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(D) The circuit consists of one $NAND$ gate,one $NOT$ gate,one $NOR$ gate,and one final $NOR$ gate.Let the output of the top $NAND$ gate be $Y_1 = \ov

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$A=1, B=1$

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(D) The circuit consists of one $NAND$ gate,one $NOT$ gate,one $NOR$ gate,and one final $NOR$ gate.Let the output of the top $NAND$ gate be $Y_1 = \overline{A \cdot B}$.Let the output of the bottom branch ($NOT$ gate followed by $NOR$ gate) be $Y_2 = \overline{\overline{A} + B}$.The final output $Y$ is the output of a $NOR$ gate with inputs $Y_1$ and $Y_2$,so $Y = \overline{Y_1 + Y_2} = \overline{(\overline{A \cdot B}) + (\overline{\overline{A} + B})}$.Testing the options:For $A=1, B=1$:$Y_1 = \overline{1 \cdot 1} = 0$$Y_2 = \overline{\overline{1} + 1} = \overline{0 + 1} = 0$$Y = \overline{0 + 0} = 1$.Thus,for $A=1, B=1$,the output $Y$ is $1$.

In the following digital logic circuit,the output $Y$ will be ' $1$ ' for inputs:

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