TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer

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  • 8/4/2019 TRIAL ADDMATE SPM 2011 Sarawak Zon a Paper 2 Answer

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    3472/2

    Matematik

    TambahanKertas 22 jam

    Sept 2011

    SEKOLAH-SEKOLAH MENENGAH ZON A KUCHING

    PEPERIKSAAN PERCUBAANSIJIL PELAJARAN MALAYSIA 2011

    MATEMATIK TAMBAHAN

    Kertas 2

    Dua jam tiga puluh minit

    JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

    Skema Pemarkahan ini mengandungi 13 halaman bercetak

    MARKING SCHEME

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    2

    ADDITIONAL MATHEMATICS MARKING SCHEME

    TRIAL SPM exam Zon A Kuching 2011 PAPER 2

    QUESTION

    NO.SOLUTION MARKS

    1

    ( )

    ( )( )

    2

    2 1

    (2 1) 2 1 3

    2 1 2 0

    x y

    y y y

    y y

    = +

    + + =

    + =

    1, 2

    2

    @

    2, 3

    y y

    x x

    = =

    = =

    5

    2

    (a)

    (b)

    (c)

    2 2 2

    2 2

    2

    ( ) 3[ 2( 2) ( 2) ( 2) ] 7

    5or 3[( 2) 4] 7 or 3[( 2) ]

    3

    3( 2) 5

    f x x x

    x x

    x

    = + + +

    +

    =

    (2, 5)min

    Shape

    (2, 5) and (0, 7)

    (1, 22) and (4, 7)

    2

    1

    3

    Solve the quadratic

    equation by using the

    factorization @ quadratic

    formula @ completing the

    square must be shown

    Eliminate orx y

    Note :

    OW 1 if the working of solving

    quadratic equation is not shown.

    5

    6

    P1

    K1

    K1

    N1

    N1

    K1

    N1

    N1

    N1

    N1

    N1

    Ox

    (2, 5)

    (1, 22)

    (4, 7)7

    y

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    3

    QUESTION

    NO.SOLUTION MARKS

    3

    (a)

    (b)

    (c)

    T10 = 100 + 9(4) or 64

    Area = 512

    2

    100 + (n 1)(4) = 4 or100

    4@ 100 + (n 1)(4) > 0

    n = 25

    2

    [ ]8640

    2(100) ( 1)( 4) or 10802 8

    nn+ =

    (n 36)(n 15) = 0

    n = 15

    3

    4(a)

    (b)xy

    5

    4=

    Number of solutions = 3

    4

    3

    Shape of sine curve

    Modulus

    Amplitude or period

    Translation

    P1

    P1

    P1

    P1

    y

    x0

    -1

    1

    3

    1 2siny x=

    2

    2

    2

    3

    xy5

    4=

    K1

    N1

    K1

    7N1

    K1

    N1

    K1

    K1

    N1

    7

    P1Sketch straight

    line correctly

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    4

    QUESTION

    NO.SOLUTION MARKS

    5

    (a) 43 2 48 4 53 8 58 11 63 9 68 4 73 2 2325fx = + + + + + + =

    2325

    40x =

    58.125=

    3

    (b)

    Q3 = 60.5 +3

    (40) 2554

    9

    = 63.28

    OR1

    1(40) 6

    450.5 58

    53

    Q

    = +

    =

    Interquartile range = 63.28 53

    = 10.28

    4

    6

    (a) BCO = 60

    = 1.047 rad

    AOB = 0.6982 rad

    3

    (b) 21 (10) (1.047)2

    or 52.35

    21 (10) (0.6982)

    2

    or 34.91

    2

    (c)Area of segmentBC= 52.35 2

    1(10) sin1.047

    2

    r= 9.237

    Area of the shaded region = 25.673

    3

    N1

    7

    K1

    N1

    P1

    Lower

    boundary OR

    3(40) 25

    549

    N1

    7

    K1

    K1

    N1

    K1

    N1

    K1

    K1

    N1

    K1

    N1

    N1

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    5

    (0.78, 1.94)

    (0, 0.30)

    x10log

    .0

    .2

    .8

    0.2 0.3 0.4 0.5

    .2

    0

    og10y Q7

    x10log 0.30 0.48 0.60 0.78 0.85 0.95

    y10log 0.93 1.30 1.57 1.94 2.07 2.31

    Correct both axes (Uniform scale) K1

    All points are plotted correctly N1

    Line of best fit N1

    0.1 0.6 0.7 0.8

    N1

    N1

    .4

    .6

    .8

    .0

    .2

    .4

    .6

    0.9 1.0

    (a) Each set of values correct (log10y must be at least

    2 decimal places) N1, N1

    log 10y = nlog 10x + log 10 (p + 1) K1

    where Y= log 10y, X= log 10x,

    m = n and c = log 10 (p + 1)

    (c) (i)X= log 10 5.6 = 0.748

    Y= 1.88 = log 10yy = 75.86 N1n = gradient

    n =1.94 0.30

    0.78 0

    = 2.103 N1

    log 10 (p + 1) = Y-interceptlog 10 (p + 1) = 0.30 K1

    p = 0.9953 N1

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    6

    QUESTION

    NO.SOLUTION MARKS

    8

    (a) (i)

    (ii)

    (b) (i)

    (ii)

    2,0

    2

    82

    12

    18

    2

    =>

    =

    =

    =

    aa

    a

    a

    aa

    3

    ++=

    +

    2)2(5,

    206

    23,

    24 yx

    D(2, 6)

    @

    Solving the equationsy = 4x 2 and2

    13

    4

    1+= xy

    D(2, 6)

    2

    0214

    5)2()0(

    22

    22

    =++

    =++

    yyx

    yx

    Get equation ofBC,y = 4x 19

    ( )

    0544

    )264)(17(4136

    4

    026413617

    021)194(4)194(

    2

    2

    2

    22

    >=

    =

    =+

    =++

    acb

    xx

    xxx

    The locus intersects the lineBC.

