函数同构

发布时间 2023-04-22 22:24:06作者: Lucius7

\(1\):若对 \(\forall x > 0\),恒有 \(a(e^{ax}+1) \ge 2 (x + \frac{1}{x}) \ln x\),则实数 \(a\) 的最小值为:\(\frac{2}{e}\)

\((e^{ax} + 1) ax \ge (x^2 +1) \ln x^2 \Rightarrow(e^{ax} + 1) \ln e^{ax} \ge (x^2 +1) \ln x^2\)

\(f(x) = (x + 1) \ln x\)\(f'(x) = \ln x + \frac{x + 1}{x}\)。令 \(g(x) = \ln x + \frac{x + 1}{x}\)\(g'(x) = \frac{x-1}{x^2}\)

\(\therefore g'(x)>0 \Leftrightarrow x>1, g'(x) < 0 \Leftrightarrow 0<x<1\)\(g(x)\)\((0, 1)\)\(\downarrow\),在 \((1, +\infty)\)\(\uparrow\)

\(\therefore g(x)_{\min} = g(1) = 2\) \(\therefore f’(x) \ge 2\)\(f(x)\)\((0, +\infty)\)\(\uparrow\)

\(\therefore e^{ax} \ge x^2 \Rightarrow a \ge \frac{2 \ln x}{x}\)

\(h(x) = \frac{2 \ln x}{x}\)\(h'(x) = \frac{2 - 2 \ln x}{x^2}\)

\(\therefore h'(x)>0 \Leftrightarrow 0<x<e, h'(x) < 0 \Leftrightarrow x > e\)\(h(x)\)\((0, e)\)\(\uparrow\),在 \((e, +\infty)\)\(\downarrow\)

\(\therefore h(x)_{\max}=h(e)=\frac{2}{e}\) \(\therefore a_\min=\frac{2}{e}\)

\(2\):已知函数 \(f(x)=e^x-a\ln(ax-a)+a(a>0)\),若关于 \(x\) 的不等式 \(f(x)>0\) 恒成立,则实数 \(a\) 的取值范围为:\(a<e^2\)

\[\begin{align} &e^{x - \ln a} > \ln(ax-a) - 1 \\ \Rightarrow &e^{x - \ln a} - \ln a > \ln(x - 1) - 1 \\ \Rightarrow &e^{x - \ln a} + (x - \ln a) > (x - 1) + \ln (x - 1) \\ \Rightarrow &e^{x - \ln a} + (x - \ln a) > e^{\ln(x - 1)} + \ln (x - 1)\\ \end{align} \]

\(g(x) = e^x + x\)\(g(x) \uparrow\)

\(\therefore x - \ln a > \ln (x - 1) \Rightarrow -\ln a > \ln (x - 1) - x\)

\(\because \ln x \le x - 1\)(当且仅当 \(x=1\) 时取等)\(\therefore \ln (x - 1) \le x - 2\)(当且仅当 \(x = 2\) 时取等)

\(\therefore \ln (x - 1) - x \le - 2\)(当且仅当 \(x = 2\) 时取等)

\(x = 2\)\(-\ln a > -2 \Rightarrow a \in (0, e^2)\)