Question 15

The given Boolean function is:

Each circuit uses a 4:1 multiplexer with select lines $S_1 = y$ and $S_0 = z$. We determine the required inputs $I_0, I_1, I_2, I_3$ as functions of $x$:

• For $(y, z) = (0, 0)$ (line 0): includes minterms $m_0 (x=0)$ and $m_4 (x=1)$. Since both are present in $f$, $I_0 = 1$.
• For $(y, z) = (0, 1)$ (line 1): includes minterms $m_1 (x=0)$ and $m_5 (x=1)$. Since both are present in $f$, $I_1 = 1$.
• For $(y, z) = (1, 0)$ (line 2): includes minterms $m_2 (x=0)$ and $m_6 (x=1)$. Here $m_2 \notin f$ and $m_6 \in f$, so when $x=0$, $f=0$ and when $x=1$, $f=1$. Thus $I_2 = x$.
• For $(y, z) = (1, 1)$ (line 3): includes minterms $m_3 (x=0)$ and $m_7 (x=1)$. Here $m_3 \in f$ and $m_7 \notin f$, so when $x=0$, $f=1$ and when $x=1$, $f=0$. Thus $I_3 = \bar{x}$ ($x'$).

Therefore, the inputs are $I_0 = 1$, $I_1 = 1$, $I_2 = x$, $I_3 = x'$, which exactly matches the circuit in Option A.