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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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

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,

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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