
The truth table for the circuit given in the figure is:

- $\begin{aligned} \begin{array}{|c c c|} \hline A & B & Y \\ \hline 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \hline \end{array} \end{aligned}$
- $\begin{aligned} &\begin{array}{|c c c|} \hline A & B & Y \\ \hline 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \hline \end{array} \end{aligned}$
- $\begin{aligned} \begin{array}{|c c c|} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} \end{aligned}$
- $\begin{aligned} \begin{array}{|c c c|} A & B & Y \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \\ \end{array} \end{aligned}$
Solution

Asked in: JEE Main 2019 (12 Apr Shift 1)