## 张祖锦2023年数学专业真题分类70天之第13天
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277、 8、 (15 分) 已知 $\displaystyle f(x)$ 在 $\displaystyle [-1,1]$ 上由三阶连续导数, $\displaystyle f(-1)=0, f(1)=1, f'(0)=0$. 求证: 存在 $\displaystyle \xi\in (-1,1)$, 使得 $\displaystyle f'''(\xi)=3$. (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) / 由 Taylor 展开,
\begin\{aligned\} 0=&f(-1)=f(0)+f'(0)(-1-0)+\frac\{f''(0)\}\{2\}(-1-0)^2+\frac\{f'''(\eta)\}\{6\}(-1-0)^3\\\\ =&f(0)+\frac\{f''(0)\}\{2\}-\frac\{f'''(\eta)\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同理,
\begin\{aligned\} 1=f(1)=f(0)+\frac\{f''(0)\}\{2\}+\frac\{f'''(\zeta)\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相减即得
\begin\{aligned\} 1=\frac\{f'''(\eta)+f'''(\zeta)\}\{6\}. \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\} f'''(\eta)\geq 3\Rightarrow f'''(\zeta)\leq 3;\quad f'''(\eta) < 3\Rightarrow f'''(\zeta) > 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
无论何种情形, 均可由连续函数的介值定理导出结论: $\displaystyle \exists\ \xi$ 介于 $\displaystyle \eta,\zeta$, 使得 $\displaystyle f'''(\xi)=3$.跟锦数学微信公众号. [在线资料](
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278、 1、 (15 分) 计算函数
\begin\{aligned\} y(x)=\sqrt\{x+\sqrt[3]\{x+\sqrt[4]\{x\}\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的导数, 并求
\begin\{aligned\} \lim\_\{x\to+\infty\}\left(\sqrt\{x+\sqrt[3]\{x+\sqrt[4]\{x\}\}\}-\sqrt\{x\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle y$ 在它的定义域上是否一致连续? (中国科学技术大学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)、
\begin\{aligned\} y'=\frac\{1+\frac\{1+\frac\{1\}\{4x^\frac\{3\}\{4\}\}\}\{3(x+x^\frac\{1\}\{4\})^\frac\{2\}\{3\}\}\}\{2\sqrt\{x+\sqrt[3]\{x+\sqrt[4]\{x\}\}\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再者,
\begin\{aligned\} &\lim\_\{x\to+\infty\}\left(\sqrt\{x+\sqrt[3]\{x+\sqrt[4]\{x\}\}\}-\sqrt\{x\}\right)\\\\ =&\lim\_\{x\to+\infty\} \frac\{\sqrt[3]\{x+\sqrt[4]\{x\}\}\}\{\sqrt\{x+\sqrt[3]\{x+\sqrt[4]\{x\}\}+\sqrt\{x\}\}\}\\\\ =&\lim\_\{x\to+\infty\} \frac\{\sqrt[3]\{1+\frac\{\sqrt[4]\{x\}\}\{\sqrt[3]\{x\}\}\}\}\{\sqrt\{1+\frac\{\sqrt[3]\{x+\sqrt[4]\{x\}\}+\sqrt\{x\}\}\{\sqrt\{x\}\}\}+1\}\cdot\frac\{1\}\{x^\frac\{1\}\{6\}\} \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\}
(2)、 为了回答最后一问, 我们先证明如下结果. 设函数 $\displaystyle g(x)$ 在 $\displaystyle [a,+\infty)$ 上一致连续, $\displaystyle f(x)$ 在 $\displaystyle [a,\infty)$ 上连续, 且 $\displaystyle \lim\_\{x\to+\infty\}[f(x)-g(x)]=0$. 则 $\displaystyle f(x)$ 在 $\displaystyle [a,\infty)$ 上一致连续. 事实上, 对 $\displaystyle \forall\ \varepsilon > 0$,
\begin\{aligned\} \exists\ X > a+1,\mathrm\{ s.t.\} x\geq X\Rightarrow |f(x)-g(x)| < \frac\{\varepsilon\}\{3\};\\\\ \exists\ 0 < \delta\_1 < 1,\mathrm\{ s.t.\} x,y\geq a,\ |x-y| < \delta\Rightarrow |g(x)-g(y)| < \frac\{\varepsilon\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &x,y\geq X,\ |x-y| < \delta\_1 \\\\ \Rightarrow &|f(x)-f(y)|\leq |f(x)-g(x)| +|g(x)-g(y)| +|g(y)-f(y)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又 $\displaystyle f$ 在 $\displaystyle [a,X+1]$ 上连续, 而一致连续,
\begin\{aligned\} \exists\ 0 < \delta < \delta\_1,\mathrm\{ s.t.\} a\leq x,y\leq X+1,\ |x-y| < \delta\Rightarrow |f(x)-f(y)| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
如此, 对 $\displaystyle \forall\ a\leq x < y,\ |x-y| < \delta$, 若 $\displaystyle a\leq x\leq X$, 则 $\displaystyle a\leq y\leq x+\delta < x+\delta\_1 < x+1\leq X+1$, 而 $\displaystyle |f(x)-f(y)| < \varepsilon$; 若 $\displaystyle x > X$, 则 $\displaystyle y > x > X$, 而也有 $\displaystyle |f(x)-f(y)| < \varepsilon$. (3)、 回到题目. 对 $\displaystyle \forall\ \varepsilon > 0, \exists\ \delta=\varepsilon^2 > 0,\mathrm\{ s.t.\} \forall\ 0\leq x < y < x+\delta$,
\begin\{aligned\} \sqrt\{y\}-\sqrt\{x\}\leq \sqrt\{y-x\} < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \sqrt\{x\}$ 在 $\displaystyle [0,+\infty)$ 上一致连续. 由第 2 步即知 $\displaystyle y$ 在其定义域 $\displaystyle x\geq 0$ 上一致连续.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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279、 6、 (15 分) 设
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}x^2\sin^2\frac\{1\}\{x\},&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\}
计算 $\displaystyle f$ 的导数 $\displaystyle f'$, 并讨论 $\displaystyle f'$ 的连续性. (中国科学技术大学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) / 当 $\displaystyle x\neq 0$ 时,
\begin\{aligned\} f'(x)=-2\sin\frac\{1\}\{x\}\cos \frac\{1\}\{x\}+2x\sin^2\frac\{1\}\{x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle x=0$ 时,
\begin\{aligned\} f'(0)=\lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x\} =\lim\_\{x\to 0\}x\sin^2\frac\{1\}\{x\}\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\}
由
\begin\{aligned\} f'\left(\frac\{1\}\{k\pi+\frac\{\pi\}\{4\}\}\right)=&-1+\frac\{4\}\{(4k+1)\pi\}\xrightarrow\{k\to+\infty\}-1\neq 0=f'(0),\\\\ f'\left(\frac\{1\}\{k\pi+\frac\{3\pi\}\{4\}\}\right)=&1+\frac\{4\}\{(4k+3)\pi\}\xrightarrow\{k\to+\infty\}1\neq 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 $\displaystyle f'$ 除了以 $\displaystyle 0$ 为第二类间断点外都连续.跟锦数学微信公众号. [在线资料](
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280、 10、 (15 分) 设
\begin\{aligned\} g\_n(x)=\mathrm\{e\}^\{x^2\}\frac\{\mathrm\{ d\}^n\}\{\mathrm\{ d\} x^n\}(\mathrm\{e\}^\{-x^2\}). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求证: $\displaystyle g\_n$ 是 $\displaystyle n$ 次多项式; (2)、 求证: $\displaystyle g\_n$ 的所有根都是实数. (中国科学技术大学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)、 用数学归纳法证明 $\displaystyle (\mathrm\{e\}^\{-x^2\})^\{(n)\}=g\_n(x)\mathrm\{e\}^\{-x^2\}$, 其中 $\displaystyle g\_n(x)$ 是 $\displaystyle n$ 次多项式. 当 $\displaystyle n=1$ 时, 结论自明. 设结论对 $\displaystyle n$ 成立, 则当 $\displaystyle n+1$ 时,
\begin\{aligned\} (\mathrm\{e\}^\{-x^2\})^\{(n+1)\} =[g\_n(x)\mathrm\{e\}^\{-x^2\}]' =[g\_n'(x)-2xg\_n(x)]\mathrm\{e\}^\{-x^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle g\_\{n+1\}(x)=g\_n'(x)-2xg\_n(x)$ 是 $\displaystyle n+1$ 次多项式. (2)、 不断分部积分知
\begin\{aligned\} &\int\_\{\mathbb\{R\}\} x^kg\_n(x)\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x =\int\_\{\mathbb\{R\}\} x^k \mathrm\{ d\} (\mathrm\{e\}^\{-x^2\})^\{n-1\}\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&-\int\_\{\mathbb\{R\}\} k x^\{k-1\}(\mathrm\{e\}^\{-x^2\})^\{(n-1)\}\mathrm\{ d\} x \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}\cdots=0, 0\leq k\leq n-1.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 设 $\displaystyle g\_n(x)$ 重数为奇数的不同实根为 $\displaystyle x\_1,\cdots,x\_m$ ($m\leq n$, 而其它实根的重数均为偶数, 所有虚根成对出现). 这表明 $\displaystyle g\_n(x)(x-x\_1)\cdots(x-x\_m)$ 在 $\displaystyle (0,+\infty)$ 上不变号, 且仅在至多 $\displaystyle n$ 个点处取值为零. 由于 $\displaystyle \mathrm\{e\}^\{-x^2\} > 0$, $\displaystyle \mathrm\{e\}^\{-x^2\}g\_n(x)(x-x\_1)\cdots(x-x\_m)$ 也满足上述性质. 于是
\begin\{aligned\} \int\_\{\mathbb\{R\}\} e^\{-x^2\}g\_n(x)(x-x\_1)\cdots(x-x\_m)\mathrm\{ d\} x\neq 0.\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle m\leq n-1$, 则由 $\displaystyle (I)$ 知 $\displaystyle \int\_\{\mathbb\{R\}\} e^\{-x^2\}g\_n(x)(x-x\_1)\cdots(x-x\_m)\mathrm\{ d\} x=0.$ 这与 $\displaystyle (II)$ 矛盾. 故 $\displaystyle m=n$, $\displaystyle g\_n(x)$ 的重数为奇数的不同实根为 $\displaystyle x\_1,\cdots,x\_n$. 这些重数也都 $\displaystyle =1$.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](
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281、 3、 设 $\displaystyle f(x)$ 为 $\displaystyle [a,b]$ 上的二阶可导函数, $\displaystyle f(a)=f(b)=0$, 并存在一点 $\displaystyle c\in (a,b)$, 使得 $\displaystyle f(c)=0$, 且 $\displaystyle f(x)$ 非常值函数. 求证: 存在 $\displaystyle \xi\in (a,b)$, 使得 $\displaystyle f''(\xi) < 0$. (中国科学院大学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 $\displaystyle \forall\ x\in (a,b), f''(x)\geq 0$, 则 $\displaystyle f$ 是凸函数,
\begin\{aligned\} a < x < b\Rightarrow&c=(1-\theta\_x)a+\theta\_x b\\\\ \Rightarrow&f(x)\leq (1-\theta\_x)f(a)+\theta\_x f(b)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f$ 非常值函数知 $\displaystyle \exists\ \xi\in (a,b)$, 使得 $\displaystyle f(\xi) < 0$. 不妨设 $\displaystyle a < \xi < c$ (若 $\displaystyle c < \xi < b$, 类似讨论可得矛盾). 进而
\begin\{aligned\} \xi < c < b\Rightarrow&\exists\ \tau\in (0,1),\mathrm\{ s.t.\} c=(1-\tau)\xi+\tau b\\\\ \Rightarrow& 0=f(c)\leq (1-\tau)f(\xi)+\tau f(b)=(1-\tau)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\}
这是一个矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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282、 8、 (15 分) 求函数 $\displaystyle f(x)=\sqrt[3]\{(x-2)(x+1)^2\}$ 的单调区间, 凹凸区间, 极值, 拐点及渐近线. (中国矿业大学(北京)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^3(x)=(x-2)(x+1)^2\left\\{\begin\{array\}\{llllllllllll\}\leq 0,&x\leq 2,\\\\ > 0,&x > 2\end\{array\}\right.\\\\ \Rightarrow&3f^2(x)f'(x)=3(x-1)(x+1)\left\\{\begin\{array\}\{llllllllllll\} < 0,&-1 < x < 1,\\\\ > 0,&x < -1\mbox\{或\} x > 1.\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 $\displaystyle f$ 在 $\displaystyle (-\infty,-1)\nearrow$, 在 $\displaystyle (-1,1)\searrow$, 在 $\displaystyle (1,+\infty)\nearrow$. 从而 $\displaystyle f$ 在 $\displaystyle x=-1$ 处取得极大值 $\displaystyle 0$, 在 $\displaystyle x=1$ 处取得极小值 $\displaystyle -\sqrt[3]\{4\}$. 再者,
\begin\{aligned\} f''(x)=-\frac\{2\}\{(x-2)^\frac\{5\}\{3\}(x+1)^\frac\{4\}\{3\}\}\left\\{\begin\{array\}\{llllllllllll\} > 0,&x < 2,\\\\ < 0,&x > 2\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 $\displaystyle f$ 在 $\displaystyle (-\infty,2)$ 上凸, 在 $\displaystyle (2,+\infty)$ 上凹, $\displaystyle (2,0)$ 为拐点. 最后, $\displaystyle \lim\_\{x\to\infty\}\frac\{f(x)\}\{x\}=1$,
