2011 maths s semester 2 sample exams solutions(1)
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Mathematics S
Semester 2
Sample Examinations
SOLUTIONS
2011
UNSW Foundation Year UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2011 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
SAMPLE A
(SOLUTIONS)
Mathematics S
Final Examination Paper
Time Allowed: 3 hours
Reading Time: 5 minutes
UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2011 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
Mathematics S Sample A Final Examination (SOLUTIONS)
1. (i) Let log832 = x
then 32 = 8x
32 = 23x
25 = 23x
5 = 3x (Equating indices.)53
= x
i.e. log8 32 =53
.
(ii) 3 1 5 5 3 1 54 3 643
2
x xx
x
− < ⇔ − < − <
− < <
− < < .
[ ][ ]
(iii) dtt
dt
t
t
t1
4
1
4
12
1
4
1
4
12
1
2
2
2 4 1
2
⌠⌡⎮
= ⌠⌡⎮
=
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
=
= −
=
−
.
(iv)
[3 significant figures.]
T = × ×⋅⋅
= ⋅
2 60 79 8
15 6
π
(v) 4
[Equating Indices]2
xx
x x
x xx x
x x
x
2
2
18
2 22 3
3 02 3 0
0 32
2 3
2
2
= ⎛⎝⎜
⎞⎠⎟
=
= −
+ =+ =
= −
−
( )
,
(
( )
vi) 22 02 1 0
0 12
2
2
x xx x
x x
x x
≥
− ≥− ≥
≤ ∪ ≥
(vii) If form a G.P., then:
If the terms are 2, 6, 18.If the terms are but all three terms are positive.So is the only solution.
x x xx
xx
xx x x
x x xx x
x xx x
xxxx
− −
−=
−
= − −
= − +
= − +
= − +∴ − − =
=== − −=
4 5 12
45 12
4 12
5 32 480 4 32 480 8 12
6 2 06 2
62 2 2 26
2
2 2
2
2
, ,
( )(5 )
( )( ), .
, , , ,
(viii)Let
thenand
By Newton's Method a second approximation will be
2D).
xf x x
f x xf
f
x xf xf x
x
3
3
2
3
2
2 11
1
2
5050
33 5 3 5 50
7 1253 5 3 3 5
36 75
3 5 7 12536 75
3 69
=
= −
=
⋅ = ⋅ −= − ⋅
⋅ = ⋅= ⋅
= −
= ⋅ −− ⋅
⋅= ⋅
( )' ( )
( ) ( )
' ( ) ( )
( )' ( )
(
(ix) If
then
x x
x x
x x
x x
x x
x xx x
14
14
14
14
2
12
12
12
12
12
12
2
1
1
2
4
2 4
6
36
2 334
− =
−⎛
⎝⎜
⎞
⎠⎟ =
− + =
+ =
+⎛
⎝⎜
⎞
⎠⎟ =
+ + =
+ =
−
−
−
−
−
−
− .6
2 7
8 7 1
8 7
3 2 52
6 2 52 2
2 17
2 1 72
2 11
72 7 1 2 1
2 1 72 14 2 1
2 1 715
2 1 7
2
8
7
7
. ( ) ( )
( ) ( )
( ) .
sin cos
cos sin .
ln
ln[ ] ln[ ]
( ) ( )( )( )
( )( )
( )( ).
(i) a Let
then
(b) Let
then
(c) Let
then
(d) Let
y xdydx
x
x
y x x
dydx
x x
y xxx x
dydx x x
x xx x
x xx x
x x
y
= −
= − −
= − −
= −
= +
=+
−⎡⎣⎢
⎤⎦⎥
= + − −
=+
−−
=− − +
+ −
=− − −
+ −
=−
+ −
= xeu x v e
u v xdydx
uv vu
x xe e
e x
xx
x
x x
x
2
2
2
2 2
2
11
1
1 1
1 2
2
2 2
2 2 2
2 2 1
−−
−
− −
−
= =
′ = ′ =
⎡
⎣⎢⎢
⎤
⎦⎥⎥
= ′ + ′
= +
= +
then
[ ] [ ]
[ ].
e
(ii) = 5
At
The slope of the tangent.
the slope of the normal = 15
Also at
The equation of the normal at (2,5) will be:-
yx
xdydx
x
x
x dydx
x y
y x
y xx y x y
−= −
= − −
=−−
= =−−
= − ⇐
∴
= =−
=
∴
− = −
− = −= − + ⇔ − + =
−
−
15 1
5 1
51
2 52 1
5
2 52 1
5
5 15
2
5 25 20 5 23 5 23 0
1
2
2
2
( )
( )
( )
,( )
.
, .
( )
.
(iii)
B
B
W
B
WB
W
WB
W
B
WB
W
BBB
BBW
BWB
BWW
WBB
WBWWWB
WWW See next page.
(a) Pr[No white cards] = Pr[all Black cards]
= Pr[BBB]
= 49
× ×
=
38
27
121
.
(b) Pr[at least 2 black cards] = Pr[2 or 3 black cards]= Pr[BBW] or Pr[BWB] or Pr[WBB] or Pr[BBB]
= 49
× × + × × + × × + × ×
=
38
57
49
58
37
59
48
37
49
38
27
1742
.
(iv) First find the point of intersection of and by solving simultaneously:
Point of intersection is Now the equation of any line is
l lx yx y
x yx y
yyy
xx
xx
x yx y k
1 2
3 4 4 02 5 9 0
6 8 8 0 16 15 27 0 2
1 2 7 35 07 35
53 4 5 4 0
3 24 03 24
88 5
4 7 13 07 4
− + =− − =
⎫⎬⎭
− + =− − =
− ⇒ + == −= −
⇒ − − + =+ =
= −= −− −
⊥ − + =+ +
................( )............( )
( ) ( )
( )
( , ).
=− − − + − + =
=+ + =
08 5 8 4 5 0
767 4 76 0
.( , ) ) ( )
..
If lies on this line then 7(
So the equation of the required line is
kk
x y
3 5 2 32
52
2 32
52
2 62
12
. cos sin sin
sin cos .
(i) (a) x x dx x x c
x x c
−⎛⎝⎜
⎞⎠⎟
⌠⌡⎮
= +
= +
cos +
+
(b) (5x − 7)3 dx∫ = (5x − 7)4
5 × 4+ c
=(5x − 7)4
20+ c.
(c) 23
23−
⌠⌡⎮
= −−−
⌠⌡⎮
= − − +2 2
3 2x
dxx
dx
x cln[ ] .
(d) 7 72
2
72
2 2
2
xe dx xe dx
e c
x x
x
∫ ∫=
= + .
(ii)
Let
Then
then or no solutions,
8 28 2
8 2
8 22 8 0
4 2 04 24 22
2
2
e e
ee
m e
mm
m mm m
m mm m
e ex
x x
xx
x
x x
− − =
− =
=
− =
− =
+ − =+ − =
= − =
= − ==
( )( ),
ln .
(iii) (a)
(b)
x dx−∫ = +
= +=
2
8 12
2 2 12
2 3234
( )( ) (8)(8)
.
[ ]
.cm 095
]9270sin9270[)80(21
sin21=Area Shaded
80
84=OB
Theorem, 'PythagorasBy .9270
107112=AOB Then,radians][10711 So,
48tan(iv)
2
2
2
22
⋅=
⋅−⋅=
−
=
+
⋅=⋅×−∠
⋅=∠
=∠
θθ
π
r
BOC
BOC
4 11
0 2 5
2 251
1 2 251 25
2 1 252 2 24 0
12 04 3 04 4 1 5
3 3 1 2
2 2
2 2
2 2
2 2
2 2
2
2
. ( , ) , ,.
( , )
( ) ( )
( )
( )( )
( )( )
.
(i) The point lies on the line so we can substitute
Also we know that is 5 units from (0,2), so
But,
h k &y x x h y kk h
h k
h k
h kk h
h hh h
h h hh hh h
h hh kh k
= + = =⇒ = +
− + − =
+ − == +
⇒ + + − =
+ − =
+ − + =
− − =
− − =− + =
∴ = ⇒ = + == − ⇒ = − + = −
(ii) (a) To find the points of intersection we solve simultaneously:
From (2) =
Now substitute (3) into (1)
So the points of intersection are (4,2) and (8,8).
x yx y
x yx y
x xx xx x
x yx y
2
2
2
8 13 2 8 0 2
3 8 212 32 8 3
12 3212 32 04 8 04 28 8
=− − =
−− =
⇒ = −
− + =− − =
= ⇒ == ⇒ =
.........( ).......( )
.......( )
( )( ),,
[ ]
( ) ( ) ( ) ( )
(b) Shaded area =
square units.
3 82 8
18
12 32
18
6 323
18
6 8 32 8 83
6 4 32 4 43
43
2
4
8
2
4
8
23
4
8
23
23
x x dx
x x dx
x x x
−−
⎡
⎣⎢
⎤
⎦⎥
⌠
⌡⎮
= − −
= − −⎡
⎣⎢
⎤
⎦⎥
= − −⎧⎨⎩
⎫⎬⎭
− − −⎧⎨⎩
⎫⎬⎭
⎡
⎣⎢
⎤
⎦⎥
=
∫
[ ]
( )
( ) ( )[ ] [ ]
(c)
cubic units.
V x dx x dx
x dx x dx
x x
=−⎡
⎣⎢⎤⎦⎥
⌠
⌡⎮−
⎡
⎣⎢
⎤
⎦⎥
⌠
⌡⎮
= − −
=−×
⎡
⎣⎢⎢
⎤
⎦⎥⎥
−⎡
⎣⎢
⎤
⎦⎥
= × − − × − −
=
∫ ∫
π π
π π
π π
π π
π
3 82 8
43 8
64
43 8
3 3 64 5
363 8 8 3 4 8
3208 4
645
2
4
82 2
4
8
2
4
84
4
8
3
4
85
4
8
3 3 5 5_
( )(iii) tan2
0
x dx x dx
x x
π π
π
π π
π
4 2
0
4
0
4
1
4 40 0
14
⌠⌡⎮
= −⌠⌡⎮
= −⎡
⎣⎢
⎤
⎦⎥
= − − −
= −
sec
tan
[tan ] [tan ]
.
(iv) and
Substitute (1) into (2) gives:+ (-4 ) = 40
=
So,
T T T Ta d a d a d a da d a d a d
a d
a aa
a d d
S
11 6 8 9
6
5 410 5 5 7 8 4010 5 25 2 15 40 2
4 15 1
22 40
20 4 20 15 163
62
2 20 5 163
40
= × 0+ =+ = + + + + =+ = + + =
− =
−
= − ⇒ − − = ⇔ =
= × − + ×⎡⎣⎢
⎤⎦⎥
= −
[ ]......( )
.................( )
( ) .
.
5. (i)
ex2dx
0
k
∫ = 1
2ex2
⎡
⎣ ⎢
⎤
⎦ ⎥
0
k
= 1
2ek2 − 2e0 = 1
2ek2 − 2 = 1
2ek2 = 3
ek2 = 3
2k2
= loge32
⎡ ⎣ ⎢
⎤ ⎦ ⎥
k = 2ln32
⎡ ⎣ ⎢
⎤ ⎦ ⎥
k = ln94
⎡ ⎣ ⎢
⎤ ⎦ ⎥ .
