Consider the 4-to-1 multiplexer with two select lines $S_1$ and $S_0$ given below. The minimal sum-of-products form of the Boolean expression for the output $F$ of the multiplexer is

From the multiplexer diagram, the select lines are $S_1 = P$ and $S_0 = Q$, and the data inputs are:

• $I_0 = 0$

• $I_1 = 1$

• $I_2 = R$

• $I_3 = \overline{R}$

The output $F$ of a 4-to-1 MUX is given by:

Substituting the inputs:

To find the minimal SOP form, we can simplify using Boolean algebra or a K-map: Expand $\overline{P}Q$ into minterms:

Now $F$ contains the minterms $m(2, 3, 5, 6)$ in terms of $(P, Q, R)$:

• $m_2 = \overline{P}Q\overline{R}$

• $m_3 = \overline{P}QR$

• $m_5 = P\overline{Q}R$

• $m_6 = PQ\overline{R}$

Groupings:

1. Grouping $m_2$ and $m_3$: $\overline{P}Q\overline{R} + \overline{P}QR = \overline{P}Q$

2. Grouping $m_2$ and $m_6$: $\overline{P}Q\overline{R} + PQ\overline{R} = Q\overline{R}$

3. The essential minterm $m_5$ ($P\overline{Q}R$) cannot combine with any adjacent 1s in the K-map, so it remains $P\overline{Q}R$.

Thus, the minimal SOP expression is:

This matches option (A).