119 views

Above question modification.

This is the actual question

Boolean expression$:(x+y)(x+\bar{y})+\overline{{(x\bar{y}+\bar{x})}}$

$(A) x$               $(B)y$           $(C)xy$               $(D)x+y$

retagged | 119 views
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0 ??
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Even I am getting 0
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I'm also getting $0$

How $x$ is possible?

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I'm getting 0 too. I dunno why answer is given x
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can you provide me shift name of GATE 2014 ECE?

Your question is wrong. I add the question, which is actual question and answer is $x$

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oh! probably during copy pasting the sign went missing
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$Alwyas$,you can type the question, it will beneficial for others also.

thanks
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yes

(x+y)(x+y') = x+xy+xy'=x(1+y+y')=x.1=x

{(x+y')+x'}'= {x+x'+y'}'={1+y'}'=1'=0

(x+y)(x+y') . {(x+y')+x'}' = x.0=0

by Boss (23.7k points)
(x+y)(x+y')+(xy).............after simplification of complement part

now just simplify

x+xy'+xy+0+xy=x ans.
by Active (2.8k points)

(x+y)(x+y')+(xy'+x')'

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

x+xy'+xy+xy     // x+xy'=x

x + xy    // x+xy = x

x

by Junior (959 points)

+1 vote