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

Let for $a \ne {a_1} \ne 0,$ $f\left( x \right) = a{x^2} + bx + c\;,g\left( x \right) = {a_1}{x^2} + {b_1}x + {c_1},p\left( x \right) = f\left( x \right) - g\left( x \right),$ If $p\left( x \right) = 0$ only for  $ x=-1 $ and $p\left( { - 2} \right) = 2$ then value of $p\left( 2 \right)$ 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 $f(x) = \cos (\log x)$, then the value of $f(x).f(4) - \frac{1}{2}\left[ {f\left( {\frac{x}{4}} \right) + f(4x)} \right]$

Domain of the function $f(x) =$ $\frac{1}{{\sqrt {\ln \,{{\cot }^{ - 1}}x} }}$ is

If $f(x) = \frac{{\alpha \,x}}{{x + 1}},\;x \ne - 1$. Then, for what value of $\alpha $ is $f(f(x)) = x$