catholic junior college - wordpress.com junior college h2 mathematics jc2 preliminary examination...
TRANSCRIPT
Page 1 of 19
CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS JC2 PRELIMINARY EXAMINATION PAPER 1 SOLUTIONS 2015
1
No.
Feedback
2
2 2
4 3 22
w z z
z z z
4 3 22 2 3 10 0z z z z
4 3 2 22 3 3 10 0z z z z z
4 3 2 22 3 10 0z z z z z
2 3 10 0w w
5 2 0w w
5w or 2w 2 5z z 2 2z z 2 5 0z z 2 2 0z z
51
2
1 4z
1 41 2
2z
11
2
2
1
2
7
i1
2
7
Page 2 of 19
2
No.
Method :
3 2
5OP
b a
Since F lies on line passing through points O and A , OF a for a lambda value.
3 23
55 5
2FP
a b a
b a
Since FP is perpendicular to OA
2
2
3 20
5 5
3 20
5 5
3 2cos
5 5
3 π 24 1 cos 1 0
5
0
0
3 5
8
5
FP OA
b a a
b a a a
b a a
Therefore 8
5OF a
Method :
2
3
Page 3 of 19
Triangle ABE is similar to triangle APF
3
5
AP AF PF
AB AE BE
cosπ
3
1
4 2
2
OE
OB
OE
OE
Then 2 1 1AE OE OA .
Then 3 3 3
15 5 5
AF AE
Hence 3 8
15 5
OF OA AF .
5
8OF a
E
2
3
Page 4 of 19
3
No.
(i) Odd day duration forms AP with 20a , 9d , 38n
38 20 38 1 9
353
U
(ii) Odd day duration forms AP with 20a , 9d , 38n
38
382 20 38 1 9 7087
2S
Even day duration form AP with 20a , 10d , 37n
37
372 20 37 1 10 7400
2S
Total no. = 7087+7400 = 14487
(iii) 75th day onwards forms GP with 353a , 0.8r , 26n
26 1
26 353 0.8
1.33
U
(iv) Duration of exercise is too low to be effective towards the end portion of the 100 days.
Page 5 of 19
4
No.
(i) R.H.S.
11 2 1 1
3r r r r r r
1
2 1 13
r r r r
1
3 13
r r
1r r =L.H.S.
(ii) 1 2 2 3 3 4 1n n
1
( 1)n
r
r r
1
11 2 1 1
3
n
r
r r r r r r
11 2 3 0 1 2
3
2 3 4 1 2 3
3 4 5 2 3 4
( 1)( 2) ( 1) ( 1)n n n n n n
1
1 2 0 1 23
n n n
1
1 23
n n n
Page 6 of 19
(iii) Let Pn be the statement
1
1( 1)( 2) 1 2 3
4
n
r
r r r n n n n
,where n
When 1n , L.H.S. 1 2 3 6
1
R.H.S. 1 2 3 4 6 L.H.S.4
1P is true.
Suppose Pk is true for some k , i.e.
1
11 2 1 2 3
4
k
r
r r r k k k k
R.T.P. 1Pk is true, i.e.
1
1
11 2 1 2 3 4
4
k
r
r r r k k k k
L.H.S. 1
1
1 2k
r
r r r
1
1 2 1 2 3k
r
r r r k k k
1
1 2 3 1 2 34
k k k k k k k
1
1 2 3 44
k k k k R.H.S.
1Pk is true.
Since 1P is true, Pk
is true 1Pkis true. By mathematical induction, Pn
is true for all n .
(iv)
1
1 2 3
11 2 3 4
5
n
r
r r r r
n n n n n
Page 7 of 19
5
No.2
(i) 1
11
22 2n nv v
1
11
22 1n nv v
1
1
2n nv v
Since 1 1constant
2
n
n
v
v
, sequence V is a GP.
(ii) Method :
1 1 2 1v u
1
11
2
n
nv
when n , 1
11
20
n
nv
.
So the limit of sequence V is 0.
Since 2n nu v , when n , 2nu .
So the limit of sequence U is 2.
Method :
Let the limit of U be L . According to the given recurrence relation, when n ,
11
2L L
So 2L , i.e. the limit of sequence U is 2.
Since 2n nv u , when n , 2 2 0nv .
So the limit of sequence V is 0.
(iii) For V , sum to infinity1
21
12
.
Page 8 of 19
(iv)
1
1( 1) 1
2 1 12 1 2 2
1 2 21
2
n
n nn
r
r
v
1 1 1
12 2 2 2 2
2
nn n n
r r r
r r r
u v v n n
When n , 1
22
0
n
,
2n ,
thus 1
n
r
r
u
, i.e., sum to infinity doesn’t exist.
Page 9 of 19
6
No.
(i) Let the angle between 1p and
2p be .
1 2 1 2
2 5 2 5 cos
4 7 4 7
40 21
c
cos
0.153 rad or 8
os
78
.8
1 2 1 2n n n n
(ii)
A vector equation of line l is
19 6
6 1 ,
0 1
r .
