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

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3 .$ If $\sum \limits_{i=1}^{n} f(i)=363,$ then $n$ is equal to

Let $N$ be the set of positive integers. For all $n \in N$, let $f_n=(n+1)^{1 / 3}-n^{1 / 3} \text { and }$ $A=\left\{n \in N: f_{n+1}<\frac{1}{3(n+1)^{2 / 3}} < f_n\right\}$ Then,

Domain of the function $f(x)\,=\,\frac{1}{{\sqrt {(x + 1)({e^x} - 1)(x - 4)(x + 5)(x - 6)} }}$

Let ${a_2},{a_3} \in R$ such that $\left| {{a_2} - {a_3}} \right| = 6$ and $f\left( x \right) = \left| {\begin{array}{*{20}{c}}
1&{{a_3}}&{{a_2}}\\
1&{{a_3}}&{2{a_2} - x}\\
1&{2{a_3} - x}&{{a_2}}
\end{array}} \right|,x \in R.$ Then the greatest value of $f(x)$ is

Greatest value of the function, $f(x) =  - 1 + \frac{2}{{{2^x}^2 + 1}}$ is