4U Cambridge 2020 Sols - Extension 2 Mathematics Paper PDF

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Extension 2 Mathematics Paper ...

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CambridgeMATHS Practice HSC Examination

Mathematics Extension 2

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

CambridgeMATHS

Practice HSC Examination

Extension 2

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

What is the negation of the statement: If a triangle has two equal sides, then it has two equal angles. A. If a triangle does not have two equal sides, then it does not have two equal angles. B. If a triangle does not have two equal angles, then it does not have two equal sides. C. A triangle has two equal sides and it does not have two equal angles. D. A triangle has two equal angles and it does not have two equal sides.

2.

Which function is a primitive of A. ln|2฀฀ − 1|

2฀฀+1 ? 2฀฀−1

B. ฀ ฀ + ln|2฀฀ − 1|

C. ฀ ฀ + 2 ln|2฀฀ − 1| D. 2฀ ฀ + ln|2฀฀ − 1|

3.

Which number is a fourth root of ฀฀? ⁄ A. ฀฀ ฀฀฀฀

⁄ B. ฀฀ 3฀฀฀฀ ⁄ C. ฀฀ ฀฀฀฀

8

8

4

⁄ D. ฀฀ 3฀฀฀฀

4

Page 2 of 29

CambridgeMATHS

4.

Practice HSC Examination

Extension 2

2 and passes through the point (3, 2). Which of the following is A line has gradient − 3 a vector equation of the line?

฀฀ 3 � , where ฀฀ ∈ ℝ A. � ฀฀� = � � + ฀฀ �−2 3 2 ฀฀ 3 3 B. � ฀฀� = � � + ฀฀ � � , where ฀฀ ∈ ℝ −2 2 ฀฀ −2 0 C. � ฀฀� = � � + ฀฀ � � , where ฀฀ ∈ ℝ 4 3 ฀฀ 0� + ฀฀ � 3 � , where ฀฀ ∈ ℝ D. � ฀฀� = � 4 −2

5.

The acceleration of a particle moving in a straight line with velocity ฀฀ is given by ฀฀󰇘 = ฀฀ 2 . Initially ฀ ฀ = 1. What is ฀฀ as a function of ฀฀? A. ฀ ฀ = 1 − ฀฀

B. ฀ ฀ = ln|1 − ฀฀| C. ฀ ฀ =

฀฀

1−฀฀

D. ฀ ฀ =

1

1−฀฀

6.

Suppose that ฀฀ is an integer and ฀ ฀ > 2. What is the largest number that is always a divisor of ฀฀3 − ฀฀? A. 2 B. 3 C. 6 D. 12

Page 3 of 29

CambridgeMATHS

7.

Practice HSC Examination

Extension 2

In the diagram below, the points ฀฀ and ฀฀ represent the complex numbers ฀฀1 and ฀฀2 respectively. In which quadrant is the point ฀฀ representing ฀฀2 − ฀฀1 ? Im

฀฀(฀฀1 )

฀฀(฀฀2 ) Re

A. 1st quadrant B. 2nd quadrant C. 3rd quadrant D. 4th quadrant

8.

Into which definite integral is ∫0 8 2฀฀ ฀฀฀฀ 1+฀฀

A. ∫0

B. ∫0

4 ฀฀

1+฀฀

8

1

1+√2฀฀

฀฀฀฀

฀฀฀฀transformed by the substitution ฀ ฀ = √2฀฀?

8 2 ฀฀฀฀ 1+฀฀

C. ∫0

D. ∫0

4

9.

1 ฀฀฀฀ 1+฀฀

The line ℓ1 has vector equation ฀฀฀ ฀ = ฀ ฀ + ฀฀(฀฀ − ฀฀) and the line ℓ2 has vector equation ฀฀฀ ฀ = (3฀ ฀ + 2฀฀ − ฀฀) + ฀฀(2฀ ฀ + ฀฀).

Which of the following statements is correct? A. ℓ1 and ℓ2 are parallel

B. ℓ1 and ℓ2 are perpendicular

C. ℓ1 and ℓ2 intersect at a point D. ℓ1 and ℓ2 are skew

Page 4 of 29

CambridgeMATHS

10.

