# Skema Jawapan & Pemarkahan Add Maths Kertas 1&2 SPM by nklye

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```									MENJAWAB SOALAN SPM
Kertas 1 dan 2
Skema Jawapan & Pemarkahan
Siri Ceramah Di:
(i) SMK Dato’ Klana Putra, Lenggeng, Negeri Sembilan,
(ii) SMK Engku Husain, Semenyih, Selangor,
(iii) SMK Dengkil, Kuala Langat, Selangor,
(iv) SMK Taman Jasmin 2, Kajang, Selangor,
(v) SMK Jalan 4, Bandar Baru Bangi, Selangor,
(vi) SMK Dato’ Ahmad Razali, Amapang, Selangor.
Q.1:

{-
(a) -1, 1 / -1 and 1 / {-1, 1}
….√(1)
reject: 1 or -1 / (-1, 1) / [-1, 1]

many-to-
(b) many-to-one /                     ….√(1)
many with one /
many → one
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):
Q.2(b):
gf(2)
(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:
Q.3:
   gh(x) = g[h(x)]
= mx2 + n – 3 ……..(P1)
= 3x2 + 7
Bandingkan:
m=3             ……...(1)
n–3=7             ………(P1)
n = 10          ………(2) (Kedua-duanya)
Q.4
Q.4:
   2x2 + p + 2 = 2px + x2
x2 – 2px + p + 2 = 0             √ ………..(P1)
    a = 1, b = - 2p, c = p +2
    b2 - 4ac > 0
   (-2p)2 – 4(1)(p + 2) > 0
4p2 – 4p – 8 > 0                √ ………..(P2)
p2 – p – 2 > 0                               x
-1       2
          Let: p  2 – p – 2 = 0

(p – 2)(p+ 1) = 0
p = 2 atau p = - 1
   Therefore: p > 2 atau p < -1 √ ..…….(3)
Q.5:
Q.5:
   (x – ⅔)(x + 4) = 0      ….…….(P1)
x2 – ⅔ x – 4x – 8/3 = 0
3x2 + 10x – 8 = 0       …………(2)

OR:
x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(P1)
3x2 + 10x – 8 = 0           …………(2)
Q.6:
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          √ …..…….(P1)
   2xlog10 3 + log10 3 = xlog10 4

   2x(0.4771) – x(0.6021) = - 0.4771   √.………..(P2)
   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 + ———          √……….(P1)
log24

log2M2 = log24 + 2log2N   √..……..(P2)

log2M2 = log24N2          √..….…(P3)
M2 = 4N2
M = √4N2
M = 2N                 √..…....(4)
Q.9:
 (y – 1) – x = (5 + 2x) – (y – 1) √....(P1)
 y – 1 – x = 5 + 2x – y + 1
         2y = 3x + 7             √….(P2)
3x + 7
          y = ———                √….(3)
2
Q.10(a):
   a = 108

ar2 = 12

ar2   =   12    √ ……….(P1)
a        108

r2 = ¹/9
r = ¹/3        √..……….(2)
Q.10(b):
108
S∞    = ———       √…….….(P1)
1 – ¹/3

108
= ———
⅔

=   162         √…..…….(2)
Q.11:

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

¹/y =   a/       +b   √ ……….(P1)
x

= - 6/4 = - 3/2    √ ……….(2)

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

 x/y - y/4 = 1
4x - 2y =        8        √..……(P1)
2y =      4x - 8
y =    4/ x – 4
2
y–7=       2(x – 2)
y =   2x + 3 √.……(3)
Q.13:
              4m + 9s                        2m + 6t        √…. (P1)
S   =                      or      t=
5                          5
P(2m, m)                Q(s, t)           R(3s, 2t)

3                  2

5s – 9s = 4m,           or 5t – 6t = 2m
- 4s = 4m,                 - t = 2m            √……(P2)
- s = m,
- 2s = - t
s = t/2             √…...…………………...(3)
Q.14:
Q.14:
   PA : PB = 2 : 1                       √..…(P1)
reject:   PA/      = 2/1
PB
   PA = 2 √(x – 5)2 + (y – 0) 2 √…..(P1) or
PB = √(x + 2)2 + (y + 3)2               √….(P1)
   PA2 = 4[(x – 5)2 + y 2 ] =
PB2 = (x + 2)2 + (y + 3)2                √....(P2) or
   4[(x – 5)2 + y 2 ] = (x + 2)2 + (y + 3)2 √…..(P2)
   3x2 + 3y2 - 44x + 6y + 87 = 0 √....(3)
Q.15
D                  C
AC = AB + BC
= AB - CB                 .........(P1)
A                 B                 = (i + 2j) – (-5i – 6j)

= 6i + 8j                 ….…(2)

