COMEDK 2025 Morning Solved Paper

Concept:For a cubic polynomial, a local minimum at a point requires the first derivative to be zero and the second derivative to be positive at that point.Explanation:Let $f(x) = a x^3 + b x^2 + c x + d$.Compute the first derivative: $f'(x) = 3a x^2 + 2b x + c$.Since there is a minimum at $x=1$, the slope must be zero: $f'(1) = 0$.Thus $3a(1)^2 + 2b(1) + c = 0$, giving $3a + 2b + c = 0$.Compute the second derivative: $f''(x) = 6a x + 2b$.For a local minimum, the second derivative must be positive: $f''(1) > 0$.So $6a(1) + 2b > 0$, which simplifies to $6a + 2b > 0$.Dividing by 2 yields $3a + b > 0$.Therefore the condition from the minima is $3a + b > 0$.Answer:Option B: $3a + b > 0$.