C3 Revision
THROUGHOUT THE PAPER:
1) Check that your calculator is in the correct mode (radians or degrees)
2) Give exact answers in terms of  , e or ln unless told otherwise (if in doubt give to 3sf as well)
Algebra
You need to be able to add, subtract, multiply, divide and simplify algebraic fractions
When simplifying algebraic fractions ALWAYS factorise first and look for the difference of 2 squares
– IF YOU CANNOT FACTORISE YOU CANNOT CANCEL
Don’t forget C2 algebraic long division and the Factor and Remainder Theorems – they may come in
useful.
The operations on algebraic fractions are exactly the same as those on normal fractions
-
to add & subtract, find the lowest common denominatoar and find equivalent fractions
to mulitply, multiply the numerators, multiply the denominators; you should cancel first, a
term from a numerator with a term from a denominator
to divide, turn the SECOND fraction upside down and multiply
Functions
A function is a mapping such that each element in the domain is mapped to exactly one element in
the range
One-one function
Many – to – one function
One – to - many mapping
Many – to – many mapping
Only one-to one mappings and many-to-one mappings are functions.
Only one-to-one functions have inverses
A function may be created from a mapping by restricting the domain of the mapping
A composite function consists of two or more functions combined eg: fg(x) – the function g(x) is
substituted into the function f(x) - work backwards from right to left
f 1  x  is the inverse function of f  x  (not to be confused with the reciprocal function
f  x 
1

1
)
f x
To find the inverse function write y  f  x  and change the subject of the formula. Remember to
write the inverse function in terms of x at the end.
The range of the original function is the same as the domain of the inverse function and vice versa
On a graph the inverse function is a reflection of the original function in the line y  x
If you are asked to sketch a function should only sketch it for the given domain
2
Exponential and log functions – from C2
You need to know the shape of exponential graphs. They all go through the point (0,1) and they are
never negative.
1
y 
2
y  3 x
y  2x
y  5x
eg:
y  a x , a  1 or
y  ax , 0  a  1
x
y  a  x , a  0 or
y  ax, 0  a  1
Logarithms are the inverse of exponentials.
All the graphs go through the point (1,0) and is
valid for positive values of x
You must learn the following laws:
ax  n

log a n  x
a 1
log a 1  0
a1  a
log a a  1
a x  a y  a x y
log a x  log a y  log a ( xy )
a x  a y  a x y
x
log a x  log a y  log a  
 y
a 
n log a x  log a x n
0
x
y
 a xy
1
log a    log a 1  log a x   log a x
x
log b x
log a x 
log b a
The domain of exponential functions is x   and the range is f  x   0
3
The exponential function y  e x is such that the gradient of the function is equal to the value of the
function itself
y  ex
Ie:
and
dy
 ex
dx
The inverse function of e x is ln x . The domain of ln x is x  ,
x  0 . The range of ln x is x  
Transformation of Graphs
f  x  a  translates by –a in the x-direction
f  x   a translates by +a in the y-direction
f  kx  stretches by a factor of
1
1
in the x-direction (multiplies x-coordinates by )
k
k
kf  x  stretches by a factor of k in the y-direction (multiplies y-coordinates by k)
f  x  reflects the whole graph in the x-axis
f   x  reflects the whole graph in the y-axis
f  x  reflects all parts of the graph below the x-axis above the x-axis
f  x  the graph is the same as the original for x  0 and this is reflected in the y-axis
These transformations may be combined and you must consider each transformation inside the
bracket first, followed by those outside the bracket
To solve equations of the form g  x   f  x  :
-
Sketch f  x  and g  x  to find the number of roots (take care with the gradients, but scales
-
need not be accurate)
Solve g  x   f  x  and g  x   f  x  (or whichever is appropriate if they do not all intersect)
To solve equations of the form g  x   f  x 
-
Sketch f  x  and g  x  to find the number of roots (take care with the gradients, but scales
-
need not be accurate)
Solve g  x   f  x  and g  x   f  x  (or whichever is appropriate if they do not all intersect)
A function should only be sketched for the given domain
4
Even and Odd Functions
Even functions – reflective symmetry in the y-axis f   x   f  x 
Eg:
f  x   x2
f  x    x 
2
 x2
 f x
Odd functions – rotational symmetry of 180o about the origin f   x   f  x 
Eg:
f  x   x3
f  x    x 
3
 x3
 f  x 
Trigonometry
1
cos 
1
cosec  
sin 
1
cos 
cot  

