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JEE MainMathematicsApplication of Derivatives

Let f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 with a > 0 . The local maximum of f(x) occurs at x = x₁ . It is given that x₁ [0, 3] and the absolute maximum value of f(x) on the interval [0, 3] is equal to f(x₁) . Then, the set of all possible values of a is

Options

  1. A(0, 3]
  2. B[ 6 5 , 3 2 ]
  3. C[ 6 5 , 3 ]
  4. D[3, )

Correct answer

C. [ 6 5 , 3 ]

Step-by-step solution

Given f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 . Differentiating with respect to x , we get: f'(x) = 6x^2 - 18ax + 12a^2 = 6(x^2 - 3ax + 2a^2) = 6(x-a)(x-2a) The critical points are x = a and x = 2a . Since a > 0 , the smaller root is a and the larger root is 2a . For a cubic with a positive leading coefficient, the local maximum occurs at the smaller root. Thus, x₁ = a . We are given that x₁ [0, 3] , which means a [0, 3] . The absolute maximum of f(x) on the interval [0, 3] must occur either at the local maximum x = a or

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