Question

The difference between greatest and least value
of $f (x) = 2 \sin x + \sin 2x, x \in \left[0, \frac{3\pi}{2}\right] $ is-

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{3\sqrt{3}}{2}-2$
• C. $\frac{3\sqrt{3}}{2}+2$
• D. None of these

Answer 1

Answer: (C)

$f(x)=2 \sin x+\sin 2 x $
$f^{\prime}(x)=2 \cos x+2 \cos 2 x=2(\cos x+\cos 2 x)$
$\therefore f^{\prime}(x)=0 \Rightarrow 2 \cos 2 x+\cos x-1=0 $
$\cos x=\frac{-1 \pm 3}{4}=-1, \frac{1}{2} $
$\therefore x=\pi, \frac{\pi}{3}$
Now, $f(0)=0, f\left(\frac{3 \pi}{2}\right)=-2$
$f(\pi)=0, f\left(\frac{\pi}{3}\right)=2 \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{3 \sqrt{3}}{2}$
$\therefore $ difference between greatest value and least value
$=\frac{3 \sqrt{3}}{2}+2$