Semiconductor Electronics question for JEE Main (JEE Main 2021 (26 Aug, Shift 1)), with solution

Answer: (D) NOR

Each NAND gate has both inputs joined, so it acts as a NOT gate: $X = \bar{A}$, $Y = \bar{B}$.

The final AND gate gives
$$Z = \bar{A}\cdot\bar{B} = \overline{A + B}$$
which is the NOR operation.

Solution by Sreeraj P, M.Sc Physics