Verified Solution • 2025–26 Session moderate

QJEEM24Madd34: JEE Main 2024 Mathematics - Class 11 JEEM24Madd34 Solution: Let . Suppose is the set of all the subsets of , then the…

Let S={1,2,3,…,10}S=\{1,2,3, \ldots, 10\} . Suppose MM is the set of all the subsets of SS , then the relation

R={(A,B):A∩B≠ϕ;A,B∈M}\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\} is :

Multiple Choice Options

  • A symmetric only
  • B reflexive only
  • C symmetric and reflexive only
  • D symmetric and transitive only

Step-by-Step Verified Solution

Let S={1,2,3,…,10}S=\{1,2,3, \ldots, 10\}

R={(A,B):A∩B≠ϕ;A,B∈M}R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}

For Reflexive,

MM is subset of ' SS '

So ϕ∈M\phi \in \mathrm{M}

for ϕ∩ϕ=ϕ\phi \cap \phi=\phi

⇒\Rightarrow but relation is A∩B≠ϕ\mathrm{A} \cap \mathrm{B} \neq \phi

So it is not reflexive.

For symmetric,

 ARB amp;A∩B≠ϕ,⇒BRAamp;⇒B∩A≠ϕ,\begin{array}{ll}\text { ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \\\Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi,\end{array}

So it is symmetric.

For transitive,

 If Aamp;={(1,2),(2,3)}Bamp;={(2,3),(3,4)}Camp;={(3,4),(5,6)}\begin{aligned}\text { If } A & =\{(1,2),(2,3)\} \\B & =\{(2,3),(3,4)\} \\C & =\{(3,4),(5,6)\}\end{aligned}

ARB\mathrm{ARB} & BRC\mathrm{BRC} but A\mathrm{A} does not relate to C\mathrm{C} So it not transitive

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