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Unit 2 - 1
...
a3x3 + a2x2 + a1x + a0
with a0, …
...


Quotient
Divisor

r
...


1

9

30

75

3

Example:

3

Find the quotient and remainder
3

2

D

2

The quotient is x + 9x + 30 and the
remainder is 75

The shaded row and column are used there only to refer to the table for explanation
...

Step 2
...


Put the contents of A1 straight down to A3
Multiply A3 by the divisor and put result in B2
Add B1 to B2 and put result in B3

To illustrate this f(3) = 33 + 6(3)2 + 3(3) – 15
= 27 + 54 + 9 – 15 = 75

Step 4
...


Multiply B3 by the divisor and put result in C2
Add C1 to C2 and put result into C3

If we divided the quadratic f(x) by x – h then
the remainder would be f(h)

Step 6
...


Multiply C3 by divisor and put result into D2
Add D1 and D2 and put result into D3

A3, B3, C3 are the coefficients of the quotient and D3 is the remainder
...


-½

2

3

-5

3

↓

-1

-1

3

2

2

-6

6

The quotient is 2x2 + 2x - 6 and
the remainder is 6
However we now have to divide
the quotient by the factor of 2 that
we took out
...


-1-

Unit 2 - 1
...


The Factor Theorem
If the remainder when dividing a polynomial
f(x) by x – h is 0 then x – h is a factor of f(x)

This is a follow on from the Remainder Theorem and is perhaps more important
and certainly useful
...
e
...

This allows us to find factors of polynomials of
any degree
...

We can then repeat the process to obtain
another factor, if one exists
...

The easiest way to use the factor Theorem is as
follows:
1
...


Evaluate f(h) until you find f(h) = 0 and
then you have a factor
...


Once you have a factor, divide the
polynomial by it using synthetic division
and obtain the polynomial quotient which
is of degree one less
...


Hence:

4
...
1

Polynomials

Finding approximate roots of the equation
f(x) = 0
The previous method using the factor theorem
will work providing the polynomial has factors
i
...
the roots are rational
...

Solving by Iteration
Recall the graph of a function
...


Show that x3 – 3x + 1 = 0 has a real root between 1 and 2

To one side of the root f(x) will be positive and
on the other side of the root, f(x) will be
negative
...


So by finding two points such that f(x) is
positive at one point and negative at the other,
you know that the root must lie between the two
points
...


Take the middle point between these two points
and depending upon whether this is positive or
negative it will tell you on which side of the
middle point the root lies
...

If you want accuracy to 1 decimal place then
you need to find the root with knowledge of the
2nd decimal place
...
5) ≈ -0
...
5 < α < 2

f(1
...
81

So 1
...
7

f(1
...
30

So 1
...
6

f(1
...
07

So 1
...
55

f(1
...
03 So 1
...
54
hence root is 1
...


-3-

Unit 2 - 1
...
24 or -1
...
to 2 d
...
)

Example:

solve 3x2 + 4x – 5 = 0 using the formula

We know the following methods for solving quadratic
equations:
1
...

2
...
Use the quadratic formula

−b ± b 2 − 4ac
x=
2a

Here we have a = 3, b = 4 and c = -5
Use: x =

4
...

we call b 2 − 4ac the Discriminant – because it
discriminates between different types of solution
...


2

−4 ± 16 + 60
6

and

x=

−4 ± 76
6

x = 0
...
12 (correct to 2 dec
...


b 2 − 4ac = 4 – 4(1)(p) = 4 – 4p for equal roots this must be zero
so 4 – 4p = 0 hence 4 = 4p and

p=1

Example:

Find the range of values for m for which 5x2 – 3mx + 5 = 0 has two real
and distinct roots
...

To touch the x axis, there must be equal roots so discriminant = 0

b 2 − 4ac = 9k2 – 4(k)(9) = 9k2 – 36k
9k2 – 36k = 0

-4-

9k(k – 4) = 0

so k = 0 or k = 4

Unit 2 - 1
...
1) with c = 3

y = x2 + 3x + 4
by substitution we get

5x + c = x2 + 3x + 4

re-arranging gives:

x2 - 2x + (4 – c) = 0

………
...
e
...
e
...

(i) length to be 10m more than breadth
(ii) area of pad to lie between 375m2 and 600m2

Let breadth of pad be x metres
...

Area of pad is x ( x + 10) and area has to be between 375 m2 and 600 m2
So we have the inequality:

375 < x(x + 10) < 600

Sketch the graph of y = x(x + 10)
We know that this graph crosses the x axis at x = 0 and x = -10

Calculate the limits for the breadth of the pad
...


See method of working opposite:
Summary:
•

y = 60 0

Form an inequality

•

Sketch the graph using ‘=’ signs

•

Which part of graph is required

•

y = 3 75

i
...
y = 375 and y = x(x + 10) ⇒ y = x2 + 10x
So solve x2 + 10x – 375 = 0

Interpret the result

0

-10

( x + 25 )(x - 15 ) = 0
so x = -25 or x = 15 ( discard negative value)

Now we need to solve y = 600 and y = x2 + 10x
i
...
solve x2 + 10x - 600 = 0
In summary with inequalities – sketch the curve
and isolate the part that is relevant
...


-5-

Unit 2 - 2

Integration

Differential Equations
An equation involving a derivative such as

dy
= 8x
dx
To solve this, we ‘undo’ the differentiation
...

y = 4x2 + c is called the anti-derivative of 8x

y = 4x2 + c

since any constant will differentiate to 0
...

Particular Solution
To narrow it down to a particular parabola, we
need more information (a boundary condition)
such as when x = 1, y = 6
...


General solution: y = 4x2 + c
But when x = 1, y = 6 so substitute to give: 6 = 4 + c
So c = 2
2

Particular Solution is: y = 4x + 2

Now we have the Particular Solution
...


If:

Their velocity after x seconds is v = 5 + 10 x m/s

Since we know that y = 0 when x = 0 then c = 0 (substitute in general soln)

If they have fallen y metres then v =
a)
b)

dy
= 5 + 10 x
dx

Find the distance y metres, they fall in x
seconds, given y = 0 when x = 0
Calculate the distance they fall in 10
seconds
...


Leibnitz’ notation
Leibnitz invented a useful notation for antiderivatives:

∫ 8 x dx = 4 x + c
In general ∫ f ( x ) dx = F ( x ) + c

The anti-derivative is called the integral and c is the constant of integration
...


The process of calculating the ant-derivative is
known as Integration
...


n ≠ −1

n

(note opposite of differentiation – which was multiply by the index, then
DECREASE the index by 1)

∫ ( f ( x) + g ( x) ) dx = ∫ f ( x) dx + ∫ g ( x) dx

Integral of a sum is the sum of the integrals
...


Examples:

Examples:
See opposite
Treat each term separately and do not forget the
constant of integration
...

(See opposite for solution)
...

Example:
Integrate:

1
2− 2
x

Integrate:

1
x−
x

−2
∫ 2 − x dx ⇒ 2 x −

⇒

x −1
+c ⇒
−1

2x +

1
+c
x

1

1

Integrate:  u − 
u


⇒

∫x−x

−

1
2

dx ⇒

2

⇒

2
∫u −2+

x2 x 2
− 1 +c ⇒
2
2

1 2
x −2 x +c
2

1
u3
u −1
du ⇒ ∫ u 2 − 2 + u −2 du ⇒
− 2u +
+c
u2
3
−1
1
1
⇒ u 3 − 2u − + c
3
u

-7-

c = 19

Unit 2 - 2

Integration

Example:
Integrate:

v3 + v
v

v3 v
∫ v + v dv ⇒

⇒

2
∫ v + 1 dv ⇒

v3
1
+ v + c ⇒ v3 + v + c
3
3

Applications:
The rate of growth per month (t) of the
population P(t) of Carlos Town is given by the
differential equation

dP
= 5 + 8t
dt

1
3

a)

Find the general solution of this equation
...

