the width are increasing at a constant rate of per second. Find the rate of increase of the area of the rectangle at the instant when and .

8

The surface area of a closed cylinder is given by , where is the height and is the radius of the base. At the time when the surface area is increasing at the rate of , the radius is , the height is and is decreasing at the rate of . Find the rate of change of radius at this time.

9

The radius of a cone is increasing at the rate of and its height is decreasing at the rate of . Find the rate of change of the volume of the cone at the instant when the radius and and the height is .

10

. Find

. Find the rate of increase of the

per second. Find the rate

A point is moving in the plane so that its coordinates are both functions of time, the coordinates of the point are , the -coordinate is increasing at the rate of

. When units per

second and the -coordinate is increasing at the rate of units per second. At what rate is the distance of the point from the origin changing at this instant?



Section 4 : More complicated areas Area above and below the -axis You know from Student Book 1, Chapter 14, that when a region is below the -axis, the value of the definite integral will be negative. Therefore, when there are parts of the curve above and below the axis, to calculate the total area you must separate out the two sections. WORKED EXAMPLE 12.15

a Find

.

b Sketch the graph of

. Hence find the area of the region enclosed between the -axis,

the curve

and the lines

and

.

Integrate and evaluate at the upper and lower limits.

a

b

You can see that the required area is made up of two parts, one above the -axis and one below it, so evaluate each of them separately.

The integral for the part of the curve below the axis is negative, but the area must be positive.

. The area of the part above the axis is found as normal.

. Now add the two areas together.

You can interpret the fact that the integral in Worked example 12.15 was zero as meaning that the area above the axis is exactly cancelled by the area below the axis.

Common error When you are asked to find an area, it is very important to sketch the graph first (if it is not given). Don’t just integrate in one go between the -limits of the region to find its area unless you are sure the curve is always totally above or totally below the -axis. If part of the region is above the -axis and part is below, you must find the area of each separately.

Area between two curves The area in the diagram is bounded by two curves with equations You can find the area by taking the area bounded by the upper curve the area bounded by the lower curve

and

.

and the -axis and subtracting

and the -axis, i.e.

You can do the subtraction before integrating so that you only have to integrate one expression instead of two. This gives an alternative formula for the area.

Key point 12.7 The area A bounded by two curves with equations below curve , is given by:

and

, where curve

is

where and are the -coordinates of the intersection points of the two curves.

Tip Make sure you subtract the lower curve from the upper curve when using this method.

WORKED EXAMPLE 12.16

Find the exact area enclosed between the curves For the intersection:

and

.

First you need to find the -coordinates of the intersections.

It may help to do a rough sketch to see the relative positions of the two curves.

Subtract the lower curve from the upper curve before integrating.

Subtracting the two equations before integrating is particularly useful when one of the curves is partly below the -axis. As long as is always above then the expression being integrated, , is always positive, so you do not have to worry about the signs of and themselves.

Rewind In Student Book 1, Chapter 14, you will have found an area such as that in Worked example 12.16 by subtracting the area under the curve from the area of a trapezium. This new method is more direct.

WORKED EXAMPLE 12.17

Find the area bounded by the curves

and

.

Sketch the graph to see the relative position of the two curves.

Intersections:

Find the intersection points; start by multiplying through by .

This is a disguised quadratic which can be factorised.

Write down the integral representing the area. You can see from the graph that the top curve is . Simplify before integrating.

Use

.

WORK IT OUT 12.3 Find the area enclosed between the line

and the curve

.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Sketching the graph shows that part of the area is below the -axis, so split it up into two parts.

Solution 2 The intersection points are at

Solution 3

and

.

The intersection points are at

and

.

Area between a curve and the -axis How can you find the area of the shaded region in this diagram, bounded by the curve and the -axis?

You know how to find the area between the curve and the -axis, so one possible strategy is to draw vertical lines to create the region labelled rectangle plus the blue rectangle.

. The required area is equal to the area of

minus the red

But there is a quicker way. You can treat as a function of , effectively reflecting the whole diagram in the line .

Key point 12.8 The area bounded by the curve

where

, the -axis and the lines

is the expression for in terms of , i.e.

and

is given by:

.

Rewind From Chapter 2, Section 4, you may have realised that this is related to inverse functions.

WORKED EXAMPLE 12.18

The curve shown has equation

.

Find the area of the shaded region.

Express in terms of .

Find the limits on the -axis.

When When



WORKED EXAMPLE 12.19

Find the exact area enclosed by the graph of

, the -axis and the line

.

Sketch the graph and identify the area required. The graph crosses the -axis at

.

The area between a curve and the -axis is given by , so express in terms of … … and then evaluate the definite integral.

Remember that

.

Notice that this value is negative; this is because the shaded area is to the left of the y-axis.

EXERCISE 12F 1

Find the area of each shaded region. You need to find the intersection points of the two curves first. a i

ii

b i

ii

c i

ii

d i

ii

e i

ii

2

Find the areas of the shaded regions. a i

ii

b i

ii

c i

ii

3

The diagram shows part of the graph of curve, the -axis and the lines

and

. The shaded region is enclosed by the .

Find the area of the shaded region. 4

Find the area enclosed between the graphs of

5

The diagram shows the graphs of

and

and

.

. Find the exact value of the area of the shaded

region.

6

Show that the area of the shaded region is .

7

The diagram shows the curve If the shaded area is

.

find the value of .

8

Find the exact value of the area enclosed by the graph of axis.

9

Find the exact value of the area between the graphs of

, the line

and

10

Find the total area enclosed between the graphs of

and

11

The area enclosed between the curve

is

and the line

. 12

The diagram shows the graph of

.

and the –

. . . Find the value of if

Find the value of a for which the areas of the red region and the blue region are equal.

Checklist of learning and understanding A convex curve has At a point of inflection,

; a concave curve has

.

and the curve changes from convex to concave, or

vice versa. Parametric equations are a way of describing a curve where both and coordinates are given in terms of a parameter (usually called or ). Each parameter value corresponds to a single point on the curve. The gradient of a curve given in parametric form is

.

The area between the -axis and a part of a curve with parametric equations is given by

, where and are the parameter values at the end-

points. The chain rule can be used to connect rates of change of two related variables; If depends on and depends on , then

. You often need the geometric

context of the question to work out how depends on . The area enclosed between two curves with equations by

is given

, where and are -coordinates of the intersection points.

The area between a curve, the -axis and the lines , where



and

and

is the expression for in terms of .

is given by

Mixed practice 12 1

A curve is defined by the parametric equations for Find the value of the parameter at the point

. Choose from these options.

A B C D 2 3

Find the coordinates of the point of inflection on the graph of A curve has parametric equations a Show that the point

.

.

lies on the curve, and the find the value of at this point.

b Find the equation of the tangent to the curve at . 4

The diagram shows the graph of with a horizontal line drawn through its intercept. Find the exact value of the area of the shaded region.

5

a Find

.

b The diagram shows the graph of the -axis and the lines and

Find the area of the shaded region.

. The shaded region is bounded by the curve, .

6

A curve is defined by the parametric equations

.

a Find the gradient at the point on the curve where

.

b Find a Cartesian equation of the curve. [© AQA 2014] 7

For the curve

.

Which statement is true? A There is a point of inflection at

.

B There is a point of inflection at

if

C There isn’t a point of inflection at

as well. if

.

D More information is required to determine whether there is a point of inflection at 8

The diagram shows the graph of

The area of the shaded region is 9

10

.

.

units. Find the value of .

The curve shown in the diagram has parametric equations for . A tangent to the curve is drawn at the point . Find the shaded area enclosed between the curve, the tangent and the -axis.

a Solve the equation

for

b The diagrams shows the curves the shaded region.

. and

. Find the exact value of the area of

11

Consider the graph of

for

.

a Show that the -coordinates of the points of inflection satisfy

.

b Use graphs to find the number of points of inflection on the graph. 12

a Find the coordinates of the stationary points on the graph of

and determine

their nature. b Prove that the graph has one non-horizontal point of inflection. 13

Prove that the function

14

The diagram shows a part of the graph of

has a stationary point of inflection at the origin. for

.

The red area is three times larger than the blue area. Find the value of .

15

The diagram shows an isosceles right-angled triangle of side the side towards point so that the area of the trapezium constant rate of .

Let

. Point is moving along is decreasing at the

.

a Write down an expression for the area of the trapezium b Show that

.

Initially point is at vertex . c Given that

, find the value of .

in terms of .

16

A curve is defined by the parametric equations a Find

.

in terms of .

b The point , where

, lies on the curve.

i Find the gradient of the curve at . ii Find the coordinates of . iii The normal at crosses the -axis at the point . Find the coordinates of . c Find the Cartesian equation of the curve in the form

, where is an integer. [© AQA 2013]

17

A particle moves in a straight line with velocity given by .



a Find the times when is instantaneously at rest. b Find the total distance travelled in the first seconds of motion, giving your answer to three significant figures. 18

Show that the area of the shaded region in the diagram is .

19

The ellipse shown in the diagram has parametric equations .

, with

a State the values of at the points marked and . b Find the area of the shaded region, and hence state the total area enclosed by the ellipse. 20

The graph shows

.

On a sketch of this graph: a mark points corresponding to a local minimum of

as

b mark points corresponding to a local maximum of

as

c mark points corresponding to a point of inflection of

21

a Show that

as .

.

b Find the coordinates of the points of intersection of the graphs .

and

c Find the area enclosed between these two graphs. d Show that the fraction of this area above the axis is independent of and state the value that this fraction takes.

13 Differential equations In this chapter you will learn how to: solve differential equations of the form write differential equations in a variety of contexts interpret a solution of a differential equation and decide whether it is realistic in the given context.

Before you start… Student Book 1, Chapter 7

You should be able to rearrange expressions involving exponents and logarithms.

1 Given that in terms of .

Student Book 1, Chapter 18

You should be able to draw force diagrams and find net forces.

2 An object of weight falls under gravity. The magnitude of the air resistance is . Find the net force on the object.

Chapter 10

You should be able to write equations involving related rates of change.

3 The rate of change of the radius, , of a sphere is . Find an expression for the rate of change of volume.

Chapter 11

You should be able to integrate, using partial fractions, and simplify the answer, using laws of logs.

4 Integrate and simplify

Chapter 11

You should be able to use integration by substitution and by parts.

5 Integrate

, write

a b

Chapter 11

You should be able to integrate, using trigonometric identities.

6 Find

.

Modelling using differential equations In Student Book 1, Chapter 15, you met problems involving velocity as the rate of change of displacement, and acceleration as the rate of change of velocity. Many natural processes can be modelled by equations involving the rate of change of some variable, such as population growth and cooling of bodies. Newton’s second law actually states that force is equal to the rate of change of momentum. To get from these rates of change to find the underlying variable involves solving differential equations. In this chapter you will look at forming differential equations with an emphasis on real-world applications

and at a method for solving a particular type of differential equation.

Section 1: Introduction to differential equations To solve a differential equation you go from an equation involving derivatives to one without. You have done this already for the case where the equation can be written in the form As an example, consider the differential equation need is integration:

.

. To solve this differential equation all that you

.

Because of the constant of integration you find that there is not just one solution to the differential equation. It could be any one of a family of solutions. All the curves have the same gradient function, so they have the same shape. The solution equation.

is called the general solution to the differential

You may also be told that the curve passes through a point, e.g.

. This is called an initial condition

and it allows you to find the constant of integration and so narrow down the general solution to the particular solution – in this case . Sometimes you will need to rearrange the equation before integrating.

WORKED EXAMPLE 13.1

Find the particular solution of the differential equation

, given that

. Write the equation in the form

.

Then integrate. Since the numerator is a multiple of the derivative of the denominator, a substitution will work.

Make the substitution, and simplify as much as possible.

Remember

; this is the general solution.

when

Use the given condition to find .

This value of gives the particular solution.

EXERCISE 13A 1

Find the general solution of each differential equation. a i ii b i ii c i ii d i ii

2

Find the particular solution of each differential equation. a i ii b i ii c i ii d i ii



Section 2: Separable differential equations In Section 1 you looked at differential equations where

depends just on . But there are situations

where the gradient depends on , or on both variables, for example,

.

You can’t solve this equation by simply integration as the right-hand side contains but if the is moved to the left-hand side first, to give

and you can then integrate both sides of the equation with respect to .

However,

so the equation becomes:

Key point 13.1 To solve a differential equation that can be written in the form

:

get all of the s on one side and all of the s on the other side by multiplication or division separate

as if it were a fraction

integrate both sides. Just as with integration by substitution, you get the same results from just splitting up

as if it were a

fraction when you separate the and ys to different sides; this leads to the expression marked *. This method of solving differential equations is called separation of variables. WORKED EXAMPLE 13.2

Solve the equation

given that

when

.

Get the term onto the left-hand side

Separate

and integrate.

Since the difference of two constants is just another constant, you only need on one side. Use the initial condition to find .



Using

.

Sometimes you need to factorise the equation first to get it into the correct form. WORKED EXAMPLE 13.3

Show that the general solution to the differential equation if

can be written as

. You can use separation of variables if you can write the equation in the form

.

Separate the variables: divide by Then integrate.

But since



and multiply by

Since is a constant, relabel it as .



WORK IT OUT 13.1 Find the general solution of the differential equation Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

Solution 2

.

Solution 3

If

depends just on you can use the fact that

.

Key point 13.2 If

then

.

WORKED EXAMPLE 13.4

Newton’s law of cooling states that the rate of change of temperature of a cooling body is proportional to the difference between the body’s temperature and the temperature of the surroundings. A cup of coffee cools in the room where the air temperature is coffee, , satisfies the differential equation

. The temperature of the

where is the time, measured in minutes. The initial temperature of the coffee is

. Find the temperature of the coffee after six minutes.

You want to integrate with respect to so take the reciprocal of both sides… So:

... and integrate.

Use the initial conditions to find c.

Substitute in the given value of before rearranging.

EXERCISE 13B 1

Find the particular solution of each differential equation, giving your answer in the form simplified as far as possible. a i ii b i ii

2

Find the particular solution of each differential equation. You do not need to give the equation for explicitly. a i ii b i ii c i ii

3

Find the general solution of each differential equation, giving your answer in the form simplified as far as possible. a i ii b i ii c i ii

4

Solve the differential equation

given that when

. Give your answer in the

form 5

.

The function when

satisfies the differential equation

. When

. Find the value of

.

6

Given that

, where is a positive constant, show that

7

Find the general solution of the differential equation

. , giving your answer in the form

. 8

Given that

and that

when

, show that

, where is a

constant to be found.

Worksheet See Support Sheet 13 for a further example of solving a separable differential equation and for more practice questions.



Section 3: Modelling with differential equations Now that you can solve various differential equations, you can look at how such equations arise in a variety of contexts. There are many situations where you know (or it is reasonable to assume) what the rate of change of a quantity depends on. You can then write down and solve a differential equation to find out how the actual quantity behaves.

Rewind You met the idea of the rate of change of a quantity in practical applications being

in

Chapter 12, Section 3. You have also met specific rates of change, such as acceleration , in Student Book 1, in mechanics.

WORKED EXAMPLE 13.5

A skydiver of mass jumps out of an airplane with zero initial velocity. The air resistance is proportional to velocity and may be modelled as . Write and solve a differential equation to find an expression for the velocity of the skydiver in terms of time. Give your answer in terms of . Forces on the skydiver:

Always start by drawing a force diagram! The weight acts downwards and the air resistance upwards.

Write Newton’s second law equation, taking downwards as positive (because that’s the direction of motion).

Since the right-hand side is in terms of , re-write the equation in terms of

.

Now integrate with respect to .

Now use the initial condition.

So:

You need to express in terms of .

Here you have used:If

then

Notice that on the penultimate line of Worked example 13.5 the modulus sign was removed. This is only justified if i.e., . Since initially , this is certainly true for some initial part of the motion. But you should ask whether eventually becomes larger than ; if it does then the equation needs to be changed. Looking at the final solution, and the solution is valid for all .

is always positive, so is in fact always less than

Focus on … There are a number of examples of modelling real life situations with differential equations in Focus on… Modelling 2.

Did you know? In fact, it.

decreases and tends to zero, so increases towards

without ever reaching

This maximum velocity is called the terminal velocity. Of course, the object might hit the ground before getting close to this velocity.

Often you have a model where some constants are unknown. You can find them by using experimental or observational information. WORKED EXAMPLE 13.6

In a simple model of a population of bacteria, the growth rate is assumed to be proportional to the number of bacteria. a Let thousand be the number of bacteria after minutes. Initially there are bacteria and this number increases to after minutes. Write and solve a differential equation to find the number of bacteria after minutes. b Comment on one limitation of this model. a

The rate of growth is

.

‘Proportional to ’ means you can write it as for some constant .

You have information about and , but not about

. So you need to solve the

differential equation before you can put in the numbers. To find the constants and , use the two given conditions: first use (i.e. when

initially

).

A second equation is needed to find so next use the condition that Using

when

.

.

So: Put and back in and rearrange the equation to get in terms of . Check that you can follow all the steps!

b This model predicts unlimited population growth, which is not realistic.

Exponential growth models often work initially, but need to be adapted for longer time periods.

Sometimes a problem has several variables and you need to use the geometric context and related rates of change to produce a single differential equation.

Rewind Remember, from Chapter 12, Section 3, that in connected rate of change problems you need to start by expressing all the information given in terms of derivatives. Then use the chain rule to link these derivatives.

WORKED EXAMPLE 13.7

A cylindrical tank with cross-sectional area

and height

is initially filled with water. The

water leaks out of the tank out of a small hole at the bottom at the rate of is the height of water in the tank after seconds.

, where

a Find an equation for

in terms of .

b Hence find how long it takes for the tank to empty. a Let be the volume of the

Define any variables not already defined in the question.

tank. Express all the given information in terms of derivatives. Use here that volume of cylinder cross-sectional area height Water leaks out at the rate of . This tells you the rate of change of volume. It is negative as the volume is decreasing. You want to find

. Use the chain rule to express this in

terms of as this appears in the previous expressions.

b

Solve the equation by integrating

.

Initially the tank is full.

The tank is empty when

.

So the tank is empty after seconds.

EXERCISE 13C 1

Write differential equations to describe each situation. You do not need to solve the equations. a i A population increases at a rate equal to five times the size of the population ii The mass of a substance

.

decreases at a rate equal to three times the current mass.

b i The rate of change of velocity is directly proportional to the velocity and inversely proportional to the square root of time. ii A population size increases at a rate proportional to the square root of the population size and to the cube root of time. c i The area of a circular stain increases at a rate proportional to the square root of the radius. Find an equation for the rate of change of radius with respect to time. ii The volume of a sphere decreases at a constant rate of of decrease of the radius. 2

. Find an equation for the rate

In a simple model of a population of bacteria, the rate of growth is proportional to the size of the population.

When the population size a Show that

is

bacteria, the rate of growth is

bacteria/minute.

.

b Initially there were bacteria. Solve the differential equation and predict the number of bacteria after minutes. 3

The mass of a radioactive substance mass is and it decays at the rate of a Show that

decays at a rate proportional to the mass. Initially the .

.

b Find the amount of time it takes for the mass to halve.   4

A particle of mass is moving with speed in a straight line on a horizontal table. A resistance force is applied to the particle in the direction of motion. The magnitude of the force is proportional to the square of the speed, so that . a Show that

.

b Find an expression for the velocity in terms of time, and hence find how long it takes for the speed to decrease to below 5

.

The population of the fish in a lake, thousand, can be modelled by the differential equation

where is the time, in years, since the fish were first introduced into the lake. Initially there were fish. a Show that the population initially increases and find when it starts to decrease. b Find the expression for in terms of . c Hence find the maximum population of the fish in the lake. d What does this model predict about the size of the population in the long term? 6

In Economics, there is a model for how the demand

for a product depends on its price

. It

states that the rate of change of with respect to is proportional to but inversely proportional to . a Explain how this model leads to the differential equation

where is a negative constant (called elasticity). b Find the general solution of this differential equation. c Sketch the graph of against , and describe how demand depends on price, in each case. i ii 7

Newton’s law of cooling states that the rate of change of temperature of the body is proportional to the difference in the temperature between the body and its surroundings. A bottle of milk is taken out of the fridge, which is at the temperature of table in the kitchen, where the temperature is

, and placed on the

. Initially, the milk is warming up at the rate of

per minute. a Show that

, where

is the temperature of the milk and is time in minutes

since the milk was taken out of the fridge. The temperature of the milk is measured to the nearest degree. b Solve the differential equation and hence find how long, to the nearest minute, it takes for the temperature of the milk to reach the kitchen temperature. 8

A particle of mass is pulled through liquid by means of a light inextensible string. The tension in the string is . The resistance force is proportional to the velocity, and equals . The particle starts from rest. a Find an expression for the velocity after time . b Describe the velocity of the particle for large values of .

9

An inverted cone has base radius

and height

The cone is filled with water at a constant rate of a Show that the height of water

. .

satisfies the differential equation

.

b Given that the cone is initially empty, find how long it takes to fill it.

10

A particle of mass

is moving at the speed of

when it enters a viscous liquid at the

point . Inside the liquid the resistance force is proportional to the velocity, and is initially equal to . a Show that the velocity of the particle satisfies the differential equation

.

b Find an expression for the velocity of the particle seconds after entering the liquid. c Find the displacement of the particle from point , and describe what happens to the displacement for large values of .

Checklist of learning and understanding A differential equation is an equation for the derivative of a function. To solve a differential equation means to find an expression for the function itself. Some differential equations can be solved by separation of variables. Write the equation in the form

and integrate both sides.

Initial conditions can be used to find the constant of integration. A differential equation often describes the rate of change of a quantity – this is the derivative with respect to time. The rate of change is often proportional to one of the variables. You may need to use given information to find the constant of proportionality. Sometimes a problem involves more than one variable and you need to use related rates of change to write a differential equation.



Mixed practice 13 1

A car depreciates at a rate proportional to its value, , at time years after it was made. is a constant. Find a possible differential equation to model the value of the car. Choose from these options. A B C D

2

A cake is placed in the oven. Its temperature,

is modelled by the differential equation

where is a positive constant and is the time, in minutes, after it was put in the oven. What is the maximum temperature the cake can reach? Choose from these options. A B C D There is no maximum. 3

Solve the differential equation in the form

4

given that

when

. Give your answer

.

Find the particular solution of the differential equation

such that

when

. 5

In a chemical reaction, the amount of the reactant

follows the differential equation

. Initially there was

of the reactant.

Show that 6

a Find

, where and are constants to be found. .

b Solve the differential equation

given that

when

. Give your answer in the form

. [© AQA 2013]

7

A population of fish initially contains fish and increases at the rate of Let be the number of fish after months.

fish per month.

In a simple model of population growth the rate of increase is directly proportional to the population size.

a Show that

.

b Solve the differential equation and find how long it takes for the number of fish to reach . c Comment on the long-term suitability of this model. An improved model takes into account seasonal variation:

d Given that initially there are

fish, find an expression for the size of the population after

months. 8

A model for a relationship between the price

of a commodity and the demand

for the

commodity states that

where is the elasticity. a Find the general solution to this differential equation. b Describe the relationship between the price and demand when c For most commodities, . 9

.

. Suggest, with a reason, what sort of commodity might have

A cylindrical tank has radius

and height

. The tank is being filled with water so that

. Initially the tank is empty. a Show that

and hence find an expression for the height of water in the

tank at time , giving your answer in the form

.

b Hence or otherwise, determine whether the tank will ever completely fill with water. 10

A ball of mass falls vertically downwards. When the velocity of the ball is resistance has magnitude . a Show that

the air

.

b The ball falls from rest. Show that

, stating the values of constants and .

c Hence describe the motion of the ball. 11

A giant snowball is melting. The snowball can be modelled as a sphere whose surface area is decreasing at a constant rate with respect to time. The surface area is at time days after it begins to melt. a Write down a differential equation in terms of the variables and and a constant , where , to model the melting snowball. b i Initially the radius of the snowball is Show that

, and days later, the radius has halved.

.

(You may assume that the surface area of a sphere is given by radius.)

, where is the

ii Use this model to find the number of days that it takes the snowball to melt completely. [© AQA 2011]

12

Consider this model of population growth:

where thousand is the population size at time months. a Suggest what the term b Given that initially

could represent. , solve the differential equation. Give your answer in the form

, where and are constants. c Hence describe what happens to the population in the long term. 13

Variables and satisfy the differential equation Given that

14

when

, show that

. .

A particle moves in a straight line. Its acceleration depends on the displacement: . a Find an expression for

in terms of and .

b Initially the particle is at the origin and its speed is

. Show that

.

c Find expressions for the displacement and velocity in terms of time.

Worksheet See Extension Sheet 13 for questions on some other types of differential equation.



14 Numerical solution of equations In this chapter you will learn how to: work with equations that cannot be solved by algebraic rearrangement find an interval that contains a root of an equation, and how to check that a given solution is correct to a specified degree of accuracy (the sign-change method) approximate a part of the curve by a tangent, and use this to find an improved guess for a solution (the Newton–Raphson method) create a sequence that converges to a root of an equation (fixed point iteration) identify situations in which the above methods fail to find a solution.

Before you start… Chapter 4

You should be able to use the term-to-term rule to generate a sequence.

1 Find the first four terms of the sequence defined by .

Chapter 2; Chapter 8

You should be able to rearrange equations involving polynomials, fractions, exponentials, logarithms and trigonometric functions.

2 Rearrange each equation into the required form. a b c

Chapter 9

You should be able to differentiate a variety of functions.

into into into

3 Differentiate each equation. a b c

What is a numerical solution? Throughout your studies of mathematics you have learnt to solve various types of equations: linear, quadratic, exponential, trigonometric. You can solve some equations by algebraic rearrangement, for example:

Sometimes you can write solutions in an exact form, such as:

or

Alternatively, you can use your calculator to find approximate solutions; for example:

But there are some equations that you can’t solve by any combination of algebraic rearrangement.

Tip You can write the solution of

in an exact form: it is just

For example, how would you solve the equation it into the form

.

? However you try rearranging it, you can’t get

.

If you have a graphical calculator or graphing software you can use them to find approximate solutions by drawing the graphs of

and

and finding their intersection.

Your calculator might have a ‘solver’ function that allows it to find approximate solutions without drawing the graph. But how does it do that?

In this chapter you will learn several different methods for solving equations numerically (this means that a solution is not found by algebraic manipulation, but by evaluating certain expressions to find increasingly accurate approximations). You should be aware that these methods can only give approximate solutions, but you can choose the level of accuracy. You will also see that in some situations these methods fail to find an answer. You will learn how to recognise such situations.

Section 1: Locating roots of a function The simplest numerical methods involve finding an interval in which the solution lies, and then improving accuracy by making this interval smaller. For example, consider the equation the equation into the form ‘equation

. It is easier to see how the method works if you rearrange ’:

A sensible way to start is to try to some values of and see if any of them give an answer close to zero.

From the table you can see that

is a solution. But you can also see that the value of the expression

changes sign between and : it is positive when means that it must equal zero somewhere between these two -values. You can see this if you draw the graph of negative, then the graph must cross the -axis.

but negative when

. This

. If the -coordinate changes from positive to

You can try to locate this solution more accurately by looking for a smaller interval. For example:

This tells you that the solution is between and . You can continue finding smaller and smaller intervals to find the solution to any degree of accuracy you want.

Key point 14.1 The sign-change rule If is a continuous function and and are numbers such that changes sign between and , then the equation has a root between and .

Tip This method produces only one solution. The equation

also has a negative

solution, between and . In general, with numerical methods you can never be sure that you have found all the solutions.

Common error Note that the converse is not true: if has a root between and then does not necessarily change sign between and . This is because there could be more than one root in the interval.

If the graph of is not a continuous line (for example, when it has a vertical asymptote) then it is possible for to change sign without the graph crossing the -axis.

WORKED EXAMPLE 14.1

Let a Show that the equation b i

Evaluate

and

has a solution between

and

.

.

ii Explain why in this case, the change of sign does not imply that between and .

has another solution

Show clearly which values of you are using.

a

There is a change of sign, so there is a root between and .

State the conclusion.

Show clearly which values of you are using.

b i

ii

has a vertical asymptote at , so the change of sign does not imply that the graph crosses the -axis.

Think about the graph: if it has an asymptote then it does not necessarily cross the -axis. is not defined for

, so the graph will

have an asymptote there.

In practice, rather than knowing that a solution lies in a particular interval, you want to express a solution you already have to a specified degree of accuracy, such as one decimal place. Consider, as an example, the equation

. By trying some values of you can find that this equation has a solution around

. But how can you check that this is in fact correct to one decimal place? The solution, when rounded to one decimal place, equals . But the numbers that round to between and , so you should look for a change of sign between those two values.

are

WORKED EXAMPLE 14.2

Show that the equation Let

has a root

, correct to one decimal place. First write the equation in the form .

.

Then look for a change of sign.

There is a change of sign, so there is a root between and . So this root is





to

You should note that you are looking at the degree of accuracy of the -value here, not the -value. For the equation from Worked example 14.2, which equals to two decimal places. However, the root is not is not equal to

to

Look at this table of values.

The change of sign actually occurs between

and

so the root is

correct to

WORK IT OUT 14.1 Find the smallest positive root of

correct to two decimal places.

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution A

As there is a change of sign, there is a root between So the root is

and

.

to

Solution B

As there is a change of sign, there is a root between So the root is

and

.

to

Solution C

As there is a change of sign, there is a root between So the root is

and

to

Focus on … Focus on… Problem solving 2 compares a numerical and analytical solution to the same problem.

EXERCISE 14A 1

Classify these equations as ‘can find exact solutions’ or ‘cannot rearrange algebraically’. a b c d e f

g h 2

Each equation has a root between lies.

and . In each case, find two integers between which the root

a b c d 3

For each equation, show that there is a root in the given interval. a i

, between and , between

ii b i

, between and , between

ii 4

and

and

For each equation, show that the given root is correct to the stated degree of accuracy. a i ii b i ii c i ii d i ii

5

a Show that the equation

has a solution between and .

b Show that this solution equals 6

The equation

correct to one decimal place.

has two solutions.

a Show that one of the solutions equals

correct to three significant figures.

b The other solution lies between positive integers and 7

A function is defined by

.

a Show that the equation b i Evaluate

and

. Find the value of .

has no solutions.

.

ii Alicia says that the change of sign implies that the equation . Explain why she is wrong. 8

Let

has a root between and

.

a i Sketch the graph of

for

.

ii State the number of solutions of the equation b i State the values of

and

.

between and .

ii George says, ‘There is no change of sign between and so the equation

has no

roots in this interval.’ Use your graph to explain why George’s reasoning is incorrect. iii Use the change of sign method (without referring to the graph) to show that the equation has two roots between and .



Section 2: The Newton–Raphson method The sign change rule allows you to show that a root of an equation lies in a certain interval, but how do you know which interval to try? In the first example, with , the table of integer values of showed that there is a solution between and . You could then look at to locate the change of sign more accurately.

This shows that the root is between

and

. You can continue like this until you make the interval as

small as you like. If you wanted to locate the root to three or four decimal places, this could take quite a long time. In some applications solutions of equations are required to even higher accuracy, and then this method becomes unfeasible, even with a fast computer. The numbers in the table suggest that the root may be much closer to than to , so maybe the next search should not start from but from or – but how do you know which number? It would be good to have a method to tell you which number to try next. There are in fact many such methods. In this section you will meet the Newton–Raphson method, which uses the tangent to the graph of to suggest where to look for the root. The diagram shows the section of the graph of for . You can see that it crosses the -axis between and . The diagram also shows the tangent to the graph at . Near this point, the tangent follows the graph closely, so it will cross the -axis near the root of the equation.

Use a graph plotter to draw the graph and the tangent and find where the tangent crosses the -axis. You will find that the value is You can now repeat the same procedure by drawing a tangent at this new value of ; this new tangent crosses the -axis at You can check that this gives the correct root to three decimal places:

Repeating one more time gives the root correct to seven decimal places – this would have taken a very long time with the original sign-change method! So far you have used a graph plotter to draw tangents and find their -intercepts. But there is a formula to calculate the -intercept of the tangent, without having to rely on the graph. You can find the equation of a tangent at any given point, and then use this equation to find where the tangent crosses the -axis.

Key point 14.2 Newton–Raphson method A sequence of approximations to the root of the equation

is given by:

This will be given in your formula book.

Rewind You met equations of tangents in Student Book 1, Chapter 13.

PROOF 8 The tangent to the graph at where .

has gradient

and it passes through the point

Therefore its equation is:

To find where this tangent crosses the -axis set

This is point

:

as shown on the diagram.

You can repeat this process to find

from

.

WORKED EXAMPLE 14.3 a Show that the equation b Starting from

has a root between and .

, use three repetitions of the Newton–Raphson method to find a better

approximation for the root. c Show that this approximation gives the root correct to three decimal places.

a

Use the sign change rule.

There is a change of sign, so there is a root between and . b

To use the Newton–Raphson formula you need . Call the next approximation

c There is a change of sign, so there is a root between and . So the root is

Use

to find

.

Use

to find

.

.

To show that the root equals to you need to show that it is between and .

.

Repeating the Newton–Raphson calculation several times, as shown in Worked example 14.3, involves using the same formula with different numbers. Every time the number you need to put into the equation is the answer from the previous calculation. Most calculators have an ANS button that you can use to carry out such repetitive calculations. Here is how you can use it for Worked example 14.3.

Worksheet See Support Sheet 14 for a further example of using the Newton-Raphson method and for more practice questions.

Start by entering the starting value of and press the

(in this case

) and press the

button again.

The answer is the value of . Pressing the button repeatedly gives you can quickly generate as many approximations to the root as you like.

EXERCISE 14B

button. Then type in

and so on. This way

EXERCISE 14B 1

For each equation, use technology to sketch the graph. Draw the tangent at approximation to the root.

and find the next

a i ii b i ii 2

For each equation, use the Newton–Raphson method, with the given starting value, to find the root correct to three significant figures. Use the change of sign method to show that your root is correct to s.f. a i ii b i ii

3

The equation has a root near . Using this value as the first approximation, use the Newton–Raphson method to find the next approximation to the root.

4

a Show that the equation

can be written as

b Given that the equation has a root near two approximations. 5

The equation

has a root near

.

, use the Newton–Raphson method to find the next

. Use the Newton–Raphson method to find the next two

approximations to the root, giving your answers to five significant figures. 6

a Show that the equation

has a root between and .

b Using as the starting value, use the Newton–Raphson method to find the next approximation to the root. Give your answer correct to two decimal places. c Show that the approximation from part b is correct to one decimal place, but not to two decimal places. 7

a Show that the equation

has a root between and .

b Use the Newton–Raphson method to find this root correct to three decimal places, and show that the root you found is correct to 8

The equation

has a root between and .

Use the Newton–Raphson method to find this root correct to three significant figures, and show that the solution you found is correct to s.f.



Section 3: Limitations of the Newton–Raphson method The Newton–Raphson method can find a root to a high degree of accuracy very quickly, so it seems to be a very good method. Unfortunately, there are some situations when it does not work. For example, try to find a root of the equation . The sign- change check confirms that this equation has a root between and .The Newton–Raphson iteration formula is:

If you start from

you get

. But you cannot divide by zero, so it is impossible to find

You could try changing the starting point slightly, for example from the root but then comes back again.

. Try it: the sequence moves away

Key point 14.3 The Newton–Raphson method does not work if the starting value is a stationary point of . If the root is close to a stationary point, the sequence may initially move away from the root.

WORKED EXAMPLE 14.4 a Show that the equation b Explain why root.

has a root between and .

is not a suitable starting point for a Newton–Raphson iteration to find this

c Use the starting value

to find the root correct to three decimal places.

a Let

Use the sign change rule.

Then

There is a change of sign, so has a root between and . The Newton-Raphson method fails when find to check.

b

, so

so there would be division by zero in the Newton–Raphson formula. Use the Newton–Raphson formula with

c

The sequence converges to .

.

Continue the sequence until you can see what the limit is. You are not being asked to write down all the iterations.

.

Another situation where the Newton–Raphson method can fail is if the function has an asymptote so that falls outside of the domain of the function. WORKED EXAMPLE 14.5

a The equation from

has a root between

and . Explain why a Newton–Raphson iteration

fails to find this root.

b Starting from carry out three iterations of the Newton–Raphson method, showing the values correct to three decimal places. In order to use the Newton–Raphson formula first find need to use the quotient rule.

a

Use the formula with

But

is not defined for so it is not possible

to find

to find

. You

.

In order to find you need to use the formula with . But has a vertical asymptote at and is not defined for , so this is impossible.

.

b

You can use graphing software to verify the result in part a of Worked example 14.5: you can see that the tangent at

crosses the -axis to the left of the vertical asymptote.

EXERCISE 14C 1

For each equation carry out one iteration of the Newton–Raphson method starting with the given value of

. In each case sketch the graph (using technology) to explain why

is not a better

approximation to the root than

.

a i ii b i ii 2

Let

.

a Find the -coordinates of the stationary points of b Show that the equation

.

has a root between and .

c The Newton–Raphson formula with

is used to find

. Explain why

may not be an

improved approximation for the root. 3

The diagram shows the curve with equation

.

Find the -coordinate of the stationary point of and hence explain, with an aid of a diagram, why a Newton–Raphson iteration with will not converge to the root of .

4

Let

.

a Write down the values of between and The equation b Taking

for which

is not defined.

has a root between and . , use the Newton–Raphson method to find

.

c Explain why Newton–Raphson iteration cannot be continued to find a better approximation to the root. 5

The diagram shows the curve with equation

.

a Find the coordinates of the turning points of The curve crosses the -axis at

.

and

, with

The Newton–Raphson method is to be used to find the roots of

. , with

.

b To which root, if any, do the successive approximations converge when: i ii c Write down the range of values of for which the Newton–Raphson iteration converges to .

6

The function

has three zeros

between

and .

a For each zero, find two integers between which it lies. b Find the exact coordinates of the stationary points of

and sketch its graph.

c Find the coordinates of the point where the tangent to the graph at The Newton–Raphson method is to be used to find root d Explain why the iteration with e Use an iteration with 7

Let

.

does not converge to .

to find correct to three significant figures.

.

a Find the coordinates of the stationary point on the graph maximum point. b Show that The equation

crosses the -axis.

and show that this is a

has no points of inflection. Hence sketch the graph of

.

has two roots.

c Show that one of the roots is between and and find two integers between which the other root lies. The Newton–Raphson iteration is to be used to find the roots. d Explain why the starting value e State the range of values of root. 8

cannot be used to find the smaller root.

for which the Newton–Raphson iteration converges to the larger

Each equation has a root near the given starting value i explain why starting with

. In each case:

does not lead to a better approximation to the root

ii use technology to investigate for which starting values Newton–Raphson method works. a

to find the root between and .

b c

to find the root between and .

d 9

to find the root between and .

A function is defined by

.

a Show that, if is any root of the equation The graph of

, then

.

is shown.

The Newton–Raphson formula with a positive value of

is used to find

.

b State the range of positive values of .

for which

is closer than

c Use a diagram to show that there is a positive value of such that approximations converge to the positive root of .

10

a Sketch the graphs of

and

for

to the positive root of

, but subsequent

. Hence show that the equation tan

has a solution, , in that interval. b Consider the Newton–Raphson iteration with the starting value Show that the iteration converges to when

but not when

c There is a value such that the iteration with your sketch and draw the tangent to the curve at

. .

converges to . Show the value on .

d When , the Newton–Raphson iteration may not converge to . Describe two different cases that can arise.



Section 4: Fixed-point iteration You saw in Section 3 that the Newton–Raphson method does not always work. There are alternative methods you can use in such situations. In this section you will learn about fixed-point iteration, which also involves creating a sequence that gets closer and closer to the root, but in a different way from the Newton–Raphson method.

Did you know? The term, fixed-point iteration, refers to the fact that the solution is a fixed point of the function – it is the value where the output equals the input.

To use fixed-point iteration the equation needs to be rearranged into the form . Suppose you have a starting guess . If you have a found a solution of the equation. Otherwise you are looking for an improved guess. Since you want to equal

, it makes sense to try

.

You can see why this works by looking at the graph. The solution of the equation is the intersection of the graphs and . Starting from the point on the -axis you can find on the -axis by using the graph . To see whether is closer to the solution than was, you need to find on the -axis. You can do this by reflecting it in the line .

In the example shown in the diagram, is closer to the solution than . If you now repeat the same process you can hope to get closer and closer to the solution. In other words, you follow this sequence: start with

You can write the general rule for the sequence as . This sequence may converge to a limit, meaning that the terms of the sequence get closer and closer to a certain number . If this is the case, both and will get closer to so the sequence equation becomes and so is a solution of the equation .

In Worked example 14.6 the sequence increased towards the limit. But it is also possible for a sequence to oscillate above and below the limiting value.

Rewind Sequence rules and convergence were covered in Chapter 4.

Key point 14.4 Fixed-point iteration To solve an equation in the form using a starting guess

:

, generate a sequence

if this sequence converges to a limit, then this limit is a solution of the equation.

WORKED EXAMPLE 14.6

The equation

has a solution between and .

a Use fixed-point iteration with the starting value to

find the first five approximations to the

solution. Give the values correct to five significant figures. b Show that this solution is correct to two decimal places. a Sequence:

Start by inputting ‘ ’ into a calculator and use to generate subsequent values.

b The solution appears to be.

To check that the solution is correct to rearrange the equation into the form and look for a sign change between and .

There is a sign change, so the solution is between and so it equals . WORKED EXAMPLE 14.7

,

WORKED EXAMPLE 14.7

The equation has a root between and . The sequence used to find an approximation to this root.

with

is

a Draw a graph to illustrate the first three approximations. b Find the root correct to a

Draw the curve and the line . Start with on the -axis. Find the corresponding point on the curve and reflect it in the line to find . Repeat to find .

b

Start with sequence.

The sequence converges to

. use

to generate the

Continue until the third decimal place stops changing.

Key point 14.5 The graphs showing fixed-point iteration are often called staircase and cobweb diagrams.

EXERCISE 14D

EXERCISE 14D 1

Use fixed-point iteration with the given starting value to find the first five approximations to the roots of each equation. Give your answers to three decimal places. a i ii b i ii c i ii

2

Use fixed-point iteration with the given starting value to find an approximate solution correct to two decimal places. a i ii b i ii

3

Use fixed-point iteration to solve each equation. Draw a graph and use technology to investigate these questions. i Does the limit depend on the starting point? ii Does starting on different sides of the root give a different limit? iii If there is more than one root, which one does the sequence converge to? Does it matter where you start? a b c d

4

The equation

has a root between and . Use fixed-point iteration to find this root

correct to two decimal places. 5

a Use the iterative formula

with

to find

. Give your answer correct to four

decimal places. b This value of is an approximation to the root of the equation . Write down an expression for and suggest the value of the root correct to two decimal places. 6

The equation

has a root between and . Use fixed-point iteration to find this

root correct to three decimal places. Show the first three approximations correct to five decimal places. 7

a Use the iterative formula

to find

and

correct to four decimal places with

. b The sequence and .

converges to the root of the equation

. Find the values of

8

The diagram shows the graphs of of the equation . The value

and . The iteration is shown on the diagram.

converges to the root

The values of and correspond to the points marked and . Which of the two values corresponds to the point ?

9

a Show that the equation

has a root between and .

b Use fixed-point iteration with c Show that your value of 10

to find the next three approximations to the root.

gives the root correct to two decimal places.

The diagram shows the graphs of

and

.

a Use the diagram to show that the iteration

converges to a root of the equation

. b Use the iteration from part a to find this root correct to two decimal places.

11

c By rewriting the equation in the form the percentage error in your approximation.

, find the exact value of the root. Hence find

The diagram shows graphs of

.

and

a Show on the diagram that the iteration b Prove that the iteration converges to

with .

converges.



Section 5: Limitations of fixed-point iteration and alternative rearrangements Fixed-point iteration seems a little easier to implement than the Newton–Raphson method, as there is no need to differentiate or rearrange the equation in any other way. It also works in some situations when the Newton–Raphson method does not. However, there are other situations when fixed-point iteration does not work because the sequence fails to converge to the root. For example, try solving the equation

for

by using the iterative formula

Sketching graphs and trying some integer values of shows that there is a root between Starting the iteration at

. and

.

gives the sequence

The sequence is clearly getting further away from the root: the sequence diverges. You can also try starting the sequence below the root, at

; the resulting sequence is

Again, the sequence does not seem to be approaching the root. You can see this on the graph by drawing the staircase diagram.

If you continue the second sequence it actually converges to the other root of the equation:

But is there any way fixed-point iteration can be used to find the first root? You can write the equation

in a different way:

The sequence based on this rearrangement,

, starting at

, is:

Looking at the graph confirms that this sequence does indeed converge to the root between and . Interestingly, starting this iteration near the other root to one converging to the positive root. Starting at

Starting with you get is not defined for

leads either to a divergent sequence or gives the sequence

and then the sequence cannot be continued further because .

Key point 14.6 Some rearrangements of the equation lead to convergent sequences, and others to divergent sequences. A divergent sequence does not find the required root. If an equation has more than one root, different rearrangements may converge to different roots.

WORKED EXAMPLE 14.8 a Show that the equation

has a root between and .

b Starting with find the next four terms of the sequence behaviour of the sequence.

and describe the

c Use a cobweb diagram to illustrate the behaviour of the sequence. d Show that the equation

can be written as

. Use this rearrangement to

find an approximate solution of the equation correct to three decimal places. a Write

.

To use the change of sign method, rewrite the equation in the form .

There is a change of sign, so there is a root between and . b

The sequence seems to diverge.

The terms of the sequence are getting further away from each other.

c

d

Make sure you show all the steps in a ‘show that ...’ , question.

Use this new arrangement to form a sequence. It makes sense to start at again.

You are not asked to show all the approximations, so just keep going until the third decimal place stops changing. In this question you are also not asked to demonstrate that the root is correct to

Is there any way you can tell whether a sequence from a particular rearrangement would converge without trying it? Looking at some staircase diagrams suggests that the answer has something to do with the gradient of the graph of

at the point where it crosses the line

.

    In the first diagram the gradient of the graph of is smaller than the gradient of and the sequence seems to converge. In the second diagram, where the sequence diverges, the graph of steeper than the line . This suggests that the sequence converges when the gradient of is smaller than one and diverges when it is greater than one. It is less obvious what happens when the gradient of

is

is negative so that the iteration produces a

cobweb diagram. It turns out that it depends on whether the gradient of .

is smaller or greater than

    Both cases are summarised in Key point 14.7.

Key point 14.7 Fixed-point iteration converges if  diverges if 

If

or

: near the root near the root.

it is impossible to tell whether the sequence will converge without trying it.

WORKED EXAMPLE 14.9

The equation are

has a root near and

. Two possible rearrangements of this equation

.

Determine which rearrangement will produce a sequence that converges to the root. Write the sequence as

The iteration converges if  root.

near the

Differentiate g.

The root is near at this point.

, so evaluate the derivative

near the root, so this iteration will not converge.

near the root, so this iteration will converge.

EXERCISE 14E 1

For each sequence, find the first five terms, the of the sequence.

and the

term, and describe the behaviour

a b c d e f g 2

Each diagram shows the graphs of happens to the iteration defined by a

b

c

d

and

and the starting value .

. Describe what

e

f

g

3

Find an alternative rearrangement for each equation. a i ii b i ii c i ii d i ii

4

Find the missing constants to rearrange each equation into the given equivalent form. a i ii b i ii c i

ii d i ii 5

Each equation has more than one root. For each root, use technology to find a rearrangement that converges to it. a b c

6

The diagram shows the graph of a function and the line roots, marked and . Which of the roots does the iteration

. The equation has three converge to when:

a b Justify your answer by showing several iterations on the diagram.

7

The iterative formula The graphs of

is used to find an approximate solution of the equation and

.

are shown.

Use the diagram to determine the behaviour of the sequence

when:

a b c

8

The Newton–Raphson iteration for solving as a fixed-point iteration for solving

is .

. This can also be considered

a Express

in terms of

.

Let be the solution of the equation, and assume that b Find

for near .

, and hence prove that the Newton–Raphson method always converges when the

starting value is sufficiently close to . 9

a Show that the equation

has a root between

and

Let and define the sequence with b and hence determine to which of the two roots the sequence 10

The function real roots,

is defined by

, and another between and . . Solve the inequality converges.

. The equation

has two positive

.

a Show that is between and , and find two consecutive integers between which lies. b The rearrangement

is used to find an approximate root of the equation

. Without carrying out the iteration, determine to which of the two roots the sequence converges. c Show that an alternative rearrangement is

and find the constants

and . Use this

rearrangement to find the root correct to three decimal places. 11

a Find, in terms of , the roots of the equation Let b Find

where

.

.

, where is the non-zero root if the equation

c The iterative formula equation root.

with

.

is used to find approximate roots of the

. Find the set of values of for which the iteration converges to the non-zero

Checklist of learning and understanding Some equations can only be solved by finding numerical approximations. Numerical methods are methods that tell you how to find an improved approximation. When an equation is written in the form (solution) between

and

you can show that it has a root

by showing that

and

have different signs.

You can use the change-of-sign method to check the accuracy of the solution by ‘unrounding’ the number. Both methods involve creating a sequence

which converges to the root of the

equation. The first term of the sequence needs to be chosen from an interval that contains the root. The Newton–Raphson method works by approximating the curve by its tangent. The equation needs to be written in the form

. The resulting sequence is given

by:

The Newton–Raphson method fails to converge if the starting point is close to a stationary point of (where ). Fixed-point iteration requires the equation written in the form

. The

iteration sequence is given by: This iteration can be represented graphically as a staircase or a cobweb diagram.

    

Sometimes the sequence diverges, meaning that the terms get further away from the required root. To get a convergent sequence you may need to rearrange the equation. If the equation has several roots, different rearrangements may converge to different roots. The iteration converges if



near the root and diverges if

.

Mixed practice 14 In this exercise you will apply numerical solutions of equations in various contexts where the resulting equation cannot be solved exactly. 1

The recurrence relation If

is used to try to find the root of

, how can the sequence

.

be demonstrated?

Choose from these options. A cobweb diagram B staircase increasing diagram C staircase decreasing diagram D divergence diagram 2

has a roots

and

where

and

Using the Newton–Raphson method, which initial values,

.

, will fail to find either of these

roots? Choose from these options. A B C D 3

The graph of

has a stationary point between and .

a Show that the -coordinate of the stationary point satisfies

.

b Use fixed-point iteration with a suitable starting point to find the -coordinate of the stationary point correct to three decimal places. 4

The diagram shows a sector of a circle with radius of the shaded region equals . a Show that

. The angle at the centre is . The area

.

b Show that the equation in part a has a root between and . c Use the Newton–Raphson method with a suitable starting point to find the value of correct to two decimal places.

5

a Sketch the graphs of and of solutions of the equation

on the same set of axes. State the number .

b Show that the equation in part a has a solution between and . c Use an iteration of the form

to find this solution correct to two

decimal places. 6

The diagram shows the graph of coordinate axes and the line .

. The shaded area is enclosed by the curve, the

a Given that the shaded area equals square units, show that

.

b Show that the equation in part a has a root between and . c Use the Newton–Raphson method to find the value of correct to three decimal places.

7

The equation

has exactly one root, . Taking as a first approximation to , use the Newton–Raphson method to find a second approximation, , to . Give your answer to four significant figures. [© AQA 2013] 8

Let

for

.

a Show that the equation b Starting from

has a root between

and

.

use the Newton–Raphson method to find

.

c Explain why this iteration cannot be continued to find the root. d Use the Newton–Raphson method, with places. 9

A curve is defined by

for

, to find the root correct to three decimal

.

a Show that the -coordinate of any stationary point on the curve satisfies the equation . One of the roots of this equation is between and . b By considering the derivative of converge to this root.

prove that the iteration

c Find an alternative rearrangement of the equation coordinate of the stationary point on the curve

does not

and use it to find the between

and

.

Give your answer correct to three decimal places. 10

a The equation

has a single root, .

Show that lies between and . b Use the recurrence relation answers to three decimal places. c The diagram shows parts of the graphs of

, with

, to find

and

and

, giving your

, and a position of

.

On a copy of the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of and on the -axis. [© AQA 2013] 11

A ball is thrown vertically upwards from the roof of a house. Its height metres above the ground after time seconds is . a Write down the height of the ball above ground level when it was released. b i Show that one possible iterative formula for finding an approximation to the maximum height reached by the ball is

.

ii Find the set of values of for which the iteration will converge to a root. iii Using the formula in part b i, find this root to two decimal places and comment on your answer. iv Find a second iterative formula for finding the maximum height. v

Given that the root is near places.

, use this second formula to find the root to two decimal

c Given that the time taken for the ball to hit the ground is between and seconds, use the Newton–Raphson method to find this time, correct to two decimal places.

Worksheet See Extension Sheet 14 for a look at Euler’s method for finding approximate solutions to differential equations.

15 Numerical integration In this chapter you will learn how to: understand why definite integration is connected to area under a curve approximate integrals that cannot be found exactly establish whether these approximations are overestimates or underestimates.

Before you start… GCSE

You should be able to calculate

1 Find the area of this shape.

the area of a trapezium.

Student Book 1, Chapter 14

You should know that a definite integral represents the area between the curve and the -axis.

Student Book 1, Chapter 16

You should know how to find distance travelled from a velocity– time graph.

2 Shade the area given by

3 Find the total distance travelled in the first seconds.

An approximation to definite integration Using definite integration to find the area between a curve and the -axis has many applications, for example, in Mechanics (finding distance travelled from a velocity–time graph) and Statistics (where the area under the normal distribution curve represents probability). Since you already know many different integration methods, you may think that in most cases the area under the curve can be found exactly. It turns out that in many real-world problems in engineering and finance, for example, it is actually not possible to integrate the function exactly (or that the integration method is very complicated). In such cases you can find an approximation for the area, using one of the methods from this chapter.

Section 1: Integration as the limit of a sum The simplest way to estimate the value of an area is to split it into rectangles. For example, you will need an approximation to the area under the curve

between

because it turns out that it is impossible to find the exact vale of the integral

and .

Fast forward You will see in Chapter 21 that the function is used when finding probabilities from a normal distribution, so it is important to be able to find the area under its graph.

In the diagram, the required area has been split into five rectangles of equal width,

. The height of

each triangle is the -coordinate of its top-right vertex; so the height of the first rectangle is the value of when

which is

. The table shows the heights of all five rectangles.

-coordinate height The total area of the five rectangles (each of width

) is

and this is an approximate value of the required area. You can see from the diagram that the actual area is a little bit larger; a lower bound for the area is . Unfortunately, this procedure does not tell you anything about how good the approximation is. To assess this, it would also be useful to have an upper bound, so that you know two numbers between which the area lies. To find an upper bound, you can use rectangles with the top edge above the curve. You can draw such rectangles by using the left end-point of each interval. The heights of the five rectangles are: -coordinate height

So the total area is

You can therefore say that the required area under the curve is certainly between can write

and

, and you

.

Key point 15.1 You can find upper and lower bounds for the area under a curve by using rectangles that lie above and below the curve. The actual area lies between the lower and the upper bounds.

WORKED EXAMPLE 15.1

The diagram shows a part of the graph of

where is in radians.

Using six rectangles of equal width, find a lower bound for the value of

.

From the graph you can see that for the lower bound (rectangles below the curve) you need to use coordinates at the right end-point of each interval.

The width of each rectangle is

.

Each rectangle has width coordinate.

and height equal to the -

So

WORKED EXAMPLE 15.2

A part of the curve with equation

is shown in the diagram.

Use four rectangles of equal width to find an upper bound for From the graph you can see that for the upper bound you now need to use -coordinates at the right end-point of each interval.

The width of each rectangle is

.

Each rectangle has width and height equal to the coordinate. So

WORK IT OUT 15.1 Three students tried to find an upper bound for the value of

, using six

rectangles. Which solution is correct? Identify the errors in the incorrect solutions.

Solution 1 Using

So

Solution 2

:

Using

:

So

Solution 3 Using

:

So

To get a more accurate approximation for the area, the upper and lower bounds need to be closer to each other. You can do this by using more rectangles of smaller width. For the original example, estimating , you can use a spreadsheet to calculate areas using more and more rectangles. Here are some of the results.

Tip Always use the graph to decide which rectangles to use.

Number of rectangles

Width



Lower bound

Upper bound



You can use a graphical calculator or graphing software to check that the actual area is

.

Key point 15.2 As the number of rectangles increases, the upper and the lower bounds approach a limit, which is the actual value of the definite integral. Using rectangles of height and width

EXERCISE 15A 1

Use five rectangles to find upper and lower bounds for the value of each integral. Use technology to draw the graph first. a i ii b i ii c i ii

2

Use a spreadsheet to find upper and lower bounds for each integral, using: i

rectangles

ii

rectangles

iii

rectangles.

In each case find the difference between the upper and lower bounds. How does the difference decrease when the number of rectangles doubles? a b

c d 3

The diagram shows part of the curve with equation

.

Using six rectangles of equal width, find upper and lower bounds for

4

The diagram shows the graph of a

.

.

Use five rectangles of equal width to find upper and lower bounds for

.

b How could the difference between the upper and lower bounds be reduced?

5

a Sketch the graph of

for

.

b Use four rectangles of equal width to find an upper bound for c If 6

rectangles were used, would the upper bound increase or decrease?

rectangles of width and

.

are used to estimate the area under the graph of

between

.

Find

7

The diagram shows a part of the graph

.

a State the exact coordinates of the maximum point on the curve. b In the part of the graph shown in the diagram, use four rectangles of equal width to find a lower

bound for



.

Section 2: The trapezium rule You saw in Section 1 that you can approximate a definite integral by using sums of areas of rectangles. However, you may need many rectangles to achieve high accuracy. You can improve this method by replacing each rectangle by a trapezium that connects the end-points of the interval. Each trapezium has an area somewhere between the upper and the lower rectangles.

The area of a trapezium is given by

, where and are the parallel sides and is the height.

Looking at the first trapezium, its height is equal to the width of the interval have lengths equal to the -coordinates of the end-points. If there are intervals, you can label the -coordinates coordinates . The areas of the trapezia are then:

and the parallel sides

and the corresponding -

When you add these together, each -coordinate appears twice, except for the first and the last one.

Key point 15.3 The area under a curve using equal intervals with end-points trapezium rule:

where

and

is given by the

.

This will be given in your formula book.

You may be asked to find an approximation by using a certain number of intervals, or by using a certain

number of ordinates. The ordinates are the -values you substitute into the formula.

Tip It is a good idea to set out the - and -values in a table. The formula then says

times

(first + last + twice the rest), where is the difference between the -coordinates.

Common error ordinates means intervals. So, for example, if you use the trapezium rule with five ordinates between and , this means four intervals each of width , not five intervals of width .

WORKED EXAMPLE 15.3

Use the trapezium rule with five equal intervals to find an approximate value of

. Give

your answer correct to three decimal places. Divide the interval from to into five equal parts. The -values start from and go up in steps of until they reach . Note that since there are five intervals, there should be six -values. The -values are calculated from . Since you want the answer correct to d.p., record the values to d.p. and round at the end. Use the formula. You should show the numbers used the in the calculation.

Tip You may have a TABLE function on your calculator which will produce the table of values. Alternatively, you can save the six numbers in memory to use in the trapezium rule calculation. Or if the expression for

is short, you can just type in the whole sum as

The actual value of the integral in Worked example 15.3 is ( d.p.), so the approximation is within of the correct value. In Section 1, rectangles were needed to get this close. This demonstrates that the trapezium rule, although slightly more complicated, is a lot more accurate than using rectangles.

Did you know? Your calculator may have a function for finding approximate values of definite integrals. It probably uses the trapezium rule.

In this case the approximation is a slight underestimate. Looking at the graph explains why this is – most of the trapezia lie below the curve.

Rewind You can see from the graph that the trapezium rule gives an underestimate when the function is concave – see Chapter 12, Section 1, for a reminder of the definition of a convex function.

Key point 15.4 The trapezium rule will: underestimate the area when the curve is concave

overestimate the area when the curve is convex.

WORKED EXAMPLE 15.4

The diagram shows the graph of a

.

Use the trapezium rule with five ordinates to estimate the value of

giving your

answer correct to three significant figures. b Explain whether your answer is an overestimate or an underestimate.

Divide the interval from to into four

a

equal parts. The -coordinates start from and increase in steps of

. The -

coordinates are found using

.

Use the trapezium rule formula.

b Since the function is convex between the trapezia are above the curve.

and

The explanation should refer to the shape of the graph.

Hence this approximation is an overestimate.

WORK IT OUT 15.2 Three students were asked to estimate the value of

using trapezium rule with five

strips. Which solution is correct? Identify the errors in the incorrect solutions. Solution 1

Solution 2

Solution 3

EXERCISE 15B 1

Use the trapezium rule, with the given number of intervals, to find the approximate value of each integral. Compare your answer to the upper and lower bounds found in question 1 of Exercise 15A. a i ii

, five intervals , four intervals

b i ii c i ii 2

, five intervals , four intervals , six intervals , three intervals

For each integral from question 1: i

use technology to find its value, correct to eight decimal places

ii use a spreadsheet to calculate trapezium rule approximations, using

and

intervals

iii find the percentage error in each estimate. How do the percentage errors decrease when you double the number of intervals? 3

For each integral, find either its exact value where possible, or an approximation using six trapezia. a b c d e f

4

a Sketch the graph of b

.

Use the trapezium rule with six ordinates to estimate the value of

. Give your

answer to two decimal places. c Explain whether your answer is an overestimate or an underestimate. 5

a

Use the trapezium rule with four intervals to find an approximate value of

.

b Describe how you could obtain a more accurate approximation. 6

The diagram shows a part of the graph of

.

The graph crosses the -axis at the point where

.

a Find the exact value of . b Use the trapezium rule with five ordinates to find an approximation for c Is your approximation an overestimate or an underestimate? Explain your answer.

.

7

A particle moves in a straight line with the velocity given by , where is measured in and in seconds. Use the trapezium rule with six strips to find the approximate distance travelled by the particle in the first seconds.

Worksheet See Support Sheet 15 for a further example of using the trapezium rule and for more practice questions. 8

The velocity, , of a particle moving in a straight line is given by shows the velocity–time graph for the particle. a The particle changes direction when

and

(with

. The diagram

). Find the exact values of and .

b Use the trapezium rule, with eight intervals, to estimate the total distance travelled by the particle during the first seconds.

Checklist of learning and understanding Some definite integrals cannot be evaluated exactly. In those cases it is possible to use rectangles to find upper and lower bounds. An upper bound for an area is a number that is larger than the area; a lower bound is a number that is smaller than the area. As the rectangles get smaller, the upper and lower bounds get closer to each other. The actual area is the limit of the sum of the rectangles. The trapezium rule is a way of using trapezia to estimate the area. You need fewer trapezia than rectangles to achieve the same accuracy. Using equal intervals with end-points

where

, the area under the curve is:

and

The trapezium rule will: underestimate the area when the curve is concave overestimate the area when the curve is convex.



Mixed practice 15 1

The trapezium rule with seven ordinates is used to find an estimate for

.

What is the width of each interval? Choose from these options. A B C D 2

The diagram shows a part of the graph of

.

Using five rectangles of equal width, find a lower bound for the value of

.

Give your answer correct to one decimal place.

3

The diagram shows a part of the curve with equation a

.

Use the trapezium rule with six ordinates, to estimate the value of

.

b State, with a reason, whether your answer is an underestimate or an overestimate. c Explain how you could find a more accurate estimate.

4

A curve , defined for by the equation , where is in radians, is sketched here. The region bounded by the curve , the -axis from to and the line is shaded. The area of the shaded region is given by

, where is in radians.

Use the trapezium rule with five ordinates (four strips) to find an approximate value for the area of the shaded region, giving your answer to three significant figures.

[© AQA 2011] 5

a Use the trapezium rule with five ordinates (four strips) to find an approximate value for giving your answer to three significant figures. b State how you could obtain a better approximation to the value of the integral using the trapezium rule. [© AQA 2012]

6

Use six rectangles of equal width to find lower and upper bounds and such that .

7

The trapezium rule with four equal intervals is used to find an estimate for If f is a decreasing function and trapezium rule if intervals are used?

.

, what will happen to the value obtained by the

Choose from these options. A Increase B Decrease C Stay the same D Depends on the particular function  . 8

The diagram shows a part of the curve with equation . Use the trapezium rule with six strips of equal width to estimate the area enclosed between the curve and the axis.

9

A particle moves in a straight line so that its velocity is given by the equation a Find the values of

.

when the velocity is zero.

b Use the trapezium rule with strips of width to find an approximate value of the distance travelled by the particle from

to

.

10

A part of the graph of

is shown in the diagram.

Use four rectangles of equal width to find a rational number such that .

11

A curve has equation

.

a Sketch the curve, showing the coordinates of any stationary points. b 12

Use four rectangles of equal width to find a lower bound for

.

The diagram shows the velocity–time graph for a particle moving in a straight line. The velocity of the particle is measured at the table.

-second intervals and the results are recorded in

Estimate the average speed of the particle during the

seconds.

Worksheet See Extension Sheet 15 for a look at how approximation methods were used in the development of calculus.



16 Applications of vectors In this chapter you will learn how to: use displacement, velocity and acceleration vectors to describe motion in two dimensions use some of the constant acceleration formulae with vectors use calculus to relate displacement, velocity and acceleration vectors in two dimensions when acceleration varies with time represent vectors in three dimension using the base vectors and use vectors to solve geometrical problems in three dimensions.

Before you start… Student Book 1, Chapter 15

You should be able to link displacement vectors to coordinates and perform operations with vectors.

1 Consider the points and . Let and Write in column vector form:

.

a b c d

Student Book 1,

You should be able to find the

Chapter 15

magnitude and direction of a vector.

Student Book 1, Chapter 16

You should understand the concepts of displacement and distance, instantaneous and average velocity and speed and acceleration.

.

2 Find the magnitude and direction of the vector

.

3 In the diagram, positive displacement is measured to the right.

A particle takes seconds to travel from to and another seconds to travel from to . Find: a the average velocity b the average speed for the whole journey. Student Book 1, Chapter 16

You should be able to use calculus to work with displacement, velocity and acceleration in one dimension.

4 A particle moves in a straight line with velocity . Find: a the acceleration when b an expression for the

.

displacement from the starting position. Student Book 1,

You should be able to use

Chapter 17

constant acceleration formulae in one dimension.

5 A particle accelerates uniformly from to distance of

while covering a in a straight line.

Find the acceleration. Chapter 12

You should be able to work with curves defined parametrically.

6 Find the Cartesian equation of the curve with parametric equations .

Why do you need to use vectors to describe motion? In Student Book 1, you studied motion in a straight line. You saw how displacement, velocity and acceleration are related through differentiation and integration:

In the special case when the acceleration is constant, you can use the constant acceleration equations:

But the real world has three dimensions, and objects do not always move in a straight line. You need to be able to describe positions and motion in a plane (such as a car moving around a race track) or in space (for example, flight paths of aeroplanes). This requires the use of vectors to describe displacement, velocity and acceleration.

Section 1: Describing motion in two dimensions When a particle moves in two dimensions, the displacement, velocity and acceleration are vectors. The distance and speed are still scalars.

Rewind Vectors were introduced in Student Book 1, Chapter 15. Remember that you can also write the vector

as

.

WORKED EXAMPLE 16.1

Points

and have position vectors

and

, where the distance is measured

in metres. A particle travels in a straight line between each pair of points. It takes seconds to travel from to and then a further seconds to travel from to . Find: a the average velocity and average speed from to b the final displacement of the particle from c the average velocity for the whole journey d the average speed for the whole journey.

a

First find the displacement from to . This is the difference between the position vectors. Remember:

Then use

.

Use Pythagoras’ theorem to find the distance Then use

.

.

The displacement from to is the difference between their position vectors. It is highlighted blue in the diagram.

b

c Total time

To get the total distance travelled, add the distance from to to the distance from to . This is highlighted red in the diagram.

d



You have already found that the total time is seconds.

Common error Notice that in Worked example 16.1 the average speed is not the magnitude of the average velocity vector. This is because the particle changes direction during the motion. You can calculate average acceleration by considering the change in velocity. WORKED EXAMPLE 16.2

A particle moves in a plane. It passes point with velocity seconds later with velocity .

and passes point

Find the magnitude and direction of the average acceleration of the particle between and . First find the acceleration vector using average acceleration

.

Then find its magnitude.

You can use any angle to define the direction as long as you clearly state where you are measuring the angle from. The angle above (or anti-clockwise from) the vector is the most common choice. Draw a diagram to make sure you find the angle you intended!

The direction is

above the vector .

Acceleration may cause a change in the direction of the velocity as well as its magnitude (speed). This means that the object will not necessarily move in a straight line. If you know how the displacement vector varies with time, you can sometimes find the Cartesian equation of the object’s path.

Rewind The displacement vector gives the parametric equations of the object’s path, with the parameter being time. See Chapter 12, Section 2, for a reminder of parametric equations.

Tip The path an object follows is also called a trajectory.

WORKED EXAMPLE 16.3

A particle moves in a plane. At time the particle is at a point and its displacement from the origin is given by the position vector a Prove that the particle moves along a parabola. b Sketch the parabola. a The coordinates of are: The two components of the position vector give the - and -coordinates of the particle’s position. You want to find in terms of , so to eliminate , substitute from the first equation into the second equation.

Hence the path of the particle is a parabola. Note that, since parabola with

b

, the object covers only a part of the .

You can also look at two particles moving in a plane and ask questions about the distance between them, and whether they ever meet. To do this you need to work with position vectors.

Key point 16.1 If a particle starts at the point with position vector and moves with constant velocity , its position vector at time is . Two particles, and , meet if

for the same value of .

Rewind In Worked example 16.4, part is an example of proof by contradiction – see Chapter 1 for a reminder.

WORKED EXAMPLE 16.4

Two particles, and , move in the same plane. starts from the origin and moves with constant velocity . a Write down the position vector of in terms of . Particle starts from the point with position vector .

and moves with constant velocity

b Prove that and never meet. c Find the minimum distance between the two particles. a

Use origin.

b Position vector of is

This time

with

.

as starts at the

The particles meet when

:

From the first equation:

You want to show that and are never in the same place at the same time. So try to find the value of when the two displacements are equal and show that this is impossible.

