Formula List and Important topics
(for SPM Additional Mathematics)
1.Functions
(a) Composite function. (b) Inverse function.
(c) Finding function ( i ) given function
f
andfg
, find functiong
. or given functiong
andgf
, find functionf
. ( ii ) given functiong
andfg
, find functionf
. or given functionf
andgf
, find functiong
. (d) Graph sketching2.
Quadratic Equations
(a)
ax
2+
bx
+
c
=
0
, roots of the quadratic equation x=α
,β
Hence,S.O.R. =
Sum of Roots =a
b
−
S.O.P. =
Product of Root =a
c
(b)
x
2−
(
New
S
.
O
.
R
)
x
+
(
New
P
.
O
.
R
)
=
0
(c)α
2+
β
2=
(
α
+
β
)
2−
2
αβ
(d) Factorisation,
ax
2+
bx
+
c
=
0
Sign for For a =1, given
p
>
q
b c
+ +
(
x
+
p
)(
x
+
q
)
−
+(
x
−
p
)(
x
−
q
)
+−
(
x
+
p
)(
x
−
q
)
−
−
(
x
−
p
)(
x
+
q
)
(e) ( i ) Two real and distinct/different roots means
b
2− ac
4
>
0
( ii ) Two real and equal/same roots meansb
2− ac
4
=
0
( iii ) Two real roots (special case) meansb
2− ac
4
≥
0
( iv ) No real roots meansb
2− ac
4
<
0
3.
Quadratic Functions
(a) Completing the square
y
=
a
(
x
+
p
)
2+
q
(b) Quadratic Inequalities( i )
y
=
ax
2+
bx
+
c
>
0
if a>0, the range of x : x<α
orx
>
β
. if a<0, the range of x :α
< x
<
β
. ( ii )y
=
ax
2+
bx
+
c
<
0
if a>0, the range of x :α
< x
<
β
if a<0, the range of x : x<
α
orx
>
β
. Two ways to solve quadratic inequalities i.e. Number line method and Graph sketching method.(c) Points of intersection between a straight and a curve.
Simultaneous Equation – equalises the two equations to form a quadratic equation
0
2
+
+
=
c
bx
ax
( i ) Intersects at two different points means
b
2− ac
4
>
0
( ii ) touches at one point @ tangent meansb
2− ac
4
=
0
( iii ) Does not intersect, always positive (a>0) @ always negative (a<0)
means
b
2− ac
4
<
0
4.
Simultaneous Equation
(a)
ax
2+
bx
+
c
=
kx
+
hy
=
m
wherea
,
b
,
c
,
k
,
h
,
m
are constants.- Separate the equation into two equations
ax
2+
bx
+
c
=
m
&kx
+
hy
=
m
- Always start from the linear equation- Substitute the linear equation into the non-linear equation and solve it.
(b) Graph – finding the points of intersection between a straight line and a curve. - Always starts from the straight line equation
- Substitute the straight line equation into the equation of the curve and solve it. (c) Daily problems
- Form two equation base on the information given (one linear and one non-linear) Always start from the linear equation
- Substitute the linear equation into the non-linear equation and solve it. 5.
Indices and Logarithm
Indices (a) x
a
N
=
,a
> N
0
,
>
0
(b)a
0=
1
,a
1=
a
(c) x xa
a
−=
1
(d) an =n a 1 eg., 3 3 1 a a = (e) m n n m m na
a
a
)
=
(
)
=
×(
(f) n m n n m ma
a
a
)
(
)
(
)
(
1 1=
=
(g) If (Left_Hand_side) (Right_Hand_side)
a
a
=
,Logarithm
(a)
log
aN
=
x
⇔
N
=
a
x (interchange form)(b)
log
a1
=
0
, (c)log
aa
=
1
(d) If
log
a(
Left
_
Hand
_
side
)
=
log
a(
Right
_
Hand
_
side
)
,Then
(
Left
_
Hand
_
side
)
=
(
Right
_
Hand
_
side
)
(Compare the values)(e) If
(
Left
_
Hand
_
side
)
>
(
Right
_
Hand
_
side
)
,Then
log
a(
Left
_
Hand
_
side
)
>
log
a(
Right
_
Hand
_
side
)
6.
Coordinate Geometry
(a) Finding area of quadrilateral.A(x1,y1) ) , (x2 y2 B D(x4,y4)
)
,
(
x
3y
3C
Area =(
1 2 2 3 3 4 4 1)
(
1 2 2 3 3 4 4 1)
2
1
x
y
y
x
x
y
x
y
y
x
y
x
y
x
y
x
+
+
+
−
+
+
+
(b) Method to find the equation of straight line.
