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

$$\begin{gathered} Given f\left(x\right)=x+\sin 2x,x\in \left[\begin{array}{cc}0,& 2\pi \end{array}\right] \\ diff w.r.to x on both sides \\ {\text{f}}^1\left(x\right)=1+2cos2x \\ for max or min f\text{'}\left(x\right)=0 \\ \Rightarrow 1+2\cos 2x=0 \\ \Rightarrow \cos 2x=\frac{-1}{2}=\cos \frac{2\pi }{3} \\ \Rightarrow 2x=\frac{2\pi }{3} \\ \Rightarrow x=\frac{\pi }{3} \\ Now f\left(0\right)=0 \\ f\left(\frac{\pi }{3}\right)=\frac{\pi }{3}+\sin \frac{2\pi }{3}=\frac{\pi }{3}+\frac{\sqrt{3}}{2} \\ f\left(2\pi \right)=2\pi +0=2\pi \end{gathered}$$

$\therefore$Maximum value of f(x) is $2\mathrm{\pi }$