If $f: R \to R$ is a continuous function such that $|f(x) - f(y)| \geqslant |e^x - e^y|$ for all $x, y \in R$,then $f(x)$ is:

Match the functions of List-$I$ with their nature in List-$II$ and choose the correct option.

$A$. $f: R \rightarrow R$ defined by $f(x) = \cos(112x - 37)$	$I$. Injection but not surjection
$B$. $f: A \rightarrow B$ defined by $f(x) = x|x|$ when $A = [-2, 2]$ and $B = [-4, 4]$	$II$. Surjection but not injection
$C$. $f: R \rightarrow R$ defined by $f(x) = (x-2)(x-3)(x-5)$	$III$. Bijection
$D$. $f: N \rightarrow N$ defined by $f(n) = n+1$	$IV$. Neither injection nor surjection
	$V$. Composite function

Let $f: R \rightarrow R$ be defined by $f(x) = x^{2} - \frac{x^{2}}{1+x^{2}}$ for all $x \in R$. Then,

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 $f: R \rightarrow R$,then the function $f(x) = x|x|$ is:

The function $f: N \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2} & , n \text{ is even} \\ -\left(\frac{n-1}{2}\right) & , n \text{ is odd} \end{cases}$ is . . . . . . .