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JEE MainMathematicsApplication of Derivatives

Let the function f(x) = 3x^4 - 32x^3 + 102x^2 - 48x + K have a local maximum value of 102 + 48 3 . The value of the constant K is

Options

  1. A6
  2. B10
  3. C25
  4. D13

Correct answer

C. 25

Step-by-step solution

Given the function f(x) = 3x^4 - 32x^3 + 102x^2 - 48x + K . Differentiating with respect to x : f'(x) = 12x^3 - 96x^2 + 204x - 48 = 12(x^3 - 8x^2 + 17x - 4) For critical points, f'(x) = 0 . By inspection, x = 4 is a root. 12(x - 4)(x^2 - 4x + 1) = 0 Equating the quadratic factor to zero, the roots are x = 4 16 - 4 2 = 2 3 . Now, find the second derivative: f''(x) = 12(3x^2 - 16x + 17) Checking the sign of f''(x) at the critical points: At x = 4 : f''(4) = 12(48 - 64 + 17) = 12 > 0 (Local minimum) At x = 2 + 3 : f''

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