2013 foundation test
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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 ~
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