Question

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

Answer: (C)

$f^{\prime }\left(x\right)=2cos2x-1$
$\Rightarrow f^{\prime }\left(x\right)=0$
$\Rightarrow 2cos2x-1=0$
$\Rightarrow cos2x=\frac{1}{2}$
$\Rightarrow 2x=\frac{\pi }{3},-\frac{\pi }{3}$
$\Rightarrow x=\frac{\pi }{6},-\frac{\pi }{6}$
Now, $f^{\prime }\left(-\frac{\pi }{2}\right)=\frac{\pi }{2}$,
$f^{\prime }\left(\frac{\pi }{2}\right)=-\frac{\pi }{2}$
$f^{\prime }\left(-\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{2}+\frac{\pi }{6}$ 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 reqd. difference
$=\frac{\sqrt{3}}{2}-\frac{\pi }{6}-\left(-\frac{\pi }{2}\right)=\frac{\sqrt{3}}{2}+\frac{\pi }{3}$