JEE Mathematics 2023 Question with Solution

Given: $S = \{1, 2, 3, \ldots, 10\}$ and $P(S)$ is the power set of $S$.

Find: Whether $R_1$ and $R_2$ are equivalence relations.

For $R_1$,

$$(A \cap B^c) \cup (B \cap A^c) = \varnothing$$

This implies $A = B$.

So we check the three properties:

1. Reflexive: For any $A \in P(S)$,

$$(A \cap A^c) \cup (A^c \cap A) = \varnothing$$

Hence $A R_1 A$.

2. Symmetric: If $A R_1 B$, then

$$(A \cap B^c) \cup (B \cap A^c) = \varnothing$$

which implies $A = B$. Therefore, $B R_1 A$.

3. Transitive: If $A R_1 B$ and $B R_1 C$, then $A = B$ and $B = C$. Hence $A = C$, so $A R_1 C$.

Therefore, $R_1$ is an equivalence relation.

For $R_2$,

$$A \cup B^c = B \cup A^c$$

Again check the three properties:

1. Reflexive: For any $A$,

$$A \cup A^c = A \cup A^c$$

Hence $A R_2 A$.

2. Symmetric: If $A R_2 B$, then

$$A \cup B^c = B \cup A^c$$

Reversing the equality gives

$$B \cup A^c = A \cup B^c$$

So $B R_2 A$.

3. Transitive: If $A R_2 B$ and $B R_2 C$, then

$$A \cup B^c = B \cup A^c$$

and

$$B \cup C^c = C \cup B^c$$

Using substitution, we get

$$A \cup C^c = C \cup A^c$$

Hence $A R_2 C$.

Therefore, $R_2$ is also an equivalence relation.

So, both $R_1$ and $R_2$ are equivalence relations. The correct option is A.

Given: The relations are defined on $P(S)$.

Find: Which of $R_1$ and $R_2$ are equivalence relations.

For $R_1$, the condition

$$(A \cap B^c) \cup (B \cap A^c) = \varnothing$$

means there is no element belonging to exactly one of $A$ and $B$. So the symmetric difference is empty, which means $A = B$. Equality is always reflexive, symmetric, and transitive.

For $R_2$, the given solution verifies reflexivity, symmetry, and transitivity directly from

$$A \cup B^c = B \cup A^c$$

Therefore, $R_2$ is also an equivalence relation.

Hence both relations are equivalence relations, so the correct option is A.