WBJEE · Maths · Sets and Relations

On the set \(R\) of real numbers we define \(x P y\) if and only if \(x y \geq 0\). Then, the relation \(P\) is

A reflexive but not symmetric
B symmetric but not reflexive
C transitive but not reflexive
D reflexive and symmetric but not transitive

Verified Solution

Answer & Solution

Correct Answer

(D) reflexive and symmetric but not transitive

Step-by-step Solution

Detailed explanation

For every real number \(x, x^{2} \geq 0\) \(\therefore\) \((x, x) \in P\) Hence, \(P\) is reflexive. Now let \((x,y) \in P \\ \Rightarrow xy \geq 0 \\ \Rightarrow yx \geq 0\) \((y, x) \in P\) Hence, \(P\) is symmetric. Again (-1,0)\(\in P\) and (0,2)\(\in P\) But (-1,2)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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