Class JEE Mathematics Sets, Relations, and Functions Q #1056
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4 Marks 2020 JEE Main 2020 (Online) 3rd September Evening Slot MCQ SINGLE
Let $R_1$ and $R_2$ be two relation defined as follows: $R_1 = {(a, b) \in R^2 : a^2 + b^2 \in Q}$ and $R_2 = {(a, b) \in R^2 : a^2 + b^2 \notin Q}$, where $Q$ is the set of all rational numbers. Then :
(A) Neither $R_1$ nor $R_2$ is transitive.
(B) $R_2$ is transitive but $R_1$ is not transitive.
(C) $R_1$ and $R_2$ are both transitive.
(D) $R_1$ is transitive but $R_2$ is not transitive.
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Correct Answer: A
Explanation
For $R_1$: Let $a = 1 + \sqrt{2}$, $b = 1 - \sqrt{2}$, $c = \sqrt[4]{8}$. $aR_1b : a^2 + b^2 = 6 \in Q$ $bR_1c : b^2 + c^2 = 3 - 2\sqrt{2} + 2\sqrt{2} = 3 \in Q$ $aR_1c : a^2 + c^2 = 3 + 2\sqrt{2} + 2\sqrt{2} \notin Q$ $\therefore$ $R_1$ is not transitive. For $R_2$: Let $a = 1 + \sqrt{2}$, $b = \sqrt{2}$, $c = 1 - \sqrt{2}$ $aR_2b : a^2 + b^2 = 5 + 2\sqrt{2} \notin Q$ $bR_2c : b^2 + c^2 = 5 - 2\sqrt{2} \notin Q$ $aR_2c : a^2 + c^2 = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6 \in Q$ $\therefore$ $R_2$ is not transitive.

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