Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$,then $2 a + 3 b = $

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 $f: R \rightarrow R$ 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 $f(x) = \frac{3x - 4}{2x - 3}$,then $f(f(f(x)))$ will be

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

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