CSsA 2006

Centre Number 

Student Number 

CATHOLIC SECONDARY SCHOOLS 

ASSOCIATION OF NEW SOUTH WALES 

2006 

TRIAL HIGHER SCHOOL CERTIFICATE 

EXAMINATION 

Mathematics 

Morning Session 

Monday 7 August 2006 

General Instructions 

• Reading time – 5 minutes 

• Working time – 3 hours 

• Write using blue or black pen 

• Board-approved calculators may be 

used 

• A table of standard integrals is 

provided separately 

• All necessary working should be 

shown in every question 

• Write your Centre Number and 

Student Number at the top of this 

page 

Total marks – 120 

• Attempt Questions 1-10 

• All questions are of equal value 

Disclaimer 

Every effort has been made to prepare these ‘Trial’ Higher School Certificate Examinations in accordance with the Board of Studies documents, 

Principles for Setting HSC Examinations in a Standards-Referenced Framework (BOS Bulletin, Vol 8, No 9, Nov/Dec 1999), and Principles for 

Developing Marking Guidelines Examinations in a Standards Referenced Framework (BOS Bulletin, Vol 9, No 3, May 2000). No guarantee or 

warranty is made or implied that the ‘Trial’ Examination papers mirror in every respect the actual HSC Examination question paper in any or all 

courses to be examined. These papers do not constitute ‘advice’ nor can they be construed as authoritative interpretations of Board of Studies 

intentions. The CSSA accepts no liability for any reliance use or purpose related to these ‘Trial’ question papers. Advice on HSC examination issues is 

only to be obtained from the NSW Board of Studies. 

1 

2602-1

Total marks – 120 

Attempt Questions 1-10 

All questions are of equal value. 

Answer each question in a SEPARATE writing booklet. 

Question 1 (12 marks) Use a SEPARATE writing booklet. 

Marks 

(a) 

Evaluate 

3 

− 38.67 × 7.2 

2 

( 11.7) − ( 1.83) 

2 

correct to 3 decimal places. 

2 

(b) 

By rationalising the denominator, simplify 

1− 

2 − 

2 

. 

8 

2 

(c) 

Solve 

1 

cosθ = − , for 0 ≤ θ ≤ 2π 

. 

2 

2 

(d) Completely factorise 4 xy+ xb+ 

8ay+ 

2ab. 2 

(e) For what values of k does the quadratic equation 4x 2 + kx + 9 = 0 have 

equal roots? 

2 

(f) Solve the equation x − 2 = 3 . 2 

2


Marks 

2 2 

(a) Sketch the region represented by x + y < 4 . 2 

(b) The function f (x) 

is given by f (x) = 

x + 1 

x 

2 

− 9 

for 

for 

x ≤ 3 

x > 3 

2 

Find f (3) 

- f (6) 

. 

(c) 

y 

• 

A(a,6) 

6 

• 

B(b,6) 

L 1 

NOT TO SCALE 

O 

x 

C ( − 2, − 4) 

• 

L 2 

L 3 

On the diagram above line L 

1 

is parallel with the x-axis and crosses the 

y-axis at 6. Lines L 

2 

and L 

3 

each pass through the origin, O, and 

intersect with the line L 

1 

at the points A (a, 6) 

and B (b, 6) 

respectively. 

The point C ( −2, 

− 4) 

lies on the line L 

3. 

(i) Show that the equation of the line L 

3 

is 2 x − y = 0 . 2 

(ii) Show that the x-ordinate of point B is 3. 1 

(iii) Find the coordinates of point A such that ∠ AOB is a right 

angle. 3 

(iv) 

Write down the distance between points A and B. Hence, or otherwise, 

calculate the area of ∆ AOB. 

2 

3


Marks 

(a) For the parabola ( x − 2) 

2 = 16y 

, find the coordinates of the: 

(i) Vertex. 1 

(ii) Focus. 1 

(b) 

Differentiate with respect to x the following expressions: 

(i) 

(ii) 

3 x log 

e 

x . 2 

2 

sin x . 2 

(c) 

Find: 

(i) 

∫ 

cos 2006x dx . 2 

(ii) 

∫ 1 0 

2 

e x 

dx . (Leave your answer in exact form). 

2 

(d) Show that the equation of the normal to the curve y = x 

3 − 5x 

at the 

point ( 1, − 4) 

is given by x − 2 y − 9 = 0. 

2 

4


Marks 

1 3 2 

(a) Consider the curve f ( x) 

= − x − x + 3x 

+ 1 . 

3 

(i) 

Find the coordinates of any stationary points and determine 

their nature. 

3 

(ii) Find any point(s) of inflexion. 2 

(iii) Sketch the curve in the domain, − 6 ≤ x ≤ 3. 2 

(iv) What is the maximum value of f (x) 

in the given domain? 1 

(b) 

P 

x 

Q 

120 

A 

80 o 

NOT TO SCALE 

S 

R 

PR and QS are straight lines intersecting at point A. Also PS = QR , 

o 

o 

∠PSA = ∠QRA 

= 80 , ∠PAQ =120 and ∠ PQA = x. 

(i) 

Copy the diagram into your writing booklet. 

(ii) Prove that ∆PSA 

is congruent to ∆QRA 

. 2 

(iii) Hence, show that x = 30 o . 2 

5


(a) (i) Write 

• 

0 .7 5 as the sum of an infinite geometric series. 

Marks 

1 

(ii) 

Hence, express 

• 

0 .7 5 as a fraction. 

