07 scss 4amp2 ans
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Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 1ai 1aii 1b
4k = Minimum value is −2 Period is 2
2
( )
( ) ( )
2
2
12 1
2 1 0
4 01 4 2 1 0
314
x mxx
m x x
b acm
m
+ = +
− − + =
− <
− − <
<
3i 3ii
VAB = VA − VB
= 5i + 6j
2 2AB
-1
V 5 6
7.81 ms (3 )sf
= +
=
5tan639.8
θ
θ
=
= °
VAB is in the direction of 039.8°
4i
( )
( )
1 1 2 21 2
1 3 2
2 2 .2 . .2 . ...
2 .2 . . 1 .2 . ...
n n n n n n
n n n
x C x C x
n x n n x
− −
− −
+ = + + +
= + + − +
4ii
( )1
3
2
.2 1. 1 .2 10
2 11 10
41
n
n
nn n
nn
−
− =−
=−
=
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 5
2 3 23x + <
( )
( )
3sin 2 3 22sin 2 33
Basic angle 0.72972 3 0.7297, 2.4119, 7.0129, 8.6950,
13.2961,14.9782,19.5793, 21.2614, 25.8625
Largest 2 3 21.2614Largest 9.13 (3sf)
x
x
x
xx
+ =
+ =
=+ =
+ ==
6i 6ii
3 2Let ( ) 6
(1) 5 05............................(1)
( 1) 6 1 43...............................(2)
, 4, 1
f x x x ax bf a b
a bf a ba b
Solving a b
= − − += − + =
− + = −− = − − + + = −+ =
= = −
( ) ( )
( ) ( )( ) ( ) ( )
3 2
2
2
Let ( ) 6 4 1
1 6 1
Sub. 2,( ) 25 2 35
5
( ) 1 6 5 1
1 2 1 3 1
f x x x x
x x px
xf x p
p
f x x x x
x x x
= − − + −
= − + +
== + ==
= − + +
= − + +
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 7i
( ) ( )8
20 17 21 10 66616
⎛ ⎞⎜ ⎟ =⎜ ⎟⎜ ⎟⎝ ⎠
7ii 2 8 0 1
3 10 5 25 16 7 5
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
7iii 2 8 0 0
3 10 5 955 16 7 0
⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
8a
5 5 3
3 2 1
40C C C+ ×=
8bi
4! 5!
2880×
=
8bii
6
35!14400
P×=
9i
( ) ( )Point is , or , 1 2
Hence area of rectangle is 1 2
1 2
Q x y x x
A x x
x x
−
= × −
= −
9ii
( ) ( ) ( )
( ) ( )
1 12 2
1 12 2
1. 1 2 2 1 22
0
1 2 . 1 213
dA x x xdx
x x x
x
−
−
= − − + −
=
− = −
=
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 9iii 1 21
3 30.192 (3sf)
A = −
=
10i
[ ] ( )( )cos 2
. sin 2 2 cos 2
2 sin 2 cos 22
d x xx x x
dxx x x
k
= − +
= − +=
10ii
[ ]
[ ]
[ ]
4 4
4 4 4
44
00
00 0
00
2 sin 2 cos 2 cos 2
1 1sin 2 cos 2 cos 22 21 sin 2 1 cos 22 2 2
1 0 0414
x x x dx x x
x x dx x dx x x
x x x
π π
π π π
ππ
− + =
= −
⎡ ⎤= −⎢ ⎥⎣ ⎦
= − −
=
∫
∫ ∫
11i 2
2
2
6
-1
42
sub. 0, 53
Hence 2 3At 3,
2 3809.858810 ms (3sf)
t
t
t
a ev e ct vcv etv e
=
= += ==
= +=
= +=
=
11ii
v (ms-1) 810 5
0 3 8 t(s)
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 11iii 3 2
0
32
0
For 0 3,
Distance tr. 2 3
3
411.429
For 3 8,1Distance tr. 5 809.857622024.644
Total distance travelled 411.429 2024.6442436.072440 m (3sf)
t
t
t
e dt
e t
t
< <
= +
⎡ ⎤= +⎣ ⎦=
< <
= × ×
=
= +==
∫
12i
( )( ) ( ) ( )
( )
( )
2
2
3 2 1 1 313 2 3 2
53 2
x xd xdx x x
x
+ − − −−⎡ ⎤ =⎢ ⎥+⎣ ⎦ +
−=
+
( )2( ) 5 2,
33 2df x x
dx x−
= ≠ −+
12ii
There is no solution of x for ( ) 0df xdx
= . Hence f does not have a turning point.
This implies that f is a one-to-one function and thus f-1 exists.
12iii
1
1( ) 03 21
Hence 1
xf xx
xf −
−= =
+=
=
12iv
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 1For ( ) ( ),
We solve ( )
1Hence 3 2
Solving, 1.26, 0.264 (3 )
f x f xf x x
p ppp sf
−==
−=
+= −
Since (p, k) is the point of intersection of f anf f-1, p = k. Hence the positive value of k is 0.264 (3sf)
13E i
ii
1 80tan50
57.99
α − ⎛ ⎞= ⎜ ⎟⎝ ⎠
= °
, 180122.0
θ α= °−= °
,
sin sin2 6
16.41957.99 16.419 41.6
β θ
β
=
= °°− ° = °
The man should steer at an angle of 41.6° with the bank upstream. 13E iii
R
R
2 2
V 6sin(180 16.419 122 ) sin122
V 4.6954
80 5094.339894.3398Hence time taken 20.1 s (3sf)4.6954
PQ
=°− °− ° °
=
= +=
= =
VW VR β
2 ms-1
Q
P
80 m
50 m
α θ VE
Q
VW VR VE
Swiss Cottage Secondary School 2007 Prelims A Maths Paper 2 Solution
Qn Answer 13 OR i
ii
55sin sin 55150 160
50.171
315 50.171 365.171
αβ
β
= °°
=
= °
°+ ° = °
The tanker’s velocity is in the bearing of 005.2°.
iii
LT
LT
LT
V 160sin(180 55 50.171 ) sin 55
V 188.5167200Time taken V
1.06 h (3sf)
=°− ° − ° °
=
=
=
80° VL(150) L α −VT VT (160) β 315° T