2009 PSPM Kedah AddMath 12 w Ans

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    3472/1 [Lihat sebelah

    SULIT

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH

    NEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    Kertas soalan ini mengandungi 17 halaman bercetak

    For Examiners use only

    Question Total MarksMarks

    Obtained

    1 2

    2 4

    3 3

    4 3

    5 3

    6 4

    7 4

    8 3

    9 3

    10 3

    11 3

    12 4

    13 3

    14 3

    15 2

    16 3

    17 4

    18 419 3

    20 2

    21 4

    22 3

    23 3

    24 3

    25 4

    TOTAL 80

    M A T E M A T I K T A M B A H A N

    Kertas 1

    Dua jam

    J A N G A N B U K A K E R T A S S O A L A N IN I

    S E H I N G G A D I B E R I T A H U

    1 This question paper consists of 25 questions.

    2. Answerall questions.

    3. Give on ly on e answer for each quest ion.

    4. Write your answers clearly in the spaces provided in

    the question paper.

    5. Show yo ur working . I t may help you to get marks .

    6. If you wish to change you r answer, cross out the work

    tha t you have done. Then write down the new

    answer.

    7. The diagrams in the questions provided are not

    drawn to scale unless stated.

    8. The marks a l loca ted for each quest ion and sub- art

    of a question are shown in brackets.

    9. A l ist of formulae is provided on pag es 2 to 3.

    10 . A boo klet of four-figure ma thematical tables is

    provided .

    .

    11 You m ay use a non-programm able scien t if iccalculator.

    12 This question paper must be handed in at the end of

    the examinat ion .

    Name : ..

    Form : ..

    3472/1

    Additional Mathematics

    Paper 1

    Sept 2009

    2 Hours

    ADDITIONAL MATHEMATICSPaper 1

    Two hours

    3 to 4.

    2kk

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    2

    BLANK PAGE

    HALAMAN KOSONG

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    3

    The following formulae may be helpful in answering the questions. The symbols given are the ones

    commonly used.

    ALGEBRA

    12 4

    2

    b b acx

    a

    - -=

    2 am an = a m + n

    3 am an = a m - n

    4 (am)

    n= a

    nm

    5 log a mn = log a m + log a n

    6 log an

    m= log a m log a n

    7 log a mn

    = n log a m

    8 logab =

    a

    b

    c

    c

    log

    log

    9 Tn = a + (n1)d

    10 Sn = ])1(2[2

    dnan

    -+

    11 Tn = arn-1

    12 Sn =r

    ra

    r

    rann

    -

    -=

    -

    -

    1

    )1(

    1

    )1(, (r 1)

    13

    r

    aS

    -

    =1

    , r

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    4

    STATISTIC

    1 Arc length,s = rq

    2 Area of sector ,A = 21

    2rq

    3 sin 2A + cos2A = 1

    4 sec2A = 1 + tan

    2A

    5 cosec2A = 1 + cot

    2A

    6 sin 2A = 2 sinA cosA

    7 cos 2A = cos2A sin

    2A

    = 2 cos2A 1

    = 1 2 sin2A

    8 tan 2A =A

    A2

    tan1

    tan2

    -

    TRIGONOMETRY

    9 sin (A B) = sinA cosB cosA sinB

    10 cos (A B) = cosA cosB m sinA sinB

    11 tan (A B) =BA

    BA

    tantan1

    tantan

    m

    12C

    c

    B

    b

    A

    a

    sinsinsin==

    13 a2 = b2 + c

    2 2bc cosA

    14 Area of triangle = Cabsin2

    1

    1 =N

    x

    2 =

    f

    fx

    3 s =N

    xx - 2)(= 2

    2

    xN

    x-

    4 s =

    -f

    xxf 2)(= 2

    2

    xf

    xf-

    5 m = Cf

    FN

    Lm

    -

    + 2

    1

    6 1

    0

    100Q

    IQ

    =

    71

    11

    w

    IwI

    =

    8)!(

    !

    rn

    nPrn

    -=

    9!)!(

    !

    rrn

    nCr

    n

    -=

    10 P(A B) = P(A)+P(B) P(A B)

    11 P (X= r) = rnrrn qpC - , p + q = 1

    12 Mean = np

    13 npq=s

    14 z =s

    m-x

    2kk

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    5

    Answer all questions.

    Jawabsemua soalan.

    1. Diagram 1 shows the relation between set A and set B.

    Rajah 1 menunjukkan hubungan antara set A dan set B.

    a) State the image of 9.Nyatakan imej bagi 9.

    b) Find the value ofx.

    Cari nilai x. [ 2 marks]

    [2 markah]

    Answer/Jawapan : (a) ..

    (b) ...

    2. Given5

    3:

    1 xxf-

    - , find the value of

    Diberi5

    3:

    1 xxf

    -- , cari nilai bagi

    (a) )3(-f ,

    (b)p if 7)( -=pf .[ 4 marks ]

    [4 markah]

    Answer/ Jawapan : (a) ..

    (b) ...4

    2

    2

    1

    Forexaminers

    use only

    Set A Set B

    49

    9

    9

    7

    3

    Diagram 1

    Rajah 1

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    3. Given that function axx +2: and 9:2 -bxxg .

    Diberi fungsi axg +2: dan 9:2 -bxxg .

    Find the value of a and ofb

    Cari nilai bagi a dan b .

    [3 marks]

    [3 markah]

    Answer/Jawapan : a =.........................b =.........................

    4. Given that the straight line 14 +=y is a tangent to the curve kxy += 2 .

    Find the value of k.

    Diberi garis lurus 14 += xy ialah tangen kepada lengkung kxy += 2 .

    Cari nilai k.

    [ 3 marks]

    [3 markah]

    Answer/Jawapan : k=......

    For

    examiners

    use only

    3

    4

    3

    3

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    5.

    4)2(3 2 --= xy

    Diagram above shows the graph of the function 4)2(3 2 --= xy . Q is the minimum pointof the curve and the curve intersects the y-axis at point P. Find the equation of the straightline PQ.

    [ 3 marks ]

    Rajah di atas menunjukkan graf bagi fungsi 4)2(3 2 --= xy . Q ialah titik minimum bagi

    lengkung itu dan lengkung tersebut bersilang dengan paksi-y di titik P. Cari persamaan garislurus PQ.

    [3 markah]

    Answer /Jawapan: ........................

    ___________________________________________________________________________

    6. Given that aand b are the roots of the quadratic equation 0792 =+- x .

    Find the value of

    Diberi a danb adalah punca bagi persamaan kuadratik 0792 =+- x . Cari nilai

    bagi

    (a) ba+(b) ab

    (c)22 ba + [ 4 marks ]

    [4 markah]

    Answer/Jawapan : (a).............................

    (b)............................

    (c).............................

    3

    5

    4

    6

    Forexaminers

    use only

    x

    y

    P

    Q

    2kk

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    7. Solve the equation:

    Selesaikan persamaan:

    xxx 51 4)8(2 =- .

    [4 marks][4 markah]

    Answer/Jawapan :x =................................

    8. The set of positive integers 2, 5, 7, 9, 11,x,y has a mean 8 and median 9. Find the

    values ofx and ofy ify >x.

    [3 marks]

    Satu set integer positif 2, 5, 7, 9, 11, x, y mempunyai min 8 dan median 9. Cari

    nilai-nilai bagi x dan y jika y > x.

    [3 markah]

    Answer/Jawapan : ...................................

    4

    7

    3

    8

    Forexaminers

    use only

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    9. Given that 2loglog 93 =+ yx . Expressy in terms ofx.

    [ 3 marks ]

    Diberi 2loglog 93 =+ yx . Ungkapkan y dalam sebutan x.

    [3 markah]

    Answer/Jawapan : ......................................

    10. The sixth and eleventh terms of an arithmetic progression are 12 and 37 respectively.Find the value of the sixteenth term of this arithmetic progression.

    [3 marks]

    Sebutan keenam dan kesebelas bagi suatu janjang aritmetik ialah 12 dan 37 masing-masing. Cari nilai bagi sebutan keenambelas bagi janjang aritmetik ini.

    [3 markah]

    Answer/Jawapan : ......

    3

    10

    Forexaminers

    use only

    3

    9

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    11. The first three terms of a geometric progression are 36, 36 p and q. If the

    common ratio is3

    1- , find the value of

    Tiga sebutan pertama suatu janjang geometri ialah 36, 36 p dan q. Jika nisbah

    sepunya ialah31- , cari nilai bagi

    (a) p ,

    (b) q.

    [ 3 marks ][3 markah]

    Answer/Jawapan: a) p = ...........

    b) q =...............................

    12. The first term of a geometric progression is a and the common ratio is r. Given that

    096 =+ ra and the sum to infinity is 32, find the value of a and of r.

    [ 4 marks ]

    Sebutan pertama bagi suatu janjang geometri ialah a dan nisbah sepunya r.Diberi 096 =+ ra dan hasil tambah hingga sebutan ketakterhinggaan ialah 32,cari nilai a dan r.

    [4 markah]

    Answer/Jawapan:a=...........

    r=...............................4

    12

    Forexaminers

    use only

    3

    11

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    13. Given that A, B )4,4(- , C )7,2( are collinear and 3AB=BC, find the coordinates of A.[ 3 marks ]

    Diberi A, B )4,4(- ,C )7,2( adalah segaris dan 3AB=BC, carikan koordinat titik A.[3 markah]

    Answer/Jawapan : ...

    14. Diagram below shows the graph of y10log against x .

    Rajah di bawah menunjukkan graf y10log lawan x.

    The variablesx andy are related by the equation2310 -= xy

    Find the value ofh and ofk.

    Pembolehubah x dan y dihubungkait dengan persamaan2310 -= xy .

    Cari nilai h dan k.[3 marks]

    [3 markah]

    Answer/Jawapan : h=

    k=....

    3

    13

    3

    14

    Forexaminers

    use only

    ),0( k

    )2,(hy10log

    x

    2kk

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    12

    15.

    In the diagram above, jniOA += 8 and jiAB 23 -= . Given

    unitsOA 10= , find

    Dalam rajah di atas, jniOA += 8 dan jiAB 23 -= . Diberi

    unitOA 10= , cari

    (a) the value ofn.

    nilai n.

    (b) coordinates of B.

    koordinat B.

    [2 marks][2 markah]

    Answer/Jawapan : (a) n =

    (b)..

    16 Given that jipa 2+= and jib --= 2 , find the value ofp if ba - is parallel

    to j .

    [ 3 marks ]

    Diberi jipa 2+= dan jib --= 2 , cari nilaip jika ba - selari dengan j .

    [3 markah]

    Answer/Jawapan :..

    2

    15

    Forexaminers

    use only

    3

    16

    y

    O

    A

    B

    2kk

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    13

    17. Solve the equation 0cotcos2 =+ x for 00 3600 x .

    [ 4 marks ]

    Selesaikan persamaan 02 =+ kotxkosx bagi 00 3600 x .

    [4 markah]

    Answer/Jawapan: ............

    18. Diagram below shows a circle with centre O.

    Rajah di bawah menunjukkan satu bulatan dengan pusat O.

    Given that the minor angle POQ is p3

    2radian and the area of the shaded

    region is212 cmp . Find the length of the minor arc PQ.

    Diberi sudut minor POQ ialah p3

    2radian dan luas sektor berlorek

    ialah212 cmp . Cari panjang lengkok minor PQ.

    [4 marks][4 markah]

    Answer/Jawapan:

    _

    p3

    2_1

    O

    P

    4

    17

    4

    18

    Forexaminers

    use only

    Q

    2kk

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    19. Find the gradient of the curve2

    5

    2

    34

    2 -+=x

    xy at the point )3,1( .

    [ 3 marks ]

    Cari kecerunan kepada lengkung

    2

    5

    2

    34

    2 -+=x

    xy pada titik )3,1( .

    [3 markah]

    Answer/Jawapan:

    20. Differentiate

    12

    14 2

    +

    -xwith respect tox .

    [2 marks]

    Bezakan12

    14 2

    +

    -

    x

    xterhadapx.

    [2 markah]

    Answer/Jawapan: ............

    3

    19

    2

    20

    For

    examiners

    use only

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    21. Given that the gradient function of a curve passing through the point (1, 2) is

    2)12(

    3

    -x+ 2x , determine the equation of the curve.

    Fungsi kecerunan bagi suatu lengkung yang melalui titik ( 1, 2) ialah

    2)12(3-x

    + 2x, tentukan persamaan bagi lengkung ini.

    [4 marks]

    [4 markah]

    Answer/Jawapan: ..

    22. Given that10

    )32( 5-=x

    y andx is increasing at the rate of 2 units per second, find the

    rate of change ofy when2

    1=x . [ 3 marks ]

    Diberi10

    )32( 5-=

    xy dan x bertambah dengan kadar 2 unit sesaat, cari kadar

    perubahan bagi y apabila2

    1=x . [3 markah]

    Answer/Jawapan: .

    4

    21

    Forexaminers

    use only

    3

    22

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    23. 90 percent of the students of Form 5 Euler passed the April mathematics test. Amongthose who passed, 20 percent score with distinction.

    90 peratus pelajar Tingkatan 5 Euler lulus ujian matematik bulan April. Antaramereka yang lulus, 20 peratus skor dengan cemerlang.

    (a) If a student of Form 5 Euler was selected at random, find the probability that hepassed the April mathematics test with distinction.

    Jika seorang pelajar dari tingkatan 5 Euler dipilih secara rawak, cari

    kebarangkalian dia lulus ujian matematik bulan April dengan cemerlang.

    (b) If 5 students of Form 5 Euler were selected at random, find the probability that

    only one of the five students selected passed the April mathematics test with

    distinction.

    Jika 5 orang pelajar dari tingkatan 5 Euler dipilih secara rawak, cari

    kebarangkalian hanya seorang daripada lima pelajar terpilih lulus ujian

    matematik bulan April dengan cemerlang.

    [3 marks]

    [3 markah]

    Answer/Jawapan: (a) ..

    (b) ..

    24. Five cards are numbered 1, 2, 3, 4 and 5 respectively. How many different odd numbers

    can be formed by using four of these five cards?

    [ 3 marks ]

    Lima kad masing-masing ditulis dengan nombor 1, 2, 3, 4 dan 5. Berapa nombor ganjil

    boleh dibentuk dengan menggunakan empat daripada lima kad ini ?[3 markah]

    Answer/Jawapan: 3

    24

    3

    23

    Forexaminers

    use only

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    25. A random variable X is normally distributed with mean 370 and standard deviation10. Find the value of

    Satu pembolehubah rawak X bertaburan normal dengan min 370 dan sisihan piawai10. Cari nilai bagi

    (a) the z-score if X = 355.skor z jika X = 355

    (b) )367(

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    SULIT3472/2

    Additional Mathematics

    Kertas 2September, 2009

    2 jam 30 minit

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

    SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICS

    Kertas 2

    Dua jam tiga puluh minit

    JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

    1. This question paper consists of three sections : Section A, Section B andSection C.

    2. Answerall questions in Section A, four questions from Section B andtwo questionsfrom Section C.

    3. Give only one answer/solution to each question.

    4. Show your working. It may help you to get your marks.

    5. The diagrams provided are not drawn according to scale unless stated.

    6. The marks allocated for each question and sub - part of a question are shown inbrackets.

    7. You may use a non-programmable scientific calculator.

    8. A list of formulae is provided in page 2 and 3.

    This question paper consists of19 printed pages and 1 blank page.

    3472/2 [Lihat sebelah

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    3472/2 [Lihat sebelah

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    2

    The following formulae may be helpful in answering the questions. The symbols given are the onescommonly used.

    ALGEBRA

    1. x =a

    acbb

    2

    42 -- 8.

    a

    bb

    c

    ca

    log

    loglog =

    2. aaa nmnm += 9. dnaT n )1( -+=

    3. aaanmnm -= 10. ])1(2[

    2dna

    nSn -+=

    4. aamnnm =)( 11.

    1-= nn arT

    5. nmmn aaa logloglog +=12.

    r

    ra

    r

    raS

    nn

    n-

    -=

    -

    -=

    1

    )1(

    1

    )1(, r 1

    6. log log loga a am

    m nn

    = - 13.r

    aS

    -=

    1, r < 1

    7. mnm ana loglog =

    CALCULUS

    1. y = uv,dx

    duv

    dx

    dvu

    dx

    dy+=

    4 Area under a curve

    = b

    adxy or

    = b

    adyx

    2. y = v

    u

    , 2v

    dx

    dvu

    dx

    duv

    dx

    dy-

    =

    5. Volume of revolution

    = b

    adxy 2p or

    = b

    adyx2p

    3.dx

    du

    du

    dy

    dx

    dy=

    GEOMETRY

    1. Distance = 2122

    12 )()( yyxx -+-4. Area of triangle

    =1 2 2 3 3 1 2 1 3 2 1 3

    1( ) ( )

    2x y x y x y x y x y x y+ + - + +

    2. Mid point

    (x , y ) =

    ++

    2,

    2

    2121 yyxx5. 22 yxr +=

    3. Division of line segment by a point

    (x , y ) =

    +

    +

    +

    +

    nm

    myny

    nm

    mxnx 2121 ,

    6.2 2

    xi yj

    ry

    +=

    +% %

    2kk

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    3

    STATISTICS

    1.N

    xx

    = 7

    =i

    ii

    W

    IWI

    2.=

    f

    fxx 8

    )!(

    !

    rn

    nPr

    n

    -=

    3.N

    xx -=2)(

    s = 22

    xN

    x- 9 !)!(

    !

    rrn

    nCr

    n

    -=

    4.

    -=

    f

    xxf 2)(s = 2

    2

    xf

    fx-

    10 P(AB) = P(A) + P(B) P(AB)

    11 P (X = r) = rnrrn qpC - ,p + q = 1

    5. m = L + Cf

    FN

    m

    -21 12 Mean , m= np

    13 npq=s

    6. 10001 = Q

    QI 14 Z = s

    m-X

    TRIGONOMETRY

    1. Arc length,s = rq 8. sin (A B ) = sinA cosB cosA sinB

    2. Area of sector, A = q2

    2

    1r

    9. cos (A B ) = cosA cosBm sinA sinB

    3. sin A + cosA = 110 tan (A B ) =

    BA

    BA

    tantan1

    tantan

    m

    4. sec A = 1 + tan A11 tan 2A =

    AA2

    tan1tan2

    -

    5. cosec A = 1 + cot A12

    C

    c

    B

    b

    A

    a

    sinsinsin==

    6. sin 2A = 2sinA cosA 13 a = b + c 2bc cosA7. cos 2A = cos A sin A

    = 2 cos A 1= 1 2 sin A

    14 Area of triangle =1

    sin2

    ab C

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    4

    Section A

    Bahagian A

    [ 40 marks ][ 40 markah ]

    Answerall questions.

    Jawab semua soalan.

    1. Solve the simultaneous equations 3 2 0x y- - = and ( )2 1 3x y - = .

    Give your answers correct to three decimal places.[5 marks]

    Selesaikan persamaan serentak 3 2 0x y- - = dan ( )2 1 3x y - = .

    Beri jawapan anda betul kepada tiga tempat perpuluhan.[5 markah]

    2.

    In the diagram, the gradient andy-intercept of the straight line PQ

    are 2 and 3 respectively.R is a point on thex-axis.

    (a) Find the value ofh and k.

    (b) Given thatPQ is perpendicular to QR, find thex-intercept ofQR.

    (c) Calculate the area ofPQR.

    [3 marks]

    [3 marks]

    [2 marks]

    Dalam rajah, kecerunan dan pintasan-y bagi garis lurus PQ masing-masing ialah 2 dan 3 . R ialah titik pada paksi-x.

    (a) Cari nilai h dan nilai k .

    (b) Diberi bahawa PQ berserenjang dengan QR, cari pintasan-x

    bagi QR.

    (c) Hitungkan luas PQR.

    [3 markah]

    [3 markah]

    [2 markah]

    x

    y

    0

    Q(4, k)

    P(h, 3)

    R

    2kk

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    5

    3.(a) Sketch the graph of 2 sin 2x= - for0 2x p .

    (b) Hence, using the same axes, sketch a suitable straight line to find

    the number of solutions for the equation sin22

    xx

    p= for

    0 2x p . State the number of solutions.

