1 Methods of Proof. 2 Consider (p (p→q)) → q pqp→q p (p→q)) (p (p→q)) → q TTTTT TFFFT FTTFT FFTFT.
· PDF fileSet P Q correctly shaded, award K1 K1 K2 P 3 Q R P Q R P Q R SULIT
Transcript of · PDF fileSet P Q correctly shaded, award K1 K1 K2 P 3 Q R P Q R P Q R SULIT
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
1449/2 (PP) Matematik Kertas 2 Peraturan Pemarkahan 17 Sept 2008
PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN.
PEPERIKSAAN PERCUBAAN SPM 2008
UNTUK KEGUNAAN PEMERIKSA SAHAJA
Peraturan pemarkahan ini mengandungi 17 halaman bercetak
MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
2
Section A [ 52 marks]
Question Solution and Mark Scheme Marks
1(a)
(b)
Note: Set P Q correctly shaded, award K1
K1
K2
3
P Q R
P Q R
P Q R
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
3 Question Solution and Mark Scheme Marks
2
3 21p q or 22 143
p q or equivalent
5 25p or 5 103
q or equivalent
OR
42
qp or 4 2q p or equivalent (K1)
5 25p or 5 103
q or equivalent (K1)
OR
1 71131 41 1 2 2 1
3
pq
or equivalent (K2)
5p
6q Note : Accept
1. 56
pq
as final answer, award N1
2.
111311 1 2 2 1
3
or equivalent seen, award (K1)
K1
K1
N1
N1
4
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
4
Question Solution and Mark Scheme Marks 3
23 10 0x x
3 5 2 0x x or equivalent
2x
53
x 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 52 0, with , 23 3
x x x
for KK2
K1
K1
N1
N1
4
4
Identify UPT or TPU
tan 2 2
75 4
UPT
or equivalent
66 8 or 66 48 '
P1
K1
N1
3
5(a)
(b)
32
y
Gradient SR = 4
*64
1y
x
or *6 4 1 c or equivalent
4 10y x or equivalent
10
N1
P1
K1
N1
N1
5
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
5
Question Solution and Mark Scheme Marks
Non-statement P1
Implication 1 : If x p > y p , then x > y K1
Implication 2 : If x > y, then x p > y p K1
6(a)
(b)
(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 14360 7
or 60 222 7360 7
14 + 90 222 14360 7
+ 60 222 7360 7
+ 7 + 7
172
3 or 157
3 or 57 33
290 22 14360 7
or 2180 22 7360 7
or 260 22 7360 7
290 22 14
360 7 2180 22 7
360 7 + 260 22 7
360 7
3083
or 21023
or 102 67
Note :
1. Accept π for K mark. 2. Correct answer from incomplete working, award KK2.
K1
K1
N1
K1
K1
N1
6
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
6
Question Solution and Mark Scheme Marks
8(a)
(b)
1 23 5
215
85
31
52
31
52
31
Note : 1 23 5 or 1 3
3 5 or
85
31 , award K1
(with condition of not more than 3 pairs)
360195 or
12065 or
2413 or 0.5417
K1
N1
K2
N1
5
9(a)
(b)
10k
3p
4 2 413 1 24 1 2 3
xy
2x
1y Note :
1. * inverse 4
matrix 2xy
or
4 213 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. 21
xy
as final answer, award N1.
4. Do not accept any solution solved not using matrices.
P1
P1
K2
N1
N1
6
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
7
Question Solution and Mark Scheme Marks
10(a) 9
P1
(b) 9 3 3 95 0 0 5
or
K1
6 11 1 25 5
or or
N1
Note: Without working, award K1N1
(c) 1 13 9 5 5 9 9 21 16 1532 2
t t or equivalent K2
12 N1 6
Note : Allow one mistake in distance expression for any two or
one correct areas for K1.
11
40 7
1 22 3 3 53 7
40 7 1 22 3 3 53 7
232 8 232 9
Note: 1. Accept π for K mark 2. Correct answer from incomplete working, award KK2
K1
K1
K1
N1
4
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
8
Question Solution and Mark Scheme Marks
12(a)
(b)
(c)(i)
(ii)
(d)
10
5
Note : K only meant for table value Graph Axes drawn in correct direction, uniform scales in 3 5 x 4 and – 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 for 3 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 of x : 1 8 1 7x
2 7 2 8x Note : 1. Allow P mark or N mark if values of y and x shown on graph
2. Values of y and x 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 of x on the graph must correct
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
9
y
x O 1 2 3 4
5
10
1 2 3 4
5
10
15
25
35
20
30
15
х
х
х
х
х
х х
х
х
22 5 8y x x
3 2y x
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
10
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) or equivalent. P3
Note :
1. P2: Rotation 90 anticlockwise or Rotation 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) or Enlargement scale factor 3.
2. P1: Enlargement.
(c) 23*198 K1
23*198198 or 198 – 22 K1
176 N1 3
OR 198 1618
OR 19889
or 219883
, award K2
12
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
11
Question Solution and Mark Scheme Marks
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 for 29 5 99 5x and 0 y 40 *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 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
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
12
0
5
10
15
20
25
30
35
40
39.5 49.5 59.5 69.5 79.5 89.5 99.5 29.5
Marks
Cumulative Frequency
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
13
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, 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.
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
14
Question Solution and Mark Scheme Marks
15(a)
Correct shape with rectangles JFGR , JKPN and square NPQR All solid lines. JF FG GQ = QR = RN > NJ Measurement correct to 0 2 cm (one way) and all angles at vertices of rectangles = 901
K1
K1
dep K1
N1 dep K1K1
3
R Q G
N P
J K F
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
15
Question Solution and Mark Scheme Marks
15(b)
Correct shape with the triangle FEV , rectangle JKPN and hexagon ABFEKJ 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
K1 dep K1
N2 dep K1K1
4
A B
E
N P
J K
F
V
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
16
Question Solution and Mark Scheme Marks
15(c)
Correct shape with rectangles BCGF, GWVU and FULK All solid lines Note : Ignore PQ P and Q joined with dotted line to form square PQGU or rectangle PQWV CW > BC > VL > VW = BK > KL > UL = LP = PV = BF = FK Measurement correct to 0 2 cm (one way) and all angles at the vertices of rectangles = 901
K1
K1
dep K1
K1 dep K1K1
N2 dep K1K1K1
5
12
U
B C
G
P
L K
F
W V
Q
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
[Lihat sebelah 1449/2(PP) SULIT
17
Question Solution and Mark Scheme Marks
16(a)
(b)(i) (ii)
(c)
(36N, 70W) Note: 70oW or (65N, W / 70 E ), award P2 36N or (180 110), award P1 50 36 60 or equivalent 5160
140 60 cos50 or equivalent 5399.42
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
http://tutormansor.wordpress.com/
SULIT 1449/2(PP)
1449/2(PP) SULIT
18
http://tutormansor.wordpress.com/