4

So,

83 ⇒

( 8)
3

4

⇒ 24 ⇒ 16

Hence the population after 8 months = 40 + 96 + 5000 = 5136
The area under a curve

y

y=x

You have calculated many areas bounded by
straight lines,

We will work out a method for calculating the
area bounded by the x-axis, the lines x = a and
x = b and the curve y = f(x)
...


It is not so easy to calculate the area bounded
by a curve
...

y = f(x)

y

y = f(x)

y

f(x)
a

f(x+h)

A(x)
h

O

y = f(x)

y

x

b

x

O

a

h
x x+h

b

O

x

a

h
x x+h

b

x

Area of shaded section

Area of shaded section

Area of shaded section

h × f(x)

A(x+h) – A(x)

h × f(x+h)

(simple rectangle)

Area up to (x+h) – Area up to x

(simple rectangle)

h × f(x)
Dividing by h (≠ 0 )

by the definition of integration
...


O

2π

π

x

2

0

The Area under a curve – A formula;
definite integrals

y = f(x)

y

The area under the curve y = f(x) from x = a
to x = b is
b

∫ f ( x) dx = [ F ( x)]

b

= F (b) − F (a)

a

A(x)

a
b

∫

O

a

x

From the last section
Let

A( x) = ∫ f ( x) dx

∫ f ( x) dx = F ( x) + c

where

and A(a) = 0

⇒

(from the diagram)

Now

A(b) = F(b) – F(a)

F(b) – F(a) is denoted by

∫ x dx
3

 x4 
= 
 4 1

∫ 2 x(3x + 1) dx
−1

so c = - F(a)

which is the area we are trying to find
...


a

limit a and upper limit b
...

Solution:
Applying the above formula:
0

3

∫ x + 3 dx

1

2

Required area =

3

x

1

 27
 1

 + 9  −  + 3
 3
 3 
1
2
= 18 − 3
= 14
units2
3
3

⇒

⇒

Example:

y
y = x2 - 1

Calculate the area of the region:
a) bounded by the x-axis, the line x = 2
and the parabola y = x2 – 1 (A)
b) enclosed by the parabola and the x-axis (B)

A

0
-1

1

B

2

x

Note:
Integration will give negative values for areas
under the x-axis, since for the shaded strip of
width h, f(x) < 0 and h > 0 so h × f(x) is
negative
...

In the example above the total area of A + B
1 1
2
would be 1 + 1 = 2 units2
3 3
3

- 10 -

Unit 2 - 2

Integration

The area between two curves

y

The area between the curves y = f(x)
and y = g(x) from x = a to x = b
y = f(x)

= the area from the x-axis to the upper curve

y = g(x)

– the area from the x-axis to the lower
curve
...


y = 2x
A

2

y = x meets y = 2x where
x2 = 2x i
...
x(x – 2) = 0
so x = 0 or x = 2

O

The area of the shaded region =

2

2

2

0

0

0

2
2
∫ 2 x dx − ∫ x dx = ∫ 2 x − x dx

2


x3 
=  x2 − 
3 0


8

=  4 −  − (0 − 0)
3


=1

1
3

BEWARE:
Remember to watch out for areas below the
x-axis
...

You must not integrate over a range which
includes POSITIVE AND NEGATIVE values
of f(x)
...
1

Calculations in 2 and 3 dimensions

Reminders:
Basic Trigonometry

SOH-CAH-TOA Sin A = Opposite

Sine Rule and Cosine Rule

Sine Rule

Hypotenuse

Adjacent
Hypotenuse

Tan A =

Opposite
Adjacent

a
b
c
=
=
Sin A
Sin B
Sin C

Cos A =

a 2 = b2 +c2 -2bc Cos A

Cosine Rule
Area of Triangle

Cos A =

Area of ∆ ABC =

Area of Triangle

b2 + c2 − a 2
2bc

1
ab Sin C
2

Related Angles – sketch them in on ASTC
quadrants to convince yourself
...