\begin\{aligned\} \lim\_\{x\to\infty\}\frac\{f(x)-x\}\{x\} =\lim\_\{x\to\infty\}\left\[\sqrt[3]\{\left(1-\frac\{2\}\{x\}\right)\left(1+\frac\{1\}\{x\}\right)\}-1\right\] =0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 只有斜渐近线 $\displaystyle y=x$.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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283、 12、 (10 分) 已知函数 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上有二阶导数, 且 $\displaystyle f(0)=f(1)=f'(0)=f'(1)=0$. 证明: 存在 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f''(\xi)=f(\xi)$. (中国矿业大学(北京)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\} &\int \mathrm\{e\}^x f''(x)\mathrm\{ d\} x=\int \mathrm\{e\}^x \mathrm\{ d\} f'(x) =\mathrm\{e\}^x f'(x)-\int \mathrm\{e\}^x f'(x)\mathrm\{ d\} x\\\\ =&\mathrm\{e\}^x f'(x)-\int \mathrm\{e\}^x \mathrm\{ d\} f(x) =\mathrm\{e\}^x f'(x)-\left\[\mathrm\{e\}^x f(x)-\int \mathrm\{e\}^x f(x)\mathrm\{ d\} x\right\]\\\\ =&\mathrm\{e\}^x[f'(x)+f(x)]+\int \mathrm\{e\}^x 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\}
知
\begin\{aligned\} \mathrm\{e\}^x[f''(x)-f(x)]=\left\\{\mathrm\{e\}^x[f'(x)+f(x)]\right\\}'. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故令 $\displaystyle F(x)=\mathrm\{e\}^x [f'(x)+f(x)]$ 后, 由题设知 $\displaystyle F(0)=F(1)=0$. 而由 Rolle 定理知
\begin\{aligned\} \exists\ \xi\in (0,1),\mathrm\{ s.t.\} 0=F'(\xi)=\mathrm\{e\}^\{\xi\}[f''(\xi)-f(\xi)] \Rightarrow 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](
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284、 3、 (15 分) 设函数 $\displaystyle F(x)=\int\_0^x \mathrm\{e\}^\{-t\}\cos t\mathrm\{ d\} t$, 计算 $\displaystyle F(x)$ 在 $\displaystyle [0,2\pi]$ 上的最大值与最小值. (中国人民大学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)=\mathrm\{e\}^\{-x\}\cos x\left\\{\begin\{array\}\{llllllllllll\} > 0,&2k\pi-\frac\{\pi\}\{2\} < x < 2k\pi+\frac\{\pi\}\{2\},\\\\ < 0,&2k\pi+\frac\{\pi\}\{2\} < x < 2k\pi+\frac\{3\pi\}\{2\}\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\}
知 $\displaystyle F$ 在 $\displaystyle \frac\{\pi\}\{2\}$ 处取得极大值, 在 $\displaystyle \frac\{3\pi\}\{2\}$ 处取得极小值. 再由
\begin\{aligned\} F(x)\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{\mathrm\{e\}^\{-x\}\}\{2\}(\mathrm\{e\}^x-\cos x+\sin x) \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\} \max\_\{[0,2\pi]\}F=&\max\left\\{F\left(\frac\{\pi\}\{2\}\right),F\left(2\pi\right)\right\\}=F\left(\frac\{\pi\}\{2\}\right) =\frac\{1+\mathrm\{e\}^\{-\frac\{\pi\}\{2\}\}\}\{2\},\\\\ \min\_\{[0,2\pi]\}F=&\min\left\\{F(0),F\left(\frac\{3\pi\}\{2\}\right)\right\\} =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\}
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)/
---
285、 4、 (15 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 内有二阶导数, $\displaystyle c$ 为 $\displaystyle (a,b)$ 内任意一点. 证明: $\displaystyle \exists\ \xi\in (a,b)$, 使得
\begin\{aligned\} \frac\{f(a)\}\{(a-b)(a-c)\}+\frac\{f(b)\}\{(b-c)(b-a)\}+\frac\{f(c)\}\{(c-a)(c-b)\}=\frac\{1\}\{2\}f''(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学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) / 考虑函数
