$A$ 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:

$f$ is a real-valued function satisfying the relation $f\left(3x + \frac{1}{2x}\right) = 9x^2 + \frac{1}{4x^2}$. If $f\left(x + \frac{1}{x}\right) = 1$, then $x =$

Given the function $f(x) = \frac{a^x + a^{-x}}{2}$,where $a > 2$. Then $f(x + y) + f(x - y) = $

If $f: R \rightarrow R$ is defined by $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$,then $\sum_{r=1}^n f(r)=$

If $f(x) = \frac{x}{x - 1} = \frac{1}{y}$,then $f(y) = $

$A$ function $f: R \rightarrow R$ is such that $f(1)=2$ and $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in R$. The area (in square units) enclosed by the lines $2|x|+5|y| \leq 4$ expressed in terms of $f(1)$, $f(2)$, and $f(4)$ is