Mathematics Advanced Sample Examination

• Reading time – 10 minutes• Working time – 3 hours• Write using black pen• Calculators approved by NESA may be used• A reference sheet is provided at the back of this paper• For questions in Section II, show relevant mathematical reasoning

and/ or calculations

Section I – 10 marks (pages 2–7)• Attempt Questions 1–10• Allow about 15 minutes for this section

Section II – 90 marks (pages 9–39)• Attempt Questions 11–38• Allow about 2 hours and 45 minutes for this section

General Instructions

Total marks: 100

Mathematics Advanced

Sample HIGHER SCHOOL CERTIFICATE EXAMINATION

NSW Education Standards Authority

– 2 –

Section I

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

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

1 What type of relation is shown?

64

5

6

77

88

A. Many-to-many

B. One-to-many

C. One-to-one

D. Many-to-one

2 At which point on this curve are the first and second derivatives BOTH negative?

x

y

O

A B

C D

A. A

B. B

C. C

D. D

– 3 –

3 What is the gradient of any line perpendicular to 3x + 2 y = 5 ?

A. 23

B. − 23

C. 32

D. − 32

4 What is the derivative of 52 x + 3 ?

A. 2 × 52 x + 3

B. (2 x + 3) × 52 x + 2

C. ln 5 × 52 x + 3

D. ln 5 × 2 × 52 x + 3

– 4 –

5 A school collected data on the reasons given by students for arriving late. The Pareto chart shows the data collected.

0

20

40

60

80

100

120

140

160

180

200

0%Slept in Family Appointment

Reasons for arriving late

Num

ber

of la

te a

rriv

als

Cum

ulat

ive

per

cent

age

Train orbus delay

Trafficproblems

Car breakdown

10%

20%

30%

40%

50%

60%

70%

80%

90%

100%

What percentage of students gave the reason ‘Train or bus delay’?

A. 6%

B. 15%

C. 30%

D. 92%

– 5 –

6 What are the values of x for which | 3 − 4 x | = 4 ?

A. x = − 74

and x = − 14

B. x = − 74

and x = 14

C. x = 74

and x = − 14

D. x = 74

and x = 14

7 The diagram shows the graph of a continuous probability density function.

x2 31O

y

1

0.2

(0.8, 0.6)

(2.5, 0.07) (3, 0.2)

Which of the following is the mode?

A. 0.07

B. 0.6

C. 0.8

D. 3

– 6 –

8 The graphs show the future values over time of $P, invested at three different rates of compound interest.

Time (years)

Futu

re v

alue

($)

P

W

X

Y

Which of the following correctly identifies each graph?

A.W

5% pa, compounding annually

X10% pa, compounding annually

Y10% pa, compounding quarterly

B.W


X10% pa, compounding quarterly

Y10% pa, compounding annually

C.W

10% pa, compounding quarterly

X10% pa, compounding annually


D.W


X10% pa, compounding quarterly


– 7 –

9 The scores on an examination are normally distributed with a mean of 70 and a standard deviation of 6. Michael received a score on the examination between the lower quartile and the upper quartile of the scores.

Which shaded region most accurately represents where Michael’s score lies?

58 64 76 8270

A.

58 64 76 8270

B.

58 64 76 8270

C.

58 64 76 8270

D.

10 Given the function y = log7 ( x x ), which expression is equal to dydx

?

A. 1

x ln 7

B. 1

ln 7 × log7 ( x x − 1 )

C. 1

x x ln 7

D. log7 x + 1

ln 7

BLANK PAGE

© 2020 NSW Education Standards Authority

– 8 –

– 9 –

HIGHER SCHOOL CERTIFICATE

EXAMINATIONSample

Please turn over

90 marksAttempt Questions 11–38Allow about 2 hours and 45 minutes for this section

Mathematics Advanced

Section II Answer Booklet

• Answer the questions in the spaces provided. These spaces provide guidance for the expected length of response.

• Your responses should include relevant mathematical reasoning and/or calculations.

Instructions

– 10 –

Question 11 (2 marks)

Find the equation of the tangent to the curve â ( x ) = x 3 + 1 at the point (1, 2) .

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................


The diagram shows a triangle with sides of length x cm, 11 cm and 13 cm and an angle of 80°.

80°

11 cm NOT TOSCALE

13 cm

x cm

Use the cosine rule to calculate the value of x, correct to two significant figures.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

3

– 11 –


A credit card requires a four-figure personal identification number (PIN) for purchases. The figures are chosen from the digits 0, 1, 2, 3, . . . , 9. Repetition is allowed and the PIN can start with any of the 10 digits.

The credit card is lost and the finder tries to guess the PIN by entering four digits.

(a) What is the probability that the four digits entered are the correct PIN?

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

(b) What is the probability that the finder will guess at least one digit in its correct position?

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

Please turn over

1

1

– 12 –


A function is given by â ( x ) = 18 x 2 − x 4 .

(a) Find the stationary points and determine their nature.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

Question 14 continues on page 13

4

– 13 –

Question 14 (continued)

(b) Sketch the curve, labelling the stationary points and axis intercepts.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

End of Question 14

2

– 14 –


Two teams play a game. There are only two possible ways to score points: hitting the red target or hitting the blue target. Hitting the red target scores R points and hitting the blue target scores B points. The results of a game are shown.

Team Number of red target hits

Number of blue target hits

Total score

Team 1 7 6 47

Team 2 5 11 47

By forming a pair of simultaneous equations, or otherwise, find the values of R and B.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................


Differentiate e sin ( px ).

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

3

2

– 15 –


Given the function â ( x ) = x 2 + 2 and g x( ) = x − 6 , sketch y = â ( g ( x ) ) over its natural domain.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

– 16 –


The diagram shows a continuous function y = â ( x ) defined in the domain −2, 10 . The function consists of a quarter of a circle centred at (0, 0) with radius 2, a straight line segment and a logarithmic function â ( x ) = ln ( x − 2) in the domain 3, 10 .

10 x987654321O

−3

−2

−1

1

2

3

4

y

−1−2−3

(a) Find the exact area bounded by the function y = â ( x ) and the x-axis in the domain −2, 3 .

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


2

– 17 –


(b) Hence, find the exact value of ln x − 2( )dx3

10⌠

⌡⎮ ,

given that ƒ x( )dx = 8ln8 −10 − π−2

10⌠

⌡⎮ .

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


A discrete random variable X has the probability distribution table shown.

X = x 11 12 13 14

P ( x ) 0.2 0.3 m 0.4

By finding the value of m , calculate the expected value and the variance of X.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

3

– 18 –


A student was asked to differentiate â ( x ) = x 2 + 4 x from first principles.

The student began the solution as shown below.

Complete the solution.

′ƒ x( ) = limh→0

ƒ x + h( ) − ƒ x( )h

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

– 19 –


The diagram shows the distances of four towns A, B, C and D from point O. The true bearings of towns A, B and D from point O are also shown.

D(318°)

A(029°)

B(125°)

C

O

N

16 km

25 k

m

18 k

m

12 kmNOT TOSCALE

198 km2

The area of the acute-angled triangle BOC is 198 km2.

Calculate the true bearing of town C from point O, correct to the nearest degree.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

3

– 20 –


A small business makes and sells bird houses.

Technology was used to draw straight-line graphs to represent the cost of making bird houses (C ) and the revenue from selling bird houses (R ). The x-axis displays the number of bird houses and the y-axis displays the cost/ revenue in dollars.

0 10 20 30 40 x

y

1000

1500

500

C

R

(a) How many bird houses need to be sold to break even?

...............................................................................................................................

(b) By first forming equations for cost (C ) and revenue (R ), determine how many bird houses need to be sold to earn a profit of $1900.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

1

3

– 21 –


The function â ( x ) = tan3 x is given.

If â ′( x ) = 3 tanm x + 3 tan2 x, find the value of m.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................


The function â ( x ) = | x | is transformed and the equation of the new function is of the form y = k â ( x + b) + c, where k , b and c are constants.

The graph of the new function is shown.

−4 O

y

x

−4

6

−6

4

2

−2

4−2 2

What are the values of k , b and c ?

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

2

– 22 –


A circle is given by the equation x 2 + y 2 + 4 x − 10 y = −16.

Find the centre and radius of this circle.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................


By drawing graphs on the number plane, determine how many solutions there are to

the equation sin x = x

5 in the domain (− , ).

x3p2pp−p O

−2

−1

1

2

y

−2p−3p

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

Questions 11–26 are worth 45 marks in total

2

3

– 23 –


The function â ( x ) = cos x is transformed to g ( x ) = 3 cos 2 x .

Describe in words how both the amplitude and period change in this transformation.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................


The graph of a function â ( x ) is shown. It has an asymptote at y = 2 .

2

O x

y

−2

4

2 4−2 6 8

(−1, 4)

(2, −2)

Using interval notation, state the domain and range of â ( x ) .

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

2

2

– 24 –


Let X denote a normal random variable with mean 0 and standard deviation 1.

The random variable X has the probability density function ƒ x( ) = 1

2πe−x2

2 ,

where –∞ < x < ∞.

The diagram shows the graph of y = â ( x ) .

