If $f(x) = x - \frac{1}{x}$, $x \neq 0$, then $3f(x) =$

If $(f(x))^2 = f(x^2) + f(1)$ holds good,then find $f(x)$.

Let $f$ be a non-zero real-valued continuous function satisfying $f(x+y) = f(x) \cdot f(y)$ for all $x, y \in R$. If $f(2) = 9$, then $f(6)$ is equal to

If $f(x+2y, x-2y) = xy$, then $f(x, y)$ is equal to

If $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y), \forall x, y \in R$ and $f(1)=10$, then, $\sum_{r=1}^n(f(r))^2=$

If $g:[-2, 2] \to R$ where $g(x) = x^3 + \tan x + \left[ \frac{x^2 + 1}{P} \right]$ is an odd function,then the value of the parameter $P$ is: