If $g(x) = 1 - \sqrt{x}$ and $f(g(x)) = 5 + 4\sqrt{x} + x$, then the value of $f(6)$ is...

If $R$ is a relation $  < $ from $A = \{1, 2, 3, 4\}$ to $B = \{1, 3, 5\}$ i.e.,$(a, b) \in R \iff a < b$,then $R \circ R^{-1}$ is

Let $f: R - \{-\frac{1}{2}\} \rightarrow R$ and $g: R - \{-\frac{5}{2}\} \rightarrow R$ be defined as $f(x) = \frac{2x+3}{2x+1}$ and $g(x) = \frac{|x|+1}{2x+5}$. Then the domain of the function $f \circ g$ is:

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

If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

If $f(x) = \frac{1 - x}{1 + x}$, then $f(f(\cos x)) = $