隱零點代換，短除因式分解，計算量偏大
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)當\(a\in(0,1]\)時，證明:\(g(x)\geq f(x)\)
解
(1)\(\ln x-ax+a\leq 0\)
記\(F(x)=\ln x-ax+a,F^{\prime}(x)=\dfrac{1}{x}-a\)
當\(a\leq 0\)時，\(F(x)\)單調遞增，而\(F(1)=0\)，從而\(x>1,F(x)>0\)不合題
當\(0<a<1\)時，\(\dfrac{1}{a}>1\)，則\(F(x)\)在\(\left(0,\dfrac{1}{a}\right)\)上增，在\(\left(\dfrac{1}{a},+\infty\right)\)上減
即\(F(x)\)在\(\left(1,\dfrac{1}{a}\right)\)上增，因\(F(1)=0\)
則\(F(x)\)在\(x\in\left(1,\dfrac{1}{a}\right)\)上正，不合題
當\(a>1\)時，同樣得到\(F(x)\)在\(\left(\dfrac{1}{a},1\right)\)上減，而\(F(1)=0\)
則\(F(x)\)在\(\left(\dfrac{1}{a},1\right)\)上正，不合題
當\(a=1\)時，因\(y=x-1\)與\(y=\ln x\)相切，合題
綜上\(a=1\)
(2)\(\ln x+a-(x-1)e^{x-a}-1\leq 0\)
記\(\varphi(x)=\ln x+a-(x-1)e^{x-a}-1,\varphi^{\prime}(x)=\dfrac{1}{x}-xe^{x-a}\)單調遞減
而\(x\to 0\),\(\varphi^{\prime}\to+\infty,x\to+\infty,\varphi^{\prime}(x)\to -\infty\)
則存在唯一的\(x_0\)使\(\varphi^{\prime}(x_0)=0\)，並且\(x=x_0\)是\(\varphi(x)\)的極大值
\(\varphi(x_0)=\ln x_0+a-(x_0-1)e^{x_0-a}-1\)
因\(\dfrac{1}{x_0}-x_0e^{x_0-a}=0\)，即\(\dfrac{1}{x^2_0}=e^{x_0-a}\)，即\(x_0+2\ln x_0=a\)
則\(\varphi(x_0)=\ln x_0+x_0+2\ln x_0-\dfrac{x_0-1}{x^2_0}-1\)
因\(\varphi^{\prime}(1)=1-e^{-a}<0,\varphi^{\prime}\left(\dfrac{1}{2}\right)=2-\dfrac{e^{\frac{1}{2}-a}}{2}>0\)
記\(\gamma(x)=3\ln x+x-\dfrac{x-1}{x^2}-1,x\in\left(\dfrac{1}{2},1\right)\)
\(\gamma^{\prime}(x)=\dfrac{x^3+3x^2+x-2}{x^3}\)
\(=\dfrac{(x+2)(x^2+x-1)}{x^3}\)
\(=\dfrac{(x+2)\left(x+\dfrac{1+\sqrt{5}}{2}\right)\left(x-\dfrac{\sqrt{5}-1}{2}\right)}{x^3}\)
則\(\gamma(x)\)在\(\left(\dfrac{1}{2},\dfrac{\sqrt{5}-1}{2}\right)\)減，\(\left(\dfrac{\sqrt{5}-1}{2},1\right)\)增
而\(\gamma\left(\dfrac{1}{2}\right)=-3\ln 2+\dfrac{1}{2}+2-1=\dfrac{3}{2}-3\ln 2<0,\gamma(1)=-1<0\)
則\(\varphi(x)_{\max}=\varphi(x_0)=\gamma(x)<0\)，得證！