Let f(x) = x^{2025} - x^{2000}, x in [0, 1] a | vedclass

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(A) Given $f(x) = x^{2025} - x^{2000}$.To find the minimum value, we find the derivative $f'(x) = 2025x^{2024} - 2000x^{1999}$.Setting $f'(x) = 0$, we

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$-81$

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(A) Given $f(x) = x^{2025} - x^{2000}$.To find the minimum value, we find the derivative $f'(x) = 2025x^{2024} - 2000x^{1999}$.Setting $f'(x) = 0$, we get $x^{1999}(2025x^{25} - 2000) = 0$.Since $x \in [0, 1]$, the critical point is $x = (\frac{2000}{2025})^{1/25} = (\frac{80}{81})^{1/25} = \alpha$.Evaluating $f(\alpha) = (\frac{80}{81})^{2025/25} - (\frac{80}{81})^{2000/25} = (\frac{80}{81})^{81} - (\frac{80}{81})^{80}$.$f(\alpha) = (\frac{80}{81})^{80} (\frac{80}{81} - 1) = (\frac{80}{81})^{80} (-\frac{1}{81}) = 80^{80} \cdot 81^{-80} \cdot (-81)^{-1} = 80^{80} \cdot (-81)^{-81}$.Comparing this with $(80)^{80}(n)^{-81}$, we get $n = -81$.

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

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