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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) √
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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
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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
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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
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Question 12 (15 marks) Use the Question 12 writing booklet.
(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
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Question 13 (15 marks) Use the Question 13 writing booklet.
(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
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Question 14 (15 marks) Use the Question 14 writing booklet.
(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
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Question 15 (15 marks) Use the Question 15 writing booklet.
(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
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Question 16 (15 marks) Use the Question 16 writing booklet.
(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
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11
2020 Trial HSC Examination
Mathematics Advanced Mathematics Extension 1 Mathematics
Extension 2
REFERENCE SHEET
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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
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1
Western Mathematics Exams
Mathematics Extension 2
SOLUTIONS
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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
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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
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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)
1 1 mark for correct answer
(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
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Question 11 2020 Solution Marks Allocation of marks
(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
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Question 11 2020 Solution Marks Allocation of marks
(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
1 1 mark for correct answer
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Question 11 2020 Solution Marks Allocation of marks
(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
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Question 12 2020 Solution Marks Allocation of marks
(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
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Question 12 2020 Solution Marks Allocation of marks
(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
1 1 mark for correct answer
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Question 12 2020 Solution Marks Allocation of marks
(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
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Question 13 2020 Solution Marks Allocation of marks
(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
1 1 mark for correct answer
(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
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Question 13 2020 Solution Marks Allocation of marks
(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
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14
Question 13 2020 Solution Marks Allocation of marks
(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
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15
Question 14 2020 Solution Marks Allocation of marks
(a) (i) 𝑝 ∙ 𝑞 2 3 3 2 4 4 16
1 1 mark for correct answer
(ii) |𝑝| 2 3 4 √29 |𝑞| 3 2 4 √29 |𝑝|
1 1 mark for correct answer
(a) (iii) 𝑐𝑜𝑠 𝜃 𝑝 ∙ 𝑞 |𝑝||𝑞|
16√29√29 𝜃 123° 29′
1 1 mark for correct answer
(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
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16
Question 14 2020 Solution Marks Allocation of marks
(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 ∴
1 1 mark for correct answer
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17
Question 14 2020 Solution Marks Allocation of marks
(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
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18
Question 15 2020 Solution Marks Allocation of marks
(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
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19
Question 15 2020 Solution Marks Allocation of marks
(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
Question 15 2020 Solution Marks Allocation of marks
(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.
1 1 mark for correct answer
(ii) 𝑘 2 3𝑖 2 𝑖 4 2𝑖 6𝑖 3𝑖 7 4𝑖
1 1 mark for correct answer
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22
Question 16 2020 Solution Marks Allocation of marks
(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