高中数学选择性必修 第二册（381题）

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知识点：第五章 一元函数的导数及其应用

参考答案：先证： \(\ln {x}>\frac {2\left ( {x-1} \right )} {x+1}\left ( {x>1} \right )，\ln {x}<\frac {2\left ( {x-1} \right )} {x+1}\left ( {0<x<1} \right )\)

不妨设\(0<{x}_{1}<\frac {1} {a}{<x}_{2}\)

\(\frac {{x}_{2}} {\frac {1} {a}}=a{x}_{2}>1⇒\ln {\left ( {a{x}_{2}} \right )}>\frac {2\left ( {a{x}_{2}-1} \right )} {a{x}_{2}+1}⇔{a}^{2}{x}^{2}_{2}+\left ( {a\ln {a-a}} \right ){x}_{2}+\ln {a+2>0}\)①

\(\frac {{x}_{1}} {\frac {1} {a}}=a{x}_{1}<1⇒\ln {\left ( {a{x}_{1}} \right )}<\frac {2\left ( {a{x}_{1}-1} \right )} {a{x}_{1}+1}⇔{a}^{2}{x}^{2}_{1}+\left ( {a\ln {a-a}} \right ){x}_{1}+\ln {a+2<0}\)②

①-②：\({a}^{2}\left ( {{x}^{2}_{2}-{x}^{2}_{1}} \right )+\left ( {a\ln {a-a}} \right )\left ( {{x}_{2}-{x}_{1}} \right )>0\)则\({x_1}{\rm{ + }}{x_2} > \frac{{1 - \ln a}}{a}\)