SUCCESS ONE HSC...SUCCESS ONE ® HSC * MATHEMATICS EXTENSION 1 Mathematics Extension 1 2012 HIGHER...

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SUCCESS ONE ®

HSC*

MATHEMATICS EXTENSION 1

Mathematics Extension 1

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 – 2 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 11–14, showrelevant mathematical reasoningand/or calculations

Total marks – 70

Pages 3–6

10 marks

• Attempt Questions 1–10

• Allow about 15 minutes for this section

Pages 7–14

60 marks

• Attempt Questions 11–14

• Allow about 1 hour and 45 minutes for this section

Section I

Section II

359

2012 HIGHER SCHOOL CERTIFICATE — EXAMINATION PAPER

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 E X T E N S I O N 1S U C C E S S O N E H S C • M A T H E M A T I C S E X T E N S I O N 1

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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 Which expression is a correct factorisation of x3 – 27?

2 The point P divides the interval from A(–2, 2) to B(8, –3) internally in the ratio 3 :2.

What is the x-coordinate of P?

(A) 4

(B) 2

(C) 0

(D) –1

3 A polynomial equation has roots α, β and γ where

α + β + γ = –2, αβ + αγ + βγ = 3 and αβγ = 1.

Which polynomial equation has the roots α, β and γ ?

(A) x3 + 2x2 + 3x + 1 = 0

(B) x3 + 2x2 + 3x − 1 = 0

(C) x3 − 2x2 + 3x + 1 = 0

(D) x3 − 2x2 + 3x − 1 = 0

(A)

(B)

(C)

x x x

x x x

x x x

−( ) − +( )−( ) − +( )−( ) +

3 3 9

3 6 9

3 3

2

2

2 + +( )−( ) + +( )

9

3 6 92(D) x x x

360


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4 Which function best describes the following graph?

5 How many arrangements of the letters of the word OLYMPIC are possible if the C andthe L are to be together in any order?

(A) 5!

(B) 6!

(C) 2 × 5!

(D) 2 × 6!

x–2

2

y

3p2

3p2

( ) sin

( ) sin

( ) sin

( )

A

B

C

D

y x

y x

yx

y

=

=

=

=

−

−

−

3 2

3

22

32

3

2

1

1

1

ssin−1

2

x

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6 A particle is moving in simple harmonic motion with displacement x. Its velocity v isgiven by

.

What is the amplitude, A, and the period, T, of the motion?

7 Which expression is equal to ?

8 When the polynomial P(x) is divided by (x + 1)(x – 3), the remainder is 2x + 7.

What is the remainder when P(x) is divided by x – 3?

(A) 1

(B) 7

(C) 9

(D) 13

v x2 216 9= −( )

sin2 3x dx⌠

⌡⎮

(A)

(B)

(C)

12

13

3

12

13

3

x x C

x x C

−⎛⎝⎜

⎞⎠⎟+

+⎛⎝⎜

⎞⎠⎟+

sin

sin

112

16

6

1

2

1

66

x x C

x x C

−⎛⎝⎜

⎞⎠⎟+

+⎛⎝⎜

⎞⎠⎟+

sin

sin(D)

(A) and

(B) and

(C) and

(D) an

A T

A T

A T

A

= =

= =

= =

=

32

34

43

4

π

π

π

dd T = 2

3

π

362


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 E X T E N S I O N 1

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9 What is the derivative of cos –1 (3x)?

10 The points A, B and P lie on a circle centred at O. The tangents to the circle at A and Bmeet at the point T, and ∠ATB = θ .

What is ∠APB in terms of θ ?

(A)

(B)

(C)

(D)

1

3 1 9

1

3 1 9

3

1 9

3

1 9

2

2

2

2

−

−

−

−

−

−

x

x

x

x

O

A P

B

T q

(A)

(B)

(C)

(D)

θ

θ

θ

θ

2

902

180

° −

° −

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Section II

60 marksAttempt Questions 11–14Allow about 1 hour and 45 minutes for this section

Answer each question in a SEPARATE writing booklet. Extra writing booklets are available.

In Questions 11–14, your responses should include relevant mathematical reasoning and/orcalculations.

Question 11 (15 marks) Use a SEPARATE writing booklet.

(a) Evaluate .

(b) Differentiate x2 tan x with respect to x.

(c) Solve .

(d) Use the substitution u = 2 – x to evaluate .

(e) In how many ways can a committee of 3 men and 4 women be selected froma group of 8 men and 10 women?

(f) (i) Use the binomial theorem to find an expression for the constant term

in the expansion of .

(ii) For what values of n does have a non-zero constant term?

2

3x

x −<

32

3

1

1

213

12

xx

−⎛⎝⎜

⎞⎠⎟

213xx

n

−⎛⎝⎜

⎞⎠⎟

2

3

x x dx25

1

2

−( )⌠

⌡⎮

1

9 20

3

+

⌠

⌡⎮

xdx

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(a) Use mathematical induction to prove that 23n – 3n is divisible by 5 for n ≥ 1.

(b) Let .

(i) Find the domain of .

(ii) Find an expression for the inverse function .

(iii) Find the points where the graphs and y = x intersect.

(iv) On the same set of axes, sketch the graphs andshowing the information found in part (iii).

(c) Kim and Mel play a simple game using a spinner marked with the numbers 1, 2,3, 4 and 5.

The game consists of each player spinning the spinner once. Each of the fivenumbers is equally likely to occur.

The player who obtains the higher number wins the game.

If both players obtain the same number, the result is a draw.

(i) Kim and Mel play one game. What is the probability that Kim wins thegame?

(ii) Kim and Mel play six games. What is the probability that Kim winsexactly three games?

1

2

1

ƒ x x( ) = −4 3

ƒ x( )ƒ − ( )1 x

2

y x= ( )ƒ

y x= ( )ƒ y x= ( )−ƒ 1

2 3

1 45

1

2

3

Question 12 continues on the following page

365


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Question 12 (continued)

(d) Let A(0, –k) be a fixed point on the y-axis with k > 0. The point C(t, 0) is onthe x-axis. The point B(0, y) is on the y-axis so that �ABC is right-angled withthe right angle at C. The point P is chosen so that OBPC is a rectangle as shownin the diagram.

(i) Show that P lies on the parabola given parametrically by

x = t and .

(ii) Write down the coordinates of the focus of the parabola in terms of k.

End of Question 12

1

A (0, –k)

C (t, 0)

PB (0, y)

y

xO

yt

k=

2

2

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(a) Write in the form , where a and b are rational.

(b) (i) Find the horizontal asymptote of the graph .

(ii) Without the use of calculus, sketch the graph , showing the

asymptote found in part (i).

(c) A particle is moving in a straight line according to the equation

x = 5 + 6 cos 2t + 8 sin 2t,

where x is the displacement in metres and t is the time in seconds.

(i) Prove that the particle is moving in simple harmonic motion by showing

that x satisfies an equation of the form .

(ii) When is the displacement of the particle zero for the first time?

1

2

yx

x=

+

2

9

2

2

yx

x=

+

2

9

2

2

2

3

x n x c= − −( )2

2sin cos22

31− ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

a b


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(d) The concentration of a drug in the blood of a patient t hours after it wasadministered is given by

C(t) = 1.4te–0.2t,

where C(t) is measured in mg/L.

(i) Initially the concentration of the drug in the blood of the patientincreases until it reaches a maximum, and then it decreases.

Find the time when this maximum occurs.

(ii) Taking t = 20 as a first approximation, use one application of Newton’smethod to find approximately when the concentration of the drug in theblood of the patient reaches 0.3 mg/L.

End of Question 13

2

3

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(a) The diagram shows a large semicircle with diameter AB and two smallersemicircles with diameters AC and BC, respectively, where C is a point on thediameter AB. The point M is the centre of the semicircle with diameter AC.

The line perpendicular to AB through C meets the largest semicircle at thepoint D. The points S and T are the intersections of the lines AD and BD withthe smaller semicircles. The point X is the intersection of the lines CD and ST.

Copy or trace the diagram into your writing booklet.

(i) Explain why CTDS is a rectangle.

(ii) Show that �MXS and �MXC are congruent.

(iii) Show that the line ST is a tangent to the semicircle with diameter AC.

A M C

D

X

S

T

B

1

2

1


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(b) A firework is fired from O, on level ground, with velocity 70 metres per secondat an angle of inclination θ. The equations of motion of the firework are

x = 70t cosθ and y = 70t sinθ – 4.9t2. (Do NOT prove this.)

The firework explodes when it reaches its maximum height.

(i) Show that the firework explodes at a height of 250 sin2θ metres.

(ii) Show that the firework explodes at a horizontal distance of250 sin 2θ metres from O.

(iii) For best viewing, the firework must explode at a horizontal distancebetween 125 m and 180 m from O, and at least 150 m above the ground.

For what values of θ will this occur?

2

3

x

y

O

150

125 180

q

1


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(c) A plane P takes off from a point B. It flies due north at a constant angle α to thehorizontal. An observer is located at A, 1 km from B, at a bearing 060° from B.Let u km be the distance from B to the plane and let r km be the distance fromthe observer to the plane. The point G is on the ground directly below the plane.

(i) Show that .

(ii) The plane is travelling at a constant speed of 360 km/h.

At what rate, in terms of α, is the distance of the plane from the observerchanging 5 minutes after take-off?

End of paper

3r u u= + −1 2 cosα

A

G

P

60°

1

B

r

u

a

N

2

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372

2012 Higher School Certifi cateWorked Answers

(Total 10 marks)

1. x3 2 27 = (x 2 3)(x2 1 3x 1 9)

Answer C

[Check by expanding.]

2. A(22, 2), B(8, 23), 3 : 2

x = kx2 1 lx1

k 1 l

= 3 3 8 1 2 3 22

3 1 2

= 4

Answer A

3. ax3 1 bx2 1 cx 1 d = 0

[Note: all the choices have a = 1.]

Now a 1 b 1 g 52ba

6 a 1 b 1 g = 2b when a = 1

But a 1 b 1 g = 22

6 2b = 22

b = 2

ab 1 ag 1 bg 5ca

6 ab 1 ag 1 bg = c when a = 1

But ab 1 ag 1 bg = 3

6 c = 3

abg 52da

6 abg = 2d when a = 1

But abg = 1

6 2d = 1

d = 21

So b = 2, c = 3 and d = 21

6 x3 1 2x2 1 3x 2 1 = 0 has roots a, b and g.

Answer B

4.

x–2

2

y

3p2

3p2

y = sin21 u has domain 21 # u # 1

Here, 22 # x # 2

6 21 # x2

# 1

So the equation is of the form y = a sin21 x2

.

When x = 2, y = 3p

2

3p

2 = a sin21 1

= a 3 p

2

6 a = 3

The equation is y = 3 sin21 x2

.

Answer C

5. If the C and L are kept together they count as a single element. But the C and L can be arranged in 2 ways.

Total arrangements = 2 3 6!

Answer D

Section ISection I


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2012 HIGHER SCHOOL CERTIFICATE — WORKED ANSWERS

6. v2 = 16(9 2 x2)

The equation is of the form v2 = n2(A2 2 x2).

A2 = 9

6 A = 3 (A > 0)

n2 = 16

6 n = 4 (n > 0)

T = 2p

n

= 2p

4

= p

2

6 A = 3 and T = p

2

Answer A

7. cos 2x = cos2 x 2 sin2 x= 1 2 2 sin2 x

2 sin2 x = 1 2 cos 2x

sin2 x = 12

(1 2 cos 2x)

3sin 23x dx = 31

2(1 2 cos 6x)dx

= 12ax 2

16

sin 6xb 1 C

Answer C

8. P(x) = (x 1 1)(x 2 3)Q(x) 1 2x 1 7

P(3) = (3 1 1)(3 2 3)Q(3) 1 2(3) 1 7= 13

6 the remainder is 13.

Answer D

9. y = cos21(3x)

dydx

= 21"1 2 (3x)2

3 3

= 23"1 2 9x2

Answer D

10.