(ii) At , also when and when
x f x x f xx f x
= = < >> >
2 0 22 0
, ' ( ) , ' ( ), ' ( )
0
x x<2 x=2 x>2 f '(x) + 0 +
∴ A horizontal inflexion point will occur at x = 2.
(ii) (b)
(iii)
When
When
Now we have to find when of 10 = 7 5.7 5 =
dMdt
kM M M e
t M MM e
t Me
e
k
k
t Me
ekt
kt
t
t
okt
okt
k
k
kt
kt
= ⇒ =
= = ⇒ =
== =
=
=
⋅ =⋅
=
= ⋅
⋅
⋅ =⋅ =⋅
=
⋅⋅
=
⋅ =
.
,
,
ln( )ln( )
ln( )ln( )
ln( )ln( )
.
0 10 10105 8
8 108
100 8 50 85
75%10
0 750 750 75
0 750 85
6 4
5
5
ππππ
ππππ
2,,04,2,02
12cosb)(
2cossincos
)sin21cos
21(2
)sin2
1cos2
1)(sin2
1cos2
1(2
]sin)4
sin(cos)4
][cos(sin)4
sin(cos)4
[cos(2LHS(a)6(i)
22
22
===
=−=
−=
−−=
+−=
xxx
xxx
xx
xxxx
xxxx
(ii) V = 2000 + 6t2 − t3 0 ≤ t ≤ 6.
dVdt
= 12t − 3t2
d2Vdt2 = 12 − 6t
(a) Put
When is maximum here.
dVdt
t t
t tt t
t d Vdt
V
= ⇒ − =
− == =
= = − < ⇒
0 12 3 0
3 4 00 4
4 12 0
2
2
2
( ), .
, ( )
(b)kl.
Vmax = + × −=
2000 6 4 42032
2 3
(c)
(d) When t = 2
(iii) f (x) = x3 ⇒ f (−x) = (−x)3
= −x3
= − f (x).∴ f (x) is odd.
(b)
(c)
is odd.
f x dx
f x x x f x x xx xf x
f x
( )
( ) cos ( ) ( ) cos( )cos [cos( ) cos ]
( ).( )
−∫ =
= ⇒ − = − −
= − − == −
∴
1
1
3 3
3
0
2 22 θ θ
[ ]
[ ]
( cos cos
( ) , ( )
.
d)
if is odd.
e x x dx e dx x x dx
e f x f x
e e
e e
x x
x
−
−
−
− −
−
− −
−
−
+ = +
=−
⎡
⎣⎢
⎤
⎦⎥ + =
⎡
⎣⎢
⎤
⎦⎥
=−
−
=−
∫ ∫ ∫
∫
2 3
1
12
1
13
1
1
2
1
1
1
1
2 2
2 2
2 2
20 0
12
2
( )
( )
71 9
13
3
1 3
13
3
13
12
12
13 6 6218
9
216
16
216
16
1
16
16
1 1
. (
sin
sin sin
.
i) dxx
dx
x
x
−
⌠⌡⎮
=−
⌠
⌡⎮
= ⎡⎣⎢
⎤⎦⎥
= ⎛⎝⎜
⎞⎠⎟
−−⎛
⎝⎜⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
= −−⎛
⎝⎜⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
=
=
− −
−
−
− −
π π
π
π
(ii)
Test Points→ x = − 4 x = −2 x = 0 (−3x−9)/(3x+5) negative positive negative
Solution x < −3 ∪ x > −123
.
See over page for alternative method
3 13 5
2
3 13 5
2 0
3 1 2 3 53 5
0
3 1 6 103 5
0
3 93 5
0
3 9 03
3 5 05
3
xx
xx
x xx
x xx
xx
xx
x
x
++
<
++
− <
+ − ++
<
+ − −+
<
− −+
<
− −= −
+ =
=−
( )
;
Critical values:-=
and
Alternative method for part (ii)
0for which values thefind and)93)(53( parabola sketch the we
3/5 and3 are roots0)93)(53(
0]10613)[53(0)]53(2)13)[(53(0)53(2)53)(13(
)53(2)53)(13(
)53(by hroughout multiply t
25313
2
2
2
<−−+=
−−<−−+
<−−++<+−++<+−++
+<++
+
<++
yxxxy
xxxxx
xxxxxx
xxx
xxx
y
x-3 -5/3
From the graph we can see that
0)93)(53( <−−+ xx when 3/5or 3 −>−< xx
(iii) ( (
(
For the term independent of
2 3 1 2 2 3 191
2 92
2
2 3 1 18 144
2 1 1 18 3
29
2 2
22
− + −⎛⎝⎜
⎞⎠⎟
= − + −⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ −
⎡
⎣⎢
⎤
⎦⎥
= − + − + −⎡⎣⎢
⎤⎦⎥
⇒ × + − × − + ×
x xx
x xx x
x xx x
x
) ) ( ) ( ) ......................
) ...................................
: ( ) ( ) ( ) ( ) ( ) (144 452) =
[ ] [ ] [ ][ ]
[ ][ ]
( )
[ ]
( )
.4
13
21
22
21
21
2)1ln1(ln
21log
2loglog2
loglog2
log2log
log2log
2log1log(iv)
4
44
424
1
1
2
12
1
11
2
1
2
2
22
2
22
2
222
+=
⎭⎬⎫
⎩⎨⎧
+−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡−−×−=⎮⌡
⌠
⎥⎦
⎤⎢⎣
⎡−=⎮⌡
⌠
⎮⌡⌠−⎮⌡
⌠=⎮⌡⌠
=−
+=
+⎥⎦⎤
⎢⎣⎡=
e
ee
eeedxxx
xxxdxxx
dxxdxxxdxddxxx
xxxxxdxd
xxxxxdxd
xxx
xxxdd
e
e
ee
e
e
e
ee
e
e
e
ee
ee
ee
( )
(v)
let u =
Now: when
when
Also:
and:
kg.
dMdt
tt
M tt
dt t
t u
t u
dut
dt
t u
tt
dt u du
u u
=−
=−
⌠⌡⎮
−
= ⇒ = − =
= ⇒ = − =
=−
= +
−⌠⌡⎮
= +⌠⌡⎮
= +⎡
⎣⎢
⎤
⎦⎥
= + − +⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
=
2
22
6 6 2 2
11 11 2 3
12 2
2
22 2
23
2
2 9 6 83
4
16 23
6
11
2
6
112
2
3
3
2
3
: .
( )
8. (i)
(ii) x x x xx x
x
x xx
x xx
xx
dx x xx
dx
x x c
2 3 2
3
3
2 2
3
2 2
22
2 0 0 02
2
22
2
22
2
22
+ + + +
+
−
⇒+
= −+
+⌠⌡⎮
= −+
⌠⌡⎮
= − + +
[ ]
ln[ ]
(iii)
6 3
6 3
6 3
6
1 36
3
1 9
10 1
6100
1 36100
1
3100
1 9100
1
1237412
1 1
1 1
2
2
2
2
x x
x x
x xt
ddt
x
x
dxdt
x
x
dxdt
x dxdt
ddt
ddt
= + =
⎛⎝⎜
⎞⎠⎟
= + ⎛⎝⎜
⎞⎠⎟
=
= ⎛⎝⎜
⎞⎠⎟
− ⎛⎝⎜
⎞⎠⎟
=
−
+× −
−
+×
= = −
=
−
+× − +
+× −
=
⋅
− −
− −
tan( ) tan
tan tan
tan tan
:
,
( ) ( )
θ α α
θ α α
θ
θ
θ
θ
and
Now differentiating implicitly with respect to
When
radians / second
= 0 0166 radians / second.
(iv) In doing this question we must use the fact that the degree of the remainderis always less than the degree of the divisor. So when we divide a polynomialby a quadratic, then the remainder could be linear (i.e.
Let the polynomial be Then andAlsowhen is divided by (
ax b
P xP P
P x x x Q x ax bP x x x
P Q x a b a bP Q x a b a b
+
− = = −
= − − + +
− −− = + − + = ⇒ − + =
= + + = − ⇒ + = −
).
( ).( ) ( ) ,
( ) ( ) ( )( ) ).
( ) ( ) ( ) ( ) ....( )( ) ( ) ( ) ( )
1 6 3 22 3
2 31 0 1 6 6 1
3 0 3 2 3 2
2
2
.....( )
..
2
6
3 24 8
2 42 4
Now solve (1) and (2) simultaneously:
+ =
The remainder is
−−
+ = −
⎫
⎬⎪
⎭⎪
− == − ⇒ =
∴ − +
a b
a ba
a bx
( )
(v) AB rr
r
rr
r r
r r
r
r
= −
=−
= = =
−=
= −
+ =
+ =
=+
=−
−= −
10
105125
55 5
15
1015
5 10
5 10
1 5 10
101 5
10 1 54
52
5 1
sin
sin
( )
( ) .
α
α
9 3 1 21 1 2 12 2 0
1 00 1
23
23
32
32
1. ( ) sin ( )(i) (a) (b)
Domain:
Range:
f x xx
xxx
y
y
= −
− ≤ − ≤− ≤ − ≤
≥ ≥≤ ≤
⎫
⎬⎪⎪
⎭⎪⎪
−× ≤ ≤ ×
−≤ ≤
⎫
⎬⎪
⎭⎪
−
π π
π π
(c) f y x f xx y
x y
y x
y x
: sin ( ) : sin (
sin ( )
sin
sin
sin .
= − ⇒ = −
= −
⎛⎝⎜
⎞⎠⎟
= −
= − ⎛⎝⎜
⎞⎠⎟
= − ⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
− − −
−
3 1 2 3 1 2
31 2
31 2
2 13
12
13
1 1 1
1
y)
(ii)
To use the graphs to solve we must find the points of intersection, A and B of the two graphs.To find A we must solve simultaneously:
= (
To find B we must solve simultaneously:
Now read off the points along the axis where =i.e. where the first graph is the second graph.
2 3 3 1
2 33 1
2 3 3 1
2 525
2 15
2 33 1
2 3 3 1
4 112 3 3 1
4 25
x x
y xy x
x x
x
x y
y xy x
x x
x yx y x y x
x x
− < +
− −= +
⎫⎬⎭
⇒ − + = +
=
= ⇒ =
= − −= − +
⎫⎬⎭
⇒ − + = − −
= − ⇒ =
− − < = +
< − ∪ >
)
( )( )
.
.
below
(iii) Let , , and be the roots of Then = + .
Now + + =
and =
also + + =
α β γα β γ
α β γ α α
αβγ βγ βγ
αβ βγ αγ α β γ βγ α
x kx mx n
k k k
n k n nk
m nk
m
k nk
m
k n kmk km n
3 2
2
2
3
3
0
22
22
2
42
8 44 8 0
+ + + =
− ⇒ = − ⇒ =−
− ⇒−
= − ⇒ =
⇒ + + = + =
+ =
+ =
− + =
.
( )
(iv) (a) (
[ is even]
(
11
1 1 2
0 1 22
33
11
0 1 22
33
11
0 22
44
+ = + + + + + + +
− = − + − + − +
+ + − = + + + +
−−
−−
x c c x c x c x c x c xx c c x c x c x c x c x n
x x c c x c x c x
nn
nn
n
nn
nn
n
n nn
n
) ................( ) ..................