2 

(b) 

Simplify 

2 

1− 

sin x 

. 

cot x 

2 

(c) 

The circle below has centre O, radius 

π 

6 cm and arc length MN = 1 cm. 

M 

O 

θ 

N 

NOT TO SCALE 

(i) Find the size of θ in radians. 1 

(ii) Hence, or otherwise, find the exact area of the sector MON. 2 

(d) 

During qualification for the 2006 World Cup, the Socceroos goalkeeper, 

Mark, defended many penalty shots at goal. In fact, the probability that 

he can stop a penalty shot at goal is 

5 

3 . During a particular match, the 

opposing team had three penalty shots at goal. 

Using a tree diagram, find the probability that: 

(i) the goalkeeper will stop all shots at goal. 2 

(ii) the goalkeeper will stop at least 1 shot at goal. 2 

6


Marks 

(a) 

A 

NOT TO SCALE 

16 cm 12 cm 

α 

β 

O 1 O 2 

20 cm 

B 

A circle with centre at O 1 and radius 16 cm intersects with another circle 

with centre at O 2 and radius 12 cm. Their points of intersection are A 

and B and the distance between their centres, O 1O2 

, is 20 cm. 

∠AO 1O2 

= α and ∠AO 2O1 

= β . 

(i) Show that ∆ O 1 

AO2 

is a right angled triangle. 1 

(ii) Find the size of the angles α and β . 2 

(iii) 

Find the shaded area enclosed by these circles. (Give your 

answer correct to 2 decimal places). 

3 

Question 6 continues on page 8 

7

Question 6 (continued) 

Marks 

(b) 

y 

NOT TO SCALE 

2 

x 

y = e 

5 

1 

O 

x 

2 

The shaded region bounded by the graph y = e 

x , the line y = 5 and 

the y-axis is rotated about the y-axis to form a solid of revolution. 

(i) Show that the volume of the solid is given by V 

5 

= ∫ log y 

y 

π 

e 

dy . 

1 

2 

(ii) 

Copy and complete the following table into your writing booklet. 

Give all answers correct to three decimal places. 

1 

y 1 2 3 4 5 

log y 0 0.693 1.099 1.609 

e 

(iii) 

Use Simpson’s Rule with five function values to approximate the 

volume of the solid of revolution V , correct to three decimal places. 

y 

3 

End of Question 6 

8


Marks 

(a) Solve the following equation: + log ( x + 7) 3 

log 

2 

x 

2 

= , for x > 0. 3 

(b) 

y 

3 

2 

y = x 

NOT TO SCALE 

A 

O 

B 

y = 3 − 2x 

2 

x 

2 

2 

The shaded region OAB is bounded by the parabolas y = x , y = 3 − 2x 

and the x-axis. Point A is the intersection of the two parabolas and point B is 

2 

the x-intercept of the parabola y = 3 − 2x 

. 

(i) Find the x-ordinates of points A and B. 2 

(ii) 

By considering the sum of two areas, show that the exact area of the 

3 

shaded region OAB is given by 2 − 2 square units. 

2 

3 


9


Marks 

(c) 

NOT TO SCALE 

r 

= 2 cm 

R = 6cm 

The playing track of a CD is made out of a number of concentric circles with 

the inner circle having a radius of 2 cm and the outer circle having a radius 

of 6 cm. The CD is rotating at 5 revolutions per second and takes 25 minutes 

to completely play. 

(i) Find the total number of revolutions for this CD. 1 

(ii) 

Find the total length of the playing track in km, correct to one 

decimal place. 

3 

End of Question 7 

10

Question 8 (12 marks) Use a SEPARATE writing booklet 

Marks 

(a) 

m 

4m 

If A = 3, 

find the value of A − 5. 2 

(b) 

A 100 mg tablet is dissolved in a glass of water. After t minutes the amount 

of undissolved tablet, U in mg, is given by the formula: 

U = 100 

kt 

e − 

, where k is a constant. 

(i) 

(ii) 

Calculate the value of k, correct to 4 decimal places, given that 2 mg 

of the tablet remain after 10 minutes. 

Find the rate at which the tablet is dissolving in the glass of water 

after 12 minutes. Give your answer correct to two decimal places. 

2 

2 

(c) 

Mr. Egan borrows $P from a bank to fund his house extensions. The term of 

the loan is 20 years with an annual interest rate of 9%. Each month, interest 

is calculated on the balance at the beginning of the month and added to the 

balance owing. Mr. Egan repays the loan in equal monthly instalments of 

$1 050. 

(i) Write an expression for the amount, A 

1 

, Mr. Egan owes immediately 

at the end of the first month. 

(ii) Show that at the end of n months the amount owing, A 

n, is given by: 

n 

n 

A = P( 1.0075) − 140000(1.0075) + 140 000 

n 

(iii) If at the end of 20 years the loan has been repaid, calculate the amount $ 

Mr. Egan originally borrowed, correct to the nearest dollar. 

1 

3 

2 

11


Marks 

(a) 

Two particles, A and B, move along a straight line so that their 

displacements, x 

A 

and x B 

, in metres, from the origin at time t seconds are 

given by the following equations respectively: 

x A 

= 12 t +5 

x B 

= 

2 3 

6t 

− t 

(i) Find two expressions for the velocities of particles A and B. 2 

(ii) Which of the two particles is travelling faster at t = 1 second? 1 

(iii) At what time does particle B come to rest? 1 

(iv) Find the maximum positive displacement of particle B. 2 

(b) 

y 

NOT TO SCALE 

A 

P 

1 

Q 

B 

O 

C 

x 

y = 1− 

x 

2 

In the diagram above P and Q are two points on the parabola y = 1 – x 2 . The 

tangents to the parabola at P and Q intersect each other on the y-axis at A 

and the x-axis at B and C respectively. Triangle ABC is an equilateral 

triangle. 

(i) Show that the gradient of the tangent at point P is equal to 3 . 1 

(ii) 

Show that the coordinates of point P are 

⎛ 3 1 ⎞ 

⎜ ⎟ 

− , . 

⎝ 2 4 ⎠ 

2 

(iii) Find the value of ∠ POQ , in radians, correct to one decimal place. 3 

12


Marks 

(a) Find the trigonometric equation for the graph below: 2 

y 

NOT TO SCALE 

2 

1 

O 

π 

2 π 

3 π 

4 π x 

-1 

-2 


13


Marks 

(b) 

y 

6 

NOT TO SCALE 

( x y) 

S , 

− 6 

T ( 0,2) 

O 

α 

R ( 4,0) 

6 

x 

− 6 

2 2 

The diagram above shows the circle + y = 36 S x, y lies 

on the circle in the first quadrant. O is the origin, R ( 4,0) 

lies on the x-axis 

and T ( 0,2) 

lies on the y-axis. The size of ∠ ROS is α radians, 

π 

where 0 < α < . 

2 

x . The point ( ) 

(i) Show that the area of triangle SOR is 12 sin α . 1 

(ii) 

Hence show that the area, A, of the quadrilateral ORST is given by: 

( 2tan 1) 

A = 6 cosα 

α + 

3 

(iii) Find the value of tan α for which the area A is a maximum. 3 

(iv) 

Hence, show that for this maximum area, the coordinates of 

⎛ 6 12 ⎞ 

point S are ⎜ 5, 5 ⎟ 

⎝ 5 5 ⎠ 

3 

End of Paper 

14

BLANK PAGE 

15

EXAMINERS 

Liviu Spiridon (convenor) 

Magdi Farag 

Kimon Kousparis 

Timothy Dimech 

LaSalle Catholic College, Bankstown 

LaSalle Catholic College, Bankstown 

Trinity Catholic College, Auburn / Regents Park 

Concord High School, Concord 

16

CATHOLIC SECONDARY SCHOOLS ASSOCIATION 

2006 TRIAL HIGHER SCHOOL CERTIFICATE EXAMINATION 

MATHEMATICS – SUGGESTED SOLUTIONS 

These marking guidelines show the criteria to be applied to responses along with the marks to 

be awarded in line with the quality of responses. These guidelines are suggested and not 

prescriptive. This is not intended to be an exhaustive list but rather an indication of the 

considerations that students could include in their responses. 

Question 1 (12 marks) 

(a) (2 marks) 

Outcomes Assessed: P3, P4 

Targeted Performance Bands: 2-3 

Criteria 

• Gives correct answer. 

• Correctly rounds to THREE decimal places. 

Mark 

1 

1 

Sample Answer 

3 

− 38.67 × 7.2 

2 

( 11.7) − ( 1.83) 

2 

= 1.277508818 

= 1.278 (3 decimal places) 

(b) (2 marks) 



Criteria 

• Correctly write the conjugate and expands to get the denominator OR the 

numerator correct 

• Gives the correct answer 

Mark 

1 

1 

Sample Answer 

1− 

2 − 

2 1− 

= 

8 2 − 

2 − 2 

= 

− 2 

= 

− 4 

1 

= 

2 

2 2 + 

× 

8 2 + 

2 + 

4 − 8 

8 

8 

8 − 

16 

DISCLAIMER 

The information contained in this document is intended for the professional assistance of teaching staff. It does not constitute advice to students. Further it is not the intention of 

CSSA to provide specific marking outcomes for all possible Trial HSC answers. Rather the purpose is to provide teachers with information so that they can better explore, 

understand and apply HSC marking requirements, as established by the NSW Board of Studies. 

No guarantee or warranty is made or implied with respect to the application or use of CSSA Marking Guidelines in relation to any specific trial exam question or answer. The CSSA 

assumes no liability or responsibility for the accuracy, completeness or usefulness of any Marking Guidelines provided for the Trial HSC papers. 

2602-2

(c) (2 marks) 



Criteria 

• Gives ONE correct answer in radians OR TWO correct answers in degrees. 

• Gives TWO correct answers in radians. 

Mark 

1 

1 

Sample Answer 

cosθ 

1 

= − 

2 

π 

Basic angle is (First Quadrant). 

3 

2π 

4π 

∴ θ = , 

0 ≤ θ ≤ 2π 

3 3 

(d) (2 marks) 



Criteria 

• Factorises in pairs, e.g. x ( 4 y + b) 

+ 2a(4 

y + b) 

. 

• Completes the factorisation into TWO brackets. 

Mark 

1 

1 

Sample Answer 

4 xy + xb+ 

8ay 

+ 2ab 

= x ( 4 y + b) 

+ 2a(4 

y + b) 

= ( x + 2a)(4 

y + b) 

(e) (2 marks) 



Criteria 

• Correctly writes down the discriminant and equates to zero. 

• Solves the equation to give correct values of k. 

Mark 

1 

1 

Sample Answer 

4x 2 + kx + 9 = 0 equal roots when ∆ = 0 

∆ = k 

2 − 144 

0 = k 

2 −144 

k = ±12 

2 

DISCLAIMER 





assumes no liability or responsibility for the accuracy, completeness or usefulness of any Marking Guidelines provided for the Trial HSC papers.

