$f:R\to R\text{ defined by }f\left(x\right)=\frac{1}{2}x\left|x\right|+\cos x+1\text{ is }$

$f\left(x\right) = \frac{x\left|x\right|}{2}+\cos x+1$

$If x\ge 0 then f\left(x\right) = \frac{x^2}{2}+\cos x +1$

$f\text{'}\left(x\right) = x - \sin x \ge 0 for all x\ge 0$ 

$if x<0 then f\left(x\right) = \frac{-x^2}{2} + \cos x +1 \Rightarrow f\text{'}\left(x\right)=-x -\sin x >0 \sin ce x is negative$

 therefore $f\text{'}\left(x\right) >0 for all x \in R \Rightarrow f is increa\sin g , hence f is one-one$ 

$and f\left(x\right)\to \infty when x\to \infty and f\left(x\right)\to -\infty when x\to -\infty$

  Range is = R ,    hence f is onto