隐零点代换，短除因式分解，计算量偏大

已知\(f(x)=\ln x-ax+a,g(x)=(x-1)e^{x-a}-ax+1\)

(1)若\(f(x)\leq 0\)，求\(a\)

(2)当\(a\in(0,1]\)时，证明:\(g(x)\geq f(x)\)

解

(1)\(\ln x-ax+a\leq 0\)

记\(F(x)=\ln x-ax+a,F^{\prime}(x)=\dfrac{1}{x}-a\)

当\(a\leq 0\)时，\(F(x)\)单调递增，而\(F(1)=0\)，从而\(x>1,F(x)>0\)不合题

当\(0<a<1\)时，\(\dfrac{1}{a}>1\)，则\(F(x)\)在\(\left(0,\dfrac{1}{a}\right)\)上增，在\(\left(\dfrac{1}{a},+\infty\right)\)上减

即\(F(x)\)在\(\left(1,\dfrac{1}{a}\right)\)上增，因\(F(1)=0\)

则\(F(x)\)在\(x\in\left(1,\dfrac{1}{a}\right)\)上正，不合题

当\(a>1\)时，同样得到\(F(x)\)在\(\left(\dfrac{1}{a},1\right)\)上减，而\(F(1)=0\)

则\(F(x)\)在\(\left(\dfrac{1}{a},1\right)\)上正，不合题

当\(a=1\)时，因\(y=x-1\)与\(y=\ln x\)相切，合题

综上\(a=1\)

(2)\(\ln x+a-(x-1)e^{x-a}-1\leq 0\)

记\(\varphi(x)=\ln x+a-(x-1)e^{x-a}-1,\varphi^{\prime}(x)=\dfrac{1}{x}-xe^{x-a}\)单调递减

而\(x\to 0\),\(\varphi^{\prime}\to+\infty,x\to+\infty,\varphi^{\prime}(x)\to -\infty\)

则存在唯一的\(x_0\)使\(\varphi^{\prime}(x_0)=0\)，并且\(x=x_0\)是\(\varphi(x)\)的极大值

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

因\(\dfrac{1}{x_0}-x_0e^{x_0-a}=0\)，即\(\dfrac{1}{x^2_0}=e^{x_0-a}\)，即\(x_0+2\ln x_0=a\)

则\(\varphi(x_0)=\ln x_0+x_0+2\ln x_0-\dfrac{x_0-1}{x^2_0}-1\)

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

记\(\gamma(x)=3\ln x+x-\dfrac{x-1}{x^2}-1,x\in\left(\dfrac{1}{2},1\right)\)

\(\gamma^{\prime}(x)=\dfrac{x^3+3x^2+x-2}{x^3}\)

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

\(=\dfrac{(x+2)\left(x+\dfrac{1+\sqrt{5}}{2}\right)\left(x-\dfrac{\sqrt{5}-1}{2}\right)}{x^3}\)

则\(\gamma(x)\)在\(\left(\dfrac{1}{2},\dfrac{\sqrt{5}-1}{2}\right)\)减，\(\left(\dfrac{\sqrt{5}-1}{2},1\right)\)增

而\(\gamma\left(\dfrac{1}{2}\right)=-3\ln 2+\dfrac{1}{2}+2-1=\dfrac{3}{2}-3\ln 2<0,\gamma(1)=-1<0\)

则\(\varphi(x)_{\max}=\varphi(x_0)=\gamma(x)<0\)，得证！