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The Boolean expression $(X+Y)(X+\overline{Y}) + \overline{(X\;\overline{Y}) + \overline{X}}$ simplifies to

  1. $X$
  2. $Y$
  3. $XY$
  4. $X+Y$
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(x+y)(x+!y)+!((x!y)+!x)

=x+xy+x!y+xy

=x
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Given boolean expression: $(X+Y)(X+\bar Y)+\overline{(X\bar Y)+\bar X}$

$\implies$ $(X.X+X\bar Y+X.Y+Y.\bar Y)+(\overline{(X\bar Y)}*\overline{\bar X})$

$\left [ \because \textrm{Demorgon’s law= $\overline{A+B}=\bar A+\bar B$, Idempotent law= A*A=A}\right]$

$\implies$ $(X+X\bar Y+XY)+(\bar X+\overline{\bar Y})*X)$

taking $X$ as common:

$\implies$ $X(1+\bar Y+Y)+(\bar X+Y)*X$

 $\left [ \because A+\bar A=1\textrm{( Complement law)}\right ]$

$\implies$ $X+X*\bar X+X*Y$

$\implies$ $X+XY$

$\implies$ $X(1+Y)$

$\implies$ $X$

Option A is correct.

Ref: Properties of Boolean Algebra

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