SULIT 3472/2 Kertas 2 2017 BAHAGIAN … MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and...
Transcript of SULIT 3472/2 Kertas 2 2017 BAHAGIAN … MATHEMATICS [P2] TRIAL SPM 2017 5 Number Solution and...
SULIT
3472/2
Additional Mathematics
Kertas 2
Ogos
2017
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KECEMERLANGAN
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2017
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN
Peraturan pemarkahan ini mengandungi 12 halaman bercetak
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
2
Number Solution and Marking Scheme Sub
Marks
Full
Marks
1 (a)
(b)
(c)
01141 k
3k
4
3opm or
3
4PQm
13
4
4
3
PQOP mm
5TP
53422 yx
06822 yxyx
K1
N1
P1
N1
P1
K1
N1
7
2
33 2 or
2
2 5(3 2 ) 9 (3 2 ) or
2 5 2 59 or 9
33 2
2
3 5 2 1 0 or 3 1 2 0
5 1,
3 2
1, 2
3
yy x x
x x x x
yx x y
x x y y
x x
y y
P1
K1
K1
N1
N1
5
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
3
Number Solution and Marking Scheme Sub
Marks Full
Marks
3 (a)
75.0,......25.20,27,3636 roror
K1
52.175.03652.11
n
n orT K1
75.0log
36
52.1log
136
52.1log75.0log1
norn
K1
12n N1
(b) 75.01
36
K1
75.01
36248
K1
336 cm
N1 7
4 (a)
(i)
(ii)
2
51505100 ..
125.5 (accept without working)
110
(20)275.5225.5(30)(10)175.5(40)125.575.5(10)
180.05
K1
N1
K1
N1
(b)
200.5 or 60 or 30
5030
605825200
..
238
P1
K1
N1
7
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
4
Number Solution and Marking Scheme Sub Marks Full
Marks
5 (a)
050125
2 2 xx
(x – 25)(x – 5) = 0
x = 25 , x = 5
Width = 25 – 5 = 20 unit
K1
K1
N1
(b) 12
5
4 x
dx
dy OR 10015
5
2 2 xy
x = 15 OR 40155
2 2 xy
y = – 40 or max depth = 40
K1
N1
N1
6
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
5
Number Solution and Marking Scheme Sub
Marks Full
Marks
6 (a)
x
x
x
xx
x
cos
sin
2cos
2sincos
sin
K1
= xxxx
xx
x
x
2cossin2sincos
cos2cos
cos
sin
=
)1cos2(sincossin2cos
cos2cos
cos
sin2
xxxxx
xx
x
x
=
)sincossin2cossin2
cos2cos
cos
sin22 xxxxx
xx
x
x
= cos 2x N1
6 (b)
Shape
Amplitude or 2 cycles or modulus
All correct
2
xy
Sketch graph
6 number of solution
P1
P1
P1
N1
K1
N1
8
Use x
x
cos
sin
or sin 2x = 2sin x cos x
or cos 2x = 2cos2
x – 1
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
6
Number Solution and Marking Scheme Sub
Marks
Full
Marks
7
(a)
(b)
9
12tan//
9
12tan PORPOQ
1.8548 rad // 1.855 rad
SQR = 9(*1.8548) // 9(*1.855) or 13.53sin9
QM
= 16.6932 // 16.695
Chord QR = 2.7*2
= 14.4
Perimeter = SQR + * Chord QR
= 31.09 → 31.098
K1
N1
K1
N1
N1
K1
N1
(c)
Area of sector OQR or Area of triangle POQ
= 8548.192
1 2 = 912
2
1
= 75.12 cm2 = 54 cm
2
2(Area of triangle *POQ) − Area of sector *OQR
2(54) − 75.12
32.88 cm2
K1
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
7
Number Solution and Marking Scheme Sub
Marks Full
Marks
8
Refer to graph
9 (a)
(i)
(ii)
(b)
OQAOAQ or APOAOP
baAQ4
3
baOP4
1
4
3
OPmOS
=
bam
4
1
4
3
banaOS
4
3
nm 14
3 or nm
4
3
4
1
443 nm and nm 3 shown
Solve simultaneous linear equations
13
12m
13
4n
K1
N1
N1
N1
N1
K1
N1
K1
N1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
8
Number
Solution and Marking Scheme Sub
Marks Full
Marks
10 (a)
52 xdx
dy
2
5052 xorx
2
55
2
52
y
4
25y
K1
N1
(b)
A1 = 54
25
A2 = 2
5
3
23 xx
21 AA
42.10//12
510//
4
125
K1
K1
K1
N1
(c) 5
0
22 5 dxxxV or dxxxx 5
0
234 2510
3
25
4
10
5
345 xxx
= 17.104//6
1104//
6
625
K1
K1
N1
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
9
Number Solution and Marking Scheme Sub Marks Full
Marks
11 (a) i)
37
7
10 )45.0()55.0()7( CxP
= 0.1665
K1
N1
(ii)
100
0
10 )55.0()45.0(C or 91
1
10 )55.0()45.0(C
)1()0(1 XPXP equivalent
= 0.9767
K1
K1
N1
(b) i)
(ii)
25.0
56.36.3
0.4364
Seen - 0.524 or 0.524
524.025.0
56.3
k
k = 3.429
K1
N1
P1
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
10
Number Solution and Marking Scheme Sub Marks Full
Marks
12 (a)
(b)
(c)
24306 2 tt
014 tt
4t and 1t
3012 t
2
5t
subtitute t into v
242
530
2
56
2
v
1ms2
113 v or 1ms
2
113 moves to the left
t < 2
5
1241151223
1 S or
2
524
2
515
2
52
23
2
5S
Total distance = 2
51 *2 SS
= 24.5
K1
K1
N1
K1
K1
N1
N1
K1
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
11
Number Solution and Marking Scheme Sub
Marks Full
Marks
13 (a) zor
yor
x 100
15
18100100
20175100
70
x = 40, y = 20, z = 120
K1
N2, 1, 0
(b)
100
461202410010125121508175 )(*)()()()(I
123.7
K1
N1
(c)
7.123*
10088006
P
RM 711.40
K1
N1
(d)
Seen orI 25.16615/18 165 or 137.5 or 110 or 132
100
4613224110105.13712165825.166 I
97133.
N1
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2017
12
Number Solution and Marking Scheme Sub Marks Full
Marks
14 Refer to graph
15 (a)
(i)
sin sin 22.18
7.71 3.46
122.73
ADC
ADC
K1
N1
(ii)
2 2 27.71 4.98 4.07 2(4.98)(4.07)cos
116.49
ABC
ABC
K1
N1
(iii)
Area of triangle ADC
49.116sin)07.498.42
1
9.070 cm2 // 9.07 cm
2
07.971.72
1 BE
BE = 2.353 cm
K1
N1
K1
N1
(b)
2 22.35 2.06
=3.125
BD
cm
P1
N1 10
END OF MARKING SCHEME
x 3 5 6 9 10 12
y10log 0.37 0.22 0.15 0.09- 0.16- 0.31-
0 2 4 6 8 10 12 14 x
0.3-
y10log
x
x
x
x
x
x
(b)
Correct axes, uniform scale and K1
one point correctly plotted
All points correctly plotted K1
Line of best fit N1
(c) qxpy 101010 logloglog P1
(i) Use p10log = y – intercept K1
p = 3.98 N1
(ii) Use q10log = gradient K1
q = 1.19 N1
(iii) y = 3.31 N1
QUESTION 8
0.4-
0.2-
0.1-
0.1
0.2
0.3
0.4
0.5
0.6
N1
10
45
( 4,38)
(a) I. y - x ≥ 10 N1
II . x ≥ 𝑦
10. or 10x ≥ y N1
III. 3000 x + 1800y ≤81 000 N1
Note: Don’t accept x : y = 1: 10
(b) Correct axes with uniform scale and 1 graph correct K1
All graphs correct N1
Correct shaded region N1
(c) (i) 33 (From point (3,30)) N1
(ii) k = 10y + 25x , let k = 250 and
max point ( 13, 23) K1
maximum allowance = 10 (23) +25 (13) K1
= RM 555 N1
R
(13, 23)
QUESTION 14