Which one of the following set of values of (X0, X1, X2, X3, X4, X5, X6, X7) will realise the Boolean function $\overline{\text{A}} + \overline{\text{A}}.\overline{\text{C}} + \text{A}.\overline{\text{B}}.\text{C}$?

To find the values of $(X_0, X_1, X_2, X_3, X_4, X_5, X_6, X_7)$, let us simplify the given Boolean function $F$:

Using absorption law $\overline{A} + \overline{A}\cdot\overline{C} = \overline{A}$:

Now, let us determine the minterms of $F(A, B, C)$:

• When $A = 0$, $F = 1$ for all combinations of $(B, C)$, which corresponds to $(A, B, C) \in \{000, 001, 010, 011\}$ ($m_0, m_1, m_2, m_3$).

• When $A = 1$, $F = 1$ only when $\overline{B}\cdot C = 1$, i.e., $B = 0, C = 1$, which corresponds to $(A, B, C) = 101$ ($m_5$).

• For all other combinations ($m_4 = 100, m_6 = 110, m_7 = 111$), $F = 0$.

From the multiplexer circuit:

• Multiplexer M3 is controlled by select line $S = B$:

  • When $B = 0$, $Q = Q_{M1}$ (output of M1)

  • When $B = 1$, $Q = Q_{M2}$ (output of M2)

• For M1 and M2, select lines are $S_1 = A$ and $S_0 = C$.

Evaluating the inputs for each combination of $(A, B, C)$:

1. For $B = 0$ (M1 selected):

   • $(A, C) = (0, 0) \implies (A, B, C) = (0, 0, 0) \implies X_0 = F(0, 0, 0) = 1$

   • $(A, C) = (0, 1) \implies (A, B, C) = (0, 0, 1) \implies X_1 = F(0, 0, 1) = 1$

   • $(A, C) = (1, 0) \implies (A, B, C) = (1, 0, 0) \implies X_2 = F(1, 0, 0) = 0$

   • $(A, C) = (1, 1) \implies (A, B, C) = (1, 0, 1) \implies X_3 = F(1, 0, 1) = 1$

2. For $B = 1$ (M2 selected):

   • $(A, C) = (0, 0) \implies (A, B, C) = (0, 1, 0) \implies X_4 = F(0, 1, 0) = 1$

   • $(A, C) = (0, 1) \implies (A, B, C) = (0, 1, 1) \implies X_5 = F(0, 1, 1) = 1$

   • $(A, C) = (1, 0) \implies (A, B, C) = (1, 1, 0) \implies X_6 = F(1, 1, 0) = 0$

   • $(A, C) = (1, 1) \implies (A, B, C) = (1, 1, 1) \implies X_7 = F(1, 1, 1) = 0$

Thus, $(X_0, X_1, X_2, X_3, X_4, X_5, X_6, X_7) = (1, 1, 0, 1, 1, 1, 0, 0)$, which corresponds to Option (C).