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Unit 1 - 1

The Straight Line
Applications:

The distance formula

( x2 − x1 )

2

+ ( y2 − y1 ) 2

derived using Pythagoras’ Theorem
...


y +y 
x +x
M 1 2 , 1 2 
2 
 2
The gradient formula

mAB =

This is a simple average of the co-ordinates of A and B
...

Use in circles to calculate radius or diameter

change in y from A to B
change in x from A to B

The gradient of AB is denoted by mAB
where θ is the anti-clockwise angle from OX to AB
For a line sloping down from left to right,
θ is obtuse angle, tan θ is negative and so gradient is negative
If the gradients of two perpendicular lines are m1 and m2, then

gradients are equal

m1 = −

Perpendicular: product of gradients = –1

1
m2

if m1 =

a
b

then

m2 = −

b
a

Lines parallel to OX and OY
The gradient of a line parallel to the x-axis is 0
The gradient of a line parallel to the y-axis is undefined

gradient = 0

gradient = undefined

Applications of: m1 × m2 = -1

• Finding gradients and equations of perpendicular lines
• Finding equations of tangents to circles (perpendicular to radius)
• Showing that an angle is a right angle
...
g
...

• Finding equations of parallel and perpendicular lines
...


Re-arranging equations
Equations of straight lines come in many forms:

Show that the gradients PQ and QR are the same
State that the two line segments PQ and QR have a common point Q

Failure to state (ii) will cost you marks !!!
Use the simple rules of algebra:
•

Change side, change sign

y = 4x – 5

y + 4x = -5

•

Multiply both sides by a constant

y + 4x + 5 = 0

2y + 8x + 10 = 0

•

Divide both sides by a constant

are all the same equation
...


Straight lines only have x and y appearing with
no higher power than 1
...
g
...
Recall the gradient formula, you need two points on
the line to obtain the gradient
...
Use the gradient formula again in this form
...


treat the equations of the lines as simultaneous equations and solve them
...


Altitude

of a triangle – the line drawn from a vertex perpendicular to the opposite side
...


at right angles to

Intersect where lines cut each other
Bisect

lines cut each other exactly in half
...

By using this condition, you can find the equation of this path which is called the locus
...


M e d ia n s

A lt it u d e s

P e rp e n d ic u la r b is e c to rs

A n g l e b is e c t o r s

Gradients
m=
m=
m = -1

m=

m=
m=
135°

negative

zero

positive

undefined

45°

lines at 45° ~ gradients +1, -1

-2-

gradients relative to m = 1

Unit 1 - 2
...


Notation:
y = f(x)

or

1

3

-2

2
3
4
5

6
9
12
15

2

4

0

0

1
-1

1

f : x → y ( f maps x to y )
Domain

Domain and Range
The domain of a function is the input – the
variable the function operates upon
...


Domain of h(x)=√x and k(x) =
√

1
x −1

We write this as:
domain of h(x) is {x : x∈ℜ : x ≥ o }
∈
and:
domain of k(x) is: {x : x∈ℜ : x ≠ 1 }
∈

The largest domain of h(x) = √x is the set of real numbers greater than or equal to
zero, since you cannot take the square root of a negative number
...


You should always be aware of the danger of
dividing by zero
...

Example:

f(x) = 3x + 1

what is

f(x+1)

Given f(x), what is f(x+1) or f(x ) or f(2x) etc
...

and simplify
...


h(x) = x2 – 3x

Example:

If

f(x) = x2 + 3x – 1

Evaluate f(-1)

f(-1) = (-1 )2 + 3(-1) – 1
If

f(x) = 3x3 – 5x + 2
3

Solution:

f(a) = 3a – 5a + 2

-3-

⇒ -3

What is f(a)

Unit 1 - 2
...
This variable is known as the ‘independent variable’ since
you can choose any value in the domain
...
This variable is known as the ‘dependent variable’, as once
you have chosen a value for x or t then f or g is determined by the function
...
f(…
...


x
x −1
x
x −1

x ]
x −1

now simplify to get f(f(x)) = x

−1

If the original function is f(x) = 2x + 3
The function takes a number, doubles it and adds 3

An inverse of a function, is a function that
‘undoes’ the operation of the original function
...

This means that for: every member of the domain, there is exactly ONE
corresponding member of the range and for every member of the range there is
exactly ONE corresponding member of the domain i
...
No duplicate values
...

A quick way to check for an inverse on a graph is to slide a ruler horizontally
down
...


-4-

Unit 1 - 2
...


