116811
The number of equivalence relations on the set \(\{1,2,3\}\) containing \((1,2)\) and \((2,1)\) is
1 3
2 1
3 2
4 None of these
Explanation:
C Equivalence relation of the set \(\{(1,2,3)\}\) containing \((1,2)\) and \((2,1)\) \(\mathrm{A}_1=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}\) \(\mathrm{A}_2=\{(1,1),(2,2),(3,3)(1,2)(2,1),(2,3)(3,1)(3\), \(2),(1,3)\}\) So, There are only two equivalence relation are possible.
AMU-2015
Sets, Relation and Function
116812
Let \(R\) and \(S\) be two equivalence relations on a non-void set \(A\). Then
1 \(\mathrm{R} \cup \mathrm{S}\) is a equivalence relation
2 \(\mathrm{R} \cap \mathrm{S}\) is equivalence relation
3 \(\mathrm{R} \cap \mathrm{S}\) is not equivalence relegation
4 \(\mathrm{R} \cup \mathrm{S}\) is not a equivalence relation
Explanation:
B Given, \(\mathrm{R}\) and \(\mathrm{S}\) be two equivalence relations on a non-void set A. For reflexive : - \(R\) and \(S\) are reflexive this means for any \(a \in A\). \(\therefore \quad(a, a) \in R\) and \((a, a) \in S\) \(\Rightarrow \quad(a, a) \in R \cap S\) \(\therefore \quad \mathrm{R} \cap \mathrm{S}\) is reflexive. For symmetric:- \((a, b) \in R \cap S\) Then, \(\quad(a, b) \in R,(a, b) \in S\) Since, \(R\) and \(S\) are symmetric. \(\therefore(b, a) \in R\) and \((b, a) \in S\) For transitive:- Let, \(\quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R} \cap \mathrm{S}\) \(\Rightarrow \quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R}\) \(\therefore \quad(a, c) \in R\), since, \(R\) is transitive. And, \((\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{S}\) This means \((a, c) \in S\) since, \(S\) is transitive. \(\therefore \quad(a, c) \in R \cap S\). So, \(\mathrm{R} \cap \mathrm{S}\) is transitive. Hence, \(\mathrm{R} \cap \mathrm{S}\) is an equivalence relation.
WB JEE-2022
Sets, Relation and Function
116813
If there are 2 elements in a set \(A\), then what would be the number of possible relations from the set \(A\) to set \(A\) ?
1 2
2 4
3 16
4 32
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) Hence, number of possible relation from set \(A\) to set \(A\) \(\Rightarrow \quad 2^{n^2}=2^{2^2}=2^4=2 \times 2 \times 2 \times 2=16\)
J and K CET-2019
Sets, Relation and Function
116814
Let \(\mathbf{X}=\{\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}\}\) and \(\mathbf{R}=\{(\mathbf{a}, \mathbf{a}),(\mathbf{b}, \mathbf{b})\), (c, \(c),(a, b),(b, a)\}\). Then the relation \(R\) on \(X\) is
1 reflexive and symmetric
2 not reflexive but symmetric
3 symmetric and transitive, but not reflexive
4 reflexive but not transitive
Explanation:
C Given, \(\mathbf{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\} \text { and }\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c}),(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a})\}\) Since, \((a, b) \in R,(b, a) \in R\) and \((a, a) \in R\) So, Relation is transitive for all \(a, b \in X\) \((a, b) \in R\) and ( \(b, a) \in R\) so relation \(R\) is symmetric The relation \(\mathrm{r}\) is not reflexive because ( \(\mathrm{d}, \mathrm{d}) \notin \mathrm{R}\) and \((\mathrm{e}, \mathrm{e}) \notin \mathrm{R}\)
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Sets, Relation and Function
116811
The number of equivalence relations on the set \(\{1,2,3\}\) containing \((1,2)\) and \((2,1)\) is
1 3
2 1
3 2
4 None of these
Explanation:
C Equivalence relation of the set \(\{(1,2,3)\}\) containing \((1,2)\) and \((2,1)\) \(\mathrm{A}_1=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}\) \(\mathrm{A}_2=\{(1,1),(2,2),(3,3)(1,2)(2,1),(2,3)(3,1)(3\), \(2),(1,3)\}\) So, There are only two equivalence relation are possible.
AMU-2015
Sets, Relation and Function
116812
Let \(R\) and \(S\) be two equivalence relations on a non-void set \(A\). Then
1 \(\mathrm{R} \cup \mathrm{S}\) is a equivalence relation
2 \(\mathrm{R} \cap \mathrm{S}\) is equivalence relation
3 \(\mathrm{R} \cap \mathrm{S}\) is not equivalence relegation
4 \(\mathrm{R} \cup \mathrm{S}\) is not a equivalence relation
Explanation:
B Given, \(\mathrm{R}\) and \(\mathrm{S}\) be two equivalence relations on a non-void set A. For reflexive : - \(R\) and \(S\) are reflexive this means for any \(a \in A\). \(\therefore \quad(a, a) \in R\) and \((a, a) \in S\) \(\Rightarrow \quad(a, a) \in R \cap S\) \(\therefore \quad \mathrm{R} \cap \mathrm{S}\) is reflexive. For symmetric:- \((a, b) \in R \cap S\) Then, \(\quad(a, b) \in R,(a, b) \in S\) Since, \(R\) and \(S\) are symmetric. \(\therefore(b, a) \in R\) and \((b, a) \in S\) For transitive:- Let, \(\quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R} \cap \mathrm{S}\) \(\Rightarrow \quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R}\) \(\therefore \quad(a, c) \in R\), since, \(R\) is transitive. And, \((\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{S}\) This means \((a, c) \in S\) since, \(S\) is transitive. \(\therefore \quad(a, c) \in R \cap S\). So, \(\mathrm{R} \cap \mathrm{S}\) is transitive. Hence, \(\mathrm{R} \cap \mathrm{S}\) is an equivalence relation.
WB JEE-2022
Sets, Relation and Function
116813
If there are 2 elements in a set \(A\), then what would be the number of possible relations from the set \(A\) to set \(A\) ?
1 2
2 4
3 16
4 32
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) Hence, number of possible relation from set \(A\) to set \(A\) \(\Rightarrow \quad 2^{n^2}=2^{2^2}=2^4=2 \times 2 \times 2 \times 2=16\)
J and K CET-2019
Sets, Relation and Function
116814
Let \(\mathbf{X}=\{\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}\}\) and \(\mathbf{R}=\{(\mathbf{a}, \mathbf{a}),(\mathbf{b}, \mathbf{b})\), (c, \(c),(a, b),(b, a)\}\). Then the relation \(R\) on \(X\) is
1 reflexive and symmetric
2 not reflexive but symmetric
3 symmetric and transitive, but not reflexive
4 reflexive but not transitive
Explanation:
C Given, \(\mathbf{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\} \text { and }\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c}),(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a})\}\) Since, \((a, b) \in R,(b, a) \in R\) and \((a, a) \in R\) So, Relation is transitive for all \(a, b \in X\) \((a, b) \in R\) and ( \(b, a) \in R\) so relation \(R\) is symmetric The relation \(\mathrm{r}\) is not reflexive because ( \(\mathrm{d}, \mathrm{d}) \notin \mathrm{R}\) and \((\mathrm{e}, \mathrm{e}) \notin \mathrm{R}\)
116811
The number of equivalence relations on the set \(\{1,2,3\}\) containing \((1,2)\) and \((2,1)\) is
1 3
2 1
3 2
4 None of these
Explanation:
C Equivalence relation of the set \(\{(1,2,3)\}\) containing \((1,2)\) and \((2,1)\) \(\mathrm{A}_1=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}\) \(\mathrm{A}_2=\{(1,1),(2,2),(3,3)(1,2)(2,1),(2,3)(3,1)(3\), \(2),(1,3)\}\) So, There are only two equivalence relation are possible.
AMU-2015
Sets, Relation and Function
116812
Let \(R\) and \(S\) be two equivalence relations on a non-void set \(A\). Then
1 \(\mathrm{R} \cup \mathrm{S}\) is a equivalence relation
2 \(\mathrm{R} \cap \mathrm{S}\) is equivalence relation
3 \(\mathrm{R} \cap \mathrm{S}\) is not equivalence relegation
4 \(\mathrm{R} \cup \mathrm{S}\) is not a equivalence relation
Explanation:
B Given, \(\mathrm{R}\) and \(\mathrm{S}\) be two equivalence relations on a non-void set A. For reflexive : - \(R\) and \(S\) are reflexive this means for any \(a \in A\). \(\therefore \quad(a, a) \in R\) and \((a, a) \in S\) \(\Rightarrow \quad(a, a) \in R \cap S\) \(\therefore \quad \mathrm{R} \cap \mathrm{S}\) is reflexive. For symmetric:- \((a, b) \in R \cap S\) Then, \(\quad(a, b) \in R,(a, b) \in S\) Since, \(R\) and \(S\) are symmetric. \(\therefore(b, a) \in R\) and \((b, a) \in S\) For transitive:- Let, \(\quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R} \cap \mathrm{S}\) \(\Rightarrow \quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R}\) \(\therefore \quad(a, c) \in R\), since, \(R\) is transitive. And, \((\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{S}\) This means \((a, c) \in S\) since, \(S\) is transitive. \(\therefore \quad(a, c) \in R \cap S\). So, \(\mathrm{R} \cap \mathrm{S}\) is transitive. Hence, \(\mathrm{R} \cap \mathrm{S}\) is an equivalence relation.
WB JEE-2022
Sets, Relation and Function
116813
If there are 2 elements in a set \(A\), then what would be the number of possible relations from the set \(A\) to set \(A\) ?
