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1992–2013

SUCCESS ONE ®

HSC*

MATHEMATICS

Mathematics

2012H I G H E R S C H O O L C E R T I F I C AT E

E X A M I N AT I O N

General Instructions

• Reading time – 5 minutes

• Working time – 3 hours

• Write using black or blue penBlack pen is preferred

• Board-approved calculators maybe used

• A table of standard integrals isprovided at the back of this paper

• In Questions 11–16, showrelevant mathematical reasoningand/or calculations

Total marks – 100

Pages 2–6

10 marks

• Attempt Questions 1–10

• Allow about 15 minutes for this section

Pages 7–18

90 marks

• Attempt Questions 11–16

• Allow about 2 hours and 45 minutes for thissection

Section I

Section II

2012 HIGHER SCHOOL CERTIFICATE—EXAMINATION PAPER

432

9781741254716 SuccessOne HSC Mathematics BOOK 2014Ed.indb 4329781741254716 SuccessOne HSC Mathematics BOOK 2014Ed.indb 432 14/01/14 3:58 PM14/01/14 3:58 PM

Exce l S U C C E S S O N E H S C • M A T H E M A T I C S

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Section I

10 marksAttempt Questions 1–10Allow about 15 minutes for this section

Use the multiple-choice answer sheet for Questions 1–10.

1 What is 4.097 84 correct to three significant figures?

(A) 4.09

(B) 4.10

(C) 4.097

(D) 4.098

2 Which of the following is equal to ?

3 The quadratic equation x2 + 3x –1 = 0 has roots α and β.

What is the value of αβ + (α + β )?

(A) 4

(B) 2

(C) –4

(D) –2

1

2 5 3−

(A)

(B)

(C)

(D)

2 5 37

2 5 3

7

2 5 3

17

2 5 317

−

+

−

+


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4 The diagram shows the graph .

Which of the following statements is true?

5 What is the perpendicular distance of the point (2, –1) from the line y = 3x + 1?

(A)

(B)

(C)

(D)

6

10

6

5

8

10

8

5

a x

y

O

(A) and

(B) and

(C)

′ ( ) > ′′( ) <

′( ) > ′′( ) >

′

ƒ ƒ

ƒ ƒ

ƒ

a a

a a

0 0

0 0

aa a

a a

( ) < ′′( ) <

′( ) < ′′( ) >

0 0

0 0

and

(D) and

ƒ

ƒ ƒ

y x= ( )ƒ


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6 What are the solutions of for 0 ≤ x ≤ 2π ?

7 Let a = ex.

Which expression is equal to ?

(A)

(B)

(C)

(D)

e

e

x

x

x

x

2

2

2

2

loge

a2( )

3 1tan x = −

(A) and

(B) and

(C) and

(D) and

2

3

4

3

2

3

5

3

5

6

7

6

5

6

11

π π

π π

π π

π ππ6


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8 The diagram shows the region enclosed by y = x – 2 and y2 = 4 – x.

Which of the following pairs of inequalities describes the shaded region in the diagram?

(A) y2 ≤ 4 – x and y ≤ x – 2

(B) y2 ≤ 4 – x and y ≥ x – 2

(C) y2 ≥ 4 – x and y ≤ x – 2

(D) y2 ≥ 4 – x and y ≥ x – 2

9 What is the value of ?

22 33

–3

–4

–2

–1

1

2

3

4

x

y2 = 4 – x

y = x – 2

1–1–2–3–4–5–6 4 5 6

y

O

1

31

4

xdx

⌠

⌡⎮

(A)

(B)

(C)

(D)

13

3

13

4

9

12

ln

ln

ln

ln


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10 The graph of has been drawn to scale for 0 ≤ x ≤ 8.

Which of the following integrals has the greatest value?

22 331 4 5 6 x7 8

y

O

y x= ( )ƒ

(A)

(B)

(C)

(D)

ƒ

ƒ

ƒ

ƒ

x dx

x dx

x dx

x

( )

( )

( )

⌠

⌡⎮

⌠

⌡⎮

⌠

⌡⎮

0

1

0

2

0

7

(( )⌠

⌡⎮ dx

0

8

y x= ( )ƒ


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Section II

90 marksAttempt Questions 11–16Allow about 2 hours and 45 minutes for this section

Answer each question in the appropriate writing booklet. Extra writing booklets are available.

In Questions 11–16, your responses should include relevant mathematical reasoning and/orcalculations.

Question 11 (15 marks) Use the Question 11 Writing Booklet.

(a) Factorise 2x2 – 7x + 3.

(b) Solve .

(c) Find the equation of the tangent to the curve y = x2 at the point where x = 3.

(d) Differentiate .

(e) Find the coordinates of the focus of the parabola x2 = 16(y – 2).

(f) The area of a sector of a circle of radius 6 cm is 50 cm2.

Find the length of the arc of the sector.

(g) Find .

2

3

23 1 2x − <

2

2

2

2

3 25

+( )e x

sec2

0

2

2

xdx

π⌠

⌡⎮


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(a) Differentiate with respect to x.

(i)

(ii)

(b) Find .

(c) Jay is making a pattern using triangular tiles. The pattern has 3 tiles in the firstrow, 5 tiles in the second row, and each successive row has 2 more tiles than theprevious row.

(i) How many tiles would Jay use in row 20?

(ii) How many tiles would Jay use altogether to make the first 20 rows?

(iii) Jay has only 200 tiles.

How many complete rows of the pattern can Jay make?

Question 12 continues on page 9

2

2x xe

−( )1 log

cos x

x2

24

62

x

xdx

+

⌠

⌡⎮

1

2

2

Row 1

Row 2

Row 3

Question 12 continues on following page


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Question 12 (continued)

(d) At a certain location a river is 12 metres wide. At this location the depth of theriver, in metres, has been measured at 3 metre intervals. The cross-section isshown below.

(i) Use Simpson’s rule with the five depth measurements to calculate theapproximate area of the cross-section.

(ii) The river flows at 0.4 metres per second.

Calculate the approximate volume of water flowing through thecross-section in 10 seconds.

End of Question 12

3

1

0.52.3

3 3 3 3

2.93.8

2.1


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(a) The diagram shows a triangle ABC. The line 2x + y = 8 meets the x and y axesat the points A and B respectively. The point C has coordinates (7, 4).

(i) Calculate the distance AB.

(ii) It is known that AC = 5 and . (Do NOT prove this.)

Calculate the size of ∠ABC to the nearest degree.

(iii) The point N lies on AB such that CN is perpendicular to AB.

Find the coordinates of N.


C (7, 4)

NOT TOSCALE

xA2x + y = 8

B

y

O

2

2

3

BC = 65



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(b) The diagram shows the parabolas y = 5x – x2 and y = x2 – 3x. The parabolasintersect at the origin O and the point A. The region between the two parabolasis shaded.

(i) Find the x-coordinate of the point A.

(ii) Find the area of the shaded region.

(c) Two buckets each contain red marbles and white marbles. Bucket A contains3 red and 2 white marbles. Bucket B contains 3 red and 4 white marbles.

Chris randomly chooses one marble from each bucket.

(i) What is the probability that both marbles are red?

(ii) What is the probability that at least one of the marbles is white?

(iii) What is the probability that both marbles are the same colour?

End of Question 13

1

1

3

y = 5x – x2

y = x2 – 3x

x

y

O

A

1

2


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(a) A function is given by .

(i) Find the coordinates of the stationary points of and determine theirnature.

(ii) Hence, sketch the graph showing the stationary points.

(iii) For what values of x is the function increasing?

(iv) For what values of k will 3x4 + 4x3 – 12x2 + k = 0 have no solution?

(b) The diagram shows the region bounded by , the x-axis, the y-axis,

and the line x = 1.

The region is rotated about the x-axis to form a solid.

Find the volume of the solid.


2

ƒ x x x x( ) = + −3 4 124 3 2

1

1

ƒ x( )

y x= ( )ƒ

yx

=+( )3

22

x1

y

O

3

3



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(c) Professor Smith has a colony of bacteria. Initially there are 1000 bacteria. Thenumber of bacteria, N(t), after t minutes is given by

N(t) = 1000ekt.

(i) After 20 minutes there are 2000 bacteria.

Show that k = 0.0347 correct to four decimal places.

(ii) How many bacteria are there when t = 120?

(iii) What is the rate of change of the number of bacteria per minute, whent = 120?

(iv) How long does it take for the number of bacteria to increase from 1000to 100 000?

End of Question 14

1

1

2

1


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(a) Rectangles of the same height are cut from a strip and arranged in a row. Thefirst rectangle has width 10 cm. The width of each subsequent rectangle is 96%of the width of the previous rectangle.

(i) Find the length of the strip required to make the first ten rectangles.

(ii) Explain why a strip of length 3 m is sufficient to make any number ofrectangles.

(b) The velocity of a particle is given by

,

where x is the displacement in metres and t is the time in seconds. Initially theparticle is 3 m to the right of the origin.

(i) Find the initial velocity of the particle.

(ii) Find the maximum velocity of the particle.

(iii) Find the displacement, x, of the particle in terms of t.

(iv) Find the position of the particle when it is at rest for the first time.


1

1

2

1

10 cm

NOT TOSCALE

x t= −1 2cos

2

2



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(c) Ari takes out a loan of $360 000. The loan is to be repaid in equal monthlyrepayments, $M, at the end of each month, over 25 years (300 months).Reducible interest is charged at 6% per annum, calculated monthly.

Let $An be the amount owing after the nth repayment.

(i) Write down an expression for the amount owing after two months, $A2.

(ii) Show that the monthly repayment is approximately $2319.50.

(iii) After how many months will the amount owing, $An, become less than$180 000?

End of Question 15

1

2

3


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(a) The diagram shows a triangle ABC with sides BC = a and AC = b. Thepoints D, E and F lie on the sides AC, AB and BC, respectively, so that CDEF isa rhombus with sides of length x.

(i) Prove that �EBF is similar to �AED.

(ii) Find an expression for x in terms of a and b.

2

2

bb

aa

x

C D

E

B

F

A



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(b) The diagram shows a point T on the unit circle x2 + y2 = 1 at angle θ from the

positive x-axis, where .

The tangent to the circle at T is perpendicular to OT, and intersects the x-axis

at P, and the line y = 1 at Q. The line y = 1 intersects the y-axis at B.

(i) Show that the equation of the line PT is

x cosθ + y sinθ = 1.

(ii) Find the length of BQ in terms of θ .

(iii) Show that the area, A, of the trapezium OPQB is given by

.

(iv) Find the angle θ that gives the minimum area of the trapezium.

02

< <θ π

x

Q

T

Pq

y

B y = 1

O

2

1

2

A = −22

sincos

θθ

3



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(c) The circle x2 + (y – c)2 = r2, where c > 0 and r > 0, lies inside the parabolay = x2. The circle touches the parabola at exactly two points locatedsymmetrically on opposite sides of the y-axis, as shown in the diagram.

(i) Show that 4c = 1 + 4r2.

(ii) Deduce that .

End of paper

1

2

c > 1

2

x

y

O

(0, c)


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2012 Higher School Certificate

Worked Answers

QUESTION 14.097 84 = 4.10 (to 3 sig. fi gs.)

Answer B (1 mark)

QUESTION 2

1

2"5 2 "3 =

1

2"5 2 "3 3

2"5 1 "3

2"5 1 "3

= 2"5 1 "3

20 2 3

= 2"5 1 "3

17Answer D (1 mark)

QUESTION 3 x2 1 3x 2 1 = 0

a 1 b = 2ba

= 23

a b = ca

= 21

a b 1 (a 1 b) = 21 1 (23)= 24

Answer C (1 mark)

QUESTION 4At x = a, the curve is increasing and concave down.

6 f 9(a) > 0 and f 0(a) < 0

Answer A (1 mark)

QUESTION 5Use (2, 21), 3x 2 y 1 1 = 0,

d = 2ax1 1 by1 1 c"a2 1 b22

= 23(2) 2 1(21) 1 1"32 1 (21)22

= 2 8"102

= 8"10

Answer C (1 mark)

QUESTION 6 "3tan x = 21

tan x = 21"3

x = 5p

6,

11p

6

Answer D (1 mark)

QUESTION 7 a = ex

a2 = (ex)2

= e2x

Taking logs of both sides:

loge (a2) = loge (e2x)= 2x loge e= 2x

Answer C (1 mark)

Section I



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2012 HIGHER SCHOOL CERTIFICATE—WORKED ANSWERS

QUESTION 8Choose a point inside the shaded region: e.g. (3, 0).

For y2 = 4 2 x, 0 = 4 2 3 ?

0 = 1 ?

But 0 # 1

6 y2 # 4 2 x

For y = x 2 2, 0 = 3 2 2 ?

0 = 1 ?

But 0 # 1

6 y # x 2 2

6 y2 # 4 2 x and y # x 2 2

Answer A (1 mark)

QUESTION 9

34

1

13x

dx = 13 3

4

1

1x

dx

= 133ln x441

= 133ln 4 2 ln 14

= 13

ln 4

Answer B (1 mark)

QUESTION 10

As 36

2

f(x) dx < 0 and

38

6

f(x) dx < | 36

2

f(x) | dx, then the greatest

value is 32

0

f(x) dx.

Answer B (1 mark)

QUESTION 11

(a) 2x2 2 7x 1 3 = (2x 2 1)(x 2 3) (2 marks)

(b) |3x 2 1| < 2

Two cases:

3x 2 1 < 2 2(3x 2 1) < 2

3x < 2 1 1 3x 2 1 > 22

3x < 3 3x > 22 1 1

x < 1 3x > 21

x > 213

6 213

< x < 1 (2 marks)

(c) y = x2

dydx

= 2x

At x = 3, gradient m = 2 3 3= 6

At x = 3, y = (3)2

= 9 6 (3, 9)

Using y 2 y1 = m(x 2 x1)

y 2 9 = 6(x 2 3)

y 2 9 = 6x 2 18

6 6x 2 y 2 9 = 0 (2 marks)

(d) y = (3 1 e2x)5

Using the function of a function rule:dydx

= 5(3 1 e2x)4.d

dx(3 1 e2x)

= 5(3 1 e2x)4. 2e2x

= 10e2x(3 1 e2x)4 (2 marks)

(e) x2 = 16(y 2 2)

which is of the form (x 2 h)2 = 4a(y 2 k)

6 vertex (0, 2), and a = 4,

As parabola is concave up then focus (0, 6).

(2 marks)

Section II



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(f) Area = 12

r2 u

50 = 12

3 62 3 u

6 50 = 18u

u = 259

Now, / = ru

= 6 3 259

= 503

6 length is 503

cm (2 marks)

(g) 3p2

0

sec2x2

dx

= c2tan

x2d p

2

0

= 2 c tan p

42 tan 0 d

= 2[1 2 0]= 2 (3 marks)

QUESTION 12

(a) (i) (x 2 1) logex

Using the product rule,

Let u = x 2 1, u9 = 1

Let v = logex, v9 = 1x

dydx

= u9.v 1 v9.u

= 1. logex 1 1x

. (x 2 1)

= logex 1 x 2 1

xc or x loge x 1 x 2 1

xd (2 marks)

(ii) cos x

x2

Using the quotient rule,

Let u = cos x, u9 = 2sin x

Let v = x2, v9 = 2x

dydx

= v.ur 2 u.vr

v2

= x2.(2sin x) 2 cos x.2x

(x2)2

= 2x2 sin x 2 2x cos x

x4

= 2x sin x 2 2 cos x

x3 (2 marks)

(b) 3 4xx2 1 6

dx = 2 3 2xx2 1 6

dx

= 2 loge(x2 1 6) 1 c

(2 marks)

(c) Arithmetic series: 3 1 5 1 7 1 ca = 3, d = 2

(i) Use n = 20,

Tn = a 1 (n 2 1)d

T20 = 3 1 (20 2 1) 3 2= 3 1 38= 41 (2 marks)

(ii) Sn = n2

[a 1 /]

= 202

[3 1 41]

= 440 (1 mark)

(iii) Sn = n2

[2a 1 (n 2 1)d] = 200n2

[2 3 3 1 (n 2 1) 3 2] = 200

n[6 1 2n 2 2] = 400

n[2n 1 4] = 400

2n2 1 4n 2 400 = 0

n2 1 2n 2 200 = 0

6 n = 22 6 "22 2 4 3 1 3 (2200)

2 3 1

= 22 6 "804

2

= 13.18, 215.18 (corr. 2 dec. pl.)

6 only 13 complete rows could be made (2 marks)

(d) (i) Area <h3

[y0 1 y4 1 4(y1 1 y3) 1 2(y2)]

= 33

[0.5 1 2.1 1 4(2.3 1 3.8) 1 2(2.9)]

= 32.8

6 area is approximately 32.8 m2

(3 marks)

(ii) As 10 3 0.4 = 4, then

Volume = 32.8 3 4= 131.2

6 volume is approximately 131.2 m3

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QUESTION 13

(a) (i) A(x, 0): subs y = 0 in 2x 1 y = 8

2x = 8

x = 4

B(0, y): subs x = 0 in 2x 1 y = 8

y = 8

6 A(4, 0), B(0, 8)

AB2 = 42 1 82

= 16 1 64= 80

AB = "80

6 distance is 4"5 units

(2 marks)

(ii)

Let /ABC = u

Using cos u = ("65)2 1 ("80)2 2 52

2 3 "65 3 "80

= 120

144.222 051c

= 120

144.222 051c= 0.832 05c

u = 33.690 067 51c6 /ABC = 34o (nearest degree)

(2 marks)

(iii) Gradient of 2x 1 y = 8 is 22

6 gradient of perpendicular = 12

equation of CN:

y 2 y1 = m(x 2 x1)

y 2 4 = 12

(x 2 7)

2y 2 8 = x 2 7

x 2 2y = 21

Solve simultaneously:

2x 1 y = 8 .............................. 1

x 2 2y = 21 .............................. 2

2 3 2: 2x 2 4y = 22 .............................. 3

1 2 3 : 5y = 10

y = 2

Subs in 1:

2x 1 2 = 8

2x = 8 2 2

2x = 6

x = 3

6 N(3, 2) (3 marks)

(b) (i) Solve simultaneously:

x2 2 3x = 5x 2 x2

2x2 2 8x = 0

2x(x 2 4) = 0

x = 0, 4

6 A has x-coordinate of 4 (1 mark)

(ii) Area = 34

0

5x 2 x2 2 (x2 2 3x) dx

= 34

0

8x 2 2x2 dx

= c4x2 22x3

3d 4

0

= 4 3 422 2(4)3

3 2 0

= 64 2 1283

= 643

6 area is 2113

units2 (3 marks)

(c) (i) P(RR) = 35

3 37

= 9

35 (1 mark)

(ii) P(at least one white) = 1 2 P(RR)

= 1 2935

= 2635

(1 mark)

0

N

u

y

x

B (0, 8)

A (4, 0)

C (7, 4)

5

2x 1 y 5 8

�� 65

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(iii) P(both same colour) = P(RR) 1 P(WW)

= 35

3 37

1 25

3 47

= 1735

(2 marks)

QUESTION 14

(a) f (x) = 3x4 1 4x3 2 12x2

(i) f 9(x) = 12x3 1 12x2 2 24x = 0

12x(x2 1 x 2 2) = 0

12x(x 1 2)(x 2 1) = 0

6 Stat. points at x = 0, 22, 1

f (0) = 3(0)4 1 4(0)3 2 12(0)2 = 0 6 (0, 0)

f (22) = 3(22)4 1 4(22)3 2 12(22)2 = 232 6 (22, 232)

f (1) = 3(1)4 1 4(1)3 2 12(1)2 = 25 6 (1, 25)

6 Stat. points at (0, 0), (22, 232), (1, 25)

f 0(x) = 36x2 1 24x 2 24

For (0, 0):

f 0(0) = 36(0)2 1 24(0) 2 24

< 0 6 Max (0, 0)

For (22, 232):

f 0(22) = 36(22)2 1 24(22) 2 24

> 0 6 Min (22, 232)

For (1, 25):

f 0(1) = 36(1)2 1 24(1) 2 24

> 0 6 Min (1, 25)

6 Max (0, 0), Min (22, 232), Min (1, 25)

(3 marks)

(ii)

(2 marks)

(iii) From (ii), the function is increasing for 22 < x < 0 and x > 1

(1 mark)

(iv) 3x4 1 4x3 2 12x2 1 k = 0

3x4 1 4x3 2 12x2 = 2k

Solution is point(s) of intersection of

y = 3x4 1 4x3 2 12x2 and y = 2k.