Practice HSC Examination

Extension 2

Which of these inequalities is false? (Do not attempt to evaluate the integrals.) A. B. C. D.

∫1

2 1

฀฀

∫฀฀3   6

1+฀฀

sin ฀฀ ฀

฀฀฀฀< ∫1

฀

21

฀฀

฀฀

< ∫฀฀3   ฀฀฀฀ 6

฀฀฀฀

1

฀฀

฀฀฀฀

∫1 ฀฀ −฀฀ ฀฀฀฀ < ∫0 ฀฀ −฀฀ ฀฀฀฀ 2

2

1

2

∫04  tan2 ฀฀ ฀฀฀฀< ∫04  tan3 ฀฀ ฀฀฀฀ ฀฀

฀฀

Page 5 of 29

CambridgeMATHS

Practice HSC Examination

Extension 2

Section II Attempt Questions 11–16 Allow about 2 hours and 45 minutes for this section

Question 11 (15 marks)

(a)

Express

23−14฀฀ 3−4฀฀

in the form ฀ ฀ + ฀฀฀฀, where ฀฀, ฀฀ ∈ ℝ.

3 + 4฀฀ 23 − 14฀฀ × [฀฀] 3 − 4฀ ฀ 3 + 4฀ ฀ 69 + 92฀฀ − 42฀ ฀ + 56 = 9 + 16 =

125 + 50฀฀ 25

= 5 + 2฀฀

(b)

2

[฀฀]

Find the square roots of −16 + 30฀฀.

2

Let (฀ ฀ + ฀฀฀฀)2 = −16 + 30฀฀.

Then (฀฀ 2 − ฀฀ 2 ) + 2฀฀฀฀฀฀ = −16 + 30฀฀. ฀฀ 2 − ฀฀ 2 = −16 and ฀฀฀฀ = 15.

Equating real and imaginary parts,

By inspection, ฀ ฀ = 3 and ฀ ฀ = 5 or ฀ ฀ = −3 and ฀ ฀ = −5. [฀฀]

So the square roots are 3 + 5฀฀ or − 3 − 5฀฀ .

Page 6 of 29

[฀฀]

CambridgeMATHS

(c)

Practice HSC Examination

Extension 2

Let ฀ ฀ = −√3 + ฀฀. (i)

Write ฀฀ in modulus-argument form. ฀ ฀ = 2 cis

(ii)

1

5฀฀ 6

[฀฀]

Hence find ฀฀ 9 in Cartesian form.

2

9

5฀฀ � 6 45฀฀ = 29 cis [฀฀] 6 15฀฀ = 512 cis 2 3฀฀ 3฀฀ = 512 �cos + ฀ ฀ sin � 2 2

฀฀ 9 = �2 cis

= −512฀฀ (d)

Find ∫ √10฀฀−฀฀2 ฀฀฀฀.

[฀฀]

1

�

√10฀฀ − 1

2 ฀฀ 2

1 ฀฀฀฀ ฀฀฀฀ = � 2 �25 − (฀฀ − 10฀ ฀ + 25) =�

�25 − (฀฀ − 5)2

= sin−1

1

฀฀ − 5 + ฀฀ 5

Page 7 of 29

฀฀฀฀

[฀฀]

[฀฀]

CambridgeMATHS

(e)

(i)

Practice HSC Examination

Find the values of ฀฀ and ฀฀ such that

2฀฀2 +3 ฀฀2 +฀฀

Extension 2

฀฀

฀฀

= 2 +฀ ฀ +฀฀+1

2฀฀ 2 + 3 = 2฀฀(฀ ฀ + 1) + ฀฀(฀ ฀ + 1) + ฀฀฀฀

Substituting ฀ ฀ = −1,

.

2

[฀฀]

2 + 3 = ฀฀(−1)

฀ ฀ = −5.

Substituting ฀ ฀ = 0, 3 = ฀฀.

So ฀ ฀ = 3 and ฀ ฀ = −5.