Recall sin-1(x), cos-1 (x) and tan-1 (x) on your calculator – these are the inverse
functions of sin x, cos x and tan x
...


−1

divide by 4 ~ hence f ( x) =

Sometimes it is more difficult to see the inverse:

Find the inverse of f ( x) =

Example:

x
4

3x − 1
2

In this case the number is multiplied by three,
one is subtracted, and the result divided by 2
...


put

y=

3x − 1
2

Summary:

then x =

2x + 1
3

3x − 1
2

2 y +1
3

1
...

3
...


Change function to y = …
...


f −1 ( x) =

2x + 1
3

This method is much easier and less prone to error
...

The graph y = f -1(x) can easily be found by
reflecting the graph of y = f(x) in the line y = x

Example:

f(x) = 4x

then

f −1 ( x ) =

x
4

y = 4x

y=x

y=

x
4

y = f(x)

Note the reflection in y = x
This technique is useful for sketching
graphs of inverse functions, and indeed
can also allow us to deduce what an
inverse function might be
...
2

Algebraic Functions and Graphs

Completing the square
of the quadratic function

Complete the square of 2x2 + 12x – 5

Example:

(

)

ensure coefficient of x2 is 1

step 1
...


2 ( x + 3) − 9

step 4
...
Arrange for the coefficient of x2 to be 1
2
...
Subtract the square of the last term taking
care to include brackets
4
...


Method summary:

2 x2 + 6x − 5

2 ( x + 3) − 18 − 5

{

2

forming the square ~ put x first and
and half the coefficient of x in last place
...


Example:

•

Intersection with x, y axes

•

turning point

•

axis of symmetry

DO NOT PLOT the graph
– you will be awarded NO MARKS for plotting

Sketching graphs of related functions
If you have the graph of a function such as
y = f(x), then you can deduce and sketch
the graph of a related function such as:
•

y = - f(x)

•

y = f(-x)

•

y = f(x ± a)

•

y = f(x) ± k

or any combination of these transformations
...

x 2 + 8 x + 4 = ( x + 4) 2 − 16 + 4 = ( x + 4) 2 − 12

Solution:

Minimum value is -12 when

if ‘a’ is negative
the function can be arranged as:
f(x) = q + a(x + p)2 and since ‘a’ is negative,
the term: a(x + p)2 will be subtracted
Maximum value = q when x = - p

Sketching the graph of quadratic
functions

⇒ 2( x 2 + 6 x + 9) − 23 ⇒ 2 x 2 + 12 x − 5

( x + 4)

2

=0

Minimum value = -12 when x = - 4

Method: To sketch the graph of y = f(x) where f(x) is a quadratic function
1
...

3
...

5
...

7
...
y-co-ordinate)
Note the turning point is on the axis of symmetry (half-way between the
two roots of the equation ~ step 2)
Find y co-ordinate of the turning point by substitution in the equation
Using these points – sketch the graph – marking in co-ordinate values
...


Given y = f(x) then: (assuming a > 0 and k > 0)
•
•
•
•
•
•

y = - f(x)
y = f(-x)
y = f(x + a)
y = f(x - a)
y = f(x) + k
y = f(x) – k

This reflects the graph in the x-axis
This reflects the graph in the y-axis
This slides the graph a units to the left
This slides the graph a units to the right
This slides the graph k units upwards
This slides the graph k units downwards

These may be combined, for example:
y = f(x + a) – k would move graph a units to left then k units down
...


-6-

Unit 1 - 2
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f (1) = 41 = 4

1
f (0) = 40 = 1
f   = 42 = 2
2
−1
f ( −1) = 4 = 0
...


Note:

The graph of the function

12

f ( x) = a x where a > 0 and a ≠ 1
The graph of the function has equation

(-3, -2)

1

y = ax
a<1
(0, 1)

(0, 1)

if a < 1
then the function is a decreasing function
...
g
...


Note that the line x = 0 is
an asymptote to the graph y = loga x

-7-

2

Unit 1 - 2
...
e
...
e
...

Example:

Sketch y = log2 (x – 3)
y = log2(x - 3)

log2 1 = 0

When plotting loga(x – 2) or similar
Choose a value of x to make (x – 2) equal to 0
Choose a value of x to make (x – 2) equal to a

⇒ (x – 3) = 1
so x = 4 giving point (4, 0)
log2 2 = 1

(5, 1)
3
(4, 0)

⇒ (x – 3) = 2
so x = 5 giving point (5, 1)
Note this confirms our previous knowledge of related functions
y = log2 (x – 3) is simply the graph of y = log2 x shifted 3 units to the right
...

