If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$ such that $f(1)=3$ and $\sum\limits_{x = 1}^n {f(x) = 120}$,find the value of $n$.

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals $-1$ and $f(0)=1$, then $f(2)=$

Let $f(x+y)=f(x)f(y)$ for all $x, y$ where $f(0) \neq 0$. If $f(5) = 2$ and $f'(0) = 3$,then $f'(5)$ is equal to

If $f(x)=x^2-2x+4$, then the set of values of $x$ satisfying $f(x-1)=f(x+1)$ is

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:

Let $f$ be a function such that $3f(x) + 2f\left(\frac{m}{19x}\right) = 5x$, $x \neq 0$, where $m = \sum_{i=1}^9 (i)^2$. Then $f(5) - f(2)$ is equal to