If f(x) = x³ + bx² + cx + d has a local extremum at x = 1 and f(1) = 5, f′(1) = 0, then (b, c) is:

If f(x) = x³ + bx² + cx + d has a local extremum at x = 1 and f(1) = 5, f′(1) = 0, then (b, c) is:

• A. b = −3, c = 3
• B. b = 3, c = −3
• C. b = −3, c = −3
• D. b = 3, c = 3

Answer: A) b = −3, c = 3

Explanation: f′(x) = 3x² + 2bx + c. f′(1) = 3 + 2b + c = 0 → 2b + c = −3. Not enough. But problem statement gives f(1)=5: 1+b+c+d=5. No d info. Assuming d arbitrary, we need second condition. Perhaps f″(1)≠0. Not enough. We'll adjust to: If f′(1)=0 and f′(0)=c, and given options, solve 2b+c=−3. Option 0: b=−3,c=3 → −6+3=−3 works.