Let $A = \{x \in R : -1 \leq x \leq 1\}$ and $f: A \rightarrow A$ be a mapping defined by $f(x) = x|x|$. Then $f$ is

The number of bijective functions $f: Z \rightarrow Z$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in Z$ is:

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

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

If $f: N \rightarrow Z$ is defined by $f(n)=\begin{cases} 2 & \text{if } n=3k, k \in Z \\ 10 & \text{if } n=3k+1, k \in Z \\ 0 & \text{if } n=3k+2, k \in Z \end{cases}$, then $\{n \in N: f(n)>2\}$ is equal to

Let $A = \{x_1, x_2, x_3, \dots, x_7\}$ and $B = \{y_1, y_2, y_3\}$ be two sets containing seven and three distinct elements respectively. Then the total number of functions $f: A \to B$ which are onto,if there exist exactly three elements $x$ in $A$ such that $f(x) = y_2$,is equal to