Let $f : R -\{0,1\} \rightarrow R$ be a function such that $f(x)+f\left(\frac{1}{1-x}\right)=1+x$. Then $f(2)$ is equal to :

Range of the function , $f (x) = cot ^{-1}$ $\left( {{{\log }_{4/5}}\,\,(5\,{x^2}\,\, - \,\,8\,x\,\, + \,\,4)\,} \right)$ is :

The number of functions $f :\{1,2,3,4\} \rightarrow\{ a \in Z :| a | \leq 8\}$ satisfying $f ( n )+$ $\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$ is

Let $R$ be the set of real numbers and $f: R \rightarrow R$ be defined by $f(x)=\frac{\{x\}}{1+[x]^2}$, where $[x]$ is the greatest integer less than or equal to $x$, and $\left\{x{\}}=x-[x]\right.$. Which of the following statements are true?

$I.$ The range of $f$ is a closed interval.

$II.$ $f$ is continuous on $R$.

$III.$ $f$ is one-one on $R$

The function $f\left( x \right) = \left| {\sin \,4x} \right| + \left| {\cos \,2x} \right|$, is a periodic function with period

Function $f(x)={\left( {1 + \frac{1}{x}} \right)^x}$ then Domain of $f (x)$ is