The truth table for the given circuit is

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(D) The given logic circuit consists of a $NOR$ gate and an $AND$ gate whose output is fed into a $NAND$ gate. Let the inputs be $A$ and $B$.$1$. The

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(D) The given logic circuit consists of a $NOR$ gate and an $AND$ gate whose output is fed into a $NAND$ gate. Let the inputs be $A$ and $B$.$1$. The output of the $NOR$ gate is $\overline{A+B}$.$2$. The output of the $AND$ gate is $A \cdot B$.$3$. These two outputs are fed into a $NAND$ gate. The final output $Y$ is given by:$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$Using De Morgan's law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$:$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$$Y = (A+B) + (\overline{A} + \overline{B})$$Y = (A + \overline{A}) + (B + \overline{B})$Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:$Y = 1 + 1 = 1$Thus,the output $Y$ is always $1$ for all combinations of inputs $A$ and $B$. Therefore,the truth table will have $1$ in all rows for output $Y$. Comparing this with the given options,option $(D)$ is correct.

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