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JEE Main 2024MathematicsSets And RelationsSymmetric Transitive And Reflexive PropertiesmediumMCQ

JEE Main 2024 — Sets And Relations Question with Solution

From: JEE Main 2024 (Online) 1st February Evening Shift

Question

Consider the relations

R_{1}

and

R_{2}

defined as

a R_{1} b \Leftrightarrow a^{2} + b^{2} = 1

for all

a, b \in R

and

(a, b) R_{2} (c, d) \Leftrightarrow

a + d = b + c

for all

(a, b), (c, d) \in N \times N

. Then :

Choose an option

Show full solutionCorrect option: C

Correct answer

COnly

R_{2}

is an equivalence relation

Step-by-step explanation

To determine if the given relations $R_{1}$ and $R_{2}$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Let's start by analysing $R_{1}$ :

Reflexivity: A relation $R$ on a set $S$ is reflexive if every element is related to itself, that is, for every $a \in S$ , the pair $(a, a)$ is in $R$ . In the case of $R_{1}$ , for an arbitrary $a \in R$ , we need to check if $a R_{1} a$ holds, which translates to checking if $a^{2} + a^{2} = 1$ . This would mean $2 a^{2} = 1$ or $a^{2} = \frac{1}{2}$ . Since this is not true for every real number $a$ , $R_{1}$ is not reflexive.

Symmetry: A relation $R$ is symmetric if whenever $a R b$ , then $b R a$ . For $R_{1}$ , if $a^{2} + b^{2} = 1$ , then it is also true that $b^{2} + a^{2} = 1$ . Thus, $R_{1}$ is symmetric.

Transitivity: A relation $R$ is transitive if whenever $a R b$ and $b R c$ , then $a R c$ . For $R_{1}$ , suppose $a R_{1} b$ and $b R_{1} c$ , this means $a^{2} + b^{2} = 1$ and $b^{2} + c^{2} = 1$ . However, if we add these two, we get $a^{2} + 2 b^{2} + c^{2} = 2$ . There is no guarantee that $a^{2} + c^{2} = 1$ . Therefore, $R_{1}$ is not transitive.

Conclusion: Relation $R_{1}$ is not reflexive or transitive, and hence it is not an equivalence relation.

Now let's consider $R_{2}$ :

Reflexivity: For any $(a, b) \in N \times N$ , it's clear that $a + b = b + a$ , which is just a reiteration of the commutative property of addition. Therefore, $(a, b) R_{2} (a, b)$ , and $R_{2}$ is reflexive.

Symmetry: If $(a, b) R_{2} (c, d)$ , meaning $a + d = b + c$ , then by reordering the terms we can similarly have $c + b = d + a$ , which means $(c, d) R_{2} (a, b)$ , so $R_{2}$ is symmetric.

Transitivity: Suppose $(a, b) R_{2} (c, d)$ and $(c, d) R_{2} (e, f)$ , i.e., $a + d = b + c$ and $c + f = d + e$ , we want to show that $(a, b) R_{2} (e, f)$ . Adding the two equations, we get $a + d + c + f = b + c + d + e$ . By rearranging and simplifying, we get $a + f = b + e$ , thus, $(a, b) R_{2} (e, f)$ .

So $R_{2}$ is reflexive, symmetric, and transitive, and therefore it is an equivalence relation.

Conclusion: According to the above examination, only $R_{2}$ is an equivalence relation. Thus, the correct answer is:

Option C: Only $R_{2}$ is an equivalence relation.

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About this question

This is a previous-year question from JEE Main 2024, covering the Sets And Relations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.