## 张祖锦2023年数学专业真题分类70天之第10天
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208、 (4)、 设函数 $\displaystyle f(x)$ 连续, 且 $\displaystyle \lim\_\{x\to 0\}\frac\{f(x)\}\{x\}=1$, $\displaystyle g(x)=\int\_0^1 f(xt)\mathrm\{ d\} t$. 求 $\displaystyle g'(x)$, 并讨论 $\displaystyle g'(x)$ 在 $\displaystyle x=0$ 处是否连续. (北京科技大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} g'(x)=\left\\{\begin\{array\}\{llllllllllll\} \frac\{1\}\{x^2\}\left\[xf(x)-\int\_0^x f(s)\mathrm\{ d\} s\right\],&x\neq 0,\\\\ \lim\_\{x\to 0\}\frac\{1\}\{x^2\}\int\_0^x f(s)\mathrm\{ d\} s =\lim\_\{x\to 0\} \frac\{f(x)\}\{2x\}=\frac\{1\}\{2\},&x=0. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \lim\_\{x\to 0\}g'(x)&=\lim\_\{x\to 0\}\left\[\frac\{f(x)\}\{x\}-\frac\{1\}\{x^2\}\int\_0^x f(s)\mathrm\{ d\} s\right\]\\\\ &=1-\lim\_\{x\to 0\}\frac\{f(x)\}\{2x\}=1-\frac\{1\}\{2\}=\frac\{1\}\{2\}=g'(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle g$ 在 $\displaystyle x=0$ 处连续.跟锦数学微信公众号. [在线资料](
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209、 (4)、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上可导, 且 $\displaystyle 0\leq f(x)\leq \frac\{x\}\{1+x^2\}, x\geq 0$. 求证: 存在 $\displaystyle \xi\in (0,+\infty)$, 使得
\begin\{aligned\} f'(\xi)=\frac\{1-\xi^2\}\{(1+\xi)^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (4-1)、 先证推广的 Rolle 定理. 设 $\displaystyle f$ 在 $\displaystyle [a,+\infty)$ 上连续, 在 $\displaystyle (a,+\infty)$ 上可微, 且 $\displaystyle \lim\_\{x\to+\infty\}f(x)=f(a)$. 证明: 存在 $\displaystyle \xi > a$, 使得 $\displaystyle f'(\xi)=0$. (Rolle 定理在无限区间上的推广) 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. (4-2)、 再证题目. 令 $\displaystyle x=0$ 得 $\displaystyle f(0)=0$. 设 $\displaystyle F(x)=f(x)-\frac\{x\}\{1+x^2\}$, 则
\begin\{aligned\} F(0)=f(0)=0, -\frac\{x\}\{1+x^2\}\leq F(x)\leq 0\Rightarrow \lim\_\{x\to+\infty\}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\}
据第 1 步即知结论成立.跟锦数学微信公众号. [在线资料](
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210、 (8)、 求 $\displaystyle f(x)=(1-x)\sqrt\{|x|\}$ 在 $\displaystyle (-1,1)$ 的极值点和极值. (大连理工大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}(1-x)\sqrt\{-x\},&-1 < x < 0,\\\\ (1-x)\sqrt\{x\},&x\geq 0\end\{array\}\right.$ 知
\begin\{aligned\} -1 < x < 0\Rightarrow f'(x)=\frac\{3x-1\}\{2\sqrt\{-x\}\} < 0, 0 < x < 1\Rightarrow f'(x)=\frac\{1-3x\}\{2\sqrt\{x\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 在 $\displaystyle x=0$ 处取得极小值 $\displaystyle f(0)=0$, 在 $\displaystyle x=\frac\{1\}\{3\}$ 处取得极大值 $\displaystyle f\left(\frac\{1\}\{3\}\right)=\frac\{2\}\{3\sqrt\{3\}\}$.跟锦数学微信公众号. [在线资料](
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211、 (9)、 求证下列函数在 $\displaystyle \mathbb\{R\}$ 上连续可导,
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}\mathrm\{e\}^\frac\{1\}\{x\},&x < 0,\\\\ x^2,&x\geq 0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle f'\_+(0)=0$,
\begin\{aligned\} f'\_-(0)=&\lim\_\{x\to 0^-\}\frac\{\mathrm\{e\}^\frac\{1\}\{x\}-0\}\{x\}\stackrel\{-\frac\{1\}\{x\}=t\}\{=\}-\lim\_\{t\to+\infty\}\frac\{t\}\{\mathrm\{e\}^t\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} -\lim\_\{t\to+\infty\}\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\}
故 $\displaystyle f$ 在 $\displaystyle x=0$ 处可导, 而在 $\displaystyle \mathbb\{R\}$ 上可导. 再由
\begin\{aligned\} x > 0\Rightarrow f'(x)=2x\Rightarrow \lim\_\{x\to 0^+\}f'(x)=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\}
\begin\{aligned\} x < 0\Rightarrow& f'(x)=-\frac\{1\}\{x^2\}\mathrm\{e\}^\frac\{1\}\{x\}\\\\ \Rightarrow& \lim\_\{x\to 0^-\}f'(x)\stackrel\{-\frac\{1\}\{x\}=t\}\{=\}-\lim\_\{t\to+\infty\}\frac\{t^2\}\{\mathrm\{e\}^t\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots =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\}
知 $\displaystyle f'$ 连续, 而 $\displaystyle f$ 连续可导.跟锦数学微信公众号. [在线资料](
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212、 (2)、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上有二阶导数, 且 $\displaystyle f(0)=f(1)=0$. 若 $\displaystyle \min\_\{[0,1]\}f=-1$, 证明: 存在 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f''(\xi)\geq 8$. (大连理工大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (2-1)、 由 $\displaystyle f(0)=f(1)=0$ 知 $\displaystyle f$ 在 $\displaystyle [0,1]$ 上的最小值 $\displaystyle -1$ 只能在 $\displaystyle (0,1)$ 中某点 $\displaystyle \xi$ 处取得. 由 Fermat 定理, $\displaystyle f'(\xi)=0.$ (2-2)、 又由 Taylor 展式,
\begin\{aligned\} f(0)&=f(\xi)+f'(\xi)(0-\xi)+\frac\{f''(\eta)\}\{2!\}(0-\xi)^2,\\\\ f(1)&=f(\xi)+f'(\xi)(1-\xi)+\frac\{f''(\zeta)\}\{2!\}(1-\xi)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f''(\eta)=\frac\{2\}\{\xi^2\}, f''(\zeta)=\frac\{2\}\{(1-\xi)^2\}.$ (2-3)、 当 $\displaystyle 0 < \xi\leq \frac\{1\}\{2\}$ 时, $\displaystyle f''(\eta)=\frac\{2\}\{\eta^2\}\geq 8$. (2-4)、 当 $\displaystyle \frac\{1\}\{2\} < \xi < 1$ 时, $\displaystyle f''(\zeta)=\frac\{2\}\{(1-\zeta)^2\}\geq 8$.跟锦数学微信公众号. [在线资料](
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213、 2、 计算题 (共 50 分, 每小题 10 分). (1)、 求使得不等式 $\displaystyle \left(1+\frac\{1\}\{n\}\right)^\{n+\alpha\} > \mathrm\{e\}$ 对所有正整数 $\displaystyle n$ 都成立的最小的数 $\displaystyle \alpha$. (电子科技大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\left(1+\frac\{1\}\{n\}\right)^\{n+\alpha\} > \mathrm\{e\}, \forall\ n\geq 1 \Leftrightarrow (n+\alpha)\ln \left(1+\frac\{1\}\{n\}\right) > 1\\\\ \Leftrightarrow&\alpha > \frac\{1\}\{\ln \left(1+\frac\{1\}\{n\}\right)\}-n \Leftrightarrow \alpha\geq \sup\_\{n\geq 1\} \left\[\frac\{1\}\{\ln \left(1+\frac\{1\}\{n\}\right)\}-n\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f(t)=\frac\{1\}\{\ln (1+t)\}-\frac\{1\}\{t\}$, 则
\begin\{aligned\} f'(t)=-\frac\{1\}\{(1+t)\ln^2(1+t)\}+\frac\{1\}\{t^2\} =\frac\{(1+t)\ln^2(1+t)-t^2\}\{(1+t)t^2\ln^2(1+t)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle g(t)=\sqrt\{1+t\}\ln(1+t)-t$, 则
\begin\{aligned\} g'(t)=\frac\{\ln(1+t)\}\{2\sqrt\{1+t\}\}+\frac\{1\}\{\sqrt\{1+t\}\}-1 =\frac\{\ln(1+t)+2-2\sqrt\{1+t\}\}\{2\sqrt\{1+t\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle h(t)=\ln(1+t)+2-2\sqrt\{1+t\}$, 则
\begin\{aligned\} &h'(t)=\frac\{1\}\{1+t\}-\frac\{1\}\{\sqrt\{1+t\}\}\leq 0, h(t)\leq h(0)=0,\\\\ &g'(t)\leq 0, g(t)\leq g(0)=0, f'(t)\leq 0, f(t)\leq f(0)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \alpha\_\{\min\}=&\lim\_\{t\to 0\}\left\[\frac\{1\}\{\ln(1+t)\}-\frac\{1\}\{t\}\right\] =\lim\_\{t\to 0\}\frac\{t-\ln(1+t)\}\{t\ln (1+t)\}\\\\ \xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\}&\lim\_\{x\to 0\}\frac\{t-\left\[t-\frac\{t^2\}\{2\}+o(t^2)\right\]\}\{t^2\}=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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214、 3、 证明题 (共 40 分, 每小题 10 分). (1)、 设函数 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 上可导且存在 $\displaystyle c\in (a,b)$, 使得 $\displaystyle f'(c)=0$. 证明: 存在 $\displaystyle \xi\in (a,b)$, 使得
\begin\{aligned\} f'(\xi)-f(\xi)+f(a)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(电子科技大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设
\begin\{aligned\} F(x)=\mathrm\{e\}^\{-x\}[f(x)-f(a)], \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle F(a)=0$,
\begin\{aligned\} F'(x)=\mathrm\{e\}^\{-x\}\left\\{f'(x)-[f(x)-f(a)]\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-1)、 若 $\displaystyle f(c)=0$, 则由 Rolle 定理,
\begin\{aligned\} \exists\ \xi\in (a,c),\mathrm\{ s.t.\} F'(\xi)=0\Rightarrow f'(\xi)=f(\xi)-f(a). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 若 $\displaystyle f(c) > 0$, 则 $\displaystyle F(c) > 0=F(a), F'(c) < 0$. 故 $\displaystyle F$ 在 $\displaystyle [a,c]$ 上的最大值只能在 $\displaystyle (a,c)$ 内某处 $\displaystyle \xi$ 取得, 而 $\displaystyle F'(\xi)=0$. (1-3)、 若 $\displaystyle f(c) < 0$, 则 $\displaystyle F(c) < 0=F(a), F'(c) > 0$. 故 $\displaystyle F$ 在 $\displaystyle [a,c]$ 上的最小值只能在 $\displaystyle (a,c)$ 内某处 $\displaystyle \xi$ 取得, 而 $\displaystyle F'(\xi)=0$.跟锦数学微信公众号. [在线资料](
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215、 4、 设 $\displaystyle y=y(x)$ 由方程
\begin\{aligned\} y^2-x+\sin y=0\left(x\geq 1\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
确定, 且 $\displaystyle y=y(x)$ 经过 $\displaystyle (\pi^2,\pi)$. 试讨论 $\displaystyle y(x)$ 在 $\displaystyle (1,+\infty)$ 上的零点个数, 并求 $\displaystyle \lim\_\{x\to+\infty\}y(x)$. (东北师范大学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](
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 若 $\displaystyle y=0$, 则代入方程知 $\displaystyle x=0$. 故当 $\displaystyle x > 1$ 时, $\displaystyle y(x)\neq 0$. 又由 $\displaystyle y=y(x)$ 经过 $\displaystyle (\pi^2,\pi)$ 及连续函数介值定理知 $\displaystyle \forall\ x > 1, y(x) > 0$. 这表明 $\displaystyle y(x)$ 在 $\displaystyle (1,+\infty)$ 上的零点个数为 $\displaystyle 0$, 且
\begin\{aligned\} x > 1\Rightarrow& y^2=x-\sin y > 0\stackrel\{y > 0\}\{\Rightarrow\}y=\sqrt\{x-\sin y\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \lim\_\{x\to+\infty\}y(x)=\lim\_\{x\to+\infty\} \frac\{\sqrt\{x-\sin y\}\}\{x\} =\lim\_\{x\to+\infty\}\sqrt\{\frac\{1\}\{x\}-\frac\{\sin y\}\{x^2\}\}\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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216、 1、 $\displaystyle y=y(x)$ 由 $\displaystyle \sin(x+y)+\ln(y-x)=0$ 确定, 求 $\displaystyle \left.\frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}\right|\_\{x=0\}$. [题目有问题, 跟锦数学微信公众号没法做哦.] (东南大学2023年数学分析考研试题) [微分 ]
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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. 毕竟 $\displaystyle \sin y+\ln y=0\Rightarrow y=?$] 方法大家都知道.跟锦数学微信公众号. [在线资料](
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217、 7、 一个分式的麦克劳林展开. [题目不全, 跟锦数学微信公众号没法做哦.] (东南大学2023年数学分析考研试题) [微分 ]
[纸质资料](
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / [题目不全, 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)/
---
218、 15、 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可导, 下凸, 且无极值点. 证明: $\displaystyle f$ 在 $\displaystyle [a,b]$ 上单调. (东南大学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)、 由 $\displaystyle f$ 下凸知 $\displaystyle f'$ 递增. 往证 $\displaystyle f'\neq 0, x\in (a,b)$. 若不然, $\displaystyle \exists\ x\_0,\mathrm\{ s.t.\} f'(x\_0)=0$, 则由 $\displaystyle f'$ 递增知 $\displaystyle f'(x)\left\\{\begin\{array\}\{llllllllllll\} < 0,&x < x\_0,\\\\ > 0,&x > x\_0.\end\{array\}\right.$ 从而 $\displaystyle x\_0$ 是 $\displaystyle f$ 的极小值点. 与题设矛盾. 故有结论. (2)、 既然 $\displaystyle f'\neq 0, x\in (a,b)$, 我们就可用反证法及导数介值定理知要么 $\displaystyle \forall\ x\in (a,b), f'(x) < 0$, 要么 $\displaystyle \forall\ x\in (a,b), f'(x) > 0$. 不论何种情形, $\displaystyle f$ 是单调的.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](
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---
219、 注: 内容包括数学分析, 常微分方程, 实变函数与复变函数 (其中前三题填空每题 10 分, 其余均为 15 分一题). 1、 填空题. (1)、 在 $\displaystyle Oxy$ 平面上给定点 $\displaystyle O(0,0), A(1,0)$, 动点 $\displaystyle P(x,y)$ 在直线 $\displaystyle y=x+1$ 上, 则当 $\displaystyle P(x,y)=\underline\{\ \ \ \ \ \ \ \ \ \ \}$ 时, $\displaystyle \angle OPA$ 取到取到最大. (复旦大学2023年分析(第6,7,8,9,10题没做)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 $\displaystyle P(x,x+1)$ 知
\begin\{aligned\} \overline\{PO\}=\left\\{-x,-x-1\right\\}, \overline\{PA\}=\left\\{1-x,-x-1\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} &|\cos\angle OPA|^2=\left|\frac\{\overline\{PO\}\cdot \overline\{PA\}\}\{|\overline\{PO\}|\cdot |\overline\{PA\}|\}\right|^2 =\frac\{(2x^2+x+1)^2\}\{(1+2x+2x^2)(2+2x^2)\}\equiv f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 (可用对数求导法)
\begin\{aligned\} f'(x)=\frac\{x(x+1)(x+2)(2x^2+x+1)\}\{(1+x^2)^2(1+2x+2x^2)^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \min \cos\angle OPA=\sqrt\{\min\left\\{f(-2), f(-1), f(0)\right\\}\} =\sqrt\{f(0)\}=\frac\{1\}\{\sqrt\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle P=(0,1)$ 时, $\displaystyle \angle OPA$ 最大, 为 $\displaystyle \frac\{\pi\}\{4\}$.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](
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---
220、 (2)、 若 $\displaystyle \mathrm\{e\}^\{x^2+y\}-x^2y=0$, 求 $\displaystyle \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}$. (广西大学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 $\displaystyle F(x,y)=$, 则由 $\displaystyle F=0$ 知
\begin\{aligned\} &F\_x+F\_y y'=0\Rightarrow y'=-\frac\{F\_x\}\{F\_y\},\\\\ &(F\_\{xx\}+F\_\{xy\}y')+(F\_\{yx\}+F\_\{yy\}y')y'+F\_yy''=0\\\\ \Rightarrow& y''=-\frac\{F\_\{xx\}+2F\_\{xy\}y'+F\_\{yy\}y'^2\}\{F\_y\} =-\frac\{F\_y^2F\_\{xx\}-2F\_xF\_yF\_\{xy\}+F\_x^2F\_y\}\{F\_y^3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
代入后并利用 $\displaystyle F\_x=2\mathrm\{e\}^\{x^2+y\}x-2xy=2x^3y-2xy$, $\displaystyle F\_y=\mathrm\{e\}^\{x^2+y\}-x^2=x^2y-x^2$ 化简 [很重要, 变成了多项式的求导哦] 知
\begin\{aligned\} y''=-\frac\{2y[-3+5x^2-4x^4+2x^2y+2x^4y+y^2-3x^2y^2]\}\{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](
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---
221、 3、 (10 分) 已知定义在 $\displaystyle \mathbb\{R\}$ 上的函数 $\displaystyle y=f(x)$ 二阶连续可导, 且满足
\begin\{aligned\} xf''(x)+x^2f'(x)=1-\mathrm\{e\}^\{-x\}, \forall\ x\in\mathbb\{R\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle x=c\ (c\neq 0)$ 是 $\displaystyle f(x)$ 的极值点, 则 $\displaystyle x=c$ 是极大值点还是极小值点? (2)、 若 $\displaystyle x=0$ 是 $\displaystyle f(x)$ 的极值点, 则 $\displaystyle x=0$ 是极大值点还是极小值点? (哈尔滨工程大学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)、 由极值条件知 $\displaystyle f'(c)=0$. 代入题中等式得
\begin\{aligned\} f''(c)=\frac\{1-\mathrm\{e\}^\{-c\}\}\{c\} > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle c$ 是极小值点. (2)、 由题设, $\displaystyle f'(c)=0$. 再由
\begin\{aligned\} f''(0)=\lim\_\{x\to 0\}f''(x)\xlongequal\{\tiny\mbox\{题设\}\} \lim\_\{x\to 0\}\left\[\frac\{1-\mathrm\{e\}^\{-x\}\}\{x\}-xf'(x)\right\]\xlongequal[\tiny\mbox\{代换\}]\{\tiny\mbox\{等价\}\} 1-0=1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle 0$ 是极小值点.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](
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---
222、 (3)、 函数 $\displaystyle f$ 在区间 $\displaystyle I$ 上可导, 且 $\displaystyle f'$ 在 $\displaystyle I$ 上处处有限, 则 $\displaystyle f'$ 在 $\displaystyle I$ 上有界. (哈尔滨工业大学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
\begin\{aligned\} f(x)=&\left\\{\begin\{array\}\{llllllllllll\}x^2\sin\frac\{1\}\{x^2\},&\in [-1,0)\cup (0,1],\\\\ 0,&x=0\end\{array\}\right.\\\\ \Rightarrow f'(x)=&\left\\{\begin\{array\}\{llllllllllll\}2x\sin\frac\{1\}\{x^2\}-\frac\{2\}\{x\}\cos\frac\{1\}\{x^2\},&x\neq 0,\\\\ 0,&x=0.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f'\left(\frac\{1\}\{\sqrt\{2k\pi\}\}\right)=-2\sqrt\{2k\pi\}\to-\infty$ 知 $\displaystyle f'$ 在 $\displaystyle [-1,1]$ 上无界.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)/
---
223、 3、 设 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上二次可微, 且对任意的 $\displaystyle h > 0$, 有
\begin\{aligned\} f(x)\leq \frac\{1\}\{2\}[f(x+h)+f(x-h)]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle f''(x)\geq 0$. (哈尔滨工业大学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\} 0\leq&\lim\_\{h\to 0\}\frac\{f(x+h)+f(x-h)-2f(x)\}\{h^2\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \lim\_\{h\to 0\}\frac\{f'(x+h)-f'(x-h)\}\{2h\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \lim\_\{h\to 0\}\frac\{f''(x+h)+f''(x-h)\}\{2\}=f''(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)/
---
224、 4、 (15 分) 若 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上有三阶导数且 $\displaystyle f(1)=0$. 设 $\displaystyle F(x)=x^3f(x)$, 证明: 在 $\displaystyle [0,1]$ 上至少存在一点 $\displaystyle \xi$, 使得 $\displaystyle F'''(\xi)=0$. (合肥工业大学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
\begin\{aligned\} F'(x)=3x^2f(x)+x^3f'(x), F''(x)=6xf(x)+6x^2f'(x)+x^3f''(x) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle F(0)=F(1)=F'(0)=F''(0)=0$. 由 Rolle 定理知
\begin\{aligned\} &\exists\ 0 < x\_1 < 1,\mathrm\{ s.t.\} F'(0)=F'(x\_1)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ 0 < x\_2 < x\_1,\mathrm\{ s.t.\} F''(0)=F''(x\_2)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}& \exists\ 0 < \xi < x\_2,\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\}
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](
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---
225、 4、 解答题. 每题 15 分, 共 60 分. (1)、 设 $\displaystyle f(x)\in C[0,1]$, 在 $\displaystyle (0,1)$ 内二阶可导, 且
\begin\{aligned\} f(0)=0;\quad f(1)=1;\quad \forall\ x\in (0,1), f''(x)\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: (1-1)、 $\displaystyle f\left(\frac\{1\}\{2\}\right)\neq \frac\{1\}\{2\}$; (1-2)、 存在唯一的 $\displaystyle \xi\in \left(\frac\{1\}\{2\},1\right)$, 使得 $\displaystyle f\left(\xi-\frac\{1\}\{2\}\right)=f(\xi)-\frac\{1\}\{2\}$. (河海大学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) / (1-1)、 用反证法. 若 $\displaystyle f\left(\frac\{1\}\{2\}\right)=\frac\{1\}\{2\}$, 则由 Lagrange 中值定理知 $\displaystyle \exists\ 0 < \xi < \frac\{1\}\{2\} < \eta < 1,\mathrm\{ s.t.\}$
\begin\{aligned\} f'(\xi)=\frac\{f\left(\frac\{1\}\{2\}\right)-f(0)\}\{\frac\{1\}\{2\}-0\}=1, f'(\eta)=\frac\{f(1)-f\left(\frac\{1\}\{2\}\right)\}\{1-\frac\{1\}\{2\}\}=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再由 Rolle 定理知 $\displaystyle \exists\ \zeta\in (\xi,\eta),\mathrm\{ s.t.\} f''(\zeta)=0$. 这与题设矛盾. 故有结论. (1-2)、 设 $\displaystyle F(x)=f\left(x-\frac\{1\}\{2\}\right)-f(x)+\frac\{1\}\{2\}$, 则
\begin\{aligned\} F\left(\frac\{1\}\{2\}\right)=&f(0)-f\left(\frac\{1\}\{2\}\right)+\frac\{1\}\{2\} =\frac\{1\}\{2\}-f\left(\frac\{1\}\{2\}\right),\\\\ F(1)=&f\left(\frac\{1\}\{2\}\right)-f(1)+\frac\{1\}\{2\} =f\left(\frac\{1\}\{2\}\right)-\frac\{1\}\{2\}. \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\left(\frac\{1\}\{2\}\right)\cdot F(1)=-\left\[f\left(\frac\{1\}\{2\}\right)-\frac\{1\}\{2\}\right\]^2 < 0$. 由连续函数介值定理, $\displaystyle \exists\ \xi\in \left(\frac\{1\}\{2\},1\right),\mathrm\{ s.t.\} F(\xi)=0$. 又由 $\displaystyle f''\neq 0$ 及导数介值定理知要么 $\displaystyle f''$ 恒小于 $\displaystyle 0$, 要么 $\displaystyle f''$ 恒大于 $\displaystyle 0$. 不妨设前一情形成立, 则
\begin\{aligned\} F'(x)=f'\left(x-\frac\{1\}\{2\}\right)-f'(x)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f''(\xi\_x)\left(-\frac\{1\}\{2\}\right) > 0\Rightarrow F\mbox\{严\}\nearrow . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而 $\displaystyle F$ 的在 $\displaystyle \left(\frac\{1\}\{2\},1\right)$ 上的零点 $\displaystyle \xi$ 是唯一的.跟锦数学微信公众号. [在线资料](
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---
226、 6、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, $\displaystyle f(0)=0, f(1)=1$. 证明: 存在 $\displaystyle \lambda,\mu\in (0,1)$, $\displaystyle \lambda\neq \mu$, 使得 $\displaystyle f'(\lambda)[1+f'(\mu)]=2$. (河南大学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) / 设 $\displaystyle F(x)=f(x)-2+2x$, 则 $\displaystyle F(0)=-2 < 1=F(1)$. 由连续函数介值定理知
\begin\{aligned\} \exists\ \xi\in (0,1),\mathrm\{ s.t.\} 0=F(\xi)\Rightarrow f(\xi)=2-2\xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 Lagrange 中值定理, $\displaystyle \exists\ 0 < \lambda < \xi < \mu < 1,\mathrm\{ s.t.\}$
\begin\{aligned\} f'(\lambda)=\frac\{f(\xi)-f(0)\}\{\xi-0\}=\frac\{2(1-\xi)\}\{\xi\}, f'(\mu)=\frac\{f(1)-f(\xi)\}\{1-\xi\}=\frac\{2\xi-1\}\{1-\xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle f'(\lambda)[1+f'(\mu)]=2$.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](
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---
227、 2、 设 $\displaystyle y=x^\{\sin x\}$, 求 $\displaystyle \mathrm\{ d\} y$. (黑龙江大学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\} &\ln y=\sin x\ln x\Rightarrow \frac\{y'\}\{y\}=\cos x\ln x+\frac\{\sin x\}\{x\}\\\\ \Rightarrow&\mathrm\{ d\} y=x^\{\sin x\}\left(\cos x\ln x+\frac\{\sin x\}\{x\}\right)\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)/
---
228、 4、 设 $\displaystyle f(x)$ 是 $\displaystyle (0,+\infty)$ 上的凸函数. 求证: $\displaystyle F(x)=\frac\{1\}\{x\}\int\_0^x f(t)\mathrm\{ d\} t$ 为 $\displaystyle (0,+\infty)$ 上的凸函数. (湖南大学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) / 由 $\displaystyle F(x)\stackrel\{t=xs\}\{=\}\int\_0^1 f(xs)\mathrm\{ d\} s$ 知对 $\displaystyle \forall\ 0 < x < y, 0\leq \theta\leq 1$,
\begin\{aligned\} &F\left((1-\theta)x+\theta y\right)=\int\_0^1 f\left(\left((1-\theta)x+\theta y\right)s\right)\mathrm\{ d\} s\\\\ =&\int\_0^1 f\left((1-\theta)xs+\theta ys\right)\mathrm\{ d\} s \leq \int\_0^1 \left\[(1-\theta)f(xs)+\theta f(ys)\right\]\mathrm\{ d\} s\\\\ =&(1-\theta)F(x)+\theta F(y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle F$ 为 $\displaystyle (0,+\infty)$ 上的凸函数.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)/
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229、 4、 已知 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上连续, $\displaystyle h > 0$, 记
\begin\{aligned\} F(x)=\frac\{1\}\{h^2\}\int\_0^h \mathrm\{ d\} s\int\_0^h f(x+s+t)\mathrm\{ d\} t, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle F''(x)$. (华东理工大学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) / 由
\begin\{aligned\} F(x)&\stackrel\{x+s+t=\tau\}\{=\}\frac\{1\}\{h^2\}\int\_0^h\mathrm\{ d\} s\int\_\{x+s\}^\{x+s+h\}f(\tau)\mathrm\{ d\} \tau \stackrel\{x+s=\theta\}\{=\}\frac\{1\}\{h^2\}\int\_x^\{x+h\} \mathrm\{ d\} \theta\int\_\{\theta\}^\{\theta+h\}f(\tau)\mathrm\{ d\} \tau \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} F'(x)=&\frac\{1\}\{h^2\}\left\[\int\_\{x+h\}^\{x+2h\}f(\tau)\mathrm\{ d\} \tau-\int\_x^\{x+h\}f(\tau)\mathrm\{ d\} \tau\right\],\\\\ F''(x)=&\frac\{[f(x+2h)-f(x+h)]-[f(x+h)-f(x)]\}\{h^2\}\\\\ =&\frac\{f(x+2h)-2f(x+h)+f(x)\}\{h^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)/
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230、 6、 设 $\displaystyle a > 1$, 且
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}x^a, &x\in \mathbb\{Q\},\\\\ 0,&x\not\in\mathbb\{Q\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
讨论 $\displaystyle f(x)$ 的可微性. (华东师范大学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)、 若 $\displaystyle a\not\in \mathbb\{Q\}$ 或 $\displaystyle a=\frac\{p\}\{q\}, (p,q)=1, q=2k$, 则 $\displaystyle f$ 的定义域为 $\displaystyle [0,+\infty)$. 对 $\displaystyle x\_0 > 0$, 取有理数列 $\displaystyle r\_n\to x\_0$, 则 $\displaystyle f(r\_n)=r\_n^a\to x\_0^a$. 取无理数列 $\displaystyle x\_n\to x\_0$, 则 $\displaystyle f(x\_n)=0$. 故 $\displaystyle f$ 在 $\displaystyle x\_0$ 处不连续, 更不可微. (2)、 若 $\displaystyle a=\frac\{p\}\{q\}, (p,q)=1, q=2k-1$, 则 $\displaystyle f$ 的定义域为 $\displaystyle \mathbb\{R\}$. 同上可知 $\displaystyle f$ 在非零点处不连续, 更不可微. 而在 $\displaystyle x\_0=0$ 处,
\begin\{aligned\} \frac\{f(x)-f(0)\}\{x\}=\left\\{\begin\{array\}\{llllllllllll\}x^\{a-1\}, &x\not\in \mathbb\{Q\},\\\\ 0,&x\not\in \mathbb\{Q\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle \left|\frac\{f(x)-f(0)\}\{x\}\right|\leq |x|^\{a-1\}$. 由 $\displaystyle a > 1$ 及迫敛性知 $\displaystyle f$ 在原点处可微 [若为定义域端点, 则就考虑单侧导数], 且 $\displaystyle f'(0)=0$.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 08:45:37
