2022考研数二第21题解析：泰勒公式、函数的单调性、拉格朗日中值定理、定积分、不等式的证明、洛必达、极限的保号性

一、题目

难度评级：

二、解析

解法 1：泰勒公式+反证法

$\textcolor{lightgreen}{\blacktriangleright}$ 首先，由泰勒公式，得：

$$
f\left(x\right) = f\left(\frac{a + b}{2}\right) + f^{\prime}\left(\frac{a + b}{2}\right)\left(x – \frac{a + b}{2}\right) + \frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}
$$

其中，$\xi$ 介于 $x$ 与 $\frac{a + b}{2}$ 之间.

于是：

$$
\begin{aligned}
\int_{a}^{b} f\left(x\right) \mathrm{~d} x & = \int_{a}^{b} \left[f\left(\frac{a + b}{2}\right) + f^{\prime}\left(\frac{a + b}{2}\right)\left(x – \frac{a + b}{2}\right) + \frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x \\ \\
& = \int_{a}^{b} f\left(\frac{a + b}{2}\right) \mathrm{~d} x + \int_{a}^{b} f^{\prime}\left(\frac{a + b}{2}\right)\left(x – \frac{a + b}{2}\right) \mathrm{~d} x + \int_{a}^{b} \left[ \frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2} \right] \mathrm{~d} x
\end{aligned}
$$

其中：

$$
\begin{aligned}
\int_{a}^{b} f\left(\frac{a + b}{2}\right) \mathrm{~d} x & = f \left( \frac{a+b}{2} \right) \int_{a}^{b} 1 \mathrm{~d} x \\ \\
& = f\left(\frac{a + b}{2}\right)\left(b – a\right) \\ \\ \\
\int_{a}^{b} f^{\prime}\left(\frac{a + b}{2}\right)\left(x – \frac{a + b}{2}\right) \mathrm{~d} x & = f^{\prime}\left(\frac{a + b}{2}\right) \int_{a}^{b} \left(x – \frac{a + b}{2}\right) \mathrm{~d} x \\ \\
& = f^{\prime}\left(\frac{a + b}{2}\right) \left[ \int_{a}^{b} x \mathrm{~d} x – \frac{a+b}{2} \cdot \left( b-a \right) \right] \\ \\
& = f ^{\prime} \left( \frac{a+b}{2} \right) \left( \frac{1}{2} x^{2} \Bigg|_{a}^{b} – \frac{b^{2} – a^{2}}{2} \right) \\ \\
& = f ^{\prime} \left( \frac{a+b}{2} \right) \left( \frac{b^{2} – a^{2}}{2} – \frac{b^{2} – a^{2}}{2} \right) \\ \\
& = 0
\end{aligned}
$$

所以：

$$
\textcolor{pink}{
\int_{a}^{b} f\left(x\right) \mathrm{~d} x = f\left(\frac{a + b}{2}\right)\left(b – a\right) + \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x
}
$$

因此：

$$
\begin{aligned}
& \ \frac{1}{b-a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x = f\left(\frac{a + b}{2}\right) + \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightblue}{
f\left(\frac{a + b}{2}\right) = \frac{1}{b-a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x } – \textcolor{orange}{ \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x
}
\end{aligned}
$$

$\textcolor{orangered}{\blacktriangleright}$ （正证法）证明由 $f^{\prime \prime}\left(x\right) \geqslant 0$ 可以推导出 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$:

若 $f^{\prime \prime}\left(x\right) \geqslant 0$，则：

$$
\begin{aligned}
& \ f^{\prime \prime}\left(\xi\right) \geqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{1}{2} f ^{\prime \prime} \left( \xi \right) \left( x – \frac{a+b}{2} \right)^{2} \geqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \begin{cases}
b \geqslant a \rightarrow \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x \geqslant 0 \\
b < a \rightarrow \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x > 0
\end{cases} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{orange}{ \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x \geqslant 0 }
\end{aligned}
$$

又由前面的推导可知：

$$
\textcolor{lightblue}{
f\left(\frac{a + b}{2}\right) = \frac{1}{b-a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x } – \textcolor{orange}{ \frac{1}{b-a} \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x
}
$$

所以：

$$
\textcolor{lightgreen}{
f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x
}
$$

综上可知，由 $f^{\prime \prime}\left(x\right) \geqslant 0$ 可以推导出 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$.

$\textcolor{orangered}{\blacktriangleright}$ （反证法）证明由 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$ 可以推导出 $f^{\prime \prime}\left(x\right) \geqslant 0$:

根据「荒原之梦考研数学」的《峰图 | 从包含关系的视角理解用反证法逆转充分条件和必要条件的证明逻辑》这篇文章可知，要证明由 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$ 可以推导出 $f^{\prime \prime}\left(x\right) \geqslant 0$, 可以从 $f^{\prime \prime}\left(x\right) \geqslant 0$ 的对立事件入手，用反证法证明：

假设存在 $x_{0}$ 使得 $f^{\prime \prime}\left(x_{0}\right) < 0$，因为 $f\left(x\right)$ 有二阶连续导数，故存在 $\delta > 0$ 使得 $f^{\prime \prime}\left(x\right)$ 在 $\left[x_{0} – \delta, x_{0} + \delta\right]$ 内恒小于零.

