# Menjawab Soalan SPM Add Math

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```					MENJAWAB SOALAN SPM
MATHEMATICS

SMK TAMAN TASIK,
AMPANG, SELAGOR

26 APRIL 2012
Q.1:

(a) -1, 1 / -1 and 1 / {-1, 1}
reject: -1 or 1 / (-1, 1) / [-1, 1]
.…..(1)
(b) many to one /
many with one/
many → 1                              …...(1)
Q.2(a):
(a)   f(x) = 4x + 2
f -1(x) = y
f(y) = x
= 4y + 2   √ -----------(1)
x = 4y + 2

x-2
y = ———        √------------(2)
4
Q.2(b):
(b) gf(2) = g[f(2)]
= g[4(2) + 2]     √………..………..(1)
= g(10)
= 102 – 3(10) – 1
= 69              √…………………..(2)

f(2) = 4(2) + 2
= 10            √ ………………….(1)
Q.3:
   gh(x) = g[h(x)]
= mx2 + n – 3   ……..(1)
= 3x2 + 7
Bandingkan:
m=3                ……..(1) or
n–3=7                 ……..(1)
n = 10             ……..(2)
Q.4:
•   2x2 + p + 2 = 2px + x2
x2 – 2px + p + 2 = 0           √ ………..(1)
•    a = 1, b = - 2p, c = p +2
•    b2 - 4ac > 0
•   (-2p)2 – 4(1)(p + 2) > 0
4p2 – 4p – 8 > 0              √ ………..(2)
p2 – p – 2 > 0
•          Let: p2 – p – 2 = 0                      x
(p – 2)(p+ 1) = 0     -1        2

p = 2 atau p = - 1

• Therefore: p > 2 atau p < -1 √ ..…….(3)
Q.5:
• (x – ⅔)(x + 4) = 0        ……….(1)
x2 – ⅔ x – 4x – 8/3 = 0
3x2 + 10x – 8 = 0          …………(2)

OR:
x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(1)
3x2 + 10x – 8 = 0           …………(2)
Q.6:

 (a) p = 3 (axis of symmetry) √.…..(1)
 (b) x = 3                    √……(1)

 (c) (3, -1)                  √……(1)
   Note: Minimum value = -1
Q.7:
• 32x + 1 = 4x
• (2x + 1)log10 3 = xlog10 4     √ …..…….(1)
• 2xlog10 3 + log10 3 = xlog10 4

• 2x(0.4771) – x(0.6021) = - 0.4771 √.….(2)
• 0.9542x – 0.6021x = - 0.4771
• 0.3521x = - 0.4771
- 0.4771
•        x =     ————
0.3521

•           = - 1.355       √ ………...(3)
Q.8:
• 2log2M = 2 + 4log4N
4log2N
2log2M = 2 + ———          √……….(1)
log24

log2M2 = log24 + 2log2N   √..……..(2)

log2M2 = log24N2          √..….…(3)
M2 = 4N2
M = √4N2
M = 2N                √..…....(4)
Q.9:

(y – 1) – x = (5 + 2x) – (y – 1) √....(1)
y – 1 – x = 5 + 2x – y + 1
2y = 3x + 7         √……...(2)
3x + 7
y = ———             √….…..(3)
2
Q.10(a):
•    a = 108

ar2 = 12

ar2 = 12      √ ……….(1)
a     108

r2 = ¹/9

r   = ¹/ 3   √..……….(2)
Q.10(b):

108
S∞ = ——— √…….….(1)
1 – ¹/ 3

108
= ———
⅔

=   162   √…..…….(2)
Q.11:

x
•    y    =   ————
a + bx
a + bx
¹/y   =   ————
x

¹/y   =   a/
x   +b   √ ……….(1)

= - 6/4 = - 3/2        √ ……….(2)

b = y-intercept
= 6           √……….(1)
Q.12:

• x/y - y/4 = 1
4x - 2y =       8       √..……(1)
2y =     4x - 8
y =    4/ x – 4
2
y–7=      2(x – 2)
y =   2x + 3 √.……(3)
Q.13:
4m + 9s                    2m + 6t √…. (1)
S   =               or        t=
3+2                       3+2
P(2m, m)             Q(s, t)          R(3s, 2t)

3                  2

5s – 9s = 4m,      or 5t – 6t = 2m
- 4s = 4m,            - t = 2m √……(2)
- s = m,
- 2s = - t
s = t/2      √……………………...(3)
Q.14:
PA : PB = 2 : 1                     √..…(1)
reject:   PA/      = 2/1
PB
PA = 2 √(x – 5)2 + (y – 0) 2 √…..(1) or
PB = √(x + 2)2 + (y + 3)2               √….(1)
PA2 = 4[(x – 5)2 + y 2 ] =
PB2 = (x + 2)2 + (y + 3)2                √...(2) or
4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(2)
3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)
Q.15
D                     C
(a) AC = AB + BC
= AB - CB                 .........(1)
A                  B                   = (i + 2j) – (-5i – 6j)

= 6i + 8j                 ….…(2)

6i + 8j
(b) Unit vector AC = ————                       ……..….(1)
√ 62 + 82

6i + 8j 3i + 4j
= ———— or ————               …..……(2)
10       5
Q.16
PQ = PO + OQ
P(1, 2)
= - OP + OQ
O                          = - (i + j) + (8i + 4j)
= 7i + 3j               ….(1)
Q(8, -4)

7i + 3j = ma + nb
= m(i + j) + n(2i – 3j)
= (m + 2n)i + (m – 3n)j   ……………….…..(2)

Compare:     m + 2n = 7            5n = 4
m – 3n = 3             n = 4/5 )
m = 27/5 )……(3)
Q.17:
   sec θ = 1/cos θ                      1
1 – t2

         = 1/t   ………(1)
t

cos (90 – θ) = sin θ       ……….(1)

=√1–t
2   ……….(2)
Q.18:
 2 sec2θ – 3tanθ = 4
 2(1 + tan2θ) – 3tanθ = 4                     …..…..(1)
2 + 2 tan2θ – 3tanθ = 4
2tan2θ - 3tanθ – 2 = 0
 (2tanθ + 1)(tanθ – 2) = 0                   ………..(2)
 tanθ = -½ atau tanθ = 2
θ = (360o – 26o 34’), (180o - 26o 34’)
   θ = 333o 26’, 153o 26’                    ………...(3)
θ = 63o 26’, (180o + 63o 26’)
   θ = 630 26’, 2430 26’                  ….....….(3)
   θ = 630 26’, 153o 26’, 2430 26’, 333o 26’. …….….(4)
Q.19:

 ̸̱ BOC = π – 1
= 3.142 – 1
= 2.142         ………..(1)

s = jθ
BC = 15 x 2.142   ………..(2)
= 32.13        ………..(3)
Q.20:
2
dy    (x – 4)(4x) – (2x + 3)(1)
— = ———————————                      ……(2)
dx            (x - 4)2
4x   …….(1)

2
2x – 16x - 3
= ———————                      ..……(3)
(x - 4)2
Q.21:
dy/dx = 4x – 3                   ………..(1)
δx = 0.01                     ………..(1)
x = 2

δy∕      ≈ dy∕dx
δx

δy ≈ dy∕dx (δx)

= [4(2) – 3](0.01)     …………(2)
= 5(0.01)
= 0.05                 …………(3)
Q.22:
3          3
   ∫ f(x) + ∫ (kx)dx = 20
1         1
3
       8 + [ kx2 ∕ 2 ] = 20 …..(1)
1

3
           [ kx2 ∕ 2 ] = 12 …..(2)
1
9k/2 – k/2 = 12,   8k/2 = 12   …..(3)
k = 3     …..(4)
Q.23:

6                      4
P           or    P           ……….(1)
4                  3

6               4
P x P OR (6 x 5 x 4 x 3) x (4 x 3 x 2) .......(2)
4           3

= 360 x 24
= 86400      …….…..(3)
Q.24:
9               5
• (a)        C       x       C                                           ……….(1)
6               4

atau 9 x 8 x 7
3x2
•   = 84                                                                 …………(2)

5           9               5       9           5       9
• (b)        C x C               +   C x C           +   C x C           …...…….(1)
5              5           4       6           3       7

= 948                                                            …………(2)
Q.25:

   0.5 – 0.225           ……….(1)
 = 0.275                 ……….(2)
TEKNIK MENJAWAB SOALAN
SPM
PAPER 1

THE END
TEKNIK MENJAWAB SOALAN
SPM
PAPER 1

THANK YOU

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