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JEE Main 202126 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2021 · 26 February · Morning Shift · Mathematics · Sets and Relations

Let R = (P, Q) | P and Q are at the same distance from the origin...

Let R=(P,Q)∣PandQareatthesamedistancefromtheoriginR = { \left(P, Q\right) | P and Q are at the same distance from the origin}bearelation,thentheequivalenceclassof(1,−1)be a relation, then the equivalence class of \left(1, - 1\right) is the set

Options

  1. AS={(x,y)∣x2+y2=1}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = 1\right\}
  2. BS={(x,y)∣x2+y2=2}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = 2\right\}
  3. CS={(x,y)∣x2+y2=2}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = \sqrt{2}\right\}
  4. DS={(x,y)∣x2+y2=4}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = 4\right\}

Correct answer

B. S={(x,y)∣x2+y2=2}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = 2\right\}

Step-by-step solution

Equivalence class of (1,−1)\left(1, - 1\right) is a circle with centre at (0,0)\left(0, 0\right)and radius =2= \sqrt{2}

⇒x2+y2=2⇒ {x}^{2} + {y}^{2} = 2

S={(x,y)∣x2+y2=2}S = \left\{\left(x, y\right) | {x}^{2} + {y}^{2} = 2\right\}

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