JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 9)
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 $\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}$,
$\mathrm{S}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}-\{0\}, 2+\frac{\mathrm{a}}{\mathrm{b}}>0\right\}$
and $\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathbb{Z}\right\}$,
$\mathrm{S}$ is transitive but $\mathrm{T}$ is not
both $\mathrm{S}$ and $\mathrm{T}$ are symmetric
neither $S$ nor $T$ is transitive
$T$ is symmetric but $S$ is not
Explanation
For relation $\mathrm{T}=\mathrm{a}^{2}-\mathrm{b}^{2}=-\mathrm{I}$
Then, $(\mathrm{b}, \mathrm{a})$ on relation $\mathrm{R}$
$\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}$
$\therefore \mathrm{T}$ 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 $(b, a) \in \mathbf{S}$ then
$2+\frac{\mathrm{b}}{\mathrm{a}}$ not necessarily positive
$\therefore \mathrm{S}$ is not symmetric
Then, $(\mathrm{b}, \mathrm{a})$ on relation $\mathrm{R}$
$\Rightarrow \mathrm{b}^{2}-\mathrm{a}^{2}=-\mathrm{I}$
$\therefore \mathrm{T}$ 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 $(b, a) \in \mathbf{S}$ then
$2+\frac{\mathrm{b}}{\mathrm{a}}$ not necessarily positive
$\therefore \mathrm{S}$ is not symmetric
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