116738
If \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\), then \(A \cap B\) consists of all multiples of
1 16
2 12
3 8
4 4
Explanation:
B Given, \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\) Then, \(A=\{4,8,12,16,20,24,28,32,36 \ldots \ldots\). and \(B=\{6,12,18,24,30,36 \ldots \ldots\}\) So, \(\mathrm{A} \cap \mathrm{B}=\{12,24,36 \ldots .\). i.e. \(A \cap B=\{x: x\) is a multiple of 12\(\}\)
UPSEE-2014
Sets, Relation and Function
116740
If \(A=\left\{(x, y): x^2+y^2 \leq 1 ; x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4 ; x, y \in R\right\}\), then
1 \(\mathrm{A}-\mathrm{B}=\phi\)
2 \(\mathrm{B}-\mathrm{A}=\phi\)
3 \(\mathrm{A} \cap \mathrm{B} \neq \phi\)
4 \(\mathrm{A} \cap \mathrm{B}=\phi\)
Explanation:
D Given, \(A=\left\{(x, y): x^2+y^2 \leq 1, x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4, x, y \in R\right\}\) From equation, draw the graph of following above set A, B - From figure we see that set \(\mathrm{A}\) inside of the circle and set \(B\) outside of the circle. So, \(\mathrm{A} \cap \mathrm{B}=\phi\)
UPSEE-2013
Sets, Relation and Function
116741
The set \(A=\{x:|2 x+3|\lt 7\}\) is equal to the set
A Given, Set \(A=\{x:|2 x+3|\lt 7\}\) Then, \(A=\{x:-7\lt 2 x+3\lt 7\}\) \(A=\{x:-7-3\lt 2 x\lt 7-3\}\) \(A=\{x:-10\lt 2 x\lt 4\}\) \(A=\{x:-5\lt x\lt 2\}\) \(A=\{x:-5+5\lt x+5\lt 2+5\}\) \(A=\{x: 0\lt x+5\lt 7\}\)So, the set \(\mathrm{A}\) is equal to set \(\mathrm{D}\)
Karnataka CET 2014
Sets, Relation and Function
116743
Set \(A\) and \(B\) have 3 and 6 elements respectively. What can be the minimum number of elements in \(A \cup B\) ?
1 18
2 9
3 6
4 3
Explanation:
C From question, No. of element in set \(A=3\) No. of element in set \(B=6\) Maximum no. of element can be in set \(A \cap B=3\) Minimum no. of element can be in set \(A \cup B\) is, \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=3+6-3\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=6\)
116738
If \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\), then \(A \cap B\) consists of all multiples of
1 16
2 12
3 8
4 4
Explanation:
B Given, \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\) Then, \(A=\{4,8,12,16,20,24,28,32,36 \ldots \ldots\). and \(B=\{6,12,18,24,30,36 \ldots \ldots\}\) So, \(\mathrm{A} \cap \mathrm{B}=\{12,24,36 \ldots .\). i.e. \(A \cap B=\{x: x\) is a multiple of 12\(\}\)
UPSEE-2014
Sets, Relation and Function
116740
If \(A=\left\{(x, y): x^2+y^2 \leq 1 ; x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4 ; x, y \in R\right\}\), then
1 \(\mathrm{A}-\mathrm{B}=\phi\)
2 \(\mathrm{B}-\mathrm{A}=\phi\)
3 \(\mathrm{A} \cap \mathrm{B} \neq \phi\)
4 \(\mathrm{A} \cap \mathrm{B}=\phi\)
Explanation:
D Given, \(A=\left\{(x, y): x^2+y^2 \leq 1, x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4, x, y \in R\right\}\) From equation, draw the graph of following above set A, B - From figure we see that set \(\mathrm{A}\) inside of the circle and set \(B\) outside of the circle. So, \(\mathrm{A} \cap \mathrm{B}=\phi\)
UPSEE-2013
Sets, Relation and Function
116741
The set \(A=\{x:|2 x+3|\lt 7\}\) is equal to the set
A Given, Set \(A=\{x:|2 x+3|\lt 7\}\) Then, \(A=\{x:-7\lt 2 x+3\lt 7\}\) \(A=\{x:-7-3\lt 2 x\lt 7-3\}\) \(A=\{x:-10\lt 2 x\lt 4\}\) \(A=\{x:-5\lt x\lt 2\}\) \(A=\{x:-5+5\lt x+5\lt 2+5\}\) \(A=\{x: 0\lt x+5\lt 7\}\)So, the set \(\mathrm{A}\) is equal to set \(\mathrm{D}\)
Karnataka CET 2014
Sets, Relation and Function
116743
Set \(A\) and \(B\) have 3 and 6 elements respectively. What can be the minimum number of elements in \(A \cup B\) ?
