Application of Derivatives — Class 12 Maths Solution

Let $f\left( x \right) = \sin 2x,\,\,0\,\, \le x\,\, \le 2\pi$
$\Rightarrow f'\left( x \right) = 2\cos 2x$

For critical points, $f'\left( x \right) = 0 \Rightarrow 2\cos 2x = 0 \Rightarrow \cos 2x = 0$
$\Rightarrow 2x = \cfrac{\pi }{2},\cfrac{{3\pi }}{2},\cfrac{{5\pi }}{2},\cfrac{{7\pi }}{2}$
$\Rightarrow x = \cfrac{\pi }{4},\cfrac{{3\pi }}{4},\cfrac{{5\pi }}{4},\cfrac{{7\pi }}{4}$

So to find the maximum and minimum values, we have to evaluate $f\left( 0 \right),\,f\left( {2\pi } \right),\,f\left( {\cfrac{\pi }{4}} \right),\,f\left( {\cfrac{{3\pi }}{4}} \right),\,f\left( {\cfrac{{5\pi }}{4}} \right),\,f\left( {\cfrac{{7\pi }}{4}} \right)$

Now, $f(0) = \sin (2 \times 0) = 0,f(2\pi ) = \sin (2 \times 2\pi ) = 0$,
$f\left( {\cfrac{\pi }{4}} \right) = \sin \left( {2 \times \cfrac{\pi }{4}} \right) = \sin \cfrac{\pi }{2} = 1$

$f\left( {\cfrac{{3\pi }}{4}} \right) = \sin \left( {2 \times \cfrac{{3\pi }}{4}} \right) = \sin \cfrac{{3\pi }}{2} = \sin \left( {\pi + \cfrac{\pi }{2}} \right), = - \sin \cfrac{\pi }{2} = - 1$

$f\left( {\cfrac{{5\pi }}{4}} \right) = \sin \left( {2 \times \cfrac{{5\pi }}{4}} \right) = \sin \cfrac{{5\pi }}{2} = \sin \cfrac{\pi }{2} = 1$
and $f\left( {\cfrac{{7\pi }}{4}} \right) = \sin \left( {2 \times \cfrac{{7\pi }}{4}} \right) = \sin \cfrac{{7\pi }}{2} = \sin \cfrac{{ - \pi }}{2} = - 1$

Therefore the maximum value of $f(x)$ is 1 at $x = \cfrac{\pi }{4}$ and $x = \cfrac{{5\pi }}{4}$.