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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. $x' + y$

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

edited
0
how x' + xy = x' + y ????
0
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

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

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

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

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

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

use K map
0
Use distribution rule  ...
= X'Y' + XY +X'Y

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

= X' + Y

hence option d is correct
+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

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