MCQ
For any two sets A and $\text{B, A - B}\cup\text{B}=\text{A}=$
- A$\text{(A - B)}\cup\text{A}$
- B$\text{(B - A)}\cup\text{B}$
- C$\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}$
- D$\text{(A}\cup\text{B)}\cap\text{(A}\cap\text{B)}.$
Solution:
$\text{(A}-\text{B)}\cup\text{(B}-\text{A)}=\text{(A}\cap\text{B}')\cup\text{(B}\cap\text{A}')$
$=[\text{A}\cup\text{(B}\cup\text{A}')]\cap[\text{B}'\cup\text{(B}\cap\text{A}')]$ [Using distribution law]
$=[\text{(A}\cup\text{B})\cap\text{(A}\cup\text{A}')]\cap[\text{(B}'\cup\text{B})\cap\text{(B}'\cup\text{A}')]$ [Using distribution law]
$=[\text{(A}\cup\text{B)}\cup\text{(U)}]\cap[\text{(U)}\cap\text{(B}'\cup\text{A}')]$ $[\text{A}\cup\text{A'= U = B}'\cup\text{B}]$
$=[\text{A}\cup\text{B}]\cap[\text{B}'\cup\text{A}']$ $\begin{bmatrix}\text{(A}\cup\text{B)}\cap\text{(U)}=\text{(A}\cup\text{B)}\\\text{ and (U)}\cap\text{(B}'\cup\text{A)}'=\text{(B}'\cup\text{A}')]\end{bmatrix}$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cap\text{B)}']$ $[\text{(A}\cap\text{B)}'=\text{B}'\cup\text{A}']$
$=[\text{A}\cup\text{B}]\cap[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}]$
$=[\text{(A}\cup\text{B)}-\text{(A}\cap\text{B)}].$
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