116788
If \(A=\{x, y, z\}, B=\{1,2\}\), then the total number of relations from set \(A\) to set \(B\) are
1 16
2 32
3 8
4 64
Explanation:
D Given, \(\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}, \mathrm{B}=\{1,2\}\) Then, \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{x}, 1),(\mathrm{y}, 1),(\mathrm{z}, 1)(\mathrm{x}, 2)(\mathrm{y}, 2),(\mathrm{z}, 2)\}\) Then number of element \(-n(A \times B)=6\) So, the total number of relations from set \(A\) to set \(B\) are \(=2^{\mathrm{n}}=2^6=64\).
MHT-CET 2020
Sets, Relation and Function
116793
If \(A\) and \(B\) are two equivalence relations defined on set \(C\), then
1 \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
2 \(\mathrm{A} \cap \mathrm{B}\) is not an equivalence relation
3 \(A \cup B\) is an equivalence relation
4 \(\mathrm{A} \cup \mathrm{B}\) is not an equivalence relation
Explanation:
A Given, A and B are two equivalence relations defined on set \(C\). So, then \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
UPSEE-2011
Sets, Relation and Function
116801
Let \(R\) be a relation on \(N \times N\) defined by \((a, b) R\) (c, \(d)\) if and only if ad \((b-c)=b c(a-d)\). Then \(R\) is
1 transitive but neither reflexive nor symmetric
2 symmetric but neither reflexive nor transitive
3 symmetric and transitive but not reflexive
4 reflexive and symmetric but not transitive
Explanation:
B Let R be relation defined by (a, b) R (c, d) \(\Leftrightarrow\) \(\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) For reflexive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{ab}(\mathrm{b}-\mathrm{a})=\mathrm{ba}(\mathrm{a}-\mathrm{b})\) \(\therefore\) It is not reflexive. For symmetric \(\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}\) \((\mathrm{a}-\mathrm{d})\) and \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})=\mathrm{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b})\)It is true Which is symmetric. For transitive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})=\mathrm{cf}(\mathrm{d}-\mathrm{e})=\mathrm{de}(\mathrm{c}-\mathrm{f})\) So, \(\operatorname{adcf}(b-c)(d-e)=b c d e(c-d)(c-f)\) \(\text { af }(b-c)(d-e)=\text { be }(a-d)(c-f)\) It is not transitive.
Shift-I
Sets, Relation and Function
116806
Let \(A=\left\{(x, y): y=e^{-x}\right\}\) and \(B=\{(x, y): y=-\) \(x\}\) Then the correct statement is :
1 \(\mathrm{A} \cap \mathrm{B}=\phi\)
2 \(\mathrm{A} \subset \mathrm{B}\)
3 \(\mathrm{B} \subset \mathrm{A}\)
4 \(\mathrm{A} \cap \mathrm{B}=\{(0,1),(0,0)\}\)
Explanation:
A We have , \(A=\left\{(x, y): y=e^{-x}\right\}, B=\{(x, y): y=-x\}\) Now, \(\quad A=(x, y)=\left(x, e^{-x}\right), B=(x, y)=(x,-x)\) Since image of \(x\) in \(A\) cannot be equal to image of \(x\) in B i.e. \(\mathrm{e}^{-\mathrm{x}} \neq-\mathrm{x}\) \(\mathrm{A} \cap \mathrm{B}=\phi\)
116788
If \(A=\{x, y, z\}, B=\{1,2\}\), then the total number of relations from set \(A\) to set \(B\) are
1 16
2 32
3 8
4 64
Explanation:
D Given, \(\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}, \mathrm{B}=\{1,2\}\) Then, \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{x}, 1),(\mathrm{y}, 1),(\mathrm{z}, 1)(\mathrm{x}, 2)(\mathrm{y}, 2),(\mathrm{z}, 2)\}\) Then number of element \(-n(A \times B)=6\) So, the total number of relations from set \(A\) to set \(B\) are \(=2^{\mathrm{n}}=2^6=64\).