(ii)

Hence find ∫

2฀฀2 +3 ฀฀2 +฀฀

[฀฀]

฀฀฀฀.

1

3 5 2฀฀ 2 + 3 ฀฀฀฀ = � � 2 + − � ฀฀฀฀ � ฀฀ 2 + ฀ ฀ ฀฀ ฀฀+1

= 2฀ ฀ + 3 ln|฀฀| − 5 ln|฀ ฀ + 1| + ฀ ฀

(f)

Find the exact value of ∫1 ฀฀ 4 ln ฀฀ ฀฀฀฀.

[฀฀]

฀฀

฀฀

� ฀฀ 4 ln ฀฀ ฀฀฀฀ 1

฀฀ ฀฀ 1 1 = � ฀฀ 5 ln ฀฀� − � ฀฀ 4 ฀฀฀฀ 5 1 1 5

=

1 5 1 1 5 ฀฀ ฀฀ − � ฀฀ � 5 5 5 1

฀ ฀ = ln ฀฀ ฀฀ ′ =

฀฀

= [฀฀฀฀]1฀ ฀ − � ฀฀฀฀ ′ 1

3

[฀฀]

1 5 1 5 ฀฀ − (฀฀ − 1) 5 25 1 (4฀฀ 5 + 1) [฀฀] = 25

=

End of Question 11

Page 8 of 29

1 ฀฀

฀฀ ′ = ฀฀ 4

1 ฀ ฀ = ฀฀ 5 5

[฀฀]

CambridgeMATHS

Practice HSC Examination

Extension 2

Question 12 (15 marks) (a)

The points ฀฀(1, −2, 3) and ฀฀(−5, 4, −1) lie on the line ℓ. (i)

Find a vector equation for ℓ.

2

1 −5 ���� = � 4 � − � −2� A direction vector of ℓ is �฀฀฀฀ −1 3 −6 [฀฀] =� 6� −4 so a vector equation for ℓ is

1 −6 ฀ ฀ = −2 � � + ฀฀ � 6 � , where ฀฀ ∈ ℝ. −4 3

(ii)

Does the point ฀฀(43, −44, 29) lie on ℓ ?

1

43 If ฀ ฀ =−44 � �, 29 −6 43 1 then � −44 � = �−2 � + ฀฀ � 6 � , 3 29 −4 so 1 − 6฀ ฀ = 43,

−2 + 6฀ ฀ = −44

and 3 − 4฀ ฀ = 29.

The first two equations are satisfied by ฀ ฀ = −7,

but the third equation isn′ t.

So the point ฀฀ doesn′ t lie on ℓ.

Page 9 of 29

[฀฀]

[฀฀]

CambridgeMATHS

(b)

Practice HSC Examination

Extension 2

It is known that 5 + 6฀฀ is a zero of the polynomial ฀฀(฀฀) = 2฀฀ 3 − 19฀฀ 2 + 112฀ ฀ + ฀฀, where ฀฀ is real. (i) Find the other two zeroes of ฀฀(฀฀). 2 5 + 6฀ ฀ = 5 − 6฀฀ is a zero of ฀฀(฀฀),

since the coefficients of ฀฀(฀฀) are real. Let the remaining zero be ฀฀ .

19 . 2 19 So (5 + 6฀฀) + (5 − 6฀฀) + ฀ ฀ = 2 19 10 + ฀ ฀ = 2 1 ฀฀=−. 2

[฀฀]

Now, the sum of the zeroes is

[฀฀]

1 So the other two zeroes are 5 − 6฀฀ and − . 2

(ii)

Find the value of ฀฀.

2

฀฀ . 2 1 ฀฀ So − (5 + 6฀฀)(5 − 6฀฀) = − 2 2 ฀ ฀ = (5 + 6฀฀)(5 − 6฀฀) The product of the zeroes is −

= 25 + 36 = 61.

[฀฀]

Page 10 of 29

[฀฀]

CambridgeMATHS

(c)

(i)

Practice HSC Examination

Extension 2

Use the result cos 3฀ ฀ + ฀ ฀ sin 3฀ ฀ = (cos ฀ ฀ + ฀ ฀ sin ฀฀)3 to find expressions for sin 3฀฀ and cos 3฀฀ in terms of sin ฀฀ and cos ฀฀.

2

cos 3฀ ฀ + ฀ ฀ sin 3฀ ฀ = cos 3 ฀ ฀ + 3 cos 2 ฀ ฀ (฀ ฀ sin ฀฀) + 3 cos ฀ ฀ (฀ ฀ sin ฀฀)2 + (฀ ฀ sin ฀฀)3 [฀฀] = cos 3 ฀ ฀ + 3฀ ฀ cos 2 ฀ ฀ sin ฀฀ − 3 cos ฀ ฀ sin2 ฀฀ − ฀฀ sin3 ฀฀

cos 3฀ ฀ = cos 3 ฀฀ − 3 cos ฀ ฀ sin2 ฀฀

Equating real and imaginary parts: sin 3฀ ฀ = 3 cos 2 ฀ ฀ sin ฀฀ − sin3 ฀ ฀ (ii)

[฀฀]

Hence show that tan 3฀ ฀ =

3 tan ฀฀−tan3 ฀฀ 1−3 tan2 ฀ ฀

.

1

sin 3฀฀ tan 3฀ ฀ = cos 3฀฀ =

3 cos 2 ฀ ฀ sin ฀฀ − sin3 ฀฀ cos 3 ฀฀ − 3 cos ฀ ฀ sin2 ฀฀

Dividing numerator and denominator by cos3 ฀ ฀ : 3 sin ฀฀ sin3 ฀฀ − cos ฀ ฀ cos 3 ฀฀ tan 3฀ ฀ = 3 cos ฀฀ 3 sin2 ฀฀ − cos 3 ฀ ฀ cos 2 ฀ ฀ =

3 tan ฀฀ − tan3 ฀฀ 1 − 3 tan2 ฀ ฀

Page 11 of 29

[฀฀]

CambridgeMATHS

(iii)

Practice HSC Examination

Extension 2

Deduce that ฀ ฀ = tan is a root of the equation ฀฀ 3 − 3฀฀ 2 − 3฀ ฀ + 1 = 0. 12 ฀฀

2

฀฀ Let ฀ ฀ =12 in the identity in (ii): ฀฀ 3 ฀฀ ฀฀ 3 tan 12 − tan 12 tan = [฀฀] ฀฀ 4 1 − 3 tan2 12

฀฀ ฀฀ ฀฀ ฀฀ = 3 tan − tan3 �since tan = 1� 12 12 12 4 ฀฀ ฀฀ ฀฀ [฀฀] +1=0 so tan3 − 3 tan2 − 3 tan 12 12 12 ฀฀ so ฀ ฀ = tan satisfies the equation ฀฀ 3 − 3฀฀ 2 − 3฀ ฀ + 1 = 0. 12

so 1 − 3 tan2

Use the substitution ฀ ฀ = √2 sin ฀฀ to show that0∫

(d) �

1

0

฀฀ 2

√2 − ฀฀ 2

=�

0

=�

0

฀฀ 4

1

฀฀฀฀

2 sin ฀฀ 2

× √2 cos ฀฀ ฀฀฀฀ √2 − 2 sin2 ฀฀

฀฀ 2 42 sin ฀฀ ฀฀ 4

√2 cos ฀฀

× √2 cos ฀฀ ฀฀฀฀

= � 2 sin2 ฀฀ ฀฀฀฀ 0

[฀฀]

[฀฀]

฀฀ 4

= � (1 − cos 2฀฀)฀฀฀฀ 0

฀฀ 1 4 sin 2฀฀] = [฀฀ − 0 2 ฀฀ 1 = − ×1 4 2 1 [฀฀] = (฀฀ − 2) 4

End of Question 12

Page 12 of 29

฀฀2

√2−฀฀2

฀฀฀฀ =4 (฀฀ − 2). 1

฀ ฀ = √2 sin ฀฀

฀฀฀฀ = √2 cos ฀฀ ฀฀฀฀ ฀฀ 0 1 ฀฀ 0 ฀฀ 4

3

CambridgeMATHS

Practice HSC Examination

Extension 2

Question 13 (15 marks)

