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

$f : R \to R$ is defined as $f(x) = \begin{cases} x^2 + 2mx - 1, & x \leq 0 \\ mx - 1, & x > 0 \end{cases}$. If $f(x)$ is one-one,then the set of values of $m$ is:

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

Let $f : N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$ for all $n \in N$. State whether the function $f$ is bijective. Justify your answer.

If $f : R \rightarrow R$ is defined as $f(x) = x - [x] + 3$,$\forall x \in R$,then $f$ is

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