First Edition © C Morris. Only authorised for use by students at Reading School.
OCR
Additional
Mathematics
FSMQ
(6993)
Revision Notes
Version 1.1 March 2008  Clive Morris (E&OE)
1
First Edition © C Morris. Only authorised for use by students at Reading School.
Index
Index...................................................................................................................................... 2
Syllabus for OCR FSMQ in Additional Mathematics (6993) ...................................... 4
Formulae............................................................................................................................... 7
Algebra.................................................................................................................................. 8
Rationalising Surds ............................................................................................................8
Manipulation of Algebraic Expressions ...........................................................................8
Addition and subtraction of polynomials .....................................................................8
Multiplication of polynomials.......................................................................................9
Division of polynomials ................................................................................................9
Remainder Theorem.........................................................................................................12
Factor Theorem ................................................................................................................13
Solution of Equations.......................................................................................................15
Solving Equations Reducing to Quadratics................................................................15
Solving Simultaneous Linear and Quadratic Equations (A Reminder)....................16
Completing the Square ................................................................................................17
Another (shorter) way of Completing the Square......................................................20
Sketching Quadratics using Completing the Square .................................................21
Finding the maximum and minimum point for a Quadratic Curve ..........................21
Solving Quadratic Equations by Completing the Square..........................................23
Solving Cubic Equations Using The Factor Theorem...............................................24
Solving cubic and cubic inequalities ..........................................................................26
Discriminant.................................................................................................................27
The Binomial Expansion .................................................................................................28
Pascal’s Triangle ..........................................................................................................28
Application to Probability – Binomial Distribution.......................................................30
Co-ordinate Geometry...................................................................................................... 32
The Straight Line..............................................................................................................32
Gradient of a Straight Line..........................................................................................32
Mid-point of a Line Segment ......................................................................................32
Length of a Line Segment ...........................................................................................33
Finding the Equation of a Straight Line .....................................................................33
Parallel and Perpendicular Gradients..........................................................................35
The Coordinate Geometry of Circles..............................................................................38
Equation of a Circle .....................................................................................................38
Finding the Centre and Radius of a Circle .................................................................39
Useful Properties in Circle Problems .........................................................................41
Finding the Equation of a Tangent to a Circle ...........................................................42
Finding the Equation of a Normal to a Circle ............................................................43
Finding the Closest Distance of a Given Point from a Circle ...................................43
When do circles meet?.................................................................................................44
Regions .............................................................................................................................45
Applications to Linear Programming .............................................................................47
2
First Edition © C Morris. Only authorised for use by students at Reading School.
Trigonometry ..................................................................................................................... 50
IGCSE Revisited ..............................................................................................................50
Applications......................................................................................................................50
Graphs of Sine, Cosine and Tangent for Any Angle .....................................................51
Trigonometric Identities ..................................................................................................54
Solving simple trigonometric equations.........................................................................55
Trigonometry and Pythagoras in 3 Dimensions.............................................................59
Angle between a line and a plane ...............................................................................59
Line of greatest slope...................................................................................................59
Angle between two planes...........................................................................................60
Calculus............................................................................................................................... 61
Differentiation ..................................................................................................................61
Notation ........................................................................................................................61
Gradient Function ........................................................................................................61
Differentiation of powers of x and constant multiples, sums and differences. ........62
Equations of tangents and normals .............................................................................63
Location and Nature of Stationary Points ..................................................................64
Sketching Curves .........................................................................................................67
Practical Maximum and Minimum Problems ............................................................68
Integration.........................................................................................................................69
Integration as the Reverse of Differentiation .............................................................69
Indefinite Integration of powers of n, constant multiples, sums and differences ....69
Finding the constant of integration using given conditions ......................................70
Definite Integrals..............................................................................................................71
Area between a curve and the x axis ..........................................................................71
Area between two curves.............................................................................................74
Application to Kinematics ...............................................................................................75
Motion in a Straight Line ............................................................................................75
SUVAT Equations (Constant Acceleration Formulae) .............................................79
Displacement-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.
3
First Edition © C Morris. Only authorised for use by students at Reading School.
Syllabus for OCR FSMQ in Additional Mathematics (6993)
Those statements in bold are in the Additional Mathematics syllabus but are not on the
IGCSE syllabus.
Algebra
Manipulation of algebraic expressions

Be able to simplify expressions including algebraic
fractions, square roots and polynomials.
The remainder theorem

Be able to find the remainder of a polynomial up to
order 3 when divided by a linear factor.
The factor theorem

Be able to find linear factors of a polynomial up to
order 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, the
use 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 2
unknowns.
Be able to solve two simultaneous equations in 2 unknowns
where 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 to
simultaneous linear equations in two unknowns.
Inequalities


Be able to manipulate inequalities.
Be able to solve linear and quadratic inequalities algebraically
and graphically.
The binomial expansion

Understand and be able to apply the binomial
expansion of (a + b)n where n is a positive integer.
Application to probability

Recognise probability situations which give rise to the
binomial distribution.
Be able to identify the binomial parameter, p, the
probability of success.
Be able to calculate probabilities using the binomial
distribution.


4
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Co-ordinate Geometry
The straight line
The co-ordinate geometry of circles



Know the definition of the gradient of a line.
Know the relationship between the gradients of parallel
and 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.

Know that the equation of a circle, centre (0,0), radius
r is x2 + y2 = r2.
Know that (x – a) 2 + (y – b)2 = r2 is the equation of a
circle 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 linear
inequalities. Be able to use graphs of linear inequalities to
solve 2-dimensional maximisation and minimisation
problems, know the definition of objective function and
be able to find it in 2-dimensional cases.

Be able to use the definitions of sin  , cos  and tan 
for 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.
Trigonometry
Ratios of any angles and their graphs



Be able to apply trigonometry to triangles with any
angles.

Know and be able to use the identity that

Know and be able to use the identity sin 2   cos 2   1 .

Be able to solve simple trigonometrical equations in given
intervals.
Be able to apply trigonometry to 2 and 3 dimensional
problems.

5
sin 
 tan 
cos 
First Edition © C Morris. Only authorised for use by students at Reading School.
Calculus
Differentiation

Be able to differentiate kxn where n is a positive integer or
0, and the sum of such functions.

Know that the gradient function

Know that the gradient of the function is the gradient of the
tangent at that point.
Be able to find the equation of a tangent and normal at
any point on a curve.
Be able to use differentiation to find stationary points on a
curve.


Integration


Be able to determine the nature of a stationary point.
Be able to sketch a curve with known stationary points.

Be aware that integration is the reverse of
differentiation.

Be able to integrate kxn where n is a positive integer or
0, and the sum of such functions.
Be able to find a constant of integration.

Definite integrals

Be able to find the equation of a curve, given its gradient
function and one point.

Know what is meant by an indefinite and a definite
integral.
Be able to evaluate definite integrals.
Be able to find the area between a curve, two ordinates
and the x-axis. Be able to find the area between two
curves.

Application to kinematics
dy
gives the gradient
dx
of the curve and measures the rate of change of y with
respect to x.

Be able to use differentiation and integration with respect
to time to solve simple problems involving variable
acceleration.

Be able to recognise the special case where the use of
constant acceleration formulae is appropriate.
Be able to solve problems using these formulae.

6
First Edition © C Morris. Only authorised for use by students at Reading School.
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 so
the only formulae you will have with you are the ones that you have taken in there in
your head!
You are advised to make your own list of things to learn from the work covered during
the Additional Mathematics course.
You are also reminded to learn the formulae from the IGCSE formula sheet including
b 2  c 2  a2
any rearrangements of formulae on it e.g. cos A 
2bc
7
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Algebra
Rationalising Surds
As well as the surds at IGCSE Level you may be asked to simplify more complicated
surds using the difference of two squares as an aid.
2
2 3
3 5
3 5

2
2 

2  3   4  2 3
3  2  3  2   3 

2
2

42 3 42 3

42 3
43
1
3  5  3  5   9  6 5  5  14  6 5  14  6
9 5
4
3  5  3  5  3   5 
2
2
5

7 3 5 7 3
5
 
2
2 2
Manipulation of Algebraic Expressions
A polynomial is an expression that only contains positive integer powers of x and
constants.
A polynomial is therefore of the form a0  a1 x  a2 x 2  a3 x3    an x n .
The numbers a1, a2 ,... are called the coefficients of x, x 2 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
(3  2 x  2 x 2  3x 3 )  (7  5 x  3x 2  x 3 )  3  7  2 x  5 x  2 x 2  3x 2  3 x 3  x 3
 10  3x  5 x 2  4 x 3
(6  2 x  4 x 3 )  (4  4 x  7 x 2 )  6  4  2 x  (4 x )  7 x 2  4 x3
 2  6 x  7 x 2  4 x3
8
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Multiplication of polynomials
This is exactly like expanding linear brackets at GCSE. Multiply each term in the
second bracket by each term in the first bracket and then simplify. Laying your work
out systematically can avoid silly slips being made as shown in the example below.
(3  4 x  2 x 3 )(1  x  x 2  x3 )  3 3x 3x2 3x3
4 x 4 x 2 4 x 3 4 x 4
2 x 3 2 x 4 2 x5 2 x6
 3  x 7 x 2 5 x 3 6 x4 2 x5 2 x6
Division of polynomials
There are 2 principal methods of dividing one polynomial by another.
Example
x into 2x 3 goes 2x 2 times
Find (2 x 3  2 x2  x  3)  ( x  2)
x into 2x 3 goes 2x 2 times
Method 1 (Long Division)
x2
2x 2  2 x  5
2 x3  2 x 2  x  3
2 x3  4 x 2
2 x2  x
2x 2 times ( x  2) is (2 x3  4 x2 )
Take (2 x3  4 x2 ) from (2 x3  2 x2 ) and bring down the x
2 x2  4 x
 5x  3
2x times ( x  2) is (2 x 2  4 x)
 5 x  10
7
That is to say
Take (2 x 2  4 x) from (2 x 2  x) and bring down the 3
( x  2) does not go into 7 so this is your remainder
(2 x 3  2 x2  x  3)  ( x  2)  (2x2  2 x  5) remainder 7
Other ways of writing this are
2 x3  2 x 2  x  3
7
 2x 2  2 x  5 
x2
x2
or
2 x 3  2 x 2  x  3  ( x  2)(2x2  2 x  5)  7
9
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Method 2 (Comparison of Coefficients)
From Method 1 it is clear that when you divide a cubic by a linear term you will obtain
a quadratic plus a constant remainder. Using the last form from above:
(2 x 3  2 x2  x  3)  ( x  2)(ax 2  bx  c)  d where d is a constant.
Expanding on the right hand side gives
(2 x 3  2 x 2  x  3)  ax3  bx 2  cx  2ax 2  2bx  2c  d
(2 x 3  2 x 2  x  3)  ax3  bx 2  2ax2  cx  2bx  2c  d
So equating coefficients a  2
b  2a  2
(comparing coefficients of x 3 )
(comparing coefficients of x 2 )
b  4  2
b2
c  2b  1 (comparing coefficients of x )
c  4 1
c5
2c  d  3 (comparing constants)
10  d  3
d 7
(2 x 3  2 x 2  x  3)  ( x  2)  (2x2  2 x  5) remainder 7
10
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Another method (related to Method 2) is to do the following.
Method 3
(2 x 3  2 x2  x  3)  ( x  2)(ax 2  bx  c)  d where d is a constant.
We can work through this is stages
In order to get the 2x 3 term a is clearly 2.
(2 x 3  2 x2  x  3)  ( x  2)(2 x2  bx  c)  d
The next stage is to look at the coefficient of x 2 .
The terms that will have an x 2 in come from x  bx and 2  2x 2 . These together must
give 2x 2 .
b  4  2
b2
(2 x 3  2 x2  x  3)  ( x  2)(2 x2  2 x  c)  d
The terms that will have an x in come from 2  2x and cx . These together must give
x .
c  4 1
c 5
(2 x 3  2 x2  x  3)  ( x  2)(2 x2  2 x  5)  d
Equating the constant terms on both sides gives
2  5  d  3
10  d  3
d 7
So
(2 x 3  2 x2  x  3)  ( x  2)  (2x2  2 x  5) remainder 7
11
First Edition © C Morris. Only authorised for use by students at Reading School.
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.
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 
Example 1
The remainder when f ( x)  x 3  2 x 2  7 x  8 is divided by ( x  3) is given by
f (3)  33  2  32  7  3  8  27  18  21  8  32 .
Example 2
The remainder when f ( x)  3 x3  2 x 2  6 x  8 is divided by ( x  2) is given by
f (2)  3  (2)3  2  (2)2  6  ( 2)  8  24  8  12  8  52 .
Example 3
The remainder when f ( x)  8 x3  4 x 2  2 x  1 is divided by (2 x  1) is given by
f ( 12 )  8  12   4  12   2  12   1 .
3
2
 8  18  4  14  1  1
 1 1 1 1
2
12
First Edition © C Morris. Only authorised for use by students at Reading School.
Factor Theorem
This is a special case of the remainder theorem.
If f (a)  0 then ( x  a ) is a factor of f ( x)
If f  ab   0 then (bx  a ) is a factor of f ( x)
That is to say dividing f ( x) by ( x  a ) or ( bx  a) respectively leaves no remainder!
Example 1
Show that x  3 and x  2 are factors of x 3  3x 2  10 x  24 .
Solution
Let f ( x )  x3  3 x 2  10 x  24
f (3)  33  3  32  10  3  24
 27  27  30  24  0
x  3 is a factor of f ( x )
x  2  x  (2)
f (2)  (2)3  3  (2)2  10  (2)  24
 8  12  20  24  0
x  2 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 (3x...) but it is better to leave this
to naturally appear as example 2 shows rather than looking for it directly.
13
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Use the factor theorem to factorise the cubic polynomial 2 x 3  9 x 2  7 x  6
Solution
Let f ( x )  2 x 3  9 x 2  7 x  6
f (1)  2  9  7  6  6  0 so ( x  1) is not a factor of f ( x)
f (1)  2  9  7  6  12  0 so ( x  1) is not a factor of f ( x )
f (2)  16  36  14  6  0 so ( x  2) is a factor of f ( x)
2 x 3  9 x 2  7 x  6  ( x  2)(2 x 2  bx  3)
In the quadratic bracket on the right the x 2 coefficient must be 2 to give 2x 3 when you
multiply out.
In the quadratic bracket on the right the constant must be 3 to give 6 when you
multiply out.
To find the value of b look at the x 2 term on the right hand side when you multiply out
and compare this with the x 2 term on the left hand side.
2  2 x 2  bx 2  9 x 2
b  4  9
b  5
2 x 3  9 x 2  7 x  6  ( x  2)(2 x2  5x  3)
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 x  3x  7 x which agrees with the left hand side.
If the quadratic factorises we can now complete the factorisation. In this case it does and
leads to
2 x 3  9 x 2  7 x  6  ( x  2)( x  3)(2 x  1)
14
First Edition © C Morris. Only authorised for use by students at Reading School.
Solution of Equations
Solving Equations Reducing to Quadratics
Example 1
By first making the substitution y  x 2 solve the equation x 4  5 x 2  36  0 .
Solution
x 4  5 x 2  36  0
( x 2 )2  5( x 2 )  36  0
y 2  5 y  36  0
( y  9)( y  4)  0
y  9  0 or y  4  0
y  9 or  4
x 2  9 or  4
x  3 since it is not possible for x 2 to be equal to  4
Example 2
4
Solve the equation 2 x  7   0
x
Solution
4
2x  7   0
x
2 x 2  7 x  4  0 (multiplying through by x)
(2 x  1)( x  4)  0
2 x 1  0 or x  4  0
x   12 or 4
15
First Edition © C Morris. Only authorised for use by students at Reading School.
Solving Simultaneous Linear and Quadratic Equations (A Reminder)
Example 1
Find where the circle x 2  y 2  25 and the line x  y  7 meet.
Solution
Where the graphs meet
x 2  (7  x )2  25 (subtituting for y in terms of x from x  y  7)
x 2  49  14 x  x 2  25 (expanding)
2 x 2  14 x  24  0
x 2  7 x  12  0 (divide through by 2 to make life easier)
( x  3)( x  4)  0
x  3 or 4
y  4 or 3
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 lose
marks.
Example 2
By solving the equations simultaneously find where the line y  5 x  6 and the curve
y  x 2  x  2 meet and comment on your answer.
Solution
Where the graphs meet
x 2  x  2  5 x  6 (equating y values)
x2  4 x  4  0
( x  2)2  0
x2
y  5 2  6  4
So since there is one repeated solution y  5 x  6 is a tangent to y  x 2  x  2 when
x  2 and y  4 i.e. at the point (2, 4) .
16
First Edition © C Morris. Only authorised for use by students at Reading School.
Completing the Square
There are quicker methods for those good with mental gymnastics but this basic routine
is always effective. Another way of approaching things is to be found at the end of this
section.
Example 1
Write x 2  4 x  3 in the form ( x  a)2  b .
Solution
x 2  4 x  3  ( x  a) 2  b
x 2  4 x  3  ( x  a)( x  a )  b
x 2  4 x  3  x 2  2ax  a 2  b
Then compare the bits on the 2 sides.
2a  4 (comparing the x terms)
a2
a 2  b  3 (comparing the numbers)
2 2  b  3
4  b  3
b  3  4  7
So
x 2  4 x  3  ( x  2) 2  7
17
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Write 7  6x  x 2 in the form a  ( x  b)2 .
Solution
7  6 x  x 2  a  ( x  b) 2
7  6 x  x 2  a  ( x  b)( x  b)
7  6 x  x 2  a  ( x 2  2bx  b2 )
7  6 x  x 2  a  b2  2bx  x 2
Then compare the bits on the 2 sides.
2b  6 (comparing the x terms)
b  3
a  b2  7 (comparing the numbers)
a  (3)2  7
a 9  7
a  16
So
7  6 x  x 2  16  ( x  3)2
18
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 3
Write 2 x 2  8 x  11 in the form a( x  b)2  c .
Solution
2 x 2  8 x  11  a( x  b)2  c
2 x 2  8 x  11  a( x  b)( x  b)  c
2 x 2  8 x  11  a( x 2  2bx  b2 )  c
2 x 2  8 x  11  ax 2  2abx  ab2  c
Then compare the bits on the 2 sides.
a  2 (comparing the x 2 terms)
2ab  6 (comparing the x terms)
2 2 b  8
b2
ab 2  c  11 (comparing the numbers)
2  22  c  11
8  c  11
c3
So
2 x 2  8 x  11  2( x  2)2  3
19
First Edition © C Morris. Only authorised for use by students at Reading School.
Another (shorter) way of Completing the Square
Since
( x  a)2  ( x  a)( x  a)  x2  2ax  a 2
we can use the fact that
x 2  2ax  ( x  a) 2  a 2
In the brackets with the x is half the number of x’s in the original expression.
Example 1
x 2  4 x  3  ( x  2)2  22  3  ( x  2)2  7
Example 2
2 x 2 16 x  7  2( x 2  8x )  7
 2( x  4)2  (4)2   7
 2( x  4)2 16  7
 2( x  4)2  25
Example 3
3  6 x  2 x 2  3  2( x 2  3x )

 3  2 ( x  32 ) 2   32 
2

 152  2( x  32 ) 2
20
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Sketching Quadratics using Completing the Square
For example the curve y  x 2  4 x  3 i.e. y  ( x  2)2  7 is the curve y  x 2
translated 2 units in the negative x direction and translated 7 units in the negative ydirection.
Therefore the vertex of the curve (in this case the lowest point) is at ( 2, 7) .
y
10
8
6
4
2
-6 -5 -4 -3 -2 -1-2
-4
-6
-8
( – 2, – 7)
-10
2
y = x + 4x – 3
1 2 3 4 5
x
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 y  2 x 2  8 x  2 and hence find the coordinates of the
maximum point on the curve.
Solution
y  2  x2  4 x   2
y  2( x  2)2  2  2  (2)2
y  2( x  2)2  6
Since the smallest value of 2( x  2) 2  6 will be when x  2 the minimum value of
y  2 x 2  8 x  2 will be y  6 when x  2 .
The minimum point will therefore have coordinates (2, 6) .
21
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Complete the square on y  3  8x  4 x 2 and hence find the coordinates of the
maximum point on the curve.
Solution
y  4  x 2  2 x   3
y  4( x  1) 2  3  4  12
y  7  4( x  1)2
Since the largest value of 7  4( x  1)2 will be when x  1 the maximum value of
y  3  8x  4 x 2 will be y  7 when x  1 .
The maximum point will therefore have coordinates ( 1, 7) .
22
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Solving Quadratic Equations by Completing the Square
Example 1
Solve the equation x 2  4 x  3  0 by first completing the square.
Solution
x2  4 x  3  0
( x  2)2  22  3  0
( x  2)2  7  0
( x  2)2  7
x 2   7
x  2  7
Sometimes you are asked to give answers in surd form (which will be exact as no
decimal approximation will have taken place) but if you have to give decimal answers
you 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 16  12 x  2 x 2  0 by completing the square.
Solution
x 2  6 x  8  0 (dividing by  2 to obtain an equation starting x2 ..)
( x  3)2  (3)2  8  0
( x  3) 2  17  0
( x  3)2  17
x  3   17
x  3  17
x  1.12 or 7.12 (3 sf)
23
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Solving Cubic Equations Using The Factor Theorem
Example 1
Show that x  1 is a factor of x 3  1 and hence factorise x 3  1 .
Solution
Let f ( x )  x 3  1
f (1)  13  1  0
x  1 is a factor of x 3  1
x 3  1  ( x  1)(ax 2  bx  c)
 ax 3  (b  a) x 2  (c  b) x  c
Comparing coefficients we have
a 1
b  a  b 1  0  b  1
c  b  c 1  0  c  1
x 3  1  ( x  1)( x2  x  1)
This cannot be factorised further as x 2  x  1 does not factorise.
24
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Factorise fully 2 x 3  5x 2  x  6 .
Solution
Let f ( x )  2 x 3  5 x 2  x  6
f (1)  2  5  1  6  2  0 so ( x  1) is not a factor
f (1)  2  5  1  6  0 so ( x  1) is a factor
You can then go through two different routes. Either take out x  1 as a factor (as in the
example on the next page) or find another factor. Pursuing this route gives
Let f ( x )  2 x3  5 x 2  x  6
f (2)  16  20  2  6  0 so ( x  2) is a factor
2 x 3  5 x2  x  6  ( x  1)( x  2)(2 x  3)
This final factor comes from observing that it must start with a 2x to give 2x 3 and
must end up with 3 to get 6 .
25
First Edition © C Morris. Only authorised for use by students at Reading School.
Solving cubic and cubic inequalities
Example
Solve the equation 2 x 3  7 x 2  3x  18  0 .
Hence solve the inequality 2 x 3  7 x 2  3x  18  0 .
Solution
Let f ( x )  2 x3  7 x 2  3 x  18
f (1)  10 so ( x  1) is not a factor of f ( x )
f (1)  12 so ( x  1) is not a factor of f ( x)
f (2)  0 so ( x  2) is a factor of f ( x)
Either by long division or by
comparing coefficients
f ( x )  ( x  2)(2 x 2  3x  9)
f ( x )  ( x  2)(2 x  3)( x  3)
f ( x )  0 when x  2 or 3 or  32
Consider the graph of y  f ( x)  2 x 3  7 x 2  3x  18.
y
y = (x – 2)(2x + 3)(x – 3)
15
10
5
-5
-4
-3
-2
-1
1
2
3
4
5
6
x
-5
-10
f ( x )  0 when the curve is on or above the x axis so f ( x )  0 when  32  x  2 or x  3.
NB See also the inequalities section of the IGCSE Revision Notes.
26
First Edition © C Morris. Only authorised for use by students at Reading School.
Discriminant
When solving the quadratic equation
ax 2  bx  c  0, a  0
You know that the solutions (if there are any) are given by the quadratic equation
formula
x
b  b2  4ac
2a
The part underneath the square root sign is called the discriminant, often given the
symbol  . So   b2  4ac .
The discriminant gives quite a lot of information about the solutions of a quadratic
equation 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
3x 2  2 x  4  0 has two solutions because   22  434  52  0 .
4 x 2  4 x  1  0 has one (repeated) solution because   42  4 41  0 .
3x 2  2 x  3  0 has no solutions because   22  433  32  0 .
6 x 2  x  2 factorises because   (1)2  462  49 which is a perfect square.
27
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The Binomial Expansion
Pascal’s Triangle
The coefficients for expanding (1  x)n come from the rows of Pascal’s Triangle.
1
1
1
1
1
1
1
1
7
2
3
4
5
6
1
3
6
10
15
21
1
1
4
10
20
35
1
5
15
35
1
6
21
The numbers in Pascal’s Triangle also come from nCr 
1
7
1
n!
using appropriate
r !(n  r )!
values of r and n.
For example the entries in the row beginning 1, 6, 15, 20, 15, 6, 1 come from
6
C0 , 6C1 , 6C2 , 6C3 , 6C4 , 6C5 , 6C6 respectively.
So we have that for positive integer values of n only.
Binomial Expansions Type 1
(1  x) n  1  nC1 x  n C2 x 2  n C3 x 3    nCr x r   xn
Where nCr 
n!
.
r !(n  r )!
n
Note that nCr is sometimes written as   (NB no fraction line!)
r
28
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Binomial Expansions Type 2
( a  x ) n  a n  n C1 a n 1 x  n C2 a n  2 x 2  nC3 a n  3 x3    n Cr a n r xr   x n
Example 1
Find the first five terms in the expansion of (1  x)10 in ascending powers of x.
Solution
(1  x)10  1  10C1 x  10C2 x 2  10C3 x3  10C4 x 4  ...
 1  10 x  45x 2  120 x 3  210 x 4  ...
Example 2
Find the terms in the expansion of (1  2 x)7 in ascending powers of x up to and
including the term in x 3 .
Solution
(1  2 x )7  1  7C1 ( 2 x )  7C2 ( 2 x )2  7 C3 ( 2 x )3  
 1  14 x  84 x 2  280 x 3  ...
Example 3
Find the binomial expansion of (3  2 x)5 and use your expansion to estimate 3.0025
correct to 1 decimal place.
Solution
(3  2 x)5  35  5C1 34 (2 x)  5C2 33 (2 x )2  5C3 32 (2 x )3  5C4 31 (2 x )4  (2 x )5
 243  810 x  1080 x 2  720 x 3  240 x4  32 x5
3.0025  (3  2  0.001)5
 243  810  0.001 (since higher power terms will not affect first dp)
 243.8 (1 decimal place)
29
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 4
Use the answer to example 2 to find the expansion in ascending powers of x up to and
including the term in x 3 of (2  3x)(1  2 x)7 .
Solution
(2  3 x)(1  2 x )7  (2  3x )(1  14 x  84 x 2  280 x3  ...)
 2  28 x  168 x 2  560 x3  3 x  42 x 2  252 x3  ...
 2  25 x  126 x 2  308 x3  ...
Application to Probability – Binomial Distribution
If X is the number of successes in n independent trials each of which has probability p
of success and probability q (  1  p ) of failure then X is said to have a Binomial
Distribution with parameters n and p and we write X  B(n, p) .
P( X  r )  n Cr p r q n  r
P( X  r )  n Cr p r (1  p)n  r
Mean of X  np
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( X  r ) n Cr p r (1  p)n  r there must be
r successes giving rise to the p r part of the formula
n  r failures giving rise to the (1  p)n  r part of the formula
The r success can occur in any of n Cr ways.
Note that the powers of p and 1  p always add up to n.
30
First Edition © C Morris. Only authorised for use by students at Reading School.
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
X  B (9, 0.15)
(a) P( X  5) 9C5  0.155  0.854  0.00499 (3 sf)
(b) P( X  3)  P( X  0 or 1 or 2 or 3)
P( X  3)  0.859 9C1  0.151  0.858  9C2  0.152  0.857  9C3  0.153  0.856  0.966 (3 sf)
(c) P( X  1)  1  P( X  0)
P( X  1)  1  0.859  0.768 (3 sf)
(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 at
least one six?
Solution
P(at least one six)  1  P(no sixes)
 1   56   0.99
n
 56 
n
 1  0.99
 56 
n
 0.01
 56 
25
 0.01048...  0.01
 56 
26
 0.008735...  0.01 (using trial and improvement)
At least 26 dice must be rolled for there to be a probability of at least one six
31
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Co-ordinate Geometry
The Straight Line
Gradient of a Straight Line
y
( x2 , y 2 )
x
Gradient 
y 2  y1
x2  x1
( x1 , y1 )
Example
Find the gradient of the line joining the points ( 3, 7) and (2, 13) .
Gradient 
13  7 20

 4
2  (3)
5
Mid-point of a Line Segment
 x  x2 y1  y2 
Mid point M is  1
,

 2
2 
y
( x2 , y 2 )
M
x
( x1 , y1 )
Example
Find the midpoint of the line joining the points ( 3, 7) and (2, 13) .
 3  2 7  (13) 
1
The midpoint is 
,
    2 , 3 
2
2


32
First Edition © C Morris. Only authorised for use by students at Reading School.
Length of a Line Segment
y
( x2 , y 2 )
Distance between ( x1 , y1 ) and ( x2 , y2 )  ( x2  x1 )2  ( y2  y1 ) 2
x
( x1 , y1 )
Example
Find the length of the line segment joining the points ( 3, 7) and (2, 8) .
Distance  (2  (3))2  (8  7) 2
 52  (15)2
 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. If
it is not you must rearrange the equation first.
Be careful that you give the answer in the required form. Sometimes a question will ask
you to give your answer a specific way e.g. in the form ax  by  c  0 where a, b and c
are integers.
There are several approaches each of which needs you to have a gradient and a point
that the line goes through. If you are given 2 points you can obviously find the gradient
from this.
33
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 1
The gradient of the line 3x  4 y  7  0 is 
3
since rearranging the equation gives
4
3x  4 y  7  0
4 y  3 x  7
3
7
y  x
4
4
Example 2
Find an equation of the straight line with gradient 3 going through the point with
coordinates (3, 7) .
Approach 1
y  3x  c
Line goes through (3, 7) so y  7 when x  3
7  3(3)  c
c  2
y  3x  2
Approach 2 (preferred)
Gradient is
y7
 3
x  (3)
y  7  3( x  3)
y  7  3x  9
y  3x  2
34
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 3
Find an equation of the straight line through the points with coordinates
(2, 5) and (  3,15) .
y  (5) 15  (5)

x2
3  2
Gradient
y  5 20

 4
x  2 5
y  5  4( x  2)
y  5  4 x  8
y  4 x  3
Parallel and Perpendicular Gradients

Two lines are parallel if and only if they have the same gradient.

Two lines with gradients m1 and m2 are perpendicular if and only if
m2  
NB If in an exam you are asked to show that
two lines are perpendicular, show that when
you multiply their gradients you get 1 .
1
i.e. m1m2  1 .
m1
Example
The equations of 5 lines are given below. Which lines are parallel to L and which lines
are perpendicular to L.
L : y  2x  5
M : 4x  2 y  8
N:
P:
2y  x  4
2x  4 y  5
R:
4x  2 y  3  0
Rearranging the lines give
L : y  2x  5
M : y  2x  4
N:
y   12 x  2
P:
y  12 x  25
R:
y  2 x  32
This shows that M and R are parallel to L and N are perpendicular to L.
35
First Edition © C Morris. Only authorised for use by students at Reading School.
Example
Find the equation of the line that is parallel to y  2 x  7 and goes through the point
( 3, 3) .
Solution
The gradient of y  2 x  7 is 2 so the equation of the required line is given by
y 3
2
x  (3)
y  3  2( x  3)
y  3  2x  6
y  2x  9
Example
Find the equation of the line that is perpendicular 3x  2 y  2  0 and goes through the
point (2, 5) giving your answer in the form ax  by  c  0 where a, b and c are
integers.
Solution
3x  2 y  2  0 can be rearranged to give y  32 x  1 and therefore has gradient
The gradient of the line perpendicular to 3x  2 y  2  0 is therefore given by
1
3
2
  23
The equation of the perpendicular line is therefore
y  (5)
2

