特别典型的一道端点效应与放缩找点

已知\(f(x)=m(x-1)^2-2x+2\ln x,m>2\)

(1)证明：函数\(f(x)\)存在单调递减区间，并求出该函数单调递减区间\((a,b)\)的长度\(b-a\)的取值范围

(2)当\(x\geq 1\)时，\(f(x)\leq 2xe^{x-1}-4x\)恒成立，求\(m\)的取值范围.

解

(1)\(f^{\prime}(x)=2m(x-1)-2+\dfrac{2}{x}=2\cdot\dfrac{mx^2-(1+m)x+1}{x}=2\cdot\dfrac{(mx-1)(x-1)}{x}\)

\(f^{\prime}(x)=0\)得\(x_1=\dfrac{1}{m},x_2=1\)，因\(m>2\)，则\(0<x_1<x_2\)

从而得到\(f(x)\)在\((x_1,x_2)\)上单调递减

则\(a=\dfrac{1}{m},b=1\)

则\(b-a=1-\dfrac{1}{m}\in\left(\dfrac{1}{2},1\right)\)

(2) 原不等式为\(m(x-1)^2-2x+2\ln x-2xe^{x-1}+4x\leq 0\)

记\(\varphi(x)=m(x-1)^2+2\ln x-2xe^{x-1}+2x\)，发现\(\varphi(1)=0\)

\(\varphi^{\prime}(x)=2m(x-1)+\dfrac{2}{x}-2e^{x-1}(1+x)+2\)，发现\(\varphi^{\prime}(1)=0\)

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

\(\varphi^{\prime\prime\prime}(x)=\dfrac{4}{x^3}-2e^{x-1}(x+3)\)，其是单调递减的

则\(\varphi^{\prime\prime\prime}(x)<\varphi^{\prime\prime\prime}(1)=-4<0\)

则\(\varphi^{\prime\prime}(x)\)单调递减

并且\(\varphi^{\prime\prime}(1)=2m-8\)

Case1 若\(m>4\)时，\(\varphi^{\prime\prime}(1)>0\)

因\(\varphi^{\prime\prime}(x)<8-\dfrac{2}{x^2}-2x(x+2)=8-\dfrac{2}{x^2}-2x^2-4x\)

取\(x=2\)，有\(\varphi^{\prime\prime}(2)<-\dfrac{17}{2}<0\)

则由零点存在定理，在\(x\in(1,2)\)中，一定存在\(x_0\)，使得\(\varphi^{\prime\prime}(x_0)=0\)

从而在\(x\in(1,x_0)\)上，\(\varphi^{\prime}(x)\)单调递增，则\(\varphi^{\prime}(x)>0\)

从而在\(x\in(1,x_0)\)上，\(\varphi(x)>\varphi(1)=0\)，不合题

Case2 若\(m\leq 4\)时，\(\varphi^{\prime}(x)<\varphi^{\prime\prime}(1)<0\)

则\(\varphi^{\prime}(x)\)单调递减，从而\(\varphi^{\prime}(x)<\varphi^{\prime}(1)=0\)

则\(\varphi(x)\)单调递减，从而\(\varphi(x)<\varphi(1)=0\)，合题

综上\(m\in(2,4]\)