    (a) Lakar graf bagi 2 sin 2y x= - untuk 0 2x p .

    (b) Seterusnya, dengan menggunakan paksi yang sama, lakar satugaris lurus yang sesuai untuk mencari bilangan penyelesaian

    bagi persamaan sin22

    xx

    p= untuk 0 2x p .

    Nyatakan bilangan penyelesaian itu.

    [4 marks]

    [3 marks]

    [4 markah]

    [3 markah]

    4. Given that ( )2 1x x - is the gradient function of a curve which passes

    through the point )1,1( -P . Find

    (a) the gradient of the tangent to the curve atP,

    (b) the equation of the curve,

    (c) the coordinates of the turning point at x = 1 . Hence determine

    whether the turning point is a maximum or a minimum point.

    [1 mark]

    [3 marks]

    [3 marks]

    Diberi ( )2 1x x - ialah fungsi kecerunan bagi suatu lengkung yang

    melalui titik )1,1( -P . Cari

    (a) kecerunan tangen kepada lengkung itu di P,

    (b) persamaan lengkung itu,

    (c) koordinat bagi titik pusingan pada x = 1 . Seterusnya tentukansama ada titik pusingan itu adalah titik maksimum atau titikminimum.

    [1 markah]

    [3 markah]

    [3 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    6

    5. A set of fifty numbers, 1 2 3 50, , ,......x x x x , has a mean of 11 and a

    standard deviation of 8.

    (a) Find

    (i) ,(ii)

    2

    (b) If each of the numbers is multiplied by 1.8 and thenincreased by 5, find the new value for the

    (i) mean,

    (ii) variance.

    [3 marks]

    [3 marks]

    Suatu set yang terdiri daripada lima puluh nombor,

    1 2 3 50, , ,......x x x x , mempunyai min 11 dan sisihan piawai 8.

    (a) Cari

    (i) ,

    (ii)2

    (b) Jika setiap nombor didarab dengan 1.8 dan ditambahdengan 5, cari nilai yang baru untuk

    (i) min ,

    (ii) varians .

    [3 markah]

    [3 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    7

    6.

    A strip of metal is cut and bent to form some semicircles. Thediagram shows the first four semicircles formed. The radius of the

    smallest semicircle is 5 cm. The radius of each subsequentsemicircle is increased by 3 cm.

    (a) If the radius of the largest semicircle is 104 cm, find the number

    of semicircles formed.

    (b) Calculate the total cost needed to form all the semicircles in (a) if

    the cost of the metal strip is RM4 per meter.[ Round off your answer to the nearest RM ]

    Satu jalur logam dipotong dan dibengkok untuk membentukbeberapa semi bulatan. Rajah menunjukkan empat semi bulatan

    pertama yang telah dibentukkan. Jejari semi bulatan yang terkecilialah 5 cm. Jejari semi bulatan yang berikutnya bertambah

    sebanyak3cm setiap satu.

    (a) Jika jejari bagi semi bulatan yang terbesar ialah 104 cm , caribilangan semi bulatan yang telah dibentuk.

    (b) Hitungkan jumlah kos yang diperlukan untuk membentuk semua

    semi bulatan dalam (a) jika kos jalur logam ialah RM4 semeter.[ Bundarkan jawapan anda kepada RM yang terdekat]

    [3 marks]

    [4 marks]

    [3 markah]

    [4 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    8

    Section B

    Bahagian B

    [ 40 marks ][ 40 markah ]

    Answerfour questions from this section.

    Jawab empatsoalan daripada bahagian ini.

    7.

    The table shows the values of two variables, x andy, obtained from

    an experiment. Variables x and y are related by the equation

    1=+yxlm , where mand lare constants.

    (a) Ploty

    1against

    1, using a scale of 2 cm to 0.1 unit on both

    axes. Hence draw the line of best fit.

    (b) Use your graph in 7(a) to find the value of

    (i) m,

    (ii) l.

    [5 marks]

    [5 marks]

    Jadual menunjukkan nilai-nilai bagi dua pembolehubah, x dan y,

    yang diperoleh daripada satu eksperimen. Pembolehubah x dan y

    dihubungkan oleh persamaan 1=+yx

    lm, dengan keadaan mdan l

    adalah pemalar

    (a) Ploty

    1melawan

    x

    1, dengan menggunakan skala 2 cm

    kepada 0.1 unit pada kedua-dua paksi. Seterusnya, lukis

    garis lurus penyuaian terbaik.(b) Gunakan graf di 7(a) untuk mencari nilai

    (i) m,

    (ii) l.

    [5 markah]

    [5 markah]

    x 1.5 2.0 2.5 4.0 5.0 10.0

    y 0.96 1.2 1.4 2.0 2.2 3.0

    2kk

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    SULIT

    9

    8.

    The diagram shows straight linex +y = 4 that intersects with thecurve y = (x 2 )2

    at points Pand Q.

    Find

    (a) the coordinates of Q,

    (b) the area of the shaded regionA,

    (c) the volume generated, in terms of , when the shaded region

    B is revolved through 360o

    about thex-axis.

    [2 marks]

    [4 marks]

    [4 marks]

    Rajah menunjukkan garis lurus x + y = 4 bersilang denganlengkung y = (x 2 )

    2pada titik P dan Q.

    Cari

    (a) koordinat Q,

    (b) luas rantau berlorek A,

    (c) isipadu janaan, dalam sebutan , apabila rantau berlorek B

    dikisarkan melalui 360opada paksi-x.

    [2 markah]

    [4 markah]

    [4 markah]

    O x

    y = (x 2 )2

    P

    QAB x + y = 4

    y

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    10

    9.

    In the diagram, PQRS is a quadrilateral. The diagonalsPR and QS

    intersect at point T. It is given PQ = 2~

    x , PS = 3~

    y and

    SR =~

    x ~

    y .

    (a) Express in terms of~

    x and~

    y :

    (i) QS

    (ii) PR .

    (b) Given that QT = m QS , PT = n PR , where m and n are

    constants, express

    (i) QT in terms of m,~

    x and~

    y ,

    (ii) PT in terms of n,~

    x and~

    y .

    (c) Using PQ = PT + TQ , find the value of m and of n.

    [10 marks]Dalam rajah, PQRS ialah sebuah sisiempat. Pepenjuru-pepenjuru

    PR dan QS bersilang di titik T . Diberi PQ = 2~

    x , PS = 3~

    y

    dan SR =~

    x ~

    y .

    (a) Ungkapkan dalam sebutan~

    x dan~

    y :

    (i) QS

    (ii) PR .

    (b) Diberi QT = m QS , PT = n PR , dengan keadaan m dan nialah pemalar, ungkapkan

    (i) QT dalam sebutan m,~

    x dan~

    y ,

    (ii) PT dalam sebutan n,~

    x dan~

    y .

    (c) Dengan mengguna PQ = PT + TQ , cari nilai m dan

    nilai n. [10 markah]

    P

    T

    S

    R

    Q

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    11

    10.

