Let R be the relation on the set R of real numbers defined by \(R = \{(a, b) : a^{2} - 3ab + 2b^{2} = 0\}\). Consider the following statements:
(A) R is reflexive.
(B) R is symmetric.
(C) R is transitive.
(D) \((6, 3) \in R\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (D) only
The condition factorises:
\(a^{2} - 3ab + 2b^{2} = (a - b)(a - 2b)\)
So \((a, b) \in R\) exactly when \(a = b\) or \(a = 2b\).
\((2, 1) \in R\), since \(2 = 2 \times 1\)
\((1, 2) \notin R\), since \(1 \ne 2\) and \(1 \ne 4\)
\((4, 2) \in R\) and \((2, 1) \in R\)
\((4, 1) \notin R\), since \(4 \ne 1\) and \(4 \ne 2\)
So R is reflexive but neither symmetric nor transitive, and it contains every pair with a = 2b.
Checking the statements
(A) R is reflexive. — Correct. \(a = a\) for every real a, so \((a, a) \in R\).
(B) R is symmetric. — Incorrect. \((2, 1) \in R\) but \((1, 2) \notin R\).
(C) R is transitive. — Incorrect. (4, 2) and (2, 1) are in R, but (4, 1) is not.
(D) \((6, 3) \in R\) — Correct. \(6 = 2 \times 3\).