Semiconductor Electronics question for JEE Main (JEE Main 2023 (29 Jan, Shift 2)), with solution

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

The upper AND gate gives $\bar A B$ and the lower one gives $A\bar B$, so
$$X=\bar AB+A\bar B=A\oplus B\ \text{(XOR)}$$
$X=1$ only when the inputs differ: $0,1,1,0$.

Solution by Sreeraj P, M.Sc Physics