    The diagram shows the cross-section of a cylindrical roller with

    centre O and radius 20 cm resting on a horizontal groundPQ.OAB is a straight line that represents the handle of the roller and

    OA :AB = 1 : 3.

    Calculate

    (a) POA in radian,

    (b) the perimeter, in cm, of the shaded region,

    (c) the area, in cm2

    , of the shaded region.

    [3 marks]

    [3 marks]

    [4 marks]

    Rajah menunjukkan keratan rentas sebuah penggelek berbentuk

    silinder dengan pusat O dan jejari 20 cm yang terletak di atas

    lantai mengufuk PQ. OAB ialah garis lurus yang mewakilipemegang penggelek itu dan OA : AB = 1 : 3.

    Hitungkan

    (a) POA dalam radian,

    (b) perimeter, dalam cm, kawasan berlorek,

    (c) luas, dalam cm2

    , kawasan berlorek.

    [3 markah]

    [3 markah]

    [4 markah]

    QP

    O

    A

    B

    2kk

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    SULIT

    12

    11(a) In a survey carried out in a school, it is found that 40% of its

    students are participating actively in co-curricular activities.

    (i) If 6 students from that school are chosen at random,calculate the probability that at least 4 students are

    participating actively in co-curricular activities.

    (ii) If the variance of the students who are active in co-curricularactivities is 288, calculate the student population of the

    school.[5 marks]

    (b) The masses of chicken eggs from a farm has a normal distribution

    with a mean of 62 g and a standard deviation of 8 g. Any egg thathas a mass exceeding 68 g is categorised as grade double-A.

    (i) Find the probability that an egg chosen randomly from the

    farm has a mass between 60 g and 68 g.

    (ii) If the farm produces 3000 eggs daily, calculate the number ofeggs with grade double-A.

    [5 marks]

    (a) Dalam satu tinjauan yang dijalankan ke atas murid-murid di sebuah sekolah, didapati 40% daripada murid-murid sekolah itu

    mengambil bahagian secara aktif dalam aktiviti kokurikulum.

    (i) Jika 6 orang murid daripada sekolah itu dipilih secara rawak,

    hitungkan kebarangkalian bahawa sekurang-kurangnya 4orang murid adalah aktif dalam aktiviti kokurikulum.

    (ii) Jika varians murid-murid yang mengambil bahagian secaraaktif dalam aktiviti kokurikulum ialah 288, hitungkan bilangan

    murid dalam sekolah itu.[5 markah]

    (b) Jisim telur ayam dari sebuah ladang adalah mengikut satu taburannormal dengan min 62 g dan sisihan piawai 8 g. Sebarang telur

    dengan jisim melebihi 68 g dikategorikan sebagai telur greddouble-A

    (i) Cari kebarangkalian bahawa sebiji telur yang dipilih secara

    rawak dari ladang itu mempunyai jisim di antara 60 g dan68 g.

    (ii) Jika ladang itu menghasilkan 3000 biji telur setiap hari,

    hitungkan bilangan telur yang mempunyai gred double-A.

    [5 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    13

    Section C

    Bahagian C

    [ 20 marks ]

    [ 20 markah ]

    Answertwo questions from this section.Jawab dua soalan daripada bahagian ini.

    12. A particle P starts from a fixed point O and moves in a straight

    line so that its velocity, v ms-1

    , is given by v = 8 + 2t t2, where

    t is the time, in seconds, after leaving O.[Assume motion to the right is positive.]

    Find

    (a) the initial velocity, in ms-1 , of the particle,

    (b) the value of t at the instant when the acceleration is 1 ms-2

    ,

    (c) the distance of P from O when P comes to instantaneousrest,

    (d) the total distance, in m, travelled by the particle P in the first

    5 seconds.

    Suatu zarah P mula dari suatu titik tetap O dan bergerak di sepanjang garis lurus. Halajunya vms

    -1 , diberi oleh

    v = 8 + 2t t2 , dengan keadaan t ialah masa, dalam saat, selepas

    melalui O.[Anggapkan gerakan ke arah kanan sebagai positif]

    Cari

    (a) halaju awal, dalam ms-1

    , bagi zarah itu,

    (b) nilai bagi t apabila pecutannya ialah 1 ms-2,

    (c) jarak P dari O apabila P berada dalam keadaan rehatseketika,

    (d) jumlah jarak yang di lalui, dalam m , oleh zarah P dalam

    5 saat yang pertama.

    [1 mark]

    [2 marks]

    [3 marks]

    [4 marks]

    [1 markah]

    [2 markah]

    [3 markah]

    [4 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    14

    13. The pie chart shows five items,A, B, C, D and E used in makingcakes. The table shows the prices and the price indices of these

    items.

    E

    D

    C

    B

    A

    30

    75

    120

    Items

    Bahan

    Price (RM)

    per kg for theyear 2003

    Harga (RM)per kg padatahun 2003

    Price (RM)

    per kg for theyear 2006

    Harga (RM)per kg padatahun 2006

    Price index for the

    year 2006 based onthe year 2003

    Indeks harga pada

    tahun 2006berasaskan tahun

    2003

    A 0.40 x 150

    B 1.50 1.65 110C 4.00 4.80 y

    D 3.00 4.50 150

    E z 2.40 120

    (a) Find the value of(i) x,

    (ii) y,(iii) z.

    (b) Calculate the composite index for the cost of making these

    cakes in the year 2006 based on the year 2003.

    (c) The total expenditure on the items in the year 2006 isRM 5000. Calculate the corresponding total expenditure

    in the year 2003.

    (d) The price of each item increases by 20 % from the year 2006to the year 2008. Find the composite index for total

    expenditure on the items in the year 2008 based on the year2003.

    [3 marks]

    [3 marks]

    [2 marks]

    [2 marks]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    15

    Carta pai menunjukkan lima bahan, A, B, C, D dan E yang

    digunakan untuk membuat sejenis kek. Jadual menunjukkan hargabahan dan nombor indeks bagi kelima-lima bahan tersebut.

    (a) Carikan nilai

    (i) x,(ii) y,

    (iii) z.

    (b) Hitungkan nombor indeks gubahan bagi kos penghasilan kekitu pada tahun 2006 berasaskan tahun 2003.

    (c) Jumlah kos bahan-bahan tersebut pada tahun 2006 ialah

    RM 5000. Hitungkan jumlah kos yang sepadan pada tahun2003.

    (d) Harga bagi setiap bahan bertambah sebanyak 20% daritahun 2006 ke tahun 2008. Cari nombor indeks gubahanbagi jumlah kos ke atas bahan-bahan tersebut pada tahun

    2008 berasaskan tahun 2003.

    [3 markah]

    [3 markah]

    [2 markah]

    [2 markah]

    2kk

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    16

    14. Use graph paper to answer this question.

    A factory produces two components, S and T for a digitalcamera, by using machines P and Q. The table shows the time

    taken to produce components S and T respectively.