MHT-CET 2020
Sets, Relation and Function
116793
If \(A\) and \(B\) are two equivalence relations defined on set \(C\), then
1 \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
2 \(\mathrm{A} \cap \mathrm{B}\) is not an equivalence relation
3 \(A \cup B\) is an equivalence relation
4 \(\mathrm{A} \cup \mathrm{B}\) is not an equivalence relation
Explanation:
A Given, A and B are two equivalence relations defined on set \(C\). So, then \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
UPSEE-2011
Sets, Relation and Function
116801
Let \(R\) be a relation on \(N \times N\) defined by \((a, b) R\) (c, \(d)\) if and only if ad \((b-c)=b c(a-d)\). Then \(R\) is
1 transitive but neither reflexive nor symmetric
2 symmetric but neither reflexive nor transitive
3 symmetric and transitive but not reflexive
4 reflexive and symmetric but not transitive
Explanation:
B Let R be relation defined by (a, b) R (c, d) \(\Leftrightarrow\) \(\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) For reflexive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{ab}(\mathrm{b}-\mathrm{a})=\mathrm{ba}(\mathrm{a}-\mathrm{b})\) \(\therefore\) It is not reflexive. For symmetric \(\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}\) \((\mathrm{a}-\mathrm{d})\) and \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})=\mathrm{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b})\)It is true Which is symmetric. For transitive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})=\mathrm{cf}(\mathrm{d}-\mathrm{e})=\mathrm{de}(\mathrm{c}-\mathrm{f})\) So, \(\operatorname{adcf}(b-c)(d-e)=b c d e(c-d)(c-f)\) \(\text { af }(b-c)(d-e)=\text { be }(a-d)(c-f)\) It is not transitive.
Shift-I
Sets, Relation and Function
116806
Let \(A=\left\{(x, y): y=e^{-x}\right\}\) and \(B=\{(x, y): y=-\) \(x\}\) Then the correct statement is :
1 \(\mathrm{A} \cap \mathrm{B}=\phi\)
2 \(\mathrm{A} \subset \mathrm{B}\)
3 \(\mathrm{B} \subset \mathrm{A}\)
4 \(\mathrm{A} \cap \mathrm{B}=\{(0,1),(0,0)\}\)
Explanation:
A We have , \(A=\left\{(x, y): y=e^{-x}\right\}, B=\{(x, y): y=-x\}\) Now, \(\quad A=(x, y)=\left(x, e^{-x}\right), B=(x, y)=(x,-x)\) Since image of \(x\) in \(A\) cannot be equal to image of \(x\) in B i.e. \(\mathrm{e}^{-\mathrm{x}} \neq-\mathrm{x}\) \(\mathrm{A} \cap \mathrm{B}=\phi\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116788
If \(A=\{x, y, z\}, B=\{1,2\}\), then the total number of relations from set \(A\) to set \(B\) are
1 16
2 32
3 8
4 64
Explanation:
D Given, \(\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}, \mathrm{B}=\{1,2\}\) Then, \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{x}, 1),(\mathrm{y}, 1),(\mathrm{z}, 1)(\mathrm{x}, 2)(\mathrm{y}, 2),(\mathrm{z}, 2)\}\) Then number of element \(-n(A \times B)=6\) So, the total number of relations from set \(A\) to set \(B\) are \(=2^{\mathrm{n}}=2^6=64\).
MHT-CET 2020
Sets, Relation and Function
116793
If \(A\) and \(B\) are two equivalence relations defined on set \(C\), then
1 \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
2 \(\mathrm{A} \cap \mathrm{B}\) is not an equivalence relation
3 \(A \cup B\) is an equivalence relation
4 \(\mathrm{A} \cup \mathrm{B}\) is not an equivalence relation
Explanation:
A Given, A and B are two equivalence relations defined on set \(C\). So, then \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
UPSEE-2011
Sets, Relation and Function
116801
Let \(R\) be a relation on \(N \times N\) defined by \((a, b) R\) (c, \(d)\) if and only if ad \((b-c)=b c(a-d)\). Then \(R\) is
1 transitive but neither reflexive nor symmetric
2 symmetric but neither reflexive nor transitive
3 symmetric and transitive but not reflexive
4 reflexive and symmetric but not transitive
Explanation:
B Let R be relation defined by (a, b) R (c, d) \(\Leftrightarrow\) \(\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) For reflexive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{ab}(\mathrm{b}-\mathrm{a})=\mathrm{ba}(\mathrm{a}-\mathrm{b})\) \(\therefore\) It is not reflexive. For symmetric \(\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}\) \((\mathrm{a}-\mathrm{d})\) and \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})=\mathrm{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b})\)It is true Which is symmetric. For transitive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})=\mathrm{cf}(\mathrm{d}-\mathrm{e})=\mathrm{de}(\mathrm{c}-\mathrm{f})\) So, \(\operatorname{adcf}(b-c)(d-e)=b c d e(c-d)(c-f)\) \(\text { af }(b-c)(d-e)=\text { be }(a-d)(c-f)\) It is not transitive.
