1 PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019 PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR (PINTAS) MATEMATIK TAMBAHAN KERTAS 2 NO SOLUTIONS MARKS 1 (a) dx x x V 4 0 2 2 ) 4 ( π[ 16x 3 3 -2x 4 + x 5 5 ] 4 0 (Integrate + limit) [ 16(4) 3 3 − 2(4) 4 + (4) 5 5 ]−0 15 512 π K1 K1K1 N1 4 6 (b) 2× 32 3 3 64 K1 N1 2 2 (a) 8 1 3 n=2 Accept answer without working for K1N1 K1 N1 2 8 b) 3 r (9 s ) 2 = 81 or (3 r ) 2 9 s = 1 3 or equivalent 3 r (3) 4s =3 4 or 3 2r-2s =3 -1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation 5 2 r 10 9 s K1 K1 K1 K1 N1 N1 6 SET 2
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1
PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019
PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR
(PINTAS)
MATEMATIK TAMBAHAN KERTAS 2
NO SOLUTIONS MARKS
1 (a) dxxxV
4
0
22 )4(
π [16x3
3-2x4 +
x5
5] 40 (Integrate + limit)
𝜋 [16(4)3
3− 2(4)4 +
(4)5
5] − 0
15
512π
K1
K1K1
N1
4
6
(b)
2 ×32
3
3
64
K1
N1
2
2 (a)
81
3
n = 2 Accept answer without working for K1N1
K1
N1
2
8
b)
3r(9s)2 = 81 or (3r)2
9s =1
3 or equivalent
3r(3)4s = 34 or 32r-2s = 3-1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation
5
2r
10
9s
K1
K1
K1
K1
N1
N1
6
SET 2
cikguJEP
cikgujep
2
NO SOLUTIONS MARKS
3
a) AD⃗⃗⃗⃗ ⃗ = AB⃗⃗⃗⃗ ⃗ + BC⃗⃗⃗⃗ ⃗ + CD⃗⃗⃗⃗ ⃗
AD⃗⃗⃗⃗ ⃗ = (q – p)a + qb
q – p = -5 or 2
3
pq
p = 13 and q = 8
K1
K1
N1
3
6
(b) 1
2 x DC x 4 = 52
p |a| = 26
2
K1
K1
N1
3
4
5x + y + (x + y) = 24 or 222 )()()5( yxyx
y = 12 – 3x
0)312(12 2 xxx
045 2 xx 0)45( xx or equivalent
5
4x or y =
5
48
48 4
5 5
192
25 or equivalent
P1
K1
K1
K1
N1
K1
N1
7
7
3
NO SOLUTIONS MARKS
5
(4.6 × 3) + (5.1 × 5) + (5.6 × 5) + (6.1 × 10) + (6.6 × 12) + (7.1 × 5) or
(4.6 × 2) + (5.1 × 7) + (5.6 × 5) + (6.1 × 12) + (6.6 × 8) + (7.1 × 6)
(4.6×3)+(5.1×5)+(5.6×5)+(6.1×10)+(6.6×12)+(7.1×5)
40 or
(4.6×2)+(5.1×7)+(5.6×5)+(6.1×12)+(6.6×8)+(7.1×6)
40
6.075 or 6.0375
(3 × 4.62) + (5 × 5.12) + (5 × 5.62) + (10 × 6.12) + (12 × 6.62) + (7.12 × 5) or
(2 × 4.62) + (7 × 5.12) + (5 × 5.62) + (12 × 6.12) + (8 × 6.62) + (7.12 × 6)
√(3×4.62)+(5×5.12)+(5×5.62)+(10×6.12)+(12×6.62)+(7.12×5)
40-(6.075)2 or
√(2×4.62)+(7×5.12)+(5×5.62)+(12×6.12)+(8×6.62)+(7.12×6)
40-(6.0375)2
0.7242 or 0.7175
ORCHARD B
K1
K1
N1
K1
K1
N1
N1
7
7
6
(a) 1cot
1cot2
2
2
2
2
2
cos1
sin
cos1
sin
2 2
2
2 2
2
cos sin
sin
cos sin
sin
K1
K1
4
NO SOLUTIONS MARKS
22 sincos
2cos
N1 3
6
(b)
ekcos
12cos
sin
1
12cos
sinsin21 2
01sinsin2 2
0)1)(sin1sin2(
1sin,2
1sin
270,150,30
K1
K1
N1
3
7
(a) 24012060 oror
3
4189.4 or
K1
N1
2
10
(b) 12OP or 60cos = OP
6
108PS or equivalent
180
142.312012PQRarcofLength or 25.14
92.45
108214.25
K1
K1
K1
K1
N1
5
(c )
120sin)12(2
1
180
142.3120)12(
2
1 22 or
or
12 128 6
2
45.88
35.628.150
K1
K1
N1
3
5
NO SOLUTIONS MARKS
8 (a)
2
95,
2
80
( 4, 7)
K1
N1
2
10
(b) )0(1)4(9)8(5)0(0)0(4)1(8)9(0)5(02
1
34
K1
N1
2
(c) m =2
1
midpoint QR ( 6,5)
y – 5 = 2
1 ( x – 6 ) or equivalent
y = 2
1 x +8
K1
K1
K1
N1
4
(d) 2
1
63
14
5
h or 8
3
14
2
1
h
h = 3
17
K1
N1
2
9
(a) p= 0.75 or q = 0.25 or
rnr
rC )25.0()75.0(15
(i) )12( XP
)15()14()13()12( XPXPXPXP
0.4 613
(ii) 75.015
11.25
P1
K1
N1
K1
N1
5
6
NO SOLUTIONS MARKS
(b)(i) 370 8
8
370380
8
370355ZP or )25.1875.1( ZP
0.8641 0.86396
(ii) 0.8641 x 3130
2704
P1
K1
N1
K1
N1
5
10
10 (a) All values of
x
1 and
y
1 correct.
x
1
0.67 0.40 0.25 0.20 0.17 0.13 0.10
y
1
0.22 0.91 1.30 1.43 1.52 1.56 1.67
N1, N1
2
10
(b) Refer graph paper
Plot log10 y against x with correct axes, uniform scales and
at least one point.
6 points plotted correctly
Line of best fit
K1
N1
N1
3
(c) hxh
k
y
1)
1(
1
(i) h
192.1
5208.0h
(ii) 5439.25208.0
k
3248.1k
P1
K1
N1
K1
N1
5
7
NO SOLUTIONS MARKS
11
(a)r = x – 5 or h = x – 2
V = π (x – 5)2 (x – 2) or equivalent
P1
N1
3
10
(b)
2
5 (1) ( 2)(2)( 5)(1)
3 ( 5)( 3)
dvx x x
dx
x x
3 ( 5)( 3) 0
3 5
x x
or
When x = 3 , V = 4π
K1
N1
K1
N1
N1
4
(c) 0.01x
9π (0.01)
0.09π
P1
K1
N1
3
12 (a) v = 3 N1 1
10
(b) (3t + 1)(t – 3) = 0
t = 3
K1
N1
2
(c) sp = 3t + 4t2 – t3
t = 3 , sp = 18 or t = 4 sp = 12
24
K1
K1
N1
3
(d) 8 – 6t = 0 or t = 4
3
Qv = 3t2 – 6t – 6
Qv = 6
3
46
3
43
2
Qv =` 3
26
K1
K1
K1
N1
4
8
NO SOLUTIONS MARKS
13 (a) 80100
80.2
x or 130100
60.2
y or z100
5
85.5
117,2,24.2 zyx
Note : All three values correct N2
Any two values correct N1
One or none values correct N0
K1
N2
3
10
(b) )26117()45130()14120()1580(
100
)26117()45130()14120()1580(
117.72
K1
K1
N1
3
(c) 72.11710035
x
RM 41. 20
K1
N1
2
(d) 100
11472.117
134.2
K1
N1
2
9
NO SOLUTIONS MARKS
14 (a) (i) 24)120)(sin)(8(
2
1BC
6.928
(ii) AC2 = 82 + 6.9282 – 2(8)(6.928) kos 120
12.94
(iii) 94.12
120sin
928.6
sin
A
27.62
K1
N1
K1
N1
K1
N1
6
10
(b)
A’
C ‘
B ‘
180 – 27.62 – 60 = 92.38
1(12.94)(6.928)sin 92.38
2
45.12
P1
K1
K1
N1
4
10
NO SOLUTIONS MARKS
15 (a) x + y ≥ 10
10x + 8 y ≤ 420
𝑥 ≤ 2𝑦
N1
N1
N1
3
10
(b)Refer to graph paper
One *straight line drawn correctly
All * straight line drawn correctly
Correct region
K1
K1
N1
3
(c) 2113 y
420 – [20(10) + 10(8)]
Seen (20 , 10 )
= 140
K1
N1
N1
N1
4
11
12
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