UNTUK KEGUNAAN PEMERIKSA SAHAJA

  

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

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Section A

  [ 52 marks]

  Question Solution and Mark Scheme Marks y

  1

  6 y = - x - 1

  4

  2 4 x

• 6 -4 -2

  2

• 2

   y = 2x + 2

• 4

  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 (a)

  P1

  2

  ÐMFH (b)

  K1

  7 tan ÐMFH = or equivalent

  3

  13 N1 28.3° or 28° 18¢

• = x or
• –0.67 Note :

  1. Accept without ‘= 0’

  4

  K1 K1 N1 N1

  3

  4

  9 ² = -

  x ( )( )

  2

  3

  2 3 = + - x x

   or equivalent

  3

  2 = x or 0.67

  3

  2

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

  3.

• × + ×
• = - or 7
• = - = = =

• = = or equivalent
• æ ö æ ö
• ç ÷ è ø è ø
• æ ö æ ö

  æ ö

  3

  11

  3

  1

  2

  3

  1 (1) (3)(7)

  7

  1

  2

  x y

  ç ÷ =

  x or y

  ç ÷ ç ÷ ç

  ÷ - æ ö

  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

  = ç ÷ ç ÷ è ø è ø

  as final answer, award N1

  K1 OR

  78

  2

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

  0 67 0 67 x x

  

with 0 67, 0 67 x x

  = × = - ×

  award Kk2

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  4

  14

  3

  2

  x y

  21 + 77 x y = or equivalent K1

  OR

  ( )( )

  ( )

  3

  1 2 1 7

  11

  2 11 3

  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.

  39

  13

  13

  3

  Question Solution and Mark Scheme Marks

  K1

  5(a)

  1

  22

  14

  14 ´ ´ ´ ´

  20

  2

  7

  2

  2 1540

  N1 K1

  1

  22

  14

  14

  1

  (b)

  20 + 20 10 AB 14 6790 ´ ´ ´ ´ + ´ ´ ´ =

  ( )

  4

  2

  7

  2

  2

  2

  25 N1 Note : 1.

  Accept p for K mark.

2. Correct answer from incomplete working, award Kk2.

  (a) benar / true P1

  6

  (b) Jika n adalah nombor negatif maka n < . P1 If n is a negative number then n < . benar / true

  (c) Set R ialah subset bagi set K P1

  5 Set R is subset of set K (d) Bilangan subset bagi suatu set yang mempunyai 7 unsur ialah K1 ₇

  2 ₇ The number of subsets in a set with 2 . ₇ n elements is N1 2 128

  =

  x = -

  (a)

  5 P1

  7

  (b)

  4 M = M = _{SR PQ} P1

  5 ^{* *} - 4 y

  4

  4 4 = - + 5 c or

  =

  ( )

  K1

  5

  5 x - -

  5

  5

  ( ) y 4 +

  8 = x

  N1

  5

  y-intercept = 8

  N1

  8 (a)

  12 P1

  1

• 12 0

  or equivalent

  (b) K1

• 10 0

  6

  or × 1 2

  N1

  2

  5 Note: Accept answer without working for K1N1

  1

  1 (c) (18 28) 12 (( t 28)(12 20) = 468 or + + - +

  ( )

  2

  2 equivalent method K2

  Note:

  1

  1

• (18 28)(12) or (( t 28)(12 20)

  2

  2 equivalent, award K1

  40 N1

  3

  6

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  K1 K1 N1 K1 K1 N1

  7

  22 360

  120 ´ ´ - ²

  7

  7

  22 360 180

  ´ ´

  3 385 or

  3

  1 128 or 128.33

  Note : 1.

  Accept π for K mark.

  6

  180 ´ ´ ₂

  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.

  14

  22 360

  9(a) (b)

  2 360 120

  14

  7

  22

  2 360

  120 ´ ´ ´ or

  7

  7

  22

  2 360 180

  ´ ´ ´ 14 + 14 + 14 +

  14

  7

  22

  ´ ´ ´ +

  7

  7

  7

  22

  2 360 180

  ´ ´ ´

  3 280 or

  3

  1 93 or 93.33 ₂

  14

  7

  22 360 120

  ´ ´ or ²

  7

2. Correct answer from incomplete working, award Kk2.

  Question Solution and Mark Scheme Marks 11 (a)

  1

  2

  1 - - æ ö ç ÷

  3 1 2 6

  6

  3 ´ - - ´

• è ø

  ( ) ( )

  1

  2 æ ö

  P2

  2 ç ÷

  15

  15 ç ÷

  6

  1 ç ÷

• ç ÷ è

  15 5 ø

  3 2 x

  9 æ öæ ö æ ö

  = (b)

  P1 ç ÷ç ÷ ç ÷

  7 è øè ø è ø

  6 1 y - -

  x

  2

  9 æ ö 1 æ öæ ö

  1 - -

  = or ç ÷ ç ÷ç ÷ _*

  y

  6

  3

  7 (3)( 1) (2)(6 - - - -

  è ø è øè ø

  x Inverse

  9 æ ö æ öæ ö

  K1 =

  ç ÷ ç ÷ç ÷

• y matrix

  7 è ø è øè ø

  1

  x = -

  N1

  3

   y = 5

  N1

  4

  6 Note:

  1 æ ö

• x

  æ ö ç ÷

  1. as final answer, award N1

  =

  3 ç ÷ ç ÷

  y

  è ø ç ÷

  5 è ø 2. Do not accept any solution solved no using matrix method. _{* *}

  inverse

  3 2 inverse

  1 æ ö æ ö æ ö æ ö

3. Do not accept or

  = = ç ÷ ç ÷ ç ÷ ç ÷

  6

• matrix matrix

  1

  1 è ø è ø è ø è ø

  K1

  12 x = -4, y = 3

  12

  (a)

  K1 x = -1.5, y = 8 Graph

  (b)

  Axes drawn in correct direction, uniform scales in

• 6<x< -0.75

  P1 and .

  0 < y < 16 All 6 points and *2 points correctly plotted or curve passes through these points and .

  K2

• 6 < x < -0.75 0 < y < 16 A smooth and continuous curve without any straight line and passes through all 9 correct points using the given scale for

  N1 and .

• 6 < x < -0.75 0 < y < 16 Note : 1. 6 or 7 points correctly plotted, award K1.

2. Ignore curve out of range.

  

Graf untuk Soalan12

Graph for Question 12

  16 .

  14

  12 .

  10

  8 .

  6 .

  . . .

  4 .

  2 .

• 4 -3 -2 -1
• 6 -5

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  P1

  (c) (i) 4.6 < y < 5.0

  P1

  (ii) -1.0 < x < = - 0.8

K1 K1

  y = 2x + 12 or garis lurus y = 2x + 12 dilukis

  (d)

N1N1

• 4.8 < x < -4.65 , -1.35 < x < -1.20

  13 (a)(i) (2 , 6)

  P2 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) : Rotation, 90° clockwise at (10 , 10) P3

  W

  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

  1 (b) V: Enlargement at centre (13 , 10), with scale factor P3

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

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

  3

  1 skala

  3 P1: Enlargement// Pembesaran ₂ (ii)

  1 æ ö

  = ´ 180 ç ÷

  K1

  3 è ø

  20 N1

  12

  Question Solution and Mark Scheme Marks 14 (a)(i) Marks Mid-point Frequency

  Markah Titik tengah Kekerapan

  20

  24.5

  2 I

• – 29

  30

  34.5

  5 II

• – 39

  40

  44.5

  8 III

• – 49

  50

  54.5

  6 IV

• – 59

  60

  64.5

  4 V

• – 69

  70

  74.5

  3 VI

• – 79

  80

  84.5

  2 VII

• – 89 Marks : (II to VII) P1

  Mid point : (II to VII) P1

  4 Frequency : (I to VII) P2

  Note : Allow two mistake in frequency for P1. (b)(i)

  40 P1

• – 49 (ii)
• _{* * * * * * *} K2 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

  4 1555 or

  30 Note: 1. Allow *midpoint for K1 311

  5 N1 or 51 or 51.83

  6

  6 Note: Correct answer from incomplete working, award Kk2 (c) Axes drawn in correct direction and uniform scale for P1 24 . 5 £ x £ 84 . 5 and 0 £ y £ 8.

  K2

• 7 points correctly plotted Note :

  4

• 5 or *6 points correctly plotted or bar passes through using at least 6 correct mid-point, award K1.

  N1 Correct bar passes all 7 correct points for using given scales 24 . 5 £ x £ 84 .

  5

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  Graf untuk Soalan14 Graph for Question 14

  frequency

  8

  7

  6

  5

  4

  3

  2

  1 24.5 34.5 44.5 54.5 64.5 74.5 84.5 Mark

  Question Solution and Mark Scheme Marks

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 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.

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  Question Solution and Mark Scheme Marks 15 (a)

  M N 3.5 cm

  P 7 cm L 3.5 cm

  J K 8 cm

  Correct shape with rectangles JKMN , JKLP and PLMN K1 All solid lines.

  K1 dep K1

  JK > KM > KL=LM

  3 Measurement correct to ± 0 2 N1 dep × cm (one way) and all angles

  K1K1 at vertices of rectangles = 90° ± 1°

  15 (b)(i) P 8 cm L

  3 cm J K

S

  3 cm 2 cm F E Q R

  4 cm 2 cm

  Correct shape with rectangle PLKJ, SQ perpendicular to EQ K1 All solid lines

  K1

  4 PL > LF > EQ > LK = KF > RF = QR = QS dep K1

  Measurement correct to ± 0 2 N2 × cm (one way) and all angles at the dep vertices of rectangles = 90° ± 1°

  K1K1

  Question Solution and Mark Scheme Marks

  5 15 (b)(ii)

  L M K 1 cm

  S

  V 2 cm

  F G 7 cm

  Correct shape K1

  All solid lines Note : Ignore *SV

  K1

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

  K1 dep

  FG > LM > MG

  K1K1

  5 N2 dep

  K1K1K1 Measurement correct to ± 0 2 × cm (one way) and all angles at

  12

  the vertices of rectangles = 90° ± 1° _{o o}

  16 (a)

  P2 105 E T)

  ∕∕ 105

  Note : _{o o o} 105 or E T, award P1 θ ∕∕ θ

  (b)(i) K1

  3900

  60 _o N1N1

  35 N // U _o (ii)

  K2 60 cos 35 4669 q ´ ´ =

  20 N1 75 105 ´

• (c) 60 ´ cos

  30 K2K1

  ( )

  600

  15.59 N1 Note :

  cos 45 + (75 105 ) , award K1

• 12

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