JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Semiconductors

The inverter fed by $A$ puts $\bar{A}$ on one rail and the inverter fed by $B$ puts $\bar{B}$ on the other. The upright inverter draws its input from the $\bar{B}$ rail, its lead hopping over the $\bar{A}$ rail without joining it, and so restores $B$. The NOR gate therefore receives $A$ and $B$, while the NAND gate receives $\bar{A}$ and $\bar{B}$.
With $A=0$ and $B=1$, the NOR gate gives $\overline{0+1}=0$ and the NAND gate gives $\overline{1\cdot 0}=1$. The AND gate at the output combines them:

The output is $x=0$.