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JEE Main202622 January 2026Morning ShiftMathematicsApplication of DerivativesActual

Let f(x)=x²⁰²⁵-x²⁰⁰⁰, x [0,1] and the minimum value of the function f(x) in the interval [0,1] be (80)⁸⁰(n)⁻⁸¹ . Then n is equal to

Options

  1. A-40
  2. B-41
  3. C-80
  4. D-81

Correct answer

D. -81

Step-by-step solution

f(x) = x²⁰²⁵ - x²⁰⁰⁰ , f'(x) = x¹⁹⁹⁹(2025x²⁵ - 2000) = 0 . Critical point: x²⁵ = 2000 2025 = 80 81 , i.e., x = ( 80 81 )^ 1/25 . f(0) = f(1) = 0 . At the critical point: f = x²⁰⁰⁰(x²⁵-1) = ( 80 81 )⁸⁰ ( 80 81 -1 ) = - 80⁸⁰ 81⁸¹ . This equals (80)⁸⁰(n)⁻⁸¹ , so n⁻⁸¹ = -81⁻⁸¹ n = -81 .

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