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#68969cae-e55f-4f86-98cd-752f88d939cb基础填空题导数与单调性导数

10. (2023春•青岛期中)已知4a=ln44a=ln4b=e1b=e^{-1}5c=ln55c=ln5,则aabbcc的大小关系为((  ))

解析
【解答】解:由题意得a=ln44a=\frac{ln4}{4}b=lnee=1eb=\frac{lne}{e}=\frac{1}{e}c=ln55c=\frac{ln5}{5}, 令f(x)=lnxx(x>0)f(x)=\frac{lnx}{x}(x>0), 则f(x)=1lnxx2f\prime (x)=\frac{1-lnx}{{x}^{2}}, 由f(x)<0f\prime (x)<0x>ex>e,即f(x)f(x)(e,+)(e,+\infty )上单调递减, 又e<4<5e<4<5,则ff(e)>f>f(4)>f>f(5),即1e>ln44>ln55\frac{1}{e}>\frac{ln4}{4}>\frac{ln5}{5}b>a>c\therefore b>a>c. 故选:DD