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 $\mathrm{T}=\left\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathbb{R}, \mathrm{a}^{2}-\mathrm{b}^{2} \in \mathrm{Z}\right\}$,

(B) T is symmetric but S is not

(A) S is transitive but T is not

(C) Neither S nor T is transitive

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Sets Relations and Functions

Among the relations

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

And

T = {(a, b) : a, b \in R, a^{2} - b^{2} \in Z}

Answer: (B)

(a, b) \in T

then

a^{2} - b^{2} \in Z

, hence

b^{2} - a^{2} \in Z

and

(b, a) \in T

, so

T

is symmetric. For

S

, take

a = - 1, b = 3

: here

2 + \frac{a}{b} = 2 - \frac{1}{3} > 0

, so

(- 1, 3) \in S

, yet

2 + \frac{b}{a} = 2 - 3 = - 1 < 0

, so

(3, - 1) \in / S .

Thus

S

fails symmetry, and

T

is symmetric while

S

is not.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option