Logic Gates question from MHT CET 2024 (16 May, Shift 2)

Answer: (c)

1. $Y = A + G(A, B)$ and the table gives $Y = 1$ in every row.
2. When $A = 0$, G must give 1 for $B = 0$ and $B = 1$.
3. NAND: $\overline{0 \cdot 0} = 1$, $\overline{0 \cdot 1} = 1$. AND and NOR fail at $(0,1)$; OR fails at $(0,0)$.
4. So the answer is (c) NAND.