6i + 8j
Unit vector AC = ————                       ……..….(P1)
√ 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   ……………….…..(P1)

Compare:   m + 2n = 7              5n = 4        .….(P1)
m – 3n = 3               n = 4/5 )..(2 either)
m = 27/5 )..(3)(both)
Q.17:
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:
Q.18:
   2 sec2θ – 3tanθ = 4
3tanθ
   2(1 + tan2θ) – 3tanθ = 4
3tanθ                     …..…..(1)
2 + 2 tan2θ – 3tanθ = 4
3tanθ
2tan2θ - 3tanθ – 2 = 0
3tanθ
   (2tanθ 1)(tanθ
(2tanθ + 1)(tanθ – 2) = 0                ………..(2)
   tanθ = -½ atau tanθ = 2
tanθ              tanθ
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       ………..(P1)

s = jθ
BC = 15 x 2.142    ………..(P2)
= 32.13         ………..(3)
Q.20:
Q.20:
2
dy     (x – 4)(4x) – (2x + 3)(1)
— = ———————————                     ……(P2)
dx             (x - 4)2
4x   …….(P1)

2
2x – 16x - 3
= ———————                     ..……(3)
(x - 4)2
Q.21:
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    …..(P1)
1

3
             [ kx2 ∕ 2 ]       = 12 …..(P2)
1

9k/2 – k/2 = 12,         8k/2 = 12   …..(P3)
k = 3     …..(4)
Q.23:
Q.23:

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

6               4
P x P = (6 x 5 x 4 x 3) x (4 x 3 x 2) ..(P2)
(4
4           3

= 360 x 24
= 86400       ……...(3
……...(3)
Q.24:
Q.24:
9               5
   (a)        C       x       C                                           ……….(P1)
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           …...….(P1)
5              5           4       6           3       7

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

    0.5 – 0.225   ……….(P1)

=   0.275         ……….(2)
PAPER 2
Q1:
x+4
x = 3y – 4   or   y = ———        ………………. √(P1)
3

Eliminate x or y:
(3y – 4)2 + y(3y – 4) – 40 = 0   …………………. √(M1)
3y2 -7y – 6 = 0
(3y + 2) (y – 3)= 0 or using quadratic equation …. √(M1)
or completing the square method.
y = - 2/3 , y = 3.     ………………………………√(A1)

x = - 6, x = 5      ………………………………√(A1)
Q3:
(a)   r = 1.05                     √….(P1)
Use T6 = ar 5
T6 = 18000 x (1.05)5
(1 05)        √….(M1)
=   RM 22 973.00
√….(A1)

(b) Use ar n-1 ≥ 36 000            √….(M1)
18000 x (1.05)n-1 ≥ 36 000
(1 05)
n ≥ 16          √….(A1)

No working minus 1 mark if answer is
correct
n
a(r – 1)
(c)   Use   S6 = ———
r-1

6
Use   S6=18000{(1.05) – 1}..√(M1)
1.05 – 1

= RM 122 434            √..(A1)
Q8:
(a)
x    1.5 3.0 4.5 6.0 7.5 9.0
log10 y 0.40 0.51 0.64 0.76 0.89 1.00

√.... (P1)
Plot graph log y against x2
log y
√ P1
1.0                                                   +
+       √ G1
0.8                                   +
√ P1
+
0.6
+
+
0.4

0.2

0.0                                                          x
0      1   2   3    4      5   6       7       8   9
(b) Plot log10 y against x
(correct axes and uniform scales)…..√(P1)
6 points plotted correctly …………..√(P1)
Lines of best fit               …………..√(G1)

(c) log10 y = log10 h + 2x log10 k …………….√(P1)
c = log10 h            .................... √(M1)
h = 1.78                   ………….√(A1)
m = 2 log10 k               ………….√(M1)
k = 1.09 – 1.12            ………….√(A1)
Q13:
P1
(a)   Use   I = —— x 100
P0
1.80
h   = —— x 100 ………………..√(M1)
1.50

= 120           ………………….√(A1)

0.90
112.5 =
112.        —— x 100 ……………… √(M1)*
k

k = 0.8 or 80 sen   ……...……………√(A1)
    I      W         IW
150     30       4500
120     45       5400
112.5
112.    15       1687.5
1687.
105     10       1050
100     12637.5
12637.    ……………√(P1)
……………

12637.5
12637.

I = —— x 100      …………√(M1)
100

=      126.38   …………√(A1)
(c) (i)             150
126.38 = —— x 100 ……√(M1)
100

189.
= 189.6     ……√(A1)

P1
(c) (ii) —— x 100 = 189.6 ……√(M1)
25

47.
= 47.39 ……√(A1)
MENJAWAB SOALAN