MHTCET 11 May 2024 Shift 1 Paper

Concept:The given circuit uses NAND gates to produce an output that matches the XOR gate truth table.Explanation:Let the intermediate gate outputs be:$C = \overline{A \cdot B}$$P = \overline{A \cdot C}$$Q = \overline{C \cdot B}$Final output: $Y = \overline{P \cdot Q}$Evaluate for all input combinations:For $A=0, B=0$: $C=1$, $P=1$, $Q=1$, so $Y=0$For $A=0, B=1$: $C=1$, $P=1$, $Q=0$, so $Y=1$For $A=1, B=0$: $C=1$, $P=0$, $Q=1$, so $Y=1$For $A=1, B=1$: $C=0$, $P=1$, $Q=1$, so $Y=0$The output $Y$ is $1$ only when $A$ and $B$ are different.This is exactly the behaviour of an XOR gate.Answer:The combination is equivalent to an X-OR gate, so option D is correct.