## 张祖锦2023年数学专业真题分类70天之第12天
---
254、 2、 (可能有误) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, $\displaystyle f(0)=f(1)=0$, $\displaystyle f(t\_0)=\alpha, t\_0\in (0,1)$. 证明:
\begin\{aligned\} \forall\ \lambda\in\mathbb\{R\},\exists\ \eta\in (0,1),\mathrm\{ s.t.\} f'(\eta)-\lambda [f(\eta)-\alpha \eta]=\alpha. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[题目有问题, 这与 $\displaystyle t\_0$ 何干? 跟锦数学微信公众号没法做哦.] (天津大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目有问题, 跟锦数学微信公众号没法做哦.]跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
255、 3、 设 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}x^2,&x\in\left\[0,\frac\{1\}\{2\}\right),\\\\ x^\alpha,&x\in [1,+\infty).\end\{array\}\right.$ 请给出 $\displaystyle f$ 在 $\displaystyle \left\[\frac\{1\}\{2\},1\right\]$ 的一个表达式, 使得 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上有连续的二阶导数. (天津大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle f(x)$ 在 $\displaystyle \left\[\frac\{1\}\{2\},1\right\]$ 上为多项式, 则
\begin\{aligned\} f\left(\frac\{1\}\{2\}\right)=\frac\{1\}\{4\}, f'\left(\frac\{1\}\{2\}\right)=1, f''\left(\frac\{1\}\{2\}\right)=2; f(1)=1, f'(1)=\alpha, f''(1)=\alpha(\alpha-1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故可取 $\displaystyle f(x)$ 为 $\displaystyle 5$ 次多项式.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
256、 7、 设 $\displaystyle f(x)$ 满足如下条件: (1)、 存在点列 $\displaystyle \left\\{x\_n\right\\}$ 严格单调递减且 $\displaystyle \lim\_\{n\to\infty\}x\_n=0$, 使得 $\displaystyle f(x\_n)=0$; (2)、 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上任意阶可导, 且存在 $\displaystyle M > 0$, 使得 $\displaystyle |f^\{(k)\}(x)|\leq M, k\in\mathbb\{Z\}\_+$. 求证: $\displaystyle f(x)\equiv 0, x\in\mathbb\{R\}$. (同济大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f(0)=\lim\_\{n\to\infty\}f(x\_n)=0$. 在 $\displaystyle [x\_\{n+1\},x\_n]$ 上应用 Rolle 定理知
\begin\{aligned\} \exists\ y\_n\in (x\_\{n+1\},x\_n),\mathrm\{ s.t.\} f'(y\_n)=0\Rightarrow f'(0)=\lim\_\{n\to\infty\}f'(y\_n)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
继续在 $\displaystyle [y\_\{n+1\},y\_n]$ 上应用 Rolle 定理, 得 $\displaystyle f''(0)=0$. 如此一直做下去知 $\displaystyle f^\{(n)\}(0)=0$. 于是
\begin\{aligned\} |f(x)|\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \left|\frac\{f^\{(n+1)\}(\xi)\}\{(n+1)!\}x^\{n+1\}\right| \leq\frac\{M|x|^\{n+1\}\}\{(n+1)!\}\xrightarrow\{n\to\infty\}0\Rightarrow f(x)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
257、 (5)、 $\displaystyle y(x)$ 由方程 $\displaystyle x=\int\_1^\{y-x\} \sin^2\left(\frac\{\pi\}\{4\}t\right)\mathrm\{ d\} t$ 确定, 求 $\displaystyle \left.\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}\right|\_\{x=0\}$. (武汉理工大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 易知 $\displaystyle y(0)=1$. 进而
\begin\{aligned\} x=\int\_1^\{y-x\} \sin^2\frac\{t\}\{4\}\mathrm\{ d\} t&\Rightarrow 1=\sin^2\frac\{\pi(y-x)\}\{4\}\cdot (y'-1)\left(\mbox\{两边关于 $\displaystyle x$ 求导\}\right)\\\\ &\Rightarrow 1=\frac\{1\}\{2\}\left\[y'(0)-1\right\]\left(\mbox\{$(x,y)=(0,1)$ 代入\}\right)\\\\ &\Rightarrow y'(0)=3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
258、 (3)、 $\displaystyle f(x)=(x^2+2x-3)^\{100\} \cos\frac\{\pi x^2\}\{4\}$, 则 $\displaystyle f^\{(100)\}(1)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &f(x)=(x+3)^\{100\}(x-1)^\{100\}\cos\frac\{\pi x^2\}\{4\}\\\\ \Rightarrow& f^\{(100)\}(1)=100! \left.(x+3)^\{100\}\cos\frac\{\pi x^2\}\{4\}\right|\_\{x=1\} =100! 4^\{100\}\frac\{1\}\{\sqrt\{2\}\}=\frac\{4^\{100\}100!\}\{\sqrt\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
259、 (6)、 $\displaystyle f(y)=\int\_1^\{y^2\}\frac\{\ln (x+y)\}\{x\}\mathrm\{ d\} x$, 则 $\displaystyle f'(y)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle 1+y > 0, y^2+y > 0$ 知 $\displaystyle f$ 的定义域为 $\displaystyle y > 0$. 此时,
\begin\{aligned\} f'(y)=&\int\_1^\{y^2\}\frac\{1\}\{x\}\frac\{1\}\{x+y\}\mathrm\{ d\} x+\frac\{\ln (y^2+y)\}\{y^2\}\cdot 2y\\\\ =&\frac\{1\}\{y\}\int\_1^\{y^2\}\left(\frac\{1\}\{x\}-\frac\{1\}\{x+y\}\right)\mathrm\{ d\} x+\frac\{2\}\{y\}\ln (y^2+y) =\frac\{3\ln y+2\ln(1+y)\}\{y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
260、 (6)、 已知方程 $\displaystyle xy=\ln(x+y)$. (6-1)、 对任意的 $\displaystyle x\in (2,+\infty)$, 存在唯一确定的函数 $\displaystyle y=f(x)$ 满足原方程; (6-2)、 $\displaystyle y=f(x)$ 关于 $\displaystyle x\in (2,+\infty)$ 连续可微; (6-3)、 证明: 对于函数 $\displaystyle y=f(x)$, 有
\begin\{aligned\} y=\frac\{\ln x\}\{x\}+O\left(\frac\{\ln x\}\{x^3\}\right)\left(x\to+\infty\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西安交通大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (6-1)、 设 $\displaystyle F=xy-\ln(x+y)$, 对任意固定的 $\displaystyle x > 2$, 由
\begin\{aligned\} F(x,0)=-\ln x < 0, \lim\_\{y\to+\infty\}F(x,y)=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及连续函数介值定理知 $\displaystyle \exists\ y\in (0,+\infty),\mathrm\{ s.t.\} F(x,y)=0$. 又由 $\displaystyle F\_y=x-\frac\{1\}\{x+y\} > x-\frac\{1\}\{x\} > 0$ 知满足 $\displaystyle F(x,y)=0$ 的 $\displaystyle y$ 是唯一的. (6-2)、 由隐函数存在定理即知 $\displaystyle y=f(x)$ 关于 $\displaystyle x\in (2,+\infty)$ 连续可微. (6-3)、
\begin\{aligned\} &xy=\ln(x+y)\leq x+y-1\Rightarrow (x-1)(y-1)\leq 0\\\\ \stackrel\{x > 2\}\{\Rightarrow\}&y < 1\Rightarrow y=\frac\{\ln(x+y)\}\{x\} < \frac\{\ln(x+1)\}\{x\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含
\begin\{aligned\} y&=\frac\{\ln(x+y)\}\{x\} =\frac\{\ln x\}\{x\}+\frac\{1\}\{x\}\ln \left(1+\frac\{y\}\{x\}\right) =\frac\{\ln x\}\{x\}+\frac\{1\}\{x\}O\left(\frac\{y\}\{x\}\right)\\\\ =&\frac\{\ln x\}\{x\}+O\left(\frac\{\frac\{\ln(x+1)\}\{x\}\}\{x^2\}\right) =\frac\{\ln x\}\{x\}+O\left(\frac\{\ln x\}\{x^3\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
261、 3、 (20 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上二阶可导, 且 $\displaystyle f(0)=f(1)=0$. 试证明: 存在 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f''(\xi)=\frac\{2f'(\xi)\}\{1-\xi\}$. (西南财经大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Rolle 定理, $\displaystyle \exists\ \eta\in (0,1),\mathrm\{ s.t.\} f'(\eta)=0$. 设 $\displaystyle F(x)=(x-1)^2f'(x)$, 则 $\displaystyle F(\eta)=F(1)=0$. 再由 Rolle 定理,
\begin\{aligned\} \exists\ \xi\in (\eta,1)\subset (0,1), \mathrm\{ s.t.\} 0=F'(\xi)\Rightarrow \frac\{2f'(\xi)\}\{1-\xi\}=f''(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
