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Q.
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
Application of Derivatives
Solution:
$f'\left(x\right)=2\,cos\,2x-1$
$\Rightarrow f'\left(x\right)=0$
$\Rightarrow 2\,cos\,2x-1=0$
$\Rightarrow cos\,2x=\frac{1}{2}$
$\Rightarrow 2x=\frac{\pi}{3}, - \frac{\pi}{3}$
$\Rightarrow x=\frac{\pi}{6}, -\frac{\pi}{6}$
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
$=\frac{\sqrt{3}}{2}-\frac{\pi}{6}-\left(-\frac{\pi}{2}\right)=\frac{\sqrt{3}}{2}+\frac{\pi}{3}$