( i ) Given the gradient of the straight line, mand 1 point A(x1,y1) )
( 1
1 m x x y
y− = − ( ii ) Given 2 points A(x1,y1)and B(x2,y2)
1 2 1 2 1 1
x
x
y
y
x
x
y
y
−
−
=
−
−
( iii ) Given x− intercept = band
y
−
intercept = c+
=
1
c
y
b
x
(c) The equation of straight line can be written in three forms ( i ) y=mx+c ( ii )
ax
+
by
+
c
=
0
( iii )+
=
1
c
y
b
x
(d) If two straight lines are parallel, then m1 =m2
(e) If two straight lines are perpendicular to each other, then m1× m2 =−1
Area = 1 1 4 4 3 3 2 2 1 1
2
1
y
x
y
x
y
x
y
x
y
x
(f) Locus of point
P
(
x
,
y
)
The general form of answer for locus is
0
2 2+
+
+
+
=
e
dy
cx
by
ax
wherea
,
b
,
c
,
d
,
e
=
constant ( i ) Distance from point A(x1,y1) is always k units. ∴AP=kk
y
y
x
x
−
+
−
=
⇒
2 1 2 1)
(
)
(
( ii ) Equidistance from two fixed points A(x1,y1) and B(x2,y2)
∴
AP
=
BP
2 2 2 2 2 1 2 1)
(
)
(
)
(
)
(
x
−
x
+
y
−
y
=
x
−
x
+
y
−
y
⇒
( iii ) Distance from two points A(x1,y1) and B(x2,y2) always in the ratio of m :n
mBP
nAP
n
m
BP
AP
=
⇒
=
∴
2 2 2 2 2 1 2 1)
(
)
(
)
(
)
(
x
x
y
y
m
x
x
y
y
n
−
+
−
=
−
+
−
⇒
Square both sides,
]
)
(
)
[(
]
)
(
)
[(
2 2 2 2 2 2 1 2 1 2y
y
x
x
m
y
y
x
x
n
−
+
−
=
−
+
−
⇒
7.Statistics
(a) Median, C f F N L m m ) ( − + = 12L
- lower boundary of median class N- total frequency,∑
f
F
- cumulative frequency before median classm
f
- frequency of median class C- width of median class (b) Find the mode from a histogramaxis
x
−
- the lower boundaries and upper boundaries of all the classesaxis
y
−
- the frequency of each class 5. (c) Cumulative Frequency curve or Ogiveaxis
x
−
- upper boundaries of classes including the class before the first class.axis
y
−
- cumulative frequencies of classes(the cumulative frequency of the class before the first class is ZERO) 10. (d) The effects on mean and variance when all the data changed uniformly
A new set of data
v
=
ku
±
h
Then, mean of
v
=k
×
(mean ofu
)±
h
standard deviation of
v
=k
×
(standard deviation ofu
) variance ofv
=k
2×
(variance ofu
)8.
Circular Measure
(a) Length of chordAB
=2
2
r
sin
θ
,θ
in unit ( O )(b) Area of triangle OAB 2
sin
θ
2
1
r
=
,θ
in unit ( O )(c) Area of the segment ACB= 2
(
θ
−
sin
θ
)
2
1
j
9.Differentiation
(a) If nax
y
=
, then=
n
ax
n−1dx
dy
(b) If nb
ax
y
=
(
+
)
, thenn
ax
b
a
dx
dy
n•
+
=
(
)
−1(c) For graph of a curve, the gradient of tangent to the curve at the point A(x1,y1),
1 m =
dx
dy
= f'(x1) when x= x1,dx
dy
= m1The gradient of the normal to curve at pointA(x1,y1),
1 2
1
m
m
=
−
because 1 2 1× m =− m(d) Maximum and minimum point
When
=
0
dx
dy
, the value of x is the x−coordinate for
- maximum point if 2
0
2<
dx
y
d
, - minimum point if 20
2>
dx
y
d
.(e) Rate of change
dt
dx
dx
dy
dt
dy
×
=
Example, volume of sphere, 3
3
4
r
V
=
π
. then,dt
dr
dr
dV
dt
dV
×
=
(f) Small changes and approximationsx
dx
dy
y
δ
δ
≈
×
Where
δ
x
=
x
new−
x
initial and the value ofdx
dy
is whenx
=
x
initialy
y
y
new=
initial+
δ
θ
r
B
A
O
C
10.
Solution of Triangles
(a) Ambiguous Case11.
Index Number
(a) Finding weighsIf a circle is given, the weightages are the simplest ratio of the angles. Example, o o o o o o=360 −(100 +60 +90 )=110 x (b) Information given
( i ) The price increased by 30% from year 2003 to year 2006 means
Price index,
100
130
2003 2006×
=
=
P
P
I
( ii ) The price decreased by 20% from year 2003 to year 2006 means
Price index, 100 80 2003 2006 × = = P P I
(c) Change of base time
If given 100 120 2003 2006 1 = × = P P I and 100 90 2003 2004 2 = × = P P I
Price Index for year 2006 based on year 2004,
3 133 100 90 100 100 120 100 100 2004 2003 2003 2006 2004 2006 . = × × = × × = × = P P P P P P I
A
'
C
'
'
C
BCC
BC
=
∠
∠
'
BC
BC
=
=
∠BAC
constantB
D
C
o60
o100
o
x
A
Items Angle WeightageA
100
o 10 B60
o 6 C90
o 9 D110
o 11B
C
12.
