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 $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{cases}$. Then for all $x$,$f(g(x))$ is equal to

If a real-valued function $f$ is defined by $f(x) = \frac{ax + \sqrt{a^2 - x^2}}{bx}$,then $f$ 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 $n \in N$ then $f$ is $\dots \dots \dots$

The function $f:[0, \infty) \rightarrow [0, \infty)$ defined by $f(x) = \frac{x}{1+x}$ is

If $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .