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

If $f(x) = \frac{3x+2}{5x-3}$,where $x \in R - \{\frac{3}{5}\}$,then:

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 $Y = \{a, b, c\}$.

Let $f : R \rightarrow R$ be defined by $f(x) = x^3 - 3x^2 + 3x - 2$. Then $f^{-1}(x)$ is given by:

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ and $g: R \rightarrow (0, 4)$ be functions defined by $f(x) = \log_e(x^2 + 2x + 4)$ and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $h(x) = (f \circ g^{-1})(x)$,where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $h(x)$ at $x = 2$ is:

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