Question

Let A = R - [2] and B = R - [1] If f : A → B is a function defined by f(x) = \(\frac{x-1}{x-2}\) show that f is one-one and onto. Find f^-1.

Answer 1

Given, A = R - {2} and B = R - {1}

f: A → B id defined as f(x) = \(\frac{x-1}{x-2}\)

(i) One-one : Let x, y e A such that f(x) = f(y)

⇒ \(\frac{x-1}{x-2}\) = \(\frac{y-1}{y-2}\) ⇒ (x -1) (y - 2) = (x - 2) (y -1)

⇒ xy - 2x - y + 2 = xy - x - 2y + 2

⇒ -2x - y = -x - 2y

⇒ 2x - x = 2y - y

⇒ x = y

So, f is one-one function.

(ii) Onto : Let y ∈ B = R - {1}. Then y ≠ 1.

The function f is onto if there exists x ∈ A such that f(x) = y

Therefore, f is onto.

Hence, function f is one-one and onto.

From equation (i), we get f^-1(y) = \(\frac{1 - 2y}{1 - y}\)

⇒ f^-1 (x) = \(\frac{1 - 2x}{1 - x}\), which is the inverse of f(x)