SPM AddMath2 Ans (Kedah)

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  • 8/14/2019 SPM AddMath2 Ans (Kedah)

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    Question Solution and marking schemeSub-

    mark

    Full

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    1. 3

    2

    yx

    += or

    2y x= 3

    2

    23 3 102 2

    y yy y

    + + + =

    0

    0

    or ( ) ( )22 2 3 2 3 1 x x x x + =

    2 2 29 6 6 2 4 40 0 y y y y y+ + + =

    or

    2 2 22 3 4 12 9 10 x x x x x + + + =

    y = 3.215 , 3.215

    or

    x = 3.107 ,

    0.107

    x = 3.107 / 3.108 , 0.107 / 0.108

    or

    y = 3.214 / 3.215 , 3.214 / 3.215

    Answer must correct to 3 decimal places. 5

    Make x or y asthe subject

    P1

    Eliminate

    x or y

    Solve quadratic

    equation23 31 0y =

    2

    31

    3y =

    or23 9 1 0x x =

    or

    using formula

    or

    completing the

    square

    N1

    K1

    K1

    N1

    j2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    2(a)

    (b)

    16 ,8 , 4 ,......

    16a = 1

    2r=

    7

    7

    116 1

    2

    11

    2

    S

    =

    =3

    314

    or 31.75

    64 ,16 , 4 ,......

    64a = 1

    4r=

    64

    11

    4

    S

    =

    = 1853

    or 85.33

    3

    3 6

    3(a) 2( ) f x x px= + + q

    =

    2 2

    2

    2 2

    p p x px q

    + + +

    =

    2 2

    2 4

    p p

    x q

    + +

    32

    p = p =6

    365

    4q + = q = 4

    N1

    K1

    User

    raS

    n

    n

    =

    1

    )1(K1

    N1

    K1

    N1

    P1

    P1

    N1

    1

    aS

    r =

    Use

    2

    2 )2b

    (

    b

    use x + bx = ( x + ) or

    32

    b

    a =use axis of symmetry

    j2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    Alternative solution

    ( )

    223 5 x px q x+ + =

    =2 6 9 5x x +

    =2

    6 4x x +

    6p =

    q = 43

    3(b)2 6 4 31x x + 0

    0

    0

    2 6 27x x

    ( 9)( 3)x x +

    3 9x 2 5

    4(a)

    tanx +cos

    1 sin

    x

    x+

    =sin

    cos

    x

    x+

    cos

    1 sin

    x

    x+

    =2sin (1 sin ) cos

    cos (1 sin )

    x x x

    x x

    + +

    +

    =2 2

    sin sin coscos (1 sin ) x x x

    x x+ +

    +

    =sin 1

    cos (1 sin )

    x

    x x

    +

    +

    =1

    cosx

    = secx3

    K1

    N1

    Use ( ) 31 0f x andfactorization

    K1

    K1

    Usesin

    tancos

    xx

    x=

    Use identity2 2

    sin cos 1x x+ =

    N1

    K1

    N1

    Comparing coefficient of x

    or constant term

    N1

    j2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    4(b)(i)

    Negative sine shape correct.

    Amplitude = 3 [ Maximum = 3 and Minimum = 3 ]

    Two full cycle in 0 x 2

    3

    4(b)(ii)

    53sin 2 2

    xx

    =

    or

    52

    xy

    =

    Draw the straight line

    5

    2

    x

    y =

    Number of solutions is 3 .

    3 9

    K1

    N1

    N1

    P1

    P1

    P1

    y

    3

    3

    2

    2

    3

    2 x

    52

    xy =

    O

    y 3sin2x=

    j2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    5 (a)

    (b)

    1220

    x=

    240x =

    The mean

    240 5 + 8 + 10 + 11 + 14 288

    25 25X

    += =

    = 11.52

    2

    212 3

    20

    x =

    2 3060x =

    The standard deviation

    ( )2 2 2 2 2

    23060 5 8 10 11 1411.52

    25

    + + + + +=

    = ( )23566

    11.5225

    = 9.9296

    = 3.151

    7 7

    K1

    xx N=

    Use

    N1

    N1

    K1

    Use formula

    22x

    xN

    =

    N1

    N1

    For the new

    K1

    Xand2

    x

    j2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    6(a)

    (b)

    (i) PR PO OR= +uuur uuur uuur

    = 6 15a b +% %

    (ii) OQ OP PQ= +uuur uuur uuur

    =3

    65

    a O+uuur

    %R

    r

    5%

    5%

    k

    k

    = 6 9a b+% %

    (i) OS hOQ=uuur uuu

    = (6 9 )h a b+% %

    (ii) OS OP PS= +uuur uuur uuur

    = 6a k PR+uuur

    %

    = ( )6 6 1a k a b+ +% %

    =(6 9 )h a b+% %

    ( )6 6 1a k a b+ +% %

    6 6 6h = 9 15h k=

    1h = 5

    3h k=

    5 13

    k k=

    3

    8k=

    5 3

    3 8h

    =

    =

    5

    8

    3

    5 8

    K1

    N1Use

    PR PO OR= +uuur uuur uuur

    or

    OQ OP PQ= +uuur r uuuruuu

    N1

    N1

    N1

    N1

    N1

    K1 Equate coefficient

    of a or b % %and

    Eliminate

    h or k

    j2k

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    Qn. Solution and Marking SchemeSub-

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    7(a)

    (b)(i)

    (ii)

    kn

    xn

    y 4log1

    log1

    log +=

    log x 0.18 0.30 0.40 0.60 0.74

    log y 0.48 0.54 0.59 0.69 0.76

    6

    410

    N1

    P1

    All points plotted

    correctly

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    Correct axes and

    scale

    Line of best-fit

    N1

    K1

    intercept

    = kn

    4log1

    n = 2 k= 1.51

    K1

    N1

    N1

    K1

    N1

    N1

    gradient

    =n

    1

    2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    0.6

    0.7

    Graph of log10y against log10x

    0.8

    log 10y

    0.5

    0.4

    0.1

    0.2

    0.3

    0.9

    0

    0.39

    0.4 0.5 0.6 0.7

    log 10 x

    0.1 0.2 0.8 0.90.3

    2k

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    8.

    (a)

    (b)

    (c)

    Solving simultaneous equation

    P( 2, 2) Q(4, 8)

    Use dxyy )( 12

    + dxx

    x )2

    4(2

    Integrate dxyy )( 12

    64

    2

    32 xx

    x+

    18

    Note : If use area of trapezium and ydx , give the marksaccordingly.

    Integrate dxx

    2

    2

    )2

    (

    =

    20

    5x

    5

    8

    3

    4

    310

    K1

    N1 N1

    K1

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    N1

    K1Use correct

    limit into

    4

    2K1

    64

    2

    32x

    xx

    +

    N1

    K1

    K1

    Use correct

    limit into2

    0

    20

    5

    x

    2k

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    9(a)

    (b)

    (c)

    (d)

    Equation ofAD :

    y 6 = 2 (x 2 )

    y =2x + 10 or equivalent

    y =2x + 10

    and

    x 2y = 0

    D(4, 2)

    p = 3 or C(8, 4)

    Substitute (8, 4) intoy = 3x + q

    q = 20

    OADOABCArea = 4

    Using formula

    0

    0

    6

    2

    2

    4

    0

    0

    2

    1= OAD

    40

    Alternative solution :

    B(10, 10)

    Using formula

    0

    0

    6

    2

    10

    10

    4

    8

    0

    0

    2

    1=OABCArea

    40

    2

    2

    3

    3

    10

    Use m = 2 and find

    equation of straight lineK1

    N1

    Solving

    simultaneous

    equations

    K1

    N1

    K1

    N1

    P1

    N1

    Find area of

    triangle

    K1

    K1

    K1

    P1

    Find area of

    parallelogram

    N1

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    2k

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    10(a)

    (b)

    (c )

    =

    10

    6sin 1 or equivalent

    = 2POQ

    radPOQ 287.1=

    Alternative solution :

    122

    = 102

    + 102

    2(10)(10) cos POQ

    +=

    )10)(10(2

    121010cos

    2221POQ

    radPOQ 287.1=

    Using (2 1.287)

    Major arc PQ

    = 10 ( 2 1.287 )

    = 49.96 cm

    Lsector= 287.1)10(2

    1 2

    Ltriangle = 287.1sin)10(2

    1 2

    = 16.35 cm2

    = 3.9149 cm

    3

    3

    4 10

    K1

    K1

    N1

    N1

    K1 Use cosine

    rule

    N1

    Use formula

    s = rK1

    N1

    K1

    Using formula

    Lsector= 21 r

    2K1

    K1

    K1Lsector - L

    Using formula

    L = absin C 21

    N1

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

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    11

    (a)(i)

    (ii)

    b)(i)

    (ii)

    p =5

    3, p + q = 1

    P( X = 0 )

    =5C0(

    5

    3)0(

    5

    2)

    5

    = 0.01024

    Using P( X 4 )= P( X = 4 ) + P( X = 5 )

    =5C4(

    5

    3)4(

    5

    2)

    1+ (

    5

    3)5

    = 0.337

    P ( 30 X 60 )

    = P (10

    3530 Z 60 3510 )

    Use P( 0.5 Z 2.5 )

    = 1 P( Z 0.5 ) P( Z 2.5 )

    = 1 0.30854 0.00621

    = 0.68525

    Number of pupils = P( X 60 ) 483

    3

    3

    2

    3

    2 10

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    K1

    N1

    K1

    N1

    Use P(X = r) = n Crprqnr,

    p + q = 1

    K1

    N1

    K1

    use

    Z =

    X

    K1

    P1

    N1

    2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

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    12.

    (a)

    (b)

    (c)

    (d)

    Subst. t= 0 intodv

    dt

    a= 15 6t15 ms

    -2

    1 134 4

    ms75 /18ms

    Integrate

    Use s = 0

    15

    2/

    17 / 7.5

    2

    S4 S3

    115

    2

    Note : If use , give the marks accordingly.4

    3vdt

    2

    2

    3

    3

    10

    K1

    N1

    0dv

    dt= K1and subst. t in v = 15t 3t

    2Use

    5[t= ]

    2

    N1

    K1

    N1

    K1

    K1

    K12 315

    2dt t t s v = =

    Subst. t = 3 or t = 4 in

    N1

    2 315

    2s t t=

    2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

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    13.

    (a)

    (b)

    (i)

    (ii)

    (iii)

    Use I = 2007

    2005

    100P

    P

    Value of m : 25, m, 80, 30 or equivalent

    120 25+130m+135 80+139 30

    Use i i

    i

    I WIW

    =

    132.1 =12025+130m+13580+139 30

    135+m

    m =65

    100150.00x

    132.1

    RM 113.55

    /I . ( . x .132 1 132 1 0 3= + )08 05

    171.73

    3

    3

    2

    2

    10

    K1

    N2, 1, 0x = 48.6y = 135z = 80

    K1

    K1

    N1

    K1

    N1

    P1

    N1

    2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

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    14.

    (a)

    (b)

    (c)

    x + y 80 or equivalent

    y 4x or equivalent

    x + 4y 120 or equivalent

    x=30

    (16,64)

    x + 4y = 120

    x + y = 80

    y = 4x

    100

    90

    80

    70

    60

    50

    40

    30

    20

    10

    10080604020 9070503010

    x

    y

    At least one straight line is drawn correctlyfrom inequalities involving x and y

    All the three straight lines are drawn correctly

    Region is correctly shaded

    (i) minimum = 23

    (ii) (16,64)

    Subst. point in the range

    in 20x + 40y

    RM2880

    3

    3

    4 10

    N1

    N1

    N1

    K1

    N1

    N1

    N1

    K1

    N1

    N1

    2k

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelahSULIT

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    15.

    (a)

    (b)

    (c)

    (d)

    120 9.3 6 sin

    2

    BCD=

    45o48 / 45.8

    o

    0.74545 4555Use cosine rule in BCD

    BD2

    = 9.32

    + 62 29.36 cos 4548

    6.685

    Use sine rule in BCDo

    sin CBD sin '

    . .

    = 45 489 3 6 685

    94

    o

    10

    4444

    qqq11111aaaaaaaaaaaaaaaaaa 14s4

    5.555555

    5555 z5555555555555555

    Sum of area:

    20 cm2

    + ABD 555 102

    58.82 cm2

    4

    2

    4 10

    Use area = ab sin c in BCD K1

    N1

    K1

    N1

    K1

    N1

    K1

    K1

    N1

    K1 Obtain ADB by using180o 85o50 BAD or equivalent

    Use area ADB = 6.685 13 sinADB

    2k