Gce Model Questions Vol1 2017 Further Mathematics Question 42

Practice objective / multiple choice question 42 from the 2017 Gce Model Questions Vol1 Further Mathematics examination.

Gce Model Questions Vol1 2017 Further Mathematics Objective / Multiple Choice Medium Difficulty
Question 42 GCE_MODEL_QUESTIONS_VOL1 • 2017 • FURTHER_MATHEMATICS • objective

The function \(f(x)=2\sin x+\sqrt{3}\cos x\) can be written in the form \(R\sin(x+\alpha)\) where \(R>0\) and \(0<\alpha<\frac{\pi}{2}\). (i) Find \(\tan\alpha\). (ii) Hence determine the maximum value of \(f(x)\) and the smallest positive \(x\) at which this maximum occurs.

Answer Options

Correct Answer A. tan\alpha=\frac{\sqrt{3}}{2}, max=\sqrt{7}, x=\frac{\pi}{2}-\arctan\left(\frac{\sqrt{3}}{2}\right)
B. tan\alpha=\frac{\sqrt{3}}{2}, max=\sqrt{8}, x=\frac{\pi}{2}-\arctan\left(\frac{\sqrt{3}}{2}\right)
C. tan\alpha=\frac{\sqrt{3}}{3}, max=\sqrt{6}, x=\frac{\pi}{2}-\arctan\left(\frac{\sqrt{3}}{3}\right)
D. tan\alpha=\frac{2}{\sqrt{3}}, max=\sqrt{5}, x=\frac{\pi}{2}-\arctan\left(\frac{2}{\sqrt{3}}\right)
Correct Answer
A
Correct Option:
tan\alpha=\frac{\sqrt{3}}{2}, max=\sqrt{7}, x=\frac{\pi}{2}-\arctan\left(\frac{\sqrt{3}}{2}\right)

About This Question

This is Gce Model Questions Vol1 2017 Further Mathematics Question 42. It is one of the objective questions from the 2017 Gce Model Questions Vol1 Further Mathematics examination.

Difficulty level: Medium .

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