It is given that A = R – {3} and B = R – {1}

f: A → B defined by

Now, let x, y ϵ A such that f(x) = f(y)

⇒ (x-2)(y-3) = (y-2)(x-3)

⇒ xy – 3x -2y + 6 = xy -3y -2x +6

⇒ -3x -2y = -3y -2x

⇒ x = y

⇒ f is one –one.

Let y ϵ B = R – {1}

Then, y ≠ 1.

The function f is onto if there exist x ϵ A such that f(x) = y

Now, f(x) = y

⇒ x – 2 = xy -3y

⇒ x(1-y) = -3y + 2

⇒ y ϵ B, there exists ϵ A such that

⇒ f is onto.

Therefore, function f is one- one and onto.