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

Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals

Statement $1$ : If $A$ and $B$ be two sets having $p$ and $q$ elements respectively, where $q > p$. Then the total number of functions from set $A$ to set $B$ is $q^P$.
Statement $2$ : The total number of selections of $p$ different objects out of $q$ objects is ${}^q{C_p}$.

Let $A=\{1,2,3,5,8,9\}$. Then the number of possible functions $f : A \rightarrow A$ such that $f(m \cdot n)=f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to $...............$.

If $f\left( x \right) = {\log _e}\,\left( {\frac{{1 - x}}{{1 + x}}} \right)$, $\left| x \right| < 1$, then $f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ is equal to

Let $c, k \in R$. If $f(x)=(c+1) x^{2}+\left(1-c^{2}\right) x+2 k$ and $f(x+y)=f(x)+f(y)-x y$, for all $x, y \in R$, then the value of $|2( f (1)+ f (2)+ f (3)+\ldots \ldots+ f (20)) \mid$ is equal to