Spm Mm Ans (kedah)

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN

PEPERIKSAAN PERCUBAAN SPM 2008 MATHEMATICS ANSWER FOR PAPER 1

1449/1

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 =

( K 1 + K 2) × 100 140

1449/2 (PP) Matematik Kertas 2 Peraturan Pemarkaha n 17 Sept 2008

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA SEKOLAH MENENGAH NEGERI KEDAH DARUL AMAN.

PEPERIKSAAN PERCUBAAN SPM 2008

MATHEMATICS Kertas 2 PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 17 halaman bercetak Section A [ 52 marks] Question

Solution and Mark Scheme

P

Q

Marks

R

1(a) K1

P

Q

R

(b)

K2

Note:

P

Q

Set P ∪ Q correctly shaded, award K1

R

3

Question 2

Solution and Mark Scheme 3 p  q  21

or

2 2 p  q  14 3

5  q  10 3

5 p  25 or

Marks K1

or equivalent

or equivalent

K1

OR

p

4q 2

or

q  42p 5  q  10 3

5 p  25 or

or equivalent

or equivalent

(K1)

(K1)

OR

 p    q



1  7 3     4  1   2   2 1   3

1 

1 1   

1



or equivalent

(K2)

p5

N1

q  6

N1

Note : Accept 1.

2.

 p   5     as final answer, award N1  q   6  1 3  or equivalent seen, award (K1)   1   2   2 1   3

1 

1 1   





1

4

Question 3

Solution and Mark Scheme

K1

3 x 2  x  10  0

 3x  5  x  2 

0

or equivalent

x  2

x

5 3

Marks

K1 N1 N1

or 1 67

4

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. 5 5  4. Accept  x   x  2   0, with x  ,  2 for KK2 3 3  4

Identify UPT or TPU tan UPT  66 8 or

5(a) (b)

y

7 5  42 2

or equivalent

6648'

K1 N1

3 2

3

N1 P1

Gradient SR = 4 y   6   4* or x 1

P1

6  4*  1  c or equivalent

K1

y  4 x  10 or equivalent

N1

10

N1

5

Question

Solution and Mark Scheme

Marks

6(a)

Non-statement

(b)

Implication 1 :

If x − p > y − p , then x > y

K1

Implication 2 :

If x > y, then x − p > y − p

K1

(c)

P1

If k is a multiple of 9, then it is multiple of 3 True

(d)

7(a)

384 is divisible by 12 or 384 boleh dibahagi tepat dengan 12

K1

60 22  2 7 360 7

K1

90 22  2  14 360 7

172 3

(b)

dep P1

Premise 2 :

14 +

P1

or

90 22 60 22  2  14 +  2 7 + 7 + 7 360 7 360 7 or

57

1 3

90 22  142 360 7

or 57 33

or

180 22 2  7 360 7

K1 N1

or

60 22 2  7 360 7

90 22 180 22 2 60 22 2  142   7 +  7 360 7 360 7 360 7 308 2 or 102 3 3

4

or 102 67

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

K1

K1

N1

6

Question 8(a)

(b)

Solution and Mark Scheme

Marks

1 2  3 5 2 15

K1 N1

 1 2     3 5 Note :

 1 3     3 5

1 2  3 5

 1 3     3 8

or

1 3  3 5

K2 1 3  , award K1 3 8

or

(with condition of not more than 3 pairs) 165 360

9(a)

(b)

or

55 120

11 24

or

or

0 458

N1

k  10

P1

p  3

P1

 x 1    y   4   1   2   3



4 2  4    3 1   2  

K2

x2

N1

y  1

N1

Note : *

 x  inverse   4  1.        y  matrix   2  award K1

or

1  4   1   2   3

*

 inverse   1 2  2. Do not accept     or  matrix   3 4 

5

*



4 2  seen,  3 1  

 inverse   1     matrix   0

 x   2    as final answer, award N1.  y   1 

3. 

4. Do not accept any solution solved not using matrices.

0 . 1

6

Question 10(a) (b)

Solution and Mark Scheme

Marks

9 93 39 or 50 05

P1 K1

6 1 or 1 or 1 2 5 5

N1

Note: Without working, award K1N1 (c)

1 1   3  9   5   t  5   9    9  21  16  t   153 or equivalent 2 2 12 Note : Allow one mistake in distance expression for any two or

K2 N1

6

one correct areas for K1.

11

40

K1

1 22   3 3 5 3 7

K1

40 

1 22   3 3 5 3 7

232 8  232 9

K1 N1

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

Question

Solution and Mark Scheme

Marks

4

12(a)

−10

K1

5

K1

2

Note : K only meant for table value (b)

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. 2. (c)(i) (ii)

(d)

P1 K2 (does not depend on P)

N1 (depends on P and K)

4

7 or 8 points correctly plotted, award K1 Ignore curve out of range P1

2 5  y  2 7 2 45  x   2 40

Identify equation y  3x  2 or equivalent

K1

Straight line y  3x  2 correctly drawn

K1

Values of x : 1 8  x   1 7 2 7  x  2 8

2

P1

Values of x on the graph must correct

N1 (dep 2nd K1)

N1 (dep 2nd K1)

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 (> 2 decimal places)

12

y

15 х

10

х

y  3x  2

х х

5 х

−4

−3

−2

−1

O

1

2

3

4 х

−5

х

−10

−15

−20

х

−25

−30 х

y  2 x 2  5 x  8

−35

x

Question 13(a)(i) (ii)

Solution and Mark Scheme

Marks

(5, 4)

P1

(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)

198 *(3) 198 

K1 198 *(3)

or 198  22

176 OR

K1 N1

3

198  198   198   16 OR 8   or 8  2  , award K2 18  9   3  12

Question

Solution and Mark Scheme

Marks

14(a)

(i)

Upper Boundary Sempadan Atas 29 5

0

Cumulative Frequency Kekerapan Longgokan 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

Upper boundary

:

(II to VIII)

Frequency

:

(II to VIII)

Cumulative frequency

:

(II to VIII)

Frequency Kekerapan

P1 P2 P1

Note : Allow one mistake in frequency for P1 P1

(ii) 74 5 (b)

Axes drawn in correct direction and uniform scale for 29 5  x  99 5 and 0  y  40

5

P1 K2

*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  x  99 5 (c)

First Quartile = 56 0  0 5

or Third Quartile = 75 5  0 5

Interquartile range = 75 5*  56.0* = 19 5  0 5

N1

4

K1 K1 N1

3 12

Cumulative Frequency 40

× ×

35

×

30

25 × 20

15 × 10

×

5 × 0× 29.5

39.5

49.5

59.5

69.5

79.5

89.5

99.5

Marks

Question 15

Solution and Mark Scheme

Marks

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 :

(ii)

Solid line :

If outside the diagram, award the N mark. If inside the diagram, award N0. 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.

Question

Solution and Mark Scheme

Marks

15(a)

R

N

J

Q

G

P

K

F

Correct shape with rectangles JFGR , JKPN and square NPQR All solid lines.

Question

K1

JF  FG  GQ = QR = RN > NJ

K1 dep K1

Measurement correct to  0 2 cm (one way) and all angles at vertices of rectangles = 901

N1 dep K1K1

Solution and Mark Scheme

Marks

3

15(b)

V

N

P

J

K

E

A

F

B

Correct shape with the triangle FEV , rectangle JKPN and hexagon ABFEKJ All solid lines

Question

K1

AB > EV > EF = JA = NP > BF = NJ = VP

K1 dep K1

Measurement correct to ± 0 2 cm (one way) and all angles at the vertices of rectangles = 901

N2 dep K1K1

Solution and Mark Scheme

Marks

4

15(c) V

W

Q

P

K

F

B

L

U

G

C

Correct shape with rectangles BCGF, GWVU and FULK All solid lines

K1

Note : Ignore PQ P and Q joined with dotted line to form square PQGU or rectangle PQWV

K1 dep K1

CW > BC > VL > VW = BK > KL > UL = LP = PV = BF = FK

K1 dep K1K1

Measurement correct to ± 0 2 cm (one way) and all angles at the vertices of rectangles = 901

N2 dep K1K1K1

5

12

Question 16(a)

Solution and Mark Scheme

(b)(i)

P3

(36°N, 70°W) Note:

70oW

or

(65°N, θ °W / 70 °E ), award P2

36°N

or

(180 − 110), award P1

 50  36   60

or

3

K1

equivalent

N1

5160

(ii) 140  60  cos 50

or

K2

equivalent

N1

5399 52

Note : Usage of cos 50 or

(c)

Marks

180  60  cos 36 or

 30  110  ,

award K1

K2

8737 38

Note : Usage of cos 36 or

 70  110  ,

5

award K1

 180  60  cos 36  * 800 10 92 or

K1 10 hours 55 minutes

N1

4

12

END OF MARK SCHEME