(iii) Since the three planes does not have common point of intersection, the line l is parallel to 3p .
6 1
1 0
1 3
6 3 0
3
a
a
a
Since 3p contains the point 1,1,1 ,
1 1
1 3
1 3
1 3 3
7
b
b
b
Page 10 of 19
(iv)
18 11
7 319
1 3
18 21 3
19
6
19
619 units
19
h AB
3n
Page 11 of 19
7
No.
(i) 2
1f( )
1x
x x
2
2
1 2f '( )
1
xx
x x
(ii)
1
2
f( )
1
x
x x
2 32 2 2
1 2 1 2 31 ( 1)( ) ..........
2! 3!x x x x x x
2 2 3 3
3
1 2 ..........
1
x x x x x
x x
(iii)
2
2
2
2
(1 2 )sin (3 )
1
f '( ) sin (3 )
1 3 3 ........
1 3 3 ......
xx
x x
x x
x x
x x
(iv) For part (ii), the validity of the expansion: 2 1x x
From G.C., 1.618 0.618x .
Hence, it is not valid to use the answer in part (ii) to find the approximation value of 1
0f( ) dx x
Page 12 of 19
8
No.
(i) 2310
4x y
2
40
3y
x
232
2A x xy
2 2
2
3 40 3 80 32
2 2 33A x x x
xx
2
d 80 33
d 3
Ax
x x
2
2
80 33 0
3
80
3
xx
xx
1/3
3 80 80
3 3x x
OR 2
2 3
d 1603
d 3
A
x x
When
1/380
3x
, 2
2
d0
d
A
x
least amount of material used when
1/380
3x
x
1/380
3
1/380
3
1/380
3
d
d
A
x -ve 0 +ve
Page 13 of 19
(ii) cos6
h
k
2
3k h
21 2 55
2 3 3
hV h h
d 10 d
d d3
V hh
t t
d 3
d 30
h
t h
or
d d d
d d d
h V h
t t V
d 1 1
d 3 10
3
h
t h
d 3
d 30
h
t h
When 1,h d 3
m /d 30
hs
t or 0.0577 m/s
Page 14 of 19
9
No.
(i) d2 1
d
xt
t
d2 1
d
yt
t
d 2 1
d 2 1
y t
x t
As 1
2t , , tangent to C tends to vertical line.
As 1
2t ,
d0
d
y
x , tangent to C tends to horizontal line.
(ii)
(iii) At 1t , 2, 0x y and
d 1
d 3
y
x
0 3 2y x
2 23 2t t t t
22 3 0t t
( 1)(2 3) 0t t
3
1 rej. point or2
t P t
Q3 15
4, 4
(iv) 22 ,x y t 2
2
x yx y t t
22
2 ,2 2
x yx yx y x y
3 1,
4 4
6, 2
2,6
1 3,
4 4
Page 15 of 19
10
No.
(i)
1
2
0
1
2
0
12
0
2
8Area d
1
24 d
1
4 ln( 1)
4ln 2 units
xx
x
xx
x
x
Page 16 of 19
(ii) Change of limits: 0, 0x
π
1,4
x
2dtan sec
d
xx
212
2
0
242
2
0
π
24
2
0
π
42
0
8π 4 1 π d
1
8tan16π π sec d
tan 1
64 tan16π d
sec
16π 64π sin d
xV x
x
p=16, q = 64, 0,4
a b
(iii) 4
0
π
4
0
2 3
1 cos216π 64 d
2
sin 216π 32π
2
116π 32π
4 2
32π 8π units
V
Page 17 of 19
11
No.
(i)
2
2
2
2
f ( ) Asymptotes: and 0
f '( ) 1
For f '( ) 0, 1 0
(rej since 0)
x a ax x y x x
x x
ax
x
ax
x
x a
x a a x
ay
2
Turning point ,2
aa
a
a a
(ii) For fg to exist, g fR D .
fD (0, )
gR (1, )
Since g fR (1, ) (0, ) D , fg exists.
2(e +1)fg : ,
e +1
x
x
ax x
since fg gD D
x
y
y = x
Page 18 of 19
(iii) Sketch
2(e +1)
e +1
x
x
ay
2
fg
1As , e 0, hence .
1
R (1 , )
x ax y
a
Alternative
Consider the graph of f ( )y x with restricted domain of (1, ) [which is gR ]. Since 0 1, then 0 1,a a so the minimum point in the graph in (i)
will not be considered under the restricted domain.
21
f (1) 11
aa
, hence fgR (1 , )a .
(iv) Greatest value of k is a .
x
y
y = 1+a
fg( )y x
x
y
1
Page 19 of 19
2
2
2
2
2 2
Let
0
( ) 4(1)( ) (Method 1)
2
4 4 or
2 2
rej
x ay
x
xy x a
x yx a
y y ax
y y a y y ax x
x a
2
2 2
2 2
2 2
0
0 (Method 2)2 2
2 2
or 2 4 2 4
rej
x yx a
y yx a
y yx a
y y y yx a x a
x a
1
21
ff
4f ( )
2
D R [2 , )
x x ax
a