Mathematics -Extension 2 4U - WME - Trial...Mathematics -Extension 2 General Instructions Reading time – 10 minutes Working time – 3 hours Write using black pen Approved calculators

Western

Mathematics

2020 TRIAL HIGHER SCHOOL CERTIFICATE EXAMINATION

Mathematics -Extension 2

General Instructions

Reading time – 10 minutes

Working time – 3 hours

Write using black pen

Approved calculators may be used

A reference sheet is provided at the back of this paper

In Questions in Section II, show relevant mathematical reasoning and/or calculations

Total marks: 100

Section I – 10 marks (pages 2 – 4)

Attempt Questions 1 – 10

Allow about 15 minutes for this section

Section II – 90 marks (pages 5 – 10)

Attempt Questions 11 – 16

Allow about 2 hours and 45 minutes for this section

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Section I

10 marks Attempt Questions 1 – 10. Allow about 15 minutes for this section.

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

1. Which expression is equal to 2𝑥 𝑒 𝑑𝑥? (A) 𝑥𝑒 𝑒 𝑑𝑥 (B) 𝑥𝑒 𝑒 𝑑𝑥 (C) 2𝑥𝑒 𝑒 𝑑𝑥 (D) 2𝑥𝑒 𝑒 𝑑𝑥

2. A student wants to prove that there is an infinite number of prime numbers. To prove this statement by contradiction, what assumption would the student start their proof with?

(A) There is only one prime number that is even.

(B) There is an infinite number of Primes.

(C) There is a finite number of Primes.

(D) All prime numbers are less than 100

3. Which of the following is the complex number 4 √3 4𝑖 ?

(A) 4𝑒 (B) 4𝑒

(C) 8𝑒

(D) 8𝑒 4. A particle is describing SHM in a straight line with an amplitude of 4 metres. Its speed is 6m/s

when the particle is 2 metres from the centre of the motion.

What is the period of the motion?

(A) √

(B) √

(C) √3𝜋 (D) √

- 3 -

5. If �̰� 42

4 is a non zero vector, then the corresponding unit vector is:

(A) �̰�

⎝⎜⎛

⎠⎟⎞

(B) �̰�

⎝⎜⎛

⎠⎟⎞

(C) �̰� 1

1

(D) �̰� 21

1

6. Which of the methods below would be suitable to use when proving the following statement?

“For any infinite sequence of integers, there will always be two numbers in that sequence that differ by a multiple of 5243.”

(A) Proof by Contradiction

(B) Proof by Induction

(C) Proof using the Pigeonhole Principle

(D) Proof using Probability Theory

7. Which vector is perpendicular to 32

4 ?

(A) 64

8

(B) 3

24

(C) 361

(D) 251

- 4 -

8. A mass of m kilograms falls from a stationary balloon at height h metres above the ground.

It experiences air resistance during its fall equal to 𝑚𝑘𝑣 , where v is its speed in metres per second and k is a positive constant.

The distance in metres of the mass from the balloon, measured positively as it falls is x.

What is the equation of motion of the mass due to gravity g?

(A) 𝑥 𝑔 𝑘𝑣 (B) 𝑥 𝑔 𝑘𝑣 (C) 𝑥 𝑔 𝑘𝑣 (D) 𝑥 𝑔 𝑘𝑣

9. If , then A and B have values of: (A) 𝐴 1,𝐵 2 (B) 𝐴 1,𝐵 2 (C) 𝐴 2,𝐵 1 (D) 𝐴 2,𝐵 1

10. The equation, in Cartesian form, of the locus of the point z if |𝑧 2𝑖| |𝑧 4| is: (A) 𝑥 2𝑦 3 0 (B) 2𝑥 𝑦 3 0 (C) 𝑥 2𝑦 3 0 (D) 2𝑥 𝑦 3 0

- 5 -

Section II

90 marks Attempt Questions 11 – 16.

Allow about 2 hours and 45 minutes for this section.

Answer each question in the appropriate writing booklet. Extra writing booklets are available. For questions in Section II, your responses should include relevant mathematical reasoning and/or calculations.

Question 11 (15 marks) Use the Question 11 writing booklet.

(a) If 𝐴 5 3𝑖 and 𝐵 3 4𝑖, evaluate the following:

(i) AB 1

(ii) 1

(iii) √𝐵 2

(b) For a complex number z, arg .

(i) Find the cartesian equation of the locus of z. 3

(ii) Describe the locus of z. 2

(c) The polynomial P 𝑥 𝑥 5𝑥 + ax + b , where a and b are real, has one root at 𝑥 3 2√2𝑖.

(i) Find the values of a and b. 2

(ii) Hence solve the equation P 𝑥 0. 1

(d) For a complex number z, shade the region of the Argand Plane in which: 3

1 |𝑧| 3 and 𝑎𝑟𝑔 𝑧 .

End of Question 11

- 6 -


(a) Evaluate:

(i) sin 𝑥 cos 𝑥 𝑑𝑥. 2

(ii) . 3

(b) (i) If

, find the values of A, B and C.

3

(ii) Hence evaluate . 3

(c) A point P, which moves in the complex plane, is represented by the equation |𝑧 4 3𝑖 | 5.

.

(i) Sketch the locus of the point P. 1

(ii) Find the value of arg z when P is in the position that maximises |z|.

1

(iii) Find the modulus of z when arg 𝑧 tan . 2

End of Question 12

- 7 -


(a) If 𝑧 √2 √6𝑖,

(i) Express z in modulus-argument form.

2

(ii) Evaluate 𝑧 . 1

(b) (i) Show that a reduction formula for 𝐼 𝑥 𝑒 𝑑𝑥 is: 𝐼 𝐼 .

2

(ii) Hence evaluate 𝑥 𝑒 𝑑𝑥 .

3

(c) (i) By considering the cases where a positive integer k is even 𝑘 2𝑥 and odd 𝑘 2𝑥 1 , show that kk 2 is always even.

2

(ii) Using the result in part (i), prove, by mathematical induction, that for all positive integral values of n , nn 53 is divisible by 6.

3

(d) For two positive real numbers a and b, prove that their arithmetic mean

is always

greater than or equal to their geometric mean √𝑎𝑏 . 2

End of Question 13

- 8 -


(a) For the vectors �̰� 234

and �̰� 32

4 :

(i) Find the scalar product. 1

(ii) Show that �̰� �̰� . 1 (iii) Find the angle between the two vectors, correct to the nearest minute. 1

(b) Find the point(s) of intersection of the line with parametric equation 𝑟 𝑖 3𝑗 4𝑘 𝑡 𝑖 2𝑗 2𝑘 and the sphere with equation 𝑥 1 𝑦 3 𝑧 4 81.

4

(c) A particle A is projected horizontally at a velocity of 50 ms-1 from the top of a tower which is 100 metres high. At the same instant, another particle B is projected from the bottom of the tower, in the same vertical plane at a velocity of 100 ms-1 with an angle of elevation of 60. Using the base of the tower as the origin, show that the particles will collide and find the co-ordinates of the point where they do so. (Use g = 10 ms-2 and give your answer to the nearest metre).

4

(d) For d, an integer where d > 1,

(i) Show that 1

(ii) Noting that show that, for a positive integer n : . . . . . . . . . . 2.

3

End of Question 14

- 9 -


(a) (i) Use De Moivre’s Theorem to express cos 5θ and sin 5θ in terms of powers of sin θ and cos θ.

2

(ii) Write an expression for tan 5θ in terms of t, where t = tan θ. 1

(iii) By solving tan 5θ = 0, deduce that: tan tan tan tan 5. 3

(b) A body of mass 1 kilogram is projected vertically upward from the ground at a speed of 10 ms-1. The particle is under the effect of both gravity and a resistance which, at any time, has a magnitude of 𝑣 , where v is the particle’s velocity at that time. Gravity is taken as 10 ms-2.

(i) Show that the equation of motion is given by: 𝑥 10 𝑣 . 1

(ii) Taking 𝑥 𝑣 , calculate the greatest height reached by the patricle. 3

(iii) Taking 𝑥 , calculate the time taken to reach the greatest height. 3

(c) 2 3𝑖 is a root of the equation 𝑥 4 2𝑖 𝑥 𝑘 0.

(i) State the other root. 1

(ii) Find the value of k. 1

End of Question 15

- 10 -


(a) A particle moves under gravity for which the resistance to its motion is directly proportional to the product of its mass and velocity (v).

(i) Show that 𝑔 – 𝑘𝑣 where k is a constant. 1

(ii) If the particle falls vertically from rest, find its terminal velocity, 𝑉 . 1

(iii) If the particle is projected vertically upwards with velocity 𝑉 show that after time t seconds :

(𝛼) its speed is 𝑉 2𝑒 1 . 𝛽 its height above its starting point is 𝑉 2 2𝑒 𝑘𝑡 .

4

3

(b) Prove that 33 – 16 – 28 11 is divisible by 85 for all positive integers 𝑛 2. 3

(c) If P = i + j + k and R = 9i + 3j + 8k, find the point Q on 𝑃�⃗� such that PQ : QR = 2 : 3. 3

End of Paper

11

2020 Trial HSC Examination

Mathematics Advanced Mathematics Extension 1 Mathematics Extension 2

REFERENCE SHEET

- 13 -

- 14 -

- 15 -

WesternMathematics

2020 Trial Higher School Certificate Examination MathematicsAdvanced

Name ________________________________ Teacher ________________________

SectionI – MultipleChoiceAnswerSheet

Allowabout25minutesforthissectionSelect the alternative A, B, C or D that best answers the question. Fill in the response oval completely. Sample: 2 + 4 = (A) 2 (B) 6 (C) 8 (D) 9

A B C D If you think you have made a mistake, put a cross through the incorrect answer and fill in the new answer.

A B C D If you change your mind and have crossed out what you consider to be the correct answer, then indicate the correct answer by writing the word correct and drawing an arrow as follows.

A B C

D

1. A B C D

2. A B C D

3. A B C D

4. A B C D

5. A B C D

6. A B C D

7. A B C D

8. A B C D

9. A B C D

10. A B C D

1

Western Mathematics Exams

Mathematics Extension 2

SOLUTIONS

2

MultipleChoiceWorkedSolutionsNo Working Answer 1 2𝑥 𝑒 𝑑𝑥

𝑢 2𝑥 𝑣′ 𝑒 𝑢′ 2 𝑣 𝑒

𝑢𝑣 𝑣𝑢′ 2𝑥 12 𝑒

12 𝑒 2

𝑥𝑒 𝑒 𝑑𝑥

A

2 Contradicting an infinite number of primes is that there is a finite number of primes

C

3 4 √3 4𝑖 in 4th quadrant therefore angle is 𝑒 𝑐𝑜𝑠 𝜋6 𝑖𝑠𝑖𝑛

𝜋6

√ Need to multiply by 8 to give desired result.

8𝑒 8 √32 𝑖2 4 √3 4𝑖

C

4 Using 𝑣 𝑛 𝑎 𝑥 6 𝑛 4 2 36 12𝑛 𝑛 3 i.e. 𝑛 √3

Periodic Time = √ √

B

5 |𝑢| 42

4 4 2 4 √36 6

𝑢

⎝⎜⎜⎛

462646 ⎠⎟⎟⎞

⎝⎜⎜⎛

231323 ⎠⎟⎟⎞

B

6 Pigeonhole principle as follows: Label each number in the sequence by the remainder it leaves upon division by 5243. There are 5243 possible labels: 0, 1, 2, …, 5242. As there an infinite number of entries in the sequence, two entries must have the same label, that is, the same remainder upon division by 5243. This means that their difference is a multiple of 5243.

C

7 Show that the dot product is zero. Test each option and find only D works. 2 3 + 5 (2) + 1 4 = 6 10 + 4 = 0 Therefore option D is perpendicular

D

4

TrialHSCExamination2020MathematicsExt2Course

Name ________________________________ Teacher ________________________

SectionI – MultipleChoiceAnswerSheet

Allowabout15minutesforthissectionSelect the alternative A, B, C or D that best answers the question. Fill in the response oval completely. Sample: 2 + 4 = (A) 2 (B) 6 (C) 8 (D) 9

A B C D If you think you have made a mistake, put a cross through the incorrect answer and fill in the new answer.

A B C D If you change your mind and have crossed out what you consider to be the correct answer, then indicate the correct answer by writing the word correct and drawing an arrow as follows.

