Find the number of all one-one functions from set $A = \{1, 2, 3\}$ 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

Let $A = \{1, 2, 3, 4, 5, 6\}$. The number of one-one functions $f: A \to A$ such that $f(1) \ge 3, f(3) \le 4$ and $f(2) + f(3) = 5$,is ————

Let $f: \mathbb{Z} \rightarrow \mathbb{Z}$ be defined by $f(x) = x^3 + 2$. Then,$f$ is . . . . . . .

If $f: R \rightarrow R$ is defined by $f(x)=2x+\sin x, x \in R$,then $f$ is

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