Derive a Boolean function f(x, y, z) that yields 1 when exactly two out of three input variables are 0.

• To derive a Boolean function f(x, y, z) that yields 1 when exactly two out of three input variables are 0, we can construct a truth table and then create the function based on the conditions.
• Step 1: Make a truth table for the variables x, y, and z.
• x | y | z | f(x, y, z) ---|---|---|----------- 0 | 0 | 0 | 0 0 | 0 | 1 | 1 0 | 1 | 0 | 1 0 | 1 | 1 | 0 1 | 0 | 0 | 1 1 | 0 | 1 | 0 1 | 1 | 0 | 0 1 | 1 | 1 | 0
• Step 2: From the truth table, identify the rows where the function yields a 1. We see that it happens when:
• - x = 0, y = 0, and z = 1 - x = 0, y = 1, and z = 0 - x = 1, y = 0, and z = 0
• Step 3: Write down the Boolean expressions that correspond to the combinations that yield 1:
• - x'y'z (x NOR y AND z) - x'yz' (x NOR z AND y) - xy'z' (y NOR z AND x)
• Step 4: Combine these with OR operations because the output is 1 for any of these conditions:
• f(x, y, z) = x'y'z + x'yz' + xy'z'
• Step 5: We can check the function by inputting different sets of x, y, and z to make sure it gives an output of 1 only when exactly two inputs are 0.

Thus, the function f(x, y, z) = x'y'z + x'yz' + xy'z' will yield 1 when exactly two out of the three input variables are 0.