Which of the following functions from $\mathbb{Z}$ to $\mathbb{Z}$ is a bijective function? (where $\mathbb{Z}$ is set of integers)

A function is bijective if it is both one-one (injective) and onto (surjective).

One-one means different inputs always give different outputs.

Onto means every element in the output set ($\mathbb{Z}$) is covered — for every integer $y$, there exists an integer $x$ such that $f(x) = y$.

Both conditions must be satisfied.

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$f(x) = x^3$

One-one: Yes — different integers always give different cubes.

Onto: No — there is no integer $x$ such that $x^3 = 2$, since $x = \sqrt[3]{2} = 1.26...$ is not an integer. So the integer $2$ has no preimage.

Not bijective (fails onto).

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$f(x) = 2x + 1$

One-one: Yes — $2x_1 + 1 = 2x_2 + 1 \Rightarrow x_1 = x_2$

Onto: No — there is no integer $x$ such that $2x + 1 = 2$, since $x = \frac{1}{2}$ is not an integer. In fact, $2x + 1$ always gives odd numbers, so no even integer is ever an output.

Not bijective (fails onto).

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$f(x) = x^2 + 1$

One-one: No — $f(1) = 1 + 1 = 2$ and $f(-1) = 1 + 1 = 2$. Two different inputs ($1$ and $-1$) give the same output.

Not bijective (fails one-one).

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$f(x) = x + 2$

One-one: Yes — $x_1 + 2 = x_2 + 2 \Rightarrow x_1 = x_2$. Different inputs always give different outputs.

Onto: Yes — for any integer $y$, pick $x = y - 2$ (which is also an integer). Then $f(y - 2) = (y - 2) + 2 = y$. Every integer in the output is covered.

Bijective.

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Therefore, $f(x) = x + 2$ is the bijective function.