Reflexive: (1,1),(2,2),(3,3) all present, yes.
Symmetric: (1,2) in R but (2,1) not in R, so not symmetric.
Transitive: (1,2) and (2,3) give (1,3), present. All transitive checks pass.
Hence R is reflexive and transitive.
CUET UG 2023 — Mathematics Algebra
Let A = {1,2,3}. Consider the relation R = {(1,1),(2,2),(3,3),(1,2),(2,3),(1,3)}. Then R is
Held on 25 May 2023 · Verified 13 Jul 2026.
reflexive only
reflexive and transitive
symmetric and transitive
neither symmetric nor transitive
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If f(x) = 2x + 3, then f⁻¹(x) is:
A relation R in the set A = {1, 2,3, 4} is given by R = {(1,1), (2,2), (1,2), (2,3), (3,4), (4,4), (1,3), (2,4), (1,4)} is
If $P(A) = \frac{3}{5}$, $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{4}$, then $P(\overline{A} | \overline{B})$ is
If corner points of the bounded feasible region are (0, 0), (3, 0) and (0, 3) and objective function is $z = 4x + 7y$, then the maximum value of $z$ is
The maximum value of a LPP $z = 3x + 4y$ subject to the constraints: $x + y \leq 6$, $x \geq 0$, $y \geq 0$ is:
Work through every CUET UG Algebra PYQ, year by year.