A Level Further Mathematics for AQA: Student Book 1 (AS/Year 1) [Student ed.] 1316644294, 9781316644430, 9781316644294, 9781316644546, 9781316644553

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Brighter Thinking

A Level Further Mathematics for AQA Student Book 1 (AS/Year 1) Stephen Ward, Paul Fannon, Vesna Kadelburg and Ben Woolley

Contents Introduction How to use this resource 1 Complex numbers 1: Definition and basic arithmetic of i 2: Division and complex conjugates 3: Geometric representation 4: Locus in the complex plane 5: Operations in modulus–argument form 2 Roots of polynomials 1: Factorising polynomials 2: Complex solutions to polynomial equations 3: Roots and coefficients 4: Finding an equation with given roots 5: Transforming equations 3 The ellipse, hyperbola and parabola 1: Introducing the ellipse, hyperbola and parabola 2: Solving problems with ellipses, hyperbolas and parabolas 3: Transformations of curves 4 Rational functions and inequalities 1: Cubic and quartic inequalities 2: Functions of the form 3: Functions of the form 4: Oblique asymptotes 5 Hyperbolic functions 1: Defining hyperbolic functions 2: Hyperbolic identities 3: Solving harder hyperbolic equations 6 Polar coordinates 1: Curves in polar coordinates 2: Some features of polar curves 3: Changing between polar and Cartesian coordinates Focus on … Proof 1 Focus on … Problem solving 1 Focus on … Modelling 1 Cross-topic review exercise 1 7 Matrices 1: Addition, subtraction and scalar multiplication 2: Matrix multiplication 3: Determinants and inverses of matrices 4: Linear simultaneous equations 8 Matrix transformations 1: Matrices as linear transformations 2: Further transformations in 2D 3: Invariant points and invariant lines 4: Transformations in 3D 9 Further applications of vectors 1: Vector equation of a line 2: Cartesian equation of a line 3: Intersections of lines 4: The scalar product

10 Further calculus 1: Volumes of revolution 2: Mean value of a function 11 Series 1: Sigma notation 2: Using standard formulae 3: Method of differences 4: Maclaurin series 12 Proof by induction 1: The principle of induction 2: Induction and series 3: Induction and matrices 4: Induction and divisibility 5: Induction and inequalities Focus on … Proof 2 Focus on … Problem solving 2 Focus on … Modelling 2 Cross-topic review exercise 2 Practice paper Formulae Answers to exercises Worked solutions for book chapters 1 Complex numbers 2 Roots of polynomials 3 The ellipse, hyperbola and parabola 4 Rational functions and inequalities 5 Hyperbolic functions 6 Polar coordinates 7 Matrices 8 Matrix transformations 9 Further vectors 10 Further calculus 11 Series 12 Proof by induction Worked solutions for Cross-topic Reviews Cross-topic review exercise 1 Cross-topic review exercise 2 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 resource will realise that it can also be incredibly frustrating, surprising and ultimately beautiful. The resource 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 are 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: factorise polynomials and solve equations which may have complex roots link between the roots of a polynomial and its coefficients use substitutions to solve more complicated equations.

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

Before you start. . . GCSE

You should be able to solve linear inequalities.

1

A Level Mathematics Student Book 1, Chapter 3

You should be able to solve quadratic inequalities.

2 Solve

.

.

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, problem-solving 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. 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 that

. Use the principle of mathematical induction to prove .

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

8

Prove that

9 10

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. Purple – challenging questions that apply the concept of the current chapter across other areas of maths. Yellow – designed to encourage reflection and discussion.

– indicates content that is for A Level students only.

– indicates content that is for AS Level students only.



1 Complex numbers In this chapter you will learn how to: work with a new set of numbers called complex numbers do arithmetic with complex numbers use the fact that complex numbers occur in conjugate pairs interpret complex numbers geometrically interpret arithmetic with complex numbers as geometric transformations represent equations and inequalities with complex numbers graphically.

Before you start... GCSE

You should be able to use the quadratic formula.

1 Solve the equation , giving your answers in simplified surd form.

GCSE

You should be able to represent a

2 Draw two points, and , apart. Shade the locus of all the points that are less than from and closer to than to .

locus of points on a diagram.

A Level Mathematics Student Book 1, Chapter 5

You should be able to solve simultaneous equations.

3 Solve these simultaneous equations.

Finding the square roots of negative numbers Until now, every number you have met has been a real number. You will have been told that it is not possible to find the square root of a negative number or equivalently, that no number squares to give a negative answer. However, in this chapter, you will learn that there is a number that squares to give , and so a whole new branch of numbers appears: complex numbers. Remarkably, these numbers have many useful applications in the real world, from electrical impedance in electronics to the Schrödinger equation in quantum mechanics!

Section 1: Definition and basic arithmetic of i There is no real number that is a solution to the equation but mathematicians have defined there to be an imaginary number that solves this equation. This imaginary number is given the symbol .

Key point 1.1

or equivalently,

In all other ways, acts just like a normal constant. WORKED EXAMPLE 1.1

Simplify these terms. a b c Use the normal laws of indices.

a

.

b

c

You need to be familiar with some common terminology. A complex number is one that can be written in the form where and are real. Commonly, is used to denote an unknown complex number and is used for the set of all complex numbers. In this definition, is the real part of and it is given the symbol and it is given the symbol . So, for example, is and

; is the imaginary part of is .

You can now do some arithmetic with complex numbers.

Tip Your calculator may have a ‘complex’ mode, which you can use to check your calculations.

Common error Remember that the imaginary part of the complex number

WORKED EXAMPLE 1.2

Given that

and

find the value of each expression.

is the real number , not .

a b c d a

Group the real and the imaginary parts.

b c

Multiply out the brackets.

Remember,

.

Use the normal rules of fractions.

d

The original purpose of introducing complex numbers was to solve quadratic equations with negative discriminants.

Focus on ... See Focus on… Problem solving 1 to find out how some cubic equations can be solved by using a formula.

WORKED EXAMPLE 1.3 a Find the value of

.

b Hence solve the equation a

. Use the standard rules of square roots: . . Use the quadratic formula.

b

Using part a.

Did you know You may be sceptical about the idea of ‘imagining’ numbers. However, when negative numbers were introduced by the Indian mathematician Brahmagupta, in the 7th century, there was just as much scepticism. For example, in Europe, it took until the 17th century for negative numbers to be accepted. Mathematicians had worked successfully for thousands of years without using

negative numbers, treating equations such as as having no solution. But once negative numbers were ‘invented’ it took only a hundred years or so to accept that equations such as also have solutions. When you look at a new number system, one of the most important questions to ask is: ‘When are two numbers equal?’

Key point 1.2 If two complex numbers are equal, then their real parts are the same and their imaginary parts are the same. Although this may seem obvious, in mathematics it pays to be careful. For example, if two rational numbers and are equal, it does not mean that

and

.

Despite its apparent simplicity, the result in Key point 1.2 has some remarkably powerful uses. One is to find square roots of complex numbers. WORKED EXAMPLE 1.4

Solve the equation Let

.

. is a complex number, so write where and are real.

,

Expand the brackets.

Now equate the real parts and equate the imaginary parts.

From (2):

Use substitution to solve these simultaneous equations.

Substituting into (1):

Multiplying through by quadratic.

leads to a disguised

There are two possible values for is real so is impossible.

. However,

Find the value of for each value.

Write the answer in the form

.

Notice that there are two numbers that have the same square; this is consistent with what you have already seen with real numbers. EXERCISE 1A

EXERCISE 1A 1

State the imaginary part of each complex number. a i ii b i ii c i ii d i ii e i ii f

i ii

2

Simplify each expression, giving your answers in the form

.

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

Simplify each expression, giving your answers in the form

.

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

Simplify each expression, giving your answers in the form a i

.

ii b i ii c i ii 5

Evaluate each expression, giving your answers in the form a i ii b i ii c i ii d i ii

6

Solve the equations, simplifying your answers. a i ii b i ii c i ii d i ii

7

Evaluate, simplifying your answers. a i ii b i ii c i ii d i ii

.

8

Find real numbers and such that: a i ii b i ii c i ii

9

By writing

, solve these equations.

a i ii b i ii 10

Find the exact values of



such that



such that

. Give your answers in the form

. 11

Find the exact values of

12

By writing

13

a Find values and such that b Hence express



.

, solve the equation

in the form

. . .

14

By writing

, solve the equation

15

Solve the equation

16

Use an algebraic method to solve the equation

.

. .

Section 2: Division and complex conjugates In Section 1, you saw that quadratic equations may have two complex roots. In Worked example 1.3, the roots were and ; the imaginary part arises from the term after the sign in the quadratic formula. These two numbers form a complex conjugate pair. They differ only in the sign of the imaginary part.

Fast forward In Chapter 2 you will use the fact that complex roots occur in conjugate pairs to factorise and solve cubic and quartic equations.

Key point 1.3 If

, then the complex conjugate of ,

So for example, if

then

, or if

then

.

At first, the concept of conjugates may not appear particularly useful, but you will need them when you are dividing complex numbers. WORKED EXAMPLE 1.5

Write

in the form

.

Multiply the numerator and the denominator by the complex conjugate of the denominator, . Use the difference of two squares in the denominator. .

Worked example 1.5 shows the general procedure for dividing by a complex number.

Key point 1.4 To divide by a complex number, write as a fraction and multiply the numerator and the denominator by the conjugate of the complex number in the denominator.

Tip This procedure should remind you of rationalising the denominator when working with surds, which is also based on the difference of two squares. You can prove that this procedure always results in a real number in the denominator. PROOF 1

Prove that

is always real.

Let

, where and are real.

Then

.

Writing and in terms of their real and imaginary parts is often the best way to prove results about complex conjugates.

.

and are real numbers.

which is real.

Worked example 1.6 shows how to use the idea of equating real and imaginary parts to solve equations involving complex conjugates. WORKED EXAMPLE 1.6

Find the complex number such that Let

.

As before, the best way to describe complex conjugates is in terms of their real and imaginary parts.

Then:

You can now equate real and imaginary parts.

So

Being able to divide complex numbers means that you can solve more complicated equations. WORKED EXAMPLE 1.7

Solve these simultaneous equations.

The best method to use here is elimination: multiply the first equation by and the second by …

Subtracting:

… and then subtract them to eliminate .

Divide by : multiply top and bottom by the complex conjugate.

Substituting into the second equation: Substitute back to find : use the second equation as it looks simpler.

So

,

.



EXERCISE 1B 1

Find the complex conjugate of each number. a i ii b i ii c i ii d i ii

2

Write in the form

. Use a calculator to check your answers.

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

Solve these equations. a b

4

Solve these simultaneous equations.

a b 5

Find the complex number, , if: a b

6

By writing

, solve these equations.

a b 7

If and are real numbers find the complex conjugate

when:

a i ii b i ii c i ii d i ii 8

Write

9

a Let

in the form

. Show all your working.

. Find, in terms of and , the real and imaginary parts of

b Find the complex number such that 10

Find real numbers and such that

11

By writing

12

Solve

13

Solve

14

If

, prove that

. , where .

. . , find the real and the imaginary parts of

answers as far as possible.



.

in terms of and , simplifying your

Section 3: Geometric representation You can represent real numbers on a number line.

To represent complex numbers you can add another axis, perpendicular to the real number line, to show the imaginary part. This is called an Argand diagram.

Did you know Using real and imaginary axes was first suggested by the land surveyor Caspar Wessel. However, it is named after Jean-Robert Argand who popularised the idea.

WORKED EXAMPLE 1.8

Represent

on an Argand diagram.

The point representing .

has coordinates

Operations with complex numbers have useful representations on Argand diagrams. Consider two numbers, coordinates

and

. On the diagram they are represented by points with

and . Their sum, , has coordinates . But you can also think of coordinates as position vectors:

So you can add complex numbers geometrically in the same way as you add vectors.

You can represent subtraction similarly, by adding the negative of the second number.

Taking a complex conjugate results in a reflection in the real axis.

WORKED EXAMPLE 1.9

Two complex numbers, and , are shown on this Argand diagram.

Represent these complex numbers on the same diagram. a b a To find , first label parallelogram.

and then draw a

is the reflection of in the real axis.

b

EXERCISE 1C 1

Represent each number on an Argand diagram. Use a separate diagram for each part. a i ii b i Let ii Let c i Let ii Let

2

and and . Represent ,

and .

. Represent , . Represent , . Represent ,

and .

,

and

and

.

.

Solve each quadratic equation and represent the solutions on an Argand diagram. a i ii b i ii

3

Each diagram shows two complex numbers, and . Copy the diagrams and add the points corresponding to , and . a i

ii

b i

ii

Modulus and argument When a complex number is represented on an Argand diagram, its distance from the origin is called the modulus. There are lots of complex numbers with the same modulus; they form a circle centred at the origin. You can uniquely describe a particular complex number by giving its angle relative to the real axis; this is called the argument. The argument is conventionally measured anti-clockwise from the positive real axis.

Focus on ... See Focus on… Modelling 1 to find out how complex numbers can be used in electronics. The symbols used for these are: or for the modulus

or for the argument.

The notation

is sometimes used for a complex number with modulus and argument .

This description of a complex number is called the modulus–argument form. The description in terms of the real and imaginary parts,

, is called the Cartesian form.

Note here that the argument is not usually stated in degrees but in an alternative measure of angle called radians.

Radians The radian is the most commonly used unit of angle in advanced mathematics.

Key point 1.5 radians You can deduce the sizes of other common angles: for example, a right angle is one quarter of a full turn, so it is

radians.

Although sizes of common angles measured in radians are often expressed as fractions of , you can also use decimal approximations: for example, a right angle measures approximately

radians.

Rewind Radians are covered in more detail in A Level Mathematics Student Book 2, Chapter 7. WORKED EXAMPLE 1.10 a Convert

to radians.

b Convert

radians to degrees.

a

is

as a fraction of a full turn.

Calculate the same fraction of

.

This is the exact answer. By evaluating

on a

calculator you can also give a decimal answer: radians ( s.f.).

b

radians is

as a fraction of a full turn.

Calculate the same fraction of

.

Tip It is worth remembering some common angles in radians. Degrees Radians

Tip When you need to evaluate trigonometric functions of angles in radians, make sure that your calculator is set to radian mode.

EXERCISE 1D

EXERCISE 1D 1

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

2

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

3

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

Converting between the modulus–argument and Cartesian forms To convert between the modulus–argument and Cartesian forms you can use a diagram and some trigonometry.

Tip Always draw the complex number on an Argand diagram before finding the argument. It is important to know which angle you need to find.

Key point 1.6

To convert from modulus–argument form to Cartesian form:

To convert from Cartesian to modulus–argument form:

The argument can be measured either between and question which is required.

or between

and . It will be made clear in the

WORKED EXAMPLE 1.11

A complex number has modulus and argument . Write the number in Cartesian form.

Use

and

.

The Cartesian form is

.

WORKED EXAMPLE 1.12

Write

in modulus–argument form, with the argument between and

.

Give your answer correct to Use

.

Before attempting to find the argument, draw a diagram to see where the complex number actually is.

Calculate the angle . Remember to use radian mode on your calculator. From the diagram, So

and



.



WORK IT OUT 1.1 Find the argument of

, where

.

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

Solution 2

Solution 3

There is another nice way of expressing complex numbers which, although strictly in Cartesian form, explicitly shows the modulus and argument.

Key point 1.7 A complex number with modulus and argument can be written as

Fast forward In Further Mathematics Student Book 2 you’ll see another way of writing complex numbers that involves the modulus and argument. When writing a number in this form, it is important to notice that there must be a plus sign between the two terms and the modulus must be a positive number; otherwise you need to draw a diagram to find the argument. WORKED EXAMPLE 1.13

Write

in the form

, with the argument between − and .

In order to draw the diagram, you need to identify the signs of the real and imaginary parts.

From the diagram:

An Argand diagram can help to identify the modulus and argument of a negative and complex conjugate of a number. WORKED EXAMPLE 1.14

A complex number and .

is shown in the diagram. The argument is measured between

On the same diagram, mark the numbers

and

. Hence write

and

in modulus–argument

form. To find , make both real and imaginary parts negative. is the reflection of in the real axis.

All three points are the same distance from the origin, so all have the same modulus. As the argument is measured between , both numbers on the diagram have a negative argument. Write both numbers in modulus–argument form.

EXERCISE 1E 1

Find the modulus and the argument (measured between a i ii b i ii c i ii d i ii

and ) for each number.

and

e i ii f

i ii

2

Find the modulus and the argument (measured between and

) for each number.

a i ii b i ii c i ii 3

Given the modulus and the argument of , write in Cartesian form. a i ii b i ii c i ii

4

Write each complex number in Cartesian form without using trigonometric functions. Display each one on an Argand diagram. a i ii b i ii c i ii

5

Write each complex number in the form a i ii b i ii

6

Find the modulus and the argument of each number. a i ii

.

b i ii c i ii d i ii e i ii 7

8

a Write

in Cartesian form in terms of surds only.

b Write

in the form

Let

and

. .

a Find the modulus and the argument of and . b Represent and on the same Argand diagram. c Find the modulus and the argument of 9

If possible.

. Comment on your answer.

, write these numbers in terms of and , simplifying your answers as far as

a b c

Rewind Part c requires a double angle formula from A Level Mathematics Student Book 2, Chapter 8. 10

If

, express the real and imaginary parts of

answer as far as possible.



in terms of , simplifying your

Section 4: Locus in the complex plane You have met various examples of representing equations and inequalities graphically. You can represent an equation involving two variables, such as , on a graph, show an inequality such as on the number line and shade a region corresponding to an inequality in two variables, such as a graph.

, on

Now that you can represent complex numbers as points on an Argand diagram, there is a way to show equations and inequalities involving complex numbers graphically.

Locus involving the modulus In Section 3, you defined the modulus of a complex number, , as its distance from the origin, , on an Argand diagram. This means that all complex numbers with the same modulus, , form a circle around the origin, with radius .

You can extend this idea to measure the distance between any two complex numbers on an Argand diagram. If you write two numbers in Cartesian form,

and

, then:

so:

Notice that the last expression is the distance between the points with coordinates

and

.

Key point 1.8 The distance between points representing complex numbers and on the Argand diagram is given by . You can now find complex numbers that satisfy equations or inequalities involving the modulus by thinking

about the geometric interpretation, rather than by doing calculations. WORKED EXAMPLE 1.15

Show on an Argand diagram a set of points satisfying the equation The points will lie on a circle, centre , radius .

.

The equation says ‘the distance between and is ’.

Key point 1.9 The equation

represents a circle with radius and centre at the point .

WORKED EXAMPLE 1.16

Shade on an Argand diagram the set of points satisfying the inequality

.

You need to write the inequality in the form . The points will lie on a circle, centre radius .

,

The inequality can be interpreted as saying, ‘the distance between and is less than ’. Remember that the dashed line means that the circle itself is not included.

You can also describe the inequality in Worked example 1.16 by using notation. The circle with centre points satisfying the inequality is

and radius has equation .

coordinates and set theory , so the set of all

Tip A locus is a set of all the points that satisfy a given condition.

WORKED EXAMPLE 1.17

Sketch the locus of points in the Argand diagram that satisfy

.

The equation says ‘the distance from to is the same as the distance from to ’. This is the perpendicular bisector of the line segment joining the points and .

Locus involving the argument The argument measures the angle that the line connecting to the origin makes with the real axis. So the three points shown on the diagram all have the same argument, .

Since the argument is measured from the positive -axis, any number on the dashed part of the line would have argument (or if the argument is negative). If you replace by

, the line shifts from the origin to the point .

Key point 1.10 The locus of points satisfying angle with the positive -axis.

is the half-line starting from the point and making

Tip A half-line is a line extending from a point in only one direction.

WORKED EXAMPLE 1.18

On the Argand diagram, sketch the locus of points that satisfy

.

You need to write the expression in brackets in the form .

The half-line starts from the point and makes a angle with the horizontal.

WORKED EXAMPLE 1.19

On a single diagram, shade the locus of points that satisfy the inequalities and

. The first inequality represents the region between two half-lines, starting from the point and making angles and

with the

vertical axis. The second inequality represents the inside of the circle with radius and centre at the origin.

The half-lines should be dashed and the circle solid. You can indicate the required region by shading.

EXERCISE 1F 1

Sketch each locus on an Argand diagram. Shade the required regions where relevant. a i ii b i ii c i ii d i ii

2

The set of points from Question 1 part ai can be described as

.

Write the sets of points for the rest of Question 1 using similar notation. 3

Sketch each locus on an Argand diagram. Shade the required regions where relevant. a i ii b i ii

4

Sketch each locus on an Argand diagram. a i ii b i ii c i ii

5

On an Argand diagram, shade the region where

.

6

On an Argand diagram, sketch the locus of points where

7

On an Argand diagram, shade the region where

8

a On the same diagram, sketch the loci of

. . and

.

b Hence find the complex number that satisfies both equations. 9 10

On an Argand diagram, shade the region where

and

On an Argand diagram, shade the locus of points that satisfy

. ,

and

. 11 12



Find the complex number that satisfies is a complex number satisfying

. . Find the maximum possible value of

.

Section 5: Operations in modulus–argument form Addition and subtraction are quite easy in Cartesian form but multiplication and division are more difficult. Raising to a large power is even harder. These operations are much easier in modulus–argument form, thanks to the following result. When you multiply two complex numbers you multiply their moduli and add their arguments.

Key point 1.11 For complex numbers and ,

The proof requires the use of compound angle formulae from the A Level Mathematics course:

Rewind Compound angle formulae are covered in A Level Mathematics Student Book 2, Chapter 8. PROOF 2

Let:

and

Start by introducing variables.

Multiply them together and group real and imaginary parts.

Now use the compound angle formulae given above. So:

This is in the form so, by comparison, you can state the modulus and argument of .

i.e.

Finish the proof with a conclusion.

and

A similar proof gives the result for dividing complex numbers. When you divide two complex numbers you divide their moduli and subtract their arguments.

Key point 1.12 For complex numbers and ,

WORKED EXAMPLE 1.20

WORKED EXAMPLE 1.20 a If

b Write

a

and

in the form

, write

in the form

.

Multiply the moduli and add the arguments.

b

Then convert to Cartesian form: .

WORK IT OUT 1.2

Find the argument of

Solution 1

Solution 2

Solution 3

.

.

EXERCISE 1G

EXERCISE 1G 1

Simplify each expression, giving your answers in the form

.

a i ii b i

ii

2

Write each of these expressions in the form

, where

a i ii b i ii

c i

ii

d i

ii

3

Write

in the form

4

Write

in the form

5

Let a Write

.

and in the form

. , without using trigonometric functions.

b Hence find the exact value of 6

.

.

Use trigonometric identities to show that .

.

Checklist of learning and understanding A complex number is of the form Apart from the fact that numbers.

, where and are real and

.

, the arithmetic of complex numbers is the same as for real

To divide by a complex number, multiply top and bottom by its complex conjugate . It is useful to represent complex numbers geometrically using an Argand diagram. Addition can be represented on a diagram in the same way as adding vectors. There are two ways of describing numbers in the Argand diagram: Cartesian form modulus–argument form . The two forms are linked by:

and

Multiplication and division are easier in modulus–argument form:

Equations and inequalities can be represented on an Argand diagram: represents a circle with centre and radius represents the perpendicular bisector of the line segment connecting points and represents a half-line starting from and making angle with the positive real axis vertical and horizontal lines are represented by equations of the form respectively.



or

,

Mixed practice 1 1

If

, find an expression for

. Choose from these options.

A B C D 2

What is

equivalent to? Choose from these options.

A B C D 3

Let

and

.

Showing all your working clearly, find in the form

:

a b 4

Express

in the form

5

a For the complex number

. , find:

i ii

, where

.

b State the modulus and argument of 6

a Solve the equation

.

.

b Represent the solutions on an Argand diagram. 7

Find the complex number such that

8

On an Argand diagram, illustrate the locus of points that satisfy the inequality .

9

It is given that

.

, where and are real numbers.

a Find, in terms of and , the real and imaginary parts of b Hence find the complex number such that

. . [©AQA 2010]

10

Find, in terms of , the complex number that satisfies both Choose from these options. A B C

and

.

D 11

Given that

12

Let and be complex numbers satisfying

and

show that

. .

a Express in terms of . b Show that, if 13

, then

Represent on an Argand diagram the region defined by the inequalities

and

. 14

Two complex numbers, and , are shown on the Argand diagram.

a Add the points representing the numbers

and

to the diagram.

b Sketch the locus of points that satisfy 15

Two loci,

and

.

, on an Argand diagram are given by:

The point represents the complex number

.

a Verify that the point is a point of intersection of b Sketch

and

and

.

on one Argand diagram.

c The point is also a point of intersection of represented by .

and

. Find the complex number that is

[©AQA 2013] 16

Let and be complex numbers such that

17

If

18

If and are complex numbers, solve the simultaneous equations:

19

Let

prove that

and

a Show that b Find the value of

and

. Find the real part of .

.

. .

in the form

, where and are to be determined exactly in surd

form.Hence find the exact values of



and

20

By considering the product

, show that

21

If

find

and

.

in its simplest form.

.

2 Roots of polynomials In this chapter you will learn how to: factorise polynomials and solve equations which may have complex roots link the roots of a polynomial and its coefficients use substitutions to solve more complicated equations.

Before you start... A Level Mathematics Student Book 1, Chapter 4

You should be able to use the factor theorem to factorise cubic polynomials.

1 a Show that

Chapter 1

You should be able to perform arithmetic with complex

2 a Expand and simplify

numbers and solve quadratic equations with complex roots.

Chapter 1

You should be able to work with complex conjugates.

is a root of

b Hence factorise

b Solve the equation

3 Given

completely.

.

, find:

a b

Using complex numbers in factorising This chapter draws together ideas from the AS Mathematics course about factorising cubics and quartics with the theory of complex numbers from Chapter 1 of this resource to enable you to find complex roots of these polynomials. It also looks at the relationship between the coefficients of a polynomial and its roots.

Section 1: Factorising polynomials You are already familiar with the link between factorising and real roots of a polynomial: for example, if is a root of the equation , then is a factor of the polynomial . This is the factor theorem (which you met in A Level Mathematics Student Book 1, Chapter 4), and it also applies to polynomials with complex roots. WORKED EXAMPLE 2.1 a Solve the equation b Hence factorise

. . Use the quadratic formula.

a

b Therefore:

Make the link between roots and factors.

You can use the same method to factorise polynomials of higher degree. WORKED EXAMPLE 2.2

Let a Show that b Hence write

. as a product of a linear factor and a real quadratic factor.

c Solve the equation

and write

as a product of three linear factors.

a

Substitute

b

As is a root, is one factor. You can find the quadratic factor by long division or comparing coefficients.

into

.

c

If

Use the formula to solve the quadratic.

State all three roots of

.

Each root has a corresponding factor

.

This means that you can now factorise some expressions that were impossible to factorise using just real numbers. A particularly useful case is this extension of the difference of two squares identity to the sum of two squares.

Key point 2.1

WORKED EXAMPLE 2.3

Factorise

. This is a difference of two squares. The first factor is a difference of two squares again. The second factor is a sum of two squares.

EXERCISE 2A

EXERCISE 2A 1

Solve the equation

and hence factorise

.

a i ii b i ii c i ii 2

Factorise each expression into linear factors. a i ii b i ii c i ii

3

Given one root of the cubic polynomial, write it as a product of a linear factor and a real quadratic factor, and hence find the other two roots. a i

; root

ii

; root

b i

; root

ii

; root

4 Show that 5

a Given that

is a factor of

and find all the solutions of the equation

is a factor of

b Factorise

, show that

.

.

completely.

6 a Show that b Factorise

is a factor of

.

and hence solve the equation

.

7 a Show that

and

are factors of

.

b Write as a product of two linear factors and a quadratic factor. Hence find all solutions of the equation . 8



Find all solutions of the equation

.

Section 2: Complex solutions to polynomial equations You know from Chapter 1 that if a quadratic equation has two complex solutions, then they are always a conjugate pair: for example, the equation has solutions and . This happens because of the in the quadratic formula.

Did you know This result is not true if the polynomial has complex coefficients. For example, the equation has solutions

and .

If you look at the cubic and quartic polynomials you factorised in Exercise 2A, you will see that the complex roots were either real or a complex conjugate pair. It can be proved that this result generalises to any real polynomial (a polynomial in which all of the coefficients are real numbers).

Key point 2.2 Complex solutions of real polynomials come in conjugate pairs. If

then

.

Focus on ... See Focus on… Proof 1 for a proof of this result. This result can be very useful when you are factorising polynomials and solving equations. If you know one complex root, then you can immediately write down another one. WORKED EXAMPLE 2.4

Given that

is a root of the polynomial

is a root so So

is also a root.

and

Complex roots come in conjugate pairs.

are factors of

. Therefore

, find the other two roots.

Link roots to factors (using the factor theorem).

is a factor. Expand and simplify.

By inspection,

Use polynomial division to find the third factor.

.

So the solutions are

and

.

In Worked example 2.4, in order to find the quadratic factor of , which is a product of the form

, you needed to expand . You get this type of expression

whenever you are expanding two brackets corresponding to a pair of complex conjugate roots, and so it is useful to know a shortcut.

Key point 2.3

In the case of

,

and so:

WORKED EXAMPLE 2.5

Given that one of the roots of the polynomial

, find the

remaining roots. is a root so So

is also a root.

and

Complex roots come in conjugate pairs.

are factors of

Therefore:

.

Link roots to factors (using the factor theorem). Multiply out the brackets (using the shortcut from Key point 2.3). If

, then:

is a factor.

and .

Factorising:

Divide by (using polynomial division or comparing coefficients).

For the roots of the second factor:

You already know the roots of the first factor. Find the roots of the second factor by factorising.

So the roots of

are:

and

.

You can also find a polynomial with given roots. WORKED EXAMPLE 2.6

Find a cubic polynomial with roots and Give your answer in the form is a root of

, so

is a root, so So

.

is a factor.

is another root.

also has factors

.

From the factor theorem, if is a root then is a factor. Complex roots occur in conjugate pairs.

and

Use the result of Key point 2.3 to expand the brackets corresponding to the complex roots. If then: and

Expand and simplify.

WORKED EXAMPLE 2.7

The polynomial numbers , , and . The other two roots are

has roots and

and

.

Hence the two quadratic factors are:

. Find the values of the real

Write down the remaining roots and use them to find factors. The complex conjugate pairs will give quadratic factors.

and:

Hence:

Multiply the two factors to get the polynomial.

So

WORK IT OUT 2.1 Find a cubic polynomial with real coefficients, given that two of its roots are and Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 The other complex root is

, so the polynomial is:

Solution 2 The other complex root is

, so the polynomial is:

Solution 3 The other complex root is

, so the polynomial is:

EXERCISE 2B 1

Find the real values of and such that the quadratic equation roots. a i ii

2

and and

b i

and

ii

and

Given one complex root of a cubic polynomial, factorise the polynomial and write down all its roots. a i

; root

ii

; root

b i

; root ; root

ii 3

Given one complex root of the quartic polynomial, find one real quadratic factor and hence find all four roots. a i

; root

ii

; root

b i

; root ; root

ii c i

; root ; root

ii 4

has the given

Given that

is one root of the equation

, find the remaining two roots.

5 a Show that

.

b Hence solve the equation 6

.

is a root of the equation

.

a Write down another complex root of the equation. b Find the remaining two roots. 7

Two roots of the equation two roots and hence write

are and . Write down the other as a product of two real quadratic factors.

8 a Show that b Write

.

as a product of two real quadratic factors.

c Hence find the remaining solutions of the equation 9

Find a quartic equation with real coefficients and roots and

. .



Section 3: Roots and coefficients When you first learned to factorise quadratic equations you were probably told to look for two numbers that add up to the middle coefficient and multiply to give the constant term. For example, factorises as because and and , add up to give the negative of the middle coefficient,

. This means that the roots, , and multiply to give the constant term,

.

However, there are infinitely many other quadratic equations with the same roots, because you can multiply the whole equation by a constant. For example, another equation with roots and is . The two roots still add up to

, which is

, and multiply to give

, which is

. This is

a particular example of a very useful general result.

Key point 2.4 If and are the roots of the quadratic

then:

PROOF 3

If a quadratic equation has roots and then:

Write the equation in factorised form.

You can multiply the whole equation by a number, and it will still have the same roots. Hence:

and

So:

Compare coefficients with

.

and

You can also find other functions of roots. This requires some algebraic manipulation. WORKED EXAMPLE 2.8

The equation

has roots and . Find the value of:

a b

. You could actually find the roots by using the quadratic formula, but Key point 2.4 provides a much quicker way to answer the question. It is often helpful to combine fractions into a single fraction.

a

b

You can get

by squaring

is a sum of two squares, so you may think that you made a mistake because the answer is negative. However, and can be complex numbers, so this is in fact possible!

The Greek letters , and are often used instead of and for the roots. WORKED EXAMPLE 2.9

The equation

has roots and .

a Write down expressions for b Find

and

in terms of .

in terms of .

a

Apply Key point 2.4.

b

You can find expansion of

from the binomial .

You need to try to express everything in terms of and

EXERCISE 2C

EXERCISE 2C 1

The equation

has roots and . Find the value of each expression in terms of .

a b c d e f 2

The equation

has roots and . Find the value of each expression in terms of .

a b c d

Cubic and quartic equations For cubic equations there are three relationships between roots and coefficients. Just as in the proof of the relationships between coefficients and roots of a quadratic equation, if the roots of the equation are and then you can write:

Expanding and comparing coefficients then gives the results.

Key point 2.5 If , and are the roots of the cubic

then:

You can again combine these with algebraic identities to find other combinations of roots. WORKED EXAMPLE 2.10

The equation Find the value of

has roots

and .

. You can find .

from the expansion of

Now use Key point 2.5.

Similar relationships between roots and coefficients can be found for polynomial equations of any degree. The expressions get increasingly complicated.

Tip To help you to remember these equations, notice that the pattern is always the same.

The sum of all the roots is related to . The sum of all possible products of two roots is related to . The sum of all possible products of three roots is related to .

The signs alternate between and .

Key point 2.6 If , , and are the roots of the quartic

, then:

WORKED EXAMPLE 2.11

The equation

Find the value of

has roots

and .

.

Combine into a single fraction.

Use Key point 2.6.

You can use information about the roots to prove facts about the coefficients of an equation.

Rewind Remember that an arithmetic sequence is one in which each successive term changes by a constant number. WORKED EXAMPLE 2.12

The roots

and of the equation

Show that

form an arithmetic sequence.

. If

and form an arithmetic sequence then .

You know three equations relating the roots to the coefficients. The most useful one seems to be Substitute from

. .

To involve and you need to use the other two relationships as well. Start with with . You know that

Finally use Substitute from , as required.

EXERCISE 2D

.

with .

.

EXERCISE 2D 1

The equation

has roots , and . Find the value of:

a b c d 2

. and are the roots of the given quartic equation. In each case, find the value of the

given expressions. a

; 

b

; 

c 3

The equation

4

The equation

and

.

and

; 

and

. .

has roots has roots

and . Find the value of

.

and . Find the value of:

a b 5

.

When two resistors of resistances total resistance in the circuit is resistance satisfies

and

are connected in series in an electric circuit, the . When they are connected in parallel, the total

.

Two resistors have resistances equal to the two roots of the quadratic equation . Find the total resistance in the circuit if the two resistors are connected: a in series b in parallel. 6

The cubic equation

has roots , and .

a Write down the value of b Show that 7

.

.

A random-number generator can produce four possible values, all of which are equally likely. The four values satisfy the equation If a large number of random values are generated, estimate their mean.

8

The equation

9

The equation

, where is a real constant, has roots

a Find an expression for

in terms of .

has roots

b Explain why this implies that the roots are not all real.



and

. Show that

. and .

Section 4: Finding an equation with given roots You can use the relationships between roots of polynomials and their coefficients to find unknown coefficients in an equation with given roots.

Rewind You can already do this by writing down the factors and expanding brackets (see Worked example 2.7), but this method is more direct. WORKED EXAMPLE 2.13

The quadratic equation values of and . The other root is

has real coefficients and one of its roots is

.

. Find the

Complex roots occur in conjugate pairs.

Then:

Use

Use

.

.

Remember:

.

WORKED EXAMPLE 2.14

A quartic equation . Find the values of and . The four roots are: ,

,

,

has real coefficients and two of its roots are and

As you know two of the complex roots, you can find the other two (their conjugates). Use

, being careful to note

how the coefficients have been labelled in this question ( is the coefficient of here).

Use

.

Use

.

You can also find a new equation with roots that are related to the roots of a given equation in some way, and you can do this without solving the equation. The strategy is to use the sum and product of roots of the first equation to find the sum and product of roots of the second equation. WORKED EXAMPLE 2.15

The quadratic equation coefficients and roots and

has roots and . Find a quadratic equation with integer

You don’t need to find and and product.

Let the equation be Then: Set

.

and .

Then:

and

.

just their sum

The coefficients of the new equation are related to the roots and . All equations with the required roots are multiples of each other, so you can set . You need to relate and . The second one is easier. Substitute

and

.

You can square

to get

Substitute

and

Substitute into

The equation is:

to

.

.

.

You want the equation with integer coefficients, so multiply through by .

WORKED EXAMPLE 2.16

The equation

has roots

For the equation

, and .

Use Key point 2.5 for the original equation.

If the equation with roots and is

and . Find a cubic equation with roots

,

Now use the roots to find the coefficients. You can set the coefficient of to be .

then: Using the first part of Key point 2.5 for the new equation.

Using the second part of Key point 2.5 for the new equation.

Using the third part of Key point 2.5 for the new equation.

The equation is

.

Fast forward You will see in Section 5 that you could also find the equation from Worked example 2.16 by using the substitution

.

WORKED EXAMPLE 2.17

The equation . For the equation

If the equation with roots then:

has roots

:

. Find a cubic equation with roots

and

Use Key point 2.5 for the original equation.

is

You can take the coefficient of

to be .

By Key point 2.5, the sum of the roots of the new equations is

.

Use the second part of Key point 2.5. Factorise.

The product of the roots is

by Key point 2.5.

So the equation is

You can also write this equation as

.

EXERCISE 2E 1

Given the roots of the equations, find the missing coefficients. a i

; roots ; roots

ii

2

b i

; roots

ii

; roots

Find the polynomial of the lowest possible order with the given roots. a i ii b i ii c i ii d i ii

3

A cubic equation has real coefficients and two of its roots are and

.

a Write down the third root. b Find the equation in the form 4

.

The quartic equation and

has real coefficients, and two of its roots are

.

a Write down the other two roots. b Hence find the values of and . 5

The equation and .

6

The equation coefficients and roots

7

has roots

The quadratic equation

has roots

and . Find a cubic equation with roots

and . Find a cubic equation with integer .

has roots and . Find a quadratic equation with roots

and . 8

9

The equation and

has roots

and . Find a quartic equation with roots

.

Let and be the roots of the equation

.

a Find the values of and

.

b Hence find a quadratic equation with integer coefficients and roots and 10

The equation

has roots

a Find the value of

.

b Find a cubic equation with roots 11

The equation and roots and .

12

a Expand

and .

has roots

and . and . Find a cubic equation with integer coefficients

.

b If and are the roots of the equation roots and .



find a cubic equation with

Section 5: Transforming equations In Section 4, you saw how to use relationships between roots and coefficients of a polynomial equation to find another equation with roots that are related to the roots of the first one. In some cases, however, there is an easier way to find this equation – by using a substitution. Suppose the original equation, with roots and , is If you set

Since

.

you get a new equation.

this equation has roots

and .

Key point 2.7 Given an equation in that has a root resulting equation in has a root

, if you make a substitution

, then the

.

WORKED EXAMPLE 2.18

The equation

has roots

and .

a Find a cubic equation with roots

and

b Hence find the value of

.

.

When , substitution

a

so make the .

Substitute for .

This is the product of the roots of the equation in .

b

Tip If you were just asked to find , you could expand and use the relationships between roots and coefficients of the original equation. You can decide for yourself whether this or the substitution method is simpler.

WORK IT OUT 2.2 The roots of the quadratic equation

are and . Find quadratic equation with

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

From the first equation: and So for the second equation: and Hence the new equation is:

Solution 2 Let

; then and are the roots for .

Make the substitution

.

Solution 3 Replace by :

A substitution is particularly useful if it transforms a difficult equation into one that you can solve easily. WORKED EXAMPLE 2.19

Use the substitution

to solve the equation

. Make the given substitution. It’s a good idea to line up terms when expanding lots of brackets.

You can solve the equation in easily by factorising.

Using

EXERCISE 2F

.

EXERCISE 2F 1

Use the given substitution to transform the equation in into a polynomial equation for . a i

; ;

ii b i

.

;

ii

.

;

c i ii 2

.

. ;

.

;

The equation

. has roots

equation with roots 3

.