Find the values of a and b
(-2, 0) lies on the curve, so

(1, 5)
(-2, 0)

0 = a log4 ( -2 + b)
so b – 2 = 1,

y = a log4(x + b)

hence

b=3

(1, 5) lies on the curve, so
5 = a log4 (1 + b),

In all cases use the two special logarithms

since b = 3

5 = a log4 (4), now log4 4 = 1

log a 1 = 0

Example:

log a a = 1

so a = 5

Sketch y = log 3  

Choose values of x as appropriate

1
x

Choose x = 1
⇒ log3 1 = 0 giving point (1, 0)

(1 /3 , 1)
(1, 0)

Choose x = 1
3
⇒ log3 3 = 1

giving point (1/3, 1)

Note that the term

1 results in a decreasing function
x

Consider what happens for large x and small x approaching zero

-8-

Unit 1 - 2
...

π radians = 180°

and

r

θ°
r
=
360° 2π r

so

θ
O

θ° =

2π radians = 360°

π

30

Use proportion based on

180

1 radian =

4

π

60

3
π

90

1 degree =

π
π
180

degrees so multiply your radians by
radians so multiply your degrees by

2π
3

135

3π
4

180

π

Note the top line of the multiplier:

tan

π
π
180

to get degrees
to get radians

180

π
π
180

gives degrees

gives radians

Recall the use of Pythagoras and SOHCAHTOA
for obtaining exact values for 30°, 45° and 60°
...
3°

Changing between degrees and radians

Radians

Degrees

360
2π

1

30°

2

2

√3

3

Maximum and minimum values of
trigonometric functions
•

Look when the trig
...


1

1
Hence sin 30° =
2

3

cos 30° =

3
2

tan 30° =

1
3

etc
...
You must be able to work in radians as well as degrees
...


•

In all cases look at when the sin or cos part of the function is at a
maximum or minimum
...

Draw in the perpendicular from base to vertex
...

Again use Pythagoras to calculate the altitude as

60°
1

You should be familiar with this table, if you cannot
memorise it, learn how to create it
...
3

Trigonometric Functions and Graphs

Angles greater than 90°
°

S

When considering angles in the 2nd 3rd and 4th quadrants, remember
the acute angle is always between the rotating arm and the x-axis
...


Sketching Trigonometric Graphs
y = a sin nx

y = a cos nx

tan 11π
6

related acute angle = π ~ tan 11π
6
6

2
p

= – tan π

1
2

= 1
3

6

2
p

a

y = a cos x

a

y = a sin x
3π

a = amplitude (max and min values of y)

2

π

n = number of waves in 360° or 2π

period of the graph is 360 ° or 2π radians
n

π

π
x

π

2

3π

2

-a

x

2

-a

n
when you have for example: cos θ = -0
...
7

Solving Trigonometric Equations

Reminder:

All these equations can ultimately be resolved
into the form
sin ( …
...
) = constant
tan (……
...


Ignore the negative sign when getting the acute angle on your calculator
...


However – first you have to get the equation into this form !
See below for strategies for the different types of equations
...
5
0 ≤ x ≤ 360°

1
1
Taking square roots gives sin x = ±
2
2
Note now there are 2 equations to solve and you will obtain 4 solutions
...
Solutions are: x = 3
...
43 radians
Note that sin x + 2 = 0 has no solutions so discard it
...
) = 0
...
) = 30° or 150° or 390° or 510° (giving four solutions)
This gives four equations: like 2x – 20 = 30, 2x – 20 = 150, etc
...
1

Introduction to Differentiation

The limit formula

Differentiation relates primarily to the
gradient of a graph or function
(generally a curve)
...


x

x+h

x

A graph is a pictorial representation of a
function
...


f ′(x) is the derivative of f(x)
...


This is a function that allows us to calculate the
gradient at any point x on the curve
...


Calculating the gradient
at a point P (x, y) on the curve y = f(x)

Example:

Find the gradient on the curve f(x) = 2x2 + 3x + 5

Solution:

Differentiate ⇒ f′(x) = 4x + 3

• Differentiate the function f(x) to get f ′(x)
• Evaluate function f ′(x) at point P(x, y)

at P(-2, 1)

Evaluate f′(-2) = 4(-2) + 3 ⇒ f′(-2) = –5
Gradient at P(-2, 1) = –5

Rules for differentiation:
f(x)

f ′(x)

xn

nxn-1

c (constant)

0

ax (a is a constant)

a

a xn

a nxn-1

f(x) + g(x)

These rules work for any power of n - positive or negative, whole number or
fractional
...