262、 4、 (15 分) 已知 $\displaystyle x+y-z=\mathrm\{e\}^z, x\mathrm\{e\}^x=\tan t,y=\cos t$, 求 $\displaystyle \left.\frac\{\mathrm\{ d\} x\}\{\mathrm\{ d\} t\}\right|\_\{t=0\}, \left.\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} t\}\right|\_\{t=0\}, \left.\frac\{\mathrm\{ d\}^2z\}\{\mathrm\{ d\} t^2\}\right|\_\{t=0\}$. (西南财经大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \left.\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} t\}\right|\_\{t=0\}=(-\sin t)|\_\{t=0\}=0$. 再者, 由 $\displaystyle x\mathrm\{e\}^x=\tan t$ 知当 $\displaystyle t=0$ 时, $\displaystyle x=0$, 且
\begin\{aligned\} (x+1)\mathrm\{e\}^x\frac\{\mathrm\{ d\} x\}\{\mathrm\{ d\} t\}=\sec^2t \stackrel\{t=x=0\}\{\Rightarrow\}\left.\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} t\}\right|\_\{t=0\}=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再者, $\displaystyle y''(0)=-1$,
\begin\{aligned\} (x+1)\mathrm\{e\}^x x'=0\sec^2t \Rightarrow&(x+2)\mathrm\{e\}^x x'+(x+1)\mathrm\{e\}^x x''=2\sec t \cdot \sec t\tan t\\\\ \stackrel\{t=x=0\}\{\Rightarrow\}&2+x''(0)=0\Rightarrow x''(0)=-2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后, 由 $\displaystyle x+y-z=\mathrm\{e\}^z$ 知当 $\displaystyle t=0, x=0, y=1$ 时, $\displaystyle z=0$, 且
\begin\{aligned\} &x+y+z+\mathrm\{e\}^z\Rightarrow x'+y'=z'+\mathrm\{e\}^z z'\Rightarrow z'(0)=\frac\{1\}\{2\},\\\\ &x''+y''=(1+\mathrm\{e\}^z)z''+\mathrm\{e\}^z z'\cdot z'\Rightarrow z''(0)=-\frac\{13\}\{8\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
263、 4、 (15 分) 证明导函数介值定理: 若 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上可导, 且 $\displaystyle f'\_+(a)\neq f'\_-(b)$, $\displaystyle k$ 为介于 $\displaystyle f'\_+(a)$ 与 $\displaystyle f'\_-(b)$ 之间的任一实数, 则至少存在一点 $\displaystyle \xi\in (a,b)$, 使得 $\displaystyle f'(\xi)=k$. (西南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 这就是导数介值定理. (1)、 先证: 若 $\displaystyle f'(a) < 0, f'(b) > 0$, 则存在 $\displaystyle \xi\in (a,b)$, 使得 $\displaystyle f'(\xi)=0$. 事实上, 由
\begin\{aligned\} f'(a)=\lim\_\{x\to a^+\}\frac\{f(x)-f(a)\}\{x-a\} < 0 < f'(b)=\lim\_\{x\to b^-\}\frac\{f(x)-f(b)\}\{x-b\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及函数极限的局部保号性知 $\displaystyle \exists\ \delta\in \left(0,\frac\{b-a\}\{2\}\right),\mathrm\{ s.t.\}$
\begin\{aligned\} \forall\ a < x < a+\delta, f(x) < f(a); \forall\ b-\delta < x < b, f(x) < f(b). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上的最小值只能在 $\displaystyle (a,b)$ 内某 $\displaystyle \xi$ 处取得. 由极值条件知 $\displaystyle f'(\xi)=0$. (2)、 再证题目. 不妨设 $\displaystyle f'\_+(a) < f'\_-(b)$. 令 $\displaystyle F(x)=f(x)-k$, 则 $\displaystyle F'(a) < 0 < F'(b)$. 由第 1 步知
\begin\{aligned\} \exists\ \xi\in (a,b),\mathrm\{ s.t.\} F'(\xi)=0\Rightarrow f'(\xi)=k. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
264、 3、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle (a,+\infty)$ 内可微, 且 $\displaystyle f(a+0)=f(+\infty)=L$. 求证: 存在 $\displaystyle \xi\in (a,+\infty)$, 使得 $\displaystyle f'(\xi)=0$. (西南交通大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle \forall\ x > a,\ f'(x)\neq 0$, 则由导数介值定理知要么 $\displaystyle \forall\ x > a,\ f'(x) > 0$, 要么 $\displaystyle \forall\ x > a,\ f'(x) < 0$. 不妨设 $\displaystyle \forall\ x > a,\ f'(x) > 0$ 成立, 则由 Lagrange 中值定理,
\begin\{aligned\} &f(a+1)-f\left(a+\frac\{1\}\{n\}\right)=f'(\xi\_n)\left(1-\frac\{1\}\{n\}\right) > 0\ (n\geq 2)\\\\ \Rightarrow& f(a+1)\geq \lim\_\{n\to\infty\}f\left(a+\frac\{1\}\{n\}\right)=f(a)\left(n\to\infty\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且 $\displaystyle f(a+2)-f(a+1)=f'(\eta) > 0$. 进一步,
\begin\{aligned\} &\left\[\forall\ x > a+2,\ f(x)-f(a+2) =f'(\zeta\_x)(x-(a+2)) > 0\right\]\\\\ \Rightarrow& \lim\_\{x\to+\infty\}f(x)\geq f(a+2) > f(a+1)\geq f(a). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这与题设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
265、 8、 (15 分) 设有界函数 $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上二次可微. 求证: 存在 $\displaystyle x\_0\in (-\infty,+\infty)$, 使得 $\displaystyle f''(x\_0)=0$. (西南交通大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 用反证法. 若 $\displaystyle \forall\ x\in\mathbb\{R\}, f''(x)\neq 0$, 则由导函数介值定理知要么 $\displaystyle \forall\ x, f''(x) > 0$, 要么 $\displaystyle \forall\ x, f''(x) < 0$. 不妨设前一情形成立 (若不然, 考虑 $\displaystyle -f$ 即可). 于是 $\displaystyle f$ 是严格凸函数, $\displaystyle f'\mbox\{严\}\nearrow $. 从而 $\displaystyle f$ 不可能是常值函数. (1)、 若 $\displaystyle f(a) < f(b)$, 则
\begin\{aligned\} &\exists\ \xi\in (a,b),\mathrm\{ s.t.\} f'(\xi)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{f(b)-f(a)\}\{b-a\} > 0\\\\ \Rightarrow&\forall\ x > \xi, f(x)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f(\xi)+f'(\eta)(x-\xi) > f(\xi)+f'(\xi)(x-\xi)\\\\ \Rightarrow&\lim\_\{x\to+\infty\}f(x)=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle f(a) > f(b)$, 则
\begin\{aligned\} &\exists\ \xi\in (a,b),\mathrm\{ s.t.\} f'(\xi)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{f(b)-f(a)\}\{b-a\} < 0\\\\ \Rightarrow&\forall\ x < \xi, f(x)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f(\xi)+f'(\eta)(x-\xi) > f(\xi)+f'(\xi)(x-\xi)\\\\ \Rightarrow&\lim\_\{x\to-\infty\}f(x)=+\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
不论何种情形, 都与题设’$f$ 有界‘矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
266、 2、 解答如下问题: (1)、 设函数 $\displaystyle y=y(x)$ 由 $\displaystyle \mathrm\{e\}^\{x^2+y\}-x^2y=0$ 确定, 求 $\displaystyle \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}$, $\displaystyle \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}$; (湘潭大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 题中方程关于 $\displaystyle x$ 求导有
\begin\{aligned\} &0=\mathrm\{e\}^\{x^2+y\}(2x+y')-2xy-x^2y' =x^2y(2x+y')-2xy-x^2y'\\\\ =&2x^3y+x^2yy'-2xy-x^2y' =x\left\[x(y-1)y'+2y(x^2-1)\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle y'=\frac\{2y(1-x^2)\}\{x(y-1)\}$, 且
\begin\{aligned\} &x(y-1)y'+2y(x^2-1)=0\\\\ \Rightarrow& (y-1)y'+x\{y'\}^2+x(y-1)y'' +2y'(x^2-1)+2y\cdot 2x=0\\\\ \Rightarrow& y''=-\frac\{2y\left\[3+2x^4+(y-2)y+x^2(y-3)(y+1)\right\]\}\{x^2(y-1)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
267、 (2)、 设 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x=t(1-\sin t),\\\\ y=t\cos t.\end\{array\}\right.$ 求 $\displaystyle \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}$, $\displaystyle \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}$. (湘潭大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}&=\frac\{y'(t)\}\{x'(t)\}=\frac\{\cos t-t\sin t\}\{1-t\cos t-\sin t\},\\\\ \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}&=\frac\{\mathrm\{ d\} \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}\}\{\mathrm\{ d\} x\} =\frac\{\frac\{y''(t)x'(t)-y'(t)x''(t)\}\{\{x'\}^2(t)\}\}\{x'(t)\}\\\\ &=\frac\{y''(t)x'(t)-y'(t)x''(t)\}\{\{x'\}^3(t)\}=\frac\{t\cos t+2\sin t-2-t^2\}\{(t\cos t+\sin t-1)^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
268、 (4)、 设 $\displaystyle I(y)=\int\_y^\{y^2\}\frac\{\sin (xy)\}\{x\}\mathrm\{ d\} x$, 则 $\displaystyle I'(y)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} I'(y)=&\int\_y^\{y^2\}\frac\{\cos(xy)\cdot x\}\{x\}\mathrm\{ d\} x +\frac\{\sin (y^2\cdot y)\}\{y^2\}\cdot 2y -\frac\{\sin (y\cdot y)\}\{y\}\cdot 1\\\\ =&\left.\frac\{1\}\{y\}\sin(xy)\right|\_\{x=y\}^\{x=y^2\}+2\frac\{\sin(y^3)\}\{y\}-\frac\{\sin(y^2)\}\{y\} =\frac\{3\sin(y^3)-2\sin(y^2)\}\{y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
269、 (5)、 设 $\displaystyle f(x)$ 可微, $\displaystyle y=f(\ln x)\mathrm\{e\}^\{f(x)\}$, 则 $\displaystyle \mathrm\{ d\} y=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (云南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mathrm\{ d\} y=y'\mathrm\{ d\} x=\left\[\frac\{f'(\ln x)\}\{x\}+f(\ln x)f'(x)\right\]\mathrm\{e\}^\{f(x)\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
270、 2、 (10 分) 设 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\},&x\neq 0,\\\\ 0,&x=0.\end\{array\}\right.$ 证明 $\displaystyle f(x)$ 在 $\displaystyle x=0$ 处存在任意阶导数. (云南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先证: 对 $\displaystyle \forall\ \alpha\in\mathbb\{R\}$, $\displaystyle \lim\_\{t\to+\infty\}\frac\{t^\alpha\}\{\mathrm\{e\}^t\}=0$. 当 $\displaystyle \alpha\leq 0$ 时,
\begin\{aligned\} \mbox\{原式\}=\lim\_\{t\to+\infty\}\frac\{1\}\{t^\{-\alpha\}\}\cdot\frac\{1\}\{\mathrm\{e\}^t\}=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle \alpha > 0$ 时, 由
\begin\{aligned\} &t > 1\Rightarrow 0 < \frac\{t^\alpha\}\{\mathrm\{e\}^t\} < \frac\{t^\{[\alpha]+1\}\}\{\mathrm\{e\}^t\},\\\\ &\lim\_\{t\to+\infty\} \frac\{t^\{[\alpha]+1\}\}\{\mathrm\{e\}^t\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{t\to+\infty\}([\alpha]+1) \frac\{t^\{[\alpha]\}\}\{\mathrm\{e\}^t\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots =0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及夹逼原理知原式 $\displaystyle =0$. (2)、 由归结原理即知
\begin\{aligned\} \lim\_\{t\to 0\}t^\alpha \mathrm\{e\}^\{-\frac\{1\}\{t^2\}\}\stackrel\{\frac\{1\}\{t^2\}=s\}\{=\}\lim\_\{s\to+\infty\}s^\{-\frac\{\alpha\}\{2\}\}\frac\{1\}\{\mathrm\{e\}^s\} \overset\{\tiny\mbox\{第1步\}\}\{=\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 用数学归纳法证明
\begin\{aligned\} x\neq 0\Rightarrow f^\{(n)\}(x)=P\_n\left(\frac\{1\}\{x\}\right) \mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle P\_n(y)$ 是 $\displaystyle y$ 的多项式. 事实上, 当 $\displaystyle n=1$ 时, $\displaystyle f'(x)=\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\} \frac\{2\}\{x^3\}$. 若结论对 $\displaystyle n$ 成立, 则
\begin\{aligned\} f^\{(n+1)\}(x)&=\left(P\_n\left(\frac\{1\}\{x\}\right)\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}\right)' =P\_n'\left(\frac\{1\}\{x\}\right) \left(-\frac\{1\}\{x^2\}\right) \mathrm\{e\}^\{-\frac\{1\}\{x^2\}\} +P\_n\left(\frac\{1\}\{x\}\right) \mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}\frac\{2\}\{x^3\}\\\\ &=\left\[-y^2P\_n'(y)+2y^3P\_n(y)\right\]\_\{y=\frac\{1\}\{x\}\}\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(4)、 再用数学归纳法证明 $\displaystyle f^\{(n)\}(0)=0$. 当 $\displaystyle n=1$ 时,
\begin\{aligned\} f'(0)=\lim\_\{x\to 0\}\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}\overset\{\tiny\mbox\{第2步\}\}\{=\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
假设结论对 $\displaystyle n$ 成立, 则
\begin\{aligned\} f^\{(n+1)\}(0)=\lim\_\{x\to 0\}\frac\{f^\{(n)\}(x)-f^\{(n)\}(0)\}\{x\} =\lim\_\{x\to 0\}\frac\{1\}\{x\} P\_n\left(\frac\{1\}\{x\}\right) \mathrm\{e\}^\{-\frac\{1\}\{x^2\}\}\overset\{\tiny\mbox\{第2步\}\}\{=\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
271、 4、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上有二阶导数, 且 $\displaystyle f(a)=f(b)=0$, $\displaystyle f'(a)\cdot f'(b) > 0$. 证明: 存在 $\displaystyle \eta\in (a,b)$, 使得 $\displaystyle f''(\eta)=0$. (云南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 不妨设 $\displaystyle f'(a) > 0, f'(b) > 0$ [否则用 $\displaystyle -f$ 代替 $\displaystyle f$ 考虑]. 由局部保号性知
\begin\{aligned\} \exists\ 0 < \delta < \frac\{b-a\}\{2\},\mathrm\{ s.t.\} \forall\ a < x\leq a+\delta, \frac\{f(x)-f(a)\}\{x-a\} > 0\Rightarrow f(x) > 0,\\\\ \forall\ b-\delta\leq x < b, \frac\{f(x)-f(b)\}\{x-b\} > 0\Rightarrow f(x) < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle [a-\delta,b+\delta]$ 上应用连续函数介值定理知
\begin\{aligned\} \exists\ \xi\in (a,b),\mathrm\{ s.t.\} f(\xi)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle f(a)=f(\xi)=f(b)=0$
\begin\{aligned\} \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ a < \zeta\_1 < \xi < \zeta\_2 < b,\mathrm\{ s.t.\} f'(\zeta\_1)=f'(\zeta\_2)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ \zeta\_1 < \eta < \zeta\_2,\mathrm\{ s.t.\} f''(\eta)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
272、 1、 选择题 (每题 5 分, 共 30 分). (1)、 设函数 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上具有连续的导数, 且 $\displaystyle f'\_+(a) > 0, f'\_-(b) < 0$, 则下列结论错误的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. 至少存在一点 $\displaystyle x\_0\in (a,b)$, 使得 $\displaystyle f(x\_0) > f(a)$
B. 至少存在一点 $\displaystyle x\_0\in (a,b)$, 使得 $\displaystyle f(x\_0)=0$
C. 至少存在一点 $\displaystyle x\_0\in (a,b)$, 使得 $\displaystyle f(x\_0) > f(b)$
D. 至少存在一点 $\displaystyle x\_0\in (a,b)$, 使得 $\displaystyle f'(x\_0)=0$ (长安大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle B$. 比如 $\displaystyle f(x)=1-x^2, x\in [-1,1]$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
273、 9、 (20 分) 证明题. (1)、 设函数 $\displaystyle f(x)$ 在 $\displaystyle (a,b)$ 内可导, 且 $\displaystyle f(a+0)=f(b-0)$. 证明: 存在 $\displaystyle \xi\in (a,b)$, 使得 $\displaystyle f'(\xi)=0$. (郑州大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F(x)=\left\\{\begin\{array\}\{llllllllllll\} f(a+0),&x=a,\\\\ f(x),&a < x < b,\\\\ f(b-0),&x=b, \end\{array\}\right.$ 则 $\displaystyle F$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 上 $\displaystyle F(x)=f(x)$, 而可导. 由 Rolle 定理知
\begin\{aligned\} \exists\ \xi\in (a,b),\mathrm\{ s.t.\} 0=F'(\xi)=f'(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
274、 2、 判断题 (每小题 3 分, 共 30 分, 判断正误并说明理由). (1)、 若函数 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 可导且无界, 则 $\displaystyle f'(x)$ 在 $\displaystyle (0,1)$ 也无界. (中国海洋大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \surd$. 用反证法. 若 $\displaystyle f'$ 有界, 设为 $\displaystyle M\geq 0$, 则
\begin\{aligned\} \forall\ x\in (0,1), |f(x)|\leq&\left|f(x)-f\left(\frac\{1\}\{2\}\right)\right|+\left|f\left(\frac\{1\}\{2\}\right)\right|\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}& |f'(\xi\_x)|\cdot \left|x-\frac\{1\}\{2\}\right|+\left|f\left(\frac\{1\}\{2\}\right)\right| \leq\frac\{M\}\{2\}+\left|f\left(\frac\{1\}\{2\}\right)\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 有界, 与题设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
275、 (3)、 若 $\displaystyle f'(x)$ 在 $\displaystyle (0,1)$ 无界, 则 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 无界. (中国海洋大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \times$. 比如 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\} x^2\sin\frac\{1\}\{x^2\},&x\neq 0\\\\ 0,&x=0 \end\{array\}\right.$ 在 $\displaystyle (0,1)$ 上可导. 由
\begin\{aligned\} &f'(x)=\left\\{\begin\{array\}\{llllllllllll\} 2x\sin\frac\{1\}\{x^2\}-\frac\{2\}\{x\}\cos\frac\{1\}\{x^2\},&x\neq 0,\\\\ \lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x-0\}=0,&x=0. \end\{array\}\right.\\\\ \Rightarrow&f'\left(\frac\{1\}\{\sqrt\{2n\pi\}\}\right)=-2\sqrt\{2n\pi\}\to-\infty\left(n\to\infty\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f'$ 无界. 但 $\displaystyle f$ 在 $\displaystyle [0,1]$ 上连续而有界.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
276、 5、 (10 分) 设 $\displaystyle f(x)$ 在 $\displaystyle 0$ 处可导, 请证明下面的极限存在并求其值:
\begin\{aligned\} \lim\_\{n\to\infty\}\frac\{f\left(\sin\frac\{1\}\{n\}\right)-f\left(\sin \left(-\frac\{1\}\{n^2\}\right)\right)\}\{\frac\{1\}\{n\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国海洋大学2023年数学分析考研试题) [微分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\lim\_\{n\to\infty\} \frac\{f\left(\sin\frac\{1\}\{n\}\right)-f(0)\}\{\sin\frac\{1\}\{n\}\}\cdot \frac\{\sin\frac\{1\}\{n\}\}\{\frac\{1\}\{n\}\}\\\\ &-\lim\_\{n\to\infty\}\frac\{f\left(\sin \left(-\frac\{1\}\{n^2\}\right)\right)-f(0)\}\{\sin\left(-\frac\{1\}\{n^2\}\right)\} \cdot\frac\{\sin\left(-\frac\{1\}\{n^2\}\right)\}\{-\frac\{1\}\{n^2\}\}\cdot \left(-\frac\{1\}\{n\}\right)\\\\ =&f'(0)\cdot 1-f'(0)\cdot 1\cdot 0=f'(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 08:46:50
