Logic Gates question from JEE Main 2025 (28 Jan, Shift 1)

Answer: (a)

1. Upper AND: $A\bar{B}$. Lower AND gets $A$ and $B$ directly: $AB$.
2. NOR output: $Y = \overline{A\bar{B} + AB} = \overline{A(\bar{B} + B)} = \bar{A}$.
3. Option (a) is a NAND gate with both inputs joined to A, which is a NOT gate: $Y = \bar{A}$.
4. So the answer is (a).