Shift-I
Sets, Relation and Function
116806
Let \(A=\left\{(x, y): y=e^{-x}\right\}\) and \(B=\{(x, y): y=-\) \(x\}\) Then the correct statement is :
1 \(\mathrm{A} \cap \mathrm{B}=\phi\)
2 \(\mathrm{A} \subset \mathrm{B}\)
3 \(\mathrm{B} \subset \mathrm{A}\)
4 \(\mathrm{A} \cap \mathrm{B}=\{(0,1),(0,0)\}\)
Explanation:
A We have , \(A=\left\{(x, y): y=e^{-x}\right\}, B=\{(x, y): y=-x\}\) Now, \(\quad A=(x, y)=\left(x, e^{-x}\right), B=(x, y)=(x,-x)\) Since image of \(x\) in \(A\) cannot be equal to image of \(x\) in B i.e. \(\mathrm{e}^{-\mathrm{x}} \neq-\mathrm{x}\) \(\mathrm{A} \cap \mathrm{B}=\phi\)
116788
If \(A=\{x, y, z\}, B=\{1,2\}\), then the total number of relations from set \(A\) to set \(B\) are
1 16
2 32
3 8
4 64
Explanation:
D Given, \(\mathrm{A}=\{\mathrm{x}, \mathrm{y}, \mathrm{z}\}, \mathrm{B}=\{1,2\}\) Then, \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{x}, 1),(\mathrm{y}, 1),(\mathrm{z}, 1)(\mathrm{x}, 2)(\mathrm{y}, 2),(\mathrm{z}, 2)\}\) Then number of element \(-n(A \times B)=6\) So, the total number of relations from set \(A\) to set \(B\) are \(=2^{\mathrm{n}}=2^6=64\).
MHT-CET 2020
Sets, Relation and Function
116793
If \(A\) and \(B\) are two equivalence relations defined on set \(C\), then
1 \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
2 \(\mathrm{A} \cap \mathrm{B}\) is not an equivalence relation
3 \(A \cup B\) is an equivalence relation
4 \(\mathrm{A} \cup \mathrm{B}\) is not an equivalence relation
Explanation:
A Given, A and B are two equivalence relations defined on set \(C\). So, then \(\mathrm{A} \cap \mathrm{B}\) is an equivalence relations
UPSEE-2011
Sets, Relation and Function
116801
Let \(R\) be a relation on \(N \times N\) defined by \((a, b) R\) (c, \(d)\) if and only if ad \((b-c)=b c(a-d)\). Then \(R\) is
1 transitive but neither reflexive nor symmetric
2 symmetric but neither reflexive nor transitive
3 symmetric and transitive but not reflexive
4 reflexive and symmetric but not transitive
Explanation:
B Let R be relation defined by (a, b) R (c, d) \(\Leftrightarrow\) \(\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) For reflexive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{a}, \mathrm{b}) \Rightarrow \mathrm{ab}(\mathrm{b}-\mathrm{a})=\mathrm{ba}(\mathrm{a}-\mathrm{b})\) \(\therefore\) It is not reflexive. For symmetric \(\Rightarrow(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}\) \((\mathrm{a}-\mathrm{d})\) and \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{a}, \mathrm{b})=\mathrm{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b})\)It is true Which is symmetric. For transitive - \((\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})=\mathrm{ad}(\mathrm{b}-\mathrm{c})=\mathrm{bc}(\mathrm{a}-\mathrm{d})\) \((\mathrm{c}, \mathrm{d}) \mathrm{R}(\mathrm{e}, \mathrm{f})=\mathrm{cf}(\mathrm{d}-\mathrm{e})=\mathrm{de}(\mathrm{c}-\mathrm{f})\) So, \(\operatorname{adcf}(b-c)(d-e)=b c d e(c-d)(c-f)\) \(\text { af }(b-c)(d-e)=\text { be }(a-d)(c-f)\) It is not transitive.
Shift-I
Sets, Relation and Function
116806
Let \(A=\left\{(x, y): y=e^{-x}\right\}\) and \(B=\{(x, y): y=-\) \(x\}\) Then the correct statement is :
1 \(\mathrm{A} \cap \mathrm{B}=\phi\)
2 \(\mathrm{A} \subset \mathrm{B}\)
3 \(\mathrm{B} \subset \mathrm{A}\)
4 \(\mathrm{A} \cap \mathrm{B}=\{(0,1),(0,0)\}\)
Explanation:
A We have , \(A=\left\{(x, y): y=e^{-x}\right\}, B=\{(x, y): y=-x\}\) Now, \(\quad A=(x, y)=\left(x, e^{-x}\right), B=(x, y)=(x,-x)\) Since image of \(x\) in \(A\) cannot be equal to image of \(x\) in B i.e. \(\mathrm{e}^{-\mathrm{x}} \neq-\mathrm{x}\) \(\mathrm{A} \cap \mathrm{B}=\phi\)
