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(D) Let the inputs be $X$ and $Y$. The top branch consists of a $NOT$ gate followed by a $NOR$ gate. The input to the $NOR$ gate is $\bar{X}$ and $Y$.

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Not possible

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(D) Let the inputs be $X$ and $Y$. The top branch consists of a $NOT$ gate followed by a $NOR$ gate. The input to the $NOR$ gate is $\bar{X}$ and $Y$. Thus,$P = \overline{\bar{X} + Y} = X \cdot \bar{Y}$.The bottom branch consists of a $NAND$ gate with inputs $X$ and $\bar{Y}$. Thus,$Q = \overline{X \cdot \bar{Y}} = \bar{X} + Y$.The final gate is a $NOR$ gate with inputs $P$ and $Q$. Thus,$R = \overline{P + Q} = \overline{(X \cdot \bar{Y}) + (\bar{X} + Y)}$.Using De Morgan's laws,$R = \overline{(X \cdot \bar{Y})} \cdot \overline{(\bar{X} + Y)} = (\bar{X} + Y) \cdot (X \cdot \bar{Y})$.Expanding this,$R = (\bar{X} \cdot X \cdot \bar{Y}) + (Y \cdot X \cdot \bar{Y}) = 0 + 0 = 0$.Since the output $R$ is always $0$ regardless of the inputs $X$ and $Y$,it is not possible to obtain a high output $(1)$ at $R$.

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$0$
$1$	$0$	$1$
$1$	$1$	$0$

The figure shows a system of logic gates. From the study of the truth table,it can be found that to produce a high output $(1)$ at $R$,we must have

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