Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals
$-4$
$30$
$-2$
$8$
Suppose $f:[2,\;2] \to R$ is defined by $f(x) = \left\{ \begin{array}{l} - 1\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{for}}\; - 2 \le x \le 0\\x - 1\;\;\;\;\;{\rm{for}}\;0 \le x \le 2\end{array} \right.$, then $\{ x \in ( - 2,\;2):x \le 0$ and $f(|x|) = x\} = $
The domain of the derivative of the function $f(x) = \left\{ \begin{array}{l}{\tan ^{ - 1}}x\;\;\;\;\;,\;|x|\; \le 1\\\frac{1}{2}(|x|\; - 1)\;,\;|x|\; > 1\end{array} \right.$ is
The graph of $y = f(x)$ is shown then number of solutions of the equation $f(f(x)) =2$ is
The range of the function $f(x) = \frac{x}{{1 + \left| x \right|}},\,x \in R,$ is
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\left( x \right) = 120,} $ find the value of $n$