Given the function πΉ = πβ² + ππ , where πΉ is a function in three Booleanβ¦
2015
Given the function πΉ = πβ² + ππ , where πΉ is a function in three Boolean variables π,π and π and πβ² = ! π , consider the following statements.
(S1) πΉ = β(4, 5, 6)
(S2) πΉ = β(0, 1, 2, 3, 7)
(S3) πΉ = β(4, 5, 6)
(S4) πΉ = β(0, 1, 2, 3, 7)
Which of the following is true?
- A.
(S1)- False, (S2)- True, (S3)- True, (S4)- False
- B.
(S1)- True, (S2)- False, (S3)- False, (S4)- True
- C.
(S1)- False, (S2)- False, (S3)- True, (S4)- True
- D.
(S1)- True, (S2)- True, (S3)- False, (S4)- False
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Correct answer: A
Correct. P' = 1 whenever P = 0, so minterms 0, 1, 2, 3 make F = 1. QR = 1 when Q = R = 1, which adds minterm 7. Thus F's ones are {0,1,2,3,7} (so F = Ξ£(0,1,2,3,7)) and its zeros are {4,5,6} (so F = Ξ (4,5,6)). Therefore the statements evaluate as: F = Ξ£(4,5,6) is false; F = Ξ£(0,1,2,3,7) is true; F = Ξ (4,5,6) is true; F = Ξ (0,1,2,3,7) is false.
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