So

Use formula for area of a triangle
...


b)

1
qr Sin (α + β )
2

p
q
=
Sin P
Sin Q
However ∠ Q = α, so substitute and then re-arrange:
Applying sine rule to ∆PQR gives:

p
q
=
Sin P
Sin α

⇒

p=

q sin P
Sin α

and from part a) we showed that sin P = sin ( α +β )

So:

p=

- 12 -

q sin (α +β )
Sin α

q
...
d
...
e
...


Unit 2 - 3
...
e
...


We have QC from part a), we have angle y
we are trying to find AC – this is a right angled triangle
– which suggests SOH-CAH-TOA - the sine ratio
...
e
...


Always start from something you know
...

If it does then you know where to start
...

– Can you find a rule or formula linking it with something on the Right Hand Side
...


- 13 -

Unit 2 - 3
...

We can use the rules listed above, by applying
them to 2-dimensional planes within the 3
dimensional solid
...


To find the angle between HB and the plane
ABCD use the perpendicular HD and form a
right angled triangle ∆ HDB

G

E

F

∠ HBD is the required angle

D

C

Calculations may involve Pythagoras and SOHCAH-TOA
A

(ii)

B

H

Angle between two planes
...


G

E

F

Then a line in each plane perpendicular to AB,
in this diagram, (BC and BG)
...


D

C

∠ DAH would also do
A

B

Some terminology:
Face diagonal – this is a diagonal across a face
...
g
...

Space diagonal – this is a diagonal linking two
vertices which are not in the same face
...
g
...

y

Co-ordinates in 2 and 3 dimensions
To fix the position of a point on a plane
(2 dimensions), you need two axes OX and OY
...


z
Q(x, y, z)
y

and three co-ordinates – x, y and z

z

Q is the point (x, y, z)
x

In 3 dimensions, we usually show the
z direction vertically
...
2

Compound Angle Formula

Reminders:
Related Angles: (Sketch the ASTC quadrants)
sin (180° – A ) = sin A

π

π

π

6

4

3

Degrees

30°

45°

60°

sin

1
2

1
2

3
2

cos

3
2

1
2

1
2

tan

(the sine of an angle is the sine of its
supplement - Recall ASTC)

1
3

1

Radians

sin (90 – A ) = cos A
sin (-A) = - sin A
cos (180 – A) = - cos A
cos (90 – A) = sin A
cos (-A) = cos A

Sin, cos, tan formulae

sin A
= tan A
cos A

3

sin 2 A + cos 2 A = 1
Radians and Degrees
π radians = 180°
Compound Angle formulae
cos (A + B) = cos A cos B – sin A sin B
cos (A – B) = cos A cos B + sin A sin B

sin (A + B) = sin A cos B + cos B sin A
sin (A – B) = sin A cos B – cos B sin A
These formulae are true for all angles A and B
whether in degrees or radians
...

then by using sin 2 A + cos 2 A = 1

cos A = ½ (1 + cos 2A)
sin2 A = ½ (1 – cos 2A)
These are all useful identities and are often
used in proofs and calculations
...
2

Compound Angle Formula

Examples:
1
...


sin 2 A = 2sin A cos A
cos 2 A = cos 2 A − sin 2 A
cos 2 A = 1 − sin 2 A
∴ sin 3 A = 2sin A cos A cos A + sin A(1 − 2sin 2 A)
∴ sin 3 A = 2sin A(1 − sin 2 A) + sin A(1 − 2sin 2 A)
∴ sin 3 A = 2sin A − 2sin 3 A + sin A − 2sin 3 A
∴ sin 3 A = 3sin A − 4sin 3 A