x2
3
3( y  5)  2( x  2)
3 y  15  2 x  4
2 x  3 y  11  0
36
3
2
.
First Edition © C Morris. Only authorised for use by students at Reading School.
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 y  k1 will therefore be of the form x  k 2
where k1 and k2 are constants and vice versa.
Example
Find the equation of the line perpendicular to x  4 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 y  3 .
37
First Edition © C Morris. Only authorised for use by students at Reading School.
The Coordinate Geometry of Circles
Equation of a Circle
The equation of a circle centre the origin and with radius r is given by
x2  y 2  r 2
The equation of a circle centre ( a, b) and with radius r is given by
( x  a) 2  ( y  b) 2  r 2
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 r 2 must be positive
Examples
x 2  y2  121 is a circle centre the origin and radius 11.
3x 2  3 y 2  147 is a circle centre the origin and radius 7 since it can be rewritten as
x 2  y 2  49 by dividing throughout by 3.
( x  2)2  ( y  3)2  36 is a circle with centre (2, 3) and radius 6.
( x  3)2  ( y  1)2  16 is a circle with centre ( 3,1) and radius 4.
38
First Edition © C Morris. Only authorised for use by students at Reading School.
Finding the Centre and Radius of a Circle
The recommended approach to determine whether a given equation is a circle and to
determine its centre and radius is simply to complete the square from first principles on
the x and y terms.
Example 1
x2  y 2  8 x  6 y  5  0
x2  y 2  8 x  6 y  5  0
( x  4)2  42  ( y  3)2  (3)2  5  0
( x  4)2  ( y  3)2  20  0
( x  4) 2  ( y  3)2  20
This is a circle centre ( 4, 3) and radius
20  4  5  2 5 .
39
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Which of the following are equations of circles and why
(i)
x 2  2 y2  6 x  8 y  36
(ii)
x 2  y 2  2 xy  6 x  12 y  11  0
(iii)
( x  y )2  ( x  y)2  50
(iv)
y 2   x2  4 x  6 y  12
(v)
x 2  y 2  10 x  2 y  50  0
Solutions
(i)
This is not a circle because the coefficients of x 2 and y 2 are not the same.
(ii)
This is not a circle because there is a term in xy .
(iii)
When this is expanded you get
x 2  2 xy  y 2  x 2  2 xy  y 2  50
x 2  y 2  25
This is a circle centre the origin and radius 5.
(iv)
When rearranged this gives
x 2  4 x  y 2  6 y  12
( x  2)2  ( y  3)2  12  22  32  25
This is a circle centre (2, 3) and radius 5.
(v)
When rearranged this gives
x 2  y 2  10 x  2 y  50  0
( x  5)2  ( y  1) 2  50  52  12  24
This is not a circle because the right hand can’t be a radius squared since it is
negative.
40
First Edition © C Morris. Only authorised for use by students at Reading School.
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
 2  (4) 4  8 
The centre of the circle is at 
,
  1, 2 
2
2 

The diameter of the circle is given by
(4  2)2  (8  (4) 2  (6)2  12 2  180  36  5  6 5
The radius of the circle is therefore
6 5
3 5
2
The equation of the circle is therefore
 x  (1)    y  2 
2
 x  1   y  2
2
2
2
 (3 5) 2
 45
41
First Edition © C Morris. Only authorised for use by students at Reading School.
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 the
radius of the circle.
Example
Find the equation of the tangent to the circle x 2  y2  6 x  8 y  0 at the point with
coordinates (6, 8) .
x2  y2  6 x  8 y  0
( x  3)2  ( y  4)2  (3) 2  42  25
Centre is (3, 4), Radius is 5
Gradient of radius to (6, 8) 
4  (8)
4

3 6
3
Gradient of tangent at (6, 8)  
1
3

4 4

3
The equation of the tangent is therefore as before
y  (8) 3

x 6
4
4( y  8)  3( x  6)
3x  4 y  50
42
First Edition © C Morris. Only authorised for use by students at Reading School.
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 equation
of 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 the
circle.
Example
In the scenario outlined above, the equation of the normal at the point with coordinates
(6, 8) on the circle with equation x 2  y2  6 x  8 y  0 is given by
y  (8)
4

x 6
3
3( y  8)  4( x  6)
4x  3 y  0
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 through
the given point, A and the centre of the circle, C, meet the circle and use this point to
calculate the required distance.
y
4
2
C
P
A
2
4
x
43

Find the centre of the circle

Find the distance AC.

Find AP  AC  radius .
First Edition © C Morris. Only authorised for use by students at Reading School.
Example
Find the point on the circle with equation ( x  2)2  ( y  7)2  5 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
(8  2)2  (19  7) 2  62  122  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 r1 and r2 are the radii of two circles, r1  r2 and d is the distance between their
centres then
(i) Circles touch externally
(ii) Circles touch internally
d  r1  r2
(iii) Circles do not intersect
d  r2  r1
(iii) Circles intersect at two distinct points
d  r1  r2
r2  r1  d  r1  r2
44
First Edition © C Morris. Only authorised for use by students at Reading School.
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 representing
the inequality x  3 you would shade the region that is NOT x  3
y
5
4
3
2
1
y
x>3
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
x2
5
4
3
2
1
y
5
4
3
2
1
y
5
4
3
2
1
y> –2
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
y
y1
y
5
4
3
2
1
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
1 2 3 4 5 x
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
5
4
3
2
1
1 2 3 4 5 x
y<x+2
-5 -4 -3 -2 -1-1
-2
-3
-4
-5
45
1 2 3 4 5 x
x – 2y  2
First Edition © C Morris. Only authorised for use by students at Reading School.
If you are not sure which way to shade, just pick a test point off the line and see whether
it satisfies the inequality. You will know which way to go then.
In the last diagram, for example, if you use the point (3, 2)
x  2 y  3  2  2  7  2 so (3, 2) does satisfy the inequality.
you get
When drawing a graph such as 3x  2 y  24 a very quick way is to find out where it
meets the axes by putting x  0 to get y  12 and by putting y  0 to get x  8 .
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 the
following inequalities. You should shade the region that is not required.
x2
y2
x  y  3
Solution
y
4
x=2
y=2
2
-6
-4
-2
0
-2
-4
x+y= 3
46
2
4
6
x
First Edition © C Morris. Only authorised for use by students at Reading School.
Applications to Linear Programming
Work on regions can be extended to encompass problems involving linear
programming.
This is where an objective function is maximised or minimised subject to certain
constraints that are usually able to be expressed as inequalities.
Consider the feasible region defined by the inequalities
x  2y  6
2x  y  8
x  2
y  3
Suppose we wish to maximise the objective function x  y .
Such maximum points usually occur at a vertex (corner) of the region. The slope of
lines of the form x  y  k indicates that in this case this would be where x  2 y  6
and 2 x  y  8 meet. By solving simultaneously this pair of equations it can be shown
that these meet at  3 13 ,1 13  . The maximum value of x  y is therefore 3 13  1 13  4 23 .
y
4
x + 2y = 6
2
x= 2
-4
-2
0
2
4
2x + y = 8
-2
-4
47
y= 3
x
First Edition © C Morris. Only authorised for use by students at Reading School.
Example
Grizelda is going to make some small cakes to sell at school and raise money for
charity. 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 kg
of 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 the
number of batches of yummy munchies, write down and simplify two
inequalities relating to the available ingredients.
(ii)
Illustrate the region satisfied by these inequalities, using the horizontal axis
for 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 the
greatest number of cakes that Grizelda can make.
48
First Edition © C Morris. Only authorised for use by students at Reading School.
Solution
(i)
300 x  200 y  2000 (restriction on flour) Using m to represent the number
3x  2 y  20
50 x  125 y  750 (restriction on butter)
2 x  5 y  30
(ii)
y
20
15
10
3x + 2y = 20
5
The coordinates (4, 4)
give the largest value of
12 x  16 y in the feasible
region i.e. 112 cakes.
(iii)
2x + 5y = 30
0
5
10
15
N = 12x + 16y
20
The objective function is N  12 x  16 y .
Remember that lines of the form 12 x  16 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 the
number of batches of each must be a whole number). Remember that this
will usually be at or near a corner of the region formed by the constraints.
The largest value of the objective function is 112 when x  4, y  4 so 112
cakes is the largest number that can be made.
49
x
First Edition © C Morris. Only authorised for use by students at Reading School.
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 use
of the sine rule to find an angle.

