If A = R – {3} and B = R – {1} and f : A → B is a mapping define

1.	If A = R – {3} and B = R – {1} and f : A → B is a mapping defined by f (x) = \(\frac{x-2}{x-3}\) , Show that f is one-one onto.
Answer» One-One Let x, y be any two elements of A, then f (x) = f (y) ⇒ \(\frac{x-2}{x-3} = \frac{y-2}{y-3}\) ⇒ (x – 2) (y – 3) = (y – 2) (x – 3) ⇒ xy – 2y – 3x + 6 = xy – 2x – 3y + 6 ⇒ – x = – y ⇒ x = y ⇒ f is one-one. Onto Let y be an element of B. Then f (x) = y ⇒ y = \(\frac{(x-2)}{(x-3)}\) ⇒ (x – 3)y = (x – 2) ⇒ xy – 3y = x – 2 ⇒ xy – x = 3y – 2 ⇒ x (y – 1) = 3y – 2 ⇒ x = \(\frac{3y-2}{y-1}\) For y ≠ 1, x = \(\frac{3y-2}{y-1}\) is a real number. Also, A = R – {3} ⇒ \(\frac{3y-2}{y-1}\) ≠ 3, because if we take \(\frac{3y-2}{y-1}\) = 3, then 3y – 2 = 3y – 3) ⇒ 2 = 3 which is not true. Hence, f is onto. ⇒ f is both one-one and onto.

If A = R – {3} and B = R – {1} and f : A → B is a mapping defined by f (x) = \(\frac{x-2}{x-3}\) , Show that f is one-one onto.

Answer»

One-One

Let x, y be any two elements of A, then

f (x) = f (y) ⇒ \(\frac{x-2}{x-3} = \frac{y-2}{y-3}\)

⇒ (x – 2) (y – 3) = (y – 2) (x – 3)

⇒ xy – 2y – 3x + 6 = xy – 2x – 3y + 6

⇒ – x = – y

⇒ x = y

⇒ f is one-one.

Onto

Let y be an element of B. Then

f (x) = y

⇒ y = \(\frac{(x-2)}{(x-3)}\)

⇒ (x – 3)y = (x – 2) ⇒ xy – 3y = x – 2

⇒ xy – x = 3y – 2 ⇒ x (y – 1) = 3y – 2

⇒ x = \(\frac{3y-2}{y-1}\)

For y ≠ 1, x = \(\frac{3y-2}{y-1}\) is a real number.

Also, A = R – {3} ⇒ \(\frac{3y-2}{y-1}\) ≠ 3, because if we take \(\frac{3y-2}{y-1}\) = 3, then 3y – 2 = 3y – 3)

⇒ 2 = 3 which is not true.

Hence, f is onto.

⇒ f is both one-one and onto.