f4 final sbp 2006 math skema p 1 & p 2
TRANSCRIPT
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SEKTOR SEKOLAH BERASRAMA PENUH BAHAGIAN SEKOLAH
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN TINGKATAN 4, 2006
MATHEMATICS PAPER 1 & PAPER 2
NOTE
This marking scheme serves as a guide only. Other method which are equivalent and
suitable may also be adapted and used in the awarding of marks.
Paper 1: 40 marks
Paper 2: 60100
×x
Total Mark: P1 + P2
THANK YOU
This marking scheme consists of 7 printed pages
PEPERIKSAAN AKHIR TAHUN INGKATAN 4, SBP 2006
SKEMA JAWAPAN MATEMATIK KERTAS 1
MARKING SCHEME
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Q Answer Q Answer
1. B 21. D
2. B 22. D
3. C 23. D
4. A 24. A
5. A 25. B
6. C 26. C
7. A 27. B
8. A 28. D
9. A 29. D
10. B 30. C
11. C 31. D
12. C 32. C
13. D 33. C
14. A 34. C
15. B 35. A
16. C 36. C
17. D 37. B
18. C 38. B
19. D 39. A
20. C 40. A
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PEPERIKSAAN PERTENGAHAN TINGKATAN 4, SBP 2006 SKEMA JAWAPAN MATEMATIK KERTAS 2
Q Marking Scheme Sub mark
Full mark
1 4p2 – 3p – 10 = 0 (4p + 5)(p – 2) = 0
p = 45
− , 2
P1 K1
N1N1
4
2 p + 4q = 26 or equivalent 4p = 24 or equivalent p = 6 q = 5
K1 K1
N1
N1
4 3 ∠PTQ or ∠QTP
tan ∠PTQ = 129
∠PTQ = 36052’ or 36.90
P2
K1 N1
4
(a) –2 P1 4
(b) -2 = -2(4) + c c = 6 y = – 2x + 6 y-intecept = 6
K1 K1 N1 P1
5
(a) 5015 or equivalent P1
(b) ×
51 50
10
K1 N1
5
(c) 60
1010 +
6020 or equivalent
K1
N1
5
6 V = 15128722
31
××××
V = 480
4807722 2 =×× h
h = 77240 or 3.12
K1
N1
K1
N1
4
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Q Marking Scheme Sub mark
Full mark
7 (a) A1 = 1414722
36090
××× and A2 = 77722
36060
×××
A1 – A2
12831
K1K1
K1
N1
(b) P1 = 147222
36090
××× or P2 = 77222
360180
×××
P1 + P2 + 14 58
K1
K1 N1
7
(a) Statement. False statement P1P1 8 (b) 3 is not a factor of 15. False P1P1
(c) Some triangles have a right angle P1 5
(a)
P1 9
(b)
P2
3
10 (a) 110 P1 (b) 20 P2 (c) 40 P2 5
(a) (i)
– 0.91
P1
(ii) –0.42 – 1 –1.42
K1 N1
(c) (i) 20 N1
11
(ii) 135 N2 6
(a) (i) 22−=
xy
y-intercept = – 2
K1 N1
(ii) 2m = 8 – 4 m = 2
K1 N1
12
(iii) mPQ= mRS =
21
y – 5 = 21 (x – 4)
2y = x + 6
P1
K1
N1
P Q R
P Q R
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Q Marking Scheme Sub mark
Full mark
(b) (i) y – 10 = 2(x – 4) or y = 2x – 8 + 10 y = 2x + 2
K1 N1
12
(ii) (0, –8) P1 (iii)
04810
−+
29
K1
N1
12
(a) (i) 10 P1 13 (ii) Identify ∠EPF
Tan ∠EPF = 107
340 59’
P1
K1
N1
(iii) ∠UQV
P2
(b)
(i) Tan ∠PKM = 105
∠PKM = 26.60 or 26057’
K1
N1
(ii) 10 tan 38o h = 5 + 10 tan 38o
12.81
K1 K1K1
N1
(a) (i) (a) or P1 (b) or P1 (ii) 5∉ A ∪ B. N2 (iii) 3n 2 + 1, n = 2, 3, 4, 5, … K1N1 (b) (i) 3 P1 (ii) 11 P1
14
(iii) 13 P2 (iv) 9 K1N1
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Q Marking Scheme Sub mark
Full mark
(a)
Column 1 – P1 Column 2 – P1 Column 3 – P2
(b) 81 N1
15
(c) 40
)2(5.95)1(5.85)4(5.75)8(5.65)10(5.55)4(5.45)3(5.35)3(5.25)5(5.15 ++++++++
or 40
2100
52.5
K2
N1
(d) Refer to the graph given 9 plotted points Joining points A straight line from point (5.5, 0) to point (15.5, 5) and (95.5, 1) and (105.5, 0)
P2 P1
P1
12
Distance (km)
Midpoint Frequency
11 – 20 15.5 5 21 – 30 25.5 3 31 – 40 35.5 3 41 – 50 45.5 4 51 – 60 55.5 10 61 – 70 65.5 8 71 – 80 75.5 4 81 – 90 85.5 1 91 – 100 95.5 2
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Q Marking Scheme Sub mark
Full mark
16 (a) 72 N1 (b) 60 – 69 N1 (c)
Column 1 – P1 Column 2 – P1 Column 3 – P1
(d) Refer to the graph given Uniform scale 8 plotted points Smooth curve
K1 P2 N1
(e) (i) 62.5
N1
(ii) 72 – 9 63
K1 N1
12
Remark:
Marks Upper boundary
Cumulative frequency
20 – 29 29.5 0 30 – 39 39.5 4 40 – 49 49.5 12 50 – 59 59.5 24 60 – 69 69.5 44 70 – 79 79.5 60 80 – 89 89.5 70 90 – 99 99.5 72
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The graph for 15(d)
15.5 25.5 35.5 45.5 55.5 65.5 75.5 Distance (km)
Freq
uenc
y
1
2
3
4
5
6
7
8
×
×
×
×
× ×
85.5
×
9
10
5.5 95.5 105.5
×
×
×
×
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The graph for 16(b)
29.5 39.5 49.5 59.5 69.5 79.5 89.5 ×
Upper boundary
Cum
ulat
ive
freq
uenc
y
10
20
30
40
50
60
70
80
×
×
×
×
×
×
99.5
×
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