Additional Maths Revision Notes

First Edition © C Morris. Only authorised for use by students at Reading School.

1

OCR

Additional

Mathematics

FSMQ

(6993)

Revision Notes

Version 1.1 March 2008 Clive Morris (E&OE)


2

Index

Index...................................................................................................................................... 2Syllabus for OCR FSMQ in Additional Mathematics (6993) ...................................... 4Formulae............................................................................................................................... 7Algebra.................................................................................................................................. 8

Rationalising Surds ............................................................................................................8Manipulation of Algebraic Expressions ...........................................................................8

Addition and subtraction of polynomials .....................................................................8Multiplication of polynomials.......................................................................................9Division of polynomials ................................................................................................9

Remainder Theorem.........................................................................................................12Factor Theorem ................................................................................................................13Solution of Equations.......................................................................................................15

Solving Equations Reducing to Quadratics................................................................15Solving Simultaneous Linear and Quadratic Equations (A Reminder)....................16Completing the Square ................................................................................................17Another (shorter) way of Completing the Square......................................................20Sketching Quadratics using Completing the Square .................................................21Finding the maximum and minimum point for a Quadratic Curve ..........................21Solving Quadratic Equations by Completing the Square..........................................23Solving Cubic Equations Using The Factor Theorem...............................................24Solving cubic and cubic inequalities ..........................................................................26Discriminant.................................................................................................................27

The Binomial Expansion .................................................................................................28Pascal’s Triangle ..........................................................................................................28

Application to Probability – Binomial Distribution.......................................................30Co-ordinate Geometry...................................................................................................... 32

The Straight Line..............................................................................................................32Gradient of a Straight Line..........................................................................................32Mid-point of a Line Segment ......................................................................................32Length of a Line Segment ...........................................................................................33Finding the Equation of a Straight Line .....................................................................33Parallel and Perpendicular Gradients..........................................................................35

The Coordinate Geometry of Circles..............................................................................38Equation of a Circle .....................................................................................................38Finding the Centre and Radius of a Circle .................................................................39Useful Properties in Circle Problems .........................................................................41Finding the Equation of a Tangent to a Circle ...........................................................42Finding the Equation of a Normal to a Circle ............................................................43Finding the Closest Distance of a Given Point from a Circle ...................................43When do circles meet?.................................................................................................44

Regions .............................................................................................................................45Applications to Linear Programming .............................................................................47


3

Trigonometry ..................................................................................................................... 50IGCSE Revisited ..............................................................................................................50Applications......................................................................................................................50Graphs of Sine, Cosine and Tangent for Any Angle .....................................................51Trigonometric Identities ..................................................................................................54Solving simple trigonometric equations.........................................................................55Trigonometry and Pythagoras in 3 Dimensions.............................................................59

Angle between a line and a plane ...............................................................................59Line of greatest slope...................................................................................................59Angle between two planes...........................................................................................60

Calculus............................................................................................................................... 61Differentiation ..................................................................................................................61

Notation ........................................................................................................................61Gradient Function ........................................................................................................61Differentiation of powers of x and constant multiples, sums and differences. ........62Equations of tangents and normals .............................................................................63Location and Nature of Stationary Points ..................................................................64Sketching Curves .........................................................................................................67Practical Maximum and Minimum Problems ............................................................68

Integration.........................................................................................................................69Integration as the Reverse of Differentiation .............................................................69Indefinite Integration of powers of n, constant multiples, sums and differences ....69Finding the constant of integration using given conditions ......................................70

Definite Integrals..............................................................................................................71Area between a curve and the x axis ..........................................................................71Area between two curves.............................................................................................74

Application to Kinematics ...............................................................................................75Motion in a Straight Line ............................................................................................75SUVAT Equations (Constant Acceleration Formulae) .............................................79Displacement-time and Velocity-time Graphs...........................................................83

These Revision Notes contain the material that is additional to the IGCSE syllabus.Material already covered in the IGCSE Revision Notes will not be repeated here.


4

Syllabus for OCR FSMQ in Additional Mathematics (6993)

Those statements in bold are in the Additional Mathematics syllabus but are not on theIGCSE syllabus.

Algebra

Manipulation of algebraic expressions Be able to simplify expressions including algebraicfractions, square roots and polynomials.

The remainder theorem Be able to find the remainder of a polynomial up toorder 3 when divided by a linear factor.

The factor theorem Be able to find linear factors of a polynomial up toorder 3.

Solution of equations Be confident in the use of brackets.

Be able to solve a linear equation in one unknown.

Be able to solve quadratic equations by factorisation, theuse of the formula and by completing the square.

Be able to solve a cubic equation by factorisation.

Be able to solve two linear simultaneous equations in 2unknowns.

Be able to solve two simultaneous equations in2 unknownswhere one equation is linear and the other is quadratic.

Be able to set up and solve problems leading to linear,quadratic and cubic equations in one unknown, and tosimultaneous linear equations in two unknowns.

Inequalities Be able to manipulate inequalities.

Be able to solve linear and quadratic inequalities algebraicallyand graphically.

The binomial expansion Understand and be able to apply the binomialexpansion of (a + b)n where n is a positive integer.

Application to probability Recognise probability situations which give rise to thebinomial distribution.

Be able to identify the binomial parameter, p, theprobability of success.

Be able to calculate probabilities using the binomialdistribution.


5

Co-ordinate Geometry

The straight line Know the definition of the gradient of a line.

Know the relationship between the gradients of paralleland perpendicular lines.

Be able to calculate the distance between two points.

Be able to find the mid-point of a line segment.

Be able to form the equation of a straight line.

Be able to draw a straight line given its equation.

Be able to solve simultaneous equations graphically.

The co-ordinate geometry of circles Know that the equation of a circle, centre (0,0), radiusr is x2 + y2 = r2.

Know that (x – a)2 + (y – b)2 = r2 is the equation of acircle with centre (a, b) and radius r.

Inequalities Be able to illustrate linear inequalities in two variables.

Be able to express real situations in terms of linearinequalities. Be able to use graphs of linear inequalities tosolve 2-dimensional maximisation and minimisationproblems, know the definition of objective function andbe able to find it in 2-dimensional cases.

Trigonometry

Ratios of any angles and their graphs Be able to use the definitions of sin , cos and tanfor any angle (measured in degrees only).

Be able to apply trigonometry to right angled triangles.

Know the sine and cosine rules and be able to apply them.

Be able to apply trigonometry to triangles with anyangles.

Know and be able to use the identity thatsin

tancos

Know and be able to use the identity 2 2sin cos 1 .

Be able to solve simple trigonometrical equations in givenintervals.

Be able to apply trigonometry to 2 and 3 dimensionalproblems.


6

Calculus

Differentiation Be able to differentiate kxn where n is a positive integer or0, and the sum of such functions.

Know that the gradient functiond

d

y

xgives the gradient

of the curve and measures the rate of change of y withrespect to x.

Know that the gradient of the function is the gradient of thetangent at that point.

Be able to find the equation of a tangent and normal atany point on a curve.

Be able to use differentiation to find stationary points on acurve.

Be able to determine the nature of a stationary point.

Be able to sketch a curve with known stationary points.

Integration Be aware that integration is the reverse ofdifferentiation.

Be able to integrate kxn where n is a positive integer or0, and the sum of such functions.

Be able to find a constant of integration.

Be able to find the equation of a curve, given its gradientfunction and one point.

Definite integrals Know what is meant by an indefinite and a definiteintegral.Be able to evaluate definite integrals.

Be able to find the area between a curve, two ordinatesand the x-axis. Be able to find the area between twocurves.

Application to kinematics Be able to use differentiation and integration with respectto time to solve simple problems involving variableacceleration.

Be able to recognise the special case where the use ofconstant acceleration formulae is appropriate.

Be able to solve problems using these formulae.


7

Formulae

You have been spoilt by not having to learn many mathematical formulae.

In the Additional Mathematics examination you are not given a formula sheet and sothe only formulae you will have with you are the ones that you have taken in there inyour head!

You are advised to make your own list of things to learn from the work covered duringthe Additional Mathematics course.

You are also reminded to learn the formulae from the IGCSE formula sheet including

any rearrangements of formulae on it e.g.2 2 2

cos2

b c aA

bc


8

Algebra

Rationalising Surds

As well as the surds at IGCSE Level you may be asked to simplify more complicatedsurds using the difference of two squares as an aid.

22

2 32 2 4 2 3 4 2 3 4 2 34 2 3

4 3 12 3 2 3 2 3 2 3

22

3 5 3 53 5 9 6 5 5 14 6 5 14 6 5 7 3 5 7 35

9 5 4 2 2 23 5 3 5 3 5 3 5

Manipulation of Algebraic Expressions

A polynomial is an expression that only contains positive integer powers of x andconstants.

A polynomial is therefore of the form 2 30 1 2 3

nna a x a x a x a x .

The numbers 1 2, ,...a a are called the coefficients of 2,x x etc.

The degree of a polynomial is the highest power of x that occurs.

A polynomial of degree 0 is a constant.

A polynomial of degree 1 is linear

A polynomial of degree 2 is quadratic.

A polynomial of degree 3 is cubic.

Addition and subtraction of polynomials

When adding or subtracting polynomials combine the terms with the same powers.

Examples

2 3 2 3 2 2 3 3

2 3

3 2 2 3

2 3

(3 2 2 3 ) (7 5 3 ) 3 7 2 5 2 3 3

10 3 5 4

(6 2 4 ) (4 4 7 ) 6 4 2 ( 4 ) 7 4

2 6 7 4

x x x x x x x x x x x x

x x x

x x x x x x x x

x x x


9

Multiplication of polynomials

This is exactly like expanding linear brackets at GCSE. Multiply each term in thesecond bracket by each term in the first bracket and then simplify. Laying your workout systematically can avoid silly slips being made as shown in the example below.

3 2 3 2 3

2 3 4

3 4 5 6

2 3 4 5 6

(3 4 2 )(1 ) 3 3 3 3

4 4 4 4

2 2 2 2

3 7 5 6 2 2

x x x x x x x x

x x x x

x x x x

x x x x x x

Division of polynomials

There are 2 principal methods of dividing one polynomial by another.

Example

Find 3 2(2 2 3) ( 2)x x x x

Method 1 (Long Division)

2

3 2

3 2

2

2

2 2 5

2 2 2 3

2 4

2

2 4

5 3

5 10

7

x x

x x x x

x x

x x

x x

x

x

That is to say

3 2 2(2 2 3) ( 2) (2 2 5) remainder 7x x x x x x

Other ways of writing this are

3 222 2 3 7

2 2 52 2

x x xx x

x x

or

3 2 22 2 3 ( 2)(2 2 5) 7x x x x x x

x into 32x goes 22x times

22x times ( 2)x is 3 2(2 4 )x x

x into 32x goes 22x times

Take 3 2(2 4 )x x from 3 2(2 2 )x x and bring down the x

2x times ( 2)x is 2(2 4 )x x

Take 2(2 4 )x x from 2(2 )x x and bring down the 3

( 2)x does not go into 7 so this is your remainder


10

Method 2 (Comparison of Coefficients)

From Method 1 it is clear that when you divide a cubic by a linear term you will obtaina quadratic plus a constant remainder. Using the last form from above:

3 2 2(2 2 3) ( 2)( )x x x x ax bx c d where d is a constant.

Expanding on the right hand side gives

3 2 3 2 2

3 2 3 2 2

3

2

(2 2 3) 2 2 2

(2 2 3) 2 2 2

So equating coefficients 2 (comparing coefficients of )

2 2 (comparing coefficients of )

4 2

2

2 1

x x x ax bx cx ax bx c d

x x x ax bx ax cx bx c d

a x

b a x

b

b

c b

3 2 2

(comparing coefficients of )

4 1

5

2 3 (comparing constants)

10 3

7

(2 2 3) ( 2) (2 2 5) remainder 7

x

c

c

c d

d

d

x x x x x x


11

Another method (related to Method 2) is to do the following.

Method 3

3 2 2(2 2 3) ( 2)( )x x x x ax bx c d where d is a constant.

We can work through this is stages

In order to get the 32x term a is clearly 2.

3 2 2(2 2 3) ( 2)(2 )x x x x x bx c d

The next stage is to look at the coefficient of 2x .

The terms that will have an 2x in come from x bx and 22 2x . These together must

give 22x .

4 2

2

b

b

3 2 2(2 2 3) ( 2)(2 2 )x x x x x x c d

The terms that will have an x in come from 2 2x and cx . These together must givex .

4 1

5

c

c

3 2 2(2 2 3) ( 2)(2 2 5)x x x x x x d

Equating the constant terms on both sides gives

2 5 3

10 3

7

d

d

d

So

3 2 2(2 2 3) ( 2) (2 2 5) remainder 7x x x x x x


12

Remainder Theorem

There is a much easier way of finding the remainder when you divide by a linear term.

This is called the remainder theorem.

Example 1

The remainder when 3 2f ( ) 2 7 8x x x x is divided by ( 3)x is given by3 2f (3) 3 2 3 7 3 8 27 18 21 8 32 .

Example 2

The remainder when 3 2f ( ) 3 2 6 8x x x x is divided by ( 2)x is given by3 2f ( 2) 3 ( 2) 2 ( 2) 6 ( 2) 8 24 8 12 8 52 .

Example 3

The remainder when 3 2f ( ) 8 4 2 1x x x x is divided by (2 1)x is given by

3 21 1 1 1

2 2 2 2

1 18 4

f ( ) 8 4 2 1

8 4 1 1

1 1 1 1

2

.

The remainder when f ( )x is divided by ( )x a is f ( )a

The remainder when f ( )x is divided by ( )bx a is f ab


13

Factor Theorem

This is a special case of the remainder theorem.

That is to say dividing f ( )x by ( )x a or ( )bx a respectively leaves no remainder!

Example 1

Show that 3x and 2x are factors of 3 23 10 24x x x .

Solution

3 2

3 2

3 2

Let f ( ) 3 10 24

f (3) 3 3 3 10 3 24

27 27 30 24 0

3 is a factor of f ( )

2 ( 2)

f ( 2) ( 2) 3 ( 2) 10 ( 2) 24

8 12 20 24 0

2 is a factor of f ( )

x x x x

x x

x x

x x

If f ( ) 0a then ( )x a is a factor of f ( )x

If f 0ab then ( )bx a is a factor of f ( )x

When looking for factors in a polynomial

Check to seek whether x or powers of x are factors

Start by looking for smaller values of a – a good strategy is check 1, then 1 ,then 2, then 2 etc.

If the coefficient of the highest power of x in the polynomial is a 2 or a 3 etc.then one of the factors will start (2 ...)x or (3 ...)x but it is better to leave this

to naturally appear as example 2 shows rather than looking for it directly.


14

Example 2

Use the factor theorem to factorise the cubic polynomial 3 22 9 7 6x x x

Solution

3 2

3 2 2

Let f ( ) 2 9 7 6

f (1) 2 9 7 6 6 0 so ( 1) is not a factor of f ( )

f ( 1) 2 9 7 6 12 0 so ( 1) is not a factor of f ( )

f (2) 16 36 14 6 0 so ( 2) is a factor of f ( )

2 9 7 6 ( 2)(2 3)

x x x x

x x

x x

x x

x x x x x bx

In the quadratic bracket on the right the 2x coefficient must be 2 to give 32x when youmultiply out.

In the quadratic bracket on the right the constant must be 3 to give 6 when youmultiply out.

To find the value of b look at the 2x term on the right hand side when you multiply out

and compare this with the 2x term on the left hand side.

2 2 22 2 9

4 9

5

x bx x

b

b

3 2 22 9 7 6 ( 2)(2 5 3)x x x x x x

As a check look at the x term which should be the same on both sides.

On the right hand side this is 2 5 3 7x x x which agrees with the left hand side.

If the quadratic factorises we can now complete the factorisation. In this case it does andleads to

3 22 9 7 6 ( 2)( 3)(2 1)x x x x x x


15

Solution of Equations

Solving Equations Reducing to Quadratics

Example 1

By first making the substitution 2y x solve the equation 4 25 36 0x x .

Solution

4 2

2 2 2

2

2

2

5 36 0

( ) 5( ) 36 0

5 36 0

( 9)( 4) 0

9 0 or 4 0

9 or 4

9 or 4

3 since it is not possible for to be equal to 4

x x

x x

y y

y y

y y

y

x

x x

Example 2

Solve the equation4

2 7 0xx

Solution

2

12

42 7 0

2 7 4 0 (multiplying through by )

(2 1)( 4) 0

2 1 0 or 4 0

or 4

xx

x x x

x x

x x

x


16

Solving Simultaneous Linear and Quadratic Equations (A Reminder)

Example 1

Find where the circle 2 2 25x y and the line 7x y meet.

Solution

Where the graphs meet

2 2

2 2

2

2

(7 ) 25 (subtituting for in terms of from 7)

49 14 25 (expanding)

2 14 24 0

7 12 0 (divide through by 2 to make life easier)

( 3)( 4) 0

3 or 4

4 or 3

x x y x x y

x x x

x x

x x

x x

x

y

So the line meets the circle at (3,4) and (4,3) .

Note that the x and y values must be paired in the final answer otherwise you may losemarks.

Example 2

By solving the equations simultaneously find where the line 5 6y x and the curve2 2y x x meet and comment on your answer.

Solution

Where the graphs meet

2

2

2

2 5 6 (equating values)

4 4 0

( 2) 0

2

5 2 6 4

x x x y

x x

x

x

y

So since there is one repeated solution 5 6y x is a tangent to 2 2y x x when

2x and 4y i.e. at the point (2, 4) .


17

Completing the Square

There are quicker methods for those good with mental gymnastics but this basic routineis always effective. Another way of approaching things is to be found at the end of thissection.

Example 1

Write 2 4 3x x in the form 2( )x a b .

Solution

2 2

2

2 2 2

4 3 ( )

4 3 ( )( )

4 3 2

x x x a b

x x x a x a b

x x x ax a b

Then compare the bits on the 2 sides.

2

2

2 4 (comparing the terms)

2

3 (comparing the numbers)

2 3

4 3

3 4 7

a x

a

a b

b

b

b

So

2 24 3 ( 2) 7x x x


18

Example 2

Write 27 6x x in the form 2( )a x b .

Solution

2 2

2

2 2 2

2 2 2

7 6 ( )

7 6 ( )( )

7 6 ( 2 )

7 6 2

x x a x b

x x a x b x b

x x a x bx b

x x a b bx x


2

2


3


( 3) 7

9 7

16

b x

b

a b

a

a

a

So

2 27 6 16 ( 3)x x x


19

Example 3

Write 22 8 11x x in the form 2( )a x b c .

Solution

2 2

2

2 2 2

2 2 2

2 8 11 ( )

2 8 11 ( )( )

2 8 11 ( 2 )

2 8 11 2

x x a x b c

x x a x b x b c

x x a x bx b c

x x ax abx ab c


2

2

2

2 (comparing the terms)


2 2 8

2


2 2 11

8 11

3

a x

ab x

b

b

ab c

c

c

c

So

2 22 8 11 2( 2) 3x x x


20

Another (shorter) way of Completing the Square

Since

2 2 2( ) ( )( ) 2x a x a x a x ax a

we can use the fact that

2 2 22 ( )x ax x a a

In the brackets with the x is half the number of x’s in the original expression.

Example 1

2 2 2 24 3 ( 2) 2 3 ( 2) 7x x x x

Example 2

2 2

2 2

2

2

2 16 7 2( 8 ) 7

2 ( 4) ( 4) 7

2 ( 4) 16 7

2( 4) 25

x x x x

x

x

x

Example 3

2 2

223 32 2

215 32 2

3 6 2 3 2( 3 )

3 2 ( )

2( )

x x x x

x

x


21

Sketching Quadratics using Completing the Square

For example the curve 2 4 3y x x i.e. 2( 2) 7y x is the curve 2y x

translated 2 units in the negative x direction and translated 7 units in the negative y-direction.

Therefore the vertex of the curve (in this case the lowest point) is at ( 2, 7) .

Finding the maximum and minimum point for a Quadratic Curve

If you have completed the square on a quadratic it is easy to decide where the maximum(or minimum) point on the curve is.

Example 1

Complete the square on 22 8 2y x x and hence find the coordinates of the

maximum point on the curve.

Solution

2

2 2

2

2 4 2

2( 2) 2 2 ( 2)

2( 2) 6

y x x

y x

y x

Since the smallest value of 22( 2) 6x will be when 2x the minimum value of22 8 2y x x will be 6y when 2x .

The minimum point will therefore have coordinates (2, 6) .

x-6 -5 -4 -3 -2 -1 1 2 3 4 5

y

-10

-8

-6

-4

-2

2

4

6

8

10

( – 2, – 7)

y = x2

+ 4x – 3


22

Example 2

Complete the square on 23 8 4y x x and hence find the coordinates of the

maximum point on the curve.

Solution

2

2 2

2

4 2 3

4( 1) 3 4 1

7 4( 1)

y x x

y x

y x

Since the largest value of 27 4( 1)x will be when 1x the maximum value of23 8 4y x x will be 7y when 1x .

The maximum point will therefore have coordinates ( 1,7) .


23

Solving Quadratic Equations by Completing the Square

Example 1

Solve the equation 2 4 3 0x x by first completing the square.

Solution

2

2 2

2

2

4 3 0

( 2) 2 3 0

( 2) 7 0

( 2) 7

2 7

2 7

x x

x

x

x

x

x

Sometimes you are asked to give answers in surd form (which will be exact as nodecimal approximation will have taken place) but if you have to give decimal answersyou can obtain them easily from here.

If you needed answers to 3 decimal places they would be 4.646 and 0.646.

Example 2

Solve the equation 216 12 2 0x x by completing the square.

Solution

2 2

2 2

2

2

6 8 0 (dividing by 2 to obtain an equation starting ..)

( 3) ( 3) 8 0

( 3) 17 0

( 3) 17

3 17

3 17

1.12 or 7.12 (3 sf)

x x x

x

x

x

x

x

x


24

Solving Cubic Equations Using The Factor Theorem

Example 1

Show that 1x is a factor of 3 1x and hence factorise 3 1x .

Solution

3

3

3

Let f ( ) 1

f (1) 1 1 0

1 is a factor of 1

x x

x x

3 2

3 2

1 ( 1)( )

( ) ( )

x x ax bx c

ax b a x c b x c

Comparing coefficients we have

1

1 0 1

1 0 1

a

b a b b

c b c c

3 21 ( 1)( 1)x x x x

This cannot be factorised further as 2 1x x does not factorise.


25

Example 2

Factorise fully 3 22 5 6x x x .

Solution

3 2Let f ( ) 2 5 6

f (1) 2 5 1 6 2 0 so ( 1) is not a factor

f ( 1) 2 5 1 6 0 so ( 1) is a factor

x x x x

x

x

You can then go through two different routes. Either take out 1x as a factor (as in theexample on the next page) or find another factor. Pursuing this route gives

3 2Let f ( ) 2 5 6

f (2) 16 20 2 6 0 so ( 2) is a factor

x x x x

x

3 22 5 6 ( 1)( 2)(2 3)x x x x x x

This final factor comes from observing that it must start with a 2x to give 32x andmust end up with 3 to get 6 .


26

Solving cubic and cubic inequalities

Example

Solve the equation 3 22 7 3 18 0x x x .

Hence solve the inequality 3 22 7 3 18 0x x x .

Solution

3 2

2

Let f ( ) 2 7 3 18

f (1) 10 so ( 1) is not a factor of f ( )

f ( 1) 12 so ( 1) is not a factor of f ( )

f (2) 0 so ( 2) is a factor of f ( )

f ( ) ( 2)(2 3 9)

f ( ) ( 2)(2 3)( 3)

f ( ) 0 when 2 o

x x x x

x x

x x

x x

x x x x

x x x x

x x

32r 3 or

3 2Consider the graph of f ( ) 2 7 3 18.y x x x x

f ( ) 0x when the curve is on or above the x axis so f ( ) 0x when 32

2 or 3.x x

NB See also the inequalities section of the IGCSE Revision Notes.

x-5 -4 -3 -2 -1 1 2 3 4 5 6

y

-10

-5

5

10

15

y = (x – 2)(2x + 3)(x – 3)

Either by long division or bycomparing coefficients


27

Discriminant

When solving the quadratic equation

2 0, 0ax bx c a

You know that the solutions (if there are any) are given by the quadratic equationformula

2 4

2

b b acx

a

The part underneath the square root sign is called the discriminant, often given the

symbol . So 2 4b ac .

The discriminant gives quite a lot of information about the solutions of a quadraticequation and whether the quadratic factorises.

Value of Information given

0 Two distinct solutions

0 One (repeated) solution

0 Solutions

0 No solutions

a perfect square The quadratic factorises

Examples

23 2 4 0x x has two solutions because 22 4 3 4 52 0 .

24 4 1 0x x has one (repeated) solution because 24 4 4 1 0 .

23 2 3 0x x has no solutions because 22 4 3 3 32 0 .

26 2x x factorises because 2( 1) 4 6 2 49 which is a perfect square.


28

The Binomial Expansion

Pascal’s Triangle

The coefficients for expanding (1 )nx come from the rows of Pascal’s Triangle.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

The numbers in Pascal’s Triangle also come from!

!( )!n

r

nC

r n r

using appropriate

values of r and n.

For example the entries in the row beginning 1, 6, 15, 20, 15, 6, 1 come from

6 6 6 6 6 6 60 1 2 3 4 5 6, , , , , ,C C C C C C C respectively.

So we have that for positive integer values of n only.

Binomial Expansions Type 1

Where!

!( )!n

r

nC

r n r

.

Note thatn

is sometimes written asr

nrC

(NB no fraction line!)

2 31 2 3(1 ) 1n n n n n r n

rx C x C x C x C x x


29

Binomial Expansions Type 2

Example 1

Find the first five terms in the expansion of 10(1 )x in ascending powers of x.

Solution

10 10 10 2 10 3 10 41 2 3 4

2 3 4

(1 ) 1 ...

1 10 45 120 210 ...

x C x C x C x C x

x x x x

Example 2

Find the terms in the expansion of 7(1 2 )x in ascending powers of x up to and

including the term in 3x .

Solution

7 7 7 2 7 31 2 3

2 3

(1 2 ) 1 ( 2 ) ( 2 ) ( 2 )

1 14 84 280 ...

x C x C x C x

x x x

Example 3

Find the binomial expansion of 5(3 2 )x and use your expansion to estimate 53.002

correct to 1 decimal place.

Solution

5 5 5 4 5 3 2 5 2 3 5 1 4 51 2 3 4

2 3 4 5

5 5

(3 2 ) 3 3 (2 ) 3 (2 ) 3 (2 ) 3 (2 ) (2 )

243 810 1080 720 240 32

3.002 (3 2 0.001)

243 810 0.001 (since higher power terms will not affect first dp)

243.8 (1 decimal plac

x C x C x C x C x x

x x x x x

e)

1 2 2 3 31 2 3( )n n n n n n n n n n r r n

ra x a C a x C a x C a x C a x x


30

Example 4

Use the answer to example 2 to find the expansion in ascending powers of x up to and

including the term in 3x of 7(2 3 )(1 2 )x x .

Solution

7 2 3

2 3 2 3

2 3

(2 3 )(1 2 ) (2 3 )(1 14 84 280 ...)

2 28 168 560 3 42 252 ...

2 25 126 308 ...

x x x x x x

x x x x x x

x x x

Application to Probability – Binomial Distribution

If X is the number of successes in n independent trials each of which has probability pof success and probability ( 1 )q p of failure then X is said to have a Binomial

Distribution with parameters n and p and we write B( , )X n p .

Conditions for use of a Binomial Distribution

A binomial distribution can be used to model a situation if

Each trial has two possible outcomes (usually referred to as success or failure)

There is a fixed number of trials

The probability of success in each trial is constant

The outcome of each trial is independent of the outcomes of all the other trials

In the formula P( ) (1 )n r n rrX r C p p there must be

r successes giving rise to the rp part of the formula

n r failures giving rise to the (1 )n rp part of the formula

The r success can occur in any of nrC ways.

Note that the powers of p and 1 p always add up to n.

P( )

P( ) (1 )

Mean of

n r n rr

n r n r

r

X r C p q

X r C p p

X np


31

Example 1

It is known that in a certain population 15% are left handed,

Find the probability that in a sample of 9 people

(a) exactly 5 are left handed

(b) at most 3 are left handed

(c) at least 1 is left handed

Find also

(d) the mean number of people who are left handed

Solution

(9,0.15)X B

(a) 9 5 45P( 5) 0.15 0.85 0.00499 (3 sf)X C

(b)

9 9 1 8 9 2 7 9 3 61 2 3

P( 3) P( 0 or 1 or 2 or 3)

P( 3) 0.85 0.15 0.85 0.15 0.85 0.15 0.85 0.966 (3 sf)

X X

X C C C

(c)

9

P( 1) 1 P( 0)

P( 1) 1 0.85 0.768 (3 sf)

X X

X

(d) Mean 9 0.15 1.35

Example 2

How many fair cubical dice must be rolled for there to be a 99% chance of obtaining atleast one six?

Solution

56

56

56

2556

2656

P(at least one six) 1 P(no sixes)

1 0.99

1 0.99

0.01

0.01048... 0.01

0.008735... 0.01 (using trial and improvement)

n

n

n

At least 26 dice must be rolled for there to be a probability of at least one six


32

Co-ordinate Geometry

The Straight Line

Gradient of a Straight Line

Example

Find the gradient of the line joining the points ( 3,7) and (2, 13) .

13 7 20Gradient 4

2 ( 3) 5

Mid-point of a Line Segment

Example

Find the midpoint of the line joining the points ( 3,7) and (2, 13) .

The midpoint is 12

3 2 7 ( 13), , 3

2 2

x

y

1 1( , )x y

M

2 2( , )x y

y

x

2 2( , )x y

1 1( , )x y

2 1

2 1

Gradienty y

x x

1 2 1 2Mid point M is ,2 2

x x y y


33

Length of a Line Segment

Example

Find the length of the line segment joining the points ( 3,7) and (2, 8) .

2 2

2 2

Distance (2 ( 3)) ( 8 7)

5 ( 15)

250

25 10

5 10

Finding the Equation of a Straight Line

The equation of a straight line is of the form y mx c where m is the gradient and c

is the y-intercept.

Remember that the equation must be in this form before you can read off the gradient. Ifit is not you must rearrange the equation first.

Be careful that you give the answer in the required form. Sometimes a question will askyou to give your answer a specific way e.g. in the form 0ax by c where a, b and c

are integers.

There are several approaches each of which needs you to have a gradient and a pointthat the line goes through. If you are given 2 points you can obviously find the gradientfrom this.

y

x

1 1( , )x y

2 2( , )x y

2 21 1 2 2 2 1 2 1Distance between ( , ) and ( , ) ( ) ( )x y x y x x y y


34

Example 1

The gradient of the line 3 4 7 0x y is3

4 since rearranging the equation gives

3 4 7 0

4 3 7

3 7

4 4

x y

y x

y x

Example 2

Find an equation of the straight line with gradient 3 going through the point withcoordinates ( 3,7) .

Approach 1

3

Line goes through ( 3,7) so 7 when 3

7 3 ( 3)

2

3 2

y x c

y x

c

c

y x

Approach 2 (preferred)

7Gradient is 3

( 3)

7 3( 3)

7 3 9

3 2

y

x

y x

y x

y x


35

Example 3

Find an equation of the straight line through the points with coordinates(2, 5) and ( 3,15) .

( 5) 15 ( 5)

2 3 2

5 204

2 5

5 4( 2)

5 4 8

4 3

y

x

y

x

y x

y x

y x

Parallel and Perpendicular Gradients

Two lines are parallel if and only if they have the same gradient.

Two lines with gradients 1m and 2m are perpendicular if and only if

Example

The equations of 5 lines are given below. Which lines are parallel to L and which linesare perpendicular to L.

: 2 5

: 4 2 8

: 2 4

: 2 4 5

: 4 2 3 0

L y x

M x y

N y x

P x y

R x y

Rearranging the lines give

12

512 2

32

: 2 5

: 2 4

: 2

:

: 2

L y x

M y x

N y x

P y x

R y x

This shows that M and R are parallel to L and N are perpendicular to L.

2

1

1m

m i.e. 1 2 1m m .

Gradient

NB If in an exam you are asked to show thattwo lines are perpendicular, show that whenyou multiply their gradients you get 1 .


36

Example

Find the equation of the line that is parallel to 2 7y x and goes through the point

( 3,3) .

Solution

The gradient of 2 7y x is 2 so the equation of the required line is given by

32

( 3)

3 2( 3)

3 2 6

2 9

y

x

y x

y x

y x

Example

Find the equation of the line that is perpendicular 3 2 2 0x y and goes through the

point (2, 5) giving your answer in the form 0ax by c where a, b and c are

integers.

Solution

3 2 2 0x y can be rearranged to give 32

1y x and therefore has gradient 32

.

The gradient of the line perpendicular to 3 2 2 0x y is therefore given by

233

2

1

The equation of the perpendicular line is therefore

( 5) 2

2 3

3( 5) 2( 2)

3 15 2 4

2 3 11 0

y

x

y x

y x

x y


37

Reminder

Remember that lines parallel to the x-axis are of the form y k where k is a constant

and lines parallel to the y-axis are of the form x k where k is a constant.

All lines perpendicular to a line of the form 1y k will therefore be of the form 2x k

where 1k and 2k are constants and vice versa.

Example

Find the equation of the line perpendicular to 4x and going through the point (2,3) .

Solution

The line must be of the form y k and since it goes through a point with y-coordinate 3

it must be 3y .


