Contents

Formulae in other document but otherwise should be all covered...

ContentsFrom additional mathematics for OCR book

Linear expressionsQuadratic expressionsCompleting the squareSimultaneous equations

Contents

This is basically GCSE if not SAT revision

When simplifying remember to: Collect like terms Remove brackets Factorise Find a common

denominator when involving fractions

When solving an equation remember to: Simplify Do the same on both

sides so that it remains the same equation

They sometimes ask to rearrange an expression in which case be careful to do so correctly

Ex 1A, B & C

Contents

This is when the highest power is 2

It often involves expanding and factorising

Before starting to solve a quadratic equation, make sure that all terms of the quadratic are on the left hand side of the equation

There are 3 ways to solve a quadratic equation:1. Factorise2. Completing the square3. Using the Quadratic

Formula Remember that the

formula is:

You use this when you are in a calculator test and cannot factorise

Ex 1D

Contents

It is used when you cannot factorise a quadratic

It is also useful when sketching a graph as it identifies the line of symmetry and the vertex

Method: X²-8x=-3*

Take the coefficient of x-8

Half it

-4 Square the answer

+16 Add this to both sides Factorise where

possible Take the square root

of both sides Add the constant to

both sides Find the answer

Method: Y = x²-8x +3*

This can be written as y=(x-4)²-13

See previous example and its answer

Therefore the line of symmetry is x=4 and the vertex is (4,-13)

Ex 1E

Contents

This is when there is more than one variable

...by substitution Suitable for when y is

the subject Take the expression for y

from the equation and substitute it in the other equation – then solve as before

...by elimination Suitable for when y is

not the subject or either equation

Multiply the equations as so that when they are subtracted/added from the other they eliminate variables

Substitute this into the first equation and thus solve

Ex 1F

Linear inequalitiesQuadratic in equalitiesAlgebraic fractionsExpressions containing a square root

Contents

Like simplifying linear expressions, you do the same to both sides

However, remember to have the inequality sign the right way and whether it is equal to or not

You may also be asked to show the answer on a number line

In this case, remember that open circles at the end of the line show that the number is not included

Closed circles mean the figure is included in your answer

Both can be used in a single answer

Ex 2A

Contents

There are two methods: Sketching a graph to

show the answer Or drawing up a table

showing the values of x But remember that if it

has terms on both sides these must be collected to one side

These quadratic inequalities will be able to be factorised

Remember to be careful in reading and working the question especially when using a graph

Ex 2B

Contents

Algebraic fractions follow the same rules as the fractions in arithmetic

The common denominator should be the lowest common multiple of the original denominators

Other than being asked to simplify an algebraic fraction you may be asked to solve an equation involving fractions

This is done in the same way as before but also having to simplify fractions

Remember that when you multiply a fraction you only multiply its numerator

Ex 2C & D

Contents

It is often easier to use surds when working with square roots to get a more accurate answer than just working out the numerical value

You should try to make the number that is under the square root sign as small as possible or as easy to work with as possible

Rationalising the denominator is an important technique to be aware of

Ex 2E

TrianglesSine ruleCosine rule

Contents

Just remember the hyp, app and adj. And that θ is used for the angle

The Trigonometric ratios are:

Remember: sinθ (etc.) will give you the ratio sin known side (etc.)will give you the angle side known X sinθ (etc.) will give you the side’s length

-1

Contents

Sin is Opposite divided by Hypotenuse.

Opposite is a helpful way of remembering it.

Contents

Tan is Opposite divided by AdjacentAn easy way to remember is it doesn’t have the hyp and opp is always on topOpposite is a helpful way of remembering it.

Contents

Cos is Adjacent divided by Hypotenuse

Opposite is a helpful way of remembering it.

Contents

It is an extension of Pythagoras’ theorem which allows it to be applied to any triangle

Contents

It is based on that fact that in any triangle the length of any edge is proportional to the sine of the angle opposite to that edge

=

=

Contents

This formula (which is cyclic) is for finding the area of a triangle when the lengths of 2 edges are known and also the size of the angle between them

=

=

=

3d work

Contents

All Silver Tea Cups

Anti-clockwise is always positive Clockwise is negative Always go from the x axis Cosine and Sine are between -1 and 1 whereas Tangent is

over 1

AllSilver

Tea Cups

SinCosTan

SinCosTan

SinCosTan

SinCosTan

+++

-+-

+--

--+

1

1

-1

-1

It is like have a circle of one unit

Contents

It is really important to draw good diagrams

There are two types: Representations of 3D

objects True shape diagrams of

2D sections in a 3D object

Contents

The two main identities that need to be learnt:

Introduction – Curves, Tangents, and NormalsGradient of a curveDifferentiationTangents and normalsStationary points and Higher Derivatives

Contents

Cord: joins two points on the curve

Tangent: touches the curve at a point of contact

Normal: perpendicular to the tangent at the point of contact

The tangent to a curve can be considered as the limit position of a chord

Curved line

Contents

As B gets closer to A we can say that B tends to A (written as BA) The gradient of the cord AB the gradient of the tangent at A E.g. y = x²

A = (xA ,yA) B = (xB, yB)

From the table, we can assume that the gradient of the tangent to the graph y = x² at A(2,4) is 4

Xa Ya = (Xa)² Xb Yb = (Xb)² Mab = Xb - Xa

2 4 3.5 12.25 5.5

2 4 3 9 5

2 4 2.5 6.25 4.5

2 4 2.25 5.0625 4.25

2 4 2.1 4.41 4.1

2 4 2.05 4.2025 4.05

2 4 2.001 4.004001 4.001

Yb - Ya

Contents

X Gradient

1 2

2 4

3 6

4 8

5 10

x 2x

X Gradient

1 4

2 32

3 108

4 256

5 500

x 4x³

X Gradient

1 3

2 12

3 27

4 48

5 75

x 3x²

y = x³y = x²

y = x⁴

Contents

The value of the gradient of the chord AB as B tends to A is called the differential coefficient of y with respect to x or the derivative of y with respect to x. The limit is denoted by the symbol (read as ‘dy by dx’)

The process of obtaining the differential coefficient or derivative of a function is called differentiation.

Note that ‘d’ has no independent meaning and must never be regarded as a factor. The complete symbol means ‘the derivative with respect to x of [previous expression]’

We may also write when y is a function of x as f’(x) or y’

dy dx

dy dx

ddx

Contents

Contents

ddx

ddx

ddx

ddx

12√x

Contents

Let y=c Graphically this is a

horizontal straight line and its gradient is zero

Therefore differentiating a constant will give you zero i.e. (c) = 0

y

x

y = c

0

ddx

Contents

(axⁿ) = a (xⁿ) = anxⁿ⁻¹ Where ‘a’ is a constant

i.e. (axⁿ) = anxⁿ⁻¹

For example: (3x⁶) = 18x⁵

ddx

ddx

ddx

ddx

Contents

We differentiate each term and then add or/and subtract the terms as necessary

For example: (x⁷ + 5x² - 3x + 4)

= 7x⁶ + 10x - 3

ddx

Contents

The gradient of the chord AB as it tends to the point A, is the value of the derivative at that point A.

We can use this to find the equation of the tangent and/or of the normal to a curve at a given point

Contents

Q. Find the equation of the tangent and of the normal to the curve:y = x² + 3x – 10 at the point (1, -6)

First differentiate the equation to give: = 2x + 3 at x = 1

Thus: m = 2 X 1 + 3 = 5

Using: y – y₁ = m(x – x₁) substitute the known values y + 6 = 5(x – 1) y = 5x – 5 – 6 y = 5x – 11 equation of tangent

Then to find the equation of the normal: m = 5 so m¹ = -⅕ y + 6 = -⅕ (x-1) use previous method but using -⅕ instead of 5 5y + 30 = -x + 1 x + 5y + 29 = 0 equation of the normal

ddx

Contents

This is basically doing the second derivativeThis is just differentiating what you already

have differentiated It can be used to find stationary points in

increasing and decreasing functions

Contents

Increasing is from A to B and from C – represented with the + This means that is

positive Decreasing is from B to C

– represented with the - This means that is

negative Stationary point are A, B

and C – represented by the zero This means that = 0

y

x

A0

B

C

++

++

++

++

++

--

--

--

-

-0

0ddx

ddx

ddx

Contents

This I where the gradient is zero They can be maximum points, minimum points, or

points of inflectionTo find stationary points:

Differentiate and find the value(s) of when this = 0 Substitute these values into the original equation to

find y To find the nature of the stationary points work out

the second derivative and then substitute the value(s) of x found before to decide if they are a min/max points or points of inflection

Contents

Maximum point

Minimum point

and

(doesn’t change sign on either said of P)

Point of inflection

dydx

dydx

dydx

d²ydx²

d²ydx²

d²ydx²

d³ydx³

= 0

= 0

= 0= 0

< 0

> 0

≠ 0

At point P

Remember to physically do and say each step in a question including saying if a certain point is a max., min. or point of inflection.