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Sets and Relations question

2024 · 27 Jan · Shift 1 · Q33
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Sets and Relations question

2024 · 27 Jan · Shift 1 · Q33

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let S={1,2,3,…,10}S=\{1,2,3, \ldots, 10\}S={1,2,3,…,10}. Suppose MMM is the set of all the subsets of SSS, then the relation R={(A,B):A∩Beqϕ;A,B∈M}\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} eq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\}R={(A,B):A∩Beqϕ;A,B∈M} is :
  1. A
    symmetric only
  2. B
    reflexive only
  3. C
    symmetric and reflexive only
  4. D
    symmetric and transitive only
View written solutionFree

Correct answer: A

  1. Given relation

Let S={1,2,3,…,10}S=\{1,2,3,\ldots,10\}S={1,2,3,…,10} and let MMM be the set of all subsets of SSS.

The relation on MMM is

R={(A,B):A∩B≠∅,  A,B∈M}.R=\{(A,B): A\cap B\neq \varnothing,\; A,B\in M\}.R={(A,B):A∩B=∅,A,B∈M}.

So, (A,B)∈R(A,B)\in R(A,B)∈R exactly when AAA and BBB have at least one common element.

We check whether RRR is reflexive, symmetric, and transitive.


  1. Reflexive?

A relation on MMM is reflexive if for every A∈MA\in MA∈M, we have (A,A)∈R(A,A)\in R(A,A)∈R.

Now,

(A,A)∈R  ⟺  A∩A≠∅  ⟺  A≠∅.(A,A)\in R \iff A\cap A\neq \varnothing \iff A\neq \varnothing.(A,A)∈R⟺A∩A=∅⟺A=∅.

But ∅∈M\varnothing\in M∅∈M since it is a subset of SSS.

For A=∅A=\varnothingA=∅,

∅∩∅=∅,\varnothing\cap \varnothing = \varnothing,∅∩∅=∅,

so

(∅,∅)∉R.(\varnothing,\varnothing)\notin R.(∅,∅)∈/R.

Hence RRR is not reflexive.


  1. Symmetric?

A relation is symmetric if

(A,B)∈R  ⟹  (B,A)∈R.(A,B)\in R \implies (B,A)\in R.(A,B)∈R⟹(B,A)∈R.

If (A,B)∈R(A,B)\in R(A,B)∈R, then

A∩B≠∅.A\cap B\neq \varnothing.A∩B=∅.

But intersection is commutative:

A∩B=B∩A.A\cap B = B\cap A.A∩B=B∩A.

So,

B∩A≠∅,B\cap A\neq \varnothing,B∩A=∅,

which means

(B,A)∈R.(B,A)\in R.(B,A)∈R.

Therefore, RRR is symmetric.


  1. Transitive?

A relation is transitive if

(A,B)∈R and (B,C)∈R  ⟹  (A,C)∈R.(A,B)\in R \text{ and } (B,C)\in R \implies (A,C)\in R.(A,B)∈R and (B,C)∈R⟹(A,C)∈R.

We test by counterexample.

Take

A={1},B={1,2},C={2}.A=\{1\},\quad B=\{1,2\},\quad C=\{2\}.A={1},B={1,2},C={2}.

Then

A∩B={1}≠∅,A\cap B = \{1\}\neq \varnothing,A∩B={1}=∅,

so (A,B)∈R(A,B)\in R(A,B)∈R.

Also,

B∩C={2}≠∅,B\cap C = \{2\}\neq \varnothing,B∩C={2}=∅,

so (B,C)∈R(B,C)\in R(B,C)∈R.

But

A∩C={1}∩{2}=∅,A\cap C = \{1\}\cap \{2\} = \varnothing,A∩C={1}∩{2}=∅,

so (A,C)∉R(A,C)\notin R(A,C)∈/R.

Hence RRR is not transitive.


  1. Conclusion
  • Not reflexive
  • Symmetric
  • Not transitive

Therefore, the relation is symmetric only.

So the correct option is:

A\boxed{\text{A}}A​
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