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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Sets and Relations

Let X=R × R . Define a relation R on X as : (a_1, b_1) R(a_2, b_2)...

Let X=R×RX=\mathbf{R} \times \mathbf{R}. Define a relation RR on XX as : (a1,b1)R(a2,b2)⇔b1=b2\left(a_1, b_1\right) R\left(a_2, b_2\right) \Leftrightarrow b_1=b_2 Statement I : R\quad \mathrm{R} is an equivalence relation. Statement II : For some (a,b)∈X(a, b) \in X, the set S={(x,y)∈X:(x,y)R(a,b)}S=\{(x, y) \in X:(x, y) R(a, b)\} represents a line parallel to y=xy=x. In the light of the above statements, choose the correct answer from the options given below :

Options

  1. ABoth Statement I and Statement II are false
  2. BStatement I is true but Statement II is false
  3. CBoth Statement I and Statement II are true
  4. DStatement I is false but Statement II is true

Correct answer

B. Statement I is true but Statement II is false

Step-by-step solution

Reflexive : (a1,b)R(a1,b1)⇒b1=b1\left(a_1, b\right) R\left(a_1, b_1\right) \Rightarrow b_1=b_1 \quad True  Symmetric : (a1,b1)R(a2,b2)⇒b1=b2(a2,b2)R(a1,b1)⇒b2=b1\begin{array}{rl}\text{ Symmetric : } & \left(a_1, b_1\right) R \left(a_2, b_2\right) \Rightarrow b_1=b_2 \\ & \left(a_2, b_2\right) R \left(a_1, b_1\right) \Rightarrow b_2=b_1\end{array} True  Transitive : (a1,b1)R(a2,b2)⇒b1=b2(a2,b2)R(a3,b3)b2=b3⇒(a1,b1)R(a3.b3)⇒ True b1=b3\begin{aligned} \text{ Transitive : } & \left(a_1, b_1\right) R\left(a_2, b_2\right) \Rightarrow b_1=b_2 \\ & &\left(a_2, b_2\right) R\left(a_3, b_3\right) b_2=b_3 \\ & \Rightarrow\left(a_1, b_1\right) R\left(a_3 . b_3\right) \Rightarrow \text{ True } b_1=b_3\end{aligned} Hence Relation RR is an equivence relation Statement-I is true. For statement - II\mathrm{II} \Rightarrow \mathrm{y}=\mathrm{b} so False
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