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(D) Let the inputs be $A, B,$ and $C$.$Gate-I$ is an $AND$ gate,and $Gate-II$ is a $NAND$ gate.The output of $Gate-I$ is $A \cdot B$.The final output

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$0, 1$

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(D) Let the inputs be $A, B,$ and $C$.$Gate-I$ is an $AND$ gate,and $Gate-II$ is a $NAND$ gate.The output of $Gate-I$ is $A \cdot B$.The final output $y$ is the $NAND$ operation of the output of $Gate-I$ and input $C$,so $y = \overline{(A \cdot B) \cdot C}$.Case $1$: When all inputs are high $(A=1, B=1, C=1)$:$y = \overline{(1 \cdot 1) \cdot 1} = \overline{1 \cdot 1} = \overline{1} = 0$.Case $2$: When all inputs are low $(A=0, B=0, C=0)$:$y = \overline{(0 \cdot 0) \cdot 0} = \overline{0 \cdot 0} = \overline{0} = 1$.Thus,the outputs are $0$ and $1$ respectively.

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

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