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JEE PHYSICS MCQ: SEMI CONDUCTOR DEVICES

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Analyzing +4 | -1
Aā‹…Bˉ+Aˉ⋅B‾\overline{A \cdot \bar{B}+\bar{A} \cdot B} is equivalent to
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Correct option: B
(A+Bˉ)+(Aˉ+B)(A+\bar{B})+(\bar{A}+B)
Option D ((A+B)ā‹…(Aˉ+Bˉ)(A+B) \cdot(\bar{A}+\bar{B})) is correct because it represents the simplified form of the given Boolean expression after applying De Morgan's law and simplification.

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