1
PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019
MODUL PINTAS
MATEMATIK TAMBAHAN KERTAS 2
NO SOLUTIONS MARKS
1 (a)
0
9
)9( dyy
y
y
92
2
81
2
810
2
81
K1
K1
K1
N1
4
6
(b) (7 × 8) − 211
3
342
3
K1
N1
2
2 (a) 23(1) − 4(1) ∙ 21 + 21 − 2
0
K1
N1
2
(b) log𝑎 𝑁 =1
2(log𝑎 24 − log𝑎 0.375 − log𝑎 729)
log𝑎 𝑁 =1
2(log𝑎
24
(0.375)(729))
log𝑎 𝑁 = log𝑎(64
729)1
2
log𝑎 𝑁 = log𝑎8
27
𝑁 =8
27
log𝑎 𝑁 = 3
K1
K1
K1
K1
N1
N1
6
10
SET 1
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2
3 (a) (i) 3a
(ii) -a + b
(iii) PA = 1
3 PQ
- 1
3 +
1
3 𝐛 (accept answer without working)
N1
N1
K1
N1
4
6
(b) 𝑂𝐵 ⃗⃗⃗⃗ ⃗⃗ ⃗⃗ = 9
5 (𝑂𝑃⃗⃗⃗⃗ ⃗ + 𝑃𝐴⃗⃗⃗⃗ ⃗)
6
5 a +
3
5𝐛
K1
NI
2
4 (30.2 𝑥 20.4) − 𝑥𝑦 = 500
2𝑥 + 2𝑦 = 43.2
𝑥 = 21.6 − 𝑦
(21.6 − 𝑦)(𝑦) = 116.08
𝑦2 − 21.6𝑦 + 116.08 = 0
𝑦 =−(−21.6)±√(−21.6)2−4(1)(116.08)
2(1)
05.10,55.11 xy or 55.11,05.10 xy
Perimeter of the ungrazed field =101.2
P1
P1
K1
K1
K1
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N1
7
7
5 (a) (62 × 18) + (88 × 23) + (16 × 28) + (13 × 33) +(11 × 38) + (10 × 43)
(62 × 18) + (88 × 23) + (16 × 28) + (13 × 33) + (11 × 38) + (10 × 43)
200
24. 33
(62 × 182) + (88 × 232) + (16 × 282) + (13 × 33) + (11 × 382) + (10 × 432)
√(62 × 182) + (88 × 232) + (16 × 282) + (13 × 332) + (11 × 382) + (10 × 432)
200− (24.33 )2
6.828
K1
K1
N1
K1
K1
N1
6
(b) Mean will reduce 5 and standard deviation unchanged N1 1 7
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3
6 (a)
t
tt
ttt
sin2
sincos
)sin(cossin2
K1
K1
N1
3
6
(b)
P1
P1
P1
3
7 (a)
275.52 104.48
8cos COP or or
1.823
K1
N1
2
(b) 10
37.76kos
or
228 2
37.7612.65 3.142
180
8.338PB or
2.65BC or PD = 12.65
8.338 2.65 7.746
18.73
K1
K1
K1
K1
N1
5
(c) 2
1(12.65) 0.6591
2 or
1(10) 7.746
2
21
(12.65) 0.65912
- 1
(10) 7.7462
14.01
K1
K1
N1
3
10
Use identity:
sin 2t = 2 sin t cos t
or cos 2t = 1- 2 sin2t
Shape: sin t or cos t
Amplitude=2
Cycle = 1 20 t
v
2
-2
t
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4
8
( 2) 3( )
7 3 2
ka
k = 4
K1
N1
2
10
7 3 4 ( 2)( ) ,
2 2b
51
18
m
)1(4
78 xy
4
39
4
7 xy
K1
K1
K1
N1
4
(c) )1)(2()3(4)7(8)8(3)2(7)8(1
2
1
or
)5(3)2(6)4(1)1(2)3(4)6(52
1
26 or 6.5
5.6
26
4 : 1
K1
K1
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N1
4
9 a) p =
2
5 or q =
3
5
(i) 1- P (X=0) – P (X=1) – P (X=2) or
1- 0.0467 – 0.1866 – 0.31104
0.4557
(ii) 1308
𝜎 = 28.01
b)(i) 13−10
4 @ 0.75 𝑠𝑒𝑒𝑛
0.7734
(ii) P ( 𝑋 > 136 ) or P ( X < 9.6)
