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GATE CSE 2023 Q44: Multiplexer inputs realising A' + A'C' + AB'C

Question

A Boolean digital circuit is composed using two 4-input multiplexers (M1 and M2) and one 2-input multiplexer (M3) as shown in the figure. X0–X7 are the inputs of the multiplexers M1 and M2 and could be connected to either 0 or 1. The select lines of the multiplexers are connected to Boolean variables A, B and C as shown.

Two 4-to-1 multiplexers: M1 with inputs X0 to X3 and M2 with inputs X4 to X7, both with select lines S1 = A and S0 = C. M1's output goes to input 0 and M2's output to input 1 of a 2-to-1 multiplexer M3 whose select line is B
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Which one of the following set of values of (X0, X1, X2, X3, X4, X5, X6, X7) will realise the Boolean function A‾+A‾ C‾+A B‾ C\overline{A} + \overline{A}\,\overline{C} + A\,\overline{B}\,C?

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Answer: C (official GATE 2023 final answer key)

Concept used: Multiplexer implementation: input XkX_k is the function value for the select combination kk.

First step: Set B=0B = 0 (M1) and evaluate the function for (A,C)=00,01,10,11(A, C) = 00, 01, 10, 11 to get X0–X3; repeat with B=1B = 1 for X4–X7.

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Source: official GATE 2023 CS question paper and final answer key, organised by IIT Kanpur. Question paper · Answer key

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