For $x \in R$,two real-valued functions $f(x)$ and $g(x)$ are such that $g(x) = \sqrt{x} + 1$ and $(f \circ g)(x) = x + 3 - \sqrt{x}$. Then $f(0)$ is equal to:

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:

Let $f, g :(1, \infty) \rightarrow \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. If the range of the function $f \circ g : [2, 4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$,then $\frac{1}{\beta-\alpha}$ is equal to

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=2x+3$ and $g(x)=x^2+7$, then the values of $x$ such that $g(f(x))=8$ are

If $g(x)=x^{2}+x-1$ and $(g \circ f)(x)=4 x^{2}-10 x+5,$ then $f\left(\frac{5}{4}\right)$ is equal to

If $f(x) = (a - x^n)^{1/n},$ where $a > 0$ and $n$ is a positive integer,then $f[f(x)] = $