明天就要高考去了，今天临时抱个佛脚
Let \(a, b\) be real numbers, and \(a>1\). Consider the function \(f(x)=a^x-bx+e^2\) (\(x\in\mathbb{R}\)).
Note:
设 \(a, b\) 为实数，且 \(a>1\)，函数 \(f(x)=a^x-bx+e^2\) (\(x\in\mathbb{R}\))
注：\(e=2.71828\cdots\) 是自然对数的底数。
The derivative of \(f(x)\) is
\[f'(x)=a^x\ln a-b\]\(f''(x)=a^x\ln^2 a>0\), so \(f'(x)\) is an strictly increasing function.
When \(b<0\), \(f'(x)>\lim_{x\to-\infty}f'(x)=0\), so \(f(x)\) is an strictly increasing function.
When \(b\ge 0\), solving \(f'(x)=0\) we can get \(x=\log_a\frac{b}{\ln a}\). Since \(f'(x)\) is increasing, we have:
| $x$ | \(x\in\left(-\infty, \log_a\frac{b}{\ln a}\right)\) | \(x=\log_a\frac{b}{\ln a}\) | \(x\in\left(\log_a\frac{b}{\ln a}, +\infty\right)\) |
|---|---|---|---|
| \(f'(x)\) | \(f'(x)<0\) | \(f'(x)=0\) | \(f'(x)>0\) |
| \(f(x)\) | \(f(x)\) is decreasing | \(f(x)\) has a local minimum | \(f(x)\) is increasing |
As a result, when \(b<0\), \(f(x)\) has a monotonically increasing interval \((-\infty, +\infty)\); when \(b\ge 0\), \(f(x)\) has a monotonically decreasing interval \(\left(-\infty, \log_a\frac{b}{\ln a}\right)\) and a monotonically increasing interval \(\left(\log_a\frac{b}{\ln a}, +\infty\right)\).
Given that \(b>2e^2>0\), the function \(f(x)\) has a local minimum at \(x=\log_a\frac{b}{\ln a}\).
Also, we have:
\[\begin{aligned} \lim_{x\to-\infty}f(x)&=\lim_{x\to-\infty}a^x-bx+e^2=\infty\\ \lim_{x\to+\infty}f(x)&=\lim_{x\to+\infty}a^x-bx+e^2=\infty \end{aligned}\]Therefore, \(f(x)\) has two distinct zeros if and only if \(f\left(\log_a\frac{b}{\ln a}\right)<0\).
Let’s simplify \(f\left(\log_a\frac{b}{\ln a}\right)\):
\[\begin{aligned} f\left(\log_a\frac{b}{\ln a}\right)&=a^{\log_a\frac{b}{\ln a}}-b\log_a\frac{b}{\ln a}+e^2\\ &=\frac{b}{\ln a}-b\cdot\left(\log_ab-\log_a\ln a\right)+e^2\\ &=\frac{b}{\ln a}-\frac{b\ln b}{\ln a}+\frac{b\ln\ln a}{\ln a} + e^2\\ &=\frac{b}{\ln a}\left(1-\ln b+\ln\ln a\right)+e^2 \end{aligned}\]For \(f\left(\log_a\frac{b}{\ln a}\right)<0\) to hold, we must ensure:
\[\lim_{b\to \left(2e^{2}\right)^+}f\left(\log_a\frac{b}{\ln a}\right)\le 0\]This implies:
\[\begin{aligned} 0&\ge \lim_{b\to {\left(2e^{2}\right)}^+}f\left(\log_a\frac{b}{\ln a}\right)\\ &=\lim_{b\to \left(2e^{2}\right)^+}\frac{b}{\ln a}\left(1-\ln b+\ln\ln a\right)+e^2\\ &=\frac{2e^2}{\ln a}\left(1-\ln\left(2e^2\right)+\ln\ln a\right)+e^2\\ &=\frac{2e^2}{\ln a}\left(\ln\ln a-1-\ln 2\right)+e^2\\ \end{aligned}\]Hence, we should have:
\[2e^2\left(\ln\ln a-1-\ln 2\right)+e^2\ln a\le 0\]This simplifies to:
\[2\ln\ln a+\ln a\le 2\ln 2+2\]Let’s define \(\phi(x)=2\ln x+x\). The inequality above is equivalent to:
\[\phi(\ln a)\le \phi(2)\]Given that \(\phi(x)\) is a strictly increasing function, we can have:
\[\ln a\le 2\Rightarrow a\le e^2\]However, this is not the final answer since we should consider the whole range of \(b>2e^2\) instead of a specific value of \(b\to \left(2e^2\right)^+\). But we have a necessary condition for \(a\) to satisfy.
From \(f\left(\log_a\frac{b}{\ln a}\right)<0\), we can get:
\[\frac{b}{\ln a}(1-\ln b+\ln\ln a)+e^2<0\]Let
\[g(a, b)=\frac{b}{\ln a}(1-\ln b+\ln\ln a)+e^2\]We have:
\[\frac{\partial g}{\partial b}=\frac{1}{\ln a}(1-\ln b+\ln\ln a)-\frac{b}{\ln a}\cdot\frac{1}{b}=\frac{\ln\ln a-\ln b}{\ln a}\]When \(a\le e^2, b>2e^2\),
\[\ln\ln a-\ln b\le \ln\ln e^2-\ln 2e^2=\ln 2-\ln 2-2=-2<0\]So \(\frac{\partial g}{\partial b}<0\) which means when \(a\le e^2\),
\[g(a, b) < g(a, 2e^2)\le 0\]So when \(a\le e^2\), \(f(x)=0\) must have two distinct solutions.
In conclusion, the range of \(a\) is \(\left(1, e^2\right]\).
When \(a=e\in(1, e^2], b>e^4>2e^2\), so \(f(x)=0\) has two distinct solutions.
Let \(x_1, x_2\) be the two solutions, and \(x_1<x_2\):
\[\begin{cases} e^{x_1}-bx_1+e^2=0 & (1)\\ e^{x_2}-bx_2+e^2=0 & (2) \end{cases}\]\((1) - (2)\), we have:
\[e^{x_1}-e^{x_2}-b(x_1-x_2)=0\Rightarrow \frac{e^{x_1}-e^{x_2}}{x_1-x_2}=b\]然后不会了啊啊啊啊啊啊啊啊啊啊啊啊啊啊啊，有空再来更