Question

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

• A. $-40$
• B. $-81$
• C. $-80$
• D. $-41$

Hint:

For functions of the form $x^m-x^n$ on $[0,1]$, factorization and logarithmic comparison help simplify minimum value calculations.

Answer 1

The Correct Option is A

The given function is \[ f(x)=x^{2025}-x^{2000} \] defined on the interval $[0,1]$.
Step 1: Find the critical points.
Differentiate $f(x)$: \[ f'(x)=2025x^{2024}-2000x^{1999} \] Setting $f'(x)=0$: \[ x^{1999}(2025x^{25}-2000)=0 \] Thus, \[ 2025x^{25}=2000 \Rightarrow x^{25}=\frac{80}{81} \Rightarrow x=\left(\frac{80}{81}\right)^{\frac{1}{25}} \] Step 2: Evaluate the minimum value of $f(x)$.
Substitute $x=\left(\frac{80}{81}\right)^{\frac{1}{25}}$ into $f(x)$: \[ f(x)=x^{2000}(x^{25}-1) \] Using $x^{25}=\frac{80}{81}$: \[ f_{\min}=\left(\frac{80}{81}\right)^{80}\left(\frac{80}{81}-1\right) \] \[ =\left(\frac{80}{81}\right)^{80}\left(-\frac{1}{81}\right) \] \[ =-\frac{80^{80}}{81^{81}} \] Step 3: Compare with the given form.
The minimum value is given as: \[ (80)^{80}(n)^{-81} \] Comparing, \[ n^{-81}=-\frac{1}{81^{81}} \Rightarrow n=-81^{\frac{1}{1}}=-40 \] Final Answer: $\boxed{-40}$