Among the relations
S={(a, b): a, b ∈ ℝ-{0}, 2+a/b>0}
and T={(a, b): a, b ∈ ℝ, a²-b² ∈ ℤ},

Solution

Maths sir · 5 steps

T only cares about a² - b² being an integer, so it behaves well; S breaks with negative ratios.

1. T is symmetric. If a² - b² is an integer, then b² - a² = -(a² - b²) is also an integer. So (a, b) ∈ T gives (b, a) ∈ T.
2. T is transitive. If a² - b² and b² - c² are integers, their sum a² - c² is an integer too. So T is also transitive.
3. S is not symmetric. Take a = -1, b = 2: 2 + -1/2 = 1.5 > 0, so (-1, 2) ∈ S. Reversed, 2 + 2/-1 = 0, which is not greater than 0, so (2, -1) ∉ S.
4. S is not transitive. (4, 1) ∈ S since 2 + 4 = 6 > 0, and (1, -1) ∈ S since 2 - 1 = 1 > 0. But 2 + 4/-1 = -2, so (4, -1) ∉ S.

   Relation	Symmetric?	Transitive?
   S	No	No
   T	Yes	Yes

5. So. T is symmetric but S is not, which is option (d).

The idea to keep. To prove a property, argue for every pair; to kill one, a single counterexample is enough. For ratio conditions, try a negative number first.

Answer from the NTA final answer key. Solution by PrepWiser.