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JEE MainMathematicsLimits

Let f: R R be a thrice differentiable function such that f(0) = 4 , f^ (0) = 0 and f^ (0) = 24 . Then the value of _ e ( _ x 0 ( ₀^ x f(t) dt 4x )^ 2 x^2 ) is equal to :

Options

  1. A2
  2. B4
  3. C8
  4. D12

Correct answer

A. 2

Step-by-step solution

Let L = _ x 0 ( ₀^ x f(t) dt 4x )^ 2 x^2 . As x 0 , by L'H 00f4pital's rule, _ x 0 ₀^ x f(t) dt 4x = _ x 0 f(x) 4 = 4 4 = 1 . Thus, L is of the form 1^ . Using the formula for 1^ limits, L = e^ P , where: P = _ x 0 2 x^2 ( ₀^ x f(t) dt 4x - 1 ) P = _ x 0 ₀^ x f(t) dt - 4x 2x^3 This is a 0 0 form. Applying L'H 00f4pital's rule and the Newton-Leibniz formula: P = _ x 0 f(x) - 4 6x^2 Since f(0) = 4 , this is again a 0 0 form. Applying L'H 00f4pital's rule: P = _ x 0 f^ (x) 12x Since f^ (0) = 0 , applying L'H 00f4pital

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