Cambridge Mathematics Advanced HSC Practice Examination solutions PDF

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Due to the implementation of the new syllabus, past paper resources that cover the new content is almost impossible to find. HOWEVER, I have collated 35+ past papers from different schools for their 2020 trials :)...

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Mathematics Advanced

General Instructions • Reading time – 10 minutes • Working time – 3 hours • Write using black pen • Draw diagrams in pencil • NESA-approved calculators may be used • The HSC Reference Sheet may be used • In Section II, show relevant mathematical reasoning and/or calculations

Total marks – 100 • Section I is worth 10 marks • Section II is worth 90 marks

Section I Attempt Questions 1–10 Allow about 15 minutes for this section Marks

1.

What will an investment of $50000   be worth in 5 years if it is invested at 6% p. a. with interest compounded monthly? A. $51262   .56 B. $66911   .28

C. $67442   .51

D. $71589   .42

2.

Which statement is true at the point ฀฀ on the curve ฀ ฀ = ฀฀(฀฀)? ฀฀

฀฀

.

A. ฀฀ ′ and ฀฀ ′′ are both positive

฀฀

B. ฀฀ ′ and ฀฀ ′′ are both negative

C. ฀฀ ′ is positive and ฀฀ ′′ is negative

D. ฀฀ ′ is negative and ฀฀ ′′ is positive 3.

The results of an English examination were normally distributed. The mean and

standard deviation were 63 and 16 respectively. If Peter’s ฀฀-score was −1.5, what

was his actual examination mark? A. 39 B. 42 C. 47 D. 57

© Cambridge University Press 2020

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4.

The line ℓ has equation ฀ ฀ + 2฀ ฀ + 3 = 0. If ฀฀ is the gradient of ℓ and ฀฀ is the acute angle between ℓ and the ฀฀ -axis, then which statement is true? A. ฀ ฀ > 0 and ฀ ฀ > 45°

B. ฀ ฀ > 0 and ฀ ฀ < 45°

C. ฀ ฀ < 0 and ฀ ฀ > 45°

D. ฀ ฀ < 0 and ฀ ฀ < 45°

5.

If ฀฀(฀฀) is a primitive of ฀฀(฀฀), which statement is true? ฀฀

A. ∫฀ ฀฀(฀฀) ฀฀฀฀ = ฀฀(฀฀) − ฀฀(฀฀ ) ฀ ฀฀

B. ∫฀ ฀฀(฀฀) ฀฀฀฀ = ฀฀(฀฀) − ฀฀(฀฀ ) ฀ ฀฀

C. ∫฀ ฀฀(฀฀) ฀฀฀฀ = ฀฀(฀฀) − ฀฀(฀฀ ) ฀ ฀฀

D. ∫฀ ฀฀(฀฀) ฀฀฀฀ = ฀฀(฀฀) − ฀฀(฀฀ ) ฀

6.

Which expression is equal to sin ฀ ฀ sec ฀ ฀ ? A. cosec ฀฀

B. cos ฀฀

C. tan ฀฀ D. cot ฀฀

7.

Consider the point ฀฀�฀฀ , ฀฀(฀฀)� on the curve ฀ ฀ = ฀฀(฀฀). If ฀฀ ′ (฀฀) = 0 and ฀฀′(฀฀) is positive on both sides of ฀฀, which statement best describes ฀฀? A. ฀฀ is a point of inflection

B. ฀฀ is a horizontal point of inflection

C. ฀฀ is a minimum turning point

D. ฀฀ is a maximum turning point

© Cambridge University Press 2020

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8.

The displacement-time graph of a particle moving in a straight line is shown below. The displacement is measured in metres and the time elapsed is in seconds. ฀฀ 15

0

฀฀

5

What is the velocity of the particle? A. 20 ms−1 B. 10 ms −1

C. 45 ms −1 D. 3 ms−1

9.

What is the gradient of the tangent to the curve ฀ ฀ = (฀฀ − 2)3 at its ฀฀-intercept? A. −8

B. −6

C. 12

D. 300

10.

Which function is a primitive of ฀฀฀฀ ฀฀ ? 2

A.

1

B.

1

2

2

฀฀ ฀฀

2

฀฀  ฀฀ ฀฀

2

C. (฀฀ 2 − 1)฀฀ ฀฀ D.

2 ฀฀ ฀฀ +1

2 −1

฀฀2 +1

© Cambridge University Press 2020

Page 4 of 27

Section II Attempt Questions 11–31 Allow about 2 hours and 45 minutes for this section

Question 11 (4 marks) The heights, in cm, of a squad of 10 basketballers are listed below in ascending order: 178, 180, 183, 183, 184, 185, 187, 187, 190, 198 (a)

What is the median height?

1

1 Median = (184 + 185) 2 [฀฀] = 184.5 (b)

Find ฀฀1 , ฀฀3 and hence the interquartile range.

1

฀฀1 = 183 and ฀฀3 = 187

so IQR = ฀฀3 − ฀฀1 = 4.

(c)

[฀฀]

Draw a box plot of the given data.

178

1

198

183 184.5 187

[฀฀] (d)

Determine whether there are any outliers amongst the heights using the criterion that an outlier is a score below ฀฀1 − 1.5 × IQR or above ฀฀3 + 1.5 × IQR.

1

฀฀1 − 1.5 × IQR = 183 − 1.5 × 4 = 177

฀฀3 + 1.5 × IQR = 187 + 1.5 × 4 = 193 So 198 is an outlier.

© Cambridge University Press 2020

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Question 12 (2 marks) Lucas has opened an investment account. He will invest $฀฀ at the end of each month

2

for 4 years at an interest rate of 0.5% per month, with interest compounded monthly. He wants the future value of the account to be $250000   . If the future value of $1 is $54.0978, find the value of ฀฀ correct to two decimal places. ฀ ฀ × 54.0978 = 250 000

so ฀฀ ≈ 4621.26.

[฀฀] [฀฀]

Question 13 (3 marks) Use the cosine rule to find the value of ฀฀ in the triangle below.

3√ 2

3

฀฀

45°

7 2

฀฀ 2 = 72 + �3√2� − 2(7)�3√2� cos 45° = 49 + 18 − 42√2 ×

= 67 − 42

1

√2

[฀฀]

[฀฀]

= 25

so ฀ ฀ = 5.

© Cambridge University Press 2020

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Question 14 (3 marks) Consider the triangle ฀฀฀฀฀฀ with ฀฀฀฀ = 20, ฀฀฀฀ = 15 and ∠฀฀฀฀฀฀ = 30°.

3

Find, correct to the nearest degree, the two possible sizes of ∠฀฀฀฀฀฀.

sin ∠฀฀฀฀฀฀ sin 30° [฀฀] = 15 20 20 × 0.5 sin ∠฀฀฀฀฀฀ = 15 2 [฀฀] sin ∠฀฀฀฀฀฀ = 3 2 2 ∠฀฀฀฀฀฀ = sin−1 or 180° − sin−1 3 3 ≈ 42° or 138°

© Cambridge University Press 2020

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Question 15 (4 marks) (a)

Solve the equation |฀฀ − 2| = 2.

฀฀ − 2 = ±2

฀ ฀ = 4 or ฀ ฀ = 0

(b)

1

[฀฀]

Sketch ฀ ฀ = |฀฀ − 2| and ฀ ฀ = 2 on the same number plane, clearly indicating

2

the coordinates of the points of intersection.

[฀฀] (one mark for the graphs, one mark for the intersection points)

(c)

Hence solve |฀฀ − 2| ≥ 2.

1 ฀฀ ≤ 0 or ฀฀ ≥ 4

© Cambridge University Press 2020

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Question 16 (3 marks) A new car depreciates in value by 15% p. a. How long will it take for the car’s value

3

to fall below 20% of its original value? Write your answer as a whole number of years. Let $฀฀ be the value of the new car, and let ฀฀ be the number of years it takes for the car to reach 20% of its original value. Then ฀฀(1 − 0.15)฀ ฀ = 0.2฀฀

[฀฀]

0.85฀ ฀ = 0.2

log10 0.85฀ ฀ = log10 0.2

฀ ฀ log10 0.85 = log10 0.2

[฀฀]

log10 0.2 ฀฀= log10 0.85 = 9.903 …

So it will take about 10 years.

© Cambridge University Press 2020

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Question 17 (3 marks) The table below shows the probability distribution of a discrete random variable ฀฀. ฀฀

0

฀฀(฀ ฀ = ฀฀)

(a)

2

฀฀

0.3

5

9

2฀฀

0.1

Determine the value of ฀฀.

1

0.3 + ฀ ฀ + 2฀ ฀ + 0.1 = 1

3฀ ฀ = 0.6

฀ ฀ = 0.2

(b)

[฀฀]

Calculate E(฀฀).

2

E(฀฀) = Σ ฀฀ × ฀฀(฀ ฀ = ฀฀)

= 0 × 0.3 + 2 × 0.2 + 5 × 0.4 + 9 × 0.1

= 3.3

© Cambridge University Press 2020

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Question 18 (4 marks) Find: (a)

(b)

(c)

฀฀

฀฀฀฀

(ln 5฀฀)

฀฀ (cos 5฀฀ ) ฀฀฀฀

฀฀

฀฀฀฀

1 1 ฀฀

1 −5 sin 5฀ ฀

฀฀

[฀฀]

((5฀฀)2 )

1 50฀฀

(d)

[฀฀]

[฀฀]

1

� �

฀฀฀฀ ฀฀

© Cambridge University Press 2020

1 ฀฀ (฀฀ −1) = −฀฀ −2 ฀฀฀฀ 1 =− 2 ฀฀

[฀฀]

Page 11 of 27

Question 19 (4 marks) Find: (a)

∫

2

√฀ ฀

฀฀฀฀

2 �

2

√฀ ฀

1

฀฀฀฀ = � 2฀฀2 ฀฀฀฀ −

1

= 2 × 2฀฀2 + ฀฀

= 4√฀ ฀ + ฀฀

(b)

[฀฀]

[฀฀]

1

∫ 7฀฀+5 ฀฀฀฀

2 �

1 1 7 ฀฀฀฀ = � ฀฀฀฀ 7฀ ฀ + 5 7 7฀ ฀ + 5

1 = ln|7฀ ฀ + 5| + ฀ ฀ 7

© Cambridge University Press 2020

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Question 20 (4 marks) Consider the function ฀ ฀ = 2 + 2 sin 4฀฀. (a)

0

Sketch the graph of the function for 0 ≤ ฀฀ ≤ ฀฀.

฀฀ 4

฀฀ 2

2

฀฀

3฀฀ 4

[฀฀] (one mark for shape, one mark for position)

(b)

Use your sketch to solve the equation sin 4฀ ฀ = 0 for 0 ≤ ฀฀ ≤ ฀฀. When sin 4฀ ฀ = 0 , ฀ ฀ = 2.

2

[฀฀]

From the graph, the line ฀ ฀ = 2 intersects ฀ ฀ = 2 + 2 sin 4฀฀ five times in the domain 0 ≤ ฀฀ ≤ ฀฀ .

฀฀ ฀฀ 3฀฀ , ฀฀. These intersection points are at ฀ ฀ = 0,  ,  , 4 2 4

© Cambridge University Press 2020

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Question 21 (3 marks) It is found that the waiting times for patients to see a doctor at a medical centre are

3

normally distributed with a mean of 17.6 minutes and a standard deviation of 5.2 minutes. Find, correct to 4 decimal places, the probability that a patient will wait more than 15 minutes. ฀฀ − ฀฀ ฀ ฀ 15 − 17.6 = 5.2

฀฀=

= −0.5

[฀฀]

฀฀(฀ ฀ > −0.5) = 1 − ฀฀(฀ ฀ < −0.5) = 1 − 0.3085 = 0.6915

© Cambridge University Press 2020

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Question 22 (4 marks)

Consider the parabola ฀฀ with equation ฀ ฀ = 2฀฀ − ฀฀ 2 and the line ℓ with equation ฀ ฀ = −฀฀. (a)

By solving simultaneously, find the points of intersection of ฀฀ and ℓ.

1

2฀฀ − ฀฀ 2 = −฀฀ ฀฀ 2 − 3฀ ฀ = 0

฀฀(฀฀ − 3) = 0

฀ ฀ = 0 or 3

So the intersection points are (0, 0) and (3, −3).

(b)

Find the area of the region enclosed by ฀฀ and ℓ.

[฀฀]

3

฀฀ (0, 0) ℓ

(3, −3)

3

Area = � �2฀฀ − ฀฀ 2 − (−฀฀)� ฀฀฀฀ 0

[฀฀]

3

= � (3฀฀ − ฀฀ 2 ฀)฀฀฀ 0

3

3฀฀ 2 ฀฀ 3 =� − � 2 3 0

[฀฀]

=�

27 − 9� − (0) 2 9 = square units 2

© Cambridge University Press 2020

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Question 23 (6 marks)

Suppose we have an arithmetic series with ฀฀6 = −10 and ฀฀21 = 50, and an infinite geometric 3

series with ฀฀1 = 32 and common ratio 4 . (a)

Find the limiting sum of the geometric series.

1

฀฀ 1 − ฀฀ 32 = 3 1−4

฀฀∞ =

= 128

(b)

[฀฀]

Show that the sum of the first ฀฀ terms of the arithmetic series is ฀฀฀ ฀ = 2฀฀2 − 32฀฀. ฀฀6 = ฀ ฀ + 5฀ ฀ = −10 … (1)

฀฀21 = ฀ ฀ + 20฀ ฀ = 50 … (2)

[฀฀]

(2) − (1):

15฀ ฀ = 60

฀ ฀ = 4 and ฀ ฀ = −30. ฀฀ ฀฀฀ ฀ = (2฀ ฀ + (฀฀ − 1)฀฀) 2 ฀฀ = �−60 + 4(฀฀ − 1)� 2 ฀฀ = (4฀฀ − 64) 2 = 2฀฀2 − 32฀฀

(c)

3

[฀฀]

[฀฀]

Find the value of ฀฀ for which the sum of the arithmetic series is three times

2

the limiting sum of the geometric series.

2฀฀2 − 32฀ ฀ = 384

฀฀2 − 16฀฀ − 192 = 0

(฀฀ − 24)(฀ ฀ + 8) = 0

[฀฀]

฀ ฀ = 24 since ฀฀ is positive.

© Cambridge University Press 2020

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Questions 24 (5 marks) In a bag there are 4 red marbles, 3 green marbles and 3 yellow marbles. Three marbles are chosen at random without replacement. Determine the probability that: (a)

all three marbles are yellow

1

3 2 1 ฀฀(฀฀฀฀฀฀) = × × 10 9 8 1 [฀฀] = 120

(b)

at least one red marble is chosen

2

1 − ฀฀(none red) = 1 −

6 5 4 × × 10 9 8 1 =1− 6 5 [฀฀] = 6

(c)

[฀฀]

one marble of each colour is chosen.

2

4 3 3 ฀฀(฀฀฀฀฀฀ in that order) = × × 10 9 8 1 [฀฀] = 20

There are 3! = 6 arrangements of ฀฀฀฀฀฀, each with the same chance.

1 So ฀฀(฀฀฀฀฀฀ in any order) = 6 × 20 3 [฀฀] =  . 10

© Cambridge University Press 2020

Page 17 of 27

Question 25 (6 marks)

The random variable ฀฀ has probability density function

1 ฀฀(฀฀) = 20 � ฀฀ for 3 ≤ ฀฀ ≤ 7 0 otherwise.

Find, correct to 2 decimal places: (a)

฀฀(3.6 ≤ ฀ ฀ ≤ 5.6)

2

฀฀(3.6 ≤ ฀฀ ≤ 5.6) = �

5.6

3.6

1 ฀฀ ฀฀฀฀ 20

[฀฀]

1 [฀฀ 2 ]5.6 3.6 40 1 = (5.62 − 3.62 ) 40 [฀฀] = 0.46 =

(b)

E(฀฀)

2

7

E(฀฀) = � ฀฀ ฀฀(฀฀) ฀฀฀฀ 3

=�

3

7

1 ฀฀ 2 ฀฀฀฀ 20

[฀฀]

1 [฀฀ 3 ]73 60 1 (73 − 33 ) = 60 79 [฀฀] = 15 =

≑ 5.27

© Cambridge University Press 2020

Page 18 of 27

(c)

Var(฀฀)