The equation

has roots

coefficients and roots 4

5

. Use a suitable substitution to find a cubic

. Find a quartic equation with integer

.

a Show that the substitution equation .

transforms the equation

b Hence solve the equation

.

a Find the value of so that the substitution into the equation

transforms the equation .

b Hence find all the solutions of the equation 6

The equation

into the

.

has roots and . Using the substitution

find the value of

, or otherwise,

.

7

The equation an equation with roots

has roots and . Using a suitable substitution, or otherwise, find and .

8

The substitution equation of the form

transforms the equation .

into the

a Find the value of . b Hence solve the equation

.

Checklist of learning and understanding Complex roots of real polynomials occur in conjugate pairs: if use this fact to factorise cubic and quartic polynomials. Coefficients of a polynomial can be expressed in terms of its roots. For a quadratic equation

For a cubic equation

with roots and :

with roots , and :

then

. You can

For a quartic equation

with roots , , and :

You can use these relationships to find: a polynomial with given roots a polynomial with roots related to the roots of another polynomial. The second of these types of problem can sometimes also be solved by making a substitution.



Mixed practice 2 1

Find the sum of the roots of the equation

.

Choose from these options. A B C D 2

The quadratic equation substitution

is transformed into an equation in by the

.

Find the root of the transformed equation in terms of the constant . Choose from these options. A B C D 3

Given that

is one root of the equation

4 . Find 5

where and .

, find the other two roots.

are real constants. Two roots of

One of the roots of the polynomial

are

is

and

.

a Write down another complex root and hence find a real quadratic factor of b Solve the equation

.

6

Find a quartic equation with real coefficients given that three of its roots are

7

The quadratic equation

and .

has roots and .

a Write down the values of b Show that

.

and

.

.

c Find a quadratic equation with integer coefficients, which has roots

and

.

[©AQA 2012] 8

Given that

9

Two roots of the cubic equation

is one root of the equation

, find the other two roots.

are

and

.

a Write down the third root. b Find the values of 10

The polynomial constants and .

11

a Show that

and . has a factor of

. Find the values of the real

.

b Let and be the roots of the quadratic equation

. Find a quadratic equation

with roots 12

and

.

The equation

has roots

cubic equation with roots 13

The equation

,

and

has roots

a cubic equation with roots 14

and .

and .

in terms of . .

c Given that

ii

to find

.

b Show that

and

and . Use a substitution of the form

has roots

a i Write down the value of

i

to find a

and .

The cubic equation

ii Express

and . Use the substitution .

and that and are real, find the values of:

d Find a cubic equation with integer coefficients which has roots

and . [©AQA 2012]

15

a A cubic equation

has roots

i Write down the values of

.

and

ii Show that

in terms of

.

b The roots , and of the equation i Show that ii Show that 16

and .

form a geometric sequence.

. .

a Show that: i ii b Given that the cubic equation i write down the values of ii show that

and

ii Find the values of

in terms of

and : and

.

c The equation i Show that

has roots

has roots

,

and

and

.

.

.

iii Hence find a cubic equation with integer coefficients and roots

and

.

3 The ellipse, hyperbola and parabola In this chapter you will learn how to: recognise and work with Cartesian equations of ellipses, hyperbolas and parabolas solve problems involving intersections of lines with those curves and find equations of tangents recognise the effects of transformations (translations, stretches and reflections) on the equations of those curves.

Before you start… A Level Mathematics Student Book 1, Chapter 3

You should be able to use the discriminant to determine the number of solutions of a quadratic equation.

1 Find the set of values of for which the equation has exactly two real roots.

A Level Mathematics

You should be able to interpret

Student Book 1, Chapter 5

solutions of simultaneous equations as intersections of graphs.

2 Find the value of for which the line is a tangent to the graph of .

A Level Mathematics Student Book 1, Chapter 5

You should be able to recognise the

A Level Mathematics Student Book 1, Chapter 5

You should be able to recognise the relationship between transformations of graphs and their equations.

graph of

, and know that it has

3 Sketch the graph of

and state

the equations of its asymptotes.

asymptotes.

4 Find the equation of the resulting curve after the graph of is: a translated units in the positive -direction b stretched vertically with scale factor .

A Level Mathematics Student Book 1, Chapter 6

You should be able to write down the equation of a circle with a given centre and radius, and to find the centre and radius from a given equation.

5 Find the centre and radius of the circle with equation .

Conic sections In A Level Mathematics Student Book 1 you met two curves with equations that involve squared terms: the parabola, such as , and the circle, such as . In this chapter you will meet some other curves with similar equations; for example: and .

All such curves can be obtained as intersections of a plane with a cone; hence they are known as conic sections. As well as having many interesting mathematical properties, they have applications in modelling planetary orbits and the design of satellite dishes and radio telescopes.



Section 1: Introducing the ellipse, hyperbola and parabola The ellipse You already know that the equation Similarly, the equation

represents a circle with radius , centred at the origin. can be rewritten as

. But what happens if the coefficients of

and

, so it represents a circle with radius

in this equation are not equal?

Use graphing software to investigate equations of this form. Here are some examples.

In each case there is a closed curve centred at the origin. The axes intercepts are most easily seen if the equation is written in the form

.

Key point 3.1 The equation and -intercepts

represents an ellipse centred at the origin, with -intercepts .

Did you know It can be shown, using the universal law of gravitation and Newton’s laws of motion, that planets follow elliptical orbits.

WORKED EXAMPLE 3.1

Sketch these ellipses, showing the coordinate axes intercepts. a b Comparing this equation to Key point 3.1, and .

a

b  

To compare this to Key point 3.1, the right-hand side needs to equal . So divide the whole equation by . Multiplying by is the same as dividing by .

You can now see that this is an ellipse with and

.

The hyperbola What happens if, in the equation of an ellipse, the Here are some examples.

term is negative?

Each of these graphs is a curve called a hyperbola. As with the ellipse, it is convenient to write the equation in the form

.

You can see several common features in all the graphs: The curve has the and -axes as lines of symmetry. This is because both and terms in the equation are squared. You can find the -intercepts, which are also the vertices of the curve, by setting the curve with equation

crosses the -axis at

and

; so, for example,

.

There is a range of values of for which the curve is not defined. For example, the curve with equation is not defined for

.

For large values of and , the curve seems to approach a straight line. Because of the symmetry of the curve, this line passes through the origin.

Tip To see why

isn’t defined for . Since

be between

is never negative,

, rewrite the equation in the form can’t be smaller than

, so can’t

and .

Key point 3.2 The equation

represents a hyperbola with -intercepts

and asymptotes

.

WORKED EXAMPLE 3.2

Sketch these hyperbolas, showing the axis intercepts and stating the equations of the asymptotes. a b Comparing this to Key point 3.2, . The  -intercepts are

and

a

The asymptotes are

.

b

The asymptotes are given by

.

First write the equation in the form .

Multiplying by is the same as dividing by . You can now see that this is a hyperbola with and

.

Hence the -intercepts are

Asymptotes:

.

The asymptotes are given by

.

You can derive the equations of the asymptotes of a hyperbola (rather than just guessing them from the graph as you did in Worked example 3.2). PROOF 4

Prove that the asymptotes of a hyperbola with equation

The asymptotes pass through the origin, so their

are

.

equations are

.

First, by symmetry of the curve, the asymptotes must pass through the origin so the -intercept is zero. The two asymptotes must also be reflections of each other in the -axis so their equations are of the form and . ‒

Consider a line of the form . Whether or not this line crosses the hyperbola depends on

The hyperbola never intersects its asymptotes, so consider the intersection of the curve with the lines of the form .

the value of .

To find any possible intersections, substitute into the equation of the hyperbola. Multiply both sides by the common denominator to clear fractions. The is positive and can’t be negative. This is only possible if the expression in the bracket on is positive.

Solutions only exist if

Therefore for

if

You can divide both sides by without affecting the direction of the inequality because is a positive number.

the line doesn’t

cross the hyperbola and if The limiting case, when

it does cross. , gives the

gradient of the asymptote. A graph shows how the limiting case gives the asymptotes.

Similarly for

to give the gradient of the

other asymptote as

. Hence the

equations of the asymptotes are

and

.

You have actually met a special case of the hyperbola before:

(or, more generally,

). It can be

shown that this is indeed a hyperbola but rotated so that the asymptotes are the coordinate axes. If is restricted to be a positive constant, then it can be written as and the equation

Fast forward

is usually written as

.

. To emphasise the symmetry between

You may be surprised that the equations

and

represent members of the

same family of curves. In Further Mathematics Student Book 2 you will see how to prove that the curves are related by rotation.

Key point 3.3 The equation

represents a hyperbola with vertices at

and

and coordinate

axes as asymptotes.

WORKED EXAMPLE 3.3

Sketch the curve with equation

, labelling the coordinates of the vertices. To use Key point 3.3 the equation needs to be in the form . This is a hyperbola with the coordinate axes as asymptotes.

It has so

so

.

The parabola In the equations of an ellipse and a hyperbola, both and terms are squared. You already know that is an equation of a parabola with a vertex at the origin. If the term is squared instead, the curve still has the same shape, but is ‘sideways’.

Key point 3.4

The equation

represents a parabola with its vertex at the origin, tangent to the -axis.

WORKED EXAMPLE 3.4

A curve has equation

.

a Sketch the curve. b Let be a point on the curve with coordinates Show that the distance of from the point a

. equals the distance of from the line

.

This is a parabola with vertex at the origin.

Sketch a graph showing the required distances.

b

The distance from a point to a vertical line equals the difference between the coordinates. You can use Pythagoras’ theorem to find the distance between two points.

Point lines on the parabola, so Hence

.

, as required.

Did you know The property illustrated in Worked example 3.4 can also be taken as the defining feature of a parabola: it is the locus of the points that are equal distances from a given point line

and a given

.

The point in Worked example 3.4 is called the focus of the parabola, and has an important special property. Horizontal rays reflected off a parabola with equation all pass through the point with coordinates

.

Satellite dishes usually have a parabolic shape.

EXERCISE 3A 1

Sketch these ellipses, showing the coordinates of any axis intercepts. a i ii b i ii c i ii d i ii

2

Write down the equation of each ellipse. a i

ii

b i

ii

c i

ii

d i

ii

3

Sketch these hyperbolas, stating the coordinates of the vertices and the equations of the asymptotes. a i ii b i ii c i ii d i

ii 4

Sketch these hyperbolas, indicating the coordinates of the vertices. a i ii b i ii c i ii

5

Find the equation of each hyperbola, giving your answers in the form a i

ii

.

b i

ii

c i

ii

d i

ii

6

Sketch each of these curves, showing the coordinates of any axes intercepts and vertices, and equations of any asymptotes. a i ii

b i ii c i ii d i ii 7

What is the equation of the curve in the diagram?

Choose from these options. A B C D 8

What are the equations of the asymptotes of the hyperbola with equation

?

Choose from these options. A B C D 9

An ellipse has equation coordinates the value of .

and

and

is a point on the ellipse. Points

. Show that the sum of the distances

and

have

does not depend on

Did you know The property illustrated in Question 9 can be used as a defining feature of an ellipse: it is the locus of points whose sum of distances from two fixed points is constant.



Section 2: Solving problems with ellipses, hyperbolas and parabolas In A Level Mathematics Student Book 1 you learnt how to find intersections of lines with parabolas and circles. You also learnt how to use the discriminant to determine the number of intersections. You can use those methods to solve problems involving ellipses and hyperbolas.

Rewind See A Level Mathematics Student Book 1, Chapter 3, for a reminder of quadratic equations and inequalities. WORKED EXAMPLE 3.5

The line

intersects the hyperbola with equation

at two points.

Find the range of possible values of . To write an equation for the intersection points, substitute into the equation of the hyperbola. Clear the fractions: multiply both sides by the common denominator. This is a quadratic equation in . Make one side equal to zero. There are two solutions:

For two intersection points, the discriminant needs to be positive.

This is a quadratic inequality for , so you can sketch the graph to see that there are two solution intervals.

You can also use the discriminant to solve problems about tangents and normals.

Rewind It is also possible to use implicit differentiation to find gradients of tangents and normals to these curves. This is covered in A Level Mathematics Student Book 2, Chapter 10. WORKED EXAMPLE 3.6

The line

, with

, is a tangent to the ellipse with equation

.

a Find the value of . The line touches the ellipse at point . b Find the equation of the normal to the ellipse at . You need to find the value of for which there is only one intersection point between the line and the ellipse. Start by writing an equation for the intersections: substitute into the equation of the ellipse.

a

This is a quadratic equation for , so make one side equal to zero. As there is one solution:

For the line to be a tangent, this equation should only have one solution. This happens when the discriminant is zero.

The question says that is positive. b Coordinates of :

You first need to find the coordinates of . The -coordinate is the solution of the quadratic equation with the value of from part a.

Gradient of normal:

The normal is perpendicular to the tangent, so their gradients multiply to ‒ .

Equation of normal:

Use normal.

for the equation of the

Rewind A normal to the curve is perpendicular to the tangent at the point of contact. See A Level Mathematics Student Book 1, Chapter 6, for a reminder of perpendicular lines. WORKED EXAMPLE 3.7

A parabola has equation gradient .

. A line passes through the point

Given that the line is tangent to the parabola, express in terms of .

, with

, and has

For the intersection with the parabola:

One solution:

Write the equation for the intersection of the line and the parabola: substitute from the equation of the line into the equation of the parabola.

If the line is a tangent to the parabola, this quadratic equation for should have only one solution so the discriminant is zero. Note that this gives a single value of for every .

EXERCISE 3B

EXERCISE 3B 1

Find the value of for which the line .

2

A hyperbola has equation

is tangent to the parabola with equation

.

a Sketch the hyperbola, stating the equations of the asymptotes. b The line with equation 3

The line

is tangent to the hyperbola. Find the possible values of .

is a tangent to the ellipse

.

Find the possible values of . 4

A parabola has equation

.

a Show that the line with equation

is tangent to the parabola and find the

coordinates of the point of contact, . b The normal to the parabola at intersects the parabola again at . Find the coordinates of . 5

The line

is tangent to the ellipse

at the point .

a Find the coordinates of . The normal to the ellipse at intersects the ellipse again at . b Find, to significant figures, the coordinates of . 6

A curve has equation

.

a Sketch the curve, showing the intercepts with the coordinate axes. b A line with equation possible values of . 7

intersects the ellipse at two distinct points. Find the range of

A hyperbola has equation

.

a Write down the equation of the asymptote with a positive gradient. b Show that every other line parallel to this asymptote intersects the hyperbola exactly once. 8

A hyperbola has equation a The line

.

is a tangent to the hyperbola. Find the possible values of .

b Show that this hyperbola has no tangents which pass through the origin.



Section 3: Transformations of curves In Sections 1 and 2 you only looked at ellipses and hyperbolas centred at the origin, and parabolas with vertex at the origin. You can use your knowledge of transformations of graphs to translate these curves.

Rewind Transformations of graphs are covered in A Level Mathematics Student Book 1, Chapter 5.

Translations You know how to apply a translation to a curve.

Key point 3.5 Replacing by

and by

results in a translation of the curve by the vector

.

WORKED EXAMPLE 3.8

The hyperbola with equation

is translated by the vector

.

Find the equation of the resulting curve and state the coordinates of its vertices. Replace by

and by

.

Clear the fractions: multiply by the common denominator .

The original hyperbola has vertices and , so the new hyperbola has vertices

The vertices of a hyperbola

are at

. The curve has been translated by

.

and

In A Level Mathematics Student Book 1 you learnt how to complete the square to find the centre and radius of a circle when the equation is written in an expanded form. You can apply the same technique to ellipses, hyperbolas and parabolas.

Rewind See A Level Mathematics Student Book 1, Chapter 3, for a reminder of completing the square. WORKED EXAMPLE 3.9 a Show that the equation represents a hyperbola, and find the coordinates of its vertices and the equations of its asymptotes. b Hence sketch the curve.

a

You want to write the equation in the form , so start by completing the squares (separately for the and terms). Remember to take out a factor of term (rather than just ).

Hence:

from each

Now put these back into the original equation.

Divide both sides by the required form.

to get the equation in

This is a hyperbola with equation translated by the vector

The vertices of the original hyperbola are at and , so the vertices of the new hyperbola are and .

The original hyperbola was translated units to the left and units up.

The asymptotes of the original hyperbola are

The asymptotes of a hyperbola are given by

and

.

The new asymptotes are:

and

b

.

. The translation replaces by .

and by

Stretches You also know how to apply horizontal and vertical stretches to curves.

Key point 3.6 Replacing by and by results in a horizontal stretch with scale factor and a vertical stretch with scale factor .

WORKED EXAMPLE 3.10

Show that the ellipse with equation

can be obtained from a circle with radius by

applying a horizontal stretch and a vertical stretch. State the scale factor of each stretch. You want to see the connection between the given ellipse and the circle , so start by making the of the equation equal to . This is the equation of the circle, with replaced by Hence the ellipse is obtained from the circle by a horizontal stretch with scale

and replaced by

Remember that a stretch with scale factor replaces by .

factor and a vertical stretch with scale factor .

Reflections Key point 3.7 Replacing with

reflects a curve in the -axis.

Replacing with

reflects a curve in the -axis.

Replacing with and with reflects a curve in the line Replacing with

and with

WORKED EXAMPLE 3.11

reflects a curve in the line

. .

.

WORKED EXAMPLE 3.11

A parabola has equation

. Find the equation of the resulting curve when this parabola is:

a reflected in the -axis b reflected in the line

.

a

Reflection in the -axis replaces by

b

Reflection in the line

.

swaps and .

WORKED EXAMPLE 3.12

Find the equation of the hyperbola shown in the diagram.

Reflect the hyperbola in the line

.

You can obtain this hyperbola from a standard hyperbola by reflecting it in the line . First find the equation of the reflected hyperbola by using its vertex coordinates and equations of asymptotes.

The asymptotes are:

The reflection swaps and ; you can use this to find the asymptotes of the reflected hyperbola. Equation of the reflected hyperbola:

The vertices are at

From the vertices:

From the asymptotes:

The equations of the asymptotes are

.

The reflected hyperbola has equation

The original hyperbola has equation

Reflection

in swaps and .

EXERCISE 3C 1

Find the equation of each curve after the given transformation. Give your answers in expanded and simplified form. a i

; translation with vector

ii

; translation with vector

b i ii c i ii

2

; translation with vector ; translation with vector ; vertical stretch with scale factor ; horizontal stretch with scale factor

d i

; reflection in the line

ii

; reflection in the line

e i

; reflection in the -axis

ii

; reflection in the -axis

Each of these curves is a translation of a standard ellipse, hyperbola or parabola. Identify the translation vector, and hence sketch each curve. a i ii

b i ii c i ii d i ii 3

An ellipse with equation

is translated by the vector

equation

. The resulting curve has

.

Find the values of and . 4

A hyperbola with equation

is translated by the vector

. Given that the

translated hyperbola passes through the origin, find the positive value of . 5

The curve

with equation

a Sketch the curve

, is translated by the vector

. The resulting curve is

, showing the coordinates of the vertices and equations of any

asymptotes. b Find the equation of 6

in the form

A parabola with equation resulting curve is

.

is stretched parallel to the -axis. The equation of the . Find the scale factor of the stretch.

7

A parabola with equation is reflected in the line . The two curves intersect at the origin and at one more point . Find, in terms of , the coordinates of .

8

Sketch the curve with equation

, stating the coordinates of the

vertices and equations of any asymptotes. 9

The curve

has equation

a Sketch the curve The curve

, showing the intercepts with the coordinate axes.

is obtained from

b Given that the point 10

.

by a stretch with scale factor parallel to the -axis.

lies on

The curve with equation

, find the value of . is stretched parallel to the -axis with scale factor . The

resulting curve is the circle with equation 11

The line

, where

. Find the positive values of and .

, is tangent to the curve with equation

.

a Find, in exact form, the possible values of . b Hence find the equations of the tangents to the curve point 12

.

The hyperbola with equation the new hyperbola is

13

The ellipse

which pass through the

is translated by the vector

. One asymptote of

. Find the equation of the other asymptote.

has equation

horizontal stretch. a State the scale factor of the stretch.

.

is obtained from a circle

, with radius , by a

.

The point

lies on

and the point lies on

. is the image of under the stretch.

b Find the coordinates of . c Find the equation of the tangent to

at .

d Hence find the equation of the normal to 14

at .

The diagram shows the hyperbola with equation . The line to the hyperbola, and crosses the asymptotes at the points and .

is a tangent

a Find the value of . b Write down the equations of the asymptotes of the hyperbola. c Find the exact distance

.

Checklist of learning and understanding The equation

represents an ellipse with axis intercepts

The equation

represents a hyperbola with vertices at

and

.

and asymptotes

. The equation represents a hyperbola with vertices axes as asymptotes. The equation line of symmetry.

and

and the coordinate

represents a parabola with the vertex at the origin and the -axis as the

You can use the discriminant of a quadratic equation to solve problems about the number of intersections of a line with a parabola, hyperbola or an ellipse, including identifying tangents. You can apply transformations to curves by changing their equations. Replacing with

and with

results in a translation with vector

.

Replacing with and with results in a horizontal stretch with scale factor and a vertical stretch with scale factor . Replacing with

results in a reflection in the -axis.

Replacing with

results in a reflection in the -axis.

Replacing with and with results in a reflection in the line Replacing with

and with

results in a reflection in the line

. ‒

.



Mixed practice 3 1

What is a possible equation of the curve shown in the diagram?

Choose from these options. A B C D 2

What is the equation of the ellipse shown in the diagram?

Choose from these options. A B C D 3

A parabola with equation

is translated by the vector

What is the equation of the resulting curve? Choose from these options. A B

.

C D 4

The line

is a tangent to the ellipse

5

The line

is tangent to the hyperbola

6

The parabola

has equation

7

, and find the coordinates

.

.

Find the equation of the tangent to

Point

is tangent to

is reflected in the -axis to form a new curve,

b Write down the equation of c

. Find the possible values of .

.

a Find the value of for which the line of the point of contact. The parabola

. Find the possible values of .

at the point

lies on the circle with equation

. .

a Show that the tangent to the circle at has equation

.

The circle is stretched horizontally to form an ellipse with equation

.

b Find the scale factor of the stretch. c Find the equation of the tangent to the ellipse at the point 8

The ellipse with equation

.

is translated by the vector

. What is the equation

of the resulting curve? Choose from these options. A B C D 9

The parabola has equation parabola at the point .

. The line with equation

is tangent to the

a Find the value of . b Find the coordinates of . The parabola

is translated by the vector

to form a new parabola,

c Find the coordinates of the points where the parabola d Sketch the parabola

The parabola

crosses the coordinate axes.

, showing the coordinates of its vertex.

e Find the equation of the tangent to 10

.

, with equation

at the point

.

, is reflected in the line

to form a new parabola,

. The two parabolas intersect at the origin and another point, . a Write down the equation of

.

b Find the coordinates of . c Use differentiation to find the equation of the tangent to d Hence find the equation of the tangent to

at .

at .

11

Point lies on the parabola with equation . For all values of , the distance of the point from the line is the same as its distance from the point . Find the value of .

12

A curve

has equation

a Sketch the curve b

.

, stating the values of its intercepts with the coordinate axes.

The curve

is translated by the vector

, where

, to give a curve

.

Given that

passes through the origin

, find the equations of the asymptotes of

.

[© AQA 2015] 13

An ellipse is shown below.

The ellipse intersects the -axis at the points and . The equation of the ellipse is . a Find the -coordinates of and . b The line

is a tangent to the ellipse, with point of contact .

i. Show that the -coordinate of satisfies the equation

.

ii. Hence find the exact value of . iii. Find the coordinates of . [© AQA 2013] 14

The hyperbola with equation

is translated by the vector

. What are the

asymptotes of the resulting curve? Choose from these options A

and

B

and

C D 15

and and

A hyperbola with equation

, where is a positive constant, is reflected in the line

. Given that the original hyperbola and its image intersect, find the range of possible values of .

16

a b

Show that the line

is a tangent to the ellipse

The line is a tangent to both the ellipse

when

.

and the ellipse

. Find the

possible equations of . 17

An ellipse has equation

a Sketch the ellipse , showing the values of the intercepts on the coordinate axes. b Given that the line with equation show that . c

The ellipse is translated by the vector

intersects the ellipse at two distinct points,

to form another ellipse whose equation is

. Find the values of the constants

and .

d Hence find an equation for each of the two tangents to the ellipse that are parallel to the line . [© AQA 2014]

4 Rational functions and inequalities In this chapter you will learn how to: solve cubic and quartic inequalities sketch graphs of the form solve inequalities of the form sketch graphs of the form

.

Before you start… GCSE

You should be able to solve linear inequalities.

1 Solve

A Level Mathematics Student Book 1, Chapter 3

You should be able to solve quadratic inequalities.

2 Solve

A Level Mathematics

You should be able to use the

Student Book 1, Chapter 3

quadratic discriminant.

3 Find the values of for which has no real solutions.

A Level Mathematics Student Book 1, Chapter 4

You should be able to sketch polynomials from their factorised form.

4 Sketch the graph of

A Level Mathematics Student Book 1, Chapter 4

You should be able to use the factor theorem to factorise polynomials.

5 Factorise

A Level Mathematics Student Book 1, Chapter 5

You should be able to describe graphically simple transformations of functions.

6 What transformation turns ?

A Level Mathematics Student Book 1, Chapter 5

You should be able to sketch graphs of reciprocal functions and know associated terminology.

7

A Level Mathematics

You should be able to simplify

.

.

.

.

What are the asymptotes of

into

?

Student Book 2,

rational functions.

Chapter 5

8 Simplify

.

Why are inequalities important? Although equations are much easier to deal with, in the real world inequalities are often more important. For example, you do not usually need a car with a fuel efficiency of exactly miles per gallon – you want a fuel efficiency of at least miles per gallon. You may want the temperature in a fridge to be at most . Unfortunately, not all of the theory you have learnt for solving equations will always work with inequalities. For example, you should know that if graphs to help you solve inequalities.

this does not mean that

. Instead you need to use

In this chapter, you will explore the graphs of more complicated functions that are the ratio of two polynomials (called rational functions).

Section 1: Cubic and quartic inequalities You already know that to solve quadratic inequalities you can sketch the graph. You can extend the same method to cubic and quartic inequalities.

Key point 4.1 To solve cubic or quartic inequalities, make one side zero and sketch the graph. Describe in terms of the regions of the graph above or below the -axis.

Tip You can also find this solution by examining the sign of each factor. For example, if

,

then the term in the first bracket is positive and the terms in the other two brackets are negative, so overall it will be positive.

WORKED EXAMPLE 4.1

Solve

. Sketch the curve of . Highlight those parts of the graph for which the -value is greater than .

So

Describe, in terms of , the highlighted sections. Notice that, since the inequality in the question is strict, the solution also involves strict inequalities.

.

If the inequality is not in a factorised form with one side zero, you may need to rearrange it and use the factor theorem. WORKED EXAMPLE 4.2

Solve

.

If

, then:

To use the factor theorem try some small integer values. Once you have found one value you could then use polynomial division to find remaining factors, but it is often quicker just to find all three factors.

So, from the factor theorem: Hence the inequality becomes:

Now sketch the graph, and highlight those parts for which the -value is less than .

or

EXERCISE 4A

Describe the highlighted region on the graph in terms of .

EXERCISE 4A 1

Solve each inequality. a i ii b i ii c i ii d i ii e i ii f

i ii

2

Find all values of that satisfy each inequality. a i ii b i ii c i ii d i ii

3

Find all values of such that

4

Solve

5

a Show that

. if

.

is a root of

.

b Hence find the exact range of values of such that 6

Solve

7

A function is given by Given that the solution to

8

9 10

given that and are positive numbers. where is

A function is given by

or

is

Find all values such that

.

, find

Find a quartic inequality for which the solution set is:

b exactly two numbers.

, and are integers. , find

.

Given that the solution to

a all real numbers

.

and .

and .



Section 2: Functions of the form If

, you know from Chapter 3 that the graph of and

forms a hyperbola, with asymptotes at

.

If you use your knowledge of transformations of functions, you can also sketch, say, is shifted one unit right and two units up. It therefore has asymptotes at and

In fact, any function of the form

will be a transformation of

, which .

, so it will also be a hyperbola.

However, it can be difficult to determine which transformations have been applied. Instead you can analyse it by finding the axes intercepts and asymptotes directly. The -intercept is when

so

The -intercept is when

. This means the numerator is zero.

This occurs when

so

.

.

The vertical asymptote will occur when the denominator is zero. This occurs when

so

.

The horizontal asymptote will occur at very large values of . This means the denominator is approximately

and the denominator is approximately

so tends

towards , which is the horizontal asymptote.

Key point 4.2 If

:

the intercepts are the asymptotes are

and and

.

WORKED EXAMPLE 4.3 a Sketch the function b Solve

where

. Label the asymptotes and intercepts.

.

a

Using Key point 4.2 you can say that the intercepts are and

. The vertical asymptote is at

and the horizontal asymptote is at . This information, along with the fact that it is a hyperbola, is enough to sketch the curve.

Add the line to the graph. The only part of the hyperbola above this line is between the intersection and the vertical asymptote.

b

If

then:

You can find the intersection point algebraically.

so

So the solution to the inequality is .

Notice that the can never equal result in division by zero.

as that would

In Worked example 4.3, you used the graph to help you solve the inequality. You can also do this algebraically, but you must take care. If you were just to multiply through by the denominator, you might be multiplying by a negative number, which would reverse the inequality. To avoid this you can use an algebraic trick.

Tip A graphical calculator does not allow you to sketch this function because it involves a parameter, . However, if you have one you can check your answer for sense, by trying to sketch the function for particular values of .

Key point 4.3 When solving inequalities involving algebraic fractions, multiply both sides by the square of the denominator. This works because square numbers are always positive, so you avoid ever having to reverse the inequality. WORKED EXAMPLE 4.4

Use an algebraic method to solve

.

Multiply through by to get rid of the fraction. Resist the temptation to multiply out brackets! Make one side zero… … then factorise by looking for a common factor – in this case

This is a quadratic inequality so sketch the graph.

So

Describe the part of the graph below the axis in terms of .

However, in the original equation, therefore:

Remember to think about the solution in the context of the original problem.

EXERCISE 4B

1

Without using a calculator, determine exact values of any asymptotes and axis intersections and sketch the graph of each function . a i ii b i ii

2

From the information given, write each function values of

by finding the

and .

a i Asymptotes

and

ii Asymptotes

3

in the form

. Root at

and

.

. -intercept at

b i Asymptote

. Root at

and -intercept

ii Asymptote

. Root at

and -intercept

. . .

Solve for . a i ii b i ii

4

Solve for , giving your answers as exact values. a i ii b i ii

5

A function

is given by

.

a Write down the vertical asymptote for The line is given by equation b i Sketch

.

and the line on the same axes.

ii Find the values of for which 6

.

A curve has equation

, where

. .

a Sketch the curve, clearly giving the equations of all asymptotes and the coordinates of all axis intercepts. b Hence solve the inequality

7

The function

is given by

.

a The line is given by the equation Given that is a tangent to The function

. , calculate all possible values of .

is given by

.

b i On the same axes, sketch

and

.

ii Find the exact values of for which 8

for some values a Assuming have:

.

and with

.

, state necessary and sufficient conditions in terms of

and for

to

i a vertical asymptote at ii a horizontal asymptote at

.

b In part a, why was the condition 9

The solution set for Find

10

required? is

or

for some values

and .

Curve has equation a A line

.

is a tangent to the curve. Find a condition for in terms of .

Two parallel lines are each tangents to the curve, touching at points . b If the distance c By differentiating

is , find

and

with

in terms of , the gradient of each of the lines.

with respect to , or otherwise, find the shortest distance between the

two branches of the curve.



and .

Section 3: Functions of the form If the top and bottom of a rational function are quadratic expressions, then the graph will no longer be a hyperbola, but you can use a similar analysis to that shown in Section 2.

Key point 4.4 If

then: when

,

when

is a solution to

vertical asymptotes are solutions to the horizontal asymptote is

.

WORKED EXAMPLE 4.5

a Sketch

.

b Hence solve the inequality a When

Find the -intercept.

.

The horizontal asymptote is

When

.

:

Vertical asymptotes occur when:

.

Consider what happens when gets very large (or just use Key point 4.4). Find the -intercepts by setting the numerator equal to zero.

Find the vertical asymptotes by setting the denominator equal to zero.

Put all the information together. It is a little unclear whether the section between the asymptotes is positive or negative. To decide this, substitute to find .

Add the line to the diagram and see where the graph is above this line.

b

When

So

Find the values of at the intersection points.

:

or

.

Describe the part of the graph above the line in terms of . Note that can’t be equal to or as there are asymptotes there.

One useful way of analysing these functions is to look at the possible -values. If you intersect these functions with you get a quadratic equation. You are not particularly interested in the value of that you get from these equations – you are more interested in whether or not a solution actually exists. To find out, use the quadratic discriminant.

Tip If you have a graphical calculator you can try to check this graph – but you will find that it is very difficult to see all the relevant features in one window. Remember that a labelled diagram does not have to be to scale.

WORKED EXAMPLE 4.6

Find the stationary points on the graph If

then:

.

You met stationary points in A Level Mathematics Student Book 1, Chapter 14. This method provides an alternative to calculus when finding stationary points.



So there are solutions when . A maximum occurs when . A minimum occurs when

You can interpret this as meaning that -values do not occur between and , so there must be a maximum at and a minimum at .

or

Put into the quadratic equation formed above to find the value of .

so

so

.

EXERCISE 4C 1

Identify all axis intercepts and vertical and horizontal asymptotes for each function Hence sketch the graph of each function. a i ii b i ii c i ii d i ii

2

By finding a condition on for the number of solutions to coordinates of each of the stationary points for each function.

, identify the exact -

a i ii b i ii c i ii 3

a Sketch the curve b Hence solve the inequality

, clearly labelling any asymptotes and axis intercepts. .

.

4

One of the asymptotes of the graph of

is

.

Find the value of and give the equation of all other asymptotes. 5

Solve the inequality

6

A curve has equation

. .

a i State the equations of any asymptotes. ii State the coordinates of any axis intercepts. b i If the line

intersects the curve, show that:

ii Hence find the coordinates of the turning points of the curve. c Sketch the curve. 7

Sketch

8

The solution to

if

. is

.

Find and . 9

is a function of the form

where is non-zero.

has no roots or vertical asymptotes, but has a maximum at , a minimum at at and a horizontal asymptote . a Show that

-intercept

.

b Given the maximum and minimum points are at

and

respectively, find

and . 10

Two particles are projected upwards at the same time. Particle begins metres off the ground with an initial velocity of ground level with an initial velocity . is the ratio of the heights above ground from ground, and . a Taking acceleration due to gravity as

until the time the first particle hits the

, find

.

b Find the maximum value of , correct to three significant figures.



. Particle begins at

Section 4: Oblique asymptotes Key point 4.4 states that the horizontal asymptote occurs at

. However, this does not work if

. In this case, you need to simplify the rational function into a polynomial plus a rational function. This means that when is very large, rather than tending towards a constant, the graph will tend towards a non-horizontal line. WORKED EXAMPLE 4.7

Find the oblique asymptotes of

.

You can also do this by polynomial long division, or by comparing coefficients.

The oblique asymptote is

.

You know this because when is very large the term is very small.

EXERCISE 4D

EXERCISE 4D 1

Identify all axis intercepts and vertical and oblique asymptotes of Hence sketch its graph.

for each function.

a i ii b i ii 2

Rewrite

in the form

By solving

.

, find the exact -coordinates of the stationary points of each curve.

a i ii b i ii 3

a Show that the function where

can be written as

and are constants to be found.

b Hence write down the oblique linear asymptote for function

.

c By finding a condition on for the number of solutions to of any stationary points of . d Write down the axis intercepts and vertical asymptotes for e Use your answers to b to d to sketch a graph of 4

A curve is given by

, where

, find the -coordinates

.

. .

a Find the equation of the oblique asymptote of . b By finding

or otherwise, show that has no stationary points.

c Sketch , labelling all asymptotes and axis intercepts.

Checklist of learning and understanding To solve cubic or quartic inequalities, take all the non-zero terms to one side, so the other side is zero, and sketch the graph. Describe in terms of the regions of the graphs above or below the -axis. If

: the intercepts are the asymptotes are

and and

.

When solving inequalities involving algebraic fractions, multiply both sides by the square of the denominator. If

then: when when

is a solution to

vertical asymptotes are solutions to the horizontal asymptote is



.

Mixed practice 4 1

State the equations of the vertical asymptotes of the curve

.

Choose from these options. A B

2

and and

C

and

D

and

Solve the inequality

Choose from these options. A

or

B

or

C D

or or

3

Solve the inequality

4

Curve is given by the equation

.

a Find all axis intercepts and asymptotes for . b Hence sketch . 5

a i Write down the equations of the two asymptotes of the curve ii Sketch the curve

.

, showing the coordinates of any points of intersection with the

coordinate axes. iii On the same axes, again showing the coordinates of any points of intersection with the coordinate axes, sketch the line . b i Solve the equation

ii Find the solution of the inequality

[© AQA 2010] 6

a Show that b

is a factor of

.

is the solution of the inequality

Find the range of possible values of . 7

The curve has equation

, where

.

a State the equations of all asymptotes of . b Sketch , labelling the coordinates of any axis intercepts c Solve the inequality

8

The function

is given by

.

a State all axis intercepts and asymptotes for the graph of

.

b Hence sketch the curve showing the asymptotes and axis intercepts. 9 10

Given that there are exactly two solutions to Curve has equation

, and has asymptotes

is a real number. The -axis intercepts of are a Find values

and

and

where

.

and in terms of .

b Given the -intercept of is 11

, find the possible values of .

, find .

Find a condition on so that function

has the whole of the real numbers as

its range. 12

The diagram shows part of the curve with equation

.

a Write down the coordinates of . b Write down the equation of any asymptotes for this curve. c Determine the coordinates of the maximum point and so find the least value such that for all . 13

A curve has equation

.

a Write down the equations of all the asymptotes of . b The curve has exactly one stationary point. The -coordinate of the stationary point is

.

i Find the -coordinate of the stationary point. ii Sketch the curve . [© AQA 2014]

14

c Solve the inequality

.

A curve has equation

.

a State the equation of the asymptote of . b The line

intersects the curve . Show that

.

c Hence find the coordinates of the stationary points of . (No credit will be given for solutions based on differentiation.) [© AQA 2015] 15

is a rational function of the form

and the function

is given by

. It is known that

has an oblique asymptote

. Find all values of for which

and .

has vertical asymptotes

5 Hyperbolic functions In this chapter you will learn how to: define the hyperbolic functions draw the graphs of hyperbolic functions

and

work with identities involving hyperbolic functions write the inverse hyperbolic functions in terms of logarithms solve equations involving hyperbolic functions.

Before you start… A Level Mathematics Student Book 1, Chapter 3

You should be able to solve quadratic equations.

1 Solve

A Level Mathematics

You should be able to interpret

Student Book 1, Chapter 5

transformations of graphs.

2 What transformation changes ?

A Level Mathematics Student Book 1, Chapter 7

You should be able to work with natural logarithms.

3 Given that

A Level Mathematics Student Book 1, Chapter 9

You should be able to complete binomial expansions of brackets with positive integer powers.

4 Expand

A Level Mathematics Student Book 1, Chapter 10

You should be able to work with the symmetries of trigonometric functions.

5 Given that

A Level Mathematics Student Book 1, Chapter 10

You should be able to work with trigonometric identities.

6 Simplify

.

to

, write in the form

.

, find

.

.

What are hyperbolic functions? Trigonometric functions are sometimes called circular functions. This is because of the definition that states: a point on the unit circle (with equation ) at an angle to the positive -axis, has coordinates .

.

In Chapter 3 you met a related curve, with equation hyperbola have coordinates



, called a hyperbola. Points on this

although can no longer be interpreted as an angle.

Section 1: Defining hyperbolic functions Although the geometric definition of hyperbolic functions gives some helpful insight, a more useful definition is related to the number e.

Fast forward In Further Mathematics Student Book 2, you will see that the trigonometric functions can also be defined in a similar way in terms of complex numbers.

Key point 5.1

You can define

by analogy with the trigonometric definition (of

).

Tip is pronounced as it reads,

is pronounced ‘

’ or ‘

’ and

is pronounced ‘

’.

Key point 5.2

There are not many special values of these functions that you need to know, but from Key points 5.1 and 5.2 you should be able to see that , and . As for trigonometric functions, you need to know the graphs of hyperbolic functions.

Did you know The function is frequently used in physics, particularly in the context of special relativity and the study of entropy.

Key point 5.3 The graph of

Key point 5.4

The graph of

Did you know You may think that the graph of looks like a parabola, but it is slightly flatter. It is called a catenary, which is the shape formed by a hanging chain.

You can see that the minimum value of

is , whereas

has no minimum (or maximum).

Fast forward In Further Mathematics Student Book 2 you will see the range of the sinh function is all real numbers and the range of the cosh function is

.

