The number of one-one functions from the set $A = \{1, 2, 3, 4, 5\}$ to itself is . . . . . . .

If the function $f:[-1,1] \rightarrow R$ is defined by $f(x) = \begin{cases} 2^x+1, & \text{for } x \in [-1,0) \\ 1, & \text{for } x=0 \\ 2^x-1, & \text{for } x \in (0,1] \end{cases}$,then in $[-1,1]$,$f(x)$ has

Show that the function $f: R_* \rightarrow R_*$ defined by $f(x) = \frac{1}{x}$ is one-one and onto,where $R_*$ is the set of all non-zero real numbers. Is the result true if the domain $R_*$ is replaced by $N$ with the co-domain remaining the same as $R_*$?

If $(x, y) \in R$ and $x, y \neq 0$,and the function $f(x, y) = \frac{x}{y}$,then this function is a/an:

$A = \{1, 2, 3, 4\}$ 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:

If $A = \{x \mid x \in N, x \leq 5\}$ and $B = \{x \mid x \in Z, x^{2} - 5x + 6 = 0\}$,then the number of onto functions from $A$ to $B$ is: