question_answer

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

A) \[\bar{A}B\]                                          

B) \[A\bar{B}\]

C) \[AB+\overline{AB}\]                

D)   \[A\overline{B}+\overline{A}B\]

Correct Answer: B

Solution :

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