    In any given week, the factory produces xunits of component S

    and y units of component T. The production of the componentsper week is based on the following constraints:

    I : MachineP operates not more than 2000 minutes.

    II : Machine Q operates at least 1200 minutes.

    III : The number of component Tproduced is not more than threetimes the number of component Sproduced.

    (a) Write three inequalities, other than x 0 and y 0, which

    satisfy all the above constraints.

    (b) Using a scale of 2 cm to 10 units on both axes, construct andshade the region R which satisfies all of the above

    constraints.

    (c) Use your graph in 14(b) to find

    (i) the maximum number of component S that could be

    produced, if the factory plans to produce only 30 unitsof component T,

    (ii) the maximum profit per week if the profit from a unit

    of component S is RM20 and from a unit of component T is RM30.

    Time taken (minutes)Masa diambil(minit)

    Component

    Komponen Machine PMesin P

    Machine QMesin Q

    S 40 15

    T 20 30

    [3 marks]

    [3 marks]

    [4 marks]

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    SULIT September, 2009

    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    17

    Gunakan kertas graf untuk menjawab soalan ini.

    Sebuah kilang menghasilkan dua komponen, S dan T bagi sesuatu kamera digital dengan menggunakan mesin P dan Q.

    Jadual menunjukkan masa yang diambil untuk menghasilkankomponen-komponen S dan T.

    Dalam mana-mana satu minggu, kilang tersebut menghasilkan x

    unit bagi komponen S dan y unit bagi komponen T. Penghasilan komponen-komponen tersebut adalah berdasarkan

    kekangan berikut:

    I : Mesin P beroperasi tidak melebihi2000 minit.

    II : Mesin Q beroperasi sekurang-kurangnya1200 minit.

    III : Bilangan komponen T yang dihasilkan tidak melebihi tigakali ganda bilangan komponen S yang dihasilkan.

    (a) Tuliskan tiga ketaksamaan, selain x 0 dan y 0 , yang

    memenuhi semua kekangan di atas.

    (b) Menggunakan skala 2 cm kepada 10 unit pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semua

    kekangan di atas.

    (c) Gunakan graf anda di 14(b) untuk mencari

    (i) bilangan maksimum bagi komponen S yang bolehdihasilkan jika kilang tersebut bercadang untuk

    menghasilkan 30 unit komponen T sahaja,

    (ii) keuntungan maksimum seminggu jika keuntungan yangdiperoleh dari satu unit komponen S ialah RM20 dan

    dari satu unit komponen T ialah RM30.

    [3 markah]

    [3 markah]

    [4 markah]

    2kk

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    18

    15. The diagram shows a cyclic quadrilateral PQRS with PQ = 9 cm,

    PR = 11 cm and QR = 7 cm.

    (a) Find PQR.

    (b) Given that PS= 6 cm, find the length ofRS.

    (c) Calculate the area ofPQRS.

    S R

    Q

    P

    11 cm

    7 cm

    9 cm

    [3 marks]

    [4 marks]

    [3 marks]

    2kk

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    19

    Rajah menunjukkan satu sisiempat kitaran PQRS denganPQ = 9 cm , PR =11 cm , dan QR = 7 cm.

    (a) Cari PQR.

    (b) Diberi PS = 6 cm, cari panjang RS.

    (c) Hitungkan luas bagi PQRS.

    [3 markah]

    [4 markah]

    [3 markah]

    END OF QUESTION PAPER

    KERTAS SOALAN TAMAT

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    SULIT JPNKd/2006/3472/1

    Additional Mathematics paper 1

    Nama Pelajar : Tingkatan 5 : .

    3472/1AdditionalMathematicsPaper1

    September 2009

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

    SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICS

    MARKING SCHEME

    Paper 1

    .

    SULIT 3472/1

    2kk

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    SULIT 3472/12

    SPM Trial Examination 2009 Kedah Darul Aman

    Marking Scheme

    Additional Mathematics Paper 1

    Question Solution/ Marking Scheme Answer Marks

    1 (a) 3

    (b) 81

    1

    1

    2

    (a) B1: 35

    3

    xor xxf 53)(

    (b) B1: 753 p or p

    5

    )7(3

    (a) 18

    (b) 2

    2

    2

    3

    B2: 934 aorb

    B1: aax )2(2

    a = 3

    b = 4

    3

    4

    B2: 410)1)(1(4)4( 2 kork

    B1 : or0142 kxx 142 xkx

    5 3

    5

    B2: 620

    )4(8

    orm

    B1: )4,2()8,0( QorP

    86 xy 3

    6

    (c) B1: )7(292

    (a) 9(b) 7(c) 67

    1

    1

    2

    7

    B3: xxx 1033

    B2 :xx or 1033 22

    B1: xxx 5213 )2()2(2

    x =2

    1 4

    3472/1 Additional Mathematics Paper 1

    SULIT

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    SULIT 3472/13

    Question Solution/ Marking Scheme Answer Marks

    8

    B2: 22 yx

    B1: 87

    119752

    yx

    12,10 yx

    13,9 yx 3

    9 B2: 2log4log 292

    3 yxoryx

    B1: 2log3log

    log2

    9log

    loglog 9

    9

    9

    3

    33 y

    xor

    yx

    2

    81

    xy 3

    10

    B2: 513 danda

    B1: 3710125 daorda 62 3

    11

    (a) B1:3

    1

    36

    36

    p

    (a) 48

    (b) 4

    2

    1

    12B3: 48

    2

    1 aorr

    B2: 32

    961

    096)1(32

    a

    aorrr

    B1: 321

    r

    a

    21,48 ra 4

    13

    B2 : 44

    374

    4

    23

    yand

    x

    B1 : 44

    374

    4

    23

    yor

    x

    ( 6, 3) 3

    14

    B2 : 23230

    22

    hor

    h

    kork

    B1 : 23log10 xy

    h =3

    4

    k= 2

    3

    15(a) 6

    (b) (11, 4)

    1

    1

    3472/1 Additional Mathematics Paper 1

    SULIT

    2kk

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    SULIT 3472/14

    Question Solution/ Marking Scheme Answer Marks

    16B2: 02 p

    B1: jip 3)2( p = 2 3

    17 B3:

    )42(330,210,270,900000

    ofoutany

    B2 : )(2

    1sin,0cos bothxx

    B1 : 0sin

    coscos2

    x

    xx

    0

    0

    00

    330

    ,270

    ,210,90

    4

    18B3 : )

    32(6 S

    B2 : 6r

    B1: 12)3

    2(

    2

    1 2 r

    12.57 cm

    ( 4 ) cm4

    19B2:

