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it is a comprehensive revision guide for further maths a level maths. its an A material
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GCSE Further
Mathematics
Revision Guide
S J Cooper
The book contains a number of worked examples covering the topics needed
in the Further Mathematics Specifications. This includes Calculus,
Trigonometry, Geometry and the more advanced Algebra.
Algebra
Indices for all rational exponents
m n mn a a a , m n mn a a a
m n mn a a , 1 0 a ,
a^2 a
1 , n a n^ a
1 , n
n a
a
, n n a a
n n
a
b
b
a
n m a n a
m n m a
Example 8 3 38 2
1 Example 3
2
1
2
1
Example^34
2 x ^2
23 3 3
2 x 4 4 8
3 x
1 2 3 2 x 3 x 6 x
2
3
x x
x
x
x x
x x x
Quadratic equations 0 2 ax bx c
Examples (i) 4 9 2 x (ii) 4 x 9 x 2 (iii) 4 9 2 2 x x
when x 32 , y 94 32 2 ^289
when x 1 , y 1 1 2 2
The geometrical interpretation here is that the straight line
2 x 3 y 8 and the parabola 2 2 y x x intersect at points
9 28 3
2
Intersection points of graphs to ‘solve’ equations. There are many
equations which can not be solved analytically. Approximate roots
to equations can be found graphically if necessary.
Example What straight line drawn on the same axes as the graph
of
3 y x will give the real root of the equation 3 0 3 x x ?
3 x x x 3 x 3
draw y 3 x
As can be seen from the sketch there is
(1,2)
0
Solutions,
0
Example
Obtain the points of intersection of the circle and the
line
Using
Hence points are ( ) (^ )
Expansions and factorisation – extensions
Example
3 2
2
2 3 2
x x x
x x
x x x x x x
Expanding
𝑦
𝑥
c
Gradient = m (= tan )
x
y
0
Geometry
Gradient/ intercept form of a straight line Equation
y mx c
Distance between two points
2 2 1
2 2 1
2 AB x x y y
say
2 1
2 1 x x
y y m
Equation of a line through ( x ’, y ’) of gradient m
Equation of a line through two points
Find the gradient using 2 1
2 1 x x
y y m
and use the formula as above.
Parallel and perpendicular lines
Let two lines have gradients (^) m 1 and m 2
Lines parallel m 1 (^) m 2
Lines perpendicular m 1 m 2 1 or 2
1
m
m
are
^1 2 x 1 x 2 , 21 y 1 y 2
General form of a straight line
ax by c 0.^ To find the gradient, rewrite in gradient/intercept
form.
2 x 3 y 12 3 y 2 x 12 y 32 x 4
gradient = 32
Gradient of perpendicular = 2
3
3
2
The Circle
Angles in semicircle is 90
Perpendicular to a chord from centre of
circle bisects the chord.
Centre, radius form of equation
(^2 ) x a y b r
Centre ( a, b ) radius = r
2 2 x y
Example Centre (1, 2) touching Equation
1 2 4
2 2 x y
General form of equation
(^2 ) x a y b r Circle centre ( ) with radius
To find centre and radius, use the method of CTS to change into
centre/radius form.
Example 2 3 3 0
2 2 x y x y
^254
2 2
(^23)
2 2
(^23) 2
(^223)
2 2
2 2
x y
x x y y
x x y y
x y x y
Centre^2 3 1 , radius = 2 5
0
(-1, 2)
P(2, 1)
Gradient CP =
gradient of tangent at P = 3
Equation
Calculus
Differentiation by rule Examples
3 5 6 1
2
2
1
2
1 2
2
1 2
1 2
1
x x x dx
d
dx
d
x dx
x d
dx
d
x
x x dx
d
dx x
d
x
x x dx
d x dx
d
Vocabulary and more notation
dx
dy is the derivative of y (with respect to x )
dx
dy is the differential coefficient of y (with respect to x ).
Example
Example^2
1 2
3 2
2 2
x x x x
x
x
x f x
_
_
_
_
To obtain coordinates of a SP. on the graph of y f ( x )
(i) Put f (^ x ) 0 and solve for x.
(iii) If (^) f (^) ( a ) 0 there will be a minimum point at x a
If f ( a ) 0 there will be a maximum point at x a
If f ^ ( a ) 0 there could be max or min or inflexion so the
second derivative rule fails. Investigate the gradient to the
immediate left and right of the stationary point. (see the + and -
signs on the diagrams in the previous section).
Example Find the stationary points on the graphs of
(i)^22
2 y x x
(ii) 3 2
3 y x x
and sketch the graphs.
(i) Here we have a quadratic function, which will have a true max
or min.
2
x dx
dy
y x x
SP at 2 x 2 0
i.e. at x 1
2 dx
d y
SP is a minimum.
(ii) 3 2 3 y x x
2 x dx
dy
For SP 3 3 0 2 x x
2 x
x 1
SPs at (1, 0) (-1, 4)
x dx
d y 2 6
2
(-1, 1)
(^0) x
y
2 Check point (0, 2)
(-1, 4)
(^0) x
y
2
(1, 0)
(2, 4)
(-2, 0)
Trigonometry
Trig ratios for 30 , 60 , 45
2 sin 30 cos 60 ^1 2 sin 45 cos 45 ^1
2
3 sin 60 cos 30 tan 45 1
tan 60 3 3 tan 30 ^1
Trig ratios for all angles NB the CAST DIAGRAM
For the sign of a trig ratio
All positive in first quadrant
Sine (only) in second quadrant
Etc…
60
30
3
1 1
2 2
45
1
1
S
T (^) C
A
Example Without using a calculator find
(i) cos 150 (ii) tan 210 (iii)
sin 240
(i) (ii) (iii)
2
3
cos 150 cos 30
3
1
tan 210 tan 30
2
3
sin 240 sin 60
Trig of Scalene triangles
Sine rule
b
A
a
sin sin
c
sin
Given AAS use it to find a second side
Given SSA use it to find a second angle (but take care to choose the
angle size appropriately – it could be acute or obtuse).
S (^) A
T^ C
30
210
S (^) A
T^ C
30
150
S (^) A
T^ C
60
A
B
C
a
b
