Mathematics · Functions

JEE Main 2026 — 22 January, Morning Shift — Question 5

Let $f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $f(x)$ in the interval $[0,1]$ be $(80)^{80}(\mathrm{n})^{-81}$ . Then n is equal to :

Option A:
-81
Correct
Option B:
-40
Option C:
-41
Option D:
-80

Answer: A

Step-by-step solution

$\mathrm{f}(\mathrm{x})=\mathrm{x}^{2025}-\mathrm{x}^{2000}$

$\mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \mathrm{x}=\left(\frac{2000}{2025}\right)^{1 / 25}=\alpha($ say $)$

$\therefore f(0)=0, f(1)=0, f(\alpha)=\left(\frac{80}{81}\right)^{80} \cdot \frac{-1}{81}=80^{80} \cdot(-81)^{-81}$

$n=-81$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Functions
Topic: Domain & range of functions