Cssa 2006 Math 2u + Solutions

45
1 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. 2602- 1

Transcript of Cssa 2006 Math 2u + Solutions

Page 1: Cssa 2006 Math 2u + Solutions

1

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.

2602-1

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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

22 83.17.112.767.38

−×−

correct to 3 decimal places. 2

(b) By rationalising the denominator, simplify

8221

−−

. 2

(c) Solve 21

cos −=θ , for πθ 20 ≤≤ . 2

(d) Completely factorise abayxbxy 284 +++ . 2

(e) For what values of k does the quadratic equation 094 2 =++ kxx have equal roots?

2

(f) Solve the equation 32 =−x . 2

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Question 2 (12 marks) Use a SEPARATE writing booklet. Marks

(a) Sketch the region represented by 422 <+ yx . 2

(b) The function )(xf is given by =)(xf

3for93for1

2 >−≤+

xxxx

Find )3(f - )6(f .

2

(c)

On the diagram above line 1L is parallel with the x-axis and crosses the y-axis at 6. Lines 2L and 3L each pass through the origin, O, and intersect with the line 1L at the points )6,(aA and )6,(bB respectively. The point )4,2( −−C lies on the line 3L .

(i) Show that the equation of the line 3L is 02 =− yx .

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

x

y

6 )6,(bB

O

•−− )4,2(C

1L

2L3L

NOT TO SCALE

)6,(aA••

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Question 3 (12 marks) Use a SEPARATE writing booklet. Marks

(a) For the parabola yx 16)2( 2 =− , find the coordinates of the:

(i) Vertex.

1

(ii) Focus.

1

(b) Differentiate with respect to x the following expressions:

(i) xx elog3 .

2

(ii) x2sin .

2

(c) Find:

(i) ∫ dxx2006cos .

2

(ii) ∫1

0

2 dxe x . (Leave your answer in exact form).

2

(d) Show that the equation of the normal to the curve xxy 53 −= at the point )4,1( − is given by 092 =−− yx .

2

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Question 4 (12 marks) Use a SEPARATE writing booklet. Marks

(a) Consider the curve 1331

)( 23 ++−−= xxxxf .

(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, 36 ≤≤− x .

2

(iv) What is the maximum value of )(xf in the given domain?

1

(b)

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

o80=∠=∠ QRAPSA , o120=∠PAQ and .xPQA =∠

(i) Copy the diagram into your writing booklet.

(ii) Prove that QRAPSA ∆∆ tocongruentis .

2

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

2

P Q

A

S R

x

80o 80o

120o

NOT TO SCALE

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Question 5 (12 marks) Use a SEPARATE writing booklet. Marks

(a)

(i) Write •57.0 as the sum of an infinite geometric series.

1

(ii) Hence, express •57.0 as a fraction.

2

(b) Simplify

xx

cotsin1 2−

.

2

(c) The circle below has centre O, radius π6

cm and arc length MN = 1 cm.

(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 53

. 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

N

O

M NOT TO SCALE

θ

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Question 6 (12 marks) Use a SEPARATE writing booklet. Marks

(a) A circle with centre at O1 and radius 16 cm intersects with another circle with centre at O2 and radius 12 cm. Their points of intersection are A and B and the distance between their centres, 21OO , is 20 cm.

α=∠ 21OAO and β=∠ 12OAO .

(i) Show that 21 AOO∆ 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

16 cm 12 cm

α O1 O2

A

B

NOT TO SCALE

20 cm β

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Question 6 (continued) Marks

(b) The shaded region bounded by the graph

2xey = , the line 5=y 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 ∫=5

1log dyyV ey π .

2

(ii)

Copy and complete the following table into your writing booklet. Give all answers correct to three decimal places.

y 1 2 3 4 5

yelog 0 0.693 1.099 1.609

1

(iii) Use Simpson’s Rule with five function values to approximate the volume of the solid of revolution yV , correct to three decimal places.

3

End of Question 6

x

NOT TO SCALE y

O

5

1

2xey =

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Question 7 (12 marks) Use a SEPARATE writing booklet. Marks

(a) Solve the following equation: ( ) 37loglog 22 =++ xx , for x > 0.

3

(b) The shaded region OAB is bounded by the parabolas 2xy = , 223 xy −= and the x-axis. Point A is the intersection of the two parabolas and point B is the x-intercept of the parabola 223 xy −= .

(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

shaded region OAB is given by 223

2 − square units.

3

Question 7 continues on page 10

x

y

O

A

B

2xy =

223 xy −=

3

NOT TO SCALE

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Question 7 (continued) Marks

(c) 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

NOT TO SCALE

cm2=r

cm6=R

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Question 8 (12 marks) Use a SEPARATE writing booklet Marks

(a) If ,3=mA find the value of 54 −mA .

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 kte − , where k is a constant.

(i) Calculate the value of k, correct to 4 decimal places, given that 2 mg of the tablet remain after 10 minutes.

2

(ii) 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

(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 ins talments of $1 050.

(i) Write an expression for the amount, 1A , Mr. Egan owes immediately at the end of the first month.

1

(ii) Show that at the end of n months the amount owing, nA , is given by:

000140)0075.1(000140)0075.1( +−= nnn PA

3

(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.

2

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Question 9 (12 marks) Use a SEPARATE writing booklet. Marks

(a) Two particles, A and B, move along a straight line so that their displacements, Ax and Bx , in metres, from the origin at time t seconds are given by the following equations respectively:

512 += tx A 326 ttxB −=

(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) In the diagram above P and Q are two points on the parabola y = 1 – x2. 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

41

,23

.

2

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

A

B C

y

x

1

O

P Q

NOT TO SCALE

21 xy −=

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Question 10 (12 marks) Use a SEPARATE writing booklet. Marks

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

Question 10 continues on page 14

π4 π2 π π3 x

y

O

1

2

-1

-2

NOT TO SCALE

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Question 10 (continued) Marks

(b)

The diagram above shows the circle 3622 =+ yx . The point ( )yxS , lies on the circle in the first quadrant. O is the origin, ( )0,4R lies on the x-axis and ( )2,0T lies on the y-axis. The size of ∠ ROS is α radians,

where 2

α << .

(i) Show that the area of triangle SOR is αsin12 .

1

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

( )1tan2cos6 += ααA

3

(iii) Find the value of αtan for which the area A is a maximum.

3

(iv) Hence, show that for this maximum area, the coordinates of

point S are

5

512

,556

3

End of Paper

( )yxS ,

α ( )0,4R 6 O

( )2,0T

6−

6

x

NOT TO SCALE

6−

y

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BLANK PAGE

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EXAMINERS

Liviu Spiridon (convenor) LaSalle Catholic College, Bankstown

Magdi Farag LaSalle Catholic College, Bankstown

Kimon Kousparis Trinity Catholic College, Auburn / Regents Park

Timothy Dimech Concord High School, Concord

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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 Mark • Gives correct answer. • Correctly rounds to THREE decimal places.

1 1

Sample Answer

( ) ( )277508818.1

83.17.11

2.767.383

22=

×−

278.1= (3 decimal places)

(b) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2-3

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

numerator correct • Gives the correct answer

1 1

Sample Answer

8282

8221

8221

++

×−−

=−−

84

168222−

−+−=

42

−−

=

21

=

2602-2

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.

Page 18: Cssa 2006 Math 2u + Solutions

2 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.

(c) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2-3

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

1 1

Sample Answer

21

cos −=θ

Basic angle is 3π

(First Quadrant).

34

,3

2 ππθ =∴ πθ 20 ≤≤

(d) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2-3

Criteria Mark • Factorises in pairs, e.g. )4(2)4( byabyx +++ . • Completes the factorisation into TWO brackets.

1 1

Sample Answer

abayxbxy 284 +++ )4(2)4( byabyx +++= )4)(2( byax ++=

(e) (2 marks) Outcomes Assessed: P2, P4 Targeted Performance Bands: 2-3

Criteria Mark • Correctly writes down the discriminant and equates to zero. • Solves the equation to give correct values of k.

1 1

Sample Answer

094 2 =++ kxx equal roots when 0=∆ 1442 −=∆ k 1440 2 −= k

12±=k

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3 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.

(f) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2-3

Criteria Mark • Gives ONE correct answer. • Gives the second correct answer.

1 1

Sample Answer

32 =−x

32 =−x or ( ) 32 =−− x ∴ 5=x or 1−=x Question 2 (12 marks) (a) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2-3

Criteria Mark • Correctly draws the circle using a dotted line. • Correctly shades the inside of the circle.

1 1

Sample Answer

x

y

2− 2

2

2−

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4 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.

Question 2 (b) (2 marks) Outcomes Assessed: P3, P4, P5 Targeted Performance Bands: 2-4

Criteria Mark • Gives correct answer ONE partial answer (e.g. ( ) 43 =f or ( ) 276 =f ) • Gives correct answer

1 1

Sample Answer

=)(xf ,9

,12 −+

xx

33

>≤

xx

41)3()3( =+=f , ( ) 27966 2 =−=f

( ) ( ) 2327463 −=−=−∴ ff

(c) (i) (2 marks) Outcomes Assessed: P4, P5 Targeted Performance Bands: 2-3

Criteria Mark • Correctly finds the gradient of line 3L .

• Correctly finds the equation of line 3L .

1 1

Sample Answer

224

2040

3==

++

=Lm

)0(20 −=− xy

02 =−∴ yx (as required)

(c) (ii) (1 mark) Outcomes Assessed: P4, P5 Targeted Performance Bands: 2-3

Criteria Mark • Correctly shows that the x-coordinate of point B is 3. 1 Sample Answer B lies on the line 6=y

6=y x26 =∴ ∴ 3=x

)6,3(B∴

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5 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.

(c) (iii) (3 marks) Outcomes Assessed: P2, P4, P5 Targeted Performance Bands: 2-4

Criteria Mark • Correctly finds the gradient of 2L which gives o90=∠AOB . • Correctly finds the equation of 2L . • Correctly finds the value of the x-ordinate of A.

1 1 1

Sample Answer

21

211

3

2−=

−=

−=

LL m

m

)0(21

0 −−=−∴ xy

xy21

−=

For 6=y ∴ x21

6 −= 12−=∴ x

( )6,12−A

(c) (iv) (2 marks) Outcomes Assessed: P2, P4, P5 Targeted Performance Bands: 2-3

Criteria Mark • Writes down the correct distance between points A and B. • Calculates the correct area of AOB∆ .

1 1

Sample Answer Distance between points A and B is 15 units.

Area of 61521

××=∆AOB

45= units 2

Page 22: Cssa 2006 Math 2u + Solutions

6 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.

Question 3 (12 marks) (a) (i) (1 mark) Outcomes Assessed: P3, P4 Targeted Performance Band: 2-4

Criteria Mark • Finds the correct vertex 1

Sample Answer

)0,2(≡Vertex

(a) (ii) (1 mark) Outcomes Assessed: P3, P4 Targeted Performance Band: 2-4

Criteria Mark • Finds the correct focus 1

Sample Answer

)4,2(≡Focus (b) (i) (2 marks) Outcomes Assessed: P7, H5 Targeted Performance Band: 2-3

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

calculation • Correctly finds the answer

1 1

Sample Answer

( ) xxxdxd

ee log33log3 +=

(b) (ii) (2 marks) Outcomes Assessed: P7, H5 Targeted Performance Band: 2-4

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

calculation • Correctly finds the answer

1 1

Sample Answer

( ) ( ) xxxxxdxd

cossin2cos.sin2sin 12 ==

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7 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.

(c) (i) (2 marks) Outcomes Assessed: H5 Targeted Performance Band: 2-4

Criteria Mark • Gives an answer of cx +2006sin • Correctly finds the answer

1 1

Sample Answer

∫ dxx2006cos = cx +2006sin2006

1

(c) (ii) (2 marks) Outcomes Assessed: H3, H5 Targeted Performance Band: 3-4

Criteria Mark

• Finds the primitive xe 2

21

but has an error in calculating the integral

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

1 1

Sample Answer

∫1

0

2 dxe x = 1

0

2

21

xe =

21

21

21

21 202 −=− eee or ( )1

21 2 −e

(d) (2 marks) Outcomes Assessed: P6, H5 Targeted Performance Band: 2-4

Criteria Mark • 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

1 1

Sample Answer

xxy 53 −= 53 2 −=∴ xdxdy

? At 21

2,1 =∴−== NT mmx

∴ equation of normal is given by )1(21

4 −=−− xy

∴ 092 =−− yx (or 29

21

−= xy )

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8 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.

Question 4 (12 marks) (a) (i) (3 marks) Outcomes Assessed: P7, H6 Targeted Performance Band: 3-5

Criteria Mark • Finds the stationary points • Finds the nature of ONE stationary point • Finds the nature of the other stationary point

1 1 1

Sample Answer

1331

)( 23 ++−−= xxxxf 32)( 2 +−−=′∴ xxxf

For stationary points ( )( ) 0310)( =+−∴=′ xxxf 1=∴ x or 3−=x

∴ the stationary points are )8,3(&32

2,1 −−

Also for the nature of the stationary points, 22)( −−=′′ xxf

At 1=x , ∴<−=′′ 04)1(f

32

2,1 is a MAXIMUM turning point

At 3−=x , )8,3(04)3( −−∴>=−′′f is a MINIMUM turning point (a) (ii) (2 marks) Outcomes Assessed: P7, H6 Targeted Performance Band: 2-4

Criteria Mark • Correctly solves the equation 0)( =′′ xf • Analyse the sign of the second derivative and gives correct answer

1 1

Sample Answer

)32

2(102222)( −=−=→=−−∴−−=′′ yxxxxf

x -1- -1 -1+

)(xf ′′ + 0 - ∴ a change in the sign of the second derivative has occurred

−−

32

2,1 is a point of inflection

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9 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.

(a) (iii) (2 marks) Outcomes Assessed: P6, H6, H7, H9 Targeted Performance Band: 3-5

Criteria Mark • Draws the correct cubic curve • Plots all important points

1 1

Sample Answer

1331

)( 23 ++−−= xxxxf

(a) (iv) (1 mark) Outcomes Assessed: H7 Targeted Performance Band: 3-4

Criteria Mark • Gives the correct answer 1

Sample Answer

19=y

• )19,6(−

)32

2,1(

x

y

O )

32

2,1( −− •

• )8,3( − • )8,3( −−

Page 26: Cssa 2006 Math 2u + Solutions

10 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.

(b) (i) (0 marks) (b) (ii) (2 marks) Outcomes Assessed: P2, H2 Targeted Performance Band: 2-4

Criteria Mark • Shows that QARPAS ∠=∠ (vertically opposite s'∠ ) or equivalent • Correctly completes proof using (AAS)

1 1

Sample Answer

QRAPSA ∠=∠ (given) QARPAS ∠=∠ (vertically opposite s'∠ )

QRPS = (given) QRAPSA ∆≡∆∴ (AAS)

(b) (iii) (2 marks) Outcomes Assessed: P2, H2 Targeted Performance Band: 3-4

Criteria Mark • Realises that PAQ∆ is isosceles & base s'∠ are equal • Gives the correct answer

1 1

Sample Answer

QAPA = (corresponding sides are equal in congruent ?’s)

PAQ∆∴ is isosceles xQPAPQA =∠=∠∴ 2x + 120 = 180o (∠ sum of a ? is 180o) x = 30o

Question 5 (12 marks) (a) (i) (1 mark) Outcomes Assessed: H5 Targeted Performance Band: 2-3

Criteria Mark • Writes the correct sum of an infinite geometric series 1

Sample Answer

...000105

10005

1005

107

57.0 +++=•

Page 27: Cssa 2006 Math 2u + Solutions

11 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.

(a) (ii) (2 marks) Outcomes Assessed: H5 Targeted Performance Band: 3-4

Criteria Mark • Writes the limiting sum formula with correct a and r • Gives the correct answer

1 1

Sample Answer

4534

181

107

1011

1005

107

1

=

+=

−+=

−=∞ r

aS

(b) (2 marks) Outcomes Assessed: P3, H5 Targeted Performance Band: 2-4

Criteria Mark • Realises that θ2cos is a substitute or equivalent • Simplifies to give the correct answer

1 1

Sample Answer

xx

cotsin1 2−

θθθ

sincoscos2

= = θθ

θcossin

cos 2 ×

θθ cossin=

Page 28: Cssa 2006 Math 2u + Solutions

12 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.

(c) (i) (1 mark) Outcomes Assessed: H5 Targeted Performance Band: 2-4

Criteria Mark • Gives the correct answer 1 Sample Answer

rl

rl =∴= θθ

661 π

πθ ==

(c) (ii) (2 marks) Outcomes Assessed: H5 Targeted Performance Band: 2-4

Criteria Mark • Writes the correct formula with necessary substitutions • Gives the correct answer in exact form

1 1

Sample Answer

θ2

21

rA =

66

21

π×

×=A

∴ Area of sector MONπ3

=⇒ A cm2

Page 29: Cssa 2006 Math 2u + Solutions

13 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.

(d) (i) (2 marks) Outcomes Assessed: H5 Targeted Performance Band: 3-4

Criteria Mark • Correctly draws a tree diagram • Gives the correct answer

1 1

Sample Answer (d) (ii) (2 marks) Outcomes Assessed: H5 Targeted Performance Band: 3-4

Criteria Mark • Uses the complementary events method (or otherwise) • Gives the correct answer with required working

1 1

Sample Answer

125117

521

52

52

521)(

3

=

−=

××−=SP

53

53

53

S

53

52

53

53

53

52

52

52

52

52

52

S

S

S

S

S

S F

F

F

F

F

F

F

12527

53

53

53

53

)(3

=

=××=∴ SP

Page 30: Cssa 2006 Math 2u + Solutions

14 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.

Question 6 (12 marks) (a) (i) (1 mark) Outcomes Assessed: P2 Targeted Performance Bands: 2-3

Criteria Mark • Correctly verifies Pythagoras’ Theorem 1 Sample Answer Triangle with sides 12 cm, 16 cm and 20 cm is right angled at A ( 222 201612 =+ , converse of Pythagoras’ Theorem) (a) (ii) (2 marks) Outcomes Assessed: P4 Targeted Performance Bands: 2-3

Criteria Marks • Correctly gives answer for α in either degrees or radians • Correctly gives answer for β in either degrees or radians

1 1

Sample Answer

2012

sin =α ∴ '5236 °=α or c644.0=α

'' 853523690 °°° =−=β or c927.0=β (using complementary angles in 21OAO∆ )

(a) (iii) (3 marks) Outcomes Assessed: H5 Targeted Performance Bands: 3-5

Criteria Marks • Correctly substitutes in the segment formula for the circle O1 • Correctly substitutes in the segment formula for the circle O2 • Correctly gives answer (to 2 d.p.)

1 1 1

Sample Answer Area enclosed between the two circles is the sum of the two segments, with angles α2 and

β2 subtended respectively at the centre.

Area = ( ) ( )927.02sin927.021221

644.02sin644.021621 22 ×−××+×−××

∴ Area = 106.2667952 cm2 ∴ Area = 106.27 cm2 (2 d.p.)

Page 31: Cssa 2006 Math 2u + Solutions

15 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.

(b) (i) (2 marks) Outcomes Assessed: H3, H8 Targeted Performance Bands: 3-4

Criteria Mark • Correctly works out 2x as a subject ( yx elog2 = )

• Correctly substitutes yx elog2 = in the volume formula and deduce the answer

1

1

Sample Answer

∫=5

1

2 dyxV y π ; Since 2xey = ∴ 2log xye = ∴ ∫=

5

1log dyyV ey π (as required)

(b) (ii) (1 mark) Outcomes Assessed: H3 Targeted Performance Bands: 2-3

Criteria Mark • Correctly completes the required value ( 386.1log =ye ) 1

Sample Answer

y 1 2 3 4 5

yelog 0 0.693 1.099 1.386 1.609

(b) (iii) (3 marks) Outcomes Assessed: H3, H5 Targeted Performance Bands: 2-4

Criteria Mark • Substitutes the correct values in the correct Simpson’s formula • Correctly calculates the answer in decimal form (e.g. 4.041) • Gives the correct answer

1 1 1

Sample Answer

Using Simpson’s Formula: ( ) ( )[ ]∫ +++++++≈b

a nyyyyyyh

dxy ...2...43 42310

( )[ ]609.1099.12386.1693.04031

+×+++×= πyV ∴ π041.4=yV

∴ 695.12=yV (3 d.p.)

Page 32: Cssa 2006 Math 2u + Solutions

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.

Question 7 (12 marks) (a) (3 marks) Outcomes Assessed: H3 Targeted Performance Bands: 3-4

Criteria Mark • 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

1 1 1

Sample Answer

( ) 37loglog 22 =++ xx ∴ ( ) 37log 2 =+xx ∴ ( ) 327 =+xx ∴ 0872 =−+ xx ∴ ( )( ) 018 =−+ xx ∴ 8−=x or 1=x ∴ solution is 1=x ( )0>x or checking the solution by substitution. (b) (i) (2 marks) Outcomes Assessed: P4 Targeted Performance Bands: 2-3

Criteria Mark • Correctly finds x-ordinate of point A • Correctly finds x-ordinate of point B

1 1

Sample Answer At point A: 22 23 xx −= ∴ 33 2 =x ∴ 1±=x ∴ 1=Ax (first quadrant )

At point B: 023 2 =− x ∴ 23

±=x ∴ 23

=Bx (first quadrant )

(b) (ii) (3 marks) Outcomes Assessed: H8 Targeted Performance Bands: 2-4

Criteria Mark • Correctly uses the sum of integrals to find area under curves • Correctly finds the primitives • Correctly finds the area using Leibnitz – Newton Formula

1 1 1

Sample Answer

( ) [ ] 23

1

310

31

023

1

22

32

331

23

−+=−+= ∫ ∫ xxxdxxdxxAAOB

∴ ( ) 223

232

323

32

23

30131

3

−=

−−

−+−=AOBA square units

Page 33: Cssa 2006 Math 2u + Solutions

17 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.

(c) (i) (1 mark) Outcomes Assessed: P4 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

750025605 =×× revolutions (c) (ii) (3 marks) Outcomes Assessed: H1, H5 Targeted Performance Bands: 3-5

Criteria Mark • Correctly realises that circles’ radii are in a arithmetic sequence • Correctly uses the formula for arithmetic series • Finds the correct answer

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:

( ) 5595.49518862222

75007500 =×+×= ππS cm =1.9 km (1 d.p.)

Page 34: Cssa 2006 Math 2u + Solutions

18 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.

Question 8 (12 marks) (a) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 2 – 4

Criteria Mark • Correctly uses index law • Find the correct answer

1 1

Sample Answer

If 3=mA ( ) 81344==∴ mA 7658154 =−=−∴ mA

(b) (i) (2 marks) Outcomes Assessed: H3, H4, H5 Targeted Performance Bands: 3 – 4

Criteria Mark • Correctly substitutes in the formula and writes –10k as a subject • Gives the correct answer

1 1

Sample Answer

U = kte−100 ke 101002 −=∴ ∴ ke 10

501 −= ke 10ln

501

ln −=∴ 10

50ln−

−=∴k

∴k = 0.3912

(b) (ii) (2 marks) Outcomes Assessed: H3, H4, H5 Targeted Performance Bands: 3-5

Criteria Mark • Correctly differentiates U • Gives the correct answer

1 1

Sample Answer

( )ktekdt

dU −−= 100

For 3912.0and12 == kt : mg/minute36.0−=dt

dU

Page 35: Cssa 2006 Math 2u + Solutions

19 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.

