Prove that the function $f: R \rightarrow R$, given by $f(x)=2 x,$ is one-one and onto.

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\} = $

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 $n$

If $f( x + y )=f( x ) f( y )$ and $\sum \limits_{ x =1}^{\infty} f( x )=2, x , y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is

If $x \in [0, 1]$, then the number of solution $(s)$ of the equation $2[cos^{-1}x] + 6[sgn(sinx)] = 3$ is (where $[.]$ denotes greatest integer function and sgn $(x)$ denotes signum function of $x$)-