Key point 5.5 The graph of

The

function has horizontal asymptotes at .

Inverse hyperbolic functions

and

:

The inverse functions of the hyperbolic functions are called

,

and

.

Rewind You saw in A Level Mathematics Student Book 2, Chapter 2, how to form the graphs of inverse functions from the original function by reflection in the line . The graphs of these functions look like this.

You can use the inverse hyperbolic functions to solve simple equations involving hyperbolic functions. For example, if then , which you can evaluate, on a calculator, as .

Tip On your calculator they might be called However, you can use the definition of

,

and

.

to derive a logarithmic form of this result. You can do this for

all three inverse hyperbolic functions.

Key point 5.6

These will be given in your formula resource. These results can all be proved in the same way. The proof for

is given here.

PROOF 5

Prove that

.

Let

Then

Let for . Take

and then look to find an expression

of both sides.

Use the definition of

.

Rearrange into a disguised quadratic in .

So

Use the quadratic formula.

Use the algebra of surds to simplify the expression.

But is a function so it can only take one value.

But So

Conventionally, you take the positive root, so this makes and .

WORKED EXAMPLE 5.1

Solve

, giving your answer in the form

for integers and .

Apply the inverse

function to find .

Use the logarithmic form

.

When you are trying to solve equations involving hyperbolic cosines, using the inverse function – unfortunately – does not give all the solutions: it just gives the positive one. Just as when taking the square root of both sides, you need to use a ‘plus or minus’ sign to get both possible solutions. This will be clear if you consider the graph of .

WORKED EXAMPLE 5.2

Given that

, express in the form

.

The expression is a disguised quadratic, so rearrange it to make one side zero and then factorise it. You could also have used the quadratic formula.

or But

or

so

has no

solutions.

Use the inverse cosh function to find . Remember that you need a plus or minus .

Use the logarithmic form

So

Use the fact that

.

.

or Simplify the second solution by rationalising the denominator to produce the required form.

So

or

You can apply the method used at the end of Worked example 5.2 to write general, so the two solutions to

for

can always be written as

EXERCISE 5A 1

Use your calculator to evaluate each expression where possible. a i ii b i ii c i ii d i ii e i ii

2

Solve each equation, giving your answers to significant figures. a i ii b i ii c i ii d i ii e i ii

3

Without using your calculator, find the exact value of each expression. a i ii b i ii c i ii d i ii e i

as

in .

ii 4

Solve the equation

5

Find and simplify the exact value of

6

Find and simplify a rational expression for

7

Solve the equation

8

Solve the equation

9

Find and simplify an expression for

. .

. . .

10

Prove that

11

Prove that

12

In the derivation of you found that two possible expressions were and . Show that their sum is zero and hence explain why the chosen expression is non-negative.



.

. .

Section 2: Hyperbolic identities Just as there is the identity linking the trigonometric functions also an identity linking the hyperbolic functions and .

and

, there is

Rewind You met the identity

in A Level Mathematics Student Book 1, Chapter 10.

Key point 5.7

This will be given in your formula resource.

Fast forward In Section 3 you will use this identity to solve equations. You can prove this identity by using the definitions of

and

given in Key point 5.1.

WORKED EXAMPLE 5.3

prove that Start from the definition of one of the hyperbolic functions. It doesn’t matter which one. It is squared in the expression so square it and simplify. Since

.

Repeat with the

term.

again. Combine the two terms and simplify.

Although the only two hyperbolic identities you are expected to know for the AS Further Mathematics course are

(Key point 5.2) and

(Key point 5.7), you may be asked to

prove other unfamiliar hyperbolic identities. To do this, always return to the definitions of the functions and follow a process similar to that in Worked examples 5.3 and 5.4.

Rewind The result in Worked example 5.3 proves that described in the introduction to this chapter.

lies on the hyperbola

, as

WORKED EXAMPLE 5.4

Prove that On the LHS use the definition of with .

and replace each

Then work from the RHS. Substitute the definitions of and . Multiply out the brackets, using the difference of two squares.

Using the rules of indices.

EXERCISE 5B 1

Prove that

2

Simplify

3

Prove that

4

Prove that

5

Prove that

6

Prove that

7

Prove that

8

Use the binomial theorem to show that

9

a Explain why

. . . . . . Hence prove that . . and

Hence show that b Write 10



Given that

.

.

in terms of

.

show that

.

.

Section 3: Solving harder hyperbolic equations When you are solving equations involving hyperbolic functions you have several options: Rearrange to get a hyperbolic function that is equal to a constant and use inverse hyperbolic functions. Use the definition of hyperbolic functions to get an exponential function that is equal to a constant and use logarithms. Use an identity for hyperbolic functions to simplify the situation to one of the two preceding options. It is only with experience that you will develop an instinct about which method will be most efficient.

Tip When you are dealing with the sum or difference of two hyperbolic functions, it is often useful to use the exponential form.

WORKED EXAMPLE 5.5

Solve

.

Use the definitions of

and

WORKED EXAMPLE 5.6

Solve

, giving your answer in the form

.

You could use the definitions of and but it is easier to use the identity

Then use the logarithmic form of

WORKED EXAMPLE 5.7

Solve

, giving your answer in logarithmic form.

.

,

The equation involves two types of function. You can use the identity to replace the term. Solve the resulting quadratic.

Use the logarithmic form of

or

WORK IT OUT 5.1 Solve Which is the correct solution? Identify the errors made in the incorrect solutions. Solution 1 Dividing by :

Solution 2 Dividing by

Solution 3

:

.

EXERCISE 5C 1

Find the exact solution to

2

Solve

3

Solve

4

Solve

5

Find the exact solution to

6

Solve

7

Solve

, giving your answers in exact form.

8

Solve

, giving your answer in logarithmic form.

9

Solve

, giving your answer in logarithmic form.

10

Solve

, giving your answer in logarithmic form.

.

, giving your answer in the form

.

, giving your answers correct to significant figures. , giving your answer in the form

, where is a rational number.

. , giving your answers in logarithmic form.

11

a Show that

and

.

b Hence find the exact solutions to the simultaneous equations. 12

Find a sufficient condition on , and for

Checklist of learning and understanding Definitions of hyperbolic functions:

Graphs of hyperbolic functions:

to have at least one solution.

Logarithmic form of inverse hyperbolic functions:

The identity



Mixed practice 5 1

Evaluate

.

Choose from these options. A Doesn’t exist. B C D 2

Solve

, giving your answer in terms of .

Choose from these options. A B C D 3

Simplify

4

Solve

5

Solve

6

Find the exact solutions to

7

a Express

. , giving your answer in terms of logarithms. , giving your answer correct to significant figures. . in the form

b Solve the equation rational number.

, where and are integers. , giving your answer in the form

, where is a

[©AQA 2008] 8

Find the exact solutions to

9

Solve the equation

. , giving your answers in exact form.

10

Solve

11

Prove that

12

Find and simplify an expression for

13

Use the binomial theorem to show that

14

a Sketch the graph of b Given that that

, giving your answers in terms of natural logarithms. for all . . .

. , use the definitions of

can be written as

ii Show that the equation Find this solution in the form a Use the definitions

in terms of and

to show

.

c i Show that the equation

15

and

has only one solution for . where is an integer. and

to show that

.

. b Solve the equation

, giving each of your answers in the form

.

[©AQA 2009] 16

a Prove the identity b If

. and

find the value of .

c Hence find the exact real solution to logarithms.

, giving your answer in a form without

17

Solve these simultaneous equations, giving your answers in exact logarithmic form.

18

Using the logarithmic definition, prove that

19

Given that

20

a Using the definition b Given that

.

, show that

.

, prove the identity and

.

, find the value of in terms of a natural logarithm.

c Hence find the real root of the equation , where and are rational numbers.

, giving your answer in the form

[©AQA 2014]

6 Polar coordinates In this chapter you will learn how to: use polar coordinates to represent curves establish various properties of those curves convert between polar and Cartesian equations of a curve.

Before you start... A Level Mathematics Student Book 2, Chapter 7

You should be able to use radians.

A Level Mathematics Student Book 1, Chapter 10

You should be familiar with graphs of trigonometric functions.

1 a Express

radians in degrees.

b State the exact value of

.

2 a Find the set of values of , between and , for which . b State the greatest possible value of .

What are polar coordinates? You are familiar with describing positions of points in the plane by using Cartesian coordinates, which represent the distance of a point from the and -axes. But you are also familiar with bearings, which determine a direction in terms of an angle from a fixed line. If you know that a point lies on a certain bearing, you can describe its exact position by also specifying the distance from the origin. For example, this diagram shows that is from on a bearing of .

Polar coordinates use a similar idea: positions of points are described in terms of a direction and distance from the origin. They can be used to describe curves that cannot easily be represented in Cartesian coordinates. In this chapter, you will learn about equations of curves such as these.

Because the distance from the origin explicitly features as a variable, polar coordinates are often used to describe quantities that vary with distance, such as the strength of the gravitational field.

Section 1: Curves in polar coordinates Polar coordinates describe the position of a point by specifying its distance from the origin (also called the pole) and the angle relative to a fixed line (called the initial line). You take the initial line to be the positive -axis, and measure the angle anti-clockwise. You write polar coordinates as , where is the distance and the angle. Angle is usually taken as lying between and . is interpreted as being the distance from the pole in the direction of the angle, so that gives the pole (irrespective of the value of ), while negative would produce a point ‘behind’ the pole. For example, point in the diagram has polar coordinates and point has polar coordinates

, point has polar coordinates

. Note that polar coordinates

,

and

would also describe points , and respectively.

WORKED EXAMPLE 6.1

Points

and have polar coordinates

and

. Find:

a the length b the area of the triangle a

, where is the pole. Polar coordinates give information about lengths and angles.

In triangle

, .

You can use the cosine rule in triangle

.

So:

b

Use area of triangle

.

In Cartesian coordinates, an equation of a curve gives a relationship between the and coordinates. Similarly, a polar equation of a curve is a relationship between and that holds for any point on the curve. WORKED EXAMPLE 6.2

WORKED EXAMPLE 6.2 a Make a table of values for the curve with polar equation

for

.

b Hence sketch the curve.

a

Use .

to calculate for various values of

b

Plot the points and join them up.

Tip If you have a graphical calculator you may be able to sketch a curve with a given polar equation.

WORKED EXAMPLE 6.3

Sketch the curve with equation

for

Sketch the graph of

is negative when: or

.

Identify the values of for which is negative.

You can also see that increases for then decreases for

,

, then repeats the

same values between and

.

You can see a similar pattern but with negative values of for

and

.

Notice that the part of the curve generated by appears in the lower right quadrant, while the part of the curve generated by appears in the upper left quadrant, because is negative for these angles.

Sometimes a polar curve will be defined with restricted to regions where .

. In these cases, the curve will not exist in

Just as in Worked example 6.3, you start by identifying values of for which simply exclude these regions as the curve isn’t defined there.

. This time, though, you

So, for example, if you had been asked to sketch for , you would get the same curve as in Worked example 6.3 but without the red and green parts, because in the regions and

.

EXERCISE 6A

1

Plot the points with the given polar coordinates.

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

For points and with given polar coordinates, find the distance triangle .

and the area of the

a i ii b i ii c i ii 3

For each equation, make a table of values (for

) and sketch the curve.

a i ii b i ii c i ii

Tip If part of a curve occurs once for certain values of when is positive and is then repeated for different values of when is negative, you do not need to indicate the repetition by drawing over the curve again to make it bolder. 4

Shade the region described by each inequality. They are given in polar coordinates. a b c

5

A curve has polar equation

for

.

a State the values of for which the curve is not defined. b Hence sketch the curve. 6

A curve has polar equation a Show that the points and , with polar coordinates b Sketch the curve. c Find the exact length of



.

and

, lie on the curve.

Section 2: Some features of polar curves When sketching curves in Cartesian coordinates you normally mark the axis intercepts, maximum and minimum points. For polar curves, there are similar features that you can deduce from the equation.

Minimum and maximum values of Since is a function of , you can use differentiation to find its minimum and maximum values.

Key point 6.1 The minimum and maximum values of occur when

.

WORKED EXAMPLE 6.4

A curve has polar equation

for

.

a Find the minimum and maximum values of , and the values of for which they occur. b Sketch the curve. The maximum and minimum values of occur

a

when When

.

:

Use the second derivative to decide which one gives the minimum and which the maximum value of . When

:

When

.

When

.

so there is a minimum at When

. : so there is a

maximum at When

. .

The minimum value could occur at the end of the domain. You have already checked , so you just need to check .

Hence the minimum value of occurs when . The maximum value of when .

occurs

As increases from to , increases from to (when , which is just below ) and then decreases to .

b

Polar equations often involve trigonometric functions. You may be able to use trigonometric graphs, rather than differentiation, to find the maximum and minimum values of . WORKED EXAMPLE 6.5

A curve has polar equation

for

.

a Find the largest and smallest values of . b Hence sketch the curve. a

Start by considering the minimum and maximum values of when , and when or

so The largest value is The smallest value is .

when

.

.

when

b

Tangents at the pole The expression

changes from positive to negative, or vice versa, when

. Each of

those values corresponds to a half-line, shown in red in the diagram. As the curve approaches each of the lines, gets closer to zero (so points on the curve gets closer and closer to the pole). This means that each of the lines

is a tangent to the curve at the pole.

Key point 6.2 For a curve with polar equation

, the line

is a tangent at the pole if

but

on one side of the line.

WORKED EXAMPLE 6.6

For the curve with polar equation

, find the tangents at the pole and hence sketch the

curve.

Sketch the graph of against to see which values of produce negative values of .

The value of

The tangents at the pole are: and

.

passes through zero when and

.

Between the tangents the value of increases from to the maximum value of and then decreases back to . Notice that the curve is actually drawn twice as varies between and .

EXERCISE 6B

EXERCISE 6B 1

For each curve, find the minimum and maximum possible value of , and the corresponding values of . Hence sketch the curve. (In all cases, .) a i ii b i ii

2

Find the equations of the tangents at the pole for each curve. Hence sketch the curve. (In all cases, .) a i ii b i ii

3

Sketch the curves given in question for

4

A curve has equation

. in polar coordinates.

a Find the maximum and minimum values of , and the corresponding values of . b Sketch the curve. 5

Consider the curve with polar equation

for

.

a Find the equations of the tangents at the pole. b State the set of values of for which the curve is not defined. c Hence sketch the curve. 6

a Sketch the curve with polar equation

.

b State the largest and smallest values of . 7

a Sketch the curve of b State the points where

8

, labelling tangents at the pole. has its maximum value.

A curve has polar equation

.

a Find the largest and smallest value of and the values of at which they occur. b Hence sketch the graph. 9

a Sketch the graph of b

10

for

.

Sketch the curve with polar equation

a Find the smallest and largest values of

for

b Sketch the curve with the polar equation 11

Sketch the curve with equation

12

Sketch the curve with equation

, for for

for

.

.

.

.



Section 3: Changing between polar and Cartesian coordinates You can use trigonometry to find the Cartesian coordinates of a point with given polar coordinates.

Key point 6.3 A point with polar coordinates

has Cartesian coordinates

.

WORKED EXAMPLE 6.7

Points and have polar coordinates

and

.

a Show points and on the same diagram. b Find the Cartesian coordinates of and . a

The first coordinate is the distance from the origin and the second coordinate is the angle. is

b For :

Use

and

is

.

and

So the Cartesian coordinates of are . For :

So the Cartesian coordinates of are .

To change from Cartesian to polar coordinates, consider the same diagram again.

.

The value of is the distance from the origin, so . It is conventional to give as a positive value. You need to be a little careful when finding the angle. Since and , you can divide the two equations to get

. However, there are two values

with the same value of

you need to consider the position of the point to decide which one is correct.

Rewind This should remind you of the modulus and argument of a complex number – see Chapter 1, Section 3.

Key point 6.4 For a point with Cartesian coordinates

the polar coordinates satisfy:

WORKED EXAMPLE 6.8

Find the polar coordinates of the points

and

.

Start by plotting the points to see which angle to use.

For :

Use

.

Find

, then add to get the two

possible values between and

.

;

Hence the coordinates of are

.

The angle for is smaller than .

.

The angle for is greater than .

For :

Hence the coordinates of are

You can now convert equations of curves between polar and Cartesian forms. WORKED EXAMPLE 6.9

Find the Cartesian equation of the curve with polar equation

.

Use

and

Substitute for curve.

Hence:

.

in the equation of the

Simplify if possible. In this case, multiply both sides by the ‘square root’ term.

WORKED EXAMPLE 6.10

Find the polar equation of the curve

. Use

and

You also know that

EXERCISE 6C 1

Each point is given in polar coordinates. Find the Cartesian coordinates. a i ii b i

. .

ii c i ii 2

Each point is given in Cartesian coordinates. Find the polar coordinates. Take

.

a i ii b i ii c i ii 3

Find the polar equation of each curve. a i ii b i ii c i ii

4

Find the Cartesian equation of each curve. a i ii b i ii c i ii

5

A curve has polar equation

.

Point , with polar coordinates

where

, lies on the curve.

a Find the value of . b Find the Cartesian coordinates of . c Find the Cartesian equation of the curve. 6

Find the polar equation of the circle .

7

Find the Cartesian equation of the curve with polar equation

8

A curve has polar equation

, giving your answer in the form

.

Show that the Cartesian equation of the curve can be written as 9

Find the Cartesian equation of the curve with polar equation: a

.

.

b

.

Checklist of learning and understanding Polar coordinates describe the position of a point in terms of its distance from the pole and the angle measured anticlockwise from the initial line. The connection between polar and Cartesian coordinates is: , For a curve with equation given in polar coordinates: when has negative values, the curve will appear to be plotted

from the angle

there may be one or more tangents at the pole, given by the values of for which



.

Mixed practice 6 1

Find the greatest distance from the pole of any point on the curve

, 

.

Choose from these options. A B C D 2 3

Find the polar equation of the curve

.

Points and have polar coordinates

and

. Find:

a the distance b the area of the triangle

.

4

Sketch the curve with polar equation

5

A curve has polar equation

,

.

. Find its Cartesian equation in the form

.

[©AQA 2007] 6

A curve has polar equation

. Find its Cartesian equation in the form

.

[©AQA 2014] 7

Find a Cartesian equation for the curve

,

.

Choose from these options. A B C D 8

A curve has polar equation

,

a Find the equations of the tangents at the pole. b Sketch the curve. c Find the Cartesian equation of the curve. 9

a Sketch the curve with polar equation

,

.

b Find the Cartesian coordinates of the point that is furthest away from the origin. 10

A curve has polar equation

for

.

a Find the equations of the tangents at the pole. b State the polar coordinates of the points at greatest distance from the pole. c Hence sketch the graph. 11

A curve is defined by the polar equation

for

.

a Sketch the curve. b Find the Cartesian coordinates of the point where the curve intersects the line

.

12

Sketch the curve with polar equation , . Indicate the equations of the tangents at the pole, and give the polar coordinates of the point where the curve crosses the initial line.

13

a Show that

can be written in the form

b A curve has Cartesian equation

.

.

Find its polar equation in the form

, given that

. [©AQA 2008]

14

The diagram shows the curve with equation

and a circle of radius . The eight

intersection points are connected to form two overlapping rectangles. Find the exact area of the shaded region.

15

The diagram shows a sketch of a curve , the pole and the initial line.

The curve has polar equation

,

a Verify that the point with polar coordinates b The circle with polar equation

lies on .

intersects at the points

i Find the polar coordinates of ii Find the area of triangle

.

and .

and . .

c Find a Cartesian equation of , giving your answer in the form

. [©AQA 2009]



FOCUS ON … PROOF 1

Roots of real polynomials In this section, you will prove that complex roots of real polynomials come in conjugate pairs. You will need to use the properties of complex conjugates.

Key point 1

for integer

All these results are proved in a similar way; here is just one of them. PROOF 6

Prove that

. The best way to describe complex conjugates is to write them in terms of their real and imaginary parts.

Start with the left-hand side.

If you get stuck, start on the right-hand side and try to meet in the middle.

Therefore

Every proof needs a conclusion.

You are now ready to prove the main result.

Key point 2 If

is a real polynomial and if

, then

.

Rewind Remember that a real polynomial is a polynomial in which the coefficients are real numbers. Key point 2 says that, if a number is a root of PROOF 7

, then so is its conjugate, .

Write a general form of the polynomial.

Let

It is helpful if you label the coefficients with subscripts that correspond to the powers. Then:

Take the complex conjugate of each term and use the results of Key point 1: . Since

is real,

.

Use the result from Proof 6. For the last term, remember that is real. So:

Use Key point 1 again:

.

Remember that you started by assuming that .

QUESTIONS 1

Prove the rest of the results in Key point 1.

2

Does the proof of Key point 2 work if is a real number? What does the result say then?

3

Where in Proof 7 did you use the fact that is a real polynomial? Does the result of Key point 2 still hold if some of the are complex numbers?

4

You already know that you can use the roots of a polynomial to write it as a product of linear factors: if the roots of are , then . However, some of the may be complex. Key point 2 implies that, once you have found all the roots of a real polynomial, you can write it as a product of real linear and quadratic factors. Can you see why this is the case? You may want to start by looking at some examples.



FOCUS ON … PROBLEM SOLVING 1

Solving cubic equations In this section, you will look at a famous historical example of problem solving. It shows you how mathematicians adapt their definitions and rules to enable them to solve a wider variety of problems.

The cubic formula You know that the quadratic equation

can be solved by applying the quadratic formula

. Some quadratic equations have no real solutions, which is the case when the discriminant

is negative.

A similar formula for the cubic equation was discovered by Cardano and Tartaglia in the 16th century. The formula is rather complicated, and here you will consider only the special case of cubic equations with no term. The equation

The quantity

has solutions given by

, where

.

plays a role similar to the discriminant of a quadratic equation: it tells you how many

solutions the equation has. Unfortunately, as you will see, there are examples where this discriminant is negative but the equation still has real solutions. This is where mathematicians needed to engage in some serious problem solving to make the formula work correctly.

Using the formula Consider the equation Here

.

, so or

When

When

So, in this case, both possible values of give the same solution for . Plotting the graph confirms that

is the only real solution of the equation.

The negative discriminant Now consider the equation

. Here

.

Then

This appears to be a dead end you are looking for real solutions, so you can’t take the square root of

.

However, you can see (by factorising) that there should be three real solutions: , any way you can make the formula work to find those three solutions?

and

. Is there

Fast forward In Further Mathematics Student Book 2 you will learn how to find cube roots of a complex number. Here, you are just checking that the three roots given are correct. Problem solving often requires perseverance and ‘thinking outside the box’. So what happens if you start working with complex numbers and carry on with the calculation? You now need

. There are in fact three different complex numbers for which the cube is i: and

When

.

:

When

:

When

, a similar calculation gives the third solution,

.

Fast forward In Further Mathematics Student Book 2 you will also use complex numbers to solve differential equations and prove trigonometric identities.

QUESTIONS 1

Apply the formula to solve the equation

2

Find the three possible values for of the equation

3

. . Confirm that these lead to the same three solutions

.

The fact that this formula uses complex numbers to find real solutions was one of the arguments that persuaded mathematicians to accept complex numbers. Did it change your opinion on whether complex numbers ‘exist’?



FOCUS ON … MODELLING 1

Complex numbers and radios You may have been told that complex numbers are used in electronics. The building blocks of electronic circuits are resistors, capacitors and inductors. Normally each of these needs to be analysed in a different way using quite tricky differential equations. For a resistor: Ohm’s law.

, where is the voltage, is the current and is the resistance. This is called

For a capacitor,

where is the capacitance.

For an inductor,

where is the inductance.

However, if you use complex numbers and assume a sinusoidal input voltage with (angular) frequency , you can treat all three in the same way, as ‘complex resistors’, using the concept of impedance, . The impedance of a resistor is , the impedance of a capacitor is

and the impedance of an inductor is

Then you can use a general version of Ohm’s law, which says that

.

This diagram shows a model of a simple circuit which can act as a radio.

The total impedance for this circuit is:

If the input voltage is

, then the resulting current is:

If you plot the modulus of the current against , your graph should look like this:

.

There is a maximum that occurs when the denominator of the expression above is minimised. This happens when Now think about what you have found. There is a natural frequency that has a higher current than any other frequency. This phenomenon is called resonance. If the input voltage has this frequency, the circuit will produce a large current that can be used to power a speaker. QUESTIONS 1

Radio has a frequency of

or

in the units required for . If an inductor has

(a fairly typical value), find the required capacitance for a circuit to ‘tune in’ to Radio . 2

a What is the current in the circuit when it is being driven at the resonant frequency? b What would happen if c Why would

3

?

be a poor modelling assumption?

Use technology to sketch the current against . Explore the effect on the graph of changing while keeping the same resonant frequency. Why might smaller values of produce a better radio?

4

a The argument of the expression for the current gives the ‘phase difference’ between the input voltage and the current – a measure of the delay between the input and the circuit responding. Find an expression for this phase difference. b What is the value of the phase difference when the circuit is at resonance?



CROSS-TOPIC REVIEW EXERCISE 1 1

Solve the equation

2

The complex number is defined by

. where is real.

a Find, in terms of , the real and imaginary parts of: i ii

.

b Show that there is exactly one value of for which

is real. [©AQA 2009]

3

The quadratic equation Express

4

has solutions and .

in terms of .

The cubic equation

, where

and are real, has roots

and .

a Write down the other root of the equation. b Find the values of 5

and .

Use the substitution

to find the exact value of the real root of the equation .

6

The parabola with equation to the -axis to form the curve a Find the equation of

is transformed by a stretch with scale factor parallel .

.

The parabola is now transformed by a stretch with scale factor form the curve . b Given that 7

and

parallel to the -axis to

are the same curve, find the value of .

A curve has equation

.

a Write down the equations of the two asymptotes to the curve. b Sketch the curve, indicating the coordinates of the points where the curve intersects the coordinate axes. c Hence, or otherwise, solve the inequality

. [©AQA 2007]

8

Solve the equation

9

Solve the equation logarithms.

10

, giving your answer in the form

.

, giving your answers in terms of natural

The Cartesian equation of a circle is

.

Using the origin as the pole and the positive -axis as the initial line, find the polar equation of this circle, giving your answer in the form . 11

a Given that

, where

, find

in terms of and .

b Sketch on an Argand diagram the locus of points satisfying 12

a By writing

, or otherwise, solve the equation

. .

13

b Solve the quadratic equation

.

A polynomial is defined by

.

a Find an integer root of the equation b Solve the equation

.

.

The integer root of

is and the complex roots are and .

c On an Argand diagram, shade the locus of points which satisfy d Calculate 14

.

.

a Show that

.

b The cubic equation roots and .

, where is a real constant, has

It is given that

.

i Find the value of . ii Show that

.

iii Find the values of and . [©AQA 2011] 15

The equation

has roots and .

a Write down the value of

and the value of

b Hence show that

.

.

c Find a quadratic equation, with integer coefficients, whose roots are

and

.

[©AQA 2013] 16

An ellipse has equation

a Sketch the ellipse , showing the coordinates of the points of intersection of the ellipse with the coordinate axes. b The ellipse is stretched with scale factor parallel to the -axis. Find and simplify the equation of the curve after the stretch. c The original ellipse, , is translated by the vector is

. The equation of the translated ellipse

.

Find the values of and . [© AQA 2009] 17

Show that

18

a Show that

. .

b Given that 19

Let a Find

, find the exact value of and

in the form

. .

.

b Find c Show 20

in the form

, where and are expressed in terms of surds.

on an Argand diagram and hence find the exact value of

.

a Two points, and , on an Argand diagram are represented by the complex numbers and respectively. Given that the points and are at the ends of a diameter of a circle

, express the equation of

b A second circle,

in the form

.

, is represented on the Argand diagram by the equation

Sketch on one Argand diagram both

and

.

c The points representing the complex numbers and lie on are such that in the form

.

and

respectively and

has its maximum value. Find this maximum value, giving your answer . [©AQA 2009]

21

The diagram shows a parabola which has equation the image of under a reflection in the line

, and another parabola which is

.

The parabolas and intersect at the origin and again at the point . The line is a tangent to both and . a i Find the coordinates of the point . ii Write down an equation for . iii Give a reason why the gradient of must be b i Given that the line .

.

intersects the parabola at two distinct points, show that

ii Find the coordinates of the points at which the line touches the parabolas and . (No credit will be given for solutions based on differentiation.) [© AQA 2011] 22

A curve has equation

.

a Find the equations of the three asymptotes of the curve. b i Show that if the line ii Given that the equation .

intersects the curve then

. has real roots, show that

iii Hence show that the curve has only one stationary point and find its coordinates. (No credit will be given for solutions based on differentiation.) c Sketch the curve and its asymptotes. [©AQA 2013] 23

Find the set of values of for which

has at least one solution. 24

The diagram shows a sketch of a curve and a circle.

The polar equation of the curve is The circle, whose polar equation is in the diagram.

, intersects the curve at the points and , as shown

a Find the polar coordinates of and the polar coordinates of . b A straight line, drawn from the point through the pole , intersects the curve again at the point . i Find the polar coordinates of . ii Find, in surd form, the length of

.

iii Hence, or otherwise, explain why the line

is a tangent to the circle

. [©AQA 2013]



7 Matrices In this chapter you will learn how to: add, subtract and perform scalar multiplication with conformable matrices use and interpret zero and identity matrices calculate the determinant of a matrix find and interpret the inverse of a matrix use matrices to solve two simultaneous equations.

Before you start… A Level Mathematics Student Book 1, Chapter 15

You should know how to add, subtract and perform scalar multiplication of vectors.

1 Calculate:

GCSE

You should be able to solve linear

2 Solve these simultaneous equations.

simultaneous equations.

What are matrices? You are already familiar with column vectors, such as

, and with the rules on how to add, subtract and

multiply vectors. A matrix is an extension of the idea of a column vector into a rectangular array of numbers.

Rewind In A Level Mathematics Student Book 2, Chapter 16, vectors are extended into dimensions. Matrices provide an efficient way of dealing with large amounts of information. They are used widely in many branches of advanced mathematics, from calculus to probability, and form the basis of many computer programs including, for example, those used in weather forecasting and quantum mechanics. In this chapter you will look at the basic techniques for working with matrices, and apply some of these to solve two linear simultaneous equations.

Section 1: Addition, subtraction and scalar multiplication A matrix (plural: matrices) is a rectangular array of elements, which may be numerical or algebraic. For example:

is a matrix with two rows and three columns, for a total of six elements.

Key point 7.1 An matrix is a rectangular array of elements with rows and columns. A matrix that has the same number of rows as columns is called a square matrix. Matrices are generally designated by a bold or underlined upper-case letter ( in print and either or when handwritten). There are two special matrices with which you will need to be familiar.

Key point 7.2 A zero matrix, denoted by Z, has every element equal to zero. An identity matrix, denoted by I, is a square matrix with on each element of the lead diagonal (upper left to lower right) and zeros everywhere else.

The

identity matrix is

, the

identity matrix is

.

Fast forward You will see the use of the identity matrix later in this chapter.

WORKED EXAMPLE 7.1

,

and

.

Write down the dimensions of the matrices , and . is a

matrix.

There are rows and columns.

is a

matrix.

There are rows and column.

is a

matrix.

There is row and columns.

Notice that a column matrix such as in Worked example 7.1 is a column vector. In some circumstances you need to flip a matrix around its lead diagonal, exchanging rows for columns. This is called transposing a matrix.

Key point 7.3

The transpose of a matrix is a new matrix,

, in which the rows of

are the columns of .

WORKED EXAMPLE 7.2

,

and

Write down the transpose matrices

.

,

and

. is a matrix so is a matrix. The first row of becomes the first column of and so on.

is a matrix so is also a matrix. The first row of becomes the first column of and so on.

is a matrix.

row matrix so

is a

column

Addition and subtraction The rules for matrix addition and subtraction are equivalent to the rules for vector addition and subtraction, with which you are already familiar.

Key point 7.4 You can only add (or subtract) two matrices with the same dimensions. Take each position in the matrix in turn and add (or subtract) the components for that position. Matrices that have the appropriate dimensions for an operation are conformable. For addition and subtraction, two matrices are only conformable if they have identical dimensions. WORKED EXAMPLE 7.3

The

matrix

For each part, determine: i a b

c

.

and ii

or explain why the sum or difference does not exist.

d The dimensions match so you can add the matrices.

a i

For each position in the matrix take the element in and add the element in .

The dimensions match so you can add the matrices.

ii

For each position in the matrix take the element in and add the element in .

b Cannot add or subtract a and a

matrix

The dimensions do not match so you cannot add or subtract.

matrix

The dimensions do not match so you cannot add or subtract.

matrix.

c Cannot add or subtract a and a matrix.

The dimensions match so you can add the matrices.

d i

For each position in the matrix take the element in and add the element in .

The dimensions match so you can add the matrices.

ii

For each position in the matrix take the element in and add the element in .

Also, just as with vectors, you can multiply a matrix by a scalar value.

Key point 7.5 To multiply a matrix by a scalar, multiply each element by that scalar.

Tip Remember that in the examination you may use a calculator to perform any operations on purely numerical matrices, without showing further working.

WORKED EXAMPLE 7.4

,

and

.

a Write down b

.

and are scalar constants; write down

c Given

and

, find the values of

.

and .

a

Multiply each element of the matrix by the scalar quantity.

b

Multiply each element of the matrix by the scalar quantity.

c

For each position in the matrix take the element in and add the element in .

Comparing each element gives a set of simultaneous equations.

and only involve unknowns and . Solve these, then substitute into and to solve for and . Substituting

into

and

EXERCISE 7A 1

a State the dimensions of each matrix. i

ii

iii

iv

b For each matrix in , write down its transpose matrix.

2



.

Calculate the resultant matrix, if there is a valid calculation. a i ii iii b i ii iii c i ii iii d i ii iii 3

Add the two given matrices where possible. a

and

b

and

c

and

d e

f

4

Given that

and and

and

,

and

, calculate:

a b c d 5

Find the values of and that satisfy each matrix equation, or state that there is no solution. a b c

d 6

Given that

and

find the matrix .

a b c d 7

and

.

Given that 8

and Given

9

.

.

, find the values of

and .

Explain why, for any two matrices and that have the same dimensions, .



and .

, find all possible values of and . and

Given 10

, find the values of

Section 2: Matrix multiplication Matrices have a variety of uses and most involve multiplication. Matrix

and matrix

To find the product

, let matrix

so that:

Multiply each pair of values together and calculate the total of these products to find So, to work out

To work out

you look at the first row of and the first column of :

you look at the first row of and the second column of :

Continuing in this fashion, you calculate each element of to find For multiplication, the left-hand matrix must have the same number of columns as the right-hand matrix has rows. If this is not the case, the matrices are not conformable for multiplication, and the product is not defined.

Key point 7.6 When an matrix

matrix and an .

matrix are multiplied together, the result is an

To find the element in row and column of : 1 take row of and column of 2 multiply each pair of corresponding values and add all the products together.

WORKED EXAMPLE 7.5

The

matrix

For each of the matrices , determine i a b

and ii

or explain why the product does not exist.

c d There is the same number of columns in as rows in so the product exists.

a i

For the top left element of the product, multiply the top row of with the left column of . Continue in the same way for each other element. There is the same number of columns in as rows in so the product exists.

ii

There is the same number of columns in as rows in so the product exists.

b i

ii Not conformable:

has columns but has rows.

c i Not conformable: has columns but

ii

has rows.

The number of columns in is different from the number of rows in so the product does not exist. The number of columns in is different from the number of rows in rows in so the product does not exist. There is the same number of columns in as rows in so the product exists.

There is the same number of columns in as rows in so the product exists.

d

In part of Worked example 7.5, you used the standard index notation itself. Similarly and onwards, for higher integer powers.

for a matrix multiplied by

Focus on ... Focus on… Modelling 2 shows how you can use matrices to represent networks. The main result that is explored involves raising these matrices to powers. Notice that, as shown in part of Worked example 7.5, matrix multiplication is not generally commutative, which means that does not necessarily equal . However, some matrices will commute with each other and, as you will see, there are some matrices that commute with all other matrices of suitable dimension.

Key point 7.7 Two matrices and commute (are commutative) if

.

Although matrix multiplication is not generally commutative, it is associative; that is, a multiple product can be calculated as or as , the result is still the same. WORKED EXAMPLE 7.6

a Calculate b Calculate a

and and

.

and explain your findings. Standard matrix multiplication.

b

The products give the same result because matrix multiplication is associative.

EXERCISE 7B 1

State whether the two matrices can be multiplied, and the dimensions of their product, when they are multiplied in the order given. a

b

c

and

and

and

d

and

e

and

f

g

h

and

and and

i

and

j

k

and

and

2

Multiply the matrices from Question 1 when possible. Use a calculator to check your answers.

3

Determine whether the matrices commute.

a

and

b

and

c

and

d

and

e

and

f 4

and

Find the values of the unknown scalars. a b c d

5

Given that

and

 work out these matrices.

a b c 6

Given that

,

and

, work out these matrices.

a b c 7

Given that

and

, complete these matrix calculations.

a b c 8

Given that

and

a b 9

. ,

a Find, in terms of :

and

, find, in terms of :

i ii

.

b Explain why

does not exist.

10

Find the value of such that the matrices

11

Prove that the matrices

and

commute for all values of

12

Given that the matrices

and

commute, find an expression for in terms of .

13

14



commute. and .

and

a Calculate

and

b Show that

.

Prove that for two

and

.

matrices and ,

.

Explain whether this is the case for matrices of other dimensions for which the product exists.



Section 3: Determinants and inverses of 2 × 2 matrices In this section you will answer the question: ‘Given two matrices, and , find a matrix such that .’

Fast forward In Further Mathematics Student Book 2 you will extend this to In the AS Level course you will focus on

matrices.

matrices.

The identity matrix In arithmetic, an identity is a number that, for a given operation, produces no changes. So, for addition in the real numbers, zero is the identity because for any real value :

For

matrix addition, the zero matrix has the property:

The identity for multiplication of the real numbers is , because, for any real value :

In matrix multiplication, the identity matrix, which was defined in Section 1, has the property:

Tip In the definition of the identity, you need to require that both because, generally, matrices don’t commute. If is a square with .

You can see this in the case of

and . This is matrix, then commutes

matrices by considering the general matrix

And similarly, you can show that This result can be generalised for multiplication of the

identity matrix with any

Focus on ... See Focus on… Proof 2 to prove that the identity matrix is unique.

Key point 7.8 The

identity matrix has the unique property that:

matrix.

for any

matrix ,

for any

matrix ,

.

Determinants When you compare real numbers, you can say that one is greater or less than another. If you want to compare vectors you can use their magnitudes. Although you cannot meaningfully write that or that

, you can compare their magnitudes, as given by the modulus of the vector.

There is a similar idea for square matrices, called the determinant, which, in many respects, can be considered the magnitude of the matrix. However, unlike a vector modulus, the determinant can take positive or negative values.

Key point 7.9 The determinant of a

matrix

is written as

or

or

, and equals

.

Tip In words, the determinant is the product of the lead diagonal elements less the product of the reverse diagonal elements.

WORKED EXAMPLE 7.7

Calculate the determinant of the matrix

.

For a matrix

You can prove some useful properties for

matrix determinants algebraically.

Key point 7.10 For two

matrices and and a scalar :

The proof of the first result is shown here. You can prove the second result in a similar way.

PROOF 8

PROOF 8

Let Then

and

.

You need to introduce some variables so you can do the calculation. First find the individual determinants of and . Use the rule for multiplying matrices to find and . Remember that they are not the same!

Expand and simplify the expression for det

.

It’s not obvious how to simplify this, so look at . Hence:

The two expressions contain the same four terms. Since is not the same as to expand det . This is again the same as

So

EXERCISE 7C

you also need .

Write a conclusion, summarising what you have proved.

EXERCISE 7C 1

Find the determinant of each matrix. a i ii b i ii c i ii d i ii

2

Find the determinants of the matrices Confirm that

3

The matrix

,

and

.

. has determinant .

Find the possible values of . 4

The matrix

has determinant .

Find the possible values of .

Inverse matrices Remember that the aim, at the start of this section, was to find a matrix such that

. Effectively,

you are trying to ‘undo’ matrix multiplication. For real numbers, you can reverse multiplication by division or, equivalently, multiplication by the reciprocal: if

then

.

The matrix equivalent of a reciprocal is called the inverse matrix.

Key point 7.11 The matrix

is the inverse of a square matrix and has the property that

You can use a calculator to find the inverse of a matrix with numerical entries, but you should also know how to use the formula.

Tip You can remember how to find

as:

swap the elements on the leading diagonal. change the signs of the elements on the reverse diagonal. divide by the determinant.

Key point 7.12 Given a matrix

with

, the inverse is

You can prove that this formula gives the inverse by checking that PROOF 9

Check that

.

Firstly multiply the second matrix by the scalar (by multiplying each element by the scalar).