Finding areas of triangles including use of the formula
1
ab sin C
2
Remember that you will not have any of the formulae and will have to have learnt
them!
Applications
You are much more likely to be asked application of trigonometry. Likely contexts
include the use of terms such as
Angle of elevation
Angle of elevation
x°
Horizontal
Angle of depression
x°
Horizontal
Angle of depression
Problems involving bearings and other real life situations are also very likely so
make sure that you revise this.
50
First Edition © C Morris. Only authorised for use by students at Reading School.
Graphs of Sine, Cosine and Tangent for Any Angle
y
y = sin x
1
0.5
-360
-270
-180
-90
0
90
-0.5
-1
Observe that
sin 0  sin180  sin 360  0
sin 90  1
sin 270  1
sin 30  sin 150  0.5
sin(180  x )  sin x 
sin(  x )   sin x
The sine graph repeats itself every 360 .
Be prepared to sketch this graph to help
you solve trigonometric equations.
51
180
270
360 x
First Edition © C Morris. Only authorised for use by students at Reading School.
y
1
y = cos x
0.5
-360
-270
-180
-90
0
90
180
-0.5
-1
Observe that
cos 0  cos360  1
cos180  1
cos90  cos 270  0
cos 60  cos300  0.5
cos120   cos 60  0.5
cos(180  x)    cos x 
cos( x )  cos x
The cosine graph repeats itself every 360 .
The cosine graph is the sine graph moved
90 to the right.
Be prepared to sketch this graph to help you
solve trigonometric equations.
52
270
360 x
First Edition © C Morris. Only authorised for use by students at Reading School.
y
10
y = tan x
8
6
4
2
-360
-270
-180
-90
0
-2
90
180
-4
-6
-8
-10
Observe that
tan 0  tan180  tan 360  0
tan 45  1
tan135   tan 45  1
tan 90  
tan 270  
tan(180  x )   tan x 
tan(  x )   tan x
The tangent graph repeats itself every 180 .
53
270
360 x
First Edition © C Morris. Only authorised for use by students at Reading School.
Trigonometric Identities
The following results are true for all values of  .
sin 2   cos 2   1
sin 
 tan 
cos 
It is also worth remembering that in a right angled triangle

z
y
90 – 
x
sin  
x
 cos(90   )
z
cos  
y
 sin(90   )
z
tan  
x
1
1


y  y  tan(90   )
 
x
54
First Edition © C Morris. Only authorised for use by students at Reading School.
Solving simple trigonometric equations
Example 1
Solve the equation sin   0.4 for 0    360 .
Solution
One value can be found from   sin 1 (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   tan 1 (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   cos 1 (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 about
  180 .
  360  101.5  258.5 (1 dp)
55
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 4
Solve sin   2 cos  where 0     360 .
Solution
sin 
2
cos 
tan   2
  63.4 or (63.4  180)
(using periodicity of tan graph)
  63.4 or 243.4
Example 5
Solve sin   2 cos   0 where 0     360 .
Solution
After a simple rearrangement this is the same as Example 4!
Example 5
Solve 2 sin 2   1 where 0     360 .
Solution
sin 2  
1
2
1
2
sin   
When sin  
1
2
  45 or 135
When sin   
1
2
  225 or 315
56
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 6
Solve 3 cos 2   5sin   5  0 for 0    360 .
Solution
3 cos 2   5sin   5  0
3 1  sin 2    5sin   5  0
3  3sin 2   5sin   5  0
3sin 2   5sin   2  0
Let y  sin 
3y2  5 y  2  0
 3 y  2  ( y  1)  0
3 y  2  0 or y  1  0
y
2
3
or y  1
sin  
2
3
or sin   1
sin  
2
3
leads to
  41.8 or 180  41.8  138.2
sin   1 leads to
  90
Example 7
Solve the equation tan  cos   1 for 0    360 .
Solution
tan  cos   1
sin 
cos   1
cos 
sin   1
  270
57
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 8
Solve the equation tan   2 sin  for 0    360 .
Solution
tan   2sin 
sin 
 2sin 
cos 
sin   2sin  cos 
sin   2sin  cos   0
sin  (1  2cos  )  0
sin   0
  0 or 180 or 360
or 1  2cos   0
cos  
1
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
58
First Edition © C Morris. Only authorised for use by students at Reading School.
Trigonometry and Pythagoras in 3 Dimensions
There are likely to be problems involving trigonometry and Pythagoras’ Theorem in 3
Dimensions and these may well be more involved and be more demanding that at
IGCSE. You are more likely to find situations where the use of the sine or cosine rules
will be more efficient.
Remember the key skills of identifying which angles you are working with and
extracting suitable triangles to work with.
You also need to be familiar with some additional terms that you might not have met
before 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 rightangled 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 if
released on the slope!
Line of Greatest Slope
59
First Edition © C Morris. Only authorised for use by students at Reading School.
Angle between two planes
This is the angle between lines in the two planes that meet at right angles to where the
two planes meet.
Angle Between the Planes
60
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Calculus
Differentiation
Notation
dy
is the (first) derivative of y with respect to x.
dx
It is the rate of change of y with respect to x.
d2 y
is the second derivative of y with respect to x.
dx 2
dy
This is the derivative of
with respect to x.
dx
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
 something  .
dx
For example if you are differentiating x 3  3x 2  3x  2 you will often see this written as
d 3
( x  3 x 2  3 x  2) .
dx
Gradient Function
dy
is also called the gradient function as it gives the gradient of a curve at the point
dx
( x , y ) when the x-coordinate of the point is substituted into it.
61
First Edition © C Morris. Only authorised for use by students at Reading School.
Differentiation of powers of x and constant multiples, sums and differences.
d
ax n   naxn1 where a is a constant.
dx
Multiply by the power and reduce the
power by 1.
d  n
m
n1
m1
where a and b are constants,
 ax  bx   nax  mbx
dx 
Remember to write functions as powers of x before you differentiate and make sure that
you simplify expressions first.
Examples
The gradient function of something
of the form mx  c is simply m from
the work covered on the gradient of
a straight line.
d
4 x 3  3x 2  4 x  7   12 x2  6 x  4

dx
The gradient function of a number
(e.g. 5) is 0 because we know that
y  5 is a straight line horizontal to
the x-axis and hence has gradient 0.
d
d
(2 x  1)(3x 2  2)   6 x3  3x2  4 x  2

dx
dx
 18 x 2  6 x  4
d  x 7  x11  d 4
8

  x  x 
3
dx  x
 dx
 4 x3  8 x7
62
You must write anything to be
differentiated as powers of x before
you begin, multiplying or dividing
out before you begin.
First Edition © C Morris. Only authorised for use by students at Reading School.
Equations of tangents and normals
The normal to a curve at a point is the line perpendicular to the tangent at that point as
shown by the diagram below.
y
Normal
Tangent
x
Example
Find the equation of the tangent and the normal to the curve y  3x 2  2 x at the point
where x  2 .
dy
 6x  2
dx
This simply means
put the value x  2
into the gradient
function.
When x  2, y  322  2 2  16
dy
dx
Equation of the tangent is
x 2
 6 2  2  14
y 16
 14
x2
y 16  14 x  28
y  14 x 12
y 16
1
Equation of the normal is

x2
14
14( y 16)  ( x  2)
14 y  x  226
63
The gradient of the
normal is 1 divided by
the gradient of the
tangent.
First Edition © C Morris. Only authorised for use by students at Reading School.
Location and Nature of Stationary Points
A stationary point is a point on a curve for which
dy
0.
dx
Stationary points can be maximum points, minimum points (these are both called
turning points) or points of inflexion (sometimes spelt inflection).
y
Maximum
Point
Point of
Inflexion
Minimum
Point
x
In order to determine the nature of a stationary point (i.e. to find out what sort of
stationary point it is) you can either
 dy 
(a) find the gradient   either side of the point, or
 dx 
 d 2 y 
(b) use the second derivative  2 
 dx 
If you use the first approach don’t go too far away from the stationary point or you
might move past another stationary point and draw an incorrect conclusion.
Conclusion
dy
to the left
dx
dy
at the point
dx
dy
to the right
dx
0
Maximum point
+ ve /
0 –––
\  ve
0
Minimum point
 ve \
0 –––
/ + ve
Point of inflection
+ ve /
0 –––
/ + ve
Point of inflection
 ve \
0 –––
\  ve
Value of
0
d2 y
dx 2
Check
dy
either side
dx
64
First Edition © C Morris. Only authorised for use by students at Reading School.
Note: The examples here are harder than you will meet in the examination but they
illustrate the principles well.
Example 1
Find the coordinates and nature of the stationary points on the curve y  x 3  2 x 2  4 x .
Solution
dy
 3x 2  4 x  4  (3 x  2)( x  2)
dx
d2 y
 6x  4
dx 2
dy
 0 when x  2 or x   23
dx
When x  2, y  8; when x   23 , y 
d2 y
dx 2
d2 y
dx 2
40
27
 8  0, so there is a minimum at  2, 8
x 2
40
 8  0, so there is a maximum at   23 , 27

