Question:
Assertion (A): A function \[ f:\mathbb{N}\to\mathbb{N} \] given by \[ f(x)=x^3+2,\quad \forall x\in\mathbb{N} \] is one-one but not onto.
Reason (R): Since, for every \(y\in\mathbb{N}\) (codomain), there does not exist \[ x=(y-2)^{1/3}\in\mathbb{N} \] (domain) such that \[ f(x)=x^3+2=y. \]
Assertion (A): A function \[ f:\mathbb{N}\to\mathbb{N} \] given by \[ f(x)=x^3+2,\quad \forall x\in\mathbb{N} \] is one-one but not onto.
Reason (R): Since, for every \(y\in\mathbb{N}\) (codomain), there does not exist \[ x=(y-2)^{1/3}\in\mathbb{N} \] (domain) such that \[ f(x)=x^3+2=y. \]
Show Hint
In Assertion-Reason questions involving functions, pay extremely close attention to the domain/codomain sets (\(\mathbb{N}, \mathbb{Z}, \mathbb{R}\)).
Watch out for quantifiers like "all" or "every"; a single counter-example makes such a statement false.
Watch out for quantifiers like "all" or "every"; a single counter-example makes such a statement false.
Updated On: Sep 11, 2026
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
• A function is one-one if \(f(x_1) = f(x_2) \implies x_1 = x_2\).
• A function is onto if the range equals the codomain.
• For \(f: \mathbb{N} \to \mathbb{N}\), all inputs and outputs must be natural numbers \(\{1, 2, 3, \dots\}\).
Step 1: Test Assertion (A) for the one-one property
Let \(f(x_1) = f(x_2)\).
\(x_1^3 + 2 = x_2^3 + 2 \implies x_1^3 = x_2^3 \implies x_1 = x_2\).
Thus, the function is one-one.
Step 2: Test Assertion (A) for the onto property
The codomain is \(\mathbb{N}\).
The range of \(f(x) = x^3 + 2\) for \(x \in \{1, 2, 3, \dots\}\) is \(\{3, 10, 29, \dots\}\).
Numbers like \(1, 2, 4, 5, \dots\) in the codomain have no pre-image in the domain.
Thus, the function is not onto. Assertion (A) is true.
Step 3: Evaluate Reason (R)
The reason states: "\(\forall y \in \mathbb{N}\) ... there does not exist \(x\)".
The universal quantifier "\(\forall y\)" (for all \(y\)) makes this statement false.
If \(y = 3\), then \(x = (3-2)^{1/3} = 1 \in \mathbb{N}\). So for \(y = 3\), an \(x\) does exist.
The reason would be true if it said "there exists some \(y\)" for which no pre-image exists, but as a universal statement, it is mathematically incorrect.
• A function is one-one if \(f(x_1) = f(x_2) \implies x_1 = x_2\).
• A function is onto if the range equals the codomain.
• For \(f: \mathbb{N} \to \mathbb{N}\), all inputs and outputs must be natural numbers \(\{1, 2, 3, \dots\}\).
Step 1: Test Assertion (A) for the one-one property
Let \(f(x_1) = f(x_2)\).
\(x_1^3 + 2 = x_2^3 + 2 \implies x_1^3 = x_2^3 \implies x_1 = x_2\).
Thus, the function is one-one.
Step 2: Test Assertion (A) for the onto property
The codomain is \(\mathbb{N}\).
The range of \(f(x) = x^3 + 2\) for \(x \in \{1, 2, 3, \dots\}\) is \(\{3, 10, 29, \dots\}\).
Numbers like \(1, 2, 4, 5, \dots\) in the codomain have no pre-image in the domain.
Thus, the function is not onto. Assertion (A) is true.
Step 3: Evaluate Reason (R)
The reason states: "\(\forall y \in \mathbb{N}\) ... there does not exist \(x\)".
The universal quantifier "\(\forall y\)" (for all \(y\)) makes this statement false.
If \(y = 3\), then \(x = (3-2)^{1/3} = 1 \in \mathbb{N}\). So for \(y = 3\), an \(x\) does exist.
The reason would be true if it said "there exists some \(y\)" for which no pre-image exists, but as a universal statement, it is mathematically incorrect.
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