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  • 7/29/2019 2013 Foundation Test

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    CATHOLIC JUNIOR COLLEGE2013 H2 MATHEMATICSFOUNDATION TEST

    NAME: _________________________________ CLASS:________ DURATION: 60 MINS

    Answer all questions.

    1 [2012/JJC/I/9]The functions f and g are defined by

    f :x a x

    x1

    , x < 0 ,

    g :x a sinx, 0 2x .(i) Explain why f1 exists. [1](ii) Define f in a similar form. [3]

    (iii) Sketch, on the same diagram, the graphs of f( )y x= ,1

    f ( )y x= and1

    f f( )y x= , giving the

    coordinates of any points where the curves cross thex- andy- axes.[3]

    (iv) Show that the composite function fg does not exist. [1]

    (v) The function h is defined by

    h :x a sinx, 2x< < .Define fh and state the range of fh. [3]

    2 [2010/RVHS/Promo/6]

    The position vectors of vertices A,B and C, relative to the origin O, are 3 4 +i k , 2 4 3 + i j k and

    11 4 9 i j k respectively.

    (i) Find a unit vector parallel to OAuuur

    . [1]

    (ii) A point P dividesACin the ratio 1 : 3. Find the position vector ofP. [2]

    (iii) Show that the points O,B and P are collinear. [2]

    (iv) OBDCforms a parallelogram. Find the position vector ofD. [2]

    3 [2008/DHS/Promo/12(b)modified]

    The lines l1 and l2 have equations

    +

    = r ,

    0

    1

    0

    1

    1

    :1

    a

    l

    and

    +

    = r ,

    1

    1

    1

    0

    1

    2

    :2l

    respectively, where a is a constant.

    Find the value ofa such that l1 is perpendicular to l2.

    For the case where 2a = ,

    [2]

    (i) find the acute angle between the lines l1 and l2. [2]

    (ii) show that the position vector of the foot of the perpendicular from the point Q(3, 2,0 ) to the

    line l1 is7 6

    5 5+i j . [3]

    (iii) find the position vector of the image 'Q ,the reflection ofQ in the line l1. [2]

    42

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    4 [2010/MI/Promo/11]

    Differentiate each of the following with respect tox.

    (a) ( )x2cosln ,[2]

    (b) xx32

    e . [2]

    5 [2011/JJC/Promo/10]

    The curve Chas equation2 2

    4 4x y xy = .

    (i) Findd

    d

    y

    xin terms ofx andy. [2]

    (ii) Find the exact coordinates of the point(s) on C at which the tangent is parallel to they-axis.

    [3]

    (iii)The tangent and normal to Cat the point P(2, 0) meet they-axis at the points TandNrespectively.

    Find the equations of the tangent and normal to Cat P and deduce the area of PTN . [6]

    ~ End of paper~

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    CATHOLIC JUNIOR COLLEGE2013 H2 MATHEMATICSFOUNDATION TEST- MARK SCHEME

    MARKS: 421. (i)

    y

    0 x

    f( )y x=

    Since any horizontal line cuts the graph of f at most once, hence f is one-one.

    Thus, f-1

    exists.

    (ii) Letx

    xy1

    =

    42

    1

    2

    2+= y

    yx

    Butx < 0,21

    42 2

    yx y = +

    Hence, f-1

    :x21 4

    2 2

    xx + ,x (- , )

    (iii)

    y y x=

    f( )y x=

    (-1,0)

    01

    f ( )y x

    = x

    (0,-1)

    1f f( )y x=

    (iv) Df : (-,0) and Rg : [-1 , 1].

    Since Rg Df , therefore fg does not exist.

    (v) f h(x) = f(sinx)

    =x

    xsin

    1sin

    Hence, f h :xx

    xsin

    1sin ,

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    2. (i) Required unit vector =

    ( )2 2

    3 31 1

    0 053 4 4 4

    OA

    OA

    = = +

    uuur

    uuur

    (ii)13

    3

    +

    +=

    OCOAOP

    =

    +

    =

    4

    31

    2

    1

    9

    4

    11

    4

    0

    3

    34

    1OP

    (iii)

    =

    3

    4

    2

    OB

    =

    =

    3

    4

    2

    4

    1

    43

    12

    1

    OP

    = OB4

    1

    Since OB is parallel to OP, and O is a common point,O,B and P are collinear.

    (iv) BD OC=uuur uuur

    OD OB OC =uuur uuur uuur

    11 2 94 4 0

    9 3 12

    OD OC OB = +

    = + =

    uuur uuur uuur

    3. l1 is perpendicular to l2

    0

    1

    1

    1

    0

    1 =

    a

    01 =+ a 1= a

    (i)

    +

    = r ,

    0

    1

    2

    0

    1

    1

    :1l

    +

    = r ,

    1

    1

    1

    0

    1

    2

    :2l

    O

    3A CP1

    B D

    CO

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    222222 )1(11012

    1

    1

    1

    0

    1

    2

    cos++++

    =

    35

    3cos 1=

    = 39.2o

    (ii) Let Fbe the foot of the perpendicular from point Q(3, 2,0 ) to the line l1.

    Since Flies on l1,

    +

    +

    =

    somefor

    0

    1

    21

    OF

    2 2

    3

    0

    QF OF OQ

    +

    = = +

    0

    0

    1

    2

    .

    0

    3

    22

    0

    0

    1

    2

    .1 =

    +

    +

    =

    QFlQF

    0344 =+++

    5

    1=

    Thus

    ( )1 7 551 6

    55

    1 27 6

    1 =5 5

    0 0

    OF

    +

    = + = +

    i j

    (iii) Let Q be the point of reflection ofQ about the line l1.

    +=

    '2

    1OQOQOF

    = OQOFOQ 2'

    =

    0

    2

    3

    05

    65

    7

    2

    =

    05

    2251

    Q

    F

    0

    1

    2

    Q

    Q

    F

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    4. (a) Let ( )xy 2cosln=

    ( )22cos

    2sin

    d

    d

    =

    x

    x

    x

    y

    xx

    x2tan2

    2cos

    2sin2==

    (b) Letx

    xy32

    e=

    xx

    x

    y xx 2e3e

    d

    d 332+=

    ( )23e3 += xx x

    5. (i) 2 8 0dy dy

    x y x ydx dx

    + =

    2 2or

    8 8

    dy y x x y

    dx y x x y

    =

    +

    (ii) For tangents // toy-axis,dy

    dxis undefined, i.e. 8 0

    8

    x y

    x y

    + =

    =

    Sub into equation ofC: ( ) ( )2 2

    2

    8 4 8 4

    1

    17

    1

    17

    =

    =

    =

    y y y y

    y

    y

    8

    17= mx

    The points are8 1 8 1

    , & ,17 17 17 17

    .

    (iii) At P(2, 0), 2dy

    dx=

    Eqn of tangent at P: ( )0 2 2 2 4y x y x = =

    Eqn of normal at P: ( )1 1

    0 2 12 2

    y x y x = = +

    Pt Tis (0, 4)PtNis (0, 1)

    Area of ( ) ( )1

    1 4 2 52

    PTN = = sq units

    ~ The End ~