JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Semiconductors

Tracing the network, the top AND gate receives $\bar{A}$ from the inverter together with $B$, giving $\bar{A}\cdot B$, while the bottom AND gate receives $A$ together with $\bar{B}$ from the second inverter, giving $A\cdot \bar{B}$. The OR gate then forms $X=\bar{A}B+A\bar{B}$, which is the exclusive-OR of the inputs. The output is $0$ when the inputs are equal and $1$ when they differ, so $X=0,1,1,0$ for the input pairs $(0,0),(0,1),(1,0),(1,1)$.