JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Semiconductors

The circuit combines two AND gates and one OR gate: one AND gate takes inputs A and B, giving $A\cdot B$; the other takes $\bar{A}$ (A inverted through a NOT gate) and B, giving $\bar{A}\cdot B$. Both outputs feed a final OR gate:


Since $A+\bar{A}=1$ for every value of A, this reduces to


So Y is 0 whenever B is 0 and 1 whenever B is 1, independent of A — giving the rows $(0,0,0),(0,1,1),(1,0,0),(1,1,1)$.