Question

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

• A. $\pi$
• B. $\sqrt{3}-\pi /3$
• C. $-\sqrt{3}+\pi /3$
• D. None of the above

Answer 1

Answer: (B)

Given, $f(x)=\sin 2x-x$
$f^{\prime }(x)=2\cos 2x-1$
$f^{\prime \prime }(x)=-4\sin 2x$
For maxima or minima, put $f^{\prime }(x)=0$
$\Rightarrow 2\cos 2x-1=0$
$\Rightarrow \cos 2x=\frac{1}{2}=\cos \frac{\pi }{3}$
$\Rightarrow \cos 2x=2n\pi \pm \frac{\pi }{3}$
$\Rightarrow x=n\pi \pm \frac{\pi }{6}$
For $n=0,x=\pm \frac{\pi }{6}$
Now, at $x=\frac{\pi }{6}$,
$f^{\prime \prime }(x)=-4\times \frac{\sqrt{3}}{2}<0$
$\therefore \text{ maxima }$
at $x=-\frac{\pi }{6}$
$f^{\prime \prime }(x)=4\times \frac{\sqrt{3}}{2}>0$
$\therefore \text{ minima }$
Now, $f\left(\frac{\pi }{6}\right)=\sin 2\left(\frac{\pi }{6}\right)-\frac{\pi }{6}=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$
$f^{\prime }\left(-\frac{\pi }{6}\right)=\sin 2\left(-\frac{\pi }{6}\right)+\frac{\pi }{6}=-\frac{\sqrt{3}}{2}+\frac{\pi }{6}$
Thus, greatest value of $f(x)=\frac{\sqrt{3}}{2}-\frac{\pi }{6}$
and least value of $f(x)=-\frac{\sqrt{3}}{2}+\frac{\pi }{6}$
Hence, required difference
$=\left(\frac{\sqrt{3}}{2}-\frac{\pi }{6}\right)-\left(-\frac{\sqrt{3}}{2}+\frac{\pi }{6}\right)$
$=\frac{2\sqrt{3}}{2}-\frac{2\pi }{6}$
$=\sqrt{3}-\frac{\pi }{3}$