Logic Gates question from MHT CET 2026 (17 Apr, Shift 2)

Answer: (d)

1. Upper gate: $P = \overline{A \cdot B}$. Lower gate: $Q = \overline{\bar{A} + B}$.
2. Final NOR: $Y = \overline{P + Q}$, so $Y = 1$ only when $P = 0$ and $Q = 0$.
3. $P = 0$ needs $A = 1, B = 1$. Then $Q = \overline{0 + 1} = 0$ as well.
4. So the answer is (d) $A = 1, B = 1$.