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

Let $f(x)=3+2x$ and $g_n(x)=(f \circ f \circ f \circ \dots \text{n times})(x)$. For all $n \in N$,if all the lines $y=g_n(x)$ pass through a fixed point $(\alpha, \beta)$,then $\alpha+\beta=$

$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=x^2$ and $g(x)=\frac{1}{x^2}$,then $x^4(f \circ g)(x)$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$, where $[x]$ is the greatest integer not exceeding $x$, then for every $x \in R, f(g(x))$ is equal to

If $f(x)=e^x$ and $h(x)=(f \circ f)(x)$, then $\frac{h^{\prime}(x)}{h(x)}=$

If $f:R \to R$ and $g:R \to R$ are given by $f(x) = |x|$ and $g(x) = |x|$ for each $x \in R$,then $\{ x \in R : g(f(x)) \le f(g(x)) \} = $