Question

Consider the following Boolean expression of a function FF :

F(P,Q)=(P‾+Q)⊕(P‾Q)F(P, Q) = (\overline{P} + Q) \oplus (\overline{P}Q)

Which of the following expressions is/are equivalent to FF ?

Choose every correct option
Official answer and solution
Official answer
A and C
From
Official GATE 2026 answer key

Solution

Expand F(P,Q)F(P, Q) into minterms or truth table values and compare with each option.

Evaluating F(P,Q)=(P‾+Q)⊕(P‾Q)F(P, Q) = (\overline{P} + Q) \oplus (\overline{P}Q) for all four input combinations (P,Q)(P, Q):

  • (0,0)(0, 0): (1+0)⊕0=1⊕0=1(1 + 0) \oplus 0 = 1 \oplus 0 = 1
  • (0,1)(0, 1): (1+1)⊕1=1⊕1=0(1 + 1) \oplus 1 = 1 \oplus 1 = 0
  • (1,0)(1, 0): (0+0)⊕0=0⊕0=0(0 + 0) \oplus 0 = 0 \oplus 0 = 0
  • (1,1)(1, 1): (0+1)⊕0=1⊕0=1(0 + 1) \oplus 0 = 1 \oplus 0 = 1

Thus F(P,Q)=P‾ Q‾+PQ=P⊙QF(P, Q) = \overline{P}\,\overline{Q} + PQ = P \odot Q, which is the XNOR function:

  • Option (A): P⊕Q‾=P⊙Q=F(P,Q)\overline{P \oplus Q} = P \odot Q = F(P, Q).
  • Option (C): P‾⊕Q=(P‾)Q‾+PQ=P‾ Q‾+PQ=F(P,Q)\overline{P} \oplus Q = (\overline{P})\overline{Q} + P Q = \overline{P}\,\overline{Q} + PQ = F(P, Q).
  • Options (B) and (D) both equal P⊕Q≠F(P,Q)P \oplus Q \neq F(P, Q).
Final answer

Options (A), (C)

Common trap

Missing that P‾⊕Q\overline{P} \oplus Q is also equal to P⊕Q‾\overline{P \oplus Q}.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question
21 of 65
Section
Computer Science and Information Technology

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