$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ is

Let $A=\{1,2,3\}, \,B=\{4,5,6,7\}$ and let $f=\{(1,4),\,(2,5),\,(3,6)\}$ be a function from $A$ to $B$. Show that $f$ is one-one.

Show that the modulus function $f : R \rightarrow R$ given by $f(x) = |x|$ is neither one-one nor onto,where $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

Consider a function $f: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $f(x) = \sin x$ and $g: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}$ given by $g(x) = \cos x$. Show that $f$ and $g$ are one-one,but $f + g$ is not one-one.

Consider the following statements:
Statement-$I$ : $A$ function $f: A \rightarrow B$ is said to be one-one if and only if $f(x) \neq f(y) \Rightarrow x \neq y$.
Statement-$II$ : $A$ relation $f: A \rightarrow B$ is said to be a function if $x \neq y \Rightarrow f(x) \neq f(y)$.
Then which one of the following is true?

If the function $f:[-1,1] \rightarrow R$ is defined by $f(x) = \begin{cases} 2^x+1, & \text{for } x \in [-1,0) \\ 1, & \text{for } x=0 \\ 2^x-1, & \text{for } x \in (0,1] \end{cases}$,then in $[-1,1]$,$f(x)$ has