The domain of the function $f(x) = \sqrt{\frac{4-x^2}{[x]+2}}$,where $[x]$ denotes the greatest integer not more than $x$,is

If the function $f: R - \{ 1, - 1\} \to A$ defined by $f(x) = \frac{x^2}{1 - x^2}$ is surjective,then $A$ is equal to

The domain of the real-valued function $f(x) = \frac{\log_2(x+3)}{\sqrt{x^2+3x+2}}$ is

Let the sets $A$ and $B$ denote the domain and range respectively of the function $f(x)=\frac{1}{\sqrt{\lceil x\rceil-x}}$ where $\lceil x \rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements
$(S1): A \cap B = (1, \infty) - \mathbb{N}$ and
$(S2): A \cup B = (1, \infty)$

If $A$ is the domain and $B$ is the range of the function $f(x) = \begin{cases} 3x-1, & x > 1 \\ x^2+1, & x \leq 1 \end{cases}$ then $A-B=$

The function $f(x) = \operatorname{sech}(x)$ on $R$ has the range