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Mathematics · Sets and Relations

JEE Main 2025 — 23 January, Evening Shift — Question 14

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II:

Let X=R×RX=R \times R. Define a relation RR on XX as: (a1, b1)R(a2, b2)⇔b1=b2\left(\mathrm{a}_{1}, \mathrm{~b}_{1}\right) \mathrm{R}\left(\mathrm{a}_{2}, \mathrm{~b}_{2}\right) \Leftrightarrow \mathrm{b}_{1}=\mathrm{b}_{2}.

Statement I: R is an equivalence relation.

Statement II: For some (a,b)∈X(\mathrm{a}, \mathrm{b}) \in \mathrm{X}, the set S={(x,y)∈X:(x,y)R(a,b)}\mathrm{S}=\{(\mathrm{x}, \mathrm{y}) \in \mathrm{X}:(\mathrm{x}, \mathrm{y}) \mathrm{R}(\mathrm{a}, \mathrm{b})\} represents a line parallel to y=x\mathrm{y}=\mathrm{x}.

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both Statement-I and Statement-II are false.

  2. Option B:

    Statement-I is true but Statement-II is false.

    Correct
  3. Option C:

    Both Statement-I and Statement-II are true.

  4. Option D:

    Statement-I is false but Statement-II is true.

Answer: B

Step-by-step solution

Reflexive: (a1,b1)R(a1,b1)⇒b1=b1True\text{Reflexive: } (a_1, b_1) R (a_1, b_1) \Rightarrow b_1 = b_1 \qquad \text{True} Symmetric: (a1,b1)R(a2,b2)⇒b1=b2True\text{Symmetric: } (a_1, b_1) R (a_2, b_2) \Rightarrow b_1 = b_2 \qquad \text{True} (a2,b2)R(a1,b1)⇒b2=b1(a_2, b_2) R (a_1, b_1) \Rightarrow b_2 = b_1 Transitive: (a1,b1)R(a2,b2)⇒b1=b2\text{Transitive: } (a_1, b_1) R (a_2, b_2) \Rightarrow b_1 = b_2 & (a2,b2)R(a3,b3)⇒b2=b3b1=b3\text{\& } (a_2, b_2) R (a_3, b_3) \Rightarrow b_2 = b_3 \qquad b_1 = b_3 ⇒(a1,b1)R(a3,b3)⇒True\Rightarrow (a_1, b_1) R (a_3, b_3) \Rightarrow \text{True}

Hence Relation R is an equivalence relation Statement-I is true.

For statement-II

⇒y=b\Rightarrow y = b

So False.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Evening Shift
Subject
Mathematics
Chapter
Sets and Relations
Topic
Relations

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