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 $
Show Answer
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} $
