SET 2 - trial.spmpaper.me Set 2/2019 Selan… · 1 peraturan pemarkahan peperiksaan percubaan spm...

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1 PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019 PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR (PINTAS) MATEMATIK TAMBAHAN KERTAS 2 NO SOLUTIONS MARKS 1 (a) dx x x V 4 0 2 2 ) 4 ( π[ 16x 3 3 -2x 4 + x 5 5 ] 4 0 (Integrate + limit) [ 16(4) 3 3 − 2(4) 4 + (4) 5 5 ]−0 15 512 π K1 K1K1 N1 4 6 (b) 32 3 3 64 K1 N1 2 2 (a) 8 1 3 n=2 Accept answer without working for K1N1 K1 N1 2 8 b) 3 r (9 s ) 2 = 81 or (3 r ) 2 9 s = 1 3 or equivalent 3 r (3) 4s =3 4 or 3 2r-2s =3 -1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation 5 2 r 10 9 s K1 K1 K1 K1 N1 N1 6 SET 2

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Transcript of SET 2 - trial.spmpaper.me Set 2/2019 Selan… · 1 peraturan pemarkahan peperiksaan percubaan spm...

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    PERATURAN PEMARKAHAN PEPERIKSAAN PERCUBAAN SPM 2019

    PROGRAM INTERVENSI TERBILANG AKADEMIK SELANGOR

    (PINTAS)

    MATEMATIK TAMBAHAN KERTAS 2

    NO SOLUTIONS MARKS

    1 (a) dxxxV

    4

    0

    22 )4(

    π [16x3

    3-2x4 +

    x5

    5] 40 (Integrate + limit)

    𝜋 [16(4)3

    3− 2(4)4 +

    (4)5

    5] − 0

    15

    512π

    K1

    K1K1

    N1

    4

    6

    (b)

    2 ×32

    3

    3

    64

    K1

    N1

    2

    2 (a)

    81

    3

    n = 2 Accept answer without working for K1N1

    K1

    N1

    2

    8

    b)

    3r(9s)2 = 81 or (3r)2

    9s=

    1

    3 or equivalent

    3r(3)4s = 34 or 32r-2s = 3-1 or equivalent r + 4s = 4 or 2r-2s = -1 or equivalent Accept any method of solving simultaneous equation

    5

    2r

    10

    9s

    K1

    K1

    K1

    K1

    N1

    N1

    6

    SET 2

    cikguJEPcikgujep

  • 2

    NO SOLUTIONS MARKS

    3

    a) AD⃗⃗⃗⃗ ⃗ = AB⃗⃗⃗⃗ ⃗ + BC⃗⃗⃗⃗ ⃗ + CD⃗⃗⃗⃗ ⃗

    AD⃗⃗⃗⃗ ⃗ = (q – p)a + qb

    q – p = -5 or 2

    3

    pq

    p = 13 and q = 8

    K1

    K1

    N1

    3

    6

    (b) 1

    2 x DC x 4 = 52

    p |a| = 26

    2

    K1

    K1

    N1

    3

    4

    5x + y + (x + y) = 24 or 222 )()()5( yxyx

    y = 12 – 3x

    0)312(12 2 xxx

    045 2 xx 0)45( xx or equivalent

    5

    4x or y =

    5

    48

    48 4

    5 5

    192

    25 or equivalent

    P1

    K1

    K1

    K1

    N1

    K1

    N1

    7

    7

  • 3

    NO SOLUTIONS MARKS

    5

    (4.6 × 3) + (5.1 × 5) + (5.6 × 5) + (6.1 × 10) + (6.6 × 12) + (7.1 × 5) or

    (4.6 × 2) + (5.1 × 7) + (5.6 × 5) + (6.1 × 12) + (6.6 × 8) + (7.1 × 6)

    (4.6×3)+(5.1×5)+(5.6×5)+(6.1×10)+(6.6×12)+(7.1×5)

    40 or

    (4.6×2)+(5.1×7)+(5.6×5)+(6.1×12)+(6.6×8)+(7.1×6)

    40

    6.075 or 6.0375

    (3 × 4.62) + (5 × 5.12) + (5 × 5.62) + (10 × 6.12) + (12 × 6.62) + (7.12 × 5) or

    (2 × 4.62) + (7 × 5.12) + (5 × 5.62) + (12 × 6.12) + (8 × 6.62) + (7.12 × 6)

    √(3×4.62)+(5×5.12)+(5×5.62)+(10×6.12)+(12×6.62)+(7.12×5)

    40-(6.075)2 or

    √(2×4.62)+(7×5.12)+(5×5.62)+(12×6.12)+(8×6.62)+(7.12×6)

    40-(6.0375)2

    0.7242 or 0.7175

    ORCHARD B

    K1

    K1

    N1

    K1

    K1

    N1

    N1

    7

    7

    6

    (a) 1cot

    1cot2

    2

    2

    2

    2

    2

    cos1

    sin

    cos1

    sin

    2 2

    2

    2 2

    2

    cos sin

    sin

    cos sin

    sin

    K1

    K1

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    NO SOLUTIONS MARKS

    22 sincos

    2cos

    N1 3

    6

    (b)

    ekcos

    12cos

    sin

    1

    12cos

    sinsin21 2

    01sinsin2 2

    0)1)(sin1sin2(

    1sin,2

    1sin

    270,150,30

    K1

    K1

    N1

    3

    7

    (a) 24012060 oror

    3

    4189.4 or

    K1

    N1

    2

    10

    (b) 12OP or 60cos = OP

    6

    108PS or equivalent

    180

    142.312012PQRarcofLength or 25.14

    92.45

    108214.25

    K1

    K1

    K1

    K1

    N1

    5

    (c )