O

A P

B

T q

/TAO = /TBO = 908 (tangents perpendicular to the radius)

/AOB = 1808 2 u (angle sum of quadrilateral is 3608)

/APB = 12

(1808 2 u) (angle at centre is

twice angle at circumference standing on same arc)

6 /APB = 908 2 u

2Answer B

QUESTION 11

(a) 33

0

19 1 x2 dx = c1

3 tan

21

x3d 3

0

= 13

tan21 1 2 13

tan21 0

= 13

3 p

4 2

13

3 0

= p

12 (3 marks)

(b) y = x2 tan x

dydx

= x2 3 sec2 x 1 (tan x) 3 2x

= x(x sec2 x 1 2 tan x) (2 marks)

(c) x

x 2 3 < 2

x

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

x(x 2 3) < 2(x2 2 6x 1 9)

x2 2 3x < 2x2 2 12x 1 18

x2 2 9x 1 18 > 0

(x 2 3)(x 2 6) > 0

Section II


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Let y = (x 2 3)(x 2 6)

y

x0 3 6

y 5 x2 2 9x 1 18

From the diagram, y > 0 when x < 3 or x > 6.

6 x

x 2 3 < 2 when x < 3 or x > 6

(3 marks)

(d) Let u = 2 2 x

du = 2dx

x = 2 2 u

When x = 1, u = 1

When x = 2, u = 0

32

1

x(2 2 x)5dx = 30

1

2 (2 2 u)u5du

= 31

0

(2u5 2 u6)du

= cu6

32

u7

7d 1

0

= 13

217

2 0

= 4

21 (3 marks)

(e) Number of ways = 8C3 3 10C4

= 11 760 (1 mark)

(f) (i) a2x3 21xb12

General term = 12Ck(2x3)ka21xb122k

= 12Ck 2k x3k(21)122k xk212

= 12Ck (21)k 2k x4k212

Constant term occurs when 4k 2 12 = 0

4k = 12

k = 3

Constant term = 12C3 (22)3

[ = 21760] (2 marks)

(ii) a2x3 21xbn

General term = nCk(2x3)ka21xbn2k

= nCk (21)n2k 2k x4k2n

Constant term occurs when 4k 2 n = 0

n = 4k

6 a2x3 21xbn

will have a non-zero

constant term when n is a multiple of 4.

(1 mark)

QUESTION 12

(a) Aim to prove that 23n 2 3n is divisible by 5 for n $ 1.

When n = 1,

23 2 31 = 8 2 3 = 5

So it is true for n = 1.

Assume true for n = k.

i.e. assume 23k 2 3k = 5m

If n = k 1 1,

23(k + 1) 2 3k + 1 = 23k+3 2 3k+1

= 23k.23 2 3k.31

= 8(23k) 2 3(3k)= (5 1 3)(23k) 2 3(3k)= 5(23k) 1 3(23k) 2 3(3k)= 5(23k) 1 3(23k 2 3k)= 5(23k) 1 3(5m)= 5(23k 1 3m)

So, if true for n = k it is also true for n = k 1 1.

It is true for n = 1, so it is true forn = 1 1 1 = 2, so it is true for n = 2 1 1 = 3 and so on.

By the process of induction, 23n 2 3n is divisible by 5 for all integers n greater than or equal to 1. (3 marks)

(b) f(x) = "4x 2 3

(i) 4x 2 3 $ 0

4x $ 3

x $ 34

The domain is all real numbers x $ 34

.

(1 mark)


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(ii) y = "4x 2 3 Range is y $ 0 Inverse function: x = "4y 2 3 x2 = 4y 2 3 4y = x2 1 3

y = x2 1 3

4

So f 21(x) = x2 1 3

4, x $ 0 (2 marks)

(iii) Graphs intersect when f(x) = x

"4x 2 3 = x 4x 2 3 = x2

x2 2 4x 1 3 = 0 (x 2 1)(x 2 3) = 0 x = 1 or x = 3So the points of intersection are (1, 1) and (3, 3). (1 mark)

(iv) y

x

y 5 f21(x)

y 5 f(x)

y 5 x

3__4

3__4

(3, 3)

(1, 1)

(2 marks)

(c) (i) P(draw) = 15

[P(Second player spins the same number

as the fi rst player did)]

P(not a draw) = 45

P(Kim wins) = 12

345

= 25

(1 mark)

(ii) P(Kim does not win) = 35

P(Kim wins exactly 3 games out of 6)

= 6C3a25b3a3

5b3

= 864

3125 (2 marks)

(d)

A (0, –k)

C (t, 0)

PB (0, y)

y

xO

(i) OBPC is a rectangle.So PC is parallel to the y-axis.So at P, x = t

m = y2 2 y1

x2 2 x1

mAC = 0 2 (2k)

t 2 0

= kt

mBC = y 2 00 2 t

= 2yt

Now AC is perpendicular to BC.

So kt

3 2yt

= 21

ky = t2

y = t 2

k

BP is parallel to the x-axis so, at P,

y = t2

k.

6 P lies on the parabola given parametrically by

x = t and y = t2

k. (2 marks)

(ii) x = t, y = t 2

k

y = x2

k ky = x2

So the parabola is of the form x2 = 4ay. Now 4a = k

a = k4

The focus of the parabola is (0, k4

).

(1 mark)


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QUESTION 13

(a) sin a2cos 21a2

3bb

Let x = cos 21a2

3b

6 cos x = 23

cos2 x = 49

sin2 x = 1 2 cos2 x

= 1 2 49

= 59

sin x = 6"5

3

But 0 # cos21 u # p for all u

6 0 # x # p

6 sin x $ 0

So sin x = "5

3

sin a2cos 21a2

3bb = sin 2x

= 2 sin x cos x

= 2 3"5

33

23

= 4"5

9 (2 marks)

(b) (i) y = 2x2

x2 1 9 y(x2 1 9) = 2x2

x2y 1 9y = 2x2

2x2 2 x2y = 9y x2(2 2 y) = 9y

x2 = 9y

2 2 yNow 2 2 y Z 0

y Z 2The horizontal asymptote is y = 2.

(1 mark)

(ii) x2 $ 0 for all values of x

6 2x2

x2 1 9 $ 0 for all values of x

f(x) = 2x2

x2 1 9 f(0) = 0

f(2x) = 2(2x)2

(2x)2 1 9

= 2x2

x2 1 9

= f(x)6 the function is even

f(x) = 2

1 19x2

6 as x S `, f(x) S 22

y

x0

y 5 2

2

y 5 2x2________

x2 1 9

(2 marks)

(c) (i) x = 5 1 6 cos 2t 1 8 sin 2tx# = 212 sin 2t 1 16 cos 2t

x$

= 224 cos 2t 2 32 sin 2t= 24(6 cos 2t 1 8 sin 2t)= 24(x 2 5)6 the particle is undergoing simple harmonic motion as the acceleration is directly proportional and of opposite sign to the displacement.

(2 marks)

(ii) If x = 0,5 1 6 cos 2t 1 8 sin 2t = 0 6 cos 2t 1 8 sin 2t = 25

r

8

610

610

cos 2t 1 810

sin 2t = 212

sin a cos 2t 1 cos a sin 2t = 212

6 sin (a 1 2t) = 212


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where sin a = 610

a = 0.643 5011…

Now t $ 0, so a 1 2t > 0

a 1 2t = 7p

6,

11p

6, …

The displacement will fi rst be negative when

a 1 2t = 7p

6

2t = 7p

6 2 0.643 5011…

= 3.021 6903…

t = 1.510 845….= 1.5 [1 d.p.]

6 the displacement of the particle will fi rst be zero after 1.5 seconds.

(3 marks)

(d) C(t) = 1.4te20.2t

(i) dCdt

= 1.4t 3 20.2e20.2t 1 e20.2t 3 1.4

= 20.28te20.2t 1 1.4e20.2t

= e20.2t(1.4 2 0.28t)

Stationary point occurs when dCdt

= 0

i.e. e20.2t(1.4 2 0.28t) = 0 e20.2t = 0 or 1.4 2 0.28t = 0 No solution 0.28t = 1.4 t = 5

If t < 5, dCdt

> 0

If t > 5, dCdt

< 0

6 the maximum occurs when t = 56 the concentration of the drug in the blood is a maximum after 5 hours.

(3 marks)

(ii) C(t) = 0.36 1.4te20.2t 2 0.3 = 0Let f(t) = 1.4te20.2t 2 0.3

fr(t) = e20.2t(1.4 2 0.28t)

t2 = t1 2 f(t1)fr(t1)

where t1 = 20

t2 = 20 2 1.4 3 20e20.2320 2 0.3

e20.2320(1.4 2 0.28 3 20)

= 22.766 798…

The concentration of the drug in the blood will be approximately 0.3 mg/L after 22 hours and 46 minutes.