IMPORTANT:

You must have the function f(x) as a polynomial, a series of powers of x
...


6x + 2

Recall rules of indices:

Indices:

Also recall meaning of fractional and negative indices
...

Both notations are equivalent
...
1

Introduction to Differentiation

Changing functions to straight line form
– or simple index form
...


•

Find the y co-ordinate of point if not
given
...

Find the y co-ordinate of the point by putting x co-ordinate into original equation
...


The derived function f ′(x) is the function
resulting from differentiating f(x)
To sketch the derived function:
Method:

Step 1
...


y = 2 (1)2 + 1 = 3

When x = 1,

Hence point is (1, 3)

Step 1
...
e
...

Note either side of the point whether gradient is positive or negative

Step 2
...

step 1

Step 2
...


Graphs of derived functions

y − a = m ( x − b)

or

y = f(x)

step 2

y = f(x)

(2, 1)
x
(-1, -1)

- 12 -

x

(2, 1)
x
(-1, -1)

x

Unit 1 - 3
...


y = x3 + 3x 2 − 9 x + 1

Differentiate to find the gradient function

•

Example:

•

Determine the nature of each stationary point
using a table of signs as shown in the
example
...

In each case you are looking either to the left or the
right of the stationary point
...
P
...
p
...
p
...
Using the factorisation in dy/dx

→ -3 → 1 →

x
(x-1)

-

↓

-

↓

+

(x+3)

-

↓

+

↓

+

dy
dx

+

0

-

0

+

– 0 +

Maximum S
...
have signs:

Differentiate to get:

+0–

Points of Inflexion have signs:

– 0 – or + 0 +

max
Hence stationary points are:

The interval on which the function is
increasing or decreasing
...

Where the gradient is positive,
the function is increasing
...


Where the gradient is negative,
the function is decreasing
...


To determine the maximum and minimum value on a closed interval:

•
•
•
•
•

Find the stationary points of the function
If any lie outside the interval, discard them
Check the S
...
of each stationary point (i
...
y co-ordinate)
Check the value of the function at each end of the interval
...


Example: Find the maximum and minimum value of y = x3 on [1, 3]

dy
= 3x2
dx

for a S
...
dy = 0

so 3x2 = 0 hence x = 0 (outside of interval)

dx

Now check ends of interval [1, 3]

y(1) = 1
3

and

y(3) = 27

Hence on the interval [1, 3], y = x has max value of 27 and min value of 1

- 13 -

Unit 1 - 3
...


Find points of intersection with x and y axes:
for intersection with y axis ~ put x = 0
for intersection with x axis ~ put y = 0 and solve the equation
...
P
...

• Behaviour for large values of + and – x
• Any useful points on the graph
...

Problem solving

•

Mathematical modelling
...


•

Look for a maximum or minimum
...

Your constraint will connect these two unknown
variables
...

Differentiate and find any stationary points –
generally there will only be one
...

The value of stationary point will cause the model to
have a maximum or minimum value
...


Find the stationary points and their nature
Differentiate and put dy/dx = 0, solve the equation to get x co-ordinates
Substitute into original equation to get y co-ordinates
Determine nature of stationary point(s) using table of signs
...
g
...
e
...
x = 0 or x = 1 (say)
Example:

A rectangle has length x cm and breadth y cm and perimeter p cm
...
Find the length and breadth of the rectangle with the smallest perimeter
...

P = 2x + 2y

This is our model with two unknowns
...

Area = xy 100 = xy so y = 100 now replace y in our equation for P
x
200 we want to find a value for x to make perimeter P a minimum
...
p
...

hence our S
...
is when x = 10
...

We should verify that this is a minimum by using the table of signs
...


Rate of Change

Velocity and acceleration

The rate of change of y with respect to x is

A point P moving along the x axis has a displacement x (OP) from the
origin O at time t
...


The acceleration of P is the rate of change of its velocity v at time t,
given by:

Note:

a=

dv
dt

When using velocity and acceleration remember that they are vectors and have direction as well as magnitude
...


- 14 -

Unit 1 - 4

Sequences

Formula for nth term
Given a formula for the nth term, we can
calculate all the terms
...
g
...


Conversely given a sequence, we can find a
formula for its nth term
...
g
...

It goes up in multiples of 4 (we add 4 on each time)
so start off with un = 4n
However it is not the 4 times table – it is offset by 1 more
So the nth term is given by un = 4n + 1
Now check to see if this generates the sequence
...

The recurrence relation is the rule for
calculating the n+1th term from the nth term
...
g
...


