If $f(x) = \sqrt{x} - 1$ and $g\{f(x)\} = x + 2\sqrt{x} + 1$,then $g(x) = $

Suppose that $g(x) = 1 + \sqrt{x}$ and $f(g(x)) = 3 + 2\sqrt{x} + x$,then $f(x)$ is

If $f(x) = \frac{4x+3}{6x-4}$,$x \neq \frac{2}{3}$ and $(f \circ f)(x) = g(x)$,where $g: R - \{\frac{2}{3}\} \rightarrow R - \{\frac{2}{3}\}$,then $(g \circ g \circ g)(4)$ is equal to

If $f(x)=2^{100} x+1$ and $g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f(g(x))=x$ is

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

If $f: R - \{\frac{3}{7}\} \rightarrow R - \{\frac{3}{7}\}$ is given by $f(x) = \frac{3x+5}{7x-3}$,then the statement which is not true is: