If the function $f\,:\,R - \,\{ 1, - 1\}  \to A$ defined by $f\,(x)\, = \frac{{{x^2}}}{{1 - {x^2}}},$ is surjective, then $A$ is equal to

Let $f(x) = cos(\sqrt P \,x),$ where $P = [\lambda], ([.]$ is $G.I.F.)$ If the period of $f(x)$ is $\pi$. then

If $f(x)$ be a polynomial function satisfying $f(x).f (\frac{1}{x}) = f(x) + f (\frac{1}{x})$  and $f(4) = 65$ then value of $f(6)$ is

Let $\mathrm{f}: N \rightarrow N$ be a function such that $\mathrm{f}(\mathrm{m}+\mathrm{n})=\mathrm{f}(\mathrm{m})+\mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in N$. If $\mathrm{f}(6)=18$ then $\mathrm{f}(2) \cdot \mathrm{f}(3)$ is equal to :

If for the function $f(x) = \frac{1}{4}{x^2} + bx + 10$ ; $f\left( {12 - x} \right) = f\left( x \right)\,\forall \,x\, \in \,R$ , then the value of $'b'$ is 

Let $R$ be the set of all real numbers and let $f$ be a function from $R$ to $R$ such that $f(x)+\left(x+\frac{1}{2}\right) f(1-x)=1$, for all $x \in R$. Then $2 f(0)+3 f(1)$ is equal to