trial kedah math spm 2013 k2 skema

7/29/2019 Trial Kedah Math SPM 2013 K2 SKEMA

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SULIT 1449/2(PP)

[Lihat halaman sebelah

1449/2(PP) SULIT

PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013

ANJURAN

MAJLIS PENGETUA SEKOLAH MALAYSIA (KEDAH)

MODUL A

MATEMATIK

KERTAS 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan pemarkahan ini mengandungi 14 halaman bercetak



SULIT 1449/2(PP)

1449/2(PP) SULIT

2

Section A[ 52 marks]

Question Solution and Mark Scheme Marks

1

2

40-2

-4

-6 -4

-2

6

4

2 x

y = - x - 1

y = 2 x + 2

Straight dotted line x = 1 correctly drawn. K1

Region correctly shaded P2 3

Note:

1 Accept solid line x = 1 for K1

2 Award P1 to shaded region bounded by two correct lines,

including part of R .(Check one vertex from any two correct lines

2(a) Ð MFH P1

3

(b)

13

7tan =Ð MFH or equivalent

K1

28.3° or 28° 18¢ N1



SULIT 1449/2(PP)


1449/2(PP) SULIT

3


3 049 2 =- x

( )( ) 02323 =+- x x or equivalent

3

2= x or 0.67

3

2-= x or – 0.67

Note : 1. Accept without ‘ = 0 ’

2. Accept two terms on the same side, in any order.

3. ( )( )0 67 0 67 x x- × + × with 0 67, 0 67 x x= × = - ×

award Kk2

K1

K1

N1

N1 4

4 14 3 2 x y+ = - or 7 21 77 x y+ = or equivalent K1

Note :Attempt to equate one of the coefficients the unknowns, award K1

OR

( )3

1 2 1 711 211 33 7 3

y x x x y or y or x or y

- - - --= - = = =

or equivalent (K1)

Note :Attempt to make one of the unknowns as the subject award K1.

3913 13 78

2 x or y- = = or equivalent K1

OR

31131

23 1

(1) (3)(7) 7 12

x

y

æ ö-æ ö æ öç ÷=ç ÷ ç ÷ç ÷ -æ öè ø è ø- -ç ÷ è ø

è ø

(K2)

Note :

Attempt to write without equation, award (K1)

1 x = - N1

4 y = N1 4

Note :

1

4

x

y

-æ ö æ ö=ç ÷ ç ÷

è ø è øas final answer, award N1



SULIT 1449/2(PP)

1449/2(PP) SULIT

4


5(a)20

2

14

2

14

7

22

2

1´´´´

K1

4

1540 N1

(b)20

2

14

2

14

7

22

2

1´´´´ + ( )

120 10 14 6790

2

AB´ + ´ ´ = K1

25 N1

Note :

1. Accept p for K mark.2. Correct answer from incomplete working, award Kk2.

6 (a) benar / true P1

5

(b) Jika n adalah nombor negatif maka 0n < .

If n is a negative number then 0n < .

benar / true

P1

(c) Set R ialah subset bagi set K

Set R is subset of set K

P1

(d ) Bilangan subset bagi suatu set yang mempunyai 7 unsur ialah72

The number of subsets in a set with n elements is 72 .

K1

72 128= N1

7 (a) 5 x = - P1

5

(b) 4

5SR PQ

M M = = P1

( )*4

4 55

c= - + or ( )

*4 4

5 5

y

x

-=

- - K1

85

4+= x y N1

y-intercept = 8 N1

8 (a) 12 P1 1

(b)12 0

10 0or equivalent

-

- K1

61 2

5or × N1 2

Note: Accept answer without working for K1N1

(c) ( )1 1(18 28) 12 (( 28)(12 20) 4682 2

t + + - + = or

equivalent method K2

Note:

1 1(18 28)(12) (( 28)(12 20)

2 2or t + - +

equivalent, award K1

40 N1 3

6



SULIT 1449/2(PP)


1449/2(PP) SULIT

5


9(a)

(b)

147

222

360

120´´´ or 7

7

222

360

180´´´

14 + 14 + 14 + 147

22

2360

120

´´´ + 77

22

2360

180

´´´

3

280 or

3

193 or 93.33

2147

22

360

120´´ or 27

7

22

360

180´´

2147

22

360

120´´ - 27

7

22

360

180´´

3

385 or

3

1128 or 128.33

Note :

1. Accept π for K mark.

2. Correct answer from incomplete working, award Kk2.

K1

K1

N1

K1

K1

N1 6

10 (a) {(P, M), (P, 4), (P, 5), (P, N), (Q, M), (Q, 4), (Q, 5), (Q, N),

(3, M), (3, 4), (3, 5), (3, N), (R, M), (R, 4), (R, 5), (R, N)}

Note :

1.

Accept 8 correct listing from not more than 16 outcomes for P1

P2

(b)(i) {(3, 4), (3, 5)} K1

16

2 or

8

1

N1

(ii) {(P, 4), (P, 5), (Q, 4), (Q, 5), (3, M), (3, N), (R, 4), (R, 5)}

16

8 or

2

1

K1

N1

NOTE :

1. Accept other method for K mark.

2. Accept answer without working from correct listing,

correct tree diagram or correct grid for K1N1.



SULIT 1449/2(PP)

1449/2(PP) SULIT

6


11 (a)

( ) ( )

1 21

6 33 1 2 6

- -æ öç ÷

-´ - - ´ è ø

1 2

15 156 1

15 5

æ öç ÷ç ÷ç ÷-ç ÷è ø

P2 2

(b)3 2 9

6 1 7

x

y

æ öæ ö æ ö=ç ÷ç ÷ ç ÷- -è øè ø è ø

P1

1 2 91

6 3 7(3)( 1) (2)(6

xor

y

- -æ ö æ öæ ö=ç ÷ ç ÷ç ÷- -- -è ø è øè ø

*9

7

x Inverse

y matrix

æ ö æ ö æ ö=

ç ÷ ç ÷ ç ÷-è ø è ø è øK1

1

3 x = - N1

y = 5 N1 4

6

Note:

1.

1

3

5

x

y

æ ö-æ ö ç ÷=ç ÷ ç ÷ç ÷è ø

è ø

as final answer, award N1

2. Do not accept any solution solved no using matrix method.

3. Do not accept*

3 2

6 1

inverse

matrix

æ ö æ ö=ç ÷ ç ÷

-è ø è ø or

*1 0

0 1

inverse

matrix

æ ö æ ö=ç ÷ ç ÷

è ø è ø

12(a)

x = -4, y = 3

x = -1.5, y = 8

K1K1

12

(b) GraphAxes drawn in correct direction, uniform scales in

-6< x< -0.75

and 0 < y < 16

.P1

All 6 points and *2 points correctly plotted or curve passes

through these points-6 < x < -0.75

and 0 < y < 16

.K2

A smooth and continuous curve without any straight line and passes

through all 9 correct points using the given scale for

-6 < x < -0.75and

0 < y < 16. N1

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

2. Ignore curve out of range.



SULIT 1449/2(PP)


1449/2(PP) SULIT

7

.

.

..

...

..

0-6 -5 -4 -3 -2 -1

2

0

4

6

8

10

12

14

16

Graf untuk Soalan12

Graph for Question 12



SULIT 1449/2(PP)

1449/2(PP) SULIT

8

(c) (i) 4.6 < y < 5.0P1

(ii) -1.0 < x < = - 0.8P1

(d ) y = 2 x + 12 or garis lurus y = 2 x + 12 dilukis

K1 K1

-4.8 < x < -4.65 , -1.35 < x < -1.20 N1N1

13(a)(i) (2 , 6) P2

12

Note: (4, 5) or (4, 5) marked , award P1

(ii) (7, 8) P2

Note: (9 , 7) or (9 , 7) marked, award P1

(b)(i)(a) W: Rotation, 90° clockwise at (10 , 10) P3

Note :

1. P2 : Rotation 90° clockwise or Rotation, at (10 , 10) / Putaran 90°

ikut arah jam atau Putaran pada (10,10)

2. P1: Rotation// Putaran

(b) V: Enlargement at centre (13 , 10), with scale factor 3

1 P3

Note:

P2: Enlargement at centre (13 , 10), or Enlargement with scale factor

3

1// Pembesaran pada (13 , 10) atau pembesaran dengan faktor

skala3

1

P1: Enlargement// Pembesaran

(ii) 21

1803

æ ö= ´ç ÷

è ø K1

20 N1



SULIT 1449/2(PP)


1449/2(PP) SULIT

9


14(a)(i)

Marks : (II to VII)

Mid point : (II to VII)

Frequency : (I to VII)

Note :

Allow two mistake in frequency for P1.

P1

P1

P2 4

(b)(i) 40 – 49 P1

4

(ii)

( ) ( ) ( ) ( ) ( ) ( ) ( )* * * * * * *

* * * * * * *

2 24.5 5 34.5 8 44.4 6 54.5 4 64.5 3 74.5 2 84.5

2 5 8 6 4 3 2

´ + ´ + ´ + ´ + ´ + ´ + ´

+ + + + + +

or 30

1555

Note: 1. Allow *midpoint for K1

6

311 or

6

551 or 51.83

Note: Correct answer from incomplete working, award Kk2

K2

N1

(c) Axes drawn in correct direction and uniform scale for

5.845.24 ££ x and 0 £ y £ 8.

*7 points correctly plotted

Note :*5 or *6 points correctly plotted or bar passes through using

at least 6 correct mid-point, award K1.

Correct bar passes all 7 correct points for using given scales5.845.24 ££ x

P1

K2

N1

4

Marks

Markah

Mid-point

Titik tengah

Frequency

Kekerapan

20 29 24.5 2 I

30 39 34.5 5 II

40 49 44.5 8 III

50 59 54.5 6 IV

60 69 64.5 4 V

70 79 74.5 3 VI

80 89 84.5 2 VII



SULIT 1449/2(PP)

1449/2(PP) SULIT

10

Graf untuk Soalan14

Graph for Question 14

8

7

6

5

4

3

2

1

0

24.5 34.5 44.5 54.5 64.5 74.5 84.5 Mark

frequency



SULIT 1449/2(PP)


1449/2(PP) SULIT

11


15 Note :

(1) Accept drawing only (not sketch).

(2) Accept diagrams with wrong labels and ignore wrong labels.

(3) Accept correct rotation of diagrams.

(4) Lateral inversions are not accepted.

(5) If more than 3 diagrams are drawn, award mark to thecorrect 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.



SULIT 1449/2(PP)

1449/2(PP) SULIT

12


15(a)

P

K

L

M N

3.5 cm

3.5 cm

7 cm

8 cm

Correct shape with rectangles JKMN , JKLP and PLMN All solid lines.

K1

3 JK > KM > KL=LM

K1dep K1

Measurement correct to ± 0 2× cm (one way) and all angles

at vertices of rectangles = 90° ± 1°

N1 dep

K1K1

15(b)(i) P L

K

F R

S

Q E 4 cm

3 cm

3 cm

2 cm

2 cm

8 cm

Correct shape with rectangle PLKJ , SQ perpendicular to EQ

All solid lines

K1

4 PL > LF > EQ > LK = KF > RF = QR = QS

K1

dep K1

Measurement correct to ± 0 2× cm (one way) and all angles at the

vertices of rectangles = 90° ± 1°

N2

dep

K1K1



SULIT 1449/2(PP)


1449/2(PP) SULIT

13


15(b)(ii)

S

G

V

M

L

K

F

1 cm

2 cm

7 cm

5

Correct shape

All solid lines

Note : Ignore *SV

K1

5

12

S and V joined with dashed line to form rectangles SVFG K1

dep K1

FG > LM > MG

K1 dep

K1K1

Measurement correct to ± 0 2× cm (one way) and all angles at

the vertices of rectangles = 90° ± 1°

N2 depK1K1K1

16(a) 105 o E ∕∕ 105 o T) Note :

105 o or θ o E ∕∕ θ oT, award P1

P2

12

(b)(i) 3900

60

35o N // U

K1

N1N1

(ii) 60 cos35 4669oq ́ ´ =

20

K2

N1(c) ( )

600

30cos6010575 ´´+

15.59

K2K1

N1

Note :

* cos45 + * (75 105 )+ , award K1



SULIT 1449/2(PP)

1449/2(PP) SULIT

14

trial kedah math spm 2013 k2 skema

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