Let $A = {set of prime numbers less than 10}; B = {factors of 36} .$ Then $A \cup B$ includes

2, 3, 4, 6, 9, 12, 18

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1, 2, 3, 4, 5, 6, 7, 8, 9

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0, 1, 2, 3, 4, 5, 6, 7, 8, 9

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1, 2, 3, 4, 5, 6, 7, 9

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Solution

The correct option is D $1, 2, 3, 4, 5, 6, 7, 9$
Thus, we have
A = { 2, 3, 5, 7 } and B = { 1, 2, 3, 4, 6, 9 }
$\Rightarrow$ $A \cup B =$ {1, 2, 3, 4, 5, 6, 7, 9}.

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Similar questions

Let U = {1,2,3,4,5,6,7,8,9},

A = {2,4,6,8} and B = {2,3,5,7}.

Verify that :

(i) $(A \cup B)^{'} = A^{'} \cap B^{'}$

(ii) $(A \cap B)^{'} = A^{'} \cup B^{'}$ .

(A \cup B)' = A' \cap B'

(A \cap B)' = A' \cup B'

U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {0, 3, 4, 7}

B = {1, 2, 8, 9}

(A U B)^{'}

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8}

C = {3, 4, 5, 6} .

(i) A^{'} (i i) B^{'} (i i i) (A \cup C)^{'}

(i v) (A \cup B)^{'} (v) (A^{'})^{'} (v i) (B - C)^{'}

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find

(i) $A^{'}$ (ii) $B^{'}$ (iii) ${(A \cup C)}^{'}$

(iv) ${(A \cup B)}^{'}$ (v) $(A)^{'}$ (vi) ${(B - C)}^{'}$

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