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MENJAWAB SOALAN SPM
ADDITIONAL MATHEMATICS
Kertas 1 dan 2
Skema Jawapan & PemarkahanSiri 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.
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Q.1:
(a) (a) --11, , 1 1 / / --1 1 and and 1 1 / {/ {--11, , 11} }
….√(1)
reject: 1 or -1 / (-1, 1) / [-1, 1]
(b) many(b) many--toto--one / one / ….√(1)
many with one /many with one /
many many → one→ one
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Q.2(a):
(a)(a) f(x) = 4x + 2f(x) = 4x + 2
f f --11(x) = y (x) = y
f(y) = xf(y) = x
= 4y + 2 = 4y + 2 √ -----------(1)
x x = = 4y + 24y + 2
x x -- 22y = y = —————— √------------(2)
44
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Q.2(b)Q.2(b)::
(b) (b) gfgf(2) = g[f(2)](2) = g[f(2)]
= g[4(2) + 2] = g[4(2) + 2] √√………..………..(1)………..………..(1)
= g(10)= g(10)
= 10= 102 2 –– 3(10) 3(10) –– 11
= 69 = 69 √√…………………..(2)…………………..(2)
f(2) = 4(2) + 2 f(2) = 4(2) + 2
= 10 = 10 √√ ………………….(1)………………….(1)
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Q.3Q.3::
gh(x) = g[h(x)]
= mx2 + n – 3 ……..(P1)
= 3x2 + 7
Bandingkan:
m = 3 m = 3 ……...(1)……...(1)
n n –– 3 = 7 3 = 7 ………(P1)………(P1)
n = 10 n = 10 ………(2) ………(2) (Kedua-duanya)
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Q.Q.44::
2x2x2 +2 + p + 2 = 2px + xp + 2 = 2px + x22
xx2 2 –– 2px + p + 2 = 0 2px + p + 2 = 0 √ ………..(P1)………..(P1)
aa = 1, b = = 1, b = -- 2p, c = p +22p, c = p +2
bb22 -- 4ac > 04ac > 0
((--2p)2p)2 2 –– 4(1)(p + 2) > 04(1)(p + 2) > 0
4p4p2 2 –– 4p 4p –– 8 > 0 8 > 0 √ ………..(P2)………..(P2)
pp2 2 –– p p –– 2 > 0 2 > 0
Let: Let: pp2 2 –– p p –– 2 = 02 = 0
(p (p –– 2)(p+ 1) = 02)(p+ 1) = 0
p = 2 atau p = p = 2 atau p = -- 11
Therefore: p > 2 atau p < Therefore: p > 2 atau p < --1 1 √ ..…….(3)
-1 2x
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Q.5Q.5::
(x (x –– ⅔⅔)(x + 4) = 0 )(x + 4) = 0 ….…….(P1) ….…….(P1)
xx22 –– ⅔ x ⅔ x –– 4x 4x –– 8/3 = 08/3 = 0
3x3x22 + 10x + 10x –– 8 = 0 8 = 0 …………(2)…………(2)
OR:
x2 – (⅔ - 4)x + (⅔)(-4) = 0 …………(P1)
3x2 + 10x – 8 = 0 …………(2)
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Q.6Q.6::
(a) p = 3 (a) p = 3 (axis of symmetry) √...(1)
(b) x = 3 √...(1)
(c) (3, -1) √...(1)
Note: Minimum value = -1
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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)
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Q.8:
2log2log22M = 2 + 4logM = 2 + 4log44NN
4log4log22NN
2log2log22M = 2 + M = 2 + —————— √……….(P1)
loglog2244
loglog22MM22 = = loglog224 + 2log4 + 2log22N N √..……..(P2)
loglog22MM22 = log= log224N4N2 2 √..….…(P3)
MM22 = 4N= 4N2 2
M = M = √4N√4N22
M = 2N M = 2N √..…....(4)
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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
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Q.10(a):
a = 108
ar2 = 12
ar2= 12 √ ……….(P1)
a 108
r2 = ¹/9
r = ¹/3 √..……….(2)
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Q.10(b):
108
S∞ = ——— √…….….(P1)1 – ¹/3
108= ———
⅔
= 162 √…..…….(2)
