$f(x)=x+\sqrt{x^2}$ is a function from $\mathbb{R}\to \mathbb{R}$. Then $f(x)$ is:

$f(x)=x+|x|$.This can be written as:
$$f(x)=\{\begin{array}{ll}x+x=2x& \text{if }x\ge 0\\ x-x=0& \text{if }x<0\end{array}$$
Injective (One-one): For all $x<0$, $f(x)=0$. For example, $f(-1)=0$ and $f(-2)=0$. Since multiple inputs give the same output, it is not injective.Surjective (Onto): The range of the function is $[0,\infty )$, but the codomain is $\mathbb{R}$. Since the range is not equal to the codomain, it is not surjective.Since it is neither injective nor surjective, it is not bijective.