GATE CS 2004 Q60 — Combinational Circuits

The output expression of a $2$-to-$1$ multiplexer with data inputs $X$ and $Y$, and control input $Z$ is given by:

We want to realize the Boolean function $f = T + R$ using this multiplexer. Let's test the given options by substituting their connections into the multiplexer equation:

• Option A: $R$ to $X$, $1$ to $Y$, $T$ to $Z$ $$F = \bar{T}R + T(1) = T + \bar{T}R$$ Using the distributive law ($A + \bar{A}B = A + B$): $$F = T + R$$ This perfectly matches the required function $f = T + R$.
• Option B: $T$ to $X$, $R$ to $Y$, $T$ to $Z$ $$F = \bar{T}T + TR = 0 + TR = TR \neq T + R$$
• Option C: $T$ to $X$, $R$ to $Y$, $0$ to $Z$ $$F = \bar{0}T + 0(R) = T \neq T + R$$
• Option D: $R$ to $X$, $0$ to $Y$, $T$ to $Z$ $$F = \bar{T}R + T(0) = \bar{T}R \neq T + R$$

Thus, only the connections in Option A correctly realize the function $f = T + R$.