−2−3 −1 0

0.2

0.1

0.3

0.4

1 2 3x

â ( x )

(a) Complete the table of values for the function given. Give your answer correct to four significant figures.

X = x 0 1 2 3

â ( x ) 0.3989 0.2420 0.004432

(b) Using the trapezoidal rule and the 4 function values in the table in part (a), show that

P −3 ≤ X ≤ 3( ) = ƒ x( )dx ≈ 0.9953−3

3⌠

⌡⎮ .

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


1

2

– 25 –


(c) The IQ (Intelligence Quotient) scores for a large population are normally distributed with a mean of 100 and a standard deviation of 15. Using the result obtained in part (b), calculate the probability of randomly selecting a person with an IQ score above 145 from this large population.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

End of Question 29

2

– 26 –


The population, P, of rabbits on an island is given by P ( t ), where t is the time in years

after the rabbits were introduced. The rabbit population changes at a rate modelled by

the function dP

dt= 30e1.25t .

Calculate the increase in the number of rabbits at the end of the first 10 years. Give your answer correct to two significant figures.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

3

– 27 –


A bid made at an auction for a real estate property, in millions of dollars, can be modelled by the random variable X with the probability density function

ƒ x( ) =k 16 − x2( )0

⎧⎨⎪

⎩⎪

1 ≤ x ≤ 4

otherwise.

(a) Show that the value of k is 127

.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


2

– 28 –


(b) Find the cumulative distribution function.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


2

– 29 –


(c) Find the probability that a bid of more than 3 million dollars will be made.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

End of Question 31

1

– 30 –


A farmer wishes to make a rectangular enclosure of area 720 m2. She uses an existing straight boundary as one side of the enclosure. She uses wire fencing for the remaining three sides and also to divide the enclosure into four equal rectangular areas of width x m as shown.

Existing boundary

x

The total length, m, of the wire fencing is given by = 5x + 720

x. (Do NOT prove

this.)

Find the minimum length of wire fencing required, showing why this is the minimum length.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

3

– 31 –


A particle is moving along the x-axis. The graph shows its velocity v metres per second at time t seconds.

O

8

4 t

v

When t = 0 the displacement x is equal to 2 metres.

On the axes below draw a graph that shows the particle’s displacement, x metres from the origin, at a time t seconds between t = 0 and t = 4. Label the coordinates of the endpoints of your graph.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

O 4 t

x

2

– 32 –


The table shows the future values of an annuity of $1 for different interest rates for 4, 5 and 6 years. The contributions are made at the end of each year.

Future value of an annuity of $1

YearsInterest rate per annum

1% 2% 3% 4%

4 4.060 4.122 4.184 4.246

5 5.101 5.204 5.309 5.416

6 6.152 6.308 6.468 6.633

An annuity account is opened and contributions of $2000 are made at the end of each year for 7 years.

For the first 6 years, the interest rate is 4% per annum, compounding annually.For the 7th year, the interest rate increases to 5% per annum, compounding annually.

Calculate the amount in the account immediately after the 7th contribution is made.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

3

– 33 –


The diagram shows the curves y = sin x and y = 3 cos x .

3y = cosx

y = sinx

2p

y

O x

1

−1

−

3

3

Find the area of the shaded region.

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

.........................................................................................................................................

4

– 34 –


An island initially has 16 100 trees. The number of trees increases by 1% per annum. The people on the island cut down 1161 trees at the end of each year.

(a) Show that after the first year there are 15 100 trees remaining.

...............................................................................................................................

...............................................................................................................................

(b) Show that at the end of 2 years the number of trees remaining is given by the expression T2 = 16 100 × (1.01)2 − 1161 (1 + 1.01).

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


1

2

– 35 –


(c) Show that at the end of n years the number of trees remaining is given by the expression Tn = 116 100 − 100 000 × (1.01)n.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

(d) For how many years will the people on the island be able to cut down 1161 trees annually?

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

End of Question 36

2

1

– 36 –


A set of bivariate data is collected by measuring the height and recording the shoe size of nine basketball players. The collected data is shown in the table and graphed in the scatterplot shown.

Shoe size 10 12 12 12 10 9 13 10 13

Height (cm) 181 190 193 196 188 175 198 192 201

170

9 10 11 12 13

175

180

185

190

195

200

205

Height(cm)

Shoe size

(a) Determine the equation of the least-squares regression line for this data.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

(b) A player with a shoe size of 11 was absent when the data was collected.

Calculate the predicted height for the missing basketball player using your answer to part (a).

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

1

1

– 37 –


A cable is freely suspended between two 10 m poles, as shown. The poles are 100 metres apart and the minimum height of the cable is 8 metres.

x(0, 0)Ground level

−50 50

10 10y

8

Pole

Cable

Pole

The height of the cable is given by y = c (e k x + e −k x), where c and k are positive constants.

