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A Boolean function $x’y’ + xy + x’y$ is equivalent to

  1. $x' + y'$
  2. $x + y$
  3. $x + y'$
  4. $x' + y$
in Digital Logic by Veteran (52.2k points) | 1.5k views

5 Answers

+19 votes
Best answer

Answer is option D.

$x'y' + x'y = x'(y+y') = x'$
$x' + xy = x' + y$

by Veteran (425k points)
edited by
0
how x' + xy = x' + y ????
+1
x' + xy = x'(y+y') + xy

           =x'y + x'y' + xy

           =x'y + x'y + x'y' + xy                   (because AB+AB+AB+.......................=AB)

           =(x'y + x'y') + (x'y + xy)

           =x'(y+y') + y(x+x')

           =x'+y
0

@lash12

how x' + xy = x' + y ????

     $x' + x.y$   [ In this case $+$ is distributed over $.$  (In boolean algebra this is also possible.]

       $x' + x.y=(x'+x).(x'+y)$

       $x' + x.y=(1).(x'+y)$         $ [\bar{A}+A=1]$

       $x' + x.y=(x'+y)= x'+y$ 

+11 votes
answer - D

use K map
by Loyal (8.6k points)
0
Use distribution rule  ...
+3 votes
= X'Y' + XY +X'Y

= X' ( Y + Y' ) + Y ( X + X')

= X' + Y

hence option d is correct
by Active (4.5k points)
+1 vote

Using K map for two variables

  y' y
x' 1 1
x 0 1

Tables for expression given looks like this from this we get:

x'+y 

 

by (93 points)
+1 vote

x'y'+xy+x'y

=x'(y'+y)+xy

=x'.1 +xy

=x'+xy

=(x'+x).(x'+y)

=1.(x'+y)

=x'+y

ans-d

by Junior (555 points)
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