    120sin)12(2

    1

    180

    142.3120)12(

    2

    1 22 or

    or

    12 128 6

    2

    45.88

    35.628.150

    K1

    K1

    N1

    3

  • 5

    NO SOLUTIONS MARKS

    8 (a)

    2

    95,

    2

    80

    ( 4, 7)

    K1

    N1

    2

    10

    (b) )0(1)4(9)8(5)0(0)0(4)1(8)9(0)5(02

    1

    34

    K1

    N1

    2

    (c) m =2

    1

    midpoint QR ( 6,5)

    y – 5 = 2

    1 ( x – 6 ) or equivalent

    y = 2

    1 x +8

    K1

    K1

    K1

    N1

    4

    (d) 2

    1

    63

    14

    5

    h or 8

    3

    14

    2

    1

    h

    h = 3

    17

    K1

    N1

    2

    9

    (a) p= 0.75 or q = 0.25 or

    rnr

    rC)25.0()75.0(15

    (i) )12( XP

    )15()14()13()12( XPXPXPXP

    0.4 613

    (ii) 75.015

    11.25

    P1

    K1

    N1

    K1

    N1

    5

  • 6

    NO SOLUTIONS MARKS

    (b)(i) 370 8

    8

    370380

    8

    370355ZP or )25.1875.1( ZP

    0.8641 0.86396

    (ii) 0.8641 x 3130

    2704

    P1

    K1

    N1

    K1

    N1

    5

    10

    10 (a) All values of

    x

    1 and

    y

    1 correct.

    x

    1

    0.67 0.40 0.25 0.20 0.17 0.13 0.10

    y

    1

    0.22 0.91 1.30 1.43 1.52 1.56 1.67

    N1, N1

    2

    10

    (b) Refer graph paper

    Plot log10 y against x with correct axes, uniform scales and

    at least one point.

    6 points plotted correctly

    Line of best fit

    K1

    N1

    N1

    3

    (c) hxh

    k

    y

    1)

    1(

    1

    (i) h

    192.1

    5208.0h

    (ii) 5439.25208.0

    k

    3248.1k

    P1

    K1

    N1

    K1

    N1

    5

  • 7

    NO SOLUTIONS MARKS

    11

    (a)r = x – 5 or h = x – 2

    V = π (x – 5)2 (x – 2) or equivalent

    P1

    N1

    3

    10

    (b)

    2

    5 (1) ( 2)(2)( 5)(1)

    3 ( 5)( 3)

    dvx x x

    dx

    x x

    3 ( 5)( 3) 0

    3 5

    x x

    or

    When x = 3 , V = 4π

    K1

    N1

    K1

    N1

    N1

    4

    (c) 0.01x

    9π (0.01)

    0.09π

    P1

    K1

    N1

    3

    12 (a) v = 3 N1 1

    10

    (b) (3t + 1)(t – 3) = 0

    t = 3

    K1

    N1

    2

    (c) sp = 3t + 4t2 – t3

    t = 3 , sp = 18 or t = 4 sp = 12

    24

    K1

    K1

    N1

    3

    (d) 8 – 6t = 0 or t = 4

    3

    Qv = 3t2 – 6t – 6

    Qv = 6

    3

    46

    3

    43

    2

    Qv =` 3

    26

    K1

    K1

    K1

    N1

    4

  • 8

    NO SOLUTIONS MARKS

    13 (a) 80100

    80.2

    x or 130100

    60.2

    y or z100

    5

    85.5

    117,2,24.2 zyx

    Note : All three values correct N2

    Any two values correct N1

    One or none values correct N0

    K1

    N2

    3

    10

    (b) )26117()45130()14120()1580(

    100

    )26117()45130()14120()1580(

    117.72

    K1

    K1

    N1

    3

    (c) 72.11710035

    x

    RM 41. 20

    K1

    N1

    2

    (d) 100

    11472.117

    134.2

    K1

    N1

    2

  • 9

    NO SOLUTIONS MARKS

    14 (a) (i) 24)120)(sin)(8(

    2

    1BC

    6.928

    (ii) AC2 = 82 + 6.9282 – 2(8)(6.928) kos 120

    12.94

    (iii) 94.12

    120sin

    928.6

    sin

    A

    27.62

    K1

    N1

    K1

    N1

    K1

    N1

    6

    10

    (b)

    A’

    C ‘

    B ‘

    180 – 27.62 – 60 = 92.38

    1(12.94)(6.928)sin 92.38

    2

    45.12

    P1

    K1

    K1

    N1

    4

  • 10

    NO SOLUTIONS MARKS

    15 (a) x + y ≥ 10

    10x + 8 y ≤ 420

    𝑥 ≤ 2𝑦

    N1

    N1

    N1

    3

    10

    (b)Refer to graph paper

    One *straight line drawn correctly

    All * straight line drawn correctly

    Correct region

    K1

    K1

    N1

    3

    (c) 2113 y

    420 – [20(10) + 10(8)]

    Seen (20 , 10 )

    = 140

    K1

    N1

    N1

    N1

    4

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