Progressions
(a)
Arithmetic Progression (A.P.)
.( i ) Method to prove a series of terms are Arithmetic Progression where exists a common difference, 1 1 − +
−
n=
n−
n nT
T
T
T
example,T
3−
T
2=
T
2−
T
1 (b)Geometry Progression (G.P.)
( i ) Method to prove a series of terms are Geometry Progression where exists a common ratio, 1 1 − + = n n n n T T T T example, 1 2 2 3 T T TT = (c) A.P. and G.P. ( i )
S
n−
S
n−1=
T
n( ii ) The sum of the first 4th terms to the first 13th terms.
3 13 13 6 5 4
T
T
T
S
S
T
+
+
+
...
+
=
−
13.Linear Law
Change the non-linear equation to linear form c mX Y = +
where
−
Y
axis−
y
new−
X
axis−
x
new − m gradient of graph − cY
−
intercept 14.Integration
(a) If
f
(x
)
dx
dy =
, thendx
x
f
dx
dx
dy
y
=
∫
(
)
=
∫
(
)
(b)
c
n
a
b
ax
dx
b
x
a
n n+
+
+
=
+
+∫
(
)
(
(
)
)
1
1(c)
Graph– equation of a curve and gradient function
If gradient function of a curve,
f
(x
)
dx
dy =
, Then the equation of the curve,dx
f
x
dx
dx
dy
y
=
∫
(
)
=
∫
(
)
( i )
∫
b =−∫
a a b f x dx dx x f( ) ( ) ( ii )∫
+
∫
=
∫
c a b a c bf
x
dx
f
x
dx
dx
x
f
(
)
(
)
(
)
( iii )∫
a
•
f
(
x
)
dx
=
a
∫
f
(
x
)
dx
example,∫
3
x
dx
=
3
∫
x
dx
15.Vector
(a) If
~a
parallel to ~b
, then ~ ~k
b
a
=
where k is a constant. (b) IfAB
=
k
BC
, thenA
,
B
and C are collinear.(c)
AB
=
OB
−
OA
(d) IfAB:BC =m:n, thenBC
n
m
AB
=
. IfAB:AC= :m m+n, thenAC
n
m
m
AB
+
=
.(e)
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
=
+
=
y
x
j
y
i
x
r
~ ~ ~ (f) If ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 1 y x u ~ and ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ = 2 2 y x v ~ , then ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ + + = + 2 1 2 1 y y x x v u ~ ~ , ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ − − = − 2 1 2 1 y y x x v u ~ ~ and ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 1 1 1 1 ky kx y x k ku ~ 16.Trigonometric Functions
(a) Quadrants
A
S
C
T
θ θ =180o+ 3 θ θ θ =360o− 4 θ θ =180o − 2I
II
III
IV
I
II
III
IV
A
B
C
m
n
(b) Graph sketching of trigonometric functions
sin
θ
,
kos
θ
andtanθ
.(c) Number of solutions
17.
Permutation and Combination
(a) Permutation – Choose with arrangement which means
arrangement does affect the number of choices
(b) Combination – Choose without involving arrangement which
means arrangement does not affect the number of choices
18.Probability
(a) Concept of Complement
)
'
)
(
A
1
P(
A
P
=
−
where)
(
)
'
)
'
(
S
A
A
n
n(
P
=
andn(
A
'
)
=
n
(
S
)
−
n
(
A
)
(b) Tree diagram – Total probability of all the branches is 1
19.Distribution of Probability
(a) Binomial distribution
( i ) Concept of Complement
)
(
)
(
)
(
)
(
)
(
3
1
P
3
1
P
2
P
1
P
0
P
X
≥
=
−
X
<
=
−
X
=
−
X
=
−
X
=
( ii )P
(
X
≥
1
)
=
1
−
P
(
X
=
0
)
and nC0 =120.
Motion on a Straight Line
(a) ( i ) Displacement ,
s
=
∫
v
dt
( ii ) Velocity,
dt
ds
v
=
;v
=
∫
a
dt
( iii ) Acceleration,
dt
dv
a
=
(b) Hidden Information( i ) Stop for a while, turn, change direction of motion ⇒ v=0
( ii ) Maximum displacement,
⇒
displacement when v=0(=
0
dt
ds
) ( iii ) Pass through the origin again ⇒ s=0
( iv ) Always move to the right ⇒ v>0
( v ) On the left side of point O, ⇒ s<0
( vi ) Particle
P
and particleQ
meet⇒
s
P=
s
Q ( vii ) Maximum velocity⇒
velocity when a=0.21.
