SPM Trial 2009 AddMath Q&A (Kedah)

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  • 8/14/2019 SPM Trial 2009 AddMath Q&A (Kedah)

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    3472/1 [Lihat sebelahSULIT

    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 22 43 34 35 36 47 48 39 3

    10 311 312 413 314 315 216 317 4

    18 419 320 221 422 323 324 325 4

    TOTAL 80

    M AT E M AT I K TA M B A H A N

    Kertas 1Dua jam

    J A N G A N B U K A K E RTA S S O A L A N IN ISEHINGGA DIBERITAHU

    1 This question paper consists of 25 questions.

    2. Answer all questions.

    3. Give onlyone answer for each question.

    4. Write your answers clearly in the spaces provided inthe question paper.

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

    6. If you wish to change you r answer, cross out the work that you have done. Then write down the newanswer.

    7. The diagrams in the questions provided are not drawn to scale unless stated.

    8. The marks allocated for each question and sub-part of a question are shown in brackets.

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

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

    .11 You m ay use a non-programm able scientific

    calculator.

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

    Name : ..

    Form : ..

    3472/1 Additional Mathematics

    Paper 1Sept 20092 Hours

    ADDITIONAL MATHEMATICSPaper 1

    Two hours

    3 to 4.

    2kk

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    BLANK PAGE HALAMAN KOSONG

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

    ALGEBRA

    12 4

    2

    b b ac x

    a

    - -=

    2 a m a n = a m + n

    3 a m a n = a m - n

    4 (a m) n = a nm

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

    6 log anm

    = log a m log a n

    7 log a mn

    = n log a m

    8 log ab =a

    b

    c

    c

    log

    log

    9 T n = a + (n 1)d

    10 S n = ])1(2[2

    d nan -+

    11 T n = arn-1

    12 S n =r r a

    r r a nn

    --=

    --

    1)1(

    1)1(

    , (r 1)

    13r

    aS

    -=

    1, r

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    STATISTIC

    1 Arc length, s = r q

    2 Area of sector , A = 212

    r q

    3 sin 2 A + cos 2 A = 1

    4 sec 2 A = 1 + tan 2 A

    5 cosec 2 A = 1 + cot 2 A

    6 sin 2 A = 2 sin A cos A

    7 cos 2 A = cos 2 A sin 2 A= 2 cos 2 A 1= 1 2 sin 2 A

    8 tan 2 A = A

    A2tan1

    tan2-

    TRIGONOMETRY

    9 sin ( A B) = sin A cos B cos A sin B

    10 cos ( A B) = cos A cos B m sin A sin B

    11 tan ( A B) = B A B A

    tantan1tantan

    m

    12C

    c B

    b A

    asinsinsin

    ==

    13 a 2 = b2 + c2 2bc cos A

    14 Area of triangle = C ab sin21

    1 x = N

    x

    2 x =

    f

    fx

    3 s = N

    x x - 2)(= 2

    2

    x N

    x -

    4 s = -

    f

    x x f 2)(= 2

    2

    x f

    x f -

    5 m = C f

    F N L

    m

    -+ 2

    1

    6 10

    100Q

    I Q

    =

    71

    11

    w I w

    I

    =

    8)!(

    !r n

    nPr n

    -=

    9!)!(

    !r r n

    nC r

    n

    -=

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

    11 P ( X = r ) = r nr r n q pC - , p + q = 1

    12 Mean = np

    13 npq=s

    14 z =s

    m - x

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    Answer all questions. Jawab semua 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 of x.Cari nilai x. [ 2 marks]

    [2 markah]

    Answer/Jawapan : (a) ..

    (b) ...

    2. Given5

    3:1

    x x f

    -- , find the value of

    Diberi5

    3:1

    x x f

    -- , cari nilai bagi

    (a) )3(- f ,

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

    [4 markah]

    Answer/ Jawapan : (a) ..

    (b) ... 4

    2

    2

    1

    For examiners

    use only

    Set A Set B

    49

    9

    x 9

    7

    3

    Diagram 1Rajah 1

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    3. Given that function a x xg + 2: and 9:2 - bx xg . Diberi fungsi a x xg + 2: dan 9:2 - bx xg .

    Find the value of a and of bCari nilai bagi a dan b .

    [3 marks][3 markah]

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

    4. Given that the straight line 14 += x y is a tangent to the curve k x y += 2 .Find the value of k .

    Diberi garis lurus 14 += x y ialah tangen kepada lengkung k x y += 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 --= x y

    Diagram above shows the graph of the function 4)2(3 2 --= x y . 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 --= x y . 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 a and b are the roots of the quadratic equation 0792 =+- x x .Find the value of

    Diberi a dan b adalah punca bagi persamaan kuadratik 0792 =+- x x . Cari nilaibagi

    (a) b a +(b) ab

    (c) 22 b a + [ 4 marks ][4 markah]

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

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

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

    3

    5

    4

    6

    For examiners

    use only

    x

    y

    P

    Q

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    7. Solve the equation:Selesaikan persamaan :

    x x x 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 of x and of y if y > 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

    For examiners

    use only

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    9. Given that 2loglog 93 =+ y x . Express y in terms of x.[ 3 marks ]

    Diberi 2loglog 93 =+ y x . 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

    For examiners

    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 is31- , 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 =+ r a 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 =+ r a dan hasil tambah hingga sebutan ketakterhinggaan ialah 32,cari nilai a dan r .

    [4 markah]

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

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

    12

    For examiners

    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, c arikan 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 variables x and y are related by the equation 2310 -= x yFind the value of h and of k .

    Pembolehubah x dan y dihubungkait dengan persamaan 2310 -= x y .Cari nilai h dan k.

    [3 marks][3 markah]

    Answer/Jawapan : h=

    k =....

    3

    13

    3

    14

    For examiners

    use only

    ),0( k

    )2,(h y10log

    x

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    15.

    In the diagram above, jniOA += 8 and ji AB 23 -= . GivenunitsOA 10= , find

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

    unit OA 10= , cari

    (a) the value of n.nilai n.

    (b) coordinates of B.koordinat B.

    [2 marks][2 markah]

    Answer/Jawapan : (a) n =

    (b)..

    16 Given that ji pa 2+= and jib --= 2 , find the value of p if ba - is parallelto j .

    [ 3 marks ]

    Diberi ji pa 2+= dan jib --= 2 , cari nilai p jika ba - selari dengan j .

    [3 markah]

    Answer/Jawapan :..

    2

    15

    For examiners

    use only

    3

    16

    y

    xO

    A

    B

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    17. Solve the equation 0cotcos2 =+ x 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 p 32

    radian and the area of the shaded

    region is 212 cmp . Find the length of the minor arc PQ.

    Diberi sudut minor POQ ialah p 32

    radian dan luas sektor berlorek

    ialah 212 cmp . Cari panjang lengkok minor PQ.

    [4 marks][4 markah]

    Answer/Jawapan :

    _

    p 32 _ 1O

    P

    4

    17

    4

    18

    For examiners

    use only

    Q

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    19 . Find the gradient of the curve25

    23

    4 2 -+= x

    x y at the point )3,1( .

    [ 3 marks ]

    Cari kecerunan kepada lengkung2

    5

    2

    34 2 -+=

    x x y pada titik )3,1( .

    [3 markah]

    Answer/Jawapan:

    20 . Differentiate12

    14 2

    +

    -

    x

    xwith respect to x .

    [2 marks]

    Bezakan1214 2

    +-

    x x

    terhadap x.

    [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

    + 2 x , 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 and x is increasing at the rate of 2 units per second, find the

    rate of change of y when21= x . [ 3 marks ]

    Diberi10

    )32( 5-= x y dan x bertambah dengan kadar 2 unit sesaat, cari kadar

    perubahan bagi y apabila21= x . [3 markah]

    Answer/Jawapan : .

    4

    21

    For examiners

    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, carikebarangkalian dia lulus ujian matematik bulan April dengan cemerlang.

    (b) If 5 students of Form 5 Euler were selected at random, find the probability thatonly one of the five students selected passed the April mathematics test withdistinction.

    Jika 5 orang pelajar dari tingkatan 5 Euler dipilih secara rawak, carikebarangkalian hanya seorang daripada lima pelajar terpilih lulus ujianmatematik 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 numberscan 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

    For examiners

    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( < X P .

    [4 marks ][4 markah]

    Answer/Jawapan : (a)

    (b)

    END OF QUESTION PAPER KERTAS SOALAN TAMAT

    4

    25

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    SULIT3472/2Additional MathematicsKertas 2September, 20092 jam 30 minit

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH

    NEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICS

    Kertas 2

    Dua jam tiga puluh mini t

    JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

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

    2. Answer all questions in Section A, four questions from Section B and two questions from 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 of 19 printed pages and 1 blank page.

    3472/2 [Lihat sebelah

    SULIT

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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. d naT n )1( -+=

    3. aaa nmnm-= 10. ])1(2[

    2d na

    nS n -+=

    4. aamnnm =)( 11.

    1-= nn ar T

    5. nmmn aaa logloglog +=12.

    r r a

    r r a

    Snn

    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 ,dxdu

    vdxdv

    udxdy += 4 Area under a curve

    = ba dx y or

    = ba dy x

    2. y = vu

    , 2vdxdv

    udxdu

    v

    dxdy

    -

    =

    5. Volume of revolution

    = ba

    dx y 2p or

    = ba dy x2p

    3.dxdu

    dudy

    dxdy =

    GEOMETRY

    1. Distance = 2122

    12 )()( y y x x -+-4. Area of triangle

    = 1 2 2 3 3 1 2 1 3 2 1 31

    ( ) ( )2

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

    2. Mid point

    ( x , y ) =

    ++

    2,

    22121 y y x x

    5. 22 y xr +=

    3. Division of line segment by a point

    ( x , y ) =

    +

    ++

    +nm

    mynynmmxnx 2121 ,

    6.2 2

    xi yj

    r x y

    +=

    +% %

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    3

    STATISTICS

    1. N

    x x

    = 7 =

    i

    ii

    W

    I W I

    2.=

    f fx

    x 8)!(

    !r n

    nPr

    n-

    =

    3. N

    x x -=2)(

    s = 22

    x N x - 9 !)!(

    !r r n

    nC r

    n-

    =

    4. -=

    f x x f 2)(

    s = 22

    x f

    fx -

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

    11 P ( X = r ) = r nr r n q pC - , p + q = 1

    5. m = L + C f

    F N

    m

    -21 12 Mean , m = np

    13 npq=s

    6. 10001 = Q

    Q I 14 Z = s

    m - X

    TRIGONOMETRY

    1. Arc length, s = r q 8. sin ( A B ) = sin A cos B cos A sin B

    2. Area of sector, A = q 221

    r 9. cos ( A B ) = cos A cos B m sin A sin B

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

    B A B A

    tantan1tantan

    m

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

    A A2tan1

    tan2-5. cosec A = 1 + cot A

    12C

    c B

    b A

    asinsinsin

    ==

    6. sin 2 A = 2sin A cos A 13 a = b + c 2 bc cos A7. cos 2 A = cos A sin A

    = 2 cos A 1= 1 2 sin A

    14 Area of triangle =1

    sin2

    ab C

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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 ]

    Answer all questions. Jawab semua soalan.

    1. Solve the simultaneous equations 3 2 0 x y- - = and ( )2 1 3 x y - = .Give your answers correct to three decimal places.

    [5 marks ]

    Selesaikan persamaan serentak 3 2 0 x y- - = dan ( )2 1 3 x y - = . Beri jawapan anda betul kepada tiga tempat perpuluhan.

    [5 markah ]

    2.

    In the diagram, the gradient and y-intercept of the straight line PQare 2 and 3 respectively. R is a point on the x-axis.

    (a) Find the value of h and k .

    (b) Given that PQ is perpendicular to QR , find the x-intercept of QR .

    (c) Calculate the area of PQR .

    [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-xbagi QR.

    (c) Hitungkan luas PQR.

    [3 markah ]

    [3 markah ]

    [2 markah ]

    x

    y

    0

    Q(4, k )

    P (h, 3) R

    j2kk

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    3472/2 Additional Mathematics Paper 2 [Lihat sebelah

    SULIT

    5

    3. (a) Sketch the graph of 2 sin 2 y x= - for 0 2 x p .

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

    the number of solutions for the equation sin22

    x x

    p = for

    0 2 x p . State the number of solutions.

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

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

    bagi persamaan sin22 x

    xp

    = untuk 0 2 x p .

    Nyatakan bilangan penyelesaian itu.

    [4 marks ]

    [3 marks ]

    [4 markah ]

    [3 markah ]

    4. Given that ( )2 1 x x - is the gradient function of a curve which passesthrough the point )1,1( -P . Find

    (a) the gradient of the tangent to the curve at P,

    (b) the equation of the curve,

    (c) the coordinates of the turning point at x = 1 . Hence determinewhether the turning point is a maximum or a minimum point.

    [1 mark ]

    [3 marks ]

    [3 marks ]

    Diberi ( )2 1 x x - ialah fungsi kecerunan bagi suatu lengkung yangmelalui 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 titik minimum.

    [1 markah ]

    [3 markah ]

    [3 markah ]

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    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 astandard deviation of 8.

    (a) Find

    (i) x ,(ii)

    2 x

    (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) x ,(ii)

    2 x

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

    (i) min ,(ii) varians .

    [3 markah ]

    [3 markah ]

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    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 thesmallest 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 numberof 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 membentuk beberapa semi bulatan. Rajah menunjukkan empat semi bulatan

    pertama yang telah dibentukkan. Jejari semi bulatan yang terkecilialah 5 cm . Jejari semi bulatan yang berikutnya bertambahsebanyak 3cm 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 semuasemi 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 ]

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    SULIT

    8

    Section B Bahagian B

    [ 40 marks ][ 40 markah ]

    Answer four questions from this section. Jawab empat soalan daripada bahagian ini.

    7.

    The table shows the values of two variables, x and y, obtained froman experiment. Variables x and y are related by the equation

    1=+ y xl m , where m and l are constants.

    (a) Plot y1

    against x1

    , 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=+ y xl m

    , dengan keadaan m dan l

    adalah pemalar

    (a) Plot y1

    melawan x1

    , 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

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    9

    8.

    The diagram shows straight line x + y = 4 that intersects with thecurve y = ( x 2 ) 2 at points P and Q.

    Find

    (a) the coordinates of Q,

    (b) the area of the shaded region A,

    (c) the volume generated, in terms of , when the shaded region B is revolved through 360 o about the x-axis.

    [2 marks ]

    [4 marks ]

    [4 marks ]

    Rajah menunjukkan garis lurus x + y = 4 bersilang denganlengkung y = ( x 2 ) 2 pada titik P dan Q .

    Cari

    (a) koordinat Q ,

    (b) luas rantau berlorek A ,

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

    dikisarkan melalui 360 o pada paksi-x.

    [2 markah ]

    [4 markah ]

    [4 markah ]

    O x

    y = ( x 2 ) 2P

    Q A B x + y = 4

    y

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    SULIT

    10

    9.

    In the diagram, PQRS is a quadrilateral. The diagonals PR 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 areconstants, 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 dannilai n. [10 markah ]

    P

    T

    S

    R

    Q

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    11

    10.

