Which one of the following circuits implements the Boolean function given below? \[ f(x,y,z) = m_0 + m_1 + m_3 + m_4 + m_5 + m_6, \] where \(m_i\) is the \(i^{\text{th}}\) minterm.
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To implement a Boolean function using a MUX, fix the select lines first and express the output for each combination in terms of the remaining variable.
Step 1: Write the minterms in binary form. For variables \((x,y,z)\), the minterms included in the function are: \[ m_0(000),\; m_1(001),\; m_3(011),\; m_4(100),\; m_5(101),\; m_6(110). \]
Step 2: Use a 4:1 MUX realization. A 4:1 multiplexer can implement a 3-variable Boolean function by taking two variables as select lines and expressing the output as a function of the remaining variable. Here, \(y\) and \(z\) are used as select lines \((s_1,s_0)\).
Step 3: Determine data inputs for each select combination. \[ \begin{array}{c|c|c} y & z & f(x,y,z) \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & x \\ 1 & 1 & x' \end{array} \] Thus, the required MUX inputs are: \[ I_0 = 1, I_1 = 1, I_2 = x, I_3 = x'. \]
Step 4: Match with the given options. Option (A) exactly corresponds to the above configuration of the 4:1 multiplexer with correct data inputs and select lines. % Final Answer
Final Answer: \[ \boxed{\text{Option (A)}} \]
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