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(A) The given circuit consists of an $AND$ gate,two $NOT$ gates,another $AND$ gate,and a $NOR$ gate.Let the inputs be $A$ and $B$. The output of the f

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$1, 1$

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(A) The given circuit consists of an $AND$ gate,two $NOT$ gates,another $AND$ gate,and a $NOR$ gate.Let the inputs be $A$ and $B$. The output of the first $AND$ gate is $A \cdot B$.The inputs to the second $AND$ gate are $\bar{A}$ and $\bar{B}$,so its output is $\bar{A} \cdot \bar{B}$.These two outputs are fed into a $NOR$ gate,so the final output $E$ is $\overline{(A \cdot B) + (\bar{A} \cdot \bar{B})}$.For $X$: $A = 0, B = 1$. $E = \overline{(0 \cdot 1) + (\bar{0} \cdot \bar{1})} = \overline{0 + (1 \cdot 0)} = \overline{0 + 0} = \overline{0} = 1$. Thus,$X = 1$.For $Y$: $A = 1, B = 0$. $E = \overline{(1 \cdot 0) + (\bar{1} \cdot \bar{0})} = \overline{0 + (0 \cdot 1)} = \overline{0 + 0} = \overline{0} = 1$. Thus,$Y = 1$.Therefore,the values of $X$ and $Y$ are $1, 1$.

In the truth table of the given circuit,the values of $X$ and $Y$ are:

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