    The diagram shows the cross-section of a cylindrical roller withcentre O and radius 20 cm resting on a horizontal ground PQ .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 cm 2 , 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 mewakili pemegang 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

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    12

    11 (a) In a survey carried out in a school, it is found that 40% of itsstudents 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 areparticipating 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 theschool .

    [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 thefarm has a mass between 60 g and 68 g.

    (ii) If the farm produces 3000 eggs daily, calculate the number of eggs 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 itumengambil 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 bilanganmurid dalam sekolah itu .

    [5 markah ](b) Jisim telur ayam dari sebuah ladang adalah mengikut satu taburan

    normal dengan min 62 g dan sisihan piawai 8 g . Sebarang telur dengan jisim melebihi 68 g dikategorikan sebagai telur gred double-A

    (i) Cari kebarangkalian bahawa sebiji telur yang dipilih secararawak 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 ]

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    13

    Section C Bahagian C

    [ 20 marks ][ 20 markah ]

    Answer two questions from this section . Jawab dua soalan daripada bahagian ini.

    12. A particle P starts from a fixed point O and moves in a straightline so that its velocity, v ms -1 , is given by v = 8 + 2t t 2, wheret 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 first5 seconds.

    Suatu zarah P mula dari suatu titik tetap O dan bergerak disepanjang garis lurus. Halajunya v ms -1 , diberi olehv = 8 + 2t t 2 , dengan keadaan t ialah masa, dalam saat, selepasmelalui 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 rehat seketika,

    (d) jumlah jarak yang di lalui, dalam m , oleh zarah P dalam5 saat yang pertama.

    [1 mark ]

    [2 marks ]

    [3 marks ]

    [4 marks ]

    [1 markah ]

    [2 markah ]

    [3 markah ]

    [4 markah ]

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    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 theseitems.

    E

    D

    C

    B

    A

    30

    75

    120

    Items

    Bahan

    Price (RM)per kg for the

    year 2003

    Harga (RM) per kg padatahun 2003

    Price (RM)per kg for the

    year 2006

    Harga (RM) per kg padatahun 2006

    Price index for theyear 2006 based on

    the year 2003

    Indeks harga padatahun 2006

    berasaskan tahun2003

    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 expenditurein the year 2003.

    (d) The price of each item increases by 20 % from the year 2006to the year 2008. Find the composite index for totalexpenditure on the items in the year 2008 based on the year2003.

    [3 marks ]

    [3 marks ]

    [2 marks ]

    [2 marks ]

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    SULIT

    15

    Carta pai menunjukkan lima bahan, A, B, C, D dan E yangdigunakan 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 kek itu pada tahun 2006 berasaskan tahun 2003 .

    (c) Jumlah kos bahan-bahan tersebut pada tahun 2006 ialahRM 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 tahun2008 berasaskan tahun 2003 .

    [3 markah ]

    [3 markah ]

    [2 markah ]

    [2 markah ]

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    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 timetaken to produce components S and T respectively.

    In any given week, the factory produces x units of component S

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

    I : Machine P operates not more than 2000 minutes.

    II : Machine Q operates at least 1200 minutes.

    III : The number of component T produced is not more than threetimes the number of component S produced.

    (a) Write three inequalities, other than x 0 and y 0, whichsatisfy 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 aboveconstraints.

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

    (i) the maximum number of component S that could beproduced, if the factory plans to produce only 30 unitsof component T,

    (ii) the maximum profit per week if the profit from a unitof component S is RM20 and from a unit of component T is RM30.

    Time taken (minutes) Masa diambil (minit)

    Component

    Komponen Machine P Mesin P

    Machine Q Mesin Q

    S 40 15

    T 20 30

    [3 marks ]

    [3 marks ]

    [4 marks ]

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    SULIT

    17

    Gunakan kertas graf untuk menjawab soalan ini.

    Sebuah kilang menghasilkan dua komponen, S dan T bagisesuatu 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 xunit bagi komponen S dan y unit bagi komponen T.Penghasilan komponen-komponen tersebut adalah berdasarkankekangan berikut:

    I : Mesin P beroperasi tidak melebihi 2000 minit.

