FP2 Revision Notes
Inequalities
There are two types of inequalities questions:
1)
Without modulus signs






2)



Rearrange the equation so that it has 0 on one side of the inequality sign
Add or subtract fractions as appropriate to achieve 1 big term
Factorise your expression completely
Find the critical values (where each factor is 0) – these are points where the graph crosses
the axes or where there is an asymptote
Draw up a table that considers all factors and work out the sign of each factor either side of
each critical value
Combine the sign to work out the sign of the function
With modulus signs
Sketch the graphs – make sure that your graphs have the correct gradient relative to each
other (ie make sure that the steeper one looks steeper). Mark the points where each graph
crosses the axes.
Solve the equations simultaneously to find points of intersection – where intersections occur
on original parts of the graph just remove the modulus brackets, where the intersection
occur on a reflected part of the graph replace the modulus brackets with a minus sign in
front of the expression.
Look at the graph carefully and use the points of intersection to solve the inequality – where
the greater than (>) function is on top.
NB: Always follow the inequality sign given EXCEPT if the question has  signs – for critical values for
terms that occur in the denominator only use > signs.
Series
Method of Differences
You need an identity with a – sign in the middle.
Substitute in 1, 2, 3,……(n-2), (n-1), n. You should find that most of the middle terms cancel out
when each line is added together. Then add together the remaining terms to get the required
expression.
1
FP2 Revision Notes
Complex Numbers
Complex Numbers can be written in three different ways:
(1) z  x  iy
(2) z  r  cos  i sin   - this is modulus-argument form
(3) z  rei - this is exponential form
r  z  x2  y2   arg z
   
When finding the argument of z and make sure that you plot the point on an Argand Diagram and
CHECK that you have taken into account the direction of the turn from the positive x-axis
(anticlockwise is positive, clockwise is negative)
If the argument is negative then z  r  cos( )  i sin( )  can be written as z  r  cos  i sin  
Multiplying and dividing complex numbers in modulus-argument form:

z1 z2  z1 z2

arg  z1 z2   arg  z1   arg  z2 


z
z1
 1
z2
z2
z 
arg  1   arg  z1   arg  z2 
 z2 
De Moivre’s Theorem – You need to be able to prove this for all integer values of n (and know that it
is true for all rational values of n)
z n   r  cos  i sin     r n  cos n  i sin n 
 When n is a positive integer use Mathematical Induction
 Show that it is true for n  0
 Show that it true for n  0
Proof:
n
Positive Integers:
n  1:
LHS   r  cos   i sin     r  cos   i sin  
1
RHS  r 1  cos 1     i sin 1      r  cos   i sin  
As LHS=RHS theorem true for n  1
n  k :  r  cos  i sin     r k  cos k  i sin k 
k
2
FP2 Revision Notes
n  k  1 :  r  cos  i sin   
k 1
  r k  cos k  i sin k    r  cos  i sin   
 r k  cos k  i sin k   r  cos  i sin  
 r k 1  cos k  i sin k  cos  i sin  
 r k 1  cos  k     i sin  k    
 r k 1  cos  k  1  i sin  k  1 
If the theorem is true when n  k it is also true when n  k  1 . As it is also true for n  1 by
Mathematical Induction it is true for all n  1 , n 

Negative Integers:
If n is a negative integer then it can be written as n   m so
LHS   r  cos   i sin   
  r  cos   i sin   
n
m
1

 r  cos   i sin   
1
 m
r  cos m  i sin m 



m
 cos m  i sin m 
1

r  cos m  i sin m   cos m  i sin m 
m
 cos m  i sin m 
r
m
 cos
2
m  sin 2 m

 cos m  i sin m 
rm
 r  m  cos m  i sin m 
 r  m  cos  m   i sin  m  
 r n  cos n  i sin n 
 RHS
Finally when n  0
LHS   r  cos  i sin     1
0
RHS  r 0  cos  0     i sin  0      1
Hence de Moivre’s Theorem is true for all integers
z  rei
The exponential form of de Moivre’s theorem is

z n  r n ein
To find trigonometric identities using de Moivre’s theorem use binomial expansions and equate real
or imaginary parts.
3
FP2 Revision Notes
As  cos  i sin    cos k  i sin k to express cos k in powers of either cos  or sin  expand
k
 cos  i sin 
k
binomially and equate real terms to cos k . To find sin k equate the imaginary
terms to sin k .
You need to be able to apply the following results:
1
 2 cos
z
1
z   2 sin 
z
1
 2 cos n
zn
1
z n  n  2 sin n
z
z
zn 
To express a power of cos  in terms of the cos of multiples of the angles use
1
1
z   2 cos
z n  n  2 cos n
z
z
For example if you want to find an expression for cos4  then as 2 cos   16 cos4  expand
4
4
2
3
1

1 1
4
31
21
 z  z   z  4z  z   6z  z   4z  z    z 


 
 
   
4
1
 z 4  4 z2  6  2  4
z
z
1
4
 z 4  4  4 z2  2  6
z
z
 2 cos 4  8 cos 2  6
So 16 cos4    2 cos  
4
4
 2 cos 4  8 cos 2  6
1
1
3
Hence cos4   cos 4  cos 2 
8
2
8
If you want to find a power if sine then do the same thing with
1
1
z   2 sin 
z n  n  2 sin n
z
z
nth roots of complex numbers
To find the nth roots of a complex number z n  a  bi

Write z n  r  cos  i sin  

Add 2k to each angle to get the expression z n  r  cos   2k   i sin   2k  


1 
   2k 
   2k
 i sin 
Apply de Moivre’s Theorem to find z  r n  cos 

n
n




Choose values of k to give answers in the required solution range
4



FP2 Revision Notes
Loci in the Complex Plane

z  z1  r is a circle centre z1 and radius r.

z  z1  z  z2 is the perpendicular bisector of the line segment joining the points z1 and z2

z  z1   z  z2 ,   0,   1 is best found using an algebraic method . Substitute in
z  x  yi and equate the moduli of both sides.

arg  z  z1    is the half-line from z1 ,angle  with a line from the fixed point z1 parallel to
the real axis.

For questions of the form arg
z  z1
  write as arg  z  z1   arg  z  z2    . Then
z  z2
  arg  z  z1  and   arg  z  z2  so that      . The locus is then the arc of a circle
standing on the chord z1 , z2 .
Angle  must be inside
the triangle.



z1
z2
If you get  inside the
triangle, draw the
lines the other side of
the chord z1 z 2

z2

z1
z1


z2

5
FP2 Revision Notes
Regions on an Argand Diagram
 Draw the locus of the points on the boundary line
 To find which region to shade in an Argand diagram, choose a point and substitute it into the
inequality. If the inequality remains true then that is the region to shade. If the inequality is
not true with that point substituted in, shade the other side of the boundary line
Transformations in the Complex Plane

The object plane is z  x  iy and the image plane is w  u  iv

a
w  z  a  ib is a translation by vector  
b 

w  kz is an enlargement by scale factor k, centre  0, 0  ( k  0 )

w  kz  a  ib is an enlargement scale factor k, centre  0, 0  followed by a translation with
a
vector  
b 

a
w  k  z  a  ib  is a translation by vector   followed by an enlargement scale factor k,
b 
centre  0, 0  .
To transform in the complex plane:




Rearrange to make z the subject
Substitute in for z into the equation
Rearrange into one of the standard loci forms
If this does not work replace w with u  iv and solve algebraically
First Order Differential Equations
These come in two forms
Separation of variables – rearrange into the form f  y 
dy
 g  x  (this will usually only occur if you
dx
have had to carry out a substitution)
Integrating Factor
d2 y
 P  x y  Q  x
dx2

Rearrange in the form



P  x dx
Find the integrating factor e 
Multiply every term in your rearranged equation by the integrating factor
Integrate both sides – the left hand side will become y  integrating factor
6
This includes the
sign of P
Don’t
forget c
on RHS
FP2 Revision Notes
Integration
In addition to the integrals given in the formula book make sure you can use the trigonometric ones
given in C3 and C4.
Basic Integral
k  x n dx k
n 1
x
n 1
k  e x dx  ke x
k
1
dx  k ln x
x
f '  x  pattern
 ax  b  pattern
n 1
k  ax  b 
n
k   ax  b  dx 
a  n  1
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
ax  b 
dx 
k
f 'x
f x
f x
dx  ke
f x
dx  k ln f  x 
k  cos xdx k sin xdx k cos ax  b dx  k sin ax  b dx k  f '  x  cos f  x  dx k sin f  x  dx



 
a
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 
Use your formula book to integrate sec 2 x and cosec 2 x
Use identities to integrate tan2 x and cot 2 x
cos2  sin2 x  1
1  tan2 x  sec 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
dv
dx
7
(1)
ln x
(2)
xn
(3)
anything else
FP2 Revision Notes
To integrate ln x :
 ln xdx   1 ln xdx
then u  ln x
parts formula thus:
dv
 1 and apply integration by
dx
du 1

vx
dx x
1
 ln xdx  x ln x   x  xdx
 x ln x   1dx
 x ln x  x
f x
 g  x  dx
To integrate fractions of the form
If f (x) is constant,
 Is it a standard form?
1
k
dx  ln  ax  b  ?
a
 ax  b 

