JEE Main 2026 (Online) 22nd January Morning ShiftQuestion 1
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 $(80)^{80}(n)^{-81}$. Then $n$ is equal to :
Differentiation and its Application
Solution
Concept. Find the critical point inside $(0,1)$ and write $f$ there as $x^{2000}(x^{25}-1)$. Formulas used. (1) $f'(x)=x^{1999}\left(2025x^{25}-2000\right)$Step 1 - Critical point. By (1), $f'=0$ in $(0,1)$ when $x^{25}=\frac{2000}{2025}=\frac{80}{81}$. $f'<0$ before it and $f'>0$ after, so it is the minimum. ($f(0)=f(1)=0$.) Step 2 - The value. $x^{2000}=\left(x^{25}\right)^{80}=\left(\frac{80}{81}\right)^{80}$, so $$f_{\min}=\left(\frac{80}{81}\right)^{80}\left(\frac{80}{81}-1\right)=-\frac{80^{80}}{81^{81}}.$$ Step 3 - Match. $-\frac{1}{81^{81}}=(-81)^{-81}$, so $f_{\min}=(80)^{80}(-81)^{-81}$ and $n=-81$. Answer: (B)