Semiconductor Electronics question for JEE Main (JEE Main 2024 (29 Jan, Shift 2)), with solution

Answer: (B) $\begin{array}{ccc} A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}$

Upper AND gate: $AB$. Lower AND gate: $\bar AB$.
$$Y = AB + \bar AB = B(A + \bar A) = B$$
So $Y$ equals $B$: $Y = 0, 1, 0, 1$ for $AB = 00, 01, 10, 11$.

Solution by Sreeraj P, M.Sc Physics