If the 2nd term is 12

i
...
u2 = 12 then u3 = u2 + 7 or u3 = 12 + 7 = 19

In general - given the first term and the recurrence relation, we can generate all
the terms of the sequence:
e
...
u1 = 5 and un+1 = 2un + 3
This will generate the sequence: 5, 13, 29, 61, …
...

e
...
13, 10, 7, 4, …
...
so: u1 = 13

Forming recurrence relationships
modelling a real-life situation
...
Each morning the number has
doubled, and the gardener picks 50 mushrooms
...

So:
un+1 = 2un – 50

Example:

There are 3 trees in Jim’s garden
...

Take un trees to be the number of trees after n days,
(This means that that original number of 3 trees is u0 ~ u0 = 3)
Then:

un+1 = un +2

Linear Recurrence Relations

These are of the form: un+1 = mun + c

(compare with y = mx + c)
...

This is an arithmetic sequence
...

e
...
un+1 = un + 2
if u1 = 3 then the sequence generated is 3, 5, 7, 9, …
...


- 15 -

Unit 1 - 4

Sequences

Geometric Sequences
If c = 0
in the recurrence relation un+1 = mun + c
then

un+1 = mun

~ each term is multiplied by a constant m

In other words the ratio of successive terms is constant
...

We are just multiplying by m each time
...
g
...
Its initial value V0 is £20

Current years price is 5% more than last year

Describe the price by a recurrence relation
...
05Vn

and

V0 = 20

Example 2:

There are 40 fish in a pond, 10% are eaten but
3 new ones are born every day
...
9un + 3 where u0 = 40

Describe this by a recurrence relation
...


Example 3:
A sequence is generated by the recurrence
relation, the nth term being un

The sequence generated is: 2, 6, 10, 14, …
...


u1 = 2

When expressing a recurrence relation – write
down the relation AND the first term u0 or u1
as appropriate
...

u1 is the first term of the recurrence relation
...
At present there are 100 but their number
doubles after each hour
...

Hn is the height after n months
...
05un

Jim is a salesman, travelling 300 km per week
...


Rn+1 = Rn + 300

- 16 -

u0 = 100

H0 = 150 cm

R0 = 9350

Unit 1 - 4

Sequences

Finite and Infinite Sequences

Examples:

A finite sequence has a finite number of terms –
there are a fixed number of terms in the
sequence
...


un = 2 −

2
...
5)n

3
...

If the nth term tends to a limiting value as n gets
very large (i
...
n tends to infinity n → ∞ ) the
sequence is convergent – it converges to a limit
...


1
n2

as n → ∞

1 → 0 and so u → 2
n
n2

as n → ∞

(0
...
Multiply the series by r, giving us:

We can write the sequence out as:
a, ar, ar2 , ar3, ………
...

The limit of Sn as n → ∞ will be S n =

a
1− r

- 17 -

a is the first term and r is the common ratio
...
Tidal
action removes 50% of the waste each week,
but a local factory discharges 8 tonnes of waste
into the loch at the end of each week
...


It seems to be levelling off at 16 tonnes
...
5un + 8
If we write down the first few terms in the sequence we find:
u0 = 10,

u1 = 13,

and so L =

u2 = 14
...
e
...
25, u4 = 15
...
5un + 8

and so L – 0
...
5

⇒

u5 = 15
...


we can say

L = 0
...
5) = 8

L=

8
0
...


c
1− m

This is an important result
...
Each morning 60% of the crop are
picked
...
Let Mn be the number ready for
picking after n days
...


Recurrence relation:

Mn+1 = 0
...
e
...
4L + 200

L – 0
...
6L = 200

Write down the recurrence relation, find the limit of
the sequence explaining what it means in the context
of this question
...
”
1
...


If you do not know the initial condition, then
you do not know whether the sequence is
decreasing down to the limit or increasing up to
the limit
...


If the multiplier is negative, then the sequence
will oscillate on both sides of the limit
...
33…
...

(In this case it will drop down to 333 since the initial condition was 1000
mushrooms)
...

Every minute 10% of the birds leave the flock
and 30 birds return
...


(60% picked ⇒ 40% left)

Recurrence relation:

Bn+1 = 0
...
e
...
9

In the long term, the number of birds in the flock will settle out at around 300
(In this case it will rise to 300 since the initial condition was 200 birds)
Example:

Consider the recurrence relation: un+1 = 0
...
5

u0 = 500

in this case the sequence drops down to this limit
...
5un + 100 and u0 = 50

i
...
the initial value is 50

then the sequence would rise up to the limit
...


- 18 -



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