tan  sin 
sec  
cos2  sin2 x  1
1  tan2 x  sec 2 x
cot 2  1  cosec 2 x
y  sec x
y  cos x
5
y  cosec x
y  sin x
y  tan x
6
y  cot x


arcsin x  sin1 x
domain 1  x  1
arccos x  cos1 x
domain 1  x  1 range 0  arccos x  
arctan x  tan1 x
domain
x 
range 
range 
y

2
2
 arcsin x 
 arctan x 
2

2
y=arcsinx
y  arcsin x
5π/8
π/2
3π/8
π/4
π/8
x
-1
-0.5
0.5
1
-π/8
-π/4
-3π/8
-π/2
-5π/8
7
y
y=arccosx
π
y  arccos x
7π/8
3π/4
5π/8
π/2
3π/8
π/4
π/8
x
-1
-0.5
0.5
1
y
y=arctanx
f(x)=pi/2
5π/8
f(x)=-pi/2
y  arctan x
π/2
3π/8
π/4
π/8
x
-π/8
-π/4
-3π/8
-π/2
-5π/8
In the formula book:
sin  A  B   sin A cos B  cos A sin B
cos  A  B   cos A cos B
tan  A  B  
sin A sin B
tan A  tan B
1 tan A tan B
Double angle formulae:
sin 2 A  2 sin A cos A
cos 2 A  cos2 A  sin2 A  2cos2 A  1  1  2 sin2 A
2 tan A
tan 2 A 
1  tan2 A
8
2 sin A cos B  sin  A  B   sin  A  B 
2cos A sin B  sin  A  B   sin  A  B 
2cos A cos B  cos  A  B   cos  A  B 
2 sin A sin B   cos  A  B   cos  A  B  
P Q 
 P Q 
sin P  sin Q  2 sin 
 cos  2 
2




P Q 
 P Q 
sin P  sin Q  2cos 
sin 


 2 
 2 
P Q 
 P Q 
cos P  cos Q  2cos 
cos 


 2 
 2 
P Q 
 P Q 
cos P  cos Q  2 sin 
sin 


 2 
 2 
To solve equtions of the form a sin   b cos  use the appropriate compound angle formulae
a sin  b cos  R sin    
R  0, 0    90o
a cos   b sin  R cos 
R  0, 0    90o
where R cos  a,

R sin  b
and
a2  b2
Yoou can use these formulae to solve equations of the form a cos   b sin   c where a, b and c
are constants.
But if c=0 ie: to solve a cos   b sin   0 you need to divide through by cos  to give
a  b tan   0
9
Differentiation
Chain Rule:
If y  f  x  
n
n 1
dy
 n f  x  f '  x 
dx
then
dy
If y  f g  x  then
 f ' g  x  g '  x 
dx
dy
1

dx  dx 
 dy 


dy dy du
and


dx du dx
dy
du
dv
v
u
dx
dx
dx
Product Rule: If y  uv then
or if y  f  x  g  x  then
dy
u
Quotient Rule: If y  then

dx
v
Or if y 
y  ex

y  ln x

y  sin x

y  cos x

y  tan x

y  cosec x

y  sec x

y  cot x

f x
g x
then
v
dy
 f 'xg x  f xg 'x
dx
du
dv
u
dx
dx
v2
dy f '  x  g  x   f  x  g '  x 

2
dx
g  x  
dy
 ex
dx
dy 1

dx x
and
y e
f x

and
y  ln f  x  

dy
 cos x
and
dx
dy
  sin x
dx
dy
 sec 2 x
dx
dy
  cosec x cot x
dx
dy
 sec x tan x
dx
dy
  cosec 2 x
dx
y  sin f  x   
dy
f x
 f 'xe  
dx
dy f '  x 

dx f  x 
dy
 f '  x  sin f  x  
dx
10