Using part (ii), no solutions if 2k < 232 i.e. k > 32

(1 mark)

(b) V = p 31

0

y2 dx

= p 31

0

c 3(x 1 2)2 d 2

dx

= p 31

0

9(x 1 2)4 dx

= 9p 31

0

(x 1 2)24 dx

= 9p c (x 1 2)23

23d 1

0

= 23p c 1(x 1 2)3 d 1

0

= 23p c 127

218d

= 23p c219216

d=

19p

72

6 volume is 19p

72 units3 (3 marks)

(c) (i) N = 1000ekt, t = 20, N = 2000

2000 = 1000e20k

e20k = 2

loge e20k = loge 2

20k = loge 2

k = loge2

20= 0.034 657 359 c = 0.0347 (to 4 dec. pl.)

(1 mark)

y

x1 2 321

25

232

2223

y 5 3x4 1 4x3 2 12x2

(1, 25)

(0, 0)

(22, 232)



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(ii) N = 1000ekt, t = 120, = 1000ek(120)

= 64 000

6 64 000 bacteria [or, using k = 0.0347, N = 64 328]

(1 mark)

(iii) N = 1000ekt

dNdt

= k. 1000ekt

= k. 64 000= 64 000k= 2218.070 978 c= 2218 (nearest whole)

6 2218 bacteria per minute [or, using k = 0.0347, 2232 per minute]

(1 mark)

(iv) N = 1000ekt, N = 100 000

100 000 = 1000ekt

ekt = 100

loge ekt = loge 100

kt = loge 100

t = loge100

k

= 132.877 1238 c= 133 (nearest whole)

6 2h 13 min (2 marks)

QUESTION 15

(a) 10, 9.6, c

(i) a = 10, r = 0.96, n = 10

Sn = a(1 2 rn)

1 2 r

= 10(1 2 0.9610)

1 2 0.96

= 83.791 841 c= 83.79 (2 dec. pl.)

6 length of the strip is 83.79 cm

(2 marks)

(ii) lim sum = a

1 2 r

= 10

1 2 0.96

= 250

As limiting sum is 250, then the strip has a maximum length of 250 cm, or 2.5 m, which means that 3 m is suffi cient.

(1 mark)

(b) (i) x# = 1 2 2 cos t

Subs t = 0,

x#(0) = 1 2 2 cos (0)= 1 2 2= 21

6 initial velocity is 21 ms-1 (1 mark)

(ii) As 21 # cos t # 1, then

22 # 2 cos t # 2

6 2 $ 22 cos t $ 22

6 3 $ 1 2 2 cos t $ 21

6 maximum velocity is 3 ms-1

OR: Maximum velocity when x$

= 0

x$

= 2 sin t = 0

t = 0, p, 2p, c x#(0) = 21 from (i)

x#(p) = 1 2 2 cos p= 1 2 2(21)= 3

6 maximum velocity is 3 ms-1 (1 mark)

(iii) x# = 1 2 2 cos t

x = t 2 2 sin t 1 c

When t = 0, x = 3,

3 = 0 2 2 sin 0 1 c

6 c = 3

6 x = t 2 2 sin t 1 3 (2 marks)

(iv) x# = 1 2 2 cos t = 0

cos t = 12

t = p

3

x (p

3) =

p

3 2 2 sin

p

3 1 3

= p

3 2 2 (

"32

) 1 3

= p

3 2 "3 1 3 (2 marks)

(c) (i) 6% pa = 0.5% per month

As 0.5% = 0.005,

A1 = 360 000 3 1.005 2 M

A2 = A1 3 1.005 2 M

= (360 000 3 1.005 2 M) 3 1.005 2 M= 360 000 3 1.0052 2 M(1 1 1.005)

(1 mark)



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(ii) A300 = 360 000 3 1.005300

2 M(1 1 1.005 1 1.0052 1 c1 1.005299)

Now, A300 = 0

Using geometric series with

a = 1, r = 1.005, n = 300

M = 360 000 3 1.005300 ÷ 1(1.005300 2 1)

1.005 2 1

= 2319.485 045

6approximately $2319.50 (2 marks)

(iii) Let An = 180 000

360 000 3 1.005n

2 2319.50 (1 1 1.005 1 c1 1.005n-1) = 180 000

360 000 3 1.005n

2 2319.50 3 1(1.005n 2 1)

0.005 = 180 000

1800 3 1.005n

2 2319.5 3 (1.005n 2 1) = 900

1800 3 1.005n

2 2319.5 3 1.005n 1 2319.5 = 900

519.5 3 1.005n = 1419.5

519.5 3 1.005n = 1419.5

1.005n =

1419.5519.5

loge 1.005n = loge c1419.5

519.5d

n loge 1.005 = loge c1419.5

519.5d

n =

loge c1419.5519.5

dloge1.005

= 201.540 811 9 c6 after 202 months the amount will be less than $180 000. (3 marks)

QUESTION 16

(a)

(i) /BFE = /FCD (corr /s, FE||CD)

/FCD = /EDA (corr /s, FC||ED)

6 /BFE = /EDA

/FBE = /DEA (corr /s, BC||ED)

6 nEBF is similar to nAED (2 /s equal)

(2 marks)

(ii) FEDA

= BFED

( matching sides of sim. ns in proportion)

x

b 2 x =

a 2 xx

x2 = (a 2 x)(b 2 x)

x2 = ab 2 ax 2 bx 1 x2

ax 1 bx = ab

x(a 1 b) = ab

x = ab

a 1 b (2 marks)

(b)

B

E

ADC

Fx

x

b 2 x

a 2 x

b

a

y

O

R

Q y 5 1

T1

u

u

Pa

b



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(i) Using xa

1 yb

= 1:

In nOPT, is OTOP

= cos u

1a

= cos u

a = 1

cos u

In nORT, is OTOR

= sin u

1b

= sin u

b = 1

sin u

6 equation is x1

cos u

1 y1

sin u

= 1

x cos u 1 y sin u = 1 ................................ 1

(2 marks)

(ii) Subs y = 1 in 1:

x cos u 1 sin u = 1

x cos u = 1 2 sin u

x = 1 2 sin u

cos u

6 Q(1 2 sin u

cos u, 1)

As B(0, 1), then BQ = 1 2 sin u

cos u

(1 mark)

(iii) Area = 12

h(a 1 b)

A = 12

3 1 3 (1 2 sin u

cos u 1

1cos u

)

= 12

3 (2 2 sin u

cos u)

= 2 2 sin u

2cos u

6 A = 2 2 sin u

2cos u units2

(2 marks)

(iv) A = 2 2 sin u

2cos u

dAdu

= 2cos u(2cos u)2 (2 2 sin u).(22sin u)

4cos 2u

= 22cos

2u 1 4sin u 2 2sin 2u

4cos 2u

= 4sin u 2 2(sin

2u 1 cos 2u)

4cos 2u

= 4sin u 2 2

4cos 2u

= 2(2sin u 2 1)

4cos 2u

= (2sin u 2 1)

2cos 2u

= 0

2sin u 2 1 = 0

sin u = 12

u = p

6 (as 0 < u <

p

2)

Test for min:

up

12p

6p

3

dAdu

< 0 0 > 0

6 u = p

6 for minimum (3 marks)

(c) (i) y = x2 ................................. 1

x2 1 (y 2 c)2 = r2 ................................. 2

6 Subs 1 in 2:

y 1 (y 2 c)2 = r2

y 1 y2 2 2cy 1 c2 = r2

y2 1 (1 2 2c)y 1 c2 2 r2 = 0

D = 0 for equal roots:

6 D = (1 2 2c)2 2 4 3 1 3 (c2 2 r2)= 1 2 4c 1 4c2 2 4c2 1 4r2 = 0

4r2 2 4c 1 1 = 0

4c = 1 1 4r2 (2 marks)

(ii) From diagram, c > r

6 4c > 4r

16c2 > 16r2

16c2 > 4(4r2)

16c2 > 4(4c 2 1) (using part i)

16c2 > 16c 2 4

16c2 2 16c 1 4 > 0

4c2 2 4c 1 1 > 0

(2c 2 1)2 > 0

2c 2 1 > 0 (since c > 0)

2c > 1

c > 12

(1 mark)



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