x  23
65
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Find the coordinates of any stationary points on the curve y  x3 ( x  1)2 and determine
their nature.
Solution
y  x 3 ( x  1)2  x3 ( x  1)( x  1)
 x 3 ( x 2  2 x  1)
 x5  2 x4  x3
dy
 5 x 4  8x 3  3 x 2
dx
d2 y
 20 x 3  24 x 2  6 x  2(10 x3  12 x2  3 x)
2
dx
dy
 5 x 4  8x 3  3 x 2  0
dx
When
x 2 (5 x 2  8x  3)  0
x 2 (5 x  3)( x  1)  0
x  0 or
d2 y
dx 2
dy
dx
or 1
 0 (no information gained)
x 0
 0.0385  0
x 0.1
dy
dx
d2 y
dx 2
3
5
 0.0225  0 so a point of inflexion at (0,0) (same sign on both sides of st pt)
x 0.1
 0.72  0 so a maximum at (0.6, 0.03456)
x 0.6
d2 y
dx 2
 2  0 so a minimum at (1, 0)
x 1
66
First Edition © C Morris. Only authorised for use by students at Reading School.
Sketching Curves
You may be expected to sketch simple curves using any of the mathematics included in
the syllabus. Useful things to bear in mind (although not all would be required on any
particular sketch, nor should you work them all out unless you are asked to) are

Where the curve meets the x-axis (i.e. when y  0 )

Where the curve meets the y-axis (i.e. when x  0 )

The coordinates of any stationary points that you might know

The general shapes of particular types of curves that you might know such as
y  x2
y  x3
y
1
x
y
1
x2
67
First Edition © C Morris. Only authorised for use by students at Reading School.
Practical Maximum and Minimum Problems
Example
A cylindrical tin has a volume of 128 cm3. Find the dimensions necessary for the tin to
have the minimum possible surface area, and find the minimum possible surface area.
r
h
Let the base radius be r and the height be h.
Surface area, S  2
 r 2  2
 rh
Circular ends
Curved surface
Volume, V   r 2 h  128
h
128
r2
S  2 r 2  2 r
 2 r 2 
128
r2
256
r
dS
256
 4 r  2
dr
r
d2S
512
 4  3
2
dr
r
dS
256
 0 when 4 r  2  0 i.e. when r 3  64 i.e. when r  4
dr
r
d2 S
dr 2
 4  8  12  0 i.e. S is a minimum value when r  4 cm
r4
So the radius, r  4 cm and the height, h 
Smin  2  42 
128
 8 cm
42
256
 96 cm 2
4
68
First Edition © C Morris. Only authorised for use by students at Reading School.
Integration
Integration as the Reverse of Differentiation
dy
 dx dx  y  c where c is an arbitrary constant of integration.
Indefinite Integration of powers of n, constant multiples, sums and differences

ax n1
c
n 1
ax n dx 
n  1 .
Remember c once you have integrated.
 a f ( x) dx  a  f ( x) dx
Make sure that you have written things
as powers of x before you integrate.
 [f ( x)  g( x)]dx   f ( x) dx   g( x) dx
Examples
3
 3x dx  8 x
7
 (x
2
8
c
 2)( x 3  1)dx   ( x 5  x3  x 2  2)dx


x6 x4 x3
   2x  c
6 4 3
x10  5 x8
dx   ( x 6  5 x 4 )dx
4
x
x7 5 x5
 
c
7
5
x7
  x5  c
7
69
First Edition © C Morris. Only authorised for use by students at Reading School.
Finding the constant of integration using given conditions
Example 1
Suppose that a curve y  f ( x) is such that
point (1, 2) .
dy
 2 x  1 and the curve passes through the
dx
Solution
y   (2 x  1) dx  x 2  x  c
2  12  1  c
since the curve passes through (1, 2)
c  4
so y  x 2  x  4
Example 2
Suppose that
dy
 3x 2  4 x  b where b is a constant.
dx
Given that y  7 when x  1 and that y  12 when x  2 , find y in terms of x.
Solution
dy
 3x 2  4 x  b
dx
y   (3x 2  4 x  b)dx
 x3 
4 x2
 bx  c
2
y  x 3  2 x 2  bx  c
7  13  2 12  b  1  c
bc 4
 (1)
12  23  2  22  b  2  c
2b  c  4  (2)
b  8
(2)  (1)
c  12
substituting for b in (1)
y  x 3  2 x 2  8 x  12
70
First Edition © C Morris. Only authorised for use by students at Reading School.
Definite Integrals
b
If  f ( x ) dx  F( x )  c then  f ( x ) dx  F(b)  F(a) .
a
Area between a curve and the x axis
y
y = f(x)
a
x
b
To find the area between the curve y  f ( x) the x-axis and the lines x  a and x  b we
calculate
b
b
a
a
 y dx   f ( x) dx  F(b)  F(a )
For example the area between the x-axis, the curve y  x 2  x  2 and the lines x  1
and x  3 is given by
3
Area   ( x 2  x  2) dx
2
y
1
y=x –x+2
10
9
8
7
6
5
4
3
2
1
3
 x3 x2

    2x
3 2
1
 33 32
  13 12

    2  3      2 1 
3 2
 3 2

9
1
1
 9  2  6   3  2  2
 8 23
-3 -2 -1
71
1
2
3
4
5 x
First Edition © C Morris. Only authorised for use by students at Reading School.
Note that sometimes areas can be calculated using more straightforward methods. For
example, when working out the area under a straight line you can use the area of a
triangle or trapezium.
Note also that using symmetry can occasionally make calculations quicker.
Example
Find the shaded area between the curve y  x 3 and the line y  4 x .
Solution
The curve and line meet when
3
y
y=x
y = 4x
10
8
6
4
2
x3  4 x
x3  4 x  0
x ( x 2  4)  0
x  0, 2
-3
-2
The area between 2 and 0 is clearly the
same as the area between 0 and 2 so we
require twice the area under the line take
away the area under the curve.
-1 -2
-4
-6
-8
-10
1
There are two ways of proceeding. The first is completely by integration.
2
2
0
0
2
Area  2 4 x dx  2 x3 dx  2  (4 x  x 3 ) dx
0
2

x4 
 16 

 2  2 x 2    2  8     0  0    8 square units
4 0
4



The second uses the fact that the area under the line is the area of a triangle.
When x  2 , y  8 .
2
Area  2  12  2  8  2 x 3 dx
0
2
x 
16 
 16  2    16  2   0  8 square units
4

 4 0
4
72
2
3 x
First Edition © C Morris. Only authorised for use by students at Reading School.
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 the
two parts separately and combine the sizes of the areas.
Example
For example consider the area between the curve y  x 2  9 , the x-axis and the lines
x  2 and x  5 .
A sketch shows that part of the area is below the x-axis and part is above.
y
2
y=x –9
12
8
4
-2 -1
-4
1 2 3 4 5 6 7 8 9 10 x
-8
3
 x3

(
x

9)
d
x


9
x
  9  27    83  18  2 32


2
3
2
3
2
The  sign is because the area is below the axis.
The actual area is 2 23 .
5
 x3

125
2
(
x

9)
d
x

 3  9 x    3  45    9  27   14 3
3

3
5
2
Area required  2 23  14 23  17 13
5
NB Had we done  ( x 2  9) dx then we would have obtained that the (incorrect) answer 12!
2
This is because the integral would have given  2 23  14 23  12.
73
First Edition © C Morris. Only authorised for use by students at Reading School.
Area between two curves
To find the area between two curves y  f ( x) and y  g( x ) which meet at x  a and
x  b which are such that f ( x)  g( x ) between these values find
b
 f ( x)  g( x) dx
a
Example
To find the area between the curves y  x 2  2 and y   x 2  6 .
y
10
8
6
4
2
-6
-4
-2
2
y=x –2
2
-2
-4
-6
-8
-10
4
x
6
2
y= –x +6
First find where the curves meet.
x2  2   x2  6
2x2  8
x  2
We therefore need to evaluate the integral of the top curve take away the bottom one.
2
Area 
   x
2
2
2

