(A) The inverse relation ${R^{ - 1}}$ is obtained by interchanging the elements of the ordered pairs in the relation $R$. Given $R = \{(1, 3), (2, 5),

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(A) The inverse relation ${R^{ - 1}}$ is obtained by interchanging the elements of the ordered pairs in the relation $R$. Given $R = \{(1, 3), (2, 5), (3, 3)\}$. By swapping the coordinates $(x, y)$ to $(y, x)$ for each pair in $R$: $(1, 3) \rightarrow (3, 1)$ $(2, 5) \rightarrow (5, 2)$ $(3, 3) \rightarrow (3, 3)$ Therefore,${R^{ - 1}} = \{(3, 1), (5, 2), (3, 3)\}$. Comparing this with the given options,the correct option is $A$.

Class 11 Mathematics Relations and Functions Relations

Medium

Let $A = \{1, 2, 3\}$ and $B = \{1, 3, 5\}$. If a relation $R$ from $A$ to $B$ is defined as $R = \{(1, 3), (2, 5), (3, 3)\}$,then find the inverse relation ${R^{ - 1}}$.

A
$\{(3, 1), (5, 2), (3, 3)\}$
B
$\{(1, 3), (2, 5), (3, 3)\}$
C
$\{(1, 3), (5, 2)\}$
D
None of these

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Relations and Functions

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Define a relation $R$ on the set $N$ of natural numbers by $R = \{(x, y) : y = x + 5, x \text{ is a natural number less than } 4; x, y \in N\}$. Depict this relationship using roster form. Write down the domain and the range.

Medium

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The solution set of $x \equiv 3 \pmod{7}$,where $x \in \mathbb{Z}$,is:

Medium

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$A$ relation on the set $A = \{x : |x| < 3, x \in Z\}$,where $Z$ is the set of integers,is defined by $R = \{(x, y) : y = |x|, x \neq -1\}$. Then the number of elements in the power set of $R$ is

DifficultJEE MAIN 2014

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Let $A = \{1, 2, 3, \ldots, 14\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$. Write down its domain,codomain,and range.

Easy

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Let $A = \{1, 2, 3, 4, \ldots, 10\}$ and $B = \{0, 1, 2, 3, 4\}$. The number of elements in the relation $R = \{(a, b) \in A \times A : 2(a - b)^2 + 3(a - b) \in B\}$ is $.........$.

DifficultJEE MAIN 2023

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Let $A = \{1, 2, 3\}$ and $B = \{1, 3, 5\}$. If a relation $R$ from $A$ to $B$ is defined as $R = \{(1, 3), (2, 5), (3, 3)\}$,then find the inverse relation ${R^{ - 1}}$.

Similar Questions

Define a relation $R$ on the set $N$ of natural numbers by $R = \{(x, y) : y = x + 5, x \text{ is a natural number less than } 4; x, y \in N\}$. Depict this relationship using roster form. Write down the domain and the range.

The solution set of $x \equiv 3 \pmod{7}$,where $x \in \mathbb{Z}$,is:

$A$ relation on the set $A = \{x : |x| < 3, x \in Z\}$,where $Z$ is the set of integers,is defined by $R = \{(x, y) : y = |x|, x \neq -1\}$. Then the number of elements in the power set of $R$ is

Let $A = \{1, 2, 3, \ldots, 14\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$. Write down its domain,codomain,and range.

Let $A = \{1, 2, 3, 4, \ldots, 10\}$ and $B = \{0, 1, 2, 3, 4\}$. The number of elements in the relation $R = \{(a, b) \in A \times A : 2(a - b)^2 + 3(a - b) \in B\}$ is $.........$.

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