一题多解,换主元、隐零点

已知函数\(f(x)=\ln x-x+a-1\)

(1)若\(f(x)\leq 0\)，求\(a\)取值范围

(2)当\(a\in(0,1]\)时，证明:\(f(x)\leq \dfrac{(x-1)e^x-xe^a}{e^a}\)

解

(1) 因\(f(1)=a-2\)，则必须有\(a\leq 2\)

现说明，\(a>2\)是所求范围

\(f^{\prime}(x)=\dfrac{1}{x}-1\)，则不难得到

\(f(x)\)在\((0,1)\)上增，在\((1,+\infty)\)上减

从而\(f(x)_{\min}=f(1)=a-2\leq 0\)

综上\(a\leq 2\)

(2) \(f(x)\leq \dfrac{(x-1)e^x-xe^a}{e^a}\)

即\((\ln x-x+a-1)e^a\leq (x-1)e^x-xe^a\)

即\((\ln x+a-1)e^a\leq (x-1)e^x\)

即\(\ln x+a-1-(x-1)e^{x-a}\leq 0\)

法一：换主元

记\(h(a)=\ln x+a-1-(x-1)e^{x-a}\)

\(h^{\prime}(a)=1+(x-1)e^{x-a}=e^{x-a}\left(e^{a-x}+x-1\right)>e^{x-a}(e^{-x}+x-1)>e^{x-a}(-x+1+x-1)=0\)

从而\(h(a)\leq h(1)=\ln x-(x-1)e^{x-1}\leq (x-1)-(x-1)e^{x-1}=(x-1)(1-e^{x-1})\leq 0\)

法二：隐零点处理

记\(\varphi(x)=\ln x+a-1-(x-1)e^{x-a}\)

\(\varphi^{\prime}(x)=\dfrac{1}{x}-xe^{x-a}\)单调递减

而\(\varphi^{\prime}\left(\dfrac{1}{2}\right)=2-\dfrac{1}{2}e^{\frac{1}{2}-a}>0,\varphi^{\prime}(1)=1-e^{1-a}<0\)

从而存在唯一的\(x_0\)使得\(\varphi^{\prime}(x_0)=\dfrac{1}{x_0}-x_0e^{x_0-a}=0\)

并且\(x=x_0\)是\(\varphi(x)\)的极大值点

从而\(\varphi(x)\leq \varphi(x_0)=\ln x_0+a-1-(x_0-1)e^{x_0-a}\)

而\(\dfrac{1}{x_0}-x_0e^{x_0-a}=0\)，则\(e^{x_0-a}=\dfrac{1}{x_0^2},a=x_0+2\ln x_0\)

则\(\varphi(x_0)=3\ln x_0+x_0-1-(x_0-1)\dfrac{1}{x_0^2}=3\ln x_0+x_0-1-\dfrac{1}{x_0}+\dfrac{1}{x_0^2}\)

因\(\ln x\leq x-1\)

则\(\varphi(x_0)\leq 3(x_0-1)+x_0-1-\dfrac{1}{x_0}+\dfrac{1}{x_0^2}\leq\dfrac{(x_0-1)(2x_0-1)(2x_0+1)}{x_0^2}<0,x_0\in\left(\dfrac{1}{2},1\right)\)

得证!