The correct answer is Option (2) → $f(x) = x + 2$ ##
One-to-One (Injective): Distinct inputs must map to distinct outputs. If $f(x_1) = f(x_2)$, then $x_1 = x_2$.
Onto (Surjective): Every integer $y$ in the codomain must have a corresponding pre-image $x$ in the domain such that $f(x) = y$
Here, $f(x) = x + 2 \Rightarrow f(x_1) = f(x_2)$
$x_1 + 2 = x_2 + 2 \Rightarrow x_1 = x_2$
Let $y = x + 2$
$x = y - 2 \in Z, \forall y \in Z$
Hence, $f(x)$ is one-one and onto.
Option 1: $f(x) = x^3$
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One-to-One: Yes, since distinct integers have distinct cubes.
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Onto: No. Consider an integer in the codomain like $y = 2$. If we set $x^3 = 2$, then $x = \sqrt[3]{2}$, which is not an integer ($x \notin \mathbb{Z}$). Because numbers like $2, 3, 4, \dots$ have no integer pre-images, this function is not onto.
Option 3: $f(x) = 2x + 1$
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One-to-One: Yes, because $2x_1 + 1 = 2x_2 + 1$ or $ x_1 = x_2$
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Onto: No. Let $y = 2x + 1$ or $x = \frac{y - 1}{2}$. This function outputs only odd integers. If we pick an even integer in the codomain like $y = 4$, then $x = \frac{4 - 1}{2} = 1.5$, which is not an integer ($x \notin \mathbb{Z}$). Thus, even numbers have no integer pre-images.
Option 4: $f(x) = x^2 + 1$
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One-to-One: No. For example, $f(1) = 1^2 + 1 = 2$ and $f(-1) = (-1)^2 + 1 = 2$. Two distinct inputs map to the same output.
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Onto: No. Since $x^2 \ge 0$, the minimum output value is $f(0) = 1$. Negative integers in the codomain (like $-5$) can never be reached.
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