Which of the following functions from mathbbZ into mathbbZ are bijections? (A) f(x) = x³ (B) f(x) = x + 2 (C)

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A bijection must be both one-to-one (injective) and onto (surjective). Among the given functions from $\mathbb{Z}$ to $\mathbb{Z}$ , only $f(x) = x + 2$ satisfies both. The correct option is (B).

Why a Bijection Proof Works

A bijection is a function that pairs every element of the domain with exactly one element of the codomain, and vice versa — no element is left out, and no element is paired twice. For a function $f: \mathbb{Z} \to \mathbb{Z}$ , we need to check two things:

Injective (one-to-one): If $f(a) = f(b)$ , then $a = b$ . No two different inputs map to the same output.
Surjective (onto): For every integer $y$ in the codomain, there is some integer $x$ in the domain such that $f(x) = y$ . Every output is actually hit.

If either condition fails, the function is not a bijection. Let’s test each option.

Option (A): $f(x) = x^3$

Is it injective? If $a^3 = b^3$ , then taking cube roots (which is a one-to-one operation on integers) gives $a = b$ . So yes, it’s injective.

Is it surjective? For $f$ to be onto, every integer $y$ must be a perfect cube. But $2$ , for example, is not a cube of any integer — there is no integer $x$ with $x^3 = 2$ . So surjectivity fails.

Watch out

A common mistake is to think $x^3$ is onto because it’s onto on the reals. But on $\mathbb{Z}$ , cubes skip many integers (like $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $9$ , …). The range is only $\{ \dots, -8, -1, 0, 1, 8, \dots \}$ , which is a proper subset of $\mathbb{Z}$ .

Conclusion: Not a bijection.
Option (B): $f(x) = x + 2$

Injectivity: If $a + 2 = b + 2$ , then subtracting $2$ gives $a = b$ . So it’s one-to-one.

Surjectivity: Given any integer $y$ , choose $x = y - 2$ , which is also an integer. Then $f(x) = (y - 2) + 2 = y$ . Every integer is hit.

Since both conditions hold, this is a bijection. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2021Set A-11 markMCQ
$A = \{ x : x \in \mathbb{R} ; x$
›Reveal solutionSolution
The function $f(x) = \frac{2x}{x-1}$ on the domain $\mathbb{R}$ minus positive integers is injective but not surjective — it misses the value $2$ entirely.

The key to this problem is understanding what the domain $A$ actually is. $A$ contains all real numbers except the positive integers $1, 2, 3, \dots$ . That means $A$ includes $0$ , negative numbers, fractions, irrationals — everything real except $1, 2, 3, \dots$ .

Now, $f(x) = \frac{2x}{x-1}$ is a rational function. Its natural point of trouble is $x = 1$ , where the denominator vanishes. But $1$ is already excluded from $A$ , so the function is defined everywhere on its domain. The question is: does every real number appear as an output? And does each output come from exactly one input?

Let’s work through it.

Check injectivity (one-to-one). Suppose $f(a) = f(b)$ for $a, b \in A$ . Then

$\frac{2a}{a-1} = \frac{2b}{b-1}$
Cross-multiply (valid since $a \neq 1$ , $b \neq 1$ ):
$2a(b-1) = 2b(a-1) \implies 2ab - 2a = 2ab - 2b$
Cancel $2ab$ :
$-2a = -2b \implies a = b$
So $f$ is injective. No two different inputs give the same output.

Check surjectivity (onto). For surjectivity, every real number $y$ must have some $x \in A$ such that $f(x) = y$ . Solve for $x$ :

$y = \frac{2x}{x-1} \implies y(x-1) = 2x \implies yx - y = 2x$
Bring $x$ terms together:
$yx - 2x = y \implies x(y - 2) = y$
If $y \neq 2$ , we can write:
$x = \frac{y}{y-2}$
So for any $y \neq 2$ , there is exactly one candidate $x$ . But we must check whether this $x$ lies in $A$ — that is, whether $x$ is a positive integer.

The problematic case is $y = 2$ . Then the equation becomes $x(2-2) = 2$ , i.e. $0 \cdot x = 2$ , which has no solution. So $2$ is never attained by $f$ .

Therefore $f$ is not surjective — the value $2$ is missing from the range. …
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Q.Which of the following functions from $\mathbb{Z}$ into $\mathbb{Z}$ are bijections?
(A) $f(x) = x^3$
(B) $f(x) = x + 2$
(C) $f(x) = 2x + 1$
(D) $f(x) = x^2 + 1$

Concept understanding — Bijection Proof

Proving a Function is a Bijection

Step 1 — prove it is one-one

Step 2 — prove it is onto

A worked template

Why it matters …

Why a Bijection Proof Works

Method: Testing a Function for Bijectivity

Steps

Common Mistakes

Q.Which of the following functions from Z\mathbb{Z}Z into Z\mathbb{Z}Z are bijections? (A) f(x)=x3f(x) = x^3f(x)=x3 (B) f(x)=x+2f(x) = x + 2f(x)=x+2 (C) f(x)=2x+1f(x) = 2x + 1f(x)=2x+1 (D) f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1

Why a Bijection Proof Works

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Q.Which of the following functions from $\mathbb{Z}$ into $\mathbb{Z}$ are bijections?
(A) $f(x) = x^3$
(B) $f(x) = x + 2$
(C) $f(x) = 2x + 1$
(D) $f(x) = x^2 + 1$

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