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2014 Higher School Certificate (HSC) Mathematics

Examination Solutions Written by: ymcaec

The sample answers or similar advice contained in this document are not intended to be exemplary or even complete answers or responses. As they are part of the working document, they may contain typographical errors, omissions, or only some of the possible correct answers.

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Section I

Question 1

2

1.644934...6

1.64 (3sf )

Answer: A

Question 2

Answer: B

Question 3

28

log 1 8

1 2

257

x

x

x

Answer: D

Question 4

2 21

2

x xe dx e c

Answer: C

3

Question 5

Equation perpendicular has gradient 3

2 .

Thus by point-gradient formula: 3 0

3 2 6 02 2

yx y

x

Answer: B

Question 6

3 28 27 2 3 4 6 9x x x x

Answer: D

Question 7

Solving it gives three solutions, however 2

is undefined, thus only two solutions.

Answer: B

Question 8

1

10 11

11

3 ; 2

3 2

3 2 3072

n

n

a x r x

T x x

T x x x

Answer: C

4

Question 9

Answer: A

Question 10

3 5 3 3(none finishes in 10 seconds)

4 6 5 8

3 5(at least one will finish in 10 seconds) 1

8 8

P

P

Answer: D

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Section II

Question 11 (a)

1 5 2

5 2 5 2 5 2

5 2

Question 11 (b)

23 2 3 2 1x x x x

Question 11 (c)

2 33

2

2

2

1 3

1 1

2 3

1

x x xd x

dx x x

x x

x

Question 11 (d)

2

1 1

33dx c

xx

Question 11 (e)

22

0 0

sin 2 cos2 2

12 1

2

2 2

x xdx

6

Question 11 (f)

2

2

2

' 4 5

( ) 2 5

(2) 3 2 2 5 2

5

( ) 2 5 5

f x x

f x x x c

f c

c

f x x x

Question 11 (g)

8 2

816

7

P r

Question 12 (a)

365

2 5 8 11 1094

3 1

1094

3652 5 8 11 1094 2 1094 200020

2

nT n

T

Question 12 (b) (i)

4 1 1

0 6 6

6 2 2

2 8 0

y

x

x y

x y

7

Question 12 (b) (ii)

2 2

3 2 0 85

1 2d

Question 12 (b) (iii)

2 2AC 6 3 3 5

1 15Area 3 5 5

2 2

Question 12 (c) (i)

Question 12 (c) (ii)

5 14 14 5 70

RG or GR19 18 19 18 171

P

Possible Outcomes

RR

RG

GR

GG

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Question 12 (d) (i)

2

2

2 8 2

2 6 0

0 or 3

3,6

x x x

x x

x

A

Question 12 (d) (ii)

3

2

0

3

3 2

0

Area 2 8 2

23 9

3

x x x dx

x x

Question 13 (a) (i)

3 sin 2 2cos 2d

x xdx

Question 13 (a) (ii)

cos 2 1

ln 3 sin 23 sin 2 2

xdx x c

x

Question 13 (b) (i)

since

kt

kt

kt

M Ae

dMk Ae

dt

kM Ae M

Thus, ktM Ae is a solution to the given equation.

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Question 13 (b) (ii)

0, 20; 300, 10

120, ln 2

300

1.984 when 1000

t M t M

A k

M t

Question 13 (c) (i)

2

3

3

3

1

1

11

1

2

1

For 0, 1 0

20 for 0

1

x tt

xt

xt

t t

x tt

Question 13 (c) (ii)

As , 1t x .

Question 13 (d) (i)

2 2 2ABC 78 11 67

AC 142 220 2 142 220 cos67

210.12AC

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Question 13 (d) (ii)

2 2 2AC BC ABcos ACB

2 AC BC

ACB 38.4679

Bearing 333

Question 14 (a)

'

''

x

x

x

y e ex

y e e

y e

For stationary points, consider ' 0y .

1

' 0

1

'' 0 minimum turning point

1,0 is a minimum turning point.

x

x

y e e

e e

x

y e

Question 14 (b) (i)

4

Question 14 (b) (ii)

2 2 6

4 62

3

k

k

11

Question 14 (c)

24

0

4

0

43 2

2

0

1

1 2

4

3 2

68

3

V x dx

x x dx

xx x

Question 14 (d) (i)

At the end of the first 8-hour: 10

3 mL remaining.

After the second dose: 40

3mL present.

Question 14 (d) (ii)

Let nA be the amount of drug after the nth dose.

1

12

23

1

10

1010 10

3 3

10 1010 10

3 9 3

10 10 110 ... 10

3 9 3

1015

11

3

n

n

A

AA

AA

A

Limiting sum of nA is 15, thus nA will not go over 15.

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Question 14 (e)

Note a maximum turning point at x = 1.

Question 15 (a) (ii)

2

2

2sin cos 2 0

2 2cos cos 2 0

cos 2cos 1 0

cos 0 or 2cos 1

3 5, , ,

2 2 3 3

x x

x x

x x

x x

x

Question 15 (b) (i)

(common)

(corresponding angles, SR EF)

||| (equiangular)

RDS FDE

RSD FED

DSR DEF

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Question 15 (b) (ii)

Corresponding sides of similar triangles are proportional, thus:

DR DS x

DF DE x y

Question 15 (b) (iii)

Ratio of areas of similar triangles is equal to the square of the ratio of the sides, thus: 2

1

1 ...(1)

A x

A x y

A x

A x y

Question 15 (b) (iv)

Similarly, it can be shown |||SEQ DEF and thus SQ y

DF x y

:

Similarly, 2 ...(2)A y

A x y

1 2

1 2

(1) (2) :

1A A

A A

A A A

Question 15 (c) (i)

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Question 15 (c) (ii)

2 ...(1)xmx e

At P, the two functions have the same gradient: 22 ...(2)xm e

2 2

2

2

2 1 0

1

2

1P ,

2

x x

x

e x e

e x

x

e

Question 15 (c) (iii)

2m e

Question 16 (a)

3

3

6sec sec 4sec 2sec0 4sec sec3 3 6 6 3

8 82 2 2

18 3 3

83

9 3

x dx

Question 16 (b) (i)

Let nA be the balance at the end of nth month.

1

2 1

2

500 1.003

500 1.01 1.003

500 1.003 500 1.01 1.003

A

A A

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Question 16 (b) (ii)

2

2

3 2 2

3

1 1

1 2 1

1.011.003

1.011.003

60

500 1.003 500 1.01 1.003

500 1.003 500 1.01 1.003 500 1.01 1.003

...

500 1.003 500 1.01 1.003 ... 500 1.01 1.003

500 1.003 1.003 1.01 1.003 ... 1.01

1500 1.003

1

n n n

n

n n n

nn

A

A

A

A

44404

Question 16 (c) (i)

12 2 10

2

12 10 2

2

5 14

x y x

y x x

y x

Question 16 (c) (ii)

2

2

2 22

2

1 38

3 5 18 4

315 3

8 4

515 3

8

xL xy

xx x

x xx x

x x

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Question 16 (c) (iii)

2

2

2

515 3

8

515 2 3

8

52 3 0

8

L x x

dLx

dx

d L

dx

For turning points, 0dL

dx .

52 3 15

8

15

52 3

8

60

24 5

x

x

2

20

60is a maximum turning point.

24 5

d L

dx

x

60 60 10,

24 5 24 5x y

to maximise L.