) ( ) [ ..................... ] ..........( )
[ ..................... ]
[ ..................... ]
[ ..................... ].
1
12 0 2
22
2
0 2 4
0 2 4
10 2 4
If we now let in equation (1) then we get:xc c c c
c c c c
c c c c
nn
n
n
nn
=
+ = + + + +
= + + + +
= + + + +−
(b)122
[These are even co - efficients in the expansion of where
Using the answer to part (a)]
r
x xc c c c
r
⎛⎝⎜
⎞⎠⎟ =
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟
+ == + + + +
=
=
=
−
∑0
6
12
0 2 4 1212 1
11
120
122
124
126
128
1210
1212
1 1
22
( )...............[
.
.]
(c)
The above expansion is the sum of the even co - efficients in the expansion( where The sum of the even co - efficients is found by using Now if then:
122
5120
5122
5124
5126
5128
51210
51212
5
1 51 1
5122
5 1 5
0
62 0 2 4 6 8 10 1
12
12 12
0
62 12
r
x xx x
x
r
r
r
r
r
⎛⎝⎜
⎞⎠⎟ =
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟ +
⎛⎝⎜
⎞⎠⎟
+ =
+ + −=
⎛⎝⎜
⎞⎠⎟ = + +
=
=
∑
∑
) .( ) ( ) .
,
[( )
2
( ) ]
[ ( ) ][ ][ ]
[ ]
[ ]..
1 5 2
6 4 26 4 23 2 2 2 22 3 2
22 3 2
11 12
12
12 12
12 12
12 12 12 12
12 12 12
11 12 12
− ÷
= + − ÷
= + ÷
= × + × ÷
=+
= +∴ = =a b and
( ) ( )
[ ]
10 1
11
22 11 1
1 22 1
11
1
11
1
11
11
0
1 1 2
2
2
2
2
2
2
2
2
2
2
2 2
. ( ) sin sin
' ( )( )
,
.
(i)
Since
Since if 0 1
=
f x x x
f xx
xxx
ddx
x xx
x
xx
x
x
xx
xx x x
x x
= + −
=−
+
−
−− −
− =−
−
=−
− −
=−
− − = ≤ ≤
−−
−=
− −
( ' ( ) ( )
] ( ).
( ) sin sin .
( ) .
i) (b) Since then must be a constant.To find the value of this constant, we can substitute any numberfrom the Domain [0 into
Let us choose
The graph of this function is shown below.
f x f x
x f x
x f
f x
=
≤ ≤
= ⇒ = + − =
=
− −
0
1
0 0 0 1 02
2
1 1 π
π
So The area of the rectangle.]f x dx( ) [
.
0
1
12
2
∫ = ×
=
π
π
(ii) y = ln(xyx yx y
dydx x y
dydx
dydx y
dydx x
dydx y x
dydx
x
ydydx
yxy x
2
2
2
1 2 1
2 1
1 2 1
1
1 2
2
)ln ln( )ln ln
.
= += +
= +⎛⎝⎜
⎞⎠⎟
− × =
−⎛⎝⎜
⎞⎠⎟ =
=−
=−
( )
Now when
So
y
xx
e xe x
dydx e e
e e
e
=
= ×=
=
=
=⎛⎝⎜
⎞⎠⎟ −
⎛⎝⎜
⎞⎠⎟
=−
=
3
3 33 9
9
93
93 2
927
3 227
2
3
3
3 3
3 3
3
ln( )ln( )
.
(iii) Sample space for throwing two dice
1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12
(a) 91
364)5( ==P
(b) 43 throw)P(second
41
369)8or 5( =∴==P
(c) To win on second throw he must first not get a result on first throw and to win on third throw he must not get a result on first two throws.
91
43
43
91
43
91 throw)3rdor 2nd, ,1st on P(win ××+×+=∴
(d) 94
4191
431
91
1==
−=
−=∞ r
aS
(iv) To answer this question we try to sketch the graph of y = P(x) and see if the
graph cuts the x − axis at 3 different points.
P(x) = 2x3 − 9x2 +12x − kP ' (x) = 6x2 −18x + 12P"(x) = 12x −18
Put P ' (x) = 0, ⇒ 6x2 −18x +12 = 0
x2 − 3x + 2 = 0(x − 2)(x − 1) = 0
x = 1, 2.
Now P"(1) = −6 (< 0) ⇒ Maximum turning point occurs here.P"(2) = 6 (> 0) ⇒ Minimum turning point occurs here.P(1) = 2 − 9 +12 − k
= 5 − k ⇒ maximum point is 1, 5 − k( )P(2) = 16 − 36 + 24 − k
= 4 − k ⇒ mimimum point is 2, 4 − k( )
We can see that when the maximum point and the minimum point are on oppositesides of the x - axis, then our function will have 3 distinct roots.
So (5 − k)(4 − k) < 0 [Opposite signs when multiplied give < 0]4 < k < 5. [See the graph below]
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Telephone: 61 2 9385 5396 Facsimile: 61 2 9662 2651 Email: foundation.year@unsw.edu.au Web: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education,
training and consulting services and a wholly owned enterprise of the University of New South Wales ABN 62 086 418 582 CRICOS 00098G
SAMPLE B
(SOLUTIONS)
Mathematics S
Final Examination Paper
Time Allowed: 3 hours
Reading Time: 5 minutes
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Mathematics S Sample B Final Examination (SOLUTIONS)
1. (i)
712127
3664)2(364
<
<−<−−<−
x
xxxxx
(ii)
3553
559512
5453206
=∴=
=−
=−
aa
a
a
(iii) (iv)
2415
)5,1( Substitute42
=−−=−
−−−+=
kk
kxxy
(v)
62log
64log)11881(log
)123(log
62
2
2
232
==
=+−=+− xx
(vi)
2,60)2)(6(
0124
124
33
813
2
2
124
3
2
2
−===+−=−−
+=
=
=+
+
xxxx
xx
xx
xx
xx
(vii)
)0,52(
5225
11515
)15ln(0)15ln(
0
∴
=
==−=−
−=−=
x
xx
ex
xxy
(viii)
337
9373813or 813
813
−==
−==−=+=+
=+
xx
xxxx
x
5 3 18
31 5 3
)1 32 (3 5
1 38 35 3
53
1
55
8
1
3 5
8 3 2
=
×=
−=
⎟ ⎟ ⎠
⎞ ⎜ ⎜ ⎝
⎛ − =
⎥⎦
⎥⎥⎤
⎢⎣
⎢⎢⎡
t∫ =t dt
2. (i)
32
32
42
)53(24
6.)53(4
)53()(
−=
−=
−
xx
xx
xdxda
25
25
5
25
)(
x
x
x
e
edxd
edxdb
=
=
1)1ln(2
11.)1ln(.2
)1ln()(
2
2
2
+++=
+++=
+
xxxx
xxxx
xxdxdc
(ii) 29)( xya −=
y
x3-3
3
ex
ex
xxx
xxxyb
1
1ln0ln1 intercept
0 asymptote vertical0domain
ln1)(
1
=
=
−==+
=∴>
+=
−
y
x1/e
(iii) (iv)
8608610
)815(220
)(2
811395)14(20
1
=×=
+=
+=
++++=+∑=
lan
rr
…
3 is normal ofgradient 31- is tangent ofgradient
31
33)4(
)1(3
)1(31
3
2
2
2
−=
−=′
−−
=
−−=′−
=
−
y
x
xyx
y
(v)
3 is ofgradient
3tan)180tan(
)180tan( is PR ofgradient
PQRPRQ Now3then tan If
0
0
−∴
−=−=−
−
==<<==<
PR
PQR
θθ
θ
θθθ
3. (i)
27106
27172
943
34
1973)
34(
))(()(973
29
19
12)34(
2)()(
133)(
34)(
0343
2233
2
222
2
−=
−=
×−=
⎟⎠⎞
⎜⎝⎛ +−=
+−+=+
=
+=
−×−−=
−+=+
−=−
==
−=−=+
=−+
βαβαβαβα
αββαβα
αβ
βα
d
cacb
aba
xx
(ii)
4or t0solution )4(3 ofgraph consider
0)4(30123
123
46
2
2
23
><−=
>−>−
−=
+−=
ttty
tttt
ttdtdA
ttA
y
t4
(iii)
xx
xxxxxx
dxda
ln11ln
11.ln.1)ln()(
=−+=
−+=−
111
10ln
lnln11
=+−×=
+−−=
⎥⎦⎤
⎢⎣⎡ −=∫
eeeee
xxxdxxee
(iv)
hundred)nearest (1340020000
202000year In (b)
places) decimal3(020.010819.0ln
10819.0ln819.0
2000016380
16380,10when 20000)(
20020.0
10
10
0
==
=
=−
=
−==
=
===
×−
−
−
eP
t
k
k
ke
e
PtPa
k
k
4. (i)
)6(sec
)6tan(
2 x
xdxd
−−=
−
(ii)
∫ ∫
∫
+−=−−=
+=
cedxxedxxe
b
cxdxxa
xxx coscoscos
2
sinsin
)(
5tan515sec)(
(iii)
25 4020°
B
DC
kmCD
CDa
6.1861.345
20cos2000160062520cos402524025)( 222
==
−+=×××−+=
Distance is shorter by (25+18.6)-40 = 3.6 km
2171
20sin402521 Area)(
km
b
=
××=
(iv)
units square436
048127
4
3
)3( Area)(
3
0
43
3
0
32
3
0
2
=
−−=
⎥⎦
⎤⎢⎣
⎡−=
−=
−=
∫
∫
xx
dxxx
dxxxa
35729
72187729
52187
766
59
69
)69(
)3( Volume)(
3
0
765
3
0
654
223
0
4
3
0
24
π
π
π
π
π
π
=
⎥⎦⎤
⎢⎣⎡ +−=
⎥⎦
⎤⎢⎣
⎡+−=
+−=
+−=
−=
∫
∫
∫
xxx
dxxxx
dxxxx
dxxxb
y
x3
5.
( ) 963(a) 22 ++=+ xxx
)
( ) ( )xPx
xxxxxxxxxx
xxxxxxxxx
xx
offactor a is 30 isremainder So,
09696
1812212112
969122496
12
2
2
2
23
23
234
2342
2
+⇒
++++
−−−−−−
+++−−+++
+−
(b)
roots) doubleboth (1or 30)1()3(
)1()3(
)12)(96()( above From
22
22
22
=−==−+
−+=
+−++=
xxxx
xx
xxxxxP
(c)
)3)(1)(1(4)3)(1(4)3)(44(
)3(4)3(4124124)(
2
2
2
23
+−+=+−=
+−=
+−+=
−−+=′
xxxxxxx
xxxxxxxP
Stationary points when 0)( =′ xP
( ) )0,3()16,1(0,1
0160311
−−===
−=−==yyyxxx
(d)
point minimum a is)0,3(32)3(point maximum a is)16,1(16)1(
point minimum a is)0,1(32)1(42412)( 2
−∴=−′′−∴−=−′′
∴=′′−+=′′
PPP
xxxP
(e)
y
x1-3
(-1,16)
(f)
622
6242
2442
8164
024,0
0)24(
024
91224129
91224
129
2
22
234
234
234
±−=
±−=
±−=
+±−=
=−+=
=−+
=−+
+−−+=−
+−−+=
−=
x
xxx
xxx
xxx
xxxxx
xxxxy
xy
(g) We can use the previous parts of the question
91291224 toequivalent is024 234234 +−>+−−+>−+ xxxxxxxx Therefore we draw on the same diagram the graphs of
line)(straight 912 andpart v) (see91224 234 +−=+−−+= xyxxxxy Note that from part vi we know that the line cuts the curve at 62 and62 −−=+−= xx and touches the curve at (double root) 0=x
f(x)
x-2-√6-2+√6
We require the x values for which 91291224 234 +−>+−−+ xxxxxi.e. the x values for which the curve is above the line. Clearly from sketch 62or62 +−>−−< xx Note This part could also be solved by drawing the curve and finding the x values for which
234 24 xxxy −+=0>y
6. (i)
(a) xx
edxd x 1
2
sin sin1
11 −
−=
−
x
xx
e
eedxdb
21
1)(sin)(
−=−
xx
x
xdxdx
dxdc
3tan63cos
3sin32
)3ln(cos2)3ln(cos)( 2
−=
−×=
=
(ii) (a) dadada 4,3, +++ form a GP
0
)1010(2
11
)102(2
11
5or reject)(00)5(05
4596343
11
2
2222
=
+−=
+=
−===+=+
++=++++
=++
dd
daS
daddad
dad
dadadadadada
dada
(b)
ddSSd
ddd
daS
1210121121
1111)2110(11
)212(222
1122
22
=−=−=
×=+−=
+=
Therefore the sum of the next 11 terms is . d121
(iii)
2222
22222222
222222 )()()(
++
++
−=
−+−=
−+−=
nn
nnnn
nnn
yx
yxyyxx
yxyyxxRHSa
(b)
),()(]andinpolynomiala[)(ie)](),(y)[x(x
assumptionusing))((),()(
(i)partfrom)()(
LHS
Proof
andinpolynomialsomeis),( where),()(provei.e.
1 when trueisstatement theProve
andin xpolynomialsomeis),(where),()(assumei.e.
when trueisstatement theAssume
trueisstatementthe1when))(,())((
1Put
22
22
222222
2222
)1(2)1(2
22
2222
yxSyxyxyxyxyyxQ
yxyxyyxQyxx
yxyyxx
yx
yxyxSyxSyxyx
kn
yyxQyxQyxyx
kn
nyxyxPyxyx
yxyxn
k
k
kkk
kk
kk
kk
nn
+=×+−++=
−+++=
−+−=
−=
+=−
+=
+=−
=
=∴+=+−=
−=−=
++
++
Therefore if the statement is true when kn = it is also true for 1+= kn . But the statement is true when . 1=nBy the process of mathematical induction the statement is true for all positive integers 1≥n
7. (i)
cx
cx
cx
dxxx
dxxxdxx
xc
cxdxx
b
ce
dxexdxexa
x
xx
+−
−=
+−××−=
+−
×−=
−−−=
−=−
+=−
+=
=
∫
∫∫
∫
∫∫
−
−
−
441
)41(12
81
21
)41(81
)41(881
)41(41
)(
2sin21
41
1)(
31
331)(
2
21
2
21
2
21
2
21
22
12
22
2
33
(ii)
6
5
x6-x
5-y
y
S R
PA Q
B
(a)
222
22
22
)
)5(36)
)6(25)
yxAB
ySB
xSA
+=
−+=
−+=
δ
β
α
(b)
2
2
2
2
2222
2222
222
51
56
102
1012
21210
21210
123625102536
)6(25)5(36
xxy
xxy
xxy
xxy
yxxxyy
yxxy
ABSASB
−=
−=
−=
+−=−
+++−+=+−+
++−+=−+
+=
(c)
59
59
518
3513
56
362026
052
56 when maximum
5252
56
2
2
2
=−=
×−×=
===−
=−∴
−=
−=
y
y
xxx
xy
dxyd
xdxdy
Therefore the maximum length of QB is 541 metres
(d)
4 when maximum isA
056
1024
56)4(
4(reject)00)4(30312
0103
560put
106
56
103
56
1053
)51
56(
2121Area
2
2
2
32
2
=∴
<−=−=′′
===−=−
=−=′
−=′′
−=′
−=
−=
=
x
A
xxxxxx
xxA
xA
xxA
xx
xxx
xy
2
max
2.31032
1064
5163
m
A
=
=
−×
=
8. (i) By t method
75373292
13.5362.292
57.262
30.1462
21
2tan,
32
2tan
21,
32
0)12)(23(026
04212
44288
41
2)11(8
4sincos8
00
00
00
2
2
22
22
2
′=′=
==
==
=−=
=−=
=−+=−+
=−+
+=−−
=+
−+−
=−
xx
xx
xx
xx
tt
tttt
tt
ttttt
tt
xx
By auxiliary angle method
73292or753
5160360or516087
83678787
3600
quadrant4th or 1st is angle and 5160 is angle basic
4961.0654)87cos(
4)87cos(65
8781tan
6518
4sincos8
00
0000
000
00
0
0
0
0
22
′′=
′−′=′+∴
′≤′+≤′
≤≤
′
=
=′+
=′+
′=
=
=+=
=−
x
x
x
x
x
x
R
xx
α
α
(ii) (a)
triangleangledright ain it put weand angle acutean is that know we0 sincetanthen tanput 1
ααα
>==−
aaa
a
1α
π/2-α
clearly
a1)
2tan( =−απ
)1(tan2
.. 1
aei −=−απ
)1(tantan
2
)1(tan2
11
1
aa
a−−
−
+=
+=
π
απ
(b)
[ ]
6
231
)31(tan3tan
31
)31(tan3tan
31
3tan31
91
11
11
1
91
1
1
91
2
π
π
=
×=
⎥⎦⎤
⎢⎣⎡ +=
⎥⎦⎤
⎢⎣⎡ −−=
=
+
−−
−−
−
−
−
∫
x
dxx
dx
(iii) (a) We solve simultaneously
parabola o tangent tis lineonceintersect only parabola and line
uslysimultaneo solve hen wesolution w one044
02
2
2
22
22
22
2
2
∴∴∴
=−=Δ
=+−
=−
=
−=
mm
mmxx
xmmx
xy
mmxy
(b)
4,00)4(
40
)0,2( sub2
2
2
===−
−=
−=
mmmm
mm
mmxy
(c) Taking we put 22 mmxy −= 4 and == mom
168 gives40 gives 0
−====
xymym
(iv) (a) )1()2(72 −++=+ xbxax
339:1Put
=∴==
aax
133:2Put −=∴−=−=
bbx
)1()2(372 −−+=+∴ xxx
cxx
dxx
dxx
dxxx
xxdxxx
x
++−−=
⎮⌡⌠
+−⎮⌡
⌠−
=
⎮⌡⌠
−+−−+
=⎮⌡⌠
−++
2ln1ln32
11
3)1)(2(
)1()2(3)1()2(
72(b)
9. (i)
15.2or 15.06
4866
12366
0163)( 2
−=
±−=
+±−=
=−+
k
k
k
kka
(b)
y
k-2·15 0·15
( ) 0)2(41 and 0
if roots real with quadratic0)2()1
2
2
≥+−+=Δ≠
=+++
kkkk
kxk(+kx
2+ − + ≥k
(c)
15.015.2only when solutions has thisabove from0)2()1(
0
1 have weintegers positive are they because that note2kc1,kbk,aput
integerspositiveeconsecutiv are,,
2
2
≤≤−=++++
=++
≥+=+==
<<
kkxkkx
cbxax
k
cbacba
but we require so therefore the equation cannot have real roots. 1≥k(ii)
)4
(cos
ORArea
)4
cos(
)4
cos()(
22
2
απ
απ
απ
−=
=
−=
−=
r
rOR
rORa
15 )0( ≠k .015 . 2 issolution 16 3 ofs from ketch
0.15 a 2.15 nd- are oots r a) ( from 01 6 3 2
0 163 2
0 841 2
0 )2 ( k k( )1 42 2+ + − − ≥k k k k
− − + ≥k k
+ − ≤k k
2= + −y k k− ≤ ≤k
2
2
2
2
)2(21
)]4
(22
[2121 sector of Area)(
r
r
r
rb
α
α
αππ
θ
=
=
−−=
=
142sin2sin14
)22
cos(14
)]22
cos(1[212
)2cos1(21cos using
)4
(cos2
)4
(cos21
square of Area21sector of Area
2
2
222
−=+=
−+=
−+=
+=
−=
−=
=
αααα
απα
απα
απα
απα
xx
rr
10. (i) 15.03.05.0)(points)2()( =×== WWPPa
0.690.31-1points)211P(
0.31
0.150.16points)2or211(
16.006.010.0
3.02.02.5.0
)or(points)211(
points2or,211,1,
21,0scoreCan)(
==<∴
=
+=
=+=
×+×=
=
P
DWWDPP
b
(ii) (a) after t seconds P has moved 2t metres
x
2t
P
R B45°
t
tx
txt
x
22
12
45cos2
45cos2
0
0
=
×=
=
=
ondskk
k
k
k
kkb
sec822
42
speeddistance time
42 length
41
2
2
diagonal oflength )()(2
22
=÷==
=
=
=
+=α
(ii)(b) )(β
21
tPR
t
tt
BRPBPR
tBR
tPB
2
2
24
2
2
2
22
222
=
=
−=
−=
=
=
from similar triangles
2
2
2
2
2
)2(2
)2(222
)2(
22)2(2
22
tkk
tkktktk
tk
tktkkQBdtd
tkkt
ARABPRQB
ARAB
PRQB
−=
−+−
=
−
×−−−=
−×
=
×=
=
9216
916
12
)4
3(
2
4
2
)8
22(
2
82put
2
2
2
2
2
2
2
2
=
×=
=
⎟⎠⎞
⎜⎝⎛ −
=
−
=
=
kk
kk
kk
k
kk
kQBdtd
kt
UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia
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SAMPLE C
(SOLUTIONS)
Mathematics S
Final Examination Paper
Time Allowed: 3 hours
Reading Time: 5 minutes
UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2011 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
Mathematics S Sample C Final Examination (SOLUTIONS)
1. i) a
a
2 2 27 17 2 7 17 46131 3= + − × × × °
∴ ≈cos
. ( significant figures) since a > 0
( ) ( )5 3 4 2 715 15
1
kkk
+ − ===
ii)
iii) 2
sec21
2tan 2 xx
dxd
=⎟⎠⎞
⎜⎝⎛
( )
lim lim
lim
x x
x
x xx
x xx
x→ →
→
+=
+
=+
=
0
2
0
0
43
434
343
iv)
v)
)42)(2(9
)8(9729
22
33
33
yxyxyxyx
yx
++−=
−=
−
vi)
( )x xx
x dx
x x k
+⎛⎝⎜
⎞⎠⎟
⌠
⌡⎮= +⌠
⌡⎮
= + +
1 1
2
2
vii)
35
654
2121area 2
π
π
θ
=
××=
= r
viii)
( ) ( )
( )
4 8
2 2
2 22 3
2 3 02 3 0
032
2
2
2
2 3
2 3
2
2
x x
x x
x x
x xx x
x x
x
=
=
=
∴ =
− =
− =
= ,
ix)
cosθ =
23
tanθ = −5
2
2. i)
satisfies4only shows Checking2634
521or521
====
+−=−−=−
xxxx
xxxx
ii) a)
∴ + =line line AB x y4 6 5
m
x y
m
AB =− −− −
= −
+ =
= − = −
3 11 523
4 6 546
23
( )
For
b) ( ) (M M
− + + − )⎛⎝⎜
⎞⎠⎟ ≡
1 52
3 12
2 1, ,
c) d)
( )
( )
bisectorlar perpendicu theofequation theis 0423
6322
2231
23=2 Using
=−−−=−
−=−
yxxy
xy
gradient and 1,
( ) ( )
( ) ( )
d
r
x yM A
AM = − − + −
= +
=
=
− + − =
1 2 3 1
9 4
13
13
2 1 13
2 2
2 2is the equation of circle
centre and through and .B
iii) y x x
y x xx
x
=
′ = +⎛⎝⎜
⎞⎠⎟
= +
ln
ln
ln
1
1
and when x e= 2 ′ = +
=y eln 2 1
3
Gradient of tangent is 3.
