Trial SPM 2013 / Kertas Percubaan SPM 2013 Kedah

1. SULIT 1449/2(PP)
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1449/2(PP) SULIT
PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013
ANJURAN
MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)
MODUL A
MATEMATIK
KERTAS 2
PERATURAN PEMARKAHAN
UNTUK KEGUNAAN PEMERIKSA SAHAJA
Peraturan pemarkahan ini mengandungi 14 halaman bercetak

2. SULIT 1449/2(PP)
1449/2(PP) SULIT
2
Section A
[ 52 marks]
Question Solution and Mark Scheme Marks
1
2
40-2
-4
-6 -4
-2
6
4
2 x
y
y = - x - 1
y = 2x + 2
Straight dotted line x = 1 correctly drawn. K1
Region correctly shaded P2 3
Note:
1 Accept solid line x = 1 for K1
2 Award P1 to shaded region bounded by two correct lines,
including part of R.
(Check one vertex from any two correct lines
2(a) ÐMFH P1
3
(b)
13
7
tan =ÐMFH or equivalent
K1
28.3° or 28° 18¢ N1

3. SULIT 1449/2(PP)
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1449/2(PP) SULIT
3
Question Solution and Mark Scheme Marks
3 049 2
=-x
( )( ) 02323 =+- xx or equivalent
3
2
=x or 0.67
3
2
-=x or 0.67
Note : 1. Accept without = 0
2. Accept two terms on the same side, in any order.
3. ( )( )0 67 0 67x x- × + × with 0 67, 0 67x x= × = - ×
award Kk2
K1
K1
N1
N1 4
4 14 3 2x y+ = - or 7 21 77x y+ = or equivalent K1
Note :
Attempt to equate one of the coefficients the unknowns, award K1
OR
( )
3
1 2 1 711 211 3
3 7 3
y xx
x y or y or x or y
- - - --
= - = = =
or equivalent (K1)
Note :
Attempt to make one of the unknowns as the subject award K1.
39
13 13 78
2
x or y- = = or equivalent K1
OR
3
1131
2
3 1
(1) (3)(7) 7 1
2
x
y
æ ö
-æ ö æ öç ÷=ç ÷ ç ÷ç ÷ -æ öè ø è ø- -ç ÷ è ø
è ø
(K2)
Note :
Attempt to write without equation, award (K1)
1x = - N1
4y = N1 4
Note :
1
4
x
y
-æ ö æ ö
=ç ÷ ç ÷
è ø è ø
as final answer, award N1

4. SULIT 1449/2(PP)
1449/2(PP) SULIT
4
Question Solution and Mark Scheme Marks
5(a)
20
2
14
2
14
7
22
2
1
´´´´
K1
4
1540 N1
(b) 20
2
14
2
14
7
22
2
1
´´´´ + ( )
1
20 10 14 6790
2
AB´ + ´ ´ =
K1
25 N1
Note :
1. Accept p for K mark.
2. Correct answer from incomplete working, award Kk2.
6 (a) benar / true P1
5
(b) Jika n adalah nombor negatif maka 0n < .
If n is a negative number then 0n < .
benar / true
P1
(c) Set R ialah subset bagi set K
Set R is subset of set K
P1
(d) Bilangan subset bagi suatu set yang mempunyai 7 unsur ialah
7
2
The number of subsets in a set with n elements is 7
2 .
K1
7
2 128= N1
7 (a) 5x = - P1
5
(b) 4
5
SR PQM M= = P1
( )
*
4
4 5
5
c= - + or
( )
*
4 4
5 5
y
x
-
=
- -
K1
8
5
4
+= xy N1
y-intercept = 8 N1
8 (a) 12 P1 1
(b)
12 0
10 0
or equivalent
-
-
K1
6
1 2
5
or × N1 2
Note: Accept answer without working for K1N1
(c) ( )
1 1
(18 28) 12 (( 28)(12 20) 468
2 2
t+ + - + = or
equivalent method K2
Note:
1 1
(18 28)(12) (( 28)(12 20)
2 2
or t+ - +
equivalent, award K1
40 N1 3
6

5. SULIT 1449/2(PP)
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1449/2(PP) SULIT
5
Question Solution and Mark Scheme Marks
9(a)
(b)
14
7
22
2
360
120
´´´ or 7
7
22
2
360
180
´´´
14 + 14 + 14 + 14
7
22
2
360
120
´´´ + 7
7
22
2
360
180
´´´
3
280
or
3
1
93 or 93.33
2
14
7
22
360
120
´´ or 2
7
7
22
360
180
´´
2
14
7
22
360
120
´´ - 2
7
7
22
360
180
´´
3
385
or
3
1
128 or 128.33
Note :
1. Accept for K mark.
2. Correct answer from incomplete working, award Kk2.
K1
K1
N1
K1
K1
N1 6
10 (a) {(P, M), (P, 4), (P, 5), (P, N), (Q, M), (Q, 4), (Q, 5), (Q, N),
(3, M), (3, 4), (3, 5), (3, N), (R, M), (R, 4), (R, 5), (R, N)}
Note :
1. Accept 8 correct listing from not more than 16 outcomes for P1
P2
(b)(i) {(3, 4), (3, 5)} K1
16
2
or
8
1 N1
(ii) {(P, 4), (P, 5), (Q, 4), (Q, 5), (3, M), (3, N), (R, 4), (R, 5)}
16
8
or
2
1
K1
N1
NOTE :
1. Accept other method for K mark.
2. Accept answer without working from correct listing,
correct tree diagram or correct grid for K1N1.

6. SULIT 1449/2(PP)
1449/2(PP) SULIT
6
Question Solution and Mark Scheme Marks
11 (a)
( ) ( )
1 21
6 33 1 2 6
- -æ ö
ç ÷
-´- - ´ è ø
1 2
15 15
6 1
15 5
æ ö
ç ÷
ç ÷
ç ÷-ç ÷
è ø
P2 2
(b)
3 2 9
6 1 7
x
y
æ öæ ö æ ö
=ç ÷ç ÷ ç ÷- -è øè ø è ø
P1
1 2 91
6 3 7(3)( 1) (2)(6
x
or
y
- -æ ö æ öæ ö
=ç ÷ ç ÷ç ÷- -- -è ø è øè ø
*
9
7
x Inverse
y matrix
æ ö æ öæ ö
=ç ÷ ç ÷ç ÷
-è ø è øè ø
K1
1
3
x = - N1
y = 5 N1 4
6
Note:
1.
1
3
5
x
y
æ ö
-æ ö ç ÷=ç ÷ ç ÷ç ÷è ø
è ø
as final answer, award N1
2. Do not accept any solution solved no using matrix method.
3. Do not accept * 3 2
6 1
inverse
matrix
æ ö æ ö
=ç ÷ ç ÷
-è ø è ø
or * 1 0
0 1
inverse
matrix
æ ö æ ö
=ç ÷ ç ÷
è ø è ø
12
(a)
x = -4, y = 3
x = -1.5, y = 8
K1
K1
12
(b) Graph
Axes drawn in correct direction, uniform scales in -6<x< -0.75
and 0 < y < 16.
P1
All 6 points and *2 points correctly plotted or curve passes
through these points -6 < x < -0.75 and 0 < y < 16. K2
A smooth and continuous curve without any straight line and passes
through all 9 correct points using the given scale for
-6 < x < -0.75 and 0 < y < 16. N1
Note : 1. 6 or 7 points correctly plotted, award K1.
2. Ignore curve out of range.

7. SULIT 1449/2(PP)
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1449/2(PP) SULIT
7
.
.
.
.
..
...
0-6 -5 -4 -3 -2 -1
2
0
4
6
8
10
12
14
16
Graf untuk Soalan12
Graph for Question 12

8. SULIT 1449/2(PP)
1449/2(PP) SULIT
8
(c) (i) 4.6 < y < 5.0 P1
(ii) -1.0 < x < = - 0.8
P1
(d) y = 2x + 12 or garis lurus y = 2x + 12 dilukis
K1 K1
-4.8 < x < -4.65 , -1.35 < x < -1.20
N1N1
13(a)(i) (2 , 6) P2
12
Note: (4, 5) or (4, 5) marked , award P1
(ii) (7, 8) P2
Note: (9 , 7) or (9 , 7) marked, award P1
(b)(i)(a) W: Rotation, 90° clockwise at (10 , 10) P3
Note :
1. P2 : Rotation 90° clockwise or Rotation, at (10 , 10) / Putaran 90°
ikut arah jam atau Putaran pada (10,10)
2. P1: Rotation// Putaran
(b) V: Enlargement at centre (13 , 10), with scale factor
3
1
P3
Note:
P2: Enlargement at centre (13 , 10), or Enlargement with scale factor
3
1
// Pembesaran pada (13 , 10) atau pembesaran dengan faktor
skala
3
1
P1: Enlargement// Pembesaran
(ii) 2
1
180
3
æ ö
= ´ç ÷
è ø
K1
20 N1

9. SULIT 1449/2(PP)
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1449/2(PP) SULIT
9
Question Solution and Mark Scheme Marks
14(a)(i)
Marks : (II to VII)
Mid point : (II to VII)
Frequency : (I to VII)
Note :
Allow two mistake in frequency for P1.
P1
P1
P2 4
(b)(i) 40 49 P1
4
(ii)
( ) ( ) ( ) ( ) ( ) ( ) ( )* * * * * * *
* * * * * * *
2 24.5 5 34.5 8 44.4 6 54.5 4 64.5 3 74.5 2 84.5
2 5 8 6 4 3 2
´ + ´ + ´ + ´ + ´ + ´ + ´
+ + + + + +
or
30
1555
Note: 1. Allow *midpoint for K1
6
311
or
6
5
51 or 51.83
Note: Correct answer from incomplete working, award Kk2
K2
N1
(c) Axes drawn in correct direction and uniform scale for
5.845.24 ££ x and 0 £ y £ 8.
*7 points correctly plotted
Note :
*5 or *6 points correctly plotted or bar passes through using
at least 6 correct mid-point, award K1.
Correct bar passes all 7 correct points for using given scales
5.845.24 ££ x
P1
K2
N1
4
Marks
Markah
Mid-point
Titik tengah
Frequency
Kekerapan
20 29 24.5 2 I
30 39 34.5 5 II
40 49 44.5 8 III
50 59 54.5 6 IV
60 69 64.5 4 V
70 79 74.5 3 VI
80 89 84.5 2 VII

10. SULIT 1449/2(PP)
1449/2(PP) SULIT
10
Graf untuk Soalan14
Graph for Question 14
8
7
6
5
4
3
2
1
0
24.5 34.5 44.5 54.5 64.5 74.5 84.5 Mark
frequency

11. SULIT 1449/2(PP)
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1449/2(PP) SULIT
11
Question Solution and Mark Scheme Marks
15 Note :
(1) Accept drawing only (not sketch).
(2) Accept diagrams with wrong labels and ignore wrong labels.
(3) Accept correct rotation of diagrams.
(4) Lateral inversions are not accepted.
(5) If more than 3 diagrams are drawn, award mark to the
correct ones only.
(6) For extra lines (dotted or solid) except construction lines, no
mark is awarded.
(7) If other scales are used with accuracy of ± 0.2 cm one way,
deduct 1 mark from the N mark obtained, for each part
attempted.
(8) Accept small gaps extensions at the corners.
For each part attempted :
(i) If £ 0 4× cm, deduct 1 mark from the N mark obtained.
(ii) If > 0 4× cm, no N mark is awarded.
(9) If the construction lines cannot be differentiated from the
actual lines:
(i) Dotted line :
If outside the diagram, award the N mark.
If inside the diagram, award N0.
(ii) Solid line :
If outside the diagram, award N0.
If inside the diagram, no mark is awarded.
(10) For double lines or non-collinear or bold lines, deduct 1
mark from the N mark obtained, for each part attempted.

12. SULIT 1449/2(PP)
1449/2(PP) SULIT
12
Question Solution and Mark Scheme Marks
15(a)
P
J K
L
MN
3.5 cm
3.5 cm
7 cm
8 cm
Correct shape with rectangles JKMN , JKLP and PLMN
All solid lines.
K1
3JK > KM > KL=LM
K1
dep K1
Measurement correct to ± 0 2× cm (one way) and all angles
at vertices of rectangles = 90° ± 1°
N1 dep
K1K1
15(b)(i)
J
P L
K
FR
S
QE
4 cm
3 cm
3 cm
2 cm
2 cm
8 cm
Correct shape with rectangle PLKJ, SQ perpendicular to EQ
All solid lines
K1
4
PL > LF > EQ > LK = KF > RF = QR = QS
K1
dep K1
Measurement correct to ± 0 2× cm (one way) and all angles at the
vertices of rectangles = 90° ± 1°
N2
dep
K1K1

13. SULIT 1449/2(PP)
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1449/2(PP) SULIT
13
Question Solution and Mark Scheme Marks
15(b)(ii)
S
G
V
M
L
K
F
1 cm
2 cm
7 cm
5
Correct shape
All solid lines
Note : Ignore *SV
K1
5
12
S and V joined with dashed line to form rectangles SVFG
K1
dep K1
FG > LM > MG
K1 dep
K1K1
Measurement correct to ± 0 2× cm (one way) and all angles at
the vertices of rectangles = 90° ± 1°
N2 dep
K1K1K1
16(a) 105 o
E 105 o
T)
Note :
105 o
or o
E o
T, award P1
P2
12
(b)(i) 3900
60
35 o
N // U
K1
N1N1
(ii) 60 cos35 4669o
q ´ ´ =
20
K2
N1
(c) ( )
600
30cos6010575 ´´+
15.59
K2K1
N1
Note :
*cos45 + *(75 105 )+ , award K1

14. SULIT 1449/2(PP)
1449/2(PP) SULIT
14