spm percubaan 2007 sbp add maths paper 2 marking scheme

Upload: chinwynncom

Post on 31-May-2018

384 views

Category:

Documents


3 download

TRANSCRIPT

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    1/17

    3472/2

    Matematik

    Tambahan

    Kertas 2

    2 jam

    Ogos 2007

    SEKTOR SEKOLAH BERASRAMA PENUH

    BAHAGIAN SEKOLAH

    KEMENTERIAN PENDIDIKAN MALAYSIA

    PEPERIKSAAN PERCUBAAN

    SIJIL PELAJARAN MALAYSIA 2007

    MATEMATIK TAMBAHAN

    Kertas 2

    Dua jam tiga puluh minit

    JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

    Skema Pemarkahan ini mengandungi 16 halaman bercetak

    MARKING SCHEME

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    2/17

    ADDITIONAL MATHEMATICS MARKING SCHEME

    TRIAL SBP 2007 PAPER 2

    QUESTION

    NO. SOLUTION MARKS

    1

    2 *

    2

    66 4

    4

    ( 6 4 ) 2

    6 6

    24 4

    y y x or x

    x x x

    or

    y y

    y

    1, 4

    2, 10

    2, 10

    1, 4

    x

    or

    y

    OR

    y

    or

    x

    5

    2(a)

    1

    3 2 2

    0

    2

    5 4 2

    4 3 0

    ( 1)( 3) 0

    1, 3

    x kx k x

    k k

    k k

    k

    3

    2

    Solve the quadratic equation

    by using the factorization @

    quadratic formula @

    completing the square

    Eliminate x or y

    Integration(all terms correctly

    integrated)

    Solve quadraticequation

    Note :OW-1 if the working of solving

    quadratic equation is not shown.

    5

    P1

    K1

    K1

    N1

    N1

    K1

    K1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    3/17

    QUESTION

    NO.SOLUTION MARKS

    (b)

    += dxxdxdy )512(xx 56

    2 +=

    2

    5

    2

    52

    2

    522

    2

    52

    )56(

    23

    3

    2

    +=

    ++=

    ++=

    +=

    xxy

    c

    cxx

    dxxxy

    5

    3

    Integrate2

    2

    d y

    dx

    Integratedy

    dx

    Find the value

    ofc

    8

    K1

    K1

    K1

    N1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    4/17

    QUESTION

    NO.SOLUTION MARKS

    3(a)

    (b)

    (c)

    3

    2@

    2

    3=

    = SRPS mm

    *

    24 ( 2)

    3

    3 2 16 0

    y x

    y x

    = +

    =

    3

    *R (4, 8)

    *Q (6, 5)

    Area =0 6 4 2 01

    1 5 8 4 12

    = 26 unit2

    3

    2

    4

    Solve simultaneous equation between SR andPR

    P1

    Use straight line formula

    Or equivalent

    * equation ofSR

    *R coordinate

    Use the formula ofpolygon

    * R and Q coordinate

    8

    K1

    N1

    K1

    N1

    N1

    K1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    5/17

    QUESTION

    NO.SOLUTION MARKS

    4

    (a)

    (b)

    (c)

    20 115cos n= @ 2sin 25 1 m

    2 2

    cos 40 cos(25 15 )

    cos 25 cos15 sin 25 sin15

    1 1m n n m

    3

    2

    sin 50 sin 2(25 )

    2sin 25 cos25

    2 1m m

    2

    21 1cos 2(12 ) 2 cos 12 12 2

    1 1cos12

    2 2

    m

    2

    5

    (a) x = 5 1

    QUESTION

    NO.SOLUTION MARKS

    5

    Use theadditional

    foemulae

    Use the double-

    angle formulae

    Use the double-angle formulae

    7

    K1

    K1

    K1

    N1

    N1

    N1

    N1

    P1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    6/17

    (b)

    (c)

    4 12 6 4 53

    13

    13

    x

    x

    x

    2

    Use any acceptable method

    60 x

    OR

    Listing method

    1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, .,5

    60 x

    Note : Accept x = 0, 1, 2, 3, 4, 5, 6

    2

    6 (a)