38

The Coordinate Geometry of Circles

Equation of a Circle

The equation of a circle centre the origin and with radius r is given by

The equation of a circle centre ( , )a b and with radius r is given by

Notes

For a circle the x and y coefficients must be the same.

There can never be an xy term in the equation of a circle.

The value of 2r must be positive

Examples

2 2 121x y is a circle centre the origin and radius 11.

2 23 3 147x y is a circle centre the origin and radius 7 since it can be rewritten as2 2 49x y by dividing throughout by 3.

2 2( 2) ( 3) 36x y is a circle with centre (2,3) and radius 6.

2 2( 3) ( 1) 16x y is a circle with centre ( 3,1) and radius 4.

2 2 2x y r

2 2 2( ) ( )x a y b r


39

Finding the Centre and Radius of a Circle

The recommended approach to determine whether a given equation is a circle and todetermine its centre and radius is simply to complete the square from first principles onthe x and y terms.

Example 1

2 2

2 2

2 2 2 2

2 2

2 2

8 6 5 0

8 6 5 0

( 4) 4 ( 3) ( 3) 5 0

( 4) ( 3) 20 0

( 4) ( 3) 20

x y x y

x y x y

x y

x y

x y

This is a circle centre ( 4,3) and radius 20 4 5 2 5 .


40

Example 2

Which of the following are equations of circles and why

(i) 2 22 6 8 36x y x y

(ii) 2 2 2 6 12 11 0x y xy x y

(iii) 2 2( ) ( ) 50x y x y

(iv) 2 2 4 6 12y x x y

(v) 2 2 10 2 50 0x y x y

Solutions

(i) This is not a circle because the coefficients of 2x and 2y are not the same.

(ii) This is not a circle because there is a term in xy .

(iii) When this is expanded you get

2 2 2 2

2 2

2 2 50

25

x xy y x xy y

x y

This is a circle centre the origin and radius 5.

(iv) When rearranged this gives

2 2

2 2 2 2

4 6 12

( 2) ( 3) 12 2 3 25

x x y y

x y

This is a circle centre (2, 3) and radius 5.

(v) When rearranged this gives

2 2

2 2 2 2

10 2 50 0

( 5) ( 1) 50 5 1 24

x y x y

x y

This is not a circle because the right hand can’t be a radius squared since it isnegative.


41

Useful Properties in Circle Problems

The diameter is twice the length of the radius.

The angle in a semicircle is a right angle.

The tangent to a circle at a point is perpendicular to the radius at that point.

The perpendicular from the centre to a chord bisects the chord.

Do not forget Pythagoras’ Theorem!

Example

Find the equation of the circle that has a diameter with endpoints (2, 4) and ( 4,8)

Solution

The centre of the circle is at 2 ( 4) 4 8

, 1,22 2

The diameter of the circle is given by

2 2 2 2( 4 2) (8 ( 4) ( 6) 12 180 36 5 6 5

The radius of the circle is therefore6 5

3 52

The equation of the circle is therefore

2 2 2

2 2

( 1) 2 (3 5)

1 2 45

x y

x y


42

Finding the Equation of a Tangent to a Circle

This can be found by using the fact that the tangent to a circle is perpendicular to theradius of the circle.

Example

Find the equation of the tangent to the circle 2 2 6 8 0x y x y at the point with

coordinates (6, 8) .

2 2

2 2 2 2

6 8 0

( 3) ( 4) ( 3) 4 25

Centre is (3, 4), Radius is 5

4 ( 8) 4Gradient of radius to (6, 8)

3 6 3

1 3Gradient of tangent at (6, 8)

4 43

x y x y

x y

The equation of the tangent is therefore as before

( 8) 3

6 4

4( 8) 3( 6)

3 4 50

y

x

y x

x y


43

Finding the Equation of a Normal to a Circle

The equation of a normal to a circle can be tackled in a similar fashion to the equationof the tangent. Remember that the equation of a normal to a circle at a particular point,P say, has the same gradient as the gradient of the line joining P to the centre of thecircle.

Example

In the scenario outlined above, the equation of the normal at the point with coordinates

(6, 8) on the circle with equation 2 2 6 8 0x y x y is given by

( 8) 4

6 3

3( 8) 4( 6)

4 3 0

y

x

y x

x y

Finding the Closest Distance of a Given Point from a Circle

Essentially this reduces to finding the coordinates of the point, P, where a line throughthe given point, A and the centre of the circle, C, meet the circle and use this point tocalculate the required distance.

y

x2 4

2

4

C

A

P

Find the centre of the circle

Find the distance AC.

Find radiusAP AC .


44

Example

Find the point on the circle with equation 2 2( 2) ( 7) 5x y that is closest to the

point with coordinates (8,19) .

Solution

The circle has centre (2,7) and radius 5 .

The distance from to the centre (2,7) to (8,19) is given by

2 2 2 2(8 2) (19 7) 6 12 180 36 5 6 5

The distance from (8,19) to the circle is therefore given by

6 5 5 5 5

When do circles meet?

If 1r and 2r are the radii of two circles, 1 2r r and d is the distance between their

centres then

(i) Circles touch externally (ii) Circles touch internally

(iii) Circles do not intersect (iii) Circles intersect at two distinct points

2 1d r r 1 2d r r

1 2d r r 2 1 1 2r r d r r


45

Regions

Solid lines are used for inequalities that include = and dashed lines otherwise.

You are usually expected to shade the regions that are excluded e.g. when representingthe inequality 3x you would shade the region that is NOT 3x

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

x > 3

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

x 2

y > – 2

y 1

x – 2y 2

x-5 -4 -3 -2 -1 1 2 3 4 5

y

-5

-4

-3

-2

-1

1

2

3

4

5

y < x + 2


46

If you are not sure which way to shade, just pick a test point off the line and see whetherit satisfies the inequality. You will know which way to go then.

In the last diagram, for example, if you use the point (3, 2) you get

2 3 2 2 7 2x y so (3, 2) does satisfy the inequality.

Questions often require you to shade regions that satisfy a number of inequalities.

Example

Use shading to show the region (often called the feasible region) that satisfies thefollowing inequalities. You should shade the region that is not required.

2

2

3

x

y

x y

Solution

When drawing a graph such as 3 2 24x y a very quick way is to find out where it

meets the axes by putting 0x to get 12y and by putting 0y to get 8x .

2

4

-2

-4

2 4 6-2-4-6 0 x

y

x = 2y = 2

x + y = 3


47

Applications to Linear Programming

Work on regions can be extended to encompass problems involving linearprogramming.

This is where an objective function is maximised or minimised subject to certainconstraints that are usually able to be expressed as inequalities.

Consider the feasible region defined by the inequalities

2 6

2 8

2

3

x y

x y

x

y

Suppose we wish to maximise the objective function x y .

Such maximum points usually occur at a vertex (corner) of the region. The slope oflines of the form x y k indicates that in this case this would be where 2 6x y

and 2 8x y meet. By solving simultaneously this pair of equations it can be shown

that these meet at 1 13 33 ,1 . The maximum value of x y is therefore 1 1 2

3 3 33 1 4 .

2

4

-2

-4

2 4-2-4 0 x

y

x + 2y = 6

2x + y = 8

x = 2

y = 3


48

Example

Grizelda is going to make some small cakes to sell at school and raise money forcharity. She has decided to make some chocolate muffins and some yummy munchies.She would like to make as many cakes as possible but discovers that she only has 2 kgof flour and 750 g of butter. She has more than enough of the other ingredients.

The cakes must be made in batches:

For 12 muffins she needs 300 g of flour and 50 g of butter

For 16 yummy munchies she needs 200 g of flour and 125 g of butter.

(i) Using x to represent the number of batches of muffins and y to represent thenumber of batches of yummy munchies, write down and simplify twoinequalities relating to the available ingredients.

(ii) Illustrate the region satisfied by these inequalities, using the horizontal axisfor x and the vertical axis for y, and shading the unwanted region.

(iii) Write down the objective function for the total number of cakes and find thegreatest number of cakes that Grizelda can make.


49

Solution

(i) 300 200 2000 (restriction on flour)

3 2 20

50 125 750 (restriction on butter)

2 5 30

x y

x y

x y

x y

Using m to represent the number

(ii)

(iii) The objective function is 12 16N x y .

Remember that lines of the form 12 16x y k are all parallel to each other.

Look for the largest value of N that satisfies the conditions from (i) and (ii)with both x and y whole numbers (this is required from the context since thenumber of batches of each must be a whole number). Remember that thiswill usually be at or near a corner of the region formed by the constraints.