Example 3
...
2

Compound Angle Formula

Example 5:
Solve sin 2θ + cos θ = 0

for 0 ≤ x ≤ 2π

Hence solutions are:
θ=

π
2

,

7π 3π 11π
,
,
6
2
6

sin 2θ + cos θ = 0
2sin θ cos θ + cos θ = 0
cos θ (2sin θ + 1) = 0
cos θ = 0 or sin θ = −

θ=

π
2

,

3π
2

or

1
2

acute θ =

Example 6:

so θ = π +

6

π
6

or 2π −

π
6

5cos 2θ − cos θ + 2 = 0

Solve correct to 1 decimal place for 0 ≤ θ ≤ 2π

π

5 cos 2 θ − sin 2 θ − cos θ + 2 = 0

5 cos 2θ - cos θ + 2 = 0

(

)

(

(

5 cos 2 θ − 1 − cos 2 θ

(

) ) − cos θ + 2 = 0

)

5 2 cos 2 θ − 1 − cos θ + 2 = 0

Hence solutions are:
θ = 0
...
4 or 2
...
2 radians
...
9, 2
...
2 or 5
...
927 rad or acute θ =

π

hence θ = 0
...
927 or

Summary of methods
Use double angle formula to expand:
sin 2x or cos 2x
Use sin 2 A + cos 2 A = 1
to switch from cos2 x to sin2 x or vice versa

You will generally get a quadratic
in cos x or sin x or a mixture
...


- 17 -

3

rad

θ =π −

π
3

or π +

π
3

rad

Unit 2 - 3
...


p
2

y = cos (x - π )
4

p
2

y = sin (x - π )
4

Graphs of:
y = sin x and y = cos x moved π to right
...


1

1

4
π

x

4

π
4

-1

-1

- 18 -

x

Unit 2 - 4

The Circle

The circle – centre O(0, 0) and radius r
x2 + y2 = r2

P(x, y)

The equation of a circle is given by the locus of
Point P

r

which describes a path at a constant distance r
from the origin
...

By Pythagoras:

x2 + y2 = r2

Hence the equation of the circle is:
x2 + y2 = r2
Application:
x2 + y2 = r2

Example:

the radius of the circle:

x2 + y2 = 64 is r = 8

Example:

Given the equation of a circle in the form

the radius of the circle:

3x2 + 3y2 = 48

first divide by 3 to get the form x2 + y2 = r2

we can write down the radius
...

Hence x2 + y2 = 25

Application:

Example:

We can check that a point lies on a circle – if it
does then it will satisfy the equation of the
circle:

Does the point R(12, -9) lie on the circle x2 + y2 = 225
LHS
RHS
225
x2 + y2
144 + 81
225
Since LHS = RHS, point R satisfies the equation,
so R lies on the circle
...

Using the distance formula we find that OR = 15, so R lies on the circle
...

By Pythagoras:

(x - a)2 + (y – b)2 = r2

Hence the equation of the circle is:
(x - a)2 + (y – b)2 = r2
Applications:

Example:

Given the equation of a circle in the form

State the centre and radius of the circle: (x - 3)2 + (y – 5)2 = 9

(x - a)2 + (y – b)2 = r2
we can write down the co-ordinates of the
centre of the circle and its radius
...

PC is the radius and PC =
So r2 = 40

( 3 − (−3) ) + ( −2 − 0 )
2

2

= 36 + 4 = 40

Hence equation of circle is: (x + 3)2 + y 2 = 40

Application:

Example:

If two circles touch, then we know that the
distance between the centres of the two circles
(using distance formula) is equal to the sums of
their radii
...


The converse also applies: If we want to find
whether two circles touch, check the distance
apart of the two centres and see if it is equal to
the sum of their radii
...


If PQ is a diameter, then the mid point of PQ is the centre
...