(c) (i) (1 mark) Outcomes Assessed: H4, H5 Targeted Performance Bands: 2-3

Criteria Mark • Correctly gives answer for 1A 1 Sample Answer

1050100

75.011 −

+= PA ( ) 10500075.11 −=∴ PA

(c) (ii) (3 marks) Outcomes Assessed: H4, H5 Targeted Performance Bands: 3-6

Criteria Mark • Correctly gives answer for 2A • Write the correct sum of the geometric series • Gives the correct answer

1 1 1

Sample Answer

[ ] 1050)0075.1(1050)0075.1(2 −−= PA

1050)10075(1050)0075.1( 2 −−= P )0075.11(1050)0075.1( 2 +−= P )0075.10075.11(1050)0075.1( 23

3 ++−= PA . . .

)0075.1...0075.11(1050)0075.1( 1−+++−= nn

n PA

= −nP )0075.1( 1050 ( )

−0075.0

10075.11 n

= −nP )0075.1( 140 000 [ ]10075.1 −n

= −nP )0075.1( 140 000 ( ) 0001400075.1 +n

Page 36: Cssa 2006 Math 2u + Solutions

20 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.

(c) (iii) (2 marks) Outcomes Assessed: H4, H5 Targeted Performance Bands: 3-4

Criteria Mark • Correctly substitutes n=240 and makes 0=nA • Gives the correct answer

1 1

Sample Answer When n = 240

( ) ( ) 0001400075.10001400075.10 240240 +−=∴ P

( ) ( ) 0001400075.10001400075.1 240240 −=∴ P

( )( )240

240

0075.10001400075.1000140 −

=∴ P

∴P = $116 702 (to the nearest dollar)

Page 37: Cssa 2006 Math 2u + Solutions

21 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.

Question 9 (12 marks) (a) (i) (2 marks) Outcomes Assessed: H4, H5 Targeted Performance Bands: 2-3

Criteria Mark • Correctly differentiate to find the velocity Av • Correctly differentiate to find the velocity Bv

1 1

Sample Answer

512 += tx A 12=∴ Av

326 ttxB −= ∴ 2312 ttvB −=

(a) (ii) (1 mark) Outcomes Assessed: H4, H5 Targeted Performance Bands: 2-3

Criteria Mark • Gives the correct answer 1 Sample Answer

At t = 1 second 12=∴ Av m/s, 9312 =−=Bv m/s ∴ particle A is faster (a) (iii) (1 mark) Outcomes Assessed: H4, H5 Targeted Performance Bands: 3 – 4

Criteria Mark • Gives the correct answer 1 Sample Answer

( ) 4or004303120 2 ==∴=−∴=−∴= ttttttvB seconds ∴Particle comes at rest at 4=t seconds

Page 38: Cssa 2006 Math 2u + Solutions

22 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.

(a) (iv) (2 marks) Outcomes Assessed: H4, H5 Targeted Performance Bands: 3 – 5

Criteria Mark • Correctly substitutes the value of t = 4 into the Bx equation • Gives the correct answer

1 1

Sample Answer For maximum displacement, 0=Bv ∴t ( 12 – 3t ) = 0 ∴t = 4 seconds ∴ 32446 32 =−×=Bx metres

(b) (i) (1 mark) Outcomes Assessed: P6 Targeted Performance Bands: 2 – 4

Criteria Mark • Gives the correct answer 1 Sample Answer

The gradient of the tangent at point P is 360tan == °

Pm (b) (ii) (2 marks) Outcomes Assessed: P3, P4 Targeted Performance Bands: 3 – 5

Criteria Mark • 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

1 1

Sample Answer

y = 1 – x2 xdxdy

2−=∴ and since the gradient of the tangent is 3

∴ 3 = - 2x 2

3−=∴ x ∴ y =

41

∴ coordinates of point P are

41

,23

Page 39: Cssa 2006 Math 2u + Solutions

23 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.

(b) (iii) (3 marks) Outcomes Assessed: P4, H5 Targeted Performance Bands: 3-6

Criteria Mark • Correctly find the length of PQ • Correctly find the length of 2PO (or 2QO ) • Gives the correct answer

1 1 1

Sample Answer

Coordinates of point Q are

41

,23

(By symmetry)

3=∴ PQ

1613

161

432 =+=PO

Using the Cosine Rule in POQ∆ : cos ( )

1613

1613

2

31613

1613 2

××

−+=∠POQ

radians6.2=∠POQ

Page 40: Cssa 2006 Math 2u + Solutions

24 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.

Question 10 (12 marks) (a) (2 marks) Outcomes Assessed: H5 Targeted Performance Bands: 2-5

Criteria Mark • Writes the correct value of the amplitude A • Find the correct value of n

1 1

Sample Answer

Since y= A sin nx ∴ A = 2

Period (T) = 3

4π (from graph) and since Period (T) =

nπ2

T

nπ2

=∴

23

=∴ n ∴ the equation is xy23

sin2=

(b) (i) (1 mark) Outcomes Assessed: P4, H5 Targeted Performance Bands: 3 – 4

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 αα sin12sin46 =××× (as required)

(b) (ii) (3 marks) Outcomes Assessed: P4, H5 Targeted Performance Bands: 3 – 6

Criteria Mark • Correctly finds the area of ? SOT • Correctly finds the area of the quadrilateral ORST • Correctly factorise to find the correct answer

1 1 1

Sample Answer

∴area of ? SOT = ½ ααπ

cos62

sin26 =

−×××

∴area of the quadrilateral ORST is:

A = 12 sina + 6 cosa

A = 6cosa(2tana + 1)

Page 41: Cssa 2006 Math 2u + Solutions

25 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.

(b) (iii) (3marks) Outcomes Assessed: P4, H5 Targeted Performance Bands: 3 – 6

Criteria Mark • 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

1 1 1

Sample Answer

A = 12 sina + 6 cosa ∴ ααα

sin6cos12 −=ddA

For maximum area 0=αd

dA∴ αα sin6cos120 −=

∴ 6sina = 12cosa 2tan =∴ α To prove the area is maximum when 2tan =α :

ααα

cos6sin122

2

−−=d

Ad and since 2tan =α ∴ from the triangle PQR

52

sin =∴ α and 5

1cos =α

∴ 05

305

16

52

122

2

<−=×−×−=αd

Ad

∴Maximum Area occurs at 2tan =α Alternative method: To prove the area is maximum when 2tan =α : Since 2tan =α ∴α = 1.1 radians

α c1 c1.1 c2.1

αddA

043.1 > 0 024.1 <−

Maximum

α 2 5

1 R Q

P

Page 42: Cssa 2006 Math 2u + Solutions

26 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.

(b) (iv) (3 marks) Outcomes Assessed: P4, H5 Targeted Performance Bands: 3 – 6

Criteria Mark • 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

1 1 1

Sample Answer

The gradient of the line OS = tan α = xy

and since 2tan =α

∴ 2=xy

∴y = 2x [ ]1

Substitute into 3622 =+ yx 364 22 =+∴ xx

∴ 5x2 = 36 5

362 =∴ x 556

56

==∴ x

∴Substitute into [ ]1 55

125

12==∴ y ∴point S is

5

512

,556

.

Page 43: Cssa 2006 Math 2u + Solutions

27 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.

BLANK PAGE

Page 44: Cssa 2006 Math 2u + Solutions

28 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.

BLANK PAGE

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2602-3