又由前面的推导可知：

$$
\textcolor{pink}{
\int_{a}^{b} f\left(x\right) \mathrm{~d} x = f\left(\frac{a + b}{2}\right)\left(b – a\right) + \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right)\left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x
}
$$

于是，若记 $a = x_{0} – \delta, b = x_{0} + \delta$, 则：

$$
\int_{a}^{b} f\left(x\right) \mathrm{~d} x = f\left(\frac{a + b}{2}\right)\left(b – a\right) + \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right) \left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x
$$

其中：

$$
\begin{aligned}
& \ f ^{\prime \prime} \left( \xi \right) < 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{1}{2} f^{\prime \prime}\left(\xi\right) \left(x – \frac{a + b}{2}\right)^{2} < 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \int_{a}^{b} \left[\frac{1}{2} f^{\prime \prime}\left(\xi\right) \left(x – \frac{a + b}{2}\right)^{2}\right] \mathrm{~d} x < 0
\end{aligned}
$$

于是：

$$
\begin{aligned}
& \ \int_{a}^{b} f\left(x\right) \mathrm{~d} x < f\left(\frac{a + b}{2}\right)\left(b – a\right) \\ \\ \textcolor{lightgreen}{ \leadsto } & \ f\left(\frac{a + b}{2}\right) > \frac{1}{b-a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x
\end{aligned}
$$

由于上面得到的 $f\left(\frac{a + b}{2}\right) > \frac{1}{b-a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$ 这个结论，和已知的 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$ 这个结论矛盾，所以 $f^{\prime \prime}\left(x_{0}\right) < 0$ 这个假设不成立，因此，其对立事件一定成立：

$$
f^{\prime \prime} \left(x\right) \geqslant 0
$$

综上可知，充分性必要性均得证.

解法 2：构造函数+函数的单调性+一点处导数的定义+洛必达+极限的保号性

$\textcolor{orangered}{\blacktriangleright}$ （正证法）证明由 $f^{\prime \prime}\left(x\right) \geqslant 0$ 可以推导出 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$:

首先，根据 $f^{\prime \prime}\left(x\right) \geqslant 0$ 构造函数 $F \left( x \right)$:

$$
\begin{aligned}
& \ f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f\left(\frac{a + b}{2}\right) – \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ F\left( x \right) = \left(x – a\right) f \left(\frac{a+x}{2}\right) – \int_{a}^{x} f \left(t\right)\mathrm{~d}t
\end{aligned}
$$

同时可知：

$$
F \left( a \right) = 0
$$

进而可知：

$$
\begin{aligned}
F^{\prime} \left(x\right) & = f\left(\frac{a+x}{2}\right) + \frac{1}{2} \left( x – a \right) f^{\prime}\left(\frac{a+x}{2}\right) – f\left(x\right) \\ \\
& = \frac{1}{2} \left(x-a\right) f^{\prime} \left(\frac{a+x}{2} \right) + f \left( \frac{a+x}{2} \right) – f \left( x \right)
\end{aligned}
$$

又由拉格朗日中值定理可知，存在 $\xi \in \left( \frac{a+x}{2}, x \right)$, 使得下式成立：

$$
\begin{aligned}
& \ f ^{\prime} \left( \xi \right) = \left[ f \left( \frac{a+x}{2} \right) – f \left( x \right) \right] \cdot \frac{1}{\frac{a+x-2x}{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f ^{\prime} \left( \xi \right) = \left[ f \left( \frac{a+x}{2} \right) – f \left( x \right) \right] \cdot \frac{1}{\frac{a-x}{2}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f \left( \frac{a+x}{2} \right) – f \left( x \right) = f ^{\prime} \left( \xi \right) \frac{1}{2} \left( a-x \right)
\end{aligned}
$$

于是：

$$
\begin{aligned}
F ^{\prime} \left( x \right) & = \frac{1}{2} \left( x-a \right) f^{\prime} \left(\frac{a+x}{2} \right) + f^{\prime} \left( \xi \right)\frac{1}{2} \left( a-x \right) \\ \\
& = \frac{1}{2} \left( x-a \right) f^{\prime} \left(\frac{a+x}{2} \right) – f^{\prime} \left( \xi \right)\frac{1}{2} \left( x – a \right) \\ \\
& = \frac{1}{2}\left( x – a \right) \left[f^{\prime}\left(\frac{a+x}{2}\right) – f^{\prime}\left(\xi\right)\right]
\end{aligned}
$$

又由于 $f^{\prime\prime}\left(x\right) \geqslant 0$, 所以 $f^{\prime} \left( x \right)$ 单调递增，于是可知：

$$
f^{\prime}\left(\frac{a+x}{2}\right) < f^{\prime}\left(\xi\right)
$$

故 $F^{\prime}\left(x\right)<0$，$F\left(x\right)$ 单调递减.

根据《什么是“不失一般性地假设”？什么是“不妨设”？》这篇文章可知，在此，我们不妨设：

$$
a > b
$$

于是可知：

$$
x \in \left( a, b \right) \leadsto x > a
$$

又因为：

$$
F \left( a \right) = 0
$$

所以：

$$
F\left(x\right) \leqslant F \left( a \right) = 0 \leadsto F\left(b\right) \leqslant 0
$$

又由前面的计算可知：

$$
F\left( x \right) = \left(x – a\right) f \left(\frac{a+x}{2}\right) – \int_{a}^{x} f \left(t\right)\mathrm{~d}t
$$

于是，将 $x = b$ 代入上面的式子，得：

$$
\begin{aligned}
& \ \left(b – a\right) f \left(\frac{a+b}{2}\right) – \int_{a}^{x} f \left(t\right)\mathrm{~d}t \leqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \left(b – a\right) f \left(\frac{a+b}{2}\right) \leqslant \int_{a}^{x} f \left(t\right)\mathrm{~d}t \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{lightgreen}{ f\left(\frac{a+b}{2}\right) \leqslant \frac{1}{b-a}\int_{a}^{b}f\left(x\right)\mathrm{~d}x }
\end{aligned}
$$

$\textcolor{orangered}{\blacktriangleright}$ （正证法）证明由 $f\left(\frac{a + b}{2}\right) \leqslant \frac{1}{b – a} \int_{a}^{b} f\left(x\right) \mathrm{~d} x$ 可以推导出 $f^{\prime \prime}\left(x\right) \geqslant 0$:

对于 $\forall \, x_{0} \in \left(-\infty,+\infty\right)$, 取 $a = x_{0} – h$, $b = x_{0} + h$, 其中 $h > 0$, 则：

$$
\begin{aligned}
& \ f\left(\frac{a+b}{2}\right) \leqslant \frac{1}{b-a}\int_{a}^{b}f\left(x\right)\mathrm{~d}x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f \left( \frac{x_{0} – h + x_{0} + h}{2} \right) \leqslant \frac{1}{x_{0}+h – x_{0} + h} \int_{x_{0}-h}^{x_{0}+h} f \left( x \right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ f \left( x_{0} \right) \leqslant \frac{1}{2h} \int_{x_{0}-h}^{x_{0}+h} f \left( x \right) \mathrm{~d} x \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\displaystyle\int_{x_{0}-h}^{x_{0}+h}f\left(x\right)\mathrm{~d}x-2f\left(x_{0}\right)h}{2h} \geqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{h > 0, \, h^{2} > 0, \, \frac{1}{h^{2}} > 0} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\displaystyle\int_{x_{0}-h}^{x_{0}+h}f\left(x\right)\mathrm{~d}x-2f\left(x_{0}\right)h}{2h} \cdot \frac{1}{h^{2}} \geqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{\displaystyle\int_{x_{0}-h}^{x_{0}+h}f\left(x\right)\mathrm{~d}x-2f\left(x_{0}\right)h}{2h^{3}} \geqslant 0 \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim \limits_{h\to 0}\frac{\displaystyle\int_{x_{0}-h}^{x_{0}+h}f\left(x\right)\mathrm{~d}x-2f\left(x_{0}\right)h}{2h^{3}} \rightarrow 0^{+} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{洛必达运算，对 } h \text{ 求导}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim\limits_{h\to 0}\frac{f\left(x_{0}+h\right)+f\left(x_{0}-h\right)-2f\left(x_{0}\right)}{6h^{2}} \rightarrow 0^{+} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \textcolor{gray}{\text{洛必达运算，对 } h \text{ 求导}} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \lim\limits_{h\to 0}\frac{f^{\prime}\left(x_{0}+h\right)-f^{\prime}\left(x_{0}-h\right)}{12h} \rightarrow 0^{+} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{1}{6} \lim\limits_{h\to 0} \frac{f^{\prime}\left(x_{0}+h\right)-f^{\prime}\left(x_{0}-h\right)}{2h} \rightarrow 0^{+} \\ \\
\textcolor{lightgreen}{ \leadsto } & \ \frac{1}{6}f^{\prime\prime}\left(x_{0}\right) \rightarrow 0^{+}
\end{aligned}
$$

综上，由极限的保号性可知：

$$
\textcolor{lightgreen}{
f^{\prime\prime}\left(x_{0}\right) \geqslant 0
}
$$

综上可知，充分性必要性均得证.

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