Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]
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Transcript of Trial spm smk_st_george_taiping_2013_maths_paper1_2_[a]
PEPERIKSAAN PERCUBAAN SPM MATHEMATICS FORM 5 SGI 2013
Kertas 1
Total = Paper 1 + Paper 2
= 40 + 100
= 140
1 C 11 D 21 B 31 B
2 B 12 C 22 A 32 A
3 C 13 D 23 C 33 D
4 D 14 B 24 B 34 A
5 C 15 C 25 B 35 D
6 D 16 D 26 D 36 A
7 A 17 B 27 D 37 B
8 A 18 C 28 B 38 C
9 C 19 D 29 C 39 D
10 B 20 B 30 B 40 A
Kertas 2
Q Solution Submarks Full
marks
1
Dotted line y = 3 drawn and Correct shaded region
Note : 1. Solid line y = 2 drawn (– 1 mark)
2. Give 1 mark if shaded region between 2 inequalities only
3M
3
2 (a) 640
= 240
(b)05
040
or equivalent
= 8 ms1(units not important)
(c)1 1
( 40 5) (240) 40 ( 11)2 2
t
t
t )11(402
1)240()540
2
1(
= 28 or equivalent
t = 15 s (units not important)
1M
1M
1M
1M
1M
1M
1M
7
3 31 4 22 21
2 3 7 2
or 5.105.105.10
7
22
3
4
2
1 or
222 14
357 2
or 35777
22
Volume =1 4 22
10.5 10.5 10.52 3 7
+
227 7 35
7
= 7815.5 or 2
17815 or
15631
2 cm
3 (units not important)
1M
1M
1M
3
x + y = 8
y = 8 – 2x
y = 2
8
2
x 0 8 4
y
4
(a)(i) False/ Palsu
(ii) True/ Benar
(b) k + 1 is not an odd integer/ k + 1 bukan satu integer ganjil
(c) 4n – 3, n = 2, 3, 4, ... or equivalent
Note : 4n – 3 only (Give 1 mark)
1M
1M
2M
2M
6
5 LNM
tan LNM = 8
6 or equivalent
36.87o
or 36o 52
’
1M
1M
1M
3
6 3x2 − 5x − 2 = 0
(3x + 1) (x − 2) = 0
x = 3
1 , 2
1M
1M
1M,1M
4
7 4m – n = 26
4m + 3n = 2
–4n = 24 or
12 ( 6) 13
2m or 4m – (-6) = 26 or
or 4m + 3(-6) = 2
m = 5, n = 6
Note : 5
6
m
n
only award 1 mark
1M
1M
1M,1M
4
8
(a) M PR = 2 and c = 4
y = 2x + 4
(b) Coordinate T : (2, 0) or x = 2
y = 2(2) + 4 or equivalent
R(2, 8)
1M
1M
1M
1M
1M
5
9
(a) Sampel space/ Ruang sampel = {(S,C), (S,O), (S,R), (S,E),
(C,S), (C,O), (C,R), (C,E),
(O,S), (O,C), (O,R), (O,E),
(R,S), (R,C), (R,O), (R,E),
(E,S), (E,C), (E,O), (E,R)}
(b)(i) {(C,S), (C,O), (C,R), (C,E)}
4
20 or
1
5
(ii){(S,C), (S,R), (C,S), (C,R), (O,E), (R,S), (R,C), (E,O)}
8
20 or
2
5
Note : Accept correct answer only if list out
1M
1M
1M
1M
1M
5
10 (a)
60 222 14
360 7 or equivalent
Perimeter = 60 22
2 14360 7
+14 + 14 + 7 + 7 or equivalent
= 56.67 or 3
256 or
170
3 cm (units not important)
1M
1M
1M
1132
22
4 3
11331
21 2
2(3) 4( ) 4 22
5
6
m
n
m
n
(b) 260 2214
360 7 or
277
22
360
60 or 290 22
7360 7
or equivalent
Area of the shaded region/ Luas kawasan berlorek
= 2 2 260 22 60 22 90 2214 7 14 7 7
360 7 360 7 360 7
or equivalent
= 136.5 or 1
1362
or 273
2 cm
2 (units not important)
1M
1M
1M
6
11
(a) Q = )4(3)5(2
1
53
42
=
53
42
2
1 or
1 2
3 52 2
(b)
8
13
23
45
y
x
8
13
53
42
2
1
y
x
2
1,3 yx
Note : 3
12
x
y
only award 1 mark
1M
1M
1M
1M
1M,1M
6
12
(a) x = –1, y = 12
x = 3, y = – 4
(b) Graph
All axis drawn with correct direction and uniform scale
7 points and 2 points* have been plot correctly
(7 or 8 points have been plot correctly get 1 mark)
Smooth curve and continue without straight line and passing through the
correct 8 points for –3 x 4.
(c) (i) 10 9y
(ii) 0.5 0.4x
(d) Identify equation y = 5x – 5
Straight line y = 5x – 5 correctly drawn
x = 0.4 ≤ x ≤ 0.5 , x = 3.75 x 3.85
1M
1M
1M
2M
1M
1M
1M
1M
1M
1M, 1M
12
13 (a) (i) (4, 9)
(ii) (4, 9) → (4, 3)
(b)(i)(a) U = Rotation, 900 clockwise, about centre (2, 4)
Note : 1. Rotation, 900 clockwise, get 2 marks
2. Rotation, about centre (2, 4), get 2 marks
3. Rotation, get 1 mark
(b) W = Enlargement, with centre (7, 4) , of scale factor 2
Note : 1. Enlargement, with centre (7, 4), get 2 marks
2. Enlargement, of scale factor 2 , get 2 marks
3. Enlargement, get 1 mark
(ii)22 A
22 120A A or equivalent
A = 40 m2
(units not important)
1M
1M, 1M
3M
3M
1M
1M
1M
12
14 (a) (i) 12
(ii) I II III
Column I (All Correct)
Column II (All Correct)
Column III (All Correct)
(iii) (3 112 12 117 26 122 24 127 12 132 8 137 6 142 3 147)
94
= 126.95
(b) All axis in correct direction, uniform scale, x-axis label as upper boundry
All 8 points* correctly plot.
(6 or 7 points* correctly plot get 1 mark)
Ogive pass through point (109.5, 0)
Smooth curve without any straight line and pass all the 8 points correctly
(c) Median = 125.5 0.5
Height
(cm)
Tinggi
(cm)
Mid
point
Titik
Tengah
Frequency
Kekerapan
Cumulative
Frequency
Kekerapan
Longgokan
Upper
Boundaries
Sempadan
atas
105 109 107 0 0 109.5
110 – 114 112 3 3 114.5
115 – 119 117 12 15 119.5
120 – 124 122 26 41 124.5
125 – 129 127 24 65 129.5
130 – 134 132 x = 12 77 134.5
135 – 139 137 8 85 139.5
140 – 144 142 6 91 144.5
145 - 149 147 3 94 149.5
1M
1M
1M
1M
1M
1M
1M
2M
1M
1M
1M
12
15 (a) Plan
or
The shape must be correct in quadrilateral form. All lines must be drawn in full.
JG = NK > GN = KJ > JI = QK
The measurement is accurate to 0.2 cm. (one way ) and the angles at all
verticals of the rectangle are 90 1
(b) (i) Viewed from A
The shape must be correct. All lines must be drawn in full.
JV > VF > IH > GH > FG, UV > VZ = YU = FG
Curve must be smooth
The measurement is accurate to 0.2 cm. (one way ) and the angles at all
verticals of the rectangle are 90 1
1M
1M
1M
1M
1M
1M
1M
IH JE GF
2 cm 4 cm
5 cm
KL NM QP 5 cm
2 cm
4 cm
IH
JE
GF
KL
NM
QP
TCW
JK
Y ZSR
UX VEL
IQ
FM
HP GN
2 cm
5 cm
4 cm
3 cm
4 cm 6 cm
(i) Viewed from B
The shape must be correct in quadrilateral form. All lines must be drawn in full.
ICGF and QWNM are two parallel straight lines
Note: Ignore straight line CW
C and W are joined dotted lines to shape a rectangular CFMW
VM = MQ > QI = TV > IC > CT > CG = WN
The measurement is accurate to 0.2 cm. (one way ) and the angles at all
verticals of the rectangle are 90 1
1M
1M
1M
1M
1M
12
16 (a) R = 70o N
(b) Q(70o S, 50
o W)
(c) 60)9070(
= 9600 n.m.
(d)(i) 600 2 =1200 n.m.
(ii) 70cos60
2600
= 58.48o or 58
o 29
’
58.48o + 50
o = 108.48
oW or 108
o 29
’W
2M
2M
2M
1M
1M
2M
1M
1M
12
3 cm
2 cm
3 cm
3 cm 5 cm
T
IJ
C
ZY
VU FE
QK
MLX
NPR
W
GHS
Graph For Question 12
-1 -2 -3 -4
y
x 1 2 3
5
10
15
20
25
-5
-10
0
(-2, 16)
(-1, 12)
(0, 2)
(1, -8)
(2, -12)
(3.5, 4.4)
(4, 22)
(3, -4)
(-3, 8)
Graph For Question 14
Graph for Question 14
(-3, -29)
50
10
60
40
20
30
70
80
90
100
0
Upper
Boundary
124.5 129.5 139.5 144.5 134.5 119.5 114.5 109.5 149.5
Cumulative
Frequency