(2 marks)

QUESTION 14

(a)

(i) /SDT = 908 ( angle in semi-circle, diameter AB)

/ASC = 908 ( angle in semi-circle, diameter AC)

6 /DSC = 908 (AD is a straight line)/CTB = 908 ( angle in a semi-circle,

diameter CB)6 /CTD = 908 ( DB is a straight line)6 /SCT = 908 ( angle sum of

quadrilateral)6 CTDS is a rectangle (quadrilateral

with all angles right angles)

(1 mark)

(ii) In nMXS, nMXCMX = MX (common side)MS = MC ( radii of circle)XS = XC (diagonals of rectangle

are equal, and bisect each other)

6 nMXS K nMXC SSS

(2 marks)

(iii) /MCX = 908 (given)/MSX = /MCX (corresponding

angles of congruent triangles)6 /MSX = 908

6 ST is a tangent to the circle (meets radius on the circumference at right angles)

(1 mark)

A M C

D

X

S

T

B


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(b) (i) y = 70t sin u 2 4.9t2

dydt

= 70 sin u 2 9.8t

Stationary point occurs when dydt

= 0

i.e. 9.8t = 70 sin u

t = 507

sin u

d2ydt2 = 29.8 (< 0)

6 the maximum height is reached when

t = 507

sin u

When t = 507

sin u,

y = 70 3 507

sin u 3 sin u 2 4.9 3 (507

sin u)2

= 500 sin2 u 2 250 sin2 u

= 250 sin2 u

6 the fi rework explodes at a height of 250 sin2 u metres. (2 marks)

(ii) x = 70t cosu

When t = 507

sin u,

x = 70 3 507

sin u 3 cos u

= 500 sin u cos u= 250 3 2 sin u cos u= 250 sin 2u

6 the fi rework explodes at a horizontal distance of 250 sin 2u metres from O.

(1 mark)

(iii) 08 < u < 908

Height $ 150 m 250 sin2 u $ 150 sin2 u $ 0.6 sin u $ 0.774 5966… (sin u > 0) u $ 50.768 479…8 08 < 2u < 1808

125 m < horizontal distance < 180 mSo 250 sin 2u > 125

sin 2u > 0.5 308 < 2u < 1508

158 < u < 758

And 250 sin 2u < 180sin 2u < 0.72

2u < 46.054…8 or 2u > 133.945…8

u < 23.027…8 or u > 66.972…8

6 the fi rework will be at the best position for viewing when 678 # u < 758.

(3 marks)

(c)

(i) In nPBG,

sin a = PGu

6 PG = u sin a

cos a = BGu

6 BG = u cos aIn nABG, by the cosine rule, AG2 = (u cos a)2 1 12

2 2 3 u cos a 3 1 3 cos 608

= u2 cos2 a 1 1 2 u cos aIn nPGA, by Pythagoras’ theorem, PA2 = PG2 1 AG2

r2 = u2 sin2 a 1 u2 cos2 a 1 1 2 u cos a= u2(sin2 a 1 cos2 a) 1 1 2 u cos a= u2 1 1 2 u cos a= 1 1 u2 2 u cos a

6 r = "1 1 u2 2 u cos a (r > 0)

(3 marks) (ii) r = "1 1 u2 2 u cos a

= (1 1 u2 2 u cos a)12

drdu

= 12

(1 1 u2 2 u cos a)2

12(2u 2 cos a)

= 2u 2 cos a

2"1 1 u2 2 u cos a

Now 360 km/h = 6 km/min

So dudt

= 6

drdt

= drdu?

dudt

= 2u 2 cos a

2"1 1 u2 2 u cos a3 6

= 3(2u 2 cos a)"1 1 u2 2 u cos a

5 minutes after take off u = 5 3 6= 30

6 drdt

= 3(2 3 30 2 cos a)"1 1 302 2 30 cos a

= 3(60 2 cos a)"901 2 30 cos a

km/min

[or 60 2 cos a

20"901 2 30 cos a km/s] (2 marks)

A

G

P

60°

1

B

r

u

a

N


14/01/14 4:08 PM

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SUCCESS ONE HSC...SUCCESS ONE ® HSC * MATHEMATICS EXTENSION 1 Mathematics Extension 1 2012 HIGHER...

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Transcript of SUCCESS ONE HSC...SUCCESS ONE ® HSC * MATHEMATICS EXTENSION 1 Mathematics Extension 1 2012 HIGHER...