当参变分离遇见洛必达

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我校2016$\thicksim$2017学年度(上期)半期高三(理科)考试第21题

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已知函数\(\textit{f}(\textit{x})=\dfrac{(2\textit{x}-1)\;\!\text{e}^{2\textit{x}}}{4}-\dfrac{\textit{m}}{2}\textit{x}^2\)和函数\(\textit{g}(\textit{x})=\textit{x}\ln\textit{x}\) (其中\(\text{e}=2.71828\cdots\)).

(1)若函数\(\textit{f}(\textit{x})\)有两个零点，求\(\textit{m}\)的取值范围；

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【分析】函数\(f(x)\)有两个零点\(\Rightarrow\)方程\(m=\dfrac{(2x-1)\text{e}^{2x}}{2x^2}\)有两根

\(\Rightarrow y=m\)与\(y=\dfrac{(2x-1)\text{e}^{2x}}{2x^2}\)有两个交点

\(\Rightarrow h(t)=\dfrac{2(t-1)\text{e}^{t}}{t^2}(t=2x)\Rightarrow h'(t)=\dfrac{2(t^2-2t+2)\text{e}^t}{t^3}\)

(注意：求导的快、准、狠的技巧)

\(\Rightarrow\)函数\(y=\dfrac{(2x-1)\text{e}^{2x}}{2x^2}\)的草图

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显然，当\(x\rightarrow -\infty\)时，\(y=\dfrac{(2x-1)\text{e}^{2x}}{2x^2}\rightarrow 0\)(洛必达法则)\(\Rightarrow m<0\)

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(还可以，\(x\rightarrow -\infty\)，\(y=\dfrac{(2x-1)\text{e}^{2x}}{2x^2}=\dfrac{2+\frac{1}{-x}}{2x\text{e}^{-2x}}\rightarrow 0\))

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【练习】已知函数\(f(x)=(x-2)\text{e}^x+a(x-1)^2\)有两个零点.

(1)求\(a\)的取值范围；(2016年新课标卷I理21题)

转载于:https://www.cnblogs.com/xuebajunlutiji/p/6045671.html