1 2
2 4
3 16
4 32
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) Hence, number of possible relation from set \(A\) to set \(A\) \(\Rightarrow \quad 2^{n^2}=2^{2^2}=2^4=2 \times 2 \times 2 \times 2=16\)
J and K CET-2019
Sets, Relation and Function
116814
Let \(\mathbf{X}=\{\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}\}\) and \(\mathbf{R}=\{(\mathbf{a}, \mathbf{a}),(\mathbf{b}, \mathbf{b})\), (c, \(c),(a, b),(b, a)\}\). Then the relation \(R\) on \(X\) is
1 reflexive and symmetric
2 not reflexive but symmetric
3 symmetric and transitive, but not reflexive
4 reflexive but not transitive
Explanation:
C Given, \(\mathbf{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\} \text { and }\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c}),(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a})\}\) Since, \((a, b) \in R,(b, a) \in R\) and \((a, a) \in R\) So, Relation is transitive for all \(a, b \in X\) \((a, b) \in R\) and ( \(b, a) \in R\) so relation \(R\) is symmetric The relation \(\mathrm{r}\) is not reflexive because ( \(\mathrm{d}, \mathrm{d}) \notin \mathrm{R}\) and \((\mathrm{e}, \mathrm{e}) \notin \mathrm{R}\)
116811
The number of equivalence relations on the set \(\{1,2,3\}\) containing \((1,2)\) and \((2,1)\) is
1 3
2 1
3 2
4 None of these
Explanation:
C Equivalence relation of the set \(\{(1,2,3)\}\) containing \((1,2)\) and \((2,1)\) \(\mathrm{A}_1=\{(1,1)(2,2)(3,3)(1,2)(2,1)\}\) \(\mathrm{A}_2=\{(1,1),(2,2),(3,3)(1,2)(2,1),(2,3)(3,1)(3\), \(2),(1,3)\}\) So, There are only two equivalence relation are possible.
AMU-2015
Sets, Relation and Function
116812
Let \(R\) and \(S\) be two equivalence relations on a non-void set \(A\). Then
1 \(\mathrm{R} \cup \mathrm{S}\) is a equivalence relation
2 \(\mathrm{R} \cap \mathrm{S}\) is equivalence relation
3 \(\mathrm{R} \cap \mathrm{S}\) is not equivalence relegation
4 \(\mathrm{R} \cup \mathrm{S}\) is not a equivalence relation
Explanation:
B Given, \(\mathrm{R}\) and \(\mathrm{S}\) be two equivalence relations on a non-void set A. For reflexive : - \(R\) and \(S\) are reflexive this means for any \(a \in A\). \(\therefore \quad(a, a) \in R\) and \((a, a) \in S\) \(\Rightarrow \quad(a, a) \in R \cap S\) \(\therefore \quad \mathrm{R} \cap \mathrm{S}\) is reflexive. For symmetric:- \((a, b) \in R \cap S\) Then, \(\quad(a, b) \in R,(a, b) \in S\) Since, \(R\) and \(S\) are symmetric. \(\therefore(b, a) \in R\) and \((b, a) \in S\) For transitive:- Let, \(\quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R} \cap \mathrm{S}\) \(\Rightarrow \quad(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{R}\) \(\therefore \quad(a, c) \in R\), since, \(R\) is transitive. And, \((\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{c}) \in \mathrm{S}\) This means \((a, c) \in S\) since, \(S\) is transitive. \(\therefore \quad(a, c) \in R \cap S\). So, \(\mathrm{R} \cap \mathrm{S}\) is transitive. Hence, \(\mathrm{R} \cap \mathrm{S}\) is an equivalence relation.
WB JEE-2022
Sets, Relation and Function
116813
If there are 2 elements in a set \(A\), then what would be the number of possible relations from the set \(A\) to set \(A\) ?
1 2
2 4
3 16
4 32
Explanation:
C Given, \(\mathrm{n}(\mathrm{A})=2\) Hence, number of possible relation from set \(A\) to set \(A\) \(\Rightarrow \quad 2^{n^2}=2^{2^2}=2^4=2 \times 2 \times 2 \times 2=16\)
J and K CET-2019
Sets, Relation and Function
116814
Let \(\mathbf{X}=\{\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}, \mathbf{e}\}\) and \(\mathbf{R}=\{(\mathbf{a}, \mathbf{a}),(\mathbf{b}, \mathbf{b})\), (c, \(c),(a, b),(b, a)\}\). Then the relation \(R\) on \(X\) is
1 reflexive and symmetric
2 not reflexive but symmetric
3 symmetric and transitive, but not reflexive
4 reflexive but not transitive
Explanation:
C Given, \(\mathbf{X}=\{\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}, \mathrm{e}\} \text { and }\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{a}),(\mathrm{b}, \mathrm{b}),(\mathrm{c}, \mathrm{c}),(\mathrm{a}, \mathrm{b}),(\mathrm{b}, \mathrm{a})\}\) Since, \((a, b) \in R,(b, a) \in R\) and \((a, a) \in R\) So, Relation is transitive for all \(a, b \in X\) \((a, b) \in R\) and ( \(b, a) \in R\) so relation \(R\) is symmetric The relation \(\mathrm{r}\) is not reflexive because ( \(\mathrm{d}, \mathrm{d}) \notin \mathrm{R}\) and \((\mathrm{e}, \mathrm{e}) \notin \mathrm{R}\)