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. $\frac{\sqrt3+\sqrt2}{2}$
• B. $\frac{\sqrt3+\sqrt2}{2}+\frac{\pi}{6}$
• C. $\frac{\sqrt3}{2}+\frac{\pi}{3}$
• D. $\frac{\sqrt3+\sqrt2}{2}-\frac{\pi}{3}$

Answer 1

The Correct Option is C

$f'\left(x\right)=2\,cos\,2x-1$ $\Rightarrow f'\left(x\right)=0$ $\Rightarrow 2\,cos\,2x-1=0$ $\Rightarrow cos\,2x=\frac{1}{2}$ $\Rightarrow 2x=\frac{\pi}{3}, - \frac{\pi}{3}$ $\Rightarrow x=\frac{\pi}{6}, -\frac{\pi}{6}$ Now, $f'\left(-\frac{\pi}{2}\right)=\frac{\pi}{2}$, $f'\left(\frac{\pi}{2}\right)=-\frac{\pi}{2}$ $f'\left(-\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}$ and $f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi}{6}$ Clearly $\frac{\sqrt{3}}{2}-\frac{\pi}{6}$ is the greatest value of $f\left(x\right)$ and its least value $=-\frac{\pi}{2}$. Hence the reqd. difference $=\frac{\sqrt{3}}{2}-\frac{\pi}{6}-\left(-\frac{\pi}{2}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{3}$