$A$ function $f: R \rightarrow R$ satisfies $f\left(\frac{x+y}{3}\right) = \frac{f(x)+f(y)+f(0)}{3}$ for all $x, y \in R$. If the function $f$ is differentiable at $x=0$,then $f$ is:

If $f(x)$ is a polynomial function satisfying $f(x) \cdot f\left(\frac{1}{x}\right)=f(x)+f\left(\frac{1}{x}\right)$ and $f(4)=257$,then $f(3)=$

If $f(x + y, x - y) = xy$,then the arithmetic mean of $f(x, y)$ and $f(y, x)$ is

If $f(x)$ is a polynomial function satisfying the condition $f(x) \cdot f(1/x) = f(x) + f(1/x)$ and $f(2) = 9$,then:

Let $f: R^{+} \rightarrow R^{+}$ be a function satisfying $f(x) - x = \lambda$ (constant),$\forall x \in R^{+}$ and $f(x f(y)) = f(x y) + x, \forall x, y \in R^{+}$. Then $\lim _{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}} - 1}{(f(x))^{\frac{1}{2}} - 1} =$

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