(a) Show that the value of c is 4.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


1

– 38 –


(b) Use the result in part (a) to show that one value of k is ln 250

.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................


4

– 39 –


(c) Hence find the area between the poles, the cable and the ground.

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

End of paper

3

© 2020 NSW Education Standards Authority

– 1 –XXXX


2020 HIGHER SCHOOL CERTIFICATE EXAMINATION

Mathematics AdvancedMathematics Extension 1Mathematics Extension 2

– 2 –

– 3 –

– 4 – © 2020 NSW Education Standards Authority


Page 1 of 26

Mathematics Advanced Sample HSC Marking Guidelines

Section I

Multiple-choice Answer Key

Question Answer

1 B 2 B 3 A 4 D 5 A 6 C 7 C 8 C 9 A

10 D

NESA Mathematics Advanced Sample HSC Marking Guidelines

Page 2 of 26

Section II

Question 11 Criteria Marks

• Provides correct solution 2 • Determines the gradient of the tangent, or equivalent merit 1

Sample answer:

ƒ x( ) = x3 +1

′ƒ x( ) = 3x2

x = 1 ƒ 1( ) = 3 × 12

= 3

∴ m = 3

y − y1 = m x − x1( ) x1, y1( ) = 1, 2( )

y − 2 = 3 x −1( )

y − 2 = 3x − 3

y = 3x −1 or 3x − y −1 = 0


• Provides correct solution with correct rounding 3 • Calculates correct value of x 2 • Attempts to use cosine rule 1

Sample answer:

x2 = 112 + 132 − 2 × 11 × 13 × cos80°

= 240.3306

∴ x = 240.366…

= 15.5027

= 16 (2 significant figures)


Page 3 of 26

Question 13 (a) Criteria Marks

• Calculates correct probability 1

Sample answer:

110

× 110

× 110

× 110

= 110 000

Question 13 (b) Criteria Marks

• Calculates correct answer, or equivalent merit 1

Sample answer:

P(at least one digit in correct position) = 1 – P(all incorrect)

1 – P(all incorrect) = 1− 910

⎛⎝

⎞⎠

4

= 343910 000


Page 4 of 26


• Provides correct solution 4 • Determines the coordinates of all three stationary points, or equivalent

merit 3

• Determines the coordinates of two stationary points, or equivalent merit 2 • Determines the x values of at least two stationary points, or equivalent

merit 1

Sample answer:

ƒ x( ) = 18x2 − x4

∴ ′ƒ x( ) = 36x − 4x3

= 4x 9 − x2( )

Stationary points occur where ′ƒ x( ) = 0.

ie 4x 9 − x2( ) = 0

∴ x = 0, ± 3

As ′′ƒ x( ) = 36 −12x2

′′ƒ 0( ) = 36 > 0 so (0, 0) is a minimum.

′′ƒ 3( ) = ′′ƒ −3( ) = −72 < 0 so 3, 81( ) and −3, 81( ) are maxima.

Hence (0, 0 ) is a (local) minimum and (3, 81) and (–3, 81) are (global) maxima.


Page 5 of 26


• Provides correct sketch with stationary points and roots labelled 2 • Determines at least two roots, or equivalent merit 1

Sample answer:

The roots of the function, given by solving x2 18 − x2( ) = 0 are (0, 0),

3 2, 0( ) and −3 2, 0( ).


Page 6 of 26


• Provides correct solution 3 • Writes both equations and attempts to solve them simultaneously, or

equivalent merit 2

• Provides one correct solution, or equivalent merit 1

Sample answer:

7R + 6B = 47 ________ 1

5R + 11B = 47 ________ 2 Multiplying 1 by 5 and 2 by 7 we have

35R + 30B = 235 ________ 3

35R + 77B = 329 ________ 4 Subtract 3 from 4 so 47B = 94 so B = 2 and on substituting this value into 1 7R = 35 so R = 5. Substituting these values into 2 we have a validity check with 5 × 5 + 11 × 2 = 47. R = 5 B = 2 Question 16 Criteria Marks

• Provides correct answer 2 • Differentiates the exponential or the trigonometric expression correctly 1

Sample answer:

y = esin π x( )

= eƒ x( ) where ƒ x( ) = sin π x( ) ′ƒ x( ) = π cos π x( )

dy

dx= ′ƒ x( )eƒ x( ) = π cos π x( )esin π x( )


Page 7 of 26


• Provides correct graph in the restricted domain 2 • Provides linear equation, draws the correct graph in the domain −∞,∞( ) ,

or equivalent merit 1

Sample answer:

ƒ x( ) = x2 + 2 g x( ) = x − 6

ƒ g x( )( ) = x − 6( )2 + 2

= x − 6 + 2

= x − 4

The domain of g(x) is x − 6 ≥ 0 or x ≥ 6, so ƒ g x( )( ) only exists for x ≥ 6.


