Logic Gates question from JEE Main 2012 (AIEEE)

Answer: (a)

1. First NAND: $N = \overline{AB}$; next: $\overline{AN}$ and $\overline{BN}$.
2. Final NAND: $Y = AN + BN = (A + B)\overline{AB} = A\bar{B} + \bar{A}B$ (XOR).
3. Rows: $00 \to 0$, $01 \to 1$, $10 \to 1$, $11 \to 0$.
4. So the answer is (a).