QJEEM24Ma6c30: JEE Main 2024 Mathematics - Class 11 JEEM24Ma6c30 Solution: Consider the relations and defined as for all and for all .…

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 $a \in \mathbf{R}$ , we need to check if $a R_1 a$ holds, which translates to checking if $a^2 + a^2 = 1$ . This would mean $2a^2 = 1$ 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 $a^2 + 2b^2 + c^2 = 2$ . 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 $(a, b) \in \mathbf{N} \times \mathbf{N}$ , 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.