If $A$ = {1, 3, 5, 7, 9, 11, 13, 15, 17}, $B$ = {2, 4, 6, 8, 10, 12, 14, 16, 18} and $N$ the set of natural numbers is the universal set, then $A^{'} \cup ((A \cup B) \cap B^{'})$ is:

ϕ

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N

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A

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B

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Solution

The correct option is B $N$
$A \cup B$ = {1, 2, 3, 4, 5……....17, 18}
$B'$ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21...}
$(A \cup B) \cap B'$ = (1, 3, 5, 7, 9, ...,17) = $A$
$A' \cup A$ = Universal Set

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If A = {1, 3, 5, ..... ,17}, B ={2, 4, 6, ..... ,18} and N, the set of natural numbers is the universal set, then $(A^{'} \cup [(A \cup B) \cap B^{'}])$ is

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If A = {3, 6, 9, 12, 15, 18, 21}
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C = {5, 10, 15, 20}

$\begin{matrix} A & B 1. B - C & A . {5, 10, 20} 2. A - C & B . {3, 9, 15, 21} 3. C - (A - B) & C . {3, 6, 9, 12, 18, 21} 4. C \cap (A - B) & D . {15} E . {2, 4, 6, 8, 12, 14, 16} \end{matrix}$

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If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, B = {2, 4, 6, 8, 10, 12, 14, 16, 18} and N the set of natural numbers is the universal set, then A′∪((A∪B)∩B′) is:

The correct option is B NA∪B = {1, 2, 3, 4, 5……....17, 18} B' = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21...} (A∪B)∩B' = (1, 3, 5, 7, 9, ...,17) = A A'∪A = Universal Set

If $A$ = {1, 3, 5, 7, 9, 11, 13, 15, 17}, $B$ = {2, 4, 6, 8, 10, 12, 14, 16, 18} and $N$ the set of natural numbers is the universal set, then $A^{'} \cup ((A \cup B) \cap B^{'})$ is:

The correct option is B $N$
$A \cup B$ = {1, 2, 3, 4, 5……....17, 18}
$B'$ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21...}
$(A \cup B) \cap B'$ = (1, 3, 5, 7, 9, ...,17) = $A$
$A' \cup A$ = Universal Set