3472-2 mt trial spm 2014_skema

Upload: rohaya-morat

Post on 11-Oct-2015

192 views

Category:

Documents


0 download

DESCRIPTION

MATEMATIK TAMBAHAN SKEMA PERCUBAAN SBP KERTAS 2 SBP

TRANSCRIPT

  • 1

    3472/2

    Matematik

    Tambahan

    Kertas 2

    Ogos 2014

    2 jam

    BAHAGIAN PENGURUSAN

    SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN

    KEMENTERIAN PELAJARAN MALAYSIA

    PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2014

    PERCUBAAN SIJIL PELAJARAN MALAYSIA

    ADDITIONAL MATHEMATICS

    Paper 2

    Skema Pemarkahan ini mengandungi 10 halaman bercetak

    MARKING SCHEME

  • 2

    No Solution and Mark Scheme Sub

    Marks

    Total

    Marks

    1

    2

    2

    6 2

    3

    6 2 6 25 4 5

    3 3

    10 7 24 0

    ( 7) ( 7) 4(10)( 24)

    2(10)

    1.94, 1.24

    0.71, 2.83

    yx

    y yy y

    y y

    y

    y

    x

    5

    5

    2(a)

    (b)

    (c)

    2( ) 2( 1) 2 K1

    2 7 6 K1

    6 5 N1

    h x x m

    m k

    m k

    12

    2k m N1, N1

    22 (4 ) 8 0x n x K1

    2

    4 4( 2)( 8) 0n K1

    4, 12n N1

    3

    2

    3

    8

    2

    2

    6 3 P1

    2

    6 3 6 35 4 5 K1

    2 2

    15 53 30 0

    ( 53) ( 53) 4(15)(30)K1

    2(15)

    xy

    x xx x

    x x

    x

    0.71, 2.83 N1

    1.94, 1.24 // 1.25 N1

    x

    y

    OR

  • 3

    3(a)

    (b)

    2 , 2 ( 1), 2 ( 2),......r r r K1

    7

    2(2 ) 6(2 ) 2162

    r K1

    15r

    length of ribbon 30 N1

    30 ,32 ,34 ,...... K1

    2d K1

    48 30 ( 1)(2 )n

    9n N1

    3

    3

    6

    4 (a)

    (b)

    (c)

    1m K1

    8y x N1

    4 5 26

    7

    x or

    4 5 22

    7

    y K1

    11, 3P

    N1

    8AB K1

    2 2

    4 4 8x y K1

    2 2 8 8 24 0x y x y N1

    2

    2

    3

    7

  • 4

    5(a)

    (b)

    y = 2 - x

    y = 3 cos 3

    2x

    2

    - 3

    0

    2

    3

    x

    y

    Shape of Cosine 0 2x P1

    Period 1.5 cycle P1

    Amplitude 3 P1

    2x

    y

    K1

    Draw line 2x

    y

    K1

    Number of solutions = 3 N1

    3

    3

    6

    6(a)

    (b)

    15.5 2 25.5 5 35.5 12 45.5 6 55.5 7 K1

    32

    38.94 N1

    Mean

    2 2 2 2 2

    215.5 2 25.5 5 35.5 12 45.5 6 55.5 7 (38.94) K1K132

    134.86 N1

    Varian

    2

    3

    8

  • 5

    (c)

    3

    332 19

    440.5 10 K1K16

    41.33 N1

    Q

    3

    7

    LAMPIRAN

    8 (a)

    (b)

    (c)

    4 K1dy

    xdx

    1

    4Nm K1

    1

    5 14

    y x

    1 21

    4 4y x N1

    1

    2

    0

    1Area= (4 5)(1) 2 3

    2x dx K1

    13

    0

    9 23

    2 3

    xx

    K1

    9 23 0

    2 3

    K1

    5

    6 N1

    5

    3

    52

    3

    2 2

    3Volume ( )

    2

    3 K1

    4 2

    5 3(5) 3 3(3) K1

    4 2 4 2

    ydy

    y y

    7 N1

    3

    4

    3

    10

  • 6

    9(a)

    (b)

    (c)

    BQ BA AQ

    Or AC AD DC

    K1

    (i) 3 6BQ x y

    N1

    (ii) 3

    52

    AC y x

    N1

    35

    2AP my mx

    K1

    3 3 6AP n x ny

    K1

    Compare and solve

    33 3

    2m n and 5 6m n K1

    3

    4m N1

    5

    8n N1

    3

    8PD BQ QD

    or 1 1

    4 2PD AC BQ

    K1

    9 17

    8 4PD x y

    N1

    3

    5

    2

    10

    10(a)

    (i)

    (ii)

    2

    10(0.75)

    7.5 N1

    10(0.75)(0.25)

    1.875 N1

    10 9 1 10 10 0

    9 10

    ( 9) ( 9) ( 10)

    (0.75) (0.25) (0.75) (0.25) K1 K1

    0.2440 N1

    P X P X P X

    C C

    2

    3

    10

  • 7

    (b)(i)

    (ii)

    1.3 1.5 2.5 1.5(1.3 2.5) K1

    0.8 0.8

    ( 0.25 1.25) K1

    0.4931 N1

    P x P z

    P z

    Total = 160

    0.4931 K1

    324 N1

    3

    2

    11(a)

    (b)

    (c)

    1 5sin8

    AOB

    K1

    0.6752AOB rad N1

    2 28 5 4FC K1

    4 1.571CD K1

    Perimeter 2 28 5 4 4 1.571 5 4 8 K1

    = 25.53 cm N1

    1

    5 6.2452

    BFO K1

    41

    4 0.67522

    Or 41

    4 1.571 0.67522

    K1

    Area 2 21 1 1

    5 6.245 4 0.6752 4 1.571 0.67522 2 2

    K1

    = 17.38 cm2

    N1

    2

    4

    4

    10

  • 8

    12(a)

    (b)

    (c)

    5 2 0 K1

    5

    2

    t

    t

    25 14 K1v t t

    2

    max

    5 55 14

    2 2

    20.25 N1

    v

    2

    2

    5 14 0 K1

    5 14 0

    ( 2)( 7) 0 K1

    7 N1

    t t

    t t

    t t

    t

    2 3514 K1

    2 3

    t ts t

    2 3 2 3

    7 9

    5(7) (7) 5(9) (9)14(7) 14(9) K1

    2 3 2 3s or s

    1 1 1Total distance 106 106 85 K1

    6 6 2

    1=127 N1

    3

    3

    3

    4

    10

  • 9

    13(a)

    2 2 26.5 3.5 2(6.5)(3.5)cos70EC K1

    6.24 cm N1

    2 10

    (b)

    sin sin 70

    3.5 6.24

    BAC K1

    Use 31.81ACD BAC K1

    180 62.74ADC K1

    117.26 N1

    4

    (c)(i)

    D' D C

    A

    'AD C must acute angle N1

    1

    (ii) ' 180 2 62.74D AD K1

    1

    ' 3.7 3.7 sin54.522

    ADD K1

    5.57 cm2 N1

    3

    14

    LAMPIRAN

  • 10

    END OF MARKING SCHEME

    15(a)

    10

    10

    12100 125 K1

    RM 9.60 N1

    Q

    Q

    2

    10

    (b)

    Use 120 P1

    120110 K1

    100

    132 N1

    3

    (c)

    12

    12

    100 121 K190

    RM108.90 N1

    Q

    Q

    2

    (d) 110(8) 136(5) 120 125(4)121 K1

    8 5 4

    2060 120121 K1

    17

    3 N1

    m

    m

    m

    m

    m

    3