2 2

Var(฀฀) = E(฀฀ 2 ) − �E(฀฀)�

79 2 = � ฀฀ ฀฀(฀฀) ฀฀฀฀− � � 15 3 7

=� =

3

2

[฀฀]

792 1 ฀฀ 3 ฀฀฀฀ − � � 15 20

7

1 79 2 7 4 [฀฀ ]3− � � 15 80

79 2 1 (74 − 34 ) − � � 15 80 284 [฀฀] = 225 =

≑ 1.26

Question 26 (3 marks) Of all the First-Year students at one of Sydney’s major universities, 80% are local

3

students of whom 10% hold a scholarship. Of the non-local First-Year students, 75% are scholarship holders. If a randomly chosen student holds a scholarship, what is the probability that he or she is a non-local student?

Let ฀฀ be the event ′ ฀฀ℎ฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀ ฀฀฀฀ ฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀฀฀.′

Let ฀฀ be the event ′ ฀฀ℎ฀฀ ฀฀฀฀฀฀฀฀฀฀฀฀฀฀ ℎ฀฀฀฀฀฀฀฀ ฀฀ ฀฀฀฀ℎ฀฀฀฀฀฀฀฀฀฀ℎ฀฀฀฀.′ ฀฀(฀฀|฀฀) =

฀฀(฀฀ ∩ ฀฀) ฀฀(฀฀)

0.2 × 0.75 0.8 × 0.1 + 0.2 × 0.75 15 [฀฀] = 23

=

© Cambridge University Press 2020

[฀฀] (one each for numerator & denominator)

Page 19 of 27

Question 27 (3 marks) Suppose that ฀฀ and ฀฀ are independent events.

3

Given that ฀฀(฀฀) = 2 × ฀฀(฀฀) and that ฀฀(฀฀ ∪ ฀฀) = 0.52, find ฀฀(฀฀). ฀฀(฀฀ ∪ ฀฀) = ฀฀(฀฀) + ฀฀(฀฀) − ฀฀ (฀฀ ∩ ฀฀ )

Since ฀฀ and ฀฀ are independent, ฀฀(฀฀ ∩ ฀฀) = ฀฀(฀฀) × ฀฀(฀฀).

So ฀฀(฀฀ ∪ ฀฀) = 2 × ฀฀(฀฀) + ฀฀(฀฀) − 2 × ฀฀(฀฀) × ฀฀(฀฀) Let ฀฀(฀฀) = ฀฀.

[฀฀]

Then 0.52 = 3฀฀ − 2฀฀ 2

13 2฀฀ 2 − 3฀ ฀ + = 0 25

50฀฀ 2 − 75฀ ฀ + 13 = 0

(10฀฀ − 13)(5฀฀ − 1) = 0

[฀฀]

1 ฀ ฀ = ฀฀(฀฀) = = 0.2 (฀฀ ≠ 1.3 since ฀฀ is a probability) 5 so ฀฀(฀฀) = 2 × ฀฀(฀฀) = 0.4

© Cambridge University Press 2020

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Question 28 (4 marks) Water is being pumped from a reservoir. The volume ฀฀ megalitres of water in the reservoir

after ฀฀ hours is decreasing at the rate of (360 − 30฀฀) ML/hour. (a)

Find ฀฀, the number of hours for which the pump is operating.

1

฀฀฀฀ = −(360 − 30฀฀) = 30฀฀ − 360 ฀฀฀฀ ฀฀฀฀ The pump stops when = 0. ฀฀฀฀ That is, when 30฀฀ − 360 = 0. That is, after ฀ ฀ = 12 hours.

(b)

[฀฀]

Given that the volume of water remaining in the reservoir after the pump stops is 2400 ML, derive a formula for the volume ฀฀ of water in the reservoir after

3

฀฀ hours of pumping, where 0 ≤ ฀฀ ≤ ฀฀.