If two vectors are equal then both components have to be equal. So you need a value of that works in both equations.

Check in the second equation:

There is not a value of which makes the two position vectors equal.

Hence for all so the particles never meet. c At time ,

The distance between the particles is the magnitude of the displacement between them, which is found by subtracting the two position vectors.

The distance between and is:

This expression has a minimum value when its square has a minimum value; so look at

to

avoid having to work with the square root. Let

You could complete the square to find the minimum value, but it is not easy with these numbers so differentiate instead.

The minimum value of is:

Don’t forget that this is the minimum value for .

Hence the minimum distance

is

EXERCISE 16A 1

Points

and have position vectors

and

metres. Find the average velocity if the particle travels: a i from to in seconds ii from to is seconds b i from to in seconds

, where distance is measured in

ii from to in seconds c i from to in seconds and then from to is seconds ii from to in seconds and then from to in seconds. 2

Find the average acceleration vector, and the magnitude of average acceleration in each case. a i The velocity changes from

to

ii The velocity changes from

to

in

in seconds.

b i A particle accelerates from rest to

in seconds.

ii A particle accelerates from rest to 3

seconds.

in

Three points have coordinates

and

seconds.

.

a A particle travels in a straight line from to in seconds. Find its average velocity and average speed. b Another particle travels in a straight line from to in seconds and then in a straight line from to in seconds. Find its average velocity and average speed. 4

A particle moves in the plane so that its displacement from the origin at time is given by the vector

.

a Find the particle’s distance from the origin when

.

b Find the Cartesian equation of the particle’s trajectory. 5

a An object’s velocity changes from

to

in seconds.

Find the magnitude of its average acceleration. b The object then moves for another

seconds with average acceleration

Find its direction of motion at the end of the 6

seconds.

A particle travels in a straight line from point , with coordinates coordinates

. The journey takes

.

, to point with

seconds and the distance is measured in metres.

a Find the average speed of the particle. The particle then takes a further seconds to travel in a straight line to point with coordinates . b Find the displacement from to . c Find the average velocity of the particle for the whole journey. d Find the average speed for the whole journey from to . Explain why this is not equal to the magnitude of the average velocity. 7

Two particles, and , move in the plane. has constant velocity displacementfrom the origin is

and its initial

. starts from the origin and moves with constant velocity

. Show that the two particles meet and find the position vector of the meeting point. 8

A particle moves in a plane so that this displacement from the origin at time vector

.

is given by the

a Find the distance of the particle from the origin when

.

b Sketch the trajectory of the particle. 9

An object moves with a constant velocity

. Its initial displacement from the origin is

. a Find the Cartesian equation of the particle’s trajectory. b Find the minimum distance of the particle from the origin. 10

A particle moves in the plane so that its displacement from the origin at time seconds is . Find the maximum distance of the particle from the origin.



Section 2: Constant acceleration equations When a particle moves with constant acceleration, you can use formulae analogous to those for onedimensional motion.

Rewind See Student Book 1, Chapter 17, for a reminder of the constant acceleration formulae.

Key point 16.2 Constant acceleration formulae in two dimensions:

These will be given in your formula book.

Notice that these formulae give displacement ( ) rather than position vector ( ). To find the position vector at time , you need to add on the particle’s initial position vector which gives . This is the same idea as in Key point 16.1, except that now the velocity isn’t constant so you need an equation of motion for rather than just

.

Fast forward The list in Key point 16.2 does not contain the vector version of the formula you study Further Mathematics, you will meet a way of multiplying vectors (called the scalar product) that enables you to extend this formula to two dimensions as well.

. If

WORKED EXAMPLE 16.5

A particle starts with initial velocity seconds its velocity is . Find:

and moves with constant acceleration. After

a the displacement from its initial position b the distance from the initial position at this time. a

Write down what you know and what you want.

Use

.

b Distance from starting position:

Distance is the magnitude of the displacement.

Since the particle does not move in a straight line, the distance from the starting point (which is measured in a straight line) is not the same as distance travelled (which is along a curve). You need to be a little careful when solving equations with vectors. If you are comparing two sides of a vector equation, corresponding components need to be equal.

WORKED EXAMPLE 16.6

A particle moves with constant acceleration velocity is

. It is initially at the origin and its initial

.

Find the time when the particle is at the point with position vector

.

Write down what you know and what you want. As the particle starts at the origin, its position vector at time t gives its displacement.

Use

First component:

The first component of the vectors gives a quadratic equation for . This will give two possible values.

Second component:

You need to check for which of these values of the second component equals

when when The particle is at seconds. WORK IT OUT

when

16.1 A particle moves with constant acceleration, starting from the origin. Its position vector at time is given by . How many times does the particle pass through the origin during the subsequent motion? Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 At the origin the position vector is zero. When or So the particle passes through the origin once, when

.

Solution 2 At the origin both components of the position vector are zero. When or When or So the particle passes through the origin twice, when

and

Solution 3 At the origin both components of the position vector are zero. When

or

When or So the particle does not pass through the origin again.

You saw in Student Book 1, Chapter 18, that Newton’s second law still applies in two dimensions: , where and are vectors and is a scalar. This means that the particle accelerates in the direction of the net force. You can now use this in conjunction with the constant acceleration formulae for vectors. WORKED EXAMPLE 16.7

A particle of mass

moves under the action of a constant force,

is at the point with position vector seconds its position vector is

moving with velocity . Find the vector .

. When

the particle

and when

Since the force is constant the acceleration is also constant, so you can use the constant acceleration equations. The displacement, , is the change in position vector.

Rearrange to find

Now use

.

EXERCISE 16B 1

In each question the particle moves with constant acceleration. Time is measured in seconds and displacement in metres. a i ii b i ii c i ii d i ii

2

, find when , find when , find when , find when , find . , find . , find . , find .

An object of mass moves under the action of the force the speed of the object at time seconds, in each case.

. The object is initially at rest. Find

a i ii b i ii 3

An object moves with constant acceleration

and initial velocity

.

Find its velocity and the displacement from the initial position after seconds. 4

A particle moves with constant acceleration vector

and its velocity is

. It is initially at the the point with position .

a Find the distance of the particle from the origin after seconds. b Find the direction of motion of the particle at this time. 5

A particle passes the origin with velocity

and moves with constant acceleration.

a Given that seconds later its velocity is

, find the acceleration.

b Find the time when the particle’s displacement from the origin is 6

An object moves with constant acceleration. When When

.

it has velocity

its displacement from the initial position is

.

.

Find the magnitude of the acceleration. 7

A particle moves with constant acceleration. Its initial velocity is displacement from the initial position is .

. seconds later, its

Find its direction of motion at this time. 8

An object moves with constant acceleration and initial velocity When its displacement from the initial position is

.

, its velocity is

.

Find the magnitude of the acceleration. 9

A particle moves with constant acceleration

. Given that its initial velocity is

, find the time when its displacement from the initial position is 10

A particle of mass the particle is

is subjected to a constant force .

.

The initial velocity of

Find the velocity of the particle after seconds. 11

A particle starts with initial velocity

and moves with constant acceleration

Prove that the speed of the particle increases with time. 12

A particle moves with constant acceleration When

.

the particle is at rest, at the point with the position vector

.

Find the shortest distance of the particle from the origin during the subsequent motion.



.

Section 3: Calculus with vectors When the acceleration is not constant you need to use differentiation and integration to find expressions for displacement and velocity. In Student Book 1, Chapter 15, you learnt how to do that for motion in one dimension. The same principles apply to two-dimensional motion: differentiating the displacement equation gives the velocity equation, and differentiating the velocity equation gives the acceleration equation. The only difference is that those quantities are now represented by vectors.

Key point 16.3 To differentiate or integrate a vector, differentiate or integrate each component separately.

WORKED EXAMPLE 16.8

A particle moves in two dimensions. Its position vector, measured in metres, varies with time (measured in seconds) as Find the speed of the particle when

. To find the velocity, differentiate the displacement vector. Do this by differentiating each component separately.

When

:

Speed is the magnitude of velocity.

When using integration with vectors, the constant of integration will also be a vector. WORKED EXAMPLE 16.9

A particle moves with acceleration

. The initial velocity is

.

Find an expression for the velocity at time . To find the velocity, integrate the acceleration vector. Do this by integrating each component separately.

When

Use the initial velocity to find .

You can include the constant within the existing vector.

So

Remember that, for two vectors to be equal, corresponding components need to be equal. WORKED EXAMPLE 16.10

A particle starts from point with velocity .

. Its acceleration is given by

Show that the particle never returns to . You need to find the displacement vector from and show that this is never for First integrate to find .  when

, so

Use the initial velocity to find

Now integrate the velocity to find the displacement vector. Initially , so

If then the component of the displacement is :

When

the component of

displacement is:

Now check if there is any value of (other than ) when . First find the time at which the component is .

is the starting point so use

and check

whether the component is

Hence for , so the particle does not return to the starting point.

WORK IT OUT 16.2 A particle starts from rest and moves with acceleration

. Find an expression

for the velocity of the particle. Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

Initially at rest means that the speed is zero. When

:

Solution 2

Initially at rest:

Solution 3

Initially at rest, so

When the force is variable (a function of ) you can still use , but because the acceleration will now be variable, you need to use calculus rather than the constant acceleration formulae. WORKED EXAMPLE 16.11

A particle of mass . Find: a the speed

starts from rest and moves under the action of the force

b the direction of motion of the particle after seconds. a

Find the velocity by integrating the acceleration. Use first.

When

When

so

:

The speed is:

Remember to find .

You can now use the given value of to find the velocity vector. Speed is the magnitude of the velocity vector.

The direction is given by the angle with the vector .

b

It is always a good idea to draw a diagram to see which angle you are looking for.

The direction of motion is above the horizontal.

EXERCISE 16C 1

For a particle with the given position vector, find expressions for the velocity and acceleration vectors. Also find the speed when . a i ii b i ii

2

For a particle moving with the given acceleration, find expressions for the velocity and position vectors. The particle is initially at the origin, and the initial velocity is given in each question part. Also find the distance from the starting point when a i ii b i ii

.

c i ii 3

A particle moves in a plane with position vector given by

.

a Find an expression for the velocity of the particle at time . b Find the speed of the particle when 4

.

A particle moves in a plane, starting from rest. Its acceleration varies according to the equation

a Find an expression for the velocity of the particle at time . b Find the displacement from the initial position after seconds. 5

The velocity of a particle, in

, moving in a plane is given by

.

a Find the initial speed of the particle. b Find the magnitude of the acceleration when

.

c Find an expression for the displacement from the initial position after seconds. 6

A particle starts from rest and moves with acceleration Find its distance from the initial position after

7

.

seconds.

The velocity of a particle moving in a plane is given by

.

Show that the particle never returns to its initial position. 8

For a particle moving in two dimensions, the displacement vector from the starting point is given by a The components of the displacement vector give parametric equations of the trajectory of the particle, . Use parametric differentiation to find the gradient of the tangent to this curve,

, when

.

b Find the velocity vector when 9

A particle of mass

. What do you notice?

moves under the action of the force

Its initial velocity is

.

.

a Show that the speed of the particle is constant. b By considering the and components of the displacement vector, show that the particle moves in a circle. 10

A particle moves in the plane, from the initial position

. Its velocity,

, at time is

given by the equation: Find the time when the particle is closest to the origin, and find this minimum distance.



Section 4: Vectors in three dimensions In the preceding sections you learnt how to use vector equations to represent motion in two dimensions. Now vector methods will be extended to enable you to describe positions and various types of motion in the three-dimensional world. To represent positions and displacements in three-dimensional space, you need three base vectors, all perpendicular to each other. They are conventionally called .

You can also show the components in a column vector:

.

Each point in a three-dimensional space can be represented by a position vector, which equals its displacement from the origin. The displacement from one point to another is the difference between their position vectors.

WORKED EXAMPLE 16.12

Points and have coordinates

and

, respectively. Write as column vectors:

a the position vectors of and b the displacement vector

.

a

The components of the position vectors are the coordinates of the point.

b

Relate

to the position vectors and exactly as you

would in two dimensions.

Did you know? Although our space is three dimensional, it turns out that many situations can be modelled as motion in two dimensions. For example, it is possible to prove that the orbit of a planet lies in a plane, so two-dimensional vectors are sufficient to describe it.

The formula for the magnitude of a three-dimensional vector is analogous to the two-dimensional one.

Key point 16.4 The magnitude (modulus) of a vector

is

The distance between points with position vectors and is

. .

WORKED EXAMPLE 16.13

Points and have position vectors

and

. Find the exact distance

. The distance is the magnitude of the displacement vector, so you need to find

first.

Now use the formula for the magnitude.

Remember that you can use vector addition and subtraction to combine displacements. WORKED EXAMPLE 16.14

The diagram shows points

such that

. Write each vector in component form. a b c

.

and

a

You can get from

to via .

b

Going from to .

c

You can get from to via from previous parts.

is the reverse of going from

EXERCISE 16D 1

Write each vector in column vector notation (in three dimensions). a i ii b i ii

2

Let

and

. Find each vector.

a i ii b i ii c i ii d i ii 3

Let a i ii b i ii c i ii

and

. Find each vector.

, using the answers

to

4

Find the magnitude of each vector in three dimensions.

5

Find the distance between each pair of points in three dimensions. a i

and

ii

and

b i

and and

ii 6

Find the distance between the points with the given position vectors. a

and

b

and

c

and

d 7

and

Given that a

, find the vector such that:

is the zero vector

b

is the zero vector

c d

.

8

Given that

9

Given that .

and

find vector such that and

, find the value of the scalar such that

10

Find the possible values of the constant such that the vector

11

Let

12

Points and are such that

and

has magnitude

. Find the possible values of such that

and

the possible values of such that 13

.

.

where is the origin. Find

.

Points and have position vectors value of for which the distance

.

and

. Find the

is minimum possible and find this minimum distance.

Worksheet See Support Sheet 16 for a further example of three-dimensional vectors and for more practice questions.



Section 5: Solving geometrical problems This chapter finishes with a review of how you can use vector methods to solve geometrical problems. You have already used these results: the position vector of the midpoint of line segment

is

if vectors and are parallel then there is a scalar so that the unit vector in the same direction as is

.

Rewind These results were introduced in Student Book 1, Chapter 15.

WORKED EXAMPLE 16.15

Points

have position vectors

Point is the midpoint of

.

.

a Find the position vector of . b Show that

is a parallelogram. Draw a diagram to show what is going on.

a

As you are given position vectors, it may help to show the origin on the diagram.For this part, you only need to look at points and

Tip Vector diagrams do not have to be accurate or to scale to be useful. A two-dimensional sketch of a three-dimensional situation is often enough to show you what’s going on.

b

In a parallelogram, opposite sides are equal and parallel, which means that the vectors corresponding to those sides are equal. So you need to show that

.

is a parallelogram.

WORKED EXAMPLE 16.16

Given vectors

and

a Find the values of and such that is parallel to . b Find the value of scalar such that

a Write Then:

for some scalar .

is parallel to vector

.

If two vectors are parallel you can write

.

If two vectors are equal then all their corresponding components are equal.

Write vector

b

in terms of and then use .

Parallel to

You can find from the first equation, but you need to check that all three equations are satisfied.

Check in the third equation:

WORKED EXAMPLE 16.17

a Find the unit vector in the same direction as

.

b Find a vector of magnitude parallel to . a Let the required unit vector be .

Find the magnitude of .

Then divide by its magnitude to produce a vector in the same direction as but of length .

b Let be parallel to and . Then

To get a vector of magnitude , multiply the unit vector by .

Tip Note that part has two possible answers, as could be in the opposite direction. To get the second answer you would take the scalar to be instead of .

The midpoint is just a special case of dividing a line segment in a given ratio. WORKED EXAMPLE 16.18

Points and have position vectors and . Find, in terms of and , the position vector of the point on such that . Always start by drawing a diagram. Since the question is about position vectors, include the origin.

You can get from to

via either or .

means that

EXERCISE 16E

1

a i Find a unit vector parallel to

.

ii Find a unit vector parallel to

2

.

b i Find a unit vector in the same direction as

.

ii Find a unit vector in the same direction as

.

Points and have position vectors a Write

Points

.

as a column vector.

b Find the position vector of the midpoint of 3

and

.

and have position vectors

vector of point such that

is a parallelogram.

and

. Find the position

4

Given that

vector

and

find the value of the scalar such that

.

5

Given that parallel to vector .

6

Points and have position vectors

and

so that 7

find the value of the scalar such that

and

Points and have position vectors

b Point lies on the line

. Point lies on the line segment

and of

such that

8

Given that parallel to vector

9

a Find a vector of magnitude parallel to

and .

. Find the coordinates of . find the values of scalars and such that

in terms of

lies on

. and

. Express the

and .

In the diagram, is the origin and points and have position vectors and . points on

and

such that

Prove that: a

is a straight line

b

is the mid-point of

.

is

.

Points and have position vectors and . Point position vector of

.

.

b Find a vector of magnitude in the same direction as

11

 is

. Find the position vector of .

a Find the position vector of the midpoint

10

is parallel to the

and

and are .

Checklist of learning and understanding Constant acceleration formulae in two dimensions:

To differentiate or integrate a vector, differentiate or integrate each component separately. Vectors in three dimensions can be expressed in terms of base vectors components.

using

The magnitude of a vector can be calculated using the components of the vector . The distance between the points with position vectors and is given by The unit vector in the same direction as is



.

.

Mixed practice 16 1

The point has position vector vector

and the point has position vector

. Find the

.

Choose from these options. A

B

C

D 2

A particle of mass

moves with constant acceleration

.

a Find the magnitude of the net force acting on the particle. When

the particle has velocity

.

b Find the speed and the direction of motion of the particle seconds later.

3

Points and have position vectors the exact distance

4

Points

and

. is the midpoint of

.

and have position vectors

a Find the position vector of the point such that b Prove that

. Find

and

.

is a parallelogram.

is a rhombus.

5

A particle moves in the plane so that its position vector at time seconds is . Find the speed of the particle when .

6

Points

and have position vectors .

a Prove that the triangle

and

is isosceles.

b Find the position vector of point such that the four points form a rhombus. 7

A particle moves with constant acceleration between the points and . At , it has velocity . At , it has velocity . It takes seconds to move from to . a Find the acceleration of the particle. b Find the distance between and . c Find the average velocity as the particle moves from to . [© AQA 2015]

8

A particle moves with velocity vector

.

Find the time at which the particle is moving parallel to the vector

.

Choose from these options. A B C D 9

A particle of mass

starts from rest and moves under the action of a constant force

. Find how long it takes to reach the speed of 10

.

A helicopter is initially hovering above the helipad. It sets off with constant acceleration , where the unit vectors and are directed east and north, respectively. The helicopter is modelled as a particle moving in two dimensions. a Find the bearing on which the helicopter is travelling. b Find the time at which the helicopter is

from its initial position.

c Explain in everyday language the meaning of the modelling assumption that the helicopter moves in two dimensions. 11

Points and have position vectors and . is the point on the line segment such that . Find the exact distance of from the origin.

12

A particle of mass

moves in the plane under the action of the force . The particle is initially at rest at the origin.

Find the direction of motion of the particle after seconds. 13

In this question, vectors and point due east and north, respectively. A port is located at the origin. One ship starts from the port and moves with velocity . a Write down the position vector at time hours. At the same time, a second ship starts

north of the port and moves with velocity

. b Write down the position vector of the second ship at time hours. c Show that after half an hour, the distance between the two ships is

.

d Show that the ships meet, and find the time when this happens. e How long after the meeting are the ships 14

apart?

A particle is initially at the point , which has position vector , with respect to an origin . At the point , the particle has velocity , and in its subsequent motion, it has a constant acceleration of and north respectively.

. The unit vectors and are directed east

a Find an expression for the velocity of the particle seconds after it leaves . b Find an expression for the position vector of the particle, with respect to the origin seconds after it leaves c Find the distance of the particle from the origin when it is travelling in a north-westerly direction. [© AQA 2013] 15

A particle has velocity vector

and is initially at the origin.

Find the particle’s position vector at time . Choose from these options. A

B

C

D 16

At time velocity

two aircraft have position vectors and

. The first moves with constant

and the second with constant velocity

.

a Write down the position vector of the first aircraft at time . Let be the distance between the two aircraft at time . b Find an expression for

in terms of . Hence show that the two aircraft will not collide.

c Find the minimum distance between the two aircraft. 17

A position vector of a particle at time seconds is given by

.

a Find the Cartesian equation of the particle’s trajectory. b Find the maximum speed of the particle, and its position vector at the times when it has this maximum speed. 18

A particle of mass

moves on a horizontal surface under the action of the net force . The particle is initially at the origin and has velocity The unit vectors and are directed east and north, respectively. Find the distance of the particle from the origin at the time when it is travelling in the northerly direction.

Worksheet See Extension Sheet 16 for questions on modelling rotation with vectors.



.

FOCUS ON… PROOF 2

Deriving the compound angle identities You are to going to demonstrate that

You have already seen trigonometric proofs that use right-angled triangles to prove results about more complicated figures. The same approach works here.

Rewind In Student Book 1, Focus on … Proof 2, you used this strategy to prove the sine and cosine rules.

PROOF 9

The sine compound angle formula will be proved first:

.

Create a triangle with angle by joining two right-angled triangles with angles and .

From triangle 1:

Now express all the other lengths in terms of and .

Similarly, from triangle 2:

You can write in two different ways.

The area of triangle is

Use

on each of the triangles

individually. The area of triangle is The area of the whole triangle is

And then on the triangle as a whole.

Therefore:

Dividing by

QUESTIONS

.

QUESTIONS 1

Is it possible to draw two right-angled triangles with the same height for any pair of acute angles and ?

2

Does the identity still hold when the angles and are not acute? Can you prove it?

3

To derive the compound angle identities for

and

, you could use the same

triangles and the cosine rule. However, there is a simpler proof which uses the relationship between

Write and use the compound angle identity for .



to prove that

and

:

FOCUS ON … PROBLEM SOLVING 2

Choosing between analytical and numerical methods Many problems you have encountered in this course can be solved in more than one way. In real-life situations you are free to choose whatever method and approach you prefer. In making your decision you should think about these questions. How difficult is the method? How efficient is it? Does it require lots of detailed or repeated calculations? How accurate is it? What level of accuracy do you actually need? Can it be easily adapted to solve other similar problems? For example, when solving an equation you have essentially two options: You can try to rearrange the equation, using rules of algebra (this is called an analytical solution), or you can use one of the iterative methods from Chapter 14 (this is a numerical solution). The former may not always be possible but when it works, it gives you an exact solution (for example, ). However, in many applications you only need an answer correct to a couple of decimal places, so you should consider whether the effort required to rearrange the equation is justified. On the other hand, your equation may have a parameter in it (for example, ) and you may want to know how changing the value of the parameter affects the solution. In this case, finding the analytical solution once (in terms of the parameter) may be more efficient than repeating the numerical calculation lots of times. Here you will look at a problem that can be investigated both analytically and numerically. You can try various approaches and decide for yourself which one suits you best.

The fishing lake problem The number of fish in a lake can be modelled by the equation

In this model, is the number of fish in year . Each year, due to natural birth and death rates, the population increases by a factor of , and die out due to lack of resources or natural causes, and a constant number, , fish are caught and removed. The question to ask is: what is the maximum number of fish that can be removed each year without causing the population to die out? QUESTIONS 1

Use a spreadsheet to investigate how the population changes with these parameter values: a

and: i ii iii

b

and: i

ii iii What is the largest number of fish that can be caught each year without causing the population to die out? Does this depend on the initial size of the population? 2

Vary the values of and slightly. How does this affect the maximum possible value of ?

You need to try lots of different values of the parameters to find the relationship between and . Is it possible to solve this problem analytically instead, to find an equation linking the three quantities? The sequence

is not one of the types you are familiar with; in fact, it is not possible

to find a general formula for . However, you are only interested in the long-term behaviour of the sequence – does it eventually decrease to zero or not? You know from Chapter 14 that, if a sequence has a limit, then this limit is a solution of the equation 3

Consider the case when a Solve the equation

and

.

. when:

i ii iii Compare the solution to what you observed in question 1. b Find the discriminant of the equation equation only has a solution when

in terms of . Hence show that the .

c Use the quadratic formula to write the two solutions in terms of . Hence show that, when , both solutions are positive. Could you tell, without doing the spreadsheet investigation, which of the two solutions the sequence will converge to?

Rewind In Chapter 14, Section 5, you learnt that the iteration if near the root.

4

a For the general case of the equation

converges to a root of

, find the discriminant in terms of

and

. b Hence show that the largest number of fish which can be taken out without causing the population to die out is



.

5

Did you find the theoretical analysis or the spreadsheet investigation easier to follow? Which one do you think gives a more reliable answer? Which helps you understand the problem better?

6

When and , the formula you found in question 4 b says that the maximum number of fish that can be take out is . However, with and , the spreadsheet shows that there are still fish in the lake after 50 years. So is the additional accuracy you get from using the formula always required?



FOCUS ON… MODELLING 2

Translating information into equations The aim of a mathematical model is to describe a real-life situation using equations, which can then be solved and used to make predictions. In this section you will look at writing differential equations – these are equations involving the rate of change of a quantity. You need to remember: the rate of change of with respect to is is proportional to means that

for some constant .

When writing differential equations, you usually have some information about particular values of the quantities involved. Sometimes you can use these to find constants in the equation (such as the in ), but sometimes you need to wait until you have solved the equation.

Tip In many examples, rate of change means change in time; however, look out for examples where this is not the case!

Rewind Solving differential equations is covered in Chapter 13. In this section you will just look at setting up equations rather than solving them.

WORKED EXAMPLE 1

The speed of an object decreases at the rate proportional to the square root of its current speed. When the speed is it is decreasing at the rate of . Using for speed and for time, write an equation to represent this information. Rate of change means the derivative with respect to time. The speed is decreasing, so write

to emphasise this.

You can use the given information to find . Remember that the rate of change is negative.

So the equation is

WORKED EXAMPLE 2

In one possible model of population growth, the rate of growth depends on two factors: it is proportional to the current size of the population, and also proportional to . (The second

factor represents seasonal breeding patterns.) When the measurements began the population size was . Using for the size of the population and for time measured in months, write a differential equation to represent this information. Both factors need to be multiplied together. Don’t forget to include the constant of proportionality. You only have information about when so you can’t find until you have solved the equation.

Rewind You know from Student Book 1 that the rate of change of velocity is acceleration (Chapter 16) and that the acceleration is proportional to the force acting on an object (Chapter 18). You could therefore use the model in Worked example 1 when there is a resistance force proportional to the square root of the speed, such as air resistance or drag when an object is moving through liquid.

QUESTIONS

QUESTIONS Write a differential equation to represent each situation. Where possible, find the values of any constants. 1

A population of a new town ( thousand) increases at a rate proportional to its size. Initially, the size of the population is and it is increasing at the rate of differential equation for the rate of change of population.

people per year. Write a

2

During the decay of a radioactive substance, the rate at which mass is lost is proportional to the mass present at that instant. Use for the mass of the substance in grams and for the time in seconds. Initially there is of the substance and the mass is decreasing at the rate of . Write a differential equation for the rate of change of mass.

3

In an electrical circuit, the voltage is decreasing at a rate proportional to the square of the present voltage. When the voltage is volts it is decreasing at a rate of volts per second.

4

Newton’s law of cooling states that the rate at which a body cools is proportional to the difference between its temperature and the temperature of its surroundings. A cup of tea is initially at and is cooling at the rate of in a room of temperature . Use for the temperature and for time (in minutes). Write a differential equation for the rate of change of temperature.

5

One of the ends of a metal rod, of length

, is heated. After a while the temperature remains

constant. However, the temperature changes along the length of the rod, decreasing at a rate proportional to the distance from the hot end. The temperatures at the two ends are and . Use for temperature and for the distance from the hot end (measured in cm). Write a differential equation for how the temperature changes along the rod. 6

The water pressure in the sea increases with depth. The pressure at depth is proportional to the density of the sea water. The density also varies with depth, and is modelled by the equation. . Write a differential equation for the rate of increase of pressure with depth.

7

A rumour spreads at a rate proportional to the square root of the number of people who have already heard it, and inversely proportional to the time it has been spreading. After minutes, people have heard the rumour and it is spreading at the rate of three people per minute. Write for the number of people who have heard the rumour and for the time, in minutes, since the rumour started. Write a differential equation to model this situation, and explain why the model needs to be modified for small values of .

8

A cylindrical tank has base radius . Water leaks out of the tank so that the rate at which the volume is decreasing is proportional to the height of the water remaining in the tank. Initially the height of water is and it is decreasing at the rate of rate of change of volume of water in the tank.



per minute. Find an equation for the

CROSS-TOPIC REVIEW EXERCISE 2 1

Show that, for small

2

Given that

3

Find the equation of the curve which has the gradient

where and are integers to be found. , show that

where is a constant to be found. and passes through the point

. 4

The graph of

5

a

has a point of inflection at (

). Find the values of and .

Use the trapezium rule with five ordinates to find an approximation to

, giving

your answer to four significant figures. b Show that

where and are constants to be found.

6

Find

7

A curve has parametric equations a Find

in terms of .

b The tangent to at the point where

cuts the -axis at the point .

Find the -coordinate of . 8

The curve

, is shown in the diagram.

Find the exact value of the shaded area, showing all of your working clearly. 9

a i Solve the equation

for

, giving your answers to the nearest

.

ii Solve the equation

for

, giving your answers to the nearest

.

b Describe a sequence of two geometrical transformations that maps the graph of onto the graph of . [© AQA 2011] 10

The diagram shows a part of the curve with parametric equations

.

a Find the values of at the points where the graph crosses the -axis. b Find the exact value of the shaded area. c Find the Cartesian equation of the curve. 11

By taking natural logarithms of both sides, or otherwise, find

12

a Show that

, given that

.

b Find the coordinates of the points on the curve equal to 13

.

where the gradient is

.

a Simplify b Hence or otherwise find

14

Triangle

is made out of a piece of elastic string. Vertices and are being pulled apart so

that the length of the base decreasing at the rate of a Show that

is increasing at the rate of . Initially, and

and the height, , is .

.

b Find an expression for in terms of . c Find an expression for the rate of change of the area of the triangle in terms of . d Find the rate at which the area of the triangle is changing when 15

A curve is defined by

and

.

.

Find the set of values of for which the curve is convex. 16

Consider the infinite geometric series

for

.

a Explain why the series converges. b Show that the sum of the series is

.

c Find the exact value of 17

.

A rectangle is drawn inside the region bounded by the curve in the diagram. The vertex has coordinates

.

and the -axis, as shown

a i Write down the coordinates of point . ii Find an expression for the area of the rectangle in terms of . b i Show that the stationary point of the area satisfies the equation ii By sketching graphs, show that this equation has one root for

. .

iii Use the second derivative to show that the stationary point is a maximum. c i The equation for the stationary point can be written as iterative formula, with three decimal places.

. Use a suitable

to find the root of the equation

correct to

ii Hence find the maximum possible area of the rectangle. 18

a Express

in the form

.

b Solve the differential equation

where

when

, expressing your answer in the form

where and are constants. [© AQA 2011] 19

The diagram shows the curves

The curve

and

.

crosses the -axis at the point and the curves intersect at the point .

a Describe a sequence of two geometrical transformations that maps the graph of the graph of .

onto

b Write down the coordinates of the point . c i Show that the -coordinate of the point satisfies the equation ii Hence find the exact value of the -coordinate of the point . d Find the exact value of the area of the shaded region bounded by the curves and the -axis.

and

[© AQA 2010] 20

Evaluate

21

a State an expression for b

. .

Hence or otherwise show that

.

22

a Use the identity

to show that

b The diagram shows part of the curve

.

.

Write down the coordinate of the point . c Find the area of the region shaded red in terms of , writing your answer in a form without trigonometric functions. d By considering the area of the region shaded blue, find 23

for

.

A function is defined by a Find

.

b Show that the stationary points of

satisfy the equation

c Hence show that the function has only one stationary point. 24

A curve has equation

.

a Find the range of values of a for which the curve has at least one point of inflection. b Given that one of the points of inflection is a stationary point, find the value of . c For this value of , find the range of values of for which the curve is concave. 25

a Sketch the graph

.

b The tangent to this graph at the point c For what range of values of does 26

a Given that

passes through the origin. Find the value of . have two solutions?

, use the quotient rule to show that

b Given that tan

c Show that, if

d A curve has equation

, use a trigonometrical identity to show that

, then

.

i Find the value of the -coordinate of each of the stationary points of the curve. ii Find

.

iii Hence show that the curve has a minimum point which lies on the -axis. [© AQA 2012]



17 Projectiles In this chapter you will learn how to: model projectile motion in two dimensions find the Cartesian equation of the trajectory of a projectile.

Before you start… Student Book 1, Chapter 15

You should be able to find the magnitude and direction of a vector.

1 Find the magnitude and direction of the vector

.

Student Book 1, Chapter 17

You should be able to use constant acceleration formulae in one dimension.

2 A particle accelerates uniformly from to while covering a distance of in a straight line. Work out the acceleration.

Chapter 16

You should be able to use constant acceleration formulae in two dimensions.

3 A particle initially has velocity and accelerates at . Find its velocity after seconds.

Chapter 8

You should know the sine double angle identity.

4 Express

Chapter 8

You should know the definition of .

5 Express

Chapter 8

You should be able to solve equations involving and .

6 Solve the equation

in terms of

in terms of

.

.

for .

Motion in two dimensions In Student Book 1, you used the constant acceleration formulae to analyse the motion of a particle that was projected in one dimension – either horizontally or vertically. In this chapter this idea is extended to look at projectiles moving in two dimensions – vertically and horizontally. This more realistic model provides a basis for analysing the motion of, for example, a bullet, a golf ball or water from a fountain.

Section 1: Modelling projectile motion When an object is projected upwards at an angle, it moves in a vertical plane along a symmetrical path.

Fast forward You will see in Section 2 how to prove that this is a parabola.

The only force acting on it is gravity so the acceleration is constant with magnitude and is directed downwards. Note that there is no acceleration horizontally as there is no force acting in this direction.

Tip Remember that a particle has no size and, importantly in the context of projectiles, it does not spin.

Key point 17.1 In projectile motion the acceleration of the particle is

.

This means that the motion can be described using the constant acceleration equations in two dimensions.

Rewind You met the constant acceleration formulae with two-dimensional vectors in Chapter 16, Section 2.

WORKED EXAMPLE 17.1

A particle is projected from point with the velocity

. Use

, giving your

final answers to an appropriate degree of accuracy. a Find the speed and direction of motion of the particle after

seconds.

b Find the distance of the particle from at this time. First find the velocity vector after .

seconds. Use

a

Speed is the magnitude of velocity. The direction of motion is the direction of the velocity vector. Draw a diagram to make sure you are finding the correct angle.

The direction of motion is

below

the horizontal. Use

b

Distance from the origin:

Distance is the magnitude of displacement.

The initial velocity may be specified by giving the speed and the angle of projection.

Key point 17.2 If a particle is projected with speed at an angle above horizontal, then the components of the velocity are: horizontally: vertically:

or as a vector:

When solving problems involving projectiles it is often useful to consider the horizontal and vertical motion separately, rather than using vectors. This means using the one-dimensional constant acceleration formulae in each direction separately. WORKED EXAMPLE 17.2

A particle is projected from ground level with speed horizontal.

at an angle of

above the

Find the height of the particle above ground level at the time when its horizontal displacement from the starting point is . Use , giving your final answers to an appropriate degree of accuracy. Horizontally:

Find the time first, using the horizontal displacement equation.

Vertically:

Use

If the particle is projected from a point above ground level then the vertical displacement can be negative, corresponding to the particle falling below the starting point.

Tip Since there is no acceleration horizontally, the horizontal component of velocity is constant throughout the motion, and so the only equation that you can use is .

WORKED EXAMPLE 17.3

A stone is thrown from the top of the cliff which is above sea level. The initial velocity has magnitude and is directed at above horizontal. Find the speed with which the stone hits the sea. Use appropriate degree of accuracy.

, giving your final answer to an

Horizontally:

The horizontal component of the velocity is constant.

Vertically:

Note that starting point.

as the stone finishes

Use velocity.

below its

to find the vertical component of the

Now find the magnitude of the velocity.

Questions often ask for the maximum height a projectile reaches, or how far horizontally from the starting point it lands.

Key point 17.3 A projectile is at its maximum height when

.

For a particle projected from ground level, to find the range (the maximum horizontal distance travelled) set .

WORKED EXAMPLE 17.4

A particle is projected from ground level with speed in the direction a Show that the maximum height the particle reaches is

above horizontal.

.

b Given that and , find the range of the projectile. Use final answer to an appropriate degree of accuracy. a Vertically:

The maximum height is reached when Use point.

, giving your

.

to find the vertical displacement at this

b Vertically:

The projectile lands when particle lands, using

. First find the time the .

corresponds to the starting position, so you want the other value of . You can now find the horizontal distance at this time.

In all of the worked examples so far the particle was projected upwards at an angle. But the same equations still apply if the particle is projected downwards. The only difference is that the vertical component of the initial velocity is negative. WORKED EXAMPLE 17.5

A small ball is thrown from a window below the horizontal.

above ground with the initial velocity

Find how long it takes to reach the ground. Use appropriate degree of accuracy.

directed at

, giving your final answer to an

Note that the vertical component of the initial velocity is negative as the particle is going downwards from the start.

Vertically:

Solve the quadratic for .

It takes

seconds

must be positive.

.

EXERCISE 17A In this exercise, unless instructed otherwise, use

, giving your final answers to an appropriate

degree of accuracy. 1

A particle is projected from a point . Find its velocity vector after seconds if its initial velocity is: a i ii b i ii c i

at

above the horizontal

ii

at

above the horizontal

d i

at at

ii 2

below the horizontal below the horizontal.

A particle is projected from the origin. Find its position vector after seconds if its initial velocity is as in question 1.

3

A small stone is projected from ground level with speed

at an angle of

above the

horizontal. a Find its height above the ground after

seconds.

b What is its speed at this time? 4

A particle is projected with initial velocity the right and vertically upwards. Find:

. The unit vectors and are directed to

a the time it takes the particle to reach its maximum height b the magnitude and direction of its velocity 5

In this question, use

seconds after projection.

, giving your final answers to an appropriate degree of accuracy

A ball is projected from a point on a horizontal plane with speed of

at an angle of elevation

. The ball returns to the plane at point . Find:

a the greatest height above the plane reached by the ball b the distance 6

.

A particle is projected with speed

at an angle above the horizontal. The greatest height

reached above the point of projection is Find, to the nearest degree, the value of .

.

7

A particle is projected from a point with speed

at an angle of elevation of

Find the length of time the particle is more than 8

A ball is hit from a point

.

above the horizontal level of .

above the ground, with speed

at an angle of elevation of

. The ball lands at the point , as shown in the diagram. Find: a the time taken for the ball to travel from to b the distance c the speed with which the particle hits the ground.

9

A ball is projected with speed at an angle of located from the point of projection.

above horizontal. A

high wall is

Determine whether the ball will clear the wall.

10

A ball is projected horizontally with speed

from the top of a

high building.

a Find the distance from the foot of the building of the point where the ball hits the ground. A second ball is projected with speed

from the foot of the building.

b Find the possible angles of projection so that the second ball hits the ground at the same place as the first ball.

Worksheet See Support Sheet 17 for an example of finding the angle and speed of projection and for more practice questions. 11

A golfer is aiming to land the ball on the green. The front of the green is the green is

long, as shown in the diagram. The ground is horizontal.

a If the golfer strikes the ball with speed

at an angle above the horizontal, show that the

horizontal distance travelled by the ball when it lands is b If

from his position and

.

, find the range of values of that will result in the golfer landing the ball on the green.

12

A film director wants to include the ‘man fired out of a cannon’ stunt in his film. For ethical reasons he uses a scale model of a cannon, which is one tenth of the real size, with a toy fired out of it. a What is the magnitude of the force of gravity that would be observed by someone watching the film? b Does the film need to be speeded up or be slowed down to correct this?



Section 2: The trajectory of a projectile So far you have only looked at how the displacement and velocity of a projectile change with time. But you can also find a relationship between the horizontal and vertical displacements. This leads to an equation describing the path (or trajectory) of the projectile.

Key point 17.4 To find an equation for the trajectory of a projectile: make the subject of the

equation

substitute this expression for into

.

Rewind This is an example of parametric equations which you met in Chapter 12, Section 2. You can apply the method of eliminating to find the Cartesian equation of the trajectory to any particle moving in two dimensions.

WORKED EXAMPLE 17.6

A particle is projected from ground level with speed that

at an angle above horizontal, such

. Let and be the horizontal and vertical displacements from the point of

projection, with measured upwards. Find an expression for in terms of and .

First find the components of the velocity.

Horizontally:

Make the subject of the horizontal equation.

Vertically:

And substitute into the vertical equation.

As you can see, the trajectory of a projectile is a parabola. It is important to remember that the equations that you have been using include gravitational acceleration, but no other force. In particular, this model of a projectile assumes no air resistance. If the air resistance is included the trajectory is no longer a parabola.

You may need to find the angle of projection in order for the particle to pass through a specific point. There will often be two possible values. WORKED EXAMPLE 17.7

A particle is projected from a point on horizontal ground, with speed

, at an angle

above the horizontal. The particle passes through a point , which is at a horizontal distance of from and a height of above the ground. Use , giving your final answer to an appropriate degree of accuracy. a Show that b Find the two possible values of . a Horizontally:

Make the subject of the horizontal component equation…

Vertically:

… and substitute into the vertical component equation.

and

b

Factorise and solve for . Note that there are two possible trajectories that pass through .

WORK IT OUT 17.1 A ball is thrown from a point , at ground level, with speed , at an angle above the horizontal. It lands at a point horizontally from on a platform that is above the ground. The ledge starts from and is long as shown.

The trajectory of the ball is given by Find the angle . Use

.

, giving your final answer to an appropriate degree of accuracy.

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1

as this is the first point of contact with the platform Solution 2

as the particle needs to be on the way down Solution 3

EXERCISE 17B In this exercise, unless otherwise instructed, use degree of accuracy. 1

, giving your final answers to an appropriate

A particle is projected horizontally with speed from the top of a high cliff. Let the origin of an coordinate system be located at the top of the cliff, with the -axis vertical and the axis horizontal in the direction of projection. Find, in terms of , the Cartesian equation of the trajectory of the particle.

2

A rugby ball is kicked at an angle of elevation of away horizontally from the kicker. a If the ball hits the crossbar, show that

towards a cross bar that is

high and

.

b If the cross bar is high and the kick is to be taken from the posts, find the minimum velocity with which the ball must be kicked to clear the bar. c State two modelling assumptions you needed to make.

3

In this question, use

, giving your final answer to an appropriate degree of accuracy.

a Show that the equation of the trajectory of a particle projected at speed elevation is

at an angle of

.

b An archer fires an arrow from a point above the ground with speed elevation of . The target is away and the centre of the target is

at an angle of above the ground.

The diameter of the target is Determine whether the arrow hits the target.

4

A particle is projected from the origin with speed through the point with position vector a Show that

at an angle of elevation . It passes

.

.

b The particle subsequently passes through the point with position vector

.

i Show that ii Find . 5

A particle is projected from a point on horizontal ground with speed at an angle of elevation . The maximum height reached by the particle is

and the particle hits the ground

from the

point of projection. Find and . 6

In this question use

, giving your final answer to an appropriate degree of accuracy.

A basketball player shoots at the hoop. The hoop is from the ground and the basketball player stands horizontally from the hoop. The ball is released from above the ground at an angle of above the horizontal. The initial speed of the ball is

.

a Given that the ball passes through the hoop, show that b Hence find the angle at which the ball was released, justifying your answer.

.

Checklist of learning and understanding In projectile motion the acceleration of q particle is If a particle is projected with speed at an angle

.

above horizontal, then the

components of the velocity are: horizontally: vertically: A projectile is at its maximum height when

.

For a particle projected from ground level, to find the range (the maximum horizontal distance travelled) set . To find an equation for the trajectory of a projectile: make the subject of the substitute this expression for into



equation .

Mixed practice 17 In this exercise, unless otherwise instructed, use degree of accuracy. 1

, giving your final answers to an appropriate

A particle is projected at a velocity of

. Find the minimum speed during its flight.

Choose from these options. A B C D 2

A particle of mass is projected horizontally off the edge of a cliff and lands in the sea below at a distance from the base of the cliff. A second particle of mass

is projected horizontally from the same point at the same speed

and lands in the sea at a distance from the base of the cliff. Which one of these statements is true? A B C D It depends on the height of the cliff. 3

A ball is thrown horizontally at

from the top of a

high building. Find:

a the time it takes to hit the ground below b the distance from the bottom of the building to the point where the ball hits the ground. 4

In this question, use accuracy.

, giving your final answers to an appropriate degree of

A projectile is launched from the point with velocity the point and is travelling with velocity

. After seconds it is at . Find:

a the value of b the distance 5

.

In this question, use accuracy. A ball is thrown from a window

, giving your final answers to an appropriate degree of

above the ground with velocity

It is caught by a child above the ground. The child stands below the window at ground level. Find: a the value of b the speed of the ball as it is caught.

.

from the point vertically

6

A particle is projected from a point on a horizontal plane with speed

at an angle

above the horizontal. It lands on the plane at the point . a i Show that ii Hence deduce that for fixed , the maximum range is achieved when b If

.

find, to the nearest degree, the two possible angles at which the particle was

projected. 7

A child stands on a bridge that passes over a river which is below. He throws a stone with speed directly at a small stationary rock which is from the base of the bridge as shown in the diagram. How far away from the rock does the stone land in the river?

8

A particle is projected from with velocity time seconds. a Given that

. It passes through the point at

, find the value of .

b At point the ball has the same speed as at . Find the time taken for the ball to travel from to .

9

A tennis ball is projected from a point with a velocity of , where and are horizontal and vertical unit vectors respectively. The ball travels in a vertical plane through , which is above the horizontal surface of a tennis court. During its flight, the horizontal and upward vertical distances of the ball from are metres and metres respectively. Model the ball as a particle. a Show that, during the flight, the equation of the trajectory of the ball is given by

b The ball hits a vertical net at a point . The net is at a horizontal distance of

from .

Determine the height of the point , above the surface of the tennis court. Give your answer to the nearest centimetre. c State a modelling assumption, other than the ball being a particle, that you need to make to answer this question. [© AQA 2014] 10

A bullet is fired horizontally from the top of a vertical cliff, at a height of metres above the sea. It hits the sea seconds after being fired, at a distance of the cliff, as shown in the diagram.

metres from the base of

a Find the initial speed of the bullet. b Find . c Find the speed of the bullet when it hits the sea. d Find the angle between the velocity of the bullet and the horizontal when it hits the sea.

[© AQA 2011] 11

In this question, use

, giving your final answers to an appropriate degree of

accuracy. A particle is projected from a point

above ground level with speed

as

shown in the diagram. Find its distance from the point of projection at the instant the particle is moving in a direction perpendicular to its initial velocity.

12

A ball is projected with speed at an angle of elevation above the horizontal so as to hit a point on a wall. The ball travels in a vertical plane through the point of projection. During the motion, the horizontal and upward vertical displacements of the ball from the point of projection are metres and metres respectively. a Show that, during the flight, the equation of the trajectory of the ball is given by

b The ball is projected from a point metre vertically below and metres horizontally from the point . i By taking

, show that satisfies the equation

ii Hence, given that and are constants, show that, for tan to have real values, must satisfy the inequality

iii Given that

, determine the minimum possible speed of projection. [© AQA 2007]

Worksheet See Extension Sheet 17 for a selection of more challenging problems.



18 Forces in context In this chapter you will learn how to: resolve forces in a given direction in order to calculate the resultant force use a model for friction determine the acceleration of a particle moving on an inclined plane.

Before you start… Student Book 1,

You should be able to add vectors

Chapter 18

and find magnitudes.

Student Book 1, Chapter 17

You should be able to solve problems involving motion with constant acceleration.

1 Three horizontal forces act on a particle. In newtons, the forces are: and . Calculate the magnitude of the resultant force and its angle from the direction . 2 A force of acts upon a particle with mass . If the particle is initially at rest, after how many seconds will its displacement be equal to ?

Improving the model In Student Book 1, Chapter 18, you learnt to add and subtract forces in vector form and to determine the angle and magnitude of the resultant force. This chapter deals with more complex situations involving strings and planes in different orientations, where the forces are not perpendicular. To solve these problems you will need to combine your knowledge of forces, vectors and trigonometry. In Student Book 1 you met problems where there is a constant resistance force acting on an object. Here the model of friction will be improved to enable more real-world situations to be modelled.

Section 1: Resolving forces When considering forces acting on a moving body, it can be useful to split each force into horizontal and vertical components. This process is called resolving the force. WORKED EXAMPLE 18.1

Two forces act on a particle in the horizontal plane. Force has magnitude and acts at angle to the positive -axis. Force has magnitude and acts in the direction of the negative -axis. Find the magnitude and direction of the resultant force. Always draw a diagram.

Form a right-angled triangle with the force as hypotenuse and the sides horizontal and vertical. In this case, since the force is in the -direction, you only need to resolve force .

Calculate horizontal and vertical components of and add to the horizontal direction The notation is a standard shorthand for ‘resolve horizontally’ and stands for ‘resolve vertically’. Draw another triangle showing the resultant force. This helps find its magnitude and direction. Note that means that the force is acting in the negative direction.

Magnitude: Direction:

is

, so the direction from the positive -axis.

Once you have found forces, you may have to work with them, using Newton’s second law

Rewind You met Newton’s second law in Student Book 1, Chapter 18.

WORKED EXAMPLE 18.2

A particle with mass moves on a horizontal plane with constant acceleration , where , acted upon by two forces in the plane, and . has magnitude

and acts at angle

anticlockwise from the positive -axis.

.

has magnitude a Show that

and acts at acute angle clockwise from the positive -axis. .

b Calculate the exact value of . a

Always draw a diagram.

Resolve vertically: no acceleration means the forces are equal. Rearrange to find

.

Use to convert into . Since the angle is acute you only need to take the positive root as for acute angles.

b

Resolve horizontally and use Newton’s second law.

You need to be able to apply previous techniques, in particular equations of constant acceleration, to problems involving objects moving under a combination of forces. WORKED EXAMPLE 18.3

A toy helicopter of mass has its rotors set so that the force provided by the engine is given by where and can be varied using a remote control, but the magnitude of is always . The unit vector is horizontal, is vertical. Use , giving your final answers to an appropriate degree of accuracy. a What is the acceleration if the helicopter is set to fly horizontally? b The helicopter accelerates at travel metres?

above the horizontal. How long would it take for the helicopter to

Always draw a diagram. You could define to be the angle the force makes with the vertical or the horizontal. Here, it is the vertical.

a

There is no vertical acceleration.

Next resolve horizontally and use Newton’s second law.

b The force acts at angle above :

The direction of the resultant force is above the horizontal, so choose this as one of the directions to resolve in. The net force in the perpendicular direction is zero. (You could solve this problem by resolving horizontally and vertically. However, resolving in the direction of motion gives simpler equations.)

Perpendicular to movement:

In direction of movement:

Constant acceleration problem: . Find .

EXERCISE 18A

Once you have found the acceleration you can use the constant acceleration formulae. Notice that, although the forces act in two dimensions, the helicopter moves in a straight line. Hence you can use the one-dimensional version of the constant acceleration formulae.

In this exercise, unless instructed otherwise, use degree of accuracy. 1

2

, giving your final answers to an appropriate

In each system, a particle on a horizontal surface is affected by two forces and . Resolve the resultant force into and components where represents due north, represents due east. a i

has magnitude

ii

has magnitude

and acts at a bearing of and acts at a bearing of

. .

b i

has magnitude bearing of .

and acts at a bearing of

has magnitude

and acts at a

ii

has magnitude bearing of .

and acts at a bearing of

has magnitude

and acts at a

A particle of mass from the horizontal.

hangs at rest attached to two light, inextensible strings, at angle and

For each system, find the tension in each string. a i ii b i ii

3

Find the resultant force acting on the object in the diagram, giving your answer in the form

4

In the diagram the particle is in equilibrium.

.

a Find the angle . b Find the force .

5

In this question, use , giving your final answers to an appropriate degree of accuracy. A mass of is being held by a string at an angle to the vertical with tension newtons and a horizontal force of .

a Show that b Find the value of . c Find the value of . 6

Three friends are pulling a load across a smooth horizontal surface. Alf pulls with a force of . Bashir pulls with a force and is positioned at an angle clockwise from Alf. Charlie pulls with force and is positioned at an angle clockwise from Bashir. Modelling the load as a particle, and given the load begins to move exactly towards Bashir, find and determine the initial acceleration of the load.

7

Two forces of magnitudes

and

act at a point .

a Given that the two forces are perpendicular to each other, find: i the angle between the resultant and the

force

ii the magnitude of the resultant force.

b It is given instead that the resultant of the two forces acts in a direction perpendicular to the force. i Find the angle between the resultant and the

force

ii Find the magnitude of the resultant.

8

A particle with mass an angle of

hangs at rest suspended by two light inextensible strings. One string is at

to the horizontal and the other is at an angle of

to the horizontal.

Find the tension in the two strings.

9

A light, smooth ring is threaded on a light, inextensible string. One end of the string is attached to a fixed point . The other end of the string is threaded through a fixed smooth ring directly below , and attached to a particle of mass A force of magnitude

.

is applied to ring at an angle

with the horizontal.

a Given the string is taut, find, in terms of , the exact value of required to maintain the system in equilibrium with angle b If angle .

and

.

and the system is in equilibrium, find the relationship between and angle

c If ring actually had mass equilibrium with angle

10

, find the new force required to hold the system in and

, and determine angle

.

A small, smooth ring of weight is threaded on a light, inextensible, taut string. The ends of the string are attached to fixed points and at the same horizontal level. A horizontal force of magnitude is applied to . In the equilibrium position the angle is a right angle, and the portion of the string attached to makes an angle with the horizontal a Explain why the tension is the same in each part of the string. b Find and .



Section 2: Coefficient of friction In Student Book 1 you met friction as a constant force opposing motion. However, friction is not always constant; it depends on: the force pushing the object into the surface how rough the object and the surface are the driving force applied to the object. If the driving force is strong enough, friction will be overcome and an object that was at rest will start to move. There must therefore be a maximum value of friction; this is called limiting friction. When friction is at its limiting value, the object is on the point of moving and is said to be in limiting equilibrium.

Key point 18.1 The maximum or limiting value of friction,

, is given by

where is the normal reaction force between the object and the surface and is the coefficient of friction.

Focus on … Focus on… Problem solving 3 asks you to decide whether values of certain quantities (including frictional force and ) are reasonable in the given contexts.

This coefficient deals with how rough the two surfaces are – for example, for an ice skate on an ice rink it is about , while for a car tyre on a road it is about . If the coefficient of friction is zero the surface is described as smooth and there is no friction force. If the driving force is not sufficiently strong to overcome the object remains at rest. In this case .

Once the object is moving, the frictional force is constant at

, the frictional force will just match it so that

.

Key point 18.2 If the resultant of the driving forces is smaller than , the object will remain at rest and the friction force, , will equal the resultant of the driving forces. If the resultant of the driving forces is larger than

WORKED EXAMPLE 18.4

, the object will move and

.

WORKED EXAMPLE 18.4

A block of mass lies at rest on a rough horizontal surface, with coefficient of friction . Attached to opposite ends of the block are two light strings, each of which is under tension. Use , giving your final answers to an appropriate degree of accuracy. a The tension in the left string is raised to i

.

Calculate the direction and magnitude of the frictional force.

ii If the block moves, determine its acceleration. b The tension in the right string is then raised to i

.

Calculate the direction and magnitude of the frictional force.

ii If the block moves, determine its acceleration.

a

Always draw a diagram. If the block moves it will be to the left so friction acts to the right.

Since any movement will be horizontal, there is no vertical acceleration. Resolving vertically gives . Maximum friction is

Calculate limiting friction

.

Resolve horizontally.

i to the right. ii The block will not move.

Friction will oppose the resultant up to a maximum of limiting friction. Since the resultant driving force is less than limiting friction, the object will not move. Friction will just balance the driving force.

b As before If the block moves it will now be to the right so friction acts to the left.

i

The resultant driving force exceeds limiting friction, so friction takes its limiting value.

If

But to the left. ii

The resultant force is

. Use

.

Notice that the frictional force acts in different directions in parts a and b of Worked example 18.4 because it must always oppose the resultant force which would drive motion.

Tip In some questions you may be uncertain initially which direction the friction acts. If you calculate a negative friction force in your answer, you have probably drawn the friction in the wrong direction. Change your diagram and adjust your equations.

Focus on … See Focus on… Proof 3 for proofs of formulae for the minimum force required to move a particle on a rough surface in various situations.

WORKED EXAMPLE 18.5

Two blocks, and , lie at rest on a rough surface, with block on top of block . The coefficient of friction between the blocks is , the mass of block is and the mass of block is . The coefficient of friction between block and the surface is . Use , giving your final answers to an appropriate degree of accuracy. A light inextensible string is attached to block , and the tension in the string is

a Show that the system remains at rest. The tension in the string is increased to b What is the acceleration of each block?

and block begins to slide on block .

.

a Treating the blocks as a single object of mass , with frictional force between and the surface being :

Since you are interested in the motion of the whole system, treat the two blocks as a single object. Consider horizontal and vertical components separately. Remember to include the normal reaction force in your diagram.

There is no movement in the vertical direction, so the net force is zero. Limiting friction:

Friction will oppose the resultant up to a maximum of limiting friction.

The tension is less than the maximal frictional force, so and the system remains at rest. b

exceeds so the blocks will move, with a net driving force .

Once the blocks are moving, the friction force equals .

Let the frictional force between the blocks be .

Since the two blocks will move with different accelerations, consider forces for each block separately, starting with block .

For the upper block, :

       (1) Limiting friction:

You can now find the limiting friction acting on .

          (2)

Since block is sliding on block , the friction is maximal and . For block :

Now look at the lower block, considering vertical components first. According to Newton’s third law, the normal reaction force acts downwards on block .

Substituting for

from (1):

Limiting friction:

Since the system is moving, friction with the surface is also limiting, so Substituting from

You can again use the fact that the block is moving to infer that friction is at the limiting level.

(2):

It may be necessary to resolve all driving forces before you can calculate the direction of the frictional force. WORKED EXAMPLE 18.6

A particle of mass rests on a rough horizontal surface, with coefficient of friction between the particle and the surface . The unit vectors and are both in the horizontal plane. Two horizontal forces act on the particle: a The frictional force b A third force

and

.

. Find and . is applied. Find the new frictional force in the form

a

.

Find the resultant driving force. Magnitude of the resultant driving force is Normal reaction equals weight since there is no vertical movement:

Calculate the magnitude of maximal friction and compare to the driving force.

Limiting friction The driving force is less than maximal friction so the particle will not move and friction will exactly counter the driving force.

The driving force is less than maximal friction, so the forces will be in equilibrium.

b

Find the overall driving force. Magnitude of the overall driving force is

Limiting friction

Calculate the magnitude of maximal friction (as found in part a) and compare to the driving force.

Friction is limiting. The friction force has magnitude in the direction of . Hence the friction force is:

The magnitude of the friction force is and its direction is opposite to the direction of the overall driving force (which has magnitude ). You need to change the magnitude of the force without changing its direction, so divide by and multiply by .

EXERCISE 18B In this exercise, unless instructed otherwise, use degree of accuracy. 1

In this question use

, giving your final answers to an appropriate

, giving your final answers to an appropriate degree of accuracy. In

each problem a block of mass is pulled along a rough horizontal surfaceby a light horizontal rope with tension newtons. The acceleration is . The coefficient of friction is . a Find when: i ii b Find when: i ii

2

In this question use , giving your final answers to an appropriate degree of accuracy. A car of mass has a coefficient of friction of with the road. What horizontal force is required to move the car?

3

A block of mass

lies in limiting equilibrium on a horizontal surface, with a horizontal force of

applied to it. Find the coefficient of friction between the block and the surface. 4

In this question use A child pulls a toybox, of mass

, giving your final answers to an appropriate degree of accuracy. , across a rough floor, using a light string tied to the box at one

end. The tension in the string is

and the string remains at an angle of

If the coefficient of friction between the box and the floor is

to the horizontal.

, what is the acceleration of the

box? 5

In this question use

, giving your final answers to an appropriate degree of accuracy.

Two particles, and , are connected by a light inextensible string.

Particle , which weighs , is placed on a rough horizontal surface. The connecting string runs horizontally to a smooth pulley at the end of the surface, then vertically downward to , which weighs

.

a The system is in limiting equilibrium. Calculate the coefficient of friction between and the surface. b A smooth ring with weight

is threaded on to the string so that it lies on particle . Calculate

the downward acceleration of . 6

Two particles, and , are connected by a taut, light, inextensible string and lie at rest on a plane, with one particle at point and the other at point . weighs

and weighs

The coefficient of friction between either particle and point is between either particle and point is .

and the coefficient of friction

When is at , a horizontal force of limiting equilibrium.

is applied to acting in direction

and the system is in

When is at , a horizontal force of once again in limiting equilibrium.

is applied to acting in direction

and the system is

Calculate 7

.

and

A car of weight

. , travelling at

, skids to a halt in

taking seconds.

a Assuming a constant braking force, find the deceleration of the car. b Assume that friction is the only force acting on the car and that

. Find the coefficient

of friction between the car and the road. 8

A particle of mass

lies on a rough horizontal surface, with the coefficient of friction between

surface and particle equal to

.

A light inextensible string is attached to the particle, and tension applied with the string at an angle above the horizontal where

.

a Given the system is in limiting equilibrium, show that the tension in the string satisfies the equation b The string will break if

for some value and find . . Find the range of values for for which the particle will be

caused to move by tension in the intact string. c Find the maximum possible acceleration for the particle.

9

In the model for friction used in this section, which of these factors affect the frictional force of a surface acting on an object? A The contact surface area between the surface and object B The speed of the object C The acceleration of the object D Lubrication between the surface and the object



Section 3: Motion on a slope If an object is on a slope (sometimes called an inclined plane) it is often convenient to resolve forces parallel or perpendicular to the slope. You use the same technique as for all resolving problems – drawing forces as the hypotenuse of a right-angled triangle with sides parallel and perpendicular to the slope. is a standard notation for ‘resolve parallel to the slope’. is a standard notation for ‘resolve perpendicular to the slope’. WORKED EXAMPLE 18.7

A block of mass

lies on a smooth slope inclined at

to the horizontal and slides down under

gravity. Calculate the component of the gravitational force acting in the direction of the slope. Use , giving your final answers to an appropriate degree of accuracy. Always draw a diagram.

Form a right-angled triangle with the force as hypotenuse and the sides parallel and perpendicular to the slope. Label relevant angles.

Gravity comes up so often in inclined plane problems that it is useful to remember these general results.

Key point 18.3 The components of gravity acting on a slope inclined at an angle to the horizontal are:

Tip A good way of remembering this is to think about what happens when is zero. Then there is no component of gravity parallel to the plane, and a component plane.

perpendicular to the

WORKED EXAMPLE 18.8

A block with mass

lies on a smooth slope inclined at

to the horizontal, and is held in

equilibrium by a string parallel to the slope, with tension . Calculate and the normal reaction force . Use

, giving your final answers to an

appropriate degree of accuracy. Always draw a diagram.

Resolve along and perpendicular to the slope.

The system is in equilibrium, so resolving in any direction should give a zero result. Since two of the three forces ( and ) are aligned to the slope, this is the preferred option. For each force not parallel or perpendicular to the slope, draw a right-angled triangle to take components in these directions.

The block is in equilibrium. Resolving perpendicular to the slope gives an equation for . Think carefully where the

angle is.

Resolving parallel to the slope gives an equation for .

Some problems about motion on a slope also involve friction. As previously, identify the direction of movement in the absence of friction, and assign friction to act in the opposite direction. WORKED EXAMPLE 18.9

A block with mass

lies on a rough surface inclined at

to the horizontal. The coefficient of

friction between block and surface is

.

A light inextensible string attached to the block passes over a smooth peg at the highest point of the surface (so that the length between block and peg is parallel to the slope) and is attached at its other end to a particle of mass which hangs freely. The system is initially held at rest and then released. Use to an appropriate degree of accuracy.

, giving your final answers

a Calculate the direction and magnitude of the frictional force between and the surface. b After the block has travelled metre, the string breaks. Calculate the total distance travelled in the subsequent

seconds.

a

Always start by drawing a force diagram.

For the particle:

Resolve separately for each component of the system, in their respective directions of movement.

For the block:

Adding the two calculations will eliminate the tension term.

Resolve perpendicular to the direction of movement of the object on the slope to find the normal reaction force. Limiting friction:

Calculate maximum possible friction .

(1) (2):

Determine whether the force driving movement in the system is sufficient to overcome friction.

Driving force for motion: The driving force is greater than maximum frictional force so the system will move. The system will be at limiting friction with acting downslope. b

Phase : Moving up the slope with string attached. Constant acceleration problem: . Find .

Phase : Moving up the slope without string.

Separate the movement into three phases: 1 Moving up the slope with string attached

2 Moving up the slope after the string breaks.

In equation (2), is now zero.

Constant acceleration problem: . Find .

Phase 3: Movement after coming to rest (friction now

3 Subsequent movement

acting up the slope to prevent sliding)

Note that the frictional force changes direction.

Total force acting down the slope is The driving force is less than maximal friction so the block will remain at rest. Total distance travelled after the string breaks is therefore equal to .

EXERCISE 18C In this exercise, unless instructed otherwise, use degree of accuracy. 1

, giving your final answers to an appropriate

A particle of mass is released from rest on a slope inclined at an angle to the horizontal, with coefficient of friction between the particle and the slope being . For each case determine the force of friction acting on the particle. a i ii b i ii

c i ii 2

A particle of mass string.

and a particle of mass

are connected by a light, inextensible

Particle lies on a slope inclined at an angle to the horizontal. The string passes from parallel to the line of greatest slope, and runs over a smooth peg at the top of the slope, then descends vertically to .

The coefficient of friction between and the slope is . For each system, find:

• the force of friction acting on • the acceleration of

and its direction

when the system is released from rest.

a i ii b i ii c i ii 3

In this question use

, giving your final answers to an appropriate degree of accuracy. A

particle of mass is projected with velocity smooth plane inclined at to the horizontal.

up the line of steepest slope of a long

a Calculate the reaction force from the plane acting on the particle. b Calculate the total distance travelled in the first seconds. 4

A child of mass

is sliding down a slide at an angle of

to the horizontal.

a Find the normal reaction of the slide on the child. b Find the acceleration if: i the slide is smooth ii there is a coefficient of friction of

between child and slide.

Worksheet See Support Sheet 18 for a further example of limiting equilibrium on a rough inclined plane and for more practice questions. 5

A block of mass

lies at rest on a rough board, with

.

Find, to the nearest degree, the minimum angle to which the board can be raised before the block begins to slide. 6

A particle is projected upwards along a line of greatest slope from the foot of a surface inclined

at

to the horizontal. The initial speed of is

and the coefficient of friction is

. The

particle comes to instantaneous rest before it reaches the top of the inclined surface. a Calculate the distance moves before coming to rest. b Calculate the time takes before coming to rest. c Find the time taken for to return to its initial position from its highest point. 7

A block of mass

lies on a smooth plane inclined at

to the horizontal. A light, inextensible

string is attached at one end to , and runs from up the slope parallel to the line of greatest slope on the plane to a smooth peg, and is attached at the other end to a particle of mass which hangs vertically below the peg, exactly

,

from the floor.

The system is released from rest. a Find the acceleration of up the slope. When hits the floor, the string breaks. b Assuming the initial distance between the block and the peg is sufficiently great that the block will not reach the peg, find the total distance travelled by the block when it instantaneously passes through its initial position. 8

A block of mass lies on a flat, rough surface. The coefficient of friction between the block and the surface is . The surface is initially horizontal, and its inclination is gradually increased. When the inclination of the surface exceeds

9

In this question use

to the horizontal, the block begins to move. Find . , giving your final answers to an appropriate degree of accuracy.

Two blocks and , connected by a light inextensible string, lie with above on the line of greatest slope of a rough plane inclined at to the horizontal. Block has mass and block has mass slope.

. The connecting string is

long, and block begins

The coefficient of friction between each block and the surface is

above the foot of the

.

The system is released from rest at time ; when block reaches the foot of the slope, its motion stops immediately. Find the time at which impacts with .

10

A particle with mass

lies on a rough plane inclined at

to the horizontal. A light,

inextensible string connects to then runs parallel with the line of greatest slope of the plane to a smooth peg, then vertically downwards, through a smooth, free ring with mass and then vertically upward to attach to a fixed point . The coefficient of friction between and the plane is

.

a By resolving forces vertically at , show that the acceleration of when the system is released from rest is related to the tension in the string by . b By resolving forces at , find an equation linking and with friction . c Find the direction and magnitude of the frictional force. d Determine whether will remain stationary, move upslope or move downslope when the

system is released from rest

11

A toy hovercraft with mass

is placed on a rough surface inclined at

to the horizontal.

Propulsion of the toy is achieved by directing two small fans, one either side of the toy, set at an angle to provide a driving force for movement. The force produced by the fans is The coefficient of friction between the toy and the surface is

.

.

a Find the acceleration of the toy up the slope if the fans are set to blow horizontally. b Find the acceleration of the toy up the slope if the fans are set to blow parallel to the slope. c The fans are set so that the direction of the driving force is

greater than the slope (i.e. at

above the horizontal). Find the value of that maximises the acceleration.

12

A block of mass

lies on a surface inclined at

to the horizontal. A light, inextensible string

is attached at one end to , and runs from up the slope parallel to line of greatest slope on the plane to a smooth peg . The string passes over the peg, through a smooth ring of mass , and is attached to a wall at Given the angle

equals

. and the system is in equilibrium, determine the possible values

for , the coefficient of friction between block and the surface. 13

A particle is projected upwards on a rough slope inclined at an angle to the horizontal. There is a coefficient of friction between the particle and the slope. The acceleration on the way up is twice the acceleration on the way down. Prove that

.

Checklist of learning and understanding To resolve a force in a given direction, draw a right-angled triangle with the force as the hypotenuse and the other sides of the triangle parallel and perpendicular to the direction of interest. When calculating motion on a slope, resolve forces parallel and perpendicular to the slope rather than vertically and horizontally. In particular the components of gravity acting on a slope inclined at an angle to the horizontal are: . The maximum or limiting value of friction, given by:

, between an object and a surface is

where is the normal reaction force between the object and the surface and is the coefficient of friction. If the resultant of the driving forces is smaller than , the object will remain at rest and the friction force, , will equal the resultant of the driving forces. If the resultant of the driving forces is larger than .



, the object will move and

Mixed practice 18 In this exercise, unless instructed otherwise, use degree of accuracy. 1

A canal boat of mass

, giving your final answers to an appropriate

is being pulled along a straight, smooth canal by a horse. The

horse has a rope with tension acting at an angle to the canal. Assuming the canal boat can only go in the direction of the canal and no other forces are acting, find the acceleration of the canal boat to three significant figures. Choose from these options. A B C D 2

The diagram shows two forces acting on a particle. a Find the component of the resultant force in the direction of the

force.

b The direction of the force is allowed to vary. Find the maximum and minimum value of the magnitude of the resultant force.

3

A block of mass horizontal.

slides freely from rest down a smooth slope inclined at

to the

What is the acceleration of the block down the slope? 4

In this question use accuracy.

, giving your final answers to an appropriate degree of

A particle is projected with speed from its starting point.

across a rough horizontal surface and comes to rest

a Calculate the coefficient of friction between the particle and the surface. b A second, identical particle is projected across the same surface and comes to rest from its starting point. Determine its initial speed. 5

A particle of mass is suspended in equilibrium by two light strings, and . The string makes an angle of to horizontal and the other string, , is horizontal, as shown in the diagram.

a Draw and label a diagram to show the forces acting on the particle. b Show that the tension in the string

is

c Find the tension in the horizontal string

. . [© AQA 2008]

6

A box, of mass , is placed on a rough slope inclined at an angle of is released from rest and slides down the slope.

to the horizontal. It

a Draw a diagram to show the forces acting on the box. b Find the magnitude of the normal reaction force acting on the box. c The coefficient of friction between the box and the slope is

.

Find the magnitude of the friction force acting on the box. d Find the acceleration of the box. e State an assumption that you have made about the forces acting on the box [© AQA 2013] 7

Three forces are in equilibrium in a vertical plane, as shown in the diagram. There is a vertical force of magnitude and a horizontal force of magnitude . The third force has magnitude newtons and acts at an angle above the horizontal. a Find . b Find .

[© AQA 2014] 8

A particle of mass

is on a smooth slope inclined at

to the horizontal. The particle is

held at rest by a force of newtons parallel to the slope, as shown in the diagram. a Draw a diagram to show all the forces acting on the particle. b Show that the magnitude of the normal reaction acting on the particle is

newtons.

c Find .

[© AQA 2010] 9

A box of mass is held at rest on a plane inclined at an angle of box is then released and slides down the plane.

to the horizontal. The

a A simple model assumes that the only forces acting on the box are its weight and the normal reaction from the plane. Show that, according to this simple model, the acceleration of the box would be , correct to three significant figures. b In fact, the box moves down the plane with constant acceleration and travels seconds By using this information, find the acceleration of the box.

metres in

c Explain why the answer to part b is less than the answer to part a. [© AQA 2009]

10

A block, of mass

, is made to move in a straight line on a rough horizontal surface by a

horizontal force of

newtons, as shown in the diagram.

Assume that there is no air resistance acting on the block. a Draw a diagram to show all the forces acting on the block. b Find the magnitude of the normal reaction force acting on the block. c The acceleration of the block is the block.

. Find the magnitude of the friction force acting on

d Find the coefficient of friction between the block and the surface. e Explain how and why your answer to d would change if you assumed that air resistance did act on the block.

[© AQA 2012] 11

A block is projected up a rough slope that makes an angle coefficient of friction is

with the horizontal. The

. Find the magnitude of the acceleration of the block during its

motion. Choose from these options A B C D 12

A block of mass is held at rest on a rough horizontal surface. The coefficient of friction between the block and the surface is . A light inextensible string which passes over a smooth peg is attached at one and to the block and at the other end to a particle of mass , which is hanging at rest. The system is released from rest. Find: a the magnitude of the frictional force acting on the block b the acceleration of the block and the tension in the string.

13

In this question use , giving your final answers to an appropriate degree of accuracy. A block with mass is pulled from rest up a smooth slope inclined at to the horizontal by a string with tension , maintained at an angle of to the horizontal. a After seconds, the block has moved

along the slope. Calculate .

b The block is allowed to come to rest again, then the tension is increased so that the block is about to lift off the slope. Calculate the minimum tension needed to achieve this. 14

A particle , mass

, is projected up a line of greatest slope of a rough plane inclined at

to the horizontal with initial velocity

. The particle comes to instantaneous rest on the

plane after travelling a Calculate the frictional force acting on , and the coefficient of friction between and the plane. b Determine the acceleration of down the plane subsequent to the instant of rest. 15

A particle of mass is projected with velocity plane inclined at to the horizontal.

up the line of steepest slope of a

a If the plane is smooth, determine the velocity of the particle after does not reach the end of the plane.

seconds, assuming it

b If the plane is rough, with coefficient of friction between particle and plane, calculate the maximum vertical height above its starting point that the particle will reach. 16

A particle is projected with speed down the line of steepest slope of an inclined rough plane. The coefficient of friction between the particle and the plane is . It takes the same time seconds for a particle of mass to travel from its start position if the plane is inclined at angle to the horizontal as it takes for a particle of mass to travel from its start position if the plane is inclined at angle to the horizontal. Show that

17

A block, of mass , is held at rest on a rough horizontal surface. The coefficient of friction between the block and the surface is . A light inextensible string, which passes over a fixed smooth peg, is attached to the block. The other end of the string is attached to a particle, of mass , which is hanging at rest. The block is released and begins to accelerate. a Find the magnitude of the friction force acting on the block. b By forming two equations of motion, one for the block and one for the particle, show that the magnitude of the acceleration of the block and the particle is . c Find the tension in the string. d When the block is released, it is it hits the peg.

metres from the peg. Find the speed of the block when

e When the block reaches the peg, the string breaks and the particle falls a further metres to the ground. Find the speed of the particle when it hits the ground.

[© AQA 2009] 18

A cyclist freewheels, with constant acceleration, in a straight line down a slope. As the cyclist moves metres, his speed increases from to . a i Find the acceleration of the cyclist. ii Find the time that it takes the cyclist to travel this distance. b The cyclist has a mass of the cyclist.

. Calculate the magnitude of the resultant force acting on

c The slope is inclined at an angle to the horizontal.

i Find if it is assumed that there is no resistance force acting on the cyclist. ii Find if it is assumed that there is a constant resistance force of magnitude acting on the cyclist.

newtons

d Make a criticism of the assumption described in part c ii. [© AQA 2012] 19

A child pulls a sledge, of mass

, along a rough horizontal surface, using a light rope. The

coefficient of friction between the sledge and the surface is . The tension in the rope is newtons. The rope is kept at an angle of to the horizontal, as shown in the diagram. Model the sledge as a particle. a Draw a diagram to show all the forces acting on the sledge. b Find the magnitude of the normal reaction force acting on the sledge, in terms of . c Given that the sledge accelerates at

, find .

[© AQA 2012] 20

Two right-angled triangular prisms of equal height, with angle of greatest slope are positioned as shown, with a smooth peg between the two highest points.

and

Block , with mass , is placed on the slope and block , with mass , is placed on the slope. The two blocks are connected by a light, inextensible string which runs parallel to the line of greatest slope of each prism and passes over the smooth peg. The coefficient of friction between block and the slope surface is and the coefficient of friction between block and the slope surface is . At time

, block is projected down the

slope with speed

.

a Calculate the acceleration of block in terms of . It is determined that

.

b Calculate whether the blocks will return to their original positions, and if so, the time at which this will occur.

21

A block of mass is dragged across a rough horizontal surface by a rope that is at an angle of to the horizontal. The coefficient of friction between the block and the surface is . a The tension in the rope is

newtons.

i Draw a diagram to show the forces acting on the block as it moves. ii Show that the magnitude of the normal reaction force on the block is correct to three significant figures. iii Find the magnitude of the friction force acting on the block.

newtons,

iv Find the acceleration of the block. b When the block is moving, the tension is reduced so that the block moves at a constant speed, with the angle between the rope and the horizontal unchanged. Find the tension in the rope when the block is moving at this constant speed. c If the block were made to move at a greater constant speed, again with the angle between the rope and the horizontal unchanged, how would the tension in this case compare to the tension found in part b? [© AQA 2013] 22

A crate, of mass , is initially at rest on a rough slope inclined at shown in the diagram.

to the horizontal, as

The coefficient of friction between the crate and the slope is . a Given that the crate is on the point of slipping down the slope, find . b A horizontal force of magnitude newtons is now applied to the crate, as shown in the diagram. i Find the normal reaction on the crate in terms of . ii Given that the crate accelerates up the slope at

, find .

[© AQA 2014] 23

A van, of mass , is towed up a slope inclined at to the horizontal. The tow rope is at an angle of to the slope. The motion of the van is opposed by a resistance force of magnitude

newtons. The van is accelerating up the slope at

.

Model the van as a particle. a Draw a diagram to show the forces acting on the van. b Show that the tension in the tow rope is

newtons, correct to three significant figures. [© AQA 2011]

Worksheet See Extension Sheet 18 for a selection of more challenging problems.



19 Moments In this chapter you will learn how to: find the turning effect of a force work with uniform rods and laminas understand and use rotational equilibrium.

Before you start… Student Book 1,

You should be able to recognise

Chapter 18

types of force acting on a particle.

1 A particle is pulled across a smooth horizontal table by a string that is parallel to the table. Draw a diagram and label all the forces acting on the particle.

Student Book 1, Chapter 18

You should understand when a particle is in equilibrium.

2 Three forces act on a particle as shown.

The particle is in equilibrium. Find the magnitude of .

Modelling rotating systems Until now you have used the particle model to analyse forces and motion. However, there are situations in which this is not appropriate. Consider, for example, closing a door. A force is applied to push the door closed, but when the door moves it does not do so in a straight line. Instead it rotates about the hinge. In this chapter you will find out how situations like this can be modelled.



Section 1: The turning effect of a force From your experience of closing doors you probably know that it is easier to push if your hand is further away from the hinge. This is because the moment – the turning effect of a force – depends upon both the force applied and the distance away from the pivot point.

Key point 19.1 The moment of a force about a point is:

where is the perpendicular distance of the line of action of the force from .

The units of a moment are newton metres

.

The moment will cause either clockwise or anticlockwise rotation about a point.

If the line of action of the force acts through , then the moment about will be zero (as the perpendicular distance from is

). So in this case there is no rotational effect from the force.

Many situations involving forces that cause rotation can be modelled using two basic shapes. A uniform lamina, which is a two-dimensional object usually in the shape of a rectangle. You might use this to model objects such as a door or a book. A uniform rod, which has just one dimension. You might use this to model a see-saw, snooker cue or a plank. In both cases uniform means that the object has the same density throughout. The key fact you need to know is where the centre of mass is for both of these shapes.

Key point 19.2 The centre of mass is the point at which the object’s weight acts. For a uniform rod, this is at the midpoint. For a uniform lamina, this is at the intersection of the diagonals.

Fast forward Although this might sound obvious, determining the centre of mass of more complex shapes can be quite difficult. You will see how to do this if you study the Mechanics option of Further Mathematics.

WORKED EXAMPLE 19.1

A uniform rod of length

has weight

.

Find the moment of its weight about the point: a b

a

The perpendicular distance from to the weight is The weight acts at the midpoint. It causes the rod to rotate clockwise about .

.

The perpendicular distance from to the weight is

.

b

The weight will cause the rod to rotate anticlockwise about .

WORKED EXAMPLE 19.2

A uniform rectangular lamina

by

has weight

.

The lamina is free to rotate in a vertical plane about . Find the moment about of the: a weight b

a

force.

The weight acts at the centre of the lamina.

The perpendicular distance from to the line of action of the weight is . The rotation will be clockwise.

b

The perpendicular distance from to the line of action of the force is . The rotation will be clockwise.

In just the same way that when several forces act on a body they combine to give an overall resultant force, the same is true of moments.

Key point 19.3 To find the resultant moment about a point, find the sum of the clockwise and anticlockwise moments separately. The resultant moment will be the difference between the two sums, in the direction of the larger moment.

Focus on … Focus on … Modelling 3 applies moment to the context of levers.

WORKED EXAMPLE 19.3

A uniform rod of length

and weight

is acted on by a

and

force as shown.

Find the resultant moment about the point .

Find the moment of each force in turn.

The

force is

from .

Find the sum of the moments that act in the same direction. Resultant moment about

The sum of the clockwise moments is greater.

EXERCISE 19A 1

Find the moment about the point of the weight of each uniform rod. a i

ii

b i

ii

c i

ii

2

Find the moment about the point of the weight of each uniform lamina. a i

ii

b i

ii

c i

ii

3

A

long uniform rod weighs

.

Find the net moment about the point in each situation. a i

ii

b i

ii

c i

ii

4

A

by

uniform rectangular lamina weighs

.

Find the net moment about the point in each situation, given that the lamina is in a vertical plane. a i

ii

b i

ii

c i

ii

d i

ii

e i

ii

5

A steering wheel is modelled as a ring of diameter

. The driver applies two clockwise forces of

tangentially at diametrically opposite sides of the steering wheel. Find the net moment about the centre of the wheel.

6

Find the resultant moment, including the direction, about the point in the diagram.

7

In this question use

, giving your final answer to an appropriate degree of accuracy.

a A diving board is modelled by a uniform plank of length

and mass

. A diver of mass

stands at one end of the plank. Find the total moment around the opposite end of the board. b Explain how you used the fact that the plank is rigid in your calculation in part a. 8

The diagram shows a regular hexagon. Find the resultant moment, including the direction, about the centre of the hexagon.

9

The diagram shows a square of side

with four equal forces acting at the corners. Prove that the

net moment is the same about any point on the interior of the square.

10

In this question use

, giving your final answer to an appropriate degree of accuracy.

A badminton racquet is modelled as a uniform rod of length

and mass

connected to a

square lamina of side length and mass . The racquet is held horizontally at a point, from the end of the rod. Find the net moment of the racquet about the point .



Section 2: Equilibrium To counterbalance an object’s weight and maintain it in equilibrium there will either be: smooth supports providing a normal reaction force, or

light strings providing a tension.

For an object to be at rest in equilibrium, the condition that there is no rotation needs to be added.

Key point 19.4 If an object is in equilibrium there is zero resultant force and zero resultant moment about any point.

Notice that whereas the resultant moment will in general be different about different points, if the object is in equilibrium, the resultant moment will be zero about any point.

Rewind Recall from Chapter 18 that ‘smooth’ means there is no friction (so here the force at the support will be normal to the surface of the object), and ‘light’ means the string has no mass.

Therefore, since you have a choice of which point to take moments about, in general it is a good idea to choose a point where at least one force acts. Since the moment of that force will be zero, it eliminates it and makes the calculation simpler. WORKED EXAMPLE 19.4

A plank of length diagram.

and mass

rests in equilibrium on two identical chairs, as shown in the

The chairs can provide a reaction force of before breaking. Assuming that the plank can be modelled as a uniform rod, determine if either chair will break. Since the plank is a uniform rod the centre of mass is metres from either end so metres from .

Taking moments around :

As the plank is in equilibrium, sum of clockwise moments sum of anticlockwise moments. Notice that taking moments about either or allows you to ignore one of the unknown forces. You could choose either point to start with.

So the chair will not break at . Vertical forces:

This exceeds the breaking force so

so the chair does not break. Equilibrium also means that forces up forces down.

so the chair does break.

the chair will break at .

WORKED EXAMPLE 19.5

A shop sign is formed from a rectangular plastic sheet

of weight

. It is held in

equilibrium by a vertical wire at and horizontal wires at and , as shown in the diagram below. If the sign can be modelled as a uniform lamina with

Take moments about :

and

, find

, and

Moments about eliminates two of the unknown forces. Since it is in equilibrium, sum of clockwise moments sum of anticlockwise moments. The centre of mass is

to the right of .

.

Horizontal forces:

Forces right forces left

Vertical forces:

Forces up forces down

If a rod is supported at two points, it can be in limiting equilibrium on the point of turning (or tilting) about one of these points. In this situation the force at the other support point will be zero. WORKED EXAMPLE 19.6

A wooden plank of mass and length is suspended horizontally from the ceiling by two ropes from each end of the plank at and . When a mass is placed on the plank at , the plank is on the point of turning about . a Find . b State any modelling assumptions you have made.

When the plank is on the point of turning about , the rope will be on the point of going slack at so .

a

Taking moments about :

Taking moment about eliminates the unknown tension.

b The plank is uniform. The mass is a particle. The ropes are light.

WORK IT OUT 19.1 A uniform rectangular lamina of width and height has a weight of and is attached to a fixed bolt at point , about which it can rotate freely. A horizontal string is attached midway along under a tension of . A downward force of is applied at a point which lies on with so that the lamina hangs in equilibrium with horizontal. Find the value of .

Which is the correct solution? Identify the errors made in the incorrect solutions. Solution A Taking moments about : Clockwise moments: Anticlockwise moments:

Solution B Taking moments about the centre of mass: Clockwise moments: Anticlockwise moments: So

Solution C Taking moments about : Clockwise moments: Anticlockwise moments: So

EXERCISE 19B 1

In each diagram a uniform rod of length wires. Find the unknown values. a i

ii

and weight

is being held in equilibrium by two

b i

ii

2

Each uniform rectangular lamina has a weight of plane. Find the unknown values. a i

ii

b i

ii

and is hanging in equilibrium in a vertical

c i

ii

3

Two children sit on a seesaw formed from a uniform rod of length child, of mass

, balanced in the middle. One

, sits on one end. How far from the other end should the other child, of mass

, sit so that the seesaw is balanced in a horizontal position? 4

A door is

wide. A perpendicular force of

is applied

from the hinge but a wedge at

the end of the door opposite the hinge is keeping it shut. Find the frictional force acting though the wedge. 5

a In a simplified model of a crane the arm mass

is modelled as a uniform rod of length

and

.

The arm is attached to the main body of the crane

from . A mass of

is suspended

from . A counter mass of can be moved along the arm to keep the crane in equilibrium. Find the distance from it should be attached. b Explain why the counter mass would not need to be placed precisely at the position found in a.

6

A plank

of mass . When a

and length rests on supports at points and where and mass is placed on the plank at the plank is on the point of tilting about

. Modelling the plank as a uniform rod and the mass as a particle, find the distance 7

A uniform rod of weight and length vertical wires at and , as shown. A weight of

is suspended horizontally from the ceiling by two

is placed on the rod at .

a Show that the rod will turn about . A weight of

is now placed on the rod at .

b Find the minimum value of

.

so that the rod is restored to equilibrium.

8

A uniform plank of length and mass rests in a horizontal equilibrium on two supports – one at the end of the plank and the other from the other end. Find, in terms of , the reaction force supplied by each support.

9

In this question use

, giving your final answers to an appropriate degree of accuracy.

A uniform plank of length painter of weight wires. 10

and mass

is standing

is suspended horizontally by wires at either end. A

from one end of the beam. Find the tension in each of the

a A pole vaulter holds a uniform pole of length and weight with one hand on the end and the other away. Find the vertical forces exerted by his hands in terms of . b State one additional assumption have you made in part a.

11

In this question use

, giving your final answers to an appropriate degree of accuracy.

A spade is modelled as a uniform rod of mass lamina of side

and mass

and length

attached to a uniform square

. A gardener holds the spade horizontally with hands

and

from the end of the rod. Find the vertical forces exerted by the gardener’s hands. 12

In this question use

, giving your final answers to an appropriate degree of accuracy.

A model for the elbow joint models the bicep muscle connecting to the horizontal forearm by a vertical tendon from the elbow joint. A mass is held in the hand from the elbow joint. If the maximum tension that can be exerted by the tendon before injury occurs is maximum mass that can be held in this way.

13

A shop sign weight

is modelled as a uniform rectangular lamina of width

, height

, find the

and

. It is resting on a support at with a horizontal wire connected at .

a Find, in terms of where necessary: i the tension in the wire ii the normal reaction at iii the friction force at . b How would your answer to part a i be different if a light rod rather than a wire were attached at ?

14

A door is modelled as a rectangular lamina of weight

with height

and width

,

supported in equilibrium by two hinges at and . If the force at is entirely horizontal, find the magnitude of the forces at and .

15

A snooker cue is formed by connecting end-to-end two uniform rods of length . One has mass and the other has mass . It balances above a point a distance from the exterior end of the

16

section. Find the value of .

In the film ‘The Italian Job’ a coach is balancing on the edge of a cliff with gold bullion at one end and a group of people at the other. Model the coach as a uniform rod of length with of gold at the end overhanging the cliff and How much of the coach can overhang the cliff before it falls?

and mass

of people at the other end.

Checklist of learning and understanding The moment of a force about a point is: where is the perpendicular distance of the line of action of the force from . The centre of mass is the point at which the object’s weight acts. For a uniform rod, this is at the midpoint. For a uniform lamina, this is at the intersection of the diagonals. To find the resultant moment about a point, find the sum of the clockwise and anticlockwise moments separately. The resultant moment will be the difference between the two sums (in the direction of the larger). If an object is in equilibrium there is zero resultant force and zero resultant moment about any point.





Mixed practice 19 1

What is the resultant moment about point on the uniform rectangular lamina of weight shown in the diagram? Choose from these options. A B C D

2

A rigid uniform rod of length and weight weight sits from support .

rests on two supports as shown. A

Find the reaction force of the support at . Choose from these options. A B C D

3

The diagram shows a uniform rod of length suspended from two wires, under tension respectively. Find the value of .

4

A uniform rod

of weight

and length

and weight lying in equilibrium, and , and set at distances and

is freely hinged at with a vertical string

attached at . The tension, newtons, in the string is sufficient to maintain the rod in a horizontal equilibrium. a Find the value of . b Find the magnitude and direction of the force provided by the hinge. 5

In this question use

, giving your final answers to an appropriate degree of

accuracy. A uniform plank, of length metres, has mass

. The plank is supported in equilibrium in a

horizontal position by two smooth supports at the points and , as shown in the diagram. A block, of mass , is placed on the plank at point .

a Draw a diagram to show the forces acting on the plank. b Show that the magnitude of the force exerted on the plank by the support at is newtons. c Find the magnitude of the force exerted on the plank by the support at . d Explain how you have used the fact that the plank is uniform in your solution. [© AQA 2010] 6

In this question use accuracy.

, giving your final answers to an appropriate degree of

A uniform plank , of length , has mass . It is supported in equilibrium in a horizontal position by two vertical inextensible ropes. One of the ropes is attached to the plank at the point and the other rope is attached to the plank at the point , where and

, as shown in the diagram.

a i Find the tension in each rope. ii State how you have used the fact that the plank is uniform in your solution. b A particle of mass

is attached to the plank at point , and the tension in each rope is

now the same. Find . [© AQA 2013] 7

Ken is trying to cross a river of width . He has a uniform plank, , of length and mass . The ground on both edges of the river bank is horizontal. The plank rests at two points, and , on fixed supports which are on opposite sides of the river. The plank is at right angles to both river banks and is horizontal. The distance is , and the point is at a horizontal distance of from the river bank. Ken, who has mass , stands on the plank directly above the middle of the river, as shown in the diagram.

a Draw a diagram to show the forces acting on the plank. b Given that the reaction on the plank at the point is the point from the nearest river bank.

, find the horizontal distance of

c State how you have used the fact that the plank is uniform in your solution. [© AQA 2011] 8

An oar is modelled as a uniform rod of length and mass with an additional mass of attached to end . The oar is hung by a single wire. How far from must this be attached if the oar is to hang horizontally?

9

In this question use accuracy.

, giving your final answers to an appropriate degree of

A uniform plank of mass and length rests on supports from and from . A mass is placed in a position to equalise the reaction forces on the two supports. Find the distance of the mass from . 10

In this question use accuracy. A uniform plank and .

, giving your final answers to an appropriate degree of

of mass

When a particle, , of weight

and length

hangs from two vertical ropes attached to

is attached to the plank rests horizontally in equilibrium.

If the tension in the rope at is three times the tension at , find: a the tension at b the distance

11

.

A uniform plank of length smooth supports as shown.

and weight

rests horizontally in equilibrium on two

a Find the reaction at . A child of weight stands on the plank at . The plank remains in equilibrium. The reactions on the plank at and are now equal. b Find the distance

.

12

A non-uniform plank has length shown in the diagram.

and weight

. It rests horizontally in equilibrium as

A particle of weight is placed on the plank at . The plank remains in equilibrium and the reaction at is . The centre of mass of the plank is a distance from . a Show that

.

The particle is now removed from and placed at . The rod remains in equilibrium and the reaction at is now b Find

.

and .

Worksheet See Support Sheet 19 for an example of non-uniform rods in equilibrium and for more practice questions. 13

A uniform plank of length and weight smooth supports and as shown.

rests horizontally in equilibrium on two

A particle of weight is attached to a point on the plank equilibrium and the reactions at and are now equal. a Show that

from . The plank remains in

.

b Hence find the range of possible values of .

14

A rectangular lamina is formed by connecting two square uniform laminas of side metres and masses and where . The rectangular lamina balances on a point a distance from the join line, as shown in the diagram. Find an expression for in terms of and .

Worksheet See Extension Sheet 19 for some questions involving laminas formed from a rectangle and a triangle. 15

A uniform rod of length

and weight

is being pushed over a roller

with negligible

friction and a support with a coefficient of friction length of the rod overhanging with .

. and are

apart. is the

Find, as a function of , the force required for the rod to move at a constant speed.

16

A car and contents is modelled by a uniform rectangular laminar (since depth will not be relevant) of mass and width . The wheels are located from the front of the car and from the rear of the car. a By taking moments about the rear wheel, show that the normal reaction of the ground on the front wheel is . b The car has front wheel drive. Assuming that the car has sufficient power and the coefficient of friction between the car and the road is , find, in terms of , the maximum acceleration of the car.

Rewind The coefficient of friction was covered in Chapter 18, Section 2.



FOCUS ON… PROOF 3

Overcoming friction A block of weight rests on rough horizontal ground. The coefficient of friction between the block and the ground is µ . You are going to derive the formulae for the minimum force required to move the block in various situations.

QUESTIONS 1

A horizontal force of magnitude acts on the block. Find, in terms of

2

and , the minimum value of required to move the block.

The force acts at a fixed angle above the horizontal. Prove that the minimum value of is

3

.

Both and can vary. a i Find the maximum value of

for

.

ii Hence find the minimum magnitude of the force required to move the box. iii Prove that this minimum magnitude is always less than . (In other words, however large is, it is always possible to move the box using a force smaller than its weight.) b Prove that the angle at which the minimal force needs to act is 4

.

The force pushes the block at an angle to the horizontal, as shown. a Prove that

.

b By considering the graph of for , or otherwise, prove that the required magnitude of the force is minimal when the force is horizontal, and state this magnitude.



FOCUS ON… PROBLEM SOLVING 3

Checking for reasonableness When solving problems in real contexts answers are rarely easy numbers to work with. It can therefore be difficult to tell whether your answer is correct. However, there are still some checks you can do, such as confirming that your calculation gives correct units and making sure that the answer is not completely unreasonable. QUESTIONS Here are proposed answers to some mechanics problems. Decide which ones are obviously wrong.



1

A car’s deceleration is

2

The tension in the cable supporting a lift is

3

The frictional force acting on a box is

4

The stone was dropped from a height of

5

The road is inclined at

6

The mass of a car is

7

The coefficient of friction between a box and the floor is

8

An athlete runs

9

The maximum height reached by the projectile is

. .

. .

to the horizontal. .

at an average speed of

.

. .

10

The two trains will meet after

11

The coefficient of friction between the skates and the ice is

12

The car accelerates at

13

The coefficient of friction between the box and the ice is

14

The ball takes

15

The journey from London to Manchester takes

seconds. .

.

seconds to fall from the tenth floor. hours.

.

FOCUS ON… MODELLING 3

Modelling with moments It is alleged that Archimedes once said, ‘Give me a place to stand and I will move the Earth’.

It is theoretically possible to move any weight if you use a sufficiently long lever. Archimedes would need to be standing on a different planet and he would also need a support for the lever.

Did you know?

The Ancient Greek mathematician Archimedes is perhaps best known for exclaiming ‘Eureka!’ on realising that a body submerged in water displaces its own volume of water, an idea now commonly known as Archimedes’ principle. However, his work extended well beyond this discovery and his work on levers. As well as designing the screw pump for raising water, compound pulleys and siege machines, he also developed the fundamentals of integral calculus 2000 years before Newton and Leibniz eventually formalised it.

QUESTIONS

QUESTIONS 1

Suppose the Earth is modelled as a particle of mass , resting on one end of the lever, and that Archimedes is a particle of mass , sitting on the other end. The lever rests on a support from the Earth. How long would the other side of the lever need to be?

2

Other than modelling the Earth and Archimedes as particles, what other assumption did you make in your calculation in question 1?

3

Is modelling the Earth as a particle realistic? What could you change in your model to make the particle assumption reasonable? (The radius of the Earth is .) How does this change your answer in question 1?

4



Using the modelling assumptions from question 1, suppose Archimedes moves his end of the lever for a year, at the average speed of . By how much would he move the Earth?

CROSS-TOPIC REVIEW EXERCISE 3 In this exercise, unless instructed otherwise, use

, giving your final answers to an appropriate

degree of accuracy. 1

Points and have position vectors a Find the vector

and

.

.

b Find the exact distance between the two points. 2

Points

and have coordinates

and

a Find the coordinates of the point so that b Show that

is a parallelogram.

is a rhombus.

c Find the coordinates of the point on the line 3

, respectively.

such that

.

A particle moves on a horizontal plane, in which the unit vectors and directed east and north respectively. At time seconds, the position vector of the particle is metres, where

a Find an expression for the velocity of the particle at time . b i Find the speed of the particle when

.

ii State the direction in which the particle is travelling when c Find the acceleration of the particle when d The mass of the particle is

.

.

.

Find the magnitude of the resultant force on the particle when

. [© AQA 2009]

4

A uniform plank of mass and length rests on two supports, and . The support is from one end of the plank and support is from the other end. The plank is horizontal. A box of mass is placed on the plank, from . Find, in terms of , the force acting on the plank at each support.

5

A non-uniform plank of mass mass rests on the plank, equilibrium, find, in terms of :

and length rests on a support at its mid-point. A particle of to the left of the support. Given that the plank is in

a the position of its centre of mass b the magnitude of the force acting on the plank at the support. 6

A particle, of mass , remains in equilibrium under the action of three forces, which act in a vertical plane, as shown in the diagram. The force with magnitude acts at above the horizontal and the force with magnitude acts at an angle above the horizontal. a By resolving horizontally, find . b Find .

[© AQA 2010] 7

Three points have coordinates

and

Find the values of and so that 8

is a straight line.

The position vectors of two particles, and , at time , are given by and

9

.

. Prove that the distance between the two particles is constant.

The constant forces

newtons and

newtons act on a particle. No

other forces act on the particle. a Find the resultant force on the particle. b Given that the mass of the particle is

, show that the acceleration of the particle is

. c At time seconds, the velocity of the particle is i When

. Show that

. when

.

ii Write down an expression for at time . iii Find the times when the speed of the particle is

. [© AQA 2010]

10

A particle has mass

and moves on a smooth horizontal plane. A single horizontal force, newtons, acts on the particle at time seconds.

The unit vectors and are directed east and north respectively. a Find the acceleration of the particle at time . b When

, the velocity of the particle is

. Find the velocity of the particle at

time . c Find when the particle is moving due west. d Find the speed of the particle when it is moving due west. [© AQA 2010] 11

In this question use accuracy.

, giving your final answers to an appropriate degree of

A uniform rectangular lamina has sides of length and and mass grams. The lamina is freely hinged at point and is held in equilibrium, with the longer side horizontal, by means of a light inextensible string attached to point . a Find the tension in the string. b Find the magnitude and direction of the force acting at the lamina at .

12

A block, of mass

, slides down a rough plane inclined at

to the horizontal. When

modelling the motion of the block, assume that there is no air resistance acting on it. a Draw and label a diagram to show the forces acting on the block. b Show that the magnitude of the normal reaction force acting on the block is three significant figures. c Given that the acceleration of the block is the block and the plane.

, correct to

, find the coefficient of friction between

d In reality, air resistance does act on the block. State how this would change your value for the coefficient of friction, and explain why. [© AQA 2008] 13

In this question use

, giving your final answers to an appropriate degree of accuracy.

Two small blocks, each of mass , are connected by a light inextensible string passing over a smooth pulley. One block rests on a rough horizontal table and it other hangs freely with the string vertical. The coefficient of friction between the block and the table is . The first block is initially

from the edge of the table when the system is released from rest.

a Find, in terms of , the acceleration of the system. b How long does it take for the first block to reach the edge of the table? c Find the magnitude and direction of the force acting on the pulley. d How did you use the modelling assumption that: i the pulley is smooth ii the string in inextensible?

14

Two particles, and , are connected by a light inextensible string which passes over a smooth peg. Particle has mass and particle has mass . Particle hangs freely with the string vertical. Particle is at rest in equilibrium on a rough horizontal surface with the string at an angle to the vertical. The particles, peg and string are shown in the diagram. a By considering particle , find the tension in the string. b Draw a diagram to show the forces acting on particle . c Show that the magnitude of the normal reaction force acting on particle is correct to three significant figures.

newtons

d Find the least possible value of the coefficient of friction between and the surface.

[© AQA 2011] 15

An arrow is fired from a point at a height of metres above horizontal ground. It has an initial velocity of at an angle of above the horizontal. The arrow hits a target at a height of metre above horizontal ground. The path of the arrow is shown in the diagram. Model the arrow as a particle. a Show that the time taken for the arrow to travel to the target is

seconds.

b Find the horizontal distance between the point where the arrow is fired and the target. c Find the speed of the arrow when it hits the target. d Find the angle between the velocity of the arrow and the horizontal when the arrow hits the target. e State one assumption that you have made about the forces acting on the arrow.

[© AQA 2011] 16

A car, of mass

, is moving along a straight horizontal road. At time seconds, the car has

speed . As the car moves, it experiences a resistance force of magnitude No other horizontal force acts on the car.

newtons.

a Show that:

b The initial speed of the car is

. Show that:

[© AQA 2011] 17

A particle moves in a straight line with acceleration initially at the origin and has velocity straight line.

18

A golfer hits a ball which is on horizontal ground. The ball initially moves with speed at an angle of above the horizontal. There is a pond further along the horizontal ground. The diagram below shows the initial position of the ball and the position of the pond.

. The particle is . Prove that the particle moves in a

a State two assumptions that you should make in order to model the motion of the ball. b Show that the horizontal distance, in metres, travelled by the ball when it returns to ground level is:

c Find the range of values of for which the ball lands in the pond. [© AQA 2008] 19

In this question use

, giving your final answers to an appropriate degree of accuracy.

A netball player takes a shot at the hoop,

above the ground. The player stands

from the

foot of the post and releases a ball at the height of , as shown in the diagram. The ball is released with speed at an angle above the horizontal. a Given that the ball passes through the hoop, show that b i Find the maximum value of

.

ii Hence find the minimum speed with which the ball can be released and still pass through the hoop.



20 Conditional probability In this chapter you will learn how to: use set notation to describe probabilities work with conditional probabilities in the context of Venn diagrams, two-way tables and tree diagrams use a formula for conditional probability.

Before you start… GCSE

You should be able to use tree diagrams to solve problems.

1 A mother has two children, who are not twins. What is the probability that they are both boys or both girls?

Student Book 1, Chapter 1

You should be able to use set notation.

2 Write out this set.

Student Book 1, Chapter 21

You should understand the basic laws of probability including the terms mutually exclusive and independent.

3 In a family having a car is mutually exclusive of having a motorbike. The probability of having a car is . The probability of having a motorbike is . What is the probability of having neither a car nor a motorbike?

Student Book 1, Chapter 21

You should understand probability distributions, including the binomial distribution.

4 What is the probability of getting heads when fair coins are tossed?

What is conditional probability? What is the probability that you will become a millionaire? You could just look at data for how many millionaires there are in the world, but this is not likely to give you a very accurate answer because it depends on lots of other factors. Where you were born, what your parents do and your attitude towards risk all change the probability. You might be glad to know that the fact that you are doing Mathematics A Level immediately increases your probability of becoming a millionaire! Information often changes probabilities. A probability that takes into account information is called a conditional probability. In reality nearly all probabilities are conditional – the probability of a patient having heart disease might change depending upon their age, the probability of a defendant being guilty might change depending upon their prior convictions, the probability of a football team winning might depend upon the team they are playing. Most people have a very poor intuition for conditional probabilities. In this chapter you will see various ways to visualise conditional probabilities and how you can use them solve problems.

Focus on … See Focus on… Problem solving 4 for the approach of using extreme values to suggest possible solutions.



Section 1: Set notation and Venn diagrams What is more likely when you roll a dice once: getting a prime number and an odd number getting a prime number or an odd number? The first possibility is restrictive – both conditions have to be satisfied. The second opens up many more possibilities – either condition can be satisfied. So the second must be more likely. These are examples of two of the most common way of combining events: intersection (in normal language ‘and’) and union (in normal language ‘or’). You can use set notation to describe the different ways of combining events.

Key point 20.1 is the intersection of and , meaning when both and is the union of happen.

happen.

and , meaning when either happens, or happens, or both

is the complement of , meaning everything that could happen other than .

Did you know? Why do mathematicians not just use simple words? One of the reasons for this is the ambiguity of everyday language. If I say that I play rugby or hockey some people may think this means I do not play both.

You can use Venn diagrams to illustrate the concepts of union and intersection.

     There is an important result that comes from looking at the Venn diagrams.

Key point 20.2

This will be given in your formula book.

You can interpret this formula as saying, ‘If you want the number of ways of getting or , take the number of ways of getting and add to that the number of ways of getting . However, you have counted the number of ways of getting and twice, so compensate by subtracting it’. If there is no possibility of and occurring at the same time, then mutually exclusive, and the formula reduces to WORKED EXAMPLE 20.1

. These events are .

WORKED EXAMPLE 20.1

A chocolate is selected randomly from a box. The probability of its containing nuts is . The probability of its containing caramel is . The probability of its containing both nuts and caramel is . What is the probability of a randomly chosen chocolate containing either nuts or caramel or both? Use the formula:

EXERCISE 20A 1

a i

and and

ii

. Find

and

ii

d i Find

if

ii Find

if

. .

. Find

and

ii

.

. Find and

c i

.

. Find

and

b i

2

. Find

. Find

. .

and and are mutually exclusive. and and are mutually exclusive.

In each question, you might find it helpful to draw a Venn diagram. a i When a fruit pie is selected at random,

and

.

contain both apples and pears. Find

.

ii In a library books are

of books are classed as fiction and century fiction.

are classed as

What proportion of the books are either fiction or from the b i

century. Half of the

century?

of students in a school play either football or tennis. The probability of a randomly chosen student playing football is

and the probability of their playing tennis is .

What percentage of students play both football and tennis? ii

in students in a school study Spanish and in study French. Half of the school study either French or Spanish. What fraction study both French and Spanish?

3

and

.

a By drawing a Venn diagram or otherwise find b Find 4

.

.

of students in a class have social media account and out of have social media account . One twentieth of students have neither of these social media accounts.

What percentage have both of these social media accounts? 5

Events and satisfy Find

6

.

.

of teams in a football league have French players and neither French nor Italian players.

have Italian players.

have

What percentage have French but not Italian players? 7

and Find

8

. Events and are independent.

.

The probability of having brown eyes and wearing glasses is . The probability of neither having brown eyes nor wearing glasses is . Given that eye colour and wearing glasses are independent, find the probability of: a having brown eyes b wearing glasses.

Conditional probability The probability of given that has happened can be written as

.

Venn diagrams provide a good way of thinking about conditional probability. If has happened then you must be in the hatched region. The probability of now happening depends upon the relative probability of the shaded region compared to the hatched region. This leads to an extremely useful formula for conditional probability.

Key point 20.3 The probability of given has happened is:

WORKED EXAMPLE 20.2

The probability that a randomly chosen resident of a city in Japan is a millionnaire is probability that a randomly chosen resident lives in a mansion is

. Only in

. The are

millionnaires who live in mansions. What is the probability of a randomly chosen individual being a millionnaire given that they live in a mansion?

Write the required probability in the ‘given’ notation and apply the formula.

WORKED EXAMPLE 20.3

and a Find b Hence find

.

. .

a

It is often a good idea to start with labelling the intersection with an unknown and writing in all the remaining information in terms of the unknown.

The probabilities sum to .

Make it clear that you know is

b

.

Use Key point 20.3.

You can use Venn diagrams for more than two groups. They can represent the number in the group as well as the probability. WORKED EXAMPLE 20.4

In a class of students, students have a bicycle, students have a mobile phone and students have a laptop computer. students have both a bike and a phone, students have both a phone and a laptop and students have both a bike and a laptop. students have none of these objects. a How many students have a bike, a phone and a laptop? b What is the probability that they have all three of the items, given that they have at least two of them?

a

Draw a Venn diagram showing three overlapping groups, and label the size of the central region as . Then work outwards. For example, the number who have a bicycle and a phone but not a laptop will be .

Continue working outwards. For example, the total of all the bicycle regions must be , so the remaining section is which is . You know there are students outside

Use the fact that there are form an equation.

and .

students in the class to

Therefore three students have a bicycle, a phone and a laptop. b There are

students who have at

least two items and three with three items. So

Use the Venn diagram to find the total number of students in overlapping regions.

.

EXERCISE 20B 1

For each question, write in mathematical notation the probability required. You should write an expression rather than a number. a Find the probability that the outcome on a dice is prime and odd. b Find the probability that a person is from either Senegal or Taiwan. c A student is studying A Levels. Find the probability that the student is also French. d If a playing card is a red card find the probability that it is a heart. e What proportion of German people live in Munich? f

What is the probability that someone is wearing neither black nor white socks?

g What is the probability that a vegetable is a potato if it is not a cabbage? h What is the probability that a ball drawn is red, given that the ball is either red or blue?

2

3

a i If

and

find

.

ii If

and

find

.

b i If

and

find

.

ii If

and

find

.

In these questions, you might find it helpful to draw a Venn diagram. a In a class of

students

take French,

take German and take neither. What is the

probability that someone who takes German also takes French? b In a survey

of people like pizza and

like lasagne.

like nether pizza nor lasagne. Find

the probability that someone likes lasagne given that they like pizza. 4

Simplify each expression where possible. a b c d e f g h i j k l m n

5

Out of

students in a college,

play football,

play badminton, and play both sports.

a Draw a Venn diagram showing this information. b How many students play neither sport? c What is the probability that a randomly chosen student plays badminton? d Given that the chosen student plays football, what is the probability that they also play badminton? 6

Out of

students in a college,

study Mathematics,

study Economics and

study neither of

the two subjects. a Draw a Venn diagram to show this information. b How many students study both subjects? c A student tells you that he studies Mathematics. What is the probability that he studies both Mathematics and Economics? 7

a In a survey,

of people are in favour of a new primary school and

are in favour of a new

library. Half of all those surveyed would like both a new primary school and a new library. What

percentage supported neither a new library nor a new primary school? b What proportion of those wanting a new primary school also wanted a new library? 8

If

and

:

a find b find 9

.

An integer is chosen at random from the first one thousand positive integers. Find the probability that the integer chosen is: a a multiple of b a multiple of both and c a multiple of given that it is a multiple of .

10

Denise conducts a survey about food preferences in the college. She asks students which of the three meals (spaghetti bolognese, chilli con carne, and vegetable curry) they would prefer to eat. She finds out that, out of the

students:

would eat spaghetti bolognese would eat vegetable curry would eat both spaghetti and curry would eat both curry and chilli would eat both chilli and spaghetti would eat all three meals would prefer not to eat any of the three meals. a Draw a Venn diagram showing this information. b How many students would eat only spaghetti? c How many students would eat chilli? d What is the probability that a randomly selected student would eat only one of the three meals? e Given that a student would eat only one of the three meals, what is the probability that they would eat curry? f 11

Find the probability that a randomly selected student would eat at least two of the three meals.

The probability that a person has dark hair is , the probability that they have blue eyes is the probability that they have both dark hair and blue eyes is .

and

a Draw a Venn diagram showing this information. b Find the probability that a person has neither dark hair nor blue eyes. c Given that a person has dark hair, find the probability that they also have blue eyes. d Given that a person does not have dark hair, find the probability that they have blue eyes. e Are the characteristics of having dark hair and having blue eyes independent? Explain your answer. 12

The probability that it rains on any given day is probability that it is neither cold nor raining is

and the probability that it is cold is .

a Find the probability that it is both cold and raining. b Draw a Venn diagram showing this information. c Given that it is raining, find the probability that it is not cold.

. The

d Given that it is not cold, find the probability that it is raining. e Are the events ‘it’s raining’ and ‘it’s cold’ independent? Explain your answer and show any supporting calculations. 13

If

14

If

15

If

16

a If

and

find find

and

.

find the maximum and minimum values of

represents a probability, state the possible values that

b Express

in terms of

c By considering an expression for



.

and

.

can take.

. show that

.

Section 2: Two-way tables Another useful way of looking at conditional probability is to use two-way tables. This lists all the possible outcomes varying along two factors.

Fast forward If you study the Statistics Option in Further Mathematics you will see a method called the chi-squared test to see if the two factors are independent.

WORKED EXAMPLE 20.5

The table shows the arrival time of a random sample of posted first class or second class. Next day

letters, as well as whether they were

Later

1st class 2nd class Find: a b c There are letters in the class and next day categories, out of letters.

a

b c

EXERCISE 20C 1

In each two-way table, find a i

ii

b i

ii

.

There are

next day letters out of



There are



next day letters.

class letters out of

class letters.

c i

ii

2

The two-way table lists the numbers of students in different year groups in a school. Year 9

Year 10

Year 11

Total

Girls Boys Total a Copy and complete the table. b Find the probability that a randomly selected student is a girl from year c Find the probability that a randomly selected girl is from year 3

.

.

The two-way table describes the additions made to coffee in a drinks machine in one day. Milk

No milk

Sugar No sugar a Find the probability of sugar being added. b Find the probability that sugar is added if milk is added. c Show that whether milk is added is independent of whether sugar is added. 4

The table shows the numbers of returned shoes to three stores in one day. Store A

Store B

Returned Unreturned a Find

5

.

b Find

.

c Find

.

The table shows a general two-way table.

Find, in terms of a b c

and :

Store C

d e 6

The table shows the results of the top three countries in the 2016 Olympics. Gold

Silver

Bronze

USA GB China A random result is chosen from amongst these

results. Find:

a b c d 7

A company makes three different sizes of T-shirt in three different colours. This table shows the sales in a week. S

M

L

White Black Green Find: a b c d e 8

.

This table gives the probabilities of different midday temperatures in different air pressures in the UK , based on long-term observations.

a For a randomly chosen day find: i ii iii 9

James investigates two identities: 1 2 Use a counterexample to show that one of these identities is incorrect.



Section 3: Tree diagrams A tree diagram is a useful way of illustrating situations where one outcome depends upon another. For example, this diagram shows the experience of a restaurant trying to predict how many portions of chips it serves.

The second probabilities all depend on the first, so they are conditional probabilities. For example:

You may have learnt that the probability of being on a branch is found by multiplying along the branch. For example, this means:

Key point 20.4 For a tree diagram, use the formula:

This will be given in your formula book.

If you wanted to find the overall probability of chips, you need to add different branches together:

Tip This is actually just a rearrangement of Key point 20.3:

WORKED EXAMPLE 20.6

If Lucas revises there is an a

chance he will pass the test, but if he doesn’t revise there is only

chance of his passing. He revises for of tests. What proportion of tests does he pass? Decide which probability is not conditional. Start the tree diagram with this event.The probability of passing the test is conditional on revision, so the revision branches have to come first.

Add the conditional event.

Identify which branches result in passing the test. Multiply to find the probability at the end of each branch.

Sometimes, you can use the information found from a tree diagram to find another conditional probability.

Tip Worked examples 20.6 and 20.7 use the terms ‘chance’ and ‘proportion’. These are just other words for probability.

WORKED EXAMPLE 20.7

If it is raining in the morning there is a

chance that Shivani will take her umbrella. If it is not

raining in the morning there is only a chance of Shivani taking her umbrella. On any given morning the probability of rain is

.

a What is the probability that Shivani takes an umbrella? b If you see Shivani with an umbrella, what is the probability that it was raining that morning? a

First draw a tree diagram.

Use the tree diagram to find relevant probabilities.

b

Since you need to find a conditional probability, use Key point 20.3.

In Worked example 20.7 you were given point 20.3 to derive a formula for this.

and found

. You can use Key

Tip Quite often in questions like this you use the answer to the first part in the second part.

Key point 20.3 says that written

but you could equally well have swapped and and .

However, and are the same thing (they are both the probability of and ). So the right-hand sides of the two expressions are equal. This leads to a formula that could be used to solve Worked example 20.7.

Key point 20.5 and

can be related by using the formula:

Focus on … See Focus on… Proof 4 for the use of conditional probability and tree diagrams in analysing a claim of guilt in a legal case.

WORK IT OUT 20.1 The probability of an acorn landing more than from the original tree is . Of those that land more than from the tree, germinate. Of those that land less than from the tree germinate. An acorn germinates. What is the probability that it landed more than

from the tree?

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution A

Work out

from a tree diagram.

Solution B

Solution C

Using a Venn diagram:

Since

:

So

EXERCISE 20D 1

In these questions you might find it helpful to write the information in a suitable tree diagram. a i

and and

ii

2

. Find . Find

. .

b i

and

. Find

.

ii

and

. Find

.

c i

and

. Find

.

ii

and

. Find

.

A class contains boys and girls. Two students are picked at random. What is the probability that they are both boys?

3

A bag contains red balls, blue balls and green balls. A ball is chosen at random from the bag and is not replaced A second ball is chosen. Find the probability of choosing one green ball and one blue ball in any order.

4

. a Illustrate this information on a tree diagram. b Find

.

c Find

.

d Find 5

Given that a b

. and

, find:

Worksheet See Support Sheet 20 for a further example on conditional probability and tree diagrams and for more practice questions. 6

A factory has two machines for making widgets. The older machine has a larger capacity, so it makes of the widgets, but are rejected by quality control. The newer machine has only a rejection rate. Find the probability that a randomly selected widget is rejected.

7

The school tennis league consists of players. Daniel has a chance of winning any game against a higher-ranked player, and a chance of winning any game against a lower-ranked player. If Daniel is currently in third place, find the probability that he wins his next game against a random opponent.

8

Box contains red balls and green balls. Box contains red balls and green balls. A standard fair cubical dice is thrown. If a six is obtained, a ball is selected from box ; otherwise a ball is selected from box . a Calculate the probability that the ball selected was red. b Given that the ball selected was red, calculate the probability that it came from box B.

9

Robert travels to work by train every weekday from Monday to Friday. The probability that he catches the other weekday is

train on Monday is . The probability that he catches the

train on any

. A weekday is chosen at random.

a Find the probability that he catches the b Given that he catches the

train on that day.

train on that day, find the probability that the chosen day is

Monday. 10

Bag contains red cubes and

blue cubes. Bag contains red cubes and blue cubes.

Two cubes are drawn at random, the first from Bag and the second from Bag . a Find the probability that the cubes are of the same colour. b Given that the cubes selected are of different colours, find the probability that the red cube was selected from Bag . 11

On any day in April there is a chance of rain in the morning. If it is raining there is a chance David will remember his umbrella, but if it is not raining there is only a chance he will remember his umbrella. a On a random day in April, what is the probability David has his umbrella? b Given that David has his umbrella on a day in April, what is the probability that it was raining?

12

A new blood test has been devised for early detection of a disease. Studies show that the probability that the blood test correctly identifies someone with this disease is , and the probability that the blood test correctly identifies someone without that disease is incidence of this disease in the general population is .

. The

The result of the blood test on one patient indicates that he has the disease. What is the probability that this patient has the disease? 13

Louise has two coins: one is a normal fair coin with heads on one side and tails on the other. The second coin has heads on both sides. Louise randomly picks a coin and flips it. The result comes up heads. What is the probability that Louise chooses the fair coin?

14

There are discs in a bag. Some of them are black and the rest are white. Two are simultaneously selected at random. Given that the probability of selecting two discs of the same colour is equal to

the probability of selecting two discs of different colour, how many black discs are there in the bag? 15

Prove that if and are independent then so are

and

.

Checklist of learning and understanding You can use set notation when describing probabilities: is the intersection of and , meaning both and

happen

is the union of and , meaning happens or happens or both happen is the complement of , meaning everything that could happen other than . From a Venn diagram: if are mutually exclusive , and the formula reduces to

. If

, then and .

is the probability of happening if has happened. This can be visualised in Venn diagrams, two-way tables or tree diagrams. You can also use the formula

.

In a tree diagram this formula is often rearranged to get: You can relate



and

by using the formula:

Mixed practice 20 1

A box of chocolates contains milk and dark. Three are chosen at random, without replacement. Given that the first chocolate chosen is milk, find the probability that all three are milk. Choose from these options. A B C D

2 Which statement is true? Choose from these options. A

and are mutually exclusive.

B

and are independent.

C

and are neither mutually exclusive nor independent.

D More information is required to decide. 3

A drawer contains red socks, black socks and white socks. Two socks are picked at random without replacement. What is the probability that two socks of the same colour are drawn?

4

Out of flies studied in an experiment have neither gene.

have the gene

and

have the gene

.

a Illustrate this information on a Venn diagram. b Find c Find 5

. .

The table shows the colour of hair and the colour of eyes of a sample of particular population.

people from a

Colour of hair Black

Dark

Medium

Fair

Auburn

Total

Blue Colour of eyes

Brown Green Total

Calculate, to three decimal places, the probability that a person, selected at random from this sample, has: i fair hair ii auburn hair and blue eyes iii either auburn hair or blue eyes but not both iv green eyes, given that the person has fair hair

v fair hair, given that the person has green eyes. [© AQA 2014] 6 a Illustrate this information on a tree diagram. b Find

.

c Find

.

d Find 7

.

The number of hours spent practising before a music examination was recorded by a teacher for each of her students. Hours practised

Male

Female

or fewer to or more a Find

.

b Two different students are randomly selected for further interviews. Find the probability that both are male. 8

Alison is a member of a tenpin bowling club which meets at a bowling alley on Wednesday and Thursday evenings. The probability that she bowls on a Wednesday evening is that she bowls on a Thursday evening is

. Independently, the probability

.

a Calculate the probability that, during a particular week, Alison bowls on: i two evenings ii exactly one evening. David, a friend of Alison, is a member of the same club. The probability that he bowls on a Wednesday evening, given that Alison bowls on that evening, is . The probability that he bowls on a Wednesday evening, given that Alison does not bowl on that evening, is . The probability that he bowls on a Thursday evening, given that Alison bowls on that evening, is . The probability that he bowls on a Thursday evening, given that Alison does not bowl on that evening, is . b Calculate the probability that, during a particular week: i Alison and David bowl on a Wednesday evening ii Alison and David bowl on both evenings iii Alison, but not David, bowls on a Thursday evening iv neither bowls on either evening. [© AQA 2013] 9

The probability that a student plays badminton is . The probability that a student plays neither football nor badminton is and the probability that a student plays both sports is . a Draw a Venn diagram showing this information. b Find the probability that a student plays badminton, but not football.

c Given that a student plays football, the probability that they also play badminton is

.

Find the probability that a student plays both badminton and football. d Hence complete your Venn diagram. What is the probability that a student plays only badminton? e Given that a student plays only one sport, what it the probability that they play badminton? 10

Only two international airlines fly daily into an airport. Pi Air has Air has

flights a day. Passengers flying with Pi Air have a

flights a day and Lambda

probability of losing their

luggage and passengers flying with Lambda Air have a probability of losing their luggage. Someone complains that their luggage has been lost. Find the probability that they travelled with Pi Air. 11

A girl walks to school every day. If it is not raining, the probability that she is late is . If it is raining, the probability that she is late is . The probability that it rains on a particular day is . On one particular day the girl is late. Find the probability that it was raining on that day.

12

The probability that a man leaves his umbrella in any shop he visits is . After visiting two shops in succession, he finds he has left his umbrella in one of them. What is the probability that he left his umbrella in the second shop?

13

a A large bag of sweets contains red and yellow sweets. Two sweets are chosen at random from the bag without replacement. Find the probability that red sweets are chosen. b A small bag contains red and yellow sweets. Two sweets are chosen without replacement from this bag. If the probability that two red sweets are chosen is that

, show

.

Ayesha has one large bag and two small bags of sweets. She selects a bag at random and then picks two sweets without replacement. c Calculate the probability that two red sweets are chosen. d Given that two red sweets are chosen, find the probability that Ayesha had selected the large bag. 14

The probability that it rains during a summer’s day in a certain town is . In this town, the probability that the daily maximum temperature exceeds is when it rains and when it does not rain. Given that the maximum daily temperature exceeded particular summer’s day, find the probability that it rained on that day.

15

Given that

and

16

Two events, and satisfy

find

.

a Find the possible values can take. b Find

in terms of .

Worksheet See Extension Sheet 20 for a selection of more challenging problems.

on a

21 The normal distribution In this chapter you will learn how to: calculate probabilities for a normally distributed random variable relate any normal distribution to the standard normal distribution calculate the value of the variable with a given cumulative probability find mean and standard deviation from information about probabilities use the normal distribution as a model use the normal distribution as an approximation to the binomial distribution

Before you start… GCSE

You should be able to solve simultaneous equations.

1 Solve these simultaneous equations.

Student Book 1,

You should be able to interpret

2 What is the frequency density of a

Chapter 20

histograms.

Student Book 1, Chapter 21

You should be able to use the binomial distribution.

group of people with masses strictly between and ? 3 The probability of rolling a six on a biased dice is . The dice is rolled four times. Find the probability of getting exactly six.

Chapter 20

You should be able to work with tree diagrams.

4 There is a chance of it raining. If it rains there is a chance I am late. If it does not rain there is a chance I am late. What is the probability that I am late?

Chapter 20

You should be able to calculate conditional probabilities.

5 Two dice are rolled – one red and one blue. If the total score is , what is the probability that the score on the red dice is ?

What is the normal distribution? You have already met discrete random variables, which you can describe by listing all possible values and their probabilities. With a continuous variable, such as height or time, it is impossible to list all values so you need a different way to describe how the probability is distributed across the possible values.

Rewind For a reminder of discrete probability distributions, see Student Book 1, Chapter 21.

To do this you can use a similar idea to a histogram: draw a curve such that the area under the curve represents probability. This curve is called the probability density function.

Rewind You met histograms in Student Book 1, Chapter 20.

There are many situations where a variable is very likely to be close to its average value, with values further away from the average becoming increasingly unlikely. You can model many such situations by using the normal distribution, which is symmetrical about the average value. Natural measurements, such as heights of people or masses of animals, tend to follow a normal distribution.

Worksheet See Extension Sheet 21 for an investigation that explores why the normal distribution occurs so often.



Section 1: Introduction to normal probabilities To specify a normal distribution fully you need to know its mean follows this distribution you use the notation .

and variance

. If a variable

The probability density function is symmetrical about the mean. The standard deviation affects the width of the curve – the larger is, the wider the curve is. Remember that the probability corresponds to the area under the graph. You know from Student Book that you can find the area under a curve by integration. However, the equation for the normal distribution curve cannot be integrated exactly, so most calculators have a builtin function to find approximate probabilities.

Common error Be careful with the notation:

is the variance, so

has standard deviation

. You may find it helpful to sketch a diagram to get a visual representation of the probability you are trying to find.

The diagrams can also provide a useful check, as you can see whether to expect the probability to be smaller or greater than

.

Tip With a continuous variable, and mean exactly the same thing, so you don’t have to worry about whether end-points should be included. This is not the case for discrete variables!

WORKED EXAMPLE 21.1

The average height of people in a town is

with standard deviation

. What is the

probability that a randomly selected resident: a is less than

tall

b is between

and

tall?

is the height of a town resident so

State the distribution used. The standard deviation is so the variance is .

,

a

State the probability to be found and use your calculator.

b

State the probability to be found and use your calculator.

Note that the domain of the normal probability density function is all real numbers – the -axis is an asymptote to the graph. This means that, in theory, a normal variable could take any real value. However, most of the data lies within three standard deviations of the mean. It is useful to remember the percentages summarised in Key point 21.1.

Key point 21.1 Approximately Approximately

of the data lie within three standard deviations of the mean. of data lie within two standard deviations of the mean.

Approximately two-thirds of the data lie within one standard deviation of the mean.

It turns out that the points of inflection of the normal distribution curve lie one standard deviation from

the mean. You can use this to estimate the standard deviation. PROOF 10

The curve corresponding to the normal distribution with mean zero and standard deviation is . Prove that the points of inflection of

occur at

.

To find a point of inflection, use the fact that . First, use the chain rule to differentiate.

Use both the chain rule and the product rule to find the second derivative.

Factorise and solve, noting that

Since

.

,

      

If a normally distributed random variable has mean , should a value of be considered unusually large? The answer depends on how spread out the variable is – and this is measured by its standard deviation. If the standard deviation were then a value around would be quite common; however, if the standard deviation were then would indeed be very unusual. The probability of a normally distributed random variable being less than a given value ( , called the cumulative probability) depends only on the number of standard deviations is away from the mean. This is called the z-score.

Key point 21.2 For mean:

the -score measures the number of standard deviations away from the

WORKED EXAMPLE 21.2

Given that

:

a work out how many standard deviations b find the value of which is

is away from the mean

standard deviations below the mean. The number of standard deviations away from the mean is measured by the -score.

a

is the variance so take the square root to get the standard deviation. is standard deviations away from the mean. Values below the mean have a negative -score.

b

If you are given a random variable µ you can create a new random variable which takes the values equal to the -scores of the values of .

Fast forward This is an extremely important property of the normal distribution, which needs to be used in situations when the mean and standard deviation of are not known (see Section 3).

Whatever the original mean and standard deviation of , this new random variable always has normal distribution with mean and variance , called the standard normal distribution: .

Key point 21.3 The probabilities of and are related by:

WORKED EXAMPLE 21.3

Let

. Write, in terms of probabilities of :

a b

a

b

You are given that

so you can calculate .

You now have two -values, so find the corresponding value for each of them.

Questions on the normal distribution can be combined with other probability facts – in particular, watch

out for questions that bring in conditional probability or the binomial distribution.

Rewind Conditional probability was covered in Chapter 20. The binomial distribution was covered in Student Book 1, Chapter 21.

Did you know? Before graphical calculators (which was not so long ago!) people used tables showing cumulative probabilities of the standard normal distribution. Because of their importance they were given special notation: . Although you do not have to use this notation, you should understand what it means.

WORKED EXAMPLE 21.4

The mass of fish caught by a trawler follows a normal distribution with mean deviation . A juvenile fish is classified as one with a mass less than .

and standard

a What is the probability that a randomly chosen fish is a juvenile? b If

fish are caught, what is the probability that there are more than ten juveniles? State any

assumptions you have to make, and comment on their validity. c If a fish is a juvenile, what is the probability that its mass is more than a Let be mass of fish so

?

State the distribution. Variance

.

Use your calculator to find the required probability. b Let be number of juveniles so .

Write down the names of all the random variables to make it obvious when a distribution has changed. Even though you write down parameters to three significant figures, use exact values in your calculations. Use the formula for conditional probability.

c

If is bigger than and .

and less than it is between

It is very common in this type of question that you can use some of the probabilities already calculated in an earlier part.

EXERCISE 21A 1

By shading the appropriate section of the normal distribution, find each probability.

a i ii b i ii c i ii d

has a normal distribution with mean

and standard deviation .

i ii e If has a normal distribution with mean

and standard deviation

.

i ii

Tip Remember that for a continuous variable, 2

is the same as

.

Find the -score corresponding to each given value of . a i ii b i ii

3

Given that variable.

, write each probability in terms of probabilities of the standard normal

a i ii b i ii c i ii 4

It is found that the lifespan of a certain brand of laptop battery follows a normal distribution with mean hours and standard deviation hours. A particular battery has a lifespan of hours. a How many standard deviations below the mean is this? b What is the probability that a randomly chosen laptop battery has a lifespan shorter than this?

5

When Ali competes in long jump competitions, the lengths of his jumps are normally distributed with mean

and standard deviation

.

a What is the probability that Ali will record a jump between

and

?

b Ali needs to jump

to qualify for the school team.

i What is the probability that he will qualify with a single jump? ii If he is allowed three jumps, what is the probability that he will qualify for the school team? c What assumptions did you have to make in your answer to part b ii? Are these likely to be met in this situation? 6

Masses of a species of cat have a normal distribution with mean and variance sample of such cats, estimate the number that will have a mass above .

7

A normal curve has points of inflexion with coordinates and deviation of this distribution.

8

Copy the diagram and mark on the approximate position of the points of inflexion. Hence estimate the mean and standard deviation of the normal curve.

9

Estimate the mean and standard deviation of the normal curve.

10

The

. Given a

. Find the mean and standard

time of a group of athletes can be modelled by a normal distribution with mean

seconds and standard deviation seconds. a Find the probability that a randomly chosen athlete will run the

in under

seconds.

b Show that, if the binomial distribution can be used, the probability that all four athletes in a team run under seconds is . c Miguel says that this means that the probability of the team breaking the school record of minutes and seconds is only . Give three reasons why this is likely to be incorrect. Do you think the real value will be greater or less than 11

If

, find:

a b c 12

If a b

, find:

?

13

The masses of apples are normally distributed with mean mass

and standard deviation

Supermarkets classify apples as medium if their masses are between

and

.

.

a What proportion of apples are medium? b In a bag of 14

apples what is the probability that there are at least eight medium apples?

The wingspans of a species of pigeon are normally distributed with mean length deviation

and standard

. A pigeon is chosen at random.

a Find the probability that the length of its wingspan is greater than b Given that its length is greater than greater than . 15

, find the probability that the length of its wingspan is

Grains of sand are believed to have a normal distribution with mean size . a Find the probability that a randomly chosen grain of sand is larger than

and variance

.

b The sand is passed through a filter that blocks grains wider than . The sand that passes through is examined. What is the probability that a randomly chosen grain of filtered sand is larger than 16

?

The amount of paracetamol per tablet is believed to be normally distributed with mean

and

standard deviation . A dose of less than is ineffective in dealing with toothache. In a trial of people suffering toothache, what is the probability that two or more of them have less than the effective dose? 17

A variable has a normal distribution with a mean that is times its standard deviation. What is the probability of the variable taking a value less than times the standard deviation?

18



If

and

find

in terms of .

Section 2: Inverse normal distribution You now know how to find probabilities given information about the variable. In real life it is often useful to work backwards from probabilities to estimate information about the data. This requires the inverse normal distribution .

Key point 21.4 For a given value of probability , the inverse normal distribution gives the value of such that .

Tip Note that many textbooks use the normal distribution: If

notation mentioned in Section 1 to write inverse , then

.

WORKED EXAMPLE 21.5

The length of men’s feet is thought to be normally distributed with mean

and variance

. A shoe manufacturer wants only of men to be unable to find shoes large enough for them. How big should their largest shoe be? If is length of a man’s foot then

Convert the information into mathematical terms.

You want to find the value of such that . You may have to convert into a probability of the form

Use the inverse normal distribution on your calculator. So their largest shoe must fit a foot long.

WORK IT OUT 21.1 Given that

and

find the value of .

Which is the correct solution? Identify the errors made in the incorrect solutions.

Solution 1

From your calculator the corresponding -score is Therefore

Solution 2

Therefore

Solution 3

EXERCISE 21B 1

a The random variable follows the normal distribution

. Find if:

i ii b The random variable follows the normal distribution

. Find if:

i ii c The random variable follows the normal distribution

. Find if:

i ii 2

IQ tests are designed to have a mean of needed to be in the top of IQ scores?

3

Rabbits’ masses are normally distributed with an average mass of and a variance of . A vet decides that the top of rabbits are obese. What is the minimum mass for an obese rabbit?

4

The amount of coffee dispensed by a machine follows normal distribution with mean standard deviation

and a standard deviation of

b Find the value of if decimal place.

of cups contain more than

of coffee.

of coffee. Give your answer to

The times taken for students to complete a test are normally distributed with a mean of and standard deviation of minutes. a Find the probability that a randomly chosen student completes the test in less than b

and

.

a Calculate the probability that the machine dispenses less than

5

. What IQ score is

minutes

minutes.

of students complete the test in less than minutes. Find the value of .

c A random sample of eight students had their time for the test recorded. Find the probability that exactly two of these students complete the test in less than minutes. 6

An old textbook says that the range of data can be estimates as times the standard deviation. If

the data is normally distributed what percentage of the data is within this range? 7

The time taken to do a maths question can be modelled by a normal distribution with mean seconds and standard deviation seconds. The probability of two randomly chosen questions both taking longer than seconds is . Find the value of .

8

The concentration of salt in a cell, , can be modelled by a normal distribution with mean and standard deviation . Find the value of such that: .

9

For a normal distribution find the ratio: a b



10

Evaluate , where standard normal distribution.

11

If

12

Most calculators have a random number generator which generates random numbers from to . These random numbers are uniformly distributed, which means that the probability is evenly spread over all possible values. How can you use these to form random numbers drawn from a normal distribution?

and

find

is the inverse normal distribution function for the

in terms of

.

Section 3: Finding unknown or One of the main applications of statistics is to determine parameters of the population given information about the data. But how can you use the normal distribution calculations if the mean or the standard deviation is unknown? This is where the standard normal distribution comes in useful: replace all the values by their -scores as they follow a known distribution, .

Tip This involves solving equations, and sometimes simultaneous equations. As the numbers usually have many decimal places you may want to use your calculator.

WORKED EXAMPLE 21.6

The random variable follows a normal distribution with standard deviation . An experiment estimated that . Estimate the mean of correct to two significant figures. Get the probability in the form

If

.

Since you do not know , convert the probability into information about . Find from your calculator.

Relate to the given -value using

.

WORKED EXAMPLE 21.7

The masses of gerbils are thought to be normally distributed. If of gerbils have a mass more than and have a mass less than , estimate the mean and the variance of the mass of a gerbil. Let be mass of a gerbil. Then

Convert the information into mathematical terms.

Get the second statement in the form

.

Use the inverse normal distribution for and relate it to the given values.

Use your calculator again to find .

Solve the simultaneous equations.

EXERCISE 21C 1

a Given that

find if:

i .

ii b Given that

find if:

i .

ii 2

Given that a i

find and if: and

ii

and

b i

and

ii

and

.

Worksheet See Support Sheet 21 for a further example of finding unknown mean and standard deviation and for more practice questions. 3

A manufacturer knows that their machines produce bolts whose diameters follow a normal distribution with standard deviation . The manager takes a random sample of bolts and finds that of them have diameter greater than . Find the mean diameter of the bolts.

4

The energy of an electron can be modelled by normal distribution with mean 12 eV and standard deviation . If of electrons have an energy above find the value of .

5

It is known that the heights of a certain plant follow a normal distribution. In a sample of plants, are less than tall and are more than tall. Estimate the mean and the standard deviation of the heights.

6

The time taken for a computer to start is modelled by a normal distribution. It is tested times and on times it takes longer than seconds. On times it takes less than seconds. Estimate the mean and standard deviation of the start-up times.

7

The actual voltage of a brand of battery is thought to be normally distributed with standard deviation and mean , where is the time, in hours, for which the battery has been used. The batteries can no longer power a lamp when they drop below . A batch of batteries is found and only

can power the lamp. Assuming that the model is correct estimate how long the

batteries have been used, assuming that they were all used for the same amount of time. 8

A scientist noticed that of temperature measurements were lower than the average. Assuming that the measurements follow a normal distribution, estimate the standard deviation.

9

The waiting time for a train is normally distributed with mean minutes. of the time the waiting time is over minutes. Find the probability that a person waits over minutes on exactly two out of three times they wait for the train.

10



The random variable models the temperature in an oven in . It follows a normal distribution where the mean value is the temperature set on the oven. The probability of being within five Celsius degrees of the temperature set is . Find the probability of being within degrees of the temperature set.

Section 4: Modelling with the normal distribution Key point 21.5 For a normal distribution you would expect: a histogram to show an approximately symmetrical distribution with only one mode and no sharp cut off summary statistics to show that nearly all of the data fall within three standard deviations of the mean.

Focus on Many real-world situations, such as the heights of people, masses of gerbils or error in experimental measurements, can be modelled by using the normal distribution. However, you should not just assume that a variable follows a normal distribution. There are various useful checks that you can use. You can explore this in more detail in Focus on … Modelling 4.

WORKED EXAMPLE 21.8

For each histogram explain why the normal distribution is not a good model for the data. a

b

a There appear to be two modes. b These data do not seem to be symmetrical and there is a sharp cut off at the top end because it is not possible to score over

.

Fast forward If you study the Statistics option of Further Mathematics you can find out about a statistical test called the chi-squared test which allows you to decide more precisely whether data follow a normal distribution.

WORKED EXAMPLE 21.9

In a sample it is found that the mean time taken to complete a puzzle is

seconds with a

standard deviation of seconds. Explain why the normal distribution would not an appropriate model to predict the time taken to complete this puzzle. There is a cut-off at zero seconds which is standard deviations below the mean. The normal distribution would therefore predict

With only the mean and standard deviation provided (no graph of the distribution), consider whether virtually all the data fall within three standard deviations of the mean.

that a significant number of people complete the puzzle in a negative amount of time which is not possible.

Many other statistical distributions can be approximated by the normal distribution. Before people had computers and graphical calculators this was very important, because calculations with the normal distribution were often much easier. For example, consider the bar charts for the binomial distribution:

The distribution on the left has a sharp cut off at and is not symmetrical, but the distribution on the right appears to have roughly the shape of a normal distribution. It turns out that it can indeed be approximated by a normal distribution, although the proof of this is beyond what you need.

Key point 21.6 If with distribution

and .

then can be approximated by the normal

Did you know? Because the binomial distribution deals with discrete variables and the normal distribution deals with continuous variables, technically you should use something called a continuity correction – this means saying that in the binomial distribution is equivalent to in the normal distribution. However, you do not need to use this in the A Level course.

WORKED EXAMPLE 21.10

A die is rolled

times. The random variable models the number of sixes thrown.

a Explain why follows a binomial distribution. b The random variable can be approximated by a normal distribution. i

Find the mean and variance of this normal distribution.

ii What properties of make this approximation valid? c Let be a random variable with the normal distribution from part b. Find d i

Find

.

.

ii Explain why this is not the same as your answer to part c.

Rewind You met the conditions for a binomial distribution to be appropriate in Student Book 1, Chapter 21.

a The events are independent with constant probability. Classify the outcomes into six or not six. b i

Use the fact that approximate distribution.

and

is the

and So the approximate normal distribution is

ii

and This means that the binomial distribution will be reasonably symmetrical, making a normal approximation valid.

Check whether

and

.

c

State the distribution of and use the calculator to find the required probability.

d i

Use the calculator to find the probability for the original binomial distribution.

ii

and do not have exactly the same distribution – is discrete and is continuous.

The normal distribution is only an approximation to the binomial, so the probabilities won’t be exactly the same.

The normal approximation is useful in binomial hypothesis tests when the value of is large, because the exact binomial probabilities can be difficult to find. WORKED EXAMPLE 21.11

A survey of a random sample of

people in a large city was conducted to test a hypothesis

about the proportion of people who cycle to work. In this sample, people cycled to work. The test concluded that there is evidence, at the significance level, that this proportion is less than . Use an appropriate normal distribution to estimate the maximum possible value of . Let be the number of people who cycle to work. Then

This is a hypothesis test for the proportion of the binomial distribution. So start by stating the distribution and the hypotheses.

The hypotheses are:

Need such that: , where

Approximate normal distribution:

is the critical value for this test. The significance level is .

You can use a normal approximation because and .

So Using this normal distribution,

Use the inverse normal distribution to find the value of .

So

is the number of people who cycle to work, so needs to be a whole number.

EXERCISE 21D 1

For each histogram, decide if the data could be modelled by a normal distribution. If they cannot, give a reason. a

b

c

d

2

The mean number of children in a family in the UK is with a standard deviation of . Use these figures to explain why a normal distribution would not be a good model of the number of children in a family in the UK.

3

The quantities of bread purchased by a random sample of

people in a one-week period were

recorded. The smallest quantity purchased was

and the largest was

.

The summary statistics for the sample are: and a Find the sample mean and standard deviation. b Explain whether the normal distribution would be a suitable model for the weekly quantity of bread purchased. 4

A psychology student asks people to estimate the value of an angle. Her results are summarised in the histogram.

a Copy and complete the frequency table. Angle, Frequency

b Hence estimate the mean and standard deviation of the data. c What features of the graph suggest that a normal distribution might be an appropriate model? d The student compares this data to another group of students. If they follow the same normal distribution, how many would you expect to estimate over ?

5

A fair coin is tossed

times.

a Find the probability that there are more than

heads, using:

i the binomial distribution. ii the normal approximation. b Find the percentage error in using the normal distribution in this situation. 6

The number of people voting for the Orange party in an election can be modelled by a binomial distribution. There are voters and the probability of each voting for Orange is . Use a normal approximation to the binomial to find the probability that the Orange party wins a majority if: a b

7

.

The random variable has the binomial distribution

.

a The distribution of can be approximated by a normal distribution. Estimate the mean and standard deviation of this normal distribution, showing your method clearly. b Hence estimate the values of and . 8

Data collected over a long period of time indicate that of children contract a certain disease. Following a public awareness campaign, a doctor conducts a survey to find out whether this proportion has decreased. The doctor uses a random sample of hypothesis test at the significance level.

children and conducts a

Use an appropriate normal distribution to find the approximate critical region for this test. 9

a Prove that if

and

then

.

Use a counterexample that the reverse is not true. b A binomial distribution 10

The random variable Using this approximation that if mean are positive.

has a probability of

of being above

is approximated by the normal distribution

. Find the value of . .

then all values of within three standard deviations of the

Checklist of learning and understanding The normal distribution models many physical situations. It is completely described once you know its mean and its variance . Calculators can provide the probabilities of being in any given range. These values are useful to know: approximately approximately

of the data lie within three standard deviations of the mean of data lie within two standard deviations of the mean.

approximately two thirds of the data lie within one standard deviation of the mean. The -score is the number of standard deviations above the mean which has a given cumulative probability. It is related to the original variable through the equation

If you know probabilities relating to a variable with a normal distribution you can use the inverse normal distribution to deduce information about the variable. You need to use the -score when the mean or the standard deviation are unknown. If distribution



with

and

then can be approximated by the normal

Mixed practice 21 1

. What is

?

Choose from these options. A B C D 2

The test scores of a group of students are normally distributed with mean

and variance

. a Find the percentage of students with scores above b What is the lowest score achieved by the top 3

.

of the students?

The masses of kittens are normally distributed with mean a Out of a group of ? b

and standard deviation

.

kittens, how many would be expected to have a mass of less than

of kittens have a mass of more than

. Determine the value of .

4

The random variable is normally distributed with a mean of and standard deviation of . By sketching a normal curve or otherwise, find the value of such that .

5

The masses,



, of babies born at a certain hospital satisfy

Find the value of such that . 6

.

of the babies have masses between

and

where

The volume, litres, of Cleanall washing-up liquid in a -litre container may be modelled by a normal distribution with a mean, , of and a standard deviation of . a Determine the probability that the volume of Cleanall in a randomly selected -litre container is: i less than

litres

ii more than litres. b Determine the value of such that

. [© AQA 2013]

7

and the interquartile range is

. Find the value of .

Choose from these options. A B C D 8

The adult female of a breed of dog has average height

with variance

.

a If the height follows a normal distribution, find the probability that a randomly selected adult female dog is more than tall. b Find the probability that in six independently selected adult female dogs of this breed exactly four are above tall.

9

Heights of trees in a forest are distributed normally with mean . a Find the probability that a tree is more than b What is the probability that among tall?

10

and standard deviation

tall.

randomly selected trees at least two are more than

It is known that the scores on a test follow a normal distribution are above

and

of the scores are below

a Show that

.

of the scores

.

.

b By writing another similar equation, find the mean and the standard deviation of the scores. 11

people were asked to estimate the size of an angle.

gave an estimate that was less

than and gave an estimate that was more than . Assuming that the data follows a normal distribution, estimate the mean and the standard deviation of the results.

12

The volume, millilitres, of energy drink in a bottle can be modelled by a normal random variable with mean

and standard deviation

.

a Find: i ii .

iii

b Determine the value of such that

.

c The energy drink is sold in packs of bottles. The bottles in each pack maybe regarded as a random sample. Calculate the probability that the volume of energy drink in at least of the bottles in a pack is between and . [© AQA 2013] 13

of students in a university are female. The discrete random variable , which is the number of female students in a group of size , is assumed to follow a binomial distribution. a Explain why, if is large, the binomial distribution can be approximated by the normal distribution and state its parameters. b A dancing club contains students. Assuming the binomial distribution is valid, use the normal approximation to find the probability that more than are female. c Are the conditions for the binomial distribution met in this situation? Explain your answer.

14

The results of an examination have a mean of of .

, a median of

and a standard deviation

a Explain why a normal distribution is a plausible model for this data. Grades are awarded in the following way: The top

get an distinction.

The next

get a merit.

The next

get a pass.

The remaining people get a fail. b Assuming a normal model, find the grade boundaries for this examination.

15

A company makes a large number of steel links for chains. They know that the force required to break any individual link is modelled by a normal distribution with mean . The company tests chains consisting of links. If any link breaks, the chain will break. A force of is applied to all of the chains and break. a Estimate the probability of a single link breaking. b Hence estimate the standard deviation in the breaking strength of the links.

16

a

of sand from Playa Gauss falls through a sieve with gaps of

, but

passes

through a sieve with gaps of . Assuming that the sand’s diameter is normally distributed, estimate the mean and standard deviation of the size of the grains of sand. b

of sand from Playa Fermat falls through a sieve with gaps of

.

of this filtered

sand passes through a sieve with gaps of . Assuming that the sand’s diameter is normally distributed, estimate the mean and standard deviation of the sand.

22 Further hypothesis testing In this chapter you will learn how to: treat the sample mean as a random variable and see how it is distributed test whether the mean of a normally distributed population is different from a predicted value test whether a set of bivariate data provides evidence for significant correlation.

Before you start… Student Book 1, Chapter 20

You should be able to interpret correlation

1 Information on height, mass, waist size and average time spent exercising per week was recorded from a random sample of adult males. Match the values of the product moment correlation coefficient with each of the sets of variables:

coefficients.

A height and mass B height and time spent exercising C waist measurement and time spent exercising. 1 2 3 Student Book 1,

You should be able to

Chapter 22

conduct hypothesis tests using the binomial distribution.

Chapter 21

You should be able to conduct calculations using the normal distribution.

2 A dice is rolled ten times and four sixes are obtained. It is claimed that the dice is biased in favour of getting a six. Test this claim at the level. 3

. Find: a b c

such that

.

Testing means and correlation coefficients A note on a cereal packet claims that it has an average mass of . A sample of packets contains a mean of of cereal. Is this evidence that the company is systematically underfilling the packets? Intuition suggests probably not – you would not expect the mean of every sample to be exactly and the result seems to be reasonably close. But how far below would the mean have to be before there was significant evidence? To answer questions like this you can use a hypothesis test.

In this chapter you will look at two different types of hypothesis test. The first is a test to see if the mean of a sample is very different from a predicted value – such as in the cereal packet example. To do this you first need to establish some theory about the distribution of sample means.

Rewind You met hypothesis testing with the binomial distribution in Student Book 1, Chapter 22.

In Student Book 1, Chapter 20, you saw that correlation coefficients could be used to describe the strength of correlation, but it was not clear how big the coefficient needed to be to have significant evidence of correlation. In Section 3 you will see how hypothesis tests can be used to decide if the sample correlation coefficient provides evidence for correlation in the population.

Section 1: Distribution of the sample mean If data follow a normal distribution then every observation is a random variable, meaning that each observation can take a different value. The histogram shows samples taken from an distribution.

Instead of looking at a single value, you can look at a sample of observations and take a mean. This might take a different value every time you do it, so it is also a random variable and given the symbol For the distribution shown, take lots of samples of size dark blue.

and create a histogram of their mean, shown in

Tip You might like to use technology to see if you can create a similar histogram.

There are two important things to note about the sample means. They are clustered around the same mean as the original data. They are less spread out. It can be shown that the sample mean of observations of a normal distribution also follows a normal distribution with parameters relating to the original distribution and .

Key point 22.1 If the original distribution was

then:

You know from Chapter 21, Section 1, that if you take a normally distributed random variable, subtract its mean and divide by its standard deviation, the new random variable is . Doing that here to

.

means that the random variable

is

. You can use this fact in hypothesis testing (see

Section 2).

Fast forward You will meet a proof of this if you study the Statistics option of Further Mathematics.

WORKED EXAMPLE 22.1

A distribution is a Find b Find

. . .

c Comment on your results in parts a and b. a

(from calculator)

b

You can use your calculator to find the probabilities for a given normal distribution. Check that your answer is reasonable – the required region is within less than one standard deviation away from the mean, so the result must be less than two-thirds. Use

with

So the standard deviation is (from calculator)

Here the required region is within just over one standard deviation of the mean, so the answer should be just over two-thirds.

c The mean of observations is much more likely to be within unit of the true mean than a single observation is.

Common error Make sure you know whether you are working with an observation, , or the mean of several observations,

EXERCISE 22A

. If you are working with

, don’t forget to divide the variance by .

1

Write down the distribution of the sample mean, given the original distribution. a i If

find

ii If b i If ii If 2

find find find

. . .

Find each probability. a i If ii If b i If ii If

3

.

find

.

find

.

find

.

find

.

is the energy (in eV) of beta particles emitted from a radioactive isotope. It is known that . is the average energy of beta particles. a Stating one necessary assumption, write down the distribution of

4

b Find

.

c Find

.

The mass of a breed of dog is known to follow a normal distribution with a mean of standard deviation of their mean is more than

5

and a

. A random sample of four dogs is measured. What is the probability that ?

The volume of apple juice in a carton follows a normal distribution with a mean of and a standard deviation of . A quality control process rejects a batch if a random sample of cartons has a mean of less than

6

along with its parameters.

. Find the probability that a batch gets rejected.

Eggs are sold in boxes of . The masses of eggs have a normal distribution with mean

and

variance . What is the minimum value of that must be chosen if the average mass of an egg must be more than in at least of boxes? 7

The lifetime of a bulb, hours, is modelled by lifetime of less than . Find the value of .

8

The length of a species of fly follows a normal distribution with mean and standard deviation . of samples of flies have a mean of more than . Find the value of .

9

The diameter of an apple has mean

.

of samples of

and standard deviation

bulbs have a mean

. A sample of apples is

chosen and their mean diameter measured. a What is the probability that the mean diameter is between b The probability that the mean diameter is between

and

and

if

?

must be at least

. What is

the smallest value of that must be chosen? c The probability of the mean diameter being between

and

is required to double to

. How

many times bigger than in part b must now be? 10

The mass of a student in a group is

where follows the

four observations find the probability that: a the total mass is less than b the heaviest student has a mass less than

.

distribution. In a sample of



Section 2: Hypothesis tests for a mean One very common decision you have to make is if a mean is different from a predicted value – for example, you may be told that the average IQ is and want to see if students in a school have above average IQ. If you only have a sample from the school it is possible that the mean is above just through chance. You can conduct a hypothesis test to see if the difference above is big enough to be significant.

Tip Make sure you distinguish between the mean of the sample population

. You use

and the true mean of the

to make a decision about .

In hypothesis testing, you assume that the conservative position – called the null hypothesis – is true and then see how likely you are to see something like the observed data. If the probability of seeing the observed data is very low, you reject the null hypothesis. To conduct a hypothesis test you need to make some assumptions about the underlying distribution. The test you will study requires the underlying distribution to be normal with known variance.

Rewind The terminology associated with hypothesis tests was introduced in Student Book 1, Chapter 22.

Key point 22.2 To test the value of a population mean, , against a suggested value,

, at significance

level : 1 Set up appropriate hypotheses, depending on the context, using one of:       2 Conduct the test with one of these methods. See if your observed mean falls into the acceptance region or the critical region. To do this write down the distribution of (using Key point 22.1). Then find the regions at the ends of the distribution which have a total probability of (the critical region). Where these regions are depends on the alternative hypothesis.

Use your calculator to find the p-value, , of the observed mean. This is the probability of getting the observed value or more extreme (according to the null hypothesis). 3 Reject the null hypothesis if the mean falls into the critical region or if

.

It is important that you put your conclusion in context and that it is not overly certain – you must show an appreciation that you have only found evidence rather than stating a certain conclusion.

Testing using the critical region The method using the critical region (or acceptance region) will be looked at first. WORKED EXAMPLE 22.2

The level of testosterone in blood is normally distributed with mean and standard deviation . After completing a race, a sprinter gives two samples with an average of . Is this sufficiently different (at testosterone level is above average?

significance) to suggest that the sprinter’s

Define the variables. µ

State the hypotheses. This is a one-tailed test because the question is only looking for evidence of high testosterone. Under

State the distribution of gave two samples.

under

. The sprinter

Find the critical region by sketching the normal curve and using the inverse normal distribution.

so

(from

calculator).

So, falls in the rejection region; reject the null hypothesis. There is evidence that the sprinter’s testosterone level is above average.

Draw a conclusion in context.

Sometimes questions only ask you to find the critical region without actually performing a test. WORKED EXAMPLE 22.3

The temperature of a water bath is normally distributed with a mean of and a standard deviation of . After the equipment is serviced it is assumed that the standard deviation is the same. The temperature is measured on five independent occasions and a test is performed at the significance level to see if the temperature has changed from . What range of mean temperatures would result in accepting that the temperature has changed? Define the variables. State the hypotheses. This is a two-tailed test because the question does not specify that you are looking for evidence that the bath is too warm or too cold. State the distribution of were taken.

under

. Five measurements

Use the inverse normal distribution to find the critical values of for the two-tailed region. The probability of being in each tail is half of the significance level, so it is .

(from calculator) by symmetry about the mean,

or

Write down the critical region.

You might be asked to work with the standard normal variable , rather than with WORKED EXAMPLE 22.4

.

WORKED EXAMPLE 22.4

Jennifer believes that her average discus throw is above . She knows that her discus throws are normally distributed with standard deviation . She decides to take an average of throws and work out the test statistic:

She then conducts a hypothesis test at the

significance level.

a Write down Jennifer’s null and alternative hypotheses. b Write down the distribution of , including the value of any parameters. c Find the critical region in terms of . d The average of Jennifer’s ten throws was

.

Find the value of and hence decide if the null hypothesis can be rejected. a

µ

This a one-tailed test, as you are looking for an increase in average distance.

µ

b

You are given that

c

Use the inverse normal distribution to find the critical value of for the one-tailed region at .

d

You are given that

so reject

so

.

. Use this to work out .

is in the critical region.

Testing using the p-value You might find the -value method more straightforward.

Tip Many calculators can calculate the -value. You should check your manual to see how to do this with yours.

WORKED EXAMPLE 22.5

Traditional light bulbs have an average lifetime of hours and a standard deviation of hours. A manufacturer claims that the lifetimes of their bulbs have the same standard deviations but that they last longer. A sample of of the manufacturer’s light bulbs has an average lifetime of hours. Test the manufacturer’s claim at the significance level. Define the variables.

State the hypotheses. State the test statistic and its distribution. Use your calculator to find the -value. Compare to the significance level and conclude. Therefore reject – there is evidence to support the manufacturer’s claim that their light bulbs last longer than hours.

WORK IT OUT 22.1 The wingspan of a species of butterflies is known to be normally distributed with mean and standard deviation . A scientist thinks he may have found a new species. The mean of a sample of six of these butterflies is . Test the scientist’s claim at the significance level, assuming the wingspans are still normally distributed with standard deviation . Which is the correct solution? Identify the errors made in the incorrect solutions. Solution A

If

then (from calculator) so accept – a new species has not been found.

Solution B

If

then

(from calculator) The -value is twice this which is . so reject – there is evidence for a new species. Solution C

If the mean is

and the standard deviation is then the critical region is

so the observed value lies in the critical region. Reject

EXERCISE 22B

.

– there is a new species.

1

Write null and alternative hypotheses for each situation. a i The average IQ, , in a school over a long period of time has been . It is thought that changing the menu in the cafeteria might have an effect upon the average IQ. ii It is claimed that the average size, , of photos created by a camera is scientist believes that this figure is inaccurate. b i A consumer believes that steaks sold in portions of

. A computer

are on average underweight.

ii A careers adviser believes that the average extra amount earned over a lifetime by people with a degree is more than the figure he has been told at a seminar. c i The mean breaking tension, , of a brake cable does not normally exceed brand claims that it regularly does exceed this value. ii The average time, , taken to match a fingerprint is normally more than computer program claims to be able to do better. 2

In each situation, it is believed that

. A new

minutes. A new

. Find the critical region in each case.

a i ii b i ii c i ii 3

In each situation, it is believed that

. Find the -value of the observed sample mean.

Hence decide the result of the test if it is conducted at the

significance level

a i ii b i ii c i ii d i ii 4

The average height of

-year-olds in England is

and the standard deviation is

.

Caroline believes that the students in her class are taller than average. To test her belief she measures the heights of students in her class. a State the hypotheses for Caroline’s test. You can assume that the heights follow a normal distribution and that the standard deviation of heights in Caroline’s class is the same as the standard deviation for the whole population. The students in Caroline’s class have average height of b Test Caroline’s belief at the 5

.

level of significance.

All students in a large school are given a typing test and it was found that the times taken to type one page of text are normally distributed with mean

minutes and standard deviation

minutes. The students were given a month-long typing course and then a random sample of students was asked to take the typing test again. The mean time was minutes. Test at

significance level whether there is evidence that the time the students take to type a page of text has decreased. 6

The national mean score in GCSE Mathematics is particular school the average of

students is

with a standard deviation of

. In a

.

a State two assumptions that are needed to perform a hypothesis test to see if the mean is better in this school than the background population. b Assuming that these assumptions are met, test at the is producing better results than the national average. 7

significance level whether the school

A farmer knows from experience that the average height of apple trees is with standard deviation . The farmer buys a new orchard and wants to test whether the average height of apple trees is different. She assumes that the standard deviation of heights is still

.

a State the hypotheses she should use for her test. b The farmer measures the heights of

trees and finds their average.

Find the critical region for the test at the c Given that the average height of the test.

level of significance.

trees is

, state the conclusion of the hypothesis

Worksheet See Support Sheet 22 for a further example of testing for the mean of a normal distribution and for more practice questions.

8

A doctor has a large number of patients starting a new diet in order to lose mass. Before the diet the mass of the patients was normally distributed with mean and standard deviation . The doctor assumes that the diet does not change the standard deviation of the masses. After the patients have been on the diet for a while, the doctor takes a sample of mean mass.

patients and finds their

a The doctor believes that the average mass of the patients has decreased following the diet. He wishes to test his belief at the level of significance. Find the critical region for this test. b State an additional assumption required in your answer to part a. c The average mass of the 9

patients after the diet was

. State the conclusion of the test.

A geologist is measuring the volume of bubbles in an underwater lava flow. He believes that the bubbles are normally distributed with variance lava flow where the mean is

, but that they are smaller than in exposed

.

a Write down the geologist’s null and alternative hypotheses. b The geologist calculates the test statistic

and rejects the null hypothesis if

.

What significance level does this correspond to? c The geologist finds a mean of

. For what values of will the geologist reject the null

hypothesis? 10

The school canteen sells coffee in cups claiming to contain coffee in a cup is normally distributed with standard deviation

. It is known that the amount of . Adam believes that on average

the cups contain less coffee than claimed. He wishes to test his belief at the

significance level.

a Adam measures the amount of coffee in randomly chosen cups and finds the average to be . Can he conclude that the average amount of coffee in a cup is less than ? b Adam decides to collect a larger sample. He finds the average to be again, but this time there is sufficient evidence to conclude that the average amount of coffee in a cup is less than . What is the minimum sample size he must have used? 11



The null hypothesis is tested and a value the alternative hypothesis is or ?

is observed. Will it have a higher -value if

Section 3: Hypothesis tests for correlation coefficients You already know that values of the correlation coefficient close to or negative correlation respectively.

represent strong positive or

However, values in between can be difficult to interpret. For example, is

evidence of significant

correlation? The answer depends on the number of data points and how certain you want to be.

Rewind See Student Book 1, Chapter 20, for a reminder of how to interpret the correlation coefficient.

You need to distinguish between two related values. The correlation coefficient of the sample has the symbol and the correlation coefficient of the underlying population has the symbol (the Greek letter rho). You can conduct a hypothesis test using to decide if there is evidence that is not zero. In a twotailed test you are looking to see if there is correlation in either direction – positive or negative – so the alternative hypothesis is In a one-tailed test you are looking for correlation in just one direction, so the alternative hypothesis would be either

or

Finding the distribution of goes beyond what you need to know, but you can find the critical values in tables that will be provided. They are calculated assuming that both variables follow a normal distribution. If the modulus of is larger than the appropriate critical value then you reject the null hypothesis. One tail Two tail

    

Did you know? The correlation coefficient used in this course is called the Pearson product-moment correlation coefficient. This is just one type of correlation coefficient – you could also use Spearman’s rank correlation coefficient or Kendall’s tau. They all have advantages and disadvantages.

WORKED EXAMPLE 22.6

The correlation coefficient between the mass and height of students in a sample of six students is . Test at the significance level whether height and mass of students from this school are positively correlated. Write down the hypotheses. You are only looking for positive correlation, so it is onetailed. One tail Two tail

Look for the critical value at the intersection of the one-tail column and the row.

The critical value from the table is State the conclusion in context. so reject . There is evidence that mass and height are positively correlated.

Did you know?

There are many examples of spurious correlations. For example, there is a very strong negative correlation between the number of pirates and global warming! Can you explain this?

WORKED EXAMPLE 22.7

Twenty students were asked about the number of hours they spent watching television each week and their results in a reading test. The correlation coefficient for their results was for evidence of correlation at the significance level.

. Test

Write down the hypotheses. You are not told to look for correlation in any particular direction, so it is two-tailed. One tail Two tail

The critical values are

Write down the critical values from the table. Remember that it is a two-tailed test so the critical region is or .

.

so do not reject . There is not significant evidence for correlation between hours watching television and results in a reading test.

EXERCISE 22C

State the conclusion in context.

1

Test each sample correlation coefficient for positive correlation at size.

significance. is the sample

a i ii b i ii 2

Test each sample correlation coefficient for correlation at

significance. is the sample size.

a i ii b i ii 3

Information for students is used to investigate the hypothesis that there is a correlation between IQ and results in a mathematics test. a Write down the null and alternative hypotheses for this investigation. b Data are collected and the -value for the correlation coefficient is conclusion of the hypothesis test at the significance level?

4

. What is the

The average speed of cars is measured at six different checkpoints at varying distances from a junction. There is a belief that in general cars get faster as they are further from the junction. a Write down the null and alternative hypotheses for this investigation. b The -value of the observed data is found to be

5

. Test the data at the

significance level.

The amount spent by a government on unemployment support is expected to be negatively correlated with the amount spent on education. Data were collected across countries in 2013. a What is the population associated with this sample? b Write down appropriate null and alternative hypotheses. c The sample correlation coefficient was found to be hypothesis test at significance?

6

. What is the conclusion of the

The correlation coefficient between the amount of water used in a town on the temperature is .

summer days and

a Jane thinks that there is a correlation between water usage and temperature on a summer day. Write down the null and alternative hypotheses that Jane should use to test her suspicion. b Conduct the test at the

significance level.

c Karl says that if people use more water the days will be warmer. Give two reasons why your hypothesis test does not support Karl’s statement. 7

The level of antibodies in blood is thought to go down as the dose of a medical drug is increased. a State appropriate null and alternative hypotheses to test this statement. b In a sample of patients the correlation coefficient between these two variables is found to be . Conduct an appropriate hypothesis test at the significance level. c What are the advantages of using a medical tests?

8

significance level rather than a

significance level for

Data are collected on the height of a cake and the temperature at which it is baked. a If a hypothesis test is conducted to test for positive correlation the -value is

. Is this

evidence of positive correlation at the

significance level?

b If the same data were instead used to test for correlation in either direction, what would be the -value? Is there evidence of correlation at the significance level? c If the correlation coefficient increases, does this increase or decrease the -value found in part a? Justify your answer. 9

It is suspected that there is correlation between the height of a tree and the total surface area of its leaves. A random sample of trees is measured and the correlation coefficient is found to be . What is the smallest value of that makes this significant at

10

Why do critical value tables for correlation coefficients start at

significance? ?

Checklist of learning and understanding The sample mean, If

, is a random variable.

, then

To test the value, , of a population mean against a suggested value, : set up appropriate hypotheses depending on the context, using one of:

then, use the distribution of and your calculator to either find the -value or set up the critical region for the given significance level, if (or if is in the critical region), reject . To test whether there is correlation between two variables: set up appropriate hypotheses depending on the context, using one of:

then look up in the tables the critical value for the given significance level and sample size if the modulus of the sample correlation coefficient, , is greater than the critical value, then reject .



Mixed practice 22 1

A random sample of people have their heights and masses measured. Which correlation coefficient is the smallest that would suggest evidence of positive correlation between height and mass, using a significance level? Choose from these options. A B C D

2

The breaking load of steel wire is known to be normally distributed with mean and standard deviation . Find the probability that the mean breaking load of a sample of such wires is between and .

3

Data are collected on HIV rates and literacy rates of

countries in 2015.

a What is the population from which this sample is drawn? b A hypothesis test is conducted to see if there is correlation between HIV rates and literacy rates. Write down appropriate null and alternative hypotheses. c The -value for the observed correlation coefficient is hypothesis test at the significance level? 4

. What is the conclusion of the

The mass of cakes produced by a bakery is known to be normally distributed with mean and standard deviation

. A new baker is employed.

a State appropriate null and alternative hypotheses to test if the mean mass of cakes has changed. b The mean of

cakes is found to be

.

i What is the -value of these data? ii What is the conclusion of the hypothesis test at the 5

significance level?

As a special promotion, a supermarket offers cartons of orange juice containing ‘ with no increase in price.

extra’

A random sample of cartons of orange juice was checked. The percentages by which the contents exceeded the nominal quantity were recorded, with the following results.

Examine whether the mean percentage by which the contents exceed the nominal quantity is less than . Use the significance level. Assume that the data are from a normal distribution with standard deviation . [© AQA 2011] 6

The results of a group of students in a test is thought to follow a distribution. The mean of a random sample of students is used to test the hypothesis against . What is the critical region at the options. A B C

significance level (given to

)? Choose from these

D 7

A test is conducted to see if the time spent revising correlates with the results in a test. a State a distributional assumption required to use tables of critical values for the correlation coefficient. b A random sample of students is surveyed. The correlation coefficient for their responses is . Conduct a hypothesis test at the significance level.

8

The time taken for a full kettle to boil is known to follow a normal distribution with mean seconds and standard deviation seconds. After cleaning, the kettle is boiled times to test if the time taken for it to boil has decreased. a Stating two necessary assumptions, find the critical region for this hypothesis test at the significance level. b The mean time is found to be

seconds. State the outcome of the hypothesis test.

c Why should there be a long delay between the ten observations for this test to be valid? 9

Paul knows the time it takes him to complete the crossword in a particular newspaper is normally distributed with mean

minutes and standard deviation

minutes.

He changes his strategy to attempt all the down clues first, and wishes to test whether this has affected his average time. To do this he finds the mean of his next

attempts and calculates the test statistic

a State the null and alternative hypothesis for the test. b If the test is conducted at the c The mean of his next

significance level, find the critical region in terms of .

attempts is

minutes.

Find the value of and hence state the conclusion of the test. d What would be the conclusion if Paul was testing at the had improved? Explain your answer fully. 10

level whether his average time

A company produces low-energy light bulbs. The bulbs are described as using watts of power. Amir, the production manager, asked Jenny to measure the power used by each of a sample of bulbs from the latest batch produced and to test the hypothesis that the mean for the batch is watts. Jenny is to carry out the test at the significance level. a What assumption must be made about the sample of to be valid? b Jenny found that the mean for the sample was was

bulbs if the result of this test is

watts and that the standard deviation

watts. Carry out the test asked for by Amir.

c When Jenny reported the conclusion of the test to Amir, he said that he had intended to ask Jenny to test whether the bulbs in the batch use more than watts on average. For the test in part b, state the effect that this new information would have on: i the alternative hypothesis ii the critical value(s) iii the conclusion. [© AQA 2014] 11

Yukun wants to test the null hypothesis

against the alternative hypothesis

. He finds the mean sample of the data, value of his data is test?

, is

. His calculator tells him that the -

. Which of these expressions defines the -value of his hypothesis

a b c d 12

The distance an athlete jumps in a long jump is known to be normally distributed with mean and standard deviation . After a change to her technique she looks at the average of jumps to see if her average distance has changed. She uses a significance level. a Write down appropriate null and alternative hypotheses for this test. b If the jumps are still normally distributed with standard deviation acceptance region in terms of . c If the mean is found to be hypothesis being rejected.

13

, find the

, find the smallest value of that would result in the null

The viewing figures of a long-running television series are given as million. In the past it was known to follow a normal distribution with a standard deviation of million. A producer wants to know if a new presenter has changed the viewing figures across episodes. He conducts a hypothesis test, assuming the viewing figures are still normally distributed with standard deviation million. The acceptance region is found to be . a Deduce the null and alternative hypotheses. b Find the significance level of this hypothesis test, giving your answer as a percentage to the nearest whole number.

Worksheet See Extension Sheet 22 for some questions to make you think about appropriate significance levels in various situations.

FOCUS ON… PROOF 4

The prosecutor’s fallacy A man is accused of robbing a jeweller’s shop and stealing diamonds. The only evidence against him is a bag of diamonds found in his car during the police investigation. The prosecution argues that the probability of the bag being in the car if the man is innocent is in and hence the probability of him being guilty is in (or ). You are going to investigate

,

whether this is a valid argument.

Investigating the argument QUESTIONS 1

Let denote the probability that the man is guilty of stealing the diamonds. Without taking into account any information (such as any findings of the police investigations), estimate the value of .

2

Denote by the event that the man is guilty, so , and let be the event that the evidence (in this case, the bag of diamonds) is found in his car. Then the prosecution’s statement says that . You also need to estimate the probability of finding the evidence if he is guilty; let’s assume this is quite likely, so set . a Complete the tree diagram. b Using the tree diagram: i find ii use the conditional probability formula to find c Prove that, when

and

in terms of .

. What does this tell you about the prosecutor’s

argument?

3

Why was it reasonable to assume that is very small? Think about what represents.

4

How do

5

Would it make sense to swap the branches on the tree diagram, so that the first ‘level’ has and ?

6

The prosecutor stated that the probability that the bag is found in the car if the man is innocent is in . Write this event using the conditional probability notation. Hence write down and interpret the event which has the probability of in .

7

In law, the phrase ‘proof beyond reasonable doubt’ is used. What do you think this means? How does this compare to mathematical proof?

and

compare to each other if is larger (say,

)?



FOCUS ON… PROBLEM SOLVING 4

Using extreme values Probability can often be counter-intuitive, and people find it difficult to evaluate their solutions. One useful strategy can be to use extreme values to make the answer more obvious.

The Monty Hall problem This game featured in a US television show. You are shown three closed doors and told that one door hides a car while the other two hide goats. You will win whatever is behind the door that you choose. The show host knows which door hides the car.

You are asked to choose a door. The host then opens one of the other two doors to reveal a goat. You are then given a choice: stick with the original door, or switch to the third one. What should you do to maximise your probability of winning the car? Most people’s intuition is that it doesn’t matter – there are two closed doors and the car is equally likely to be behind either of them. But this does not take into account the fact the host knows where the car is, so would never open that door. QUESTIONS 1

Imagine that instead there are doors and you initially choose door . The host opens other doors, leaving out door . What would you do?

of the

The fact that the host knew which door to leave closed gives you additional information, so the probability has changed from the original equally likely for each door. 2

Suppose you play the game times. Call the door hiding the car door . Your initial choice of the door is random. If you picked door then the host can choose whether to open door or door ; assume the host chooses randomly. If you initially picked Door or , then the host has no choice about which one to open.

Copy and complete this table showing the possible outcomes, assuming the car is behind Door . You choose

Host opens

Use the table to find the probability that you win the car if you switch.

Two children in the garden There are two children in the garden. One of them is a girl. What is the probability that both of them are girls? You may assume that both genders are equally likely and that the genders of the two children are independent of each other – so, for example, they are not twins. 3

a Which of these arguments do you find most convincing? i Both genders are equally likely, and the genders of the two children are independent, so the probability that the second one is also a girl is . ii The two genders are independent, so the probability of two girls is iii The options for the two genders are

and

.

. Hence the probability that both are girls

is . iv The options for the two genders are

and

. Hence the probability of two girls is

. v

You already know that one child is a girl, so there are fewer options to choose from. Hence the probability of two girls is more than .

b Now suppose there are probability that all

children in the garden and of them are girls. Do you think that the

are girls is bigger or smaller than

?

c The table shows possible outcomes for two children. Copy and complete the probabilities and hence find the probability that both children are girls, given that at least one of them is a girl. First child boy Second child



girl boy

girl

FOCUS ON… MODELLING 4

When can you use the normal distribution? The normal distribution is commonly used as a model in many applications. However, it is important to be aware that this model is not always suitable. This section focuses on the properties of the normal distribution, which you should consider when deciding whether to use it as a model. WORKED EXAMPLE 1

The marks for a group of students on a Statistics exam are summarised in the histogram.The mean mark is and the standard deviation of the marks is .

a Estimate the number of students whose marks are more than two standard deviations away from the mean. b Hence state, with a reason, whether a normal distribution could be used to model the distribution of the marks.

a

No students had marks more than two standard deviations from the mean. b For a normal distribution, the number of students with marks more than two standard deviations from the mean should be around . Hence the normal distribution does not seem to be a good model for these marks. QUESTIONS

On the histogram, there are no data values below

or above

For a normal distribution,

. of the

data should be within two standard deviations of the mean.

QUESTIONS 1

The table summarises heights of a group of Height (cm)

schoolchildren.

Frequency

a Draw a histogram to represent these data (you may want to use technology to do this). b Hence explain whether a normal distribution would be a suitable model for these heights. 2

a An old textbook says that the range of the data is about six times the standard deviation. For a normal distribution, what percentage of the values is contained in this range? The box plot summarises results of a discus throw competition (length, measured in metres). b Assuming the data follow a normal distribution, use the box plot to estimate its mean and standard deviation. Find the inter-quartile range for this normal distribution. c Hence state, with a reason, whether a normal distribution is a suitable model for the lengths of the throws.

3

4

The mean, median and standard deviation for two sets of data are given. For each set of data decide, based on this information, whether a normal distribution would be a suitable model. a

;

;

b

;

;

In social science research, subjects are often asked to rank their opinions on a five-point scale, called the Likert scale (for example: strongly disagree, disagree, no opinion, agree, strongly agree). For the purpose of statistical analysis, these responses are sometimes translated into numbers (e.g. strongly disagree , strongly agree ). Give a reason why data measured on a Likert scale should not be modelled using a normal distribution.

5

a A law firm has found that the average length of phone calls made by the employees is minutes. Give a reason why a normal distribution might not be a suitable model for the length of phone calls. b It is often said that many naturally occurring measurements approximately follow a normal distribution. Discuss whether a normal distribution would be an appropriate model in each situation. i The mass of red squirrels in a forest. ii The heights of all parents and children at a nursery school open day. iii The number of children in a family.

6

The diagram shows three cumulative frequency curves. Which curves show:

a a symmetrical distribution b a normal distribution?



CROSS-TOPIC REVIEW EXERCISE 4 1

Asher has a large bag of sweets, half of which are red. He rolls a fair six-sided dice once. If the dice shows or , he randomly picks two sweets from the bag. If the dice shows any other number, he randomly picks three sweets from the bag. Find the probability that Asher picks at least one red sweet.

2

Elsa is investigating whether there is any correlation between the average daily temperature and the daily amount of rainfall. a State suitable null and alternative hypotheses for her test. Elsa collects the data for a random sample of days and calculates that the correlation coefficient between the average temperature and the amount of rainfall is . b Conduct a hypothesis test at the

3

level of significance. State your conclusion in context.

It is known that the heights of a certain type of rose bush follow a normal distribution with mean and standard deviation . Larkin thinks that the roses in her garden have the same standard deviation of heights, but are taller on average. She measures the heights of rose bushes in her garden and finds that their average height is . a State suitable hypotheses to test Larkin’s belief. b Showing your method clearly, test at the level of significance whether there is evidence that Larkin’s roses are taller than average.

4

The lifetime of a certain type of lightbulb, hours, is modelled by the distribution given that .

.It is

a Find the value of . b Find the probability that, in a sample of than hours.

randomly selected lightbulbs, at least

last more

c By considering the range of values that contains nearly all values of , discuss whether the normal distribution is a reasonable model for the lifetime of the lightbulbs. 5

During June 2011, the volume, litres, of unleaded petrol purchased per visit at a supermarket’s filling station by private-car customers could be modelled by a normal distribution with a mean of and a standard deviation of . a Determine: i ii iii b Given that during June 2011 unleaded petrol cost £ per litre, calculate the probability that the unleaded petrol bill for a visit during June 2011 by a private-car customer exceeded £ . c Give two reasons, in context, why the model is unlikely to be valid for a visit by any customer purchasing fuel at this filling station during June 2011. [© AQA 2012]

6

A data set consists of four numbers: and . Find the value of for which the standard deviation of the data is the minimum possible.

7

Two events and are such that diagram, or otherwise, find:

and

. By use of a tree

a b c 8

.

Theo repeatedly rolls a fair dice until he gets a six. a Show that the probability of him getting this six on the third roll is b

.

is the probability of getting his first six on the th roll. Find an expression for in terms of .

c Prove algebraically that 9

Roger is an active retired lecturer. Each day after breakfast, he decides whether the weather for that day is going to be fine , dull or wet . He then decides on only one of four activities for the day: cycling , gardening , shopping from day to day may be assumed to be independent.

or relaxing

. His decisions

The table shows Roger’s probabilities for each combination of weather and activity. Fine

Weather Dull

Cycling Activity

Gardening Shopping Relaxing

a Find the probability that, on a particular day, Roger decided: i that it was going to be fine and that he would go cycling; ii on either gardening or shopping; iii to go cycling, given that he had decided that it was going to be fine; iv not to relax, given that he had decided that it was going to be dull; v

that it was going to be fine, given that he did not go cycling.

b Calculate the probability that, on a particular Saturday and Sunday, Roger decided that it was going to be fine and decided on the same activity for both days. [© AQA 2013] 10

The masses of bags of sugar are normally distributed with mean .

and standard deviation

a Find the probability that a randomly chosen bag of sugar weighs more than b Find the probability that in a box of

.

bags there are at least two that weigh more than

c Darien picks out bags of sugar from a large crate at random. What is the probability that he has to pick up exactly bags before he finds one that weighs more than 11

?

A machine, which cuts bread dough for loaves, can be adjusted to cut dough to any specified set weight. For any set weight, grams, the actual weights of cut dough are known to be approximately normally distributed with a mean of grams and a fixed standard deviation of grams. It is also known that the machine cuts dough to within

grams of any set weight.

a Estimate, with justification, a value for . b The machine is set to cut dough to a weight of

grams.

.

As a training exercise, Sunita, the quality control manager, asked Dev, a recently employed trainee, to record the weight of each of a random sample of such pieces of dough selected from the machine’s output. She then asked him to calculate the mean and the standard deviation of his recorded weights. Dev subsequently reported to Sunita that, for his sample, the mean was standard deviation was grams.

grams and the

Advise Sunita on whether or not each of Dev’s values is likely to be correct. Give numerical support for your answers. c Maria, an experienced quality control officer, recorded the weight, y grams, of each of a random sample of pieces of dough selected from the machine’s output when it was set to cut dough to a weight of grams. Her summarised results were as follows. and Explain, with numerical justifications, why both of these values are likely to be correct. [© AQA 2013] 12

A random variable has a normal distribution with mean and standard deviation . a Use your calculator to find

correct to five decimal places.

The exact value of the probability is given by:

b Use the trapezium rule with six strips to estimate the value of this integral. Give your answer correct to five decimal places. c Find the percentage error in using the trapezium rule to estimate this probability. 13

The random variable has binomial distribution distribution of .

. The line graph shows the probability

The distribution of can be approximated by a normal distribution. a Use the graph to estimate the mean and standard deviation of the normal distribution.Explain clearly how you arrived at your answer. b Hence estimate the values of and . 14

A student is investigating whether there is any correlation between the amount of time spent revising and marks gained on a test. She uses information from a sample of six tests. In order to have a larger sample, she collects data from two of her friends. She finds that the correlation coefficient between hours spent revising and the percentage mark on the test is . She therefore suggests that there is negative correlation between

the amount of time spent revising and the test marks. a State the null and alternative hypotheses for her test. b Test the hypotheses at the

level of significance and interpret the conclusion in context.

The scatter graph of her data is shown here.

c Suggest one possible interpretation of the data and the result of the hypothesis test. 15

A company manufactures bath panels. The bath panels should be deep, but a small amount of variability is acceptable. The depths are known to be normally distributed with standard deviation . a In order to check that the mean depth is , Amir takes a random sample of bath panels from the current production and measures their depths, in millimetres, with the following results.

Test whether the current mean is

, using the

significance level.

b Isabella, a manager, tells Amir that, in order to check whether the current mean is is necessary to take a larger sample. Amir therefore takes a random sample of size the current production and finds that the mean depth is . Test whether the current mean is significance level.

, it from

, using the data from this second sample and the

[© AQA 2011] 16

Daniel and Theo play a game with a biased coin. There is a probability of of the coin showing a head and a probability of of it showing a tail. They take it in turns to toss the coin. If the coin shows a head the player who tossed the coin wins the game. If the coin shows a tail, the other player has the next toss. Daniel plays first and the game continues until there is a winner. a Write down the probability that Daniel wins on his first toss. b Calculate the probability that Theo wins on his first toss. c Calculate the probability that Daniel wins on his second toss. d Show that the probability of Daniel winning is . e State the probability of Theo winning.

f

17

They play the game with a different biased coin and find that the probability of Daniel winning is twice the probability of Theo winning. Find the probability of this coin showing a head.

In a large typesetting company the time taken for a typesetter to set one page is normally distributed with mean minutes and standard deviation minutes. A new training scheme is introduced and, after all the typesetters have completed the scheme, the times taken by a random sample of typesetters are recorded. The mean time for the sample is minutes. You may assume that the population standard deviation is unchanged. a Test, at the

significance level, whether the mean time to set a page has decreased.

b It is required to redesign the test so that the probability of incorrectly rejecting the null hypothesis is less than when the sample mean is . Find the smallest sample size needed. 18

A normal distribution can be represented by a curve, as shown in the diagram. The equation of the curve is a Find the -coordinates of the points of inflection of the curve. The random variable follows this normal distribution. b Use the trapezium rule with four strips to estimate

. Give your answer to

three significant figures. c Is it possible to tell, without doing any further calculations, whether the answer in part b is an over-estimate or an under-estimate? Explain your answer.



PRACTICE PAPER 1 2 hours, 100 marks

A Level Mathematics: Paper 1 1

If

then

is equivalent to which one of these?

A

[1 mark]

B C D 2

The graph of graph?

is translated two units to the right. What is the equation of the resulting

Choose from these options. A

[1 mark]

B C D 3

Find the coordinates of the centre, , and the radius, , of the circle with equation . Choose from these options. A

[1 mark]

B C D 4

Find the derivative of

.

Choose from these options. [1 mark]

A B C D 5

Solve the equation

.

[4 marks]

Give your answer in terms of e. 6

Find an approximate expression for the terms in and above.

7

The triangle in the diagram has area

when is small enough to neglect

. The angle is obtuse.

[3 marks]

Find the length of 8

correct to one decimal place.

Find the equation of the normal to the curve Give your answer in the form

9

a Solve the inequality

[5 marks] at the point where

, where

. [6 marks]

and are integers.

.

[3 marks]

b A sequence is given by

for

. [4 marks]

Prove that the sequence is increasing. 10

11

The sector shown in the diagram has perimeter .

Find, in terms of , the largest possible area of the sector.

[6 marks]

a Express

[3 marks]

in the form

, where and are integers.

b Hence, or otherwise, find the binomial expansion of including the term in

[6 marks]

up to and

.

c Find the range of values of for which the binomial expansion of

[1 mark]

is

valid. 12

a Three consecutive terms in an arithmetic sequence are values of . b Prove that there is no value of for which geometric sequence.

13

Use a substitution to find

. Find the possible [5 marks]

are consecutive terms of a

in the form ln

[3 marks] where and are rational

[7 marks]

numbers. 14

The diagram shows a part of the graph with equation

a Show that the -value of the maximum point of the curve lies between

and

.

[4 marks]

b Use three rectangles of equal width to find values of and such that:

[3 marks]

c Use the trapezium rule with six strips to find an approximate value of

[3 marks]

. State, with a reason, whether your answer is an overestimate or an underestimate. 15

A curve is defined by the implicit equation a Find an expression for

. [5 marks]

in terms of and .

b Hence prove that the curve has no tangents parallel to the -axis. 16

Consider the equation

[3 marks]

, where is measured in radians.

a By means of a sketch, show that this equation has only one solution.

[3 marks]

b Show that this solution lies between and .

[2 marks] , where and [2 marks]

c Show that the equation can be rearranged into the form are constants to be found.

d Hence use a suitable iterative formula to find an approximate solution to the equation 17

The polynomial

correct to three decimal places. is defined by

a Use the factor theorem to show that b Hence express

.

[2 marks]

is a factor of

.

[2 marks] [2 marks]

as a product of three linear factors.

c i Show that the equation .

can be written as

ii Hence find all solutions of the equation , giving your solutions to the nearest degree.

[3 marks]

, where

in the interval [8 marks]

PRACTICE PAPER 2 2 hours, 100 marks

A Level Mathematics: Paper 2 Section A 1

What is a possible equation of this graph?

Choose from these options. [1 mark]

A B C D 2

Let be the smallest positive solution of the equation cos

, where is in radians.

Find the next smallest positive solution in terms of . Choose from these options. [1 mark]

A B C D 3

The diagram shows the graphs of

and

.

[2 marks]

a Find the coordinates of points and . b Shade the region determined by the inequalities

and

c State the smallest integer value of that satisfies both inequalities.

.

[1 mark] [1 mark]

4

Points and have position vectors

and

a Find the position vector of the midpoint of b Point has position vector

.

.

[2 marks]

. It is given that

. [3 marks]

Find the value of . 5

Find the exact coordinates of the point of inflection on the graph of

6

Prove by contradiction that

7

a Show that the equation

[5 marks]

.

is an irrational number.

[4 marks]

can be written in the form

State the values of the constants

.

and .

b Hence solve the equation

[3 marks]

for [4 marks]

giving your answers to the nearest degree. 8

Find

, giving your answer in the form

9

The circle shown in the diagram has equation

a Show that the point b The line

.

[5 marks] .

[2 marks]

lies on the circle.

is tangent to the circle at . [5 marks]

Find the exact value of the area shaded in the diagram. 10

The diagram shows the curves with equations

and

.

[1 mark]

a Write down the coordinates of point . b The curves intersect at the point . i Show that the -coordinate of satisfies the equation ii Hence find the exact coordinates of . c Find the exact value of the area of the shaded region.

. [4 marks] [6 marks]

Section B 11

A small box of weight is pulled along a rough horizontal floor by means of a light inextensible string. The string is horizontal and the tension in the string is . The box moves with a constant velocity. Find the coefficient of friction between the box and the floor, correct to one decimal place. Choose from these options. [1 mark]

A B C D 12

A particle moves in the plane with constant acceleration. Its velocity changes from during seconds.

to

Find the magnitude of the acceleration. Choose from these options. [1 mark]

A B C D 13

A cyclist moves along a straight horizontal road. The velocity–time graph of her journey is shown in the diagram.

a Calculate the acceleration of the cyclist during the first seconds.

[2 marks]

b Show that the cyclist returns to the starting point after

[3 marks]

seconds.

[1 mark]

c State the average velocity of the cyclist for the whole journey. 14

In this question use

, giving your final answers to an appropriate degree of accuracy.

A non-uniform plank of mass and length rests on two supports, and . The supports are located from the ends of the plank, marked and on the diagram.

Given that the reaction force acting on the plank at the support is a find the reaction force acting on the plank at

: [1 mark]

[3 marks]

b find the distance of the centre of mass of the plank from the end marked 15

Four forces act on a particle in a horizontal plane, as shown in the diagram. The unit vectors and are directed east and north, respectively.

a The particle is in equilibrium. Find the force b The force

in the form

[2 marks]

.

is suddenly removed. Given that the mass of the particle is

grams, find:

i the magnitude of the acceleration of the particle ii the angle the acceleration vector makes with the direction of the vector . 16

[5 marks]

a A particle moves in a straight line with constant acceleration . The initial velocity of the particle is . At time seconds, the velocity of the particle is and its displacement from the initial position is

.

i By considering the velocity–time graph, write down an equation relating ii Hence show that

and . [4 marks]

.

b A car is moving along a straight horizontal road when the driver applies the brakes. The car decelerates at a constant rate of

. In the next seconds is travels

. [3 marks]

Find its speed at the end of the seconds. 17

In this question use

, giving your final answers to an appropriate degree of accuracy.

A child kicks a ball from ground level at an angle above the horizontal. The speed of projection is . The ball is modelled as a particle and air resistance can be ignored. a At time seconds the ball’s displacement from the point of projection is

.

[4 marks]

Show that b The angle of projection is . The ball passes over a tall fence which is from the point of projection. Find the minimum possible value of . c How would the answer change if your model included air resistance? 18

[3 marks]

[1 mark]

Two particles, and , are connecting by a light inextensible string which passes over a smooth pulley. Particle has mass and lies on a rough plane. The plane is inclined at to the horizontal, and the coefficient of friction between and the plane is µ . Particle has mass and hangs freely below the pulley.

The system is in equilibrium.

19

a Show that

[5 marks]

b Obtain another similar inequality for .

[2 marks]

c In the case when remains in equilibrium.

[2 marks]

, find the range of values of for which the system

A particle of mass moves in a straight line. At time seconds the velocity of the particle is and the force acting on the particle is newtons, where . When the velocity of the particle is . Find the velocity of the particle after seconds.



[7 marks]

PRACTICE PAPER 3 2 hours, 100 marks

A Level Mathematics: Paper 3 Section A 1

The diagram shows the graph of the function and . The area labelled equals

What is the value of

. The graph crosses the -axis at the points and the area labelled equals .

?

Choose from these options. A B C D [1 mark] 2

Find the largest possible domain of the function

.

Choose from these options. A B C

or

D

and [1 mark]

3

The first term of an arithmetic sequence is Find the sum of the first

and the

term is

.

terms. [3 marks]

4

The polynomial

has factors

and

.

Find the values of and . [5 marks] 5

a Write

in the form

. [3 marks]

b A function is defined by i Prove that

is an increasing function.

for

.

[2 marks] ii Does

have an inverse function? Explain your answer. [2 marks]

iii Find the range of values of for which

is convex. [3 marks]

6

a Express

in the form

, where

.

Give the value of correct to three decimal places. [3 marks] b Hence find the minimum value of Give your answer in the form

. . [3 marks]

7

The diagram shows a part of the graph with equation

One of the roots of the equation

.

lies between and .

a The Newton–Raphson method is used to find an approximation to the root at . i Taking the first approximation to be

, find the second approximation,

.

Give your answer to three significant figures. ii Illustrate the relationship between

and

on a copy of the diagram. [5 marks]

b The point is the minimum point on the curve. i Find the exact -coordinate of . ii Explain why an iteration starting at would not converge to root of the equation . [4 marks] 8

A student is investigating the number of friends people have on a large social networking site. He collects some data on the percentage of people who have friends and plots this graph.

The student proposes two possible models for the relationship between and . Model :

    Model :

To check which model is a better fit, he plots the graph of against , where The graph is approximately a straight line with equation

and

.

.

a Is model or model a better fit for the data? Explain your answer. [2 marks] b Find the values of and . [3 marks] 9

A curve is given by parametric equations axis at point , and is a maximum point on the curve.

, for

. The curve crosses the -

a Find the exact coordinates of . [4 marks] b i Find the values of at the points and . ii Find the area of the shaded region. [6 marks]

Section B 10

a A university wants to investigate ways to encourage more students to apply for its courses. It takes a list of all schools in the UK and randomly selects ten to visit. The university asks students at each of the ten schools to complete questionnaires. What type of sampling is this? Choose from these options. A

Cluster sampling

B

Stratified sampling

C

Simple random sampling

D

Quota sampling [1 mark]

b In the questionnaire the number of students who said they were considering applying to the university in each school, , is summarised as:

Which of these is the best estimate of the standard deviation between the schools of the percentage applying to the university? Choose from these options. A B C D [1 mark] 11

In a survey of families the total number of takeaways a family buys each week and their total food budget has a correlation coefficient of . a A test is carried out using the following hypotheses:

Show that at the

significance level

is not rejected. [2 marks]

b Decide if each of these statements are supported by the test, explaining you reasoning. i The number of takeaways is independent of the total food budget. [2 marks] ii There is no correlation between the total number of takeaways and their total food budget. [2 marks] 12

The two-way table shows the number of people in a survey with a special food requirement along with their gender.

Male

Female

Vegan Vegetarian Pescetarian No requirement a Find (male vegan). [1 mark] b Find (male vegan). [2 marks] c Find (male   vegan). [2 marks]

d Two vegans are chosen without replacement. Find the probability that they are both male. [2 marks] e Use appropriate figures from this table to provide a counterexample to the identity

[2 marks] 13

There are

girls at a school.

of them describe themselves as having blonde hair.

a Use the binomial distribution to find the probability that a randomly selected group of contains exactly

girls

blonde girls. [2 marks]

b A club contains girls of whom have blonde hair. Use the binomial distribution to test at significance whether this is more than the expected number of girls with blonde hair. State your null and alternative hypotheses and your -value. [4 marks] c Give two reasons why the binomial distribution may not be an appropriate model for the number of girls with blonde hair in this club. [2 marks] 14

The distribution of the number of books, , borrowed by members of a library follows this distribution.

a Find the value of . [1 mark] b Find the probability that a randomly chosen member has borrowed at least one book. [1 mark] c The librarian proposes this model for the probability of a book begin overdue: If a member has borrowed books, the probability that he has an overdue book is P(overdue) . Find the probability that a randomly chosen member has an overdue book. [3 marks] 15

Two events, and , satisfy: a Find

. [3 marks]

b Show that and are not independent. [2 marks] 16

Alex keeps a record of the number of calories consumed each day over the course of several years. The mean is kilocalories and standard deviation kilocalories. The lowest value is and the highest value is . a Give two reasons why the numerical information given supports a normal model for the data. [2 marks] b Assuming that a normal model holds, find the probability that Alex eats over a day.

kilocalories in

[1 mark] c Stating a necessary assumption, find the probability that Alex eats over

kilocalories on at

least days in a week. [4 marks] d After attending a healthy eating course Alex finds that he eats kilocalories in a week. Test at the significance level if his average daily calorie intake is different after the healthy eating course, assuming that it still comes from a normal distribution with standard deviation . State your null and alternative hypotheses. [4 marks] e Jane eats more than kilocalories on of days and less than kilocalories on days. Assuming a normal distribution, estimate the value of the mean and variance.

of

[4 marks]

FORMULAE

Pure mathematics Binomial series

where

Arithmetic series

Geometic series

Trigonometry: small angles For small angle ,

Trigonometric identities

Differentiation

Integration

( constant;

where relevant)

Numerical solution of equations The Newton-Raphson iteration for solving

:

Numerical integration The trapezium rule:

, where

Mechanics Constant acceleration  

Probability and statistics Probability  

Standard discrete distributions Distribution of X

P(X = x)

Binomial B (n, p)

Mean

Variance

np

np (1 − np)

Sampling distributions  For a random sample mean and variance

of independent observations from a distribution having

:

is an unbiased estimator of

, where

For a random sample of n observations from

Critical values of the product moment correlation coefficient The table gives the critical values, for different significance levels, of the product moment correlation coefficient, , for varying sample sizes, .

One tail Two tail

10% 20%

5% 10%

2.5% 5%

1% 2%

0.5% 1%

n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 60 70 80 90

0.8000 0.6870 0.6084 0.5509 0.5067 0.4716 0.4428 0.4187 0.3981 0.3802 0.3646 0.3507 0.3383 0.3271 0.3170 0.3077 0.2992 0.2914 0.2841 0.2774 0.2711 0.2653 0.2598 0.2546 0.2497 0.2451 0.2407 0.2366 0.2327 0.2289 0.2254 0.2220 0.2187 0.2156 0.2126 0.2097 0.2070 0.2043 0.2018 0.1993 0.1970 0.1947 0.1925 0.1903 0.1883 0.1863 0.1843 0.1678 0.1550 0.1448 0.1364

0.9000 0.8054 0.7293 0.6694 0.6215 0.5822 0.5494 0.5214 0.4973 0.4762 0.4575 0.4409 0.4259 0.4124 0.4000 0.3887 0.3783 0.3687 0.3598 0.3515 0.3438 0.3365 0.3297 0.3233 0.3172 0.3115 0.3061 0.3009 0.2960 0.2913 0.2869 0.2826 0.2785 0.2746 0.2709 0.2673 0.2638 0.2605 0.2573 0.2542 0.2512 0.2483 0.2455 0.2429 0.2403 0.2377 0.2353 0.2144 0.1982 0.1852 0.1745

0.9500 0.8783 0.8114 0.7545 0.7067 0.6664 0.6319 0.6021 0.5760 0.5529 0.5324 0.5140 0.4973 0.4821 0.4683 0.4555 0.4438 0.4329 0.4227 0.4132 0.4044 0.3961 0.3882 0.3809 0.3739 0.3673 0.3610 0.3550 0.3494 0.3440 0.3388 0.3338 0.3291 0.3246 0.3202 0.3160 0.3120 0.3081 0.3044 0.3008 0.2973 0.2940 0.2907 0.2876 0.2845 0.2816 0.2787 0.2542 0.2352 0.2199 0.2072

0.9800 0.9343 0.8822 0.8329 0.7887 0.7498 0.7155 0.6851 0.6581 0.6339 0.6120 0.5923 0.5742 0.5577 0.5425 0.5285 0.5155 0.5034 0.4921 0.4815 0.4716 0.4622 0.4534 0.4451 0.4372 0.4297 0.4226 0.4158 0.4093 0.4032 0.3972 0.3916 0.3862 0.3810 0.3760 0.3712 0.3665 0.3621 0.3578 0.3536 0.3496 0.3457 0.3420 0.3384 0.3348 0.3314 0.3281 0.2997 0.2776 0.2597 0.2449

0.9900 0.9587 0.9172 0.8745 0.8343 0.7977 0.7646 0.7348 0.7079 0.6835 0.6614 0.6411 0.6226 0.6055 0.5897 0.5751 0.5614 0.5487 0.5368 0.5256 0.5151 0.5052 0.4958 0.4869 0.4785 0.4705 0.4629 0.4556 0.4487 0.4421 0.4357 0.4296 0.4238 0.4182 0.4128 0.4076 0.4026 0.3978 0.3932 0.3887 0.3843 0.3801 0.3761 0.3721 0.3683 0.3646 0.3610 0.3301 0.3060 0.2864 0.2702

One tail Two tail n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 60 70 80 90

100

0.1292

0.1654

0.1966

0.2324

0.2565

100

Answers 1 Proof and mathematical communication BEFORE YOU START 1 a b 2–4 Proof. EXERCISE 1A 1 For example, 2–5 Proof. 6 For example,

.

7 a–c Proof. 8 a–b Proof. c Its diagonals are perpendicular (so it must be a kite). 9 a–c Proof. 10 a b Proof. 11 Proof. EXERCISE 1B 1–12 Proof. EXERCISE 1C 1 a b Line . 2 a Proof. b 3 a b Line . 4 Line . 5 Line Line Line 6 If

only for

. It should be

doesn’t imply a minimum the suggested factorisation is not necessarily true.

7 Line does not follow from line . MIXED PRACTICE 1 1 C

.

2–4 Proof. 5 Line is not equivalent to line . Line does not necessarily follow from line . Line is not equivalent to line . 6 C 7–10 Proof. 11 a Line has an incorrect sign in the bracket. Line has a missing . Line is equivalent to line as given, but earlier errors mean that the final solution may not be a solution to the original equation, and/or there may be other solutions. b Proof. 12 Line . 13–15 Proof.

2 Functions BEFORE YOU START 1 a b 2 a b 3 4

or

5 a b 6 EXERCISE 2A 1 a Function; many-to-one. b Mapping. c Mapping. d Function; one-to-one. e Function; one-to-one. f

Function; one-to-one.

2 a i Yes ii Yes b i Yes ii Yes c i No ii No 3 a i One-to-one. ii Many-to-one. b i Many-to-one. ii Many-to-one. c i Many-to-one. ii One-to-one. 4 a i Yes ii Yes b i No ii No

c i No ii Yes 5 A is correct. Yes, as

.

EXERCISE 2B 1 a Domain: ; range: b Domain: ; range: c Domain:

; range:

d Domain:

; range:

2 a i ii b i

or

ii c i ii d i ii e i

or

ii f

i ii

3 a i ii b i ii c i ii d i

or

ii 4 a i ii b i ii c i

or

ii d i ii

or or

5 a b 6 7 8 9 a

b 10

or

11

or

12 a i ii b WORK IT OUT 2.1 Solution 2 is correct. EXERCISE 2C 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii

4 a b 5 a i ii b i ii c i ii d i ii 6 7 8 a b 9 a b 10 a

is not always

b 11 WORK IT OUT 2.2 Solution 1 is correct. EXERCISE 2D 1 a i ii b i ii c i ii d i ii e i ii

.

f

i ii

g i ii h i ii 2 a

b

c

d

3 Each function is one-to-one so domain of a i

; domain of

ii

; domain of

b i

; domain of

c i

; domain of

; range of

b 5 6 a .

7 8 Proof. 9 a

b Domain: c 10 a

.

; range of

.

; domain of

; range:

; range:

.

; range of

4 a

b Domain:

.

; range of

; domain of

ii

.

; range of

; domain of

.

of

.

; range of

; domain of f:

d i

and range of

; range of

ii

ii

of

; range of

. .

.

b Proof. 11

for

12 a b 13 a

b WORK IT OUT 2.3 Solution 3 is correct. EXERCISE 2E 1 a b c d 2 a

b

c

3 a b

has a turning point, so is not one-to-one.

c 4 5 MIXED PRACTICE 2 1 B 2 a b 3 a b 4 5 a b c Reflection in the line d i ii

.

iii e No solutions. 6 a b i ii c i ii 7 a Not one-to-one. b c d 8 a b c 9 a i ii iii iv v b

can be , which is not in the domain of .

c i ii Reflection in the line

.

iii iv 10 a b

c Range of is d 11 a i

; range of is

.

ii iii

b i ii 12 a b c 13 a Proof. b c 14 a b 15 a Proof. b Rotation

about the origin.

c–d Proof. e Reflection in the -axis. f–g Proof.

3 Further transformations of graphs BEFORE YOU START 1 a

b

2 a b 3 a b WORK IT OUT 3.1 Solution 1 is correct. EXERCISE 3A 1 a i

ii

b i

ii

c i

ii

d i

ii

e i

ii

2 a i

; vertical stretch of scale factor and translation

(in either order).

ii

; stretch of scale factor relative to

and translation

b i

; stretch of scale factor relative to

, reflection in

and translation

; stretch of scale factor relative to

, reflection in

and translation

.

. ii . 3 a i ii b i ii

c i ii d i ii 4 a i

; translation translation

. ; translation

ii translation b i

then reflection in -axis; or reflection in -axis then

. ; translation

scale factor relative to ii

then reflection in -axis; or reflection in -axis then

; translation scale factor relative to

then stretch of scale factor relative to then translation

.

then stretch of scale factor relative to then translation

5 a i ii b i ii c i ii 6 a b c d 7 a b c Horizontal stretch with scale factor .

; or stretch of

.

; or stretch of

8 a

b

9 a

b

c

10 11 12 13 14 a Vertical stretch with scale factor ; horizontal stretch with scale factor . b

c

EXERCISE 3B

1 a i

ii

b i

ii

c i

ii

d i

ii

2 a i ii

b i ii c i ii 3 a i ii b i ii 4 a i

or

ii

or

b i

or

ii 5 a i ii b i ii 6 a i ii b i ii 7 a

b

or

8

9

10

WORK IT OUT 3.2 Solution 1 is correct. EXERCISE 3C 1 a i ii b i ii c i ii d i

ii e i ii f

i ii

2 a i ii b i ii c i ii d i ii

or

e i ii f

i ii

3 a b 4 a b 5 a

b No solutions. 6 7

8 MIXED PRACTICE 3 1 a

b

2 B 3 4 Translation by 5 a

b 6 a b 7 D

and vertical stretch with scale factor .

or

8

9 a Translation by b Translation by c Translation by

and vertical stretch with scale factor . and translation by and vertical stretch with scale factor .

10 a Vertical stretch with scale factor ; reflection in the -axis; translation units up.

b c 11 a Translation by b

c i ii 12 a Translation b 13

; vertical stretch, scale factor ; translation

.

14



4 Sequences and series BEFORE YOU START 1 a b 2 3 4 5 6 EXERCISE 4A 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii d i ii 3 a i Increases, converges to

.

ii Decreases, diverges. b i Periodic, period . ii Periodic, period . 4 a b 5 a b Converges to (and oscillates). 6

7 a Proof. b 8 a b 9 a b 10 Proof. EXERCISE 4B 1 a i ii b i ii c i ii 2 a i

ii

b i

ii

c i

ii 3 4 a Proof. b 5 6 a b EXERCISE 4C 1 a i ii b i ii

c i ii d i ii e i ii 2 a i ii b i ii 3 a b 4 5 6

.

7 8 a Proof. b

pages.

EXERCISE 4D 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii 3 4 a b 5 6 7

8 Proof. 9 10 11 12 13 a b 14 15 a b 16 EXERCISE 4E 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii c i ii 3 a b 4 a b 5 a b 6 7 a Proof.

or or

b 8 9 10 11 12 EXERCISE 4F 1 a i ii b i ii c i ii d i

or

ii

or

2 a i ii b i ii 3 a b 4 a b 5

or

6 7 a b 8 a b 9 a Proof. b 10 a Proof. b EXERCISE 4G 1 a i

ii b i ii c i Divergent. ii Divergent. d i ii e i Divergent. ii 2 a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii h i ii i

i ii

3 4 5

or

6 a b

7 a b 8 9 a b 10 11 a b 12 a b 13 a b 14 a Proof. b 15 Proof. WORK IT OUT 4.1 Solution C is correct. EXERCISE 4H 1 a £ b £ 2 a £ b 3 a b £ c i ii 4 a b 5 a

£

b 6 a b 7 Proof. 8 a b c Physical interactions haven’t been considered.

9 a Proof. b c MIXED PRACTICE 4 1 B 2 3 4 a b 5 6 a b c i ii 7 A 8 A 9 10 a b c 11 12 a b 13 14 15 a b 16 a Proof. b 17 a Proof. b c 18 Proof. 19 20

21 22 a b c d Proof. e 23 a Proof. b c

5 Rational functions and partial fractions BEFORE YOU START 1 2 3 4 WORK IT OUT 5.1 Solution 2 is correct. EXERCISE 5A 1 a i ii b i ii 2 3 4 a b 5 a Proof. b 6 Proof; 7 Proof. 8 9 a b 10 WORK IT OUT 5.2 Solution 2 is correct. EXERCISE 5B 1 a i ii b i ii c i ii

d i ii e i ii f

i ii

g i ii 2 a i ii b i ii c i ii d i ii e i ii 3 a i ii b i ii c i ii d i ii 4 a i

ii b i ii

5 6 7 a b

or

8 9 Quotient: ; remainder: 10 Quotient:

; remainder: .

11 12 13 a Proof. b 14

.

15 Proof. 16 a b Proof. EXERCISE 5C 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii 3

.

4 5 6 7 8 a Proof. b 9 10 11 a b 12 Discussion. EXERCISE 5D 1 a i ii b i ii c i ii d i ii 2 3 4 5 a Proof. b c 6 a Proof. b

7 a b 8 a Proof. b 9 MIXED PRACTICE 5 1 C 2 a b 3 a Proof. b c d 4 5 6 7 8 9 10 a b i Proof. ii iii 11 A 12 a b 13 a b 14 a Proof. b

c d 15 a Proof. b c 16 a b 17 a b c 18 a b 19 a b 20 21 22 23 24 Proof. 25 a b Proof.

6 General binomial expansion BEFORE YOU START 1 2 3 4 EXERCISE 6A 1 a i ii b i ii c i ii d i ii 2 3 4 a b 5 a Proof. b c d 6 a b c i ii 7 8 9 10

WORK IT OUT 6.1 Solution B is correct. EXERCISE 6B 1 2 3 4 5 a b c 6 a b c 7 8 a b c 9 10

is not real for values convergence.

MIXED PRACTICE 6 1 D 2 3 4 a b 5 6 B 7 B 8 a b 9 a

so even some cunning rearrangement does not allow

b c 10 11

or

12 a i ii b 13 a b c 14 a b 15 16 17 a b Proof. c d e 18 a b i ii c Proof.

Focus on ... proof 1 Arithmetic series proof

Geometric series proof Multiply through by .

1

so can’t divide by it on last line;

.

2 Yes. 3 a b

Focus on ... Problem solving 1 1 2 3 4

Focus on ... Modelling 1 1 a i ii iii

b The maximum possible rate (asymptote, can’t be reached). 2 3 a Rational function. b Exponential function.

Cross-topic review exercise 1 1 Proof. 2 3 4

or

5 a b i Decreasing so one-to-one. ii c

6

7 a

b

or

c

or

8 a i ii Proof. b 9 a b Proof. 10 a b 11 12 a b

c Domain:

or

; range:

and

.

13 14 a–b Proof. c d Underestimate, as the kitten will grow more slowly as it gets older. 15 a b i ii Expansion is only valid for 16 17 a No b c 18 a b 19 a b c Proof. d

and

is not this interval.

7 Radian measure BEFORE YOU START 1 2 3 4 5 WORK IT OUT 7.1 Solution 2 is correct. EXERCISE 7A 1 a

b

c

d

2 a i ii b i ii c i ii d i ii 3 a i ii b i ii c i ii d i ii 4 a i ii b i ii c i ii d i ii 5 a i

ii

b i

ii

6 a b c d 7 a b c d 8 a b

c d 9 a i ii b i ii c i ii 10 a b c 11–12 Proof. 12 13 WORK IT OUT 7.2 Solution 1 is correct. WORK IT OUT 7.3 Solution 3 is correct. EXERCISE 7B 1 a i ii b i ii 2 a i ii b i ii c i ii 3 a i ii b i

ii c i ii d i ii 4 a b c 5 a i ii b i ii c i ii d i ii 6 a i ii b i ii c i ii d i ii 7

8

9 10 a Proof. b 11 12 Proof. 13 14 a Proof. b c EXERCISE 7C 1 a Amplitude: ; period: b Amplitude: ; period: c Amplitude: ; period: d Amplitude: ; period: 2 a i

ii

b i

ii

c i

ii

d i

ii

3 a b 4 a b 5 6

7 a

b Two intersections. c

solutions.

8 9 a

b c 10 a b 11 a b c 12 a b c EXERCISE 7D 1 a b 2 a b

3 a b 4 a b 5 6 a b 7 8 9 10 11 12 13 14 15 16 17 18 19 20

or

21

radians

22

radians

EXERCISE 7E 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii

4 Proof. 5 a Proof. b c EXERCISE 7F 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii 4 a b 5 a b 6 a b i ii 7 8 9 10 a

b 11 12 13 a b i Proof. ii Because for small (positive or negative) MIXED PRACTICE 7 1 C 2 a b 3 4 5 a b c 6 7 a b 8 a b i ii 9 B 10 11 a b c 12 a b c 13 a b

.

c

14 15 a

is a square.

b c d 16 17 a

and b

and 18 19 20 a Proof. b

21 a b 22 a

, right angle between a tangent and a radius.

b c d e 23 a b Proof. c

is a rectangle, because there are right angles at and , and

is parallel to

.

8 Further trigonometry BEFORE YOU START 1 a b 2 3 a b WORK IT OUT 8.1 Solution 3 is correct. EXERCISE 8A 1 a b c d 2 a b c d 3 a b 4 a Proof. b 5 a b c 6 a b 7 8 a Proof. b

9 a Proof. b EXERCISE 8B 1 a i ii b i ii c i ii 2 a i ii b i ii 3 a b c d 4 a

 

b c d 5 a–d Proof. 6 7–8 Proof. 9 a b 10 a b 11 a b 12 a i –ii Proof.

b 13 14–15 Proof. EXERCISE 8C 1 i ii 2 i ii 3 i ii 4 i ii 5 a b Vertical stretch with scale factor 6 a b 7 a b 8 a b Minimum: 9 10 EXERCISE 8D 1 a i ii b i ii c i ii 2 a i ii b i ii c i

; maximum:

; translation

units to the left.

ii d i ii 3 4 a i ii b i ii c i ii d i ii 5 a i ii b i ii c i ii d i ii 6 a i ii b i ii c i ii d i ii 7 Proof. 8 9 10–12 Proof.

13 a Proof. b 14 a Proof. b 15 a Proof. b 16 17 18 MIXED PRACTICE 8 1 C 2 a Proof. b 3 a b 4 a b c d 5 a Proof. b c 6 a b c d e f 7 a b 8 B 9 a Proof. b

10 a b Proof. c d 11 12 a Proof. b 13 a b i ii 14 a b Proof. c 15 a i ii Proof. iii b–c Proof. d e

9 Calculus of exponential and trigonometric functions BEFORE YOU START 1 2 3 4 EXERCISE 9A 1 a i ii b i ii c i ii d i ii e i ii f

i ii

2 a i ii b i ii c i ii d i ii e i ii f

i ii

3 4 5 6 7 8 9 10 11 12 Tangent:

;

normal: 13 14

local maximum;

local minimum.

15 Proof. 16 17 a

local maximum.

b

local minimum

18 a

million litres.

b WORK IT OUT 9.1 Statement 4 is correct. WORK IT OUT 9.2 Solution 2 is the correct method, but the value of the shaded area is EXERCISE 9B 1 a i ii b i ii c i ii d i ii e i

.

ii f

i ii

2 a i ii b i ii c i ii d i ii e i ii f

i ii

3 a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii 4 5 6

7 8 a b 9 10 11 a b 12 13 a Proof. b 14 Proof. 15 MIXED PRACTICE 9 1 D 2 3 i

4

ii 5 6 7 i

8

ii 9 C 10 11

local maximum; local minima;

12 13 14 a i ii b i

hours

local minimum. local maxima.

ii c i ii 15 16

hours

10 Further differentiation BEFORE YOU START 1 a b c d 2 a b 3 a b 4 a b 5 a b 6 a b EXERCISE 10A 1 a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii

h i ii i

i ii

j

i ii

k i ii l

i ii

m i ii n i ii 2 a i ii b i ii   c i ii d i ii e i ii f

i ii

g i ii h i ii 3 a i ii b i

ii c i ii d i ii 4 5 6 7 8 9 a b 10 11 12 13 a b 14 a Left post. b Proof. c 15 a b c

WORK IT OUT 10.1 Solution 2 is correct. EXERCISE 10B 1 a i ii

b i ii c i ii d i ii 2 a i ii b i ii 3 4 5 6 7 8 a b 9 10 11 a b c 12 a b

c

is even.

WORK IT OUT 10.2 All solutions are correct. EXERCISE 10C 1 a i

ii b i ii

c i ii d i ii 2 3 4 5

local maximum.

6 7 8 Proof. EXERCISE 10D 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i

ii d i ii 3 a b 4 a i ii b i ii c i ii d i ii 5 6 a Proof. b 7 a Proof. b 8 9 10 11 12 13 a b Proof. c EXERCISE 10E 1 a b 2 3 4 a Proof. b

5 a b Proof. 6 a–b Proof. c 7 a b c Proof. MIXED PRACTICE 10 1 D 2 A 3 a i ii b 4 5 a Proof. b 6 7 8 a b i ii 9 10 11 a b Proof. 12 13 a

seconds

b 14 a b c

local minimum;

local maximum.

d

15 a Proof. b 16 a b 17

and

18 a b Proof. c

local maximum;

19 a Proof. b 20 a b c

local minimum.

11 Further integration techniques BEFORE YOU START 1 a b 2 a b 3 a b 4 5 WORK IT OUT 11.1 Solutions B and C are correct. EXERCISE 11A 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii c i ii d i ii 3 a i

ii b i ii c i ii d i ii 4 a b c d e f 5 6 7 8 EXERCISE 11B 1 a i ii b i ii c i ii d i ii e i ii f

i ii

g i

ii h i ii i

i ii

2 3 4 5 6 EXERCISE 11C 1 a i ii b i ii 2 a i ii b i ii 3 a i ii b i ii c i ii 4 a i ii b i ii c i

ii 5 6 a Proof. b 7 8 9 EXERCISE 11D 1 a i ii b i ii c i ii d i ii 2 3 EXERCISE 11E 1 a b c d 2 a b 3 a b 4 Discussion. 5 6 7

8 a Proof. b 9 10 a Proof. b WORK IT OUT 11.2 They are all right, with different

EXERCISE 11F 1 a b c d e 2 a b c d e 3 a i ii b i ii 4 a i ii b i ii 5 a i ii b i ii

:

c i ii 6 7 a Proof. or, alternatively,

b 8 9 a Proof. b 10 11 12 a Proof. b 13 a b Proof. EXERCISE 11G 1 a b c No d e No f g 2 a b c d 3 a i ii b i ii c i ii d i

ii e i ii 4 a i ii b i ii 5 c e 6 a b 7 8 9 a b 10 MIXED PRACTICE 11 1 C 2 3 4 5 6 a b 7 8 a b 9 A 10 11 a

or

b 12 13 a b 14 15 16 a b 17 18 19 20 21 a b Proof. 22 a b c

12 Further applications of calculus BEFORE YOU START 1 a b 2

or

3 4 a b c WORK IT OUT 12.1 Solutions 2 and 3 are correct. EXERCISE 12A 1 a i Decreasing, convex. ii Concave. iii Increasing, convex. b i Increasing ii Increasing, concave. iii Increasing, convex. 2 a

b

3

4

5 6 Proof. 7

or

8 9 10

maximum;

11 Proof. 12 13 14

EXERCISE 12B 1 a i

ii

minimum.

b i

ii

c i

ii

2 a i ii b i

ii c i ii 3 a i ii b i ii c i ii d i ii e i ii 4 a b 5 a b

; the distance along the line.

c WORK IT OUT 12.2 Solution 2 is correct. EXERCISE 12C 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii

3 4 5 6 a b 7 8 Proof. 9 Proof. 10 a b Proof. EXERCISE 12D 1 a i ii b i ii c i ii 2 a b 3 a

and

b 4 a Proof. b 5 a b c EXERCISE 12E 1 a i ii b i ii 2 a i ii b i ii

c i ii 3 a i ii b i ii c i ii 4 5 6 7 8 9 10

units per second

WORK IT OUT 12.3 Solution 3 is correct. EXERCISE 12F 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii c i ii

3 4 5 6 Proof. 7 8 9 10 11 12 MIXED PRACTICE 12 1 D 2 3 a b 4 5 a b 6 a b 7 D 8 9 10 a b 11 a Proof. b 12 a

local minimum;

b Proof. 13 Proof. 14 15 a b Proof.

local maximum.

c 16 a b i ii iii c 17 a

and

seconds

b 18 Proof. 19 a b 20

21 a Proof. b c d

and

13 Differential equations BEFORE YOU START 1 2

downwards.

3 4 5 a b 6 EXERCISE 13A 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii d i ii WORK IT OUT 13.1 Solution 2 is correct. EXERCISE 13B 1 a i ii

b i ii 2 a i ii b i ii c i ii 3 a i ii b i ii c i ii 4 5 6 Proof. 7 8 EXERCISE 13C 1 a i ii b i ii c i ii 2 a Proof. b

; after

minutes .

3 a Proof. b 4 a Proof. b

5 a When

; decreases from

years.

b c d It will decay to zero. is the rate of change. ‘Proportional to and inversely proportional to ’ means that

6 a

, so

.

is negative because demand decreases as price increases, so the rate of change negative. b c i

Demand is inversely proportional to price. ii

Demand is independent of price. 7 a Proof. b 8 a b The velocity approaches

.

9 a Proof. b 10 a Proof. b c MIXED PRACTICE 13 1 C

; it approaches

.

is

2 B 3 4 5 6 a b 7 a Proof. b c Not suitable, as it predicts indefinite growth. d 8 a

where

b The demand decreases as the price increases. c Luxury goods; the demand increases with price. 9 a b It will never fill, as

for all .

10 a Proof. b c The velocity increases towards

as increases.

11 a b i Proof. ii 12 a Decrease in size due to, for example, competition for food. b c Increases with the limit of 13 Proof. 14 a b Proof. c

.

14 Numerical solution of equations BEFORE YOU START 1 2 a b c 3 a b c WORK IT OUT 14.1 Solution C is correct. EXERCISE 14A 1 a Exact solution. b Cannot rearrange. c Exact solution. d Exact solution. e Cannot rearrange. f

Exact solution.

g Cannot rearrange. h Cannot rearrange. 2 a

and

b

and

c

and

d

and

3−5 Proof. 6 a Proof. b 7 a Proof. b i ii

is not continuous (has an asymptote) at

.

8 a i

ii b i ii

changes sign twice.

iii Use, for example, EXERCISE 14B 1 a i ii b i ii 2 a i ii b i ii 3 4 a Proof. b 5 6 a Proof. b c Proof. 7 a Proof. b 8 EXERCISE 14C 1 a i ii b i ii

close to stationary point close to stationary point. ; on the other side of an asymptote. ; outside the domain.

and

2 a

b Proof. c It is close to a stationary point. 3

; the tangent crosses the -axis further away from the root, so subsequent values of increase.

4 a b , so falls on the far side of a discontinuity.

c 5 a b i

ii Not converge. c 6 a

and

and

b

c d It will converge to . e 7 a

and

b

c

and

d

which is outside of the domain of

e 8 a i Too close to stationary point at ii b i Too close to stationary point at ii c i

, so lies in a different region.

ii d i

, so lies in a different region.

ii 9 a Proof. b c Proof. 10 a

b Proof.

and

where

.

.

c

d Does not converge, or converges to another root (in a different interval). EXERCISE 14D 1 a i ii b i ii c i ii 2 a i ii b i ii i

ii

iii

a

No

No

N/A

b

No

No

N/A

c

Yes

Yes

Convergence to upper root only. Convergence for

lower root.

d

Yes

Yes

Convergence to lower root only. Convergence for

upper root.

3

4 5 a b 6 7 a b 8 9 a Proof. b c 10 a Proof.

b c 11 a

b Proof. EXERCISE 14E 1 a

(large, positive); diverges (increases without a limit)

b

(large, negative); diverges (decreases without a limit).

c

converges to

d

converges to

e

(decreases)

oscillates between and .

f

converges (increases to

g

undefined after

2 a Converges b Converges c Converges to B d Converges to A. e Converges to C. f

(and oscillates).

Converges to B.

g Converges to B 3 a i ii b i ii c i ii d i ii 4 a i ii

.

).

b i ii c i ii d i ii 5 These rearrangements are not unique – there are alternatives which will converge to the various roots. a lower root: upper root:

;

b lower root:

;

upper root: c lower and upper roots: central root: 6 a b 7 a Converges to P. b Converges to P. c Converges to S. 8 a b 9 a Proof. b 10 a

; the positive root. and

b c 11 a

or

b c MIXED PRACTICE 14 1 A 2 B 3 a Proof. b 4 a−b Proof.

;

c 5 a

solutions;

b Proof. c 6 a−b Proof. c 7 8 a Proof. b c

is on the other side of the asymptote.

d 9 a−b Proof. c 10 a Proof. b c Proof. 11 a b i Proof. ii Proof. iii

seconds. Not suitable in context, the required solution must be positive.

iv v c

seconds seconds

15 Numerical integration BEFORE YOU START 1 2

3 WORK IT OUT 15.1 Solution 2 is correct. EXERCISE 15A 1 a i

Lower bound:

; upper bound:

Lower bound:

; upper bound:

ii

b i

Lower bound:

; upper bound:

ii

Lower bound:

; upper bound:

Lower bound:

; upper bound:

c i

ii

Lower bound:

; upper bound:

2 a i Difference ii Difference iii Difference b i Difference ii Difference iii Difference c i Difference ii Difference iii Difference d i Difference ii Difference iii Difference In general, doubling the number of rectangles from between the upper and lower bounds. 3 Lower bound: 4 a Lower bound:

; upper bound: ; upper bound:

to

or from

to

halves the difference

b More rectangles. 5 a

b c Decrease. 6 7 a b WORK IT OUT 15.2 Solution 3 is correct. EXERCISE 15B 1 a i ii b i ii c i ii 2 i and ii See table in Worked solutions. iii Usually error decreases by a factor of to . 3 a Approximation b c d Approximation e f

4 a

b c Concave curve: underestimate. 5 a b More trapezia; exact integration is possible. 6 a b c Concave curve: underestimate. 7 8 a b MIXED PRACTICE 15 1 D 2 3 a b Concave curve: underestimate. c More intervals. 4 5 a b More intervals. 6 7 B 8 9 a b 10

and

11 a

b 12

16 Applications of vectors BEFORE YOU START 1 a b c d 2

from horizontal.

3 a b 4 a b 5 6 EXERCISE 16A 1 a i ii b i ii c i ii 2 a i ii b i ii 3 a b 4 a b 5 a b

from horizontal.

6 a b c d

; the particle changes direction during the motion.

7 8 a b

9 a b 10 WORK IT OUT 16.1 Solution 3 is correct. EXERCISE 16B 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii 3 4 a b

from horizontal.

5 a b 6 7

from horizontal.

8 9 10 11 Proof. 12 WORK IT OUT 16.2 Solution 2 is correct. EXERCISE 16C 1 a i

ii

b i

ii

2 a i

ii

b i

ii

c i

ii

3 a b 4 a

b 5 a b c 6 7 Proof. 8 a b 9 a−b Proof. c EXERCISE 16D 1 a i

ii

b i

ii

2 a i

ii

b i

ii

c i

ii

d i

ii 3 a i ii b i ii c i ii 4 5 a i ii b i ii 6 a b c d 7 a b c d

8

9 10 11 12 13 EXERCISE 16E

1 a i

ii

b i

ii

2 a

b

3

4 5 6

7 a b 8 9 a

b

10 11 a−b Proof. MIXED PRACTICE 16 1 B 2 a b 3

from the horizontal.

4 a b Proof. 5 6 a Proof. b 7 a b c 8 D 9 10 a b c It stays at a constant height. 11 12

below the horizontal

13 a b c Proof. d e

hours

14 a b c 15 A 16 a b c 17 a b 18

Focus on … Proof 2 1 Yes, by varying the base (or hypotenuse) accordingly. 2 Yes. 3 Proof.

Focus on … Problem solving 2

1 a i Tends to ii Stays at iii Dies out after

years

b i Same as in a. is the largest value without the population dying out. This doesn’t depend on the initial size of the population. 2 Increases with , decreases with 3 a i ii iii No solutions b c 4 a b Proof. 5−6 Discussion.

Focus on … Modelling 2 1 2 3 4 5 6 7

; this model predicts infinite initial rate of growth. or

8

Cross-topic review exercise 2 1 2 3 4 5 a b 6 7 a

b 8 9 a i ii b Translate

and then stretch scale factor parallel to -axis.

10 a b c 11 12 a Proof. b 13 a b 14 a Proof. b c d 15 16 a b Proof. c 17 a i ii b i −iii Proof. c i ii 18 a b 19 a Stretch scale factor parallel to -axis. Translation b c i Proof. ii

.

d 20 21 a b Proof. 22 a Proof. b c d 23 a b−c Proof. 24 a b c 25 a

b c 26 a−c Proof d i ii iii Proof.

17 Projectiles BEFORE YOU START 1

at

from vector

2 3 4 5 6 EXERCISE 17A 1 a i ii b i ii c i ii d i ii 2 a i ii b i ii c i ii d i ii 3 a b 4 a b 5 a b 6 7 8 a b c

below the horizontal.

9 Yes; height is

at that point.

10 a b

or

11 a Proof. b

or

12 a Ten times normal. b Slowed down by a factor of

.

WORK IT OUT 17.1 Solution 2 is correct. EXERCISE 17B 1 2 a Proof. b c No air resistance; rugby ball is a particle (has no size and doesn’t spin). 3 a Proof. b Yes 4 a Proof. b i Proof. ii 5 6 a Proof. b

; it must be falling when it goes through the hoop.

MIXED PRACTICE 17 1 B 2 C 3 a b 4 a b 5 a b 6 a i −ii Proof. b 7 8 a b

9 a Proof. b c No air resistance or the ball doesn’t spin or no loss of energy. 10 a b c d 11 12 a Proof. b i −ii Proof. iii

18 Forces in context BEFORE YOU START 1

; magnitude

; angle

from direction .

2 EXERCISE 18A 1 a i ii b i ii 2 Answers given to

.

a i ii b i ii 3 4 a b 5 a Proof. b c 6 7 a i ii b i ii 8 9 a b c 10 a They are two parts of the same string. b EXERCISE 18B 1 a i ii b i ii

2 3 4 5 a b 6 7 a b 8 a b c 9 C and D only. EXERCISE 18C 1 Answers given to

.

a i ii b i ii c i ii 2 Answers given to 3 a i ii

. up the slope.

down the slope.

b i  

down the slope;

ii

down the slope;

c i

up the slope;

ii

up the slope;

4 a b 5 a b i ii 6 7 a b c

up the slope. . . down the slope.

8 a b 9 10 11 a Proof. b c Up the slope with magnitude d Down the slope, with acceleration

. .

12 a b c 13 14 Proof. MIXED PRACTICE 18 1 B 2 a b 3 4 a b 5 a

b Proof. c 6 a

b c d e For example: No air resistance force acting. No other forces acting on the box. They are the only forces that act. No turning effect (due to forces). Forces are concurrent.

7 a b 8 a

b Proof. c 9 a Proof. b c The acceleration is reduced because of air resistance or the fact that there is friction. 10 a

b c d e Less friction, so a smaller coefficient of friction. 11 C 12 a b 13 a b 14 a

µ

b 15 a b 16 Proof. 17 a b Proof. c d e

down the slope.

18 a i ii b c i ii d The air resistance force will increase (vary or change) with speed. 19 a

b c 20 a

(in direction of moving up the slope)

b Yes;

seconds after the start of movement.

21 a i

ii Proof. iii iv b c The same. 22 a b i ii 23 a

b Proof.

19 Moments BEFORE YOU START 1

2 EXERCISE 19A 1 a i ii b i ii c i ii 2 a i

.

ii

.

b i

.

ii

.

c i ii 3 a i

. .

ii b i ii

.

c i

. .

ii 4 a i

. .

ii b i

.

ii

.

c i

.

ii

.

d i

.

ii

.

e i

.

.

ii 5

.

6

.

7 a b Since there is no bending, the diver is 8

from the other end.

clockwise.

9 Proof. 10 WORK IT OUT 19.1 Solution C is correct. EXERCISE 19B 1 a i ii b i ii 2 a i ii b i ii c i ii 3 4 5 a b The crane attachment would be able to handle some net moment. 6 a 7 a Proof. b 8 9 10 a

and

b The working in a assumes that the pole is rigid and horizontal. 11 12 13 a i ii

b

iii

could be bigger than

14 At

.

; At

15 16 MIXED PRACTICE 19 1 A 2 B 3 4 a b

upwards.

5 a

b Proof. c d The weight of the plank acts through its centre. 6 a i ii The weight of the plank acts through its centre. b 7 a

b c The weight of the plank acts through its centre. 8 9 10 a b 11 a b 12 a Proof. b

13 a Proof. b 14 15 16 a Proof. b

Focus on … Proof 3 1 2 Proof. 3 a i ii iii Proof. b Proof. 4 a Proof. b

Focus on … Problem solving 3 The answers that are clearly wrong are:

.

Focus on … Modelling 3 1 2 The gravitational acceleration on Archimedes’s planet is also

.

3 The support needs to be further away from the point where the Earth is resting on the lever. If this distance is increased to , the length of the other end would need to be . 4

Cross-topic review exercise 3 1 a b 2 a b Proof. c 3 a b i ii West. c

d 4 5 a

to the right of the support.

b 6 a b 7 8 Proof. 9 a b Proof. c i Proof. ii iii

or

10 a b c d 11 a b 12 a

b Proof. c d There is less friction so the coefficient of friction must be less. 13 a b c

below the horizontal (to the left).

d i  The tension is the same on both sides. ii The two blocks have the same speed and acceleration. 14 a

b

c Proof. d 15 a Proof. b c d e The weight is the only force acting OR no air resistance. 16 a Proof. b Proof. 17 Proof. 18 a It is a particle/no air resistance/lift forces act on the ball. b Proof. c 19 a Proof. b i ii

20 Conditional probability BEFORE YOU START 1 2 3 4 EXERCISE 20A 1 a i ii b i ii c i ii d i ii 2 a i ii all b i ii 3 a b 4 5 6 7 8 a

or

b EXERCISE 20B 1 a b c d e f

g h 2 a i ii b i ii 3 a b 4 a b c d e f g h i j k l m n 5 a

b c d 6 a

b c 7 a b 8 a b 9 a b c 10 a

b c d e f 11 a

b c d e No: 12 a

b

c d e No: 13 14 15 16 a b c Proof. EXERCISE 20C 1 a i ii b i ii c i ii Year 9

2 a Girls Boys Total b c 3 a b c Proof. 4 a b c 5 a

Year 10

Year 11

Total

b c d e 6 a b c d 7 a b c d e 8 a i ii iii 9 For example, numbers on a dice with WORK IT OUT 20.1 Solution A is correct. EXERCISE 20D 1 a i ii b i ii c i ii 2 3

‘even’ and

‘prime’ shows is wrong.

4 a

b c d 5 a b 6 7 8 a b 9 a b 10 a b 11 a b 12 13 14

or

15 Proof. MIXED PRACTICE 20 1 C 2 B 3

4 a

b c 5 a i ii iii iv v 6 a

b c d 7 a b 8 a i ii b i ii iii iv 9 a

b c

d

e f 10 11 12 13 a b Proof. c d 14 15 16 a b

21 Normal distribution BEFORE YOU START 1 2 3 4 5 EXERCISE 21A 1 a i ii b i ii c i ii d i ii e i ii 2 a i ii b i ii 3 a i ii b i ii c i ii 4 a b 5 a b i ii c Assumes that they are independent, but Ali might be getting tired, OR he might get more determined. 6

7 Mean ; standard deviation . 8 Mean

; standard deviation .

9 Mean

; standard deviation .

10 a b Proof. c Best athletes likely to be chosen for the team; not independent as will depend on the weather/track/how well other athletes are doing; all four going under seconds is a sufficient but not necessary condition. 11 a b c 12 a b 13 a b 14 a b 15 a b 16 17 18 WORK IT OUT 21.1 Solution 1 is correct. EXERCISE 21B 1 a i ii b i ii c i ii 2 3 4 a b 5 a b

c 6 7 8 9 a b 10 11 12 Take

, where is a uniform continuous distribution across

.

EXERCISE 21C 1 a i ii b i ii 2 a i ii b i ii 3 4 5 6 7 8 9 10 EXERCISE 21D 1 a Yes. b No – not symmetrical. c No – two modes. d No – not a characteristic bell curve. 2 Three standard deviations below the mean would be predict impossible results.

children, so the normal model would

3 a b All data lie within this range, so the normal distribution appears to be suitable, assuming the data are symmetrically distributed.

4 a b Mean

; standard deviation

.

c Symmetric, bell-shaped curve. d About

students.

5 a i ii b 6 a b 7 a b For example:

.

8 9 a Proof; For example b 10 Proof. MIXED PRACTICE 21 1 A 2 a b 3 a b 4 5 6 a i ii b 7 B 8 a b 9 a b 10 a Proof. b 11 12 a i ii iii

.

b c 13 a For

and

.

b c Probably not independent. 14 a Within standard deviations of the mean are scores from to , which are all possible. The mean is about the same as the mean, suggesting a symmetrical distribution. b Distinction 15 a b 16 a b

; Merit

; Pass

.

22 Further hypothesis testing BEFORE YOU START 1 2

; reject

3 a b c EXERCISE 22A 1 a i ii b i ii 2 a i ii b i ii 3 a

; assuming independence of emissions.

b c 4 5 6 7 8 9 a b c 10 a b WORK IT OUT 22.1 Solution B is correct. EXERCISE 22B 1 a i ii b i ii

c i ii 2 a i

or or

ii b i ii c i ii 3 a i

; reject

.

; do not reject

ii b i

; reject

.

ii

; reject

.

c i

.

; do not reject

.

ii

; do not reject

.

d i

; do not reject

.

ii

; do not reject

4 a b

; do not reject

5

; reject

.

.

6 a The scores of students in the school follow a normal distribution. The standard deviation is still . The

students were randomly selected from the cohort in the school.

b

; do not reject

.

7 a b c Reject

or .

8 a b The post-diet masses follow a normal distribution. c Reject

.

9 a b c 10 a No b 11 Higher -value if

as you have the probability from both tails.

EXERCISE 22C 1 a i Significant evidence.

ii Significant evidence. b i No significant evidence. ii No significant evidence. 2 a i No significant evidence. ii No significant evidence. b i Significant evidence. ii Significant evidence. 3 a b Reject

.

4 a b Do not reject

.

5 a i Countries in 2013. ii iii Reject

.

6 a b Reject

.

c No – correlation does not imply causation and the test was not for positive correlation. 7 a b Reject

.

c Less likely to get a false positive, which would result in a useless drug being used. 8 a Significant evidence. b

; no significant evidence.

c Decrease, since increased value of r means it is less likely to be a chance result from a population with zero correlation. 9 10 With two data points there is always a perfectly fitted straight line. MIXED PRACTICE 22 1 C Many people think that -values are a direct measure of how likely do not fall into this trap. 2 3 a Countries in 2015. b c Do not reject

.

4 a b i ii Do not reject 5

; reject

. .

is to be true. Make sure you

6 D 7 a Both variables must be normally distributed. b Reject

.

8 a Still normally distributed. Still with a standard deviation of seconds; b Reject

.

.

c They would not be independent – the water needs to cool down to the original temperature. 9 a b

or

c

; do not reject

d Critical region now

. so now reject

.

10 a Random sample. b

; do not reject

c i

becomes

ii Becomes iii Reject

.

. .

.

11 D 12 a b c 13 a b

Focus on … Proof 4 1 Very small 2 a

b i ii c Proof; the argument is flawed. 3 There are not a large number of diamond thieves in the population. 4 If 5 Yes.

, then

.

6 7 Discussion.

Focus on … Problem solving 4 1 Discussion.

2





You choose

Host opens



3 a Discussion. b Larger, because you already know that there are girls.

c



First child girl

boy

girl Second child

boy

Focus on … Modelling 4 1 a

b No, not symmetrical. 2 a b c It could be a suitable model; the interquartile range is (which is close to looks symmetrical.

) and the box plot

3 a Yes b No 4 It is discrete, whereas normal is continuous. 5 a It is likely that the distribution would not be symmetrical, as the minimum possible length of a phone call is , but the maximum can be more than

minutes.

b i Yes ii No; bimodal. iii No; discrete. 6 a Curves and . b Curve .

Cross-topic review exercise 4 1 2 a b Do not reject and rainfall.

; there is insufficient evidence of correlation between average daily temperature

3 a b Reject

; there is evidence that the rose bushes are taller

.

4 a b c Not reasonable; predicts nearly all bulbs will last more than

hours.

5 a i ii iii b c Other fuels; other vehicles; other types of customer; minimum purchase (policy); purchases in integer/fixed £ . 6 7 a b c 8 a Proof. b c Proof. 9 a i ii iii iv v b

10 a b c 11 a b

so mean unlikely to be correct. so standard deviation unlikely to be correct.

c

12 a b c 13 a µ b 14 a b Reject . Significant evidence of negative correlation; as the amount of time increases the test marks decrease. c The two sections of the scatter graph represent two different students; for each student there is positive correlation, but one student got better marks despite spending less time on revision. 15 a Do not reject b Reject

.

.

16 a b c d Proof. e f 17 a Reject

; sufficient evidence that the mean time has decreased

b 18 a

and

b c No, because some of the trapezia are above and some are below the curve.

Practice Paper 1 1 C 2 C 3 A

.

4 D 5 6 7 8 9 a

or

b Proof. 10 11 a b c 12 a

or

b Proof. 13 14 a Proof. b c

; underestimate.

15 a b Proof. 16 a, b Proof. b c 17 a Proof. b c i Proof. ii

Practice Paper 2 1 C 2 C 3 a

b

c 4 a b 5 6 Proof. 7 a b 8 9 a Proof. b 10 a b ii c 11 D 12 B 13 a b Proof. c 14 a b 15 a b i ii 16 a i ii Proof. b 17 a Proof. b c

would need to be larger.

18 a Proof. b c 19

Practice Paper 3 1 B 2 C 3 4 5 a b i Proof. ii Yes; it is increasing and therefore one-to-one. iii 6 a b 7 a i ii

b i ii The tangent does not cross the -axis. 8 a Model ; taking logs gives

which is of the form

b 9 a b i ii 10 a A b B 11 a b i No – there may be a non-linear relationship. ii No – this sample does not provide significant evidence of correlation.

.

12 a b c d e 13 a b

Do not reject

; this is not significantly above the

expected number. c Might not be independent; club not representative of whole school. 14 a b c 15 a b Proof. 16 a Mean is around middle of range. Range is about standard deviations. b c

, assuming independence.

d

Do not reject intake.

e

. No significant change in energy

Gateway to A Level – GCSE revision A. Expanding brackets and factorising You need to be able to multiply together (or expand) sets of brackets, and simplify the resulting expression by collecting together like terms. To expand two sets of brackets, multiply each term in the first bracket by each term in the second bracket.

Worked example 1 Expand and simplify Solution

Comments Multiply each term in the second bracket by . Multiply each term in the second bracket by .

Add the results together, collecting like terms (

and ).

If there are three brackets, it is a good idea to expand two of them and then multiply the result by the third set of brackets.

Worked example 2 Expand and simplify Solution

Comments Expand any two brackets, for example the second and third. Then multiply by the first bracket as in Worked example 1.

Add the results together, collecting like terms ( ).

and

You also need to be able to factorise expressions, which means putting brackets in and writing the sum or difference as a product. You can often do this by finding a factor common to each term in the sum.

Worked example 3 Factorise Solution

Comments The largest factor of: and is and

is

and is

So the largest factor in both terms is

.

Sometimes you can factorise an expression by first splitting it up into two parts and factorising each one separately. This sometimes reveals a common factor in both parts.

Worked example 4 Factorise Solution

Comments is a factor of the first two terms and of the second two. You can now take out a factor of

.

Note Gateway to A Level Section B looks at factorising quadratic expressions into two brackets using a method similar to this.

EXERCISE A

EXERCISE A 1 Simplify: a b c d e f 2 Expand and simplify: a b c d e f g h i 3 Factorise: a b c d e f 4 Factorise: a b c d e f



Gateway to A Level – GCSE revision B. Factorising quadratics You need to be able to factorise quadratic expressions of the form If the coefficient of is , look for a factorisation of the form that their product is and they add up to .

. . The numbers and are such

Worked example 1 Factorise Solution

Comments Look at factors of .

and

. You need two negative numbers to give

add up to

.

If the coefficient of is not , you need to adapt this procedure slightly. First look for two numbers that multiply to and add up to . Then split the middle term and factorise in pairs.

Worked example 2 Factorise Solution

Comments You need two numbers that multiply to and add up to . This means you need one positive and one negative number, with the positive number being larger.

The two numbers are

and

Look at factors of

:

Split the middle term: Factorise in pairs: The first two terms have a common factor and the last two terms have a common factor . Finally, take out the common factor

.

An alternative method is to simply look for numbers that work. If the coefficients are small prime numbers this can be quite quick, but otherwise the method shown in Worked example 2 is more efficient.

Worked example 3 Factorise

Solution

Comments The only way to factorise is

.

The missing numbers in brackets are and . One is positive one is negative. Try possible combinations until you find one that gives the middle term . Before you use one of these methods, you should check whether there is a common factor that can be taken out of all three terms. For example:

A special example of factorising a quadratic is the difference of two squares:

Worked example 4 Factorise Solution

Comments is the square of

EXERCISE B

and

is the square of

EXERCISE B Factorise these expressions: 1 a b c d 2 a b c d 3 a b c d 4 a b c d e f g h



Gateway to A Level – GCSE revision C. Solving quadratic equations You have met two ways of solving quadratic equations: Factorising Using the formula

.

Worked example 1 Solve the equation:

Solution

Comments For both methods, the equation needs to be

.

You are looking for two numbers that add up to and multiply to give . It must be a positive and a negative number. Make each bracket

.

You can also factorise by taking out common factors (rather than using two sets of brackets).

Worked example 2 Solve the equation:

Solution

Comments Both terms have a common factor This now has two factors:

and

. Make each factor

.

If the equation doesn’t seem to factorise, use the formula. (You can also use the formula for equations that do factorise: you will then get rational numbers as answers.)

Worked example 3 Solve the equation:

Solution

Comments The equation needs to

.

The equation doesn’t seem to factorise, so use the formula with . Be careful:

is

.

or

(to s.f.)

You usually round the answers to three significant figures.

EXERCISE C 1 Solve by factorising: a b c d e f g h i j 2 Solve by using the formula, giving your answers to s.f: a b c d e f 3 Solve these equations. Check whether you can factorise first, and if you can’t then use the formula. a b c d e f g h



Gateway to A Level – GCSE revision D. Types of numbers There are many different types of number and it is very important that you know some of the labels applied to them: Natural numbers Integers Rational numbers Irrational numbers Real numbers

The rational numbers are any numbers that can be written as a fraction of two integers. Irrational numbers cannot be written as a fraction of two integers. The main irrational numbers that you know at the moment are surds and anything involving . The real numbers comprise all numbers you have met so far. If no other indication is given, assume that you are working with real numbers. EXERCISE D 1 Tick the boxes to indicate which sets of numbers the numbers are in. The first one has been done for you.

Number ✔

✔

✔

✔



































































Gateway to A Level – GCSE revision E. Angles, polygons and bearings Parallel lines

Parallelograms and rhombuses Parallelogram:

opposite sides are equal opposite angles are equal adjacent angles add up to diagonals bisect each other Rhombus:

A rhombus is a type of parallelogram in which all four sides are equal. There are two further useful facts

about a rhombus: the diagonals are perpendicular to each other the area is equal to

, where and are the lengths of the diagonals.

Bearings Bearings are a way of describing directions by specifying the angle measured clockwise from the north.

Worked example The bearing of from is Solution

. Find the bearing of from . Comments Draw a diagram and label required angles.

Angles

and add up to

(see ‘parallel lines’).

EXERCISE E

EXERCISE E 1 Find the angles at the four vertices of this rhombus:

2 The bearing of from is and is directly south of .

Find the bearing of:



a

from

b

from .

. The distance from to is the same as the distance from to ,

Gateway to A Level – GCSE revision F. Functions A function is a rule that assigns to each input value a unique output value. To evaluate a function at a particular value, you substitute the given value into the algebraic expression. Although you will be allowed to use your calculator in examinations, it is important that you are confident about rules of arithmetic so that you can apply them to algebraic expressions.

Worked example 1 If

, find

a b c Solution

Comments Substitute

Substitute

into

.

into

.

Be careful with negatives: and Substitute

into

.

Make sure you are comfortable with the arithmetic of fractions.

You also need to simplify functions when the input is itself a function of .

Worked example 2 If

, find

a b c Solution

Comments Substitute

into

.

Make sure you square all of Substitute

into

:

.

Expand:

Substitute

into

.

Be careful with negatives: and EXERCISE F 1 Without using a calculator, evaluate each function at the given values. a

at i ii iii

b

at i ii iii

c

at i ii iii at

d i ii iii

at

e i ii iii

at

f i

ii iii 2 Simplify each function for the given inputs. a

at i ii iii

b

at i ii iii

c

at i ii iii

d

at i ii iii at

e i ii iii



Gateway to A Level – GCSE revision G. Rules of indices You need to be able to evaluate positive, negative and fractional powers.

Worked example 1 Evaluate: a b

c

Solution

Comments Write the exponent form as repeated multiplication.

Evaluate in parts. , so evaluate

first …

… and then take the reciprocal.

The power means cube root. You should also know the rules for working with powers:

Worked example 2 Simplify: a b Solution

Comments Add powers when multiplying. Subtract powers when dividing. Multiply powers. Subtract powers when dividing.

EXERCISE G

1 Are these statements true or false? a

is always larger than

b c d e f g h 2 Evaluate without a calculator: a b c d 3 Evaluate the following without a calculator, leaving your answers as a fraction where appropriate: a i ii b i ii c i

ii

d i ii e i ii f

i

ii

g i ii h i

ii

4 Simplify: a b c d 5 a Write b Write



in the form in the form

Gateway to A Level – GCSE revision H. Surds A surd (also called a root or a radical) is any number of the form whole numbers.

where

and are fractions or

The most important rule of surds deals with their product:

There is no equivalent for the sum. In general:

You can collect square roots of the same number together, so, for example: . One very useful tool for simplifying square roots is to take out any square factors of the number being square rooted. For example: .

Worked example Write Solution

in the form

. Comments Treat the expression as two brackets.

Simplify. Then put it into the required form.

EXERCISE H

EXERCISE H 1 Write in the form

, where is a whole number:

a b c d e f 2 Write in the form

where is a whole number:

a b c d e f 3 Write in the form a b c d e f



:

Gateway to A Level – GCSE revision I. Linear equations and inequalities The most important thing to remember when solving equations is that you have to do the same thing to both sides. A good tactic is to get all the unknowns on one side and everything else on the other side and then divide. Do not expect all the answers to be whole numbers.

Worked example 1 Solve:

Solution

Comments

Multiply by :

Getting rid of fractions is a good idea. Remember to multiply the whole of each side, not just the terms you are most interested in.

Add

:

Subtract

:

Get all terms containing on one side and everything else on the other. Try to get a positive coefficient of . Make the subject.

You can solve linear inequalities in a similar way, with one important exception: when multiplying or dividing by a negative number, you must reverse the inequality sign.

Worked example 2 Solve the inequality:

Solution

Comments

Multiply by :

Getting rid of fractions is a good idea.

Subtract :

Get all terms containing on one side and everything else on the other.

Subtract :

Divide by

and remember to reverse the inequality sign.

The issue of changing the inequality sign can be avoided if you always try to get a positive coefficient for . Worked example 3 demonstrates a different method of solving the same inequality.

Worked example 3 Solve the inequality:

Solution

Comments Multiply through by to get rid of the fraction.

Add

:

Get all terms containing on one side and everything else on the other.

Subtract :

Try to get a positive coefficient of .

Divide. Conventionally is written first, so you need to rewrite this. EXERCISE I 1 Solve for : a b c d e f 2 Solve these inequalities: a b c d



Gateway to A Level – GCSE revision J. Simultaneous equations If you have two equations and two unknowns then you can often solve them to find just one set of values that works in both equations. There are two algebraic methods for doing this: elimination and substitution. 1 Elimination You can add or subtract entire equations (and multiples of them) to entirely eliminate one of the variables.

Worked example 1 Solve the simultaneous equations:

Solution

Comments Labelling the equations with numbers in brackets to show clearly how you are manipulating the equations. Look for a way to multiply one equation to match the coefficient of a variable in another.

Substituting into (1):

2 Substitution You can rearrange one of the equations to make one of the variables the subject and then substitute this into the other equation. Again, you will have formed an equation with just one variable. The advantage of this method is that you can use it with some non-linear simultaneous equations.

Worked example 2 Solve the simultaneous equations:

Solution

Comments Rearrange the simpler equation to make one variable the subject. Substitute into the second equation.

When When

Use

to find for each solution for .

,

EXERCISE J 1 Find the solution of these simultaneous equations. Use whichever method you prefer: a

and

b

and

c

and

d

and

e f

and and

2 Solve these simultaneous equations: a b



and and

c

and

d

and

Gateway to A Level – GCSE revision K. Direct and inverse proportion Proportion is another word for fraction it is a way of comparing one group to the whole quantity. So the proportion of vowels in the alphabet is

.

If two variables are ‘in proportion’ then if you divide one by the other the result is always the same. i.e. if is proportional to then

a constant. You usually just rewrite this as

. (Sometimes you use

the phrase ‘direct proportion’ for this type of relationship.)

Worked example 1 If is proportional to , and when

, what is the value of when

Solution

Comments

If is proportional to then When

?

,

.



so:

Use the given values to find .

So



There are other types of proportional relationships. For example: is proportional to

means that

is inversely proportional to means that

.

Worked example 2 Given that is inversely proportional to . Solution

, and that

when

, find an expression for in terms of

Comments Use the given information to find .

EXERCISE K

EXERCISE K 1 a If is proportional to and when

, find a formula for in terms of .

b If is proportional to and when

, find a formula for in terms of .

c If is proportional to and when

, find when

.

d If is proportional to and when

, find when

.

e If is proportional to and when f If is proportional to

and when

find a formula for in terms of . , find when

2 a Given that is inversely proportional to , and that . b



is inversely proportional to and

when

c Given that is inversely proportional to when .

and that

d If is inversely proportional to , and

when

.

when

, find the value of when

. Find an expression for in terms of . when

, find the value of

, find an expression for in terms of .

Gateway to A Level – GCSE revision L. Using coordinates You can describe any point in a plane relative to an origin using two numbers, which you call coordinates. The -coordinate describes how far to the right the point is from the origin and the -coordinate describes how far up a point is from the origin. A negative coordinate means the point is to the left or down respectively. You can calculate a gradient between two points, giving a measure of the steepness of the line connecting the two points. If the gradient is positive it increases from left to right. If it is negative it decreases from left to right. If the two points have coordinates and then:

Key point Gradient

There are also simple formulae for the point exactly half-way between two coordinates and the distance between two points.

Key point Midpoint

Tip Notice that this is just the average of the coordinates of the two points.

Key point Distance between two points

Tip This formula follows from Pythagoras’ theorem.

Worked example Point has coordinates

. Point has coordinates

a

What is the gradient of the line connecting and ?

b

What are the coordinates of the midpoint of and ?

c

What is the distance between and ?

Solution

.

Comments Use the formula for gradient.

a

Use the formula for midpoint.

b

Use the formula for distance. c

EXERCISE L 1 Find the gradient of the line connecting: a b c

and and and

2 Find the midpoint of these points: a b c

and and and

3 Find the distance between the two points: a b c



and and and

Gateway to A Level – GCSE revision M. Straight line graphs If you are given a rule connecting the -coordinate and the -coordinate, only some points on the plane will satisfy them. If the rule is of the form then those points will lie in a straight line with gradient and the line will meet the -axis at If you are given the gradient and one point the line passes through, you can use that information to find the value of If two lines have the same gradient then they are parallel.

Worked example A line has equation

Find the equation of a line parallel to this line through the point

Solution

Comments Rearrange into the form find the gradient.

So the gradient is The required line also has gradient

and passes

Parallel lines have the same gradient.

through the point :

You know and one pair of values.



EXERCISE M

to

EXERCISE M 1 Find the gradient and -intercept of these straight lines: a b c d e f 2 Find the equation of a line: a through

with gradient

b through

with gradient

c through

with gradient

3 Find in the form a a line parallel to b a line parallel to c a line parallel to



the equation of: through through through

Gateway to A Level – GCSE revision N. Circle theorems You need to know this terminology:

        

These theorems about angles in circles can be useful in solving problems about diameters and tangents: 1

The angle in a semicircle is

2

Opposite angles in a cyclic quadrilateral (a quadrilateral with all four corners on the circumference of a circle) add up to

3

A tangent meets its radius at right angles.

EXERCISE N

EXERCISE N 1 Find the value of the angle marked in these diagrams, giving reasons for your answers: a

b

c



Gateway to A Level – GCSE revision O. Pythagoras and trigonometry There is a convention for labelling triangles with the angles in capital letters opposite the side with the same letter but in lower case. In a right-angled triangle the longest side is always opposite the right angle and it is called the hypotenuse. These rules only apply to right-angled triangles:

Key point

Pythagoras’ theorem:

Trigonometric ratios:

Pythagoras’ theorem also works in reverse – if you have a triangle with with the right angle at

then it is right-angled

If you know the lengths in a right-angled triangle you can then use these to find the angles. To do this you have to ‘undo’ one of the trigonometric ratios to get just using the or buttons on your calculator.

Worked example 1 Find the length in this diagram:

Solution

Comments First decide which trig ratio to use. The relevant sides are the side opposite the angle and the hypotenuse – this means sin.

Worked example 2 Find the angle in this diagram:

Solution

Comments First decide which trig ratio to use. The relevant sides are the side opposite the angle and the side adjacent to the angle – this means tan. To find you need to undo the tangent operation.

EXERCISE O 1 Find the unknown lengths to :

2 Find the unknown angles to :

   

   

3 A ship is lighthouse.



east and

north of a lighthouse. Find the bearing of the ship from the

Gateway to A Level – GCSE revision P. Area and volume You need to know how to find the areas of common shapes: Area of a parallelogram Area of a triangle

, where is the base, is the height

Area of a trapezium Area of a circle

, where is the base, is the height

where and are the parallel sides, is the height , where is the radius.

You also need to be able to find volumes of these objects: Volume of a pyramid Volume of a cuboid Volume of a cylinder

, where is the length, is the width, is the height , where is the radius, is the height

Volume of a cone

, where is the radius, is the height

Volume of a sphere

, where is the radius.

You sometimes need to calculate curved surface areas: Area of the curved surface of a cylinder Area of the curved surface of a cone side.

where is the radius, is the height where is the radius and is the length of the sloping

Worked example 1 Find the area of this shape:

Solution

Comments

Quadrilateral with two parallel sides: trapezium

Identify the shape.

Find what is needed for the formula. Quote formula, then substitute the values and calculate.

Worked example 2 A cylinder has height Solution

and curved surface area

Find its volume.

Comments Use the formula for the curved surface area to find the radius.



Now use radius and height to find the volume.

EXERCISE P 1 Find the area of each of these shapes: a

b

c

d

2 Find the volume of each of these symmetrical shapes:

a

b

c

d





Gateway to A Level – GCSE revision Q. Working with formulae A formula is a set of instructions for finding one variable if you know some others. If you treat a formula as an equation you can change the subject of the formula from one variable to another. There are a few useful ideas you need to bear in mind when you are trying to do this: If the new subject occurs in any squares or square roots, isolate these before undoing them. Get all the terms involving the new subject on one side and everything else on the other side. Then factorise.

Worked example 1 Make the subject of the formula Solution

Comments Get rid of fractions and expand brackets. Get everything involving on one side, everything else on the other side, then factorise.

It is good practice to have the subject on the left-hand side.

     

Another important tool is substituting one formula into another.

Worked example 2 and Solution

find in terms of Comments Simplify the expression.

EXERCISE Q

EXERCISE Q 1 a If b If c If

find when find when find when

and

2 Make the subject of each of these formulae: a b c d e f g h i 3 Write in terms of , giving your answer in a form without brackets: a b c d e f



Gateway to A Level – GCSE revision R. Vectors You can represent vectors either by drawing arrows or by using the column vector notation.

Worked example 1

a b

Write the vector

as a column vector.

Represent the vector

Solution

on the grid.

Comments The top number shows the number of units to the right. The bottom number shows the number of units up. The vector goes unit down, so it’s a negative number. The top number is negative, so the vector goes to the left and up.

You can add vectors by joining the endpoint of one vector to the start of another. You can use this to express vectors in terms of other vectors.

Worked example 2 The line

is parallel to

Express the vector

and

in terms of vectors and

Solution

Comments Going from to is in the same direction but twice the distance as from

  

EXERCISE R

to

You can get from

to by going from

to and then from to

EXERCISE R 1 Write these as column vectors:

2 a

is parallel to

and

b

is parallel to

and

Express in terms of and : i ii c Express



in terms of and

Express



in terms of and

is parallel to

and

Gateway to A Level – GCSE revision S. Travel graphs For an object moving in a straight line, you can represent the displacement, velocity and acceleration on a travel graph. These all have time on the horizontal axis. For a displacement–time graph: the vertical axis shows the displacement from the starting point the gradient gives the velocity. For a velocity–time graph: the vertical axis shows velocity the gradient gives acceleration the area under the graph gives distance travelled.

Worked example 1

Use this displacement–time graph to find: a

the distance travelled by the object between

b

the speed of the object when

Solution   The displacement changes from , so the distance travelled is

  

Worked example 2

and

Comments to

The object is always moving in the same direction, so the distance is the difference between the initial and the final displacements. The speed is the gradient of the graph. You need to look at the part of the graph between and

Use this velocity–time graph to find: a

the acceleration of the object between

b

the length of time during which the object is at rest

c

the distance travelled during the first seconds.

and

Solution

Comments The acceleration is the gradient of

  

the graph.   The object is at rest between

and

, so for

The object is at rest when

seconds.   

EXERCISE S

The distance is the area under the graph. This is a triangle with base and height

EXERCISE S 1 a Use this distance–time graph to find:

i

the velocity of the object during the first

ii

between which times the object is at rest.

seconds

b Is the object moving towards or away from the starting point when 2 Use this speed–time graph to find:

a the acceleration of the object when b the distance travelled in the first



seconds.

?

Gateway to A Level – GCSE revision T. Statistical diagrams You should be able to interpret different types of statistical diagrams. Pie charts Pie charts represent proportions by sizes of sectors. Bar charts Bar charts represent frequencies by the heights of bars. Pictograms Frequencies are proportional to the number of icons displayed in a pictogram. Stem-and-leaf diagrams In a stem-and-leaf diagram, the last one (or occasionally two) significant figures of each data item is listed from a ‘stem’ of all previous significant figures. Line graphs A line graph is normally used with time along the -axis and the values recorded along the -axis. Consecutive data points are joined by a straight line, representing an assumption that the data changes at a constant rate in the time interval.

Worked example This pie chart and bar chart show the types of TV shows watched by girls and boys in a class.

a In which group did a greater proportion of students watch soap operas? b Did more boys than girls watch sport? Solution a Soap is the least popular category for boys but the most popular for girls so a greater proportion of girls watch soaps.

b You cannot answer this because the pie chart shows proportions not

Comments

numbers.

EXERCISE T 1 This pie chart represents the favourite colours of

-year-old students.

  a How many students have each favourite colour? i

yellow

ii

blue

iii red b In a survey of the favourite colour of Colour

-year-olds, the results were:

Number

Blue Yellow Red In a pie chart, what angle would the sector representing each of these colours take up? i

blue

ii

yellow

iii red c Draw bar charts showing the results for



-year-olds and

-year-olds.

Gateway to A Level – GCSE revision U. Statistical measures Mean, median and mode Mean, median and mode are all measures of the centre of a group of data. The mean is the total of all the data divided by the number of data items. The median is the value of the middle item when all the data is arranged in order. If there are two middle items, the mean of these two is taken. The mode is the value of the most common data item. There may be more than one mode, or no mode at all. Range Range is a measure of the spread of the data. The range is the difference between the smallest and the largest data value.

Worked example For the data a

mean

b

median

c

mode

d

range

find its:

Solution

Comments

a

Add up all the data. data items

Count how many data items there are. Find the ratio of these.

b

Arrange the data in order. With data items, the rd item is in the middle.

Decide which is the middle data item.

Median is c

Mode is

Decide which data item has occurred most often.

d

The range is

Identify the smallest and the largest data values.

EXERCISE U

EXERCISE U 1 Find: i

the mean

ii

the median

iii the mode iv

the range

of these data: a b c d e f



Gateway to A Level – GCSE revision V. Probability One way to find probabilities is to list all possible outcomes. This can be in the form of a list, table or grid. The list of all possible outcomes is called a sample space. This method only works if all the outcomes are equally likely.

Worked example 1 a

List the sample space for the sexes of a two-child family, assuming no twins.

b

Hence find the theoretical probability that the two children are a boy and a girl.

Solution

Comments

a

A table is a systematic way to list the possibilities.

First child

Second child

boy

boy

boy

girl

girl

boy

girl

girl

There are four outcomes, and two of them are a boy and a girl.

b

Tip Don’t forget that since you can distinguish between the boy first–girl second and girl first– boy second cases; you must count them separately in the sample space.

Worked example 2 What is the probability of getting a sum of when two fair dice are rolled? Solution

Comments Dice A

Dice

Draw a sample space grid diagram showing all possible totals when two dice are rolled.

B

There are

items in the sample space,

Count how many items are in the sample space.

each of equal probability. There are sums of , therefore the

Count how many sums of there are.

probability of a sum of is

Tip A ‘ ’ on dice and a ‘ ’ on dice has to be counted as a separate event from a ‘ ’ on dice and a ‘ ’ on dice However, a ‘ ’ on dice and a ‘ ’ on dice is only one event. This often causes confusion.

EXERCISE V 1 List the sample space for: a a fair six-sided dice b rearrangements of the word ‘RED’ c the sexes of a three-child family d a six-sided dice with three ‘ ’s and the numbers from to 2 Two fair six-sided dice numbered to are rolled. By drawing probability grid diagrams find the probability that: a the sum is b the product is greater than or equal to c the product is

or

d the largest value is



Gateway to A Level – GCSE revision W. Tree diagrams Listing all possible outcomes only helps calculate probabilities when the outcomes are equally likely. When this is not the case, tree diagrams can help find the probabilities instead. The rule for calculating probabilities from a tree diagram is: multiply along branches add between branches.

Worked example 1 In a particular species of animal, the probability of a female offspring is Find the probability that there is a male and a female.

A female has two offspring.

Solution

Comments



There are two possible outcomes, but they are not equally likely. So listing all possible combinations in a table wouldn’t give the correct answer. In a tree diagram, the first set of branches represents possible outcomes for the first offspring, and the second set the possible outcomes for the second offspring.

There are two ways to get a male and a female offspring. Multiply along branches and add between branches. The probabilities on the two sets of branches don’t have to be the same; the probabilities can change depending on the outcome of the first event.

Worked example 2 A bag contains red balls and green balls. Two balls are chosen at random and not replaced. Find the probability that the two balls are different colours. Solution

Comments



The first ball is chosen out of The second ball is chosen out of ; but the probability of it being red depends on the colour of the first ball.

There are two ways to get one red and one green ball.

  

EXERCISE W 1 The probability that Daniel is late for school on any given day is is late on exactly one out of two consecutive days. 2 A box contains triangular tokens and

Find the probability that he

square tokens.

a Two tokens are picked at random and not replaced. Find the probability that they are both square tokens. b How does the answer change if the first token is replaced in the box? 3 The probability that it is windy on any given day is If it is windy, Rosie walks to work with probability and cycle with probability If it is not windy, Rosie cycles with probability and walk with probability What is the probability that Rosie cycles to work on a randomly selected day?



Gateway to A Level – GCSE revision X. Sequences A sequence is a list of numbers or terms, for example, The first term is usually denoted by

, the second term by

, ... and so on.

You should be familiar with two ways of defining a sequence: by a position-to-term rule (also called ‘the formula for the th term’), for example, by a term-to-term rule, for example,

.

Worked example 1 a b

Find the tenth term of the sequence defined by

.

Find the second and third terms of the sequence defined by

Solution

Comments

a

Substitute

b

Use

.

into the formula.

with

to link

to

.

Now substitute in as given in the definition of the sequence. Use

with

to link

... and substitute in

to

...

.

A particular type of sequence you have met is a linear or arithmetic sequence. This is one in which there is a common difference between terms. For example: ... is an arithmetic sequence with common difference ... is an arithmetic sequence with common difference The th term formula of an arithmetic sequence is always of the form difference.

. , where is the common

You may have used the method shown in Worked example 2 to find the th term formula.

Worked example 2 Find a formula for the th term of: a

, ...

b

, ... Solution

Comments

a

Common difference is so so

where Taking

.

, you can see that you need to

subtract from to get

b

Common difference is

where

.

.

so so

Taking to

, you can see that you need to add

to get

.

Tip At A Level you will use a slightly different method to find the th term formula.

EXERCISE X

EXERCISE X 1 Find the first five terms of each sequence defined by a position to term rule. a b c d e f g h 2 Find

and

for each sequence defined by a term-to-term rule.

a b c d e f 3 Find the th term formula for each arithmetic sequence. a

, ...

b

, ...

c

...

d

, ...

e f



...

Gateway to A Level – GCSE revision Y. Adding algebraic fractions To add algebraic fractions, always find the lowest common denominator first.

Worked example 1 Write

as a single fraction.

Solution

Comments You need to find a common denominator before you can add fractions. One that will always work is the product of the two denominators – in this case

.

You can create equivalent fractions with this denominator by multiplying top and bottom of the first fraction by

and

multiplying top and bottom of the second fraction by

.

Once you have a common denominator you can add fractions by just adding the numerators.

Worked example 2 Write Solution

as a single fraction. Comments To find the common denominator, you need to factorise both denominators first. In this case, multiplying the two denominators together gives

.

However, you can find a lower common denominator by using

.

Remember that the minus sign applies to both and in the second fraction!

Worked example 3 Solve Solution

Comments

Add together the fractions on the LHS in the normal way: here the common denominator is

.

Multiply through by the common denominator to remove the fraction. Rearrange and solve the quadratic equation.

Tip In Worked example 3, instead of adding the fractions you could multiply through the entire equation by

EXERCISE Y

at the beginning.)

EXERCISE Y 1 Write each expression as a single fraction. a i ii b i ii c i ii d i ii e i ii f

i ii

2 Solve these equations. a b c d



Gateway to A Level – GCSE revision Z. Exact values of trigonometric functions You can find the exact values of sin, cos and tan of that contain these angles.

and

by constructing right-angled triangles

Worked example 1 Find the exact values of

and

Solution

. Comments Draw a right-angled triangle that has a angle (which makes it isosceles). You can choose any length for the sides of the triangle. Let

.

Use Pythagoras‘ theorem to find

.

Now use the definitions of sin, cos and tan.



Rationalise the denominator of



:

The results for other values are summarised in this table.

Not defined

You need to be able to use these values in more complicated expressions.

Worked example 2 Show that Solution

. Comments Use

and

Rationalise the denominator of

. .

Create a common denominator and add the fractions.

EXERCISE Z Do not use your calculator in this exercise. 1 Evaluate each expression, simplifying as far as possible. a b c d 2 Show that: a b c d



.

Gateway to A Level – GCSE revision answers A. Expanding brackets and factorising 1 a b c d e f 2 a b c d e f g h i 3 a b c d e f 4 a b c d e f

Gateway to A Level – GCSE revision answers B. Factorising quadratics 1 a b c d 2 a b c d 3 a b c d 4 a b c d e f g h

Gateway to A Level – GCSE revision answers C. Solving quadratic equations 1 a b c d e f g h i j 2 a b c d e f 3 a b c d e f g h

Gateway to A Level – GCSE revision answers D. Types of numbers 1 a b c d e f g h

Gateway to A Level – GCSE revision answers E. Angles, polygons and bearings 1 2 a b

Gateway to A Level – GCSE revision answers F. Functions 1 a i ii iii b i ii iii c i ii iii d i ii iii e i ii iii f

i ii iii

2 a i ii iii b i ii iii c i ii iii d i ii iii e i

ii iii

Gateway to A Level – GCSE revision answers G. Rules of indices 1 a False b False c True d True e False f False g True h False 2 a b c d 3 a i ii b i ii c i ii d i ii e i ii f

i ii

g i ii h i ii 4 a

b c d 5 a b

Gateway to A Level – GCSE revision answers H. Surds 1 a b c d e f 2 a b c d e f 3 a b c d e f

Gateway to A Level – GCSE revision answers I. Linear equations and inequalities 1 a b c d e f 2 a b c d

Gateway to A Level – GCSE revision answers J. Simultaneous equations 1 a b c d e f 2 a b



or or

c

or

d

or

Gateway to A Level – GCSE revision answers K. Direct and inverse proportion 1 a b c d e f 2 a b c d

Gateway to A Level – GCSE revision answers L. Using coordinates 1 a b c 2 a b c 3 a b c

Gateway to A Level – GCSE revision answers M. Straight line graphs 1 a b c d e f 2 a b c 3 a b c

Gateway to A Level – GCSE revision answers N. Circle theorems 1 a



(right-angled triangle)

b

(opposite angles in cyclic quadrilateral are complementary)

c

(radius meets tangent at right-angle, angle in a semicircle is

)

Gateway to A Level – GCSE revision answers O. Pythagoras and trigonometry 1 a b c d e f 2 a b c d e f 3

Gateway to A Level – GCSE revision answers P. Area and volume 1 a b c d 2 a b c d

Gateway to A Level – GCSE revision answers Q. Working with formulae 1 a b c 2 a b c d e f g h i 3 a b c d e f

Gateway to A Level – GCSE revision answers R. Vectors 1 a b c d 2 a b i ii c

Gateway to A Level – GCSE revision answers S. Travel graphs 1 a i ii between

and

b Towards the starting point 2 a b

Gateway to A Level – GCSE revision answers T. Statistical diagrams 1 a

i. ii. iii.

b i ii iii c



Gateway to A Level – GCSE revision answers U. Statistical measures 1 a i ii iii iv b i ii iii iv c i ii iii iv d i ii iii None iv e i ii iii None iv f

i ii iii iv



Gateway to A Level – GCSE revision answers V. Probability 1 a b c d 2 a b c d

Gateway to A Level – GCSE revision answers W. Tree diagrams 1 2 a b 3

Gateway to A Level – GCSE revision answers X. Sequences 1 a b c d e f g h 2 a b c d e f 3 a b c d e f

Gateway to A Level – GCSE revision answers Y. Adding algebraic fractions 1 a i ii b i ii c i ii d i ii e i ii f

i ii

2 a b c d

Gateway to A Level – GCSE revision answers Z. Exact values of trigonometric functions 1 a b c d 2

Chapter 1 Worked solutions for book chapters 1 Proof and mathematical communication Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 1A 1 When

it is again the case that

(as it is for

for any integer ).

2 by definition of displacement, so

So

has only one solution, at

.

.

Therefore the particle never again has zero displacement from its original position. 3 a

b If

then it is still the case that

, so the implication does not hold in the opposite

direction. 4 5

Tip Since the question mentions multiples of , it is a good idea to consider any integer n as belonging to one of four groups. It is either a multiple of , one more than a multiple of , two more than a multiple of or three more than a multiple of . Note that the last group could also be thought of as one less than a multiple of . For any integer , there are four cases to consider: Case 1:

which is a multiple of . Case 2:

which is one more than a multiple of . Case 3:

which is a multiple of . Case 4:

which is one more than a multiple of . So any square number is either a multiple of or one more than a multiple of . 6 The statement is false. For example: Mean So

, mode but

.

, median

, contrary to the given statement.

7 a

so By the cosine rule in triangle

b In right-angled triangle c

:

:

so

8

a

b

So

and are parallel and of equal length, which is a sufficient condition to show that is a parallelogram.

c If So

is a rectangle then

is perpendicular to

Then

9 a For example: ln

ln

ln

are perpendicular.

.

is perpendicular to

That is, the diagonals of

ln

and

.

must be perpendicular.

b ln

ln

ln

ln

Taking each side as an exponent of e:

, since then

, so division by

is valid here without restriction.

c The steps are valid in both directions as long as any arguments of the logarithm lie within its domain. However, if, for example,

and

, then although

, the statement ln

ln ln is problematic, because you require that a logarithm only takes a positive value as its argument.

Tip If you have studied further mathematics, you will know that the logarithm of a negative number can be calculated as a complex number. In this example, ln ln and ln ln ln so you still have ln ln ln . However, even if you allow that the arguments can be negative, you must still require that they are non-zero, so remains a problem. Although

at the origin, you cannot calculate ln 

as any finite

value. 10 a b

is prime if it has no factors other than itself and . From part a you know that can only be prime if Then

and

are factors.

or

If

then

which is not prime.

If

then

which is prime.

If Thus 11 If

then

must be greater than , so

. .

which is not prime.

is prime if and only if

.

then

for an integer such that

So the sum of the probabilities is

.

Consider now the expansion of the binomial expression

.

The binomial theorem gives that this equals

.

Rewriting as , this is the same expression as the sum of the probabilities.

EXERCISE 1B 1 Suppose that is odd. Then

for some integer . so

is odd.

.

You can conclude that

.

For even , suppose that is odd. As already shown, that would mean that contradicting the initial statement, meaning that cannot be odd. You can conclude that if 2 Suppose that

must be odd,

is even, must also be even.

where and are integers with no common factors.

Squaring both sides: This means that is a multiple of so must also be a multiple of . You can then write: Substituting this into

, so :

.

This means that is a multiple of , so must be a multiple of . But it has been shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect. Therefore

cannot be written as , so is not rational.

3 Suppose that there is a finite number of even numbers. This finite list can then be ordered so that Then the largest even number is But

.

.

would also be even and clearly greater than

(The same argument could be made using

, so is not in the list.

.)

Therefore there are more than even numbers. This contradicts the initial proposition. Conclude that there are infinitely many even numbers.

Tip Be careful to ensure that the contradiction applies directly to the initial assumptions. In the proof in question , you assume a finite number of even numbers and produce a contradiction showing that there must be more. Alternatively, you could assume a largest even number and show that there must be a larger one. Either approach is fine, but you must not mix them!

4 Suppose that there exists a rational number ( and coprime integers) and an irrational number such that their sum Then

is rational.

for some coprime and .

Tip Two integers are coprime if their highest common factor is .

But then

which is rational, since

and are integers.

This contradicts the initial supposition. Therefore there is no example of a rational and an irrational with a sum that is rational. Conclude that the sum of a rational and irrational number is always irrational. 5 Suppose that integers and are both odd and

is even.

odd means that

for some integer .

odd means that

for some integer .

Then So is one more than an even number and is therefore odd, which contradicts the initial statement. Conclude that if

is even for integers and , at least one of and must also be even.

6 Suppose that

where and are coprime integers.

Cubing both sides: This means that is a multiple of so must also be a multiple of . You can then write:

, so

Substituting this into

.

:

.

This means that is a multiple of , so must be a multiple of . But it has been shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect. Therefore

cannot be written as , so is not rational.

7 Suppose that

where and are coprime integers,

Taking each side as a power of gives

.

.

is an integer, so raising each side to the power No integer power of is a multiple of , so

can only be true if

.

This contradicts the original assertion. Therefore 8

cannot be written as , so is not rational.

is a composite integer, so has at least one prime factor where Suppose has no prime factors less than or equal to Therefore

.

.

Since is a factor of not equal to , there is an integer Then

.

so

such that

.

.

If is prime, this contradicts the supposition that no prime factor is less than or equal to If is not prime, then it must have a prime factor

.

.

But if is a factor of then it must also be a factor of . Since

, this again contradicts the statement that no prime factor is less than

.

Conclude that a composite integer must have at least one prime factor less than or equal to 9 Suppose that at most two dates are within each month. Then at most

dates are chosen, which conflicts with the initial information.

Conclude that in at least one month there must be at least three dates chosen. 10 Suppose Then Then

for some integers and . .

is a multiple of and so must be a multiple of .

.

for some integer . Substituting:

.

The left side is a multiple of but the right side is not. Conclude that there are no integers and such that . 11 a Substituting

into the cubic:

Multiplying through by : b If and are both odd then each term on the left side must be odd (products of odd values are themselves odd). So

and

for some integers

and .

Substituting: The left side is odd but the right side is zero, which is even. Conclude that there can be no solution with and both odd. c If one of and is odd and the other even then two of the terms on the left side must be even and the third must be odd. Again, the left side (even even odd) must be odd, but this is inconsistent with the right side equalling zero. The only possibility then is that both and are even, but this means that and have a common multiple and are therefore not coprime. If there is a rational solution to the cubic then that rational solution must have a simplest form, where numerator and denominator are coprime. The working shows that they cannot be coprime, which contradicts the assertion. Conclude that there are no rational solutions to 12 Suppose that a triangle with sides

and has

. .

By the cosine rule, the angle is such that

.

Comparing it to the initial statement, you see that Reject

or

Therefore

.

, since then the shape is not a triangle, but a line. .

Since is an internal angle of a planar triangle, The only solution to

in that interval is

. .

Tip Using trigonometry can feel like a circular argument, since proofs of many trigonometric laws refer to some properties of right-angled triangles, such as this one. In such a question, you should feel free to use standard results unless specifically told otherwise.

EXERCISE 1C

EXERCISE 1C 1 a b In line the double implication is incorrect. From line , the correct equivalent line would be , and this explains the introduced solution . 2 a If

then both

and

b The correct symbol is solution.

so

is a solution to the original equation.

(if), which shows that

is a solution but not necessarily the only

3 a b In line the implication sign is incorrect, and should be an ‘only if’ . When multiplying both sides of the equation by in line , a false solution is introduced into the work. In fact, noticing that the numerator and denominator have a common factor faster solution would show:

in line , a

4 In line the implication sign is incorrect, and should be an ‘only if’ . When combining two logs using the rule , the rule assumes that both and are known to be positive. Since you have no such information here, a forward implication sign should be used, or (as in this case) a false solution may be introduced. 5 From line to line Andrew takes the positive square root but omits the negative root. Line should read , or the statement ‘so is a stationary point’ (it is acceptable to mention only since that is the focus of the question, but there should not be a suggestion that it is unique). Line is an error in understanding; when taking the derivative, the function expression should be used, not the derivative of a value (which is always zero). Having supposedly found that the second derivative is zero, Andrew makes the incorrect assertion in line that this shows that the point is a minimum on the curve. Line onwards should read:

so

is a minimum.

6 It is always necessary to compare all coefficients to ensure that all criteria are found and are consistent. Only the term is evaluated. Clearly the coefficients of

and

Comparing coefficients of

:

are consistent so check

However, this is not necessarily true if

(as that would fix

So, the solution given for the other factor being 7 Line is false:

Mixed practice 1

to complete the solution.

shows that is even, not .

.

is only true if

.

1

is a solution but not necessarily the only one. (Answer C)

2 If is even then this product is even. If is odd then way, the product is even. 3 Suppose that

is even, so the product is even. Either

where and are integers with no common factors.

Squaring both sides: This means that is a multiple of so must also be a multiple of . You can then write: Substituting this into

, so

.

:

.

This means that is a multiple of , so must be a multiple of . But you have shown that both and are multiples of , so they share a factor of . This contradicts the original assertion, so it must be incorrect. Therefore

cannot be written as , so is not rational.

4 Suppose that there are only a finite number of square numbers. Let the total number of squares be . Then the squares can be listed in increasing order square number. But

, so that

is a square number, which must be greater than

So there is a square larger than

, since

is the largest .

, which means that there are more than squares.

This contradicts the initial statement. Conclude that there must be infinitely many square numbers. 5 Moving from line to line , both sides are multiplied by

.

Whenever two sides of an equation are multiplied by a function , the roots of may be introduced as additional solutions, if not already solutions of the original problem, and the implication sign has to be one way. Line should read either: . or (better) . Then from line to line contains the more critical error; both sides are divided by . Whenever two sides of an equation are divided by a common factor g , the roots of g should be noted as possible solutions of the problem to be carried forward in the working. Line should read (continuing from 2b): .

or

Tip In fact, in this case, multiplying by cos x does not introduce false solutions, but this does not change the fact that the working should have noted the possibility and excluded it. Then (even disregarding the omission of the solutions from there is a solution from the given interval of missing in line so lines and are not equivalent. Lines and (continuing from 3. given in this solution) should read:

or (for 6 For the given numbers, 7 Suppose that

)

is not prime for

. (Answer C)

where and are coprime integers,

Taking each side as a power of gives

.

.

is an integer, so raise each side to the power No integer power of five is a multiple of , so

. can only be true if

.

This contradicts the original assertion. Therefore log2 cannot be written as , so is not rational. 8

If

is a parallelogram then

so that

Since the displacement vector from to is a multiple of the displacement vector from to , it follows that and are collinear. 9 If is even then a multiple of .

is also even, so this is a product of two even numbers and must be

for some integer .

If is odd then be odd:

is also odd, so this is a product of two odd numbers and must also

for some integer .

Tip You could also check this by an exhaustive list of the four combinations of a and b being even and odd. By rearranging the equation you can reduce the number of relevant cases to only , saving working.

10 a

Tip It may be hard to make a coherent proof of something which seems obvious; break down your argument carefully and ensure that you do not use any circular reasoning.

Let

be a polynomial of order given by

First, note

with integer coefficients .

that the product of two integers is an integer

In consequence Also note that

is closed under multiplication).

an integer to an integer power is also an integer. the sum of two integers is an integer

is closed under addition).

Then for an integer : By By

is an integer for each value . is an integer for each value .

By repeated application of b

, the finite sum

is an integer.

is a polynomial with non-integer coefficients which nonetheless always gives an integer result for an integer value .

11 a Line has an incorrect implication sign (or lacks an additional condition). It should read either: . or, more usefully when working out the solution: .

and

(The condition ‘and

’ can be then added to each line up to .)

Line : an incorrect sign has been used in the bracket when completing the square. Carrying through from 1b: 4. Line : as it stands the implication sign should be the other way round: correcting for the incorrect sign and carrying through from the new line : 5.

. However,

and

The final line should then use the full information from to give only one solution: 6.

Tip To guarantee the final values are actually solutions of the original equation, you need symbols on every line of working. It is often clearer to work with this in mind by adding conditions as necessary as was done in 1b and subsequently here, rather than using the approach of 1a and then looking for places where solutions may have been lost or incorrect ‘solutions’ generated. b

12 Line is an error; by line is divisible by . So you can conclude that which does not indicate it is non-prime.

is divisible by ,

Tip Simply removing the from line will in fact give a valid proof − the working shows that for any r, there is a sequence of consecutive non-primes, from which can be deduced that an arbitrarily long sequence of consecutive composite (nonprime) numbers exists. 13 Let

for integers

and .

Suppose that and are both odd. Then

and

for some integers and . .

Clearly this is both even and not a multiple of . Now consider : If is odd then is odd − but

has been shown to be even.

If is even then for some so that shown not to be a multiple of . No integer exists that can equal

and is a multiple of , but

has been

, contradicting the initial statement.

Conclude that the initial statement must be incorrect and that at least one of or must be even. 14 a For right-angled triangle

But

with a right-angle at :

, so

Hence b Then:

So the product of all the LHS from these four lines equals , and is rational. 15

Tip There are many ways to prove this result, using a variety of trigonometric facts. Here are three simple methods, but there are many more. Method 1: Reducing to a single trigonometric function

Between

and

By this working, between

and

Method 2: Squaring and using known features

Between

and

and: From

is non-negative and

is non-positive, and in consequence:

  and

between

Taking square roots and using

:

and between

and

.

Method 3: Calculus Let At

and at

Taking the derivative of

. :

has a stationary point where In the interval .

so

.

, there is a single solution to this, at

, which corresponds to

Since is continuous in the interval, and has a single stationary point greater than the two interval end points, that stationary point represents a maximum and therefore is never less than the minimum of the two end points. Hence

in the interval.

Chapter 2 Worked solutions for book chapters 2 Functions Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 2A 5 ‘Square root’ is a function and so has a single output value for any input value. The square root of is defined as . Statement A is correct. A function could be defined as the negative square root; if consistently output the negative root of its input. .

then function

will

Tip This is a very important issue, frequently misunderstood. The solution to the equation substituted.

is

, since either value leads to a true statement when

The reason for writing the line of working is that the square root is specified as positive, so it is important to write to indicate both solutions.

EXERCISE 2B 5 a

b 6

ln Domain: Require the argument of the logarithm to be positive. so Range: Range of logarithm is the whole real line .

7 Domain: Require the argument of the root to be non-negative. ln  8

so

or

You cannot have division by zero, so

, , .

The square root only takes non-negative values. Require The domain of

is

.

or, in interval notation,

9 a This is a negative quadratic with roots at and

.

b A square root requires a non-negative argument. You require: From the graph: 10

ln takes only positive values, so you require: or

11 You cannot have division by zero:

.

The square root only takes non-negative values. You require either and Domain of

and or

is

or

or

and

and

.

.

, or, in interval notation,

.

12 a A square root only takes non-negative values. Require that takes only positive values. Require that i ii b

EXERCISE 2C

the domain is

.

.

.

the function has no domain, or, in other words, its domain is the empty set, .

EXERCISE 2C 6

If

then

So

or

.

7

8 a b 9 a Domain of

: You require non-negative values for

b Over the domain

,

takes all values except

.

and so the range of fg is

10 a The range of f is

.

The range of f lies within the domain of g, so gf is a valid composition. The range of g is

.

Values lie within the range of g but not within the domain of f, so fg is not a valid composition for the full domain of g. b You require the range of g to be limited to , or, alternatively,

so restrict the domain to .

11 By observation,

EXERCISE 2D 4 a b Since

, then (because f is stated as being one-to-one and so .

5

is well defined)

.

6 a

, or, equivalently, b The range of f is The domain of f is

so the domain of

is

so the range of

is

. .

7

8 Suppose that

, defined over the range of f.

That is, for any element in the range of h, element in the domain of h, . Then, by definition,

. Similar working shows that for any

.

9 a The graph of the inverse function is the original graph, reflected through the line

.

Tip Always be alert for axes with different scales; the scale is the same in this case, so is the usual line.

b The domain of The range of c

is the range of f:

.

is the domain of f (since f is one-to-one):

.

Tip Remember that the solutions to will lie on the line if the functions are always increasing. This is not the case for a decreasing function; for example, the function is self-inverse, with every value in the domain being a solution to , but only for does . Since fis an increasing function, The solutions are

10 a

,

.

will have solutions on the line

.

b 11

(Select the negative root since can only take non-positive values.)

f has an asymptote at

(as

or, in interval notation, Hence the domain of

and is always decreasing, so the range of f is

,

. is

.

Tip The domain is an essential part of the definition of a function. If you are asked to find a function, always specify its domain as part of your answer for anything other than the entire real line. 12 a

The range of f is , so the domain of

is also .

b 13 a The graph of the inverse function is the original graph, reflected through the line

b Since is an increasing function, all solutions of will also lie on the line . From the graph, this appears to be at

EXERCISE 2E

, intersections of the two lines,

, which you can show algebraically: so

.

EXERCISE 2E 3 a b The function has a turning point at with ln for which .

ln and is continuous. Therefore there are values and

The function is not one-to-one across its domain, so

is not defined.

c For ln , the function f is strictly decreasing the domain ln .

and so the function is one-to-one over

4

So You require that

.

Comparing these confirms that

.

Tip Alternatively, multiply out and rearrange to find for all values of , from which

. So .

5 You require that f is one-to-one for the inverse to be defined, and so there can be at most one stationary point in the cubic.

Tip Remember that a single stationary point on a cubic (as for example in ) will be a horizontal inflexion, which does not prevent it being one-to-one. However, if there are two stationary points, they must be a local maximum and a local minimum, and the function will not be one-to-one.

For there to be at most one real root for this quadratic, the discriminant

.

. The positive quadratic is less than zero between the roots, so .

Mixed practice 2 1 The reciprocal is a self-inverse function. (Answer B) 2 a

The range of f is b

so the domain of

is

.

, or, in interval notation,

The range of f is

, so the domain of

3 a The reflection of

in the line

is

.

gives the graph of

so is

.

b 4

(domain Then

, range

,

.

with the same domain and range as

For

restricted to the same domain as (domain

.

, the composite function

, range

is defined:

.

(selecting the positive root in accordance with the domain) 5 a b

for

c The graphs of a function and its inverse are reflections of each other in

.

d i Since

, choose the positive root.

ii The domain of is

so the range of

iii The range of is

is

so the domain of

is

. .

e so there are no solutions to this equation, since the roots are and 6 a The range of is

.

.

b i ii c i

ii (Select the positive root as must lie in the domain of f.)

Tip In such cases as this, you must explicitly reject the invalid solution – in an examination there is often an accuracy mark awarded for doing so. 7 a

is not one-to-one, because

b Let

for all and the domain is given as all real values.

But

so

Rewriting as a function of :

c The range of

d

is the domain of

only for

:

or

or (not defined) or The only solution is

.

8 a b (Select the positive root as the domain of is

c The domain of

is the range of f. Since

.)

, the domain of

is

.

9 a i ii The range of is

.

iii iv v b The value

is in the range of but not in the domain of , so

is not defined.

c i

Note that

is self-inverse.

ii The graphs of a function and its inverse are reflections of each other in iii

is self-inverse so the domain and range of all be the same value set. The domain of

iv The range of

is is

and the domain and range of

. .

10 a b It has a vertex at

.

, it is a positive quadratic, with -intercept at

.

must

c The range of d

is

. The range of

. Range of on a domain The range of

is

is

is

.

.

11 a i So ii The range of

is the domain of

iii The -intercept is

.

, the horizontal asymptote is

b i

ii

12 a b c

13 a

b Substituting for with :

c Subtracting

:

:

.

14 a The domain of than 2.

is

, so restrict the domain of g so that the range excludes values less

For the excluded interval

.

The positive quadratic is less than zero between the roots.

So

,

b The range of for domain

is

The range of for domain

is

.

.

So the range of the composite is

.

15 a So

for any value .

Hence, by definition,

is an odd function.

b It has rotational symmetry of c Let

about the origin.

. Then

By the definition given,

is an odd function.

d So the modulus is an even function. e An even function has reflective symmetry through the line f Let

(the

).

. Then

By the definition,

is an even function.

g Since a multiple of an odd function must be odd and a multiple of an even function must be even, as defined must be odd and But

.

.

must be even.

.

Any function is therefore the sum of an even function and an odd function.

Chapter 3 Worked solutions for book chapters 3 Further transformations of graphs Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 3A

6 a Vertical translation by :

becomes

:

Vertical stretch s.f. :

becomes

:

b Vertical stretch s.f. :

becomes

:

Vertical translation by :

becomes

c Horizontal stretch s.f. :

:

Replace with :

Horizontal translation by

:

Replace with

:

d Horizontal translation by

:

Replace with

:

Horizontal stretch s.f. :

Replace with :

7 a Horizontal stretch s.f. :

Replace with :

Vertical stretch s.f. :

becomes

:

Because the function is of a power of , a stretch along one axis is equivalent to a stretch along the other axis. For any scale factor

, a horizontal stretch with scale factor will be reversed by a vertical stretch with .

b Horizontal translation by Vertical stretch s.f. :

Replace with becomes

c Vertical translation by : To replace with

:

becomes

: :

:

requires a horizontal stretch with scale factor .

Tip It is a defining quality of the exponential curve that a horizontal translation is equivalent to a vertical stretch. A logarithm, being the inverse function to an exponential, has its graph and consequent properties reflected through ; for a logarithm, a horizontal stretch is equivalent to a vertical translation. 8 The graph connects a

to

to

: This is a translation The new graph connects

b

b

.

and horizontal stretch, factor relative to the to

to

. The new graph connects

: Vertical asymptote

, intercept

: Vertical asymptote This is a translation

to (

) to (

, followed

) to

.

.

, intercept

.

.

: This is a vertical stretch, scale factor , with reflection through the by a translation

9 a

to

to

.

and a vertical stretch, scale factor , relative to the

.

: Vertical asymptote

c

This is a translation

10

, intercept

.

and horizontal stretch, scale factor relative to the

: This is a translation

, replace with

.

The function becomes

.

Reflection through the

: multiply the function by

The function becomes 11

.

.

: Translation by

Reflection through

: replace with

: replace with

, add to function.

.

Horizontal stretch, scale factor : replace with

.

Comparing coefficients: 12 Reflection through

Translation by

: Replace with

: Replace with

.

, add to function.

Horizontal stretch, factor : Replace with .

.

Comparing coefficients: : : : 13 Vertical stretch, scale factor : multiply the function by .

Translation by

: replace with

, add to function.

Horizontal stretch, scale factor : replace with

.

14 a Replace with : horizontal stretch, scale factor . Multiply the function by : vertical stretch, scale factor . b The graph has asymptote

and intercept

.

After the two transformations, the new graph will have the same asymptote and an intercept at .

c From the previous graph, replace with The new graph will have asymptote

: translation by and intercept

. .

EXERCISE 3B 7 a The graph of

crosses the

at

.

b The graph in part a is translated units upwards.

8 The graph of

9

crosses the

at (

).

10

EXERCISE 3C

Tip You could solve the problems in this exercise either by a direct algebraic method (square the modulus and rearrange) or by drawing the graphs to determine which part of the modulus graph is involved in the intersection, and thereby eliminating the modulus altogether. If the problem involves an inequality you might find a graph useful. Both methods are shown for the first question, then the graphical method alone is shown for subsequent inequality questions. 3 Method 1: Graphical approach a

The line At

:

crosses the modulus graph once on each side.

At

:

b From the graph,

for

Method 2: Algebraic approach a Squaring both sides:

b Testing an intermediate value: when , the inequality required lies between the two boundary values.

is true, so the region

for 4 a

intersects both arms of At

:

so

At

:

so

.

b From the graph, the required region lies outside the interval

.

for 5 a

b There are no intersections; there is no solution to

.

6

The graph of is

lies above the graph of

for all values of , so the solution set for

.

7

Both intersections lie in the left arm of :

so

:

so

(intersection on

at

.

.

.

From the graph, the inequality

has solution set

8 Squaring: That is, for

, which is

so , the solution is

.

.

Mixed practice 3 1 a

to

:

Translation Points

b

and vertical stretch, factor .

and

on original will shift to

and

.

to Horizontal stretch, scale factor , reflection through the Points

and

on original will shift to

and

2 Squaring:

so

or

(Answer B)

3 Original curve: Translation

: replace with

.

Vertical stretch, scale factor : multiply the function by .

4

is transformed to Replace with

: translation

.

followed by translation .

.

Multiply the function by : vertical stretch, scale factor . 5 a

b There is one intersection on each arm of the modulus function. At

:

so

At

:

so

From the graph, the inequality

is satisfied between these two intersection values:

.

6

a Both intersections are on the right arm of the function At the first intersection point:

so

At the second intersection point: b From the graph, 7

Replace with Replace with

8

so

outside the interval

is transformed to : translation

.

:

or

.

. produces

.

: horizontal stretch, scale factor , produces

. (Answer D)

Both intersections lie on the left arm of At

:

so

At

:

so

.

From the graph, the inequality is satisfied outside the interval 9 a

is transformed to Replace with

:

or

.

.

: translation

.

Multiply the function by : vertical stretch, scale factor . b

is transformed to Replace with Add

: translation

.

to the function: translation

c

.

.

is transformed to

.

This is achieved by part b followed by part a: translation 10 a

followed by a vertical stretch, scale factor 3.

is transformed to

.

Multiply the function by : vertical stretch, scale factor . Multiply the function by

: reflection in the

Add to the function: translation

.

.

Tip The stretch and reflection can be taken in either order but the specified translation must occur last. Alternative correct answers would include: vertical stretch, factor , translation reflection through translation (in either order).)

b

, translation

then reflection through then vertical stretch, factor

then reflection through x-axis and vertical stretch, factor

At

:

so

At

:

so

c From the graph, the inequality holds in the interval enclosed by these two intersection values: . 11 a

is transformed to Replace with

b

: translation

has asymptote So

. .

and intercept

has asymptote

.

and intercept

. It also has an intercept at , ln .

takes the previous graph and reflects it in the This graph will have the same asymptote It will have axis intercepts at

and will pass through

and

and so is a translation

.

ii Translation

: replace with

Reflection in the

12 a

is transformed to Start from

.

.

: multiply the function by

.

, descending.

must be the left-most turning point on

If the only transformation prior to the reflection in the to

.

.

c i The image of the first (left-most) turning point the image graph .

takes

then translates it

.

is a translation, the translation

Replace with

: translation

.

The new curve: Multiply the function by : vertical stretch, scale factor . The new curve: Subtract from the function: translation

.

Tip Alternatively, translation

b When When

. , ln

so

The axis intercepts are 13

14



so

followed by vertical stretch, factor .

and

.

and or

.

Chapter 4 Worked solutions for book chapters 4 Sequences and series Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 4A 4 a

,

b

so ,

5 ,

a b If

,

,

then or

are the stable (equilibrium) values for this sequence.

From this working (and continuing on GDC), the sequence approaches oscillating.

as

,

Tip You can formally demonstrate this by taking

So the difference

changes from to

where is ‘small’:

.

For small , this will have a different sign (so the sequence oscillates either side of 3) and the distance in each sequential term will be smaller:

This shows that as long as does not take values strictly between and , each subsequent term in the series will be both closer to and on the other side of . 6

, The sequence limit (if finite) should be an equilibrium:

7

, a

so

b gives a sequence which decreases from

.

so

.

,

8 a

The sequence cycles through these five values, so

.

b 9 a If the series converges, it is to some value where

.

Tip

b

There are several approaches here. You could answer this question using observation on the formula and direct reasoning, or you could use algebra and find the range of values of for which gives with . This more analytical approach is not covered explicitly in the course but is acceptable in an examination. Method 1: Argument on the function Irrespective of the sign of

, all further terms in the sequence must be positive, since .

Moreover, the sequence for will be the same as the sequence for , so the convergence criteria will be symmetrical about .

for any value of

So , for any value ; the sequence is always increasing (so a start value greater than or less than must give rise to positive terms getting progressively further from .) However, if

⩽

then

, so the increasing sequence is bounded

above by and so converges. Method 2: Analysis of the error term Suppose

.

Then For consistent convergence, require that (Equality would imply that value .) For this to be the case, (squaring):

.

, which is already shown only to occur at the equilibrium

The solution to this is that

.

So progressive convergence occurs if

lies between

and

10

So

for all . The sequence is increasing.

EXERCISE 4B 3 ,

,

,

,

4 a ,

,

b

so 5

6 a

b The sum of every six consecutive terms in the sequence is zero.

EXERCISE 4C

, that is,

.

EXERCISE 4C 3 a

so

b Solve

:

The first 4

terms are less than

, the

st term is greater than

.

, so

5

,

Let

=

.

Then so 6

,

Let



.

Then so The ladder has 7

,

rungs. ,

,

Let be the common difference between consecutive terms. (1) (2) (3) Also, Substituting from

and

:

so 8 a The first pages are numbered with single digits for a total of digits. The

th and

th pages are each numbered with two digits.

The total for the first

pages is

digits.

b For the first pages: pages at digit per page, total For pages

:

For pages

There are There are in total

EXERCISE 4D

pages at digits per page, total :

pages at digits per page, total

three-digit pages. pages

EXERCISE 4D 3

,

so

(

is an integer, so reject the first factor. 4 a

,

so so

b

,

5

:

so

,

6

: 7

so

,

(

:

so

,

8 The odd numbers form an arithmetic sequence with

9

,

.

, For the largest possible sum, you require there to be all positive values in the sequence (and no negative values). Since

10

, it follows that

, so the largest possible sum is

,

: so The roots of

are

The least positive such that 11

and

.

is therefore

.

Tip If you are given a formula for calculate .

, then by substituting

you can immediately

Rearranging: Substituting

into

Then from

:

:

12

Substituting

into

:

so The smallest sector is

.

13 a The multiples of form an arithmetic sequence with

and

is the greatest multiple not greater than

.

, and is the

nd term in the

sequence. × b The sum of the integers from to arithmetic series with :

inclusive is the .

th triangular number (or

The difference between the two sums is the sum of all positive integers up to multiples of .

that are not

14 Rearranging:

.

Substituting:

All terms in the sequence are positive so

,

.

So 15 a

The multiple

is an arithmetic series with

,

and

.

The sum of the logarithms is therefore b 16

so

.

and

Tip You can approach this by considering this as either the sum of two simpler series or as the difference of two simpler series. Both methods are given here. Method 1: Sum of two series

You can consider this as the sum of two series:

for the

is the sum of values with form

.

is the sum of values with form

.

: a series with

,

,

: a series with

,

,

Method 2: Difference of two series Let be the sum of all three-digit multiples of Let

.

be the sum of all three digit multiples of

.

: a series with : a series with

EXERCISE 4E 3 a

: so

b So 4 a : so b 5 a

: Substituting into

:

so

or b 6

(1) : Then So

(2)

, ,

, ,

7 a

The negative sign indicates that is an even number, so

.

b ,

8

This describes a geometric sequence with common ratio

and first term

.

So Require the least integer such that

So the least integer is

.

.

is the first term greater than million. 9 This is a geometric sequence with first term

Require that

and common ratio

.

:

Tip when

so dividing through causes a reversal in the inequality.

The first term less than

is

.

10

Cancelling

Either

11

and factorising from the LHS:

, or, cancelling

,

Tip In sequences and series, letters , and normally have standard meanings, so take extra care in questions that use these letters in other ways and do not use any letter for more than one meaning! { , , } is an arithmetic progression so:

and

{ , , } is a geometric progression so: and

Substituting gives So

into

:

which contradicts the requirement

.

,

12

The common difference of the arithmetic sequence is

.

But

The geometric sequence is So sequence.) Then

. (Reject so

since this clearly does not lie in the (increasing) arithmetic .

EXERCISE 4F 3 a Comparing with the general form for a term from a geometric sequence

:

b

4 a

b

5

Comparing to the general term for the series sum

:

So

, ,

Using your GDC, there are two solutions:

or

.

6

:

Cancelling and cross-multiplying:

Reject

, since this will not give the sums stated.)

Using your GDC, there is only one root to this:

.

So

Tip Here is an alternative approach, using the summation formulae:

:

So Again you would need a calculator to find the root of this quartic, at the root at

which arises from cancelling the (

7 a

: : : b From

8 a

:

(1) (2)

factors.

, rejecting

: So b Require the smallest integer such that

.

So So the least integer is

.

9 a

: Rearranging: b Factorising: (

(

is a false solution introduced by cancelling the ( The only true solution is therefore

10 a

and

denominators in

, for which

and

.

.

so

That is, the sum of the second terms is a factor

greater than the sum of the first terms.

b

So

or

or

By inspection, this has a root (

(

, and factorising gives:

.

Since the quadratic factor has no roots (discriminant is negative), the only solution is From part a, the ratio of the first .

terms to the sum of the next

terms is

.

which is

EXERCISE 4G If you are told that the sum to infinity has a finite value, you can assume that information.

without further

If you are not told that the series converges, you should always explicitly show that answer when evaluating or using .

as part of your

3

:

4

:

5

:

6 a

: but you know that is positive so reject the negative value.

So

,

b

, 7 a

:

Each term of the series is positive so reject the negative solution.

Conclude: b 8

:

Each term of the series is positive so reject the negative solution. Conclude:

, so

.

9 , a Convergence criterion: (that is, | | < b

:

10

:

11 Comparing features:

and

a Convergence criterion:

b

so

and

Require the least integer such that

. .

Require at least seven terms for a sum greater than

.

12

(Replacing the dummy variable with to release for use as the common ratio.) f(

is an infinite geometric series with

and

.

a When

,

so

and the series will converge.

b When

,

so

and the series will not converge to a finite value.

is infinite. 13 a Convergence criterion: so b

so 14 a This is an infinite geometric series with

e–x and

.

Rearranging: Multiplying through by e : b

(no real solution) or

15

where, for convergence, But

ln 3

.

so

Then the convergence condition is

.

Then So

, as long as the sum to infinity is positive.

Tip Negative would result in the condition requirement that a .

EXERCISE 4H

, which contradicts the

1

Tip Be careful to define the terms you use; in finance questions it will often be critical whether you consider as the value at the start of year or the end of year n. If you are defining your own variables, always state this clearly at the start of the solution. a Let

represent the balance at the start of year .

This is a geometric sequence:

.

The interest for the sixth year is £

.

b The balance after six years is the balance at the start of the seventh year,

The balance after six years is £ 2 a Let

.

.

be Lars’ salary in the th year.

Ths is an arithmetic sequence:

.

In the twentieth year his salary will be £

.

b Require So

The roots of this positive quadratic are

and

.

He will have earned more than £ million after 22 years. 3 Let

be the balance at start of year . £

a After full years the balance is the same as at the start of year b The balance at the end of 5 years: £

.

£

c i ii The balance will exceed £ 4 Let

after

whole years.

be the number of seats in row .

This is an arithmetic sequence: a Require

.

:

The roots of this positive quadratic are

and

.

10 rows are required for there to be in excess of b

seats.

, The percentage of seats in the first rows is

5 a The balance at the start of year is £ £

. .

£ £

: £

.

b The balance at the end of month is £ Require £

It takes 6 Let

:

months.

be the number of miles run on day .

This is an arithmetic sequence: a Require

.

:

The roots of this positive quadratic are The total distance exceeds b Require

miles after

.

days.

:

He runs more than 7 Let

and

miles on day

.

and be the amount paid in by Aaron and Blake respectively at the end of the th month. is an arithmetic progression with

. Aaron’s balance is the sum

. is a geometric progression with

. Blake’s balance is the sum

Require the least integer such that

.

.

Let: and and both value of such that

. and g(

are increasing functions for all

, so there can be at most one

.

, , So,

and

Therefore the first time Blake has more than Aaron is after 8 Let

months.

be the height the ball rises on the th bounce, that is, after hitting the ground times.

This is a geometric sequence:

.

a b Total distance travelled at the end of bounce is:

The ball hits the ground for the th time at the end of bounce .

c The model predicts a height of . is sufficiently small that it is likely to be absorbed by model inaccuracies and physical interactions not considered in this simple model. 9 The account balance at the beginning of year is given by £

.

, a So

b The pattern shows that

is the sum of a geometric sequence with

,

.

c Require

:

After 29 years of saving, Samantha will have accumulated at least £

Mixed practice 4 1

so

So

r

2

):

3

:

4 a

:

(Answer B)

.

b

so the sum to infinity converges to a finite value.

5

, If

then

6 a Then

and

.

b c i

ii 7

:

(Answer A)

Tip

8

Take care when the sum is from zero instead of one.

,

,

This is an infinite geometric series:

.

(Answer A) 9 This is a geometric sequence:

.

Require the least integer such that

.

The least integer satisfying this condition is 10 Let

.

be the number of bricks in row , where row is the top row. ,

This is an arithmetic sequence with a b So

,

.

c

(Reject the solution

, in context.)

So the total number of bricks

.

11 Let £

be the amount after years in plan A and £

the amount after years in plan B.

Using your GDC: the intersection of the two graphs is at 12 a

, so for the first 19 years,

.

2:

So

or –3 b If

then

Since If

and

.

, the series does not converge as then

Since

and

. .

, the series converges as

:

13 This is the sum of two infinite geometric series: In

where

,

and

,

.

,

and

,

.

The sum to infinity

In

where

The sum to infinity The total value is therefore

.

14 The least multiple of greater than

Let

is

be an arithmetic sequence with

Then the sum of the multiples of between

15 a

,

,

and the greatest less than or equal to

, and

and so is:

,

A positive quadratic is less than zero between the roots.

.

is

.

b

Tip This is not obviously either arithmetic or geometric, but with some manipulation of the sum you can form it into a sum of a known series and prove the convergence. Alternatively, you can assume that there is convergence from the phrasing of the question and use the stability property to shortcut to the value. Method 1: Manipulation and proof of convergence For So

is the sum of a geometric term and a geometric series.

The series has

,

.

Hence Given

, as

,

Method 2: Assume convergence, use stability property If the series converges to a value then the iteration

will mean that

. Rearranging: Given

,

16 a

If the sequence tends to .

as

then (substituting

the iteration gives

Rearranging: Substituting b From

into

:

.

:

17 a

b : c From part a: Require such that

Integer

:

Then 18

,

Then

The rational function

is greater than zero for all values except

–2 < , so is positive for all positive integers . Consequently, each term is more than greater than the previous one, and the sequence must be increasing. 19 Let the geometric sequence be common difference .

and have common ratio , and the arithmetic sequence have

,

2:

2

(constant sequence) or

.

20 Let the geometric sequence be and have common ratio , and the arithmetic sequence have common difference and first term .

): ar( (or

which produces the constant zero sequence, for which the common ratio is

not defined.) To factorise this cubic, note that if then the two sequences are constant and the conditions would be met trivially, so is a factor. (

(

The quadratic term has roots

.

If the sum to infinity converges to a finite value then criteria. 21 Let the sequence be identified as

, where

so only the root

meets the

Comparing this to the general form for a term of an arithmetic sequence: seen to be an arithmetic sequence with

,

this is

.

Then

22 a b

integers The final integer on the th line is the th triangle number:

c The first integer on the th line must be

.

less than the final integer:

d This is an arithmetic sequence of consecutive values from

to

:

,

e Using your GDC: 23 a Consider the mortgage as held in one account ( ) and the payments into a separate account ( ). The mortgage account then just rises at its interest rate: is a geometric sequence with

,

. .

At the end of three years the mortgage account stands at The payments account works as previous payments and then a new £

.

, since each year interest is added to the payment is made.

Therefore is the geometric series with

,

.

At the end of three years, the payments account stands at

.

So, after three years, the outstanding mortgage balance is Balance at years b Continuing the pattern: Balance at years

c Require mortgage balance at years to be less than zero.

So the mortgage is paid off after 21 years.

.

.



Chapter 5 Worked solutions for book chapters 5 Rational functions and partial fractions Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 5A

Tip When proving factors for a polynomial expression, it is often useful to define the expression as first, so that you can express the factor theorem more easily, citing . 2 Let

.

Then By the factor theorem, if

then

So

is a factor of

for some integer

.

and .

Comparing coefficients: : : : :

.

3 Let

.

By the factor theorem, if

4 a By the factor theorem, if

b So

is a factor of

then

is a factor of

for some integer

then

.

.

and .

Tip Never reuse an unknown constant in a different context within the same question. You already have a value for a from the first part, so you must use a different letter for the quadratic coefficient of the other factor.

Comparing coefficients: : : : :

is consistent.

5 a Let

.

By the factor theorem, if b

then

is a factor of

for some integers

.

and .

Comparing coefficients: : : : :

.

6 By the factor theorem, if So

then

is a factor of

for some integers

.

and .

Comparing coefficients: : : : :

.

has solutions 7 Let If

If

9 a

.

. is a factor of

8 Let

or

then, by the factor theorem,

.

. is a factor of

, then, by the factor theorem,

By the factor theorem, if

and

are both factors,

.



(2) (1): Substituting into (1): b So

for some integers

and .

Expanding:

Comparing coefficients: : : : :

.

:



for so, by the factor theorem,

10

so, by the factor theorem, So

is a factor of for some integers

Comparing coefficients: : : : :

.

:

.

for

EXERCISE 5B

5

. and .

is a factor of is a factor of

. .

6

7 a b

or

or

8

Tip

9

Shown here are the long-division method, a gradual fraction rearrangement method equivalent to long division, and two approaches using a predicted general solution where unknowns are found either by substitution or by comparing coefficients. Make sure you understand how each works, then choose the one you prefer. Method 1: Long division

 

Quotient is , remainder is

.

Method 2: Converting improper rational function to polynomial plus proper rational function

Quotient is , remainder is

.

Method 3: General solution specified by substitution and Substituting Substituting

are both of degree :

:



for some constants and .

(1) (2)

Substituting (2) into (1): Quotient is , remainder is

.

Method 4: General solution specified by comparing coefficients and

are both of degree

Comparing coefficients: : : Quotient is 2, remainder is 25. 10

Tip Using method 2 from the solution to question 9.

Quotient is

, remainder is .

for some constants and .

11

Tip Using method 2 from the solution to question 9.

12 Let

.

Then

.

By the factor theorem, if

then

So

is a factor of

for some integers

.

and .

Comparing coefficients: : : : :

.

Then 13 a Let

.

Then

.

By the factor theorem, if

then

b So

is a factor of

for some integers

.

and .

Comparing coefficients: : : : :

.

Then 14

is a quadratic and

is linear.

So

for some integers

and .

You know that the remainder is . Comparing coefficients: : : :

The quotient 15

. where is the quotient function and is the remainder function.

Tip The remainder must be a lower order than If

then

So

is divisible by

so is a constant term in this case.

.

16 a Total distance: Total time:

hours

Mean speed over journey total distance ÷ total time b

So either or

(constant speed throughout the journey)

(the speed in the second section is double the speed in the first section).

EXERCISE 5C

3 Multiplying through by the RHS denominator:

Tip As detailed in the chapter, from here you can either compare coefficients or substitute convenient values. Be careful if you choose the values method if your general form is wrong you will still get a solution that seems to work. Comparing coefficients if you are thorough and compare all the coefficients will alert you to an error. Method 1: Substitute

Tip Try to pick either ‘easy’ values such as for simple simultaneous equations or deliberately select values of that eliminate one or other of the unknowns. Substitute

:

Substitute

:

Method 2: Compare coefficients Comparing coefficients: : : Substituting in (1): Then from (1):

(1)

4

for some constants and . Multiplying through by the LHS denominator: Substituting

:

Substituting

:

5

for some constants

and .

Multiplying through by the LHS denominator:

Substituting

:

Substituting

:

Substituting

:

6

for some constants

and .

Multiplying through by the LHS denominator:

Substituting

:

Substituting Substituting

: :

7 for some constants

and .

Multiplying through by the LHS denominator:

Substituting Substituting

: :

Substituting Substituting

: :

8 a b

for some constants and . Multiplying through by the LHS denominator:

Substituting

:

Substituting

:

9

for some constants and . Multiplying through by the LHS denominator: Substituting

:

Substituting

:

10

for some constants and . Multiplying through by the LHS denominator: Substituting Substituting

: :

11 a So b

for some constants and . Multiplying through by the LHS denominator:

12

Substituting

:

Substituting

:

Tip This question highlights the dangers of the substitution method for completing partial fractions. The question is answered first with substitution and then with the safer method of comparing variables. Method 1: Substitute Suppose

for some constants and .

Multiplying through by the LHS denominator: Substituting

:

Substituting

:

You might conclude incorrectly that

as shown by this working.

However, testing a third substitution shows this is not valid. For example, with , which is clearly false.

, you get that

The problem is that the proposed form

is not valid.

If you compare coefficients, it is immediately apparent that the proposed form is incorrect. Method 2: Compare coefficients Multiplying through by the LHS denominator: Comparing coefficients:

:

No values of or will make this true, so the proposed form is invalid. Solution In general, for any rational function, first rewrite (as in the solution to question 8) as the sum of a polynomial and a proper rational function, where the numerator has a lower degree than the denominator.

If the proper rational function has a denominator that factorises then you can split it into partial fractions.

Tip Within the scope of this specification, all such factors will be linear though if there are repeated factors you need a modified approach, as seen in Section 4 of this Chapter. for some constants and . Multiplying through by the LHS denominator: Substituting

:

Substituting

:

EXERCISE 5D

2

for some constants

and .

Multiplying through by the LHS denominator: Comparing coefficients: : : :

3

for some constants Multiplying through by the LHS denominator:

and .

Comparing coefficients: : : :

4

for some constants Multiplying through by the LHS denominator: Comparing coefficients: :



(1)

:



(2)

:



(3)

(3)

× (2):

Substituting in (1):

5 a Let

.

Then By the factor theorem, if b

then

is a factor of

for some constants

.

and .

Comparing coefficients: : : : :

Consistent.

c

for some constants Multiplying through by the LHS denominator:

Comparing coefficients: :



(1)

:



(2)

:



(3)

2 × (3) (2): Substituting in (1):

and .

and .

6 a Let

.

Then

.

By the factor theorem, if b

then

is a factor of

for some constants

.

and .

Comparing coefficients: : : : :

Consistent.

Then

.

Multiplying through by the LHS denominator:

Comparing coefficients: : : : 3 × (3)

× (2):

Substituting in (1):

7 a

is an improper rational function with numerator of degree 4 and denominator of degree 3, so you can rewrite it as the sum of a degree 1 polynomial and a proper rational function with numerator of lower degree than the denominator. Suppose

for some constants

Multiplying through by the LHS denominator:

Comparing coefficients: : : : : :

.

and .

Since all results are consistent, you can write b

for some constants

as and .

Multiplying through by the LHS denominator: Comparing coefficients: : : :

8 a So

and

.

b

for some constants

and .

Multiplying through by the LHS denominator: Comparing coefficients: :



(1)

:



:



(2) (3)

(3) (2):

9

for some constants Multiplying through by the LHS denominator:

Comparing coefficients: :



: :

(1)



(2) (3)

and .

.

Tip Even if this still holds true. In some other circumstances you might need to specify because repeated factors in the denominator must be treated differently. For example, consider this problem. Write

in partial fractions.

For distinct and , the solution is . This solution fails if

where , because then m and n are

infinite. For

, the solution is

.

Remember that if you are finding partial factors with an unknown value in the problem, set conditions or take separate cases to ensure that the constants you find cannot have a zero denominator.

Mixed practice 5 1

(Answer C)

2 a

b Let

.

Then

(

That is,

×

The prime factors of

. are

and

.

3 a Let Then

By the factor theorem, if

then

b So

for some integers

Expanding and comparing coefficients: : : : :

c



when

.

.

is a factor of and .

.

d

4

for some constants and . Multiplying through by the denominator of the LHS: Comparing coefficients: :



: (2)

(1)

(2)

× (1):

5

for some constants

and .

Multiplying through by the denominator of the LHS:

Comparing coefficients: :



:



:



(1) (2) (3)

From (3): (2)

× (1):

6

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients: : :

7

for some constants Multiplying through by the denominator of the LHS:

Comparing coefficients:

and .

: : :

8

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients: :



(1)

:



Substituting

(2)

into (2):

9

for some constants

and .

Multiplying through by the denominator of the LHS: (

(

(

Comparing coefficients: :



(1)

:



: (2) (4)

(2)

× (3):

(3)

(4)

× (1):

10 a

for some quotient function

and remainder .

The remainder will be a constant, as it must have a lower degree than the (linear) dividing factor . Then

.

But

.

So the remainder on division by

equals .

Tip You can also answer this question using standard long division or equating coefficients.

b i ii By the factor theorem, if So

then for some integers

is a factor of

.

and .

Expanding and comparing coefficients: : : : :

Consistent.

iii 11

for some constants and .

Tip The only solution of the correct form is A, so no further working is needed, but substituting gives and substituting gives , from which . (Answer A) 12 a

for some constants and . Multiplying through by the denominator of the LHS: Comparing coefficients: : :



(1) (2)

2 × (2) (1):

b

has numerator of degree 3 and denominator of degree 2 so you can express it as the sum of a linear function and a proper rational function.

c Applying the result from part a:

13 a

b

for some constants and . Multiplying through by the denominator of the LHS: Comparing coefficients: :



: (2)

(1)

(2)

× (1):

14 a Let

.

Then

.

By the factor theorem, if

then

b So

is a factor of

for some integers

.

and .

Expanding and comparing coefficients: : : : :

Consistent.

for

c

.

d for some constants Multiplying through by the denominator of the LHS:

Comparing coefficients: :



(1)

: : (2) (3): (4) (1):

15 a



(2)



(3)

(4)

and .

By the factor theorem, if

then

b So

is a factor of

for some constants

.

and .

Comparing coefficients: : : : :

Consistent.

for c for some constants

, and .

Multiplying through by the denominator of the LHS:

Comparing coefficients: :



:

(1)



:

(2)

(1)

× (3):

(4)

× (2):

(3)

(4)

16 a for some constants

and .

Multiplying through by the denominator of the LHS:

Comparing coefficients: : : : :

b

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients:

:



(1)

: (2)



(2)

× (1):

17 a By the factor theorem, if



is divisible by

and

, then

(1)

Substituting into (1):

.

b So

for some integers

and .

Expanding:

Comparing coefficients: : : : :

Consistent.

:

Consistent.

c

when

.

18 a

for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients: : : (2) (1):

b



(1)

(2)

.

19 a

Quotient is , remainder is .

b for some constants and . Multiplying through by the denominator of the LHS:

Comparing coefficients: :



: (2)

20 If

(1)

(2)

× (1):

then

Otherwise,

for some constants

Multiplying through by the denominator of the LHS:

Comparing coefficients: : : :

Tip This general form simplifies to when

21

.

for some constants and . Multiplying through by

:

and .

Comparing coefficients: : : : 22

for some constants and . Expanding and comparing coefficients: : : :

23 Given q So

q

r

r

then

q

q

q

.

24 Multiplying through by the denominator of the RHS:

Comparing coefficients: If Coefficient of 25 a

b

:

is linear then the coefficient of

on the RHS is zero.

Chapter 6 Worked solutions for book chapters 6 General binomial expansion Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 6A

Tip As with the standard binomial expansion, it is always wise to write out an expansion in its most general format first, bracketing elements where appropriate. Only after you have the essential structure written down clearly should you perform any multiplication, division or application of indices needed to calculate coefficients.

2

3

4 a

b Require

or, equivalently,

.

5 a

b Require

or, equivalently,

.

c d Let

in the expansion from part c, which is within the convergence interval.

Then

.

6 a

b Require

or, equivalently,

.

c i Let

in the expansion in part a, which is within the convergence interval in part b.

Then ii Let

in the expansion in part a, which is within the convergence interval in part b.

Then 7 The cubic term has the form So

and

.

8

Comparing coefficients:

So

,

From ( ): 9

Comparing coefficients:

.

So

,

From ( ): 10 Then

Comparing coefficients: Consistent with the first coefficient of the binomial Consistent with the second coefficient

.

Consistent with the third coefficient

.

EXERCISE 6B

1

2

For convergence, require that

or, equivalently,

.

3

4

5 a

for some constants and . Multiplying through by the denominator on the LHS:

Substituting

so

Substituting

so

.

b c

d Require

and

; the more stringent requirement is

or, equivalently,

so this is the convergence criterion for the whole expansion. 6 a

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

,

b

,

Tip It is perfectly reasonable to find the three expansions for the three rational functions in part a and sum them, and this is the intention in the question. It is slightly less work to amalgamate the second and third first, since their denominators consist of the same linear factor to different powers. Recombining the second and third fractions from part a:

,

c Require

and

; the more stringent requirement is

, so this is the convergence

criterion for the whole expansion. 7

Tip You can approach this question either using partial fractions and adding two expansions or by direct multiplication of the two expansions. Both methods are shown here; in this case, direct multiplication is not particularly arduous but there are many circumstances where the small amount of preparatory work establishing the partial fractions is preferable to the multiplication, particularly if more terms are required. Method 1: Direct multiplication

Method 2: Partial fractions for some constants and . Multiplying through by the denominator on the LHS:

Substituting

so

Substituting

so

8 a

b The first expansion is valid for , the second is valid for , which is a more restrictive condition. The whole expansion is therefore valid only for , equivalent to . c Substituting

:

9

Comparing coefficients:

so so

and

so

,

10

is not real for values convergence.

so even some cunning rearrangement does not allow

Tip If you study further mathematics, you might be interested to consider the complex rearrangement available:

This is indeed convergent for any complex such that

Mixed practice 6

.

1 A:

is valid for

B: C:

.

is valid for all ( is a positive integer so the expansion is finite). is valid for all ( is a positive integer so the expansion is finite).

D:

is valid for

or, equivalently,

(Answer D) 2

The expansion is valid for

, equivalent to

.

3

The expansion is valid for 4 a

so

, equivalent to ,

.

b

5

6 The quadratic term is 7 A: B: C: D: (Answer B)

. (Answer B)

.

.

8 a

b This expansion is valid for

, equivalent to

so substitute

9 a

.

, which is within the convergence interval.

for some constants , and . Multiplying through by the denominator on the LHS:

Comparing coefficients:

,

,

b

c The first expansion is valid for , equivalent to condition is 10

, equivalent to

. The second expansion is valid for

. The sum is valid where both these are valid; the more restrictive

. for some constants and .

Multiplying through by the denominator on the LHS:

Substituting

so

Substituting

The first expansion converges as long as

and the second expansion converges as long as

. The latter is the stronger condition so for the sum to converge, so require that which is equivalent to

.

11

Comparing coefficients:

Substituting

so

Substituting ,

or

so

and (

,

12 a i

ii Replacing with

b

in the expansion from part i:

,

13 a

b

Tip You can jump from the second line of working to the end line if you recognise that the expansion is identical to the one in part a, replacing with

c The expansion in part b is valid for

, equivalent to

.

.

Tip Always establish the convergence criterion for your expansion. You want to find and must be certain that the most obvious substitution is valid. lies in the interval of convergence for the expansion. Substituting

:

14 a

The product of the two expansions is

so

.

b Substituting

:

So 15

Comparing coefficients:

Substituting ( ) into ( ):

so

, Then from ( ): 16

Tip Be careful here – when the coefficient of turns out to be , you need to extend the expansion to to get three terms. At that point you need to go back and continue the expansion structure to include the fourth term. If you got as your third term you probably neglected to check this. 17 a

,

,

The expansion is valid for

.

b

c

,

,

The sum converges for

, equivalent to

d Using the expansion from part b with e

:

.

Using the expansion from part a with

:

18 a

The expansion is valid for:

b

where i

. so

So ii

is

greater than

.

so

So

is

greater than

.

c The th term in the expansion has the form:

That is, every term is positive, so the sequence increases the more terms are included in any approximation: .

Chapter 7 Worked solutions for book chapters 7 Radian measure Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 7A

11

12 13 14

So

EXERCISE 7B

7

so Primary solution: Secondary solution: Periodic solutions lie outside the required interval.

8

9 Let

then

Primary solution: Periodic solutions:

10 a

This is a quadratic in sin  ; using the quadratic formula:

Since

has range

, reject the negative root.

b Primary solution: Secondary solution: 11 or

Tip If you divide both sides of an equation by an expression (here sin or tan ) you must also allow for the possibility that the expression equals zero. If you multiply both sides by an expression (here cos ), you might introduce a false solution where that expression equals zero. or

so

or

.

12 Since cos has the real numbers as its domain and arccos has only the interval any value outside that interval will be a counterexample. Let

. Then arccos (cos

arccos

.

as its range,

13 Let

. Then

.

Primary solution: Secondary solution: Periodic solutions: Further periodic solutions lie outside the interval for .

14 a For example:

gives

In this case,

. .

Tip The false idea being disproved by counterexample here is that division ‘passes through’ the inversion of a function that for a function follow that

b Let

it should

; this is generally not the case!

. Then

.

So c

EXERCISE 7C

3 a

has range

so has range

.

Low tide has depth metres and high tide has depth b High tide occurs when

metres.

.

so So high tide occurs at hours after midnight a.m.) and at p.m.) each day. 4 a Amplitude b

seconds Time for five complete oscillations is

5 Amplitude so

.

seconds.

hours after midnight

6 Amplitude Period is

so the maximum is one quarter period

before

.

° 7 a

b From the graph, there are two intersections in this interval. c The two functions each have a whole number of periods in the interval shown. In an interval four times as large there will be four times as many intersections: solutions to the equation . 8 Centre of oscillation is at Amplitude is so 9 a Amplitude , roots at

b Maximum

and

, minimum at

; minimum

c The graph is translated by

and maximum at

.

. . Maximum is

; minimum is

.

10 a Period Amplitude equals the radius:

.

b If the particle starts at and moves upwards initially, it will pass the lowest point threequarters of the way through its first rotation, after seconds. 11

a The cosine ranges from Least height is

to so has range

.

above ground and greatest height is

above ground.

b The period is c The ball will move from the lowest to the highest position in half a period: 12 a The vertical distance below the centre is

.

So the vertical distance above the ground is b One rotation At

.

radians) takes minutes.

so is directly proportional to .

c Substituting into the equation in part a: Require

: when

So

for

, i.e.

.

, a duration of minute

seconds or

seconds.

EXERCISE 7D 5 6 a b 7 The angle in the major sector So the minor angle

.

8 9 10 The angle in the major sector So the minor angle

.

11 12 so

13 Then

14 The perimeter is composed of three arcs, each with radius So

and angle

.

15 16

17

so so

18

19

So the area of the shaded region is 20 Perimeter Area Substituting

into

:

Multiplying both sides by and rearranging:

or 21 Let the minor angle be , so the major angle is

.

Minor area: Major area: : radians 22 On the cone, slant height The perimeter of the base

EXERCISE 7E

, which is the radius of the sector. , which is the arc length of the sector.

EXERCISE 7E 4 so 5 a By the cosine rule in triangle

:

b Using the cosine rule in triangle

:

c The area of the shaded region is the sum of two segments: radius , angle

and radius , angle

radians.

Area of the shaded region

EXERCISE 7F 4 a b Let equal

. Then using part a: for small

5 a b From part a: So

6 a So Ignoring terms in

and higher, the approximation is

b i Percentage error equals

.

ii Percentage error equals 7

. for small angles .

.

Using a binomial expansion for the second bracketed expression:

8

for small angles . Using a binomial expansion for the second bracketed expression:

9

so

and

10 a Expanding, using a binomial approximation:

b

Tip is already quite large enough to make the approximation fairly inaccurate. See Worked solution for a boundary on the approximation having a or lower error. 11 Expanding the second bracket, using a binomial approximation:

12

Expanding the second bracket, using a binomial approximation:

13 a For small : Require that

, for which the solution is

, so the upper bound is

. For small : Require that bound is

, for which the solution is

, so the upper

.

For small : Require that

, for which the solution is

, so the upper bound is

. b i Approximating

with :

But the binomial expansion of this expression is (to the first term containing a power of , since is small):

So, approximating for small

.

ii For small values of is a positive value, lying either in the first or fourth quadrant, so it is inappropriate to include a symbol.

Mixed practice 7 (Answer C)

1 2 a Amplitude is

metres

b 3 Primary solution: Secondary solution: Solutions are 4

, alternative primary solution: , alternative secondary solution

5 a Using angles on a straight line: b

c

6

7 a Segment area sector area triangle area

b 8 a b i ii Perimeter is

and

.

9 arcsin is a function, so for any argument there is a single output value. (Answer B) 10 There is a maximum at

and the curve passes through

, so the period is .

so Amplitude 11 a The bridge has zero height at either end: The river width is therefore b Require that a rectangle

.

. by

fits under the curve.

when Let

,

Primary solution: Secondary solution: So

or

The width between these is barge.

.

metres, which is the maximum width of the

c Centring the barge under the bridge, it extends

either side of the midpoint at

Height of the bridge at the ends of the barge cross-section: 12 a One lap takes a single period. b Circle radius is

metres, so its circumference is

. . .

.

c 13 a b Let

, then

sin so solutions are: Solutions for are: Coordinates are c

14 Using a binomial expansion and ignoring terms in

15 a

is a rhombus since each side is a radius in length. If

b

and higher:

then

must be a square, so each internal angle is

, the diagonal of a square with side .

c Area of sector

, it is a quarter circle.

d The overlap consists of two segments, each subtending the square :

16

.

because

, or (equivalently) two sectors less

is a tangent to the circle.

So by Pythagoras’ theorem: Hence the area of the triangle

is

The area of the shaded region equals the area of triangle

17 a

has domain

and range

.

less the area of the sector.

The end points are b

and

becomes Reflecting in the

The endpoints are 18 Let

.

. and translating by

and

. Then

is equivalent to reflecting in the line

. and

Primary solution: Secondary solution: Periodic solutions in the given interval: So Then 19

(Ordered: 20 a

(Reject this second solution as it is outside the range of cos.)

.

, so

b

and

Primary solution: Secondary solution: Periodic solutions: Solutions:

or

21 a Using a binomial expansion and ignoring terms in

and higher:

b

22 a

, since

is tangent to the circle and there is a right angle between a tangent and a

radius. b By the same reasoning,

and hence

is a rectangle, so

.

c So

d

so

and

e

23 a

Tip The reverse is not true: b Let

arccos so that

Then So c Using parts a and b: Since

, use only the positive root.

only if lies in the range of arcsin .



Chapter 8 Worked solutions for book chapters 8 Further trigonometry Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 8A 4 a

b

5 a b

c

Primary solutions: These are not in the required interval. Periodic solutions:

6 a This function has a maximum value , occurring at of at which there is a maximum is

, so the smallest positive value

.

b This function has a maximum value , occurring at value of at which there is a maximum is . 7



for small values of .

, so the smallest positive

8 a

b Let

.

Then from part a:

is the only solution in the given interval. 9 a

b Using part a,

so

or

(for which there are no real solutions)

so

EXERCISE 8B 6 If

then

When

:

When

:

Solutions are: 7 a Set

.

b Set

Set

Tip You are allowed to define your own constants or variables if doing so will make working clearer or faster to write.

Since is known to be positive

, then

8 From the double angle formulae:

.

and

Then 9 a So

b Then For

, solutions are

.

10 a

b

From part a:

Then Dividing numerator and denominator by

:

11 a b

Tip It is at least as fast in this case to start again as to convert the answer from part a, using , since you can reuse the first line of working. 12 a i ii

b

13 or

(no solutions in the given interval)

and

14 a

b

15 a From the double angle formula: Let

.

Then

so

b

Restrict to lie within the domain of

, so

.

EXERCISE 8C 5 a Require Require

b

so and

. ; is in the first quadrant and

so

.

to Replace with

: translation

Multiply the function by

: vertical stretch, scale factor

.

6 a Require Require

b

so and

. ; is in the first quadrant and

so

.

has range

7 a Require Require

so and

b so

. ; is in the first quadrant and

so

.

The smallest positive value of is

.

8 a Require

so

Require

b For

and

. ; is in the first quadrant and

:

the maximum occurs when the argument is a multiple of the minimum occurs when the argument is

:

9 2 sin Let So

and

Then

and lies in the first quadrant with

so

so Primary solution:

so

Secondary solution:

so

10 So Let

so

.

Primary solution: Secondary solution: Periodic solutions: So the solutions are:

. .

EXERCISE 8D

7

so

: .

.

8

Primary solutions: Secondary solutions: Periodic solutions: Taking the solutions in the required interval:

or

or

9 or

or

10

11

12 Dividing the numerator and denominator by

:

13 a But

.

Substituting:

:

b

or : :

or

So the solutions are

.

14 a

Multiplying numerator and denominator by

:

b or : : So the solutions are:

.

15 a

b Primary solution: Periodic solution: So the solutions are

or

.

16 Using the first few terms of a binomial expansion to approximate this:

17

Let

so :

sin

with further periodic solutions :

so the solutions are 18 Let But

. Then

.

so

Therefore The inverse to the function

is the arccosine of the reciprocal of .

Mixed practice 8 1

for some and

So

for some and so has maximum value

(Answer C) 2 a Setting b

:

.

Let

.

: : The solutions are

.

3 a b Using similar working: If the two expressions are equal then

So

, which occurs at

.

for integer .

Within the given interval, the solutions are 4 a Diameter

and angle

Therefore b

.

(angle in a semicircle).

and

.

is a right-angled triangle so

.

c Using the sine rule for area: d So 5 a Setting

:

b tan c Let

so that tan

. Call

Then using part a: So This quadratic has roots Since Therefore, 6 a Amplitude b Amplitude is

c

.

, tan must be positive. . and period

so seconds

.

tan

and

so lies in the first quadrant and

so

.

d Amplitude is , the period remains at seconds. e Require

: for some integer so least positive occurs when

f Require

.

:

Let

with

so that

.

Primary solution:

Secondary solution: So the first two solutions are

or

.

7 a

and

so is in the first quadrant and

b

so

so Let

. Then

and

Primary solution: Secondary solution: No further periodic solutions exist in the required interval. The solutions are: The solutions are:

or

. or

.

.

.

8

(Answer B) 9 a Setting

:

b

Using the formula in part a and the double angle formula for sin,

Either

or, dividing by

Since

, require solutions for

:

:

.

But is not a solution of the original equation as this would make the denominator of the LHS zero ( ).

So, the solutions are

.

Tip It is important to check that the values you get are actually solutions of the original equation. Here appeared to be correct, but on checking back you can see that it isn’t a valid solution of the original equation.

10 a Let

.

By inspection,

so, by the factor theorem,

So

for some integers

is a factor of

.

and .

Expanding and comparing coefficients: : :

so

:

so

:

Consistent.

Tip

b

There are other approaches. Here is a direct approach, using compound and double angle formulae for . Compound angle formula: Double angle formula:

Multiplying numerator and denominator by

:

c d Suppose for values that satisfy the equation. Then Let

or

. Since this is a single period, there should be three

and so

or

.

. Then using the formula from part b, for any of the values :

Rearranging gives

, for which the solutions, from part a are: .

As noted, within the period of

to

there are three values for which

.

From the working, the tangent of each of these must correspond to one of the roots of the cubic . Since

is an increasing function in the interval

tan

and

, it follows that .

11 Since

:

The range of is the real line excluding the interval real solutions for .

so the first of these roots has no

Primary solution: Secondary solution: Further periodic solutions lie outside the required interval. The solutions are:

or

.

12 a

So b

; if

then

and

.

(Reject negative root as noted.) Primary solution: Secondary solution: Further periodic solutions lie outside the given interval. The solutions are:

.

13 a

and

so lies in the first quadrant and

so

.

b i The maximum value for f occurs with the minimum positive denominator. (The denominator is always non-zero.) So require

ii

, at which

so

14 a

for all

b Let

.

Then So

. for

.

c Using the double angle formula: or 15 a i

.

ii LHS, being a squared expression, is never negative for real so

iii The equality will hold when the initial expression

, when

b Let the angle of inclination to the base of the picture be ; then The angle of inclination to the top of the picture is Using the compound angle formula

. .

with

Multiplying numerator and denominator by

c From part a i:

, and

.

and

:

so

d From part a iii: equality occurs for e It is clear from context that

;

is an increasing function within this domain.

Consequently the greatest value of corresponds with the greatest value of When

radians.

.

:

Chapter 9 Worked solutions for book chapters 9 Calculus of exponential and trigonometric functions Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 9A

3 4 5 6 Require that

:

Checking the answers for validity: reject So the solution is

as it is not within the domain of g.

.

Tip Always check for validity of solutions in any question containing ln or a square root, since the working can give rise to solution values outside the domain of the original function. 7

8

9 Require the gradient to be :

The point on the curve is

, gradient is .

The tangent has equation

.

10 Require the gradient to be :

If

then

.

The point on the curve is

, gradient is .

The tangent has equation

11

.

 

12

The gradient of the tangent is , the gradient of the normal is

The tangent has equation

:

or, equivalently,

.

The normal has equation

:

or, equivalently,

.

13

Require that

14 Stationary points occur where

.

So

The stationary points are

and

.

is a local maximum. is a local minimum. 15 Stationary points occur where

:

When The stationary point is

.

.

16 Stationary points occur where

:

The stationary point is

.

, so the point is a minimum. The range of f is

.

17 a

The stationary point is

.

is a local maximum. b

The stationary point is

.

is a local minimum. 18 a

has range

so the minimum volume is

million litres.

b Since all water flow is through the dam, water flow will equal the change in lake volume.

You can assume that the hydroelectric dam can generate electricity when flow is in either direction, so that there is generation at all times except when flow is zero. Maximum flow occurs when Hence maximum flow in the first days occurs at

EXERCISE 9B

4 5

.

6

7 8 a b Area is

(below the

).

9 The curve passes through

:

10

11 a b The curve passes through

12 The curve crosses the

:

when

From your calculator, this is at

. .

13 a

The curve has roots at b Area of the shaded region

and

.

14 The value of the integral is independent of . 15 The gradient of the normal equals

gradient of the tangent equals

Require that

Mixed practice 9 (Answer D)

1 2

The equation of the tangent with gradient

3

4i

ii

5

6 Area of the shaded region

through

is

.

7

8 i ii 9 There is a stationary point when

: (Answer C)

10

11 Stationary points occur where

:

or The stationary points are

and

.

is a local maximum. is a local minimum. 12 Stationary points occur where

Using the chain rule:

local minimum local maximum local maximum local minimum

:

13

The gradient of the normal at

is

.

The equation of normal is

.

14 a i The initial population is

bacteria.

ii From your calculator: The population reaches

million after

hours.

b i ii

The rate of growth is million per hour after

hours.

c i The second derivative gives the rate of change of the growth rate. In this case the second derivative is always positive and increasing so the growth rate will always be increasing at an ever growing rate. ii Minimum number of bacteria will occur either at a boundary point or when

The stationary point at

has a lower bacterial population than the initial condition.

so this is a minimum point on the curve. According to the model, the minimum number of bacteria is 15 The graphs intersect when

.

The enclosed area consists of two parts: (i) below (ii) below

for for

:

.

.

16



Chapter 10 Worked solutions for book chapters 10 Further differentiation Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 10A

4

The tangent at

5

is given by:

so

The normal at

has gradient

and is given by:

6 Require that

But the domain of

:

is

, so the only solution is

7 The gradient is zero at a stationary point:



Stationary points have coordinates

.

8 The gradient is zero at a stationary point: Stationary points have coordinates 9 a

.

.

At b At 10 Stationary points occur where

:

since Stationary points have coordinates 11

and

.

The gradient is zero at a stationary point: The stationary point has coordinates

.

12 Require that So

:

.

13 a b since 14 a

and Since

, the post at

(the left post) is taller.

b The stationary point occurs where

:

Classifying the point:

The stationary point is therefore a minimum. So the minimum height occurs when

.

c

15 a Using the trigonometric identity for

:

b The stationary point occurs where Using the identity for cos

.

:

This is a quadratic in cos  . Using the quadratic formula:

or

or

or

Stationary points are

.

c Knowing the roots and the stationary points for the oscillating function, you can sketch it (very roughly). You can use additional points as well:

EXERCISE 10B 3 Let

4 Let

Let

5 Let

Stationary points occur where

:

6 Let

Stationary points occur where

7 Let

:

. Then

and

Using the chain rule: Using the product rule:

8 a Let

b 9 Let

Stationary points occur where

:

The stationary point lies at

.

10 Let

11 a b Using the chain rule: Using the product rule: From part a: c Stationary points occur where

:

since The stationary point is at

.

.

12 a Using the product rule:

Stationary points occur where

(The solution In the interval

requires

:

and the solution

requires

, the only stationary point is

.)

.

Tip You do not have to show this in your answer, but you should satisfy yourself that this must lie in the given interval so:

b

: there is a double root at : there is a triple root at

. .

This is a positive polynomial of order . The stationary point is at

.

The turning point is from towards .

c The curve will finish with a positive gradient for any

.

If the second root is repeated an odd number of times ( is odd), the curve will be negative in the interval and the stationary point will be a minimum. If is even, the curve will be positive in

and the stationary point will be a maximum.

The stationary point is maximum if (and only if) is even.

EXERCISE 10C

EXERCISE 10C 2 Let

The normal at

has gradient

.

The normal equation is given as

3 Let

Stationary points occur where The stationary points are at

. and

.

4 Let

Require that 5 Let

.

Stationary points occur where Let

.

The stationary point at 6 Let

.

.

is a local maximum.

Require that

:

(since the denominator is always non-negative)

Checking for validity in the function, For

is increasing.

7 Let

8

is not in the domain.

.

has a local maximum at Let

. Then

So

for small

.

. ; there is a stationary point at

Also,

and

. , so

is a local

minimum.

Tip It might seem better to use second derivative analysis to determine the nature of the stationary point, but there are good reasons not to do this; it is possible for a local maximum to have a zero second derivative, such as in the curve , so any proof would require contingencies for this circumstance, and in any case calculating the second derivative of

requires multiple uses of chain rule, product rule

and quotient rule and is unnecessarily complicated.

EXERCISE 10D

5 Using implicit differentiation: At 6 a Using implicit differentiation: b At The equation of the tangent is 7 a When

and

b Using implicit differentiation:

. so lies on the curve.

The gradient of the normal at (

is

The equation of the normal is 8 Using implicit differentiation: Substituting At the point

: , the gradient is .

9 Using implicit differentiation: Substituting

:

The equation of the tangent is

.

10 Using implicit differentiation: Substituting At the point

, the gradient is infinite; that is, the tangent is vertical.

The tangent at

is given as

11

Using implicit differentiation:

12 Using implicit differentiation: The stationary point occurs where Substituting into the curve equation: The stationary point is at

.

13 a This is restricted to The curve is

.

.

Using implicit differentiation: At

so is a tangent at

The gradient at

is

.

.

has equation b For the intersection of and , substituting

c One (repeated) root must be

Comparing coefficients:

since is a tangent at this point.

.

Substituting into the equation for : The point of intersection is

EXERCISE 10E 1 a b Using implicit differentiation:

3

:

.

intersects again when

2

into

–1.

(on the lower part of curve ).

When

:

So 4 a

for all , so there are no turning points. The cubic is therefore one-to-one and so must have an inverse.

b When

, by examination of the function.

The gradient of the inverse at Let

is the reciprocal of the gradient at

. Then

5 a Let

. Then

so

.

b

6 a So f(

is an increasing function.

b

so this point is on the graph.

c At the point

on the graph of f(

, the gradient is

. Therefore, at the point this:

on the graph of

, the gradient is the reciprocal of

.

7 a b 8

is an increasing function if Assuming of fat (

( .

for all values of in the domain.

is defined, you know that its gradient at

is the reciprocal of the gradient

Since the reciprocal of a positive value is also positive, wherever the gradient of it is positive, so is also an increasing function.

is defined,

Tip If you require that is a strictly increasing function – that is, the gradient is greater than zero and never equal to zero – then you know that the inverse exists (f must be one-to-one) and you need not be concerned about the possibility of a stationary point on . However, if you consider the example of , you can see that this restriction is not needed for the result to stand.

Mixed practice 10 1

(Answer D) 2 When

(Answer A)

3 a i ii Using the chain rule:

b Using the product rule:

4

5 a Substituting

into the LHS:

So point does lie on the curve. b Using implicit differentiation:

When

:

6

When

, the gradient is

The tangent at

) has gradient

. so has equation

.

Axis intercepts of this tangent line are Area of the right-angled triangle

and

is

. .

7

8 a Using the product rule: b i When

e,

So the equation of the tangent is ii The intersection with the

is when

. :

9

10 Using implicit differentiation:

At the stationary point, Substituting into the curve equation: The stationary point is at

.

11 a b 12 The minimum velocity will either be achieved at one of the end-points of the motion (

or

or at a point during the motion where

.

Using the second derivative to determine the nature of each stationary point:

When

:

When

:

Then finding the velocity at this minimum point: When

,

Checking this against the initial velocity: When

. so the minimum velocity is

Tip There is no need to check the velocity at once you know that minimum point for the velocity, as the function must be increasing for 13 a The particle is instantaneously at rest when

is a .

.

Using the product rule:

, i.e. the particle is instantaneously at rest at

seconds.

b The maximum speed will either be achieved at one of the end-points of the motion (

or

or at a point during the motion where

Using the product rule:

Substituting into

:

when

.

Checking the end-points: when when

.

Therefore the maximum speed is

.

14 a Vertical asymptote where denominator equals zero: b Using the quotient rule:

.

The stationary point occurs where

The stationary points are

.

) and , –1).

Tip

c

Second derivative analysis is possible to establish the nature of the stationary points but it is easier, in this case, to assess the sign changes of the first derivative. If doing this, always remember to include consideration of any vertical asymptotes, as behaviour of the curve may change either side.

–ve Slope From the table,

0

+ve

+ve

0

−

/

/

−

is a local minimum and

is a local maximum.

d

15 a Let

.

Using the chain rule: b When The equation of the tangent at

is

16 a For g( b At

, g'(

so g is a strictly increasing function.

is therefore one-to-one and has an inverse. has gradient .

Therefore at

has gradient .

17 Using implicit differentiation:

Stationary points occur where

:

.

–ve

Substituting back into the original equation:

Since

, the coordinates of the stationary point are

and

.

18 a Using implicit differentiation: Stationary points occur where Therefore

.

. Substituting into the original equation:

Stationary points are at

and

.

b Using implicit differentiation: At stationary points,

so this reduces to

.

c

At , 4), At (–2, –4),

so

) is a local maximum.

so (–2, –4) is a local minimum.

19 a Starting from the identity

and dividing through by

Using implicit differentiation:

b Using the chain rule on the result from part a: If

then

The gradient when

20 a

. is .

The gradient of the normal at

is

The equation of the normal at

is

. .

:

b

The stationary point occurs where

, so at

Tip

c

You can approach this problem directly and without reference to the work in parts a and b, using a substitution . However, part a offers a faster route, which avoids integration and substitution, as shown here. Either approach would be acceptable.

Let

If

and

then

Since the integral of

is

:

Chapter 11 Worked solutions for book chapters 11 Further integration techniques Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 11A

5

6 The graph lies above the

throughout.

Tip Watch out for graphs that dip below the in case you need to split the integral into multiple regions to find the total area enclosed.

7 The curve intersects the

8 If

then

EXERCISE 11B

.

at

and

.

EXERCISE 11B

Tip It can often be helpful in questions of this sort to assign a letter to equal the integral (typically I is used) – this makes referring to the integral easier in the later working.

2 Let

. Then

so

Let

3 Let

.

.

. Then

Let

so

.

.

4 Let

. Then

Let

.

5 Let

. Then

Let

.

Let

. Then

so

.

so

.

so

.

6

EXERCISE 11C

5 Let Let

. Then

so .

.

6 a Let

. so by the factor theorem,

is a factor of

.

b Let

. Then

Let

.

so

.

Tip Any time you integrate a function of the form

, the result will be ln

, and it is acceptable to show no intermediate steps of working – though you should justify this step.

7 Let

. Then

Let

so .

8 Let

. Then

Let

9 Let

.

so

.

.

. Then

Let

EXERCISE 11D

so .

.

EXERCISE 11D

2 Let

.

Integrating by parts: set

3 Let

and

so that

and

.

.

Integrating by parts: set

ln

and

so that

and

.

EXERCISE 11E

4 If you include an unknown constant when integrating

, the result is:

So the constant cancels out in the final answer, leaving only the standard unknown constant (here using . 5 Let

.

Integrating by parts: set

6

.

and

so that

and

.

Tip Remember that usually you differentiate the polynomial and integrate the other function but, when dealing with a logarithm, you must always differentiate the logarithm to simplify the integrand in future steps of the integration.

Integrating by parts: set

ln and

so that

and

.

7 Let Integrating by parts: set

and

Now integrating by parts again: set

8 a Let

. Then

b Let

so

so that

and

and

so that

.

and

.

.

.

Integrating by parts: set

9 Let

. Then

Substituting bounds: when

and

so that

so

and

.

. and when

.

Tip In this question, substituting for the bounds is faster than keeping the bounds in terms of and substituting back, but either approach will give the same answer.

Let

.

Integrating by parts: set

. and

so

and

.

10 a Integrating by parts: set

cos and so

Now set

sin and

so that

and

.

. so that

and

.

Tip Remember that as soon as you eliminate the last of the integration signs you must insert an unknown constant. b Adding to both sides of the equation from part a and halving:

Tip but both are unknown constants. It is best to use a different letter for the constant rather than using c in both lines.

EXERCISE 11F

6 Using the identity

:

7 a b

Tip The full method is shown here. However, you should not need to undertake a substitution technique to integrate either of these terms, and should be able to move directly to the answer with reference to a formula in the formula book.

Let

. Let

Let

tan . Then

. Let

so

cos . Then

.

so

.

So

Tip Because tan and sec are closely interlinked in calculus problems, you will often find that sec is preferred to cos in such results; this answer could also be given as .

8

Require that

:

This has infinitely many solutions for , since the curve three times for the primary curve (though for each other periodic repeat of the curve.

The least positive solution is

is not a solution of the original problem)and once

.

9 a Let

. Then

.

b Let

. Then

so

intersects the line

.

10 Let

11 Let

Then

. Then

so

.

.

so

.

12 a b

Let

. Then

so

13 a The full circle has equation

.

In this region, both and are positive, so rearranging for : b Area of shaded region Let

cos  . Then

Bounds of integral: when

so

. and when

.

EXERCISE 11G

6 a

for some constants and . Multiplying through by the denominator on the LHS: Comparing coefficients:

Tip You can compare coefficients or substitute trial values of ; the advantage of the coefficients method is that if you have made an error in your assertion of the partial fraction form (for example, missed out a term) then you will be unable to solve the resultant coefficient equations consistently, but substitution – unless you try further values to check – will not give you this safeguard. : : :

b

Tip The indefinite integral will be valid but any definite integral for which the interval included either or both of the roots of the denominator and would be ill-defined.

7

for some constants and .

Tip The roots of the denominator lie outside the integral interval, so the function is defined throughout the interval of integration. Multiplying through by the denominator on the LHS: Comparing coefficients: : : :

8

9 a

for some constants and . Multiplying through by the denominator on the LHS: Comparing coefficients: : : :

b

Tip The roots of the denominator and lie outside the integral interval, so the function is defined throughout the interval of integration.

10

for some constants

and .

Tip The roots of the denominator and lie outside the integral interval, so the function is defined throughout the interval of integration. Multiplying through by the denominator on the LHS: Comparing coefficients: : : : From

:

Then from

:

Substituting into

:

Mixed practice 11 Tip

1

Given the integral alone, you would try a substitution , observing that the derivative of that substitution is present in the integrand. Alternatively, since this is a multiple choice, you could try differentiating each function and rejecting the ones which do not produce the required derivative form. Method 1: Integrate with substitution Let

. Then

so

.

(Answer C) Method 2: Differentiate each of the multiple choice answers. A: B: C:

D: (Answer C) 2

3 Let

.

Integrating by parts: set

and

so that

and

.

4

However, the function

has a discontinuity at

, and any integral with bound at

is infinite, so the negative solution is not valid for the original problem.

5

6 a The function to be integrated has the form So b

Tip If you want to show more working you can perform a substitution a linear function to a power, this is sufficient.

7 Let

.

Integrating by parts: set

and

8 a b Let

.

so that

and

.

but for

Substituting

:

Then

.

9 Let

.

Integrating by parts: set

and

so that

and

.

(Answer A) 10 Let

.

Integrating by parts: set

and

so that

and

.



Tip

11

In general, simplifying a function of this sort would involve reducing as much as possible the number of instances of . However, this actually makes the subsequent calculations slightly harder. Method 1 is more thorough, method 2 has slightly easier integration. Method 1 a First, multiplying numerator and denominator by

b Let Let

. Then . Then

. so

.

and then factorising:

Partial fractions:

for some constants and .

Multiplying through by the denominator on the LHS: Comparing coefficients: : :

Method 2 a b Let

. Then

Let

. Then

. so

.

Tip Notice that the two answers given differ only by a constant, which can be accounted for within the logarithm.

12 Let

. Then

Let

so

.

.

13 a

for some constants and . Multiplying through by the denominator on the LHS: Comparing coefficients: : : :

b

14 Let

.

Substitution: Let

. Then

so

.

Tip Using integration by parts could lead to difficulties… Integrating by parts: set

and

so that

and

. Unfortunately, with I on both sides of the equation (rather than some multiple on the RHS), this does not lead to a straightforward integral.

15 Let

. Then

Let

so

.

.

16 a

for some constants and . Multiplying through by the denominator on the LHS:

Comparing coefficients: : : : : : b

.

Tip The argument of the logarithms has an even power so the modulus signs are redundant and can be dropped.

17 Let

. Then

Let

.

so

Integral bounds: when

.

and when

.

Tip If you have an integral with mixed variables, you should label the bounds as to which variable they apply.

Tip You don’t need this level of detailed working in an examination answer; it is given here for clarity. As long as your working shows how you get from each line of working to the next, it is sufficient.

18

When

and

:

for all

so

for all .

Therefore, the particle never changes direction. So, distance travelled in first seconds is given by Integrating by parts for

:

So,

Distance travelled is 19

Tip Although it would be a perfectly valid approach, using the standard partial fractions approach is unnecessarily long-winded if the denominator is just a repeated factor.

20

for some constants and . Multiplying through by the denominator on the LHS: Comparing coefficients: : :

For the function to be defined over the integral interval, . (This is also necessary for the individual logarithms to have real values.) That being the case, and are both positive, so you can drop the modulus signs.

21 a Let Then

. Let so

Limits of the integration: when

 sin  . . and when

.

b The internal area of the ellipse is twice the area of the upper half. Rearranging the equation

for non-negative:

so Therefore the area of the upper half of the ellipse is given by

.

From part a: Therefore, the total area of the ellipse is

.

22 a b Let

c

sin

cos . Then

so

.

.

Chapter 12 Worked solutions for book chapters 12 Further applications of calculus Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 12A

4

   At a point of inflection, So

.

and the point of inflection is

5

.

   At a point of inflection, So

6

.

. The points of inflection are

and

.

   At a point of inflection, But

7

so

.

at all points of inflection.   

The curve is convex where

.

The curve is convex for 8

or

.

   At a point of inflection, So

; since

. , at both of these points the second derivative changes sign so

they are points of inflection. so at points of inflection, The points of inflection are

and

. .

9 Using the chain rule: In the given interval, the derivative only equals zero at

, so this is the only stationary point.

So at

.

At a point of inflection,

.

Since is an odd function (so for all , it follows that the first derivative must be even and, if non-zero, the second derivative must be odd, and therefore is also a point of inflection.

Tip Alternatively, try values either side of such as to show that the second derivative passes from negative to positive through . Remember that it is not sufficient to show that the second derivative is zero.

10 Stationary points occur at

and

.

so there is a local minimum at

.

preventing immediate classification. To determine whether it is a maximum, minimum or horizontal inflection, use the sign of the first derivative on either side.

shape From the table it is clear that there is a local maximum at

.

11

At a point of inflection

:

For this to be a stationary point of inflection,

12

:

   The curve is concave where so

so the curve is concave for

.

13

   The curve is concave where Since

.

is positive for all , require that

.

The positive quadratic is less than zero between the roots. The roots of

are

The solution is 14 A: Local minimum: f′( B: Local maximum: f′(

. .

with f′( with f′(

and f′( and f′(

C: Inflection points of f are stationary points of f′(

EXERCISE 12B

for small positive . for small positive .

, which are not horizontal inflections in f'(

.

EXERCISE 12B 4 a

 m

b When

(launch) or

(hitting the ground).

At when

5 a

When

so the -intercept is

b When

and

.

When

and

.

The distance between these points is

.

.

is in fact equal to the distance travelled along the path.

Tip This specific example does not prove the generality, but you can do that by finding the distance between the point at and . When

and

When

. and

.

The distance between these points is

.

c Substituting

: the Cartesian equation is

.

EXERCISE 12C

3

and

so

When

and

The normal at ,

.

) has gradient

The equation of the normal at ,

4

and When

π,

),

) has equation

so

π,

and

The tangent at (π, –1) is 5 At

.

so

–1. .

. .



.

and

so

So the tangent at

) has gradient

.

The equation of the tangent is

When

.

is

When

. is

.

(

.

).

The area of triangle OMN is 6 a At

),

so

and

. so

So the normal has gradient . The normal at

) has equation

.

b Substituting the parametric equations into the normal equation:

at the point

) so the point where the normal meets the curve again is at

which is at the point 7

and

.

so

.

The gradient of the normal is At

.

, the point on the curve is

) so the normal has equation

.

Substituting the parametric form: The source point of the normal is at So is

so is at

.

).

8 The tangent parallel to the

occurs when

.

for all values of so there is no point at which the tangent is parallel to the 9

and At

so

, the gradient is

The tangent has equation

At

so

At

, so

The coordinates are:

. . .

.

.

10 a

at . and

so

At , the gradient of the curve is

so the gradient of the normal is

The equation of the normal is

When b

so

is (

so

is

.

.

is

.

.

EXERCISE 12D 2 a

where ( When

(

–1,

:

or –1.

. When

.

Coordinates of the points where the curve crosses the b

So the area is 3 a

when when

b When

, below the

.

. ln 2. and when

ln 2,

ln 2.

are

) and

).

4

a For the area , endpoints are at

:

or

.

b

5 a

(1)

:

Checking, this gives the desired values for and . b For the normal at : so When So the normal at has gradient

Substituting Let

: the line crosses the and

The curve has a root where

c

and hence has equation

and

at

.

.

. at point

so

).

is the Cartesian equation of the curve.

Tip If you don’t see this immediately, you can still find the solution using the method of substitution:

so

from which

. Substituting

into the parametric equation for y: . So

. Or you can

substitute both t and , but this requires that you rationalise the denominator and cancel terms to get the same result.

EXERCISE 12E 4

radius (cm),

area (

  

),

time (seconds)

  

so

Then When 5

, the rate of change of the area is

side length (cm),

area (

  

),

  

.

time (seconds) so

Then When 6

, the rate of change of the side length is

radius (cm),

volume (

   So when 7

and

, the rate of change of the volume is width (cm),

  

8

time (seconds)

  

length (cm),

So when

),

.

area (

),

.

time (seconds)

  

and

radius (cm),

, the rate of change of area of the rectangle is

height (cm),

area (

),

.

time (seconds)

so

so When 9

radius (cm),

and

, the rate of change of the radius is

height (cm),

volume (

),

time (seconds)

.

and so When

10

and

:

and

At (

):

EXERCISE 12F 3

(

(

Roots at and . To calculate the shaded area, you need to compensate for the area below the axis being deducted from the total:

4 Intersections of the two curves occur when

.

±2 The line lies above the quadratic so the difference equation is

.

5

6 The oblique line bounding the shaded area passes through points ( . The line has equation

.

and

).

Tip An alternative approach would be to integrate between and and subtract that from the area of the trapezium. Ensure that your working is clearly laid out and any valid method will be acceptable. 7

so

So 8

so

.

9 Intersections occur where So

(

Intersections at

. (

and

. .

has no discontinuities in the interval The line overlies the reciprocal in the interval

10 Intersections occur where

So

(

2

so the difference function is

.

, which has solutions at

The difference equation is

.

2

and

.

.

11 Intersections occur where

, at

or

.

So 12 Let the blue area equal and the red area equal .

The height of the rectangle made up of the blue and red regions is

. The width is

.

You can evaluate as the area of the rectangle less the area .

If

:

There is a trivial solution where cubic. Factorising: (

equivalent to

(

The roots are The diagram depicts

so the solution is

Mixed practice 12 1 Require

At At At

, cos

: A is not correct. : B is not correct. : C is not correct.

.

, so

must be a factor of the

At

and

2

: D is correct. (Answer D)

   At a point of inflection,

3 a When

so

and

b

and the coordinates of the point of inflection are

so lies on the curve at

.

so At

so the equation of the tangent at is

4 The line has equation symmetrical about

, and meets the curve again at

.

, since the quadratic is

.

The difference function between the line and the curve is

.

5 a b There is a root at

.

To calculate the shaded area, you need to compensate for the area below the axis being deducted from the total:

6 a

so When

.

.

b Substituting into the equation for :

Tip There is no need to make the subject of this equation. The question doesn’t ask for the answer in a particular form, just for a Cartesian equation. Note also, that it’s a good idea to avoid making t the subject of the y equation as the first step, as this involves the complication of the positive and negative square root.

.

7 A point of inflection requires that the gradient is at a local maximum or minimum; that is, the second derivative must be zero and additionally the second derivative must be passing through zero, either from positive to negative or from negative to positive. A gives a necessary condition for a point of inflection, but is not sufficient. B is invalid that would be a necessary (but still not sufficient) condition for a horizontal inflection. C is invalid that would be a necessary (but still not sufficient) condition for a non-horizontal inflection. D is correct. (Answer D) 8

so At

. .

At

.

So





.

Tip

9

You can integrate this using the method of parameters or by finding the Cartesian equation and integrating directly. You can also either calculate the trapezium area under the tangent and deduct the area under the curve or calculate the triangle area above the tangent and subtract this from the area between the curve and the -axis. Each approach is shown here. Method 1: Parameters and area between the curve and the At

),

so so

The gradient of the tangent at

is

The equation of the tangent is The triangle above the tangent has area

. (



.

Method 2: Cartesian equation and area between the curve and the

The area of the shaded region is the area of the trapezium under the curve between and .

),

),

),

) less the area

10 a

 sin   cos 

or cos 

or b Need the area between the

for

11 a

and

for

and the area between

and

.

   Points of inflection occur when

passes through zero.

b From the diagram, there are four values of in the interval

for which

, and since

in none of the cases are the two curves tangent, each must represent a point of inflection.

12 a Stationary points occur where The stationary points are

so

) and

so

or

.

.082).

.082) is a local maximum.

so classify by determining gradient either side.

So

) is a local minimum.

b At a point of inflection,

. or

.6

You have already established that there is a turning point at

The curve changes from convex to concave or vice versa at So at 13

.

.

, there is a non-horizontal point of inflection.

so the curve does pass through the origin. so

. There is therefore a stationary point at the origin. so

as long as

. There is therefore a point of inflection at the origin,

. so

. Therefore there is a point of inflection

at the origin. 14 The upper right vertex of the white rectangle enclosed by the coloured zones is ( rectangle has vertices at the origin, ( ), ( and ) and hence has area

) so the inner .

The upper right vertex of the coloured zones is ( ) so the outer rectangle containing the areas of interest has vertices at the origin, ( ), ( ) and ) and hence has area . Let the blue area be and the red area .

Subtracting: So if

then

.

15 a The area of the trapezium is the difference in the areas of the two triangles.

b The derivative of the area with respect to time is So

.

           

Tip

c

To solve either use the method of separating variables or take the reciprocal and integrate with respect to . Method 1 Separating variables for At

, so

Doubling and rearranging:

: .

and, given cannot exceed

so

.

Method 2 so Integrating with respect to : When

so

.

(Take the positive root since cannot exceed

.)

so 16 a

so

b i When ii When

ln ,

and

so is

iii The gradient of the normal at is . The equation of the normal at is

This intersects the

when

.

:

–50.

is ( c

So when cos 

17 a

in

; tan 

b

Using repeated integration by parts:

So

at

seconds

.

18 The curves are

and

.

Intersections occur when (

.

(

or

–2.

The line is to the right of the quadratic so the difference equation is

.

Tip You could alternatively reflect the entire graph in the line and solve the more routine integration with respect to . If you instead rotate the graph by , you will find it is exactly the same problem as you encountered in Exercise 12F question 6. This would be an unusual method and you should always accompany your working with a fully labelled diagram.

19 a At :

so

  At :

so

b The shaded region is a quarter of the ellipse.

So the total ellipse area is

.

20 Local minimum: f′(

and the gradient on the graph of f′(

is positive.

Local maximum: f′(

and the gradient on the graph of f′(

is negative.

Inflection points: Stationary points of f′(

that are not horizontal inflections.

21 a

b Intersections occur when So So

ax

. or

or

Intersections are at c The difference function is

and

. ax

(

ax

.

d

So the ratio of the enclosed area lying above the



is

.

Chapter 13 Worked solutions for book chapters 13 Differential equations Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 13B

4 Separating variables:

Substituting so

: and the curve uses the positive branch.

5 Separating variables:

where Substituting

:

and the curve uses the positive branch:

When 6 Separating variables:

where 7 Separating variables: The left integral can be approached using the method of partial fractions.

Multiplying through: ( Substituting Substituting

: :

(

.

8 Separating variables:

Substituting

,

:

However:

EXERCISE 13C 2 a Rate of population growth is proportional to population size.

When

b When When

: , the model predicts

3 a Rate of decay is proportional to mass.

When

b When

:

Half-life is such that seconds 4 a

b Separating variables: When = Require that After

:

seconds, the speed has decreased below

.

5 a So the population increases initially. For The population decreases after 5.71 years. b Separating variables: At

c

:

Completing the square in the exponent: So the maximum population is The maximum population is

d As model.

fish.

. In the long term, the fish population drops towards zero, according to the

6 a Rate of change of with respect to

is proportional to

.

for some constant of proportionality is expected to be negative because demand would normally decrease as price increases, so the rate of change

is negative.

b Separating variables: where

c i

ii

: Demand is inversely proportional to price.

: Demand is constant, independent of price.

7 a The temperature of the milk remains no greater than .

, so the temperature difference is

Rate of temperature change is proportional to the temperature difference: When

so

b Separating variables: When

Since temperature measurement is to the nearest degree, need to find when

It takes

.

minutes (to the nearest minute) for the milk to reach room temperature.

8 a Resistance force

.

Acceleration

.

Separating variables:

Tip You know from the differential equation that v cannot exceed if it starts from less than , so you do not need to use a modulus sign in the logarithm of ( ).

and

b As

so, for large values of , the velocity approaches

9 a Volume of cone is

.

.

You know that in this cone, the filled volume will always be a cone with the same ratio of height and radius as the large cone: . So Using implicit differentiation: the fill rate is

Rearranging: b Separating variables: : When

,

It takes approximately

seconds to fill the cone.

10 a Resistance force Acceleration b Separating variables:

Tip Since the only force in the system resists motion, the velocity will never be negative and you do not need to use a modulus sign in the logarithm of .

c

As so for large values of , the displacement from approaches

.

Mixed practice 13 1 From the question, rate of value change is (negative and) proportional to . A: Rate is proportional to but positive. B: Rate is inversely proportional to and positive. C: Rate of value change is (negative and) proportional to . D: Rate is inversely proportional to and negative. (Answer C) 2 Temperature increase falls to zero as the temperature approaches climb towards (though never reach) . The upper bound for the temperature is 3 Separating variables: where

. (Answer B)

; the temperature can

4 Separating variables:

5 Separating variables: where

6 a Substituting:

b Separating variables:

When

:

7 a Growth rate is proportional to population: When b

Tip For variety, instead of separating variables you can manipulate the derivative, because there is only one variable in the expression for the rate.

When Rearranging: When

:

It takes 34.7 months (just under 3 years) for the fish population to reach model.

, according to the

c The model has the population growing exponentially without limit; at some stage, limitations on space, food resources etc. will constrain the growth, and the model will be increasingly inaccurate. d Separating variables:

8 a Separating variables:

Tip You don’t need to include modulus signs as you do not consider negative price or demand. where b When , demand is inversely proportional to the root of the price; if the price quadruples, the demand halves. c

means that as the price rises, demand also rises. This is rarely the case for anything but luxury items, where rarity and prestige become relevant in the purchase decision.

9 Volume of a cylinder of radius a Using implicit differentiation:

and height m is

.

.

Separating variables:

b As

but there is no finite time at which

10 a Resultant forces:

, so the tank will never completely fill.

b Separating variables: where

c The velocity increases towards

as increases.

11 a b i

ii When

so it takes

days for the snowball to melt completely.

12 2 causes a slowing in growth with the square of the population; the square of the a The term population will be proportional to the number of interactions within the population, and so may reflect competition for resources or space.

b Separating variables: Using the method of partial fractions on the right side:

Substituting

:

Substituting

:

where

c As

; the population increases towards

as time progresses.

Tip This is an example of a logistic growth model, which is used in many population models for simple growth with an externally imposed population limit.

13 Separating variables:

14 a Using the chain rule: b Separating variables: When

c Separating variables:

Substituting into the expression for :

Chapter 14 Worked solutions for book chapters 14 Numerical solution of equations Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 14A 1 a Can factorise the quadratic exact solutions available. b Can’t rearrange algebraically

in exponent and base position.

c Can rearrange using trigonometric identities exact solutions available. d Can rearrange using logarithms exact solutions available. e Can’t rearrange algebraically

in argument of trigonometric function and outside as well.

f Can rearrange using exponential exact solutions available. g Can’t rearrange algebraically

in argument of two dissimilar functions.

h Can’t rearrange algebraically cubic-solving equation not available within this course. 5 a For

:

The cubic equation is continuous so

must have a root between and .

b

Change of sign indicates there must be a root of correct to d.p.

between

and

, which would give

6 Let a

Change of sign indicates there must be a root of give to s.f.

between

b From the diagram there is a root between and , so

.

and

, which would

7 a For a rational function to equal zero, the numerator must equal zero. for all so

cannot equal zero so has no roots.

b i ii The change of sign principle cannot be applied where there is a discontinuity in the interval. Since is a vertical asymptote, the change of sign from to does not imply a root. 8 a i

ii b i ii

A graph can cross the axis and cross back. In fact, for a continuous graph, George’s comment could be modified to say that there must be an even number of roots in the interval – in this case, two.

iii

so by the change of sign principle there must be a root between and , and another between and .

EXERCISE 14B 3

4 a Multiplying through by :

b

5

6 a

The only discontinuity is at between and . b

so the change of sign principle shows that there is a root

c

There must be a root between

and

The change of sign does not lie between decimal places.

, which is equal to

and

to d.p.

so the root does not equal

to two

Tip Be careful of your phrasing here – see Exercise 14A Q8b ii for why, without a graph, you should not declare that there is no root in the interval because of a lack of signchange! 7 a

By the change of sign principle, since between and .

is continuous in this interval, there must be a root

8

Test solution:

By the change of sign principle there is a root between d.p. 9

and

, which is equal to

to

This sequence of values is heading away from the known solution; you need a better start value.

Test the solution:

By the change of sign principle there is a root between s.f.

and

, which is equal to

EXERCISE 14C 2 a

at stationary points.

b

By the change of sign principle, there is a root between and . c

is close to the stationary point and so may be expected to have a tangent close to horizontal, which will intersect the -axis a long distance from the interval.

to

Tip You are not expected actually to use the value, simply to predict the result. If you choose to check the suggestion, you will find

is on the other side of the stationary point at , so is both a lot further from the point and also unlikely to return to the desired root. 3

There is a stationary point at , so any value will have a tangent with negative gradient, so that the iterative sequence increases, rather than decreasing towards the root between and .

4 a

is not defined at (due to the logarithm) or odd multiples of (due to the tan function), where

.

b

c This value is on the far side of a discontinuity of the graph (also on the far side of a turning point) and so continuing the iteration will not be expected to return to the root value close to . 5 a

There are turning points at The turning points are

and and

. .

b i Start values less than

converge to the first root

ii Start values at a turning point c Start values greater than

.

or do not converge.

converge to the third root

.

6 a

Turning points are at So b

and . . : there is a stationary point at (

: there is a stationary point at

). .

c

The tangent at

crosses the

at (

).

d Because is close to a stationary point, its tangent is close to horizontal. As seen, results in below the stationary point at and so would be expected to converge to instead of . e

Test the solution.

By the change of sign principle, there is a root between to s.f.

and

, which is equal to

7 a There is a stationary point at is due to the logarithm.)

so

. (Reject the negative point as outside the domain of f, which

) is a maximum.

b There would be a point of inflection where

, which has no solutions since

. So there can be no points of inflection.

c As seen on the graph, there is a root between and .

Tip You know that as and there must be a root between and .

so by the change of sign principle

There must also be a root between and . d

For , the next value from the Newton-Raphson iteration is below the lower limit of the function domain, so cannot be used. e For 8 a i

, the iteration will converge to the greater root.

Stationary points occur at interval

and

.

is likely to be too close to the stationary point, so will lead to , which is the locality of the root.

ii Limits for where if

being outside the

for which there is convergence are: then

and if

From your calculator:

then

.

.

b i

A stationary point occurs at

.

is likely to be too close to the stationary point, so will lead to domain, with . ii Require Limits for

being outside the

such that for which there is convergence are:

.

c i

Stationary points occur at odd multiples of , the curve has discontinuities at multiples of . leads to

, which is in a different region of the curve.

ii You can solve this equation algebraically:

No need for an iterative method! Limits for

for which there is convergence are:

.

d i

Stationary points occur at multiples of , the curve has discontinuities at odd multiples of . leads to

which is beyond the discontinuity at

, so this sequence

will not converge to the root between and (in fact it diverges). ii Limits for

for which there is convergence are:

9 a Using the quotient rule:

.

So b

when

, but at both such points,

as well.

and are both positive for all values greater than the root (at so for the sequence will directly converge to the root, and therefore will be closer to the root than . There is a discontinuity at

.

for all values between and the root, so for converge to the root, and again will be closer to the root than and are both negative for from the positive root. In all, the positive values of where . c

10 a

b

for which

so

the sequence will directly .

will be less than

and therefore further

is closer to the positive root are

and

The sequence converges to a root in the given interval for

but not for

c The requirement on is that the tangent at must meet the

.

at , so that, for all values

between and , the tangent meets the axis at a point to the left of and continues on a convergent sequence.

From your calculator, the approximate solution to this is

.

d As seen with = , if , the next term may lie on the far side of a discontinuity and the sequence may not converge at all, or may converge to a different root.

EXERCISE 14D 3 a

i There is only one intersection point; all staircase diagrams will lead to it, irrespective of starting point. ii The same result comes from either side of the root. iii N/A b

i There is only one intersection point; all staircase diagrams will lead to it, irrespective of starting point. ii The same result comes from either side of the root. iii N/A c

i There are two intersection points, and with

.

The greater root will be approached by any staircase diagram with a starting point greater than . ii A starting term less than yields no solution (the iterated value decreases until it drops below , where the function is not defined). iii This iteration will never converge to root . d

i There are two intersection points, and with

.

The lesser root will be approached by any staircase diagram with a starting point less than . ii A starting term greater than yields no solution (the iterated value increases forever). iii This iteration will never converge to root . 4

Test the solution. For

:

By the change of sign principle, there is a root between two decimal places. 5 a

and

, which is equal to

to

b The iteration is converging to a root of The root appears to be equal to

.

to two decimal places.

6

Test the solution. For

:

By the change of sign principle, there is a root between d.p.. 7 a

and

, which is equal to

to

b The sequence converges to a root such that:

. 8 The iteration pattern involves starting at to find . This will occur at , with

, rising to the curve, moving horizontally to the line

at .

9 a has a discontinuity at

.

By the change of sign principle, there is a root between and . b

c

to two decimal places. Check the solution.

By the change of sign principle, there is a root between two decimal places.

and

, which is equal to

to

10 a

For any greater than , fixed point iteration will converge to the greater solution, as illustrated for . b

The root is

to two decimal places.

c At the root,

.

The upper root, towards which the iteration converged, is Percentage error is

.

11 a

The sequence converges towards b At the root,

(approximately).

.

and clearly the iteration converges to the positive root.

EXERCISE 14E 5 a

There are two roots and

.

For starting values greater than , iteration will converge to .

For starting values less than , the iteration will converge to . b

roots and

.

For starting values greater than , iteration will converge to .

For starting values less than , the iteration will converge to .

Tip This second arrangement of the equation has infinitely many solutions, but because the problem posed was in terms of the function arctan, the domain for solutions is restricted to values which can equal , which is the interval

or

c

can be rearranged to: for the iteration or (and many others).

for the iteration

.

There are three roots

and

.

For starting values between and , iteration will converge to .

For starting values greater than , the iteration will converge to . For starting values between approximately and , the iteration will converge to (iterations with starting values much less than will jump into the upper iteration towards . 6 Iteration moves vertically to the curve

and horizontally to

.

a b 7 Iteration moves vertically to the curve

a Converges to . b Converges to . c Converges to . 8 a b Using the quotient rule:

and horizontally to

.

since

;

has a stationary point at the root.

9 a Let

By the change of sign principle, there is a root between

and

.

By the change of sign principle, there is a root between and . b

is always positive, and since for (and can be shown to be less than for ) convergence is to the upper (positive) root. 10 a

By the change of sign principle, there is a root between and .

By the change of sign principle, there is a root between and . b

In the vicinity of the upper root c

is consistently below , so the iteration will converge to .

³

⁴

⁴ ⁴ ⁴

⁴

Test the solution.

By the change of sign principle, there is a root between to d.p. 11 a

b

c For convergence to , require that

.

and

, which is equal to

Mixed practice 14 1

The iteration converges in an oscillatory pattern, which will produce a cobweb diagram. (Answer A) 2

By the Newton Raphson method: Since

is undefined for

, the starting point

(Answer B) 3 a

At the stationary point, b Use the iteration

so .

will not generate

.

Test the solution:

By the change of sign principle, there is a stationary point between equal to to d.p.

and

, which is

4 a The required area is

b Let

So, by the change of sign principle, there is a root between and . c

Test the solution:

By the change of sign principle, there is a stationary point between equal to to two decimal places. 5 a There are two solutions

b Let

with

.

and

, which is

By the change of sign principle, there is a stationary point between and . c

Using the iteration

:

Test the solution:

By the change of sign principle, there is a stationary point between equal to to two decimal places.

and

6 a

b Let

By the change of sign principle, there is a stationary point between and . c

, which is

Test the solution:

By the change of sign principle, there is a stationary point between equals to d.p.

and

, which

7 Let

8 a There is a discontinuity at

, which is not in the domain.

Tip Always find any discontinuities first, then you can be confident in using the change of sign rule.

By the change of sign principle, there is a root between b

and

.

c

which is on the far side of the discontinuity at ; the method fails because the iteration will not converge to the desired root, but will grow without limit, since and for and tends towards as tends to infinity, so each term will be greater than the previous one, increasingly so.

d

Test the solution:

By the change of sign principle, there is a stationary point between is equal to to d.p. 9 a

At a stationary point,

b Let Then

:

and

, which

between criterion is that

and so the iteration will not converge (the convergence in the vicinity of the root).

c Normally, taking the inverse relationship would result in a root. rearranges to The iteration

converges to a root… but not the one you want (it converges to

. This is because the range of arctan is

.

To get the value you want, you need to amend the formula:

This rearranges to Using the iteration

. :

Test the solution: For

By the change of sign principle, there is a stationary point between equal to to d.p.

and

, which is

10 a Let

, which is continuous in its domain

.

By the change of sign principle, there is a stationary point between and . b

c

11 a At The ball was released from

about the ground.

b i

Maximum height is reached when

:

So a possible iterative formula is ii The domain of the iteration is

. , since you require the argument of the logarithm to be

positive. At

the height is positive and increasing and at

Hence there is no root in the interval

so it is still positive.

and so any positive root must be beyond the

domain of the iteration. iii

The root is

( d.p.).

(This is not a valid solution as must be positive). iv You are seeking a solution to An alternative rearrangement is

. , giving the iteration:

v

Test the solution:

By the change of sign principle, the root lies between seconds to d.p. c Seeking a solution to

.

By the Newton–Raphson iteration:

and

, so is equal to

3. Test the solution:

By the change of sign principle, the solution lies between seconds, correct to two decimal places.

and

, so is equal to

Chapter 15 Worked solutions for book chapters 15 Numerical integration Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 15A 2

Number of rectangles Integral

Bound Lower

a

Upper Difference Lower

b

Upper Difference Lower

c

Upper Difference Lower

d

Upper Difference In every case, doubling the number of rectangles from difference between the upper and lower bounds. 3



Total Lower bound: 4 a

, upper bound:

to

or from

to

halves the



Total Lower bound:

, upper bound:

b To reduce the gap between the bounds, divide the area into more strips and approximate with more rectangles. 5 a

b



Total Upper bound c If rectangles were used, the bounds would move closer to the true value, so the upper bound would decrease. 6 The limit is the definition of the integral over the interval.

7 a The maximum point will occur at

.

Tip

b

Two possible interpretations could be made here. For the lower bound, you require rectangles that lie beneath the curve; since the curve is symmetrical, it would be sensible to fit four rectangles beneath the first half of the curve (the rising portion, so that the upper left corners of the rectangles lie on the curve) and double this. Alternatively, you could fit four rectangles beneath the entire curve, with the first two attached at the upper left to the curve and the second two attached at the upper right. Here the first of these is shown, which gives the more accurate estimate. Note that the first interval is divided into five, not four, because the left-most rectangle attaches to the curve at and has zero area, so is ignored.



Total The lower bound on the total area under the curve is twice this:

EXERCISE 15B 2 i, ii

.

Number of intervals Integral

Exact

error

error

error

error

error

error iii The percentage errors shrink rapidly, particularly (as in part a i) when they begin very high, due to a rapidly changing gradient. In most cases, doubling strips appears to shrink the error by a factor of between and , though this is not an absolute rule. 3 a Trapezium rule approximation:

Total Total Approximate area: b Substituting

so

and

:

c

d Trapezium rule approximation:



Total Total Approximate area: e

f

4 a There is a vertical asymptote at

.

b Trapezium rule approximation:

Total Total Approximate area: c The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. 5 a Trapezium rule approximation:

Total Total Approximate area: b A more accurate approximation could be found by increasing the number of trapezia in the interval.

Tip You could obtain an exact value by using a substitution and integration by parts.

6 a

when

so

b Trapezium rule approximation:

Total Total Approximate area: c The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. 7 Trapezium rule approximation for

:

Total Total Approximate area: The distance travelled in the first seconds is approximately

.

8 a The changes of direction occur at the roots of the function (as changes sign). and b Total distance travelled equals total enclosed area. Require Approximate the integral for

with two strips.

Approximate the integral for

with six strips.

Total Total Approximate area:

Total Total Approximate area: Total enclosed area (distance travelled) is approximately

Mixed practice 15 1

ordinates means strips, for a total width of ( Each strip must be wide. (Answer D)

2



Total Approximate area: 3 a

.

.

Total Total Approximate area: b The curve is concave so the trapezium rule will give an underestimate; all the trapezia will lie beneath the curve, with the oblique edges being chords under the curve. c A more accurate estimate could be obtained by using more intervals.

Tip You could also integrate the function exactly, using integration by parts. 4

Total Total Approximate area: 5 a

Total Total Approximate area: b Using more strips ( intervals, for example) will give a more accurate approximation. 6



Total Lower bound: 7

, upper bound:

so the curve is convex; the trapezium rule will provide an overestimate. More ordinates will improve the approximation, so will be expected to decrease the estimate towards the true integral value. (Answer B)

Tip It is possible to design a convex function artificially so that a better approximation is available from four strips than from five (for example) so it is not necessarily true that more intervals automatically gives a better approximation, but doubling the intervals will always improve the approximation for a function that is consistently convex or consistently concave in the interval. 8 Roots are where

.

or Approximating:

Total Total Approximate area: 9 a Roots of

:

at

and

at

and

The product has roots at

and

.

b Approximating:

     

      Total Total Approximate area:

Tip You can calculate this area exactly, using the identity .

10 Upper bound for h



The upper bound for the integral is given by where:

11 a Stationary points occur where

:

so the only stationary point is at

, coordinate

The curve exists for all values of and is an even function so will be symmetrical about

.

b Since the curve is symmetrical, you should use the four rectangles on only one side of the and double the result.

Total The lower bound for half the integral is

; the lower bound for the entire integral is

12 Using trapezium rule to estimate the area under the curve, which will give total distance:

Total

.

,

Total Approximate total distance covered is Estimate of average speed is

. .

Chapter 16 Worked solutions for book chapters 16 Applications of vectors Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 16A

3 a

For a straight path, the average speed is the modulus of the average velocity.

b Average velocity is

, since there is no overall displacement.

Tip You might prefer explicitly to split the answer into components as rather than use the zero vector 0.

4 a so its distance from the origin at b

and Substituting

5 a

. :

is

Magnitude of average acceleration is

.

b End velocity is

.

This direction is between

and

anticlockwise from i, so the exact angle is given by

anticlockwise from i. 6 a Distance

equals

b c

d Distance

equals

This is not equal to the modulus of the average velocity because points and are not collinear. The average velocity only takes account of start and end points and , irrespective of the route taken. The average speed is calculated for total distance travelled and so must be calculated by summing the distance travelled in each leg of the journey. The average speed must be greater than or equal to the average velocity, with equality only occurring if the path is a straight line for the entire journey. 7

The particles will meet if there is a solution

:

So The particles meet at

seconds after starting.

8 a

Distance when b

is

so Substituting into

:

This is a negative quadratic with vertex

and -intercept

. Roots are at

.

9 a So

and

Eliminating :

. is the trajectory.

b Find the shortest distance from the origin to the straight line.

Tip There are many methods to calculate this. A couple are given here, but you will already be familiar with other approaches. Method 1: Closest point Let be the closest point on the line. Then

is perpendicular to the trajectory.

The trajectory

so

has gradient

lies at the intersection of

and

has gradient and hence equation

.

.

Substituting: Distance Method 2: Vector distance For a line with vector equation The line has equation

, the closest distance to the origin is always so

10 Distance from the origin at time is where:

Since

has maximum value

has maximum value of

The maximum distance of the particle from the origin is

.

.

.

EXERCISE 16B

3

. Find v and s.

4 a

. Find r.

So the distance from the origin at time

is

b The direction of movement is at angle above the horizontal. The vector is in the first quadrant so is between and .

5 a

. Find a.

b

. Find .

:

so

.

Checking for consistency: : :

is consistent. is consistent.

The particle’s displacement from the origin is 6

when

seconds.

. Find a.

Magnitude of the acceleration is 7 . Find v.

The angle of v anticlockwise from i is between

and

(second quadrant) so is given by

. 8 . Find .

is a consistent solution to this pair of simultaneous equations.

Magnitude of the acceleration is

.

9

So require :

A check shows that this is a valid solution for both

and

.

The particle’s displacement from its initial position is 10

seconds after it starts moving.

so

So 11 So As increases, speed increases. 12

and

Distance, , from the origin at time is given by:

Need to minimise

.

Differentiating with respect to and setting equal to zero:

Finding at these two values of : When

:

When

:

So the minimum distance from the origin is

.

Tip The time that minimises also minimises so it wasn’t necessary to square root and then carry out the more complicated differentiation on the expression for . The minimum could have occurred at even if that hadn’t been a stationary point. Always check the endpoints for possible global maximum or minimum points.

EXERCISE 16C

3 a

b When

so the speed is

.

4 a

So

b

So When 5 a

Initial speed b

The magnitude of the acceleration when c So 6

is

.

Then:

So the displacement from the initial position is At

.

:

So the distance from the start position is

.

7

The particle returns to its original position when

:

So

.

when

or

and

when

or

There is therefore no time after when both particle never returns to its original position.

and

at the same time; that is, the

8 a

so so

At

, s

and

b So at As expected, this vector has gradient , since the calculations for v are the same as the ones to find the gradient by parametric means.

which has gradient

calculated as

9 a so

, and the parametric method for finding gradient is also

The speed of the particle is

.

So the speed of the particle is constant. b Then So that So the path of the particle is a circle of radius , centred units to the right position. 10

The distance to the origin is where:

From this completed square form, minimum So minimum is

when

EXERCISE 16D

8

9

and Require such that

:

is

when .

.

of the starting

is a consistent solution for each element of this vector equation. 10 Require that

so

.

Tip Take great care when taking a common factor out of a modulus sign; if the factor is an expression rather than a value you must keep it in its own modulus and ultimately allow that factor to take either sign in calculations like this.

11 Require that

:

12

Require

13

so that

2

:

Minimum

occurs for

, when

.

EXERCISE 16E

2 a

b

Midpoint M of AB is given by

3 Require that

.

:

So

Then

is the position vector of .

4 Require that

:

       From the component: Substituting So

.

into the i component:

.

.

This is consistent for all three components in

.

Tip In this case, since the i component gives the same equation as the k component there are only two different equations for the two unknowns p and k. In problems where there are fewer unknowns than equations, be careful to check that the solution is consistent with all the information available. 5 Require that

Components of i and k equal zero for 6

.

7 a

b Either If

or is the midpoint of then has coordinates

So has coordinates

: :

From

:

From

:

9 a Require that

.

.

8

                         

:

.

So the two vectors of magnitude parallel to

b Require that

are

.

:

So the two vectors of magnitude parallel to

are

.

10

11 a

Then:

and:

Two vectors are scalar multiples of each other if they are parallel. Lines

and

are parallel and both pass through and so

b From a,

, so is the midpoint of

Mixed practice 16 1 (Answer B)

.

and lie on a straight line.

2 a

The magnitude of the force is therefore:

b Constant acceleration:

. Find v.

Speed Direction of motion is in the first quadrant so it is at angle

anticlockwise

from direction i. 3

So

must be half this distance:

4 a Require that

:

So

Tip

b

If you are familiar with geometric properties, and have studied scalar products in the Further Mathematics course, you could prove this by demonstrating that the diagonals are perpendicular. You are not limited by the specification in this course when writing your answer if you know a rigorous method from other work.

So

is a parallelogram with adjacent sides of equal length, so is a rhombus.

Tip You do not need to find all the side lengths; it is already established that parallelogram, so the working shown is sufficient to show it is a rhombus.

is a

5

Then

so the speed is

.

6 a

So

So

So so

is isosceles.

b For the points to form a rhombus, must lie opposite (so that the two pre-existing rhombus sides shared with triangle are equal length). Then require that

:

Then

7 a

b

The distance between and is therefore given by:

c

8 Require that v is parallel to

:

for some .

(2)

: (Answer D)

9

so The particle will move in a straight line so reduce the problem to a one-dimensional situation.

Constant acceleration problem:

. Find .

10 a Since the helicopter moves with constant acceleration from rest, the direction of movement will be equal to the direction of the acceleration. Bearing

Tip Remember that for a bearing, you must calculate the angle from the vertical (north) rather than the usual angle over the horizontal. The formula is therefore .

b Since all movement is in a straight line, you can reduce the problem to a linear scenario. The magnitude of acceleration Constant acceleration problem:

. Find .

seconds c The helicopter is flying horizontally (at a constant height). 11

12

so

The particle begins at rest: so So Then Position is in the fourth quadrant so angle is between horizontal.

and

anticlockwise from the

The angle is The particle’s movement is directed

below the horizontal.

13 a

b

c The distance between the ships is where

When d From part c, movement.

.

. so

when

; the ships meet 2 hours after the start of the

e The ships start apart and move at constant velocity, so it will take a further 2 hours after they meet to once again be apart. 14 a Constant acceleration:

b

c Require that v is parallel to

:

:

15

So

Then

(Answer A)

16 a Let the two aircraft be and .

b

The aircraft are at all times at least c At

apart so will not collide.

, the aircraft are at minimum separation:

.

17 so

a

(This is an ellipse with horizontal semi-axis and vertical semi-axis .) b For speed :

So the maximum value of occurs when When sin 18

±1, cos

so

, at which times

.

.

so

Require to find such that

for some :

Taking the initial position as the origin, require r

:

Then The distance of the particle from the origin is

.

Chapter 17 Worked solutions for book chapters 17 Projectiles Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 17A

3 a , ( s.f.) b When

,

Velocity has magnitude

( s.f.).

4

a Maximum height occurs when

.

( s.f.) b When Velocity has magnitude and direction 5 a

Completing the square for :

( s.f.). ( s.f.) (

below the horizontal).

The greatest height achieved is

( s.f.).

b From the symmetry of the completed square, the maximum height is achieved when so the particle returns to the plane after twice that time, at Substituting into the formula for :

.

( s.f).

6

Tip There are many ways to approach this problem. Remember that you can use any known properties of quadratics to give you a quick approach to an answer. Alternatively, you could find the value of t at which the particle returns to the ground and then evaluate at the midpoint of the flight, since the vertex of a quadratic lies midway between the roots. The vertex of the quadratic equals . Completing the square for :

So the maximum height attained is

.

So ( s.f.) 7

Require that

:

This quadratic has roots

and

.

The positive quadratic is less than zero between the roots. The particle is more than 8

above the horizontal level of for

( s.f.).

a When

:

, which has positive root

b When

( s.f.).

(

c

so

9

When

so

.

The ball will clear the wall by more than

.

10 a

When

,

so

The ball hits the ground b For the second ball,

( s.f.) from the foot of the building. where is the angle at which the ball is hit.

Require that:

From

:

(ball is at ground level when it is hit) or

Substituting into

:

( s.f.) 11 a

If the initial velocity is at angle of inclination then:

The ball lands when

:

(at the strike) or Substituting into , at landing:

b Require that

:

or or 12 a What would be observed would be the acceleration due to gravity, since you can’t see the actual forces involved, only their effects. In both cases the acceleration due to gravity is clearly , but in the scale model it will appear to be ten times this so the gravitational force acting will appear to be ten times the normal force acting. Justification: In freefalling

from rest,

so

.

In the film, a scaled-down object would fall but it would still be observed with speed so where is the acceleration due to gravity observed.

b To address this, the film would need to be slowed down by a factor of appear to fall with an appropriate acceleration.

so that objects

Justification: To freefall metre from rest,

so

In the film, a scaled-down object would fall To give the effect that this normal speed.

EXERCISE 17B

1

. from rest in

is actually a full

.

, the film must be run at about one-third

Substituting

into the equation for :

2 a

Substituting

:

b To clear the bar, require that

when

:

( s.f.) c The equation models the ball as a particle, so neglects both the ball’s dimensions and its propensity to spin. Spin, in conjunction with air resistance (also ignored in the model), could cause the path to deviate significantly both within and away from the vertical plane of the model. 3

a Substituting

:

b Relative to the height of the initial position of the arrow, when require .

, to hit the target would

Tip Alternatively, use the modified trajectory equation look for are using.

and

. Either way, make clear what frame of reference you

and Using the trajectory equation from a:

When

( s.f.)

This is within the interval

required for the arrow to strike the target.

4

a Substituting

:

The trajectory passes through

:

So b i Substituting

into

:

So :

ii Substituting into

:

( s.f.) 5

Substituting

into the equation for :

The maximum height occurs at half the distance to the landing,

, so:

:

( s.f.) Substituting

into

:

( s.f.) 6

a Require that the trajectory passes through the position

From

:

:

.

Substituting into

:

Rearranging:

Multiplying through by

b Solving this quadratic in

and using

:

:

( s.f.) The ball was thrown at angle rising, at the point given.

Mixed practice 17 1

since, to go through the hoop, it must be dropping, rather than

The speed of the particle is therefore where:

This is in completed square form, and at The minimum value of is 2 In both cases Therefore

is at minimum .

. (Answer B) so the trajectory paths are identical.

. (Answer C)

Tip This might seem counterintuitive but the difference you would experience – if you were the one throwing the particles – would be the energy expended to give the particles their initial velocities. You would have to put more effort into throwing the heavier particle, but having done so it would follow the same path if the initial velocities were the same. 3 Taking the origin of the system as the point of projection:

a The ball hits the ground when

.

( s.f.) b When

( s.f.).

The ball hits the ground

away from the foot of the building.

4

a If

then

( s.f.).

b Taking the origin of the system as the point of projection:

At ( s.f.) 5

Taking the origin of the system as the point of projection:

a Relative to the point of projection, the ball is caught at ( At the catch,

so

Then

).

. .

(taking the positive root). b At the catch,

so

.

Speed 6 a i Taking the origin of the system as the point of projection:

When

(at ) or

(at )

When

so

ii Maximum value of b

by varying will occur when

, at

°.

so or ° or

7 The initial velocity u has

and is at angle

Taking the origin of the system as the point of projection:

The stone hits the water when

.

(taking the positive root) At

.

.

The stone falls

( s.f.) short of its target.

8 a Taking the origin of the system as the point of projection:

At

(moment of projection) or

( s.f.).

b If the ball has the same speed at then, by the symmetry of the parabola, will be at the same elevation as and an equal distance away from the midpoint of the trajectory (both in distance and in time). Completing the square for :

The midpoint of the trajectory occurs at

so the ball is at at

( s.f.).

9 a Taking the origin of the system as the point of projection:

Substituting

into the equation for gives the trajectory:

b When

( s.f.).

The origin point is above the surface of the court, so above the surface of the court.

represents a point

Tip Always take care to relate your frame of reference – your origin point and axis directions – to the context of the model. c In the model air resistance is ignored.

Tip Ignoring air resistance should always be your first choice of model limitation in this sort of question, besides treating the object as a particle. Issues of spin are significant in conjunction with air resistance and irregular shape and, as such, are secondary to these, though often acceptable in an examination. You could also cite the fact that energy (gravitational potential plus kinetic) is preserved throughout this model (none is lost as thermal energy, for example), but this is again related to the absence of a resistance force.) 10 a Taking the origin of the system as the point of projection:

When

.

b When

. ( s.f.)

c so

( s.f.)

d Angle of the bullet when

is

( s.f.).

The bullet enters the sea at an angle of

below horizontal.

11 Taking the origin of the system as the point of projection:

Require such that

When

(

current direction is perpendicular to original direction):

, the particle is at position

so its distance from the point of

projection is Given all the input values have been given to significant figure, you should give your final answer to the same accuracy. The particle has travelled

.

12 a

Substituting

Since

:

:

b i Taking the projection point as the origin of the motion, the trajectory passes through ( Substituting

ii This is a quadratic in For real root(s), require

into the equation in part a, with

:

, with discriminant

.

:

iii

This is a quadratic inequality in

, with roots:

A positive quadratic is greater than zero outside the roots so (given . ( s.f.)

must be positive)

).

Chapter 18 Worked solutions for book chapters 18 Forces in context Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 18A 3 Resolving forces in the unit directions:

The resultant force is ( 4 a Resolving forces in equilibrium:

b

5

a Forces are in equilibrium. Resolving forces:

Taking

:

b

:

c From ( :

.

6

Resolving perpendicular to : because all motion (and therefore resultant force) is in the direction of .

Resolving parallel to :

7 a

Taking the

force as having direction and

as having direction , the sum of the forces is

. i The angle of from the direction is

.

ii The magnitude of is b Let the

force be

, with magnitude

i The angle between and ii

.

is

.

.

8

Resolving forces at theparticle:

Substituting from ( :

From ( : 9 a

If

then with

Equilibrium: Resolving at the particle

Resolving at :

.

With

:

From ( :

so

From ( :

b From ( :

So From ( :

So : Requiring

restricts the angles so that

c Resolving at with

.

:

( still gives ( : So

Then

and

10 a They are two parts of the same string threaded through a smooth ring. b

so Resolving forces at :

Tip To solve these two equations for and , you can either use substitution to derive equations for and , then by taking the ratio or sum of squares find or , or you can use double angle formulae.) Method 1: Substitution From ( :

Substituting into ( :

Then from ( : :

Method 2 From ( :

Then from ( :

:

From ( :

EXERCISE 18B 2

Resolving for the car:

Limiting friction so Require Then The horizontal force must exceed

for the car to move.

3

Resolving for the block:

Limiting friction so

4

Resolving for the box:

Limiting friction so

5 a

Resolving for particle in equilibrium:

Resolving for particle in equilibrium and limiting friction:

At limiting friction,

b Now the weight at is

.

The force of friction is already maximal at

.

Resolving for :

Tip Be careful here: each particle.

so you need to divide the weight by to find the mass of

Resolving for :

:

6 For each particle the normal reaction force equals the weight, since all other forces are horizontal. at :

Friction opposing movement of is

and friction opposing movement of is

.

Resolving horizontally for the whole system (so that tension forces can be ignored as they will cancel):

at :

Friction opposing movement of is Resolving horizontally as previously:

:

Substituting into ( :

and friction opposing movement of is

.

7 a Constant acceleration:

. Find .

b Normal reaction from the road must equal the weight. Limiting friction:

µ

But also

8

a Resolving forces in equilibrium:

Limiting friction: So

Then

and

So

and

so

. lies in the first quadrant (

and

.

Substituting into ( :

b Require

with the particle moving so require

Solutions to

are

and

.

So

c Set

with the particle moving.

Resolving forces vertically gives the same result:

at limiting equilibrium:

both positive)

Resolving forces horizontally:

Limiting friction:

Maximum occurs for

at which

9 The frictional force equals the lesser of µ and the resultant driving force acting in the contact plane. A is not a feature of the model used since the object is assumed to be a particle and therefore has no dimension. If the surface area were taken into consideration it would affect µ . B plays no part in the model used, so doesn’t affect the frictional force. C relates directly to the resultant of driving and frictional forces, so does affect the frictional force. D will affect µ and so does affect the frictional force, albeit the model used doesn’t give you a way of quantifying this a different value of µ would just be given for a lubricated surface.

EXERCISE 18C 3

a Resolving perpendicular to the slope:

b Taking the positive direction of movement as upslope (so that acceleration will be negative): Resolving along the slope:

Constant acceleration:

. Find .

4 Modelling the child on the slide as a particle on a slope:

a Resolving perpendicular to the slope:

b Resolving along the slope:

Limiting friction:

µ

i Smooth slide: µ

ii µ

5

Require that the system be at limiting equilibrium. Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

So the angle at which limiting equilibrium occurs for is . 6 Take the dominant direction of motion as being upslope throughout the question, so that acceleration is negative.

Moving up the slope:

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

Constant acceleration:

.

a b c Moving down the slope:

as previously. Resolving along the slope:

Limiting friction:

µ

Constant acceleration:

. Find .

7

a Resolving forces for along the slope:

Resolving forces for vertically:

:

b During the initial motion up the slope: Constant acceleration problem:

. Find .

During the second part of the motion up the slope, reduces to so that

Constant acceleration problem:

The total distance travelled up the slope is position is twice that, at . 8

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

µ

. Find .

so the total travelled when it passes the initial

9 Initial motion: Take the two particles as a single body with mass

, so only external forces apply.

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

and

After travelling two metres under constant acceleration: , find .

New system only involves the upper block. and

are unchanged and there is now no tension.

To travel a further one metre under constant acceleration: , find

(Reject the negative solution to the quadratic.)

Tip The acceleration on block is constant throughout the motion. If you are sufficiently confident in your understanding of the system to assert this, you can shorten the working considerably, solving the constant acceleration problem with and for the total metres travelled by . 10

Tip Take great care here the acceleration of the ring will not be the same as the acceleration of the particle. a

Resolving at :

For the ring, the vertical acceleration is given by

:

b Resolving for the particle away from the slope:

Resolving for the particle up the slope, taking friction as acting downslope and acceleration as upslope:

Importantly, you have to note that the acceleration of the ring will be half that of the particle, because every two units of distance the particle moves up the slope will permit the ring to lower by only one unit of distance, since the string suspending it between the peg and the ceiling will lengthen by one unit on each side. .

c Using the equations from parts a and b to eliminate :

The overall force excluding friction is to move the particle downslope, so that in fact friction will act upslope. Correcting for this, and reorienting the accelerations:

The frictional force is acting up the slope to oppose a net force of

.

Limiting friction is Friction is therefore maximal (and the system will move) and friction acts up the slope with magnitude . d

will move down the slope, with acceleration

.

11 a

Resolving perpendicular to the slope:

Resolving up the slope:

Friction is opposing a force of Maximal friction equals limiting friction.

b

. , so the hovercraft will move up the slope under

Resolving perpendicular to the slope:

Resolving up the slope:

Friction is opposing a force of Maximal friction equals limiting friction.

. , so the hovercraft will move up the slope under

c

Resolving perpendicular to the slope:

Resolving up the slope:

Assuming limiting friction (which must apply since the object can move even with

You need to differentiate, so should adjust to using radians instead of degrees. Let

where is measured in radians.

Differentiating and setting the derivative equal to zero to find the stationary point:

:

12

Considering horizontal forces on , it is clear that the inclination of the string either side must be the same (or the ring would slide until this was the case). Resolving vertically at :

Resolving perpendicular to the slope for :

Resolving down the slope:

Substituting for :

Friction cannot exceed µ

, so:

13 For the upward movement:

Resolving along the slope:

For the downward movement:

In both cases, resolving perpendicular to the slope:

There is movement, so the system moves with limiting friction: Cancelling and setting

µ .

:

:



Mixed practice 18 1 Force in the direction of motion is so

.

(Answer B)

2 a b The maximum would occur if the

and

forces were in the same direction:

The minimum would occur if the

and

forces were in the opposite direction:

3

Resolving along the slope:

. .

Tip In the absence of friction there is no need to resolve perpendicular to the slope as you have no need of the normal reaction force. Also, the mass of the block is not relevant, as the acceleration down a smooth slope at inclination is always just . 4 a Resolving vertically:

Resolving horizontally:

Limiting friction:

µ

Constant acceleration:

. Find .

Then b Constant acceleration:

5 a

b Resolving vertically:

c Resolving horizontally:

6 a

. Find .

b Resolving perpendicular to the slope:

c The box moves so friction is maximal (limiting friction):

µ

.

d Resolving along the slope:

e It is assumed that these forces are constant and the only ones affecting the system. Other forces (such as air resistance) can be ignored/are negligible. No rotational effect of forces. 7 Take as vertically upward and as horizontally right in the diagram. For equilibrium,

.

a b 8 a

b Resolving perpendicular to the slope:

c Resolving along the slope:

9

a Resolving along the slope:

b Constant acceleration problem:

. Find .

c There must be an additional resistance force affecting the movement, such as friction between the box and the slope. Air resistance would also be a possible factor (or if the system is immersed in a fluid other than air, resistance to movement provided by that fluid). 10 a

b No net force in the vertical direction:

c Resolving in the direction of motion:

Given

:

d Limiting friction (the block is moving):

µ

e With part of the resistive force accounted for by air resistance, the contribution from friction would be reduced. In the working in part d, the numerator would be reduced and therefore the value of µ would be reduced. 11

Resolving perpendicular to the slope:

Resolving along the slope:

Limiting friction:

The magnitude of the acceleration is

(Answer C)

12 a Resolving vertically for the block:

Resolving vertically for the particle:

Resolving horizontally for the block:

: The force against which friction is acting is Friction will therefore be the lesser of

.

and µ (limiting friction).

µ

The particle will move with limiting friction

.

Tip You cannot necessarily assume that µ . You must first determine the resultant driving force against which the friction will be acting, since the friction cannot exceed that driving force. The question clearly implies that limiting friction will apply but you should not assume so unless it is explicitly stated that the system moves when released. b From ( : From ( : 13 a Constant acceleration problem:

. Find .

Resolving along the slope:

b

is increased until

.

Resolving perpendicular to the slope:

Substituting

:

14 Take the dominant direction of movement as down the slope throughout the question (so that there is positive acceleration but in the first part of the movement both initial velocity and displacement are negative).

a Constant acceleration problem:

Acceleration is

. Find .

directed down the slope.

Resolving perpendicular to the slope:

Resolving parallel to the slope (downwards):

Maximal friction (friction is limiting since the particle is moving)

µ .

b In the downward movement, resolving parallel to the slope (downwards):

Since maximal equals

, there will be downward movement with limiting friction:

15 Take the dominant direction of movement as downslope throughout the question (so that there is positive acceleration but in the first part of the movement both initial velocity and displacement are negative).

Resolving perpendicular to the slope:

Resolving down the slope:

a Smooth plane:

Constant acceleration problem:

After

seconds, the velocity is

. Find .

directed down the slope.

b Rough plane; the particle will move under limiting friction:

Constant acceleration problem:

. Find .

