116790
The relation \(R\) defined on the set of natural numbers as \(\{(\mathbf{a}, \mathbf{b})\) : a differs from \(b\) by 3\(\}\) is given
1 \(\{(1,4),(2,5),(3,6), \ldots\}\)
2 \(\{(4,1),(5,2),(6,3), \ldots\}\)
3 \(\{(1,3),(2,6),(3,9), \ldots\}\)
4 None of the above
Explanation:
B Given, The relation \(\mathrm{R}\) defined on the set of natural number as \(\{(\mathrm{a}, \mathrm{b})\) : a differs from b by 3\(\}\) can be also written as. \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}-\mathrm{b}=3\}\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}=\mathrm{b}+3\}\) \(\mathrm{R}=[\{\mathrm{b}+3, \mathrm{~b}\}, \mathrm{b} \in \mathrm{N}]\) \(\mathrm{R}=[\{\mathrm{n}+3, \mathrm{n}\} \mathrm{n} \in \mathrm{N}\}]\) \(\mathrm{O}_1\) If \(\mathrm{n}=1,2,3,4 \ldots\). so, the relation becomes \(R=\{(4,1),(5,2),(6,3) \ldots \ldots .\}\)
VITEEE-2012
Sets, Relation and Function
116791
If \(R\) be a relation from \(A=\{1,2,3,4\}\) to \(B=\{1,3,5\}\) such that \((a, b) \in R \Leftrightarrow a\lt b\), then \(\mathbf{R O R}^{-1}\) is
1 \(\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}\)
2 \(\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}\)
3 \(\{(3,3),(3,5),(5,3),(5,5)\}\)
4 \(\{(3,3),(3,4),(4,5)\}\)
Explanation:
C Given, \(\mathrm{A}=\{1,2,3,4\}\) \(\text { and } \quad B=\{1,3,5\}\) Such that, \((a, b) \in R \Leftrightarrow a\lt b\) Then, So, \(\mathrm{R}=\{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)\}\) and \(\mathrm{R}^{-1}=\{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)\}\) For composition \(\mathrm{ROR}^{-1}\), we will pickup an element of \(\mathrm{R}^{-1}\) first then of \(\mathrm{R}\). Eg. \((3,1) \in R^{-1},(1,3) \in R \Rightarrow(3,3) \in \operatorname{ROR}^{-1}\) Hence, \(\operatorname{ROR}^{-1}=\{(3,3),(3,5)(5,3),(5,5)\}\)
VITEEE-2011
Sets, Relation and Function
116792
If \(R\) be a relation defined as a \(R\) b if \(f|a-b|>\) 0 , then the relation is
1 reflexive
2 symmetric
3 transitive
4 symmetric and transitive
Explanation:
D Given, \(\mathrm{R}\) be a relation defined as \({ }_{\mathrm{a}} \mathrm{R}_{\mathrm{b}}\) if \(\mathrm{f}|\mathrm{a}-\mathrm{b}|>0\) Then, checking for the relation- (a) Reflexive :- Consider a be an arbitrary element \(\therefore|\mathrm{a}-\mathrm{a}|=0\) which shows \(\mathrm{a} \notin \mathrm{R}\) Then, it is not reflexive relation on \(R\). (b) Symmetric : - \(|\mathrm{a}-\mathrm{b}|>0 \Rightarrow|\mathrm{b}-\mathrm{a}|>0\) \(\Rightarrow \mathrm{aRb}=\mathrm{bRa}\) \(\text { Since, } |\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}| \text { Then, }\) Since, \(\quad|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|\) Then, \(\quad \mathrm{R}\) is symmetric. (c) Transitive : - \(|\mathrm{a}-\mathrm{b}|>0,|\mathrm{~b}-\mathrm{c}|>0 \Rightarrow|\mathrm{a}-\mathrm{c}|>0\)Therefore, \((a, c) \in R \quad\) Then, \(\mathrm{R}\) is transitive. So, the relation is symmetric and transitive.
VITEEE-2008
Sets, Relation and Function
116794
The relation \(R\) defined on as \(R=\{(1,1),(2,2),(3,3),(1,3)\}\) is
1 equivalence
2 not symmetric
3 not reflexive
4 not transitive
Explanation:
B Given, \(\mathrm{A}=\{1,2,3\}, \quad \mathrm{R}=\{(1,1),(2,2),(3,3),(1,3)\}\) Then, check relations - (a) Reflexive : - \((1,1) \in R \Rightarrow{ }_1 R_1\) Then, \(\mathrm{R}\) is reflexive. (b) Symmetric : - \((1,3) \in \mathrm{R} \Rightarrow(3,1) \notin \mathrm{R}\) \({ }_1 \mathrm{R}_3 \nRightarrow{ }_3 \mathrm{R}_1\) Then, \(\mathrm{R}\) is not symmetric. (c) Transitive :- \((1,1) \in \mathrm{R},(3,3) \in \mathrm{R} \Rightarrow(1,3) \in \mathrm{R}\) \({ }_1 \mathrm{R}_1,{ }_3 \mathrm{R}_3 \Rightarrow{ }_1 \mathrm{R}_3\) Then, \(R\) is transitive. So, the relation \(\mathrm{R}\) is not symmetric but reflexive and transitive.
116790
The relation \(R\) defined on the set of natural numbers as \(\{(\mathbf{a}, \mathbf{b})\) : a differs from \(b\) by 3\(\}\) is given
1 \(\{(1,4),(2,5),(3,6), \ldots\}\)
2 \(\{(4,1),(5,2),(6,3), \ldots\}\)
3 \(\{(1,3),(2,6),(3,9), \ldots\}\)
4 None of the above
Explanation:
B Given, The relation \(\mathrm{R}\) defined on the set of natural number as \(\{(\mathrm{a}, \mathrm{b})\) : a differs from b by 3\(\}\) can be also written as. \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}-\mathrm{b}=3\}\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}=\mathrm{b}+3\}\) \(\mathrm{R}=[\{\mathrm{b}+3, \mathrm{~b}\}, \mathrm{b} \in \mathrm{N}]\) \(\mathrm{R}=[\{\mathrm{n}+3, \mathrm{n}\} \mathrm{n} \in \mathrm{N}\}]\) \(\mathrm{O}_1\) If \(\mathrm{n}=1,2,3,4 \ldots\). so, the relation becomes \(R=\{(4,1),(5,2),(6,3) \ldots \ldots .\}\)
VITEEE-2012
Sets, Relation and Function
116791
If \(R\) be a relation from \(A=\{1,2,3,4\}\) to \(B=\{1,3,5\}\) such that \((a, b) \in R \Leftrightarrow a\lt b\), then \(\mathbf{R O R}^{-1}\) is
1 \(\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}\)
2 \(\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}\)
3 \(\{(3,3),(3,5),(5,3),(5,5)\}\)
4 \(\{(3,3),(3,4),(4,5)\}\)
Explanation:
C Given, \(\mathrm{A}=\{1,2,3,4\}\) \(\text { and } \quad B=\{1,3,5\}\) Such that, \((a, b) \in R \Leftrightarrow a\lt b\) Then, So, \(\mathrm{R}=\{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)\}\) and \(\mathrm{R}^{-1}=\{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)\}\) For composition \(\mathrm{ROR}^{-1}\), we will pickup an element of \(\mathrm{R}^{-1}\) first then of \(\mathrm{R}\). Eg. \((3,1) \in R^{-1},(1,3) \in R \Rightarrow(3,3) \in \operatorname{ROR}^{-1}\) Hence, \(\operatorname{ROR}^{-1}=\{(3,3),(3,5)(5,3),(5,5)\}\)
VITEEE-2011
Sets, Relation and Function
116792
If \(R\) be a relation defined as a \(R\) b if \(f|a-b|>\) 0 , then the relation is
1 reflexive
2 symmetric
3 transitive
4 symmetric and transitive
Explanation:
D Given, \(\mathrm{R}\) be a relation defined as \({ }_{\mathrm{a}} \mathrm{R}_{\mathrm{b}}\) if \(\mathrm{f}|\mathrm{a}-\mathrm{b}|>0\) Then, checking for the relation- (a) Reflexive :- Consider a be an arbitrary element \(\therefore|\mathrm{a}-\mathrm{a}|=0\) which shows \(\mathrm{a} \notin \mathrm{R}\) Then, it is not reflexive relation on \(R\). (b) Symmetric : - \(|\mathrm{a}-\mathrm{b}|>0 \Rightarrow|\mathrm{b}-\mathrm{a}|>0\) \(\Rightarrow \mathrm{aRb}=\mathrm{bRa}\) \(\text { Since, } |\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}| \text { Then, }\) Since, \(\quad|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|\) Then, \(\quad \mathrm{R}\) is symmetric. (c) Transitive : - \(|\mathrm{a}-\mathrm{b}|>0,|\mathrm{~b}-\mathrm{c}|>0 \Rightarrow|\mathrm{a}-\mathrm{c}|>0\)Therefore, \((a, c) \in R \quad\) Then, \(\mathrm{R}\) is transitive. So, the relation is symmetric and transitive.
VITEEE-2008
Sets, Relation and Function
116794
The relation \(R\) defined on as \(R=\{(1,1),(2,2),(3,3),(1,3)\}\) is
1 equivalence
2 not symmetric
3 not reflexive
4 not transitive
Explanation:
B Given, \(\mathrm{A}=\{1,2,3\}, \quad \mathrm{R}=\{(1,1),(2,2),(3,3),(1,3)\}\) Then, check relations - (a) Reflexive : - \((1,1) \in R \Rightarrow{ }_1 R_1\) Then, \(\mathrm{R}\) is reflexive. (b) Symmetric : - \((1,3) \in \mathrm{R} \Rightarrow(3,1) \notin \mathrm{R}\) \({ }_1 \mathrm{R}_3 \nRightarrow{ }_3 \mathrm{R}_1\) Then, \(\mathrm{R}\) is not symmetric. (c) Transitive :- \((1,1) \in \mathrm{R},(3,3) \in \mathrm{R} \Rightarrow(1,3) \in \mathrm{R}\) \({ }_1 \mathrm{R}_1,{ }_3 \mathrm{R}_3 \Rightarrow{ }_1 \mathrm{R}_3\) Then, \(R\) is transitive. So, the relation \(\mathrm{R}\) is not symmetric but reflexive and transitive.
116790
The relation \(R\) defined on the set of natural numbers as \(\{(\mathbf{a}, \mathbf{b})\) : a differs from \(b\) by 3\(\}\) is given
1 \(\{(1,4),(2,5),(3,6), \ldots\}\)
2 \(\{(4,1),(5,2),(6,3), \ldots\}\)
3 \(\{(1,3),(2,6),(3,9), \ldots\}\)
4 None of the above
Explanation:
B Given, The relation \(\mathrm{R}\) defined on the set of natural number as \(\{(\mathrm{a}, \mathrm{b})\) : a differs from b by 3\(\}\) can be also written as. \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}-\mathrm{b}=3\}\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}=\mathrm{b}+3\}\) \(\mathrm{R}=[\{\mathrm{b}+3, \mathrm{~b}\}, \mathrm{b} \in \mathrm{N}]\) \(\mathrm{R}=[\{\mathrm{n}+3, \mathrm{n}\} \mathrm{n} \in \mathrm{N}\}]\) \(\mathrm{O}_1\) If \(\mathrm{n}=1,2,3,4 \ldots\). so, the relation becomes \(R=\{(4,1),(5,2),(6,3) \ldots \ldots .\}\)
VITEEE-2012
Sets, Relation and Function
116791
If \(R\) be a relation from \(A=\{1,2,3,4\}\) to \(B=\{1,3,5\}\) such that \((a, b) \in R \Leftrightarrow a\lt b\), then \(\mathbf{R O R}^{-1}\) is
1 \(\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}\)
2 \(\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}\)
3 \(\{(3,3),(3,5),(5,3),(5,5)\}\)
4 \(\{(3,3),(3,4),(4,5)\}\)
Explanation:
C Given, \(\mathrm{A}=\{1,2,3,4\}\) \(\text { and } \quad B=\{1,3,5\}\) Such that, \((a, b) \in R \Leftrightarrow a\lt b\) Then, So, \(\mathrm{R}=\{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)\}\) and \(\mathrm{R}^{-1}=\{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)\}\) For composition \(\mathrm{ROR}^{-1}\), we will pickup an element of \(\mathrm{R}^{-1}\) first then of \(\mathrm{R}\). Eg. \((3,1) \in R^{-1},(1,3) \in R \Rightarrow(3,3) \in \operatorname{ROR}^{-1}\) Hence, \(\operatorname{ROR}^{-1}=\{(3,3),(3,5)(5,3),(5,5)\}\)
VITEEE-2011
Sets, Relation and Function
116792
If \(R\) be a relation defined as a \(R\) b if \(f|a-b|>\) 0 , then the relation is
1 reflexive
2 symmetric
3 transitive
4 symmetric and transitive
Explanation:
D Given, \(\mathrm{R}\) be a relation defined as \({ }_{\mathrm{a}} \mathrm{R}_{\mathrm{b}}\) if \(\mathrm{f}|\mathrm{a}-\mathrm{b}|>0\) Then, checking for the relation- (a) Reflexive :- Consider a be an arbitrary element \(\therefore|\mathrm{a}-\mathrm{a}|=0\) which shows \(\mathrm{a} \notin \mathrm{R}\) Then, it is not reflexive relation on \(R\). (b) Symmetric : - \(|\mathrm{a}-\mathrm{b}|>0 \Rightarrow|\mathrm{b}-\mathrm{a}|>0\) \(\Rightarrow \mathrm{aRb}=\mathrm{bRa}\) \(\text { Since, } |\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}| \text { Then, }\) Since, \(\quad|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|\) Then, \(\quad \mathrm{R}\) is symmetric. (c) Transitive : - \(|\mathrm{a}-\mathrm{b}|>0,|\mathrm{~b}-\mathrm{c}|>0 \Rightarrow|\mathrm{a}-\mathrm{c}|>0\)Therefore, \((a, c) \in R \quad\) Then, \(\mathrm{R}\) is transitive. So, the relation is symmetric and transitive.
VITEEE-2008
Sets, Relation and Function
116794
The relation \(R\) defined on as \(R=\{(1,1),(2,2),(3,3),(1,3)\}\) is
1 equivalence
2 not symmetric
3 not reflexive
4 not transitive
Explanation:
B Given, \(\mathrm{A}=\{1,2,3\}, \quad \mathrm{R}=\{(1,1),(2,2),(3,3),(1,3)\}\) Then, check relations - (a) Reflexive : - \((1,1) \in R \Rightarrow{ }_1 R_1\) Then, \(\mathrm{R}\) is reflexive. (b) Symmetric : - \((1,3) \in \mathrm{R} \Rightarrow(3,1) \notin \mathrm{R}\) \({ }_1 \mathrm{R}_3 \nRightarrow{ }_3 \mathrm{R}_1\) Then, \(\mathrm{R}\) is not symmetric. (c) Transitive :- \((1,1) \in \mathrm{R},(3,3) \in \mathrm{R} \Rightarrow(1,3) \in \mathrm{R}\) \({ }_1 \mathrm{R}_1,{ }_3 \mathrm{R}_3 \Rightarrow{ }_1 \mathrm{R}_3\) Then, \(R\) is transitive. So, the relation \(\mathrm{R}\) is not symmetric but reflexive and transitive.
116790
The relation \(R\) defined on the set of natural numbers as \(\{(\mathbf{a}, \mathbf{b})\) : a differs from \(b\) by 3\(\}\) is given
1 \(\{(1,4),(2,5),(3,6), \ldots\}\)
2 \(\{(4,1),(5,2),(6,3), \ldots\}\)
3 \(\{(1,3),(2,6),(3,9), \ldots\}\)
4 None of the above
Explanation:
B Given, The relation \(\mathrm{R}\) defined on the set of natural number as \(\{(\mathrm{a}, \mathrm{b})\) : a differs from b by 3\(\}\) can be also written as. \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}-\mathrm{b}=3\}\) \(\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}, \mathrm{b} \in \mathrm{N}, \mathrm{a}=\mathrm{b}+3\}\) \(\mathrm{R}=[\{\mathrm{b}+3, \mathrm{~b}\}, \mathrm{b} \in \mathrm{N}]\) \(\mathrm{R}=[\{\mathrm{n}+3, \mathrm{n}\} \mathrm{n} \in \mathrm{N}\}]\) \(\mathrm{O}_1\) If \(\mathrm{n}=1,2,3,4 \ldots\). so, the relation becomes \(R=\{(4,1),(5,2),(6,3) \ldots \ldots .\}\)
VITEEE-2012
Sets, Relation and Function
116791
If \(R\) be a relation from \(A=\{1,2,3,4\}\) to \(B=\{1,3,5\}\) such that \((a, b) \in R \Leftrightarrow a\lt b\), then \(\mathbf{R O R}^{-1}\) is
1 \(\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\}\)
2 \(\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\}\)
3 \(\{(3,3),(3,5),(5,3),(5,5)\}\)
4 \(\{(3,3),(3,4),(4,5)\}\)
Explanation:
C Given, \(\mathrm{A}=\{1,2,3,4\}\) \(\text { and } \quad B=\{1,3,5\}\) Such that, \((a, b) \in R \Leftrightarrow a\lt b\) Then, So, \(\mathrm{R}=\{(1,3)(1,5)(2,3)(2,5)(3,5)(4,5)\}\) and \(\mathrm{R}^{-1}=\{(3,1)(5,1)(3,2)(5,2)(5,3)(5,4)\}\) For composition \(\mathrm{ROR}^{-1}\), we will pickup an element of \(\mathrm{R}^{-1}\) first then of \(\mathrm{R}\). Eg. \((3,1) \in R^{-1},(1,3) \in R \Rightarrow(3,3) \in \operatorname{ROR}^{-1}\) Hence, \(\operatorname{ROR}^{-1}=\{(3,3),(3,5)(5,3),(5,5)\}\)
VITEEE-2011
Sets, Relation and Function
116792
If \(R\) be a relation defined as a \(R\) b if \(f|a-b|>\) 0 , then the relation is
1 reflexive
2 symmetric
3 transitive
4 symmetric and transitive
Explanation:
D Given, \(\mathrm{R}\) be a relation defined as \({ }_{\mathrm{a}} \mathrm{R}_{\mathrm{b}}\) if \(\mathrm{f}|\mathrm{a}-\mathrm{b}|>0\) Then, checking for the relation- (a) Reflexive :- Consider a be an arbitrary element \(\therefore|\mathrm{a}-\mathrm{a}|=0\) which shows \(\mathrm{a} \notin \mathrm{R}\) Then, it is not reflexive relation on \(R\). (b) Symmetric : - \(|\mathrm{a}-\mathrm{b}|>0 \Rightarrow|\mathrm{b}-\mathrm{a}|>0\) \(\Rightarrow \mathrm{aRb}=\mathrm{bRa}\) \(\text { Since, } |\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}| \text { Then, }\) Since, \(\quad|\mathrm{a}-\mathrm{b}|=|\mathrm{b}-\mathrm{a}|\) Then, \(\quad \mathrm{R}\) is symmetric. (c) Transitive : - \(|\mathrm{a}-\mathrm{b}|>0,|\mathrm{~b}-\mathrm{c}|>0 \Rightarrow|\mathrm{a}-\mathrm{c}|>0\)Therefore, \((a, c) \in R \quad\) Then, \(\mathrm{R}\) is transitive. So, the relation is symmetric and transitive.
VITEEE-2008
Sets, Relation and Function
116794
The relation \(R\) defined on as \(R=\{(1,1),(2,2),(3,3),(1,3)\}\) is
1 equivalence
2 not symmetric
3 not reflexive
4 not transitive
Explanation:
B Given, \(\mathrm{A}=\{1,2,3\}, \quad \mathrm{R}=\{(1,1),(2,2),(3,3),(1,3)\}\) Then, check relations - (a) Reflexive : - \((1,1) \in R \Rightarrow{ }_1 R_1\) Then, \(\mathrm{R}\) is reflexive. (b) Symmetric : - \((1,3) \in \mathrm{R} \Rightarrow(3,1) \notin \mathrm{R}\) \({ }_1 \mathrm{R}_3 \nRightarrow{ }_3 \mathrm{R}_1\) Then, \(\mathrm{R}\) is not symmetric. (c) Transitive :- \((1,1) \in \mathrm{R},(3,3) \in \mathrm{R} \Rightarrow(1,3) \in \mathrm{R}\) \({ }_1 \mathrm{R}_1,{ }_3 \mathrm{R}_3 \Rightarrow{ }_1 \mathrm{R}_3\) Then, \(R\) is transitive. So, the relation \(\mathrm{R}\) is not symmetric but reflexive and transitive.
