Question

If $Y=\overline{\overline{A}B+\overline{B}C+A\overline{C}}+CD$, then dual of Y is

• A. $(A+\overline{B})(B+\overline{C})(C+D)$
• B. $(A\overline{B}+B\overline{C}+\overline{A}C)(C+D)$
• C. $(A+\overline{B})(\overline{B}+C)(A+\overline{C})(C+D)$
• D. $(A\overline{B}+\overline{B}C+\overline{A}C)\left(CD\right)$

Answer 1

The correct option is B $(A\overline{B}+B\overline{C}+\overline{A}C)(C+D)$

$Y=\overline{\overline{A}B+\overline{B}C+A\overline{C}}+CD$

$=\left(\overline{\overline{A}B}\right)\left(\overline{\overline{B}C}\right)\left(\overline{A\overline{C}}\right)+CD$

$Y=(A+\overline{B})(B+\overline{C})(\overline{A}+C)+CD$

$DualofY=(A\overline{B}+B\overline{C}+\overline{A}C)(C+D)$