Question

The difference between the greatest and least values of the function $ f(x) = \sin2x-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, $f(x) =\sin 2 x-x$
$f^{\prime}(x) =2 \cos 2 x-1 $
$f^{\prime \prime}(x) =-4 \sin 2 x$
For maxima or minima, put $f^{\prime}(x)=0$
$\Rightarrow 2 \cos 2 x-1=0 $
$\Rightarrow \cos 2 x=\frac{1}{2}=\cos \frac{\pi}{3} $
$\Rightarrow \cos 2 x=2 n \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} $
$=\sqrt{3}-\frac{\pi}{3}$