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(D) The circuit consists of an $AND$ gate followed by a $NAND$ gate. Let the three inputs be $A, B, C$. The first $AND$ gate takes two inputs (say $A$

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$0,1$

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(D) The circuit consists of an $AND$ gate followed by a $NAND$ gate. Let the three inputs be $A, B, C$. The first $AND$ gate takes two inputs (say $A$ and $B$) and produces an output $X = A \cdot B$. The $NAND$ gate takes $X$ and the third input $C$ to produce the final output $Y = \overline{X \cdot C} = \overline{A \cdot B \cdot C}$.Case $1$: All inputs are high $(A=1, B=1, C=1)$.$Y = \overline{1 \cdot 1 \cdot 1} = \overline{1} = 0$.Case $2$: All inputs are low $(A=0, B=0, C=0)$.$Y = \overline{0 \cdot 0 \cdot 0} = \overline{0} = 1$.Thus,the outputs are $0$ and $1$ respectively. Therefore,the correct option is $D$.

The output $Y$,when all the three inputs are first high and then low,will respectively be

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