Question

Find the maximum value of $\sqrt{3}\sin x+3\cos x$ .

• A. $\sqrt{3}$
• B. $2\sqrt{3}$
• C. $\sqrt{3}+1$
• D. $4\sqrt{3}$

Answer 1

The correct option is B $2\sqrt{3}$

$\sqrt{3}\sin x+3\cos x$

$=\sqrt{3}(\sin x+\sqrt{3}\cos x)$.

$=2\sqrt{3}(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x)=2\sqrt{3}\sin (x+\frac{\pi }{3}).$

Since, $-1\le \sin x\le 1$

Clearly it is max. when $x+\pi /3=\pi /2.i.e.x=\pi /6$ and Max. value is $=2\sqrt{3}\sin \frac{\pi }{2}=2\sqrt{3}\cdot 1=2\sqrt{3}.$
Ans: B