Let a relation $R$ be defined by $R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}$. Then ${R^{ - 1}}oR$ is

Define $f(x) = \begin{cases} 1 + x, & 0 \leq x \leq 2 \\ 3 - x, & 2 < x \leq 3 \end{cases}$. If $f \circ f(x)$ is discontinuous at $a$ and $b$ in $[0, 3]$ and $a < b$, then $2 a + 3 b = $

Two functions $f: R \rightarrow R, g: R \rightarrow R$ 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:

Let $f(x) = \sin^{-1} x$ and $g(x) = \frac{x^2 - x - 2}{2x^2 - x - 6}$. If $g(2) = \lim_{x \to 2} g(x)$,then the domain of the function $f \circ g$ 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

Let $f, g$ and $h$ be functions from $R$ to $R$. Show that:
$\begin{cases} (f+g)oh = foh + goh \\ (f \cdot g)oh = (foh) \cdot (goh) \end{cases}$