# 2024年考研数二第21题解析：证明绝对值式子小于XX，需要“两头围堵”

## 二、解析

### 第 (1) 问 | 解法一

$$|f(x) – f(0)(1-x)-f(1) x| \leq \frac{x(1-x)}{2}$$

$$\textcolor{springgreen}{- \frac{x(1-x)}{2}} \leq f(x) – f(0)(1-x)-f(1) x \leq \textcolor{orangered}{\frac{x(1-x)}{2}}$$

\begin{aligned} g(x) \\ & = f(0)(1-x) + x f(1) \\ & = f(0) – xf(0) + x f(1) \end{aligned}

$$g^{\prime}(x) = -f(0) + f(1) \Rightarrow \text{常数}$$

$$\begin{cases} g(0) = f(0) \\ g(1) = f(1) \end{cases}$$

$$F(x)=f(x)-g(x)-\frac{x(1-x)}{2}, \quad x \in(0,1)$$

$$\textcolor{orangered}{ F(x) \leq 0 }$$

$F(x)$ 在闭区间 $[0,1]$ 端点处的取值情况：

\begin{aligned} F(0) \\ & = f(0) – g(0) – 0 \\ & = f(0) – f(0) \\ & = 0 \end{aligned}

\begin{aligned} F(1) \\ & = f(1) – g(1) – 0 \\ & = f(1) – f(1) \\ & = 0 \end{aligned}

【一阶导】

\begin{aligned} F^{\prime}(x) \\ & = f^{\prime} (x) – g^{\prime} (x) – \frac{1}{2} (1 – 2x) \\ \\ & = f^{\prime} (x) + x + \text{常数} \end{aligned}

【二阶导】

\begin{aligned} F^{\prime \prime}(x) \\ & = f^{\prime \prime}(x) + 1 \\ \\ & \underrightarrow{\left|f^{\prime \prime}(x)\right| \leq 1 \quad } \quad F^{\prime \prime}(x) \geq 0 \end{aligned}

$$\textcolor{orangered}{ F(x) \leq 0 }$$

$$\textcolor{springgreen}{ f(x)-f(0)(1-x)-f(1) x \leq \frac{x(1-x)}{2} }$$

$$– \frac{x(1-x)}{2} \leq f(x) – f(0)(1-x)-f(1) x$$

$$F(x)=f(x)-g(x)+\frac{x(1-x)}{2}, \quad x \in(0,1)$$

$$\begin{cases} F(0)=0 \\ F(1)=0 \end{cases}$$

$$F^{\prime \prime}(x)=f^{\prime \prime}(x)-1$$

$$\left|f^{\prime \prime}(x)\right| \leq 1$$

$$F^{\prime \prime}(x) \geq 0$$

$$\textcolor{orangered}{ F(x) \geq 0 }$$

$$\therefore f(x)-f(0)(1-x)-f(1) x \geq-\frac{x(1-x)}{2}$$

$$|f(x)-f(0)(1-x)-f(1) x| \leq \frac{x(1-x)}{2}$$