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) 6 40 = 240 (b) 0 5 0 40 or equivalent = 8 ms 1 (units not important) (c) 1 1 ( 40 5) (240) 40 ( 11) 2 2 t t t ) 11 ( 40 2 1 ) 240 ( ) 5 40 2 1 ( = 28 or equivalent t = 15 s (units not important) 1M 1M 1M 1M 1M 1M 1M 7 3 3 1 4 22 21 2 3 7 2 or 5 . 10 5 . 10 5 . 10 7 22 3 4 2 1 or 2 22 14 35 7 2 or 35 7 7 7 22 Volume = 1 4 22 10.5 10.5 10.5 2 3 7 + 22 7 7 35 7 = 7815.5 or 2 1 7815 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
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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),