Angle in a semi-circle is a right angle
A tangent to a circle is at right angles (90°) to
the radius (or diameter)

Example:

Look for bisected chords (right angles to radius
or diameter)

The small circle centre C has equation (x + 2)2 + (y + 1)2 = 25
The large circle, centres A and B, touch the small circle
and AB is parallel to the x-axis
...


Look for isosceles triangles
...


Solution:
a)

Given three points P, Q, R, you can find the
equation of the circle passing through them
all
...
Find equation of perpendicular
...
Find equation of perpendicular
...


For circle centred on C: co-ordinates of C are (-2, -1) and radius is 5
Hence: A is (-7, -1) and B is (3, -1) and radii of both circles are the
same ( = AB) which is the diameter of circle at C
radii of Circle A and circle B is 10
b)
Equation of circle centre A is: (x + 7)2 + (y + 1)2 = 100
Equation of circle centre B is: (x - 3)2 + (y + 1)2 = 100

Distance from centre to P or Q or R will give
radius
...


- 21 -

Unit 2 - 4

The Circle

The general equation of a circle:
x2 + y2 + 2gx + 2fy + c = 0

g2 + f 2 − c

with centre (-g, -f) and radius
2

2

provided that g + f – c > 0
Note that the coefficients of x2 and y2 must be 1
...

Can we show that: x2 + y2 + 2gx + 2fy + c = 0 represents a circle ?
x2 + 2gx + y2 + 2fy = - c

Re-arranging we get:

(x + g)2 – g2 + (y + f)2 – f2 = -c

Now complete the square

(x + g)2 + (y + f)2 = g2 + f2 – c

Strategies:

thus:

Given a circle in this form – we can find the centre
and radius

we may choose to write this as:

We can then continue using strategies stated
previously
...


this gives us: 2g = -4 so g = -2 and 2f = 8 so f = 4 and c = -12
Condition for a circle is: g2 + f2 – c > 0
and the coefficients x2 and y2 are equal to 1
(-2)2 + 42 – (-12) = 4 + 16 + 12 = 32 which is > 0
so it is the equation of a circle
...

Join PQ and QR
...

Where they meet at C is the centre of the circle
...


x2 + y2 = 4
2

Centre (0, 0) and radius = 2

2

x + y – 8x + 6y + 24 = 0

so -g = 4

2f = 6 so -f = -3

so radius = √(16 + 9 – 24) = 1

g2 + f 2 − c

radius =

2g = -8

Centre (4, -3) and radius 1

Hence
sketch

A

B

(4, -3)

The shortest distance between the circumferences will be along the line
joining their centres
...


Find the co-ordinates of the points of intersection of

Generally you will get two points of
intersection
...


all we have to do is solve the equations simultaneously

Re-arrange ( 1 ) to give x = 5y + 7 and substitute into ( 2 )
⇒

25y2 + 70y + 49 + y2 + 10y + 14 – 2y – 11 = 0

⇒

or if the line misses the circle altogether, in
which case there will be no points of
intersection
...


Find the values of k for y = x + k to be a tangent

For example, by considering the quadratic
equation which results from a simultaneous
equation solution

to the circle x2 + y2 = 8

Solution:
The line and circle intersect where

We can deduce that:

x2 + (x + k)2 = 8
(quadratic equation from simultaneous substitution)

Line meets the circle
in two distinct points
real and distinct roots

b2 – 4ac > 0

at one point only (tangent)
equal roots

b2 – 4ac = 0

at no point
no real roots

b2 – 4ac < 0

i
...


x2 + x2 + 2kx + k2 = 8

⇒

2 x2 + 2kx + k2 – 8 = 0

for a tangent we require only one solution i
...
equal roots
so

b2 – 4ac = 0
⇒ 4k2 – 4(2)(k2 – 8) = 0
⇒ -4k2 – 64 = 0

(divide throughout by 4)

⇒ -k – 16 = 0

⇒ k2 = 16

2

Hence k = ± 4

- 24 -

( giving tangents y = x ± 4 )



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