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    1

    SPM TRIAL EXAM 2012

    Marking Scheme

    Additional mathematics Paper I

    Number Solution and marking scheme

    Sub

    Marks

    Full

    Marks

    1 (a) {p, r, s}

    (b) {a , b, c, d}(c) Many to One

    1

    11

    3

    2(a)

    2

    7

    B1: 2x 5 = 2 atau f(x) =2

    5+x

    2 2

    3 (a)5(b) 42 +x

    B2: ( ) 7653 =+ xxf B1 : ( ) 53 + xf

    1

    3

    4

    4

    p = 7 , q = 10 ( both )

    B2: p = 7 or q = 10

    B1:33

    7 p= or33

    10 q= or 01073 2 = xx

    or 3(1)2

    +p(1)+q = 0, 3(10/3)2

    +p(10/3) + q = 0

    3 3

    5 h =2 andk= 3

    B2 : h = 2 atau k= 3

    B1 : 1

    3

    =h

    or 922 =++ kh

    3 3

    6

    45 x

    B2: ( ) 0)5(4 + xx or

    B1 : x2 +x 20 0

    3 3

    4-5

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    2

    7

    pr

    12+ or

    pr

    rp +2

    B3:mm 3

    3

    2

    2

    log

    3log2log

    2log +

    B2: log m2 + 2logmmm 3

    2

    3

    2

    2

    log

    3log

    log

    2log+3or

    B1 : log 2 + log32

    or 2logmmlog

    2log3 or or

    mlog

    3log

    4 4

    8 158 += px

    B2: )32(43 += px

    B1: 32 or )32(42 +p

    3 3

    9 a) 3b) 360

    B1: [ ]18

    2(3) (18 1)22

    +

    1

    2

    3

    10 13=a , 4=d (both)

    B2: 13=a or 4=d

    B1: 52 =+ da or 157 =+ da

    3 3

    11 (a) r=x

    2

    (b) 31

    B1:2

    2

    18

    1

    x

    x

    =

    1

    2

    3

    12a) qpx

    x

    y+= 2

    b) 2=p , 8=q B2 : 2=p or 8=q

    B1:42

    04

    =p or q+= )4)(2(0 or 4 =p(2) + q

    1

    3

    4

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    3

    13 2,12

    B2: 110

    )1(3

    4

    73=

    hh or form equation using Pythagoras

    theorem

    B1:4

    73

    hor

    10

    )1(3

    h of find the lengths of AB, BC and AC using

    distance formula

    3 3

    14(a)

    w

    1

    (b) 212 ww B1:

    21 w or cos2w

    1

    2

    3

    15 26.57 ,116.57 , 206.57 , 296.57 B3: 26.57 and 116.57 B2: )2)(tan1tan2( + xx

    B1: 02tan3tan2 2 =+ xx

    4 4

    16(a)13(b)k = 13

    B1:

    +

    ++25

    112 kor jik )25()112( ++++

    1

    2

    3

    17 (a) 4a + 4b(b) ba 42 +

    B1: )44(6 baa ++

    12

    3

    18 8

    B2: 9.42)3()3()3(3.13.1 =+++++ rrrrrr

    B1: r3.1 or )3(3.1 r

    3 3

    19

    5

    12

    B2:

    +0

    12

    )2(32

    2

    B1:

    2

    0

    2 1

    32

    +x

    x

    3 3

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    4

    20 122 + aa

    B2: ))4()4(()( 22 ++ aa

    B1: xx +2

    3 3

    21 k= 8

    B3: (k+1)(k8) = 0

    B2:

    22 2 22 5 2 56

    3 3

    k k+ + + + =

    B1:3

    52 kx

    ++= or 2222 52 kx ++=

    4 4

    22 1 2( , )3 3

    B2 : x =1

    3, y = 2()(32)

    B1 : 12x 4 = 0

    3 3

    23 (a) 1(b)i) 5040

    ii) 288

    B1: 2 3! 4!

    1

    12

    4

    24(a)

    15

    4

    (b) 35

    B1 :3 2 2 1 3 1 2 2

    1 or5 3 5 3 5 3 5 3

    + +

    1

    2

    3

    25(a) 53 (b)

    625

    144

    B1:

    32

    2

    5

    5

    2

    5

    3

    C

    1

    2

    3

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    5

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    1

    SPM TRIAL EXAM 2012

    Marking Scheme

    Additional Mathematics Paper 2SECTION A

    Question Important Steps Marks

    1 y = 3x 4 1

    5x2 4x(3x 4) + (3x 4)2

    or 2x= 9

    21

    8x + 7 = 0

    )2(2

    )7)(2(4)8()8( 2 =x

    1

    x = 2.707, 1.293 1y = 3(2.707) 4 , y = 3(1.293) 4

    = 4.121 = 0.121 1TOTAL 5

    2(a)

    Change base of logarithm :4

    2

    2

    log (1 2 )

    log 4log (1 2 )

    xx

    = or

    equivalent

    1

    Use n logx = logx n : 2 log 2 ( 1 2x ) = log2 ( 1 2x ) 12

    Solve : (2x + 5 ) = ( 1 2x ) 12

    x = 221 , 1

    x = 21

    1(b)

    31 1

    TOTAL 6

    3(a)

    8 1

    (b)Use Tn 1= a + (n 1 ) d : 8 + ( 22 1 ) ( 3 )

    Use Sn ])([ dnan 122 += : 2 [ 2 ( 8) ( 1) (3) ]n n + 1

    Solve : ])()()([ 31822

    + nn = 55 1

    n = 10 1

    (c)T 11328 1

    TOTAL 7

    4

    (a)x = 3

    1

    (b)5

    5

    9105.9

    +=median

    1

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    2

    Median = 10.5 1

    (c) All midpoints are correct. 1

    11

    20

    220

    20

    )22(2)17(4)12(5)7(6)2(3==

    ++++=x 1

    =++++= 3150)22(2)17(4)12(5)7(6)2(3222222fx

    or f(x-x )21

    = 7302

    2

    20

    220

    20

    3150

    = or

    20

    7302 = 1

    = 36.5 1

    TOTAL 8

    5(a)

    y

    xy

    2

    =

    x

    Shape1

    Max/min1

    Oneperiod

    1

    Completefrom 0 to

    2

    1(b)

    Equation xy

    2= 1

    Straight line xy

    2= 1

    2 solutions 1

    TOTAL 7

    6

    (a)

    3

    2

    QSm = 1

    )6(2

    31 = xy 1

    82

    3= xy 1

    (b) Q(0, 8) 1

    [(x 6)2 + (y 1)2 [(6 0)] or 2 + (1+8)2 1]

    x2 12x + 36 +y2 1 2y + 1 = 117

    x2

    +y2

    1 12x 2y 80 = 0TOTAL 7

    22

    32

    3

    -3

    O

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    3

    SECTION B

    7(a)

    x 1 1.5 2 2.5 3 3.5 4log10 0.04y 0.18 0.28 0.40 0.52 0.62 0.76

    1

    (b)Plot log10 1y againstx [ correct axes and uniform scales ]All 7 points plotted correctly 1Line of best fit 1

    (c) (i) 3.72 1

    (ii)22101010 hk xy

    loglog)(log = 1

    Use :log210 hc = 20

    210 .

    log=

    h 1

    h = 2.5 1(iii)

    Use :log210 km = 240

    210 .

    log=

    k 1

    k = 3.0 1TOTAL 10

    8(a)

    = dxxy 2 1

    cx

    y +

    =2

    2 2 1

    92 += xy 1

    (b)3

    2

    0( 9)x dx + 1

    33

    0

    93

    xx

    +

    1

    )10)(10(2

    1

    32

    0( 9)x dx + or 50

    32

    0( 9)x dx +

    or )10)(10(21

    33

    0

    93x x +

    or 50

    33

    0

    93x x +

    1

    = 32 1

    (c) Volume = dyy)9( 1

    92

    0

    92

    yy

    1

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    4

    81

    2or equivalent

    1

    TOTAL 10

    9(a)

    Angle AOB = 6.5/5 1= 1.3 rad. 1

    AnglePOQ = 0.8667 rad. 1(b) MN= 5 sin(0.8667 rad.) = 3.811 cm

    or ON = 5 cos( 0.8667 rad.) = 3.2367 cm

    1

    Length of arcPQ = 6 0.8667 = 5.2002 1

    Perimeter = 3.811 + 5.2002 + 1 + (63.2367) 1

    = 12.77 cm 1(c) Area of sectorOPQ = 62 1 0.8667

    Area of shaded region = 15.60 (3.811)(3.2367) 1

    = 9.433 1TOTAL 10

    10

    (a) (i)RTPRPT +=

    )148(2

    1xyRT += 1

    )148(2

    18 xyyPT ++=

    = yx 47 +

    1

    (ii)RS RP PS= +

    )14(3

    18 xy += 1

    yx 83

    14= 1

    (b) (i) yhxhPM 47 += 1

    (ii)RMPRPM +=

    )83

    14(8 yxky +=

    1

    = ykxk )88(3

    14+ 1

    (c)14

    73

    h k= or hk 488 = 1

    2

    1=h

    1

    4

    3=k

    1

    TOTAL 10

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    11(a)(i)

    37

    7

    10 )55.0()45.0()7( CXP == 1

    = 0.07460 1

    (ii)

    P(X= 0, 1, 3)= 10C0(0.45)0(0.55)10 + 10C1(0.45)1(0.55)9 + 10C2(0.45)2(0.55) 18

    = 0.09956 1

    (b) (i)

    6.0 7.2 8.1 7.2( )

    1.2 1.2P z

    < < OR ( 1 0.75)P z < < 1

    = 2266.01587.01 1

    = 0.6147 1

    (ii) 8.06048

    2.12.7 ==

    > tzp 1

    842.02.1

    2.7=

    t 1

    190.61896.6 ort = 1

    TOTAL 10

    SECTION C

    12(a)

    a = 1.4 0.6dv

    tdt = 1

    = 1.4 0.6(2)= 0.2 1

    (b)1.4t0.3t2 1+ 0.5 = 0

    (3t+1)(t 5) = 0 or using quadratic formula 1t= 5 1

    (c)s = 2(1.4 0.3 0.5)t t dt + = 0.7t

    2 0.1t3 1+ 0.5t+ c integrate

    At t= 0,s = 0 c = 0 finding c

    or +++

    5

    0

    10

    5

    225.03.04.15.03.04.1 dtttdttt

    limits

    1

    When t= 5,s = 7.5 m, when t= 10,s = 25 mor substitute t=0, 5, 10 in [0.7t2 0.1t3

    1+ 0.5t]

    Total distance = 7.5 x 2 + 25 or 7.5 + |25 7.5| 1= 40 m 1

    TOTAL 10

    13(a)

    158100

    130x = 1

    = 121.54 1

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    6

    (b)

    150100 107.14

    140A = 1

    121.54100 90.03

    135B = 1

    120 100 109.09110

    C = 1

    123100 102.5

    120D = 1

    (c)

    12

    )25.102()209.109()503.90()314.107( +++=I 1

    = 99.56 1

    (d)56.99

    100

    1202013

    =I 1

    = 119.47 1

    TOTAL 10

    14(a) (i)

    sin sin 30

    12 7

    ACB =

    1

    ACB = 59 1

    (ii) 2 2 24 11.47 12cos

    2(4)(11.47)AKB

    + = 1

    cos AKB = 0.0388AKB = 87.78 or 8747

    1

    (iii) ABC= 911

    AreaABC= (7)(12) sin 91or Area ofAKB = (4)(11.47) sin 87.78

    1

    Area of quadrilateral= AreaABC+ Area ofAKB= 41.99 + 22.92

    1

    = 64.91 cm 12

    (b)(i)

    1

    (ii) ACB = 121

    1

    TOTAL 10

    B

    AC

    12 cm

    7 cm

    30o

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    7

    15(a)(i)

    x +y 10 or equivalent 1y x 4 or equivalent 1x 2y or equivalent 1

    (b)Draw correctly one straight line from the inequalities 1Draw correctly two more straight line from the inequalities 1RegionR correctly shaded 1

    (c)(i)Maximum point ( 3 , 7 ) 1RM [ 10(3) + 25(7) ] = RM 205 1

    (ii)Minimum point (2 , 6 ) 1RM [ 10(2) + 25(6) ] = RM 170 1

    TOTAL 10

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    8

    GRAPH FORQUESTION 7

    00.5 1 1.5 2 2.5 3 3.5 4

    x

    log10 y

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    0.1

    0.2

    ( 0 , 0.2 )

    ( 4 , 0.76 )

    3.2

    0.57

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    9

    GRAPH FORQUESTION 15

    01

    1

    x

    y

    2 3 4 5 6 7 8 9

    2

    3

    4

    5

    6

    7

    8

    9

    10y x = 4

    x +y = 10

    x = 2y

    R

    ( 3 , 7 )

    10x + 25y = 150( 2 , 6 )

    y = 6