6. Application of derivatives — Maths STD 12 Science — Question

==> $f'(x) = \cos 2x + \cos x$

and $f''(x) = - 2\sin 2x - \sin x = - (2\sin 2x + \sin x)$

For maximum or minimum value of $f(x)$, $f'(x) = 0$

$\cos 2x + \cos x = 0$==> $\cos x = - \cos 2x$

==> $\cos x = \cos (\pi \pm 2x)$

$\therefore x = \pi \pm 2x$ or $x = \frac{\pi }{3},\,\, - \pi $

Now $f''\,\left( {\frac{\pi }{3}} \right) = - 2\sin \frac{{2\pi }}{3} - \sin \frac{\pi }{3} $

$= - 2\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 3 }}{2} = - \frac{{3\sqrt 3 }}{2} = - ve$

Hence $f(x)$ is maximum at $x = \frac{\pi }{3}$.