Question

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

• A. $\frac{\sqrt{3}+\sqrt{2}}{2}$
• B. $\frac{\sqrt{3}+\sqrt{2}}{2}+\frac{\pi }{6}$
• C. $\frac{\sqrt{3}}{2}+\frac{\pi }{3}$
• D. $\frac{\sqrt{3}+\sqrt{2}}{2}-\frac{\pi }{3}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}+\frac{\pi }{3}$

$f^{\prime }\left(x\right)=2\cos 2x-1$

$\therefore f^{\prime }\left(x\right)=0\Rightarrow 2\cos 2x-1=0$

$\Rightarrow \cos 2x=\frac{1}{2}\Rightarrow 2x=\frac{\pi }{3},-\frac{\pi }{3}\Rightarrow x=\frac{\pi }{6},-\frac{\pi }{6}$

Now, $f(-\frac{\pi }{2})=\frac{\pi }{2},f^{\prime }\left(\frac{\pi }{2}\right)=-\frac{\pi }{2}$

$f^{\prime }(-\frac{\pi }{6})=\frac{\sqrt{3}}{2}+\frac{\pi }{6}$ and $f^{\prime }\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$

Clearly $\frac{\sqrt{3}}{2}-\frac{\pi }{6}$ is the greatest value of $f\left(x\right)$ and its least value $=-\frac{\pi }{2}$

Hence the required difference $\frac{\sqrt{3}}{2}-\frac{\pi }{6}(-\frac{\pi }{2})=\frac{\sqrt{3}}{2}+\frac{\pi }{3}$