Universal set,
$U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}$
$A = {x | x^{2} - 5 x + 6 = 0}$
$B = {x | x^{2} - 3 x + 2 = 0}$
what is $(A \cap B)^{'}$ equal to

{1, 3}

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{1, 2, 3}

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{0, 1, 3}

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{0, 1, 2, 3}

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Solution

The correct option is C ${0, 1, 3}$
Given $U = {x | x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}$

$x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0$
$x^{2} (x^{3} - 6 x^{2} + 11 x - 6) = 0$
$x^{2} (x^{2} - 5 x + 6) (x - 1) = 0$
$x^{2} (x - 3) (x - 2) (x - 1) = 0$
Hence $U = {0, 1, 2, 3}$

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

$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
Hence $A = {2, 3}$

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

$x^{2} - 3 x + 2 = 0$
$(x - 2) (x - 1) = 0$
$B = {1, 2}$

$A \cap B = {2}$
$(A \cap B)^{'} = {0, 1, 3}$

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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 is an integer root of the equation x^{5} - 6 x^{4} + 11 x^{3} - 6 x^{2} = 0}

B = {x | x is a real root of the equation a x^{5} + 2 a x^{3} + 2 b x^{2} + b = 0, a, b \in R such that the given equation have maximum number of real roots.}

P = {x / x^{2} + 14 x + 40 = 0}

Q = {x / x^{2} - 5 x + 6 = 0}

R = {x / x^{2} + 17 x - 60 = 0}

X = {- 20, - 10, - 4, 2, 3, 4}

P \cup Q

6 x^{2} - 13 x + 6 = 0

\sqrt{6} x^{2} - 5 x + \sqrt{6} = 0

24 x^{2} - 17 x + 3 = 0

x^{2} + 2 x + 4 = 0

x^{2} + x + 1 = 0

x^{2} - 3 \sqrt{3} x - 30 = 0

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