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Q.11:
x
y = ————
a + bx
a + bx
¹/y = ————x
¹/y = a/x + b √ ……….(P1)
a = gradient
= - 6/4 = - 3/2 √ ……….(2)
b = y-intercept
= 6 √……….(1)
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Q.12:
xx/y /y -- yy/4 = 1/4 = 1
4x 4x -- 2y = 8 2y = 8 √..……(P1)
2y = 4x 2y = 4x -- 88
y = y = 44//22x x –– 44
gradient, m = 2 m = 2 √..……(P2)
y y –– 7 = 2(x 7 = 2(x –– 2) 2)
y = 2x + 3 y = 2x + 3 √.……(3)
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Q.13:
4m + 9s4m + 9s 2m + 6t2m + 6t √…. (P1)
P(2m, m) P(2m, m) Q(s, t)Q(s, t) R(3s, 2t)R(3s, 2t)
3 23 2
5s 5s –– 9s = 4m, 9s = 4m, or 5t 5t –– 6t = 2m6t = 2m
-- 4s = 4m, 4s = 4m, -- t = 2m t = 2m √……(P2)
-- s = m, s = m,
- 2s = - t
s = t/2 √…...…………………...(3)
S = or t =5 5
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Q.Q.1414::
PA : PB = 2 : 1 PA : PB = 2 : 1 √..…(P1)
reject: PA/PB = 2/1
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)
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Q.15
A B
C
AC = AB + BC
= AB - CB .........(P1)
= (i + 2j) – (-5i – 6j)
= 6i + 8j ….…(2)
6i + 8jUnit vector AC = ———— ……..….(P1)
√ 62 + 82
6i + 8j 3i + 4j
= ———— or ———— …..……(2)
10 5
D
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Q.16
P(1, 2)
O
Q(8, -4)
PQ = PO + OQ
= - OP + OQ
= - (i + j) + (8i + 4j)
= 7i + 3j ….(1)
7i + 3j = ma + nb
= m(i + j) + n(2i – 3j)
= (m + 2n)i + (m – 3n)j ……………….…..(P1)
Compare: m + 2n = 7
m – 3n = 3
5n = 4 .….(P1)
n = 4/5 )..(2 either)
m = 27/5 )..(3)(both)
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Q.Q.1717::
sec sec θθ = = 11//cos cos θθ
= = 11//t t ………(1) ………(1)
cos (90 cos (90 –– θ) = sin θ θ) = sin θ ……….(1)……….(1)
= √ 1 = √ 1 –– tt22 ……….(2)……….(2)
1 – t21
t
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Q.18Q.18:: 2 sec2 sec22θθ –– 3tan3tanθθ = 4= 4
2(1 + tan2(1 + tan22θθ) ) –– 3tan3tanθθ = 4 = 4 …..…..(1)…..…..(1)
2 + 2 tan2 + 2 tan22θθ –– 3tan3tanθθ = 4= 4
2tan2tan22θθ -- 3tan3tanθθ –– 2 = 02 = 0
(2tan(2tanθθ + 1)(tan+ 1)(tanθθ –– 2) = 0 2) = 0 ………..(2)………..(2)
tantanθθ = = --½ ½ atau atau tantanθθ = 2= 2
θθ = (360= (360oo –– 2626o o 34’), (18034’), (180o o -- 2626o o 34’)34’)
θθ = 333= 333oo 26’, 15326’, 153o o 26’ 26’ ………...(3)………...(3)
θθ = 63= 63o o 26’, (18026’, (180o o + 63+ 63o o 26’)26’)
θθ = 63= 630 0 26’, 24326’, 2430 0 26’ 26’ ….....….(3)….....….(3)
θθ = 63= 630 0 26’, 15326’, 153o o 26’, 24326’, 2430 0 26’, 33326’, 333oo 26’.26’. …….….…….….(4)(4)
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Q.19:
BOC = BOC = ππ –– 1 1
= = 33..142 142 –– 1 1
= = 22..142 142 ………..(P1)
s = jθ
BC = 15 x 2.142 ………..(P2)
= 32.13 ………..(3)
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Q.Q.2020::
dy (x dy (x –– 4)(4x) 4)(4x) –– (2x(2x22
+ 3)(1)+ 3)(1)—— = = —————————————————————— ……(P2)dx (x dx (x -- 4)4)22
4x …….(P1)
2x2x22
–– 16x 16x -- 33= = —————————————— ..……(3)
(x (x -- 4)4)22
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Q.21Q.21::
dy/dx = 4x dy/dx = 4x –– 3 3 ………..(1)
δx = 0.01 δx = 0.01 ………..(1)
x = 2x = 2
δyδy∕∕δxδx ≈ ≈ dydy∕∕dxdx
δyδy ≈ ≈ dydy∕∕dxdx (δx)(δx)
