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(C) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. Let the inputs be $A$ and $B$.$1$. The top $AND$ gate receives in

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(C) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate. Let the inputs be $A$ and $B$.$1$. The top $AND$ gate receives inputs $\bar{A}$ and $B$. Its output is $Y_1 = \bar{A} \cdot B$.$2$. The bottom $AND$ gate receives inputs $A$ and $\bar{B}$. Its output is $Y_2 = A \cdot \bar{B}$.$3$. The final $OR$ gate combines these outputs: $X = Y_1 + Y_2 = \bar{A}B + A\bar{B}$.This is the Boolean expression for an $XOR$ gate.The truth table for an $XOR$ gate is:- If $A=0, B=0$,then $X = 0 \cdot 0 + 0 \cdot 1 = 0$.- If $A=0, B=1$,then $X = 1 \cdot 1 + 0 \cdot 0 = 1$.- If $A=1, B=0$,then $X = 0 \cdot 0 + 1 \cdot 1 = 1$.- If $A=1, B=1$,then $X = 0 \cdot 1 + 1 \cdot 0 = 0$.Thus,the correct truth table is the one where $X=1$ only when $A$ and $B$ are different,which corresponds to option $C$.

For the given logic gates combination,the correct truth table will be

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