Universal set $U = {x : x^{5} - {6 x}^{4} + {11 x}^{3} - 6 x^{2} = 0}$ , $A = {x : x^{2} - 5 x + 6 = 0}$ and $B = {x : x^{2} - 3 x + 2 = 0}$ . Then $(A \cap B)^{'}$ is

1, 2, 3

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0, 1, 3

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2, 3

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0, 1, 2

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Solution

The correct option is B $0, 1, 3$
$U = {x : x^{5} - {6 x}^{4} + {11 x}^{3} - 6 x^{2} = 0} \Rightarrow x^{5} - {6 x}^{4} + {11 x}^{3} - 6 x^{2} = 0 \Rightarrow x^{2} (x^{3} - 6 x^{2} + 11 x - 6) = 0 \Rightarrow x^{2} (x - 1) (x - 2) (x - 3) = 0 \Rightarrow x = 0, 1, 2, 3 \Rightarrow U = {0, 1, 2, 3} A = {x : x^{2} - 5 x + 6 = 0} \Rightarrow x^{2} - 5 x + 6 = 0 \Rightarrow (x - 2) (x - 3) = 0 \Rightarrow A = {2, 3} B = {x : x^{2} - 3 x + 2 = 0} \Rightarrow x^{2} - 3 x + 2 = 0 \Rightarrow (x - 1) (x - 2) = 0 \Rightarrow B {1, 2} \Rightarrow (A \cap B) = {2} U = {0, 1, 2, 3} \Rightarrow (A \cap B)^{'} = {0, 1, 3}$
So option $B$ is correct.

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Similar questions

U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}

A = {x | x^{2} - 5 x + 6 = 0}

B = {x | x^{2} - 3 x + 2 = 0}

(A \cap B)^{'}

U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}

A = {x | x^{2} - 5 x + 6 = 0}

B = {x | x^{2} - 3 x + 2 = 0}

(A \cap B)^{'}

A = {x : x^{2} - 5 x + 6 = 0}, B = {2, 3}

C = {2, 3, 3}

A = B = C

A = {x : x^{2} - 3 x + 2 = 0}

B = {x : x^{2} + 4 x - 5 = 0}

A - B

The sets A = ${4, 5, 6}$ and B = ${x : x^{2} - 5 x - 6 = 0}$ are disjoint sets.

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