If $f: A \rightarrow B$ and $g: B \rightarrow C$ are functions such that $g \circ f: A \rightarrow C$ is onto,then a necessary condition is:

Let $Q$ be the set of all rational numbers in $[0,1]$ and $f:[0,1] \rightarrow [0,1]$ be defined by $f(x) = \begin{cases} x & \text{for } x \in Q \\ 1-x & \text{for } x \notin Q \end{cases}$. Then,the set $S = \{x \in [0,1] : (f \circ f)(x) = x\}$ is equal to

Find $gof$ and $fog$,if $f: R \rightarrow R$ and $g: R \rightarrow R$ are given by $f(x) = \cos x$ and $g(x) = 3x^2$. Show that $gof \neq fog$.

If $f: R \rightarrow R$ is defined by $f(x) = \frac{x}{x^{2}+1}$,find $f(f(2))$.

Let $f, g :(1, \infty) \rightarrow \mathbb{R}$ be defined as $f(x) = \frac{2x+3}{5x+2}$ and $g(x) = \frac{2-3x}{1-x}$. If the range of the function $f \circ g : [2, 4] \rightarrow \mathbb{R}$ is $[\alpha, \beta]$,then $\frac{1}{\beta-\alpha}$ is equal to

Find the function $g(t)$ if $f(t)=3t-2$ and $(g \circ f)^{-1}(t)=t-2$.