If $[x]$ denotes the greatest integer $\leq x$, then the domain of the function $f(x)=\sqrt{\frac{4-x^2}{[x]+2}}$ is

If $f: R-\{2\} \rightarrow R$ is a function defined by $f(x)=\frac{x^2-4}{x-2}$,then the range is

The real-valued function $f(x) = \frac{\operatorname{cosec}^{-1} x}{\sqrt{x - [x]}}$,where $[x]$ denotes the greatest integer less than or equal to $x$,is defined for all $x$ belonging to:

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 the domain of the function $f(x) = \frac{\sqrt{x^2-25}}{4-x^2} + \log_{10}(x^2+2x-15)$ is $(-\infty, \alpha) \cup [\beta, \infty)$,then $\alpha^2 + \beta^3$ is equal to:

The largest interval lying in $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ for which the function $f(x) = 4^{-x^2} + \cos^{-1}\left( \frac{x}{2} - 1 \right) + \log(\cos x)$ is defined is: