JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Semiconductors

The inverter on the lower input presents $\bar{B}$ to the OR gate, whose output is $A+\bar{B}$. The inverter that follows gives

by De Morgan's theorem. The output is high only for $A=0$ with $B=1$, and low for the other three input combinations. The circuit is therefore equivalent to an AND gate fed by $B$ directly and by $A$ through an inverter.