1. NOTES AND FORMULAE ADDITIONAL MATHEMATICS FORM 4
1. FUNCTIONS 2 6 The axis of symmetry is x – p = 0 or
(a) Arrow Diagram fg(x) = f( )= 2 x=p
x x
(c) Quadratic inequalities
2. QUADRATIC EQUATION (x – a)(x – b)  0
2
(a) ax + bx + c = 0 Range x  a, x  b
2
x =  b  b  4ac (x – a)(x – b)  0
2a Range a  x  b
Sum of roots: 4. INDICES AND LOGARITHM
Domain = set A = {−2, 1, 2, 3}
b
Codomain = set B = {1, 4, 8, 9}    n
Range = set of images = {1, 4, 9} a (a) x=a  loga x = n
Object of 4 = −2, 2 Product of roots: Index Logarithm
Image of 3 = 9 c Form Form
2  
In function notation f : x  x a
(b) Types of Relation (b) Logrithm Law
(b) Equation from the roots: 1. logaxy = logax + logay
2
x - (sum of roots)x + product of 2. loga x = logax – logay
roots = 0 y
n
One-to-one One-to-many 3. loga x = n logax
3. QUADRATIC FUNCTION 4. loga a = 1
(a) Types of roots 5. loga 1 = 0
2
b - 4ac > 0  2 real and
6. loga b = log c b
distinct/different roots.
Many-to-one Many-to-many 2 log c a
b - 4ac = 0  2 real and equal
roots/two same roots. 7. loga b = 1
(c) Inverse Function 2
b - 4ac < 0  no real root. log b a
Function f maps set A to set B 2
-1 b - 4ac  0  got real roots.
Inverse function f maps set B to y y y
set A. 5. COORDINATE GEOMETRY
Given f(x) = ax + b, let y = ax + b (a) Distance between A(x1, y1) and
y b -1 xb B(x2, y2)
x= f :x 0 x
a a 0 AB = ( x 2  x1 ) 2  ( y 2  y1 ) 2
x
(d) Composite Function 0 x (b) Mid point AB
2 2 2
fg(x) means the function g followed b - 4ac > 0 b - 4ac = 0 b - 4ac < 0 M  x1  x 2 , y1  y 2 
by the function f.  
 2 2 
2 (b) Completing the square
E.g. f : x  3x – 2, g : x  2 (c) P which divides AB in the ratio m : n
x y = a(x - p) + q
a +ve  minimum point (p, q)
a –ve  maximum point(p, q)
Prepared by Mr. Sim Kwang Yaw 1

2. m : n By Histogram :
x
 fx Frequency
P B(x2 , y2 )
f
A(x 1 , y1 )
For ungrouped data with frequency.
P  nx1  mx 2 ny1  my 2   fx
 ,  x
i
 nm nm  f
(d) Gradient AB For grouped data, xi = mid-point
m = y 2  y1
x 2  x1 (b) Median
The data in the centre when 0 Mode Class boundary
m =  y-intercept arranged in order (ascending or
x-intercept descending).
(e) Equation of straight line Measurement of Dispersion
(a) Interquartile Range
Formula Formula :
(i) Given m and A(x1, y1) 1
nF 1
M=L+ 2
C Q1 = L  4 n  F1  C
y – y1 = m(x – x1)
fm 1
f Q1
(ii) Given A(x1, y1) and L = Lower boundary of median 3
class. Q3 = L  4 n  F3  C
B(x2, y2) 3
f Q3
n = Total frequency
y  y1 y 2  y1
 F = cumulative frequency before the
x  x1 x 2  x1 median class Ogive :
fm = frequency of median class Cumulative frequency
(a) Area of polygon C = class interval size
L = 1 x1 x2 x3 ......... x1
By Ogive 3
__
2 y1 y 2 y 3 y1 Cumulative Frequency 4
n
(g) Parallel lines
m1 = m2 n
1
__ n
(h) Perpendicular lines 4
m1  m2 = -1. n
__
2
0 Q Q 3 Upper boundary
6. STATISTICS 1
Measurement of Central Tendency Interquartile range = Q3 – Q1
(a) Mean
0 Median Upper boundary (b) Variance, Standard Deviation
x
x
n 2
(c) Mode Variance = (standard deviation)
For ungrouped data
Data with the highest frequency
Prepared by Mr. Sim Kwang Yaw 2

3. 2 d (xn) = nxn-1 9. SOLUTIONS OF TRIANGLES
 
 ( x  x) (c) (a) The Sine Rule
n dx
sin A sin B sin C
d (axn) = anxn-1   , or,
=  x2 x
2 (d) a b c
n dx
a b c
For ungrouped data  
(e) Differentiation of product sin A sin B sin C
2 d (uv) = u dv + v du The Ambiguous Case
 
 f ( x  x)
dx dx dx
f
2
=  fx  x 2
(f) Differentiation of Quotient
f d  u  v du  u dv
For grouped data    dx 2 dx
dx  v  v
Two triangles of the same
7. CIRCULAR MEASURE (g) Differentiation of Composite measurements can be drawn given
(b) Radian  Degree Function C, AC and AB where AB < AC.
0 d (ax+b)n = n(ax+b)n-1 × a
 =   180
r
dx (b) The Cosine Rue
 2 2 2
a = b + c – 2bc cos A
(c) Degree  Radian (h) Stationary point 
dy = 0
b 2  c2  a 2
 =    rad cos A =
o
dx
180 Maximum point: 2bc
(d) Length of arc dy = 0 and d 2 y < 0
s = r 10. INDEX NUMBER
dx dx 2 (a) Price Index
(e) Area of sector p1
2
Minimum point: I  100
A = 1 r  = 1 rs 2 po
2 2 dy = 0 and d y > 0
p0 is the price in the base year.
(f) Area of segment dx dx 2
2
A = 1 r ( – sin ) (b) Composite Index
2 (i) Rate of Change
dy dy dx
  I
 IW
8. DIFFERENTIATION
(a) Differentiation by First Principle
dt dx dt W
dy had  y I = price index
 (j) Small changes: W = weightage
dx x  0  x dy
d (a) = 0  y  . x
(b) dx
dx
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