    2)1(2

    3)1(8

    B1:2

    2

    38

    xx

    dx

    dy

    2

    13

    3

    20 B1: 12 xy

    or

    2

    2

    )12(

    )14(2)8)(12(

    x

    xxx

    dx

    dy 2 2

    21

    B3 : c

    12

    32

    B2:2

    2

    2

    )12(321

    xand

    x

    B1:2

    2

    2

    )12(321

    xor

    x

    2

    5

    )12(2

    3 2

    x

    xy

    4

    3472/1 Additional Mathematics Paper 1

    SULIT

    2kk

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    SULIT 3472/1

    3472/1 Additional Mathematics Paper 1

    SULIT

    5

    Question Solution/ Marking Scheme Answer Mark

    22

    B2 : 210

    2)32(54

    x

    dt

    dy

    B1 : 210

    2)32(5 4

    dt

    dxor

    x

    dx

    dy

    32 units per

    second

    3

    23(b) B1:

    411

    5)82.0()18.0(C (a)0.18

    (b)0.40691

    2

    24 B2: 34

    13

    PP

    B1: 34

    13

    PorP

    72 3

    25(a) B1 :

    10

    370355

    (b) B1 :10

    370367

    (a) 1.5

    (b) 0.3821

    2

    2

    END OF MARKING SCHEME

    2kk

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    SULIT JPNKd/2006/3472/2

    Nama Pelajar : Tingkatan 5 : .

    3472/2Additional

    Mathematics

    September 2009

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

    SEKOLAH MENENGAHNEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICS

    Paper 2

    .

    MARKING SCHEME

    SULIT 3472/2

    2kk

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    2

    MARKING SCHEME

    ADDITIONAL MATHEMATICS PAPER 2

    SPM TRIAL EXAMINATION 2009

    N0. SOLUTION MARKS

    1

    2

    2

    2

    3 23 2 (3 2 1)

    0 6 6 3

    2 2 1 0

    2 ( 2) 4(2)( 1)

    2(2)

    1.366 0.366

    3(1.366) 2 3( 0.366) 2

    2.098 3.098

    y xx x

    x x

    x x

    x

    x or x

    y y

    = -= - -

    => = - -

    - - =

    - - -=

    = = -

    = - = - -

    = = -

    P1K1 Eliminate y

    K1 Solve quadratic

    equation

    N1

    N1

    5

    2

    (a)

    (b)

    (c)

    ( )

    2 3

    ( , 3) 3 2 3

    3

    (4, ) 2 4 3

    11

    y x

    P h h

    h

    Q k k

    k

    = +

    - - = +

    = -

    = +

    =

    OR

    3 32 3

    0

    32 11

    4 0

    hh

    kk

    += = -

    --

    = =-

    1 2

    12

    2

    11 1

    4 2

    1: 13

    2

    0 26

    m m

    y

    x

    QR y x

    y x

    = = -

    -= -

    -

    = - +

    = =

    OR

    11 0 1

    4 2

    111 2

    2

    26

    x

    x

    -= -

    -

    = - +

    =

    x-intercept = 26

    Area

    3 4 26 31

    3 11 0 32

    133 78 12 286

    2

    1385

    2

    192.5

    - -=

    - -

    = - - + -

    = -

    =

    K1 Use equation or

    gradient

    N1

    N1

    P1 For 21

    2m = -

    K1 Use equation or

    gradient

    N1

    K1 Use formula and

    find the area

    triangle

    N1

    8

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    3

    N0. SOLUTION MARKS

    3

    (a)

    (b) 2 sin 2 22

    xx

    p- = -

    or

    22

    xy

    p= -

    Draw the straight line 22

    yp

    = -

    Number of solutions = 4 .

    P1 Negative sineshape correct.

    P1 Amplitude = 1

    [ Maximum = 3

    and Minimum =

    1 ]

    P1 Two full cycle in

    0 x 2p

    P1 Shift up the graph

    N1 For equation

    K1 Sketch the

    straight line

    N1

    7

    3

    12

    2

    xy

    p= -

    y

    2

    p 2p xO p

    2 sin 2x= -

    2

    2

    3p

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    4

    N0. SOLUTION MARKS

    4

    (a)

    (b)

    (c)

    ( ) ( )2

    1 1 1 2dy

    dx= - - - = -

    3 2

    4 3

    4 3

    ( )

    4 3

    1 11

    4 3

    5

    12

    5

    4 3 12

    y x x dx

    x xy c

    c

    c

    x xy

    = -

    = - +

    = + +

    =

    = - +

    ( )2

    22

    2

    2

    2

    1 0

    0@1

    3 2

    1 1 0

    1 1 5 1

    4 3 12 3

    dy

    x xdx

    x

    d yx x

    dx

    d yx

    dx

    y

    = - ==

    = -

    = = >

    = - + =

    11,

    3

    Minimumpoint.

    N1

    K1 Integrate gradient

    function

    K1 Substitute (1,1)into equation y

    N1

    K1 Find2

    2

    d y

    dx

    N1 N1

    7

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    5

    N0. SOLUTION MARKS

    5

    (a)

    (i)

    (ii)

    (b)

    (i)

    (ii)

    11 55050

    xx= => =

    2

    2

    2

    11 850

    50(64 121)

    9250

    x

    x

    - =

    => = +

    =

    New mean = 11 1.8 5 24.8 + =

    New variance =2

    (8 1.8) 207.36 =

    K1 Use formula of

    mean and/or

    standard deviation

    N1

    N1

    K1 Find new mean

    and/or new

    variance

    N1

    N1

    6

    6

    (a)

    (b)

    ( )

    5

    3

    5 1 3 104

    1 33

    34

    l r

    a

    d

    n

    n

    n

    p

    p

    p

    p p p

    =

    =

    =

    + - =

    - =

    =

    ( ) ( )3434

    2 5 33 32

    S p p = +

    1853

    5821

    58.21

    cm

    m

    p=

    =

    =

    Total cost

    4 58.21

    232.8

    233

    RM

    RM

    RM

    =

    =

    =

    P1 Value of a and/or d

    K1 Use Tn = 104p

    N1

    K1 Find S34

    N1

    K1 RM4 S34

    N1

    7

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    6

    N0. SOLUTION MARKS

    7

    (a)

    (b)(i)

    (ii)

    ll

    m 111+-=

    xy

    y

    11.04 0.83 0.71 0.50 0.45 0.33

    10.67 0.5 0.4 0.25 0.2 0.1

    l

    1=y-intercept

    l = 5

    l

    m- = gradient

    m = 6.25

    P1

    N1

    N1

    K1 for correct axes

    and scale

    N1 for all points

    plotted correctly

    N1 for line of

    best-fit

    K1 for y-intercept

    N1

    K1 for gradient

    N1

    10

    y

    1

    x

    10.2

    0

    2kk

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    7

    N0. SOLUTION MARKS

    8

    (a)

    (b)

    (c)

    (x 2 )2 = 4 x

    Q(3, 1 )

    A = [ ]dxxx ---3

    0

    2)2()4(

    = dxxx -3

    0

    2)3(

    =3

    0

    32

    32

    3

    -

    xx

    =2

    9

    Note : If use area of trapezium and dxy , give marks accordingly.

    V = -2

    0

    4

    )2( dxxp

    =2

    0

    5

    5

    )2(

    -xp

    = p

    5

    32

    K1 Solve for x

    N1

    K1 use

    dxyy - )( 12

    K1 integrate

    correctly

    K1 correct limit

    N1

    K1 integrate

    dxy2

    p

    K1 integrate

    correctly

    K1 correct limit

    N1

    10

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    8

    N0. SOLUTION MARKS

    9

    (i)

    (ii)

    (b)

    (i)

    (ii)

    (c)

    QS= QP + PS

    = 2~

    x + 3~

    y

    PR = PS + SR

    =~

    x + 2~

    y

    QT = m QS

    = m ( 2~

    x + 3~

    y )

    = 2m~

    x + 3m~

    y

    PT = n PR

    = n (~

    x + 2~

    y )

    = n~

    x + 2n~

    y

    PQ = PT + TQ

    2~

    x = n~

    x + 2n~

    y + 2m~

    x 3m~

    y

    = (n + 2m)~

    x + (2n 3m)~

    y

    n + 2m = 2

    2n 3m = 0

    m =7

    4

    n =7

    6

    K1 for using vector

    triangle

    N1

    N1

    N1

    N1

    K1 for substituting

    & grouping into

    components

    K1 for equating

    coefficients

    correctly

    K1 for eliminating

    m or n

    N1

    N1

    10

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    9

    N0. SOLUTION MARKS

    10

    (a)

    (b)

    (c)

    1

    1cos

    4

    1cos

    4

    POA

    POA-

    =

    =

    = 75.52 @75 31"o o

    = 1.318 rad.

    Arc PA = 20 ( 1.318 ) = 26.36

    PQ2

    = 802 20

    2

    PQ = 77.46

    Perimeter

    = 60 + 26.36 + 77.46

    = 163.82 cm

    Area OPQ = ( ) ( )1

    20 77.46 774.62

    =

    Area sector POA = ( ) ( )21

    20 1.318 263.62

    =

    Area of shaded region

    = 774.6 263.6

    = 511 cm2

    K1 Use correctly

    trigonometric

    ratio

    N1

    K1 Use s rq=

    K1 Use Pythagoras

    Theorem

    K1

    N1

    K1 Use formula

    1

    2A bh=

    K1 Use formula

    21

    2A r q=

    K1

    N1

    10

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    10

    N0. SOLUTION MARKS

    11

    (a)

    (i)

    (ii)

    (b)

    (i)

    (ii)

    p = 0.4 q = 0.6 n = 6

    P ( )4X

    ( ) ( ) ( )

    ( ) ( ) ( ) ( ) ( ) ( )4 2 5 1 6 06 6 6

    4 5 6

    4 5 6

    0.4 0.6 0.4 0.6 0.4 0.6

    0.13824 0.036864 0.004096

    0.1792

    P X P X P X

    C C C

    = = + = + =

    = + +

    = + +

    =

    2

    288npqs = =

    ( )( )288

    12000.4 0.6

    n = =

    62m= 8s =

    ( )

    ( )

    ( ) ( )

    60 68

    60 62 68 62

    8 8

    0.25 0.75

    1 0.25 0.75

    1 0.4013 0.2266

    0.3721

    P X

    P Z

    P Z

    P Z P Z

    <

    = - -

    =

    ( ) ( )68 0.75 0.22663P X P Z > = > =

    0.22663 3000 = 679.89

    = 679 @ 680

    P1 Value of p and/or q

    AND p + q =1

    K1 Use P(X = r)

    =n

    Cr prq

    nr

    N1

    K1

    N1

    K1 Use Z =s

    m-X

    K1

    N1

    K1

    N1

    10

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    11

    N0. SOLUTION MARKS

    12

    (a)

    (b)

    (c)

    8 ms-1

    dva

    dt

    = =02 2t = 1

    t = 12

    8 + 2t t2 = 0

    (t 4 ) (t + 2) = 0

    t = 4 t = 2 (not accepted)

    s v dt vdt + 4 50 4

    =t t

    t t t t + - + + -

    4 53 3

    2 2

    0 4

    8 83 3

    = ( ) + - - + + - - + -

    64 125 64 32 16 0 40 25 32 16

    3 3 3

    = + -80 103 3

    = 30 m

    N1

    K1

    N1

    K1

    K1

    N1

    (for t = 4 only)

    K1

    (for and 4 5

    0 4

    )

    K1

    (for integration)

    K1

    (for summation)

    N1

    10

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    12

    N0. SOLUTION MARKS

    13

    (a)

    (b)

    (c)

    (d)

    .

    x= 150 1000 4

    (or formula finding y /z)

    x = RM 0.60

    y = 120

    z = RM 2.00

    45o

    ( ) ( ) ( ) ( ) ( )x x x x xI

    + + + += 150 30 110 90 120 75 150 120 120 45360

    =46800

    360

    = 130

    P03 = ( )100

    5000130

    = RM 3846.2

    / .I x08 03 130 1 2 (or 130 + 130x0.2)

    = 156

    N1

    N1

    N1

    P1

    K1

    N1

    K1

    N1

    K1

    N1

    10

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    13

    N0. SOLUTION MARKS

    14

    (a)

    (b)

    (c)

    (d)

    40x + 20y 2000 or

    2x + y 100

    15x +30y 1200 or

    x + 2y 80

    y 3x

    (20, 60)

    (35, 30)

    2x + y = 100y = 3x

    x + 2y = 80

    y = 30

    100

    90

    80

    70

    60

    50

    40

    30

    20

    10

    10080604020 9070503010

    x

    y

    At least one straight line is drawn correctly from inequalities

    involving x and y.

    All the three straight lines are drawn correctly

    Region is correctly shaded

    35

    Maximum point (20, 60)

    Maximum profit = 20(20) + 30(60)

    = RM 2200

    N1

    N1

    N1

    K1

    N1

    N1

    N1

    N1

    K1

    N1

    10

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    N0. SOLUTION MARKS

    15

    (a)

    (b)

    (c)

    112

    = 92

    + 72

    2(9)(7)cosPQR

    cosPQR =9

    126

    PQR = 85o 54

    PSR = 180o 85o 54

    = 94o 6

    '

    sin sinPRS = 094 66 11

    PRS = 32o 57

    \ RPS = 180o 94o 6 32o 57= 52o 57

    ' 'sinsin

    oo

    RS = 1152 57

    94 6

    RS = 8.802 cm

    Area = ' '( )( )sin ( )( . )sino

    o +1 19 7 85 54 6 8 802 94 6 2 2

    = 31.42 + 26.34

    = 57.76

    K1

    K1

    N1

    P1

    K1

    K1

    N1

    K1, K1

    (for using

    area= absinc

    and summation)

    N1

    2kk