Let A = R – {3} and B = R – {1}. Consider the function f : A → B

1.	Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by \(f(x) = (\frac{x-2}{x-3})\). Show that f is one-one and onto and hence find f–1.
Answer» One - one : Let x₁, x₂ ∈ A Now, f (x₁) = f (x₂) ⇒ \(\frac{x_1\,-\,2}{x_1\,-\,3}=\frac{x_2\,-\,2}{x_2\,-\,3}\) ⇒ (x₁ - 2) (x₂ - 3) = (x₁ - 3) (x₂ - 2) ⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁ x₂ - 2x₁ - 3x₂ + 6 ⇒ - 3x₁ - 2x₂ = - 2x₁ - 3x₂ ⇒ - x₁ = - x₂ ⇒ x₁ = x₂ Hence, f is one - one function. Onto : Let y = \(\frac{x\,-\,2}{x\,-\,3}\) ⇒ xy - 3y = x - 2 ⇒ xy - x = 3y - 2 ⇒ x(y - 1) = 3y - 2 ⇒ \(x = \frac{3y\,-\,2}{y\,-\,1}\) .....(i) From above it is obvious that ∀ y except 1, i.e, ∀ y ∈ B = R - {1}∃ x ∈ A Hence, f is onto function. Thus, f is one - one onto function. If f^-1 is inverse function of f then, \(f^{-1}(y) = \frac{3y\,-\,2}{y\,-\,1}\) [from (i)]

Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by \(f(x) = (\frac{x-2}{x-3})\). Show that f is one-one and onto and hence find f–1.

Answer»

One - one :

Let x₁, x₂ ∈ A

Now,

f (x₁) = f (x₂)

⇒ \(\frac{x_1\,-\,2}{x_1\,-\,3}=\frac{x_2\,-\,2}{x_2\,-\,3}\)

⇒ (x₁ - 2) (x₂ - 3) = (x₁ - 3) (x₂ - 2)

⇒ x₁x₂ - 3x₁ - 2x₂ + 6 = x₁ x₂ - 2x₁ - 3x₂ + 6

⇒ - 3x₁ - 2x₂ = - 2x₁ - 3x₂

⇒ - x₁ = - x₂

⇒ x₁ = x₂

Hence, f is one - one function.

Onto :

Let y = \(\frac{x\,-\,2}{x\,-\,3}\)

⇒ xy - 3y = x - 2

⇒ xy - x = 3y - 2

⇒ x(y - 1) = 3y - 2

⇒ \(x = \frac{3y\,-\,2}{y\,-\,1}\) .....(i)

From above it is obvious that ∀ y except 1,

i.e, ∀ y ∈ B = R - {1}∃ x ∈ A

Hence, f is onto function.

Thus, f is one - one onto function.

If f^-1 is inverse function of f then,

\(f^{-1}(y) = \frac{3y\,-\,2}{y\,-\,1}\) [from (i)]