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

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\}$ 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

Let $f: R \rightarrow R$ be defined as $f(x) = 2x - 1$ and $g: R - \{1\} \rightarrow R$ be defined as $g(x) = \frac{x - 1/2}{x - 1}$. Then the composition function $f(g(x))$ is:

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$,then the domain of the function $f \circ g$ is .... .

If $f$ is the greatest integer function and $g$ is the modulus function,then $(gof)\left( -\frac{5}{3} \right) - (fog)\left( -\frac{5}{3} \right) = $