    (b)

    r = 0.8

    0

    8

    8

    47.270

    8.01

    )8.01(65

    =

    =S

    3

    4

    6

    Use the mean formula

    Use formula of Sn for GP

    Use the formula of Tn with

    a = -11 and d = 6

    5

    K1

    N1

    K1

    K1

    K1

    P1

    N1

    N1

    N1

    K1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    7/17

    7

    [ 11 ( 1)6] [ 11 ( 1 1)6] [ 11 ( 2 1)6] 93

    33 18 93

    18 126

    7

    11 ( 7 1)6

    25, 31, 37

    n n n

    n

    n

    n

    T

    + + + + + + + =

    + =

    =

    =

    = +

    OR

    Listing method

    -11, -5, 1, 7, 13, 19, 25, 31, 37,

    25, 31, 37

    OR

    (m) + ( m + 6 ) + ( m + 6 + 6 ) = 93

    m = 25

    25, 31, 37

    Note : Accept answer without working, but OW -1

    7

    1 2 93n n nT T T

    N1N1

    Use d = 6

    1 2 93n n nT T T

    7

    K1

    K1

    N1

    N2,1,0

    N1

    K1

    N1

    N1

    N1

    K1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    8/17

    QUESTION

    NO.SOLUTION MARKS

    7

    (a)

    (b)(i)

    (ii)

    (iii)

    x 42.50 38.50 32.01 25.50 13.00 7.98

    log y 0.15 0.20 0.30 0.40 0.58 0.65

    log 10 y = (log 10 p) x - log 10 q

    Refer to the graph.

    m = * log 10 p

    p = 0.9667 0.05

    3.23

    10

    QUESTION SOLUTION MARKS

    8

    All values of log10 y correct

    (accept correct to 2 decimal places)

    Plot log 10 y against x

    6 points mark correctly

    Line of best fit

    c = - log 10 q

    q = 0.17 0.05

    10

    N1

    P1

    K1

    N1

    K1

    N1

    N1

    N1

    K1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    9/17

    NO.

    8(a)

    (b)

    (c)

    (d)

    BOC = 1.942 rad

    0 '1801.942 x = 111 15

    1

    0 '= cos 55 385

    = 2.822 cm

    x

    x

    2

    0 '= sin 55 385

    = 4.127 cm

    y

    y

    The length of BC = 2 x 4.127

    = 8.254 cm

    The length of the arc BC

    =s r

    5 x 1.942

    = 9.71

    Therefore the perimeter of the shaded region8.254 + 9.71

    = 17.964 cm

    4

    Area =21

    2r

    21 (5) (1.942)

    2

    = 24.28

    Area of triangle =

    1

    x 8.254 x 2.8222 = 11.65

    Therefore the area of the shaded region24.28 -11.65

    = 12.63

    3

    QUESTION SOLUTION MARKS

    9

    Use trigo to find BC

    or1

    2BC

    *

    BOC

    KK

    Find area of triangle or

    sector

    * BOC 10

    K1

    N1

    N1

    K1

    K1

    K1

    N1

    N1

    K1

    K1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    10/17

    NO.

    9

    (a)

    (b)(i)

    (ii)

    (c)

    (d)

    += APOAOP

    ~~ca

    2

    1+=

    1

    OBOQ =

    )ba(~~

    +=

    CPCQ =

    )OPCO(OQCO +=+

    COOPCOOQ -+=

    ~~~~

    c)ca(c +2

    1++-=

    ~~

    c)(a2

    -1+=

    3

    From (b), = and2

    -1=

    2

    -1=

    3

    2==

    3

    2

    3=

    18

    OBCofArea

    227= unitOBCofArea

    2

    54= unitramlogparalleofArea

    3

    QUESTION

    NO.SOLUTION MARKS

    10

    Compare the coefficient of

    ~

    a and~

    c

    Solve

    simultaneous

    equation

    10

    Use CQ = CO OQ@ CP = CO OP

    N1

    N1

    K1

    N1

    K1

    K1

    K1

    N1

    N1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    11/17

    10

    (a)

    (b)

    (c)

    2 20 33

    2

    2

    2

    xx

    x

    k

    2

    22

    10

    32

    0

    ( 3)

    33

    L x dx

    xx

    21 14

    72 3

    L

    Area = L 1 + L 2

    = 25

    4

    1

    7

    3

    72

    3

    32

    ( 3)V

    yy

    y dx

    221

    (2) 33

    V

    Volume = V1 + V2

    = 12

    4

    QUESTION

    NO.SOLUTION MARKS

    11

    Solve the simultaneous

    equation

    Integrate and use the

    limit 0 and 2

    Find the area of

    triangle

    L 1 +L2

    Integrate and use thelimit 3 and 7

    Find the volume of

    cone

    V1 + V2

    10

    N1

    N1

    N1

    K1

    K1

    K1

    K1

    K1

    K1

    K1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    12/17

    11

    (a) (i)

    (ii)

    (b) (i)

    (ii)

    np = 6

    npq = 2.4

    q = 0.4*p = 0.6* n = 10

    ( 2) ( 0) ( 1) ( 2)P x P X P X P X

    = 10C0(0.6)0(0.4)10 + 10C1(0.6)

    1(0.4)9

    + 10C2(0.6)2(0.4)8

    = 0.01230

    5

    P ( 280 < X < 350) = )Z(P25

    325-350315) = )Z(P25

    325-320>

    = P(Z > -0.2)

    = 1-P( Z < - 0.2)

    = 0.5793

    The percentage of battery that has a life-span

    of more than 315 hours is 57.93 %

    5

    QUESTION

    NO.SOLUTION MARKS

    12

    Both p and q

    * n and p

    ( 0) @ ( 1) @ ( 2) P X P X P X

    ( 0) ( 1) ( 2 P X P X P X

    KK

    10

    K1

    N1

    N1

    K1

    K1

    K1

    K1

    K1

    N1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    13/17

    12

    (a)

    (b)

    (c)

    a = 2t 7 = 0

    t = 3.5 s

    22

    7

    2

    tv t c

    10 = (0)2 7(0) + c

    c = 10

    v = t2 7t+ 10

    v = (3.5)2 7(3.5) + 10

    = - 2 .25 ms- 1

    3

    v = t2 7t+ 10 < 0

    (t 2)(t 5) < 0

    2 < t < 5

    3

    3 27

    103 2

    t t s t c

    s = 0, t = 0 , c = 0

    3 2

    7 103 2t ts t

    3 22 7(2)10(2)

    3 2s or

    3 25 7(5)10(5)

    3 2s

    =26

    3=

    25

    6

    Total distance =26

    3+

    26 25

    3 6

    = 796

    m

    OR

    4

    QUESTION

    NO.SOLUTION MARKS

    13

    Integrate and

    substitute t = 3.5

    N1

    K1

    K1

    K1

    K1

    K1

    K1

    K1

    N1

    N1

    Use a = 0

    Use v < 0

    Integrate v dt

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    14/17

    2 52 2

    0 2( 7 10) ( 7 10)

    26 93 2

    79

    6

    S t t dt t t dt

    13

    (a)

    (b)

    (c)

    (d)

    06

    50x 100 = 125

    P

    P06 = RM 40

    2

    125 4 120 80 5 150 3

    12012 2

    p p

    p

    *p = 3

    2

    07 x 100 = 12015

    P

    P07 = RM 18

    2

    I C =80

    x 115 = 92100

    125 4 120 3 92 5 150 6

    18I

    = 123.33

    4

    14

    Answer for question 14

    (a) I.170x y

    II. 80x y

    III. 2 20y x

    (b) Refer to the graph,

    1 graph correct

    3 graphs correct

    Correct area

    (c) max point ( 50,120 )

    i) k = 100x + 80y

    Max fees = 100(50) + 80(120)

    = RM 14, 600

    ii) 60 150y 10 20 30 40 50 60 700 8020401401201001601808060

    Use the limit

    2 5

    0 2and

    Integrate v

    2 5

    0 2

    See 115

    *p

    10

    10

    K1

    K1

    K1

    K1

    K1

    K1

    K1

    K1

    N1

    N1

    N1

    N1

    P1

    N1

    N1

    N1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    15/17

    15

    (50,120)

    1

    N1

    N1

    N1

    K1

    N1

    K1

    N1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    16/17

    15

    (a) (i.)

    (a) (ii)

    (b) (i)

    (b) (ii)

    82 = 52 + (AC)2 - 2 (5)(AC) cos 60

    (AC)2 5 (AC) 39 = 0

    AC =

    2( 5) ( 5) 4(1) 39

    2 1

    AC = 9.227 , ( - 4. 227 )

    EC = 9.227 8.5 = 0.727 cm

    OR

    8 5

    sin 60 sin c

    c = 32.77

    8sin 87.32 sin 60

    AC

    AC = 9.227 cm

    EC = 9.227 - 8.5 = 0.727 cm

    1

    (8.5)(15.6) sin 542

    54.54

    180 54.54 125.46AED

    5

    7tan

    10

    = 34.99 @ 35

    AE = AF = 149

    2 2 28 ( 149) ( 149) 2 149 149 cos

    cos = 0.7852

    = 38.26 @ 38 15 @ 38 16

    5

    16

    Use cos rule inABC

    Use sin rule in ABC

    Use cosrule in

    AFE

    10

    K1

    K1

    N1

    P1

    N1

    N1

    N1

    P1

    N1

    K1

    K1

    K1

    K1

  • 8/14/2019 SPM Percubaan 2007 SBP Add Maths Paper 2 Marking Scheme

    17/17

    17

    0 5 10 15 20 25 30 35 40

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.7

    0.8

    x

    x

    x

    x

    x

    x

    yLog10

    x

    (a)

    x 42.50 38.50 32.01 25.50 13.00 7.98

    Log y 0.15 0.20 0.30 0.40 0.58 0.65

    Answer for question 7