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