= [4(2) = [4(2) –– 3](0.01) 3](0.01) …………(2)
= 5(0.01)= 5(0.01)
= 0.05 = 0.05 …………(3)
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Q.22:
3 3 33
∫ f(x) + ∫ (kx)dx = 20 ∫ f(x) + ∫ (kx)dx = 20 1 11 1
33
8 + [ kx8 + [ kx2 2 ∕ 2∕ 2 ] = 20 ] = 20 …..(P1)11
33
[ kx[ kx2 2 ∕ 2∕ 2 ] = 12 ] = 12 …..(P2) 11
9k/2 – k/2 = 12, 8k/2 = 12 …..(P3)
k = 3 …..(4)
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Q.Q.2323::
6 46 4
P or P P or P ….(1)4 34 3
6 46 4
P x P = P x P = ((6 6 x x 5 5 x x 4 4 x x 33) x () x (4 4 x x 3 3 x x 22)) ..(P2)4 34 3
= = 360 360 x x 2424
= = 86400 86400 ……...(……...(33))
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Q.Q.2424::99 55
(a) C x C (a) C x C ……….(P1)66 44
atauatau 9 x 8 x 79 x 8 x 7
3 x 23 x 2
= 84 = 84 …………(2)
55 9 5 9 5 99 5 9 5 9 (b) C x C + C x C + C x C (b) C x C + C x C + C x C …...….(P1)
5 5 4 6 3 75 5 4 6 3 7
= 948 = 948 …………(2)
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Q.25:
0.5 0.5 –– 0.225 0.225 ……….(P1)……….(P1)
= 0.275 = 0.275 ……….(2) ……….(2)
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PAPER 2Q1:
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)
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Q3:
(a) (a) r = r = 11..05 05 √….(P1)
Use Use TT6 6 = ar= ar 55
TT6 6 = = 18000 18000 x (x (11..0505))5 5 √….(M1)
= RM 22 973.00√….(A1)
(b) Use (b) Use arar n-1 ≥ ≥ 36 000 36 000 √….(M1)
18000 18000 x (x (11..0505))n-1 ≥ ≥ 36 000 36 000
n ≥ n ≥ 1616 √….(A1)
No working minus No working minus 1 1 mark if answer is mark if answer is
correctcorrect
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a(ra(rnn –– 11))
(c) (c) Use Use S6 = ———
r - 1
Use Use S6=18000{(1.05)6 – 1}..√(M1)
1.05 – 1
= RM 122 434 √..(A1)
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Q8:
(a)(a)
xx 1.5 3.0 4.5 6.0 7.5 9.01.5 3.0 4.5 6.0 7.5 9.0
loglog10 yy 0.40 0.51 0.64 0.76 0.89 1.000.40 0.51 0.64 0.76 0.89 1.00
√.... (P1)
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Plot graph log y against x2
log y
x0 1 2 3 4 5 6 7 8
0.2
0.4
0.6
0.8
1.0
0.0
+
+
+
+
+
√ P1
√ P1
√ G1
9
+
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(b)(b) Plot Plot log10 y against against xx
(correct axes and uniform scales)(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)
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Q13:PP11
(a) (a) Use Use I = —— x 100 P0
11..8080
h = —— x 100 ………………..√(M1)1.50
= 120 ………………….√(A1)
00..9090
112112..5 5 = —— x 100 ……………… √(M1)*k
k = 0.8 or 80 sen ……...……………√(A1)
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I W IWI W IW
150 30 4500150 30 4500
120 120 45 540045 5400
112112..5 15 16875 15 1687..55
105 10 1050105 10 1050
100 12637100 12637..55 …………………………√(P1)
1263712637..55
I = —— x 100 …………√(M1)100
= 126.38 …………√(A1)
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(c) (i) (c) (i) 150150
126.38 = —— x 100 ……√(M1)
100
= = 189189..6 6 ……√(A1)
PP11
(c) (ii)(c) (ii) —— x 100 = 189.6 ……√(M1)
25
= = 4747..39 39 ……√(A1)
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MENJAWAB SOALAN
ADDITIONAL MATHEMATICS
SPM
PAPER 1 & 2
THE ENDTHE END