Let $f: R \rightarrow R$ be defined by $f(x)=3 x^2-5$ and $g: R \rightarrow R$ by $g(x)=\frac{x}{x^2+1}$,then $g \circ f$ is

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 the function $f(x) = \frac{1}{2} - \tan \left( \frac{\pi x}{2} \right)$ for $-1 < x < 1$ and $g(x) = \sqrt{3 + 4x - 4x^2}$,then the domain of the composite function $(g \circ f)(x)$ is:

If $f(x)$ and $g(x)$ are two functions with $g(x)=x-\frac{1}{x}$ and $f \circ g(x)=x^3-\frac{1}{x^3}$,then $f^{\prime}(x)$ is equal to

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that $f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$ and $g(n) = n - (-1)^n$. Then $fog$ is

If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=x-[x]$ and $g(x)=[x]$ for $x \in R$,where $[x]$ is the greatest integer not exceeding $x$,then for every $x \in R, f(g(x))$ is equal to