add maths trial exam p2 marking scheme 2013 set b.pdf

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  • 7/27/2019 Add Maths Trial Exam P2 Marking Scheme 2013 set B.pdf

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    2

    Nama Pelajar : Tingkatan 5 : .

    3472/2

    Additional

    Mathematics

    Sept 2013

    PROGRAM PENINGKATAN PRESTASI AKADEMIK SPM 2013

    ADDITIONAL MATHEMATICS

    Paper 2

    (SET B)

    .

    MARKING SCHEME

    SULIT 3472/2

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    3

    MARKING SCHEME

    ADDITIONAL MATHEMATICS PAPER 2

    N0. SOLUTION MARKS

    1 3x y

    or 3y x

    2 2(3 ) (3 ) 6y y y 2(3 ) ( 1) 6x x x 22 7 6 0y y 22 5 3 0x x

    (2 9)( 1) 0y y (2 3)( 1) 0x x

    3

    2x and 1x (both)

    3

    2y and 2y (both

    P1K1 Eliminate x/y

    K1 Solve quadratic equation

    N1

    N1

    5

    2

    (a)

    (b)

    2

    2

    [ ( )] [(2 1) 4]

    (2 1) 4

    3, 12 11

    1

    f g x f x

    m x m n

    m n

    n

    ( ) 5

    1( )5

    3, 5

    f x p x y

    x p y

    p q

    K1(find composite function)

    N1

    N1

    K1(find inverse function)

    N1N1

    7

    3

    (a)

    (b)

    y =x

    draw the straight line y =x

    Number of solutions = 4

    P1 cos shape correct.

    P1 Amplitude = 6 [ Maximum = 3

    and Minimum = -3 ]

    P1 2 full cycle in 0 x 2 orP1 reflection the graph

    N1 For equation

    K1 Sketch the straight line

    N1

    6

    2

    3

    -3

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    4

    4

    (a)

    (i)

    (ii)

    (iii)

    (b)

    2( )

    3

    4 2

    OQ OC CR

    x y

    1( )

    2

    3 2

    BQ BO OC

    x y

    5

    BR BQ QR

    x y

    .

    BR hOC

    cannot find h

    not parallel

    K1

    N1

    K1

    N1

    K1

    N1

    K1 find h

    N1

    8

    5

    (a)

    (b)

    i)6

    10

    60

    x

    x

    ii)

    22 2

    2

    7 610

    850

    x

    x

    new mean 3(6) 5 23

    new standard deviation 3(7) 21

    P1

    K1

    N1

    K1 N1

    K1 N1

    7

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    5

    6

    (a)

    (b)

    2 2 21 1 1, , ,...2 8 32

    p p p

    2 2

    2 2

    1 18 321 1

    2 8

    p p

    p p

    1

    4r

    (i)

    11 253200( )

    4 128

    8

    n

    n

    (ii)

    3200

    114

    24266

    3

    S

    K1

    K1

    N1

    K1K1

    N1

    K1

    N1

    8

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    6

    7

    (a)

    (b)

    (c)

    (i)

    (ii)

    (iii)

    x 1 2 3 4 5 6

    10log y 0.26 0.41 0.53 0.66 0.80 0.94

    10 10 10log log logy x h k h

    10log h = *gradient

    h = 1.37

    10logk h = *y-intercept

    = 0.88

    y = 3.98

    N1 6 correct

    values of log y

    K1 Plot10

    log y vs

    x.

    Correct axes &

    uniform scale

    N1 6 points plotted

    correctly

    N1 Line of best-fit

    P1

    K1

    N1

    K1

    N1

    N1

    10

    10log y

    0.12

    0x

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    7

    N0. SOLUTION MARKS

    8

    (a)

    (b)

    (c)

    21(6) 26.28

    2

    = 1.46 rad

    6(1.46)AD

    S = 8.76 cm

    Perimeter = 8.76 + 5 + 5 + 8

    = 26.76 cm

    Area of triangle = 17.89 cm2

    Area of rectangle = 40

    Area of the shaded region = 17.89 + 4026.28= 31.61 cm

    2

    K1

    N1

    K1 Use s r

    N1

    K1

    N1

    K1

    K1

    K1

    N1

    10

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    8

    N0. SOLUTION MARKS

    9(a)

    (b)

    c)

    2

    216

    xx

    dx

    dy

    At turning point, (p, 4) ,0

    dx

    dy

    02

    162

    p

    p

    16p3

    = 2

    8

    13 p ,2

    1p

    y = 16x 2x-2 dx

    cx

    x

    2

    2

    16 2

    = cx

    x 2

    8 2

    At point A (2

    1, 4 ) ,

    4 = 8( )2

    + 4 + c

    c = -2

    The equation of the curve is , 22

    82

    xxy

    32

    24

    16xdx

    yd

    At point A(2

    1, 4) , 048)8(416

    2

    2

    dx

    yd

    Thus,

    4,

    2

    1A is a minimum point

    P1

    K1

    N1

    K1 use

    dxyy )( 12

    K1 integrate

    correctly

    K1

    N1

    K1

    K1

    N1

    10

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    9

    N0. SOLUTION MARKS

    10

    (a)

    (b)

    (c)

    (d)

    1

    68

    yx

    Thus,A( 0,6) and B( 8, 0)

    LetD (x,y)

    CD : DE= 1:3

    x =

    84

    32

    4

    10321

    y =

    74

    28

    4

    10321

    ThusD 7,8

    4

    3

    8

    6

    ABm

    Thus ,

    3

    4

    CEm

    y - 6 = 03

    4x

    y = 63

    4

    x

    Area ofAOB=

    6706

    0880

    2

    1

    28562

    1

    N1N1

    K1

    K1

    N1

    K1

    N1

    K1 (use area formula)

    K1N1

    10

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    10

    N0. SOLUTION MARKS

    11

    (a)

    (i)

    (ii)

    (b)

    (i)

    (ii)

    X = Students have a laptop

    p =10

    7= 0.7 , n = 5

    q = 3.010

    3

    10

    71

    P(X= 5) = 5055 7.03.07.05 c

    = 0.1681

    P( X< 3) = P( X =0) + P( X = 1) + P( X= 2)

    = 32

    2

    41

    1

    50

    0 3.07.053.07.053.07.05 ccc

    = 0.1631

    X =weights of pencil box

    X 20,250N

    P(X>265) =

    20

    250265ZP

    =P(Z > 0.75)

    = 0.2266

    P(X < w ) = 0.3

    P(Z > 20

    250w= 0.3

    From table,

    20

    250w= -0.524

    W= 239.52 g

    K1 Use P (X=r ) =

    rnr

    r

    n qpC

    N1

    K1

    K1 Use P (X=r ) =

    rnr

    r

    n qpC

    N1

    K1 Use Z =

    X

    N1

    K1

    K1

    N1

    10

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    11

    N0. SOLUTION MARKS

    12 (a)

    (b)

    (c)

    (d)

    vinitial= 5 m s-1

    v= t26t +5

    dt

    dva

    =2t -6

    a = 0

    t = 3

    v = (3)2-6(3) + 5

    = -4 ms-1

    v < 0

    t26t+5 < 0(t-1) (t-5) < 0

    The range of values of t

    1 < t < 5

    5

    1

    23

    1

    0

    23

    5

    1

    2

    1

    0

    2

    1

    0

    5

    1

    533

    533

    )56()56(

    ttt

    ttt

    dtttdttt

    vdtvdt

    )1(5)1(33

    1)5(5)5(3

    3

    50)1(5)1(3

    3

    12

    3

    2

    3

    2

    3

    =13

    N1

    K1

    K1

    N1

    K1

    N1

    K1 for

    1

    0

    vdt

    K1 (for

    Integration;

    either one)

    K1 (for use and

    summation)

    N1

    10

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    12

    N0. SOLUTION MARKS

    13 (a)

    i)

    ii)

    iii)

    (b)

    i)

    (ii)

    oSU 65cos)7)(8(2)7()8( 222

    SU = 8.10 cm

    Area of triangle SVU= O65sin)8)(7(2

    1

    = 25.38 cm2

    SUTo

    sin

    6

    75sin

    1.8

    Sin SUT= 0.7155

    SUT=45.680

    TSU=180O-75

    O-45.68

    O=59.32

    O

    STS=61.36O

    oo SinSin

    SS

    32.59

    6

    36.61

    '

    = 6.123cm

    K1

    N1

    K1

    N1

    K1

    N1

    P1

    P1

    (for 59.32o

    OR

    61.36o

    K1

    N1

    10

    T S

    S

    U

    6 cm

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    13

    N0. SOLUTION MARKS

    14

    (a)

    (b)

    (c)

    i)

    ii)

    x + y 5

    10x +20y 150 or x + 2y 15

    x 2y or y 2

    x

    At least one straight line is drawn correctly from inequalities involvingx and y.

    All the three straight lines are drawn correctly. Region is correctly shaded.

    3 kg

    Minimum point (5,0)

    Minimum cost = 10(5) + 20(0)

    = RM 50

    N1

    N1

    N1

    K1

    N1

    N1

    N1

    N1

    K1

    N1

    10

    R

    7.5

    5

    5 15

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    14

    N0. SOLUTION MARKS

    15 (a)

    i)

    ii)

    (b) (i)

    (ii)

    1005.162

    20.511 xp

    P11 = RM 3.20

    100140

    15012

    13 xI

    = 107.14

    30+10+20+40 ( can be seen )

    14540201030

    )40135()20()10150()305.162(

    XxXXX

    X = 136.25

    The usage of 136.25

    56100

    25.13613 xp

    = RM 76.3

    K1

    N1

    K1

    N1

    P1

    K1

    N1

    P1

    K1

    N1

    10

    END OF MARKING SCHEME