The furthest point the particle reaches is turns around and moves back downslope. The high point is

.

up the slope from the original position. It then more elevated than the starting point.

16

Each scenario involves a constant acceleration problem with the same with the same . The specific values of and are therefore irrelevant.

and and therefore

Let the inclination of the slope be and the two scenarios be numbered and . and Resolving perpendicular to the slope:

Resolving down the slope:

Since in both cases the particle is moving, friction is limiting so

You know that the two accelerations are equal:

17 a Resolving vertically for the block:

µ

.

Resolving vertically for the particle: (1) Resolving horizontally for the block: (2) :

(3)

The force against which friction is acting is Friction will therefore be the lesser of

. and µ (limiting friction).

µ

The particle will move with limiting friction

.

b From part a: c From ( : d Constant acceleration problem:

. Find .

e At the moment the string breaks,

.

The particle falls under gravity so

.

Constant acceleration problem:

18 a Constant acceleration problem:

. Find .

.

i

ii b c

Resolving down the slope:

i If (resistance force) equals zero then

ii If (resistance force) equals a constant

.

then

.

d Air resistance force varies with speed; at

this can be expected to be significant.

19 a

b Resolving vertically:

c Resolving horizontally for the block:

Limiting friction (the sledge is moving):

.

µ

Then

20 a With block heavier and on the steeper slope, anticipate that the net force (and acceleration) will be in favour of moving downwards. With travelling downwards initially, friction will act to slow the movement.

For block , resolving perpendicular to the slope:

Resolving up the slope:

Friction is limiting while the blocks move so

µ

.

For block , resolving perpendicular to the slope:

Resolving down the slope:

Friction is limiting while the blocks move so

µ

.

Substituting and adding ( to ( to eliminate :

b When the blocks come to rest, forces generating movement will remain the same but the forces opposing movement will swap direction.

From block :

From block :

: Maximum frictional forces are:

Total frictional force to oppose motion is

.

Therefore the blocks will accelerate in the direction opposite to the initial velocity and will pass again through their original positions. Initial movement:

Constant acceleration problem:

. Find and .

The blocks come to instantaneous rest after

seconds,

from their start positions.

Subsequent movement:

Constant acceleration problem:

. Find .

seconds Total time from the start:

seconds.

The blocks are back in their initial positions

seconds after the start of the movement.

21 a i

ii Resolving vertically:

iii The block is moving so friction is maximal (limiting):

µ

iv Resolving horizontally:

b Reducing so that horizontal forces are in equilibrium:

Tip Be careful here if you reuse then you are neglecting the fact that affects the normal reaction force and thereby will change as changes.

c Since the calculation in part b is independent of the speed at which the block is moving, the tension needed to oppose the frictional force and maintain a constant speed will be the same regardless of that speed (assuming resistances which are functions of speed, such as air

resistance, can be disregarded). 22 a

Resolving perpendicular to the slope:

Resolving in the direction of the slope in limiting equilibrium:

Limiting friction so

µ :

b i Resolving perpendicular to the slope:

ii Resolving in the direction of the slope upwards:

Limiting friction so

µ (now acting down the slope) where

23 a

b Resolving in the direction of the slope upwards:

.



Chapter 19 Worked solutions for book chapters 19 Moments Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions.

EXERCISE 19A

EXERCISE 19A 5 Moment sum of Clockwise moment: 6 Moment sum of Clockwise moment: Anticlockwise moment: Resultant moment:

clockwise

7 a Assuming the plank remains rigid: Moment sum of Moment

(1 s.f.)

b In asserting that the perpendicular distance between the diver’s weight and the pivot point is constant at , and that the perpendicular distance between the weight acting at the board’s centre of mass is a constant , you require that the board remains exactly horizontal and rigid. 8 Perpendicular distance from the force to the centre is

.

Tip This is clear for the points at the top and bottom, and since it is a regular hexagon must also be the case for the other side – don’t start doing trigonometry unnecessarily! Moment sum of Clockwise moment 9 Let the bottom left corner of the square be the origin, and any point in the square be defined by its coordinates relative to that corner. For the point (

, the four forces have perpendicular distances

and

.

Moment sum of Clockwise moment Hence, the clockwise moment is the same for every point as the moment is independent of and . 10 Moment sum of Distance from centre of rod to is Distance from centre of lamina to is

. .

EXERCISE 19B 3 Let the distance from the end to the point where the second child sits be Taking moments about the pivot point of the seesaw:

.

4 Taking moments about the vertical line of the hinges:

where is the perpendicular frictional force acting through the wedge.

5 a Let the counter mass be metres from . Taking moments about the point of support with the crane’s upright:

The counter mass should be

away from

b The point of attachment with the crane upright should have tolerance for some net moment if the crane is to be usable in practice. 6

Let distance

be given by .

The plank is on the point of tipping about so must lie between and as shown, so that the forces are in equilibrium and the reaction is zero. Taking moments about :

So 7 a Let the tension in the wires be

and

The total weight being supported is

.

N so

.

Suppose that the rod remains in equilibrium. Then taking moments about : so

s.f.)

That is, to maintain equilibrium, there would need to be a thrust of Since a wire cannot provide thrust, the rod will rotate about .

through the wire at .

b Including the weight at and taking moments about again:

For So

to be minimum,

(i.e. the rod is on the point of rotating about ).

.

8 Let the reaction force at the end be 1 and the reaction force The total upward force must equal the weight of the plank:

from the other end be 2. 2

.

Taking moments about the end with the support:

so 9 Let the tension in the wire nearer the painter be 1 and the tension in the further wire be 2. The total tension must equal the total weight carried:

2

.

Taking moments about the end nearer the painter:

so 10 a Let the downward force at the end of the pole be and the upward force at the other handhold be The total upward force must equal the weight of the pole so

.

Taking moments about the end of the pole being held:

So b The working in part a assumes that the pole is rigid and horizontal. 11 Let the vertical force nearer the handle be and the force nearer the blade be . Then the total

must equal the total weight of the spade:

.

Taking moments about the point of :

s.f.) So

s.f.)

12 Require equilibrium in moments. Taking moments at the elbow joint:

s.f.) 13 a i Let the tension the wire be . Taking moments at : clockwise moment anticlockwise moment

ii For equilibrium of forces: the vertical component of the force at (the normal reaction) equals

.

iii For equilibrium of forces: the horizontal component of the force at (the friction) equals

.

b A rod can be under tension or compression but a wire can only be under tension; since , in a system with a wire must be positive so (the resting point is on the same side as the wire tether). In a system with a rod, can take any value between and .

14 Let the horizontal component of the force at the upper hinge be Taking moments at the upper hinge, with the lower hinge weight away:

Vertical forces have equal magnitude:

(away from the door).

below it and the line of action of the

.

So magnitude of force at is

.

Horizontal forces have equal magnitude:

So magnitude of force at is 15 Let the heavy rod be

.

and the lighter rod be

The balance point is where

.

Moment in favour of descending:

)

3.5gx s.f.) 16 Let the distance from the people end to the edge of the cliff point be metres. It is assumed that the coach will not slip! For equilibrium, .

Taking moments about the edge of the cliff:

of the coach overhangs the cliff. So

s.f.) of the coach overhangs the cliff.



Mixed practice 19 1 Taking moments about : clockwise anticlockwise 2 Taking moments about :

(Answer B) 3 Taking moments about the left end of the rod:

(Answer A)

4 a Taking moments about :

b Resolving the system vertically shows that the vertical component of the force at is . There are no horizontal forces in the system. The force at is therefore

upwards.

5 a

b Taking moments about : 3.2 × × c The total of the two reaction forces must equal the total weight of the system:

So

s.f.)

d It is assumed that the centre of mass is at the midpoint of the plank, as will be the case for a uniform plank, so the weight of the plank acts through its centre. 6 a i Let the tensions in the ropes be

and

.

The total tension must equal the weight of the plank:

.

Taking moments about : 2 ×

×

So So

s.f.) and

s.f.)

ii It is assumed that the centre of mass is at the midpoint of the plank, as will be the case for a uniform plank, so the weight of the plank acts through its centre. b The total tension now equals tension . Taking moments about :

Cancelling :

or

. If the tension in the two ropes is equal then each has

7 a Ken stands at the point . The distance from to the river bank is so from the middle of the plank to is and is .

b Taking moments about :

Cancelling by throughout: .

is

from the river bank.

c The centre of mass is assumed to lie at the midpoint of the plank. 8 Let the distance from to the wire be . Taking moments about the point of connection of the wire:

9 Total load on the supports is

.

Therefore if equal, the normal reaction at each support is

.

Let the distance of the mass from be . Taking moments about the support nearer to :

10 a Let the tension in the rope at be given as so the tension in the rope at is The total tension must equal the total weight of the plank and the particle. So Then

. and

, the tension at , equals

.

b Let distance CB be . Taking moments about :

11 a Taking moments about , with reaction at given as

:

.

b Let distance AD be . The sum of the two reactions must equal the total weight of the plank and the child, If the two reactions are equal then

.

.

Taking moments about again:

12 a Taking moments about after the weight is added at : 125(

)

Multiplying through by :

b With the weight at , again taking moments about : 125(

(1)

)

:

The weight is

and the distance is

.

13 a Let each of the normal forces at and be . Taking moments about :

The supports are holding the plank and weight so 2R Substituting

into

.

:

xW (

b Since

must be finite (and given you have no information as to the strength of the plank), the

equation in part a implies that ⩽

.

Tip The plank will bend significantly – and then break – for high values of W, so in fact will have a maximum value somewhat lower than

, but that is a topic for

material sciences and beyond the scope of this specification. 14 Taking moments about the pivot point, requiring equilibrium: ( So

mg( –

Tip You can check an answer for sense even when the question has abstract values. In this case, you can check that if m then the pivot point should be at the join line and if then the pivot point should be in the centre of the M block ( , both of which can be seen from the answer. 15 If the overhang is then the distance from to the centre of mass is also , since and are 3 m apart, for a rod of length . Let the normal force at be

.

Taking moments about :

So

for

If the rod is moving then friction at is limiting and frictional force

NS.

For the rod to move at constant speed, the force moving the rod must exactly counter this frictional force, so the rod must be pushed with force . 16 a Let the normal reaction of the ground to the rear wheel be ground to the front wheel be .

and the normal reaction of the

Taking moments about the rear wheel:

b At maximum acceleration, friction will be limiting (the car would slip rather than be provided greater horizontal force) Maximum friction

×

ma so The greatest acceleration available is

 s−2.

Chapter 20 Worked solutions for book chapters 20 Conditional probability Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 20A

Tip Each of these questions could be answered using any of a variety of methods. For example: abstract algebra, Venn diagram illustration, tables or tree diagrams. A range of methods is given here but all ultimately use the same logic and working.) 3 a Taking

, then:

Since the total in the diagram must equal :

b 4

or 5 Taking

, then:

Since the total in the diagram must equal :

Then:

6 Let be the proportion of teams containing French players and be the proportion of teams containing Italian players. Taking

, then:

Since the total in the diagram must equal :

Then 7

or so

.

and are independent so Hence Then 8 a Putting the information in a table: Glasses

Not glasses

Total

Brown Not brown Total Eye colour and glasses-wearing are independent, so the value in each inner cell must be the product of the total for that row and the total for that column.

P(brown)

and P(not glasses)

P(brown not glasses) P(brown) P(not glasses)

or P(brown)

or

b Similarly, P(glasses)

or

EXERCISE 20B n( ) means the number of items in set .

Tip In several of these questions, a complex Venn diagram is used to illustrate an answer. Preliminary calculations for populating the diagram are given in the worked solution, but you don’t usually need to show them in an examination. 5 a

b The number playing neither sport equals c d 6 a

only) only)

b From the diagram:

.

c

7 a so so Then

b 8 so so Then

a b 9 Numbers selected from: to a Since

, there are

multiples of .

(multiple of ) b Since

, there are

multiples of

.

(multiple of and ) c 10 a Labelling as Bolognese, as chilli and as vegetable curry:

b From the diagram: c d e f

11 a

b From the diagram: c d e

so blue eyes and dark hair are not independent.

12 a

b .

c d e The two events are not independent. 13

14 Let Then

and

15 The maximum value of

is

, when is included within .

The minimum value of

is

, when

Tip Be careful, it is an easy mistake to say that the smallest the intersection could be is zero; they cannot be mutually exclusive because . Then

16 a Any probability must be non-negative and no greater than :

b

c

EXERCISE 20C 2 a Ensuring that all row totals and column totals are achieved (including the grand total): Year 9

Year 10

Year 11

Total

Girls Boys Total b

c

3 Extending the table in the question: Milk Sugar No sugar Total a

No milk

Total

b

c

so milk and sugar are independent; knowing the presence or absence of milk gives no information as to whether sugar is present.

4 Expanding the table: Store A

Store B

Store C

Total

Returned Unreturned Total a

b

c

5 Expanding the table: Total

Total a b c d e 6 Expanding the table: Gold USA GB China Total a b c

Silver

Bronze

Total

d 7 Expanding the table: S

M

L

Total

White Black Green Total a b c

d

e 8 Expanding the table: Total ℃ ℃ to

℃

℃

Total a i ii iii

9

is incorrect. For example, suppose James rolls a fair

-sided dice with sides numbered to

Let be ‘even number’ and be ‘prime number’. Total

Total

But On the other hand, is true. If you know is the case then either or

must be true so

.

EXERCISE 20D

Tip In all these questions a tree diagram, populated with relevant values, is sufficient preliminary working. The tree diagrams have been given in these answers, together with full algebraic working such as would be needed for an answer if a tree diagram were not drawn. 2

3

Tip In a question like this, you don’t need to draw a ‘full’ tree diagram, since only a few results are of interest. Thus, after blue in the first draw, all that is of interest is whether the second ball is green or not; G/G′ branches are sufficient there is no need for three branches B/G/R. Similarly, only B/B′ are needed after G and a first draw of R need not be further detailed.

4 a

b c d 5

a

b

Alternatively:

6

7

8

a

b

9

a

b

10

a

b

11

a

b

12

is the event that a patient has the disease. is the event that the patient tested positive for the disease.

Tip This is a very real problem in medical tests; for a condition which is very rare, a test regime has to be incredibly accurate to be reliable. As shown here, despite a seemingly very predictive test, the probability of a positive test result being a false positive is much more likely than a true positive because the incidence of the disease ( ) is so much lower than the probability of an incorrect test on a healthy patient ( ). 13

is the event of picking the fair coin. is the event of flipping heads.

14

Let the number of black counters be .

or 15 and are independent if and only if

This, by definition, means that

and

.

are independent.

Mixed practice 20 1

Tip You could recast the question as starting after the first choice; what is the probability of drawing two milk chocolates from a box containing five milk and four dark, avoiding the conditional altogether. 2 There is enough information to populate a Venn diagram completely:

so and are not mutually exclusive. and (Answer B). 3

4 a

If is the number with both genes, then:

b c 5 a i ii iii

iv v

so and are independent.

6 a

b c d 7 Adding totals to the table: Hours practised

Total

Total a b 8 a Let be the event that Alison bowls on Wednesday and Thursday. i

be the event that she bowls on

The evenings are independent so:

ii

b Let and respectively. i ii

iii iv

9 a

be the events that David bowls on a Wednesday and on a Thursday

b c

d

e 10

11

12 Let

mean that he left the umbrella in the first shop, that he left it in the second shop and be that he kept the umbrella in each case. Clearly terminates the events and so needs no branches from it in the probability tree, so the three possible event paths are or or .

13 a b

(reject c

as not valid in context.)

d

Tip

14

The diagram is the clearest and fastest way to show working, but the full algebraic working is also given here.

15

16 a

If

:

So Then when b



for .

, with

when and are mutually exclusive,

Chapter 21 Worked solutions for book chapters 21 The normal distribution Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions. EXERCISE 21A 4 a b 5 a b i ii c It is assumed that his probability of success is identical on each attempt, and that the attempts are independent of each other. It might be speculated that he would improve with the practice or get increasingly determined (so his chance of failure would decrease) or that he might get tired or dispirited (so his chance of failure would increase).

Tip You are free to argue in whichever direction you think reasonable, having established what the assumptions are. 6

So the expected number in a sample of 7 Points of inflection are at

.

standard deviation from the mean.

So the mean is the midpoint: µ Standard deviation

is

.

.

8

Mean is at the mode: µ

.

Standard deviation is distance from inflection to mean:

.

9 Mean is at the mode: µ

.

Standard deviation is distance from inflection to mean:

.

10 a b Assuming the runners are selected at random from the group: Let be the number of runners in the team of four who run under

c The real probability is likely to be greater than

seconds.

.

To beat a total time of does not require all four runners to beat a time, it requires their mean to beat . If one or more runs the distance in well under , they can still beat the record even if the rest run relatively slowly. The athletes chosen for the team are likely to be at the upper end of the distribution rather than randomly selected, so their team mean time should be less than and their standard deviation also smaller. Unless the data relates to the exact conditions under which a race will be run (weather, track, competition) then the conditions encountered will cause the experienced mean to be different. 11 a

b

c 12 a

b

13 a of apples are medium. b Let be the number of medium apples in a bag of ten.

14 a

b

15 a b

16

Let be the number of people in the sample of

who receive a less than effective dose.

17

18

and By symmetry about µ

for any .

Then:

EXERCISE 21B 2

is the score in an IQ test.

3

is the mass of a rabbit.

4

is the amount of coffee dispensed (

a b

).

5

is the time to take a test.

a b (

)

c Let be the number of students in the sample who complete the test in less than

minutes.

6

of the data is within this range.

Tip This assumes that the range is centred on the mean, which while not necessarily the case is a valid assumption. 7

is the time taken to do a maths question (seconds).

8

is concentration (%).

Distribution is symmetrical about µ so Then

. so

Tip Did you try sketching a graph? What is the probability of being below 9 a Normal distribution is symmetrical about the mean, so median = mean.

b

Then So 10 Let

, so that

.

?

Then But

, by the symmetry about as shown in the diagram.

So

So 11

12 If is a uniform continuous distribution across To transform to a normal

EXERCISE 21C 3

is the diameter of a bolt.

4

is the energy of an electron (eV).

, take

then

for .

.

5

is the height of a plant (cm).

and

:

Substituting into (1): 6

is the time taken to start a computer (seconds).

and

:

Substituting into (1):

Tip This calculation puts the mean less than two standard deviations above zero for a distribution that can only take positive values, and therefore it might not be appropriate to call it ‘normal’. 7

is the voltage of a battery (V).

So

Estimate that the batteries have been used for 3.23 hours (or 193.5 minutes). 8

is the measured temperature (°C).

Standard deviation is 9

.

is the time waited for a train (minutes).

Let be the number of times out of three that a person waits over

minutes.

10 Let be the temperature in the oven (°C) and the set temperature be

Using the symmetry of the distribution:

.

.

So

Approximately

of the time, the oven temperature will be within

of the set temperature.

EXERCISE 21D 2 For a normal distribution to be appropriate, ideally three standard deviations either side of the mean should be possible values. However, µ children, so the normal model would have a substantial proportion of families predicted to have a negative number of children. 3 a Sample standard deviation b

All data lie within this range, so the normal distribution appears to be suitable, assuming the data are symmetrically distributed. 4 a Reading from the graph:

Angle Frequency

b Taking xi as the midpoint of each category:

Tip A parameter with a ‘hat’ on is the best estimate for that parameter. So could also be written as , since it is the best estimate for available. You might also write here as , the standard deviation of the sample. If you study statistics at a higher level you will also encounter , the estimate of the population standard deviation based on a sample. c The graph is symmetrical and bell-shaped. d Using these estimated parameters, estimate over . 5 Let be the number of heads in

so expect

out of

students to

throws.

a i

ii

b

Tip As mentioned in the chapter, this should properly be calculated using the ‘continuity correction’, whereby you would interpret the probability of more than heads as , since the event is equivalent to when read from a continuous scale. which is much closer to the binomial being estimated, with an error of . The continuity correction is not required in the A Level specification. 6 Let be the number of people voting for this party. can be approximated by a b 7 a The line of symmetry of the graph is approximately

.

The points of inflection of the graph (when made continuous by joining the tops of the bars to form a smooth curve) are at approximately and , a distance of from the mean. Since the points of inflection of a normal curve are one standard deviation from the mean, .

b mean: variance:

So

and

.

8 Let be the number of children out of

contracting the disease.

approximated by H0: H1:

(one-tailed test)

Use significance level Critical zone is

.

where

So

The critical region is therefore 9 a If

and

.

then summing these together:

.

. However, if, for example b

and

, it is true that

is approximated by



The solution

gives

but

.

.

(1)

and sorelates to

.

Tip The solution gives and relates to false solution introduced when squaring both sides of equation (1). 10 Standard deviation If

then

, a

since .

So,

i.e. the mean is greater than three standard deviations above zero, so all values of within three standard deviations of the mean must be positive.

Mixed practice 21 1 A continuous distribution has 2

is a test score;

for any . (Answer A) .

a

of students have scores above

.

b By the symmetry of the distribution, the mean score lower of students. 3

is the mass of a kitten (kg).

is the boundary between the upper and

).

a So the expected number of kittens out of

with a mass less than

Tip An expectation, like a mean, need not be a realisable value. b

4 Require

:

So

5

. Since

6 a i

, require that

:

is

.

ii b Interval symmetrical about the mean so require

7 The UQ is µ

and the LQ is µ

where

.

. If the IQR is

µ

then

.

So

(Answer B) 8

is the height of an adult female of the breed described.

.

a b Let be the number of dogs, in a random sample of females, that are more than

9

is the height of a tree in the forest (m).

.

a b Let be the number of trees, in a random sample of

10

is a test score;

.

a Rearranging:

(1)

Rearranging:

(2)

b

:

so

Substituting into (1): 11

is an angle estimate (degrees).



(1)

:

(2) so

Substituting into (1):

, that are more than

tall.

tall.

12

is the volume of drink in a bottle (

.

a i ii Using the fact that the interval is symmetrical about the mean:

iii For a continuous distribution, the probability of having any specific value is zero so:

b

so

c Let be the number of bottles in a pack which contain between

and

of drink.

From part a ii, Then 13

is the number of female students in a group of size .

a The binomial can be approximated by the normal if can be approximated by a normal

and

. Since

, for

.

b c A binomial distribution requires a constant probability for each ‘trial’, that is, the gender of each student attending must be independent of all others, which is probably not valid in this case. 14 a Range of scores included within three standard deviations from the mean is to , all of which are possible scores. The mean and median are very close, suggesting the distribution is symmetrical.

Tip The distribution could still be far from normal – it could for example be bimodal, but you cannot tell any more detail from the statistics given. There are no strong reasons the distribution should not be normal. b

is the score of a student in the examination; For distinction boundary : For merit boundary

:

For pass boundary : 15

).

so so

. .

so

.

is the breaking force for a chain (kN)

a Let be the probability of one link breaking under

Then the probability of a -link chain breaking is

b

force.

.

16 a

is the diameter of a Playa Gauss grain of sand (mm).

so (1) so (2) :

so

From (1): b

is the diameter of a Playa Fermat grain of sand (mm).

so (3)

Then

and so (4) :

From (3):

so

Chapter 22 Worked solutions for book chapters 22 Further hypothesis testing Worked solutions are provided for all levelled practice and discussion questions, as well as cross-topic review exercises. They are not provided for black coded practice questions.

Tip In these worked solutions, it is assumed that calculations on a stated normal distribution can be made accurately, either by using a calculator or by using the formula , so the working often moves directly between probabilities and -values. Refer to Chapter 21 for further detail in working.

EXERCISE 22A 3 a Assuming the energy values of particles are independent:

b



c 4

is the mass of a dog of the given breed (

).

so

5

is the volume in a carton of apple juice ( so

6

is the mass (grams) of an egg. so Require that

:

Then

.

must be at least 7

.

is the lifetime (hours) of a bulb. so

so

).

8

is the length (

) of a fly.

so so

9

is the diameter (

.

) of an apple.

so a

because this is a symmetrical interval about the mean. :

b Require

so

Must choose at least

.

c Require

so

Must choose 10

.

.

times as many apples.

is the mass (in kg) of a student. So a

Tip

b

With a little care, you can adjust your working to consider the least or greatest value in a group. The calculations for second or third greatest/least are much more arduous. If the heaviest has a mass less than

then all of them must have a mass less than

Assuming the sample students have independent masses:

EXERCISE 22B 4

is the height ( a

) of an 18-year-old in England.

µ µ

(one-tailed)

.

b

, test at

significance level.

Do not reject there is insufficient evidence at the significance level to show that the class mean height is significantly greater than that of the general population of 18-year-olds. 5

is the time (minutes) taken in the writing test. µ

(one-tailed)

µ

, test at

Reject there is evidence at the decreased after the typing course. 6

significance level.

significance level to show that the mean typing time has

is a score in Mathematics GCSE. a Assume that: the scores in the school follow a normal distribution the standard deviation in the school is also the b

students chosen were randomly selected from the cohort in the school.

µ µ

(one-tailed) , test at

significance level.

Do not reject there is insufficient evidence at the significance level to show that the school mean is significantly greater than the national mean. 7

is the height (metres) of an apple tree. a

µ µ

(two-tailed)

b

, test at

significance level (

The critical region will lie outside

in each tail)

where is such that

.

The critical region is c 8

lies within the critical region so reject the new orchard differs significantly from

there is evidence that the mean height of trees in  m.

is the mass of a patient. a

µ µ

(one-tailed) , test at

Critical value is such that

significance level. .

Critical region is

.

b The test assumes that the post-diet masses also follow a normal distribution. c 9

is within the critical region of the test so reject at significance: There is significant evidence that the diet has reduced the mean mass of patients. is bubble volume

a

.

µ

(one-tailed)

µ

b The significance level is c Require that

. so

.

For the geologist to reject the null hypothesis, he would have needed a sample size 10

is the amount (

.

) of coffee in a cup.

µ

(one-tailed)

µ

Test at

significance level.

a

Cannot reject .

there is insufficient evidence that the mean volume in a cup is less than

b Require

:

The minimum sample size is

.

11 The -value represents the probability of finding the observed value or ‘more extreme’. With a two-tailed test, ‘more extreme’ means further away from the mean in either direction, whereas for a one-tailed test it means further away only on one side. For

, the -value will be

For greater.

, the -value will be

EXERCISE 22C 3 a (two-tailed test)

. , which will be

b

so reject results in a Mathematics test.

there is evidence of a significant correlation between IQ and

4 a (one-tailed test) b

so do not reject there is insufficient evidence to show that car speed is positively correlated with distance from the junction.

5 a The population being sampled is ‘countries in 2013’. b (one-tailed test) c

, test at

significance.

From the table, the critical region for

is

.

Reject at significance − there is significant evidence of a negative correlation between unemployment support and amount spent on education. 6 a (two-tailed test) b

, test at

significance.

From the table, the critical region for

is

.

Reject at significance − there is significant evidence of a correlation between water use and temperature. c Correlation does not imply causation (and it would be hard to see how water use could drive daily temperature anyway − the reverse causal link would seem more probable!) Furthermore, the test only showed that a correlation exists, not the direction of that correlation. A one-tailed test with (using fresh data, to avoid data-snooping) would be needed to find evidence that there is a positive correlation. 7 a (one-tailed test) b

, test at

significance.

From the table, the critical region for

is approximately

.

Reject at significance − there is significant evidence of a negative correlation between dose and antibody level. c A more strict significance level reduces the probability of a false positive and therefore helps prevent ineffective therapies being mistakenly classed as effective.

Tip There is always a pay-off. Reducing the false positives will potentially increase the probability of a false negative, which, in context, would mean discarding an effective therapy because it was not shown to be significantly effective in a drug trial. Where a drug is only therapeutic for a subset of patients this is particularly problematic: most medical conditions are multi-factorial; if drug will cure patients with a particular rare gene, but is ineffective in all other patients, then it may be discarded at an early stage of trials if this linkage is not detected, since it will not have a sufficiently significant benefit in general trials. 8 a

so this is significant evidence of positive correlation.

b The -value for a two-tailed test would be twice that for a one-tailed test. there would not be significant evidence of correlation.

so

c If increases then it becomes less likely that this is a chance result from a population with The -value therefore decreases.

.

9 (two-tailed test) From the value table,

is significant for a two-tailed

significance test for

.

10 With only two data points there is always a perfectly fitted straight line, with positive gradient ( ) or negative gradient ( ). Zero gradient ( ) has zero probability if the dependent variable is normally distributed. As such, the value of cannot have any useful means of showing whether the two variables are correlated.

Mixed practice 22 1 (one-tailed test) . Test at

significance.

From the table, the critical value is For a positive correlation, require rejecting . (Answer C) 2

. so C is the least of the given values that would result in

is the breaking load of steel wire. Then

3 a The population is ‘countries in 2015’. b (two-tailed test) c

. Test at

significance.

The critical region is Do not reject rates. 4

so

does not lie within the critical region.

there is insufficient evidence of correlation between HIV rates and literacy

ismass (grams) of cakes. a

µ µ

(two-tailed test)

b i is the -value for the data. ii 5

so do not reject there is insufficient evidence to conclude that the mean cake mass of the new baker differs from the previous mean.

is the percentage increase over nominal quantity. Then µ µ

(one-tailed test)

, test at

Reject quantity.

significance level.

there is evidence that the volume is significantly less than

above nominal

6 (one-tailed test)

Critical value is such that So

. . (Answer D)

7 a For the Pearson moment correlation coefficient for critical values to be used, both variables (times spent revising and test results) must be normally distributed. b (two-tailed test)

Tip You might assume that revision would be be linked with improved test results, indicating a one-tailed test. However, the question phrasing is neutral and since it could equally be argued that weaker students would spend more time revising, you should take a two-tailed test. Whatever you choose, be clear so that your working can be assessed on the basis of your interpretation. The critical region for

in a

Since , reject spent revising and test results. 8

significance two-tailed test is

.

there is evidence of a significant correlation between time

is the time (seconds) taken for a full kettle to boil. a Assume that after cleaning the distribution of times remains normal with the same standard deviation seconds. µ µ

(one-tailed test) , test at

significance level.

Critical region is

where

So the critical region of the test is b

so reject in boiling time.

.

.

there is evidence that cleaning the kettle caused a significant reduction

c Since a single device is being used in a heating experiment, both it and any contents must return to room temperature between trials. 9

is the time to complete a crossword. a

µ µ

b

(two-tailed test)

two-tailed significance levels for a -test are

.

c

. which does not lie in the critical region. Do not reject there is insufficient evidence to conclude that the new strategy changes the time taken to complete a crossword.

d If checking solely for improvement, the test would have been one-tailed and the critical region . Under that scenario, the value observed would lie in the critical region, so would be rejected and the conclusion made that there was evidence of a significant improvement in time. 10 a The sample must be randomly selected from the batch (not all from the same machine or time period as independence would be in question). b

is the power rating (watts) of a bulb. µ

(two-tailed test)

µ

. Test at

significance level.

Do not reject there is insufficient evidence that the latest batch has a different underlying mean than is asserted in the description. c i

µ

(one-tailed test)

ii The critical region

would become

, where:

and That is, the value changes from

to

.

iii The -value is less than so for a one-tailed test at significance, the conclusion becomes rejection of there is evidence that the latest batch has a significantly higher underlying mean than is asserted in the description. 11 The -value is the probability of the observed result or more extreme, calculated under the assumption that is true. (Answer D) 12

is the distance (metres) the athlete jumps in a long jump a (two-tailed test) , test at

b

significance level.

The acceptance region is

where is given by:

So

So the acceptance region is

.

Or, alternatively, the acceptance region is c Require

.

to lie beyond the acceptance region:

so the least value of resulting in a rejection of 13

is .

is the viewing figure (millions) for the series. a The null hypothesis mean must lie at the midpoint of the acceptance region.

(two-tailed test) b The acceptance region can be rewritten as But

. .

The rejection region to the right of the acceptance region represents and therefore the rejection region to the left of the acceptance region must also represent . P(reject) To the nearest integer percentage, the test has significance level

.

Cross topic review exercise 1 worked solutions Worked solutions are provided for all levelled practice and discussion questions, as well as cross topic review exercises. They are not provided for black coded practice questions.

Cross topic review exercise 1 1 Proof by contradiction: is rational so

for some integers and .

Suppose that is rational. Then Then

where

and

is irrational.

for some integers and .

are integers. By definition,

is rational which contradicts the initial

conditions. Conclude that for rational and

irrational, cannot be rational.

2 By the formula for the sum of terms:

Comparing the coefficient in

:

So 3 The mean value equals the sum of the terms divided by :

4

Comparing coefficients:

If If Consistent with both Solution:

,

or

,

5 a The domain of f is restricted to

and

.

. .

Since the range of the square root is the positive reals, the range of is b i The function is one to one, so ii

is defined.

.

Domain of c

range of

is negative for

6 For

,

For

,

The graph of

. ; that is,

, and is not defined for

so

for

so so has a corner at

so the derivative is not defined.

7 a

b There is one solution on each arm of the graph:

c From the graph,

outside the solutions in part :

or 8 a i ii By the factor theorem, if b From part a i, and .

then

is also a factor of

Expanding: Comparing coefficients:

so

is a factor of

. for some integers

so so Consistent. Consistent. So

9 a Completing the square:

The range of

is

.

b If , and for some (so that the three expressions given represent consecutive terms of an arithmetic sequence with difference ) then:

So But, as shown in part

is outside the range of

for real values of .

Therefore the three expressions cannot represent consecutive terms of a real arithmetic sequence. 10 a This is a geometric sequence with first term

, common ratio

.

Tip The standard formulae must be adjusted if the sequence is numbered with the first term instead of . You can check the validity of your formulae, using or .

b 11 This is a geometric series with common ratio Geometric series converges for

.

.

So

The expression inside the modulus sign is never less than disregarded.

Quadratic has roots Positive quadratic is less than zero between the roots. Problem solution is

12 a

so the modulus sign can be

, b Vertical asymptote at

, horizontal asymptote at

Axis intercepts are

Range is

and

.

.

.

c Require the domain such that

; from the graph, this is

The range of h for this restricted domain is

,

So the range of

.

is

,

or

.

.

13 For equal roots, require discriminant

14 a The amount eaten in week

.

is

.

(Geometric progression with

and

)

So the amount of food eaten in the fourth week is

grams.

b The sum over weeks is:

Require

so

.

c Then

The bag will last for

full weeks.

d The model may work for a while but percentage growth rate is likely to decline as the kitten ages so the kitten will probably eat less than predicted and the bag will consequently last longer, so part is an underestimate. 15 a

for some constants and . Multiplying both sides by the denominator on the LHS:

Comparing coefficients:



,

b i

The expansion converges for

so

The expansion converges for

.

so

.

The sum of these two expansions gives the expansion for f( ):

ii The second expansion is only convergent for approximation is not likely to be accurate for

.

is not in this interval so this

.

16

Tip The large pi notation is for products, just as the large sigma notation is used for sums.

17 a

so the function is not one-to-one.

b c The derivative is unaffected by the vertical translation and has its sign reversed by the reflection.

18 a

reflected in

becomes

reflected in

becomes

The image of is b After a

. .

.

rotation anticlockwise, a point

As shown in part then, a reflection in .

has image

.

rotation is equivalent to a reflection in

followed by a

Reflection in

is the transformation from a function to its inverse.

Reflection in

: replace with

.

The new graph has equation 19 a If a polynomial

.

when divided by a polynomial for some polynomial

;

has a remainder

then:

is the polynomial analogue of a ‘remainder’

when dividing integers. Rearranging:

.

Substituting in this question , so for some constants and . So

is called

and the remainder

is linear

.

b Using the product rule:

c The remainder and

d For the remainder to be equal to zero for all values of (so that for all



Substituting

into

:

Substituting

into

:

is a factor of

):

Cross topic review exercise 2 worked solutions Worked solutions are provided for all levelled practice and discussion questions, as well as cross topic review exercises. They are not provided for black coded practice questions.

Cross topic review exercise 2 1 For small values of : and So for small :

Then and 2

So

3

4

The inflection occurs when the second derivative is zero: so so 5 a Trapezium rule approximation:

Total Total Approximate area: b Let Using the substitution Then

:

so

, 6 7 a

and

b At

, the gradient is

and the point on the curve has coordinates

The tangent has equation

At

so

.

8

Integrating by parts: set

9 a i

so

and

so

and

.

Primary solution: Secondary solution: Periodic solution: Solutions in the given interval are ii Given

:

Let so (reject as outside the range of cosec) or

From part :

or

Since only solutions for solutions.

.

are required, you don’t need to find any further

or or b f( ) cosec  Replace with

Replace with

: translation

: horizontal stretch, scale factor

10 a Graph crosses the

when

(parallel to

:

b Integrating by parts: set

and

so

c Eliminating :

11

Using implicit differentiation and the product rule:

and

)

Tip Alternatively, use the chain rule. 12 a Let

cosec 

(sin 

Using the chain rule:

b

Reject the first root as it is outside the range of cos  Within the given interval,

for

13 a

So b Set

14 a Let

and

:

( ).

From the question: Integrating: Substituting

b

and .

:

.

so h

c So d At

and

At 15

sin

Second derivative is positive for convex regions.

.

so the point with this gradient is

.

Require that

:

sin So

or

16 a The common ratio cos  , which for criterion for a geometric series to converge.

has value such that |

, which is the

b The first term

and

so

Using the double angle formula

with

:

c

17 a i By the symmetry of the sine graph, is

.

ii b i Let be the area.

At a stationary value for area,

:

Dividing through by cos  :

ii

There is only one intersection of .

and

tan  , so only one root of

iii

At the stationary point, cos

because the point lies in

.

Using the result in part b i,

at the stationary point so

Negative second derivative indicates that the value is a local maximum. c i Try iteration

xn

0

1

2

3

4

5

…

1

0.519

0.811

0.65

0.744

0.691

17

18

0.710

0.710

Test the solution: Let

By the change of sign principle, the solution is

to d.p.

Tip Not unexpectedly, this iteration is slow. For faster convergence, it would be better to use the Newton-Raphson method: Let

xn

so

0

1

2

3

4

1

0.777

0.713

0.710

0.710

You would use the same sign test to demonstrate the validity of this solution. ii For 18 a

for some constants Multiplying through by the denominator on the LHS:

Comparing coefficients: :

: 3 ×

× ×

So b Separating variables and integrating:

and .

When

.

Substituting:

and

with ex

19 a Initial function: Replace with

: horizontal stretch, scale factor

The function becomes

(parallel to

).

.

Subtract 1: vertical translation by one unit downwards, translation The function becomes

.

At

).

.

b

c i

so is ( At , so

Collecting terms and multiplying through by

ii Factorising this quadratic in

or

:

:

(Reject since

for all real .)

is ( d The difference function is

20 The expression is the binomial expansion of Using the identity

21 a This is the sum of the first

:

terms of a geometric series with

and

Tip

b

There are many ways to get to this result but here are two tidy variations. Method 1 Let Then So From part :

Method 2 From part : The LHS is a finite order polynomial, so can be differentiated term by term:

Using the quotient rule on the LHS:

22 a But sin(arcsin  ) b

so

has -coordinate so has -coordinate arcsin 

c d You can find the blue shaded area by subtracting the result in part c from the rectangle area arcsin  .

But you can also find the blue area by finding the area to the left of the curve:

Equating these results and changing the dummy variable in the integration from to :

(0 <

since must be positive and within the range of sin )

23 a

b Stationary points occur where f'( )

Using the double-angle formula for

:

and multiplying through by

:

c

The only solution in the interval

is

.

So the function only has one stationary point. 24 a

)ex

(

For a point of inflection, require the second derivative to equal zero:

For real roots, the discriminant must be non-negative:

b Inflection is also a stationary point, so for the value require:

:

so

Then c For

,

The curve is concave when the second derivative is negative, that is for 25 a Asymptote b

and axis intercept (

.

so the tangent at ( , ln  ) has gradient The equation of the tangent is

. .

Require that this passes through the origin. Substituting

:

c From the nature of the graph, any line through the origin with positive gradient less than will intersect the graph twice, while any line with negative gradient will intersect the graph once and any line with gradient greater than will not intersect the graph at all. For two intersections: 26 a Using the quotient rule:

b

c

tan  Using the result from part a and noting that the derivative of 1 is zero: as before. From part b,

so

Taking the reciprocal:

d i Stationary points occur when

.

Using :

or ii

iii When When

so there is a local minimum when so there is a local maximum when

When So the minimum point is at (

) which lies on the

.

. .

Cross topic review exercise 3 worked solutions Worked solutions are provided for all levelled practice and discussion questions, as well as cross topic review exercises. They are not provided for black coded practice questions.

Cross topic review exercise 3 1 a b

2 a Require

so has coordinates

b

.

and

.

Two adjacent sides of the parallelogram are equal length so it is a rhombus. or

c

so that has coordinates

.

3 a b i At ii At

, the movement is solely in the negative direction: due west.

c

so

d At

, the magnitude of the acceleration is so the resultant force at that moment is

4 Let

and

be the reaction forces at and respectively.

Equilibrium of forces in the plank means that total reaction force equals total weight: quilibrium of forces in the plank means that total reaction force equals total weight:

Taking moments about the contact point :

5 a Suppose the centre of mass is

to the right of the support.

Taking moments about the point of contact with the support:

So

; the centre of mass is

b Total weight of the system is

to the right of the support.

, to be exactly equalled by the reaction at the support.

6 a

b

7

For

to be a straight line,

so

.

and therefore

and

8 So

for all values of .

9 a Resultant force b

so

c i so Then ii iii The square of the speed

or

when

.

seconds.

10 a b so Then c Require component of velocity to be zero and component negative.

From At time d

, which also satisfies

.

, the particle is moving due west. so the speed is

.

11 a Let the tension in the string be . Taking moments about in equilibrium: so b All other forces acting on the system are vertical, so in equilibrium the force at must also be vertical. If the force acting at is upwards then resolving vertical forces:

From part a,

(upwards)

.

12 a

b Resolving perpendicular to the slope: so c Resolving parallel to the slope, taking acceleration as directed down the slope:

The block is moving so friction is maximal:

so

d Air resistance would act in the same direction as friction to oppose the motion. The total resistive force would still have to be for the block to be decelerating at so, since some of this force would now be due to air resistance, less of it would be due to friction. Therefore the coefficient of friction would be lower (as the normal reaction would be the same). 13

a Resolving vertically for the hanging block:

Resolving for the supported block: so

Maximal friction is equal .

which is less than

b Constant acceleration problem:

so the blocks will move and friction will

. Find .

which gives

, so

c The two parts of the string each affect the pulley with force , so the resultant is at horizontal and of magnitude .

so the resultant force on the pulley has magnitude

to the

.

d i There is no friction at the pulley so the tension is the same on both sides of the pulley. ii The acceleration of the suspended block downwards equals the acceleration of the supported block horizontally. 14

a Resolving vertically for particle : so b

c Resolving vertically for : so d Resolving horizontally for : so Friction is limited to µ so the minimum value µ is given by

.

15 a

Known point on the trajectory as the arrow hits the target:

where is the horizontal

distance from the firing point to the target. So From the -component of this equation,

, which has positive root

b Substituting into the -component:

c So d

so the velocity has angle

relative to the

horizontal. e Throughout the working it is assumed that the only force affecting the arrow is gravity; no account is made of air resistance or wind. 16 a

so from which (The force is negative because, being a resistance, it is directed against the direction of movement.)

b Separating variables:

so

and hence

.

17

so

so

and then

Then at all times,

, so the trajectory is along the straight line given by that equation.

18 a Model the ball as a particle (no effect of spin or width to be considered) and ignore all forces except gravity (no account taken of air resistance or friction interacting with rotation). b

Let the ball return to ground

from its start point. Then for some :

From the -component of this equation, the ball returns to ground level when so c

.

Using the double-angle formula

.

and

Require such that

so

19 a

Known point on the trajectory as the ball passes through the hoop: So From the -component of this equation,

.

Substituting into the -component and rearranging:

b i Let and Then

and lies in the first quadrant so so has maximum value

ii



From

:

. . .

Cross topic review exercise 4 worked solutions Worked solutions are provided for all levelled practice and discussion questions, as well as cross topic review exercises. They are not provided for black coded practice questions.

Cross topic review exercise 4 1

2 a (two-tailed test) b Critical values for two-tailed test with are . so do not reject there is insufficient evidence to show that there is linear correlation between average daily temperature and rainfall. 3

is the height (µ a

;

of a rose bush.

) so µ µ

(one-tailed test)

b so reject

at

significance level; there

is evidence that Larkin’s rose bushes have a population mean height significantly greater than . 4 a and so b Let be the number of light bulbs in a sample of (

which last more than

hours.

)

c Although the mean is many standard deviations above zero (so the model does not suffer from the problem of a significant proportion being predicted to have a negative life-expectancy), the fact that the mean is so far above zero is also a problem – in this model there is a vanishingly small probability of a bulb failing in early life, which is contrary to real life experience. The model predicts that nearly all bulbs will have a lifetime between and hours ( standard deviations above and below the mean). 5

( a i

)

ii iii b Let be the bill for the fuel;

c The model is specified as applying to unleaded petrol for private-car customers only. There is no reason to believe it would be more widely applicable: For example: trucks and other large commercial vehicles might be expected to have a far greater fuel capacity different fuel types might typically be purchased in greater or lesser volume depending on the fuel business customers might have required fuel increments, minimum purchases or fixed volume deals customers might fill fuel cans in addition to or instead of filling a vehicle’s fuel tank. 6 Minimum standard deviation will use the same value as minimum variance.

From this completed square, variance is minimised by

.

7

a b

c

8 a b c

is the infinite sum of a geometric series with

9 a i P(fine and cycle)

and

.

ii P(garden or shop) iii iv v

b 10 Let be the mass (in grams) of a bag of sugar. Then

(

)

a b Let be the number of bags in a box of packed with randomly selected bags). Then

(

which weigh more than

(assuming boxes are

)

c P(fourth weighs more than

, first three weigh less than

)

11 a The normal distribution is almost entirely within three standard deviations of the mean, so estimate that three standard deviations equals , since the machine always cuts within of the desired mean.

b

so the recorded mean is lower than the tolerance of the machine suggests is possible for even a single piece of dough, and so is unlikely to be correct. is far bigger than the suggested standard deviation of , and is unlikely to be correct; it is certainly incompatible with the knowledge that the machine cuts to within grams of the set weight.

c The mean observed weight is

grams, which is within

The standard deviation of the sample of

is

expectations of accuracy for the machine ( 12 a

is a standard normal, so is equivalent to .

b Trapezium rule approximation:

of

, so is a plausible value.

which is again in line with the ) and so is plausible.

Total Total Approximate area: c 13 a The mode of the graph appears to be between the mean.

and

, so

would be the best estimate of

The graph becomes undetectable below and above ; if this interval is approximately six standard deviations (about a midpoint which is , close to the estimate of the mean from the central symmetry) then the standard deviation would be . Alternatively, looking at the points of inflection in the graph (when made continuous by joining the tops of the bars to form a smooth curve), these would be at approximately and , at a distance of approximately from the estimate of the mean, again giving a standard deviation approximately . b Mean np is approximately

and variance np(

The ratio of the two values gives

so

) is approximately and then

. .

14 a (one-tailed test) b For data points, the critical value for significance is so an observed value is significant; reject and conclude that there is evidence of negative correlation so that more revision is associated with worse test scores. c Taking two such dissimilar individuals’ data and treating it as from a single source has resulted in a false result; from the scatter graph it appears that a student with high attainment revised less on each occasion than a student with low attainment but (critically), for each student, longer revision time was associated with an improved score. The disparity between the students was substantially greater than the improvement due to revision time, so the amalgamated data hides this detail, which is clear in the graph. 15 a

is the depth (mm) of a bath panel. (µ

) so

µ µ

(two-tailed test)

Observed mean

Do not reject different from b

there is insufficient evidence to show that the current mean is significantly . .

and H1 are as in part a. Observed mean

Reject there is now evidence at the significantly different from .

significance level to show that the current mean is

16 Use the notation that is the event of Daniel winning on his th throw and is the event of Theo winning on his th throw, with being a throw of heads so that is a throw of tails. a b c

d P(Daniel wins) e P(Theo wins)

P(Daniel wins)

f Let be the probability of throwing heads. Require that P(Daniel wins)

(Reject

as a false solution introduced when multiplying through by the

denominator of the RHS in (1).) 17

Let be the time taken to type a page (mins). a

(µ

) so

µ µ

(one-tailed test)

Observed mean

Reject

, test at

significance.

there is evidence to show that the mean is significantly lower than

minutes.

b Require sample size such that Then

So require a sample size of at least

.

18 a The curve corresponding to the standard normal distribution with mean zero and standard deviation σ is

. Points of inflection are at

Comparing this to the given curve:

± . .

The curve has been translated units in the positive -direction so the mean is . Therefore, the points of inflection are at b

, i.e. at

and

.

Trapezium rule approximation:

Total Total Approximate area: c Given the inflection is at , which lies within the area being estimated, some of the trapezia will lie above and some below the curve. In consequence, without further working it is not possible to say whether part b will be an over- or under-estimate.

Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

1. Proof and mathematical communication Support Sheet

1−3 Proof.

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2. Functions Support Sheet

1 2 3 4

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3. Further transformations of graphs Support Sheet

1 2 3 4

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4. Sequences and series Support Sheet

1 a b c d

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5. Rational functions and partial fractions Support Sheet

1 2 3 4

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6. General binomial expansion Support Sheet

1 2 3 4

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7. Radian measure Support Sheet

1 2 3 4

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8. Further trigonometry Support Sheet

1 a b 2 a b 3 a b

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9. Calculus of exponential and trigonometric functions Support Sheet

1 2 3

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10. Further differentiation Support Sheet

1 2 3 4

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11. Further integration techniques Support Sheet

1 2 3 4

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12. Further applications of calculus Support Sheet

1 2 3 Proof.

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13. Differential equations Support Sheet

1 2 3

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14. Numerical solutions of equations Support Sheet

1 a Proof. b

,

,

c Proof. 2 a Proof. b

,

c Proof. 3 a Proof. b c

is a stationary point.

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15. Numerical integration Support Sheet

, 1 2 3 4

,

,

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16. Applications of vectors Support Sheet

1

2

3 4

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17. Projectiles Support Sheet

1 2 a b 3

( s.f.),

( s.f.)

4

( s.f.),

( s.f.)



Worksheet answers All answers to ‘fill in the blanks’ activities are provided in the order the blanks appear, reading from the top left line by line to the bottom right.

18. Forces in context Support Sheet

1 2 3 4

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19. Moments Support Sheet

, , 1 2 3 4

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20. Conditional probability Support Sheet , , ,

1 a b c 2 a b c 3 a b c 4 a b c

,

,

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21. The normal distribution Support Sheet , ,

,

1 2 3 a Proof. b 4

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22. Further hypothesis testing Support Sheet

insufficient insufficient mean plates 1 Insufficient evidence that the mean length has changed 2 Sufficient evidence that the lifetime is longer 3 Sufficient evidence of a decrease 4 Insufficient evidence that his mean is different

. .

. .

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1. Proof and mathematical communication Extension Sheet 1 a b Proof. Use the fact that

.

2 a Proof. is always divisible by . b Proof. Consider the numbers

.

3 a b Any digit other than . c Proof. Compare the th digits of and

.

d Proof (by contradiction: start by assuming that the numbers have been put into a list).

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2. Functions Extension Sheet 1 a

b

c

d

2 3 4 a

for some

b

for some

c d

for some

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3. Further transformations of graphs Extension Sheet 1 a

b

,

2 3 4 5 6

,

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4. Sequences and series Extension Sheet 1 a b It approaches . c 2 It approaches

.

3 It doesn’t approach a limit. 4 Investigation.

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5. Rational functions and partial fractions Extension Sheet 1 a b c Proof. 2 a Proof. Notice that b

, which is the golden ratio!

c 3 a Proof. b Proof. c

4 It converges to .

.

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6. General binomial expansion Extension Sheet 1 a b c d e 2 a b c 3 4 Proof. 5 a b c

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7. Radian measure Extension Sheet 1 2 a b Proof. c i ii 3 4 a b 5 a b c

because

lies on circles with centres and .

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8. Further trigonometry Extension Sheet 1 2 Proof. 3 Proof. 4 a Proof. b c Proof. 5

,

,

6 Proof. 7 a Proof. b c

,

,

,

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9. Calculus of exponential and trigonometric functions Extension Sheet 1 2 One if

or if

. Two if

. None otherwise.

3 a Proof. b Proof. c Proof. 4 a b c

for odd , zero for even .

d

e f Proof. Set

and use the first two terms.

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10. Further differentiation Extension Sheet 1 a–c Proof. 2 a Proof. b Proof. Consider the second derivative. 3 a b c

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11. Further integration techniques Extension Sheet 1 2 3 Proof. 4 a Proof. b ,

5 6 a Proof. b c

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12. Further applications of calculus Extension Sheet 1 About

chocolate bars.

2 a b t-shirts; pair of shorts. 3 a b computer games; DVDs. 4 a b 5 a b c



bikes bikes

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13. Differential equations Extension Sheet 1 a b 2 3

; oscillation with frequency

.

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14. Numerical solutions of equations Extension Sheet 1 a b Proof. c 2 a 1.68 b



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15. Numerical integration Extension Sheet 1 a b Proof. c

d e Because 2 a Proof; b c d 3 a b green; yellow c Proof. d Proof. 4 Discussion. 5 Discussion. 6 Discussion.

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16. Applications of vectors Extension Sheet 1 a b Proof. c d e 2 a rθ is the length of the arc that has been in contact with the ground. b c

d

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17. Projectiles Extension Sheet 1 Proof. 2 3 a Proof. 4 5 6 Skimming is faster.

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18. Forces in context Extension Sheet 1 a b 2 a b 3 4 5



( s.f.)

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19. Moments Extension Sheet 1 a Proof. b Discussion. Consider areas on each side of a median. c 2 a b c

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20. Conditional probability Extension Sheet 1 a b c 2 AAB

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21. The normal distribution Extension Sheet This is an investigation. These answers indicate approximately what results you should get. 1 Mean

; variance

2 Mean

; variance

3 Mean

; variance

4 Mean

; variance

5 Investigation.

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22. Further hypothesis testing Extension Sheet 1 a b Discussion. 2 Discussion. 3 Discussion. 4 a b c d Discussion.

Working with the large data set AQA have provided you with a data set listing the amount of food purchased in different regions of England in the years from 2001 to 2014. You can find the large data set on the AQA website. You can use this chapter to apply the ideas studied in the book chapters and to explore important themes surrounding data and statistics whilst you familiarise yourself with this large data set.

Learning about the data set Every data set has its own peculiarities. These questions are designed to encourage you to think critically about exactly what is meant by some of the data in the data set. QUESTIONS 1 In the Yorkshire and The Humber data there is a value for ‘Other food and drink’ of 02. Explore how this value is found from other values within the spreadsheet.

in 2001–

Tip It is not just the sum of all the values in the ‘Other food and drink’ section. Do you think this is a valid calculation? 2 Give some examples of ‘Other takeaway food brought home’. Use the data to decide whether your answers are reasonable. Why might that be? 3 For 2001–02 the values of ‘Milk and milk products excluding cheese’ in the different regions are shown in this table. Region

Amount (ml)

North East

1913

North West

2103

Yorkshire and Humber

2105

East Midlands

2144

West Midlands

1989

East

2062

London

1838

South East

1960

South West

2190

a Find the mean of the ‘Milk and milk products excluding cheese’ purchased across all of these regions. b The mean amount of ‘Milk and milk products excluding cheese’ purchased across all of England is Explain why this is different to your answer in part a. Why could you have predicted that it would be less than your answer to part a?

A reminder about statistical diagrams You must be able to interpret a range of diagrams including: pie charts bar charts histograms cumulative frequency diagrams box-and-whisker plots. You might find it helpful to explore how to create these (and other) diagrams using technology, and to think about when each type of diagram is useful. QUESTIONS 4 These pie charts show the proportions of cheese purchased in the North East.

a Summarise the main changes in purchasing behaviour shown by these pie charts. b What changes are not illustrated by these pie charts? 5 Draw a pie chart to show the proportion of soft drinks purchased in London which were low calorie in 2011. 6 These three bar charts display the same data, illustrating the changes in the amount of ‘Milk and milk products excluding cheese’ purchased.

a Which region tended to purchase the least milk and milk products? Which chart illustrates this best? b Is there a significant change in the pattern of milk and milk products purchased over time? Which chart illustrates this best? c Why might the second chart be considered misleading? 7 This histogram shows the milk and milk products purchased in all different regions of England across all years given.

a Approximately how many of the data items are between

and

?

b Using technology, create a histogram for the same data with class widths of size c What are the advantages and disadvantages of the histogram in part b compared to the histogram in part a? 8 This cumulative frequency diagram shows the amount of sugar and preserves purchased across all regions and all years.

a How many data items are included in this sample? b Use the diagram to estimate the median amount of sugar and preserves purchased. c Sketch and label a box-and-whisker plot for these data. 9 These box-and-whisker plots show the weekly sugar and preserves purchased in seven different regions in the years from 2001 to 2014.

a Which region has the largest range of values? b Which region has the lowest median value? c By looking at the original data set, determine which region has the box-and-whisker plot labelled ???. d This dot plot shows the values of weekly sugar and preserves purchased in each year for one of the regions displayed in the box-and-whisker plots. Which area is being represented? Explain your answer.

10 Suggest a reason for choosing each of these diagrams to illustrate a data set. a Pie chart b Bar chart c Cumulative frequency diagram d Histogram e Box-and-whisker plot

Standard deviation and interquartile range You can use a spreadsheet to find the standard deviation of data. A common syntax might look like this.

Common error This command gives the estimate of the standard deviation of the whole population based upon treating the data as a sample. This is usually more useful to statisticians. If you are interested in the standard deviation of the data you need to use the command ‘STDEVP’. You can also use spreadsheet commands to find the range and the interquartile range. A common syntax might look like this.

QUESTIONS

QUESTIONS 11 Consider the data on ‘Takeaway sauces and mayonnaise’ for the East Midlands. a The values shown on the spreadsheet are summarised in this table. Value (g) Frequency By inputting this information into your calculator, find an estimate for the standard deviation of the population. b Use the spreadsheet data to find an estimate of the standard deviation of the population. c Explain why your answers to parts a and b are very different. 12 a Find the best estimate for the standard deviation of the amount of yoghurt and fromage frais purchased each week in the North East. b Find the best estimate for the standard deviation of the amount of yoghurt and fromage frais purchased each week in London. c Hence compare the distribution of these two populations. 13 a Find the interquartile range for the amount of confectionery purchased in the West Midlands. b Find the interquartile range for the amount of confectionery bought in the East Midlands. c Hence compare the distribution of these two populations. 14 a Find the interquartile range for the amount of fats purchased in the North West. b Find the best estimate for the standard deviation of the amount of fats purchased in the South East. c Hence compare the distribution of these two populations. 15 a Find the best estimate for the standard deviation of the amount of soft drinks purchased in the South East. b Find the best estimate for the standard deviation of the amount of confectionery purchased in the South East. c Asher says that since the standard deviation of soft drinks is larger than the standard deviation of confectionery, there was greater variation in the amount of soft drinks purchased in this period. Do you agree? d Elsa suggests that a better measure to compare this variation is to use:

What are the advantages and disadvantages of this measure? 16 a Daniel is studying the purchase of soft drinks, concentrated, not low calorie in the East Midlands. He believes that there is a greater spread towards the beginning of the period studied than at the end of the period studied. To gather evidence for this he looks at the standard deviation for the first three years and the last three years. Does this support his belief? b Alessia is also studying the purchase of soft drinks, concentrated, not low calorie in the East Midlands. She believes that there is a smaller spread towards the beginning of the period studied than at the end of the period studied. To gather evidence for this she looks at the standard deviation for the first five years and the last five years. Does this support her belief? c Comment on your answers to parts a and b.

Scatter diagrams and correlation

You can use a spreadsheet to find the correlation coefficient between two variables. A common syntax might look like this.

QUESTIONS 17 This scatter diagram shows the correlation between confectionery purchased and soft drinks purchased in England in the different years covered by the data set.

a Which of these values is the correlation coefficient for these data? A B C D b How would you describe the relationship between these two variables? 18 a Use technology to plot a scatter diagram showing the relationship between the amount of soft drink, not concentrated, not low calorie and the amount of soft drink, not concentrated, low calorie purchased in London over time. b What is the correlation coefficient for these data? c Conduct a test at the

significance level to decide if this is evidence of positive correlation.

d Suggest a third factor which could explain the relationship between these two variables. 19 Milk and milk products purchased per week in the South West is shown on this line chart.

a Describe the trend shown. b Can you calculate the correlation coefficient between these variables? Can you interpret the correlation coefficient? c The graph suggests that midway between the 2006 data point and the 2007 data point the amount of milk purchased is about Is there a valid interpretation of this fact or is it a misleading use of the graph?

Did you know? There is a whole area of statistics called time series with tools for dealing with situations like this.

Outliers and cleaning data There are two standard definitions for deciding if something is an outlier. An outlier is any number more than

interquartile ranges away from the nearest quartile.

An outlier is anything more than two standard deviations away from the mean. Once you have identified an outlier you must investigate and use your judgement to decide if it is an error (a clear mistake, for example putting a decimal point in the wrong place) or just an anomaly. If it is an anomaly, then you can choose whether or not to include it depending upon exactly what question you are trying to answer.

QUESTIONS

QUESTIONS 20 This diagram shows the amount of sugar and preserves purchased in the South West in the period studied.

a Show that there must be an outlier in this data. b Is this outlier likely to be an error? Explain your answer. c Using the original data set calculate the standard deviation and interquartile range of the amount of sugar and preserves purchased in the South West. d Remove the outlier and recalculate the standard deviation and interquartile range. e Use your calculations to explain whether it is better to calculate the standard deviation or interquartile range of data with outliers. 21 a Add up the total mass of cheese, fats, sugar and preserves, bread, other food and drink and confectionery purchased in England in the year 2001–02. b Assuming three meals each day, days each week, use your answer to part a to work out the mass of food per meal on average. c Comment on your answer to part b.

Sampling and hypothesis testing Whenever you are working with real statistics you have to take into account the fact that your data is only a sample of the whole population you are interested in. You need to understand the sampling method to understand what population the sample represents. To make any decisions about the population you also need to take into account the fact that you might just have an unusual sample. The method for doing this is hypothesis testing. Many spreadsheets can take random samples from data. For example, using the data analysis ToolPak in Excel produces this layout.

QUESTIONS

QUESTIONS 22 This data set comes from the Living Costs and Food Survey. About households are visited at different times of the year and each household member over the age of six is asked to keep a diary of the food they eat for two weeks. Many more households are invited to take part, but some decline. a Explain why this survey is not a simple random sample of the population. b The data covers the years from 2001 to 2014. What sampling method has been used and on which population? c Each family is surveyed for two weeks, chosen randomly. What sampling method has been used and on which population? d New families are asked to participate until there are enough households from all different social groups and geographic regions. What type of sampling method is this? 23 It is thought that in of the years in the sample, the total amount of sugar and preserves purchased per week is less than a Use a binomial test at in: i ii

significance to determine if this is the case using data from England

the first years all the years in the sample.

b Select random samples of size from the data. In how many of these random samples is there sufficient evidence to reject the hypothesis that in of the years the total amount of sugar and preserves purchased per week is less than ? 24 The long-term amount of confectionery purchased per week is It is assumed to be normally distributed.

with standard deviation

The government is considering introducing a sugar tax which they hope will decrease the amount of confectionary purchased. a i

ii

In the year after the sugar tax is introduced, the average amount of confectionery purchased is What is the probability of getting this value or lower if the population still has a normal distribution with a mean of and standard deviation ? Would this be sufficient evidence at the changed the behaviour?

significance level to say that the sugar tax has

b Create random samples of size from the data for total confectionery purchased in England. How many of these samples would provide sufficient evidence of a decrease in the mean from using a significance level?

Statistical problem solving In real life you can use data to answer questions. However, you rarely have the perfect data to answer your questions so you have to make inferences from the data you do have. For each question, suggest a statistical method which could be used to answer or illustrate it, and describe the assumptions you are making.

QUESTIONS

QUESTIONS 25 How much sugar does the average person eat? 26 Is the popularity of margarine reducing the amount of butter purchased? 27 Was there a lot of confectionery eaten during the London 2012 Olympic Games? 28 Are London’s eating habits very different from the rest of the country?



Working with the large data set – answers AQA have provided you with a data set looking at the purchasing of different types of food in different parts of the country in different years. You can find the large data set on the AQA website. 1 It seems to be the sum of all the categories labelled in column of ‘Other food and drink’ (not double counting the numbers which are breakdowns of bigger categories). However, this adds to which is normally not possible. It certainly should not have units of . However, as long as the same calculation is done everywhere it might allow some measure of comparison across regions and years. 2 You might initially think of things like hamburgers or fish and chips. However, with a value of less than per person each week this seems unlikely. There are several obvious categories such as ‘meat’ and ‘vegetables’ which are missing from the data. Most fast food might be categorised there, so this might be more specialised things, for example ‘fried seaweed’. 3 a

( s.f.)

b The figure for England is a weighted average, taking into account the population of each region. Since by far the most populous regions in England are London and the South East, these will have a large effect on the overall average and they have low values of milk purchased. 4 a In 2014 a greater proportion of the cheese purchased was soft natural cheese. A smaller proportion of processed cheese or ‘other UK or foreign equivalent’ cheese was purchased. b The actual amount of cheese eaten is not shown in these pie charts. 5

6 a Least milk and milk products purchased in London. Best shown by second or third chart. b No significant change over time. Best shown by first chart. c The vertical axis starting at distorts the impression given of milk and milk products purchased in London compared to in the other regions. 7 a b

c There is perhaps too much detail here, which makes the overall picture harder to see. However, if fine detail is needed this might be better, particularly at the edges. 8 a b About c

9 a Yorkshire and The Humber b London c East d South West. The highest value is just under

and the lowest value is between

and

.

10 a Shows proportions of each category. b Quickly shows a value associated with each category. c Easy to read off the median and quartiles of grouped data. d Displays the frequencies of grouped continuous data. e Illustrates key statistics (median, quartiles, range) and allows distributions to be quickly compared. 11 a

( s.f.)

b

( s.f.)

c The spreadsheet displays data rounded to the nearest whole number, but the calculations are based on the true values. Most of the data is actually fairly close to . 12 a

( s.f.)

b

( s.f.)

c There has been greater variation in the purchasing of yoghurt and fromage frais in London. 13 a

( s.f.)

b

( s.f.)

c There is a greater variation in the purchasing of confectionery in the East Midlands. 14 a

( s.f.)

b

( s.f.)

c You cannot compare IQR with standard deviation. 15 a b

( s.f.) ( s.f.)

c It is not valid to compare these two standard deviations. They are for different items, with different units. d This approach does overcome the problem of different units and scaling, but it can cause a problem if the mean has a very small value; in other situations, the mean might be zero, in which case this measure would be undefined. 16 a First three years: b First five years:

; last three years: ; last five years:

. Supports Daniel’s belief.

. Supports Alessia’s belief.

c Particularly with small sample sizes choosing slightly different samples can have a big effect on the conclusion. You should be suspicious of seemingly arbitrary choices made regarding the sample. 17 a b Virtually no correlation. 18 a

b

( s.f.)

c Critical value for one-tailed test: correlation.

. There is significant evidence of positive

d For example, increasing population. 19 a The amount of milk and milk products purchased is generally decreasing. b You cannot calculate the correlation coefficient as the -axis is categorical data. Even if you transformed it into numerical data (for example by saying 2001−02 could be written as ) there would be some issues since the -axis is not a random variable so it is not really appropriate to talk about the correlation. Certainly it would not be possible to do the standard hypothesis test on the correlation coefficient. c If it is assumed that the changes in milk purchasing is linear between the two samples you

could interpret the /week as the average amount of milk purchased half way between the two sampling periods. However, it is likely that seasonal variations make the assumption of linear change unlikely. 20 a From the data, UQ is ; IQR is . So anything above of above this limit, so this is an outlier.

is an outlier. There is one value

b This is by far the largest value in any region, so it is suspicious but it is not a clear error. It is a plausible value (not a misplaced negative sign or decimal point). So you can treat it as an anomaly, but it is not necessarily an error. c IQR

,

( s.f.)

d IQR

,

( s.f.)

e The standard deviation is far more affected by the presence of the outlier, so the IQR is a more stable measure when there are outliers. 21 a b

( s.f.)

c This is far too little for an average meal. This is probably due to missing data, such as meat, fish, fruit, vegetables and cereals. 22 a Not all households are equally likely to be included in the survey, those who decline are not represented and there is no information on the basis for selection in the first place to judge whether it is a random survey of the wider population invited to participate. b Households from 2001 to 2014; opportunity sample over all years. c Two week periods of household behaviour within the calendar year; simple random sample of behaviour. d Quota sampling. 23 a i Sufficient evidence that this is not the case ii Insufficient evidence that this is not the case

. .

b This is a powerful method called simulation. In probably two of the sufficient evidence to reject the hypothesis. 24 a i

samples there will be

so

ii In a significance two-tailed test for change, change.

so there is sufficient evidence of

b Only up to three of these these samples are likely to be extreme enough to suggest the mean is below – even though (as you can see from the data) the mean is indeed below , at 25 You might choose to use the data for England and take an average (perhaps mean) of total sugar intake over all the years. This assumes that the amount of sugar purchased is the same as the amount eaten, so this will probably be an overestimate. 26 You might look at the correlation between butter and margarine purchased, but this might be influenced by population growth. It might be better to use a line graph to see how margarine and butter purchasing changes over time. 27 You might use a box-and-whisker plot to illustrate the general distribution of confectionery in London, with an arrow highlighting the 2012 figure. The problem is that the data is annual and although the Olympics might have had an impact it will be diluted across the year. It is also not clear whether the data refer to confectionery bought by families living in London, or all confectionery that was consumed in London, so it might be ignoring tourists. 28 This is very tricky! You might produce a multiple bar chart showing the average of each major food type over time for each region. There are some very advanced statistical methods such as logistic factor analysis to determine the answer to this type of question. However, is ‘amount of food purchased’ really the same as ‘eating habits’? You are ignoring, for example, how food is prepared,

where and when it is eaten, how much is thrown away, how much is spent on it.

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