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Given μ = x...

Question

Given $μ = {x / x is a integer, 0 \leq x \leq 10}$
$A = {x / the remainder is 1 when x is divided by 2}$
$B = {x / the remainder is 2 when x is divided by 3}$
Find the set $(A \cup B)^{'}$

(A \cup B)^{'} = {0, 1, 2, 4, 6, 10}

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(A \cup B)^{'} = {0, 1, 2, 3, 4, 6, 10}

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(A \cup B)^{'} = {0, 1, 2, 4, 5, 6, 10}

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(A \cup B)^{'} = {0, 1, 2, 4, 6, 9, 10}

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Solution

The correct option is A $(A \cup B)^{'} = {0, 1, 2, 4, 6, 10}$
$A -$ no' divided by 'z' leaves remainder 1
& $0 \leq x \leq 10$
$∴ A = {3, 5, 7, 9}$
$B -$ no' divided by '3' leave remain 2
$∴ B = {5, 8}$
$\Rightarrow (A \cup B) = {3, 5, 7, 8, 9}$ & $∴ (A \cup B) ∴ {0, 1, 2, 4, 6, 10}$

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Similar questions

A = ⎡ ⎢ ⎣ \begin{matrix} 1 & - 1 & 1 2 & 1 & - 3 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

(10) \times B = ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 2 - 5 & 0 & α 1 & - 2 & 3 \end{matrix} ⎤ ⎥ ⎦

B

A

α

A = {1, 3, 5}, B = {2, 4, 6}

C = {0, 2, 4, 6, 8}

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{0, 1, 2, 3, 4, 5, 6}

ϕ

{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

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

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)^{'}

{(x + \frac{1}{x} + 1)}^{6} = a_{0} + (a_{1} x + \frac{b_{1}}{x}) + (a_{2} x^{2} + \frac{b_{2}}{x^{2}}) + . . . + (a_{6} x^{6} + \frac{b_{6}}{x^{6}})

a_{0}

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)^{'}

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