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

If $f(x)$ and $g(x)$ are two real valued functions such that $f(x)=3x-2$ and $g(x)=x^2+2$, then $[(g \circ f)+(f \circ g)](x) = $

Let $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Define $f: S \rightarrow S$ as $f(n) = \begin{cases} 2n, & \text{if } n = 1, 2, 3, 4, 5 \\ 2n - 11, & \text{if } n = 6, 7, 8, 9, 10 \end{cases}$. Let $g: S \rightarrow S$ be a function such that $f \circ g(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\ n - 1, & \text{if } n \text{ is even} \end{cases}$. Then $g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5))$ is equal to

If $f: R \rightarrow R$ and $g: R^{+} \rightarrow R$ are such that $g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$, then a possible choice for $f$ and $g$ is

Let $f(x) = ax + b$ and $g(x) = cx + d$. The condition $f(g(x)) = g(f(x))$ holds for all $x$ if and only if ...

$f: R \rightarrow R$ and $g: R \rightarrow R$ are two functions such that $f(x)=x^2$ and $g(x)=\frac{1}{x^2}$,then $x^4(f \circ g)(x)$ is equal to