Upload
simpor
View
243
Download
0
Embed Size (px)
Citation preview
8/14/2019 Spm Mm Ans (Kedah)
1/18
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH
NEGERI KEDAH DARUL AMAN
PEPERIKSAAN PERCUBAAN SPM 2008 1449/1MATHEMATICS
ANSWER FOR PAPER 1
1 C 11 D 21 C 31 C
2 B 12 C 22 A 32 C
3 D 13 A 23 D 33 A
4 B 14 D 24 D 34 D
5 C 15 C 25 B 35 B
6 B 16 A 26 B 36 D
7 A 17 B 27 D 37 A
8 C 18 C 28 A 38 B
9 D 19 D 29 B 39 C
10 B 20 A 30 A 40 A
ANALISIS
A 10
B 10
C 10
D 10
NOTA: MARKAH PELAJAR = 100140
)21(
+KK
8/14/2019 Spm Mm Ans (Kedah)
2/18
1449/2 (PP)
Matematik
Kertas 2
Peraturan
Pemarkaha
n
17 Sept2008
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH
NEGERI KEDAH DARUL AMAN.
PEPERIKSAAN PERCUBAAN SPM 2008
UNTUK KEGUNAAN PEMERIKSA SAHAJA
MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN
8/14/2019 Spm Mm Ans (Kedah)
3/18
Peraturan pemarkahan ini mengandungi 17 halaman bercetak
Section A
[ 52marks]
Question Solution and Mark Scheme Marks
1(a)
(b)
Note:
SetPQ correctly shaded, award K1
K1
K2 3
P Q R
PQ
R
PQ
R
8/14/2019 Spm Mm Ans (Kedah)
4/18
Question Solution and Mark Scheme Marks
2 3 21p q or2
2 143
p q or equivalent
5 25p or 5 103 q or equivalent
OR
4
2
qp
or 4 2q p or equivalent (K1)
5 25p or5
103 q or equivalent (K1)
OR
1711
31 4
1 1 2 2 13
p
q
or equivalent (K2)
5p
6q
Note :
Accept
1. 56
pq
as final answer, award N1
2.
111
31
1 1 2 2 13
or equivalent seen, award (K1)
K1
K1
N1
N1 4
8/14/2019 Spm Mm Ans (Kedah)
5/18
Question Solution and Mark Scheme Marks
3 23 10 0x x
3 5 2 0x x or equivalent
2x
5
3x or 1 67
Note : 1. Accept without = 0
2. Accept three terms on the same side, in any order.
3. Do not accept solutions solved not using factorisation.
4. Accept 5 5
2 0, with , 2
3 3
x x x
for KK2
K1
K1
N1
N1 4
4 Identify UPT or TPU
tan2 2
7
5 4UPT
or equivalent
66 8 or 66 48'
P1
K1
N1 3
5(a)
(b)
3
2y
Gradient SR = 4
*6 41
y
x
or *6 4 1 c or equivalent
4 10y x
or equivalent
10
N1
P1
K1
N1
N1 5
8/14/2019 Spm Mm Ans (Kedah)
6/18
Question Solution and Mark Scheme Marks
6(a)
(b)
Non-statement P1
Implication 1 : Ifxp >yp , thenx >y K1
Implication 2 : Ifx >y, thenxp >yp K1
(c) If k is a multiple of 9, then it is multiple of 3 P1
True dep P1
(d) Premise 2 : 384 is divisible by 12
or
384 boleh dibahagi tepat dengan 12 K1 4
7(a)
(b)
90 222 14
360 7 or
60 222 7
360 7
14 +90 22
2 14360 7
+60 22
2 7360 7
+ 7 + 7
172
3 or
157
3 or 57 33
290 2214
360 7 or 2
180 227
360 7 or 2
60 227
360 7
290 22 14360 7
2180 22
7360 7
+ 260 22
7360 7
308
3 or
2102
3 or 102 67
Note :
1. Accept for K mark.2. Correct answer from incomplete working, award KK2.
K1
K1
N1
K1
K1
N1 6
8/14/2019 Spm Mm Ans (Kedah)
7/18
Question Solution and Mark Scheme Marks
8(a)
(b)
1 2
3 5
2
15
1 2 1 3 1 3
3 5 3 5 3 8
Note :1 2
3 5 or
1 3
3 5 or
1 3
3 8 , award K1
(with condition of not more than 3 pairs)
165
360 or
55
120 or
11
24 or 0 458
K1
N1
K2
N1 5
9(a)
(b)
10k
3p
4 2 41
3 1 24 1 2 3
x
y
2x
1y
Note :
1.
*inverse 4
matrix 2
x
y
or
4 21
3 14 1 2 3
seen,
award K1
2. Do not accept
*inverse 1 2
matrix 3 4
or
*inverse 1 0
matrix 0 1
.
3.2
1
x
y
as final answer, award N1.
4. Do not accept any solution solved not using matrices.
P1
P1
K2
N1
N1 6
8/14/2019 Spm Mm Ans (Kedah)
8/18
Question Solution and Mark Scheme Marks
10(a) 9 P1
(b)
9 3 3 9
5 0 0 5or
K1
6 11 1 25 5or or N1
Note: Without working, award K1N1
(c) 1 1
3 9 5 5 9 9 21 16 1532 2
t t or equivalent K2
12 N1 6Note : Allow one mistake in distance expression for any two or
one correct areas for K1.
11 40
1 223 3 5
3 7
40 1 22
3 3 53 7
232 8 232 9
Note:
1. Accept for K mark
2. Correct answer from incomplete working, award KK2
K1
K1
K1
N1 4
Question Solution and Mark Scheme Marks
8/14/2019 Spm Mm Ans (Kedah)
9/18
12(a)
(b)
(c)(i)
(ii)
(d)
10
5
Note : Konly meant for table value
Graph
Axes drawn in correct direction, uniform scales in 3 5 x 4and 34 y 11
7 points and *2 points correctly plotted or curve passes through
these points for 3 5 x 4
Smooth and continuous curve without any straight line and passes
through all 9 correct points using the given scales for3 5 x 4
Note : 1. 7 or 8 points correctly plotted, award K1
2. Ignore curve out of range
2 5 2 7y
2 45 2 40x
Identify equation 3 2y x or equivalent
Straight line 3 2y x correctly drawn
Values ofx : 1 8 1 7x
2 7 2 8x
Note : 1. Allow P mark or N mark if values ofy andx
shown on graph
2. Values ofy andx obtained by computations, award
P0 or N0
(> 2 decimal places)
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
Values ofx on the
graph must correct
8/14/2019 Spm Mm Ans (Kedah)
10/18
y
O 1 2 3 4
5
10
1234
5
10
15
25
35
20
30
15
22 5 8 y x x
3 2y x
8/14/2019 Spm Mm Ans (Kedah)
11/18
Question Solution and Mark Scheme Marks
13(a)(i) (5, 4) P1
(ii) (5, 2) P2 3
Note : (1, 0) or (5, 2) marked on diagram, award P1
(b)(i) W: Rotation 90 anticlockwise at centre ( 1, 2) orequivalent. P3
Note :
1. P2: Rotation 90 anticlockwise orRotation at centre ( 1, 2),
2. P1: Rotation.
(ii) V: Enlargement centre ( 5, 5) with scale factor 3 P3 6
Note :
1. P2: Enlargement centre ( 5, 5) orEnlargement scale factor 3.
2. P1: Enlargement.
(c)198
*(3)K1
198198 198 22
*(3)or K1
176 N1 3
OR198
1618
OR198
89
or 2198
83
, award K2
12
Question Solution and Mark Scheme Marks
8/14/2019 Spm Mm Ans (Kedah)
12/18
14(a)
(b)
(c)
(i)
Upper boundary : (II to VIII)
Frequency : (II to VIII)
Cumulative frequency : (II to VIII)
Note:Allow one mistake in frequency for P1
(ii) 74 5
Axes drawn in correct direction and uniform scale for29 5 99 5x and 0 y 40
*8 points correctly plotted
Note :
*6 or *7 points correctly plotted or curve passes through usingat least 6 correct upper boundary, award K1
Smooth and continuous curve without any straight line passes all
8 correct points for using given scales 29 5 99 5x
First Quartile = 56 0 0 5 or Third Quartile = 75 5 0 5
Interquartile range = * *75 5 56.0
= 19 5 0 5
P1
P2
P1
P1
P1
K2
N1
K1
K1
N1
5
4
3
12
Upper
Boundary
Sempadan
Atas
Frequency
Kekerapan
Cumulative
Frequency
Kekerapan
Longgokan
29 5 0 0 I
39 5 2 2 II
49 5 3 5 III
59 5 8 13 IV
69 5 10 23 V
79 5 11 34 VI
89 5 4 38 VII
99 5 2 40 VIII
8/14/2019 Spm Mm Ans (Kedah)
13/18
0
5
10
15
20
25
30
35
40
39.5 49.5 59.5 69.5 79.5 89.5 99.529.5
Marks
Cumulative
Frequency
8/14/2019 Spm Mm Ans (Kedah)
14/18
Question Solution and Mark Scheme Marks
15 Note :
(1) Accept drawing only (not sketch).
(2) Accept diagrams with wrong labels or without 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, nomark 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 theactual lines:
(i) Dotted line : If outside the diagram, award the Nmark.
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 1mark from the N mark obtained, for each part attempted.
Question Solution and Mark Scheme Marks
8/14/2019 Spm Mm Ans (Kedah)
15/18
15(a)
Correct shape with rectanglesJFGR ,JKPNand
square NPQR
All solid lines.
JFFGGQ = QR =RN >NJ
Measurement correct to 0 2 cm (one way) and all anglesat vertices of rectangles = 901
K1
K1
dep K1
N1 depK1K1
3
Question Solution and Mark Scheme Marks
R Q G
N P
J K F
8/14/2019 Spm Mm Ans (Kedah)
16/18
15(b)
Correct shape with the triangleFEV, rectangleJKPNand
hexagonABFEKJ
All solid lines
AB>EV>EF=JA =NP>BF= NJ= VP
Measurement correct to 0 2 cm (one way) and all angles at the
vertices of rectangles = 901
K1
K1dep K1
N2 dep
K1K1 4
Question Solution and Mark Scheme Marks
A B
E
N P
J K
F
V
8/14/2019 Spm Mm Ans (Kedah)
17/18
15(c)
Correct shape with rectanglesBCGF, GWVUandFULK
All solid lines
Note : IgnorePQ
Pand Q joined with dotted line to form squarePQGUorrectanglePQWV
CW>BC> VL > VW=BK>KL > UL =LP=PV=BF=FK
Measurement correct to 0 2 cm (one way) and all angles atthe vertices of rectangles = 901
K1
K1
dep K1
K1 dep
K1K1
N2 dep
K1K1K1
5
12
U
B C
G
P
LK
F
WV
Q
8/14/2019 Spm Mm Ans (Kedah)
18/18
Question Solution and Mark Scheme Marks
16(a)
(b)(i)
(ii)
(c)
(36N, 70W)
Note: 70oW or (65N, W/ 70E), award P2
36N or (180
110), award P1
50 36 60 or equivalent
5160
140 60 cos50 or equivalent
5399 52
Note : Usage of cos 50 or 30 110 , award K1
180 60 cos36 or 8737 38
Note : Usage of cos 36 or 70 110 , award K1
180 60 cos36 *
800
10 92 or 10 hours 55 minutes
P3
K1
N1
K2
N1
K2
K1
N1
3
5
4
12
END OF MARK SCHEME