If $R$ denotes the set of all real numbers,then the function $f: R \rightarrow R$ defined by $f(x)=|x|$ is

If the function $f: R \rightarrow R$ is defined by $f(x) = (x^{2} + 1)^{35}, \forall x \in R,$ then $f$ is

If $f(x) = \begin{cases} [x], & -3 < x \leq -1 \\ |x|, & -1 < x < 1 \\ |[x]|, & 1 \leq x < 3 \end{cases}$ then the set $\{x : f(x) \geq 0\}$ is equal to

Let $R$ denote the set of all real numbers and $R^{+}$ denote the set of all positive real numbers. For the subsets $A$ and $B$ of $R$,define $f: A \rightarrow B$ by $f(x) = x^2$ for $x \in A$. Match the items in Column-$I$ with the items in Column-$II$.

Column-$I$	Column-$II$
$A$. $f$ is one-one and onto,if	$1$. $A = R^{+}, B = R$
$B$. $f$ is one-one but not onto,if	$2$. $A = B = R$
$C$. $f$ is onto but not one-one,if	$3$. $A = R, B = R^{+}$
$D$. $f$ is neither one-one nor onto,if	$4$. $A = B = R^{+}$

The set $A$ has $4$ elements and the set $B$ has $5$ elements. Then,the number of injective mappings that can be defined from $A$ to $B$ is:

$A$ function $f: R - \{ 0 \} \rightarrow R$ is defined as $f(x) = \begin{cases} x^2 + 3x - 7, & x > 0 \\ h(x), & x < 0 \end{cases}$ If $f(x)$ is an odd function, then $h(x) =$