(f) (2 marks) 



• Gives ONE correct answer. 

• Gives the second correct answer. 

Criteria 

Mark 

1 

1 

Sample Answer 

x − 2 = 3 

x − 2 = 3 or − ( x − 2 ) = 3 

∴ x = 5 or x = −1 


(a) (2 marks) 



Criteria 

• Correctly draws the circle using a dotted line. 

• Correctly shades the inside of the circle. 

Mark 

1 

1 

Sample Answer 

y 

2 

− 2 

2 

x 

− 2 

3 

DISCLAIMER 






Question 2 

(b) (2 marks) 

Outcomes Assessed: P3, P4, P5 


Criteria 

• Gives correct answer ONE partial answer (e.g. f ( 3 ) = 4 or ( 6 ) = 27 

• Gives correct answer 

f ) 

Mark 

1 

1 

Sample Answer 

f (x) = 

x + 1, 

x 

2 − 

9, 

x ≤ 3 

x > 3 

2 

f ( 3) = (3) + 1 = 4 , f ( 6) = 6 − 9 = 27 

∴ f 

( 3) − f ( 6) = 4 − 27 = −23 

(c) (i) (2 marks) 



Criteria 

• Correctly finds the gradient of line L 

3. 

• Correctly finds the equation of line L 

3. 

Mark 

1 

1 

Sample Answer 

m 

0 + 4 4 

= = 

0 + 2 2 

L 

= 

3 

y − 0 = 2( x − 0) 

∴ 2 x − y = 0 (as required) 

2 

(c) (ii) (1 mark) 



Criteria 

Mark 

• Correctly shows that the x-coordinate of point B is 3. 1 

Sample Answer 

B lies on the line y = 6 

y = 6 ∴ 6 = 2x 

∴ x = 3 

∴ B(3,6) 

4 

DISCLAIMER 






(c) (iii) (3 marks) 



Criteria 

• Correctly finds the gradient of L2 

which gives 

• Correctly finds the equation of L 

2 

. 

• Correctly finds the value of the x-ordinate of A. 

o 

∠AOB = 90 . 

Mark 

1 

1 

1 

Sample Answer 

m 

L2 

−1 

−1 

1 

= = = − 

m 2 2 

L3 

1 

∴ y − 0 = − ( x − 0) 

2 

1 

y = − x 

2 

1 

For y = 6 ∴ 6 = − x ∴ x = −12 

2 

A ( −12,6) 

(c) (iv) (2 marks) 



Criteria 

• Writes down the correct distance between points A and B. 

• Calculates the correct area of∆ AOB . 

Mark 

1 

1 

Sample Answer 

Distance between points A and B is 15 units. 

1 

Area of ∆AOB = × 15 × 6 

2 

= 45units 2 

5 

DISCLAIMER 







(a) (i) (1 mark) 


Targeted Performance Band: 2-4 

Criteria 

Mark 

• Finds the correct vertex 1 

Sample Answer 

Vertex ≡ (2,0) 

(a) (ii) (1 mark) 



Criteria 

Mark 

• Finds the correct focus 1 

Sample Answer 

Focus ≡ (2,4) 

(b) (i) (2 marks) 

Outcomes Assessed: P7, H5 


Criteria 

• Correctly uses the product rule of differentiation but has ONE mistake in 

calculation 

• Correctly finds the answer 

Mark 

1 

1 

Sample Answer 

d 

3 xlog 

e 

x = 3 + 3log e 

dx 

( ) x 

(b) (ii) (2 marks) 



Criteria 

• Correctly uses the chain rule of differentiation but has ONE mistake in 

calculation 


Mark 

1 

1 

Sample Answer 

d 

dx 

2 

1 

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

x 

6 

DISCLAIMER 






(c) (i) (2 marks) 

Outcomes Assessed: H5 


Criteria 

• Gives an answer of sin 2006x + c 


Mark 

1 

1 

Sample Answer 

∫ 

1 

cos 2006x dx = sin 2006x + c 

2006 

(c) (ii) (2 marks) 

Outcomes Assessed: H3, H5 


Criteria 

1 

• Finds the primitive e 2x but has an error in calculating the integral 

2 

• Correctly applies the Newton-Leibnitz formula to obtain the correct answer 

in exact form 

Mark 

1 

1 

Sample Answer 

∫ 1 0 

2 

e x dx = 

⎡1 

⎢ 

⎣2 

1 

2x 

⎤ 

e = 

⎥ 

⎦ 

0 

1 2 0 2 

2 

1 

e or ( e 2 

−1) 

1 

− e 

2 

1 

= e 

2 

1 

− 

2 

2 

(d) (2 marks) 



Criteria 

• Correctly finds the gradient of the normal 

• Correctly substitutes the values for x and y into the point/gradient formula to 

find the equation of the normal 

Mark 

1 

1 

Sample Answer 

dy 

y = x 

3 − 5x 

∴ = 3x 

2 − 5 

dx 

? At 

x 

1 

1 , m T 

= −2∴ 

m = 

2 

= 

N 

1 

∴ equation of normal is given by y − −4 = ( x −1) 

2 

∴ x − 2 y − 9 = 0 (or 

1 9 

y = x − ) 

2 2 

7 

DISCLAIMER 







(a) (i) (3 marks) 



Criteria 

• Finds the stationary points 

• Finds the nature of ONE stationary point 

• Finds the nature of the other stationary point 

Mark 

1 

1 

1 

Sample Answer 

1 3 2 

2 

f ( x) 

= − x − x + 3x 

+ 1 ∴ f ′( 

x) 

= −x 

−2x 

+ 3 

3 

f ′( x) 

= 0 ∴ 1− 

x x + 3 = ∴ x = 1 or x = −3 

For stationary points ( )( ) 0 

⎛ 2 ⎞ 

∴ the stationary points are ⎜1 , 2 ⎟ & ( −3 , − 8) 

⎝ 3 ⎠ 

Also for the nature of the stationary points, f ′ ( x) 

= −2x 

− 2 

At x = 1 , f ′ (1) = −4 

< 0 

⎛ 2 ⎞ 

∴ ⎜1 , 2 ⎟ is a MAXIMUM turning point 

⎝ 3 ⎠ 

At x = −3, f ′′( −3) 

= 4 > 0 ∴( 

−3 , − 8) 

is a MINIMUM turning point 

(a) (ii) (2 marks) 



Criteria 

• Correctly solves the equation f ′′( x) 

= 0 

• Analyse the sign of the second derivative and gives correct answer 

Mark 

1 

1 

Sample Answer 

2 

f ′ ( x) 

= −2x 

− 2 ∴ − 2x 

− 2 = 0 → x = −1 

( y = −2 

) 

3 

x -1 - -1 -1 + 

f ′′(x) 

+ 0 - 

∴ a change in the sign of the second derivative has occurred 

∴ 

⎛ 2 ⎞ 

⎜−1 , − 2 ⎟ is a point of inflection 

⎝ 3 ⎠ 

8 

DISCLAIMER 






(a) (iii) (2 marks) 

Outcomes Assessed: P6, H6, H7, H9 


• Draws the correct cubic curve 

• Plots all important points 

Criteria 

Mark 

1 

1 

Sample Answer 

1 3 2 

f ( x) 

= − x − x + 3x 

+ 1 

3 

•(− 

6,19) 

y 

2 

( 1, 2 ) 

3 

• 

O 

2 

( − 1, − 2 ) • 

3 

x 

•( − 3, − 8) 

•( 3, − 8) 

(a) (iv) (1 mark) 



Criteria 

Mark 

• Gives the correct answer 1 

Sample Answer 

y = 19 

9 

DISCLAIMER 






(b) (i) (0 marks) 




Criteria 

• Shows that ∠ PAS = ∠QAR 

(vertically opposite ∠ ' s ) or equivalent 

• Correctly completes proof using (AAS) 

Mark 

1 

1 

Sample Answer 

∠ PSA = ∠QRA (given) 

∠ PAS = ∠QAR (vertically opposite ∠ ' s ) 

PS = QR 

(given) 

∴ ∆PSA ≡ ∆QRA (AAS) 

(b) (iii) (2 marks) 



Criteria 

• Realises that ∆ PAQ is isosceles & base ∠' 

s are equal 


Mark 

1 

1 

Sample Answer 

PA = QA 

(corresponding sides are equal in congruent ?’s) 

∴ ∆PAQ is isosceles ∴ ∠PQA = ∠QPA 

= x 

2x + 120 = 180 o (∠ sum of a ? is 180 o ) 

x = 30 o 


(a) (i) (1 mark) 



Criteria 

Mark 

• Writes the correct sum of an infinite geometric series 1 

Sample Answer 

• 

7 5 5 5 

0 .75 = + + + ... 

10 100 1000 10000 

10 

DISCLAIMER 









Criteria 

• Writes the limiting sum formula with correct a and r 


Mark 

1 

1 

Sample Answer 

a 

S∞ 

= 

1− 

r 

= 

7 

10 

7 

= 

10 

34 

= 

45 

+ 

+ 

5 

100 

1− 

1 

10 

1 

18 

(b) (2 marks) 



Criteria 

• Realises that cos 2 θ is a substitute or equivalent 

• Simplifies to give the correct answer 

Mark 

1 

1 

Sample Answer 

2 

1− 

sin x 

cot x 

cos 2 θ 

= = 

cosθ 

sin θ 

sin θ 

cos 2 θ × 

cosθ 

= sin θ cosθ 

11 

DISCLAIMER 






(c) (i) (1 mark) 



Criteria 

Mark 


Sample Answer 

l 

l = rθ 

∴ θ = 

r 

1 π 

θ = = 

6 6 

π 




Criteria 

• Writes the correct formula with necessary substitutions 

• Gives the correct answer in exact form 

Mark 

1 

1 

Sample Answer 

1 A = r 

2 

θ 

2 

2 

1 ⎛ 6 ⎞ π 

A = × ⎜ ⎟ × 

2 ⎝π 

⎠ 6 

3 

∴ Area of sector MON ⇒ A = cm 2 

π 

12 

DISCLAIMER 






(d) (i) (2 marks) 



• Correctly draws a tree diagram 


Criteria 

Mark 

1 

1 

Sample Answer 

3 

5 

S 

3 

5 

2 

5 

S 

F 

3 

5 

2 

5 

3 

5 

2 

5 

S 

F 

S 

F 

3 

5 

3 

5 

3 

5 

2 

5 

2 

5 

2 

5 

2 

5 

F 

S 

F 

S 

F 

S 

F 

∴ P ( S) 

= 

3 

5 

3 

× × 

5 

3 

5 

⎛ 3⎞ 

= ⎜ ⎟ 

⎝ 5⎠ 

3 

= 

27 

125 

(d) (ii) (2 marks) 



