$f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{\sqrt{x}}$,then:

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

$[x]$ represents the greatest integer function. Let $g(x) = 1 + x - [x]$ and $f(x) = \begin{cases} -3, & x < 0 \\ 0, & x = 0 \\ 5, & x > 0 \end{cases}$. Then $f(g(x))$ is:

Let $f$ and $g$ be two functions defined by $f(x) = \begin{cases} x+1, & x < 0 \\ |x-1|, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x+1, & x < 0 \\ 1, & x \geq 0 \end{cases}$. Then $(g \circ f)(x)$ is

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 $f: R \to R$ is defined by $f(x) = (x + 1)^2$ and $g: R \to R$ is defined by $g(x) = x^2 + 1$,then $(fog)(-3)$ equals: