Question:
Let \(R\) be a relation in \(\mathbb{N}\) defined by \(\{(x,y): x + 3y = 10, x,y \in \mathbb{N}\}\). Then the range of \(R\) is
Let \(R\) be a relation in \(\mathbb{N}\) defined by \(\{(x,y): x + 3y = 10, x,y \in \mathbb{N}\}\). Then the range of \(R\) is
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Always ensure both variables satisfy given domain (here $\mathbb{N}$).
Updated On: Apr 30, 2026
- $\{1,2,3,4\}$
- $\{2,3,4\}$
- $\{1,3,4\}$
- $\{1,2,3\}$
- $\{1,3\}$
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The Correct Option is D
Solution and Explanation
Concept:
Range of a relation is the set of all possible values of $y$.
Step 1: Form equation
\[ x + 3y = 10 \]
Step 2: Find natural number solutions
\[ x = 10 - 3y \] For $x \in \mathbb{N}$: \[ 10 - 3y > 0 \Rightarrow y = 1,2,3 \]
Step 3: Range
\[ \text{Range} = \{1,2,3\} \] Final Conclusion:
Option (D)
Step 1: Form equation
\[ x + 3y = 10 \]
Step 2: Find natural number solutions
\[ x = 10 - 3y \] For $x \in \mathbb{N}$: \[ 10 - 3y > 0 \Rightarrow y = 1,2,3 \]
Step 3: Range
\[ \text{Range} = \{1,2,3\} \] Final Conclusion:
Option (D)
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