The relation $R$ is defined as $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then $R^{-1} o R$ is:

Let $S, T, U$ be three non-void sets and $f: S \rightarrow T, g: T \rightarrow U$ be functions such that $g \circ f: S \rightarrow U$ is surjective. Then,

Let $f(x) = \frac{\alpha x}{x+1}$,$x \neq -1$. If $f(f(x)) = x$,then the value of $\alpha$ is . . . . . . .

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to:

The composite mapping $fog$ of the maps $f:R \to R$,$f(x) = \sin x$,and $g:R \to R$,$g(x) = x^2$ is:

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