# 构造函数的技巧：什么样的式子求导可能会产生 1 阶导和 0 阶导？

## 一、题目

A. $f(-1)>2$

C. $f(1)>2 \mathrm{e}^{2}$

B. $f(-1)<\frac{2}{\mathrm{e}^{2}}$

D. $f(1)<2 \mathrm{e}^{2}$

## 二、解析

$$f^{\prime}(x)<2 f(x) \Rightarrow f^{\prime}(x)-2 f(x)<0$$

$$\left[e^{x} f(x)\right]^{\prime} = e^{x}\left[ \textcolor{orangered}{f^{\prime}(x)} + \textcolor{springgreen}{f(x)} \right]$$

$$\left[e^{-2 x} f(x)\right]^{\prime}=e^{-2 x}\left[ \textcolor{orangered}{f^{\prime}(x)} \textcolor{springgreen}{- 2 f(x)} \right]$$

$$\varphi(x)=e^{-2 x} f(x)$$

$$\varphi^{\prime}(x)=e^{-2 x}\left[f^{\prime}(x)-2 f(x)\right]$$

$$\begin{cases} e^{-2 x}>0 \\ f^{\prime}(x)-2 f(x)<0 \end{cases}$$

$$\varphi^{\prime}(x)<0$$

$$-1 \textcolor{springgreen}{<} 0 \textcolor{springgreen}{<} 1 \Rightarrow$$

$$\varphi(-1) \textcolor{orangered}{>} \varphi(0) \textcolor{orangered}{>} \varphi(1)$$

$$\varphi(0)=1 \times f(0)=1 \times 2=2$$

$$\varphi(-1)>2>\varphi(1) \Rightarrow$$

$$e^{2} f(-1)>2>e^{-2} f(1)$$

$$\begin{cases} \textcolor{springgreen}{f(-1)>\frac{2}{e^{2}}} \\ f(1)<\frac{2}{e^{-2}} \Rightarrow \textcolor{springgreen}{f(1) < 2 e^{2} } \end{cases}$$