If $f\left( x \right) = {\log _e}\,\left( {\frac{{1 - x}}{{1 + x}}} \right)$, $\left| x \right| < 1$, then $f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is equal 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

$( S 1): A \cap B =(1, \infty)-N$ and

$( S 2): A \cup B=(1, \infty)$

The number of one-one function $f :\{ a , b , c , d \} \rightarrow$ $\{0,1,2, \ldots ., 10\}$ such that $2 f(a)-f(b)+3 f(c)+$ $f ( d )=0$ is

The domain of ${\sin ^{ - 1}}\left[ {{{\log }_3}\left( {\frac{x}{3}} \right)} \right]$ is

If $f(x) = \log \frac{{1 + x}}{{1 - x}}$, then $f(x)$ is

If $f(x) = \cos (\log x)$, then the value of $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$