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PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2010 1449/1 MATHEMATICS ANSWER FOR PAPER 1
1 B 11 C 21 C 31 D
2 A 12 D 22 A 32 B
3 D 13 A 23 A 33 B
4 B 14 B 24 B 34 D
5 C 15 C 25 B 35 C
6 B 16 B 26 D 36 B
7 C 17 D 27 A 37 D
8 C 18 C 28 D 38 C
9 A 19 A 29 A 39 D
10 B 20 A 30 C 40 A
ANALISIS
A 10
B 11
C 10
D 9
NOTA: MARKAH PELAJAR = 100140
)21( ×+ KK
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SULIT 1449/2(PP)
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1449/2 (PP) Matematik Kertas 2 Peraturan Pemarkahan 19 September 2010
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH
NEGERI KEDAH DARUL AMAN.
PEPERIKSAAN PERCUBAAN SPM 2010
UNTUK KEGUNAAN PEMERIKSA SAHAJA
Peraturan pemarkahan ini mengandungi 19 halaman bercetak
MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN
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SULIT 1449/2(PP)
1449/2(PP) SULIT
2
Section A [ 52 marks]
Question Solution and Mark Scheme Marks
1(a)
(b)
Note : P ∪ R correctly shaded, award K1
K1
K2
3
P R
Q
P R
Q
P R
Q
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3
Question Solution and Mark Scheme Marks
2
24 3 7 0− − =x x
( ) ( )1 4 7 0+ − =x x or equivalent
OR
2( 3) ( 3) 4(4)( 7)
2(4)
− − ± − − −=x (K1)
1= −x
7 31 1 75
4 4= ⋅x or or
Note : 1. Accept without ‘= 0’.
2. Accept three terms on the same side, in any order.
3. Accept correct answer from the correct term without factorisation for Kk2.
K1 K1
N1
N1
4
3
3
3 212
p q− = or 3 2 18p q+ = or equivalent
Note: Attempt to equate the coefficient of one of the unknowns award K1
4 12q− = or 4 32p = or equivalent OR
14 2p q= + or 14
2
pq
−= or equivalent (K1)
2(9 )
3
qp
−= or 3
92
q p= − or equivalent (K1)
Note: Attempt to make one of the unknowns as the subject, award K1
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1449/2(PP) SULIT
4
Question Solution and Mark Scheme Marks
OR
( ) ( )
1 2141
33 911 1 2 22
p
q
= − − −
or equivalent (K2)
8p =
3q = − Note : Accept
1. 8
3
p
q
= −
as final answer, award N1
2.
( ) ( )
1 2141
33 911 1 2 22
− − −
or equivalent seen, award (K1)
N1
N1
4(a)
(b)
PTQ∠ or QTP∠
2 2
12tanθ
6 8=
+ or equivalent
50 19⋅ ° or 50 11'° or 50 12'°
P1
K1
N1
3
5(a)(i)
(ii)
(b)
False // Palsu
True // Benar
If 2 0x > , then 0 or 0x x> < //
Jika 2 0x > , maka 0 atau 0x x> <
P1
P1
P1
4
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5
Question Solution and Mark Scheme Marks
(c)
If 0 or 0x x> < , then 2 0x > //
Jika 0 atau 0x x> < , maka 2 0x >
5 and 9 are odd numbers // 5 dan 9 adalah nombor ganjil
P1
K1
5
6(a)
(b)
0 2 5x= − − or equivalent
5
2x = − or
12
2− or 2 5− ⋅
mQR = 2−
6
4
y
x
− =−
*mQR or 6 = *mQR (4) + c or equivalent
2 14y x= − + or equivalent
K1
N1
P1
K1
N1
5
7(a)
(b)
p = 14 q = – 4
2 3 13
4 1 5
x
y
− =
1 3 131
4 2 2(2)(1) ( 3)(4)
x
y
= −− −
or
*
13
2
x inverse
y matrix
=
2x =
3y = −
P1
P1
P1
K1
N1
N1
2
4 6
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SULIT 1449/2(PP)
1449/2(PP) SULIT
6
Question Solution and Mark Scheme Marks
Note:
1. Do not accept *
2 3 1 0
4 1 0 1
inverseor
matrix
=
.
2. 2
3
x
y
= −
as final answer, award N1.
3. Do not accept any solutions solved not using matrix method.
8
21 22
3 5 63 7
× × ⋅ ×
222
5 47
× ×
2 21 22 223 5 6 5 4
3 7 7 × × ⋅ × + × ×
2739
7 or
2391
7 or 391 3⋅
Note :
1. Accept π for K mark. 2. Correct answer from incomplete working, award Kk2.
K1
K1
K1
N1
4
9(a)
(b)(i)
{(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)}
Note :
1. Accept correct tree diagram or grid for P2
2. Accept 10 correct listings for P1
{(C,S), (C,O), (C,R), (C,E)} 4
20 or
1
5
P2
K1
N1
2
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7
Question Solution and Mark Scheme Marks
(ii)
{(S,C), (S,R), (C,S), (C,R), (O,E), (R,S), (R,C), (E,O)} 8
20 or
2
5
Note : Accept answer without working from correct listing, tree diagram or grid.
K1
N1
4
6
10(a)
(b)
90 22
2 14360 7
× × × or 120 22
2 7360 7
× × ×
90 22
2 14360 7
× × × + 120 22
2 7360 7
× × × + 14 + 7 + 7 or equivalent
194
3 or
264
3 or 64 67⋅
290 2214
360 7× × or 2120 22
7360 7
× ×
290 22
14360 7
× × − 2120 227
360 7× × or equivalent
308
3 or
2102
3 or 102 7⋅
Note :
1. Accept π for K mark. 2. Correct answer from incomplete working, award KK2.
K1
K1
N1
K1
K1
N1
6
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SULIT 1449/2(PP)
1449/2(PP) SULIT
8
Question Solution and Mark Scheme Marks
11(a)
(b)
(c)
45915.3
30=
11 28
10
− or
17
10
−
7
110
− or 1 7− ⋅
Note : Without working, award K1N1
( ) ( ) ( )1 111 28 10 11 12 11 8 459
2 2v
× + × + × + × + × =
or equivalent method Note:
1. ( )111 28 10
2× + × or ( )11 12× or ( )1
11 82
v× + ×
or equivalent, award K1
2. Accept 195 or 132 for K1 22
P1
K1
N1
K2
N1
6
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9
Question Solution and Mark Scheme Marks
12(a)
(b)
(c)(i)
(ii)
(d)
7
2
Note : K only meant for table value. Graph Axes drawn in correct direction, uniform scales in 3 5 4x− ⋅ ≤ ≤ and 8 28y− ≤ ≤ . All 7 points and *2 points correctly plotted or curve passes through these points for 3 5 4x− ⋅ ≤ ≤ and 8 28y− ≤ ≤ . A smooth and continuous curve without any straight line and passes through all 9 correct points using the given scale for 3 5 4x− ⋅ ≤ ≤ . Note : 1. 7 or 8 points correctly plotted, award K1.
2. Ignore curve out of range.
10 0 11 0y⋅ ≤ ≤ ⋅
3 20 3 10x− ⋅ ≤ ≤ − ⋅
Identify equation 2 9y x= − + or equivalent. Straight line 2 9y x= − + correctly drawn.
Values of x : 2 65 2 55x− ⋅ ≤ ≤ − ⋅
3 05 3 15x⋅ ≤ ≤ ⋅ Note : 1. Allow N marks if values of x shown on the graph. 2. Values x obtained by calculation, award N0.
K1
K1
P1
K2 (does not
depend on P)
N1
(depends on P and K )
P1
P1
K1
K1
N1
(dep 2nd K1)
N1 (dep 2nd K1)
2
4
2
4
12
Note:
1. Allow P mark if value of x and y are shown on the graph.
2. Value of x and y obtained by calculation, award P0.
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SULIT 1449/2(PP)
1449/2(PP) SULIT
10
y = – 2x + 9
22 3 7y x x= − − ××××
××××
××××
××××
××××
××××
××××
×××× ××××
– 2 – 1 O 1
20
15
10
5
– 5
– 10
2 3 4 x
25
30
y
– 3 – 4
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11
Question Solution and Mark Scheme Marks
13(a)(i) (6, 1) P2
Note: (6, 1) marked or (1, 4) seen or (1, 4) marked, award P1
(ii) (8, 5) P2 4
Note: (8, 5) marked or (3, 8) seen or (3, 8) marked, award P1
(b)(i)(a) W: Reflection in the line x = 1 P2
Note :
1. P1 : Reflection // Pantulan.
(b) V: Enlargement at centre K or (6, 10) with scale factor3
2 P3
Note :
1. P2 : Enlargement at centre K or (2, 1) or
Enlargement scale factor 3
2
2. P1 : Enlargement.
(ii) * 23
882
×
K1
* 23
88 882
× −
K1
110 N1 8
12
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1449/2(PP) SULIT
12
Question Solution and Mark Scheme Marks
14(a)
(b)
Frequency : (I to VIII)
Cumulative frequency : (I to VIII)
Upper Boundary : (I to VIII)
Note : Allow one mistake in frequency for P1
* * * * * * *
* * * * * * *
( 34 5 2) ( 44 5 3) ( 54 5 8) ( 64 5 10) ( 74 5 11) ( 84 5 4) ( 94 5 2)
2 3 8 10 11 4 2
⋅ × + ⋅ × + ⋅ × + ⋅ × + ⋅ × + ⋅ × + ⋅ ×+ + + + + +
Note :
1. Allow *midpoint for K1 2. Al low two mistakes for the multiplication of midpoint
and *frequency, for K1 65 75⋅ Note :
Correct answer from incomplete working, award Kk2
e.g. 2630
65 7540
= ⋅ (without table)
P2
P1
P1
K2
N1
4 3
Marks Markah
Frequency Kekerapan
Cumulative Frequency Kekerapan Longgokan
Upper Boundary Sempadan
Atas
20 – 29 0 0 29.5 I
30 – 39 2 2 39.5 II
40 – 49 3 5 49.5 III
50 – 59 8 13 59.5 IV
60 – 69 10 23 69.5 V
70 – 79 11 34 79.5 VI
80 – 89 4 38 89.5 VII
90 - 99 2 40 99.5 VIII
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13
Question Solution and Mark Scheme Marks
(c)
(d)
Axes drawn in correct direction and uniform scale for 29 5 99 5x⋅ ≤ ≤ ⋅ and 0 ≤ y ≤ 40 Note :
Do not accept wrong label *8 points correctly plotted Note :
*6 or *7 points correctly plotted or curve passes through using at least 6 correct upper boundary, award K1 Smooth and continuous curve without any straight line passes through 8 correct points for using given scales 29 5 99 5x⋅ ≤ ≤ ⋅ Any true statement from the ogive
Note :
Do not accept answer without an ogive
P1
K2
N1
K1
4
1
12
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SULIT 1449/2(PP)
1449/2(PP) SULIT
14
0
5
10
15
20
25
30
35
40
29.5 39.5 49.5 59.5 69.5 79.5 89.5
××××
Marks
Frequency
45
××××
××××
××××
××××
××××
××××
99.5 ××××
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15
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.
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1449/2(PP) SULIT
16
Question Solution and Mark Scheme Marks
15(a)
Correct shape with pentagon ABHGF All solid lines. BA > AF > HG > BH Measurement correct to ± 0 2⋅ cm (one way) and all angles at vertices of rectangles = 90° ± 1°
K1
K1 dep K1
N1 dep K1K1
3
F
G H
B A
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17
Question Solution and Mark Scheme Marks
15(b)(i)
Correct shape with rectangles PNML, FPLG, GLJH and FPJH All solid lines
FN > FH > HJ = FP > MN = FG > LJ = GH = LM
Measurement correct to ± 0 2⋅ cm (one way) and all angles at the vertices of rectangles = 90° ± 1°
K1
K1 dep K1
N2 dep K1K1
4
N M
L J
H G F
P
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1449/2(PP) SULIT
18
Question Solution and Mark Scheme Marks
15(b)(ii)
Correct shape with rectangles QNPD and ADEF All solid lines Note : Ignore straight line KG K and G joined with dashed line to form rectangles ADKG and GKEF
QN = QA > AF > EF = AD > KE = GF > NP = QD >
EP = AG = DK
Measurement correct to ± 0 2⋅ cm (one way) and all angles at the vertices of rectangles = 90° ± 1°
K1
K1 dep K1
K1 dep K1K1
N2 dep K1K1K1
5
12
N P
Q
G K
F E
D A
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19
Question Solution and Mark Scheme Marks
16 (a)
(b)
(c)
(d)
(75°S, 100°E) // (75°S, 100°T) Note: 1. 100°E or 100°T , award P2
2. 100° or E or T , award P1
6420
60 or 107�
6420
7560
−
32�S
( )80 20 60 cos75+ × × ° or 100 60 cos75× × °
Note: ( )80 20+ or 100 or using cos75° correctly, award K1
1552 91⋅
180 60
550
× or equivalent
Note: ( )*180 60
550
×, award K1
19 64⋅ hours or 19 hours 38 minutes
END OF MARK SCHEME
P3
K1
K1
N1
K2
N1
K2
N1
3
3
3
3
12
SMKDarulaman j*k
http://edu.joshuatly.com/ http://www.joshuatly.com/