Let f (x )=x^2025-x^2000 ,x ∈ [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

Question

Accepted Answer

Correct answer: D. (d) f (x )=x^2025-x^2000 ,f^ ' (x )=x^1999(2025x^25-2000)=0. Critical point: x^25= 20002025= 8081, i.e., x= ( 8081 )^1/25. f (0 )=f (1 )=0. At the critical point: f=x^2000(x^25-1)= ( 8081 )^80( 8081-1)=- 80^8081^81. This equals (80 )^80 (n )^-81, so n^-81=-81^-81 ⇒n=-81.

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

Solution

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