Faith Anglo-Saxon Academic School Complex (FASAC), Yaoundé
First continuous assessment
Pure mathematics
Upper sixth science / Upper sixth Arts
26th September, 2024
Time Allowed: 100 minutes
Answer all questions
1. The arg 𝑧 if 𝑧 = −√3 − 𝑖 is: 5. 𝑧 = −3 + 3𝑖, in the form 𝑟(cos 𝜃 + 𝑖 sin 𝜃)
𝜋 is:
(A) 6
𝜋 𝜋
(B)
7𝜋 (A) 3 [cos (− ) + 𝑖 sin (− )]
4 4
6 3𝜋 3𝜋
(C)
5𝜋 (B) 3 [cos ( 4 ) + 𝑖 sin ( 4 )]
6
5𝜋 3𝜋 3𝜋
(D)− (C) 3√2 [cos ( 4 ) + 𝑖 sin ( 4 )]
6
𝜋 𝜋
(D)3√2 [cos (− 4 ) + 𝑖 sin (− 4 )]
2. The quadratic equation whose roots
3 + 4𝑖 is: 1
6. If cos 𝜃 = − 2, the general solution is:
(A) 𝑥 2 + 25 = 0 𝜋
(B) 𝑥 2 + 6𝑥 + 25 = 0 (A) 2𝑛𝜋 ±
3
(C) 𝑥 2 + 6𝑥 − 25 = 0 (B) 2𝑛𝜋 ±
2𝜋
3
(D)𝑥 2 − 6𝑥 + 25 = 0 2𝑛𝜋
(C) 2𝑛𝜋 + 3
2𝑛𝜋
(D)2𝑛𝜋 − 3
3. The product of 3 − 2𝑖 and its conjugate is:
(A) 5 − 12𝑖
(B) 13 − 12𝑖 7. The greatest value of 5 sin 𝜃 + 12 cos 𝜃 is:
(C) 13 (A) – 13
(D)12𝑖 (B) 13
(C) 17
(D)– 17
𝑧
4. If 𝑧1 = 3 − 3𝑖 and 𝑧2 = 1 + 𝑖, |𝑧1 | is:
2
(A) 2√2 sin 5𝑥−sin 𝑥
(B) 3 8. Simplifying sin 4𝑥−sin 2𝑥 gives
(C) 6 (A) 2 cos 𝑥
(D)None of these (B) 2 cos 2𝑥
(C) 2 sec 𝑥
(D)2 sin 𝑥
By EGBE VICTOR E. JUNIOR
𝜋
9. sin (𝑥 + ) − sin (𝑥 − ) =
𝜋 10. If 𝐴 = 𝑡𝑎𝑛−1 5 + 𝑡𝑎𝑛−1 (−3), then tan 𝐴 =
6 6 1
(A) − cos 𝑥 (A)
2
4
(B) sin 𝑥 (B) − 7
(C) cos 𝑥 1
(C)
(D)− sin 𝑥 8
1
(D)− 7
Section B: Answer all the questions, showing all working clearly
1. .
(a) Find, in radians, the solution of the equation sin 3𝜃 + sin 5𝜃 = 0 in the interval
0≤𝜃≤𝜋 (4 marks)
(b) Express 𝑓(𝜃), where 𝑓(𝜃) = √3 cos 𝜃 − sin 𝜃 in the form 𝑓(𝜃) = 𝑅 cos(𝜃 + 𝜆), where
𝑅 > 0 and 𝜆 is an acute angle. Hence, find in degrees, the general solution of the equation
√3 cos 𝜃 − sin 𝜃 = √2.
3
Find also the maximum and minimum values of 𝑔(𝜃) = 𝑓(𝜃)+4 (8 marks)
2. .
√3−𝑖
(a) Given that 𝑧 = , calculate |𝑧| and arg 𝑧 (5 marks)
√3+𝑖
(b) Given that 𝑧1 = 1 + √3𝑖 and 𝑧2 = −1 − 𝑖
(i) Express 𝑧1 and 𝑧2 in the form 𝑟(cos 𝜃 + 𝑖 sin 𝜃), where 𝑟 ∈ ℝ and −𝜋 < 𝜃 ≤ 𝜋.
𝑧 3 𝑧 3
(ii) Evaluate |𝑧1 | and arg (𝑧1 ) (5 marks)
2 2
(c) If 𝑧 = 𝑥 + 𝑦𝑖 and |𝑧 − 3| = 2|𝑧 + 6𝑖|, show that 𝑧 lies on the circle 𝑥 2 + 𝑦 2 + 2𝑥 + 16𝑦 +
45 = 0 (3 marks)
By EGBE VICTOR E. JUNIOR
