The sum of all the local minimum values of the twice differentiable function f : R → R defined by f(x) = x³ - 3x² - 3f''(2)/2x + f''(1) is :

Solution

Maths sir · 4 steps

Differentiate twice first: the unknown terms vanish, so f''(2) and f''(1) become plain numbers.

1. Find the constants. The terms with f''(2) and f''(1) are only a linear term and a constant, so they disappear after two derivatives. That leaves f''(x) = 6x - 6, which gives f''(2) = 6 and f''(1) = 0.
2. Rewrite the function. Substitute back: the coefficient of x is -(3 × 6)/2 = -9 and the constant is 0. So f(x) = x³ - 3x² - 9x.

   f'(x) = 3x² - 6x - 9 = 3(x-3)(x+1)

3. Classify each point. At x = 3, f''(3) = 12 > 0, so this is a local minimum. At x = -1, f''(-1) = -12 < 0, so this is a local maximum.
4. So. The only local minimum value is f(3) = 27 - 27 - 27 = -27. The sum of all local minimum values is -27, option (c).

The idea to keep. When a function's formula contains its own derivatives, differentiate enough times that those terms vanish. Then evaluate them as numbers and solve normally.

Answer from the NTA final answer key. Solution by PrepWiser.