Verified Solution • 2025–26 Session moderate

QJEEM25Ma9993: JEE Main 2025 Mathematics - Class 11 JEEM25Ma9993 Solution: Let . Define a relation R on X as : Statement I: is an…

Let X=R×R\mathrm{X}=\mathbf{R} \times \mathbf{R} . Define a relation R on X 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(\mathrm{a}, \mathrm{b}) \in \mathrm{X} , the set⁡S={(x,y)∈X:(x,y)R(a,b)}\operatorname{set} \mathrm{S}=\{(x, y) \in \mathrm{X}:(x, y) \mathrm{R}(\mathrm{a}, \mathrm{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 :

Multiple Choice Options

  • A Both Statement I and Statement II are true
  • B Statement I is true but Statement II is false
  • C Both Statement I and Statement II are false
  • D Statement I is false but Statement II is true

Step-by-Step Verified Solution

Statement - I :

Reflexive : (a1,b1)R(a1,b1)⇒b1=b1\left(a_1, b_1\right) R\left(a_1, b_1\right) \Rightarrow b_1=b_1 \quad True

 Symmetric : amp;(a1, b1)R(a2, b2)⇒b1=b2amp;(a2, b2)R(a1, b1)⇒b2=b1}\left.\begin{array}{rl}\text { Symmetric : } & \left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow \mathrm{b}_1=\mathrm{b}_2 \\ & \left(\mathrm{a}_2, \mathrm{~b}_2\right) \mathrm{R}\left(\mathrm{a}_1, \mathrm{~b}_1\right) \Rightarrow \mathrm{b}_2=\mathrm{b}_1\end{array}\right\} True

 Transitive: amp;(a1,b1)R(a2,b2)⇒b1=b2amp;&(a2,b2)R(a3,b3)b2=b3}b1=b3\left.\begin{array}{ll}\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\end{array}\right\} b_1=b_3

⇒(a1, b1)R(a3⋅ b3)⇒ True \Rightarrow\left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_3 \cdot \mathrm{~b}_3\right) \Rightarrow \text { True }

Hence Relation RR is an equivence relation Statement-I is true.

For statement - II ⇒y=b\Rightarrow \mathrm{y}=\mathrm{b} so False

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