Math K12 Sabah Tial 09 Answer

JAWAPAN KERTAS 1 Matematik EXCEL 2 SPM 2009

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

A D B A C A A D A D D B D B C D A B D A

21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

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C C A C A D B A A D C A C B C B C D C D

1

Question 1

Solution and Mark Scheme

Marks

y = x-3

x O

y  3

y= -2x+4 The line y  3 correctly drawn (doesn’t matter dotted or solid).

K1

The region correctly shaded (line must be dotted).

P2

Note : Award P1 to shaded region bounded by 2 correct lines (Check one vertex from any two correct lines) 3 2

-6p+12q = -60 or -5p+10q= -50 or 12p-10q =78 or

K1

equivalent Note : Attempt to equate the coefficient of one of the unknowns, award K1 7q= -21 or 6 p  24 or equivalent OR

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K1

2 2

P =2q+10 or q 

1 39 5 6 39 p  5 or p   q or q  p  or 2 6 6 5 5

equivalent

K1

Attempt to make one of the unknowns as the subject, award K1 7q= -21 or 6 p  24 or equivalent

K1

OR  p  5 2  10  1       q   1 5    6  2   6 1  39 

K2

Note : Award K1 if *

 p   inverse   10   1 2  p   10  1.        or        q   matrix   39   6 5  q   39  *

 inverse   1 2  1 0 2. Do not accept   =  or   0 1  matrix   6 5  p4

N1

q  3

N1

Note :  p  4       as final answer, award N1  q   3 

4

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3

3

6x2-5x-4=0

K1

(2x+1) (3x-4) =0 or equivalent

K1

1 2

N1

x = -0.5 or

x=

-

4 3

N1

Note : 1. Accept without “=0” 2. Accept three terms on the same side, in any order 3. Do not accept solutions solved not using factorization 4 4

1 22 2   7  30 3 7

K1

1 4 22 3   7 2 3 7

K1

1 22 2 1 4 22   7  30 +    73 3 7 2 3 7

K1

2 2258 cm 3 3

N1

4

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4 5(a) (b)

True // Benar

P1

If x 2  4 , then p  2 // Jika x 2  4 , maka p  2

P1 P1

False // Palsu (c)

12-7 n 2 , n  1, 2, 3........

K2

Note : 12-7 n 2 seen, award K1 5 6(a)

M MN =

P1

2 3

(b)

y4

(c)

M PQ = M MN =

N1 2 3

P1

y0  M PQ * or o   M PQ  (2)  c or equivalent x2 y

2 4 x or equivalent 3 3

K1

N1

5 7

Identify DRC or CRD 11 tan DRC = or equivalent 17 32.91 or 3254’

P1 K1 N1

3

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5 Question 8

Solution and Mark Scheme (a) m = -8 k = 4(-5) – 8(-3) k=4

P2

(b)  x  5 3   7  1       y  4(5)  8(3)  8 4   11  1 x    y   2   3 1 x 2 y3

K2

N1 N1

Note: x  inverse  7   5 3  1 1.       or   seen, award K1. 4(5)  8(3)  8 4   y  matrix  11 *

 inverse   4 3   inverse   1 0  2. Do not accept    or   .  matrix   8 5   matrix   0 1  *

Marks P1

*

1  x   3.    2 as final answer, award N1  y  3   4. Do not accept any solutions solve not using matrices.

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7

6

Question 9

(a) 13 - 6 = 7

Solution and Mark Scheme

Marks P1

(b) 0  14 14  20  13 7

K1 N1

=-2 Note: Accept answer without working for K1N1. (c) 1 (v  14)(6)  (14  7)  221 or equivalent method 2

K2

Note: 1 (v  14)(6)  (14  7), award K1 2

10

v = 27

N1

( A, D), ( A, 4), ( A,5), ( B, D), ( B, 4), ( B,5), (3, D), (3, 4), (3,5)

P1

(a) (3,5)

K1

(b)

1 9

N1

( A, D), ( A, 4), ( A,5), ( B, 4), (3, 4)

K1

5 9

N1

Note: Accept answer without working for K1N1

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6

5

7 11

(a)

240 22  2 7 360 7

7 +

47

or

180 22 7  2  360 7 2

240 22 180 22 7  2 7 +  2  360 7 360 7 2

1 142 cm or cm or 47.33 cm 3 3

180 22  7  (b)    360 7  2 

2

240 22 2  7 360 7 1001 12

or

or

–

83

5 12

240 22 2  7 360 7

180 22  7     360 7  2 

or

K1

K1

N1

K1

2

83.42 cm2

K1

N1 6

Note: 1. Accept  for K mark. 2. Correct answer from incomplete working, award KK2

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8 Question (a) 12

Solution and Mark Scheme x y

(b)

(i) (ii)

–1 –2

2.5 20.75

Axes drawn in correct direction with uniform scales in –2  x  5 and –25  y  25, 6 points and 2 points* correctly plotted or curve passes through these points for –2  x  5.

