Verified Solution • 2025–26 Session moderate

QJEEM23Ma20a1: JEE Main 2023 Mathematics - Class 11 JEEM23Ma20a1 Solution: Let denote the power set of . Define the relations and on as if …

Let P(S)P(S) denote the power set of S={1,2,3,….,10}S=\{1,2,3, \ldots ., 10\} . Define the relations R1R_{1} and R2R_{2} on P(S)P(S) as AR1 B\mathrm{AR}_{1} \mathrm{~B} if (A∩Bc)∪(B∩Ac)=∅\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{c}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{c}}\right)=\emptyset and AR2 B\mathrm{AR}_{2} \mathrm{~B} if A∪Bc=B∪Ac,∀A,B∈P(S)\mathrm{A} \cup \mathrm{B}^{\mathrm{c}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{c}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S}) . Then :

Multiple Choice Options

  • A only R2R_{2} is an equivalence relation
  • B both R1R_{1} and R2R_{2} are not equivalence relations
  • C both R1R_{1} and R2R_{2} are equivalence relations
  • D only R1R_{1} is an equivalence relation

Step-by-Step Verified Solution

amp;S={1,2,3,……10}amp;P(S)= power set of Samp;AR1B⇒(A∩B‾)∪(A‾∩B)=ϕamp;R1 is reflexive, symmetric amp; For transitive amp;(A∩B‾)∪(A‾∩B)=ϕ;{a}=ϕ={b}∴A=Bamp;(B∩C‾)∪(B‾∩C)=ϕ∴B=Camp;∴A=C equivalence. \begin{aligned} & \mathrm{S}=\{1,2,3, \ldots \ldots 10\} \\\\ & \mathrm{P}(\mathrm{S})=\text { power set of } \mathrm{S} \\\\ & \mathrm{AR}_1 \mathrm{B} \Rightarrow(\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi \\\\ & \mathrm{R}_1 \text { is reflexive, symmetric } \\\\ & \text { For transitive } \\\\ & (\mathrm{A} \cap \overline{\mathrm{B}}) \cup(\overline{\mathrm{A}} \cap \mathrm{B})=\phi ;\{\mathrm{a}\}=\phi=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & (\mathrm{B} \cap \overline{\mathrm{C}}) \cup(\overline{\mathrm{B}} \cap \mathrm{C})=\phi \therefore \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \text { equivalence. }\end{aligned} JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Sets and Relations Question 53 English Explanation 1

R2≡A∪B‾=A‾∪B\mathrm{R}_2 \equiv \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B}

R2→\mathrm{R}_2 \rightarrow Reflexive, symmetric

For transitive :

JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Sets and Relations Question 53 English Explanation 2

amp;A∪B‾=A‾∪B⇒{a,c,d}={b,c,d}amp;{a}={b}∴A=Bamp;B∪C‾=B‾∪C⇒B=Camp;∴A=C∴A∪C‾=A‾∪C∴ Equivalence \begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overline{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\\\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\\\ & \mathrm{B} \cup \overline{\mathrm{C}}=\overline{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \\\\ & \therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overline{\mathrm{C}}=\overline{\mathrm{A}} \cup \mathrm{C} \therefore \text { Equivalence }\end{aligned}

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