A B C

D

1. A B C D

2. A B C D 3. A B C D 4. A B C D 5. A B C D 6. A B C D 7. A B C D 8. A B C D 9. A B C D 10. A B C D

5

Question 11 2020 Solution Marks Allocation of marks

(a) (i) 𝐴𝐵 5 3𝑖 3 4𝑖 = 15 20i + 9i 12𝑖 15 11𝑖 12

= 27 11i

1 1 mark for correct answer

(a) (ii)


(a) (iii) √𝐵 = √3 4𝑖 √𝜔 √3 4 𝑖

Let iyxA (x and y real) A = x2 – y2 + 2xyi x2 – y2 = 3 . . . . . . . . . . . . .① 2xy = 4 22222222 4 yxyxyx = 3 4 = 25 𝑥 𝑦 5 . . . . . . . . . . . . .② ① + ② 2𝑥 8 → 𝑥 2 ② ① 2𝑦 2 → 𝑦 1 Since 2xy = 4, the signs of x and y are opposite √𝐵 2 𝑖

2 2 marks for finding one or both of the correct roots 1 mark for working including setting up and solving simultaneous equations, leading to an incorrect answer or equivalent merit

6


(b) (i) z + 1 is represented by the vector joining (-1,0) to (x, y) and z - 1 is represented by the vector joining (1,0) to (x, y).

As such arg(z+1) will be less than arg(z-1), so Let 𝑧 𝑥 𝑖𝑦 ∴ 𝑧 1 𝑥 1 𝑖𝑦 𝐿𝑒𝑡 𝑎𝑟𝑔 𝑧 1 𝜃 ∴ 𝑧 1 𝑥 1 𝑖𝑦 𝑎𝑟𝑔 𝑧 1 𝛼

𝑡𝑎𝑛 𝜃 𝑦𝑥 1 𝑎𝑛𝑑 𝑡𝑎𝑛 𝛼 𝑦

𝑥 1 𝛼 𝜃 𝜋4 𝑡𝑎𝑛 𝛼 𝜃 1

𝑡𝑎𝑛 𝛼 𝑡𝑎𝑛 𝜃

1 𝑡𝑎𝑛 𝛼𝑡𝑎𝑛 𝜃 1 𝑦𝑥 1

𝑦𝑥 1

1 𝑦𝑥 1 .𝑦

𝑥 1 1

𝑥𝑦 𝑦 𝑦𝑥 𝑦𝑥 𝑦 1 1

2𝑦 𝑥 𝑦 1 Cartesian Equation is

𝑥 𝑦 2𝑦 1 0

3 3 marks for correct equation 2 marks for substantial progress toward correct equation, such as using expansion for tan 𝜃 𝛼 or equivalent merit 1 mark for initial working relevant to question or equivalent merit

7


(b) ii) 𝑥 𝑦 2𝑦 1 𝑥 𝑦 2𝑦 1 1 1

𝑥 𝑦 1 2 This is the equation of a circle, Centre 0, 1 , Radius √2 units. However, the condition 𝑎𝑟𝑔 is only met by values of z above the x‐axis, ie the locus of z is the major arc of the circle above the x axis, 𝑧 1

2 2 marks for correct for correct centre, radius and description 1 mark for working including completing the square or similar merit Diagram not required, provided for illustration purposes only.

(c) (i) P 𝑥 𝑥 5𝑥 + ax + b has one root 3 2√2𝑖 Therefore another root is 3 2√2𝑖 ∴ 𝑃 𝑥 is divisible by 𝑥 3 2√2𝑖 3 2√2𝑖 𝑥

3 2√2𝑖 3 2√2𝑖 i.e. Divisible by 𝑥 6𝑥 17 Using sum of the roots, the other root is 1 other factor is 𝑥 1

∴ 𝑃 𝑥 𝑥 6𝑥 17 𝑥 1 𝑥 𝑥 6𝑥 6𝑥 17𝑥 17 𝑥 5𝑥 11𝑥 17 ∴ 𝑎 11,𝑏 17

2 2 marks for correct values of a and b 1 mark for significant working toward values of a and b.

(c) (ii) 𝑃 𝑥 𝑥 5𝑥 11𝑥 17 0 From part i) roots are 𝑥 3 2√2𝑖, 𝑥 3 2√2𝑖 and x = 1

𝑥 1, 3 2√ 2𝑖


8


(d)

3 3 marks for correct graphs and shading 2 marks for one incorrect graph with otherwise correct graph and shading or equivalent merit 1 mark for one correct graph for circle or for argument with incorrect or incomplete graph and shading, or equivalent merit

9


(a) (i) 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑥 1 𝑐𝑜𝑠 𝑥 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑥 Let 𝑢 𝑐𝑜𝑠 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑢 𝑠𝑖𝑛 𝑥 𝑑𝑥 1 𝑢 𝑢 𝑑𝑢 𝑢 𝑢 𝑑𝑢 𝑐 cos 𝑥 cos 𝑥 𝑐

2 2 marks for correct integral 1 mark for correct substitution or similar merit

(ii) √ Let 𝑢 𝑥 2 𝑎𝑛𝑑 𝑠𝑜 𝑑𝑢 𝑑𝑥 √ 𝑠𝑖𝑛 √ 𝑐 sin √ 𝑐

3 3 marks for correct integral 2 marks for significant progress toward integral 1 mark for attempt to form square or similar merit

(b) (i) 4𝑥 10

2 𝑥 𝑥 2 𝐴

2 𝑥 𝐵𝑥 𝐶𝑥 2

∴ 4𝑥 10 𝐴 𝑥 2 𝐵𝑥 𝐶 2 𝑥 If x = 2, 18 = 6A, A = 3 Equating coefficients of 𝑥 : A B 0

3 𝐵 0 𝐵 3

Equating constants: 2𝐴 2𝐶 10

6 2𝐶 0 2𝐶 4 𝐶 2

3 3 marks for three correct values. 2 marks for significant progress toward finding the three values 1 mark for initial evaluation of A or similar merit

10


(b) (ii)

𝑑𝑥

𝑑𝑥 𝑑𝑥

3 𝑑𝑥

𝑑𝑥 2 dx = 3 ln (2 𝑥 + ln 𝑥 2 √ tan √ c

3 3 marks for correct integral 2 marks for significant progress toward correct integral 1 mark for breaking into separate fractions or similar merit

(c) (i) Circle centre (4, 3) radius 5

|𝑧 4 3𝑖 | 5

1 1 mark for correct sketch of circle

(c) (ii) Maximum value of |𝑧| is 10 (when z lies along the diameter) So since this passes through the centre (4, 3).

𝑎𝑟𝑔 𝑧 𝜃 𝑡𝑎𝑛 34


11


(c) (iii) When 𝑎𝑟𝑔 𝑧 𝑡𝑎𝑛 continuing the ray to the circle, gives z = 9 + 3i.

So |𝑧| 9 3 90 |𝑧| √90 3√10 Alternately the point z can be found algebraically

𝑧 lies on 𝑥 4 𝑦 3 25 𝑧 lies on 𝑦 𝑥3 since arg 𝑧 tan

13

𝑥 4 𝑥3 3 25 𝑥 8𝑥 16 𝑥9 2𝑥 9 25 9𝑥 72𝑥 144 𝑥 18𝑥 81 225 10𝑥 90𝑥 0 𝑥 𝑥 9 0 𝑥 0 or 𝑥 9 so 𝑥 9 and 𝑦 3

2 2 marks for correct answer 1 mark for any correct reasoning leading to incorrect answer

12


(a) (i) Z √2 √6𝑖 𝑟 √2 √6 tan 𝜃 √√ √2 6 tan 𝜃 √3 √8 𝜃 2√2

∴ 𝑧 2√2 𝑐𝑜𝑠 π3 𝑖 𝑠𝑖𝑛 π3

2 2 marks for correct modulus and argument 1 mark for significant working toward modulus and argument

(a) (ii) 𝑧 2√2 𝑐𝑖𝑠 16√2 𝑐𝑖𝑠 3 16√2 𝑐𝑖𝑠 𝜋 16√2


(b) (i) 𝐼 𝑥 𝑒 𝑑𝑥

𝑢 𝑥 𝑣′ 𝑒 𝑢′ 𝑛𝑥 𝑣 𝑒

𝐼 𝑢𝑣 𝑣𝑢′ 𝐼 𝑥 13 𝑒

13 𝑒 𝑛𝑥

𝐼 𝑥 𝑒3 𝑛𝑥

3 𝑒 𝐼 𝑥 𝑒

∴ 𝐼 𝑥 𝑒3 𝑛3 𝐼

2 2 marks for correct derivation 1 mark for correct use of integration by parts with errors in derivation

13


(b) ∴ 𝐼 𝑥 𝑒3 𝑛3 𝐼

𝑥 𝑒 𝑑𝑥 𝑥 𝑒3 33 𝑥 𝑒 𝑑𝑥

𝑥 𝑒 𝑑𝑥 𝑥 𝑒3 23 𝑥 𝑒 𝑑𝑥

23 𝑥 𝑒 𝑑𝑥

23

𝑥 𝑒3

23

13 𝑥 𝑒 𝑑𝑥

𝑒 𝑒

∴ 𝑥 𝑒 𝑑𝑥

𝑥 𝑒3 𝑥 𝑒

3 2𝑥𝑒

9 2

27 𝑒 𝐶

3 3 marks for correct integral 2 marks for substantial progress toward integral or equivalent merit 1 mark for initial working relevant to integral or equivalent merit

(c) (i) If k is even, i.e xk 2 , then

m

xxxx

xxkk

222

2422

2

2

22

If k is odd, i.e 12 xk , then

m

xxxx

xxxxxkk

21322

26412144

1212

2

2

2

22

2 2 marks for showing both results 1 mark for proving only one, or equivalent merit

14


(c) (ii) Assume that nn 53 is divisible by 6 for kn i.e 𝑘 5𝑘 6𝑝 where p is an integer. Now when 1 kn

16666

2366366

63366335

55133151

2

2

23

233

mpmp

*mpkkp

kkpkkkk

kkkkkk

* from i) above

151 3 kk is divisible by 6 if true for kn , then also true for 1 kn , but since true for 1n , by induction is true for all integral values, 1n

3 3 marks for correct and complete proof 2 marks for substantial progress in proof with either an error or incomplete statements or equivalent merit 1 mark for initial working relevant to the proof or equivalent merit

(d) Prove that √𝑎𝑏 if a, b 0

𝑎 𝑏 𝑎 2𝑎𝑏 𝑏 𝑎 2𝑎𝑏 𝑏 4𝑎𝑏 𝑎 𝑏 4𝑎𝑏 If 𝑎 𝑏, Then 𝑎 𝑏 4𝑎𝑏 𝑎 𝑏 2√𝑎𝑏 √𝑎𝑏 If a = b then √𝑎𝑏 Hence √𝑎𝑏 if a, b 0

2

2 marks for correct and complete proof 1 mark for significant working toward proof

15


(a) (i) 𝑝 ∙ 𝑞 2 3 3 2 4 4 16


(ii) |𝑝| 2 3 4 √29 |𝑞| 3 2 4 √29 |𝑝|


(a) (iii) 𝑐𝑜𝑠 𝜃 𝑝 ∙ 𝑞 |𝑝||𝑞|

16√29√29 𝜃 123° 29′


(b) 𝑟 𝑖 3𝑗 4𝑘 𝑡 𝑖 2𝑗 2𝑘 𝑥 1 𝑡 𝑦 3 2𝑡 𝑧 4 2𝑡 Now 𝑥 1 𝑦 3 𝑧 4 81 1 𝑡 1 3 2𝑡 3 4 2𝑡 4 81

𝑡 2𝑡 2𝑡 81 9𝑡 81 𝑡 9 𝑡 3

Points are: 1 3, 3 2 3 , 4 2 3 4, 9, 2 1 3, 3 2 3 , 4 2 3 2, 3, 10

4 4 marks for two correct points 3 marks for substantial progress toward solving the equations simultaneously or equivalent merit 2 marks for writing parametric equations with some progress toward answer or equivalent merit 1 mark for writing parametric equations or other initial or other limited working relevant to question

16


(c) 50 m/s A B Tower 100 m/s 100 100 sin 60 100m B 60 60 100 cos 60 Particle A Particle B 𝑥 0 𝑥 0 𝑥 𝑐 50 𝑥 𝑐 100 𝑐𝑜𝑠 60 𝑥 50 𝑥 50 𝑥 50𝑡 𝑐 𝑥 50𝑡 𝑐 For both when t = o x = 0 therefore c = 0 𝑥 50𝑡 𝑥 50𝑡 𝒚 𝐠 𝒚 𝐠 𝑦 𝑔𝑡 + c * 𝑦 𝑔𝑡 𝑐 𝒚 𝒈𝒕 (t=0, 𝑦 0, 𝑐 0 𝑦 𝑔𝑡 100 𝑠𝑖𝑛 60 𝑦 𝑔𝑡 𝑐 𝒚 𝟓𝟎√𝟑 𝒈𝒕 t = 0, y = 100 c = 100 y 50√3𝑡 𝑔𝑡 + c 𝒚 𝟏𝟎𝟎 𝟏𝟐𝒈𝒕𝟐 when t = 0, y = 0 c = 0 y = 𝟓𝟎√𝟑𝒕 𝟏𝟐 𝒈𝒕𝟐 As x = 50t for both particles, they will collide if there is a solution to solving the heights simultaneously. Particles at the same height when

𝟏𝟎𝟎 𝟏𝟐𝒈𝒕𝟐 𝟓𝟎√𝟑𝒕 𝟏𝟐𝒈𝒕

𝟐

100 50√3𝑡 𝑖. 𝑒. 𝑡 10050√3

2√3 seconds

When 𝑡 √

𝑥 50 √ 𝑦 100 10 √ 𝑥 58 m nearest m 𝑦 93 m nearest m

4 4 marks for showing collision and finding co-ordinates of point of collision 3 marks for substantial progress toward finding co-ordinates of point of collision or equivalent merit 2 marks for restricted progress including finding equations for one particle correctly or equivalent merit 1 mark for initial or limited working relevant to question

(d) (i) If 𝑑 1 then 𝑑 𝑑 – 1 Also 𝑑 𝑑 𝑑 1 ∴


17


(d) (ii) Given From (i) For is undefined, so use 1 Then Therefore

. . . . . . . .

1 1 12 12

13

13

14 . . . . . . . . .. 1

𝑛 2 1

𝑛 1 1

𝑛 1 1𝑛

1 1 12 12

13

13

14

14 . . . . . . . . ..

2 0 0 0 ⋯ 0 0 2 As 𝑛 is a positive integer , 0 and 2 2

∴ 1 12 1

3 . . . . . . . 1𝑛 2

3 3 marks for correct proof 2 marks for substantial progress toward correct proof 1 mark for some correct working relevant to proof

18


(a) (i) 5 5 5cos i sin cos i sin

LHS =

5 4 3 2

2 3 4 5

5 10

10 5

cos cos isin cos isin

cos isin cos isin isin

Equating Real and Imaginary Parts

5 3 2 45 10 5cos cos cos sin cos sin 4 2 3 55 5 10sin cos sin cos sin sin

2 2 marks for expressions for both sin5θ and cos5θ 1 mark for only one of sin5θ or cos5θ correct or equivalent merit

(a) (ii)

4 2 3 5

5 3 2 45 105

10 5cos sin cos sin sintan

cos cos sin cos sin

Dividing everything by 5cos 3 5

2 45 1051 10 5t t ttan

t t

where t tan

1 1 mark for dividing to get tan 5θ

(a) (iii) If 5 0tan

...............,5

6,,5

4,5

3,5

2,5

,0

...........,5,4,3,2,,05

Also if 5 0tan

0105..05101

105 5342

53

tttei

ttttt

i.e. 4 210 5 0t t t

ofRoots 4 210 5 0t t must be ,

54,

53,

52,

5

t

Product of Roots of quartic:

2 3 4 55 5 5 5

etan tan tan tana

i.e. 2 3 4 5

5 5 5 5tan tan tan tan

3 3 marks for correct derivation of required result 2 marks for substantial progress toward derivation of required result 1 mark for initial working relevant to question or equivalent merit

(b) (i) x+ m = 1 𝑥 𝑔 𝑣 0 g = 10 𝑥 10 𝑣

1 1 mark for correct derivation

19


(b) (ii) 𝑥 10 𝑣 𝑣 𝑣 100 10 𝑥 5𝑙𝑛 𝑣 100 + c when x = 0, v = 10 0 5 𝑙𝑛 10 100 𝑐 0 = 5 ln 200 + c ∴ 𝑐 5 𝑙𝑛 200 𝑥 5 𝑙𝑛 When v = 0, 𝑥 5 𝑙𝑛 5 𝑙𝑛 5 𝑙𝑛 1 𝑙𝑛 2 5ln 2 Maximum Height = 5 ln2 metres.

