Given that A = R - {3} , B = R - {1} Consider the function f : A → B defined by f (x) = $\left(\frac{x-2}{x-3}\right)$ Let x, y ∊ A such that f (x) = f (y) ⇒ $\frac{x-2}{x-3}$ = $\frac{y-2}{y-3}$ ⇒ (x - 2) (y - 3) = (y - 2) (x - 3) ⇒ xy - 3x - 2y + 6 = xy - 3y - 2x + 6 ⇒ - 3x - 2y = - 3y - 2x ⇒ 3x - 2x = 3y - 2y ⇒ x = y ∴ f is one - one Let y ₹ B = R - {1} Then, y ≠ 1. The function f is onto if there exists x ∊ A such that f (x) = y. Now, f (x) = y ⇒ $\frac{x-2}{x-3}$ = y ⇒ x - 2 = y (x - 3) ⇒ x - 2 = xy - 3y ⇒ x - xy = 2 - 3y ⇒ x (1 - y) = 2 - 3y ⇒ x = $\frac{2-3y}{1-y}$ ∊ A [y ≠ 1] ... (1) Thus, for any y ∊ B, there exists $\frac{2-3y}{1-y}$ ∊ A such that r $\left(\frac{2-3y}{1-y}\right)$ = $\frac{\frac{2-3y}{1-y} - 2}{\frac{2-3y}{1-y} - 3}$ = $\frac{2-3y-2+2y}{2-3y-3+3y}$ = $\frac{-y}{-1}$ = y ∴ f is onto. Hence, the function is one-one and onto. Therefore, $f^{-1}$ exists. Consider equation (1). x = $\frac{2-3y}{1-y}$ ∊ A [y ≠ 1] Replace y by x and x by $f^{-1}$ (x) in the above equation, we have, $f^{-1}$ (x) = $\frac{2-3x}{1-x}$ , x ≠ 1