JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Semiconductors

Trace the two signal paths of the circuit. The lower path feeds $A$ and $B$ into a NAND gate, producing $\overline{A\cdot B}=\bar{A}+\bar{B}$, which is carried on to the output gate. The upper path delivers $A$ itself to the output gate. The output gate takes the AND of these two, so $Y=A\cdot \overline{A\cdot B}=A(\bar{A}+\bar{B})=A\bar{B}$. This is $1$ only when $A=1$ and $B=0$, and $0$ otherwise, giving outputs $0,0,1,0$ for $(A,B)=(0,0),(0,1),(1,0),(1,1)$.