Let $A$ be a set containing $10$ distinct elements. Then the total number of distinct functions from $A$ to $A$ is:

The function $f: N \rightarrow N$ defined by $f(x) = \begin{cases} x+1, & x \text{ is odd} \\ x-1, & x \text{ is even} \end{cases}$ is . . . . . . .

If $f(x) = (\frac{3}{5})^x + (\frac{4}{5})^x - 1$,$x \in R$,then the equation $f(x) = 0$ has

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 $n \in N$ then $f$ is $\dots \dots \dots$

Show that the function $f : R \rightarrow \{ x \in R : -1 < x < 1 \}$ defined by $f(x) = \frac{x}{1+|x|}$ for all $x \in R$ is a one-one and onto function.

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