GATE CS 2010 Q9 — Combinational Circuits

A 4-to-1 multiplexer with select lines $S_1 = P$ (MSB) and $S_0 = Q$ (LSB) has the following output expression:

From the given circuit diagram, the input values are:

• $I_0 = R$
• $I_1 = \overline{R}$
• $I_2 = \overline{R}$
• $I_3 = R$

Substituting these inputs into the multiplexer equation:

Now, group the terms containing $R$ and $\overline{R}$:

Using the standard Boolean identities for XOR ($\oplus$) and XNOR ($\odot$):

• $\overline{P}Q + P\overline{Q} = P \oplus Q$
• $\overline{P}\overline{Q} + PQ = P \odot Q = \overline{P \oplus Q}$

Substituting these back into the expression for $f$:

Let $X = P \oplus Q$. The equation simplifies to:

Thus, the output Boolean expression is $P \oplus Q \oplus R$, which corresponds to Option B.