Criteria 

• Uses the complementary events method (or otherwise) 

• Gives the correct answer with required working 

Mark 

1 

1 

Sample Answer 

⎡2 

2 2 ⎤ ⎛ 2 ⎞ 

P ( S ) = 1− 

⎢ 

× × = 1− 

⎜ ⎟ 

5 5 5 ⎥ 

⎣ ⎦ ⎝ 5 ⎠ 

3 

117 

= 

125 

13 

DISCLAIMER 







(a) (i) (1 mark) 

Outcomes Assessed: P2 


Criteria 

Mark 

• Correctly verifies Pythagoras’ Theorem 1 

Sample Answer 

2 2 2 

Triangle with sides 12 cm, 16 cm and 20 cm is right angled at A ( 12 + 16 = 20 , converse 

of Pythagoras’ Theorem) 




Criteria 

• Correctly gives answer for α in either degrees or radians 

• Correctly gives answer for β in either degrees or radians 

Marks 

1 

1 

Sample Answer 

12 

sin α = ∴ 

20 

' 

α = 36 ° 52 or 

α 

c 

= 0.644 

° ° ' ° ' 

β = 90 − 36 52 = 53 8 or 

c 

β = 0.927 (using complementary angles in ∆ AO 1 

O2 

) 

(a) (iii) (3 marks) 



Criteria 

• Correctly substitutes in the segment formula for the circle O 1 

• Correctly substitutes in the segment formula for the circle O 2 

• Correctly gives answer (to 2 d.p.) 

Marks 

1 

1 

1 

Sample Answer 

Area enclosed between the two circles is the sum of the two segments, with angles 2 α and 

2 β subtended respectively at the centre. 

1 1 2 

2 

2 

2 

Area = × 16 ( 2 × 0.644 − sin 2 × 0.644) + × 12 ( 2 × 0.927 − sin 2 × 0.927) 

∴ Area = 106.2667952 cm 2 ∴ Area = 106.27 cm 2 (2 d.p.) 

14 

DISCLAIMER 






(b) (i) (2 marks) 



Criteria 

• 

2 

Correctly works out x as a subject ( x 

= log y ) 

• Correctly substitutes x 

= log 

e 

y in the volume formula and deduce the 

answer 

e 

Mark 

1 

1 

Sample Answer 

∫ 

5 

V = x 2 y 

π dy ; Since 

1 

x 2 

y = e ∴ 

log 

2 

e 

y = x ∴ V 

5 

= ∫1 log y 

y 

π 

e 

dy (as required) 

(b) (ii) (1 mark) 



Criteria 

Mark 

• Correctly completes the required value ( log 

e 

y =1. 386 ) 1 

Sample Answer 

y 1 2 3 4 5 

log y 0 0.693 1.099 1.386 1.609 

e 




Criteria 

• Substitutes the correct values in the correct Simpson’s formula 

• Correctly calculates the answer in decimal form (e.g. 4.041) 


Mark 

1 

1 

1 

Sample Answer 

h 

∫ 0 1 3 

2 4 

+ 

a 

3 

b 

Using Simpson’s Formula: y dx ≈ [ y + 4( y + y + ...) + 2( y + y + ...) 

] 

1 

V 

y 

= π × [ 0 + 4( 0.693 + 1.386) 

+ 2 × 1.099 + 1.609] 

∴ V 

y 

= 4.041π 

3 

∴ V = 12. 695 (3 d.p.) 

y 

y n 

15 

DISCLAIMER 







(a) (3 marks) 



Criteria 

• Correctly uses the properties of logarithms 

• Correctly uses the definition of logarithms to obtain a quadratic equation 

• Solves the quadratic equation, indicating the correct answer 

Mark 

1 

1 

1 

Sample Answer 

3 

x + log ( x + 7) 3 ∴ log 2 

x ( x + 7) = 3 ∴ ( x + 7 ) = 2 

log 

2 

2 

= 

∴ ( x + 8 )( x −1) = 0 ∴ x = −8 

or x = 1 ∴ solution is = 1 

by substitution. 

2 

x ∴ x + 7x 

− 8 = 0 

x ( > 0) 

x or checking the solution 

(b) (i) (2 marks) 



Criteria 

• Correctly finds x-ordinate of point A 

• Correctly finds x-ordinate of point B 

Mark 

1 

1 

Sample Answer 

At point A: 

x 

2 

2 

= 3 − 2x 

∴ 3 2 = 3 

x ∴ x = ± 1 ∴ x = 1 (first quadrant) 

A 

At point B: 3 − 2x 2 = 0 ∴ 

3 

x = ± ∴ 

2 

3 

x 

B 

= (first quadrant) 

2 




Criteria 

• Correctly uses the sum of integrals to find area under curves 

• Correctly finds the primitives 

• Correctly finds the area using Leibnitz – Newton Formula 

Mark 

1 

1 

1 

Sample Answer 

A AOB 

3 

1 

2 

1 

2 

2 1 3 ⎡ 2 

= ∫ x dx + ∫ ( − 2x 

) dx = [ x ] 0 + 3x 

− x 

0 

1 

3⎤ 

3 ⎥ ⎦ 

3 

1 ⎡⎛ 

3 2 3 ⎞ 2 ⎤ 3 

⎢ 

⎞ 

A ⎜ 

⎟ ⎥ 

AOB 

⎜ ⎟ 

square units 

3 ⎢⎜ 

2 3 2 ⎟ ⎝ 3 ⎠⎥ 

2 

⎣⎝ 

⎠ ⎦ 

⎛ 

∴ = ( 1 − 0) + 3 − − 3 − = 2 − 2 

16 

DISCLAIMER 





assumes no liability or responsibility for the accuracy, completeness or usefulness of any Marking Guidelines provided for the Trial HSC papers. 

3 

⎢ 

⎣ 

3 

1 

3 

2

(c) (i) (1 mark) 


Targeted Performance Bands: 2 

Criteria 

Mark 

• Correctly shows the number of revolutions the track will rotate 1 

Sample Answer 

Record is played at 5 revolutions per second for 25 minutes therefore will rotate 

5 × 60× 

25 = 7500 revolutions 




Criteria 

• Correctly realises that circles’ radii are in a arithmetic sequence 

• Correctly uses the formula for arithmetic series 

• Finds the correct answer 

Mark 

1 

1 

1 

Therefore there will be 7500 concentric circles whose radii increases in an arithmetic 

sequence from 2 cm to 6 cm. The length of the track record can be found using the sum of the 

arithmetic sequence with 7500 terms: 

S 

7500 

= π 

2 

( 2π 

× 2 + 2 × 6) 188495. 5595 

7500 

= 

cm =1.9 km (1 d.p.) 

17 

DISCLAIMER 







(a) (2 marks) 


Targeted Performance Bands: 2 – 4 

• Correctly uses index law 

• Find the correct answer 

Criteria 

Mark 

1 

1 

Sample Answer 

m 

If = 3 

A ∴ ( m ) = 3 4 = 81 

4 

4 

A ∴ A 

m − 5 = 81 − 5 = 76 

(b) (i) (2 marks) 

Outcomes Assessed: H3, H4, H5 


Criteria 

• Correctly substitutes in the formula and writes –10k as a subject 


Mark 

1 

1 

Sample Answer 

U = 

100 e −kt 

− 

∴ 2 = 100e 10k ∴ 

50 

1 − 

= e 10k 

1 

∴ ln 

50 

− 

= ln e 10k 

∴k = 

− ln 50 

−10 

∴k = 0.3912 


Outcomes Assessed: H3, H4, H5 


• Correctly differentiates U 


Criteria 

Mark 

1 

1 

Sample Answer 

dU 

dt 

−kt 

( e ) 

= −k 

100 

dU 

For t = 12 and k = 0. 3912 : = −0.36 mg/minute 

dt 

18 

DISCLAIMER 






(c) (i) (1 mark) 



Criteria 

Mark 

• Correctly gives answer for A 1 

1 

Sample Answer 

⎛ 0.75 ⎞ 

A 1 = P ⎜1 

+ ⎟ − 1050 ∴A 

1 

= P( 1.0075) −1050 

⎝ 100 ⎠ 




Criteria 

• Correctly gives answer for A 

2 

• Write the correct sum of the geometric series 


Mark 

1 

1 

1 

Sample Answer 

A 

[ (1.0075) −1050 ](1.0075) 

1050 

2 

= P 

− 

= P (1.0075) 

= P (1.0075) 

2 

2 

−1050(10075) 

−1050 

−1050(1 

+ 1.0075) 

3 

2 

A 

3 

= P(1.0075) 

−1050(1 

+ 1.0075+ 

1.0075 ) 

. 

. 

. 

A 

n 

= P(1.0075) 

n 

− 1050(1 + 1.0075 + ... + 1.0075 

n−1 

) 

n 

= P ( 1.0075) 

− 1050 

n 

= P 1.0075) 

− 

⎡ 

⎢ 

⎣ 

n 

( −1) ⎤ 

⎥⎦ 

11.0075 

0.0075 

n 

( 140 000 [ 1.0075 

−1] 

n 

= P 1.0075) 

− 

n 

( 140 000 ( 1 .0075) + 140000 

19 

DISCLAIMER 






(c) (iii) (2 marks) 



Criteria 

• Correctly substitutes n=240 and makes A 

n 

= 0 


Mark 

1 

1 

Sample Answer 

When n = 240 

∴0 

= P 

∴ P 

240 

240 

( 1.0075) − 140000( 1.0075) + 140 000 

240 

240 

( 1.0075) = 140 000( 1.0075) −140 

000 

240 

( ) 

( 1.0075) 240 

140 000 1.0075 −140 000 

∴ P = 

∴P = $116 702 (to the nearest dollar) 

20 

DISCLAIMER 







(a) (i) (2 marks) 



Criteria 

• Correctly differentiate to find the velocity v 

A 

• Correctly differentiate to find the velocityv 

B 

Mark 

1 

1 

Sample Answer 

x A 

= 12 t + 5 v = 12 

∴ A 

x B 

2 3 

= 6t 

− t ∴ v B 

= 12t 

− 3t 

2 

(a) (ii) (1 mark) 



Criteria 

Mark 


Sample Answer 

At t = 1 second v = 12m/s, v = 12 −3 

= 9 m/s 

∴ A 

∴ particle A is faster 

B 

(a) (iii) (1 mark) 



Criteria 

Mark 


Sample Answer 

v B 

( 4 − t) = 0∴t 

= 0 or = 4 

2 

= 0 ∴12t 

− 3t 

= 0∴ 

3t 

t 

seconds 

∴Particle comes at rest at t = 4 seconds 

21 

DISCLAIMER 






(a) (iv) (2 marks) 



Criteria 

• Correctly substitutes the value of t = 4 into the x 

B 

equation 


Mark 

1 

1 

Sample Answer 

For maximum displacement, v = 0 ∴t ( 12 – 3t ) = 0 ∴t = 4 seconds 

2 3 

∴ x = 6× 

4 − 4 = 32 metres 

B 

B 

(b) (i) (1 mark) 



Criteria 

Mark 


Sample Answer 

° 

The gradient of the tangent at point P is m = tan 60 = 3 

P 




Criteria 

• Correctly finds the gradient of the tangent to the curve and equates to the 

gradient obtained from (i) 

• Correctly finds the coordinates of point P 

Mark 

1 

1 

Sample Answer 

y = 1 – x 2 dy 

∴ = −2x 

and since the gradient of the tangent is 3 

dx 

∴ 3 = - 2x ∴ x − 3 

= 

2 

∴ y = 1 ⎛ 3 1 

∴ coordinates of point P are 

4 ⎟ ⎞ 

⎜ 

− , 

⎝ 2 4 ⎠ 

22 

DISCLAIMER 









Criteria 

• Correctly find the length of PQ 

2 

• Correctly find the length of PO (or 


2 

QO ) 

Mark 

1 

1 

1 

Sample Answer 

Coordinates of point Q are 

∴ PQ = 3 

⎛ 

⎜ 

⎝ 

3 1 ⎞ 

, ⎟ 

2 4 

⎠ 

(By symmetry) 

PO 

2 

3 1 13 

= + = 

4 16 16 

Using the Cosine Rule in 

∆ POQ 

: cos 

∠POQ 

( 3) 

13 13 

+ − 

2 

= 

16 16 

13 13 

2× 

× 

16 16 

∠POQ = 2.6 radians 

23 

DISCLAIMER 







(a) (2 marks) 



Criteria 

• Writes the correct value of the amplitude A 

• Find the correct value of n 

Mark 

1 

1 

Sample Answer 

Since y= A sin nx ∴ A = 2 

Period (T) = 

4π 2π 

(from graph) and since Period (T) = 

n 

3 

2π 

∴ n = 

T 

3 

3 

∴ n = ∴ the equation is y = 2sin x 

2 

2 

(b) (i) (1 mark) 



Criteria 

Mark 

• Correctly finds the area of ? SOR 1 

Sample Answer 

In ? SOR, OS = 6 (radius of the circle) and OR = 4 

∴ area of ? SOR = ½ 

× 6 × 4× 

sin α = 12sin 

α (as required) 




Criteria 

• Correctly finds the area of ? SOT 

• Correctly finds the area of the quadrilateral ORST 

• Correctly factorise to find the correct answer 

Mark 

1 

1 

1 

Sample Answer 

⎛ π ⎞ 

∴area of ? SOT = ½ × 6 × 2 × sin ⎜ − α ⎟ = 6 cosα 

⎝ 2 ⎠ 

∴area of the quadrilateral ORST is: 

A = 12 sina + 6 cosa 

A = 6cosa(2tana + 1) 

24 

DISCLAIMER 






(b) (iii) (3marks) 



Criteria 

• Correctly finds the derivative of A 

• Correctly finds the value of tan a 

• Correctly shows that the area is a maximum when tan a = 2 

Mark 

1 

1 

1 

Sample Answer 

dA 

A = 12 sina + 6 cosa ∴ 

dα 

=12 cosα 

− 6sin α 

dA 

For maximum area = 0 ∴0 = 12cosα 

− 6sin 

α 

dα 

∴ 6sina = 12cosa ∴tan α = 2 

To prove the area is maximum whentan α = 2 : 

2 

d A 

= −12 sin α − 6 cosα 

2 

dα 

and since tan α = 2 ∴ from the triangle PQR 

2 

∴sin α = and 

5 

cos α = 

2 

d A 2 1 30 

∴ = −12× 

− 6× 

= − < 0 

2 

dα 5 5 5 

1 

5 

R 

5 

α 

1 

P 

2 

Q 

∴Maximum Area occurs at tan α = 2 

Alternative method: To prove the area is maximum whentan α = 2: 

Since tan α = 2 ∴α = 1.1 radians 

α 

dA 

dα 

c 

1 

c 

1 .1 

c 

1 .2 

1 .43 > 0 0 − 1 .24 < 0 

Maximum 

25 

DISCLAIMER 






(b) (iv) (3 marks) 



Criteria 

• Correctly finds the gradient of the line OS and equates to 2 to get the 

equation y = 2x 

• Correctly substitutes into the equation of the circle to get the x-ordinate 

of point S 

• Correctly finds the y-ordinate of point S 

Mark 

1 

1 

1 

Sample Answer 

y 

The gradient of the line OS = tan α = and since tan α = 2 

x 

y 

∴ = 2 

x 

∴y = 2x [ 1 ] 

2 2 

Substitute into x + y = 36 ∴ x 

2 + 4x 

2 = 36 

∴ 5x 2 36 6 6 

= 36 ∴ x 2 = ∴ x = = 5 

5 

5 5 

12 12 

∴Substitute into [ 1 ] ∴ y = = 5 ∴point S is 

⎛ 6 12 ⎞ 

⎜ 5, 5 ⎟ . 

5 5 

⎝ 5 5 ⎠ 

26 

DISCLAIMER 






BLANK PAGE 

27 

DISCLAIMER 






BLANK PAGE 

28 

DISCLAIMER 






2602-3

CSsA 2006

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?