GATE CS 2002 Q2.2 — Combinational Circuits

To find the function $f(x, y, z)$ implemented by the multiplexer, we analyze the inputs, selection lines, and the enable input from the circuit diagram:

1. Select Lines:
   • $A_1 = x$
   • $A_0 = y$
2. Data Inputs:
   • $I_0 = 0$
   • $I_1 = 0$
   • $I_2 = 0$
   • $I_3 = 1$
3. Enable Input ($EN$): The enable input has an active-low bubble and is connected to the variable $z$. Therefore, the multiplexer is active only when $EN = 0$, which corresponds to $z' = 1$.

The general output expression for a 4-to-1 multiplexer with an active-low enable input is:

Substituting the given values into the expression:

Thus, the function implemented by the circuit is $x y z'$, which corresponds to Option A.