    II : Mesin Q beroperasi sekurang-kurangnya 1200 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 , yangmemenuhi semua kekangan di atas.

    (b) Menggunakan skala 2 cm kepada 10 unit pada kedua-dua paksi, bina dan lorek rantau R yang memenuhi semuakekangan 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 dandari satu unit komponen T ialah RM30 .

    [3 markah ]

    [3 markah ]

    [4 markah ]

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    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 of RS.

    (c) Calculate the area of PQRS .

    S R

    Q

    P

    11 cm

    7 cm

    9 cm

    [3 marks ]

    [4 marks ]

    [3 marks ]

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    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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    Additional Mathematics paper 1

    Nama Pelajar : Tingkatan 5 : .3472/1

    Additional Mathematics Paper1September 2009

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH

    NEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICSMARKING SCHEME

    Paper 1

    .

    SULIT 3472/1

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    Question Solution/ Marking Scheme Answer Marks

    8B2: 22y x

    B1: 87

    119752 y x

    12,10 y x

    13,9 y x 3

    9 B2: 2log4log 292

    3 y xor y x

    B1: 2log3log

    log2

    9loglog

    log 99

    9

    3

    33 y

    xor

    y x

    281

    x y 3

    10B2: 513 d and a

    B1: 3710125 d aor d a 62 3

    11

    (a) B1:31

    3636 p

    (a) 48

    (b) 4

    2

    1

    12B3: 48

    21

    aor r

    B2: 32

    961

    096)1(32

    aa

    or r r

    B1: 321 r

    a

    21,48 r a 4

    13

    B2 : 4437

    44

    23 yand

    x

    B1 : 4437

    44

    23 yor

    x

    ( 6, 3) 3

    14

    B2 : 23230

    22 hor

    hk

    or k

    B1 : 23log 10 x y

    h =34

    k = 2

    3

    15(a) 6

    (b) (11, 4)

    1

    1

    3472/1 Additional Mathematics Paper 1 SULIT

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    SULIT 3472/14

    Question Solution/ Marking Scheme Answer Marks16 B2: 02 p

    B1: ji p 3)2( p = 2 3

    17 B3:

    )42(330,210,270,90 0000 of out any

    B2 : )(21

    sin,0cos both x x

    B1 : 0sincos

    cos2 x x

    x

    0

    0

    00

    330

    ,270

    ,210,90

    4

    18B3 : )

    32(6 S

    B2 : 6r

    B1: 12)32

    (21 2r

    12.57 cm( 4 ) cm

    4

    19B2:

    2)1(2

    3)1(8

    B1:22

    38

    x x

    dxdy

    2

    13

    3

    20 B1: 12x y or

    2

    2

    )12(

    )14(2)8)(12(

    x

    x x xdxdy

    2 2

    21

    B3 : c123

    2

    B2:2

    22

    )12(3 21 xand

    x

    B1:2

    22

    )12(3 21 xor

    x

    25

    )12(23 2x

    x y 4

    3472/1 Additional Mathematics Paper 1 SULIT

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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(5 4xdt dy

    B1 : 210

    2)32(5 4

    dt dx

    or x

    dxdy

    32 units persecond

    3

    23(b) B1: 411

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

    (b) 0.4069

    1

    2

    24 B2: 34

    13

    PP

    B1: 34

    13 Por P

    72 3

    25(a) B1 :

    10370355

    (b) B1 :10

    370367

    (a) 1.5

    (b) 0.3821

    2

    2

    END OF MARKING SCHEME

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    SULIT JPNKd/2006/ 3472/2

    Nama Pelajar : Tingkatan 5 : . 3472/2

    Additional Mathematics

    September 2009

    PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUASEKOLAH MENENGAH

    NEGERI KEDAH DARUL AMAN

    PEPERIKSAAN PERCUBAAN SPM 2009

    ADDITIONAL MATHEMATICS

    Paper 2

    .

