Let A = R – {3},

Given that,

In order to prove that f is one-one, it is sufficient to prove that f(x₁)=f(x₂) ⇒ x₁=x₂∀ x₁, x₂ ∈ A .

Let f(x₁)=f(x₂)

⇒

⇒ (x₁-2)(x₂-3) = (x₂-2)(x₁-3)

⇒ x₁x₂-3x₁-2x₂+6 = x₁x₂-3x₂-2x₁+6

⇒ -3x₁-2x₂ = -3x₂-2x₁

⇒ -3x₁+2x₁ = -3x₂+2x₂

⇒ (-3+2) x₁ = (-3+2)x₂

⇒ x₁ = x₂

∴ f is one-one.

f is onto if every element of B is the f-image of some element of A.

let f(x) = y

⇒

⇒ x-2 = y(x-3)

⇒ x-2 = xy-3y

⇒ x-xy = -3y+2

⇒ x(1-y) = -3y+2

⇒

⇒

Thus, for each y ∈ B there exists such that f(x) = y.

Hence, f is onto.

Since, f is one-one and onto therefore f is bijective.