Mathematics - Application of Derivative Question with Solution | TestHub

Given, $f\left(x\right)=x+\sin 2x$

$\Rightarrow f^{\text{'}}\left(x\right)=1+2\cos 2x$

For maximum or minimum value, $f^{\text{'}}\left(x\right)=0$

$\Rightarrow 1+2\cos 2x=0\Rightarrow \cos 2x=-\frac{1}{2}$

$\Rightarrow 2x=2n\pi \pm \frac{2\pi }{3}$

$\Rightarrow x=n\pi \pm \frac{\pi }{3}$

$\Rightarrow x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}$

Find $f\left(0\right), f\left(\frac{\pi }{3}\right),f\left(\frac{2\pi }{3}\right), f\left(\frac{4\pi }{3}\right), f\left(\frac{5\pi }{3}\right), f\left(2\pi \right)$

$\Rightarrow f\left(0\right)=0,f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+\frac{\sqrt{3}}{2},f\left(\frac{2\pi }{3}\right)=\frac{2\pi }{3}-0.8$

$\Rightarrow f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+0.8,f\left(\frac{5\pi }{3}\right)=\frac{5\pi }{3}-0.8$

$\Rightarrow f\left(2\pi \right)=2\pi +0=2\pi$

$\therefore$ Maximum value of $f\left(x\right)=2\pi$