3 3 marks for correct answer 2 marks for substantial progress toward correct answer such as finding equation for x or equivalent merit 1 mark for some correct working relevant to question

(b) (iii) 𝑥 10 𝑣 𝑑𝑣𝑑𝑡

110 𝑣 100 𝑑𝑡

𝑑𝑣 101

𝑣 100 𝑡 1010 𝑡𝑎𝑛

𝑣10 𝑐

𝑡 𝑡𝑎𝑛 𝑣10 𝑐 When v = 10, t = 0

0 𝑡𝑎𝑛 1010 𝑐 𝑐 𝑡𝑎𝑛 1

𝑐 𝜋4 ∴ 𝑡 𝑡𝑎𝑛 𝑣10

𝜋4

When v = 0

𝑡 𝑡𝑎𝑛 010 𝜋4

𝑡 seconds

3 3 marks for correct answer 2 marks for substantial progress toward correct answer such as finding equation for t without constant or equivalent merit 1 mark for some correct working relevant to question

20


(c) i 𝑥 4 2𝑖 𝑥 𝑘 0. The roots are 2 3𝑖 and 𝛽

∴ 2 3𝑖 𝛽 4 2𝑖1 4 2𝑖 𝛽 4 2𝑖 2 3𝑖 𝛽 2 𝑖

The other root is 2 + i.


(ii) 𝑘 2 3𝑖 2 𝑖 4 2𝑖 6𝑖 3𝑖 7 4𝑖


22


(iii) () 𝑉 2𝑒 – 1 Where x is height above starting point 𝑥 𝑉 𝑒 𝑡 𝑐 When t = 0, x = 0 ∴ 𝑐

∴ 𝑥 𝑉 2𝑘 𝑒 𝑡 2𝑉𝑘

𝑉 𝑒 𝑡 𝑉 2 2𝑒 𝑘𝑡

3 3 marks for correct derivation of required result 2 marks for substantial progress toward required result 1 mark for initial integration or other limited working relevant to question

(b) Now, 33 – 16 – 28 11 33 – 16 28 11

33 16 . . . . 28 11 . . . . 17 . . . 17 . . . Hence 33 – 16 – 28 11 is divisible by 17. Also,

33 – 16 – 28 11 33 28 16 11 33 28 . . . 16 11 . . . 5 . . . 5 . . . Hence 33 – 16 – 28 11 is divisible by 5. Thus 33 – 16 – 28 11 is divisible by 17 and by 5. Therfore 33 – 16 – 28 11 is divisible by 17 5 = 85 as 17 and 5 are prime numbers.

3

3 marks for correct and complete proof 2 marks for substantial progress in proof with minor omission or errors 1 mark for some initial statements relevant to proof

(c) 𝑃𝑄 25 𝑃𝑅 𝑃𝑅 𝑟 𝑝 𝑃𝑄 𝑞 𝑝 5𝑃𝑄 2𝑃𝑅 5 𝑞 𝑝 2 𝑟 𝑝 5𝑞 5𝑝 2𝑟 2𝑝 5𝑞 2𝑟 3𝑝 ∴ 5𝑞 2 9𝑖 3𝑗 8𝑘 3 𝑖 𝑗 𝑘 18𝑖 6𝑗 16𝑘 3𝑖 3𝑗 3𝑘 21𝑖 9𝑗 19𝑘 𝑄 215 𝑖

95 𝑗

195 𝑘

𝑖. 𝑒.𝑄 215 ,95 ,

195

3

3 marks for correct co-ordinates of Q 2 marks for substantial progress toward correct answer with minor error or incomplete solution 1 mark for initial working relevant to question

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