This is the required result:

.

as required. A similar calculation shows that .

In the formula for the inverse of , the expression at the bottom of the fraction is the determinant of . This means that if

, then the matrix has no inverse.

Key point 7.13 A matrix with

is called singular and has no inverse.

WORKED EXAMPLE 7.8

For each matrix, find the inverse matrix or establish that there is no inverse. Where there is an inverse, verify that the product with the original matrix is a b

c d

a

Find det ; if it is non-zero, then exists. Swap the elements on the lead diagonal. Change the sign of the elements on the reverse diagonal. Divide by the determinant of the original matrix. Check the answer by multiplication to find

b

Find exists.

; if it is non-zero, then

Check the answer by multiplication to find

c is singular and has no inverse. d

Find ; if it is zero, then not exist. Find exists.

does

; if it is non-zero, then

Check the answer by multiplication to find

Using the inverse You can use the inverse matrix to ‘undo’ matrix multiplication. Because matrix multiplication is generally not commutative, you need to be careful to multiply the matrices in the correct order.

Tip If is singular (i.e. if

), then there is no solution as the inverse matrix,

, doesn’t

exist. If is a non-singular matrix and if

, then you can multiply both sides by

However, if you want to solve the equation and

, you need to multiply by

on the left.

on the right, so that

are next to each other.

WORKED EXAMPLE 7.9

,

.

a Find such that

.

b Find such that

. You need to multiply through by find and . Calculate

a

and then use it to find

.

You need to cancel from the left-hand side of the product, so you must multiply by on the left on both sides of the equation. Then use and

b

in order to

and

.

You need to cancel from the right-hand side of the product, so you must multiply by on the right on both sides of the equation. Then use and

and

.

You will often need to use the inverse of a product of two matrices. With real numbers, the reciprocal of a product is the product of the reciprocals:

. Because matrix multiplication is not commutative,

you need to be a little more careful. WORKED EXAMPLE 7.10

For non-singular square matrices and : a show that b write down

exists in terms of

and

.

a If and are non-singular, then det and det are both non-zero. Since

, it follows that is also non-zero.



is non-singular and has an inverse.

b By definition of an inverse:

Multiply by inverses on the left of each side to find . Multiply by

on the left.

Now multiply by

on the left.

Key point 7.14 The inverse of a product consists of the inverses of the individual matrices, but listed in the opposite order:

EXERCISE 7D 1

Find the inverses of the matrices from Exercise 7C, Question 1. State any values of for which the inverse does not exist.

2

Find all possible values of for which these matrices are singular. a i ii b i

ii c i ii d i ii 3

Given that

 and  

, find the matrix such that:

a b c d e f 4

If

5

Given that

  find the matrix such that  and 

.

:

a show that is non-singular b find

.

6

If

, find the matrix such that

7

Find the values of for which the matrix

8

Let

. is singular.

.

a Show that is non-singular for all values of . b Find 9

in terms of .

Given that

10

, find the matrix such that

and are non-singular matrices with inverses a Show that b Simplify

is the inverse of

and

.

.

.

11

Given that and are non-singular matrices, simplify

12

Given that and are non-singular matrices, simplify: a

.

.

b

.

13

Prove that, if the non-singular matrices and commute, then so do

14

For non-singular square matrix , find in terms of det : a



b

where is a

matrix

c

where is an integer.

and

.

Section 4: Linear simultaneous equations You already know how to solve simultaneous equations by elimination and substitution. In this section, you will learn how to represent two simultaneous equations as a matrix equation, which you can then solve using the methods from Section 3. In Further Mathematics Student Book 2 you will see this technique extended to systems of three equations, where it can prove more useful than solving by elimination or substitution.

Fast forward In Further Mathematics Student Book 2 you will see the same idea applied to three simultaneous equations with a

matrix.

When you multiply a column vector by a square matrix you get another column vector. For example:

The two components of this new vector look like expressions that appear in simultaneous equations. So, for example, the system of equations:

can be written as

This is now an equation of the form

, where

so you can solve it by finding

.

WORKED EXAMPLE 7.11

Use a matrix method to these simultaneous equations.

Rewrite the problem as a matrix multiplication of the form

Find the inverse of the square matrix.

Sometimes simultaneous equations don’t have a unique solution. For example, the system doesn’t have any solutions, while the system

has infinitely many.

Tip Remember that you are allowed to perform all numerical matrix multiplication on your calculator. The system will have no unique solution whenever the determinant is zero (as it isn't possible to find the inverse matrix).

Fast forward You will see in Further Mathematics Student Book 2 that it is particularly useful with systems of three equations to use the determinant to establish whether or not the system has a unique solution.

Key point 7.15 Two simultaneous equations can be written as a matrix problem:

where contains the coefficients of the system. If det

, there is no unique solution to the system.

WORKED EXAMPLE 7.12

Find the values of for which the simultaneous equations:

do not have a unique solution. Write the simultaneous equations as a matrix equation. Since there isn’t a unique solution, the determinant is zero. Find an expression for the determinant and solve the equation for .

EXERCISE 7E

EXERCISE 7E 1

By rewriting each set of simultaneous equations as a matrix problem, solve for and . a i ii b i ii c i ii

2

Find the values of for which these simultaneous equations do not have a unique solution.

3

a Find the value of for which these simultaneous equations do not have a unique solution.

b Assuming that the equations have a unique solution, find the solution in terms of . 4

This system of equations has a unique solution.

a Find the set of possible values of . b Find the solution in terms of .

Checklist of learning and understanding An matrix is a rectangular array of numerical or algebraic elements with rows and columns. Addition (and subtraction) of conformable matrices and involves adding (or subtracting) elements of matching position. Scalar multiplication of a matrix is equivalent to multiplying each element by the scalar. When an matrix

matrix and an .

matrix are multiplied together, the result is an

To find the element in row and column of : take row of and column of multiply each pair of corresponding values and add all the products together. In general, matrices are not commutative, i.e. The zero matrix has all elements equal to . The identity matrix is a square ( and elsewhere. For suitably sized ,

,

The inverse matrix of satisfies

) matrix with elements equal to on the lead diagonal for any matrix . .

A matrix is singular if its determinant is zero. Otherwise it is non-singular. Only non-singular matrices have inverse matrices. For two non-singular matrices and ,

.

Inverse matrices can be used to solve equations. If

then

.

If

then

.

The determinant of a

matrix

The inverse of a non-singular

is matrix

is

Two simultaneous equations can be written as a matrix problem: contains the coefficients of the system. If det



, there is no unique solution to the system.

, where

Mixed practice 7 1

The matrix is given by Find

.

. Choose from these options.

A B C D 2

and Find, in terms of , the matrix

. Choose from these options.

A B C D 3

Matrices , and are given by: , If

4

and

, find

.

and .

Matrices and are given by

and

where is a constant.

a Given that and are commutative, find . b Show, by choosing matrices and , that matrix multiplication is not always commutative. 5

and

6

Matrix is given by

. Given that

, find .

.

a Given that is singular, calculate . b Given that is non-singular, express 7

in terms of .

The matrices and are given by , a Calculate the matrix b Show that c Show that

.

is of the form , where is an integer and is the .

identity matrix.

[©AQA 2008] 8 a i Find the value of for which ii Given that b 9

When

is singular.

is non-singular, find

, use

a Given that

, the inverse of

, in terms of .

to solve the simultaneous equations , prove that

.

and b Find . 10

It is given that

and that is the

a Show that

identity matrix.

for some integer .

b Given further that

, find the integer such that [©AQA 2010]

11

Let a Determine

.

b Show that

for all real values of .

c Find the value of for which the matrix

is singular. [©AQA 2012]

12

a Show that for two

matrices and , where neither nor is the zero matrix , if

then both and must be singular. b Find two matrices and , where no element of either matrix is zero, for which 13 14

Matrix and are two a Show that if

. Given that is non-singular and

, find and .

matrices. , then

b Give an example of two matrices and for which

.

. but

.

8 Matrix transformations In this chapter you will learn how to: interpret matrices as linear transformations in two and three dimensions find a matrix representing a combined transformation find invariant points and invariant lines of a linear transformation.

Before you start… GCSE

You should understand simple transformations: rotation, reflection, enlargement.

1 Triangle

is given by . Find the image of under each transformation.

a Rotation origin.

anticlockwise about the

b Reflection through the line

.

c Enlargement centred at the origin, scale factor . A Level Mathematics Student Book 1, Chapter 15

You should be able to use position vectors to represent points in and dimensions.

2 Write down the position vectors of points and .

A Level Mathematics Student Book 1, Chapter 10

You should be able to solve trigonometric equations.

3 Find the value of in

Chapter 7

You should be able to multiply conformable matrices.

4 Calculate:

Chapter 7

You should know how to find determinants and inverses of matrices.

5

and

For the matrix

for which .

find:

a det b

Chapter 7

You should be able to identify a singular matrix.

6 Find the value of so that the matrix is singular.

Why are matrices useful?

In Chapter 7 you learned about the structure and arithmetic of matrices, how to calculate inverses and determinants and how to solve equations involving matrices. In this chapter, you’ll use these techniques to describe geometric transformations with matrices. This allows visual information to be recorded mathematically, which is an essential technique in applications such as computer game design.

Section 1: Matrices as linear transformations You are familiar with transformations in two dimensions, such as reflections, rotations or enlargements, where the points (making up the object) are moved under a consistent rule to new (image) points. If you represent the points by their position vectors, then this rule can be written as a matrix equation. For example, consider the rotation through anticlockwise about the origin, applied to the triangle with vertices , and , as shown in the diagram.

The image points have coordinates

and

.

Look at how the transformation affects the position vectors:

It looks as if the rule can be written as:

This can be expressed as a matrix equation:

The matrix

represents a

anticlockwise rotation about the origin. You can use this matrix

to find the image of any other point after the rotation. For example, the point the point

would be rotated to

.

You can identify a transformation represented by a matrix by looking at the image of the unit square, which is the square with coordinates WORKED EXAMPLE 8.1

A transformation is given by

,

,

and

.

Find the image of the unit square under transformation and show this on a coordinate grid. Hence describe the transformation. Label the vertices of the unit square , , . Multiply the position vector of each vertex by the transformation matrix.

and

Looking at the images of and helps identify the line of reflection.

The transformation represents a reflection in the line .

Rather than considering all four points on the unit square, you can find the matrix to represent a given transformation by just looking at For a transformation

and

.

:

and

.

Notice that a matrix transformation never changes the position of the origin, as

.

Key point 8.1 The first column of a transformation matrix is the image of image of

and the second column is the

.

WORKED EXAMPLE 8.2

Find the matrix representing the enlargement with scale factor and centre at the origin.

and

The images of the unit vectors give the columns of the transformation matrix.

So the matrix is

Using this method you can find a matrix for any transformation in two dimensions. The most common transformations are summarised in Key point 8.2.

Key point 8.2 Transformation matrices Rotation about the origin through the given angle

Reflection in the given mirror line

scale factor Enlargement with centre at the origin

Tip Note that by convention a positive angle means the rotation is anticlockwise. If the rotation is clockwise this will either be stated or a negative angle will be given

Fast forward In Section 2 rotations are generalised to any angle about the origin, and reflections to any mirror line through the origin.

The determinant In some cases it can be difficult to distinguish between a rotation and a reflection. In Worked example 8.1 it could have been difficult to tell whether it was a reflection in the line or a rotation through .

The image of the unit square is the same in both cases, but the points are arranged in a different order. Reading anticlockwise around the shape, the square is mapped to by the reflection, and to by the rotation.

Fast forward In Further Mathematics Student Book 2, you’ll see: rotation preserving the orientation reflection reversing the orientation of the shape. A usfeul way of distinguishing between the two is to use the determinant of the transformation matrix A. Table 8.1: Summary of the determinant value for transformation matrix A Rotation

Reflection

det

det

Enlargement det

The transformations that reverse the orientation have a negative determinant. Rotations and reflections do not change the size of the shape, while enlargements do.

Successive transformations When two transformations are applied in succession, the overall result is another transformation. You can find its matrix as shown here. Suppose the original point (the object) has the position vector

Fast forward In Further Mathematics Student Book you’ll see how the determinant of a matrix representing an enlargement relates to the area of the image of a shape. If the matrix representing the first transformation is , then the first image is second transformation with matrix , the final image is . Since matrix multiplication is associative, this is the same as combined transformation is .

. After applying the

. Hence the matrix representing the

Key point 8.3 For two transformations and , if represents followed by , then transformation is given by .

Common error

Notice that the matrix for the transformation that is applied first is written on the right.

WORKED EXAMPLE 8.3

represents a reflection in the line through

and

represents a rotation

about the origin.

Transformation

is defined as followed by , and

is the image of the point

under

is defined by followed by .

and is the image of under

.

a Find the coordinates of and . b Describe the transformations

and

.

followed by is represented by

a

.

has coordinates followed by is represented

has coordinates b

Looking at the columns of vector is reversed.

, the unit vector is unchanged and the

is a reflection in the axis leaves the vector unchanged and reverses the direction of

is a reflection in the axis

Inverse transformations You can also use a matrix to reverse the effect of a transformation. If a transformation has matrix , and if the image of the object under this transformation is , then you can recover the original object by using the inverse matrix:

.

Key point 8.4 If a linear transformation is represented by a matrix , then its inverse transformation is represented by the matrix . Notice that all the transformations that you have met so far have had non-zero determinants, so their inverse matrices exist. When you have two successive transformations, the inverse transformations are applied in reverse order. This corresponds to the inverse of a product of two matrices:

WORKED EXAMPLE 8.4

Point is rotated coordinates

about the origin, and then reflected in the -axis. The final image has .

Find the coordinates of . The matrix for the

rotation is:

and for the

You are looking for the inverse of the combined transformation. Start by writing down the individual transformation matrices.

reflection:

The combined transformation is:

Find the combined transformation. Remember that the second transformation is written on the left.

You know that where .

,

are the coordinates of

You can multiply on the left by the inverse matrix.

Hence:

The inverse of

is

.

So:

The coordinates of are

.

EXERCISE 8A 1

Draw the image of the unit square under the transformation represented by each matrix. a b c d

2

Each diagram shows the image of the unit square under a linear transformation. Write down the matrix representing the transformation. a

b

c

d

3

By considering the determinant, classify each transformation as a rotation, reflection or neither. Where appropriate, state the angle of rotation or the equation of the line of symmetry. a b c d e f

4

Find the image of the given point after the given transformation (or sequence of transformations). a i Point

; reflection in the line

ii

; reflection in the -axis

b i Point

; rotation

about the origin

ii Point

; rotation

about the origin

c i Point ii Point d i Point ii Point 5

; an enlargement with scale factor followed by a rotation ; reflection in the line ; rotation

about the origin

followed by an enlargement with scale factor

about the origin followed by an enlargement with scale factor

; enlargement with scale factor followed by a reflection in the

axis

You are given a transformation (or a sequence of transformations) and the coordinates of the image. Find the coordinates of the original object. a i Rotation ii Rotation

about the origin; image about the origin; image

b i Reflection in the -axis; image ii Reflection in the line

; image

c i Enlargement with scale factor followed by a reflection in the line ii Reflection in the

axis followed by an enlargement with scale factor ; image

d i Rotation through

about the origin followed by a reflection in the -axis; image

ii Reflection in the line 6

; image

followed by a rotation through

a Find the matrix for the resulting transformation when a rotation followed by a rotation

about the origin; image

about the origin is

about the origin. Describe the transformation.

b Repeat part a for some other pairs of rotations, including combining a rotation with itself. What do you notice? c Find the matrix for the resulting transformation when a reflection in the line by a reflection in the

is followed

axis. Describe the transformation.

d Repeat part c for other pairs of reflections. What do you notice? 7

What is the inverse transformation of: a reflection in the line b rotation through

8

about the origin?

The triangle with vertices

,

and

is rotated through

anticlockwise about

the origin. Find the coordinates of the vertices of the image. 9

The triangle with vertices

and

is reflected in the line

.

Find the coordinates of the vertices of the image. Show the original and the image triangle on a grid. 10

Transformation is defined as a reflection in the line

followed by an enlargement with

scale factor . a Find the matrix representing . b The image of the point under the transformation has coordinates

. Find the

coordinates of . c The rectangle with vertices transformation . Find the area of the image.



,

,

and

is transformed using the

Section 2: Further transformations in There are five types of basic transformation that can be produced by matrices. In Section 1, you worked with enlargements and special cases of rotations and reflections. In this section, you will extend your understanding to general rotations and reflections and also meet stretches.

Rotations The origin can never move under a matrix transformation, so it has to be the centre of rotation.

Tip Remember that the columns of the transformation matrices correspond to the images of and

this allows you to write down the matrix for each of these transformations.

Rewind Check that using

and

gives the three rotation matrices from Section 1.

A rotation through angle (anticlockwise) about the origin moves the point and

to

to the point

, so the matrix for this transformation is

.

WORKED EXAMPLE 8.5

Find the image of the point

Reflections

about the origin.

Use the general expression to find the rotation matrix.

The matrix is:

The image of

after a rotation through

is:

Since the origin cannot be affected by a transformation matrix, all lines of reflection must pass through the origin. Such lines have equations of the form . The gradient can also be defined by the angle the line makes with the -axis: if this angle is then the gradient of the line is . Considering the images of the two unit vectors, it can be shown that the matrix for the reflection is .

WORKED EXAMPLE 8.6

Line has equation

.

a Find the angle that the line makes with the -axis. b Write down the matrix that describes a reflection in the line

.

a

The line is at an angle b Use

(so

above the -axis. ):

The special lines from Section 1 are also covered by this general formula. For example, the matrix from Worked example 8.1 corresponds to

(since

and

), which

is the line

.

Stretches A stretch with scale factor parallel to the -axis leaves the -axis unchanged (this is called the invariant axis). Hence its matrix is

.

Similarly, a stretch with scale factor parallel to the -axis leaves the -axis invariant and has matrix

.

WORKED EXAMPLE 8.7

Transformation is a stretch with scale factor with the -axis invariant. Transformation is a stretch with scale factor with the -axis invariant. Transformation is the result of followed by . a Find the matrix representing . b The triangle with vertices its image on a grid.

and

is transformed using . Draw triangle

a

The combined matrix is order!

be careful about the

b

Multiply each position vector by the transformation matrix.

and

Enlargements An enlargement (centred at the origin) is a combination of a vertical and horizontal stretch with the same scale factors. Its matrix is

, where is the scale factor.

The table in Key point 8.5 summarises all the 2D linear transformations.

Key point 8.5 linear transformations Transformation Rotation about , angle (anticlockwise) Reflection in the line

Stretch, scale factor , parallel to the -axis Stretch, scale factor , parallel to the -axis

Matrix

Determinant

Enlargement, centre , scale factor The matrices for rotation and reflection will be given in your formula resource.

Tip Note that only the matrices for rotation and reflection are given in the formula resource.

EXERCISE 8B 1

Find the angle of (anticlockwise) rotation represented by each matrix. a

b

c

d

2

Find the equation of the line of symmetry for the reflection represented by each matrix.

a

b

c

d

3

Describe each stretch. a b c d

4

Write down a matrix to represent each transformation in two dimensions.

a Rotation through

about the origin

b Stretch with scale factor parallel to the -axis c Reflection in the -axis d Stretch with scale factor parallel to the -axis e Reflection in the line f

Rotation through

clockwise about the origin

g Enlargement with scale factor centred at the origin 5

Find the image of the point

under each transformation.

a Reflection in the line b Rotation through

clockwise about the origin

c Reflection in the line d Stretch with scale factor parallel to the -axis e Rotation through f

about the origin

Enlargement with scale factor centred at the origin

g Stretch with scale factor parallel to the -axis 6

For each transformation, find the point that has image

.

a Reflection in the line b Rotation through

about the origin

c Reflection in the line d Enlargement with scale factor with the centre at the origin e Stretch with scale factor parallel to the -axis f

Rotation through

about the origin

g Stretch with scale factor parallel to the -axis 7

Find the image of the point

8

The vertices of a rectangle are , , and under the stretch with scale factor parallel to the -axis. a Find the coordinates of b Find the ratio (area of

9

a Find a

under a rotation through

): (area of

vertices with coordinates original triangle.

.

after reflection.

matrix representing a stretch with scale factor

b A stretch with scale factor

is the image of

).

matrix representing a reflection in the line

a Write down the

.

and .

b Find the coordinates of the point that has image 10

about the origin.

parallel to the -axis.

parallel to the -axis is applied to a triangle. The image has and

. Find the coordinates of the vertices of the



Section 3: Invariant points and invariant lines You already know that every linear transformation maps the origin to itself; you say that is an invariant point. Some transformations, such as rotations, have no other invariant points: all other points are moved by the transformation. In this section, you will learn how to determine whether a transformation has any invariant points. Consider a reflection in the line

If a point with position vector

; it is represented by matrix

.

is invariant, then its image will have the same position vector:

This is a system of linear equations:

You can see that the equations are the same, and equivalent to

.

This means that every invariant point must lie on the line . This makes sense: the points on the reflection line don’t move and all the other points do. You say that is a line of invariant points.

Key point 8.6 An invariant point is any point that is unaffected by the transformation, which means that the image of the point is the point itself There may be infinitely many invariant points, forming a line of invariant points. In that case, this equation will have infinitely many solutions of the form

WORKED EXAMPLE 8.8

Determine whether

You are looking for points

has any lines of invariant points.

such that:

Invariant points satisfy

Rewrite this as a system of equations.

Both equations are equivalent to or Hence

. is the line of invariant points.

Check whether both equations are, in fact, the same. If so, they give the line of invariant points.

WORKED EXAMPLE 8.9

Show that the transformation represented by the matrix

has no invariant points

other than the origin. Invariant points satisfy:

The only solution is

Write the condition for invariant points and turn it into simultaneous equations.

,

Solve the equations.

. Hence the only invariant point is

.

Consider again the reflection in the line . All the points on the reflection line are invariant. Any other point moves under the reflection. However, it stays on the same line perpendicular to the reflection line. This means that this perpendicular line, considered as a whole, is invariant.

Key point 8.7 An invariant line is a line for which the image of any point on is also on (though is not necessarily the same point). Any line of invariant points is an invariant line. However, the opposite is not true; there can be an invariant line that contains non-invariant points (for example, any line perpendicular to the line of reflection). WORKED EXAMPLE 8.10

Show that

is an invariant line for the reflection in

Any point on the line coordinates of the form Its image is:

has .

.

You are going to find the image of a general point on the line

You already know the matrix representing the reflection in

Then:

To check whether this image lies on the same line, you need to check that

Hence

also lies on the line , and so this line is

invariant.

WORKED EXAMPLE 8.11

Show that an enlargement has an infinite number of invariant lines through the origin.

The matrix for an enlargement is

.

Invariant lines:

Lines through the origin have the form Write an equation for the image of any point on the line.

Hence

for all , so every line

through the origin is invariant.

If line is invariant, then the coordinates of the image also satisfy

WORKED EXAMPLE 8.12

A transformation is represented by the matrix

.

a Find the equations of invariant lines through the origin. b Show that there are no lines of invariant points. a The image of

For an invariant line:

:

Any line through the origin has the form

If the line is invariant, the image also lies on the line.

If the whole line is invariant, then this is true for all

The invariant lines are .

and

If a point is invariant, then

b For an invariant point:

Solve to find

.

Hence the origin is the only invariant point.

EXERCISE 8C 1

For each matrix, find any lines of invariant points and any other invariant lines through the origin. a b c d

2

R is a reflection through the line is a

.

rotation anticlockwise about the origin.

is a stretch, scale factor , parallel to the -axis. In each case, find any lines of invariant points and any other invariant lines through the origin. a

followed by

b c d e f 3

Transformation is given by a Given that

.

is an invariant line under transformation , find the value of .

b For this value , determine whether there is a second invariant line.

4

Transformations and are given by

.

a Show that they have a common invariant line.

5

b Find any invariant lines of the product

.

Transformation is given by

.

a Find any invariant lines of and determine whether there is a line of invariant points. b Write down the equations of any invariant lines of . 6

is a reflection in the line

.

is a clockwise rotation about the origin by angle . Find a condition on for which 7

Transformation is given by

is an invariant line of

.

.

a Find any invariant lines of and determine whether any are lines of invariant points. Transformation is given by b Given that 8

is an invariant line of the product

Matrix line

. , find the value of .

represents a three-dimensional linear transformation, for which the is invariant.

a By considering the product

, find the values of and .

b For these values of and : i find 9

or show that is singular describe the image of the line

,

.

For this question, two distinct lines are said to be twinned under a transformation if each line is the image of the other. a Under transformation

, line

is twinned with

.Calculate the values

and . b For transformation



, find the line that is twinned with

.

Section 4: Transformations in When you move from the plane to three-dimensional space, some of the basic transformations need to be described in a different way. For example, while in two dimensions a rotation is about a point, in three dimensions it is about an axis of rotation. Similarly, you can’t reflect a three-dimensional object in a line, but you can use a plane of reflection.

In two dimensions, it was useful to consider the image of the unit square. In three dimensions, you can look at the unit cube. This is the cube with edge length one unit, with one vertex at the origin and edges parallel to the coordinate axes. Three of the edges are given by the unit vectors,

,

and

.

The three columns of a transformation matrix give the image positions for these three unit vectors, and so outline the three edges of the image of the unit cube. In each of the two types of transformation you will consider, at least one of the unit direction vectors will be unaffected. By fixing the row and column of that direction vector in the matrix, you can use your understanding of transformations to define the matrix.

Reflections In this course, you only need to consider three possible planes of reflection: the and the plane. The

plane contains the and axes. Every point in this plane

has -coordinate equal to zero, so this plane can also be described as the

plane. The other two planes are the

and the

planes.

plane, the

plane

WORKED EXAMPLE 8.13

Transformation represents a reflection in the plane Write down the

.

matrix for . Sketch the unit cube being reflected in the plane

Images of unit vectors:

The vector has its direction reversed. The unit vectors and lie in the plane of reflection, so are unaffected by the reflection.

Hence:

Rotations

The images of the unit vectors give the columns of the transformation matrix.

You only need to work with rotations about one of the coordinate axes. The axis of rotation is unaffected by the transformation, so one of the columns in the transformation matrix will be just the unit vector. To find the rest of the matrix, you need to think about the other two unit vectors being rotated in two dimensions.

WORKED EXAMPLE 8.14

Transformation represents the rotation

about the -axis. Find the

matrix for .

The axis of rotation is along direction , so

The unit vector lies along the axis of rotation so is unaffected by the rotation. This means that the second row and column are as shown in red.

A

The and vectors are rotated in the the remaining four elements form the rotation.

rotation about the origin in the plane has matrix:

Filling in the gaps with this rotation matrix gives:

plane. Thus matrix of a

You need to be a little careful: the anticlockwise rotation goes from towards , so the image of the unit vector has both components positive, while the image of has one negative component. This means that the negative entry (from ) is in the first column rather than the third.

The three 3D rotation matrices you need to know are summarised in Key point 8.8.

Tip Remember that a positive angle means the rotation is anticlockwise.

Key point 8.8 Rotation about the given axis -axis

-axis

-axis

These matrices will be given in your formula resource.

The three 3D reflection matrices you need to know are summarised in Key point 8.9.

Key point 8.9 Reflection in a given plane.

Other transformations As in two dimensions, combining transformations corresponds to multiplying their matrices. WORKED EXAMPLE 8.15

Find the matrix of the resulting transformation when a rotation through followed by a rotation through about the -axis. Let represent the rotation about the -axis and the rotation about the axis. Then:

about the -axis is

Start by writing down the two standard rotation matrices.

and

The combined transformation is:

The combined transformation corresponds to the matrix product.

This doesn’t seem to be any of the three standard rotations, and you don’t need to be able to interpret such a compound transformation. However, in some simple cases, you can recognise the transformation by considering the images of the three unit vectors. WORKED EXAMPLE 8.16

Describe transformation given by

Vector is unaffected by :

.

Find which of the first, second or third row and column match the identity matrix; reduce the matrix to a and identify the transformation in the plane.

The transformation in the plane shows that is a stretch parallel to the -axis, scale factor . Looking at the images of the unit vectors confirms that this is a stretch parallel to the axis.

EXERCISE 8D

EXERCISE 8D 1

Write down a matrix to represent each transformation in three dimensions. a Rotation through

about the -axis

b Rotation through

about the -axis

c Reflection in the plane d Reflection in the plane 2

Find the image of the point a Rotation through

under each transformation in three dimensions.

about the -axis

b Rotation through

about the -axis

c Reflection in the plane d Reflection in the plane 3

Point is reflected in the plane

. The coordinates of the image are

. Find the

coordinates of . 4

Transformation is a reflection in the plane

. Transformation is a rotation through

about

the -axis. Transformation

is followed by .

a Write down the matrices representing and . b Find the matrix representing 5

represents a rotation through , . Find the image of triangle

6

. Hence describe the transformation

.

clockwise about the -axis. Triangle

is given by

,

under transformation .

are the coordinates of the image of point after a rotation through

about the -axis.

Find the coordinates of the object point . 7

Find the coordinates of the image when the point

8

A rotation through

is rotated

clockwise about the -axis.

about the -axis is followed by the reflection in the plane

.

Find the matrix representing the resulting transformation. 9

A rotation through

about the -axis is followed by a rotation through

clockwise about the -

axis. Find the matrix representing the resulting transformation. 10

Point is rotated through about the -axis and then reflected in the plane of the image are . Find the coordinates of .

11

Transformation represents a axis.

rotation about the -axis followed by a

. The coordinates

rotation about the -

a Find matrix . b Show that

.

Checklist of learning and understanding A matrix can represent a linear transformation: the image of a point with position vector is

.

The first column of a transformation matrix is the image of

and the second column is the

image of For two transformations and , if represents followed by , then transformation is given by

.

If a linear transformation is represented by a matrix , then its inverse transformation is represented by the matrix . Two-dimensional linear transformations include: rotation about the origin reflection through line stretch parallel to the -axis or -axis enlargement centred at the origin. Three-dimensional linear transformations include: rotation about the -axis, -axis or -axis reflection through the plane

,

or

.

An invariant line is one whose image coincides with the original line. An invariant point is one whose image is the same point. The origin is always an invariant point. Invariant points may form a line of invariant points.



Mixed practice 8 1

A reflection is represented by the matrix

.

State the equation of the line of invariant points. Choose from these options. A B C D 2

is a rotation clockwise through is the matrix

about the -axis. .

. Calculate matrix .

a

b Describe the transformation represented by . 3

For matrices and , find any invariant lines of the form are lines of invariant points.

and determine whether any

a b 4

a Describe the single transformation given by each matrix. i ii b Write down the

5

matrix describing a rotation through

The matrix

about the -axis.

represents a rotation.

a State the axis of rotation. b Find the angle of rotation, giving your answer to the nearest degree. [©AQA 2014] 6

Find the transformation represented by the matrix

.

Choose from these options. A Rotation anticlockwise about B Rotation anticlockwise about C Reflection in the line D Reflection in the line 7

Transformation is a

rotation clockwise about the origin and transformation is a stretch,

scale factor , parallel to the -axis. a Write down the

matrices for and .

b Matrix is such that 8

. Find det .

The matrices and are defined by:

Describe fully the geometrical transformation represented by each of the following matrices. a b c d e [©AQA 2010] 9

The transformation represented by matrix

has line of invariant points

.Find and . 10

Under transformation , the images of points a By considering b

and

and

.

or otherwise, calculate the matrix representing .

is the result of an enlargement, scale factor where

are

, and a transformation with matrix

.

i Find . ii Describe the transformation . 11

a Write down the

matrix corresponding to each transformation:

i a reflection in the line ii a stretch parallel to the -axis of scale factor . b Hence find the matrix corresponding to the combined transformation of a reflection in the line followed by a stretch parallel to the -axis of scale factor . c

The matrix is defined by i Show that

.

, where is a constant and is the

identity matrix.

ii Show that the matrix corresponds to a combination of an enlargement and a reflection. State the scale factor of the enlargement and state the equation of the line of reflection in the form . [©AQA 2014] 12

Transformation is given by matrix a A rectangle

with

. and

is transformed by .

i Write down the coordinates of . ii Find the image

of

under transformation .

b i Write down the equation of the line of invariant points for . Determine whether there is a second invariant line for transformation and, if so, find its

ii equation. 13

Matrices and are given by: , a Find, in terms of , the invariant lines of . b Given that

:

i find the possible values of ii show that and have a common invariant line iii for each value of , find the equations of any other invariant lines of . 14

Matrices and are given by

and

.

a Given and commute under multiplication, find and . b For a transformation given by matrix , where

,

:

i find the matrix T, and state the equation of the invariant line ii describe the second linear transformation that comprises 15

a The matrix is defined by

.

.

Given that the image of the point under the transformation represented by is find the value of and the value of . b The matrix is defined by i Show that

.

, where is an integer and is the

identity matrix.

ii Describe the transformation represented by the matrix as a combination of two geometrical transformations. iii Find the matrix

.

,

9 Further applications of vectors In this chapter you will learn how to: write an equation of a straight line in three dimensions, both using vectors and using coordinates find the intersection point of two lines calculate an angle between two vectors or two straight lines (using the scalar product) decide whether two lines are parallel or perpendicular solve problems involving distances between points and lines.

Before you start… A Level Mathematics Student Book 1, Chapter 15

You should know how to describe vectors using components in two and three dimensions.

A Level Mathematics Student Book 1, Chapter 10

1 a Write b

using and base vectors.

as a three-dimensional column vector.

You should recognise when two vectors are equal.

2 Find the values of and such that

You should recognise when two vectors are parallel.

3 Find the values of and such that the vectors and are parallel.

You should be able to solve trigonometric equations to find an unknown angle, working in both degrees and radians.

4 Find two possible values of such that .

.

Describing lines in three dimensions Many situations involve lines in three-dimensional space: for example, representing flight paths of aeroplanes or describing the motion of a character in a computer game. In order to solve problems involving lines in space, you need a way of deciding whether a point lies on a given straight line. Suppose you are given two points, and . These two points determine a unique straight line (a ‘straight line’ means a line extending indefinitely in both directions). How can you check whether a third point lies on the same line, or describe all points on this line? You can use vectors to answer this question.

Section 1: Vector equation of a line Consider, as an example, the straight line passing through points . Then:

,

and another point

This means that the line is parallel to . But, since they both contain the point , they must be the same straight line: in other words, lies on the line .

Rewind Remember that a scalar is just a number. You can now characterise all the points on the line

. Using the above idea, a point lies on

if

and are parallel. But you know that this can be expressed using vectors by saying that some value of the scalar .

So

Rewind See A Level Mathematics Student Book 1, Chapter 15, for a reminder of vector algebra. But you also know that

, where and are the position vectors of and , respectively.

for

This means that

is the position vector of a general point on the line

words, has coordinates different points on the line; for example, .

The line is parallel to the vector

. In other

for some value of . Different values of correspond to corresponds to point , to point and to point

, so this vector determines the direction of the line. The expression

for the position vector of is usually written as

, so it is easy to identify the direction

vector.

Fast forward In Further Mathematics Student Book 2 you will see that there is more than one possible form for the vector equation of a line.

Key point 9.1 The vector equation of a line is: where: is the position vector of a general point on the line is the direction vector of the line is the position vector of any known point on the line.

WORKED EXAMPLE 9.1

Write down a vector equation of the line passing through the point vector

in the direction of the

.

The equation of the line is , where is the position vector of point on the line and is the direction vector.

You can also use the vector equation for lines in two dimensions. For example, a line with direction vector has gradient , because an increase of unit in produces an increase of units in . In three dimensions, you cannot use a single number to replace the direction vector, as you can to describe the gradient in two dimensions.

For example, for a line with direction vector

, an increase of units in produces an increase of

units in and an increase of unit in .

Worked example 9.2 shows how to find a vector equation of the line when two points on the line are known. WORKED EXAMPLE 9.2

Find a vector equation of the line through points

and

.

The line passes through

A direction vector for the line is given by the vector

(or

).

The vector equation is

.

If instead of you had used the position vector of point in the formula, then you would have got the equation

.

This equation represents the same line, but the values of the parameter corresponding to particular points will be different. For example, with the first equation point has equation point has and point has

and point has

, while with the second

.

The direction vector is not unique either, since you are only interested in its direction and not its magnitude. Hence

or

could also be used as direction vectors for the line in Worked example

9.2, as they are all in the same direction. So yet another form of the equation of the same line would be . With this equation, point has

and point has

.

When there is more than one line in a question, you should use different letters for the parameters.

Tip The most commonly used letters for the parameter in the equation of a line are (lambda) and (mu) or and .

WORKED EXAMPLE 9.3

a Show that the equations

and

represent the same straight

line. b Show that the equation

represents a different straight line.

You need to show that the two lines have parallel direction vectors (the lines will be parallel) and one common point (then they will be the same line). Two vectors are parallel if one is a scalar multiple of the other. a The direction vectors are parallel. Show that on the first line.

lies

You know that the point lies on the first line.

lies on the second line; now check it

Find the value of which gives the first coordinate.

Checking the other coordinates:

This value of must give the other two coordinates.

lies on the line. Hence the two lines are the same. Check whether the direction vectors are parallel. b So the line is parallel to the other two. Check whether lies on the first line. Find the value of which gives the first coordinate. Checking the other two coordinates:

This value of must give the other two coordinates.

does not lie on the line. Hence the second line is not the same as the first line.

WORK IT OUT 9.1 Find a vector equation of the line that passes through the points

and

.

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

Solution 2

Solution 3

Sometimes, you know that a point lies on a given line, but you do not know its precise coordinates. Worked example 9.4 shows you how to work with a general point on the line (with an unknown value of ). WORKED EXAMPLE 9.4

WORKED EXAMPLE 9.4

Point

lies on the line with equation

.

Find the possible positions of a point on the line such that

.

You know that lies on the line, so it has the position vector

for some

value of .

You can express vector in terms of and then set its magnitude equal to .

It is easier to work without the square root, so square the magnitude equation.

Now find the position vector of , using .

EXERCISE 9A 1

Find the vector equation of each line in the given direction through each given point. a i Direction ii Direction

b i Point

ii Point

; point ; point

; direction

; direction

c i Point

; direction

ii Point

; direction

d i Direction ii Direction

; point ; point

2

Find the vector equation of each line through the two given points. a i

and

ii

and

b i ii 3

and and

Decide whether or not each given point lies on the given line. a i Line

; point

ii Line

; point

b i Line

; point

ii Line 4

Determine whether the two equations describe the same straight line. a

b

5

; point

and

and

c

and

d

and

A line passes through the point

and has direction vector

.

a Write down the vector equation of the line. b Point has coordinates c Find the exact distance 6

. Show that lies on the line. .

a Find the vector equation of the line through the points with coordinates b Determine whether the point with coordinates

7

Line has vector equation .

Find the vector equation of the line that passes through the point the line with equation

9

lies on this line. . Find the vector equation of the line parallel

to which passes through the point 8

which is parallel to

.

a Show that the points

and

and

lie on the line with equation

. b Find the coordinates of the point on the line such that

.

.

10

a Find the vector equation of line through points b Point lies on and

11

and

.

. Find the possible coordinates of .

a Write down the vector equation of the line through the point

parallel to the vector

. b Calculate the magnitude of the vector

.

c Find the possible coordinates of point on such that



.

Section 2: Cartesian equation of a line In Section 1, you worked with vector equations of lines in both two and three dimensions. You already know that, in two dimensions, you can also write the equation of a straight line in the form . In this section, you will look at how the two forms of the equation are related. WORKED EXAMPLE 9.5

A straight line passes through the point

and has gradient

.

Find the vector equation of the line. Draw a gradient triangle to identify the direction vector: units to the right and units down.

The direction vector is

The line goes through point

The vector equation of the line is

so the

equation is:

Key point 9.2 In two dimensions, a line with gradient has direction vector

.

WORKED EXAMPLE 9.6

A straight line has equation

.

Write the equation of the line in the form The line has gradient and passes through the point

.

. You can find the gradient from the direction vector. The vector equation also shows the coordinates of one point on the line. Use

.

.

The equation of the line in the form

is called a Cartesian equation. This means that the

equation is in terms of - and -coordinates, rather than in terms of position vectors.

Tip The Cartesian equation of a line in two dimensions can be written as or .

,

There is another way to change from a vector to a Cartesian equation. The key is to realise that the position vector gives the coordinates of a point on the line:

.

WORKED EXAMPLE 9.7

A line has vector equation

.

By expressing and in terms of , write the equation of the line in the form

.

The position vector is related to coordinates: .

From (1):

Express from the first equation and substitute into the second.

Substituting into (2):

In three dimensions, there is no equivalent of the gradient – you cannot use a single number to replace the direction vector. But you can still use the method from Worked example 9.7 to find a Cartesian equation of the line. WORKED EXAMPLE 9.8

A line has vector equation

. Express and in terms of .

Write

.

Express from the first equation and substitute into the second and third.

From (1):

Substituting into (2) and (3):

It is possible to combine these two equations. Make the subject of all three equations and equate them to each other.