    2

    3

    K1

    K1

    K1

    N1

    N1

    N1

    K1

    N1

    K1

    K1

    N1

    K1

    10

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    7

    QUESTION

    NO.SOLUTION MARKS

    9(a) 1

    2OP x y= +

    1

    (b)

    (i)

    (ii)

    OQ x y = +

    ( )1

    2

    11

    2

    OQ OL LQ

    x LP

    x LM MP

    x y x

    x y

    = +

    = +

    = + +

    = +

    = +

    4

    x : = 1 1

    2

    y : =

    = 1 1

    2

    = =2

    3

    3

    (c)Area of triangle OLM=

    3

    2 24 = 36

    Therefore area of parallelogram OLMN= 72

    2

    P1

    10

    K1

    K1

    N1

    K1

    K1

    N1

    K1

    N1

    N1

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    8

    QUESTION

    NO.SOLUTION MARKS

    10(a)

    3 9 3

    r h hr= =

    2

    9

    dV h

    dh

    =

    2

    0.89

    h dh

    dt

    =

    2

    0.89h dh

    dt=

    4

    (b)

    (i)

    (ii)

    h = 1 and k= 1

    Area of the shaded region

    = ( ) ( )1 0

    3 3

    0 1

    y y dy y y dy

    +

    =

    14 2

    04 2

    y y

    +

    0

    1

    24

    24

    yy

    = ( )1 1

    0 04 2

    +

    ( )41 10 0

    4 2

    =2

    1

    Note: OW 1 once only for correct answer without showing theprocess of intergration.

    6

    N1

    K1

    K1

    K1

    K1

    N1

    N1

    K1

    K1

    10

    N1

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    9

    QUESTION

    NO.SOLUTION MARKS

    11(a)

    (i)

    (ii)

    (b)

    (i)

    (ii)

    0 550( 0) (0.2) (0.8)

    0.3277

    P X C= =

    =

    ( 2)P X< = 0 5 1 45 50 1(0.2) (0.8) (0.2) (0.8)C C+

    = 0.7373

    5

    ( 0.811)P Z> @R(0.811)

    = 0.7913 @ 0.79132

    P(0.811

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    10

    QUESTION

    NO.SOLUTION MARKS

    12

    (a) Initial velocity 8=Av 1

    (b)

    4

    12

    42

    55

    2

    5

    2

    5

    052

    2

    =

    +

    =

    =

    =+=

    Bv

    t

    tdt

    dv

    3

    (c) ( 2)( 4) 0

    ( 1)( 4) 0

    4

    A

    B

    v t t

    v t t

    t

    = =

    = =

    =

    3

    (d)

    3

    26

    )2(8)2(33

    )2(

    833

    83

    23

    23

    2

    =

    ++=

    +=

    += tt

    t

    dtttsA

    3

    N1

    K1

    K1

    K1

    N1

    K1

    K1

    N1

    10

    K1

    N1

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    11

    QUESTION

    NO.SOLUTION MARKS

    13

    (a)

    (b)(i)

    (ii)

    A : 12510060

    08 =P

    08 RM75P =

    2

    (125 4) (120 ) (80 5) 150( 3)120

    12 2

    n n

    n

    + + + +=

    +

    1440 240 1350 270n n+ = +

    n = 3

    06

    RM30100 120

    P =

    RM25=

    3

    (c)

    18

    )6150()580()3138()4125(

    138)15.0120(120

    06/09

    +++=

    =+

    I

    = 123

    3

    K1

    K1

    N1

    K1

    N1

    K1

    N1

    10

    N1

    K1

    K1

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    12

    QUESTION

    NO.SOLUTION MARKS

    14

    (a)

    (b)

    (c)

    18 28

    sin38 sin QSR=

    QSR = 180 7316

    = 10644

    3

    QRS = 3516

    2 2 228 18 2(28)(18) cos 35 16

    @ Sine Rule

    16.88

    PS

    QS

    = +

    =

    3

    2 2 226 11 16.88 2 11 16.88 cos PQS= +

    PQS = 13639

    Area of triangle PQR = 12

    1128sin 17639

    = 8.999

    2

    10

    K1

    N1

    K1

    N1

    K1

    N1

    K1

    N1

    K1

    N1

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    13

    Answer for question 15

    2 4 6 8 10 12 140 16

    2

    4

    14

    12

    10

    16

    8

    6 (15, 6)

    R

    (a) I. 4832 + yx

    II. 4885 + yx

    III. xy 3

    (b) Refer to the graph,

    1 or 2 graph(s) correct3 graphs correct

    Correct area

    (c) ii) max point (15, 6)

    k= RM(500x + 300y)

    Maximum Profit = RM 500(15) + RM 300(6)

    = RM 7500

    (ii) 8 units

    10

    N1

    N1

    N1

    N1

    N1

    N1

    K1

    N1

    K1

    N1

    x

    y