简单的参变分离

已知函数$ f(x)=\dfrac{\ln x}{x^2}$

(1)讨论\(f(x)\)的最值；

(2)若函数\(g(x)=e^x+x^4f(x)-x^2-ax\)有两个零点，求\(a\)的范围

解

(1)\(f(x)=\dfrac{1-2\ln x}{x^3}\)

得\(f^{\prime}\left(\sqrt{e}\right)=0\)

则\(f(x)\)在\(\left(0,\sqrt{e}\right)\)上增，在\(\left(\sqrt{e},+\infty\right)\)上减

从而\(f(x)_{\max}=f(\sqrt{e})=\dfrac{1}{2e}\)

(2)\(g(x)=e^x+x^2\ln x-x^2-ax=e^x-ax+x^2(\ln x-1)=0\)

即\(a=\dfrac{e^x}{x}+x(\ln x-1)\)

记\(\varphi(x)=\dfrac{e^x}{x}+x(\ln x-1),\varphi^{\prime}(x)=\dfrac{e^x(x-1)}{x^2}+\ln x\)

不难发现\(0<x<1\)时，\(\varphi^{\prime}(x)<0,x>0\varphi^{\prime}(x)>0,x=1,\varphi^{\prime}(1)=0\)

则\(\varphi(x)\)在\((0,1)\)上减，在\((1,+\infty)\)上增，\(\varphi(x)_{\min}=\varphi(1)=e+1\)

则使得\(y=a\)与\(y=\varphi(x)\)有两个交点，要有\(a>e+1\)