• Determines correct area, or equivalent merit 2 • Calculates area of one region, or equivalent merit 1

Sample answer:

A0 = 14× π × 22 = π units2

A1 = 12× 3 × 2 = 3 units2

Atotal −2, 3( ) = 3 + π( ) units2


Page 8 of 26


• Provides correct solution 2 • Writes an equation that correctly equates the integral and the area, or

equivalent merit 1

Sample answer:

ƒ x( )dx−2

10⌠

⌡⎮ = ƒ x( )dx

−2

3⌠

⌡⎮ + ƒ x( )dx

3

10⌠

⌡⎮

(under x-axis)

8ln8 −10 − π = −3 − π + ln x − 2( )dx3

10⌠

⌡⎮

8ln8 −10 + 3 − π + π = ln x − 2( )dx3

10⌠

⌡⎮

ln x − 2( )dx3

10⌠

⌡⎮ = 8ln8 − 7


Page 9 of 26


• Calculates correct values of m, E(X) and Var(X) 3 • Calculates m and expected value, or equivalent merit 2 • Determines correct value of m, or equivalent merit 1

Sample answer:

0.2 + 0.3 + m + 0.4 = 1

m + 0.9 = 1

m = 0.1

X = x 11 12 13 14

P(x) 0.2 0.3 0.1 0.4

E X( ) = ∑ xP x( ) = 11 × 0.2 +12 × 0.3 +13 × 0.1+14 × 0.4

E X( ) = 12.7

Var X( ) = E X 2( ) − E X( )[ ]2

= ∑ x2P x( ) − μ2

= 112 × 0.2( ) + 122 × 0.3( ) + 132 × 0.1( ) + 142 × 0.4( ) −12.72

Var(X) = 1.41


Page 10 of 26


• Provides correct solution 2 • Correctly substitutes into the given formula and expands the numerator

correctly, or equivalent merit 1

Sample answer:

′ƒ x( ) = limh→0

ƒ x + h( ) – ƒ x( )h

= limh→0

x + h( )2 + 4 x + h( ) − x2 + 4x( )h

= limh→0

x2 + 2hx + h2 + 4x + 4h − x2 − 4x

h

= limh→0

2hx + h2 + 4h

h

= limh→0

h 2x + h + 4( )h

= limh→0

2x + h + 4

= 2x + 4


• Provides correct solution 3 • Correctly calculates angle BOC 2

• Attempts to use the area of a triangle formula A = 12

absinC 1

Sample answer:

198 = 12× 16 × 25 × sin ∠BOC( )

198 = 200 sin ∠BOC( )

sin ∠BOC( ) = 198200

= 81.9°

∴Bearing of C from O is 125 + 81.9 = 207° (nearest degree)


Page 11 of 26


• Provides correct answer 1

Sample answer:

20 bird houses Question 22 (b) Criteria Marks

• Provides correct solution 3 • Provides correct formulae for R and C, or equivalent merit 2 • Provides correct formulae for R or C, or equivalent merit 1

Sample answer:

R = 40x

C = 500 +15x

∴1900 = 40x − 500 +15x( )1900 = 25x − 500

2400 = 25x

x = 240025

= 96

∴ 96 bird houses must be sold to earn a profit of $1900.


Page 12 of 26


• Provides correct solution 2 • Makes progress towards the solution 1

Sample answer:

ƒ x( ) = tan3 x = tan x( )3

′ƒ x( ) = 3 tan x( )2 sec2 x( )= 3tan2 x.sec2 x but sec2 x = 1+ tan2 x

= 3tan2 x 1+ tan2 x( )= 3tan4 x + 3tan2 x

compared to 3tanm x + 3tan2 x⎡⎣ ⎤⎦

∴ m = 4


• Provides correct solution 2 • Obtains the correct value of k, b or c 1

Sample answer:

The function ƒ x( ) = x is translated to the left 2 units, reflected through the x-axis,

translated down 1 unit and multiplied by 3. ie k = –3, b = 2 and c = –1.


Page 13 of 26


• Provides correct centre and radius 2 • Provides correct centre or radius, or equivalent merit 1

Sample answer:

x2 + 4x + 4 + y2 −10y + 25 = 13

x + 2( )2 + y − 5( )2 = 13

Centre at −2, 5( ).

Radius is 13 units.