฀ ฀ = �(30฀฀ − 360) ฀฀฀฀

= 15฀฀ 2 − 360฀ ฀ + ฀฀

When ฀ ฀ = 12, ฀ ฀ = 2400

so 2400 = 15(12)2 − 360(12) + ฀฀ so ฀ ฀ = 2400 − 2160 + 4320

[฀฀]

[฀฀]

= 4560

so ฀ ฀ = 15฀฀ 2 − 360฀ ฀ + 4560

© Cambridge University Press 2020

[฀฀]

Page 21 of 27

Question 29 (6 marks) Spiro is a property developer. He is going to borrow $5 million from the Acme State Bank at an interest rate of 3.6% pa. He intends to make an annual repayment of $฀฀ at the end of each year for 10 years. Interest is calculated at the beginning of each year based on the amount still owing and then added to the account. Let  ฀$฀฀฀ be the amount owing after the ฀฀th repayment. (a)

Write down expressions for ฀฀1 and ฀฀2 , and hence show that ฀฀3 = 5 × 106 × 1.0363 − ฀฀ (1 + 1.036 + 1.0362 ).

2

฀฀1 = 5 × 106 × 1.036 − ฀฀

฀฀2 = ฀฀1 × 1.036 − ฀฀

= 5 × 106 × 1.0362 − 1.036฀฀ − ฀฀

[฀฀]

฀฀3 = ฀฀2 × 1.036 − ฀฀

= 5 × 106 × 1.0363 − 1.0362 ฀฀ − 1.036฀฀ − ฀฀

= 5 × 106 × 1.0363 − ฀฀(1 + 1.036 + 1.0362 )

(b)

[฀฀]

Hence find the annual repayment, correct to the nearest dollar.

3

฀฀10 = 0

so 5 × 106 × 1.03610 − ฀฀(1 + 1.036 + 1.0362 + ⋯ + 1.0369 ) = 0

5 × 106 × 1.03610 [฀฀] so ฀ ฀ =  . 1 + 1.036 + 1.0362 + ⋯ + 1.0369 The denominaor is a geometric series with ฀ ฀ = 1, ฀ ฀ = 1.036, ฀ ฀ = 10 and its sum is ฀฀10 =

So ฀ ฀ =

฀฀(฀฀ 10 − 1) 1.03610 − 1 = ฀฀ − 1 0.036

5 × 106 × 1.03610 × 0.036 ≈ 604241   1.03610 − 1

 .

so the annual repayment is $604241   , to the nearest dollar.

© Cambridge University Press 2020

Page 22 of 27

(c)

Calculate, correct to the nearest dollar, the amount owing after 5 years.

1

฀฀5 = 5 × 106 × 1.0365 − ฀฀ (1 + 1.036 + 1.0362 + 1.0363 + 1.0364 ) =5×

106

×

≈ 2 720470  

1.0365

1.0365 − 1 − ฀฀ × 0.036

[฀฀]

so Spiro owes $2 720470   , to the nearest dollar, after 5 years.

Question 30 (10 marks) Consider the function ฀฀(฀฀) defined by ฀฀(฀฀) = 8฀฀  ฀฀ −0.4฀฀ . (a)

Show that ฀฀ ′ (฀฀ ) = (8 − 3.2฀฀ ฀)฀ −0.4฀฀ .

1

By the product rule,

฀฀ ′ (฀฀) = 8฀฀ −0.4฀฀ + 8฀฀(−0.4฀฀ −0.4฀฀) = 8฀฀ −0.4฀฀ − 3.2฀฀  ฀฀ −0.4฀฀ = (8 − 3.2฀฀)  ฀฀ −0.4฀฀

(b)

[฀฀]

Hence find the coordinates of the stationary point and determine its nature.

3

Stationary points occur when ฀฀ ′ (฀฀) = 0, that is, when (8 − 3.2฀฀)  ฀฀ −0.4฀฀ = 0.

8 − 3.2฀ ฀ = 0  ฀(฀ −0.4฀฀ > 0 for all real ฀฀) 8 ฀ ฀ = = 2.5 3.2

When ฀ ฀ = 2.5, ฀ ฀ = 8(2.5)(฀฀ −1) = 20฀฀ −1. So (2.5, 20฀฀ −1) is a stationary point.

[฀฀]

By the product rule again,

฀฀ ′′ (฀฀ ) = −3.2  ฀฀ −0.4฀฀ − 0.4(8 − 3.2฀฀ ฀)฀ −0.4฀฀ = 1.28฀฀  ฀฀ −0.4฀฀ − 6.4  ฀฀ −0...