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Best answer

- $f(x,y,z) = m_0+m_1+m_3+m_4+m_5+m_6$
- $S_1 = y,~S_0 = z$

Draw a small table as following and mark the min-terms in $f.$

$$\begin{array}{cccc}& \bf{I_0} &\bf{I_1}& \bf{I_2}& \bf{I_3}\\

\bar{x}& \color{Red} 0& \color{Red} 1& 2 & \color{Red} 3\\

x& \color{Red} 4& \color{Red} 5& \color{Red} 6& 7 \end{array}$$

$\begin{array}{ll}I_0 = (\bar{x} + x ) = 1 &\text{(both rows are marked)} \\I_1 = (\bar{x} + x ) = 1 & \text{(both rows are marked)}\\I_2 = x&\text{(only row corresponding to $x$ is marked)}\\I_3 = \bar{x} & \text{(only row corresponding to $\bar{x}$ is marked)}\end{array}$

**A is correct.**