Question

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

• A. $\frac{3\sqrt{3}}{8}$
• B. $\frac{\sqrt{3}}{8}$
• C. $\frac{3}{8}$
• D. $\frac{1}{2\sqrt{2}}$

Answer 1

Answer: (A)

$\frac{3\sqrt{3}}{8}$

Given, $f\left(x\right)={\cos }^2\frac{x}{2}\sin x$
$\Rightarrow f\left(x\right)=\frac{1}{2}(1+\cos x)\sin x$
$\Rightarrow f\left(x\right)=\frac{\sin x}{2}+\frac{\sin 2x}{4}$
Therefore, $f^{\prime }\left(x\right)=\frac{\cos x}{2}+\frac{\cos 2x}{2}$
$=\frac{2{\cos }^{2}x+\cos x-1}{2}$
$=\frac{(2\cos x-1)(\cos x+1)}{2}$
For maxima or minima, $f^{\prime }\left(x\right)=0$
$\Rightarrow \cos x=\frac{1}{2},\cos x=-1$
$\Rightarrow x=\frac{\pi }{3},\pi$
$f\left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{8}=\frac{3\sqrt{3}}{8}$ $=$ Maximum value
$f\left(0\right)=f\left(\pi \right)=0$ $=$ Minimum value
Required difference $=\frac{3\sqrt{3}}{8}$