Which of the following functions from $z$ to $z$ is a bijective function? (where $z$ is set of integers)

$f(x) = x+2$

The correct answer is Option (4) → $f(x) = x+2$

Let $f: \mathbb{Z} \to \mathbb{Z}$

A function is bijective if it is both:

• Injective (one-one): $f(a) = f(b) \Rightarrow a = b$
• Surjective (onto): For every $y \in \mathbb{Z}$, there exists $x \in \mathbb{Z}$ such that $f(x) = y$

Option 1: $f(x) = x^3$

• Injective: Yes, since $x^3$ is strictly increasing on $\mathbb{Z}$
• Surjective: Yes, for any $y \in \mathbb{Z}$, $\sqrt[3]{y} \in \mathbb{Z}$ if and only if $y$ is a perfect cube ⇒ not all integers are cubes ⇒ not onto

⇒ ❌ Not bijective

Option 2: $f(x) = 2x + 1$

• Injective: Yes, linear function with slope ≠ 0
• Surjective: No, image is only odd integers ⇒ not all integers covered

⇒ ❌ Not bijective

Option 3: $f(x) = x^2 + 1$

• Injective: No, since $f(1) = f(-1) = 2$
• Surjective: No, output is ≥ 1 ⇒ not all $\mathbb{Z}$

⇒ ❌ Not bijective

Option 4: $f(x) = x + 2$

• Injective: Yes
• Surjective: Yes, for any $y \in \mathbb{Z}$, $x = y - 2 \in \mathbb{Z}$

⇒ ✅ Bijective