C4 Revision
Partial Fractions
An improper algebraic fraction is one where the highest power in the numerator is greater than or
equal to the highest power in the denominator
Before splitting into partial fractions check
(1) Is the fraction improper?
(2) Is the denominator factorised completely?
(3) Is there a repeated factor in the denominator?
Linear Factors:
Repeated Factors:
A
ax  b
A
 ax  b 
2

B
or
 ax  b 
A B

x2 x
Improper Fractions:
If the highest powers in the numerator and denominator are equal:px 2  qx  r
B
C
 A

ax  b cx  d
 ax  b  cx  d 
If the power in the numerator is 1 more than the highest power in the denominator:px 3  qx 2  rx  s
C
D
 Ax  B 

ax  b cx  d
 ax  b  cx  d 
If the difference between the powers is any greater use long division to work out the ‘whole partt’ and
the remainder,
To solve either use substitution (the same value of x on both sides) or equate coefficients (the number
of x’s on the LHS = the number of x’s on the RHS)
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Binomial Expansion:
Use the formula given on the C2 page in the formula book
1  x 
n
 1  nx 
n  n  1
1 2
x 2  ........ 
n  n  1 .....  n  r  1
1  2  ......  r
x r ........
 x  1, n  
This only works when the first term in the bracket is 1. If is not you need to factorise – but make sure
that you raise the factor to the power of n.
The expansion is only valid when the modulus of the x term in the expansion is less than 1.
If you need to use the expansion to find an approximation, equate your bracket with the term you
need to approximate and solve to find out what value of x you need.
BEWARE:
(1)
(2)
(3)
(4)
signs – the x-term includes the sign
Expand carefully showing the values you substitute in – don’t try and simplify as
you go along
Make sure you square, cube etc the whole of the x-term (including the sign)
If you have factorised make sure that you multiply back in at the end.
Vectors
Unit vectors have magnitude 1
x
 
a  xi  yj  zk   y 
z 
 

a  x 2  y 2  z2
A unit vector in the direction of a vector a is aˆ 
a
a
The position vector of a point is the vector from the origin to the point.
If A and B have position vectors a and b respectively then AB  b  a
The vector equation of a line through a point P and parallel to a vector q is r  p  tq where p is the
position vector and q is the direction vector.
 p1 
 q1 
 
 
If p  p1i  p2 j  p3k   p2  and q  q1i  q2 j  q3k   q2 
p 
q 
 3
 3
 p1   q1   p1  tq1 
    

Then r  p1i  p2 j  p3k  t  q1i  q2 j  q3k    p2   t  q2    p2  tq2 
 p   q   p  tq 
3
 3  3  3
This is the
general form
for any point
on the line
Where r is the position vector of a point on the line
2
Scalar (dot) product
The scalar product is defined as a.b  a b cos 
 a1 
 b1 
 
 
If a  a1i  a2 j  a3k   a2  and b  b1i  b2 j  b3k   b2 
a 
b 
 3
 3


Then a.b  a1b1  a2b2  a3b3
If a and b is perpendicular then a.b  0
If a.b  a b then a and b are parallel but look at the vectors – they will be multiples of each other.
For equations of lines it is the direction vectors that are multiples if the lines are parallel.
To find the angle between two vectors cos  
a.b
a b
To find the angle between two lines or a line and a displacement or position vector, use cos  
a.b
a b
with the direction vector of each line
To find if two lines intersect, equate their I, j, and k components to produce simultaneous equations.
Check your answer in the third equation.
Differentiation and Coordinate Geometry
To differentiate implicitly, differentiate each x term with respect to x and each y term with respect to
dy
yand multiply by
.
dx
You can use the chain rule, product rule and quotient rule as in C3, just remember that every time you
dy
differentiate a y term, multiply it by
dx
LEARN:
y  ax
ln y  ln a x
ln y  x ln a
1 dy
 ln a
y dx
dy
 y ln a
dx
dy
 a x ln a
dx
When a  e this becomes
dy
 e x ln e  e x
dy
3
y  loga x
x
dx
dy
dx
dy
dy
dx
 ay
 a y ln a
When a  e this becomes
dy 1 1
 
dx x ln e
1

x
 x ln a
1
dy
dx
1

x ln a
1
1
 
x ln a

Parametric Equations:
These define the x and y coordinates in terms of a third variable usually t or 
To find the cartesian equation (the equation in x and y) eliminate the parameter using simultaneous
equations. If the parametric equations are trigonometric then you usually need to use one of the
following identities.
cos2  sin2 x  1
sin 2 A  2 sin A cos A
1  tan x  sec x
cos 2 A  cos2 A  sin2 A  2cos2 A  1  1  2 sin2 A
2 tan A
tan 2 A 
1  tan2 A
2
2
cot 2  1  cosec 2 x
Parametric differentiation
dy
dy dy dt


 dt
dx dt dx dx
dt
or
dy
dy dy d d



dx d dx dx
d
You can then find gradients, stationary points (where the gradient is zero) equations of tangents and
normals as in any other question.
REMEMBER: to find the equation of any straight line you need the gradient and a point
NB
To change between sin, cos and tan, put all the information in a right-angled triangle and use
Pythagoras’ Theroem to find the missing side (usually in algebraic form)
4
Integration:
k  x n dx k
n 1
x
n 1
k  e x dx  ke x
k
f '  x  pattern
 ax  b  pattern
n 1
k  ax  b 
n
k   ax  b  dx 
a  n  1
Basic Integral
n 1
f  x 
k  f '  x  f  x  dx k 
n 1
n
k f 'xe
k ax b 
e
a
1
k
k
dx  ln  ax  b 
a
 ax  b 
k  e
1
dx  k ln x
x
k  cos xdx k sin xdx
ax  b 
dx 
k
k
k  cos  ax  b  dx  sin  ax  b  dx
a
f 'x
f x
f x
dx  ke
f x
dx  k ln f  x 
k  f '  x  cos f  x  dx k sin f  x  dx
In addition you need learn the following:
1
1  cos 2x dx
2
1
2
 cos xdx  2  1  cos 2x  dx
1
 sin x cos x  2  sin 2xdx
 sin
2
xdx 
cos2  sin2 x  1
Use you formula book to integrate sec 2 x and cosec 2 x
1  tan2 x  sec 2 x
Use identities to integrate tan2 x and cot 2 x
cot 2  1  cosec 2 x
Integration by Parts:
dv
du
 u dx dx  uv   v dx dx

b
a
u
b
dv
du
b
dx  uv a   v
dx
a
dx
dx
To decide which term is u and which is
To integrate ln x :
 ln xdx   1 ln xdx
dv
dx
(1)
ln x
(2)
xn
(3)
anything else
then u  ln x
du 1

dx x
formula thus:
dv
 1 and apply integration by parts
dx
vx
1
 ln xdx  x ln x   x  xdx
 x ln x   1dx
 x ln x  x
5
To integrate by substitution – you are usually given the substitution
You must remember to substitute in for all the original variables including dx and the limits
Volumes of revolution:
V    y 2dx
simplify as far as possible before integrating.
Parametric Integration:
To find an area
A   ydx
To find a volume V    y 2dx
You then need to substitute in your parameter for y, dx and the limits
Trapezium Rule (given in the formula book)

b
a
ydx  12 h ( y0  yn )  2  y1  y2  ...........  yn1  where h 
ba
n
n is the number of intervals (not the number of x coordinates in the table)
A better way to work out h is to use the distance between the x coordinates.
Separation of Variables:
GENERAL SOLUTION ;- the solution with the constant of integration (+c)
PARTICULAR SOLUTION:- the solution with the constant of integration evaluated
If your differential equation contains 2 variables you need to rearrange the equation into the form
dy
f y 
 g x
dx
Then  f  y  dy   g  x  dx
A Rate of Change is written as
If
dy
 kt then y  Aekt
dt
d?
where the ? is replaced by the appropriate variable.
dt
and if
dy
 kt then y  Aekt
dt
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