The largest value of the objective function is 112 when 4, 4x y so 112

cakes is the largest number that can be made.

The coordinates (4, 4)

give the largest value of12 16x y in the feasible

region i.e. 112 cakes.

5

10

15

20

5 10 15 200 x

y

3x + 2y = 20

2x + 5y = 30

N = 12x + 16y


50

Trigonometry

IGCSE Revisited

You obviously need to be familiar with all the work from IGCSE to do with

Trigonometry and Pythagoras in right angled triangles.

This includes finding sides and angles.

Sine and cosine rule.

This includes finding sides and angles and may include the ambiguous case for useof the sine rule to find an angle.

Finding areas of triangles including use of the formula1

sin2

ab C

Remember that you will not have any of the formulae and will have to have learntthem!

Applications

You are much more likely to be asked application of trigonometry. Likely contextsinclude the use of terms such as

Angle of elevation

x°Horizontal

Angle of elevation

Angle of depression

x°Horizontal

Angle of depression

Problems involving bearings and other real life situations are also very likely somake sure that you revise this.


51

Graphs of Sine, Cosine and Tangent for Any Angle

Observe that

sin 0 sin180 sin 360 0

sin 90 1

sin 270 1

sin 30 sin150 0.5

sin(180 ) sinx x

sin( ) sinx x

The sine graph repeats itself every 360 .

Be prepared to sketch this graph to helpyou solve trigonometric equations.

0.5

1

-0.5

-1

90 180 270 360-90-180-270-360 0 x

y

y = sin x


52

0.5

1

-0.5

-1

90 180 270 360-90-180-270-360 0 x

y

y = cos x

Observe that

cos 0 cos360 1

cos180 1

cos90 cos 270 0

cos 60 cos300 0.5

cos120 cos 60 0.5

cos(180 ) cosx x

cos( ) cosx x

The cosine graph repeats itself every 360 .

The cosine graph is the sine graph moved90 to the right.

Be prepared to sketch this graph to help yousolve trigonometric equations.


53

Observe that

tan 0 tan180 tan 360 0

tan 45 1

tan135 tan 45 1

tan 90

tan 270

tan(180 ) tanx x

tan( ) tanx x

The tangent graph repeats itself every 180 .

2

4

6

8

10

-2

-4

-6

-8

-10

90 180 270 360-90-180-270-360 0 x

y

y = tan x


54

Trigonometric Identities

The following results are true for all values of .

It is also worth remembering that in a right angled triangle

2 2sin cos 1

sintan

cos

x

yz

90 –

sin cos(90 )

cos sin(90 )

1 1tan

tan(90 )

x

z

y

z

x

yy

x


55

Solving simple trigonometric equations

Example 1

Solve the equation sin 0.4 for 0 360 .

Solution

One value can be found from 1sin (0.4) 23.6 (1 dp) .

From the symmetry of the sine graph there will also be a solution at

180 23.6 156.4 (1 dp)

Example 2

Solve the equation tan 1.2 for 180 180 .

Solution

One value can be found from 1tan ( 1.2) 50.2 (1 dp) .

Since the tangent graph repeats itself every 180 there will also be a solution at

180 ( 50.2) 129.8 (1 dp)

Example 3

Solve the equation cos 0.2 for 180 360 .

Solution

One value can be found from 1cos ( 0.2) 101.5 (1 dp) .

Another can be found from the symmetry of the cosine graph about 0 i.e.

101.5 (1 dp)

There is a third that can be found using the symmetry of the cosine graph about180 .

360 101.5 258.5 (1 dp)


56

Example 4

Solve sin 2cos where 0 360 .

Solution

sin2

cos

tan 2

63.4 or (63.4 180) (using periodicity of tan graph)

63.4 or 243.4

Example 5

Solve sin 2cos 0 where 0 360 .

Solution

After a simple rearrangement this is the same as Example 4!

Example 5

Solve 22sin 1 where 0 360 .

Solution

2 1sin

2

1sin

2

1When sin

2

45 or 135

1When sin

2

225 or 315


57

Example 6

Solve 23cos 5sin 5 0 for 0 360 .

Solution

2

2

2

2

2

23

23

23

3cos 5sin 5 0

3 1 sin 5sin 5 0

3 3sin 5sin 5 0

3sin 5sin 2 0

Let sin

3 5 2 0

3 2 ( 1) 0

3 2 0 or 1 0

or 1

sin or sin 1

sin leads to

41.8 or 180 41.8 138.2

sin 1 leads to

y

y y

y y

y y

y y

90

Example 7

Solve the equation tan cos 1 for 0 360 .

Solution

tan cos 1

sin

cos

cos 1

sin 1

270


58

Example 8

Solve the equation tan 2sin for 0 360 .

Solution

tan 2sin

sin2sin

cos

sin 2sin cos

sin 2sin cos 0

sin (1 2cos ) 0

sin 0

0 or 180 or 360

or 1 2cos 0

1cos

2

60 or 300

Example 9

Solve the equation cos 2 0.3 for 0 360 .

Solution

If 0 360

0 2 720

cos 2 0.3

2 72.5423.. or 360 72.5423.. or 360 72.5423.. or 360 (360 72.5423..)

2 72.5423.. or 287.4576.. or 432.5423.. or 647.4576..)

36.3 or 143.7 or 216.3 or 323.7


59

Trigonometry and Pythagoras in 3 Dimensions

There are likely to be problems involving trigonometry and Pythagoras’ Theorem in 3Dimensions and these may well be more involved and be more demanding that atIGCSE. You are more likely to find situations where the use of the sine or cosine ruleswill be more efficient.

Remember the key skills of identifying which angles you are working with andextracting suitable triangles to work with.

You also need to be familiar with some additional terms that you might not have metbefore just in case they come up.

Angle between a line and a plane

You should drop a perpendicular line down from the line to the plane to form a right-angled triangle.

Plane

Line

Angle between line and plane

Line of greatest slope

This is the steepest line down a slope. It is essentially the path a ball would take ifreleased on the slope!

Line of Greatest Slope


60

Angle between two planes

This is the angle between lines in the two planes that meet at right angles to where thetwo planes meet.

Angle Between the Planes


61

Calculus

Differentiation

Notation

d

d

y

xis the (first) derivative of y with respect to x.

It is the rate of change of y with respect to x.

2

2

d

d

y

xis the second derivative of y with respect to x.

This is the derivative ofd

d

y

xwith respect to x.

The first derivative of f ( )x is written as f '( )x and the second derivative is written as

f ''( )x

When differentiating something you will often see it written as d

somethingdx

.

For example if you are differentiating 3 23 3 2x x x you will often see this written as

3 2d( 3 3 2)

dx x x

x .

Gradient Function

d

d

y

xis also called the gradient function as it gives the gradient of a curve at the point

( , )x y when the x-coordinate of the point is substituted into it.


62

Differentiation of powers of x and constant multiples, sums and differences.

1d

dn nax nax

x where a is a constant.

1 1d

dn m n max bx nax mbx

x where a and b are constants,

Remember to write functions as powers of x before you differentiate and make sure thatyou simplify expressions first.

Examples

3 2 2d4 3 4 7 12 6 4

dx x x x x

x

2 3 2

2

d d(2 1)(3 2) 6 3 4 2

d d

18 6 4

x x x x xx x

x x

7 11

4 8

3

3 7

d d

d d

4 8

x xx x

x x x

x x

The gradient function of somethingof the form mx c is simply m fromthe work covered on the gradient ofa straight line.

The gradient function of a number(e.g. 5) is 0 because we know that

5y is a straight line horizontal to

the x-axis and hence has gradient 0.

You must write anything to bedifferentiated as powers of x beforeyou begin, multiplying or dividingout before you begin.

Multiply by the power and reduce thepower by 1.


63

Equations of tangents and normals

The normal to a curve at a point is the line perpendicular to the tangent at that point asshown by the diagram below.

x

yNormal

Tangent

Example

Find the equation of the tangent and the normal to the curve 23 2y x x at the point

where 2x .

2

2

d6 2

d

When 2, 3 2 2 2 16

d6 2 2 14

d

16Equation of the tangent is 14

2

16 14 28

14 12

16 1Equation of the normal is

2 14

14( 16) ( 2)

14 226

x

yx

x

x y

y

x

y

x

y x

y x

y

x

y x

y x

This simply meansput the value 2x into the gradientfunction.

The gradient of thenormal is 1 divided bythe gradient of thetangent.


64

Location and Nature of Stationary Points

A stationary point is a point on a curve for whichd

0d

y

x .