  2 x
2
2
 6    x 2  2   dx
 8  dx
2
 2

   x3  8 x 
 3
 2
 
16
3
Note that if you had subtracted the integrals the
wrong way round you would have obtained the
answer 21 13 .
Note also that because of the symmetry of both
curves we could have found instead
2
2   x 2  6    x 2  2   dx
 16     16
16
3
0
 21 13
74
First Edition © C Morris. Only authorised for use by students at Reading School.
Application to Kinematics
Motion in a Straight Line
You need to be able to apply your knowledge of differentiation to motion in a straight
line.
The standard notation is
t
s
ds
dt
dv
a
dt
v
time
distance
velocity
acceleration
Maximum/minimum velocity occurs when maximum of graph i.e.
a0.
dv
 0 i.e. when
dt
Remember also that
s   v dt
v   a dt
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 its
value.
75
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 1
The distance of a particle, s metres, from a fixed point O after time t seconds is given by
the formula s  t 3  t 2  5t  2 .
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) v 
ds
 3t 2  2t  5
dt
(ii) a 
dv
 6t  2
dt
(iii) When t  2 , v  3  22  2  2  5  12  4  5  3 m/s
(iv) When t  2 , a  6  2  2  10 m/s 2
(v)
v  3t 2  2t  5  0
(3t  5)(t  1)  0
3t  5  0 since t  1  0 gives a negative time which is impossible
t  1 23 seconds
76
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
A particle moves such that its displacement s metres after time t seconds is given by
s  t3  t2  t 1.
(a)
Determine the value of t for which the velocity is 0.
(b)
Show that its acceleration is never 0.
Solution
v
(a)
ds
 3t 2  2t  1  (3t  1)(t  1)
dt
When v  0 , t  13 since t must be positive.
(b)
a
dv
 6t  2
dt
The only value of t that gives a  0 is t   13 and since time is always positive
the acceleration can never be 0.
Example 3
A particle has acceleration given by a  9t 2  1 where t is the time since the particle
started moving. Find the velocity in terms of t given that its initial velocity is 7 ms-1.
Solution
a  9t 2  1
v   (9t 2  1) dt 3t 3  t  c
When t  0, v  7 so 7  3  02  0  c
c7
v  3t 3  t  7
77
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 4
The velocity of a particle is given by v  t 2  t at time t seconds. Find the distance
moved in the 3rd second (i.e. between 2 and 3 seconds).
Solution
s    t 2  t  dt
s
t3 t2
 c
3 2
8 4
14
When t  2, s    c   c
3 2
3
When t  3, s 
27 9
27
 c 
c
3 2
2
 27
  14

Distance travelled    c     c   8 56 metres
 2
  3

Another alternative approach is to use a definite integral.
3
s    t 2  t  dt
2
3
t3 t2 
s  
 3 2 2
 27 9   8 4 
s       
 3 2 3 2
Distance travelled  8 56 metres
78
First Edition © C Morris. Only authorised for use by students at Reading School.
SUVAT Equations (Constant Acceleration Formulae)
These are for motion with constant acceleration in a straight line (horizontal or vertical)
only.
v  u  at
s  displacement
v 2  u 2  2as
u  initial velocity
s  ut  12 at 2
v  final velocity
uv
t
2
a  acceleration
s
s  vt  12 at 2
t  time
Remember:

As these are vector quantities they have both magnitude and direction, so the signs
of the quantities in these equations can be positive or negative. Choose a direction to
be 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 quantity
that you are trying to calculate.

If there are two unknowns to find it may be that you need to form and solve a pair of
simultaneous equations.

If a particle is projected up and then falls down again you can look at the motion in
two stages, but this is not required, as long as you are careful with directions.
Other useful results are
Average Speed 
Average Velocity 
Total Distance
Total Time
Total Displacement
Total Time
When projecting a particle vertically, at maximum height the velocity will be zero.
79
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 1
A car begins to accelerate at 0.5 ms-2 for a distance of 500 m. At the end of this the car
is travelling at 30 ms-1, calculate the initial speed of the car.
Solution
We know:
so use:
a = 0.5
s = 500
v = 30
u=?
v2 = u2 + 2as
302  u 2  2  0.5  500
u 2  30 2  2  0.5  500
u 2  400
u  20 ms-1
Example 2
A particle is projected upwards with a velocity of 30 ms-1. What is the maximum height
it will reach?
Solution
Because the particle is only acting under its own weight, its acceleration will be due to
gravity.
Because it reaches maximum height the final velocity at that height will be zero.
We therefore know:
so use:
s=?
u = 30 ↑
v=0
a = - 9.8 ↑
v2 = u2 + 2as
0  302  2  ( 9.8)  s
19.6 s  900
s  45.9 m (3 sf)
80
First Edition © C Morris. Only authorised for use by students at Reading School.
Example 3
Daisy is cycling at a steady speed of 6 ms 1 when she comes to a hill which causes her
to slow down at a rate of 0.5 ms2 .
(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 2 ms1 .
Solution
(a)
u6
v0
a  0.5
Using v 2  u 2  2as
0 2  62  2  0.5  s
0  3 s
s  36
She travels 36 metres up the hill
(b)
u6
v2
a  0.5
Using v  u  at
2  6  0.5  t
0.5t  4
t
4
8
0.5
She takes 8 seconds to slow down to 2 ms2
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First Edition © C Morris. Only authorised for use by students at Reading School.
Example 4
A particle is projected with a velocity of 20 ms-1. For how long will it be above a height
of 16m?
Solution
We know:
u = 20 ↑
a =  9.8 ↑
s = 16 ↑ i.e. from the floor
t=?
so use:
s = ut + ½ at2
16 = 20t – 4.9t 2
0 = 4.9t 2 – 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 and
on the way down during its motion.
The time above 16m will be 2.99  1.09  1.9 seconds .
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First Edition © C Morris. Only authorised for use by students at Reading School.
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-time
graphs).
(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 into
account and distance if only the total of the magnitudes of the areas are
considered.
Example 1
Displacement
(metres)
4
A
-6
B
C
D
2
4
5
F
6.5
8.5
time
(seconds)
E
Between points A and B on the graph the particle is travelling with a constant velocity
of 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 constant
velocity –4 ms-1 i.e. in the opposite direction to its initial motion. Between D and E the
particle travels 6 m away from its starting point in the opposite direction, still with
constant velocity –4 ms-1. For the final part of the journey between E and F the particle
travels with constant velocity 3 ms-1 in its original direction until it returns to its starting
point.
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First Edition © C Morris. Only authorised for use by students at Reading School.
Example 2
Velocity
-1
(ms )
4
The total of areas ABCD and
DEF represents the distance
travelled.
C
B
D
A
2
4
-6
5
6.5
F
8.5
time
(seconds)
E
At the point on the graph marked A, the particle starts at rest. Between points A and B
on the graph the particle is travelling with a constant acceleration of 2 ms-2 away from
its starting point. Between B and C the particle is travelling with constant velocity
4 ms1 away from its starting point.
Between C and D the particle decelerates with deceleration 4 ms-2 until at D it is
instantaneously stationary. It continues to decelerate with deceleration 4 ms-2 (and travel
away from its starting point in the opposite direction) until at E it is instantaneously
travelling at –6 ms-1 away from its starting point. It then accelerates with acceleration 3
ms-2 (N.B. an object travelling with negative velocity and positive acceleration is
slowing down) until at F it is stationary again.
The total distance travelled can be calculated by evaluating areas ABCD and DEF.
Area ABCD 
Area DEF 
(5  2)
 4  14
2
(This is a trapezium)
3.5  6
 10.5
2
Total distance travelled  14  10.5  24.5 metres
The displacement from its
original position needs to
take into account the
direction of the motion i.e.
you need the area of
ABCD  the area of DEF
 14  10.5  2.5 metres
Notice that Examples 1 and 2 are quite different even though the graphs look the same!!!
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