Question

Let A = R-{3} and B = R-{1}. consider the function f: A → B defined by f (x) = \(\frac{x-2}{x-3}\). Is f one-one and onto? Justify your answer.

Answer 1

f : A → B, where A = R- {3}, B = R – {1}, f is defined by

(x1 – 2)(x2 – 3) = (x2 – 2)(x1 – 3)
or x1x2 – 3x1 – 2x2 + 6 = x1x2 – 2x1 – 3x2 + 6
i.e. – 3x1 – 2x2 = -2x1 – 3x2
or -x1 = -x2
⇒ x1 = x2
∴ f is one-one.
(b) Let y = \(\frac{x-2}{x-3}\) ⇒ xy – 3y = x – 2
or x(y – 1) = 3y – 2
⇒ x = \(\frac{3 y-2}{y-1}\)
⇒ For every value of y, except y = 1, there is a pre-image
⇒ x = \(\frac{3 y-2}{y-1}\)
⇒ f is onto.
Thus, f is one-one and onto.