JEE Main 202622 Jan, Morning shiftMathematicsEasyLimit, Continuity and Differentiability420× since 2002Let f(x)=x2025−x2000f(x)=x^{2025}-x^{2000}f(x)=x2025−x2000, x∈[0,1]x\in[0, 1]x∈[0,1] and the minimum value of the function f(x)f(x)f(x) in the interval [0,1][0, 1][0,1] be (80)80(n)−81(80)^{80}(n)^{-81}(80)80(n)−81. Then nnn is equal toAOption A. −40-40−40BOption B. −80-80−80COption C. −81-81−81DOption D. −41-41−41CheckSkip · show solution
