JEE Main 12 Jan 2019 Morning Question 23

Question: The output of the given logic circuit is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

Options:

A) $ \bar{A}B $

B) $ A\bar{B} $

C) $ AB+\overline{AB} $

D) $ A\overline{B}+\overline{A}B $

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Answer:

Correct Answer: B

Solution:

  • $ Y=\overline{\overline{A.(\overline{AB})}.(\overline{AB}+B)} $ (by applying De Morgan theorem) $ Y=A.(\overline{A}+\overline{B})+(\overline{\overline{AB}}+\overline{B}) $ $ Y=A.(\overline{A}+\overline{B})+(AB.\overline{B}) $ $ (\because B.\overline{B}=0) $ $ Y=A.\overline{A}+A\overline{B},Y=A\overline{B} $