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END OF QUESTION PAPER
MARKING SCHEME
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ADDITIONAL MATHEMATICS 3472/1
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KERTAS SOALAN TAMAT
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PKBS 3 / TRIAL SPM 2010
SUB TOTAL
NO MARK SCHEME
MARKS MARKS
1 (a) many to one 1 2
(b) f(x) = x2 1
2 1
k= , p = -3 [both]
2
3 3
1
B2: k = or p = -3
2
x−6 x− p
B1 : or
2 k
3 (a) 2x -7 2 4
B1: g(x) + 4 = 2x -3
(b) 2 2
B1: 2(x + 4) -7 = 5
4 10 8 2 2
3 x 2 + 10x - 8 = 0 or x + x− =0
2
3 3
2 2
B1: x 2 -( -4 + ) x + (-4 × ) = 0 OR
3 3
2
( x + 4)( x − ) = 0 or (x + 4)(3x – 2) = 0
3
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5 (a) 2 1 3
(b) 5 1
(c) x = -2 or x + 2 = 0 1
6 x < 2 , x > 3 (both) 4 4
B3 :
2 3
B2 : (x – 2)(x – 3) > 0
B1 : x 2 - 5x + 6 > 0
7 2 3 3
B2 : 3x = 9
B1 : 3x (32 − 3) = 54
8 4+h–k 3 3
B2 : log 3 81 + log3 x − log3 y
B1 : log 81x − log y or log 81 + log x
9 4m 2 2 2
p= or p = 1.333 m 2
3
2m p
B1: =
3 2m
10 .(a) 6 2 4
8
B1: [ 2(5) + (8 − 1)d ]
2
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(b) 59
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B1: 5 + (10 – 1)6 2
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11 1 2 4
.(a) r= or 0.5 , a = 96
2
B1: 24(1+ r) = 36
2
(b) 192
96
B1: 1
1−
2
12 1 p 1 4
.(a) = +q
y x
(b) p = -2 , q = 11 (both)
3
B2: p = -2 or q = 11
7 −3
B1: p = OR 7 = 2p +q or 3 = 4p + q
2−4
13 2 3 3
h= k
3
10
(1)(k ) + (2)(6h) (1)(2h) + (2)( k)
B2: = 3k or 3 = 4h
2 +1
2 +1
10
(1)(2h) + (2)( k)
B1: (1)(k ) + (2)(6h) 3
or
2 +1 2 +1
14 0.784 3 3
B2: 6(2π − θ ) = 33
B1: 2π − θ or 3600 - θ
15 3 2 2
B1 : or =
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16 (a) -3 + 4 1 3
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3 4
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(b) − i + j
5 5 2
B1 : (−3) 2 + 42
17 k 2 3
(a) −
1− k 2
B1 : 1 − k 2 or see in diagram
(b) - 2k 1 − k 2 1
18 90 , 210 , 330 4 4
B3: 90o, 210o
B2: (2sin A + 1)(sin A − 1) = 0
B1: sin A + 2(1 − sin 2 A) = 1
19 2 3 3
or 0.6667
3
B2 : 18 p − 4(3) = 0
dy
B1 : = 18 p − 4 x
dx
20 28h 3 3
B2 : ∂y ≈ [8(3) + 4] × h
dy
B1 : = 8 x + 4 or δx =h
dx
21 (a) 4 1 4
(b) -7 3
B2 : 2(4)-[3(6)-3(1)]
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B1 : ∫ 2 g ( x)dx − ∫ 3dx or 2∫ g ( x)dx or 3x
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22 p 2 + 20 p2 3 3
k= or k = +4
5 5
25k
B2: p2 = –( 20 )2
5
B1 : x = 20 or ∑x 2
= 25k or σ = p
23 (a) 495 1 4
(b) 12 3
B2 :
5
C5 ×3 C2 ×4 C1
B1 :
5
C5 ×4 C1 or
4
C1 ×3 C2 or 5
C5 ×3 C2
24 20 5 3 3
or or 0.2778
72 18
 2 1  3 2  4 3
B2 :  ×  +  × +  × 
 9 8 9 8  9 8
 2 1  3 2  4 3
B1 :  ×  or  ×  or  × 
 9 8 9 8 9 8
25 (a) 0.76 2 4
B1 : P(z > k) = 0.5 – 0.2764
(b) 49.38 2
50.9 − µ
B1 : 0.76 =
2
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