1 18
2 9
3 6
4 3
Explanation:
C From question, No. of element in set \(A=3\) No. of element in set \(B=6\) Maximum no. of element can be in set \(A \cap B=3\) Minimum no. of element can be in set \(A \cup B\) is, \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=3+6-3\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=6\)
116738
If \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\), then \(A \cap B\) consists of all multiples of
1 16
2 12
3 8
4 4
Explanation:
B Given, \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\) Then, \(A=\{4,8,12,16,20,24,28,32,36 \ldots \ldots\). and \(B=\{6,12,18,24,30,36 \ldots \ldots\}\) So, \(\mathrm{A} \cap \mathrm{B}=\{12,24,36 \ldots .\). i.e. \(A \cap B=\{x: x\) is a multiple of 12\(\}\)
UPSEE-2014
Sets, Relation and Function
116740
If \(A=\left\{(x, y): x^2+y^2 \leq 1 ; x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4 ; x, y \in R\right\}\), then
1 \(\mathrm{A}-\mathrm{B}=\phi\)
2 \(\mathrm{B}-\mathrm{A}=\phi\)
3 \(\mathrm{A} \cap \mathrm{B} \neq \phi\)
4 \(\mathrm{A} \cap \mathrm{B}=\phi\)
Explanation:
D Given, \(A=\left\{(x, y): x^2+y^2 \leq 1, x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4, x, y \in R\right\}\) From equation, draw the graph of following above set A, B - From figure we see that set \(\mathrm{A}\) inside of the circle and set \(B\) outside of the circle. So, \(\mathrm{A} \cap \mathrm{B}=\phi\)
UPSEE-2013
Sets, Relation and Function
116741
The set \(A=\{x:|2 x+3|\lt 7\}\) is equal to the set
A Given, Set \(A=\{x:|2 x+3|\lt 7\}\) Then, \(A=\{x:-7\lt 2 x+3\lt 7\}\) \(A=\{x:-7-3\lt 2 x\lt 7-3\}\) \(A=\{x:-10\lt 2 x\lt 4\}\) \(A=\{x:-5\lt x\lt 2\}\) \(A=\{x:-5+5\lt x+5\lt 2+5\}\) \(A=\{x: 0\lt x+5\lt 7\}\)So, the set \(\mathrm{A}\) is equal to set \(\mathrm{D}\)
Karnataka CET 2014
Sets, Relation and Function
116743
Set \(A\) and \(B\) have 3 and 6 elements respectively. What can be the minimum number of elements in \(A \cup B\) ?
1 18
2 9
3 6
4 3
Explanation:
C From question, No. of element in set \(A=3\) No. of element in set \(B=6\) Maximum no. of element can be in set \(A \cap B=3\) Minimum no. of element can be in set \(A \cup B\) is, \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=3+6-3\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=6\)
116738
If \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\), then \(A \cap B\) consists of all multiples of
1 16
2 12
3 8
4 4
Explanation:
B Given, \(A=\{x: x\) is a multiple of 4\(\}\) and \(B=\{x: x\) is a multiple of 6\(\}\) Then, \(A=\{4,8,12,16,20,24,28,32,36 \ldots \ldots\). and \(B=\{6,12,18,24,30,36 \ldots \ldots\}\) So, \(\mathrm{A} \cap \mathrm{B}=\{12,24,36 \ldots .\). i.e. \(A \cap B=\{x: x\) is a multiple of 12\(\}\)
UPSEE-2014
Sets, Relation and Function
116740
If \(A=\left\{(x, y): x^2+y^2 \leq 1 ; x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4 ; x, y \in R\right\}\), then
1 \(\mathrm{A}-\mathrm{B}=\phi\)
2 \(\mathrm{B}-\mathrm{A}=\phi\)
3 \(\mathrm{A} \cap \mathrm{B} \neq \phi\)
4 \(\mathrm{A} \cap \mathrm{B}=\phi\)
Explanation:
D Given, \(A=\left\{(x, y): x^2+y^2 \leq 1, x, y \in R\right\}\) and \(B=\left\{(x, y): x^2+y^2 \geq 4, x, y \in R\right\}\) From equation, draw the graph of following above set A, B - From figure we see that set \(\mathrm{A}\) inside of the circle and set \(B\) outside of the circle. So, \(\mathrm{A} \cap \mathrm{B}=\phi\)
UPSEE-2013
Sets, Relation and Function
116741
The set \(A=\{x:|2 x+3|\lt 7\}\) is equal to the set
A Given, Set \(A=\{x:|2 x+3|\lt 7\}\) Then, \(A=\{x:-7\lt 2 x+3\lt 7\}\) \(A=\{x:-7-3\lt 2 x\lt 7-3\}\) \(A=\{x:-10\lt 2 x\lt 4\}\) \(A=\{x:-5\lt x\lt 2\}\) \(A=\{x:-5+5\lt x+5\lt 2+5\}\) \(A=\{x: 0\lt x+5\lt 7\}\)So, the set \(\mathrm{A}\) is equal to set \(\mathrm{D}\)
Karnataka CET 2014
Sets, Relation and Function
116743
Set \(A\) and \(B\) have 3 and 6 elements respectively. What can be the minimum number of elements in \(A \cup B\) ?
1 18
2 9
3 6
4 3
Explanation:
C From question, No. of element in set \(A=3\) No. of element in set \(B=6\) Maximum no. of element can be in set \(A \cap B=3\) Minimum no. of element can be in set \(A \cup B\) is, \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=3+6-3\) \(\mathrm{n}(\mathrm{A} \cup \mathrm{B})=6\)
