JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Semiconductors

Trace the signals through the gates. The circuit forms $A\cdot \overline{AB}$ on one branch and $AB\cdot \bar{B}$ on the other, then combines them with an OR. Expand $\overline{AB}$ by De Morgan's rule as $\bar{A}+\bar{B}$, so the first branch is $A(\bar{A}+\bar{B})=A\bar{A}+A\bar{B}=A\bar{B}$, since $A\bar{A}=0$. The second branch $AB\bar{B}$ vanishes because $B\bar{B}=0$. The OR of $A\bar{B}$ with $0$ is just $A\bar{B}$, so the output of the circuit is $A\bar{B}$.