The relation R on the set of real numbers defined by R = (a, b): a ≤…

The correct answer is Option (3) → (B) and (C) only

Each property must hold for all real numbers, so one counterexample is enough to show that it fails.

R is reflexive only if \(a \le a^2\) for every real a. This fails for \(a = \frac{1}{2}\):

    \(\displaystyle \left(\frac{1}{2}\right)^2 = \frac{1}{4} \lt \frac{1}{2}\)

R is symmetric only if \((a, b) \in R\) always gives \((b, a) \in R\). Take a = 1, b = 2:

    \((1, 2) \in R\) since \(1 \le 2^2\)
    \((2, 1) \notin R\) since \(2 \gt 1^2\)

R is transitive only if \((a, b) \in R\) and \((b, c) \in R\) always give \((a, c) \in R\). Take a = 5, b = 3, c = 2:

    \((5, 3) \in R\) since \(5 \le 3^2\)
    \((3, 2) \in R\) since \(3 \le 2^2\)
    \((5, 2) \notin R\) since \(5 \gt 2^2\)

So R is not reflexive, not symmetric and not transitive.

Checking the statements

(A) Reflexive — Incorrect. \(\left(\frac{1}{2}, \frac{1}{2}\right) \notin R\), since \(\frac{1}{2} \gt \frac{1}{4}\).

(B) Not symmetric — Correct. \((1, 2) \in R\) but \((2, 1) \notin R\).

(C) Neither reflexive nor transitive — Correct. Both properties fail.

(D) Transitive — Incorrect. \((5, 3), (3, 2) \in R\) but \((5, 2) \notin R\).

NCERT (2026-27 edition):
Types of Relations