Question

A function f: `RR → RR` defined as f(x) = x² − 4x + 5 is: 

Options

• injective but not surjective

• surjective but not injective

• both injective and surjective

• neither injective nor surjective

Answer 1

neither injective nor surjective

Explanation:

Given, f: `RR → RR`

f(x) = x² − 4x + 5

One-One (Injective)

f(x₁) = f(x₂)

⇒ x₁² − 4x₁ + 5 = x₁² − 4x₂ + 5

⇒ x₁² − x₂² = 4(x₁ − x₂)

⇒ (x₁ − x₂) (x₁ + x₂) = 4(x₁ − x₂)

⇒ x₁ + x₂ = 4

⇒ x₁ = 4 − x₂

Thus, f(x) is not an injective mapping.

onto surjective

let y = x² − 4x +5

⇒ y = (x − 2)² + 1

⇒ y − 1 = (x − 2)²

⇒ x = `sqrt((y - 1)) + 2`

Thus, for any value of y < 1, x ∉ R. So, we don't have a pre-image for all y ∈ R in x ∈ R.

Thus, f(x) = x² − 4x + 5 is not surjective.