0.1841 x 145 or 0.4602 x 145
13: 33
P1
K1
N1
N1
N1
K1
N1
K1
K1
N1
5
5
10
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5
10 (a) All values of x and log10y correct
x 1 2 3 4 5 6
log 10 y 0.26 0.43 0.61 0.78 0.96 1.05 (b) refer graph
N1
K1
K1N1
4
(c) (i) y incorrect =11.22 and yactual valu e= 13.49
(ii) pkxy logloglog 1010
175.0log k (0.17 - 0.175)
50.1k (1.48 - 1.50)
85.0log p
p = 7.079
N1
P1
K1
N1
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N1
6
10
11 (a)A = πr2 + 2πrh + 2 πr2
A = 3πr2 + 2πrh
V = πr2h + 2
3πr3
K1
N1
N1
3
10
(b) 3πr2 + 2πrh = 20π
220 3
2
rh
r
V = πr2220 3
2
r
r
+ 2
3πr3
V = 10πr - 5
6 πr3
K1
K1
K1
N1
4
(c)
2
2
510
2
510 (1.5) 0.4
2
1.75
dvr
dr
K1
K1
N1
3
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6
12 (a)(i) 2t – 6 = 0
v = (3)2 – 6(3) + 8
v = -1
(ii) (t – 2)(t – 4) = 0
𝑡3
3− 3𝑡2 + 8𝑡
(43
3− 3(4)2 + 8(4)) - (
23
3− 3(2)2 + 8(2))
4
3
K1
K1
NI
K1
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N1
7
10
(b)
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3
13
(a) 𝑥
3.50x100 = 125 or
6
𝑦x100 = 110 or
5.50
4.00x100 = 𝑧
𝑥 = 4.38
𝑦 = 5.45
𝑧 = 137.5
K1
N1
N1
N1
4
10
(b) (120x5) + (125x3) + (110x4) + (137.5x1)
(120x5) + (125x3) + (110x4) + (137.5x1)
13
119.42
K1
K1
N1
3
(c) 119.42x115
100
𝑝
40x100 = 137.33
RM 54.93
K1
K1
N1
3
Shape graph ∪
Graph intersect x-axis at 2 and 4
3 < 𝑡 ≤ 7
v
t
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7
14
(a) (i) 4.72 = 6.52 + 52 – 2(6.5)(5) cos A
45.99
(ii) 02.88sin
5
99.45sin
BD
3.598
K1
N1
K1
N1
4
10
P1
K1
N1
N1
4
1( ) (6.5)(4.7)(sin 38.09)
2c
9.423
K1
N1
2
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8
15 (a) 𝑥 ≥ 50
𝑦 ≥ 180
24 𝑥 + 8 𝑦 ≤ 8000 or 3 𝑥 + 𝑦 ≤ 1000
𝑥 + 𝑦 ≤ 800
N1
N1
N1
N1
4
10
(b) Refer to graph paper
One *straight line drawn correctly
All * straight line drawn correctly
Correct region
K1
K1
N1
3
(c) (i) 180 ≤ x ≤ 390
(ii) Pmax = 30 x + 10 y
30 ( 100 ) + 10 (700 )
1000
N1
K1
N1
3
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9
QUESTION 10
Plot log10 y against x with correct axes,
uniform scales and at least one point K1
6 points plotted correctly K1
Line of best fit N1
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10
QUESTION 15
1000
900
0 100 200 300 400 500 600 700
100
200
300
400
500
600
700
800
800
R
100 200 300 400 500 600 700
100
200
300
400
500
600
700
800
800
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