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 $f: R \rightarrow R$ is defined by $f(x)=3^{-x}$. Observe the following statements about it:
$I$. $f$ is one-one
$II$. $f$ is onto
$III$. $f$ is a decreasing function
Out of these, the true statements are:

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

$A$ function from $A = \{x : -1 \leq x \leq 1\}$ to itself which is not a bijection is

The function $f: N \rightarrow Z$ defined by $f(n) = \begin{cases} \frac{n}{2} & , n \text{ is even} \\ -\left(\frac{n-1}{2}\right) & , n \text{ is odd} \end{cases}$ is . . . . . . .

$A$ function $f$ from the set of natural numbers $\mathbb{N}$ to the set of integers $\mathbb{Z}$ is defined by $f(n) = \begin{cases} \frac{n-1}{2}, & \text{if } n \text{ is odd} \\ -\frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$. The function $f$ is: