Verified Solution • 2025–26 Session moderate

QJEEM23Mac599: JEE Main 2023 Mathematics - Class 11 JEEM23Mac599 Solution: Among the relations and ,

Among the relations

\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}

and T={(a,b):a,b∈R,a2−b2∈Z}\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\} ,

Multiple Choice Options

  • A S\mathrm{S} is transitive but T\mathrm{T} is not
  • B both S\mathrm{S} and T\mathrm{T} are symmetric
  • C neither SS nor TT is transitive
  • D TT is symmetric but SS is not

Step-by-Step Verified Solution

For relation T=a2−b2=−I\mathrm{T}=\mathrm{a}^{2}-\mathrm{b}^{2}=-\mathrm{I}

Then, (b,a)(\mathrm{b}, \mathrm{a}) on relation R\mathrm{R}

⇒b2−a2=−I\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}

∴T\therefore \mathrm{T} is symmetric

\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}

2+\frac{\mathrm{a}}{\mathrm{b}}>0 \Rightarrow \frac{\mathrm{a}}{\mathrm{b}}>-2, \Rightarrow \frac{\mathrm{b}}{\mathrm{a}}<\frac{-1}{2}

If (b,a)∈S(b, a) \in \mathbf{S} then

2+ba2+\frac{\mathrm{b}}{\mathrm{a}} not necessarily positive

∴S\therefore \mathrm{S} is not symmetric

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