同构问题,越复杂越有思路
已知函数\(f(x)=(\ln x-2x+a)\ln x\)
\((1)\) 当\(a=2\)求\(f(x)\)的单调性
\((2)\) 若\(f(x)\leq \dfrac{e^x}{x}-x^2+ax-a\),求实数\(a\)取值范围.
\((1)\) \(a=2,f(x)=(\ln x-2x+2)\ln x\)
\(f^{\prime}(x)=\dfrac{2\ln x}{x}-2\ln x-2+\dfrac{2}{x}=\dfrac{2(1-x)(1+\ln x)}{x}\)
从而\(f(x)\)在\((0,e^{—1}),(1,+\infty)\)上单调递减,\((e^{-1},1)\)单调递增
\((2)\) 由题\(\ln ^2x-2x\ln x+x^2\leq \dfrac{e^x}{x}+a(x-1-\ln x)\)
即\((\ln x-x)^2\leq \dfrac{e^x}{x}+a(x-1-\ln x)\)
即\((x-\ln x)^2\leq e^{x-\ln x}+a(x-\ln x-1)\)
记\(x-\ln x=t,t\in[1,+\infty)\)
即\(t^2\leq e^t+a(t-1)\)
即\(a\geq \dfrac{t^2-e^t}{t-1}\)
记\(h(t)=\dfrac{t^2-e^t}{t-1},h^{\prime}(t)=-\dfrac{(t-2)(e^t-t)}{t-1}\)
则\(h(t)_{\max}=h(2)=4-e^2\)
从而\(a\geq 4-e^2\)