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JEE Main 2020 (Online) 3rd September Evening Slot
Subjects
Mathematics
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Competency Based
Knowledge Based
2020
JEE Main 2020 (Online) 3rd September Evening Slot
Full Paper Analysis & Solutions
Mathematics
Multiple Choice Questions
1
APPLY
MCQ_SINGLE
4 Mark(s)
2020
JEE Main 2020 (Online) 3rd September Evening Slot
#1056
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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AI Explanation
MEDIUM
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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