(a)

−3 −5 4 −9�� = √14. Two spheres have vector equations �฀฀ − � �� = 2√14 and �฀฀ − �7 Prove that the spheres touch each other at a single point. 10

3

The centres of the spheres are ฀฀1 (−3, −5, 10) and ฀฀2 (−9, 4, 7), and the radii are ฀฀1 = 2√14 and ฀฀2 = √14.

We must show that the distance between the centres ฀฀1 ฀฀2 is equal to the sum of the radii ฀฀1 + ฀฀2 .

฀฀1 ฀฀2 = �(−9 + 3)2 + (4 + 5)2 + (7 − 10)2

[฀฀]

= √36 + 81 + 9

[฀฀]

= √126

= ฀฀1 + ฀฀2 , as required. = 3√14

(b)

[฀฀]

Prove by contradiction that log 5 13 is an irrational number. Assume that log 5 13 is a rational number. ฀฀ Then log 5 13 can be written in the form , ฀฀ ฀฀ where ฀฀, ฀฀ ∈ ℤ, ฀฀ ≠ 0 and is in lowest terms. ฀฀ ฀฀ So log5 13 = ฀฀ ฀฀

5฀ ฀ = 13

฀฀ ฀ ฀

�5 ฀฀ �

= 13฀฀

5฀ ฀ = 13฀฀

[฀฀]

which is impossible due to the uniqueness of the prime factorisation of a positive integer.

So the assumption that log5 13 is rational is false,

so log 5 13 is irrational.

[฀฀]

Page 13 of 29

3

[฀฀]

CambridgeMATHS

(c)

Practice HSC Examination

Extension 2

Suppose that ฀฀ and ฀ ฀ + 1 are positive integers, neither of which is divisible by 3. Prove that ฀฀3 + (฀ ฀ + 1)3 is divisible by 9. Since neither ฀฀ nor ฀ ฀ + 1 is divisible by 3, they must be

of the form 3฀ ฀ + 1 and 3฀ ฀ + 2 respectively, where ฀฀ ∈ {0, 1, 2, … }.

So ฀฀3 + (฀ ฀ + 1)3 = (3฀ ฀ + 1)3 + (3฀ ฀ + 2)3

3

[฀฀]

= 27฀฀ 3 + 27฀฀ 2 + 9฀ ฀ + 1 + 27฀฀ 3 + 54฀฀ 2 + 36฀ ฀ + 8 = 54฀฀ 3 + 81฀฀ 2 + 45฀ ฀ + 9 = 9(6฀฀ 3 + 9฀฀ 2 + 5฀ ฀ + 1)

which is divisible by 9, since ฀฀ ∈ ℤ and so 6฀฀ 3 + 9฀฀ 2 + 5฀ ฀ + 1 ∈ ℤ. (d)

[฀฀]

[฀฀]

harmonic motion about the origin. The motion has amplitude 0.2m   and period

A particle is projected from the origin with positive velocity and moves in simple

4 seconds. Find, leaving answers in exact form: (i)

the initial speed of the particle, ฀฀=

2฀฀ =4 ฀ ฀ ฀฀ so ฀ ฀ = . 2

[฀฀]

The speed is at its maximum when ฀ ฀ = 0, so the initial speed is ฀฀ ฀฀฀฀ = × 0.2 2 ฀฀ [฀฀] ms−1. = 10

Page 14 of 29

2

CambridgeMATHS

(ii)

Practice HSC Examination

Extension 2

the speed of the particle after 1.5 seconds,

2

The displacement is of the form ฀ ฀ = ฀ ฀ sin ฀฀฀฀ ,

since the particle is initially at ฀ ฀ = 0 with ฀ ฀ > 0.

