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

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

If non-zero real numbers $b$ and $c$ are such that $min \,f\left( x \right) > \max \,g\left( x \right)$, where $f\left( x \right) = {x^2} + 2bx + 2{c^2}$  and $g\left( x \right) = {-x^2} - 2cx + {b^2}$$\left( {x \in R} \right)$; then $\left| {\frac{c}{b}} \right|$ lies in the interval

If $f$ is an even function defined on the interval $(-5, 5)$, then four real values of $x$ satisfying the equation $f(x) = f\left( {\frac{{x + 1}}{{x + 2}}} \right)$ are

Domain of the function $f(x) = {\sin ^{ - 1}}\left( {\frac{{2 - |x|}}{4}} \right) + {\cos ^{ - 1}}\left( {\frac{{2 - |x|}}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{{2 - |x|}}{4}} \right)$ is