$f(x) = x + \sqrt{x^2}$ is a function from $R \to R$,then $f(x)$ 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	$4$. $A = B = R^{+}$

$A = \{1, 2, 3, 4\}$ and $B = \{1, 2, 3, 4, 5, 6\}$ are two sets, and the function $f: A \rightarrow B$ is defined by $f(x) = x + 2$ for all $x \in A$. Then the function $f$ is:

The function $f(x) = \frac{e^{2x} - 1}{e^{2x} + 1}$ is

Show that an onto function $f: \{1, 2, 3\} \rightarrow \{1, 2, 3\}$ is always one-one.

If $f(x) = \sin([\pi^2]x) - \sin([-\pi^2]x)$,where $[x]$ denotes the greatest integer function $\leq x$,then which of the following is not true?