(adding 4 + 25 = 29 to both sides to complete the squares)


• Provides correct number of solutions with correct graphs 3 • Provides correct number of solutions from incorrect graphs, or equivalent

merit 2

• Draws one correct graph and one incorrect graph, or equivalent merit 1

Sample answer:

Draw y = sin x and y = x

5

Check some points to see how high y = x

5 is.

x = 2π y = 2π5

x = −1.5π y = 1.5π5

= 1.256 = 0.94

(below sine graph)

∴ There are four solutions (four points of intersection).


Page 14 of 26


• Provides correct explanation 2 • Provides one correct transformation, or equivalent merit 1

Sample answer:

The period is halved and the amplitude is multiplied by 3. Question 28 Criteria Marks

• Provides correct domain and range 2 • Provides correct domain or range, or equivalent merit 1

Sample answer:

Domain: −1,∞[ )

Range: −2, 4[ ] Question 29 (a) Criteria Marks


Sample answer:

X = x 0 1 2 3

ƒ x( ) 0.3989 0.2420 0.0540 0.004432


Page 15 of 26


• Provides correct answer or correct numerical expression 2 • Attempts to use the trapezoidal rule with at least two correct numbers 1

Sample answer:

P −3 ≤ X ≤ 3( ) = ƒ x( )dx−3

3⌠

⌡⎮

= 2 ƒ x( )dx0

3⌠

⌡⎮

= 2 × 12

0.3989 + 0.004 + 2 0.2420 + 0.0540( )[ ]

= 0.9953

Question 29 (c) Criteria Marks

• Provides correct solution 2 • Makes progress towards the solution 1

Sample answer:

μ = 100

θ = 15

z = 145 −10015

= 3

Therefore, P(person with IQ > 145) = 0.00235

= 0.235%


Page 16 of 26


• Provides correct solution 3 • Provides correct numerical expression 2 • Writes correct integral, or equivalent merit 1

Sample answer:

dP

dt= 30e1.25t

Increase in the number of rabbits over first 10 years is given by:

30e1.25t dt0

10⌠

⌡⎮

= 30e1.25t

1.25

⎡⎣⎢

⎤⎦⎥0

10

= 24 e12.5 − e0( )= 24 268 377.2865 −1( )

= 6 440 070.877

∼ 6 400 000 rabbits (correct to 2 significant figures)


Page 17 of 26


• Provides correct solution 2 • Provides correct equation stating that the integral equals one, or

equivalent merit 1

Sample answer:

k 16 − x2( )dx = 11

4⌠

⌡⎮

16x − x3

3

⎡⎣⎢

⎤⎦⎥1

4

= 1k

16 × 4 –43

3

⎛⎝⎜

⎞⎠⎟− 16 − 1

3⎛⎝

⎞⎠ = 1

k

27 = 1k

∴ k = 127


Page 18 of 26


• Provides correct solution 2 • Provides correct integral, or equivalent merit 1

Sample answer:

ƒ x( ) =1

2716 − x2( )

0

⎧

⎨⎪

⎩⎪

1 ≤ x ≤ 4

otherwise

F t( ) = ƒ x( )dx−∞

t⌠

⌡⎮

= ƒ x( )dx1

t⌠

⌡⎮

= 127

16 − x2( )dx1

t⌠

⌡⎮

= 127

16x − x3

3

⎛⎝⎜

⎞⎠⎟1

t

= 127

16t − t3

3−16 + 1

3

⎛⎝⎜

⎞⎠⎟

= 127

16t − t3

3− 47

3

⎛⎝⎜

⎞⎠⎟

= 181

48t − t3 − 47( )

∴F x( ) =

01811

⎧

⎨⎪⎪

⎩⎪⎪

x < 1

48x − x3 − 47( ) 1 ≤ x ≤ 4

x > 4


Page 19 of 26


• Provides correct solution 1

Sample answer:

1− F 3( ) = 181

48 × 3 − 33 − 47( )⎛⎝

⎞⎠

= 1− 0.864197…( )

= 0.135802…

∼ 0.1358 (4 significant figures).


