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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Application of Derivatives

Let f(x)=x^ 2025 -x^ 2000 , x ∈[0,1] and the minimum value of the...

Let f(x)=x2025−x2000,x∈[0,1]f(x)=x^{2025}-x^{2000}, x \in[0,1] and the minimum value of the function f(x)f(x) in the interval [0,1][0,1] be (80)80(n)−81(80)^{80}(n)^{-81}. Then nn 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−x2000f(x) = x^{2025} - x^{2000}, f′(x)=x1999(2025x25−2000)=0f'(x) = x^{1999}(2025x^{25} - 2000) = 0.
Critical point: x25=20002025=8081x^{25} = \frac{2000}{2025} = \frac{80}{81}, i.e., x=(8081)1/25x = \left(\frac{80}{81}\right)^{1/25}.
f(0)=f(1)=0f(0) = f(1) = 0. At the critical point:
f=x2000(x25−1)=(8081)80(8081−1)=−80808181f = x^{2000}(x^{25}-1) = \left(\frac{80}{81}\right)^{80}\left(\frac{80}{81}-1\right) = -\frac{80^{80}}{81^{81}}.
This equals (80)80(n)−81(80)^{80}(n)^{-81}, so n−81=−81−81⇒n=−81n^{-81} = -81^{-81} \Rightarrow n = -81.
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