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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.

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Given; 

f: A → B, A = R – {3} and B = R – {1}

f(x1) = f(x2) ⇒ \(\frac{x_1-2}{x_1-3}\) = \(\frac{x_2-2}{x_2-3}\)

⇒ (x1 – 2)(x2 – 3) = (x2 – 2)(x1 – 3)

⇒ x1x2 – 2x2 – 3x1 + 6 = x2x1 – 2x1 – 3x2 + 6

⇒ -2x2 – 3x1 = – 2x1 – 3x2 ⇒ x1 = x2

Hence one-one

Let y ∈ B ⇒ f(x) = y ⇒ \(\frac{x_1-2}{x_1-3}\) = y

⇒ x – 2 = xy – 3y

⇒ x – xy = 2 – 3y

⇒ x(1 – y) = 2 – 3y 

⇒ \(\frac{2-3y}{1-y} = y\) ∈ A =

Hence onto. Therefore objective.

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