Class JEE Mathematics Sets, Relations, and Functions Q #1056
KNOWLEDGE BASED
APPLY
4 Marks 2020 JEE Main 2020 (Online) 3rd September Evening Slot MCQ SINGLE
Let $R_1$ and $R_2$ be two relation defined as follows: $R_1 = {(a, b) \in R^2 : a^2 + b^2 \in Q}$ and $R_2 = {(a, b) \in R^2 : a^2 + b^2 \notin Q}$, where $Q$ is the set of all rational numbers. Then :
(A) Neither $R_1$ nor $R_2$ is transitive.
(B) $R_2$ is transitive but $R_1$ is not transitive.
(C) $R_1$ and $R_2$ are both transitive.
(D) $R_1$ is transitive but $R_2$ is not transitive.
Prev Next
Correct Answer: A
Explanation
For $R_1$: Let $a = 1 + \sqrt{2}$, $b = 1 - \sqrt{2}$, $c = \sqrt[4]{8}$. $aR_1b : a^2 + b^2 = 6 \in Q$ $bR_1c : b^2 + c^2 = 3 - 2\sqrt{2} + 2\sqrt{2} = 3 \in Q$ $aR_1c : a^2 + c^2 = 3 + 2\sqrt{2} + 2\sqrt{2} \notin Q$ $\therefore$ $R_1$ is not transitive. For $R_2$: Let $a = 1 + \sqrt{2}$, $b = \sqrt{2}$, $c = 1 - \sqrt{2}$ $aR_2b : a^2 + b^2 = 5 + 2\sqrt{2} \notin Q$ $bR_2c : b^2 + c^2 = 5 - 2\sqrt{2} \notin Q$ $aR_2c : a^2 + c^2 = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 6 \in Q$ $\therefore$ $R_2$ is not transitive.

More from this Chapter

MCQ_SINGLE
Let $X = R \times R$. Define a relation R on X as: $(a_1, b_1) R (a_2, b_2) \Leftrightarrow b_1 = b_2$ Statement I: $R$ is an equivalence relation. Statement II: For some $(a, b) \in X$, the set $S = \{(x, y) \in X : (x, y)R(a, b)\}$ represents a line parallel to $y = x$. In the light of the above statements, choose the correct answer from the options given below:
NUMERICAL
Let $A=\{1,2,3,4\}$ and $\mathrm{R}$ be a relation on the set $A \times A$ defined by $R=\{((a, b),(c, d)): 2 a+3 b=4 c+5 d\}$. Then the number of elements in $\mathrm{R}$ is ____________.
MCQ_SINGLE
Consider the following two binary relations on the set $A = {a, b, c}$: $R_1 = {(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)}$ and $R_2 = {(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)}$. Then:
MCQ_SINGLE
Let $A = {x \in (0, \pi) - {\frac{\pi}{2}} : \log_{(2/\pi)} |\sin x| + \log_{(2/\pi)} |\cos x| = 2}$ and $B = {x \ge 0 : x(x-4) - 3|x-2| + 6 = 0}$. Then $n(A \cup B)$ is equal to :
NUMERICAL
Let $A=\{1,2,3,4,5,6,7\}$. Define $B=\{T \subseteq A$ : either $1 \notin T$ or $2 \in T\}$ and $C=\{T \subseteq A: T$ the sum of all the elements of $T$ is a prime number $\}$. Then the number of elements in the set $B \cup C$ is ________________.
View All Questions