Stationary points can be maximum points, minimum points (these are both calledturning points) or points of inflexion (sometimes spelt inflection).

x

yMaximumPoint

MinimumPoint

Point ofInflexion

In order to determine the nature of a stationary point (i.e. to find out what sort ofstationary point it is) you can either

(a) find the gradientd

d

y

x

either side of the point, or

(b) use the second derivative2

2

d

d

y

x

If you use the first approach don’t go too far away from the stationary point or youmight move past another stationary point and draw an incorrect conclusion.

Value of2

2

d

d

y

xConclusion

d

d

y

xto the left

d

d

y

xat the point

d

d

y

xto the right

0 Maximum point + ve / 0 ––– \ ve

0 Minimum point ve \ 0 ––– / + ve

Point of inflection + ve / 0 ––– / + ve

Point of inflection ve \ 0 ––– \ ve

0 Checkd

d

y

xeither side


65

Note: The examples here are harder than you will meet in the examination but theyillustrate the principles well.

Example 1

Find the coordinates and nature of the stationary points on the curve 3 22 4y x x x .

Solution

23

2

2

2

23

4023 27

2

2

2

2402

3 272

d3 4 4 (3 2)( 2)

d

d6 4

d

d0 when 2 or

d

When 2, 8; when ,

d8 0, so there is a minimum at 2, 8

d

d8 0, so there is a maximum at ,

d

x

x

yx x x x

x

yx

x

yx x

x

x y x y

y

x

y

x


66

Example 2

Find the coordinates of any stationary points on the curve 3 2( 1)y x x and determine

their nature.

Solution

3 2 3

3 2

5 4 3

4 3 2

23 2 3 2

2

4 3 2

2 2

2

35

2

2

0

( 1) ( 1)( 1)

( 2 1)

2

d5 8 3

d

d20 24 6 2(10 12 3 )

d

dWhen 5 8 3 0

d

(5 8 3) 0

(5 3)( 1) 0

0 or or 1

d0 (no information gained)

d

d

d

x

y x x x x x

x x x

x x x

yx x x

x

yx x x x x x

x

yx x x

x

x x x

x x x

x

y

x

y

0.1

0.1

2

2

0.6

2

2

1

0.0385 0

d0.0225 0 so a point of inflexion at (0,0) (same sign on both sides of st pt)

d

d0.72 0 so a maximum at (0.6,0.03456)

d

d2 0 so a minimum at (1,0)

d

x

x

x

x

x

y

x

y

x

y

x


67

Sketching Curves

You may be expected to sketch simple curves using any of the mathematics included inthe syllabus. Useful things to bear in mind (although not all would be required on anyparticular sketch, nor should you work them all out unless you are asked to) are

Where the curve meets the x-axis (i.e. when 0y )

Where the curve meets the y-axis (i.e. when 0x )

The coordinates of any stationary points that you might know

The general shapes of particular types of curves that you might know such as

2y x

3y x

1y

x

2

1y

x


68

Practical Maximum and Minimum Problems

Example

A cylindrical tin has a volume of 128 cm3. Find the dimensions necessary for the tin tohave the minimum possible surface area, and find the minimum possible surface area.

Let the base radius be and the height be .r h

2

Circular ends Curved surface

2

2

2

2

2

2

2

2 3

3

2

2

Surface area, 2 2

Volume, 128

128

1282 2

2562

d 2564

d

d 5124

d

d 2560 when 4 0 i.e. when 64 i.e. when 4

d

d

S r rh

V r h

hr

S r rr

rr

Sr

r r

S

r r

Sr r r

r r

S

2

4

2

2 2min

4 8 12 0 i.e. S is a minimum value when 4 cmd

128So the radius, 4 cm and the height, 8 cm

4

256S 2 4 96 cm

4

r

rr

r h

r

h


69

Integration

Integration as the Reverse of Differentiation

dd

d

yx y c

x where c is an arbitrary constant of integration.

Indefinite Integration of powers of n, constant multiples, sums and differences

Examples

7 833 d

8x x x c

2 3 5 3 2

6 4 3

( 2)( 1)d ( 2)d

26 4 3

x x x x x x x

x x xx c

10 86 4

4

7 5

75

5d ( 5 )d

5

7 5

7

x xx x x x

x

x xc

xx c

Remember c once you have integrated.

Make sure that you have written thingsas powers of x before you integrate.

1

d 11

nn ax

ax x c nn

.

f ( ) d f ( ) da x x a x x

[f ( ) g( )]d f ( ) d g( ) dx x x x x x x


70

Finding the constant of integration using given conditions

Example 1

Suppose that a curve f ( )y x is such thatd

2 1d

yx

x and the curve passes through the

point (1, 2) .

Solution

2

2

2

(2 1) d

2 1 1 since the curve passes through (1, 2)

4

so 4

y x x x x c

c

c

y x x

Example 2

Suppose that 2d3 4

d

yx x b

x where b is a constant.

Given that 7y when 1x and that 12y when 2x , find y in terms of x.

Solution

2

2

23

3 2

3 2

3 2

3 2

d3 4

d

(3 4 )d

4

2

2

7 1 2 1 1

4 (1)

12 2 2 2 2

2 4 (2)

8 (2) (1)

12 substituting for in (1)

2 8 12

yx x b

x

y x x b x

xx bx c

y x x bx c

b c

b c

b c

b c

b

c b

y x x x


71

Definite Integrals

If f ( ) d F( )x x x c then f ( ) d F( ) F( )b

a

x x b a .

Area between a curve and the x axis

To find the area between the curve f ( )y x the x-axis and the lines x a and x b we

calculate

For example the area between the x-axis, the curve 2 2y x x and the lines 1x

and 3x is given by

32

1

33 2

1

3 2 3 2

9 1 12 3 2

23

Area ( 2) d

23 2

3 3 1 12 3 2 1

3 2 3 2

9 6 2

8

x x x

x xx

b

y

xa

y = f(x)

x-3 -2 -1 1 2 3 4 5

y

1

2

3

4

5

6

7

8

9

10

y = x2

– x + 2

d f ( ) d F( ) F( )b b

a a

y x x x b a


72

x-3 -2 -1 1 2 3

y

-10-8-6-4-2

2468

10

y = x3

y = 4x

Note that sometimes areas can be calculated using more straightforward methods. Forexample, when working out the area under a straight line you can use the area of atriangle or trapezium.

Note also that using symmetry can occasionally make calculations quicker.

Example

Find the shaded area between the curve 3y x and the line 4y x .

Solution

The curve and line meet when

3

3

2

4

4 0

( 4) 0

0, 2

x x

x x

x x

x

There are two ways of proceeding. The first is completely by integration.

2 2 23 3

0 0 0

242

0

Area 2 4 d 2 d 2 (4 )d

162 2 2 8 0 0 8 square units

4 4

x x x x x x x

xx

The second uses the fact that the area under the line is the area of a triangle.

When 2x , 8y .

231

2

0

24

0

Area 2 2 8 2 d

1616 2 16 2 0 8 square units

4 4

x x

x

The area between 2 and 0 is clearly thesame as the area between 0 and 2 so werequire twice the area under the line takeaway the area under the curve.


73

A warning!

You are strongly advised to draw a sketch of any curve before you find the area.

This is because areas below curves are negative.

If a curve has part above the x-axis and part below the x-axis you need to consider thetwo parts separately and combine the sizes of the areas.

Example

For example consider the area between the curve 2 9y x , the x-axis and the lines

2x and 5x .

A sketch shows that part of the area is below the x-axis and part is above.

33 32 8 2

3 3

2 2

23

55 32 125 2

3 3

3 3

2 2 13 3 3

( 9) d 9 9 27 18 23

The sign is because the area is below the axis.

The actual area is 2 .

( 9) d 9 45 9 27 143

Area required 2 14 17

NB Had we done

xx x x

xx x x

52

2

2 23 3

( 9) d then we would have obtained that the (incorrect) answer 12!

This is because the integral would have given 2 14 12.

x x

x-2 -1 1 2 3 4 5 6 7 8 9 10

y

-8

-4

4

8

12

y = x2

– 9


74

Area between two curves

To find the area between two curves f ( )y x and g( )y x which meet at x a and

x b which are such that f ( ) g( )x x between these values find

f ( ) g( ) db

a

x x x

Example

To find the area between the curves 2 2y x and 2 6y x .

First find where the curves meet.

2 2

2

2 6

2 8

2

x x

x

x

We therefore need to evaluate the integral of the top curve take away the bottom one.

22 2

2

22

2

2

3

2

16 163 3

13

Area 6 2 d

2 8 d

28

3

16 16

21

x x x

x x

x x

x-6 -4 -2 2 4 6

y

-10

-8

-6

-4

-2

2

4

6

8

10

y = – x2

+ 6

y = x2

– 2

Note that if you had subtracted the integrals thewrong way round you would have obtained theanswer 1

321 .

Note also that because of the symmetry of bothcurves we could have found instead

2

2 2

0

2 6 2 dx x x


75

Application to Kinematics

Motion in a Straight Line

You need to be able to apply your knowledge of differentiation to motion in a straightline.

The standard notation is

t times distance

d

d

sv

t velocity

d

d

va

t acceleration

Maximum/minimum velocity occurs when maximum of graph i.e.d

0d

v

t i.e. when

0a .

Remember also that

d

d

s v t

v a t

Don’t forget the constant of integration when integrating an indefinite integral –you will often need to use information you are given in the question to find itsvalue.


76

Example 1

The distance of a particle, s metres, from a fixed point O after time t seconds is given by

the formula 3 2 5 2s t t t .

Find formulae, in terms of t, for

(i) the velocity

(ii) the acceleration

Use your answers to find

(iii) the velocity after 2 seconds

(iv) the acceleration after 2 seconds

(v) the time at which the velocity is 0

Solution

(i) 2d3 2 5

d

sv t t

t (ii)

d6 2

d

va t

t

(iii) When 2t , 23 2 2 2 5 12 4 5 3 m/sv

(iv) When 2t , 26 2 2 10 m/sa

(v) 2

23

3 2 5 0

(3 5)( 1) 0

3 5 0 since 1 0 gives a negative time which is impossible

1 seconds

v t t

t t

t t

t


77

Example 2

A particle moves such that its displacement s metres after time t seconds is given by

3 2 1s t t t .

(a) Determine the value of t for which the velocity is 0.

(b) Show that its acceleration is never 0.

Solution

(a) 2d3 2 1 (3 1)( 1)

d

sv t t t t

t

When 0v , 13

t since t must be positive.

(b)d

6 2d

va t

t

The only value of t that gives 0a is 13

t and since time is always positive

the acceleration can never be 0.

Example 3

A particle has acceleration given by 29 1a t where t is the time since the particlestarted moving. Find the velocity in terms of t given that its initial velocity is 7 ms-1.

Solution

2

2 3

2

3

9 1

(9 1) d 3

When 0, 7 so 7 3 0 0

7

3 7

a t

v t t t t c

t v c

c

v t t


78

Example 4

The velocity of a particle is given by 2v t t at time t seconds. Find the distancemoved in the 3rd second (i.e. between 2 and 3 seconds).

Solution

2

3 2

56

d

3 2

8 4 14When 2,

3 2 3

27 9 27When 3,

3 2 2

27 14Distance travelled 8 metres

2 3

s t t t

t ts c

t s c c

t s c c

c c

Another alternative approach is to use a definite integral.

3

2

2

33 2

2

56

d

3 2

27 9 8 4

3 2 3 2

Distance travelled 8 metres

s t t t

t ts

s


79

SUVAT Equations (Constant Acceleration Formulae)

These are for motion with constant acceleration in a straight line (horizontal or vertical)only.

2 2

212

212

displacement

2 initial velocity

final velocity

acceleration2

time

v u at s

v u as u

s ut at v

u vs t a

s vt at t

Remember:

As these are vector quantities they have both magnitude and direction, so the signsof the quantities in these equations can be positive or negative. Choose a direction tobe positive and then quantities in the opposite direction are negative. Slowing down,retardation and deceleration are all terms for negative acceleration.

Usually g, the acceleration due to gravity is taken as 9.8 ms-2 (positive ↓).

In order to use these equations successfully keep in mind:

You should choose the equation that fits the information you know and the quantitythat you are trying to calculate.

If there are two unknowns to find it may be that you need to form and solve a pair ofsimultaneous equations.

If a particle is projected up and then falls down again you can look at the motion intwo stages, but this is not required, as long as you are careful with directions.

Other useful results are

Total DistanceAverage Speed

Total Time

Total DisplacementAverage Velocity

Total Time

When projecting a particle vertically, at maximum height the velocity will be zero.


80

Example 1

A car begins to accelerate at 0.5 ms-2 for a distance of 500 m. At the end of this the caris travelling at 30 ms-1, calculate the initial speed of the car.

Solution

We know: a = 0.5s = 500v = 30u = ?

so use: v2 = u2 + 2as

2 2

2 2

2

-1

30 2 0.5 500

30 2 0.5 500

400

20 ms

u

u

u

u

Example 2

A particle is projected upwards with a velocity of 30 ms-1. What is the maximum heightit will reach?

Solution

Because the particle is only acting under its own weight, its acceleration will be due togravity.

Because it reaches maximum height the final velocity at that height will be zero.

We therefore know: s = ?u = 30 ↑v = 0a = - 9.8 ↑

so use: v2 = u2 + 2as

20 30 2 ( 9.8)

19.6 900

45.9 m (3 sf)

s

s

s


81

Example 3

Daisy is cycling at a steady speed of 16 ms when she comes to a hill which causes her

to slow down at a rate of 20.5 ms .

(a) How far up the hill does she travel before coming to a rest?

(b) How long does it take her speed to be reduced to 12 ms .

Solution

(a)

2 2

2 2

6

0

0.5

Using 2

0 6 2 0.5

0 3

36

She travels 36 metres up the hill

u

v

a

v u as

s

s

s

(b)

2

6

2

0.5

Using

2 6 0.5

0.5 4

48

0.5

She takes 8 seconds to slow down to 2 ms

u

v

a

v u at

t

t

t


82

Example 4

A particle is projected with a velocity of 20 ms-1. For how long will it be above a heightof 16m?

Solution

We know: u = 20 ↑a = 9.8 ↑s = 16 ↑ i.e. from the floort = ?

so use: s = ut + ½ at2

16 = 20t – 4.9t2

0 = 4.9t2 – 20t + 16 use the quadratic formula

t = 1.09 s

t = 2.99 s (3 s.f.)

Two answers were expected because the particle will reach 16m both on the way up andon the way down during its motion.

The time above 16m will be 2.99 1.09 1.9 seconds .


83

Displacement-time and Velocity-time Graphs

These are often useful as tools in your armoury to tackle problems involving constant(straight lined velocity-time graphs) or variable acceleration (curved velocity-timegraphs).

(t, s) [Displacement-time]

The gradient of this graph represents velocity.

(t, v) [Velocity-time]

The gradient of this graph represents acceleration.

The area under this graph represents displacement when taking direction intoaccount and distance if only the total of the magnitudes of the areas areconsidered.

Example 1

Between points A and B on the graph the particle is travelling with a constant velocityof 2 ms-1 away from its starting point. Between B and C the particle is stationary.

Between C and D the particle travels back towards its starting point with constantvelocity –4 ms-1 i.e. in the opposite direction to its initial motion. Between D and E theparticle travels 6 m away from its starting point in the opposite direction, still withconstant velocity –4 ms-1. For the final part of the journey between E and F the particletravels with constant velocity 3 ms-1 in its original direction until it returns to its startingpoint.

time(seconds)

Displacement(metres)

A

B C

D

E

F

4

-6

52 8.54 6.5


84

time(seconds)

Velocity

A

B C

D

E

F

4

-6

52 8.54 6.5

(ms-1

)

The displacement from itsoriginal position needs totake into account thedirection of the motion i.e.you need the area ofABCD the area of DEF

14 10.5 2.5 metres

Example 2

At the point on the graph marked A, the particle starts at rest. Between points A and Bon the graph the particle is travelling with a constant acceleration of 2 ms-2 away fromits starting point. Between B and C the particle is travelling with constant velocity

14 ms away from its starting point.

Between C and D the particle decelerates with deceleration 4 ms-2 until at D it isinstantaneously stationary. It continues to decelerate with deceleration 4 ms-2 (and travelaway from its starting point in the opposite direction) until at E it is instantaneouslytravelling at –6 ms-1 away from its starting point. It then accelerates with acceleration 3ms-2 (N.B. an object travelling with negative velocity and positive acceleration isslowing down) until at F it is stationary again.

The total distance travelled can be calculated by evaluating areas ABCD and DEF.

(5 2)Area ABCD 4 14 (This is a trapezium)

2

3.5 6Area DEF 10.5

2

Total distance travelled 14 10.5 24.5 metres

Notice that Examples 1 and 2 are quite different even though the graphs look the same!!!

The total of areas ABCD andDEF represents the distancetravelled.

Additional Maths Revision Notes

Documents

Transcript of Additional Maths Revision Notes