Which of the foll

A function is bijective iff it is one-one and onto.

Option A. f (x) = x³

Let f(x₁) = f(x₂)

⇒ x₁³= x₂³

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = x³

⇒ x = y^1/3 but y^1/3∉ Z

⇒ f is not onto

Thus, f is not bijective.

Option B. f (x) = x + 2

Let f(x₁) = f(x₂)

⇒ x₁+2 = x₂+2

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = x + 2

⇒ x = y – 2

⇒ for each y ∈ Z there exists x ∈ Z (domain) such that f(x) = y.

⇒ f is onto

Thus, f is bijective.

Option C. f (x) = 2x + 1

Let f(x₁) = f(x₂)

⇒ 2x₁+1 = 2x₂+1

⇒ x₁= x₂

⇒ f is one one

Let f(x) = y, y ∈ Z

⇒ y = 2x + 1

⇒ y - 1 = 2x

⇒

We observe that if we put y=0, then .

Thus, y = 0 ∈ Z does not have pre image in Z (domain)

⇒ f is not onto.

Thus, f is not bijective.

Option D. f (x) = x² + 1

let f(x₁) = f(x₂)

⇒ x₁² + 1 = x₂² + 1

⇒ x₁² = x₂²

⇒ x₁ = ± x₂

⇒ x₁= x₂ and x₁= - x₂

For e.g., f(-1) = |-1| = 1 and f(1) = |1| = 1

⇒ f is not one-one.

Since, f is not one one it cannot be bijective.