    MARKING SCHEME

    SULIT 3472/2

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    2

    MARKI NG SCHEM EADDITIONAL M ATHEMATICS PAPER 2

    SPM TRIAL EXAMINATI ON 2009

    N0. SOLUTION M ARK S

    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 x x x

    x x

    x x

    x

    x or x

    y y

    = -= - -=> = - -

    - - =

    - - -=

    = = -= - = - -

    = = -

    P1K1 Eliminate y

    K1 Solve quadraticequation

    N1

    N1

    52

    (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

    03

    2 114 0

    hh

    k k

    + = = --- = =-

    1 21

    22

    11 14 2

    1: 13

    20 26

    m m

    y x

    QR y x

    y x

    = = -

    - = --

    = - +

    = =

    OR

    11 0 14 2

    111 2

    226

    x

    x

    x

    - = --

    = - +

    =

    x-intercept = 26

    Area

    3 4 26 313 11 0 32

    133 78 12 286

    21

    3852192.5

    - -=

    - -

    = - - + -

    = -

    =

    K1 Use equation orgradient

    N1

    N1

    P1 For 212

    m = -

    K1 Use equation orgradient

    N1

    K1 Use formula andfind the areatriangle

    N1

    8

    2kk

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    N0. SOLUTION M ARK S3

    (a)

    (b) 2 sin 2 22 x

    xp

    - = -

    or

    22 x

    yp

    = -

    Draw the straight line 22 x

    yp

    = -

    Number of solutions = 4 .

    P1 Negative sineshape correct.

    P1 Amplitude = 1[ M aximum = 3and Minimum =1 ]

    P1 Two full cycle in0 x 2p

    P1 Shift up the graph

    N1 For equation

    K1 Sketch thestr aight l ine

    N1

    7

    3

    1 2 2 x

    yp

    = -

    y

    2p 2p xO p

    2 sin 2 y x= -

    2

    23p

    2kk

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    N0. SOLUTION M ARK S4

    (a)

    (b)

    (c)

    ( ) ( )21 1 1 2dydx

    = - - - = -

    3 2

    4 3

    4 3

    ( )

    4 31 1

    14 35

    12

    54 3 12

    y x x dx

    x x y c

    c

    c

    x x y

    = -

    = - +

    = + +

    =

    = - +

    ( )2

    22

    2

    2

    2

    1 00@1

    3 2

    1 1 0

    1 1 5 1

    4 3 12 3

    dy x xdx

    x

    d y x x

    dx

    d y x

    dx

    y

    = - ==

    = -

    = = >

    = - + =

    11,

    3

    Minimum point.

    N1

    K1 Integrate gradientfunction

    K1 Substi tute ( 1,1)into equation y

    N1

    K1 Find2

    2d y

    dx

    N1 N1

    7

    2kk

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    N0. SOLUTION M ARK S5

    (a)(i)

    (ii)

    (b)(i)

    (ii)

    11 55050

    x x= => =

    22

    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 ofmean and/orstandard deviation

    N1

    N1

    K1 Find new meanand/or new

    varianceN1

    N1

    66

    (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 T n = 104 p

    N1

    K1 Find S 34

    N1

    K1 RM4 S34

    N1

    7

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    N0. SOLUTION M ARKS7

    (a)

    (b)(i)

    (ii)

    l l m 111 +-=

    x y

    y1 1.04 0.83 0.71 0.50 0.45 0.33

    x1

    0.67 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 axesand scale

    N1 for all pointsplotted corr ectly

    N1 for l ine ofbest-fit

    K1 for y-intercept

    N1

    K1 for gradient

    N1

    10

    y1

    x10.2

    0

    2kk

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    N0. SOLUTION M ARKS

    8(a)

    (b)

    (c)

    ( x 2 ) 2 = 4 xQ(3, 1 )

    A = [ ]dx x x ---3

    0

    2)2()4(

    = dx x x -3

    0

    2 )3(

    =3

    0

    32

    323

    - x x

    =29

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

    V = -2

    0

    4

    )2( dx xp

    =2

    0

    5

    5)2(

    - xp

    = p

    5

    32

    K1 Solve for xN1

    K1 use

    dx y y - )( 12K1 integrate

    correctly

    K1 correct limit

    N1

    K1 integrate

    dx y 2p

    K1 integratecorrectly

    K1 correct limit

    N1

    10

    2kk

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    N0. SOLUTION M ARKS9(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 )

    = 2 m~

    x + 3 m~

    y

    PT = n PR

    = n (~

    x + 2~

    y )

    = n~

    x + 2 n~

    y

    PQ = PT + TQ

    2~

    x = n~

    x + 2 n~

    y + 2 m~

    x 3 m~

    y

    = ( n + 2 m)~

    x + (2 n 3m)~

    y

    n + 2 m = 2

    2n 3m = 0

    m =74

    n =76

    K1 for using vectortriangle

    N1

    N1

    N1

    N1

    K1 for substitut ing& grouping intocomponents

    K1 for equatingcoefficientscorrectly

    K1 for eliminating m or n

    N1

    N1

    10

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    N0. SOLUTION M ARKS10(a)

    (b)

    (c)

    1

    1cos

    41

    cos

    4

    POA

    POA -

    =

    =

    = 75.52 @75 31"o o

    = 1.318 rad.

    Arc PA = 20 ( 1.318 ) = 26.36

    PQ 2 = 80 2 202

    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 cm 2

    K1 Use correctlytrigonometricratio

    N1

    K1 Use s r q =

    K1 Use Pythagoras

    Theorem

    K1

    N1

    K1 Use formula12

    A bh=

    K1 Use formula

    212

    A r q =

    K1

    N1

    10

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    N0. SOLUTION MARKS11(a)(i)

    (ii)

    (b)(i)

    (ii)

    p = 0.4 q = 0.6 n = 6

    P ( )4 X

    ( ) ( ) ( )

    ( ) ( ) ( ) ( ) ( ) ( )4 2 5 1 6 06 6 64 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 68 628 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 qAND p + q =1

    K1 Use P(X = r)= n C r prqnr

    N1

    K1

    N1

    K1 Use Z =s

    m - X

    K1N1

    K1

    N1

    10

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    N0. SOLUTION M ARK S12(a)

    (b)

    (c)

    8 ms -1

    dva

    dt

    = =0

    2 2t = 1

    t = 1 2

    8 + 2t t 2 = 0

    (t 4 ) (t + 2) = 0

    t = 4 t = 2 (not accepted)

    s v dt v dt + 4 5

    0 4

    =t t

    t t t t

    + - + + -

    4 5 3 3 2 2

    0 4

    8 8 3 3

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

    64 125 64 32 16 0 40 25 32 16

    3 3 3

    = + -80 10 3 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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    N0. SOLUTION M ARK S13(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 x I

    + + + += 150 30 110 90 120 75 150 120 120 45 360

    =46800

    60

    = 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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    N0. SOLUTION MARKS14(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 = 100 y = 3x

    x + 2y = 80

    y = 30

    100

    90

    80

    70

    60

    50

    40

    30

    20

    10

    10080604020 9070503010x

    y

    At least one straight line is drawn correctly from inequalitiesinvolving 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 M ARK S15(a)

    (b)

    (c)

    11 2 = 9 2 + 7 2 2(9)(7)cos PQR

    cosPQR =9

    126

    PQR = 85o 54

    PSR = 180 o 85 o 54

    = 94 o 6

    '

    sin sinPRS =0

    94 6 6 11

    PRS = 32 o 57

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

    ' 'sinsin

    oo

    RS = 11 52 57

    94 6

    RS = 8.802 cm

    Area = ' '( )( )sin ( )( . )sinoo +1 1 9 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= absincand summation)

    N1

    2kk