( )
( )
( ) ( )
log log
log
,
2 2
22
2 4
2
6 4
6 4
6 26 16 0
8 2 08 2
0 2
x x
x x
x xx xx x
xx x
+ + =
∴ + =
+ =
+ − =
+ − == −
> ∴ =but only.
iv) 3. i) a)
b) original is 2+= xy inverse is 2+= yx
22
27
2 += yx 2 −= xy )( 21 −=− xxf 29)3(1 =−=−f
ii) a) 2716)( =FP (b)
11740
2615
2716)( =×=FFP
2−
y
x
2
c) )()()()( JFnJnFnFUJn ∩−+=
)(221627 JFn ∩−+= 112738)( =−=∩ JFn
2711)( =∩ JFP
iii) ( )[ ]
[ ]
Vx
dx
x
=+
⌠⌡⎮
= +
= −
=
=⋅⋅
π
π
π
π
13 2
33 2
38 5
385
0 492
1
2
12ln
ln ln
ln
.
Volume is 0.492 cubic units (to 3 decimal places) iv)
t
t
ttttt
tt
xx
xx
122
211
21
21
tan1
sin1
cotcosec
22
22
=
=
−++=
−+
+=
+=
+
4. i) a)
y
x2
3
b)
41)2()2(
]2[
2 area
11
11
1
1
++−=
−−−+−=
+−=
+=
−
−−
−
−∫
ee
ee
xe
dxe
x
x
ii) y
( )2cos3 1 −= − xy π3
x
{ }{ }π30: : Range
31: :Domain ≤≤=≤≤=
yyYxxX
1 3
iii)
α
α
α
α
−=−−
−=
+−−−+−
−=
′−=
′−=
+−=′
−−+−=
21
2
1124126222482
)2()2(2
)()(
11123)(6116)(
0
001
2
23
ff
xfxf
xx
xxxfxxxxf
iv)
Let the roots be arara ,,
Product of roots 81
1623 −=−=a
21
−=a
Now roots are 2
,21,
21 rr
−−−
441
0)4)(14(04174134444
131
162261
1626
221
21roots of Sum
2
2
2
2
−=−=
=++=++
−=++
−=++
×−=++
=−−−=
rr
rrrr
rrr
rrr
rrr
rr
Roots are 81,
21,2 − (both values of r give same roots)
5.
i) a) 2
1
91
3sin
xx
dxd
−=⎟
⎠⎞
⎜⎝⎛ −
b)
21
21
)2ln()2ln()22ln(
−−
+=
−−+=⎥⎦⎤
⎢⎣⎡
−+
xx
xxdxd
xx
dxd
c) ( ) xxx ex
exedxd
21
21 2
1
==−
ii) a) ′ = +y x3 12 At A, x = −1 ′ = − +y 3 1 12( ) Gradient is 4
3 1 43 3
11
2
2
2
xxxx
+ =
=
== ±
b) A is the point where x = −1 When x = 1, y = 4 ∴ B(1, 4) ∴ Equation : y x
y xy x
− = −− = −
=
4 4 14 4 4
4
( )
c) y x
y x xx x x
x x
=
= + +
= + +
− + =
4 12 2
4 23 2 0
3
3
3
( )( )
Test x = −2 ( ) ( )
( , )− − − + = − + + =∴ − −
2 3 2 2 8 6 2 02 8
3
C is Alternative solution : x x3 3 2 0− + = has roots α α β, , 2 0
2 32 3
1 22 22 2
11
2
2
2
3
3
12
α β
α αβ
α ββ α
α α
α
ααβ
+ =
+ = −
= −⇒ = −
− = −
− = −
==
∴ = −
( )( )( )
( )( )
and substituting into (3)2
C is (−2, −8) and B is (1, 4)
( )
( )
Area
square units.
= − +
= − +⎡
⎣⎢⎤
⎦⎥
= − +⎛⎝⎜
⎞⎠⎟ − − + −
⎛⎝⎜
⎞⎠⎟
=
−
−
∫ x x dx
x xx
3
2
1
4 2
2
1
3 2
432
2
14
32
2164
122
4
634
d) 6. (i) (a) When , 0=t 61021 ×⋅=N ∴ 61021 ×⋅=A so tkeN 61021 ×⋅=
When 5=t , 61081 ×⋅=N ∴ ke566 10211081 ×⋅=×⋅
divide both sides by 61021 ×⋅
ke551 =⋅ 51ln5 ⋅=k
51ln51
⋅=k
(3 d.p.) 0810 ⋅=k
(b) when 14=t 14081061021 ×⋅×⋅= eN
61073 ×⋅= (c) When 61053 ×⋅=N te 081066 10211053 ⋅×⋅=×⋅
divide both sides by 61021 ×⋅ te 0810922 ⋅=⋅ 922ln0810 ⋅=⋅ t 2313 ⋅≈t After 13 days there were insects in the colony. 61053 ×⋅
(d) tedtdN 0810610210810 ⋅×⋅×⋅=
tedtdN 081097200 ⋅=
When 8=t
8081097200 ×⋅= edtdN
5596185818 ⋅= (2sig.fig) 190000= The rate at which the colony of insects was increasing after
8 days is 190000 insects per day.
ii) a) yx
=+2
1
b)
solutions 2 so twiceintersectsClearly graph.other of on top Draw)(graph. from 0or2,0When )(
1 isDomain axis. abovewhen 01
2)(
3xyyyx
xxx
=<><
−>−>+
γβ
α
iii)
General term is kkk xxC )
21( 3
2020 −−
kkkk xC )
21()1(42020 −−
put 0420 =− k 5=k
term is 21484)
21()1( 550
520 −=−xC
7.
i) a) 0> x
b) 2
11xx
y −=′
0112 =−
xx
012 =−
xx
1=x 1=y
32
21xx
y +−=′′
1)1( =′′ytherefore is a minimum turning point )1,1(therefore minimum value is 1=y c) 0=′ ′y
02132 =+−
xx
023 =+−
xx
2=x is a possible point of inflection
x 1 2 3 y ′′ 1 0 -1/27
Clearly concavity changes so there is a point of inflection when 2=x d)
y
x
(1,1)
2
ii) a) Stationary points when ′ =f x( ) 0, i.e. at x = 0, 4 Test x = 0
x 0− 0 0+ ′f + 0 +
So x = 0 is a horizontal point of inflection Test x = 4
x 4− 4 4+ ′f + 0 −
So x = 4 is a maximum turning point. b) f decreasing when x( ) ′ <f x( ) 0, i.e. x > 4 c) Concave down when gradient of ′f x( ) curve is negative, i.e. x < 0 or x > 3
8. i) a)
∫ ∫∫ +−=−=−
= cxxdxxdxxdxx )2sin21(
212cos1
21
22cos1sin 2
b)
Cxx
dxxx
dxx
+−=
⎮⌡⌠
+−=⎮⌡
⌠+
−
3tan3
991
91
22
2
ii) a) 3)109(1−
b) 99.0)109(1 >− n
n)9.0(01.0 > nlog(0.9)<log(0.01)
9.0log01.0log
>n NB reversal of inequality as dividing by a negative
7.43>n Ans 44=n
iii)
P
θ
6 cm 6 cm
Q R
dtd
dtdA
A
A
APQR
θθ
θ
θ
cos18
sin18
sin6621
of Area
=
=
×××=
=Δ
When area is 9 we have
6
21sin
sin189
sin36219
πθ
θ
θ
θ
=
=
=
××=
Also cm
radians second
2dAdt
s
ddt
ddt
ddt
ddt
ddt
=
∴ = ⋅
= ⋅
=
=
∴ =
3
3 186
318 3
2
3 9 3
13 3
39
/
cos
/
π θ
θ
θ
θ
θ
)4
sin(23cos3sin3
41
33tan
231833R(a)i)(
.922
π
παα
+=+
=∴==
==+=
xxx
π2 period23 Amplitudeb)( ==
(c)
y
x
2
4
6
-2
-4
-6
-π/4 7π/43π/4
(π/4,3√2)
(5π/4,-3√2)
ii) (a) 222 )( rryx =−+
2222 2 rryryx =+−+ 22 2 yyrx −= (b) Rotate circle in (a) about y-axis for hy ≤≤0
Volume = ∫ −h
dyyyr0
22π
= hyry 0
32 ]
3[ −π
= ]3
[3
2 hrh −π
iii) a) ⎮⌡⌠=
4
01 tan
π
θθ dI
θθθ
π
d⎮⌡⌠= 4
0 cossin
4
0
cosln
π
θ ⎥⎦
⎤⎢⎣
⎡−=
1ln2
1ln +−=
21
2ln−
−= 2ln21
=
b) 2−+ nn II
⎮⌡⌠ += −4
0
2tantan
π
θθθ dnn
⎮⌡
⌠⎥⎦
⎤⎢⎣
⎡+= −4
0
22 1tantan
π
θθθ dn
⎮⌡⌠= −4
0
22 sectan
π
θθθ dn
11
1tan 4
0
1
−=⎥
⎦
⎤⎢⎣
⎡−
=−
nn
nπ
θ
c) When , 3=n21
131
13 =−
=+ II
so 2ln21
21
3 −=I
When , 5=n41
151
35 =−
=+ II
So 2ln21
412ln
21
21
41
5 +−=+−=I
Let andx y
x y
= =
∴ = =
− −cos tan
cos tan
1 135
12
35
12
10. i)
So
sin , sin
cos
x y
y
= =
=
45
15
25
sin cos tan sin( )
sin cos cos sin
− −+⎡⎣⎢
⎤⎦⎥= +
= +
= × + ×
= +
=
1 135
12
45
25
35
15
85 5
35 5
11 525
x y
x y x y
TC x C x
TC x C x
TC x C x
n n
n n
n n
21 1
32
22
2
43
33
3
1 2 3
2 2 4
3 2 5
=+
=
=+
=
=+
=
ii) a)
( ) ( ) ( ) ( )
x y xC C
xC
xC
nx
x x xn
x
n n n nn n
n n nnn
n
2 2 0 1 2 2
0 2 1 3 2 4 2
2 3 4 2
2 3 4 2
= + + + ++
⎡
⎣⎢⎤
⎦⎥
= + + + ++
⎡
⎣⎢
⎤
⎦⎥+
b)
19
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )[ ]( )
ddx
x y x x xn x
n
x x x x
x x
n n nnn n
n n nnn n
n
2 0 1 2 2 31
0 1 22
2
2
3
3
4
4
2
2
1
= + + + ++
+
= + + + +
= +
+( )
( )
c) d) ∫
∫( )∫
( ) ( )
x y x x dx u x du dx
u u du
u u du
un
un
K
x yx
nx
nK
x
n
n
n n
n n
n n
2
1
2 1
22 1
1 1
1
2 11
21
1
0
= + = + =
= −
= −
=+
−+
+
=++
−++
+
=
+
+ +
+ +
( )
( )
and using then
So
Let
( ) ( )
( ) ( )
Kn n
x yx
n nx
n nxn
xn
n n
n n
=+
−+
∴ =++
−+
−++
++
=+ −
+−
+ −+
+ +
+ +
11
12
12
12
11
11
1 12
1 11
22 1
2 1
( ) ( )
e) If x = 1 in the series for x in part (b), then y2n
r
r
n Cr +=
∑ 20 .
Substitute for x = 1 in the answer to part (d) 2 1
22 1
12 2 1 2 2 2
1 22 2 2 2 2 2 2 1
1 22 1
1 2
22 1
1 2
2 1
2 2 1 1
1 1 1 1
1
0
1
n n
n n n n
n n n n
n
nr
r
n n
n nn n n
n nn n
n nnn n
Cn
nn n
+ +
+ + + +
+ + + +
+
=
+
−+
−−
+
=+ − − − − + +
+ +
=+ − − +
+ +
=+
+ +
∴+
=+
+ +∑
. . .( ) ( )
. . . .( ) ( )
.( ) ( )
.( ) ( )
2n
UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia
Telephone: 61 2 9385 5396 Facsimile: 61 2 9662 2651 Email: foundation.year@unsw.edu.au Web: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education,
training and consulting services and a wholly owned enterprise of the University of New South Wales ABN 62 086 418 582 CRICOS 00098G
SAMPLE D
(SOLUTIONS)
Mathematics S
Final Examination Paper
Time Allowed: 3 hours
Reading Time: 5 minutes
UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2011 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
Question 1
( ) ( )
106436
2615(i) 22
=+=
++−−=d
(ii)
( ) ( )( )
θθθ
θθ
tancossin
180cos180siniii
=
−−
=−+
( ) ( )
{ }22:Domain04
4lniv2
2
<<−=>−
−=
xxx
xy
( )24
1cos41sinv 11 π
=⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ −− (property)
( )
2
41
25
lim
by bottom & topdivide4
25limvi
2
3
33
3
−=
+
−=
→+−
∞→
∞→
x
x
xxx
x
x
x
( ) ( ) ( )
( )( )
16solution real no 42
2242
2or4024082
2let
8222vii
2
2
==−=
=−−=−
=−==−+=−+∴
−=
=−+−
xxx
xx
mmmmmm
xm
xx
y
x ( )
( )
122
1223
3,1 Substitute22
62viii
3
3
2
−+=∴
−=++=
++=
+=
xxy
cc
cxxy
xdxdy
2 8−
Question 2
( ) ( ) ( ) 22 55 2ai xx xeedxd −− −=
( ) ( )( ) ( )
xxxx
xxxx
xx
dxd
2
2
cossincoscos
)sin(1coscos
b
+
−−=⎟
⎠⎞
⎜⎝⎛
( ) ( ) ( ) ( )
( ) ( )( )( ) 0322
0242
0 roots, real noFor 022ai
2
2
<−+<+−+
<Δ=++++
kkkkk
kxkxk
322 >∪−< kk
( ) ( ) ( )
)2( and 2or
0 and 02 then allfor positive is
22)( Ifb
32
2
>∪−<−>
<Δ>+
++++=
kkk
kx
kxkxkxf
3
2>∴ k ( ) xexxf 2)(iii =
( )2
2
2
.2.)()a(
xxe
exxexfx
xx
+=
+=′
points stationaryat 0)( =′ xf
( ) 02solution no
02or 0 2
=+=+=∴
xxxxex
2402or 0
eyyxx==
−==∴
( ) ( )
( )( )24
222
222)(
2
2
2
++=
+++=
+++=′′
xxe
xxxe
xeexxxf
x
x
xx
point. turningminimum a is 0)(0,
.0)(,0At ∴
>′′= xfx
point. turningmaximum a is ),2(
.0)2()(,2At
24
2
e-
exfx∴
<−=′′−= −
( )figs) sig (2 1073
)100()100(b40
1002
−
−
×⋅=
−=− ef
( )c
2−
322−
2−>k :
:)2( 32>∪−< kk
+− −
322−
y
),2( 24e
-
x 0
Question 3
(i) (a) ⎮⌡⌠ +−=− cxdxx 65 )32(
121)32(
(b) cxx
dxxx
x
+++=
⎮⌡⌠
+++
)4(ln4
12
2
2
(c) dxxx⎮⌡
⌠++ 106
12
cx
dxx
++=
⎮⌡⌠
++=
− )3(tan1)3(
1
1
2
(ii) (a) ⎟⎠⎞
⎜⎝⎛= −
2cos 1 xy
12
1 ≤≤−x
22: ≤≤− xD π≤≤ yR 0: (b)
(iii) 0452 23 =+−− kxxx
(a) 25
=++ βαα
αβ 225−=∴ (*)
22
2 −=++ αβαβα 22 −=+∴ αβα (**) Substitute (*) into (**).
222522 −=⎟
⎠⎞
⎜⎝⎛ −+ ααα
245 22 −=−+ ααα
0253 2 =−− αα 0)2()13( =−+ αα
2,31
−=α
23,
619
−=β
Roots are 6
19,31,
31
−−
or 23,2,2 −
(b) 2
2 k−=βα
βα 22−=∴ k
2719
619
312
2
−=⎟⎠⎞
⎜⎝⎛
⎟⎠⎞
⎜⎝⎛−−=∴ k
or 1223)2(2 2 =⎟⎠⎞
⎜⎝⎛−−=k
y π
2π
x2 2−
Question 4
(i) ⎟⎟⎠
⎞⎜⎜⎝
⎛→ 2
2
0
2sinlimθ
θθ
⎟⎠⎞
⎜⎝⎛ ××=
→4
22sin
22sinlim
02 θθ
θθ
θ
4411 =××= Alternative solution
2
2
0
)cossin2(limθ
θθθ→
= ⎟⎟⎠
⎞⎜⎜⎝
⎛××
→θ
θθ
θ
22
2
0cossin4lim
4114 =××= (ii) (a) tkeII −= 0
kIekIdtdI tk −=−= −
0
(b) Substitute 021 II = , 2=t
into tkeII −= 0
ke 250 −=⋅
k=−⋅2
50ln
3470 ⋅=k
(c) te 3470
10001 ⋅−=
t=⋅− 3470
10001ln
...90719 ⋅=t seconds 919 ⋅ (iii) (a) kk
k nmC −55
(b) Coefficient of is 2x 14
1523
25 11 nmCnmC ×+×
nmnm 423 510 +=Coefficient of is 3x
5 232
323
5 11 nmCnmC ×+×
2332 1010 nmnm +=
So, 2332423 1010510 nmnmnmnm +=+ 324 105 nmnm =
5
1032
4
=nmnm
22
2
=nm
Question 5 (i)
(b) 811
31 4
=⎟⎠⎞
⎜⎝⎛
(c) ⎟⎠⎞
⎜⎝⎛×⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛
32
31
31 34
271
31 3
=⎟⎠⎞
⎜⎝⎛=
(d) 91
31 2
=⎟⎠⎞
⎜⎝⎛
(ii) (a) 12sin2 =x
212sin =x
∴base angle = 6π
6
,6
2 πππ−=∴ x )220( π≤≤ x
i.e. 6
5,6
2 ππ=x
125,
12ππ
=∴ x )0( π≤≤ x
Points of intersection are
⎟⎠⎞
⎜⎝⎛ 1,12π and ⎟
⎠⎞
⎜⎝⎛ 1,
125π
(b) Area ⎮⌡⌠ −=
125
12
)12sin2(π
πdxx
125
12
2cosπ
π⎥⎦⎤
⎢⎣⎡ −−= xx
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −−−⎟
⎠⎞
⎜⎝⎛ −−=
126cos
125
65cos ππππ
122
3125
23 ππ
++−=
3
3 π−=
...684850 ⋅=
(iii) (a) x
x 1tan =
01tan =−x
x
Let x
xxf 1tan)( −=
0220)80( <⋅−=⋅f 0150)90( >⋅=⋅f ∴ root between and 80 ⋅ 90 ⋅
(b) x
xxf 1tan)( −=
22 1sec)(
xxxf +=′
22
2
)90(190sec
15090
⋅+⋅
⋅−⋅=x
860 ⋅=
C C
C
C
I I
I
I
31(a)
31
31
32
31
32
32
32
Question 6
(i) m
m21
245tan+−
=°
m
m21
21+−
=
121
2=
+−
mm or 1
212
−=+−
mm
122 +=− mm mm 212 −−=− 3−=m 13 =m
31
=m
(ii) (a) )2sin( 1 xxdxd −
2
1
4122sin1
xxx
−×+×= −
2
1
41
22sinx
xx−
+= −
(b) From part (a) we have
211
41
22sin)2sin(x
xxxxdxd
−+= −−
211
41
2)2sin(2sinx
xxxdxdx
−−=∴ −−
21
43
241
43
21
24
21
)41(41
641
)41(841
21sin
41
41
22sin
2sin So
41
0
21
2
41
0
21
21
41
02
41
0
1
41
0
1
−+=⎟⎟⎠
⎞⎜⎜⎝
⎛−+=
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡−
+×=
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎮⎮⌡
⌠
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−−−−⎟
⎠⎞
⎜⎝⎛=
⎮⎮⌡
⌠⎟⎟⎠
⎞⎜⎜⎝
⎛
−−⎥
⎦
⎤⎢⎣
⎡=
⎮⌡
⌠⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
−
−
ππ
π x
dxxx
dxx
xxx
dxx
⎪⎭
⎪⎬
⎫
=
=
2
1(a)(iii)
xy
xy
2
1x
x =
4
1x
x =
15 =x 1=x )1,1(
(b) 23
1
21
0
21
=⎮⌡⌠+⎮⌡
⌠ −k
dxxdxx
231
32
1
1
0
23
=⎥⎦⎤
⎢⎣⎡−+⎥
⎦
⎤⎢⎣
⎡ k
xx
2311
32
=⎥⎦⎤
⎢⎣⎡ +−+
k
231
321
−+=k
611
=k
6=k
Question 7
(i) (a) rraS
n
n −−
=1
)1(
32
311200 ⎟
⎟⎠
⎞⎜⎜⎝
⎛⎟⎠⎞
⎜⎝⎛−
=
n
⎟⎠⎞
⎜⎝⎛ −= n3
11300
(b) r
aS−
=∞ 1300
311
200=
−=
(c) 00010311300300 ⋅<⎟⎠⎞
⎜⎝⎛ −− n
0001031300 ⋅<× n
00010
3003⋅
>n
00000033 >n
0000003ln3ln >n
3ln
0000003ln>n
5713 ⋅>n Smallest value 14=n (ii) (a)
⎭⎬⎫
=−=ex
xy ln(b)
1ln −=−= ey )1,( −∴ e
xy ln−= xy ln=− xe y =− yex 22 −=
⎮⌡⌠ −=
−
−0
1
22 )( dyeeV yπ
0
1
22
21
−
−⎥⎦⎤
⎢⎣⎡ += yeyeπ
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +−−⎟
⎠⎞
⎜⎝⎛ += 22
21
210 eeπ
⎥⎦⎤
⎢⎣⎡ −+= 22
21
21 eeπ
⎥⎦⎤
⎢⎣⎡ += 2
21
21 eπ
)1(2
2e+=π
(c) y
1x
e
1−
y
x1
Question 8
( ) ( ) ( )
( )
2
2
2
2
12
121
12
tan21
2sec
21
2tanai
tdtdx
t
x
xdxdt
tt
+=∴
+=
⎟⎠⎞
⎜⎝⎛ +=
=
=α
( )
2
2
22
2
2
2
122
1121
11
11
cossin1
tt
tttt
tt
tt
xx
++
=
+−+++
=
+−
++2
+=
++β
( )
( )
1ln3
11ln
1ln
11
222
31
6tan,
3
0,0
122
12
Integral
cossin1b
31
0
31
0
31
0
31
02
2
3
0
−⎟⎟⎠
⎞⎜⎜⎝
⎛+=
⎥⎦⎤
⎢⎣⎡ +=
⎮⌡⌠
+=
⎮⌡⌠
+=
⎪⎩
⎪⎨⎧
===
==
⎮⎮⎮
⌡
⌠
+++=
⎮⌡⌠
++
t
dtt
dtt
tx
tx
tt
tdt
xxdx
ππ
π
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
311ln
( )( )
3
2
2
274
94
3131 Now
3212
3
ianglessimilar tr Usingaii
y
yy
yxV
yx
xy
=
××=
=
=∴
=
( ) 43431Capacity Totalb =××=
3822
272
27
421
274
3
3
3
⋅≈
=
=
×=∴
y
y
y
( )
m./min.169at rising level water So
169
4941
2712c 2
=
××=∴
⋅=
dtdy
dtdydtdyy
dtdV
1
3 2x
y
Question 9
( ) ( )686429
1411
1412
1413ai =××
( )
686257
6864291
1411
1412
14131b
=
−=××−
( ) ( )
( )( )
πθππθ
θθ
θθθθ
θθ
==
−==
=+−=+−
=+
or 3
5,3
1cosor21cos
01cos1cos20cos1cos2
0cos2cosaii2
( ) ( ) ( )αbii
yyyxP
yxyx
2cos222
cos
cos
+=+=
=
=
θ
θ
θ
( ) 60cos12is,That =+ θy
( )
( )
( )
( )2
2
2
2
2
cos12sin450
2sin21.
cos1900
sincos.cos130
sincos
sin
sin221
θθ
θθ
θθθ
θθ
θ
θβ
+=
+=
+=
=
=
=
y
xy
xyA
( )γ
( ) ( ) ( )( )4
2
cos1
sin.cos12.2sin4502cos900.cos1
θ
θθθθθθ +
−+−+=
ddA
( ) ( )[ ]
( )( )( )
( )( )
( )( )3
3
3
4
cos1cos2cos900
cos12cos9002cos900
cos1sin2sincos2cos9002cos900
cos1sin2sin900cos12cos900cos1
θθθθ
θθθθ
θθθθθθ
θθθθθ
++
=
+−+
=
+++
=
++++
=
θ
y y
( )
3 .
2But
,3
5,
3 (a), From
0cos2cos when 0
πθ
πθ
πππ
θ
θθθ
δ
=∴<
=
=+=ddA
θ 4π
3π
2π
θddA
+ 0 −
θ x
MAX
- 10 -
( ) ( ) ( )n
n
xnn
xn
xnn
x
22
2
22
22
12
02
21aiii
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟⎠
⎞⎜⎜⎝
⎛=
+
Question 10 ( ) ( )
⎢⎢⎢⎢⎢⎢
⎣
⎡
⎟⎠⎞
⎜⎝⎛ +=
+
=
===
+=
3
333tan
32
93
3sin32
cos3sin3ai
πα
α
πR
x
xx
( )
1223,
125
49,
43
3
,3
233
For
22
3sin
63
sin32
6cos3sin3b
ππ
πππ
ππππ
π
π
=
=+
+≤+≤
=⎟⎠⎞
⎜⎝⎛ +
=⎟⎠⎞
⎜⎝⎛ +
=+
x
x
x
x
x
xx
( ) ( ) cxbcxabaxxP −−+−+= )()()(Let aii 23
0)1( Then,=
−−+−+= cbcabaP
)( offactor a is )1( xPx −∴ (b) ))(1()( 2 cbxaxxxP ++−=∴
( )
( ) 131212
32
123
112
012
31
,6 and 3 Ifb
12212⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=+
== nx
( ) 231212
32
123
112
012
31
,6 and 3 If
12212⎟⎟⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛=−
=−= nx
Adding 1 and 2 gives
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=+ 1221212 3
1212
32
120
12224
( )
( )1222
122
2243
1212
32
120
12
1211
1212
1212122
+=
+=
+=⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛∴
0
)(
)()(1
2
2
23
23
2
ccxccx
bxbx
xbcbx
axax
cxbcxabaxx
cbxax
−−
−
−+
−
−−+−+−
++(c) Subtracting 2 from 1 gives
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=− 1131212 3
1112
33
123
112
224
( )122
2243
1112
33
123
112
1211
1212113
−=
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛∴
If then or . 0)( =xP 1=x 02 =++ cbxax ( )
31223
1112
33
121
12 1211102 −=⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛∴
For 3 distinct real roots, must have 2 distinct roots where
02 =++ cbxax1≠x .
So conditions are and and 042 >− acb 0≠++ cba 0≠a .
UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia
Telephone: 61 2 9385 5396 Facsimile: 61 2 9662 2651 Email: foundation.year@unsw.edu.au Web: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education,
training and consulting services and a wholly owned enterprise of the University of New South Wales ABN 62 086 418 582 CRICOS 00098G
SAMPLE E
(SOLUTIONS)
Mathematics S
Final Examination Paper
Time Allowed: 3 hours
Reading Time: 5 minutes
UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Copyright © 2011 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner.
Question 1 (i)
figs) sig. (3 851848343551
37cos32294
cos2222
⋅=⋅=
°×××−+=
−+=
…a
Abccba
(ii)
( )
32
3cos2π
=
=
p
xxf
(iii)
( )
352193
1923
20
−=−+=
+=−==
daTda
(iv)
( ) xxdxd 2cos22sin =
(v)
( )( )( )2
2
2
1
2121
+=
++=
++
xe
xxexexe
x
x
xx
(vi)
( )( ) 03
62 23
=−−++=
PkxxxxP
( ) ( ) ( )
5153
06318273323 23
−==−
=−−+−−−+−+−
kk
kk 06 =
(vii)
44
=
⎟⎠⎞
⎜⎝⎛=
=
r
r
rlππ
θ
(viii)
40
040
2
===−
=Δ
mmmm
(ix)
1312cos
135sin
−=
=
θ
θ 5
12
13
θ
Question 2 (i) 13 += xy Inverse:
( ) 31
3
3
1..
1
1
−=
−=
+=
− xxfei
xy
yx
(ii) 0132 23 =+−+ xxx
31111
3
111
=++∴
−=−=++
++=++
γβα
αβγαγβγαβ
αβγαγβγαβ
γβα
(iii)
x
xxx
xxx3
1893
133(a)3
32
+−
+−−
3183
313
22
3
−+
+=−+−
∴x
xxx
xx
(b) (α) VA: 3±=x (β) OA: xy 3=
(iv)
(a) Let 2
tan At =
2tan
22
1111
111
111
cos1cos1
2
2
2
22
22
2
2
2
2
At
ttttt
tttt
AA
=
=
=
−+++−+
=
+−
+
+−
−=
+−
(b) Let °= 30A
( )2
2
32
3232
3232
231
231
30cos130cos115tan
−=
+−
×+−
=
+
−=
°+°−
=°
( )positive is 15tan
3215tan°
−=°∴
Question 3 (i) (a)
421
cos42
1sin(b)
1
1
π
π
==
=
==
=
−
−
yx
xy
yx
xy
⎟⎠
⎞⎜⎝
⎛∴
4,
21 π lies on both curves
2211
12
11
1sin(c)
1
2
1
=
−==
−=
= −
mx
xdxdy
xy
2211
12
11
1cos
2
2
1
−=
−−==
−−=
= −
mx
xdxdy
xy
( )( )( )2370
22221
22tan
′°==
−+−−
=∴
θ
θ
(iii) y
x
π
π2
– π2
1 – 1
{ }{ } units cubic 13
2
132
4tan
3tan
2
2tan21
2sec
8
6
6
8
2
+=
−−=
⎭⎬⎫
⎩⎨⎧
⎟⎠⎞
⎜⎝⎛−−=
⎥⎦⎤
⎢⎣⎡=
⎮⌡⌠=
−
−
π
π
πππ
ππ
π
π
π
x
dxxV
(iii)(a)
264
41651
6
=+=
=
⋅=
+
xxx
xx
P
Q
R
S
T
6
x 6
1.5
∴tip of shadow is 8m from pole
P
Q
R
S
T
6
y – x x
1.5
y
(b)
m/s 32
5034
34
3444
41
=
⋅×=×=
=
=−
=−
dtdx
dtdy
xy
yxyy
xy
Question 4 (i)
( )[ ]
( )( )
xxx
xxxx
xx
xxdxd
xx
dxd
ee
e
++
=
+−+
=
+−=
+−=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛+
2
2
2
21
121
12
1loglog
1log
(ii)
( )⎮⌡
⌠−−
−
1
123
23 dxx
x
xududxuxdxduuxxu
23324,12,1
3:limits
−=−−==−=−===−=
432ln2
232ln2
434ln2
3ln2
32
32
2
4
2
42
2
42
−=
⎟⎠⎞
⎜⎝⎛ +−⎟
⎠⎞
⎜⎝⎛ +=
⎥⎦⎤
⎢⎣⎡ +=
⎮⌡⌠ −=
−⋅⎮⌡⌠ −
=
uu
duuu
duuu
(iii) ⎟⎠⎞
⎜⎝⎛ + −−
21tan
53cossin 11
21tan
53cos
21tan
53coslet 11
==
== −−
βα
βα
α
5
3
4
β
5
2
1
( )
25511
5511
51
53
52
54
sincoscossinsin
=
=
⋅+⋅=
+=+ βαβαβα
(iv)
θ
θ
sin22
sin(a)
=
=
RS
RS
θθθθθ
θ
2sin4cossin8
sin2.cos4 Areacos2(b)
====
A
AOS
4
22
12sin when4 is max (c)
πθ
πθ
θ
=
=
=A
Question 5
(i) 61
61
3
3sinlim
23sin
lim00
=⋅=→→ h
h
h
h
hh .
