Class JEE Mathematics Sets, Relations, and Functions Q #1065
KNOWLEDGE BASED
APPLY
4 Marks 2018 JEE Main 2018 (Online) 16th April Morning Slot MCQ SINGLE
Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x, y) \in N \times N : 2x + y = 10\}$ and $R_2 = \{(x, y) \in N \times N : x + 2y = 10\}$. Then :
(A) Range of $R_1$ is $\{2, 4, 8\}$.
(B) Range of $R_2$ is $\{1, 2, 3, 4\}$.
(C) Both $R_1$ and $R_2$ are symmetric relations.
(D) Both $R_1$ and $R_2$ are transitive relations.
Prev Next
Correct Answer: B
Explanation
For $R_1$, $2x + y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are:
$x = 1, y = 8 \implies (1, 8)$
$x = 2, y = 6 \implies (2, 6)$
$x = 3, y = 4 \implies (3, 4)$
$x = 4, y = 2 \implies (4, 2)$
Therefore, $R_1 = \{(1, 8), (2, 6), (3, 4), (4, 2)\}$.
Hence, the range of $R_1$ is $\{2, 4, 6, 8\}$.
$R_1$ is not symmetric.
$R_1$ is not transitive as $(3, 4), (4, 2) \in R_1$, but $(3, 2) \notin R_1$.
For $R_2$, $x + 2y = 10$ and $x, y \in N$, the possible values for $x$ and $y$ are:
$x = 8, y = 1 \implies (8, 1)$
$x = 6, y = 2 \implies (6, 2)$
$x = 4, y = 3 \implies (4, 3)$
$x = 2, y = 4 \implies (2, 4)$
Therefore, $R_2 = \{(8, 1), (6, 2), (4, 3), (2, 4)\}$.
Hence, the range of $R_2 = \{1, 2, 3, 4\}$.
$R_2$ is not symmetric and transitive.

More from this Chapter

NUMERICAL
The number of elements in the set $\left\{n \in \mathbb{N}: 10 \leq n \leq 100\right.$ and $3^{n}-3$ is a multiple of 7$\}$ is ___________.
NUMERICAL
4 Let $A=\{1,2,3\}$. The number of relations on $A$, containing $(1,2)$ and $(2,3)$, which are reflexive and transitive but not symmetric, is _________.
NUMERICAL
Let $A=\{1,2,3, \ldots \ldots \ldots \ldots, 100\}$. Let $R$ be a relation on $\mathrm{A}$ defined by $(x, y) \in R$ if and only if $2 x=3 y$. Let $R_1$ be a symmetric relation on $A$ such that $R \subset R_1$ and the number of elements in $R_1$ is $\mathrm{n}$. Then, the minimum value of $\mathrm{n}$ is _________.
MCQ_SINGLE
Let $R_1 = \{(a, b) \in N \times N : |a - b| \le 13\}$ and $R_2 = \{(a, b) \in N \times N : |a - b| \ne 13\}$. Then on N :
NUMERICAL
Let $S=\left\{p_1, p_2 \ldots, p_{10}\right\}$ be the set of first ten prime numbers. Let $A=S \cup P$, where $P$ is the set of all possible products of distinct elements of $S$. Then the number of all ordered pairs $(x, y), x \in S$, $y \in A$, such that $x$ divides $y$, is ________ .
View All Questions