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 :

If $f({x_1}) - f({x_2}) = f\left( {\frac{{{x_1} - {x_2}}}{{1 - {x_1}{x_2}}}} \right)$ for ${x_1},{x_2} \in [ - 1,\,1]$, then $f(x)$ is

Let $f(x) = (1 + {b^2}){x^2} + 2bx + 1$ and $m(b)$ the minimum value of $f(x)$ for a given $b$. As $b$ varies, the range of $m(b)$ is

The range of function $f : R \rightarrow  R$, $f(x) = \frac{{{{(x\, + \,1)}^4}}}{{{x^4} + \,1}}$ is

If $f( x + y )=f( x ) f( y )$ and $\sum \limits_{ x =1}^{\infty} f( x )=2, x , y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is

Let $f$ be a function satisfying $f(xy) = \frac{f(x)}{y}$ for all positive real numbers $x$ and $y.$ If $ f(30) = 20,$ then the value of $f(40)$ is-