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In Boolean Algebra, dual of the Boolean expression is equal to .
Assertion: In Boolean Algebra, dual of the Boolean expression $(A+B)'.1$ is equal to $0$.
Reason: In Boolean Algebra, the complement of an OR operation is equal to the AND operation of complement of the individual variables.
- aBoth A and R are true, and R is the correct explanation of A.
- bBoth A and R are true, and R is not the correct explanation of A.
- cA is true, but R is false.
- dA is false, but R is true.
Answer
Answer
AICorrect option: d
Answer: (d) A is false, but R is true.
The dual of (A+B)'.1 is formed by swapping + with . and 1 with 0: (A.B)' + 0, which simplifies to A'+B' (equivalently (A.B)') - a contingency, not the constant 0, so A is false. R correctly states De Morgan's law, (A+B)' = A'.B', so R is true.
From ISC Computer Science - Competency Focused Practice Questions (CISCE, August 2024), question 22.
Check your working with the Boolean algebra solver: the steps law by law, the K-map and the logic circuit for any expression.