Let $f, g: R \rightarrow R$ be defined as: $f(x)=|x-1|$ and $g(x)=\begin{cases} e^x, & x \geq 0 \\ x+1, & x \leq 0 \end{cases}$. Then the function $f(g(x))$ is

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 = $

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = \begin{cases} x+2, & x>0 \\ 2-x, & x \leq 0 \end{cases}$ and $g(x) = \begin{cases} x^2-2x-2, & 1 \leq x < 2 \\ x-7, & x \geq 2 \\ x+5, & x < 1 \end{cases}$,then $\lim _{x \rightarrow 0} g(f(x))$

Let $f(x)=\sqrt{x^{2}-3x+2}$ and $g(x)=\sqrt{x}$ be two given functions. If $S$ is the domain of $f \circ g$ and $T$ is the domain of $g \circ f$,then:

Given the functions $f: R \rightarrow R$ defined by $f(x) = 2x^2 - 5$ and $g: R \rightarrow R$ defined by $g(x) = \frac{x}{x^2 + 1}$,find the composite function $(g \circ f)(x)$.

Consider the function $f: R \rightarrow R$ defined by $f(x)=\frac{2x}{\sqrt{1+9x^2}}$. If the composition of $f$,$\underbrace{(f \circ f \circ \ldots \circ f)}_{10 \text{ times }}(x) = \frac{2^{10}x}{\sqrt{1+9\alpha x^2}}$,then the value of $\sqrt{3\alpha+1}$ is equal to: