2013 foundation test

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


2/6

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 ,


4/6

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 ~

2013 foundation test

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