Rewind Writing , and in terms of gives the parametric equation of the line. These are covered in A Level Mathematics Student Book 2, Chapter 12.

Key point 9.3 To find the Cartesian equation of a line given its vector equation:

write

in terms of , giving three equations

make the subject of each equation equate the three expressions for to get an equation of the form

.

Sometimes a Cartesian equation cannot be written in this form, and you need to make a slight adjustment to the procedure. WORKED EXAMPLE 9.9

Find the Cartesian equation of the line with vector equation

.

You need an equation involving , and . Remembering that

, you can express in terms of , and .

Equate the expressions for from the first two equations. However, the third equation does not contain , so you have to leave it as a separate equation. It will look neater if you rewrite the equation without ‘fractions within fractions’.

You can reverse this procedure to go from a Cartesian to a vector equation. A vector equation is convenient if you need to identify the direction vector of the line, or to solve problems involving intersections of lines.

Tip The Cartesian equation can sometimes be read from the vector equation: if the vector equation is

then the Cartesian equation is

. However, if any of

the components of the direction vector is , you should work through the procedure described in Key point 9.3.

Key point 9.4 To find a vector equation of a line, from a Cartesian equation in the form

:

set each of the three expressions equal to express , and in terms of write

to obtain in terms of .

You can adapt this procedure even when the Cartesian equation is in a different form, as in Worked example 9.10. WORKED EXAMPLE 9.10

Find a vector equation of the line with Cartesian equation

, and hence write

down the direction vector of the line. You need to introduce a parameter . As the two expressions involving and are equal, you can set them both equal to . Now express , and in terms of .

The vector equation is an equation for in terms of . You usually separate the expression into a part without and a part involving .

Now identify the direction vector. The direction vector is

or

,

.

You can change the magnitude of the direction vector so that it does not contain fractions, in this case by multiplying by a scalar.

If you want to check whether a given point lies on a given line, you can do this by substituting the numbers for the , and coordinates. WORKED EXAMPLE 9.11

Determine whether the point

Substituting

,

,

lies on the line with equation

:

The point does not lie on the line.

EXERCISE 9B 1

Write down the Cartesian equation of each line. a i ii

b i

ii

c i

ii 2

Find a vector equation of each line. a i ii

.

If the point lies on the line, the coordinates should satisfy the Cartesian equation. This means that all three expressions should be equal.

The third equality is not satisfied.

b i ii 3

Write down a vector equation of each line. a i ii b i ii c i ii

4

Determine whether each pair of lines is parallel, the same line, or neither. a

and

b

and

c

and

d

and

5

Find a vector equation of the line

6

Determine whether the point with coordinates

. lies on the line with equation

. 7

a Find the Cartesian equation of the line with vector equation

.

b Find the unit vector in the direction of the line. 8

Line has Cartesian equation

. Point

lies on the line.

a Find the values of and . b Point also lies on the line and the distance



. Find the possible coordinates of .

Section 3: Intersections of lines Suppose two lines have vector equations and . If they intersect, then there is a point that lies on both lines. Remembering that the position vector of a point on the line is given by the vector , this means that you need to find the values of and which make

.

WORKED EXAMPLE 9.12

Find the coordinates of the point of intersection of the lines

and

.

You need to make

.

If two vectors are equal, then all their components are equal.

You know how to solve two simultaneous equations in two variables. Pick any two of the three equations. Subtract the first from the third to eliminate .

The values of and you have found must also satisfy the second equation.

Check in (2):

So the lines intersect. The position of the intersection point is given by the vector (or as they should be the same).

The lines intersect at the point

.

Tip You might be able to use your calculator to solve simultaneous equations. In a plane, two different straight lines either intersect or are parallel. However, in three dimensions it is possible to have lines that are not parallel but do not intersect, like the red and the blue lines shown in the diagram.

Such lines are called skew lines. When two lines are skew, it is not possible to find values of and such that

.

WORKED EXAMPLE 9.13

Show that the lines

and

do not intersect.

Make the two position vectors equal and try to solve the three equations to find and .

You can find and from the first two equations. Check in (3):

The values found must also satisfy the third equation.

The two lines do not intersect.

This tells us that it is impossible to find and to make .

It is also possible to check whether a given line crosses one of the coordinate axes. In Worked example 9.14, you use the Cartesian equation of a line. WORKED EXAMPLE 9.14 a Find the coordinates of the point where the line with equation

intersects the

-axis. b Show that the line does not intersect the -axis. a Substitute

,

,

:

A point on the -axis has coordinates . Substitute these into the Cartesian equation.

Solve for .

The point of intersection is b Substitute

,

.

:

The line does not intersect the -axis.

EXERCISE 9C

A point on the -axis has coordinates

The first equality is not satisfied.

.

EXERCISE 9C 1

Determine whether each pair of lines intersect and, if they do, find the coordinates of the point of intersection. a i

and

and

ii

b i

and

and

ii

c i

and

ii

and

2

Show that the lines with equations are skew.

3

Show that the lines

and

and

intersect, and find the

coordinates of the intersection point. 4

Determine whether the line with equation

5

The line with equation

6

Show that the lines with equations

crosses the -axis.

crosses the -axis. Find the value of .

,

form a triangle, and find its area.



and

Section 4: The scalar product Angles between lines The diagram shows two vectors, and , directed away from the intersection point. The angle between these vectors, , can be calculated using the scalar product (also called the dot product).

Key point 9.5 The scalar product of two vectors,

, is defined as

where is the angle between the vectors and . This definition leads to a formula for finding the scalar product from the components of the two vectors.

Key point 9.6 If

and

, then:

You can now find the angle between two vectors. One of the most common uses is to find the angle between two lines, which you can see, from the diagram, is the same as the angle between their direction vectors.

Key point 9.7 The angle between two lines is equal to the angle between their direction vectors.

Common error

When taking the scalar product of two vectors to find the angle between them, you need both vectors to be directed away from the angle; otherwise the angle you find will be

.

WORKED EXAMPLE 9.15

Find the acute angle between lines with equations

and

.

Rearrange the formula from Key point 9.5 Take and to be the direction vectors of the two lines.

Find the dot product

Now use the formula to calculate the angle.

The angle found is obtuse but the question asked for the acute angle.

WORKED EXAMPLE 9.16

Given points

,

and

, find the size of angle

, in degrees.

It is always helpful to draw a diagram to identify which vectors you need to use.

Let

.

You can see that the required angle is between vectors and . Note that you need both vectors to be directed away from the angle you want to find. and

.

Find the dot product.

Use

WORK IT OUT 9.2

Find the angle between the lines

and

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

Solution 2

.

Solution 3

Perpendicular lines The formula in Key point 9.5 is very convenient for checking whether two vectors are perpendicular. If , then , so the numerator of the fraction in the formula must be zero. This means that you don’t even need to calculate the magnitudes of the two vectors.

Key point 9.8 Two vectors and are perpendicular if

.

Two lines are perpendicular if their direction vectors satisfy

.

WORKED EXAMPLE 9.17

Given that

to

and

find the value of the scalar such that

is perpendicular

.

Two vectors are perpendicular if their scalar product is equal to .

You need to find the components of and then form an equation. So:

WORKED EXAMPLE 9.18

in terms of

WORKED EXAMPLE 9.18

Prove that the lines

and

are perpendicular.

In order to identify the two direction vectors, you need to rearrange the equations into the form .

Use the scalar product to show that perpendicular.

and

are

Hence the two lines are perpendicular.

Distance from a point to a line You can use perpendicular lines to find the shortest distance from a point to a line.

Focus on ... See Focus on ... Problem solving 2 for alternatives to the method in Worked example 9.19 for finding the shortest distance between skew lines.

WORKED EXAMPLE 9.19

Line has equation

and point has coordinates

a Find the coordinates of point on so that

.

is perpendicular to .

b Hence find the shortest distance from to . c Find the coordinates of the reflection of the point in . Draw a diagram. The line needs to be perpendicular to the direction vector of .

a

You know that lies on , so its position vector is given by the equation for .

You can now find the value of for which the two lines are perpendicular.

Using this value of in the equation of the line gives the position vector of . has coordinates

b

c

. The shortest distance from a point to a line is the perpendicular distance, in other words, the distance

The reflection lies on the line and As they are also in the same direction,

.

. .

So

.

The final part of Worked example 9.19 illustrates the power of vectors: as they contain both distance and direction information, just one equation and that

was able to express both the fact that

is on the line

.

Distance between lines The approach used in Worked example 9.19 to find the shortest distance from a point to a line can also be used to find the shortest distance between two parallel lines. You can just pick any known point on the second line to play the role of the point A.

However, a different approach is needed to find the shortest distance between two skew lines. You can’t now pick a fixed point on the second line; instead you need a general point on the first line, , and a general point on the second line, . The distance between and will be shortest when perpendicular to both lines.

is

Focus on ... See Focus on ... Problem solving 2 question 2 for alternatives to the method in Worked example 9.20 for finding the shortest distance between skew lines.

WORKED EXAMPLE 9.20

Find the shortest distance between the skew lines :

and :

.

Let be a general point on and a general point on .

The position vector of a general point on a line is given by the equation of that line.

Then

and

You can then find a general vector, connecting and

If

is perpendicular to both lines then

,

This connecting vector will be shortest when it is perpendicular to both lines. The dot product of with the direction vector of each line must be zero.

and

Solving (1) and (2) simultaneously:

Substitute these values of and back into the expression for .

The distance is the length of

EXERCISE 9D 1

Calculate the scalar product for each pair of vectors.

.

a i

and

ii b i ii

and

c i

and

ii 2

and

and

Calculate the angle between the pairs of vectors from Question 1. Give your answers in radians.

3

The angle between vectors and is . Find the exact value of a i

and

ii

and

b i

and

ii c i ii 4

and and and

Which pairs of vectors are perpendicular? a i

and

ii

and

b i ii 5

in each case.

and and

Find the acute angle between each pair of lines, giving your answers in degrees. a i

and

ii

and

b i

and

and

ii

6

a The vertices of a triangle have position vectors

,

and

.

Find, in degrees, the angles of the triangle.

7

b Find, in degrees, the angles of the triangle with vertices



Points and have position vectors

. Find the angle between

and

and

and

.

.

Given four points with coordinates 8

acute angle between

and

9

The line has equation

,

,

and

, find the

. and the line has equation

.

a Find, to 1 decimal place, the acute angle between and . The line has equation

.

b Show that and are perpendicular. 10

Four points have coordinates a Show that b When

,

and

.

is a parallelogram for all values of . find the angles of the parallelogram.

c Find the value of for which 11

,

is a rectangle.

The vertices of a triangle have position vectors

,

and

.

a Show that the points are the vertices of a right-angled triangle. b Calculate the other two angles of the triangle. c Find the area of the triangle. 12

Line has equation and

13

and point has coordinates

. Point lies on

is perpendicular to . Find the coordinates of .

The line passes through the point with coordinates coordinates .

and the point with

a Find the vector equation of . The line passes through the point with coordinates b Find the distance

and is parallel to .

.

c Find the exact value of

, where is the angle between

and .

d Hence find, to 3 significant figures, the distance between and . 14

Two lines with equations

and

intersect at point

. a Find the coordinates of . b Find, in degrees, the acute angle between the two lines. Point has coordinates

.

c Show that lies on . d Find the distance e Hence find the shortest distance from to the line . 15

Find the shortest distance of the line with equation

16

Find the exact distance between the lines

17

The line has equation a Find the coordinates of point

from the origin.

and

and point has coordinates on such that

b Show that the point

.

is perpendicular to .

lies on .

c Find the coordinates of point on such that 18

.

.

Two lines have vector equations :

and :

The point on and the point on are such that

is perpendicular to both lines.

a Show that

.

.

b Find a second equation linking and . c Hence find the shortest distance between and , giving your answer as an exact value.

Checklist of learning and understanding A vector equation of a straight line has the form and is the direction vector.

, where is one point on the line

To find the equation of the line through two points with position vectors and , use the direction vector . To find the Cartesian equation of a line given its vector equation: write

in terms of , giving three equations

make the subject of each equation equate the three expressions for to get an equation of the form

.

To find the intersection of two lines, solve simultaneous equations. If there is no solution, the lines are skew. The scalar product (or dot product) of two vectors can be calculated in two ways:

Two vectors are perpendicular if

.

The angle between two vectors can be found from



.

Mixed practice 9 1

The points , and have position vectors

,

and

respectively. Find the size of the angle

.

Choose from these options. a b c d 2

Find the vector equation of the line passing through points

3

Two lines have equations

and

.

and

.

Given that the lines intersect: a find the coordinates of the point of intersection b find the acute angle between the lines. 4

Show that the lines with equations are skew.

5

The vector and the vector perpendicular, find the possible values of .

6

Find the acute angle between the skew lines

and

. Given that and are

and

. 7

a Find a vector equation of the line with Cartesian equation

.

b Determine whether the line intersects the -axis. c Find the angle the line makes with the -axis. 8

a Find the coordinates of the point of intersection of the lines with Cartesian equations and

.

b Show that the line with equation

passes through the intersection

point from part a. 9

Four points have coordinates Determine whether the lines

10

and and

Three points have coordinates

intersect. ,

a Find the value of for which the triangle

and

.

has a right angle at .

b For the value of in part a, find the coordinates of point on the side . 11

.

Two lines are given by Cartesian equations:

such that

a Show that and are parallel. b Show that the point

lies on .

c Find the coordinates of the point on such that

is perpendicular to the two lines.

d Hence find the distance between and , giving your answer to significant figures. 12

The points , and have coordinates

,

and

respectively.

The line passes through and has equation

.

a Show that the point lies on the line . b Find a vector equation of the line that passes through points and . c The point lies on the line through and such that the angle the coordinates of .

is a right angle. Find

d The point lies on the line through and such that the area of triangle is three times the area of triangle . Find the coordinates of the two possible positions of . [© AQA 2013] 13

The points and have coordinates

and

respectively. The angle

is ,

where is the origin. a i Find the vector

.

ii Show that

.

b The point is such that . Find a vector equation of .

. The line is parallel to

c The point lies on such that angle

and passes through the point

. Find the coordinates of . [© AQA 2009]

14

Find the shortest distance from the point

to the line with equation

. 15

A line passes through the points

and

.

a Find a vector equation for . A line has equation

.

b Find the exact distance between and . 16

Two lines have equations

and

intersect at point . a Show that b

lies on .

is a point on such that

. Find the possible coordinates of .

and

17

Two lines are given by a

and

and intersect at . Find the coordinates of .

b Show that the point

lies on .

c Find the coordinates of point d Find the area of the triangle 18

.

on such that

is perpendicular to .

.

a Show that the lines : not intersect.

and :

b Points and lie on and respectively, such that i Show that

do

is perpendicular to both lines.

.

ii Find a second equation for and . iii Hence find the shortest distance between and , giving your answer as an exact value. 19

The coordinates of the points and are

and

The line has equation

a i Find the vector

respectively.

.

.

ii Calculate the acute angle between . b The point lies on where

and the line , giving your answer to the nearest

. The line passes through and is parallel to

.

i Find a vector equation of line with parameter . ii The diagram shows a symmetrical trapezium . The point lies on line . The length of

, with angle

is equal to the length of

equal to angle

.

Find the coordinates of . [© AQA 2011]

10 Further calculus In this chapter you will learn how to: find the volume of a shape formed by rotating a curve around the -axis or the -axis find the mean value of a function.

Before you start … GCSE

You should know the formula for the volume of a cylinder.

A Level Mathematics Student Book 1, Chapter 14

You should know how to find the definite integral of a polynomial.

A Level Mathematics Student Book 1, Chapter 16

You should know that displacement is found by

.

1 Find the exact volume of a cylinder with base radius and height .

2 Evaluate

.

3 Find the displacement in the first seconds of a particle with velocity .

What else can you do with calculus? You have already seen several applications of calculus, such as finding tangents and normals to curves, optimisation, finding areas and converting between displacement, velocity and acceleration in kinematics. In this chapter, you will see two further applications – finding volumes and finding the mean value of a function.

Section 1:Volumes of revolution In A Level Mathematics Student Book 1, Chapter 14, you saw that the area between a curve and the -axis from

to

is given by

, as long as

. In this section, you will use a similar formula to

find the volume of a shape formed by rotating the curve about either the -axis or the -axis. If a curve is rotated about the -axis or the -axis, the resulting shape is called a solid of revolution and the volume of that shape is referred to as the volume of revolution.

Key point 10.1 When the curve

between

of revolution is given by When the curve

between

of revolution is given by

and

is rotated

about the -axis, the volume

is rotated

about the -axis, the volume

. and .

The proof of these results is very similar. The proof for rotation about the -axis is given here. PROOF 10

The solid can be split into small cylinders. Draw an outline of a representative function to illustrate the argument.

The volume of each cylinder is

.

The radius of each cylinder is the -coordinate and the height is . You are starting at and stopping at . It is only approximate because the volume of revolution is not exactly the same as the total volume of the cylinders.

The total volume is approximately:

However, as you make the cylinders smaller the volume gets more and more accurate. The sum then becomes an integral. You can leave out of the integration and multiply by it at the end.

WORKED EXAMPLE 10.1

The graph of

,

, is rotated

about the -axis.

Find, in terms of , the volume of the solid generated.

Use the formula:

.

Evaluate the definite integral.

To find the volume of revolution about the -axis you will often have to rearrange the equation of the curve to find in terms of .

Common error Remember that the limits of the integration need to be in terms of and not .

WORKED EXAMPLE 10.2

WORKED EXAMPLE 10.2

The part of the curve

between

and

is rotated

about the -axis. Find the

exact value of the volume of the solid generated. Find the limits in terms of .

Express in terms of .

Use the formula

, substituting in

.

You might also be asked to find a volume of revolution of an area between two curves.

From the diagram you can see that the volume formed when the region R is rotated around the -axis is given by the volume of revolution of minus the volume of revolution of .

Tip Remember that many calculators can find definite integrals.

Common error Make sure that you square each term within the brackets and do not make the mistake of

squaring the whole expression inside the brackets: the formula is not

.

Key point 10.2 The volume of revolution of the region between curves

where

is above

and the curves intersect at

and

is:

and

WORKED EXAMPLE 10.3

Find the volume formed when the region enclosed by about the -axis. For points of intersection:

and

is rotated through

First find the -coordinates of the points where the curves meet, by equating the RHS of both equations and solving. This will give you the limits of integration.

Sketch the graphs in the region concerned. is above

Apply the formula.

Expand and simplify.

Then evaluate the definite integral.

Did you know There are also formulae to find the surface area of a solid formed by rotating a region around an axis. Some particularly interesting examples arise if you allow one end of the region to tend to infinity; for example, rotating the region formed by the lines

,

and the -axis results in

a solid called Gabriel’s horn or Torricelli’s trumpet.

Areas and volumes can also be calculated using what are called improper integrals, and it ensues that it is possible to have a solid of finite volume but infinite surface area!

EXCERCISE 10A 1

The part of the curve for is rotated volume of revolution formed in each case. a i

;

ii

;

b i

c i ii 2

, , ;

,

;

ii ; ;

about the -axis. Find the exact

, , ,

Find the exact volume of revolution formed when each curve, for through radians about the -axis. a i ii b i ii c i ii

3

The part of the curve for

is rotated

about the -axis.

Find the exact volume of revolution formed in each case. a i

, is rotated

ii b i ii c i ii 4

The part of the curve

for

is rotated

about the -axis.

Find the exact volume of revolution formed in each case. a i ii b i ii c i ii 5

The diagram shows the region, , bounded by the curve line

, the -axis and the

.

a Find the coordinates of the point A where the curve crosses the -axis. This region is rotated about the -axis. b Find the exact volume of the solid generated. 6

The curve

, for

, is rotated through

about the -axis.

Find the volume of revolution generated, correct to 7

The part of the curve

between

and

is rotated through

radians

about the -axis. Find the exact volume of the solid generated. 8

The curve volume is

, for

, is rotated through

about the -axis. The resulting

.

Find the value of . 9

The region enclosed by the curve

and the -axis is rotated

about the -

axis. Find an expression for the volume of revolution formed. 10

The part of the curve

between

and

is rotated through

about the -axis. The volume of the resulting solid is

radians

.

Find the exact value of . 11

a Find the coordinates of the points of intersection of curves

and

.

b Find the volume of revolution generated when the region between the curves and 12

is rotated through

The region bounded by the curves

about the -axis.

and

is rotated through

about

the -axis.Find the volume of the resulting shape. 13

a Find the coordinates of the points of intersection of the curves and

b The region between the curves and

is rotated through

about

the -axis. Find the volume of the solid generated. 14

By rotating the circle radius is given by

15

around the -axis, prove that the volume of a sphere of .

By choosing a suitable function to rotate around the -axis, prove that the volume of a circular cone with base radius and height is

16



.

Find the volume of revolution when the region enclosed by the graphs of is rotated through about the line .

,

and

Section 2: Mean value of a function Suppose an object travels between looks like this.

and

with a velocity given by

. Its velocity–time graph

Its average velocity can be found from:

Suppose, instead, the object has velocity given by

. Then you can compare the two velocity–time

graphs. The formula

would give the same average velocity for the two graphs, which

can’t be correct because the red curve is underneath the blue line everywhere other than at the end points.

You need a measure of average that takes into account the value of the function everywhere. One possibility is to use

.

You can then use the fact that total distance is the integral of velocity with respect to time. For the blue line this gives:

For the red curve this gives:

This process can be generalised for any function.

Key point 10.4 The mean value of a function

between and is:

WORKED EXAMPLE 10.4

Find the mean value of

between and . Use the formula for the mean value of a function:

Notice that varies between and around seems reasonable.

EXCERCISE 10B 1

Find the mean value of each function between the given values of . a i ii b i ii c i ii

, so a mean of

2

Find the mean value of each function over the domain given. a i ii b i ii c i ii

3

The velocity of a rocket is given by in metres per second.

where is time, in seconds, and is velocity,

Find the mean velocity in the first seconds. 4

The mean value of the function

for

is zero.

Find the value of . 5

for a

.

is the mean value of

between and . Find an expression for

in terms of

. b Given that

find an expression for in terms of .

6

Show that the mean value of

between and is inversely proportional to .

7

An alternating current has time period . The power dissipated by the current through a resistor is given by . Find the ratio of the mean power of one complete period to the maximum power.

8

The mean value of

between and is .

Prove that the mean value of 9

a Sketch the graph of

between and is

.

.

b Use the graph to explain why the mean value of the function between and is less than the mean of and . c Hence prove that, if 10

If

is the mean value of

,

.

for

and

, then

.

Either prove this statement or disprove it using a counterexample.

Checklist of learning and understanding The volume of a shape formed by rotating a curve about the -axis or the -axis is known as the volume of revolution. When the curve between volume of revolution is given by

and

is rotated

about the -axis, the

When the curve between volume of revolution is given by

and

is rotated

about the -axis, the

The volume of revolution of the region between curves

where

is above

and the curves intersect at

The mean value of a function



between and is:

and

and

is:

.

Mixed practice 10 1

Find the volume of revolution when the curve around the -axis.

for

is rotated through

Choose from these options. A B C D 2

Find the mean value of

between and .

Choose from these options. A B C D 3

The curve between and is rotated through shape has a volume of .

about the -axis. The resulting

Find the value of . 4

For

, the mean value of is equal to the mean value of

.

Find the value of . 5

The mean value of

from to is .

Find the value of . 6

The curve

with

, is rotated through

about the -axis.

Find the volume of revolution generated, correct to 7

The diagram shows the curve withequation

, where

.

Calculate the volume of the solid generated when the region bounded by the curve shown and the coordinate axes is rotated through about the -axis, giving your answer in terms of. [© AQA 2009] 8

The diagram shows the curve with equation

for

The shaded region is bounded by the curve and

.

, the -axis and the lines

.

Find the exact value of the volume of the solid formed when the region is rotated through about the -axis. [© AQA 2009] 9

for a

is the mean value of

between and . Find an expression for

in terms of

. b Given that 10

find an expression for in terms of .

The region bounded by the curve the -axis.

and the -axis is rotated one full turn about

Find, in terms of , the resulting volume of revolution. 11

Prove that the mean value of between and is the arithmetic mean of and .

12

The diagram shows the curve

and the line

.

a Show that the two graphs intersect at (e, 1). The shaded region is rotated through

about the -axis.

b Find the exact value of the volume of revolution. 13

The region enclosed by about the line .

and the line

is rotated through

Find the exact value of the resulting volume. 14

The part of the curve between axis. The volume of revolution formed is

and

is rotated

about the -

.

Find the value of k. 15

Consider two curves with equations

and

a Find the coordinates of the points of intersection of the two curves. about the -axis. Write down b The region enclosed by the curves is rotated through an integral expression for the volume of the solid generated. c Evaluate the volume, giving your answer to the nearest integer. 16

a The region enclosed by

and

is labelled .

Draw a sketch showing . b Find the volume when is rotated through

about the -axis.

c Hence find the volume when is rotated through

about the -axis.

11 Series In this chapter you will learn how to: use given results for the sums of integers, squares and cubes to find expressions for sums of other series use a technique called the method of differences to find expressions for the sum of terms of a series use given results for infinite series expansions of functions such as and , to find series for more complicated functions understand for which values of these infinite series are valid.

Before you start… GCSE

You should be able to use the th term formula to generate terms of a sequence.

1 A sequence is defined by

.

Find the first three terms.

GCSE

You should be able to simplify expressions by factorising.

2 Simplify

A Level Mathematics Student Book 1, Chapter 9

You should be able to use the binomial expansion on for positive integer .

3 Expand

.

.

Summing sequences If you sum the terms of a sequence, you get a series. The formula for the th term of a sequence is useful because it allows you to find any term you wish without having to find all the terms that have come before. In the same way, it is useful to have a formula for the sum of the first terms of a sequence. In this chapter, you will look at methods for finding formulae for these sums and at forming infinite series for some common functions such as sine, cosine and the exponential function.

Section 1: Sigma notation The sum of a sequence up to a certain point is called a series. You often use the symbol sum of the first terms of a sequence.

to denote the

Tip The expression

is the standard notation for the th term of a sequence. So in the sequence , and so on.

Instead of

, you will often see exactly the same idea expressed in a shorter form,

using sigma notation:

Tip Don’t be put off by this notation. If you are in any doubt, try writing out the first few terms in full.

Key point 11.1

There is nothing special about the letter here; you could use any letter, but and are the most usual. Note that sometimes sigma is written without the and above and below it – you may just see the first and last values. WORKED EXAMPLE 11.1

Find the value of

. Substitute the starting value, be summed, .

, into the expression to

You’ve not reached the end value, so put in

.

You’ve still not reached the end value, so put in You’ve reached the end value, so stop and evaluate.

WORKED EXAMPLE 11.2

Write the series

in sigma notation.

.

The general term is

Describe each term of the series using a general term in the variable .

.

The series starts at The series ends at

. .

Note the first value of . Note the final value of . Summarise in sigma notation.

EXERCISE 11A 1

Evaluate each expression. a i

ii

b i

ii

c i

ii 2

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



Section 2: Using standard formulae In general, it is difficult to find a formula for the first terms of a series, but if you know the formula for a few ‘standard’ series, you can use them to establish formulae for many other series. You can use the three standard formulae in Key point 11.2 without proof – unless you’re asked to prove them with the method used in Section 3!

Fast forward The formulae in Key point 11.2 will be proved in Section 3 of this chapter. You will also see in Chapter 12, Section 2, how these results can be proved more directly by means of a technique called induction.

Key point 11.2 Formulae for the sums of integers, squares and cubes:





The second and third formulae will be given in your formula resource.

Did you know The formula for the sum of the first integers,

, is the th term formula for the

sequence of triangular numbers:

Before you use these results, notice how you can split up sums and take out constants. For example:

where

.

Common error

Remember that a constant, , summed times is

and not just . For example,

and

not .

Key point 11.3 You can manipulate series in several ways.



where is a constant.

WORKED EXAMPLE 11.3

a Use the formula for

b Hence find

to show that

.

.

a

You need to rearrange the expression into a form to which you can apply the standard formulae. Start by splitting up the sum. Then take out of the first sum as a factor.

and Notice that it’s always a good idea to factorise first. In this case only factorises, but in more complicated examples this will avoid having to expand and then factorise a higher order polynomial later. b

You can only use the formula in part a if the sum starts from . Therefore, work out the sum of the first terms and subtract the sum of the first 7 terms.

Now use the formula .

WORKED EXAMPLE 11.4

with

and

WORKED EXAMPLE 11.4

a Use the formulae for

,

and

to show that .

b Hence find an expression for

a

, simplifying your answer fully.

Expand the brackets.

Split up the series into separate sums. Take out constants.

Substitute in the standard formulae. Simplify the first two terms. Now factorise as many terms as possible. Note that this is much easier than expanding everything first.

b

Substitute for in the formula found in part a. Simplify and factorise a from the second bracket.

WORK IT OUT 11.1 Given that

, find an expression for

.

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





Solution 2





Solution 3





EXERCISE 11B

In this exercise, you can assume the formulae for 1

,

and

.

Evaluate each expression. a i

ii

b i

ii 2

Find a formula for each series, giving your answer in its simplest form. a i

ii

b i

ii 3

Show that

4

Show that

5

a Find an expression for

.

.

.

b Hence find the least value of such that 6

a Show that b Hence evaluate

7

Show that

8

a Show that

b Hence find a formula for 9

a Show that b Hence find, in the form

10

.

.

, where is an integer to be found.

.

, fully simplifying your answer.

. , the exact value of

Show that the sum of the squares of the first odd numbers is given by is an integer to be found.



.

. , where

Section 3: Method of differences If you are investigating a series, start by writing out a few terms to see if any patterns develop. For example, for the series

:

You can see that each term shares a common element with the next; in the first term, this element is positive and in the next it is negative. Therefore, when you complete the sum, these common elements will cancel out. This cancellation continues right through to the th term.

In fact, because:

you have just shown that:

which is the result for the sum of the first integers that you met in Section 2. This process for finding a formula for the sum of the first terms of a sequence is called the method of differences.

Tip The series won’t always take exactly this form, so always write out several terms to see how the cancellation occurs.

Key point 11.4 Method of differences If the general term of a series,

WORKED EXAMPLE 11.5

, can be written in the form

, then:

WORKED EXAMPLE 11.5

a Show that b Hence show that

a

. .

Use the binomial expansion to expand the cubed brackets.

Simplify to give the result required. b

Sum both sides of the result from a. The RHS is a difference, so you expect cancellation. Write out the first few terms and the last couple of terms .

Everything cancels except the terms shown. For the LHS, remember that

.

Make the expressions for the LHS and RHS equal. You now need to make the subject. Start by expanding the RHS and then simplify.

Factorise the RHS.

Finally, divide by .

Sometimes the cancellation occurs two terms apart. WORKED EXAMPLE 11.6

a Show that

.

Tip In an A Level question you could be expected to split

into partial

fractions rather than being given the identity to prove.

b Find an expression for c Hence find

.

.

a

b

Sum both sides of the result in a.

Writing out several terms shows that the cancellations in the series occur two terms apart.

Continue the pattern of cancellation for the last few terms .

This leaves part of the first two terms and part of the last two. You could put this all over a common denominator and combine into one fraction but, as the question doesn’t specifically require this, there is no need to do anything else.

c Let

As the denominator tends to fractions tend to zero.

in the result in part b.

, these

EXERCISE 11C 1

2

3

a Show that

.

b Hence show that

.

a Show that

.

b Hence show that

.

a Show that

.

b i Hence show that

.

ii Find

4

a Express b

.

Use the method of differences to show that

a Show that b Hence, find

7

.

a Show that b

6

in partial fractions.

Use the method of differences to show that

c Find 5

.

.

. .

a Show that b Find an expression for c Hence show that

8

Use the method of differences to find

9

a Show that

. . . .

b Use the method of differences to show that

where , and are constants to be found. c Find 10

a Use the method of differences to find b Hence, prove that the series



.

diverges.

Section 4: Maclaurin series You can write many common functions as infinite series, called Maclaurin series. You need to be aware of the results in this section, which are given in the formula resource.

Did you know Maclaurin series are named after the 18th-century mathematician Colin Maclaurin, who also developed some of Newton’s work on calculus, algebra and gravitation theory.

Tip Don’t overlook the information on the values of for which these series are valid; this is a very important part of each result.

Key point 11.5 Maclaurin series for some common functions and the values of for which they are valid: all

all all

These will be given in your formula resource. You can use these standard results to find Maclaurin series of more complicated functions.

Fast forward You will see where these results come from in Further Mathematics Student Book 2.

Rewind Note that the last result in Key point 11.5 is the binomial expansion, which is covered in A Level Mathematics Student Book 2, Chapter 6. WORKED EXAMPLE 11.7

a Use the Maclaurin series for

to find the first four terms in the series for

.

b State the values of for which the series is valid.

a

Substitute

into the series for

:

Expand and simplify.

b Valid for all

.

Both

and

are valid for all

.

This process can be more complicated if it involves finding two separate Maclaurin series and then combining them. WORKED EXAMPLE 11.8

a Use the Maclaurin series for

and to find the series for

as far as the term in

.

b State the values of for which your series is valid.

a

Start by replacing

with its series.

Now split this into a product of two terms… …and form the series for each of them. For substitute

into the series for .

Expand term by term and simplify.

b Valid for all

The series for both and .

are valid for all

WORKED EXAMPLE 11.9

a Find the first three terms in the Maclaurin series for b Hence find the Maclaurin series up to the term in

for

. .

c State the interval in which the expansion is valid.

a

You know the series expansion for so you need to write in this form. Start by factorising .

Separate the , using

Then substitute

into the series of

Expand and simplify.

Again, you need everything in the form of .

b

First, use the laws of logs. You know the series expansion for the second term from part a. For the first term, substitute series for .

So:

into the

Now put both series together.

c Since

is valid when

:

Find the interval of validity separately for each function.

is valid when This is when is valid when This is when Therefore,

is valid when

For both to be valid, you need the smaller interval.

.

EXERCISE 11D 1

Find the first three non-zero terms of the Maclaurin series for each expression. a i

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

By first manipulating it into an appropriate form, find the first three non-zero terms of the Maclaurin series for each expression. a i ii b i ii c i ii

3

By combining Maclaurin series of different functions, find the series expansion as far as the term in for each expression. a i ii b i ii c i ii

4

Find the Maclaurin series for

5

Find the Maclaurin series as far as the term in

6

Show that

7

a Find the first two non-zero terms of the Maclaurin series for b Hence find the Maclaurin series of

8

a By using the Maclaurin series for in .

and state the interval in which the series is valid. for

. . .

up to and including the term in

.

, find the series expansion for

b Hence find the first two non-zero terms of the expansion of

up to the term

.

c Use your result from b to find the first two non-zero terms of the series for

.

9

10

a Find the first four terms of the Maclaurin series for

.

b Find the equation of the tangent to

at

.

a Find the Maclaurin series for

, stating the interval in which the series is valid.

b Use the first three terms of this series to estimate the value of used.

, stating the value of

Checklist of learning and understanding

The formulae for the sums of integers, squares and cubes:

You can manipulate series in these ways:

where is a constant. Method of differences If the general term of a series,

, can be written in the form

, then:

Maclaurin series for some common functions and the values of for which they are valid include: •

all

• •

all

•

all

•



Mixed practice 11 1

Find an expression for

.

Choose from these options. A B C D 2

Use the formulae for

3

Use the formulae for

4

a Show that

and

and

to show that

.

to find the value of

.

, where and are integers to be found.

b Hence find the value of

. [©AQA 2013]

5

a Show that

.

b Hence show that 6

, where and are integers to be found.

a Write down the expansion of

in ascending powers of up to and including the term in

b Hence, or otherwise, find the term in

.

in the expansion, in ascending powers of , of

. [©AQA 2013] 7

Find the set of values of for which the Maclaurin series of the function

is valid.

Choose from these options. A

All

B C D 8

a Show that

.

b Hence find the sum of

9

a Show that rational number and b

.

can be expressed in the form

, where is a

and are integers.

Show that there is exactly one positive integer for which

.

[©AQA 2010] 10

a Show that

.

b Hence show that 11

.

a Show that

.

b Find c Hence find 12

. .

a Given that b

, show that

.

Use the method of differences to show that

. [©AQA 2013]

13

a Find the first three terms in the expansion, in ascending powers of , of

.

b Write down the first three terms in the expansion, in ascending powers of , of c Show that, for small values of ,

.

. [©AQA 2010]

14

a Express

in partial fractions.

b Hence find an expression for 15

Use the method of differences to show that are constants to be found.



.

where , and

12 Proof by induction In this chapter you will learn how to: use the principle of induction to prove that patterns continue forever apply proof by induction to series apply proof by induction to matrices apply proof by induction to divisibility problems apply proof by induction to inequalities adapt the method to solve problems in a range of other contexts.

Before you start … GCSE

You should be able to use laws of indices.

1 a Given that , express in terms of . b Simplify

Chapter 7

You should be able to multiply matrices.

2 Given that

Chapter 11

You should be able to work with sigma notation for series.

3 Evaluate

A Level Mathematics Student Book 1, Chapter 9

You should be able to work with factorials.

4 Simplify

.

find

.

.

What is induction? Mathematicians and scientists are very interested in patterns. In science, you can observe patterns and conjecture a general rule; this is called inductive reasoning. There is no way to prove that this rule is correct, and the pattern may not continue forever. In mathematics, you have so far used deductive reasoning, where you start from known facts and use logic to derive new results, but sometimes a potential new result is based on observations of a pattern. In that case, you need a way of proving that the pattern continues indefinitely. One of the most powerful ways of doing this is a method called induction.

Section 1: The principle of induction Sequences often produce interesting patterns. Consider adding up consecutive odd numbers:

But you can’t keep checking forever to be sure that this pattern continues. Suppose that you have checked (by direct calculation) that the pattern continues up to the number; so you know that the sum of the first odd numbers is

To check that the pattern continues to the first numbers from to

th odd

odd numbers, you do not have to add all the odd

. You can use the result you already have, so

Building upon the previous work you have done, rather than starting all over again, is called an inductive step. In this example it seems fairly straightforward but in other problems this can seem very difficult.

Did you know This example is the first documented example of proof by induction, by the Italian mathematician Francesco Maurolico in 1575.

Now suppose that you have checked the pattern for the first odd numbers. This means that

To check that the pattern still holds for the first odd numbers, you can use the result for the first odd numbers (which you assume has already been checked):

Therefore the pattern still holds for the first

odd numbers.

In summary, for the pattern, ‘The sum of the first odd numbers equals the pattern holds for the first case (when

’, you have found:

)

if  you can check the pattern up to some whole number , then it follows that it will also hold for

.

Does this prove that the pattern continues forever? Yes! You know that it holds for

; but if it holds for any number , then it holds for

for . But if it holds for it therefore holds for reach any number , however large.

, so it therefore holds

and so on. You can continue this process to

The reasoning described in the last paragraph is called the Principle of Mathematical Induction.

Key point 12.1 To prove a statement (or rule) about a positive integer : 1

Prove that the initial case is true.

2

Assume that the th case is true.

3

Show that if the proposition is true for , it is also true for

4

Write a conclusion.

.

The hardest step is undoubtedly . To do this you need to make a link between one proposition and the next – the inductive step. The exact way you do this depends upon the type of problem. In the following sections you will see how to apply proof by induction in various contexts.

Section 2: Induction and series In Chapter 11 you used the method of differences to find the formula for the sum of the first terms of certain sequences. To do this, however, you needed a way of splitting the general term of the sequence into a difference. In this section, you will prove results for a far wider class of series (including the formulae you assumed for and ) by induction.

Rewind See Chapter 11, Section 1, for a reminder of sigma notation.

Tip You must lay your proof out clearly, with the structure given in Key point 12.1, to show that you are following the correct logic.

Key point 12.2 When using induction to prove a result about series, use:

WORKED EXAMPLE 12.1

Prove by induction that

for all

For

:

Show that the statement is true for the starting value (in this case, ).

So, the result is true for

.

Assume that the result is true for

Let

.

:

:

State the assumption for

Consider

and relate it to .

.

by using

Substitute in the result for (assumed to be true). Combine this into one fraction and simplify. It is always a good idea to take out any common factors.

Show that this is in the required form by separating out in each place it occurs. So, the result is true for The result is true for it is also true for the result is true for all

, and if it is true for . Therefore,

Make sure you write a conclusion.

, by induction.

Common error In the conclusion, do not write: ‘The statement is true for and true for must be correct: ‘If the statement is true for , then it is true for .’

EXERCISE 12A

.’ The logic

EXERCISE 12A 1

Prove by induction that, for all

2

Prove by induction that, for all integers

3

Using mathematical induction, prove that, for all positive integers,

4

Prove by induction that, for all integers

5

Use mathematical induction to show that, for all integers

6

Prove by induction that, for all

7

Using mathematical induction, prove that, for all integers

8

Prove by induction that, for all positive integers,

9

Prove, using mathematical induction, that, for all

10



,

,

,

,

,

Prove by induction that, for all integers

,

,

,

Section 3: Induction and matrices With matrices, the inductive step involves relating one power to the next one.

Rewind See Chapter 7, Section 2, for a reminder about matrix multiplication.

Key point 12.3 When using induction to prove a result about powers of matrices, use:

WORKED EXAMPLE 12.2

Use mathematical induction to prove that

When

:

.

Show that the statement is true for the starting value (in this case, ).

So the result is true for

.

Assume that the result is true for

Let

for all integers

:

:

State the assumption for

Consider

and relate it to

.

using

.

Substitute in the result for true).

(assumed to be

Multiply the two matrices together and simplify. You must show sufficient detail of your calculation.

Show that this is clearly in the required form by separating out in each place it occurs. Hence the result is true for

.

The result is true for , and if it is true for it is also true for . Therefore, the result is true for all , by induction.

EXERCISE 12B

Always write a conclusion.

EXERCISE 12B 1

Use mathematical induction to prove that, for all positive integers ,

2

Prove by induction that, for all

,

Tip Remember that the symbol is used to denote the set of all integers (whole numbers) and the set of positive integers.

3

Let

4

Let

5

Given that

6

Using mathematical induction, show that, for all

7

Prove by induction that, for all integers

8

Matrix A is given by

. Use induction to prove that, for all

. Prove by induction that, for all positive integer powers ,

a Find the matrix

, prove by induction that, for all

,

,

,

,

. Let and be real numbers. , where I is the

identity matrix.

b Use mathematical induction to prove that , for all integers



,

.

,

Section 4: Induction and divisibility Number theory is an important area of pure mathematics: it is concerned with properties of natural numbers. One of the important tasks in number theory is studying divisibility.

Tip Note that, in this example, the first value of is . Consider the expression

. Induction does not have to start from

for

Looking at the first few values of :

It looks as if

is divisible by for all values of . You can use mathematical induction to prove this.

Key point 12.4 When using induction to prove that an expression is divisible by an integer : write the assumption as

for some integer

isolate any term in the expression for assumption.

that appears in

and substitute in from the

WORKED EXAMPLE 12.3

The expression Prove that

So

is defined by

for all integers .

is divisible by for all integers

. Show that the statement is true for the starting value (in this case, ).

is divisible by .

Assume that

is divisible by :

for some Consider

State the assumption for

.

.

:

Consider and relate it to by isolating, say, . Substitute in the result for (assumed to be true). Simplify, working towards taking out a factor of .

So

is divisible by . is divisible by , and if is divisible by , then so is . Therefore is divisible by for all integers ,

Write a conclusion.

by induction.

Sometimes it can be more difficult to manipulate the expression for

to find the factor you need.

WORKED EXAMPLE 12.4

Use mathematical induction to prove that, for all positive integers , When

:

is divisible by

.

Show that the statement is true for the starting value (in this case, ).

which is divisible by

so the statement is true for

. Assume that the statement is true for

:

State the assumption for

.

for some integer . Let

Consider and relate it to by isolating, say, .

:

Substitute in the result for (assumed to be true). Work towards finding a common factor of . Factorise from the second and third terms and see what is left. which is divisible by .

so the statement is true for

The statement is true for and if it is true for then it is also true for . Therefore the statement is true for all , by induction.

EXERCISE 12C

Write a conclusion.

EXERCISE 12C 1

Prove by induction that

is divisible by for all

.

2

Prove by induction that

is divisible by for all

.

3

Using mathematical induction, prove that

4

Use induction to prove that

is divisible by

5

Show by using induction that

is divisible by for all integers

6

Using mathematical induction, prove that

7

Use induction to show that

8

Prove, using mathematical induction, that .

9

Prove by induction that

10



is divisible by for all for all integers

. .

is divisible by for all integers

is divisible by

is divisible by

.

for all

.

.

is divisible by

for all positive integers

for all positive integers .

Prove that the sum of the cubes of any three consecutive integers is divisible by .

Section 5: Induction and inequalities One of the most surprising uses of mathematical induction is in proving inequalities. For example, consider the inequality . Trying the first few integer values of shows that this inequality is satisfied for . In fact, by looking at the graphs of and you can see that the inequality is satisfied for all real numbers greater than approximately . Note that the inequality cannot be solved algebraically. However, you can use induction to show that it holds for all integers .

WORKED EXAMPLE 12.5

Use induction to prove that the inequality For

:

Show that the statement it true for the starting value (in this case, ).

, so the inequality holds for Assume that the inequality holds for

Let

holds for all integers

. State the assumption for

:

Consider

:

.

and relate it to

Substitute in the result for be true).

.

(assumed to

Working backwards, you need the RHS to be , so separate the term. Use the fact that : then is greater than , so it is definitely greater than . , so the inequality holds for The inequality holds for and if it holds for , then it also holds for . Therefore, it holds for all integers by induction.

. Write a conclusion.

Many interesting inequalities involve the factorial function. For the inductive step you often need to make a link between and

Tip In order to get the expression for into the final form that you need, it is often useful to work backwards from the result. Just make sure this doesn’t interrupt the logic of your proof.

Key point 12.5 When using induction to prove a result involving factorials, use:

WORKED EXAMPLE 12.6

Prove that When

for all integers

.

:

Show that the statement is true for the starting value (in this case, ).

so the inequality holds for Assume the result is true for

Let

. State the assumption for

:

Consider to using

:

and relate it

Substitute in the result for (assumed to be true).    (since

)

so the inequality holds for .The statement is true for and if it is true for then it is also true for . Therefore, it is true for all by induction.

EXERCISE 12D

.

Working backwards, you need the RHS to be , so divide through by . Write a conclusion.

EXERCISE 12D 1

Prove by induction that

2

Prove by induction that

3

Use mathematical induction to show that

4

Prove, using induction, that

5

Using mathematical induction, prove that

6

Prove by induction that

7

Use mathematical induction to prove that

8

a Solve the inequality

for all for

. for

for all

for all

.

. for all

.

for all

.

.

.

b Prove by induction that 9

.

for

.

Use mathematical induction to prove that

for all integers

.

Checklist of learning and understanding In mathematics, no matter how many times a rule works this will never be enough to prove it. Mathematical induction provides a logically rigorous way of showing that a pattern will continue forever. The steps of a mathematical induction proof are: 1 Prove that the initial case is true. 2 Assume that the  th case is true. 3

Show that if the proposition is true for , it is also true for

4

Write a conclusion.

.

The inductive step depends on the type of problem.



Problem type

Inductive step

Series

Use

Powers of matrices

Use

Divisibility

Substitute from

Factorials

Use

Anything else

There will probably be a hint in the previous part of the question.

into

Mixed practice 12 Tip Some questions in this exercise are different from all of the examples in this chapter. You need to be able to adapt familiar methods to new contexts! 1

Use mathematical induction to prove that, for all positive integers ,

2

Prove that

3

Prove by induction that

4

Use mathematical induction to prove that

5

Prove by induction that for positive integer ,

6

Prove by induction that, for all

7

Use mathematical induction to show that

8

Prove by induction that

9

Use induction to prove that, for all integers

is divisible by for all

.

for all integers

is a multiple of

for all

.

is a multiple of

for all

.

,

is divisible by

10

Prove by induction that, for all

11

Prove that, for all

12

Prove by induction that, for all integers

,

.

for all integers

.

,

,

has the form

where , are positive integers.

,

[© AQA 2008] 13

a Given that

, show that

where is an integer. b Prove by induction that

is divisible by for all integers

. [© AQA 2011]

14

The expression

is given by

a Show that b Prove by induction that

.

can be expressed in the form is a multiple of

, where is an integer.

for all integers

. [© AQA 2015]

15

Prove by induction that, for any matrix M,

for all positive integers .

(You may use the result that 16

for any two matrices and .)

Prove by induction that

for all integers

. [© AQA 2009]

17

a Show that, for any two complex numbers and , b Prove by induction that

18

.

for positive integer .

De Moivre’s theorem for complex numbers states that:

Use induction to prove de Moivre’s theorem for all

.

Rewind The result in Question 18 is an extension of the rule for multiplying complex numbers in modulus–argument form, which you learnt in Chapter 1, Section 5.

Fast forward You will use de Moivre’s theorem in Further Mathematics Student Book 2.

FOCUS ON … PROOF 2

Proving properties of identity and inverse matrices Throughout Chapters 7 and 8 you used various properties of identity and inverse matrices without really checking that they must be true. But in mathematics, all properties you use must be proved. This is particularly important when working with structures such as matrices, which have many properties similar to numbers but also some that are different. (For example, matrix multiplication is not commutative, and not all non-zero matrices have an inverse.) This means that our intuition about what ‘works’ may be wrong. Key point 7.8 stated that the identity matrix has the unique property that for every matrix . It is reasonable to ask: How can you be sure that there is only one matrix with this property? Would it be possible to have two different identity matrices? You are going to prove that the identity matrix is indeed unique. PROOF 11

If is a matrix such that Consider the product

By the definition of ,

for every matrix , then One way to prove that directly.

You are proposing that

Then

State your conclusion.

so

is to use the identity property

is the identity matrix so .

But by the proposed property of ,

.

for any conformable matrix

for any conformable matrix .

The statement in Key point 7.8 requires that both and . You know that matrix multiplication is generally not commutative ( for some matrices). So it is sensible to ask: Is there some other matrix , such that but ? (In advanced mathematics, you would say that is a ‘left identity’ but not a ‘right identity’.) It turns out that this is not the case. You are going to prove this result. PROOF 12

If is a matrix such that If

for all , then also

for all , then:

. Multiplication by , by definition, does not change a matrix.

(by the definition of the identity matrix ) Group the matrices as you like, but do not change the order. (by the proposed property of )

Use the given property of in its product with .

Hence if

, for all

then also

.

State your conclusion.

QUESTIONS 1

Use arguments similar to those in proofs 11 and 12 to prove these properties of inverse matrices. a Every non-singular matrix has a unique inverse. b If is a non-singular matrix with inverse then .

, and if is another matrix such that

,

c If in a non-singular matrix, then its ‘right inverse’ and its ‘left inverse’ are the same, so if then also 2

.

In proofs 11 and 12, it was stressed several times that you cannot simply ‘cancel’ a matrix from both sides of an equation. In other words, does not necessarily imply that . a Find a counterexample to prove that

is not true.

b Find an example of two non-zero matrices whose product is zero. (Hence it is not true that or



.)

FOCUS ON … PROBLEM SOLVING 2

Alternative approaches to calculating distances In Worked example 9.19, you looked at this problem.

Line has equation

and point has coordinates

.

Find the shortest distance from to . You used this strategy to solve the problem.

Strategy 1 Write down the position vector of a general point on the line (in terms of ):

To get the shortest distance,

must be perpendicular to the line. This means that

,

which gives an equation for :

Find and hence find the coordinates of . Calculate the distance

.

You can solve this problem in several different ways. Strategies 2 and 3 give two alternative solutions: decide for yourself which is the simpler.

Strategy 2 This also starts by expressing the vector this expression. Show that

in terms of , but you then use algebra or calculus to minimise

.

Either by completing the square, or by using differentiation, find the minimum value of Hence show that the required minimum distance

Strategy 3

is

.

.

This strategy uses a geometrical approach. You find the minimum distance directly, without finding the coordinates of (or the value of ).

The point

lies on the line. The shortest distance, marked , equals

Find the exact value of

.

Use the scalar product to show that Hence show that

.

and find the exact value of

.

.

QUESTIONS 1

Strategy 3 can be simplified slightly if you introduce the concept of projection. Consider a line with direction vector through the point with position vector , and a point off the line with position vector .

Let be the foot of the perpendicular from to the line. The length of

is the projection of

onto the line. a Use the right-angled triangle

to express the length

b Use the scalar product to write cos in terms of c Hence show that

and .

and .

.

d Use this formula for the given example. Hence use distance from to the line. 2

in terms of

The lines with equations

and

to find the shortest

are skew.

Find the shortest distance between them. Try solving the problem in three different ways. i Use the scalar product, as in Strategy 1. Let be a point on the first line and a point on the second line. The shortest distance is achieved when is perpendicular to both lines. a Write

in terms of and µ .

b Use the scalar product to obtain two equations for and µ . c Hence find the minimum distancess

.

ii Starting from part a, show that

.

How could you find the minimum value of this expression? iii You can use the scalar product to find a vector perpendicular to the direction vectors of both lines. This vector will then be parallel to

.

a Use the scalar product to form two equations and hence find a vector, b Point lies on the first line and point perpendicular to both lines, is the projection of

lies on the second line. Since is onto the line with direction . Adapt

the argument from Question 1 to show that the length of that the shortest distance between the two lines is

.

is

. Hence show

.

Fast forward In Further Mathematics Student Book 2 you will see how you can use the vector product to find a vector perpendicular to two other vectors.



FOCUS ON … MODELLING 2

Counting paths in networks: choosing the right representation Describing the situation There are many situations where you are interested in connections between places or people. Examples include road systems and social networks. For many purposes you can model such a situation as a network (or graph). This is a system of points (nodes or vertices) connected by lines (edges).

Did you know One of the most famous network models is the map of the London Underground (the Tube). This model only shows which nodes are connected by an edge; it ignores things like shape of the road, the exact position of towns, or the nature of ‘friendship’ in a social network (close friend, family, etc.) The model can be refined in several ways: the edges can have numbers associated with them (representing, for example, length, time or cost); sometimes edges can be directed (for example, in a network of one-way streets). There can also be more than one edge connecting two nodes; for example, there may be both a motorway and an A-road connecting two cities, or two people may be friends on several social networking sites. The network in this diagram has nodes (labelled

) and edges (labelled

).

Selecting a mathematical representation Although there are some problems that can be solved efficiently by using pictures (for example, in geometry and some probability), using equations is a far more common way in which to represent mathematical models. Furthermore, most network problems deal with very large networks and need to be solved using a computer; this means that you need to represent the information in a way that can be used in a computer program. A useful way of representing a network is as a matrix. Each row and column corresponds to a node, and each entry is the number of edges connecting two nodes. This is called the adjacency matrix for the network. For example, the network shown can be represented by this matrix.

The circled number shows that there are two edges between and . (Notice that since the edges are not directed the matrix is symmetric.) You will consider networks in which:

there can be more than one edge connecting two nodes a node can be connected to itself the edges are not directed; this means that if is connected to then is also connected to there are no numbers on the edges; you are only interested in the number of edges between two nodes, not their length or shape.

Using the model A path between two nodes is a sequence of edges starting at the first node and ending at the second. For example, one path between and is ; this path has length (it consists of three edges). A path can use the same edge more than once; for example,

is a path of length

between and . You are going to solve this problem. How can you find the number of paths of a given length between two given nodes? For example, in the network shown, there are six paths of length between and .

QUESTIONS A To get a feel for the problem, answer these questions. 1

2

List all the paths of length between: a

and

b

and .

List all the paths of length between: a

and

b

and .

Even listing all the paths of length seems reasonably complicated. How would you count all the paths of length ? What would you do if your network had nodes?

Tip You can actually prove this result by induction. To count the number of paths of length between and , think about using paths of length to get from to another node, and then taking the final step from that node to . The way you add up the numbers of different paths is the same as the way you multiply matrices. It turns out that, with the matrix representation, it is possible to prove an extremely useful result. The number of paths of length n between two nodes is given by the corresponding entry in the matrix

.

For example,

, which shows that there are paths of length between and (as you

have seen). QUESTIONS B 3

By representing the network as an adjacency matrix, find the number of paths:

a of length between and b of length between and . 4

Find the number of paths:

a of length between and b of length between and . 5

Can you adapt the model to answer the question about the number of paths in a directed network, such as this one?

Find the number of paths of length : a from to b from to c from to .



CROSS-TOPIC REVIEW EXERCISE 2 1

Find the determinant of

2

Find the value of for which the matrix

3

Consider the matrix a Find

. is singular.

.

.

is a square with vertices transforms into .

and

. The transformation with matrix

b Find the area of . 4

and are

non-singular matrices.

Simplify 5

.

a Find a vector equation of the line b The line

has equation

and

.

.

Find the point of intersection of 6

passing through the points

and

The position vectors of three points

.

and relative to an origin are given respectively by:

, and .

7

a Find the angle between

and

b Find the area of triangle

.

.

The shape of a vase can be modelled by rotating the curve with equation between and completely about the -axis.

The vase has a base. Find the volume of water needed to fill the vase, giving your answer as an exact value. [© AQA 2013] 8

The shaded region in the diagram is bounded by the curve with equation axis and the lines and .

, the -

This region is rotated through 9

about the -axis. Find the exact volume of the solid formed.

The matrix represents reflection in the plane rotation about the -axis. a By considering the images of the unit vectors

. The

matrix represents a

and , find:

i ii

.

b Find the single matrix that represents reflection in the plane the -axis. c Hence find the image of the point 10

Line passes through points

followed by rotation about

under this combination of transformations. and

.

a Find the vector equation of . b Given that intersects the line with equation . 11

Determine whether the lines

, find the value of

µ

and

intersect or are

skew. 12

The coordinates of the points and are The line has equation

and

respectively.

.

a Find the distance between and . b Find the acute angle between the lines

and . Give your answer to the nearest degree.

c The points and lie on such that the distance coordinates of .

is equal to the distance

. Find the

[©AQA 2008] 13

Use the formulae for

and

to show that

where and are constants to be found. 14

a Show that

, stating the value of the constant .

b Hence show that

.

c Find the value of 15

Prove by induction that

16

a Show that

. is divisible by for all positive integers .

b Prove by induction that, for all positive integers ,

[© AQA 2010] 17

The transformation represented by the matrix

has an invariant line

with

. a Find the value of . b Show that this is not a line of invariant points. 18

The equation of a straight line is

.

Find the shortest distance between and the point with coordinates to decimal places. 19

The line has Cartesian equation

. Give your answer

.

a Find a vector equation for . The line has vector equation

.

b Show that and do not intersect. c i Find points and on and respectively such that is

perpendicular to both lines.

ii Hence find the shortest distance between and . Give your answer in exact form. 20

The mean value of the function

between and is .

Find the value of . 21

a Find, up to the term in

, the Maclaurin series for

.

b Find the set of values for which the expansion is valid. c By evaluating the series in at an appropriate value of , find a rational approximation to . 22

a Express

in the form

b Prove by induction that, for all integers

where and are constants. ,

[©AQA 2014] 23

A linear transformation is represented by the matrix

.

Line has equation

. Line is the image of under the transformation

.

Find the value of such that is perpendicular to . 24

a The transformation of three-dimensional space is represented by the matrix . i Write down a vector equation for the line with Cartesian equation . ii Find a vector equation for the image of under , and deduce that it is a line through the origin. b The plane transformation is represented by the matrix is the line with equation i Find, in the form ii Deduce that lines.

, and

is the image of

, the Cartesian equation for

is parallel to

. under .

.

and find, in terms of , the distance between these two

[©AQA 2011] 25

The matrix is given by a Show that

. .

b Prove by induction that for

,

c The matrix represents a linear transformation. The image of the point under this transformation has coordinates . Find the coordinates of .

PRACTICE PAPER

1 hour 30 minutes, 80 marks 1

and Find

.

Choose from these options. A

[1 mark]

B C D 2

Find the equation of the asymptote of the curve Choose from these options. A

[1 mark]

B C D 3

Two complex numbers are given by your answer in the form , calculate:

and

. Showing your method clearly, and giving

a b 4

[2 marks] .

[3 marks]

Two vectors are given by

and

, where

.

a Show that and are perpendicular for all values of . b Find the value of for which 5

The cubic equation Use the formulae for

7

The region bounded by the curve rotated through

is parallel to the vector has roots

6

and

.

[3 marks]

and . Find the value of

[5 marks]

.

to show that

[6 marks]

.

, the -axis and the lines

and

is [5 marks]

about the -axis.

Show that the volume generated is 8

[2 marks]

, where

Show that the lines with Cartesian equations

and are integers to be found. and

[5 marks]

are skew. 9

Consider the matrix

.

a Find the value of for which is a singular matrix.

[3 marks]

b Given that is such that is non-singular, find 10

The complex number is given by

in terms of .

[2 marks]

.

a Find the modulus and the argument of

[2 marks]

b On an Argand diagram represent the locus of points satisfying the inequalities

[4 marks]

and 11

A curve has polar equation

.

a Find its Cartesian equation in the form b The curve intersects the line

.

[4 marks] [3 marks]

.

Find the value of at the points of intersection. 12

a Show that the Maclaurin series of the function

up to the term in

b A student claims that this series is valid for all

is

. Show by means of a

[2 marks]

counterexample that he is wrong. The hyperbola has equation 13

.

a Find the equations of the asymptotes to . The line

[6 marks]

[2 marks]

is a tangent to .

b i Show that ii Hence find the possible values of . 14

[5 marks]

A linear transformation is represented by the matrix A combined transformation consists of the reflection in the represented by .

. plane followed by the transformation

a Find the matrix representing the combined transformation. b Find the image of the point

[3 marks]

under the combined transformation.

c Show that the only invariant point under the combined transformation is 15

Prove by induction that

is divisible by

for all

.

[1 mark] .

[3 marks] [7 marks]

FORMULAE

Further Pure Mathematics Differentiation

Integration

Complex numbers

The roots of

are given by

for

Matrix transformations Anticlockwise rotation through about :

Reflection in the line

:

The matrices for rotations (in three dimensions) through an angle about one of the axes are:

for the -axis

for the -axis

for the -axis

Summations

Maclaurin’s series

for all for all

for all

Vectors The resolved part of in the direction of is

The vector product

If is the point with position vector

, then

the straight line through with direction vector (Cartesian form) or

has equation (vector product form)

the plane through and parallel to and has vector equation

Area of a sector (polar coordinates)

Hyperbolic functions

Conics Ellipse

Parabola

Hyperbola

Standard form

Parametric form Asymptotes



none

none

Further numerical integration The mid-ordinate rule: where Simpson’s rule: where

and is even

Numerical solution of differential equations For

and small , recurrence relations are:

Euler’s method: For

:

Euler’s method: Improved Euler method:

, where

Arc length (Cartesian coordinates)

(parametric form)

Surface area of revolution



(Cartesian coordinates)

(parametric form)

Pure mathematics Binomial series

where

Arithmetic series

Geometic series

for

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

Answers to exercises 1 Complex numbers BEFORE YOU START 1 2

3 EXERCISE 1A 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 3 a i ii b i ii c i ii d i ii e i ii 4 a i ii b i ii c 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 a i ii b i ii c i ii d i ii 8 a i ii b i ii c i ii 9 a i ii b i ii

10 11

or

12 13 a b 14 15 16 EXERCISE 1B 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 a b 7 a i ii b i ii c i ii d i ii 8 9 a b 10

11 Proof. 12 13 No solutions. 14 EXERCISE 1C 1 a i

ii

b i

ii

c i

ii

2 a i

ii

b i

ii

3 a i

ii

b i

ii

EXERCISE 1D 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 ii b i ii c i ii d i ii WORK IT OUT 1.1 Solution 1 is correct. EXERCISE 1E 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 3 a i ii b i ii c i ii 4 a i

ii

b i

ii

c i

ii

5 a i ii b i ii 6 a i ii b i ii

c i ii d i ii e i ii 7 a b 8 a b

c

9 a b c 10 EXERCISE 1F 1 a i



ii

b i

ii

c i

ii

d i

ii

2 a iii b i ii c i ii d i ii 3 a i

ii

b i

ii

4 a i

ii

b i

ii

c i

ii

5

6

7

8 a

b 9

10

11 12 WORK IT OUT 1.2 Solution 2 is correct. EXERCISE 1G 1 a i ii b i ii 2 a i ii b i ii c i ii d i ii 3 4 5 a i ii 6 Proof. MIXED PRACTICE 1

1 2 3 a i ii 4 5 a i ii b 6 a b

7 8

9 a b 10 11 Proof. 12 a b proof.

;

13

14 a, b

15 a Proof. b

c 16 17 Proof. 18 19 a Proof. b 20 Proof. 21

;

2 Roots of polynomials BEFORE YOU START 1 a Proof. b 2 a b 3 a b EXERCISE 2A 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

;

4 Proof; 5 a Proof. b 6 a Proof. b

7 a Proof. b

;

, , 8

,

or

, or

WORK IT OUT 2.1 Solution 3 is correct. EXERCISE 2B 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 5 a Proof. b 6 a b 7

;

8 a Proof. b c 9 EXERCISE 2C 1 a b c d e

f 2 a b c d EXERCISE 2D 1 a b c d 2 a b c 3 4 a b 5 a b 6 a b Proof. 7 8 Proof. 9 a b

; but if

were all real, the sum of their squares would be positive.

(This is an example of proof by contradiction.) EXERCISE 2E 1 a i ii b i ii 2 a i ii

b i ii c i ii d i ii 3 a b 4 a b 5 6 7 8 9 a

;

b 10 a b 11 12 a i ii WORK IT OUT 2.2 Solution 2 is correct. EXERCISE 2F 1 a i ii b i ii c i ii 2 3 4 a Proof. b 5 a b

6 7 8 a b MIXED PRACTICE 2 1 D 2 C 3 4 5 a

;

b 6 7 a

;

b Proof. c 8 9 a b 10 11 a Proof. b 12 13 14 a i ii b Proof. c i ii d 15 a i ii Proof. b i–ii Proof. 16 a i–ii  Proof. b i ii Proof. c i Proof.

ii iii

3 The ellipse, hyperbola and parabola BEFORE YOU START 1 2 3

4 a b 5 EXERCISE 3A 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

ii

b i

ii

c i

ii

d i

iii

4 a i

ii

b i

ii

c i

ii

5 a i ii b i ii c i

ii

d i

ii

6 a i Vertices asymptotes

;

ii Vertices asymptotes

b i Vertex

ii Vertex

;

c i

ii

d i Vertices

ii Vertices

7 8 9 Proof. EXERCISE 3B 1 2 a

b 3 4 a Proof; b 5 a b

6 a

b 7 a b Proof. 8 a b Proof. EXERCISE 3C 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 4 5 a

b 6 7 8

Vertices

9 a Vertices

b 10 11 a b 12 13 a b c d 14 a

; asymptotes

; asymptotes

b c MIXED PRACTICE 3 1 2 3 4 5 6 a b c 7 a Proof. b c 8 9 a b c d

e 10 a b c d 11

12 a

b 13 a

is at

b i Proof. ii iii 14 15 16 a Proof. b 17 a

b Proof. c d

; is at

4 Rational functions and inequalities BEFORE YOU START 1 2

or

3 4

5 6 Translation units to the right. 7

and

8 EXERCISE 4A 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

or

iii d i

or

ii

or

3

or

4

or

5 a Proof. b

,

6

or

7

;

8

;

;

;

9 10 a For example, b For example, EXERCISE 4B 1 a i

ii

b i

,

,

;

;

,

,

,

;

,

ii

2 a i ii b i ii 3 a i ii b i ii 4 a i ii b i ii 5 a

,

;

b i

ii 6 a

b 7 a b i

ii

or

or

8 a i ii b If , then the numerator is a multiple of the denominator, and function. 9

;

;

is a degenerate rational

10 a b c EXERCISE 4C 1 a i

;

;

ii ,

b i

;

,

ii

c i

ii

;

,

;

;

,

,

d i

;

ii

2 a i ii b i No stationary points. ii No stationary points. c i ii

;

3 a

b 4 5 6 a i ii b i Proof. ii c

7

8 9 a Proof. b

10 a b EXERCISE 4D 1 a i

ii

;

;

b i

ii

,

;

;

,

,

2 a i

;

ii

; none

b i

;

ii

;

3 a Proof; b c d e

4 a b Proof. c

; ;

MIXED PRACTICE 4 1 2 3 4 a

,

b

5 a i ii, iii

b i ii 6 a Proof. b 7 a

;

,

b

c 8 a b

9 10 a b 11 12 a b c 13 a b i

ii

c 14 a b Proof. c 15

5 Hyperbolic functions BEFORE YOU START 1 2 Vertical stretch factor three and horizontal translation one to the left. 3 4 5 6 EXERCISE 5A 1 a i ii b i ii c i ii d i ii e i Not possible. ii Not possible. 2 a i ii b i ii c i ii d i No solution. ii No solution. e i ii 3 a i ii b i ii c i

ii d i Doesn’t exist. ii Doesn’t exist. e i ii 4

or

5 6 7 8 9 10–12  Proof. EXERCISE 5B 1 Proof. 2 3–8 Proof. 9 a Proof. b 10 Proof. WORK IT OUT 5.1 Solution 3 is correct. EXERCISE 5C 1 2 3 4 5 6 7 8 9 10 11 a Proof.

b 12 MIXED PRACTICE 5 1 2 3 4 5 6 7 a b 8

or

9 10 11 Proof. 12 13 Proof. 14 a

b Proof. c i Proof. ii 15 a Proof. b 16 a Proof. b c

17

or

18 Proof. 19 Proof. 20 a Proof. b c

.

6 Polar coordinates BEFORE YOU START 1 a b 2 a b EXERCISE 6A 1 a–d

2 a i Distance

; area

ii Distance

; area

b i Distance

; area

ii Distance

; area

c i Distance

; area

ii Distance

; area

3 a i

ii

b i

ii

c i

ii

4 a

b

c

5 a b

6 a Proof.

b

c EXERCISE 6B 1 a i Maximum

ii Maximum

b i Maxima

; minimum

; minimum

and

;  

minima

ii Maxima minima

2 a i

ii

and

and and

;

b i

ii

3 a i

ii

b i

ii

4 a Maximum

; minimum

b

5 a b c

6 a

b 7 a

and

b Maximum 8 a Largest b

9 a

at when

and ; smallest

when

.

b

10 a Maximum is b

11

; minimum is

12

EXERCISE 6C 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 5 a b c 6 7 8 Proof. 9 a b MIXED PRACTICE 6 1 2 3 a b 4

5 6 7 8 a

and

b

c 9 a

b 10 a b c

and

11 a

b 12

13 a Proof. b 14 15 a Proof. b i ii c

Focus on ... Proof 1 1–4 Proof.

Focus on ... Problem solving 1 1 2 3 Discussion.

,

Focus on ... Modelling 1 1 2 a b The resonant current would grow without limit. c Even the wires will have some resistance. 3 The width of the resonant peak is smaller, so there will be less interference from stations with slightly different frequencies.

4 a

b

Cross-topic review exercise 1 1 2 a i ii b Proof. 3 4 a b 5 6 a b 7 a b

c 8 In

9 ln or 10 11 a b

12 a

b 13 a b c

d 14 a Proof. b i ii Proof. iii 15 a b Proof. c

;

16 a

b c 17 Proof. 18 a Proof. b 19 a b c 20 a b

c 21 a i ii iii Proof. b i Proof. ii 22 a b i Proof. ii Proof;

.

c

23 24 a

;

b i ii iii Proof.

7 Matrices BEFORE YOU START 1 2 EXERCISE 7A 1 a i ii iii iv b i ii

iii

iv

2 a i ii

iii

b i ii Not possible. iii c i ii

iii

d i ii

iii

3 a b Not possible c Not possible. d e Not possible. f

4 a b c d 5 a

,

b

,

c

,

d No solution. 6 a

b

c

d

7 8

or

9 10 Switching rows for columns before or after addition makes no difference; the same elements are added and the sum ends in the same position either way. EXERCISE 7B 1 a Yes; b No

c No d Yes; e Yes; f

Yes;

g Yes; h Yes; i

Yes;

j

Yes;

k Yes; 2 a b Not possible c Not possible. d e f g h

i

j

k 3 a No b Yes c Yes d Yes e No f 4 a b c d 5 a

No

b c 6 a b c

7 a

b

c

8 a b 9 a i ii b Dimension mismatch. 10 11 Proof. 12 13 a

b Proof. 14 Proof; the result always applies for matrices and where the product EXERCISE 7C 1 a i ii b i ii c i ii d i

exists.

ii 2

; Proof.

3

or

4

or

EXERCISE 7D 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 c d e f 4 5 a

, ,

b 6 7 8 a b 9 10 a Proof. b 11 12 a b 13 Proof. 14 a b c EXERCISE 7E 1 a i ii b i ii c i ii 2 3 a b 4 a b MIXED PRACTICE 7 1 2 3 4 a

b Proof; for example:

5 6 a b 7 a b Proof; c Proof. 8 a i ii b 9 a Proof. b 10 a Proof; b 11 a b Proof. c 12 a Proof. b For example: Importantly. the second column of must be a multiple of the first, and the upper row of must be a multiple of the lower. 13 14 a Proof. b For example:

8 Matrix transformations BEFORE YOU START 1 a b c 2 3 4 a b 5 EXERCISE 8A 1 a

b

c

d

2 a b c d 3 a Reflection;

.

b Reflection;

.

c Neither. d Reflection; e Rotation; f

Neither.

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

;

rotation about the origin.

b A rotation through followed by a rotation through is equivalent to a rotation through c

rotation about the origin.

d Two successive reflections are always equal to a rotation, of double the angle between the reflection lines, in the same direction as from the first line to the second. (In part c, to the axis is a rotation, so the combination is the same as a rotation). 7 a Reflection in the same line. b Rotation

clockwise about the origin.

8 9

.

10 a b c

square units

EXERCISE 8B 1 a b c d 2 a b c d

or

3 a Stretch with scale factor , parallel to -axis.

.

b Stretch with scale factor , parallel to -axis. c Stretch with scale factor d Stretch with scale factor 4 a

b c d

e

f

g 5 a b c d e f g 6 a b c d e f g 7 8 a b

, parallel to -axis. , parallel to -axis

9 a b 10 a b EXERCISE 8C 1 a Line of invariant points b Invariant lines

; invariant line

and

.

c Invariant lines d Line of invariant points

; invariant line

2 a Line of invariant points

; invariant line

.

b None. c Line of invariant points d Invariant lines

; invariant line

.

.

e No invariant lines. f

Line of invariant points

; invariant line

3 a b No; is singular 4 a b

and

5 a

; Neither is a line of invariant points.

b 6 7 a Line of invariant points

; invariant line

b 8 a b i

so

ii Any point 9 a b EXERCISE 8D 1 a

does not exist. will have image at the origin.

.

b

c

d

2 a b c d 3 4 a

;

b 5 6 7

8

9 10 11 a b Proof. MIXED PRACTICE 8 1 2 a b Reflection in the plane 3 a Line of invariant points b Invariant line

. ; invariant line

.

4 a i Enlargement, scale factor −0.5. ii Stretch parallel to the y-axis, scale factor 4. b

5 a

-axis

b 6 7 a b 8 a Rotation about . b Reflection in

.

c Rotation about . d Identity transformation. e Reflection in 9

.

or

10 a b i ii Stretch parallel to the x-axis, scale factor

.

11 a i ii b c i Proof; ; line of reflection

ii 12 a i ii b c 13 a

;

b i ii iii For 14 a b i

:

; for

:

, ;the x-axis is invariant.

.

ii Enlargement with scale factor . 15 a b i Proof; ii Enlargement scale factor and rotation iii

about origin.

9 Further applications of vectors BEFORE YOU START 1 a b 2 3 4

and

WORK IT OUT 9.1 Solution 3 is correct. EXERCISE 9A 1 a i ii

b i

ii

c i ii

d i

ii

2 a i ii

b i

ii 3 a i Yes ii Yes

b i Yes ii No 4 a No b No c Yes d No 5 a b Proof. c 6 a

b No 7

8 9 a Proof. b 10 a

b 11 a

b c EXERCISE 9B 1 a i ii b i ii c i ii 2 a i ii

b i

ii

3 a i

ii

b i

ii

c i

ii 4 a Neither. b Parallel. c Neither. d Same line. 5 6 No 7 a

b 8 a b EXERCISE 9C 1 a i ii b i No intersection. ii No intersection. c i ii No intersection. 2 Proof.

3 Proof; 4 No 5 6 Proof; area is WORK IT OUT 9.2 Solution 2 is correct. EXERCISE 9D 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 No ii Yes b i Yes ii No 5 a i ii b i ii 6 a b

square units.

7 8 9 a

radians

b Proof. 10 a Proof. b c 11 a Proof. b c 12

13 a

b c d 14 a b c Proof. d e 15 16 17 a b Proof. c 18 a Proof. b c MIXED PRACTICE 9 1 2

3 a

b 4 Proof. 5 6

7 a

b Yes, at c 8 a b Proof. 9 No intersection (skew lines). 10 a b 11 a Proof. b Proof. c d 12 a Proof. b

c d

or

13 a i ii Proof. b

c 14

15 a

b 16 a Proof.

b

or

17 a b Proof. c d 18 a Proof. b i Proof. ii iii 19 a i ii b i

ii

10 Further calculus BEFORE YOU START 1 2 3 EXERCISE 10A 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

5 a b 6 7 8 9 10 11 a b 12 13 a b 14 Proof. 15 Proof. Use 16 EXERCISE 10B 1 a i ii b i ii c i ii 2 a i ii b i ii c i ii 3 4 5 a

.

b 6 Proof. 7 8 Proof. 9 a

b Curve is concave up. c Proof. 10 Not true. For example: MIXED PRACTICE 10 1 2 3 4 5 6 7 8 9 a b 10 11 Proof. 12 a Proof b 13 14 15 a

and

between

and

.

b c 16 a

b c

11 Series BEFORE YOU START 1 2 3 EXERCISE 11A 1 a i ii b i ii c i ii 2 a i

ii

b i

ii

c i

ii WORK IT OUT 11.1 Solution 2 is correct. EXERCISE 11B 1 a i ii b i ii 2 a i ii b i ii 3–4  Proof.

5 a b 6 a Proof. b 7 Proof; 8 a Proof. b 9 a Proof. b 10 Proof; EXERCISE 11C 1–2 Proof. 3 a−b i Proof. ii 4 a b Proof. c 5 Proof. 6 a Proof. b 7 a Proof. b c Proof. 8 9 a Proof. b c 10 a b Proof. EXERCISE 11D 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 i ii b i ii c i ii 4

;

5 6 Proof. 7 a b 8 a b

c 9 a b 10 a b

; Convergence is for ;

MIXED PRACTICE 11 1 2 Proof. 3 4 a Proof; b 5 a Proof. b Proof; 6 a b 7 8 a Proof. b 9 a Proof; b 10 a, b Proof. 11 a Proof. b c 12 a, b Proof. 13 a b c Proof. 14 a b 15 Proof;

12 Proof by induction BEFORE YOU START 1 a b 2 3 4 EXERCISE 12A 1–10 Proof. EXERCISE 12B 1–7 Proof. 8 a b Proof. EXERCISE 12C 1–10 Proof. EXERCISE 12D 1–7 Proof. 8 a

or

b Proof. 9 Proof. MIXED PRACTICE 12 1–12 Proof. 13 a Proof; b Proof. 14 a Proof; b Proof. 15–18 Proof. FOCUS ON ... PROOF 2 1, 2 Proof. FOCUS ON ... PROBLEM SOLVING 2 1 a b c Proof.

where is in the direction from to .

d 2 a i

ii

iii b Proof. c i ii Proof. FOCUS ON ... MODELLING 2 1 a b

2 a b

3 a b 4 a b 5 a b c CROSS-TOPIC REVIEW EXERCISE 2 1 2 3 a b 4 5 a b

6 a b 7 8

9 a i

ii

b c 10 a b 11 Skew. 12 a b c 13 Proof; 14 a Proof; b Proof. c 15 Proof. 16 a, b Proof. 17 a b Proof. 18 19 a b Proof. c i ii 20 21 a



b c 22 a b Proof. 23 24 a i

ii

b i ii Proof; 25 a, b Proof. c PRACTICE PAPER 1 2 3 a b 4 a Proof. b 5 6 Proof. 7 Proof; 8 Proof. 9 a b 10 a

b

11 a b 12 a Proof. b For example, 13 a b i Proof. ii

14 a b c Proof. 15 Proof.

Worked solutions for book chapters 1 Complex numbers 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 10 Comparing real and imaginary parts: Re: Im: Substituting: so 11 Expanding: Comparing real and imaginary parts:

or Im: Solutions are

or

12    Let Then Comparing real and imaginary parts: Re: Im: Substituting:

13 a Comparing real and imaginary parts: Re: Im:

Substituting:

b

14 Expanding: Comparing real and imaginary parts:

Substituting:

is real so discard the second factor.

Solution: 15 Expanding: Comparing real and imaginary parts:

Substituting:

Reject the introduced solution

.

Solution: 16 Expanding: Comparing real and imaginary parts:

Substituting:

is real so discard the second factor.

Solution:

EXERCISE 1B 8

Tip Two different methods are shown here; using complex conjugates or setting the solution equal to a complex number and then multiplying through and resolving to find and . Either method would be acceptable in an examination. Method 1: Multiplying by the complex conjugate

Method 2: Setting an unknown answer and multiplying through to compare parts Let Then Expanding: Comparing real and imaginary parts: Re: Re: Substituting:

Hence 9a

so

b Comparing real and imaginary parts:

Solution: 10

11

for

12 Let

for

13 Let

for

Require this to equal zero. This is not possible, since no value of will make the imaginary parts match. No solution for . 14

So

EXERCISE 1E

7a b

(Select angle in the second quadrant: So 8a

)

(Select angle in first quadrant:

(Select angle in fourth quadrant:

)

)

b

c

Select angle in first quadrant: Notice that

and

Tip For complex numbers, multiplication produces a complex number with modulus equal to the product of the moduli and argument equal to the sum of the arguments, so that You have seen this in question 8 and you will meet it formally in Section 5. 9a b c 10 Let

Using

for

:

So

EXERCISE 1F 5 In words: ‘The distance between and the point is greater than (the locus consists of points outside the circle with radius ) and less than or equal to (the locus also lies within the circle of radius , including its circumference)’.

6 In words: ‘The distance between and the point is equal to the distance between and the point ’ (the locus is the perpendicular bisector of and ).

7 In words: ‘The locus is all points between the two half-lines (but not including either) starting at the origin, with angles and

.

8a In words: ‘The distance between and the point is equal to the distance between and the point ’ (the locus is the perpendicular bisector of and ).

In words: ‘The locus is the half-line starting at the origin (not included), with angle

b The half line is the positive part of

.

.

The perpendicular bisector is the line

.

The intersection is clearly at the point ( complex number .

) on the Argand plane, which corresponds to the

9 In words: ‘The distance between and the point the circle of radius , centred at .

is less than ’, which describes the interior of

In words: ‘The locus is all points in the first quadrant lying between the half-line starting at the origin (not included), with angle and the positive real axis’.

10 This is equivalent to the Cartesian equation

on the Argand plane.

This is equivalent to the Cartesian equation

on the Argand plane.

In words: ‘The locus is all points in the first quadrant and all those in the second quadrant to the right of line .

11 In words: ‘The distance between and the point is equal to the distance between and the point ’, which is the perpendicular bisector of the line connecting and . This is the Cartesian line

in the Argand plane.

In words: ‘The distance between and the point is equal to the distance between and the point ’, which is the perpendicular bisector of the line connecting and . This is the Cartesian line The intersection of

and

in the Argand plane. is the point (

) which corresponds to

12 In words: ‘The distance between and the point the circle with radius and centre ).

equals ’ (the locus is the circumference of

The question requires the line of greatest argument to pass through a point in this locus; this will be the tangent to the left of the circle: the line

EXERCISE 1G

, which has angle from the positive real axis.

3

4

5a

b

6

But

since

has a period

and, for the same reason,

Mixed practice 1 1 Select angle in the second quadrant since (Answer C) 2 Let

so that

Then But since multiplication is commutative for complex numbers,

(Answer B)

3a b 4

5a i ii

Tip so add to the result of the arctan.

( s.f.) b ( s.f.) 6 a Using quadratic formula:

b

7 Let Then Comparing real and imaginary parts:

Solution: 8 In words: ‘The distance between and the point is greater than (outside the circle radius centred at ) and less than or equal to (within or on the circumference of the circle with radius

centred at

)’.

9a

So

and

b Require that so

and

,

Therefore 10

so the line from to is inclined at a

angle above the positive horizontal; that is,

for some positive value .

11

,

12 a

b Let

Then, if

for

, it follows that

13 So the line from to is at an angle anticlockwise from the positive horizontal greater than less than .

and

So lies at least units distant from on the Argand diagram.

14 a Find

by treating the complex numbers as position vectors.

is the reflection of a in the real axis. b In words: ‘The distance between and the point is equal to the distance between and the point ’ (the locus is the perpendicular bisector of and ).

15 a Substituting

so lies on

into the loci:

.

Tip Real part so lies on b

so add to the arctan. . is a circle centred at



with radius

is a half-line starting at (but not including)

positive horizontal (in Cartesian form, the line

with angle

for negative ).

from the

c The centre of the circle with locus

is

, which also lies on

.

Tip You can see this by considering the Cartesian equation for Therefore and are opposite ends of a diameter and so

16 Let

for

So Re 17 Using

Substituting:

:

. ,

18

19 a so

and arg

Tip so do not add .

Tip so do not add .

Then b Comparing this expression with the result from a: and

Tip There can be alternative expressions for this sine and cosine using nested surds instead of a sum of surds. Show that the expression nested surd equivalent form for 20 Let Then

But since 21 Let

,

is equivalent to

and find a



Worked solutions for book chapters 2 Roots of polynomials 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 4

So by the factor theorem, ( So

) is a factor of

.

.

Expanding and comparing coefficients:

so so is consistent with

Tip Since comparing coefficients gives rise to four equations for three unknowns, you will always have a check on your working. Use this check, even if you don't write it down in an examination.

Completing the square on the quadratic factor:

So the solutions to

are

5 a Let Then By the factor theorem, if ( so b So

) is a factor of

then

. .

Tip Use a different letter as the coefficient. NEVER reuse a letter to represent a new unknown constant within the same question! Expanding and comparing coefficients: so

so so is consistent

The quadratic has no real roots, but completing the square on the quadratic factor:

Using this you can find the complex factors:

6 a By the factor theorem,

is a factor of

.

b So Expanding and comparing coefficients:

so so is consistent

The roots of this quartic are

or

.

7a

By the factor theorem,

and

are both factors of

b So

Expanding and comparing coefficients:

so so is consistent is consistent

Completing the square on the quadratic factor:

Factorising into linear complex roots:

Solution to 8

a quadratic in Factorising:

.

or so

or

EXERCISE 2B 4 If

is a factor of this (By the factor theorem, if

then

is a root)

is a factor as well (Complex roots come in conjugate pairs).

The original function is a cubic, so the final factor is a real linear factor. So Expanding: Comparing coefficients:

so is consistent is consistent

The remaining two roots of the cubic are

and

.

5a b Complex roots of polynomials with real coefficients occur in conjugate pairs so root.

Expanding: Comparing coefficients:

is consistent is consistent

when

or

6 a Complex roots occur in conjugate pairs so b Expanding: Comparing coefficients:

so so is consistent

is also a root.

must also be a

is consistent

The remaining two roots are 7 Complex factors come in conjugate pairs. If and

are roots then so are

and

.

8 a b Complex factors come in conjugate pairs, so if (

) is a factor then so is (

).

Expanding:

Comparing coefficients:

so is consistent is consistent

c So

has solutions (given) and also

and

9 Complex roots come in conjugate pairs. So, if and The quartic is therefore

For example,

EXERCISE 2D 3 For a cubic Product of roots Sum of roots For the given cubic: Product of roots: Sum of roots:

4 For a cubic

are roots then so are

and

.

Product of roots Sum of root product pairs Sum of roots For the given cubic: Product of roots: Sum of root product pairs: Sum of roots: a b 5 For a quadratic Product of roots Sum of roots For the given quadratic: a Sum of roots: b Product of roots:

So 6 a Product of roots: b Sum of roots: Therefore Substituting into the answer to part (a): Rearranging: 7 Sum of roots: The sample mean will tend towards the mean of the roots for large sample size tending to infinity. Estimate of the mean: 8 Product of roots:

Sum of roots:

Substituting

:

9 Sum of roots: Sum of root product pairs: Product of roots: a b

is real and non-zero (if then the equation is not cubic and there are fewer than three roots, which contradicts the question). The right-hand side of the equation is therefore negative, in which case the left-hand side, which is the sum of the squared roots, must also be negative. This is not possible if all three are real values. Therefore, not all the roots are real. There must be a single real root and a conjugate pair of complex roots.

EXERCISE 2E 3 Complex roots for a polynomial with real coefficients will always occur in conjugate pairs. a The third root must be

.

b The cubic must have general form Expanding:

4 Complex roots for a polynomial with real coefficients will always occur in conjugate pairs. a The other two roots are b Sum of roots:

Product of roots

5 For a cubic Product of roots Sum of root product pairs Sum of roots For the given cubic:

and

.

Product of roots: Sum of root product pairs: Sum of roots: Then for the required cubic: Product of roots: Sum of root product pairs: Sum of roots: The required cubic is therefore (fixing

):

6 For the given cubic: Product of roots: Sum of root product pairs: Sum of roots: Then for the required cubic

:

Product of roots:

Sum of root product pairs:

Sum of roots: The required cubic is therefore (taking

to keep all integer coefficients):

7 For the given quadratic: Product of roots: Sum of roots: Then for the required quadratic

:

Product of roots: Sum of roots: The required quadratic is therefore (taking

8 For the given quartic: Sum of roots: Sum of product pairs:

to keep all integer coefficients):

Sum of product triples: Product of roots: For the required quartic

:

Sum of roots:

Sum of product pairs: Sum of product triples: Product of roots: The required quartic is therefore (taking

):

9 a Product of roots: Sum of roots:

b For required quadratic

:

Product of roots: Sum of roots: The required quartic is therefore (taking

to ensure integer coefficients):

10 Sum of roots: Sum of paired products of roots: Product of roots: a b For the required cubic

:

Sum of roots: Sum of paired products of roots

Product of roots: The required cubic is therefore (taking

to ensure integer coefficients):

Tip You can find the reciprocal root polynomial more simply by reversing the order of the coefficients. Consider Then (dividing by

):

where

.

If is a root of the cubic, then

must be a root of the cubic.

Here, the original polynomial was reciprocal roots can be immediately written as

, so the polynomial with the .

11 For the given cubic: Sum of roots: Sum of paired products of roots: Product of roots: For the required cubic

:

Sum of roots: Sum of paired products of roots: Product of roots: The required cubic is therefore (taking

to ensure integer coefficients):

12 a

b For the given cubic: Sum of roots: Sum of paired products of roots: Product of roots: For the required cubic

:

Sum of roots:

Sum of paired products of roots:

Product of roots: The required cubic is therefore (taking

EXERCISE 2F 2 Let Then

.

):

3 Let

.

Then

4 a Let Substituting into

b

:

so Then

or or

5 a Substituting

Require

are solutions to the original cubic.

:

so

Then

, as also required.

b

Then, since

, the solutions to the original cubic are

(repeated root) or . 6 Substituting

:

If the roots of this quadratic are and : Product of the roots: But the roots and are related to the original roots by the same transformation; so

Tip Unless the method is specified in a question, if you can see a rigorous alternative to the standard approach which seems faster, use it. In this case, a more direct route may be simpler. From the original quadratic: Sum of roots:

Product of roots: Then

7 Let

; then

Substituting:

Tip To simplify, isolate the root (or a multiple of it) on one side of the equals sign and then square.

This quadratic has roots 8 a Substituting

For

and

.

, and requiring that the terms in

to have zero coefficient,

and

have zero coefficients:

.

Then the quartic simplifies to b Factorising: or Then

gives the solutions to the original quartic as

or

.

Mixed practice 2 1 Sum of roots of equation

equals

.

So the sum of roots of the given equation equals , since the coefficient of (Answer D) 2 The equation in has repeated root . If

then

(Answer C)

3 Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root, then

Expanding: Comparing coefficients:

is also a root.

is .

is consistent is consistent. The cubic factorises as The roots are

(given),

and .

4 Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root then

is also a root.

5 a Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root then

is also a root.

The product of the two complex linear roots is therefore also a factor of b Then

for some linear factor

.

Comparing coefficients:

so is consistent is consistent. Then

, which has roots

6 Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root then

The quartic factorises as Expanding:

7 a Sum of roots: Product of roots: b

c For the new quadratic: Sum of roots:

is also a root.

.

Product of roots: The required quadratic is therefore (taking

to ensure integer coefficients):

8 Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root, then

is also a root.

The cubic factorises as

for some and .

Expanding: Comparing coefficients:

so is consistent is consistent. has roots

(given),

and

.

9 a Complex roots of polynomials with real coefficients occur in conjugate pairs. If

is a root, then

is also a root.

b Cubic factorises as Expanding:

. 10 Complex factors of polynomials with real coefficients occur in conjugate pairs. If

is a factor, then

The third factor must be linear:

is also a factor. .

So Expanding:

Comparing coefficients:

11 a Using binomial expansion on

:

Rearranging:

b From the original quadratic: Sum of roots: Product of roots: For the required quadratic of the form

:

Sum of roots: Product of roots: Taking

, the required quadratic is

12 Substituting

:

13 Let Substituting:

This cubic has roots which are half the roots of the original cubic. 14 a i ii

Sum of roots: Product of roots:

b

But from a i, this means that So

Tip Alternatively, by substituting each of the roots into the given cubic:

Summing gives:

Using a i and ii, this reduces to c i

.

Real coefficients means that complex roots occur in conjugate pairs so one root is The sum of the roots equals zero, so the remaining root must be

ii

is the sum of product pairs of the roots:

is the negative product of the three roots:

.

d Using substitution: let Then 15 a i

so

is the new cubic with the required roots.

Sum of roots: Product of roots:

ii

Writing the cubic in factorised form:

Expanding:

Comparing the coefficient of

:

Therefore b Assuming the roots are given in order, so that

:

Let the common ratio of the sequence , , be , so that i

ii

Then the product of the roots,

, so

The sum of the roots, So The sum of the product pairs of the roots, So But then:

16 a i

Therefore ii

Therefore b i

Sum of roots: Product of roots:

ii

Writing the cubic in factorised form and expanding:

Comparing the coefficients of Therefore c Sum of roots:

:

and

.

Sum of product pairs of roots: Product of roots: i

From a i:

ii

From a ii:

iii

Required cubic has the form Sum of roots: Sum of product pairs of roots: Product of roots: Required cubic (taking



to ensure integer coefficients) is

Worked solutions for book chapters 3 The ellipse, hyperbola and parabola 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

7 Ellipse general equation is Horizontal semi-major axis Vertical semi-minor axis (Answer C) 8 Asymptotes to So the asymptotes are

are

.

. (Answer C)

9 If lies on the ellipse then

.

Similarly, Now, since

, and both so that

and are positive, it follows that , which is independent of .

and

EXERCISE 3B 1 The intersection of a line and the parabola will have a single repeated root when the line is tangent. Substituting:

Discriminant

2 a Asymptotes are

forms a quadratic equation with zero discriminant.

b The intersection of a line and the hyperbola will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

3 The intersection of a line and the ellipse will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

4 a Substituting the line equation into the parabola:

There is a single repeated root at

, so the line is tangent to the parabola at

b At , the gradient of the parabola is , so the normal has gradient Normal has equation

.

or

Substituting into the parabola equation and isolating the known factor

So has coordinates

.

:

.

5 a The intersection of a line and the ellipse will have a single repeated root when the line is tangent. Substituting:

So has coordinates

.

b Normal has gradient Substituting

at so has equation and isolating the known factor

So has coordinates

, or

:

.

6 a Horizontal semi-major axis Vertical semi-minor axis

b Substituting the line equation into the ellipse equation:

For two distinct real roots, require the discriminant

7 a Positive gradient asymptote has equation b For a line with equation

for non-zero , the intersection with the hyperbola occurs when

This reduces to a linear equation

, which for

has a single solution.

8 a The intersection of a line and the hyperbola will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

b A line passing through the origin has equation Substituting

into the hyperbola equation gives

or

for some constant . , which has no roots and so the line

cannot be a tangent as it does not have any points in common with the hyperbola. Substituting

into the hyperbola equation:

For the line to be a tangent, this quadratic must have discriminant

This can only be true if , but the lines hyperbola, and cannot therefore be tangent to it.

.

are known to be the asymptotes of the

EXERCISE 3C 3 Original equation: Completing the square for the new equation: So

4 After the translation, the equation is Substituting

for the curve to pass through the origin:

so the positive value of is 5a

has vertices at After translation by and .

and

and the two axes are asymptotes.

, the vertices are at

and

b

6 Horizontal stretch with scale factor replaces with

.

The new curve is therefore So 7 Reflection through the line

switches and in the equation.

The second parabola is therefore

and the asymptotes are

Substituting

to find the intersection:

so

(or

The point has coordinates

) from which .

8 Completing the square:

The hyperbola

is centred at the origin and has asymptotes

The given curve is this hyperbola under a translation and asymptotes

9 a Hyperbola with vertices

, alternatively given as

and asymptotes

b Vertical stretch with scale factor replaces with so

If

lies on the curve then

, so has centre

has equation

and vertices

, vertices and

so

from which

Since no reflection is noted in the description of the transformation,

so

.

10 Horizontal stretch with scale factor replaces with so the new curve has equation: or If this is the same equation as matches):

, then comparing coefficients (since the

term already

11 a The intersection of the line and the ellipse will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

so b The tangents to the ellipse in part a have equations Translating the ellipse by

gives the new curve (replacing with

).

Applying the same translation to the tangent lines gives the new tangent lines with equations . 12 Asymptotes of the original hyperbola are

.

After the translation, these asymptotes have equations and Comparing with the given asymptote equation: so Therefore the other asymptote has equation 13 a

has equation Horizontal stretch with scale factor replaces with so Dividing through by

can also be written as

and comparing with the given form of

:

so b Appling the transformation, which replaces with

, maps

to

c Tangent of a circle at is perpendicular to the radius from the centre to : has gradient

so tangent has gradient

The equation of the tangent at is d The equation of the normal to

at is

so

.

Reversing the horizontal stretch replaces with and passes through (

so the normal to

) so has equation

at is parallel to

.

Tip Take care – although the normal to a circle will always go through its centre, the same is not true of an ellipse; the stretch will correctly adjust the gradient of the line so that it remains the gradient of the normal at the given point but will fix the line at the origin, rather than at the point on the perimeter as desired. 14 a The intersection of a line and the hyperbola will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

so

From the diagram, the line has positive gradient so b The centre of the hyperbola is at asymptotes have equations c Intersection of line

and the gradients of the asymptotes are and

so the

and the asymptotes:

or or So has coordinates

and has coordinates

.

Mixed practice 3 1 Hyperbola with axes as asymptotes: equation of the form 2 Ellipse general equation is

(Answer D)

, equivalent to

Horizontal semi-minor axis Vertical semi-major axis Equation is 3 Translation New equation is

(Answer A) : Replace by

and replace by

.

(Answer D)

4 The intersection of a line and the ellipse will have a single repeated root when the line is tangent. Substituting:

Discriminant

forms a quadratic equation with zero discriminant

5 The intersection of a line and the hyperbola will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant

6 a The intersection of a line and Substituting:

will have a single repeated root when the line is tangent.

forms a quadratic equation with zero discriminant

Discriminant

So from The point of contact is when b Reflection in the The equation of

, so is at

: Replace with is

c The tangent described is the reflection of the line found in part a so has equation 7 a The tangent to a circle at a point is perpendicular to the radial line centre. For

, is the origin and so

The tangent therefore has gradient

where is the circle

has gradient and so has equation

Multiplying by and rearranging: b Horizontal stretch with scale factor replaces with The new curve is therefore

. Comparing with

there is no reference to a reflection, assume c The image of under the stretch is tangent line after the stretch. Replacing with 8 Translation

:

so

,

so

.

and so the equation of the tangent will be the initial

or

: Replace with

and replace with

New curve has equation Multiplying through by and expanding: 9 a The intersection of a line and Substituting:

. Since

(Answer B)

will have a single repeated root when the line is tangent.

forms a quadratic equation with zero discriminant

Discriminant

b From (1): The point of contact is when c Translation

, so is at

: Replace with

has equation

and replace with

, so

crosses the When

.

at

,

and

, so

crosses the

at

d

e

is the image of under the translation, so the tangent at the translation: so

10 a Reflection through

: exchange and in the equation

has equation b Require c For

,

so has coordinates

.

so at the gradient is .

Equation of the tangent at is

d Reflecting back through

11 Let be the point

to find the tangent on

.

Distance between and the line

is

So Squaring and comparing coefficient of :

.

at :

will be the tangent at after

12 a

b Translation by

with

curve is shifted to the right and must pass through the origin so

has asymptotes Translation by

: replace with

Asymptotes of 13 a When

,

is at b i

ii

are so

or

and is at

Substituting

into the ellipse equation:

For the line to be tangent, there must be a single repeated root to quadratic ( ) so the discriminant must be zero. Discriminant so

iii

(selecting

)

With this value for , from ( ): Multiplying through by : , which has repeated root When

,

so has coordinates

14 Asymptotes of the original curve are with

(or equivalently

) Translation

and replace with

New curve has asymptotes or

(Answer B)

15 The two curves will intersect as long as the line Substituting

into the hyperbola equation:

intersects the curves.

: Replace

This will give a solution for as long as the denominator is positive, which (given is positive) means that . 16 a The intersection of a line and the ellipse will have a single repeated root when the line is tangent. Substituting:

forms a quadratic equation with zero discriminant

Discriminant Since

, cancelling gives

Expanding and collecting terms: Cancelling b Using part a, if has equation

so

then

and

So has four possible equations:

17 a

b Substituting

into the ellipse equation:

If there are two distinct intersections, there must be two roots to ( ) so the discriminant

c Translation by

: replace with

Expanding:

Comparing with the given equation:

and replace with

.

d The tangents to the original ellipse parallel to Translating these by vector



:

are

, from part a.

Worked solutions for book chapters 4 Rational functions and inequalities 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 3

is equivalent to Let



has roots

and

A positive cubic is less than zero to the left of the first root and between the second and third roots.

Solutions are 4 A positive cubic with roots , and will be greater than zero between the first two roots and to the right of the greatest root.

Solutions to

or

5 a Let so So by the factor theorem, constants , and . Expanding:

is a root of is a factor of

. so

for some

Comparing coefficients:

so

has roots

b

where So the solutions are 6 The quartic factorises as is always positive. So the roots are

.

Then 7 Let Then the positive cubic

must have a root at

, and for the given inequalities to be true.

So

and

8 Let Then

So 9

must have a root at and a repeated root at for the given inequalities to be true.

and and since all the given powers of are positive, there is no problem squaring: and the original problem only allows for

(due to the square root).

From the graph, the solution is 10 a A positive quartic will have a global minimum and a negative quartic will have a global maximum so for every quartic an inequality valid for the entire real line is possible.

A simple example is

for any

For example, b A quartic with two distinct equal extrema (two distinct equal minima or maxima) can be used. For example, so

has mimina at

and

has solutions

EXERCISE 4B 5 a Vertical asymptote when the denominator is zero: b i

Horizontal asymptote Axis intercepts at

ii

Intercepts of and

and

:

lies above the line for

and

6 a Vertical asymptote where denominator is zero: Horizontal asymptote at the limit of the ratio of lead terms of numerator and denominator: When When

, numerator is zero so

b The curve passes

in only one place.

gives Solution to 7 a is a tangent to

is so the equation for their intersection should give a repeated root.

Discriminant of this quadratic

b i

For

:

Vertical asymptote

horizontal asymptote

Axis intercept For

:

Vertical asymptote Axis intercepts

ii

, horizontal asymptote and

The graphs intersect where

:

From the graph,

8a i ii

for:

Vertical asymptote where denominator equals zero: Horizontal asymptote at the ratio of the coefficients:

b If then the numerator is a multiple of the denominator and function.

is a degenerate rational

Tip The degenerate rational is not exactly the same as the constant function because strictly speaking the degenerate rational is still undefined at the vertical asymptote , although in many circumstances, this would be ignored, and the continuous constant function would be used. 9

If is positive, then the graph should look something like this (the line may intersect the lower or upper branch of the hyperbola, or neither).

Depending which branch of the hyperbola this intersects, the solution would have the form (intersects neither, or is a tangent to lower)

or

,

(tangent to upper branch)

or

(intersects upper branch)

or

(intersects lower branch),

where and are the two intersection points, if such exist. If is negative, then the graph would look something like this:

This version of

must intersect the line on the upper branch of the hyperbola because it must

have a negative horizontal asymptote, so the solution will be of the form Comparing with the given solution set, it can only match branch with

or

(intersecting lower

).

So the upper intersection value is

,

and

is the lower intersection value.

and The quadratic factorises as so the lower root is 10 a Tangent to the curve: the intersection quadratic will have repeated roots

Repeated roots in (*): Determinant

b

The points and lie where quadratic (*) is a perfect square.

intersects the curve, with as given in a, so that

lies on the upper tangent line so has coordinates

.

lies on the lower tangent line so has coordinates

.

The distance

is then given by

c The derivative equals zero at max or min distance, which occurs for

Tip Reject as it isn’t valid in this context, since a negative gradient will not give tangent lines. At

, the distance

EXERCISE 4C

3a Vertical asymptote where denominator is zero: Horizontal asymptote at the ratio of the lead terms of numerator and denominator: When No -axis intercept.

b The curve passes gives Solutions to

in two locations: so are

or

4 Vertical asymptote is a root of the denominator.

Comparing:

and

.

The vertical asymptotes are therefore

(given) and

.

The horizontal asymptote is the ratio of the quadratic coefficients:

.

5 Checking the denominator: the vertical asymptotes Horizontal asymptote is Numerator has roots -axis intercept is

has roots

and

, so these are

Intersections of

and

, where

Solution is 6a i

Vertical asymptote where denominator is zero: No real solutions so there are no vertical asymptotes. Horizontal asymptote at the ratio of the lead terms of numerator and denominator:

ii

When

,

and when

Axis intercepts are b i

Intersection with

, numerator is zero so

and :

so at the intersections, Given there is at least one solution to this quadratic, the discriminant

ii

Turning point occurs when a horizontal line is tangent to the curve, so the discriminant equals zero. From (*): When When

, ,

so so

. Turning point . Turning point

c

7 Vertical asymptotes at

and

Horizontal asymptote Axis intercepts at

and

and since

8 Denominator Vertical asymptotes

and

Horizontal asymptote is Numerator

has discriminant

, so has no roots.

Only axis intercept is therefore

The only finite enclosed interval which can be the solution to the problem lies to the right of the asymptote . When

,

.

9 a No vertical asymptotes: denominator has no roots.

Tip Negative discriminant

.

No roots: numerator has no roots.

Tip Negative discriminant -intercept at

.  

Horizontal asymptote

 

has a single root for

Substituting

and

and



Single root: Discriminant equals zero

Substituting

and



Single root: discriminant equals zero   Then

and

together give

b Curve passes through Substituting and using

Substituting into

. and

.

Curve passes through

Substituting into If

. Substituting and using

gives the consistent result that then

So

and

.

. .

10 a Using

Then

b Fixing and solving for :

When is maximum, there will be only one value of , so the discriminant of this quadratic will be zero, and

For value .

will have the form

, the coefficient of is negative, so

Whereas, for Since you require that

EXERCISE 4D 3a

Tip

would factorise as

would factorise as

for some positive

for some positive value .

within the model, the maximum value of must equal

Either perform algebraic long division or equate the original function with the desired form, multiply through and compare coefficients. Both methods are shown here. Method 1: Algebraic long division – extract multiples of the denominator from the numerator

Method 2: Compare coefficients

Multiplying through by

Comparing coefficients: so so so

b c

Where there is a single repeated root, the rational function has a turning point. Determinant

Solving for to find the position of this maximum and minimum value: square for the given value of .

So the stationary points are: and

d

so the -axis intercept is The numerator of -axis intercepts are

has roots and

and

must be a perfect

Vertical asymptote is e

4a The oblique asymptote is b

.

Tip Two methods are given – using differentiation and using quadratic analysis. Method 1: Differentiation Using the Quotient Rule:

So there is no value for which

, and hence there are no stationary points.

Method 2: Quadratic analysis Suppose

.

Then

Since this is a quadratic equation, for each value there may be solution would indicate a stationary point. A single solution occurs where discriminant

or solutions where a single

.

No such values exist and so there cannot be any stationary points.

c

Mixed practice 4 1 Vertical asymptotes are at the roots of the denominator: (Answer A) 2 Positive cubic with roots at

or

and .

Positive cubic is less than zero for values of below the first root and between the second and third roots. or

(Answer A) 3 Positive quartic with roots at

and .

Positive quartic with four roots is greater than zero for values of less than the first root, greater than the fourth root or between the second and third roots. So

for

and

.

4a Denominator

so vertical asymptotes are

Numerator When

and

so roots are , so

is the -axis intercept.

Horizontal asymptote is the ratio of the

terms:

b

5a i

Vertical asymptote is at the root of the denominator: Horizontal asymptote is at the limit of the ratio of the lead terms of numerator and denominator:

ii iii

b i

Intersection occurs when

or ii

Using the solution to part b i and the graph in part a iii: when

or

6 a Let Then By the factor theorem, since

must be a factor of

.

b Factorising: The quadratic factor If If

then

, in which case

then

If If

has at least one real root when has solution

, in which case

then then

has only a single root at

, in which case

is

then

or

.

.

has solution

has three distinct roots and the solution to

So if the solution to

, so for

. has solution

.

consists of two regions.

.

7 a Vertical asymptotes are at the roots of the denominator: Horizontal asymptote is at the limit of the ratio of the lead terms of numerator and denominator: b

c From part b, the solution to If

consists of two regions.

then so

or

So the solution to

is

or

8 a Numerator

has roots and

.

and

so the axis intercepts are

Denominator so

has vertical asymptotes

Ratio of

and

coefficients gives horizontal asymptote

b

9 Graph of Denominator

so vertical asymptotes are

Horizontal asymptote is the ratio of the

coefficients:

-axis intercept is

Solving

Two roots: discriminant

Positive quadratic is greater than zero outside the roots so However, since it but the branch for

is a horizontal asymptote, and the branch of the curve for crosses it once, there can only be a single solution for

The true solution set is therefore 10

or

and

does not cross .

a Horizontal asymptote:

is the ratio of the

Vertical asymptotes: and

coefficients so

must be roots of the denominator,

so

Axis intercepts:

and

must be roots of the numerator,

so

b When

If

then the curve degenerates to

and does not have an asymptote at

so

. 11 Suppose that

.

Require that there is a solution for for any value . First check the special case

has a solution: this is at

.

A problem arises if the denominator is also equal to zero when

So cannot equal or

for there to be a solution to

.

Excluding these two values, rearrange to give

For a solution to all values of , require that the discriminant of

non-negative for all values :

This is a positive quadratic in ; to be non-negative for all , there must be at most one root, so the discriminant of this quadratic must be less than or equal to zero:

The initial working shows that the limits of the interval cannot be included in the solution, so 12 a Roots of the numerator are and so is b Denominator gives the horizontal asymptote c

Solve

.

so curve has vertical asymptote

Ratio of the

terms

Single root: discriminant

so

becomes

and is 13 a Vertical asymptotes

and

Horizontal asymptote b i ii

When

,



c Intersections of

and

:

Solution to the inequality: 14 a Denominator

has no roots so there are no vertical asymptotes.

Horizontal asymptote occurs at the ratio of the

coefficients:

b Intersect line and curve:

If line and curve intersect, then there must be at least one root of this quadratic. Discriminant

c

Positive quadratic is less than zero between the roots Substituting the boundary values into (*) to find stationary points: When

,

so

.

is a stationary point. When

,

so

.

is a stationary point. 15

has oblique asymptote

so

Rearranging:

where

Then

for some unknown constant . is a different unknown constant.

has vertical asymptotes

The vertical asymptotes must be roots of the denominator of

so

Then comparing coefficients:

Substituting: If

then

The first quadratic has a negative discriminant and so has no roots.

Tip On the graph, there is no intersection of

The second has roots At



.

Worked solutions for book chapters 5 Hyperbolic 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 5A

4

5

6

7

8

Reject

as it is outside the range of

9 10 Let

so that

.

Quadratic in :

Tip Select the positive root since you require that Taking natural logarithm of both sides: Hence 11 Let

, as required. so that

.

.

Taking natural logarithm of both sides: Hence

, as required.

12

Tip This is a proof that is an even function, where, for any given value of the range, the corresponding values in the domain come in pairs. The range of

is

For

and

Therefore,

and so

, so the domain of

is

.

. .

This first possible expression is shown to be non-negative (and so the other is non-positive, since they must sum to zero).

EXERCISE 5B

1 2

3

Tip Some identities are easier to prove if you start with the

4

Tip

and work towards the

.

Here the result from Worked example 5.3 has been used to simplify the numerator. 5

Tip The identities established in question and Worked example 5.4 have been used.

Dividing numerator and denominator by 6 Therefore 7

8

9a

so Similarly,

so

Adding these together:

b

10

, then

EXERCISE 5C 1 Using the definitions of cosh and sinh:

gives

So

and therefore

2 So 3

or So

4 Using the definitions of

and

:

5 Using the definitions of sinh and cosh:

So Multiplying through by

:

Reject negative solution. So 6 Using the definitions of sinh and cosh:

Multiplying through by and factorising:

So, 7 Using the identity

8 so Reject negative root. So Therefore,

:

Tip Since

, solutions can always be

given in the form 9 So Therefore 10 So Therefore So

11 a

so

Therefore,

and

, as required.

b Then So Expanding:

Multiplying through by and rearranging:

Solutions:

and

or

and

.

Tip Since the equations are symmetrical, it shouldn’t surprise you that the answers are also symmetrical. 12 Multiplying through by

and rearranging:

Quadratic in has at least one real root if the discriminant

:

Tip Since the quadratic can be expressed as , where both and must be positive, the symmetry line of the quadratic will always have a positive value for ex, so that, if there is at least one root, there must be at least one positive root and there is indeed a real solution for .

Mixed practice 5 1 So

(Answer C)

2 So

(Answer D)

3

4 So Therefore

Tip If you calculate the solution without using the logarithm formula for arcosh, you might get the result as your final answer instead. Is this a problem? Evaluate the two answers on your calculator and explain the result.

5

So

6

( So

as it is outside the range of

)

7a So b Multiplying through by and factorising:

So Reject negative solution. So 8 Multiplying through by and rearranging:

so 9 Rearranging and factorising: (rejecting the negative root as it is outside the range of or 10 Using the identity

: ( reject

as it is outside the range of cosh )

So 11

for all

12 Let So Then

13

)

14 a

b

So Therefore c i Dividing by

:

So the equation becomes So ii

, as required.

Factorising: So

(reject because the range of

or Then, using the formula found in :

15 a

b

is

)

16 a

So b If

 then 

c

17

so

From ( ):

.

Substituting into ( ): , so , so Solutions are 18 The logarithmic formula for Therefore:

is

19

Tip You could also answer this question using a hyperbolic identity from the A Level Further Mathematics course: Using

and taking tanh of both sides of the initial equation:

Rearranging:

20 a

So b From :

So c

.



Worked solutions for book chapters 6 Polar coordinates 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 4a

b

c

is the locus of points no more than units from the pole, which is the interior of the circle, centred at the origin, with radius (including the circumference).

is the locus of points between and units (inclusive) from the pole, which is the interior of an annulus, centred at the origin, with inner radius and outer radius . Both circumferences are included in the locus.

is the locus of points in the wedge of space between the (non-included) line at angle (this is the positive axis).

) anticlockwise to the (included) line at angle (this is the negative -

5 a Question requires values of for which for b

and

so the curve is not defined for these values of .

Tip Sketch using a table of values or by previous experience, using known properties of period and symmetry to minimise the work needed. Using a table of values:

Then evaluating further values using periodic property

6 a When

and symmetry property

and when

so both and lie on the curve. b Using a table of values:

Tip Can you convert the locus formula into a parametric Cartesian equation and prove that it is a circle? c

EXERCISE 6B

4a at

and

.

Maximum and minimum values occur at

and

, respectively.

b

5 a Tangent occurs when for b The curve is restricted to In the intervals give c

and

, the function would

so the curve is not defined for these values of .

Tip Sketch using a table of values or by previous experience, using known properties of period and symmetry to minimise the work needed. Using a table of values:

Then evaluating further values using periodic property and symmetry property

6a

b at

and

Maxima at

and

; minima at

7 a Tangents at the pole occur when so Then

.

and

b Maximum at

and

Minimum at So maximum 8a

and occurs at

has range

.

so the maximum value of is at

, and the minimum value is at

. b

Tip Sketch using a table of values or by previous experience, using known properties of symmetry to minimise the work needed. Using a table of values:

Then evaluating further values using symmetry property

9a

b

10 a at is the maximum value of . is the minimum value of . b

11

12

EXERCISE 6C 5a

b c

6

Since

gives

, the first solution is included in the second, so the equation is:

7

8

9a

b

Mixed practice 6 1 Minimum value of

is

so maximum is

(Answer D)

2

Tip The solution

is included within this, at

3 a Cartesian coordinates are

, so you do not need to state it separately

and

Distance b Area

4

(using sine rule for area):

Tip With no guidance, you may approach this question either by using a graphical calculator, or by a table of points which you plot and join, or by attempting to transform into a Cartesian equation. Be warned that it is unusual to have to transform a question into a simple Cartesian equation unless you are actively steered in that direction!

Method 1: Table of values

Tip Use

and repeat the shape for the second period of the function, rotated

.

Method 2: Transform to Cartesian equation

Tip As perhaps expected, although you can transform into a Cartesian equation, it does not particularly help! In this case, a calculator or join-the-dots approach is best. 5

6

7

8a

when

or

The tangents at the pole are

and

b

c

9a

b Maximum occurs when

, so has polar coordinate

Cartesian coordinates for this point are 10 a Tangents occur when

:

.

.

for b

; solutions are

is maximal when

, which occurs at

Polar coordinates of the points are c

11 a

b When

The point is 12

.

and

and and .

.

Tangents at the pole when

Curve crosses the initial line when

, i.e. at the point

.

13 a

b From part a:

Using the quadratic formula (and only selecting the positive root for

):

Tip Alternatively, at the second line in this working, use the fact that both sides:

14 Intersections:

to square root

so

Solutions:

One of the vertices, at

, has Cartesian coordinates

.

By symmetry, each of the two large rectangles has long side of length

and short side length .

Total shaded area is therefore equal to the central square (side length ) and four small rectangles, of dimension

.

Total shaded area 15 a When Hence b i

is indeed a point on .

The circle intersects when

is ii

and is

.

Cartesian coordinates of the points are: ,

,

The triangle is isosceles, and the line of symmetry is the The height is and the base length is The area of triangle c



is

.

.

.

Worked solutions for book chapters 7 Matrices 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

7 From the lower left-hand element: From the lower right-hand element: From the upper left-hand element: From the upper right-hand element: ,

,

so so so so

,

8

or Substituting into Solution is

when

; when

or

9 Upper left and lower right elements give simultaneous equations for and :

Substituting these values, the lower left and upper right elements give simultaneous equations for and :

10 Using algebra: Let

.

Then, using the standard notation that and th column:

means the element of matrix which lies in the th row

Tip Each element of is the sum of the corresponding elements in and . So

Tip The element in the ith row and jth column of the transpose is the element in the jth row and ith column of the original matrix. But

and

So Since all the elements match, Hence Using direct verbal reasoning: ‘Switching rows for columns before or after addition makes no difference; the same elements are added and the sum ends in the same position either way.’

EXERCISE 7B

5 a b c

6 a b c 7

Tip Parts and are shown with full working. This is included to give a detailed check on the method of multiplying matrices by hand. Since you are permitted to solve numerical

matrix multiplication by calculator in an examination, the answer alone would be sufficient.

a

b

c Using your calculator:

8 a b 9 a i ii

b No such product exists. In a matrix product must have the same number of columns as has rows. So,for a product to exist, must have as many rows as it has columns. It must be a square matrix. In this question, is a

matrix, not a square matrix.

10

The upper right-hand element gives: so All other elements are consistent. 11

The products are the same, in whichever order the matrices are written. The two matrices commute under multiplication.

Tip Each of these matrices (if not the zero matrix) represents a combination of an enlargement and a rotation, and, as such, the order of implementation would not be important.

12

For these two products to be equal, the upper right-hand elements must match (which will cause the lower left-hand elements also to match).

Tip Unless each of these matrices represents a combination of an enlargement and a reflection. For two reflections to commute, they must act through the same line, and so it follows that each must be a multiple of the other.

13 a

b From part a:

Close inspection shows that these are the same, verifying that associative.

matrix multiplication is

Tip Observing the patterns, you can see that each element of the triple product is formed of the sum of all possible element products , where and can take values or independently of each other. More succinctly (and in a manner which immediately extends to showing that all matrices are associative under multiplication: From the definition of matrix multiplication of two

Notice the central expression in this working, described 14 Algebraically: Let

matrices:

, which is the pattern

Then

by definition of matrix products.

So But

, again by definition.

So These are the same products as in Hence

.

for all and

This result always applies for matrices and where the product

EXERCISE 7C 3

or 4

or

EXERCISE 7D

4

so If

5 a

Non-zero determinant means that is non-singular. b

exists.

6

so If

then

7 , so 8 a

for all values of . Non-zero determinant, so is non-singular for all values of .

b 9 so If

then

10 a

. Therefore, by the definition of inverses,

is the inverse of

.

b 11 Taking the inverse of a product gives the inverse of each component of the product, in reverse order.

12 a

Tip Multiplication is associative. b 13

and commute So

.

Inverses exist, since neither nor is singular.

Left-multiplying and right-multiplying both sides by

:

Simplifying: Now left-multiplying and right-multiplying by

:

Simplifying: This is the property that

and

commute, as required.

14 a Since the determinant of a product is the product of the determinants:

Therefore, b If

then so

c Applying the rule that the determinant of a product is the product of the determinants: for positive integer . For negative integer

When

so

,

which is still equal to

So for any integer ,

EXERCISE 7E

2

There is no unique solution when this determinant

.

There are no solutions when:

There are no solutions when

or

3 a There is no unique solution if the determinant of the

matrix is zero.

b

4 a There is a unique solution if the determinant of the

matrix is non-zero.

Expanding about the first column:

b

Mixed practice 7 so

1

(Answer B)

2 (Answer C) 3

,

,

From Substituting into

:

Substituting into

:

Substituting into

:

4 a

if

so

b Commutativity is the exception rather than the rule, and it is likely that any randomly chosen pair of matrices will not commute (or pick and from part a with ), but an easy example is:

5

so so

6 a If is singular then

so

b If is non-singular then 7

, a b c

8 a i

Det For

singular,

so

ii b

so

9 a Left-multiply by

Then

and right multiply by

:

b

10 a So b So for

11 a b

Tip The negative sum of two squared values cannot be positive. c Det so require

and

:

is the only solution. 12 a Proof by contradiction: Suppose that is non-singular; then

exists.

so But any matrix multiplied by yields , from which that neither nor is the zero matrix .

, which contradicts the requirement

A similar argument shows that cannot be non-singular. b Since it follows that at least one of and is singular. is singular when some values

, so a general form for a singular

matrix is

for

and .

A simple example with no zero elements might be So, if

and

, you can see that

Importantly, the second column of must be a multiple of the first, and the upper row of must be a multiple of the lower. 13 because is non-singular.

From the upper right-hand element: so the determinant From the upper left-hand element: so

so 14 a Using components: Let

and

‒ ‒

b

‒

‒

‒

‒

Tip Most matrices whose determinants are values of the same magnitude and opposite sign would not sum to make a singular matrix. Try any matrix you like for , this is not a rare property.

For example:

for which

Seek a matrix for which For example: Then But



, as required.

Worked solutions for book chapters 8 Matrix transformations 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

6 a Rotation through angle (anticlockwise) has matrix For

, the matrix is

and for

.

, the matrix is

The composed transformation is This is the matrix representing a

.

. rotation about the origin.

b In the general case, a rotation through angle followed by a rotation through angle would result in a transformation:

So, the overall effect is a single rotation through angle c Reflection through

is given by matrix

Reflection through

is given by matrix

.

. .

followed by produces the transformation with matrix This is a rotation through

.

about the origin.

d A pair of successive reflections is always equal to a rotation, of double the angle between the reflection lines, in the same direction as from the first line to the second.

Tip In part rotation.

to the -axis is a

rotation, so the combination is the same as a

You can show this by taking the general case. A reflection through the line with angle anticlockwise from the positive horizontal is given by

So a reflection through line with angle followed by a reflection through the line with angle results in a transformation with matrix

This is a rotation through angle

.

7 a A reflection is self-inverse, so the inverse of a reflection through b Rotation

is the same reflection.

clockwise about the origin.

8 Rotation through

about the origin is given by matrix

If is the matrix consisting of the position vectors of the columns of .

The image points are 9 Reflection through

and

and , then the images

and

.

is given by matrix

.

Enlargement scale factor is given by matrix

.

b If has position vector then

so

.

.

c The area of the original rectangle is det

and

are

is given by matrix

The image points are

has coordinates

are

.

If is the matrix consisting of the position vectors of , and then the images the columns of .

10 a Reflection through

and

,

The area of the image will be the modulus of the product:

EXERCISE 8B

square units

EXERCISE 8B 7 Rotation

has matrix

So the image of

.

is

.

8 a A stretch with scale factor parallel to the

has matrix

So the vertices of the image are

:

and

.

so the ratio of the areas is .

b

9 a The angle of the line

is

The matrix for reflection through a line with angle above the horizontal axis is . If

then so

Tip Select positive root since the image of

b

should be above the

.

because a reflection is always self-inverse. If

then

The object point is 10 a

A stretch with scale factor

parallel to the

has matrix

.

b Then, using the matrix from part a,

So the vertices of the original triangle are

EXERCISE 8C

,

and

.

Tip In the working shown for finding an invariant line through the origin, the general equation is used, as in the worked examples. This does not allow for the possibility of being an invariant line. If you prefer to use a more general equation , you will avoid this issue but the working will look more complicated. Since this special case is easy to test for, it is fine to use . Be aware that any matrix with as an invariant line will have a zero as the upper right-hand entry, having general form

3 a If

.

is an invariant line, then

So Then

In fact, is a line of invariant points, since the image of the general point is the same point rather than a different point on the same line. b det

so there can only be one invariant line.

The matrix represents a projection onto the line

.

Tip If you are not confident to make this assertion, you can show algebraically that there is no other invariant line through the origin: Suppose that

For

is an invariant line; the general point is

to be an invariant line:

So the two apparent invariant lines are and . However, a quick check shows that any point on the line is always mapped to the origin, so this is a degenerate invariant line. 4 a Let

For

be an invariant line; the general point is

to be an invariant line:

The invariant lines are

and

.

For

to be an invariant line:

The invariant lines are

and

The two transformations have a common invariant line b

will be an invariant line of the product

For

For

For

and

be an invariant line; the general point is

.

to be an invariant line:

The invariant lines are For

.

to be an invariant line:

The invariant lines are 5 a Let

.

:

and , so this is not a line of invariant points.

:

, so this is also not a line of invariant points.

b The same lines must be invariant for transformation performed any multiple of times. The invariant lines are 6

and

A reflection through a line with angle anticlockwise from the The line

has angle

is given by matrix

Tip This is a clockwise rotation by . Require

to be invariant under

.

has matrix

.

is therefore an invariant line of . But a reflection only has two invariant lines through the origin: the reflection line itself, which is a line of invariant points, and the line perpendicular to that. Therefore either

or

First option:

Second option:

7 a Let

be an invariant line; the general point is

For

to be an invariant line:

The invariant lines are

and

For

so this is not a line of invariant points.

:

For

:

so this is a line of invariant points.

b Require

to be an invariant line so

8 a

For

to be an invariant line,

From the third element, Then b i

so

. and

Expanding down the first column:

is singular so

.

does not exist.

so

.

ii

A general point on the line

,

has position vector

, so the image of any point of the line is the origin. The whole line is mapped to the origin. 9 a If

is twinned with

, then

and also

.

and Substituting

into

and

into

, and cancelling :

b Any point on

is mapped to a point on

To verify that this is a twinning transformation: So any point on

is mapped to

, as required.

EXERCISE 8D 3 A reflection through

changes the sign of the coordinate but nothing else.

is 4 a

,

b The first and second columns of are just the vectors and respectively; the third column is . Therefore this is a reflection in the plane . 5

Image points are 6 Rotation

.

about the

so

is given by

.

is given by 7 Rotation

about the

. is given by

.

Image point has position vector

is given by

8 Rotation

.

.

about the

is given by

Reflection in the plane

is given by

.

.

followed by is given by

,

which is a reflection through the plane 9 Rotation

anticlockwise about the

Rotation

clockwise (which is

. is given by

.

anticlockwise) about the

followed by is given by

10 Rotation

.

about the

is given by

Reflection in the plane

is given by

.

.

followed by is given by

Tip This is a reflection in the plane

.

If has position vector , then Since

is given by

is a reflection, it is self-inverse and

so .

.

has coordinates

.

11 a Rotation

about the

is given by

Rotation

about the

is given by

.

.

b

Mixed practice 8 1 Testing each option: , so

is a line of invariant points.

, so

is an invariant line but not a line of invariant points.

and

, so

and

are images of each

other and neither is invariant. (Answer ) 2

,

a



b T represents a reflection in the plane

.

3 a For a point on the line

:

For

to be invariant:

For

:

so

is a line of invariant points

is an invariant line but not a line of invariant points since only the identity transformation has more than one. b For a point on the line

:

For

to be invariant:

For

:

so

is an invariant line but not a line of invariant points.

4 a i

represents an enlargement, scale factor

.

Tip This could also be seen as a rotation through about the origin and an enlargement with scale factor , but the question specifies that a single transformation should be described. represents a stretch with scale factor , parallel to the

ii

b A rotation through

about the origin in two dimensions has matrix

A rotation through

about the

5 a Rotation about the

in three dimensions will therefore have matrix

.

b The rotation matrix in the so

.

plane is

and

.

This is negative cosine so The rotation is 6

.

, so the transformation involves a reflection. Test for invariant lines: , so , so

is not invariant and cannot be the line of reflection. is a line of invariant points and is therefore the line of

reflection. (Answer ) ,

7 a b det So if

and det

.

, then det

det

So det det and therefore det 8 a

has the form

for

so represents a rotation b

about the origin.

has the form

for

so represents a reflection in the line

where

c

represents a rotation

d

equals the identity transformation (reflections are self-inverse).

about the origin.

e

which represents a reflection in

.

9 For

to be a line of invariant points of :

From

:

Then from so

:

or

10 a

so

Therefore

b i

det so

(reject

)

ii so represents a stretch, scale factor

, parallel to the

.

11 a i ii b

followed by :

c i so ii

has determinant factor .

So

so the enlargement must have determinant

and therefore scale

for

.

is an enlargement, scale factor

and a reflection in the line

12 a i so has coordinates ii

b i

so lies on the line of invariant points, which therefore has equation

ii For

to be invariant:

is also an invariant line. 13 a For

so

to be an invariant line of :

or

Invariant lines are b i

and

det det det

so ii

iii

14 a

Test

det A

or in :

so

is also an invariant line of .

For

, the other invariant line of is

For

, the other invariant line of is

.

If and commute then upper right element).



so

(from the upper left element) and

(from the

The remaining two elements are consistent with this solution. b i since

, where det

,

.

This has (the ) as a line of invariant points (and so an invariant line), since any point is mapped to itself. ii i.e.



, where is an enlargement about the origin with scale factor .

15 a

so

b i

so and then

ii so represents an enlargement scale factor together with a rotation iii



about the origin.

Worked solutions for book chapters 9 Further 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 9A 5 a b When which is the position vector of . so

c

6 a The direction vector of the line is

and the line passes through

so has

equation b The point on the line with -coordinate At So

,

will occur for

.

which is not the position vector of the point. does not lie on the line.

7 The direction vector of is

so a parallel line passing through has equation

.

8 The direction vector of the given line is

so the new line has equation

. Or equivalently, 9 a When When

.

on the line,

, which is the position vector of .

on the line,

, which is the position vector of .

Hence both and lie on the line. b

is at

and is at

.

If then is located either at coordinates

(so that

or at

at which has

10 a The direction vector of the given line is

so the line has equation

. b

lies at If

and lies at then lies at

so has coordinates

or

11 a

b c

which is a distance of five times the direction vector. So lies at

, and therefore has coordinates

or

.

EXERCISE 9B 5 Let

.

Tip Although there is no need to use instead of using or as the parameter, this can avoid any need for fractions or rearrangement later. Then

so

6 Substituting

.

and

into the line equation:

This simplifies to Since this is false, the point does not lie on the line. 7 a

has Cartesian equation

b Direction vector

so

and the unit direction vector is

Tip

.

The notation of a ‘hat’ on a vector to indicate a unit vector is standard though not used in the Student Book. 8 a For

to lie on : and

so

b In vector form, the line has equation

The direction vector

has magnitude

So travelling a distance

along the line (in either direction) from

vector

or

has coordinates

or

gives a position

. .

EXERCISE 9C 2

Tip Both lines have been given with the same parameter. To set up the simultaneous equations, you have to change these to non-matching letters or the solution cannot be calculated!. Intersection of the two lines:

From



Then from Substituting into

which is false.

The two lines have direction vectors

and

which are not multiples of each

other so the lines are not parallel and, by the working, have no intersection point. They are therefore, by definition, skew. 3 Intersection of the two lines:

so

Then from Substituting into

so which is true so the two lines do indeed intersect, at the point

Tip You can use the intersection point to make a quick visual check of the validity of your answer against the initial equations. 4 Line has equation

.

General point of intersection with the

is

.

Substituting into the line equation: Since this can never be true for any value of , the line does not pass through the 5 Point of intersection with the

is

.

Substituting into the line equation: Multiplying through by So

:

for the line to intersect the

6 Calling the three lines

.

and

, at point

respectively:

Intersection of and

From



From



Substituting into is true so the lines intersect at point Intersection of and

From



From



Substituting into is true so the lines intersect at point Intersection of and

From Substituting into is true so the lines intersect at point

.

.

.

So the lines form a triangle

with

and

Tip Finding the area of a triangle from vectors can be made easier by using the scalar product (see Section 4 of this chapter) but in the absence of that technique you can use standard geometry. The side lengths are

,

and

By the cosine rule:

Then area

square units

EXERCISE 9D

7

The angle between

and

,

8

The angle between

is given by:

.

and

is given by:



9 a The direction vectors of the two lines are

and

So the angle between the lines is given by:

The acute angle between the lines is therefore b The direction vector of is

radians

.

so the two lines are perpendicular. 10 a

and

so

is a parallelogram.

b When



and

The angle between

and

is given by:

So the angles of parallelogram c For

are

to be a rectangle, require that

and

. :

so

11

,

and

so

a

So

b c

Tip In the solutions to questions to the simplest approach is often to assign unknowns to the values which you need to determine. You can then use one or more standard equations which describe the geometry you are given. Solving the equations will give the values for the unknowns.

12

for some value

so

Since

is perpendicular to , require

Then , so has coordinates

13 a

.

so line is given by

b

c

so distance

has direction vector

since is parallel to .

Tip The sign may be positive or negative, according to whether is the acute or obtuse angle between and d The distance between the and equals 14 a At ,

so Substituting into



Substituting into



is consistent.

The two lines intersect at

b

and

.

The angle between the lines is given by:

c At



, which is the position vector of

so lies on .

d

so e Let be the closest point to on line . Then

forms a triangle with

,

and

Using standard triangle trigonometry in right-angled triangle 15 The closest point on the line to the origin is . .

Require that

.

Then

so

and

The distance 16 The distance between two parallel lines with direction vector , passing through points with position vectors and respectively is where is the angle between

and is given by

Here,

so

and

and therefore

Therefore the distance between the lines is

17 a

lies on the line so

If

b When

and

,

c For

for some value .

has coordinates which is the position vector of

require that

It follows that to . lies at of .

form an isosceles triangle, with base lying along .

and lies at

, so it follows that must lie at

and has position vector

Then

b Since

is perpendicular to :

is also perpendicular to

.



so

for

is perpendicular to be the midpoint

.

has position vector

Since

so lies on .

must lie at the midpoint of the base of that triangle, since

So has coordinates

c

.

is perpendicular to then require

so

18 a

.

.

So

and since

is perpendicular to both lines, distance

is the shortest distance

between the lines.

MIXED PRACTICE 9

1

and

Angle

(Answer B)

2 The direction vector of the line is a multiple of the vector connecting the points: so use The equation of the line passing through

3 a

with direction vector is

Tip Two methods are shown here. In the first, a direct solution to the simultaneous equations is found. In the second, an arbitrary vector is chosen, by inspection, which is perpendicular to one of the direction vectors. By taking the scalar product through the whole equated positions, you collapse the simultaneous equations immediately to find the parameter of the intersection, if there is one. Method 1:

so Intersection point is at

which corresponds to

and

Method 2: At the intersection point,

Taking the scalar product with

, which, by inspection, is perpendicular to the first line:

so The intersection point is at

, which corresponds to

and

.

b If the direction vectors of the two lines are

and

, then the angle between the lines is given

by

and

4

so

Tip As for question , two methods are shown here. In the first, a direct solution to the simultaneous equations is looked for. In the second, two vectors are chosen, by inspection, which are perpendicular to one of the direction vectors. By taking the scalar product through the whole equated positions, you collapse the simultaneous equations immediately to find the parameter of the intersection, if there is one. If both give the same parameter value, then the lines do intersect. Geometrically, this method is equivalent to looking at the intersection from different directions; if the point where the lines appear to touch is the same from every perspective then the lines do indeed intersect. Method 1:

Since this is a falsehood, there can be no intersection and the lines are skew. Method 2: At the intersection point,

Taking the scalar product with

Taking the scalar product with

, which, by inspection, is perpendicular to the first line:

, which, by inspection, is also perpendicular to the first line:

This inconsistency in the value of shows that the two lines are skew. 5 For p and q to be perpendicular, require that

Therefore

or

6 The two direction vectors are

The angle between the lines is given by

and

So the acute angle between the lines is 7 a

b Intersection with

,

Require such that

This simultaneous equation has consistent solution

, so the line does intersect the

, at

.

c Angle between direction vectors

and

is .

8 a Parametrised form of the first line: Parametrised form of the second line: Lines intersect when

These are consistent for The point of intersection is b By observation, at 9 Line

. .

, the line passes through

has equation

and line

. has equation

For an intersection, require values of and µ for which the resultant position vector is the same in both lines. By inspection, for the -coordinates to match, require that cannot match. The lines cannot therefore intersect, and are skew. 10 a

, For angle

to be a right angle,

µ but, if

µ then the -coordinates

b

so has coordinates

.

11 a The lines are given by

and

so both have direction vector

.

passes through point and passes through point and since is not a multiple of , the two lines have the same direction but do not intersect so are parallel but not the same. b Substituting

and

into the equation for :

so satisfies the equation and does lie on . c

on has coordinates

so

.

Require that So so Therefore, has coordinates

d

so

12 a When on line .

the point defined on is

b

c

so line

has equation

has position vector

so

which is the position vector of , so lies

for some µ

.

Require that so µ has coordinates d

.

will be the height of any triangle For the triangle µ

with on line

to have area three times that of µ

. , it follows that

.

lies at µ

so must lie at µ

has coordinates

.

or

.

13 a i

ii and

b

has position vector

so the line with direction

c

passing through has equation

has position vector

for some .

Require that

so

.

has coordinates

.

14 Let be the closest point on the line to

Require that

.

is perpendicular to the line so

.

Then so

Then

15 a The direction vector

so the line equation is

b The two lines are parallel. The distance between two parallel lines with direction vector , passing through points with position vectors and respectively is where is the angle between

and is given by

Here,

and

so

and therefore

Therefore the distance between the lines is

16 a At

so lies on

b Intersection occurs when

is valid

Tip The lines have the same specified point, which is clearly therefore the intersection. The working, although standard, is unnecessary in this question. Require that

form an isosceles triangle with

.

17 a At , From



From



From



so is valid.

Tip Always use all three of the equations to ensure that the solution obtained from two of them is valid. In this case, the question states that the lines do intersect, so this check serves to reveal errors in working rather than determining whether or not the lines are skew, but is nonetheless worthwhile. The two lines do intersect, at

.

b When

so

does lie on .

c

Require that So

d

18 a Supposing point lies on both lines:

From the coefficient: Substituting into the coefficient:

so

Substituting also into the coefficient: and

are incompatible so there can be no such point .

So, the two lines are skew. b i

Point has position vector

and has position vector

So

ii

iii

Since

is perpendicular to :

Since

is also perpendicular to :

so

so



So

and since

is perpendicular to both lines, distance

is the shortest

distance between the lines.

19 a i

ii

The angle between

and is where

so b i

has coordinates

so is given by

.

ii Require such that

so

or

The larger value would produce a parallelogram with and so has coordinates



.

so for the trapezium,

Worked solutions for book chapters 10 Further 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 10A 5 a When

so has coordinates

b

6 When

and when

7 8 So

from which

9 Roots of the curve are at

10 So 11 a

and therefore and

intersect where

Intersection points are b In the interval

. so:

.

12 Intersection where Intersections are at

so and

, and in the interval

13 a Squaring both sides:

so intersections are at

and

.

b Recasting the equations so that is a function of : and

, with

14 Only need the upper semicircle:

, from

to

.

15 Taking the apex of the cone at the origin, require a straight line graph such that when so the function to be used is

16 The intersection of

and

is at

.

.

The volume described is equivalent to the volume of revolution when the region beneath between and is rotated around the -axis.

EXERCISE 10B

3 4 Mean value of

between and is:

So Given

. the only solution is

5 a Mean value of

.

between and is:

b

so

6 Mean value of

between and is:

7 Maximum power is

.

Average power over a single period is

So the ratio 8 Let

is

where:

.

.

The mean value of

between and is

Then the mean value of

between and is:

9a

b The curve is concave up (the gradient is negative and the second derivative positive) for all so the curve between and lies entirely below the chord connecting and and in consequence, the mean value of the function is less than the midpoint of that chord. c From part b, mean value of

mean of

and

10 If the function is increasing or decreasing on the entire interval then the statement will be true, but if it has a local maximum of minimum in the interval it may not be. For a simple counterexample, choose a function for which is predominately less, say, than on the interval but due to a having a local minimum in the interval it then increases so that as required. For example: and but

for

,

So,

but it is not true that

.

Mixed practice 10 1

(Answer )

2

(Answer )

3

so

4

and and If

then

5

6

7 Recasting the equation so that is a function of :

8 9a b 10 The curve has roots at Taking

so and

.

By symmetry, if

then the volume will have the same positive value, so in general, the volume

of revolution is 11 Average value of

between and is:

which is the arithmetic mean of and . 12 a Let

and

Then

so the curves do intersect at

b The two curves are

.

and

The volume of revolution of the shaded region is the sum of two volumes of revolution; the curve for and the curve for .

13 Translating the curve down by unit so that the volume is rotated around the -axis:

which is bounded by the -axis between Then the volume of revolution is:

14

and

.

so the positive solution is 15 a

and

.

Intersections occur when so Solutions are

and

b In the interval

, so the intersection points are

and

.

so the volume of revolution is given by:

c 16 a

Intersection points of

and

are at

and

.

b , so the volume of revolution is the same about either axis; c The region is symmetrical about the volume of revolution about the -axis is therefore also .



Worked solutions for book chapters 11 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 11B

In this Exercise, the formulae for

,

and

can be assumed.

3

4

5a b Require Positive root of is So the least (positive) value of satisfying the inequality is 6a

b

7

So

.

8a

b Substituting

instead of :

9a

b General term is

10 The sum of the squares of the first odd numbers and the first even numbers is the sum of the squares of the first integers. So the sum of the first odd numbers equals

So

.

EXERCISE 11C 1a b Using the method of differences:

2a b Using the method of differences:

3 a

b i

Using the method of differences:

ii

4 a Let

for some constants and

Multiplying through by the denominator of the LHS:

Substituting Substituting So b

c From the working in part b,

Taking the limit as

:

5a b Using the method of differences:

6a b Using the method of differences:

Therefore, as tends to infinity, the value of the sum tends to . 7a b Using the method of differences:

c From the working in part b:

8 Using partial fractions: Suppose Multiplying through by the denominator on the LHS:

Substituting:

So Then, using the method of differences:

for some constants

and .

9a

b Using part a:

where

and

c

10 a Using the method of differences:

b So the series does not converge.

EXERCISE 11D 4

Convergence is for 5

6

so

7a

b

8a

b c

9a

b Using the expansion in a: and So Then the tangent at

is

10a

Convergence is for restricted such that So b Require So

and also

Therefore

so

Using a:

Mixed practice 11 1

(Answer A) 2

3

4a

So

and

b

5a

b So 6a

,

b

So the

term is

.

7 Requires convergence for the Maclaurin series for which is for

which is

.

So the convergence criterion is

, in other words

8a

b

9a So

,

b Require

So the only positive integer solution for is 10 a b Using the method of differences:

11 a

and also for the Maclaurin series

.

. (Answer C)

b

c Taking the limit as

in the result from part b:

12 a

b Using the method of differences:

13 a

b Replacing with

:

c

Tip Ignore higher powers of , which can be disregarded for small . 14 a Let

for some constants and .

Multiplying through by the denominator of the LHS:

Comparing coefficients:

So b Using the method of differences:

15 Partial fractions: Let

for some constants

Multiplying through by the denominator of the LHS:

Substituting Substituting Substituting So

where

and

and .

Worked solutions for book chapters 12 Proof by induction 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

1 Proposition:

Base case,

for all

.

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 2 Proposition:

Base case,

for all

:

Assume the proposition is true for

Working towards

, by the principle of mathematical induction. .

so the proposition is true for

Inductive step: so that

.

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 3 Proposition:

Base case,

for all

:

it is also true for

.

, by the principle of mathematical induction.

.

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 4 Proposition:

Base case,

for all

:

.

so the proposition is true for

Assume the proposition is true for

so that

:

.

, by the principle of mathematical induction.

Inductive step:

Working towards

it is also true for

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 5 Proposition:

Base case,

for all

it is also true for

.

, by the principle of mathematical induction. .

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 6 Proposition:

Base case,

for all

:

it is also true for

, by the principle of mathematical induction. .

so the proposition is true for

Inductive step: Assume the proposition is true for

Working towards

.

so that

:

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 7 Proposition:

Base case,

.

, by the principle of mathematical induction.

for all

.

:

so the proposition is true for

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 8 Proposition:

Base case,

it is also true for

, by the principle of mathematical induction.

for all

.

:

so the proposition is true for

Inductive step: Assume the proposition is true for

Working towards

.

so that

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 9 Proposition:

Base case,

for all

.

, by the principle of mathematical induction. .

:

so the proposition is true for

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 10 Proposition:

Base case,

for all

:

, by the principle of mathematical induction. .

so the proposition is true for

Inductive step: Assume the proposition is true for

.

so that

.

Working towards

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for , by the principle of mathematical induction.



EXERCISE 12B

1 Proposition: Base case,

for all positive integers . : so the proposition is true for

.

Inductive step: Assume that

for some positive integer .

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 2 Proposition: Base case,

it is also true for

, by the principle of mathematical induction.

for all positive integers . : so the proposition is true for

Inductive step:

.

.

Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 3 Proposition: Base case,

it is also true for

.

, by the principle of mathematical induction.

for all positive integers . : so the proposition is true for

.

Inductive step: Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 4 Proposition: Base case,

it is also true for

.

, by the principle of mathematical induction.

for all positive integers . : so the proposition is true for

.

Inductive step: Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 5 Proposition: Base case,

.

, by the principle of mathematical induction.

for all positive integers . : so the proposition is true for

.

Inductive step: Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 6 Proposition: Base case,

it is also true for

, by the principle of mathematical induction.

for all positive integers . :

.

so the proposition is true for

.

Inductive step: Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 7 Proposition: Base case,

.

, by the principle of mathematical induction. for all positive integers .

:

so the proposition is true for Inductive step:

Assume that

for some positive integer .

Working towards

:

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

8a b Proposition: Base case,

for all positive integers . : so the proposition is true for

Inductive step: Assume that

for some positive integer .

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

EXERCISE 12C 1 Proposition:

is divisible by for all

.

it is also true for

.

, by the principle of mathematical induction.

Base case,

:

, so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 2 Proposition: Base case,

is divisible by for all :

.

, by the principle of mathematical induction. .

, so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 3 Proposition:

is divisible by for all

Base case,

.

, by the principle of mathematical induction. .

, so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 4 Proposition: Base case,

is divisible by :

it is also true for

.

, by the principle of mathematical induction.

for all integers

.

, so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that for some integer :

for some integer .

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 5 Proposition: Base case,

is divisible by for all :

.

, by the principle of mathematical induction. .

, so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

for some integer .

Working towards

for some integer :

But is the product of two consecutive integers, one of which must be even, and so it itself an even number; for some integer .

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 6 Proposition: Base case,

is divisible by for all :

.

, by the principle of mathematical induction. .

, so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

for some integer .

for some integer :

But is the product of two consecutive integers, one of which must be even, and so it itself an even number; for some integer .

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 7 Proposition:

is divisible by

for all

it is also true for

.

, by the principle of mathematical induction. .

Base case,

:

, so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 8 Proposition: Base case,

is divisible by

it is also true for

.

, by the principle of mathematical induction.

for all

.

:

, so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 9 Proposition: Base case,

is divisible by :

it is also true for

.

, by the principle of mathematical induction.

for

. , so the proposition is true for

Tip The question only asks for a proof over positive integers, but is also true, and is easy to see as true without a calculator. There is no harm in proving more than required as long as you prove everything that is required! Inductive step: Assume the proposition is true for Working towards

So the proposition is true for Conclusion:

so that for some integer :

.

for some integer .

The proposition is true for

and if true for

Hence the proposition is true for all integers 10 Proposition:

it is also true for

.

, by the principle of mathematical induction.

is divisible by for all

.

Tip The proof required is for all integers, but a proof for non-negative can easily be converted into a proof for negative using symmetry. Start by proving the proposition for Base case,

:

:

, so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

for some integer .

for some integer :

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers You can extend this proof to all For

But

, let

it is also true for

, by the principle of mathematical induction.

:

.

so it has already been shown that

Therefore

is a multiple of .

is a multiple of when

as well.

Therefore the proposition is true for all integers .

EXERCISE 12D 1 Proposition:

for all

Base case,

so the proposition is true for

Inductive step: Assume the proposition is true for Working towards

.

:

so that

.

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 2 Proposition:

it is also true for

.

, by the principle of mathematical induction.

for all

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

.

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 3 Proposition:

it is also true for

.

, by the principle of mathematical induction.

for all

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

.

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 4 Proposition:

for all

it is also true for

, by the principle of mathematical induction.

.

Base case,

so the proposition is true for

Inductive step: Assume the proposition is true for Working towards

So the proposition is true for Conclusion:

so that

:

.

.

.

.

The proposition is true for

and if true for

Hence the proposition is true for all integers 5 Proposition:

for all

it is also true for

.

, by the principle of mathematical induction.

.

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

.

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 6 Proposition:

for all

it is also true for

.

, by the principle of mathematical induction.

.

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so

for

for

So Then

.

:

has roots So

so that

for gives

for

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 7 Proposition:

for all

it is also true for

.

, by the principle of mathematical induction.

.

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

:

has a single root at So

for

So Then

for gives

.

for

so

for

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 8 a Quadratic

has roots

it is also true for

.

, by the principle of mathematical induction. .

Positive quadratic is greater than zero outside the roots. So

for

b Proposition:

or

for all

.

Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

.

:

But by part a:

Then (*) gives:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 9 Proposition:

for all

it is also true for

.

, by the principle of mathematical induction.

.

Tip For all integers

is equivalent to for all integers

Base case,

.

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

.

Working towards

has a single positive root at So Then

so for

gives

So the proposition is true for

for .

for

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

Mixed practice 12 1 Proposition: Base case,

for all positive integers . :

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

.

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 2 Proposition:

is divisible by for all

Base case:

it is also true for

.

, by the principle of mathematical induction. .

, so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 3 Proposition:

for all

Base case:

it is also true for

, by the principle of mathematical induction.

. , so the proposition is true for

Inductive step: Assume the proposition is true for Working towards

:

.

so that

.

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 4 Proposition: Base case,

is divisible by :

it is also true for

.

, by the principle of mathematical induction.

for all , so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 5 Proposition:

Base case,

it is also true for

.

, by the principle of mathematical induction.

for

.

:

, so the proposition is true for

. Inductive step: Assume the proposition is true for

so that

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 6 Proposition: Base case,

it is also true for

, by the principle of mathematical induction. for all

:

.

. , so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 7 Proposition: Base case,

is divisible by

for all

:

it is also true for

.

, by the principle of mathematical induction. . , so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 8 Proposition:

is divisible by

for

Base case:

it is also true for

.

, by the principle of mathematical induction. . , so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

for some integer :

So the proposition is true for

for some integer .

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

9 Proposition:

Base case,

for all

.

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 10 Proposition:

Base case,

it is also true for

.

, by the principle of mathematical induction.

for all

:

.

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 11Proposition:

it is also true for

.

, by the principle of mathematical induction.

, where and are positive integers, for all

.

Base case,

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for .

so that

Working towards

for some positive integers and :

Since and are positive integers, So the proposition is true for

for some positive integers and

and

are also positive integers.

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 12 Proposition:

Base case,

.

, by the principle of mathematical induction.

for all integers

.

:

so the proposition is true for

. Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers

.

, by the principle of mathematical induction.

13 a

So

.

b Proposition: Base case, Inductive step:

is divisible by for all :

. so the proposition is true for

.

Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers

.

, by the principle of mathematical induction.

14 a

So

.

b Proposition:

is divisible by

Base case,

for all

.

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

15 Given result: Proposition: Base case,

for all positive integers . :

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

So the proposition is true for

:

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

16 Proposition:

Base case,

for all integers

.

:

so the proposition is true for

. Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers 17 a Let

and

Then

for real

and

it is also true for

.

, by the principle of mathematical induction.

and .

.

So, b Proposition: Base case,

for all

.

:

so the proposition is true for

Inductive step: Assume the proposition is true for Working towards

So the proposition is true for

so that

:

.

.

.

Conclusion: The proposition is true for

and if true for

it is also true for

Hence the proposition is true for all integers 18 Proposition: Base case,

.

, by the principle of mathematical induction.

for all

.

:

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

Tip the final step in this working uses the angle formulae

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

Worked solutions for Cross-topic Reviews 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 Let

so

Then Comparing real and imaginary parts: Re:

so

Im:

so

2a i Re

and Im

ii Re b If

and Im is real then Im

so

3

4 a Polynomials with real coefficients will have complex roots in conjugate pairs, so root. b Therefore the cubic expression can be factorised as case, , since the coefficient of is .

and 5 Substituting:

so Therefore

is the only solution for is the only solution for .

6 a Horizontal stretch scale factor : Replace with Curve

has equation

b Vertical stretch scale factor : Replace with

is also a and, in this

Curve

has equation

Comparing:

so

Tip Reject the negative root as this would properly be described as a reflection as well as a stretch. 7 a Vertical asymptote where denominator is zero: Horizontal asymptote is the ratio of the coefficients of b

c From the graph,

for

8

9 Factorising: so

or

Then 10

or and

.

Substituting into the Cartesian equation:

Dividing through by

:

11 a so b

: Distance from equals the distance from

.

The locus is the perpendicular bisector of the line connecting to

.

12 a Let

.

Then Comparing real and imaginary parts:

From (2): Substituting into (1):

so

Then

Tip Select positive root since

So

must be positive.

and

So b Completing the square:

13 a Testing some values, clearly need

is a root of b Factorising: Comparing coefficients:

so so is consistent

So Using the quadratic formula for the remaining two (complex) roots: for

or

c Require the locus of points with distance from the origin at least but less than the distance to , which is a ring ('annulus') around the origin with inner radius and outer radius .

d

and So

14 a Using binomial expansion:

b i Substituting

:

ii The coefficient of is the negative sum of the roots:

so iii The constant coefficient equals the negative product of the roots so

. 15 a Sum of the roots Product of the roots b

c For the new quadratic:

.

Then (taking the 16 a When

,

coefficient as .

for integer coefficients throughout) the new quadratic is

and when

,

b Vertical stretch scale factor : Replace with . New curve equation: c Translation by

: Replace with

and with

New curve equation: Multiplying through by

and expanding:

Comparing coefficients with

:

17

18 a b Method 1: Find

and hence

so

for which the only real solution is

Then Method 2: Use identities

Dividing by

and using

:

so

and

So 19 a The product of complex numbers has modulus equal to the product of the moduli and argument equal to the sum of the arguments so b

, so

c

From the diagram,

and

So 20 a The centre of the circle is at the midpoint of and : The radius of the circle is half the distance between and :

So the locus of b

is

is a circle centred at

with radius .

Tip Be careful when sketching a diagram with circles. Make sure that each circle correctly intersects, is tangential to, or does not touch each axis. So, in this case, must touch the real axis but not reach the imaginary axis. c The greatest separation distance will occur with and on the line connecting the centres of the circles. and

so

which has length Then the length separating and equals 21 a i At

so

has coordinates ii Reflecting through

and hence

.

. : Interchange and in the equation

has equation iii The diagram is, by construction, symmetrical through the line must be perpendicular to the line and hence have gradient (Alternatively: If the contact point on is be

then by symmetry, the contact point on must

and the gradient between these two points is

b i Substituting

, so a common tangent .

)

into the equation for and requiring two solutions:

has two real roots so discriminant

So ii From part b i, the tangent must have equation with .

so as to have a single intersection

Substituting into So the intersection with is at

and the intersection with is, by symmetry,

22 a Vertical asymptotes where the denominator is zero:

or

Horizontal asymptote at the ratio of the leading term of numerator and denominator, when they have equal order: b i If

is on the curve then

so ii If the quadratic has at least one real root, then the discriminant

iii From part ii, there are no roots for so the critical values

and

, may be turning points.

:

At

, there are no solutions because

At

so

has no solution. (repeated root)

is the only stationary point.

Tip This argument is sufficient, given the known shape of a rational function which is the ratio of a quadratic and a quadratic. However, in general, there may be other stationary points than simply at the boundaries of regions with zero solutions to . c

23

Let

to form a quadratic in :

For at least one solution, discriminant

:

Tip There are several other ways to approach this question; using calculus or compressing the LHS into a single function are two possible alternatives that employ Level Further Mathematics techniques. Method 2: Calculus Let Then

so the single turning point occurs when

so

Since the graph of tends to infinity as tends to be at least one solution to for

, this will be a minimum, and so there will .

Giving the solution exactly:

(negative because

matches the sign of ) and

So the minimum value of Therefore

is

has at least one solution for

Method 3: Combining the functions Let Since

,

let

and

, so that

and so so Then Range of

is

, so

has at least one solution for

.

24 a The intersections occur where so So the intersections are b i

and

must have angle above the initial line. When

ii

, so has polar coordinates ,

and

so, by the cosine rule:

iii Then Therefore Hence

is the hypotenuse of right-angled triangle and therefore

.

is a tangent to the circle centred at passing through .

Worked solutions for Cross-topic Reviews 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 2 Matrix is singular if 3a b Area of is so the area of its image will be 4 So

, the identity matrix.

5a

is the direction of the line

so the vector equation is

b The lines intersect where

so so Then from (2):

so

Checking consistent solution in (1): The lines intersect at the point

.

and

6a The angle between

and

is given by

b 7 Rewriting the curve with

as a function of :

8

9a i

ii

does not affect or but transforms into

.

does not affect but transforms to and to

.

b

c so the image has coordinates 10 a

.

so the line direction

and the line equation is given by

b At the intersection,

Tip Try taking the scalar product with vector through this whole equation. Can you see why this gives a very fast route to the solution?

so Then from (2): Then from (3):

so

11 The parameterised version of the first line is Substituting this into the second line equation: , equivalent to Resolving the first equals sign gives so , which does not lead to a consistent solution since is not true. The two lines must therefore be skew since they are not parallel (the denominators are not in constant ratio with ). 12 a

b has direction vector

and

has direction

The angle between them is given by

c If lies on then it has position vector

for some .

Then

So either 13

(so that

) or

, so that has coordinates

.

and

So

,

Tip You can also do this by a change in summation and substitution into the only:

formula

14 a So b Using method of differences:

c

As

, this sum tends to .

15 Proposition:

is divisible by for integer

Base case,

.

so the proposition is true for

.

Inductive step: Assume the proposition is true for Working towards

so that

for some integer .

for some integer :

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

16 a b Proposition: Base case,

for :

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

.

Working towards

:

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

17 a For

to be an invariant line:

or

(reject as require

) so

b A general point on the line is mapped to a different point so is not a line of invariant points.

by the transformation,

18 Let be

and be a point on the line , which has direction

If is the nearest point on the line to then

is perpendicular to .

, so

Then

19 a

has direction vector

with position vector

and passes through point

So has vector equation b Suppose the lines intersect at point . Then:

Comparing coefficients: For Substituting into coefficient for

so

Substituting into coefficient for

so

Since these contradict, there is no such point and therefore the two lines do not intersect. c i Point has position vector

and has position vector

So Since

is perpendicular to :

Since

is also perpendicular to :

so Substituting Substituting

ii Then

20

So

,

.

into (1): is into (2): is

so the shortest distance between the lines is

Quadratic in

:

Rejecting the false solution

, the only solution is

so

21 a

b Valid for equivalent to c Require

so

Substituting

from which

into the expansion in part a:

22 a b Proposition: Base case,

for :

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that .

Working towards

.

So the proposition is true for

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers

it is also true for

.

, by the principle of mathematical induction.

23 so the equation of is For to be perpendicular to , require the scalar product of the directions to be zero: so

24 a i

is given by

ii Under the transformation, a general point of is mapped to

:

So the image of is the line through the origin with direction

b i

has vector equation so

has general point

So (eliminating the introduced parameter ): is the equation of ii The gradient of both

and

, which, in the required form, is

is , so these lines are parallel.

The lines have a vertical separation of

.

The perpendicular separation has gradient

so

Then

25 a

b Proposition: Base case,

so the proposition is true for

.

Inductive step: Assume the proposition is true for

so that

Working towards

:

So the proposition is true for

.

.

Conclusion: The proposition is true for

and if true for

Hence the proposition is true for all integers c Let have coordinates

.

it is also true for

.

, by the principle of mathematical induction.

So, point has coordinates

.

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