Question

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

• A. $\pi$
• B. 0
• C. $\frac{\sqrt{3}}{2}+\frac{\pi }{3}$
• D. $-\frac{\sqrt{3}}{2}+\frac{2\pi }{3}$

Answer 1

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

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

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

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

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

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

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

and its least value is $-\frac{\pi }{2}$

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