Question: Assuming $A$ to be the MSB, the Boolean function $f(A, B, C, D) = \Sigma(2, 6, 8, 9, 11, 14) + d(10)$ simplifies to (A) $AB + C'D$ (B) $A' + BCD$ (C) $AB' + CD'$ (D) None of the other options

1. **State the problem:** Simplify the Boolean function $$f(A, B, C, D) = \Sigma(2, 6, 8, 9, 11, 14) + d(10)$$ where $A$ is the most significant bit (MSB). 2. **List minterms and don't care:** - Minterms: 2, 6, 8, 9, 11, 14 - Don't care: 10 3. **Convert minterms and don't care to binary:** - 2 = 0010 - 6 = 0110 - 8 = 1000 - 9 = 1001 - 10 = 1010 (don't care) - 11 = 1011 - 14 = 1110 4. **Group minterms by number of 1s:** - Group 1 (one 1): 2 (0010), 8 (1000) - Group 2 (two 1s): 6 (0110), 9 (1001), 10 (1010) - Group 3 (three 1s): 11 (1011), 14 (1110) 5. **Combine minterms to simplify:** - Combine 8 (1000) and 9 (1001): $A B' C' D + A B' C' D' = A B' C'$ (since $D$ changes) - Combine 9 (1001) and 11 (1011): $A B' C D + A B C D = A C D$ - Combine 6 (0110) and 14 (1110): $A' B C D' + A B C D' = B C D'$ (since $A$ changes) - Combine 2 (0010) and 6 (0110): $A' B' C D + A' B C D = A' C D$ 6. **Use don't care 10 (1010) to further simplify:** - 10 (1010) can combine with 14 (1110) to simplify $B C D'$ term. 7. **Final simplified expression:** - From combinations, the minimal sum of products is $$AB' + CD'$$ 8. **Match with options:** - Option (C) is $AB' + CD'$ **Answer:** (C) $AB' + CD'$