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[函数] $h(x)=a\mathrm e^x+x^2-x-x\ln x<0$ 恒成立求实数$a$的范围

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isee Posted 2021-9-16 01:49 |Read mode
Last edited by isee 2021-9-16 08:50题:$h(x)=a\mathrm e^x+x^2-x-x\ln x<0,x>0$恒成立,求实数 $a$ 的范围.





















这个导数压轴题,应该是高三模拟卷上的,比较棘手.

由$h(1)=a\mathrm e<0$ 知 $a<0$ ,熟知 $\mathrm e^x\geqslant x+1$ ,于是
\begin{align*}
&a\mathrm e^x+x^2-x-x\ln x<0\\
&\iff \frac {-a\mathrm e^x}{x}+\ln x-x+1>0\\
&\iff \frac {\mathrm e^{\ln (-a)}\mathrm e^x}{\mathrm e^{\ln x}}+\ln x-x+1>0\\
&\iff \mathrm e^{x-\ln x+\ln (-a)}+\ln x-x+1>0\\
\iff \mathrm e^{\color{red}{x-\ln x+\ln (-a)}}+\ln x-x+1
&\geqslant {\color{red}{x-\ln x+\ln (-a)}}+1+(\ln x-x+1)\\
&=\ln (-a)+2>0\\
&\iff a<-\frac 1{\mathrm e^2}.
\end{align*}

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oreoes Posted 2021-9-18 15:38
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 Author| isee Posted 2021-9-18 15:53
回复 2# oreoes

全分参,而且都将$x-\ln x$作为整体的全分参,也是厉害的。

另外:论坛的公式代码说明:forum.php?mod=viewthread&tid=5  (本论坛的最厉害的地方)

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