Marks K1 K1

2

P1 K2

Note: 1. 6 or 7 points correctly plotted, award K1 (iii)

(c) (i) (ii) (d)

Smooth and continuous curve without any straight line passes through all 8 correct points using the given scale for –2  x  5

N1

16  y  17

P1

–1.65  y  –1.55

P1

Identify equation y   x  18 or  3 x 2  11x  12   x  18

K1

Straight line y   x  18 correctly drawn

K1

4

2

Values of x : 0.55  x  0.65

N1

3.35  x  3.45

N1

4

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. 12

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9 Graph For Question 12 Graf untuk Soalan 12 y 25 y = –3x2+ 11x +12

×

×

20

×

×

16.5 15

y = –x + 18

×

× 10

× 5

–1.6 –2

–1

×

0

0.6

1

2

3

3.4

x 4

5

–5

× –10

–15

–20

× –25

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10

13

Question (a) (i) (ii)

Solution and Mark Scheme

Marks

m=1

P1

(–3, 6)

P1

(–4, 0)

P2

(a) 4

(b) Note: 1. Point (–4, 0) marked on diagram, award P1. (b) (i) (a)

Enlargement centre (4, –2). with scale factor Pembesaran pusat (4, –2) dan faktor skala

1 or 2

P3

1 . 2

Note: 1. P2 : Enlargement centre (4, –2) or Enlargement scale factor Pembesaran pusat (4, –2) or Pembesaran faktor skala

1 or 2

1 . 2

. 2. P1 : Enlargement or Pembesaran

(b)

Rotation 900 anticlockwise at centre (2, –3) or Putaran 900 lawan arah jam pada pusat (2, –3).

P3

Note: 1. P2 : Rotation 900 anticlockwise or Rotation at centre (2, –3) or Putaran 900 lawan arah jam or Putaran pada pusat (2, –3). 2. P2 : Rotation or Putaran.

(ii)

  1 2  1      Area ABCD or 43.5 ÷ 3 or   2  

K1

2

1 area of EFGH =    area of heptagon ABCD 2

N1

14.5

8 12

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11

Question 14

Solution and Mark Scheme

(a)

Class Midpoint interval Titik Selang tengah kelas 60 – 64 65 – 69 70 – 74 75 – 79 80 – 84 85 – 89 90 – 94

(b)

(c) (i)

(ii)

(d)

62 67 72 77 82 87 92

Frequency Kekerapan

5 9 8 5 6 4 3

Marks

Upper boundary Sempadan atas

64.5 69.5 74.5 79.5 84.5 89.5 94.5

Class interval : all the answers are correct Midpoint : all the answers are correct Frequency : all the answers are correct Upper boundary : all the answers are correct

P1 P1 P2 P1

5  62  9  67  8  72  5  77  6  82  4  87  3  92 40 3 74.75 or 74 4 Note: 1. Allow two mistakes for K1. i.e. mid point wrongly copied or wrong multiplication 2. Incomplete working followed by correct answer, award KK2 2990 i.e. = 74.75 40

K2

5

N1

3

Axes drawn in correct direction with uniform scales for 59.5  x  P1 94.5 and 0  y  9 and axis x labeled correctly with either midpoints or lower and upper boundaries 7 bars* drawn correctly according to the values in the table

K2

22

P1

3

1 12

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12 15(a)

Correct shape with pentagon ABCDE. All solid lines.

AB > AE > ED > DC > CB Measurements correct to  0.2 cm (one way) and angles at vertices A and B of pentagon are 90  1 .

K1

K1 dep K1

N1 dep K1K1 3

15(b)(i)

Correct shape with rectangles LMSR, CDIH and EDJI. All solid lines. LM > CH > LR = ED = DC. Measurements correct to  0.2 cm (one way) and C , D, E , H , I , M , L, R and S  90  1

K1 K1 dep K1

N2 dep K1K1 4

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13 15(b)(ii)

Correct shape with rectangles FPSM and AFJE. All solid lines.

K1

Note : Ignore CH and DI.

C and H joined with a solid line and D and I joined with dotted line, to form rectangle CHJE and DIFA.

PS > AF > AE > PF > AD = DC = CE, PF = AC. Measurements correct to  0.2 cm (one way) and All angles at the vertices of rectangles = 90  1

K1 dep K1 K1 dep K1K1

N2 dep K1K1K1

5 12

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14 Question 16(a)

100E

or

Solution and Mark Scheme 100T

Marks P2

Note : 100 or E or T award P1 2 16(b)(i)

6138  x  cos 47 or 9000  80 60

K1

70 E or 70T

N1

Note : If 70 without E or T 16(b)(ii)

4560 60

N0

3 K1

or 76

4560  47 60

K1

29 N

N1

or 29U

Note: If 29 without N or U 16(c)

K1

6138 or 9000 cos 47

N0

3

(180 – 29 – 47)  60

K1

6240

N1 2

16(d)

6138  4560  *c 900

K1

18.82

N1

Accept : 18 hours 49 minutes or 18.8 for N1

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2 12