If $f: R \rightarrow R$ is defined by $f(x)=|x|$,then

The relation $R$ on the set of natural numbers is defined by $R = \{(a, b) : a = 2b\}$. Then ${R^{-1}}$ is:

Let $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ be a one-one and onto function given by $f(1) = a$,$f(2) = b$,and $f(3) = c$. Show that there exists a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $X = \{1, 2, 3\}$ and $Y = \{a, b, c\}$.

If $f(x) = \frac{x}{1 + x}$,then ${f^{-1}}(x)$ is equal to

Let $f: R \rightarrow R$ be defined by $f(x) = 2x + 6$,which is a bijective mapping. Then $f^{-1}(x)$ is given by:

Let the function $f$ be defined by $f(x) = \frac{2x + 1}{1 - 3x}$,then $f^{-1}(x)$ is