Q 12-14-075JEE MainJEE Main 2024 (9 Apr, Shift 2)Medium
In the truth table of the circuit shown, the values of $X$ and $Y$ are:
$$\begin{array}{c|c|c} A & B & E \\ \hline 0 & 0 & 0 \\ 0 & 1 & X \\ 1 & 0 & Y \\ 1 & 1 & 0 \end{array}$$
Answer: (B) 1, 1
$$E = \overline{AB + \bar A\bar B}$$
For $A = 0, B = 1$: $AB = 0$ and $\bar A\bar B = 0$, so $E = \overline{0} = 1$: $X = 1$.
For $A = 1, B = 0$: again both terms are 0, so $Y = 1$.
Solution by Sreeraj P, M.Sc Physics