Page 20 of 26


• Provides correct solution 3 • Finds length at stationary point, or equivalent merit 2

• Finds

d

dx, or equivalent merit 1

Sample answer:

Stationary points occur when

0 = d

dx

= 5 − 720

x2

720

x2= 5

x2 = 144

x = 12 (x is length so ignore – 12)

d2

dx2= 1440

x3

at x = 12d2

dx2= 1440

123> 0

so minimum at x = 12

= 5 ×12 + 72012

= 120 m


Page 21 of 26


• Draws correct graph with correct endpoints 2 • Calculates the correct end value x or provides an equation for x, or

equivalent merit 1

Sample answer:

Equation of velocity: v = −2t + 8 m = − 84

m = −2⎛⎝

⎞⎠

Equation of displacement x = −2t + 8( )dt⌠

⌡⎮

x = −t2 + 8t + c

when t = 0, x = 2 ∴c = 2 ∴ x = −t2 + 8t + 2

Concave down parabola max at −2t + 8 = 0

2t = 8

t = 4

x = −42 + 8 × 4 + 2

= 18 m


Page 22 of 26


• Provides correct answer or correct numerical expression 3 • Provides correct value just before the 7th investment is made, or

equivalent merit 2

• Identifies correct value from the table, or equivalent merit 1

Sample answer:

Total after 6 years = 6.633 × 2000

= $13 266

After 7th year interest, total = 13 266(1 + 0.05)1

= $13 929.30

∴ Final total = 13 929.30 + 2000

= $15 929.30 Question 35 Criteria Marks

• Provides correct solution 4 • Writes a correct integral to determine the area, or equivalent merit 3 • Correctly determines the x-values of the points of intersection, or

equivalent merit 2

• Attempts to solve an equation to find the x-values of the points of intersection, or equivalent merit 1

Sample answer:

Need the x-values of the points of intersection first. y = sin x , y = 3 cos x sin x = 3 cos x

sin x

cos x= 3

Note: cos x ≠ 0 at these points of intersection

tan x = 3

∴ x = π3

, π + π3

x = π3

,4π3


Page 23 of 26

Shaded area = sin x − 3 cos x( )dxπ3

4π3⌠

⌡⎮

= −cos x − 3 sin x⎡

⎣⎢⎢

⎤

⎦⎥⎥ π

3

4π3

= cos x + 3 sin x⎡

⎣⎢⎢

⎤

⎦⎥⎥ 4π

3

π3

(swap the limits to change sign)

= cosπ3+ 3 sin

π3

⎛⎝

⎞⎠ − cos

4π3

+ 3 sin4π3

⎛⎝

⎞⎠

= 12+ 3 × 3

2

⎛⎝⎜

⎞⎠⎟− − 1

2⎛⎝

⎞⎠ + 3 × − 3

2

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

= 12+ 1

2+ 3

2+ 3

2

= 4 units2



Sample answer:

T1 = 16 100 × 1.01 − 1161 = 15 100 Question 36 (b) Criteria Marks

• Provides correct solution 2 • Makes progress towards solution 1

Sample answer:

T2 = T1 × 1.01 − 1161

= (16 100 × 1.01 − 1161) × 1.01 − 1161 = 16 100 × (1.01)2 − 1161(1 + 1.01)


Page 24 of 26


• Provides correct solution 2 • Writes a correct expression for T2 in expanded form, or equivalent merit 1

Sample answer:

Using this recursive rule we find T3 = 16 100 × (1.01)3 − 1161(1 + 1.01 + 1.012) Similarly Tn = 16 100 × (1.01)n − 1161(1 + 1.01 + 1.012 + · · · + 1.01n−1) Using the sum to n terms of a geometric sequence we find

Tn = 16 100 × (1.01)n − 11611.01n −11.01−1

⎛⎝⎜

⎞⎠⎟

Hence Tn = 16 100 × (1.01)n − 116 100 (1.01n − 1) So Tn = 116 100 − 100 000 × (1.01)n Question 36 (d) Criteria Marks

• Provides correct answer, or correct numerical expression 1

Sample answer:

0 = 116 100 − 100 000 × (1.01)n

So 1.01n = 116100100 000

= 1.161

n = ln1.161ln1.01

15 years (as n > 15 there are still some trees remaining after 15 years.)


Page 25 of 26


• Provides correct equation 1

Sample answer:

The least-squares regression line is y = 4.85x + 136. Question 37 (b) Criteria Marks

• Substitutes 11 into their equation from part (a) 1

Sample answer:

x = 11 y = 4.85x +136

= 4.85 ×11+136 = 189.35

= 189 cm (to nearest cm)


Page 26 of 26



Sample answer:

Using y = c ekx + e−kx( ) at (0, 8) we have 8 = c e0 + e−0( ) so c = 82= 4


• Provides correct solution 4 • Correctly writes the equation as a quadratic equation and makes progress

towards solving it 3

• Attempts to solve a quadratic equation to find k, or equivalent merit 2 • Substitutes (50, 10) and c = 4 into the equation, or equivalent merit 1

Sample answer:

Using y = c ekx + e−kx( ) at (50, 10) we have 10 = 4 e50k + e−50k( ) and so with u = e50k we

may write u − 2.5 + u−1 = 0 or u2 − 2.5u + 1 = 0 or 2u2 − 5u + 2 = 0. Factorising we have

(2u − 1)(u − 2) = 0 so 2 = e50k and hence k = ln 2( )50

with the other possibility, 12= e50k

giving the value of –k. Question 38 (c) Criteria Marks

• Provides correct solution 3 • Makes progress towards, evaluating the correct integral 2 • Writes a correct integral, or equivalent merit 1

Sample answer:

A = 2 4 ekx + e−kx( )dx0

50⌠

⌡⎮ which is

8k

ekx − e−kx( )050

= 400ln2

2 − 12

⎛⎝

⎞⎠

So the area is 600ln2

m2 ≈ 865 m2.


