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(B) The circuit consists of two $OR$ gates followed by a $NAND$ gate. Let the outputs of the two $OR$ gates be $Y_1$ and $Y_2$. Then $Y_1 = A + B$ and

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(B) The circuit consists of two $OR$ gates followed by a $NAND$ gate. Let the outputs of the two $OR$ gates be $Y_1$ and $Y_2$. Then $Y_1 = A + B$ and $Y_2 = A + B$. However, looking closely at the diagram, the top gate is a $NOR$ gate and the bottom is an $OR$ gate. Let's re-evaluate: The top gate is a $NOR$ gate, so its output is $Y_1 = \overline{A+B}$. The bottom gate is an $OR$ gate, so its output is $Y_2 = B+B = B$. The final gate is a $NAND$ gate, so the output $Y = \overline{Y_1 \cdot Y_2} = \overline{\overline{(A+B)} \cdot B} = (A+B) + \bar{B} = A + B + \bar{B} = A + 1 = 1$. Thus, the output $Y$ is always $1$ regardless of the inputs $A$ and $B$.

For the following circuit and given inputs $A$ and $B$, choose the correct option for output $Y$.

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