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Solution


Q.

Let f(x)=x2025-x2000, 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         [2026]
1 -80  
2 -81  
3 -41  
4 -40  

Ans.

(2)

f(x)=x2025-x2000

f'(x)=0⇒x=(20002025)1/25=α (say)

∴f(0)=0,  f(1)=0, f(α)=(8081)80.-181=8080·(-81)-81