(ii) (a) cxdxx
+⎟⎠⎞
⎜⎝⎛=
+−∫ 2
tan21
41 1
2 .
(b) ( ) cxdxx
x++=
+∫ 22 9ln
21
9 .
(c) ( ) ( )dxxxedxex xx ∫∫ +=+
22
1
cxe x ++=22
1 22
(iii) ( ))5log4log2 22 ++= yy ( ) 45loglog 2
22 =+− yy
45
log2
2 =⎟⎟⎠
⎞⎜⎜⎝
⎛+yy
165
2
=+yy
080162 =−− yy . ( )( ) 0420 =+− yy . 4or20 −== yy
But requires , y2log 0>y
20=∴ y is the only solution. (iv) (a) (calc.). ( ) 6320150 1 ⋅≈−=⋅ −eg . ( ) ( ) 2505050 2 ⋅=⋅=⋅h ( ) ( )5050 ⋅>⋅∴ hg .
(b) y
x1 – 1 – 2
1
2
– 1
– 2
y=h(x)
y=g(x)
(c) 2 solutions. (d) ( ) 122 −+= − xexxf ( ) xexxf 222 −−=′∴ . If 901 ⋅=x
then ( )( )9.0
90902 ffx′⋅
−⋅=
( ) ( )
( ) ( )902
9022
29.0219090 ⋅−
⋅−
−−+⋅
−⋅=e
e
9170 ⋅= to 3 sig.figs.
Question 6
(i) ( )( ) ( ) 14
!2!2!
!1!2!1
=−
+−−
nn
nn ( )1>n
( )( )( )
( )( )( ) 14
2!2!21
!2!21
=−
−−+
−−−
nnnn
nnn
( ) 142
11
1=
−+
− nnn
2822 2 =−+− nnn 0302 =−+ nn
( ) ( ) 065 =+− nn
∴Solution 5=n ( )1>n
(ii) (a) ( ) 86434422 ==+=R
3
34
34tan παα =∴==
⎟⎠⎞
⎜⎝⎛ −=+∴
3cos8sin34cos4 πxxx
(b) from part (a), 73
cos8 =⎟⎠⎞
⎜⎝⎛ −
πx
i.e. 87
3cos =⎟
⎠⎞
⎜⎝⎛ −
πx
∴for 3
233
ππππ−≤⎟
⎠⎞
⎜⎝⎛ −≤− x
..5053603
⋅±=−πx (calc.)
..5053603
⋅±=πx
(3s.f.) 5515420 ⋅⋅= or
(iii) (a) 4040,0 =∴== ANt keNt 1040110110,10 =∴==
40
110ln10 =k
10104
11ln101
⋅=⎟⎠⎞
⎜⎝⎛=k (3d.p.)
(b) After 6 weeks, 427 days. 6 =×=t
2782..87278140 421010 ≈⋅== ×⋅eN .
(c) ⇒= 300N te 101040300 ⋅=
⎟⎠⎞
⎜⎝⎛=⋅
40300ln1010 t
⎟⎠⎞
⎜⎝⎛
⋅=
430ln
10101t
...94919 ⋅=
i.e. after approx. 20 days.
(c) EITHER, when 110 ,10 == Nt
Hence ,
11.111101010 =×⋅== kNdtdN
Virus spreading at a rate of 11 people per day after 10 days.
OR, tedtdN 1010401010 ⋅×⋅= .
when , 10=t
101010401010 ×⋅×⋅= edtdN
..092211⋅=
Say 11 people/day
Question 7 (i) B 1 2 3 4 5 6 1 1,1 2,1 3,1 4,1 5,1 6,1 2 1,2 2,2 3,2 4,2 5,2 6,2 L 3 1,3 2,3 3,3 4,3 5,3 6,3 4 1,4 2,4 3,4 4,4 5,4 6,4 5 1,5 2,5 3,5 4,5 5,5 6,5 6 1,6 2,6 3,6 4,6 5,6 6,6
(a) 61
366= .
(b) 125
3615
= .
(c) 43
369
3618
3618
=−+ .
(ii) (a) L.H.S. = xxx 2sincoscos − ( )xx 2sin1cos −= xx 2coscos= x3cos= = R.H.S.
(b) ⎮⌡⌠ 3
0
3cosπ
dxx
⎮⌡⌠ −=
3
0
2sincoscosπ
dxxxx
3
0
3
3sinsin
π
⎥⎦
⎤⎢⎣
⎡−=
xx
(continued…)
i) (b) (continued)
(i
⎟⎟⎠
⎞⎜⎜⎝
⎛−−
⎟⎟⎟⎟
⎠
⎞⎛ sin 3 π
⎜⎜⎜⎜
⎝
−=3
0sin0sin3
33
sin3π
323
23
3
⎟⎟⎠
⎞⎜⎜⎝
⎛
−= 0−
833
= .
(b) Expansion =
(iii) (a) 27189 229
219
209
xx ⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛
( ) ⎥⎦
⎤⎢⎣
⎡+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛+ …27189 2
29
219
209
2 xxx
∴Coeff. of x =
89 219
2209
⎟⎟⎠
⎞⎜⎜⎝
⎛×−⎟⎟
⎠
⎞⎜⎜⎝
⎛
4096−= .
(c)
ives the coefficient of
.
Hence
when
( ) ( ) ( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛+−⎟⎟
⎠
⎞⎜⎜⎝
⎛+ …… 6354 2
69
259
xxax
g 6x
34 269
259
⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−= a
0269
259 34 =⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛− a
088416126 =××+×− a 3=a .
i.e. when
Question 8
Given x
xy elog=
(a) 0>x (b) 1=x
(c)
2
2
log1
log1
xx
x
xx
xy
e
e
−=
−⎟⎠⎞
⎜⎝⎛
=′
stationary points put 0=′y
ey
exxe
1
0log1
=
==−∴
3
4
4
2
log23
log22
2)log1(1
xx
xxxxx
x
xxx
xy
e
e
e
+−=
+−−=
−−⎟⎠⎞
⎜⎝⎛ −
=′′
368)0718,(2
point turningmaximum a is 1,
023at
3
⋅⋅
⎟⎠⎞
⎜⎝⎛∴
<+−
=′′
=
ee
ey
ex
(d) when 0=′′y
23
23log
3log2
ex
x
x
e
e
=
=
=
x 23
e< 23
e 23
e> y ′′ __ 0 +
∴concavity changes
∴⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
23
23
2
3,e
e is a point of inflection
)3350,484( ⋅⋅
(e)
y )1,(e
e
),( 23
232
3 −
ee x
1
(f) 2log
1
=∫ dxx
xke
22
)(log
1
2
=⎥⎦
⎤⎢⎣
⎡∴
k
e x
4)1(log)(log 22 =−∴ ee k 4)(log 2 =ke
2log ±=ke
1350,3972 ⋅⋅≈= ±ek
(g) x
xy e )(log −−=
x
xe
−−
=)(log
)( xf −= )1,(
ee− y
),( 23
232
3 −
− ee x
-1
Question 9 (i) (a) xx sin12cos −=
xx sin1sin21 2 −=−∴
0sinsin2 2 =−∴ xx 0)1sin2(sin =−xx
21sinor 0sin == xx
65,
6,2,,0 ππππ=x
(b) (c) 1sin2cos ≤+ xx
xx sin12cos −≤∴
65
6ππ
≤≤ x or ππ 2≤≤ x
or 0=x
(ii) (a) (α)
49
Prime (win)
Even (lose)
Even (lose)
Prime (win)15
410
510
110
59
π6 π 2π
y
x
(β)
94
94
101
104wins)(
=
×+=P
(b) P(win after 3 selections)
0040104
101
101
⋅=
××=
(c) P(win second game)
...........101
104
101
104
104
.................104
101
101
104
101
104
2
+⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+=
+⎟⎠⎞
⎜⎝⎛ ××+⎟
⎠⎞
⎜⎝⎛ ×+⎟
⎠⎞
⎜⎝⎛=
(i.e. an infinite geometric series)
94
1011
104
=
−=
Same probability as the first game
1
2
– 1
5π6
Question 10 (ii) (b) ⎮
⌡
⌠
−
−1
02
2
42 dxxxx
(i) 000805052 22 =++ yxyx
( )( )
( )
42
333
43223
62
4200sin23223
21sin2
42421
21sin2
424
11
1
0
21
221
1
0
21
21
02
2
−+=
−+−×=
⎟⎟⎠
⎞⎜⎜⎝
⎛+−−⎟⎟
⎠
⎞⎜⎜⎝
⎛+−=
⎥⎥⎦
⎤
⎢⎢⎣
⎡−+−−=
⎮⌡
⌠−−+⎮
⌡
⌠
−=
−−
−
−
π
π
xxxx
dxxxdxx
x
)I(010054 =+⎟⎠⎞
⎜⎝⎛ ++∴
dxdyy
dxdyxyx
When , 150=x
( )( )
20or)reject(3502035
07001500003575050000805075000045
2
2
2
=−=∴=−+∴=−+∴
=−+∴
=++
yyyy
yyyy
yy
Sub into (I) 20,150 == yx
5514
2750700
07502700
00002150205600
−=
−=∴
=+∴
=+⎟⎠⎞
⎜⎝⎛ ++
dxdy
dxdy
dxdy
dxdy
(iii) (a) ( )nx+1
nxnn
xn
xnn
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= 2
210
(b) Integrating both sided wrt x:
(ii) (a) ⎟⎠⎞
⎜⎝⎛ −−− 21 4
21
21sin2 xxx
dxd
( ) ( )
( )
2
2
2
2
2
22
2
22
2
4
42
242
44422
4
4
2
221
242124
21
24
2
x
xx
xx
xxx
xx
x
xxxxx
−=
−=
−
+−−=
−+
−−
−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
= −×−
−×+−−−
( )1
1 1
++ +
nx n
Cnx
nnxnxn
xn n
++⎟⎟
⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
+
132210
132
Let 1
10+
=⇒=n
Cx
( )1
11 1
+−+
∴+
nx n
132
11
231
121
0+
⎟⎟⎠
⎞⎜⎜⎝
⎛+
++⎟⎟⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛= nx
nn
nx
nx
nx
n
(c) Substitute : 1−=x
( )⎟⎟⎠
⎞⎜⎜⎝
⎛+
−++⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
+− +
nn
nnnn
n
n
11
231
121
011 1
Divide by 1− :
( )1
11
123
112
10 +
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
−++⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛nn
nn
nnn n
UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia
Telephone: 61 2 9385 5396 Facsimile: 61 2 9662 2651 Email: foundation.year@unsw.edu.au Web: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education,
training and consulting services and a wholly owned enterprise of the University of New South Wales ABN 62 086 418 582 CRICOS 00098G
UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia
Telephone: 61 2 9385 5396 Facsimile: 61 2 9662 2651 Email: foundation.year@unsw.edu.au Web: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education,
training and consulting services and a wholly owned enterprise of the University of New South Wales ABN 62 086 418 582 CRICOS 00098G
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