$\text{ Let the function }f:R\to R\text{ be defined by }f\left(x\right)=2x+\sin x\text{ for }x\in R\text{ . Then }f\text{ is }$

$\text{ Given that }f\left(x\right)=2x+\sin x,x\in R$

$\begin{array}{r}\text{ or }{f}^{\mathrm{\prime }}\left(x\right)=2+\cos x\\ \text{ But }-1\le \cos x\le 1\\ \text{ or }1\le 2+\cos x\le 3\\ \therefore f^{\mathrm{\prime }}\left(x\right)>0\mathrm{\forall }x\in R\end{array}$

$$\begin{gathered} \text{ Therefore, }f\left(x\right)\text{ is strictly increasing and, hence, one-one. } \\ \text{ Also, as }x\to \mathrm{\infty },f\left(x\right)\to \mathrm{\infty },\text{ and }x\to -\mathrm{\infty },f\left(x\right)\to -\mathrm{\infty }\text{ . Therefore, } \end{gathered}$$

$\text{ Range of }f\left(x\right)=R=\mathrm{co}\text{-}\mathrm{domain}\text{ of }f\left(x\right)$

$\text{ Hence, }f\left(x\right)\text{ is onto. }$

$\text{ Thus, }f\left(x\right)\text{ is one-one and onto. }$