Then the velocity is given by ฀ ฀ = ฀฀󰇗 = ฀฀฀฀ cos ฀฀฀฀ . So when ฀ ฀ = 1.5,

฀฀ ฀฀ 3 × cos � × � 2 2 2 ฀฀ 3฀฀ = × cos 4 10 1 ฀฀ ×− = 10 √2 ฀฀ . =− 10√2

฀ ฀ = 0.2 ×

[฀฀]

฀฀√2 ฀฀ ms−1 . or So the speed when ฀ ฀ = 1.5 is 20 10√2 (iii)

[฀฀]

when the particle first reaches a point 0.1m   from the origin. ฀ ฀ = ฀ ฀ sin ฀฀฀฀

When ฀ ฀ = 0.1,

0.1 = 0.2 sin

sin

฀฀ 1 ฀฀= 2 2

฀฀ ฀฀ 2

[฀฀]

The first positive solution occurs when ฀฀ ฀฀ ฀฀= 2 6 1 [฀฀] ฀฀=. 3 So the particle is 0.1m from the origin for the first time after 

1 seconds.   3

End of Question 13 Page 15 of 29

2

CambridgeMATHS

Practice HSC Examination

Extension 2

Question 14 (15 marks) (a)

Sketch the subset of the complex plane determined by the relation 1 11 += . ฀ ฀  ฀  ฀ 2

Observe that ฀฀ ≠ 0. Let ฀ ฀ = ฀ ฀ + ฀฀฀฀.

1 1 1 + = ฀ ฀ + ฀฀฀฀ ฀฀ − ฀฀฀฀ 2

2(฀฀ − ฀฀฀฀) + 2(฀ ฀ + ฀฀฀฀) = (฀ ฀ + ฀฀฀฀)(฀฀ − ฀฀฀฀) 4฀ ฀ = ฀฀ 2 + ฀฀ 2

(฀฀ 2 − 4฀ ฀ + 4) + ฀฀ 2 = 4 [฀฀]

(฀฀ − 2)2 + ฀฀ 2 = 4

(and (฀฀, ฀฀) ≠ (0, 0). )

0

2

[฀฀]

4

[฀฀] for sketch (must have open circle at origin)

Page 16 of 29

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CambridgeMATHS

(b)

Practice HSC Examination

1 Let ฀฀ and ฀฀ be the points on the line ฀ ฀ = �

Extension 2

3 −2 � + ฀฀ � � corresponding to ฀ ฀ = 1 and −1(1, 0, 2).1 ฀ ฀ = −1 respectively, and let ฀฀ be the point (i)

2

����� onto �฀฀฀฀ ���� . Find the projection of ฀฀฀฀

2

฀฀ is the point (5, −1, 0) and ฀฀ is (−1, 3, −2). 1 −4 5 �฀฀฀฀ ���� = � 0� − � −1 � = � 1 � 2 2 0

−1 −6 5 �฀฀฀฀ ���� = � 3 � − � −1 � = � 4 � −2 −2 0 ����� proj��฀฀฀฀ ��� ฀฀฀฀= � =�

[฀฀]

����� ⋅ �฀฀฀฀ ��� ฀฀฀฀ ���� � ฀฀฀฀ �฀฀฀฀ ���� ⋅ �฀฀฀฀ ���

24 + 4 − 4 −6 �� 4 � 36 + 16 + 4 −2

−18 ⁄7 3 −6 =  � 4 � or � 12 ⁄7 � 7 −2 −6⁄ 7

Page 17 of 29

[฀฀]

CambridgeMATHS

(ii)

Practice HSC Examination

Hence find the perpendicular distance ฀฀ from ฀฀ to the line ฀฀฀฀. ฀฀

������ �฀฀฀฀

฀฀

��� � − �proj������฀฀฀฀ ��� � ฀฀ 2 = �฀฀฀฀ ฀฀฀฀ 2

2

= ((−4)2 + 12 + 22 ) − ��−

= 21 − =

72 7

18 2 12 2 6 2 � + � � + �− � � 7 7 7

[฀฀]

75 7

so ฀ ฀ =

√7

5√3

or

5√21 units. 7 ฀฀ − 1

For a complex number z, arg � ฀฀ + 1� = (i)

2

฀฀

������� |proj�����  ฀฀฀฀| ฀฀฀฀

฀฀

(c)

Extension 2

[฀฀] ฀฀ 4

Find the cartesian equation of the locus of z. z + 1 is represented by the vector joining (-1,0) to (x, y) and z - 1 is ฀฀ − 1 ฀฀ arg � � = arg(฀฀ − 1) − arg(฀ ฀ + 1) = ฀฀ + 1 4

represented by the vector joining (1,0) to (x, y).

Let ฀฀ = ฀฀ + ฀฀฀฀

As such arg(z+1) will be less than arg(z-1), so ∴ ฀฀ + 1 = ฀฀ + 1 + ฀฀฀฀

฀฀฀฀฀฀ ฀฀฀฀฀฀ (฀฀ + 1) = ฀฀

∴ ฀฀ − 1 = ฀฀ − 1 + ฀฀฀฀ ฀฀฀฀฀฀ (฀฀ − 1) = ฀฀ ฀฀ ฀฀ ฀฀฀฀฀฀ ฀฀ = ฀฀฀฀฀฀ ฀฀฀฀฀฀ ฀฀ = ฀฀ + 1 ฀฀ − 1 Page 18 of 29

3

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Practice HSC Examination

฀฀ ฀฀ − ฀฀ = 4 ฀฀฀฀฀฀ ( ฀฀ − ฀฀) = 1

3 marks for correct equation

฀฀฀฀฀฀ ฀฀ − ฀฀฀฀฀฀ ฀฀ = 1 1 + ฀฀฀฀฀฀ ฀฀฀฀฀฀฀฀ ฀ ฀

2 marks for substantial progress toward correct equation, such as using expansion for tan(฀฀ − ฀฀ ) or equivalent merit

฀฀ ฀฀ ฀฀ − 1 − ฀ ฀ + 1 = 1 ฀฀ ฀฀ 1 + . ฀฀ − 1 ฀ ฀ + 1

฀฀฀฀ + ฀฀ − ฀฀฀฀ + ฀฀ = 1 ฀฀ 2 + ฀฀ 2 − 1

Cartesian Equation is

(ii)

Sketch the locus of z.

Extension 2

2฀฀ = ฀฀ 2 + ฀฀ 2 − 1

1 mark for initial working relevant to question or equivalent merit

฀฀ 2 + ฀฀ 2 − 2฀฀ − 1 = 0 ฀฀ 2 + ฀฀ 2 − 2฀ ฀ = 1

2

฀฀ 2 + ฀฀ 2 − 2฀฀ + 1 = 1 + 1 ฀฀ 2 + (฀฀ − 1)2 = 2

This is the equation of a circle, Centre (0, 1), Radius √2 units.

However, the condition ฀฀฀฀฀฀฀฀+1 � �= ฀฀−1

฀฀ 4

is only met by values of z above the x-

to, ฀฀ ≠ ±1(must have open circles at ± 1

axis, ie the locus of z is the major arc of the circle above the x axis, not equal

Page 19 of 29

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(d)

Find

�

Practice HSC Examination

Extension 2

฀฀฀฀ ฀฀ �฀฀ (1 + ฀฀ )฀฀฀฀ 1 1 1 Let ฀ ฀ = then = − 2 → −฀฀ 2 ฀฀฀฀ = ฀฀฀฀ , also ฀ ฀ = ฀฀ ฀฀ ฀฀฀฀ ฀฀ 1

�

1

฀฀�฀฀(1 + ฀฀)

−฀฀ 2 ฀฀฀฀ ฀฀฀฀ = � ฀฀ �฀฀(1 + ฀฀ ) = −�

= −�

1

1 1 ฀฀ � + 2 ฀ ฀ ฀฀

√1 + ฀ ฀ 1

฀฀฀฀

฀฀฀฀

= −2√1 + ฀ ฀ + ฀฀

= −2�1 +

1 + ฀฀ ฀฀