1. 2
The following formulae may be helpful in answering the questions. The symbols given
are the ones commonly used.
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang
diberi adalah yang biasa digunakan.
ALGEBRA
− b ± b 2 − 4ac log c b
1 x= 8 log a b =
2a log c a
2 a × a = a m+ n
m n
9 Tn = a + (n – 1)d
3 a m ÷ a n = a m-n
10 n
Sn = [ 2a + (n – 1) d ]
4 ( a m ) n = a mn 2
5 loga mn = loga m + loga n 11 Tn = ar n−1
6 m 12 a ( r n − 1) a (1 − r n )
loga = loga m – loga n Sn = = ,r≠1
n r −1 1− r
7 loga mn = n loga m 13 a
S∞ = , r <1
1−r
CALCULUS / KALKULUS
1 dy dv du 4 Area under a curve
y = uv, =u +v
dx dx dx Luas di bawah lengkung
b
du dv = ∫y dx or (atau)
2 u v −u a
y = , dy
v = dx 2 dx
dx v b
= ∫ x dy
a
5 Volume generated / Isipadu janaan
dy dy du
= ×
3 dx du dx
b
= ∫ πy dx or ( atau)
2
a
b
= ∫ πx dy
2
a

2. 3
STATISTICS / STATISTIK
1
x=
∑x 7
I=
∑W I i i
N ∑W i
2 ∑ fx 8 n n!
x= Pr =
∑f ( n − r )!
3 9 n!
σ= ∑ ( x − x) 2
= ∑x 2
−x
2
n
Cr =
( n − r )! r!
N N
4
∑ fx 2 10 P(A∪B) = P(A) + P(B) – P(A∩B)
∑
2
f ( x − x) 2
−x
σ= = ∑f 11 P(X = r) = n C r p r q n − r , p + q = 1
∑f
5  1 N −F  12 Mean / Min , µ = np
m=L+ 2 C
 f 
 m  13 σ = npq
6 Q1 14 X −µ
I= × 100 Z=
Q0 σ
GEOMETRY / GEOMETRI
1 Distance / Jarak 4 Area of triangle / Luas segitiga
= ( x2 − x1 )2 + ( y2 − y1 )2 1
= ( x1 y 2 + x 2 y 3 + x3 y1 ) − ( x 2 y1 + x3 y 2 + x1 y 3 )
2
2 Midpoint / Titik tengah 5 r = x2 + y2
 x1 + x 2 y1 + y 2 
(x, y) =  , 
 2 2  ∧ xi + y j
6
r=
x2 + y 2
3 A point dividing a segment of a line
Titik yang membahagi suatu tembereng garis
 nx1 + mx 2 ny1 + my 2 
(x, y) =  , 
 m+n m+n 

3. 4
TRIGONOMETRY / TRIGONOMETRI
1 Arc length, s = r θ 8 sin (A ± B) = sin A cos B ± cos A sin B
Panjang lengkok, s = j θ sin (A ± B) = sin A kos B ± kos A sin B
1 2
2 Area of sector, A = r θ 9 cos (A ± B) = cos A cos B  sin A sin B
2
1 2
Luas sektor, L = j θ kos (A ± B) = kos A kos B  sin A sin B
2
3 sin 2 A + cos 2 A =1 10 tan (A ± B ) = tan A ± tan B
sin 2 A + kos 2 A =1
1  tan A tan B
11 2 tan A
4 sec 2 A = 1 + tan 2 A tan 2 A =
1 − tan 2 A
sek 2 A = 1 + tan 2 A
5 cosec 2 A = 1 + cot 2 A 12 a b c
= =
sin A sin B sin C
kosek 2 A = 1 + kot 2 A
6 sin 2A = 2 sin A cos A 13 a 2 = b 2 + c 2 – 2bc cos A
sin 2A = 2 sin A kos A
a 2 = b 2 + c 2 – 2bc kos A
7 cos 2A = cos2 A – sin2 A 14 Area of triangle / Luas segitiga
= 2 cos 2 A – 1 1
= ab sin C
= 1 – 2 sin 2 A 2
kos 2A = kos2 A – sin2 A
= 2 kos 2 A – 1
= 1 – 2 sin 2 A