Page 1 of 2

Mathematics Advanced Sample HSC Mapping Grid Section I

Question Marks Content Syllabus outcomes

Targeted performance

bands

1 1 MA-F1 Working with Functions MA11-2 2-3

2 1 MA-C3 Applications of Differentiation MA12-6 2-3


4 1 MA-C2 Differential Calculus MA12-6 3-4

5# 1 MA-S2 Descriptive Statistics and Bivariate Data Analysis MA12-8 3-4


7 1 MA-S3 Random Variables MA12-8 4-5

8# 1 MA-M1 Modelling Financial Situations MA12-2 5-6

9# 1 MA-S3 Random Variables MA12-8 5-6


Section II



bands

11 2 MA-C1 Introduction to Differentiation MA11-5 2-4

12# 3 MA-T1 Trigonometry and Measure of Angles MA11-3 2-4

13 (a)# 1 MA-S1 Probability and Discrete Probability Distributions MA11-7 2-3

13 (b)# 1 MA-S1 Probability and Discrete Probability Distributions MA11-7 3-4

14 (a) 4 MA-C3 Applications of Differentiation MA12-3 2-5

14 (b) 2 MA-C3 Applications of Differentiation MA12-3 3-5

15# 3 MA-F1 Working with Functions MA11-1 3-5



18 (a) 2 MA-C4 Integral Calculus MA12-7 2-4

18 (b) 2 MA-C4 Integral Calculus MA12-7 3-5

19 3 MA-S1 Probability and Discrete Probability Distributions MA11-7 3-6

20 2 MA-C1 Introduction to Differentiation MA11-5 3-5

21# 3 MA-T1 Trigonometry and Measure of Angles MA11-3 3-5


Page 2 of 2



bands

22 (a)# 1 MA-F1 Working with Functions MA11-1 2-3

22 (b)# 3 MA-F1 Working with Functions MA11-1 3-5


24 2 MA-F2 Graphing Techniques MA12-1 3-5


26 3 MA-F2 Graphing Techniques MA-T1 Trigonometry and Measure of Angles

MA12-1; MA11-3 3-6

27 2 MA-T3 Trigonometric Functions and Graphs MA12-1; MA12-5 2-3


29 (a) 1 MA-S3 Random Variables MA12-8 2-3

29 (b) 2 MA-S3 Random Variables MA12-8 3-5

29 (c) 2 MA-S3 Random Variables MA12-8; MA12-10 3-5

30 3 MA-E1 Logarithms and Exponentials MA-C4 Integral Calculus

MA12-3; MA12-7 3-5

31 (a) 2 MA-S3 Random Variables MA12-8 3-5

31 (b) 2 MA-S3 Random Variables MA12-8 4-6

31 (c) 1 MA-S3 Random Variables MA12-8 5-6

32 3 MA-C3 Applications of Differentiation MA12-3 2-5

33 2 MA-C4 Integral Calculus MA12-7 4-6

34# 3 MA-M1 Modelling Financial Situations MA12-2 2-5

35 4 MA-C4 Integral Calculus MA12-7; MA12-10 3-6

36 (a) 1 MA-M1 Modelling Financial Situations MA12-4 2-3

36 (b) 2 MA-M1 Modelling Financial Situations MA12-4 3-5

36 (c) 2 MA-M1 Modelling Financial Situations MA12-4 4-6

36 (d) 1 MA-M1 Modelling Financial Situations MA12-4 5-6

37 (a)# 1 MA-S2 Descriptive Statistics and Bivariate Data Analysis MA12-8 3-4

37 (b)# 1 MA-S2 Descriptive Statistics and Bivariate Data Analysis MA12-8 3-4

38 (a) 1 MA-E1 Logarithms and Exponentials MA12-1 3-4

38 (b) 4 MA-E1 Logarithms and Exponentials MA12-1; MA12-10 3-6

38 (c) 3 MA-C4 Integral Calculus MA12-1; MA12-7; MA12-10

4-6

# - denotes a question that could be common to Mathematics Standard 2 HSC examination.

Documents

Mathematics Advanced Sample Examination