Pls give it's solution now.

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You can also do it by elimination.

xy=0. Either x or y or both has to be 0. Option C ruled out.

x+y+z = 1. You cannot have a combination that has all 0s or two 1s for x,y,z. Option D ruled out.

xz+w = 1. B says z and w both are 0. Which is not possible. Option B ruled out.

The answer is A :)

Answer 1

$x+y$  $\Rightarrow$ $x \ OR \ y$

$xy$  $\Rightarrow$ $x \ AND \ y$

 

$x + y + z = 1$

$xy = 0$

$xz + w = 1$

$xy + z'w' = 0$

$xy = 0$    $\rightarrow$ $xy = { \{01,10,00\} }$

$option \ C$ is eliminated  $\because$ $x = 1$ , $y = 1 \ and \  x.y = 1  \ but  \ we \  need \ 0 $

 

$xy + z'w' = 0$

$0 + z'w' = 0$           $\because$ $ xy = 0 $

$ z'w' = 0$  $\rightarrow$  $zw = { \{01,10,11\} }$

$option \ B$ is eliminated  $\because$ $z = 0 , w = 0 \ \ \ \ \ \ \ z' = 1 , w' = 1  \  \ \ \ \ So ,  z'w' = 1.1 = 1 \neq 0 $

 

 $option \  A$

$x = 1,  y = 0 , z = 1,  w = 1$

$1.$  $x + y + z = 1$    $\Rightarrow$    $1 + 0 + 1 = 1$

$2.$  $x.y = 0$    $\Rightarrow$    $1.0 = 0$

$3.$  $xz + w = 1$    $\Rightarrow$    $1.1 + 1 = 1$

$4.$  $xy + z'w' = 0$   $\Rightarrow$    $1.0 + 0.0 = 0$

 

$option \  D$

$x = 0, y = 1, z = 1, w = 0$

$1.$  $x + y + z = 1$  $\Rightarrow$  $0 + 1 + 1 = 1$

$2.$  $xy = 0$   $\Rightarrow$  $0.1 = 0$

$3.$  $xz + w = 1$  $\Rightarrow$  $0.1 + 0 \neq 1$

 

$option \  A $ is correct

Answer 2

option a is right