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MHT CET 2024 (3 May, Shift 2) · Logic Gates

Logic Gates question from MHT CET 2024 (3 May, Shift 2)

Q56MHT CET 2024 · 3 MAY · SHIFT 2Easy

Logic Gates

The Boolean expression for 'X-OR' gate C=(A⊕B)C = (A \oplus B) is equal to

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Answer: (d)

  1. XOR is 1 only when the inputs differ: A=0,B=1A = 0, B = 1 or A=1,B=0A = 1, B = 0.
  2. These two cases are A‾⋅B\overline{A} \cdot B and A⋅B‾A \cdot \overline{B}.
  3. A⊕B=A‾⋅B+A⋅B‾A \oplus B = \overline{A} \cdot B + A \cdot \overline{B}.
  4. So the answer is (d).
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