The range of the function $f(x) = \frac{1}{\sqrt{x-[x]}}$ is

Let $[x]$ denote the greatest integer not more than $x$. If $A$ and $B$ are the domains of the functions $f(x)=\frac{x-[x]}{\sqrt{|x|-x}}$ and $g(x)=\frac{x-[x]}{\sqrt{|x|+x}}$ respectively,then

Let $D$ be the domain of the function $f(x) = \sin^{-1} \left(\log_{3x} \left(\frac{6+2 \log_3 x}{-5x}\right)\right)$. If the range of the function $g: D \rightarrow R$ defined by $g(x) = x - [x]$ (where $[x]$ is the greatest integer function) is $(\alpha, \beta)$,then $\alpha^2 + \frac{5}{\beta}$ is equal to

Let $f = \left\{ \left(x, \frac{x^2}{1+x^2} \right) : x \in R \right\}$ be a function from $R$ into $R$. Determine the range of $f$.

$A$ real-valued function $f: A \rightarrow B$ defined by $f(x) = \frac{4-x^2}{4+x^2}$ for all $x \in A$ is a bijection. If $-4 \in A$,then $A \cap B =$

If a real valued function $f:[-1,2] \rightarrow B$ defined by $f(x) = \begin{cases} 1-x, & -1 \leq x \leq 1 \\ x-1, & 1 < x \leq 2 \end{cases}$ is a surjection,then $B=$