If $f(x) = \frac{1 - x}{1 + x}$, then $f(f(\cos x)) = $

Consider functions $f$ and $g$ such that the composite function $g \circ f$ is defined and is one-one. Are $f$ and $g$ both necessarily one-one?

Let $f(x) = x^3$ and $g(x) = 3^x$,then the quadratic equation whose roots are solutions of the equation $(f \circ g)(x) = (g \circ f)(x)$ (for $x \neq 0$) is

Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then, $(f \circ g)(\pi) + (g \circ f)(e)$ 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 $f(x)=\sqrt{x^{2}-3x+2}$ 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: