The domain of the function $f(x) = \frac{{{{\sin }^{ - 1}}(x - 3)}}{{\sqrt {9 - {x^2}} }}$ is

Let $A= \{1, 2, 3, 4\}$ and $R : A \to A$ be the relation defined by $R = \{ (1, 1), (2, 3), (3, 4), ( 4, 2) \}$. The correct statement is

If $\,\,f(x) = \left\{ {\begin{array}{*{20}{c}}
  {3 + x;\,\,\,\,\,x \geqslant 0} \\ 
  {2 - 3x;\,\,\,\,\,x < 0} 
\end{array}} \right.$ then $\mathop {\lim }\limits_{x \to 0} f(f(x))$ is equal to -

If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$, for all $x,\;y \in R$ and $f(1) = 7$, then $\sum\limits_{r = 1}^n {f(r)} $ is

The largest interval lying in $\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)$ for which the function, $f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {\frac{x}{2} - 1} \right) + \log \left( {\cos x} \right)$  is defined is

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\} = $