Virtual Gate Test Series: Digital Logic - Multiplexer

Question

Answer 1

Input for $4\times1$ multiplexer is:

 

$I_0=\bar C\bar D+C\bar D=\bar D(C+\bar C)=\bar D$

$I_1=I_2=0$

$I_3=\bar C D+C D= D(C+\bar C)= D$

$\therefore F=\bar A\bar BI_0+\bar A BI_1+ A\bar BI_2+ A BI_3$

$F=\bar A\bar B\bar D+\bar A B.0+ A\bar B.0+ A BD$

$F=\bar A\bar B\bar D+ A BD$

Option $(D)$ is correct.

Answer 2

C'D'+CD'=D'

C'D'+CD=D

A'B'D'+ABD this is answer

So d. None of these