Question

Maximum value of $f\left(x\right)=x+\sin 2x,x\in [0,2\pi ]$ is

• A. $\pi$
• B. $2\pi$
• C. $3\pi$
• D. $\pi /2$

Answer 1

The correct option is A $\pi$

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

For maximum-

$⟹f^{\prime }\left(x\right)=1+2\cos 2x=0$

$⟹\cos 2x=\frac{-1}{2}$

$⟹2x=\frac{2\pi }{3},\frac{4\pi }{3},\frac{8\pi }{3},\frac{10\pi }{3}$

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

$f^{\prime \prime }\left(x\right)=-4\sin 2x<0$

$⟹\sin 2x>0$

$2x=\frac{2\pi }{3},\frac{8\pi }{3}$

$⟹x=\frac{\pi }{3},\frac{4\pi }{3}$

Now, we got two point of local maxima.

$f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+\sin \frac{2\pi }{3}$

$⟹f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+\frac{\sqrt{3}}{2}$

And, $f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+\sin \frac{8\pi }{3}$

$⟹f\left(\frac{4\pi }{3}\right)=\frac{4\pi }{3}+\frac{\sqrt{3}}{2}$

Hence overall maximum is at $x=\frac{4\pi }{3}$ and has value $\frac{4\pi }{3}+\frac{\sqrt{3}}{2}.$