How a 4×1 MUX works. It passes exactly one of four data inputs to the output, chosen by the two select lines; the general expression is a sum of the four minterms of the select variables, each ANDed with its data input:

\(Y=\overline{S_1}\,\overline{S_0}\,I_0+\overline{S_1}S_0 I_1+S_1\overline{S_0}I_2+S_1 S_0 I_3\)

Connections from the figure. Select lines \(S_1=A,\ S_0=B\); data inputs \(I_0=I_1=C\) and \(I_2=I_3=\overline{C}\). This is the standard trick of using a MUX as a function generator: two variables drive the selects, and the third variable (in true or complemented form) is applied to the data lines.

Substitute.

\(Y=\overline{A}\,\overline{B}\,C+\overline{A}B\,C+A\overline{B}\,\overline{C}+A B\,\overline{C}\)

Group and simplify. The first two terms share \(\overline{A}C\) and the last two share \(A\overline{C}\):

\(Y=\overline{A}C(\overline{B}+B)+A\overline{C}(\overline{B}+B)\)

Using the complement law \(B+\overline{B}=1\):

\(Y=\overline{A}C+A\overline{C}=A\oplus C\)

Why B disappears. Look at the pairing of the data inputs: I₀ and I₁ carry the same value, and so do I₂ and I₃. Those pairs differ only in S₀ = B, so whichever way B goes the same value is selected — B has no influence on the output at all. Recognising this immediately reduces the 4×1 MUX to a 2×1 MUX selected by A, with C on one input and \(\overline{C}\) on the other, which is precisely an XOR of A and C.

Check the options. Options 3 and 4 involve B, which has just been shown to cancel, so both are impossible. Option 1, \(\overline{A\oplus C}\), is the XNOR — it would arise if the data inputs were swapped (\(I_0=I_1=\overline{C}\), \(I_2=I_3=C\)). The connections as given yield C when A = 0 and \(\overline{C}\) when A = 1, i.e. XOR.

Hence, the output is y = A ⊕ C.