If $f(x)=|x-1|+|x-2|+|x-3|$ for $2 < x < 3$,then $f$ is

Let $Z$ denote the set of integers. Define $f: Z \rightarrow Z$ by $f(x) = \begin{cases} \frac{x}{2}, & x \text{ is even} \\ 0, & x \text{ is odd} \end{cases}$. Then $f$ is:

Let $f: R \rightarrow R$ be defined by $f(x) = 5^{-|x|} + \operatorname{sgn}(5^{-x})$, where $\operatorname{sgn}(x)$ denotes the signum function of $x$. Then $f$ is

$f(x)=\frac{x}{e^x-1}+\frac{x}{2}+2 \cos ^3 \frac{x}{2}$ on $R-\{0\}$ is

Show that the modulus function $f : R \rightarrow R$ given by $f(x) = |x|$ is neither one-one nor onto,where $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

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