Is it of the form k 

f  x 
Is it of the form k  f '  x  f  x  dx k 
n 1
Can you use partial fractions?
n 1

n
?
If f(x) is a polynomial (not constant) then consider the following:

Is f(x) = g’(x)? then use
f 'x
 f  x  dx
n 1




f  x 
If g ( x )  h  x   and f  x   h '  x  then use k  f '  x  f  x  dx k 
n 1
Is the fraction improper? Then divide
Can you use Partial Fractions?
g 'x hx
Can you split up the fraction into the form 

dx
g x g x
n
n
Second Order Differential Equations

d2 y
dy
 b c  f  x
2
dx
dx
2
Solve the auxiliary equation am  bm  c  0 to give solutions  , 

Find the complementary Function
To solve the differential equation a
If  ,  are REAL and DISTINCT then CF is y  Ae x  Be x
If  ,  are EQUAL(ie    ) then CF is y  e x  Ax  B 
If  ,  are COMPLEX

 p  qi 
then CF is y  e px  A cos qx  B sin qx 
Find the Particular Integral
If f  x  is a polynomial try y   x2   x  
If f  x  is a power of e try y  ekx where k is the power of e on the RHS
If f  x  is trigonometric try y   cos kx   sin kx where k is the multiple of e on RHS
8
FP2 Revision Notes


Differentiate the PI and substitute it into the differential equation. Equate like terms and
find the unknown values in the PI
The General Solution is given by y  CF+PI
If you need to find the particular solution, substitute in the values given at the start of the
question.
Differential Equations using Substitutions
If you are given a substitution then you must remember that all variables are functions of x and not
constants (unless told otherwise)



y
, rearrange to get y  zx and differentiate using the product rule.
x
dz
So if f  x   z, g  x   x then f '  x  
and g  x   1
dx
dy
dz
Then
zx
dx
dx
dz
If you have to differentiate again use the product rule on the x term to get
dx
2
2
2
d y dz
d z dz
d z
dz

x 2 
 x 2 2
dx2 dx
dx
dx
dx
dx
Most other substitutions proceed in a similar fashion and you should rearrange to get the
required first or second order differential equation
For the substitution z 
The most challenging substitution is x  eu - so here it goes!
dx
 eu
du
dy dy du
dy dy 1
dy
By the chain rule
so

 u  eu


dx du e
du
dx du dx
dy dy
This can be rearranged to get x
(1)

dx du
dy
dy
To get the second order terms start from
and apply the product rule:
 eu
dx
du
dy
f  x   eu
g  x 
du
du
d 2 y du
f '  x   e u
g ' x   2
dx
du dx
As
du
 eu
dx
Hence
f '  x   e 2u
g '  x   eu

d2 y
d2 y 
dy
 e  u  e  u 2   e 2u
2
dx
du 
du

 e 2u
d2 y
dy
 e 2u
2
du
du
As e2u  x2 this becomes
9
d2 y
du 2
FP2 Revision Notes
d 2 y d 2 y dy
(2)


dx2 du 2 du
If this is far too complicated learn results (1) and (2).
x2
Maclaurin and Taylor Series



Maclaurin series is centred on the origin and gives a better approximation close to the origin
Taylor series are centred on a point a and gives a better approximation in the region of a.
Both are infinite series and can be continued as far as necessary to get a suitable
approximation
 For both of these expansions differentiate the function given the required number of terms
and substitute 0 for a Maclaurin series or a for a Taylor series. Then substitute all values in
the appropriate expansion given in the formula book
NB: Take care with your value of a in Taylor series




If you are finding the expansion of sin x in powers  x   then a  
3
3






If you are finding the expansion of sin  x   then a 
3
3

You can solve differential equations by replacing f  a  with the value of y for an initial value of x,
f '  a  with the value of
dy
at the initial value of x.
dx
Polar Coordinates

Polar coordinates are given in the form  r ,  where r is the distance of the point from the

origin and  is the angle (measured in radians) from the initial line (the positive x axis).
x  r cos ,
y  r sin 








 y
 
To sketch a polar curve find the points where the trig function is either 0, 1 or -1.
Integration of polar functions gives the area between two half lines and the curve
1  2
To find the area A   r 2 d
2  1
d
To find points where there are tangents parallel to the initial line solve
r sin   0
dx
d
To find points where there are tangents perpendicular to the initial line solve
r cos  0
dx
Equations of tangents perpendicular to the initial line are r cos  x and find the value of x
to substitute in (this will be in polar form)
Equations of tangents parallel to the initial line are r sin   y and find the value of y to
x2  y 2  r 2
  tan 1  
x
substitute in (this will be in polar form)
10