Question 35
GCE_MODEL_QUESTIONS_VOL1
•
2019
•
FURTHER_MATHEMATICS
•
objective
Solve the equation \(2\sin^{2}x - \sqrt{3}\sin x - 1 = 0\) for \(x\) in the interval \(0 \le x < 2\pi\). Then give the sum of all solutions (in radians).
Answer Options
A.
Solutions: \(x=\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\) and \(x=\pi-\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\); sum of solutions = \(\pi\) radians.
B.
Solutions: \(x=\pi+\arcsin\!\left(\dfrac{\sqrt{11}+\sqrt{3}}{4}\right)\) and \(x=2\pi-\arcsin\!\left(\dfrac{\sqrt{11}+\sqrt{3}}{4}\right)\); sum of solutions = \(3\pi\) radians.
C.
Solutions: \(x=\pi-\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\) and \(x=2\pi+\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\); sum of solutions = \(3\pi\) radians.
Correct Answer
D.
Solutions: \(x=\pi+\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\) and \(x=2\pi-\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\); sum of solutions = \(3\pi\) radians.
Correct Answer
D
Correct Option:
Solutions: \(x=\pi+\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\) and \(x=2\pi-\arcsin\!\left(\dfrac{\sqrt{11}-\sqrt{3}}{4}\right)\); sum of solutions = \(3\pi\) radians.
About This Question
This is
Gce Model Questions Vol1
2019
Further Mathematics
Question 35.
It is one of the
objective
questions from the
2019
Gce Model Questions Vol1
Further Mathematics
examination.
Difficulty level:
Medium
.