Which one of the following circuits implements the Boolean function given below? $$f(x, y, z) = m_0 + m_1 + m_3 + m_4 + m_5 + m_6\text{, where } m_i \text{ is the } i^{\text{th}} \text{ minterm.}$$

The given Boolean function is:

In all given $4 \times 1$ multiplexers, the select lines are $s_1 = y$ and $s_0 = z$. The output of a $4 \times 1$ multiplexer with select lines $y$ and $z$ is:

Let us determine the required inputs $I_0, I_1, I_2, I_3$ as functions of $x$:

1. For $yz = 00$ ($I_0$): Minterms corresponding to $yz = 00$ are $m_0$ ($x=0, y=0, z=0$) and $m_4$ ($x=1, y=0, z=0$). Since both $m_0$ and $m_4$ are present in $f$, $I_0 = 1$.

2. For $yz = 01$ ($I_1$): Minterms corresponding to $yz = 01$ are $m_1$ ($x=0, y=0, z=1$) and $m_5$ ($x=1, y=0, z=1$). Since both $m_1$ and $m_5$ are present in $f$, $I_1 = 1$.

3. For $yz = 10$ ($I_2$): Minterms corresponding to $yz = 10$ are $m_2$ ($x=0, y=1, z=0$) and $m_6$ ($x=1, y=1, z=0$). Since $m_2$ is absent and $m_6$ is present, $I_2 = x$.

4. For $yz = 11$ ($I_3$): Minterms corresponding to $yz = 11$ are $m_3$ ($x=0, y=1, z=1$) and $m_7$ ($x=1, y=1, z=1$). Since $m_3$ is present and $m_7$ is absent, $I_3 = x^{\prime}$.

Thus, the required data inputs are:

This configuration corresponds to Option (A).