Let $f, g: R \rightarrow R$ be two functions defined as $f(x)=|x|+x$ and $g(x)=|x|-x$ for all $x \in R$. Then $(f \circ g)(x)$ for $x < 0$ is

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:

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Then $R \circ S = $

For every real number $x \neq -1$, let $f(x) = \frac{x}{x+1}$. Define $f_1(x) = f(x)$ and for $n \geq 2$, $f_n(x) = f(f_{n-1}(x))$. Then the product $f_1(-2) \cdot f_2(-2) \cdot \ldots \cdot f_n(-2)$ is equal to:

Let $f(x) = 1 - x$, $g(x) = \frac{1}{1 - x}$, and $h(x) = \frac{1}{x}$ be three functions, for $x \neq 0, 1$. If a function $F(x)$ satisfies $f(F(h(x))) = g(x)$, then which of the following is true?

If $f(x) = |\cos x|$ and $g(x) = [x]$,then $gof(x)$ is equal to