The function $f(x) = \log (x + \sqrt {{x^2} + 1} )$ is:

Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$. Then $f$ is:

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	$4$. $A = B = R^{+}$

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

Let $f : R \to R$ be defined by $f(x) = \frac{ax^2 + ax + b}{ax + b}$. Then:

Show that the Signum Function $f: R \rightarrow R$,given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$ is neither one-one nor onto.