Find the number of all onto functions from the set $\{1, 2, 3, \ldots, n\}$ to itself.

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

The function $f: C \rightarrow C$ defined by $f(x) = \frac{ax + b}{cx + d}$ for $x \in C$,where $ad - bc \neq 0$,reduces to a constant function if:

The number of functions $f$ from the set $A = \{x \in N: x^{2}-10x+9 \leq 0\}$ to the set $B = \{n^{2}: n \in N\}$ such that $f(x) \leq (x-3)^{2}+1$ for every $x \in A$ is:

Let $A = \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), (4, 2) \}$. The correct statement 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: