Applications of Derivatives
Minimum Value — High-Degree Polynomial on [0,1]
nta_pyq_2026_jan
Grade 12
Question:
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
Step-by-Step Solution
Key Concept: $f'(x)=x^{1999}(2025x^{25}-2000)=0$ in $(0,1)$ at $x^{25}=80/81$, i.e., $x=(80/81)^{1/25}$. $f(0)=f(1)=0$ so minimum is at the interior critical point.
Minimum at $x=(80/81)^{1/25}$: $(80)^{80}(-81)^{-81}$. $n=-81$.
Correct Answer: 4