\begin\{aligned\} F(x)=&f(x)-\frac\{f(a)\}\{(a-b)(a-c)\}(x-b)(x-c) -\frac\{f(b)\}\{(b-a)(b-c)\}(x-a)(x-c)\\\\ & -\frac\{f(c)\}\{(c-a)(c-b)\}(x-a)(x-b), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} &F(a)=F(b)=F(c)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ a < \eta < b < \zeta < c,\mathrm\{ s.t.\} F'(\eta)=F'(\zeta)=0\\\\ \stackrel\{\mbox\{Rolle\}\}\{\Longrightarrow\}&\exists\ \eta < \xi < \zeta,\mathrm\{ s.t.\} 0=F''(\xi)\\\\ &=f''(\xi)-\left\[\frac\{2f(a)\}\{(a-b)(a-c)\}+\frac\{2f(b)\}\{(b-c)(b-a)\}+\frac\{2f(c)\}\{(c-a)(c-b)\}\right\]. \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)/
---
286、 5、 (15 分) 解答如下问题. (1)、 设函数 $\displaystyle f$ 在 $\displaystyle x=0$ 附近有三阶导数, 且其带带皮亚诺型余项的三阶麦克劳林展开式为
\begin\{aligned\} f(x)=1+x+3x^2+4x^3+o(x^3), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
试写出 $\displaystyle f$ 的反函数 $\displaystyle g$ 的带皮亚诺型余项的三阶麦克劳林展开式. (中南大学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\} g(x)=&\frac\{1\}\{f(x)\} =\frac\{1\}\{1+x+3x^2+4x^3+o(x^3)\}\\\\ =&1-(x+3x^2+4x^3)+(x+3x^2)^2-x^3+o(x^3)\\\\ =&1-x-2x^2+x^3+o(x^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)/
---
287、 (2)、 设 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上有定义, 且对任何 $\displaystyle x,y\in (0,+\infty)$, 都有
\begin\{aligned\} f(xy)=f(x)+f(y). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle f'(1)$ 存在, 求 $\displaystyle f'(x)$. (重庆大学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) / 令 $\displaystyle x=y=1$ 知 $\displaystyle f(1)=0$. 于是
\begin\{aligned\} f'(x)=&\lim\_\{y\to x\}\frac\{f(y)-f(x)\}\{y-x\} =\lim\_\{y\to x\}\frac\{f\left(x\cdot\frac\{y\}\{x\}\right)-f(x)\}\{y-x\} =\lim\_\{y\to x\}\frac\{f\left(\frac\{y\}\{x\}\right)\}\{y-x\}\\\\ =&\lim\_\{y\to x\}\frac\{f\left(\frac\{y\}\{x\}\right)-f(1)\}\{\frac\{y\}\{x\}-1\}\cdot \frac\{1\}\{x\} =\frac\{f'(1)\}\{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)/
---
288、 (4)、 求函数 $\displaystyle f(x)=\int\_1^\{x^2\}(x^2-t)\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t$ 的单调区间与极值. (重庆大学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 $\displaystyle f(x)=x^2\int\_1^\{x^2\}\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t-\int\_1^\{x^2\}t\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t$ 知
\begin\{aligned\} f'(x)=&2x\int\_1^\{x^2\}\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t+x^2\mathrm\{e\}^\{-x^4\}\cdot 2x-x^2\mathrm\{e\}^\{-x^4\}\cdot 2x\\\\ =&2x\int\_1^\{x^2\}\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t \left\\{\begin\{array\}\{llllllllllll\} > 0,&x < 0,\\\\ < 0,&0 < x < 1,\\\\ > 0,&x > 1.\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 $\displaystyle f$ 在 $\displaystyle (-\infty,0)$ 上严增, 在 $\displaystyle (0,1)$ 上严减, 在 $\displaystyle (1,+\infty)$ 上严增. 进而 $\displaystyle f$ 在 $\displaystyle x=0$ 处取得极大值
\begin\{aligned\} f(0)=\int\_1^0 (-t)\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t=\frac\{\mathrm\{e\}-1\}\{2\mathrm\{e\}\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在 $\displaystyle x=1$ 处取得极小值 $\displaystyle f(1)=0$.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)/
---
289、 4、 (10 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, $\displaystyle f(0)=0$, 且对任意 $\displaystyle x\in (0,1)$ 有 $\displaystyle f(x)\neq 0$. 证明: 对任意正整数 $\displaystyle n$, 存在 $\displaystyle \xi\in (0,1)$, 使得
\begin\{aligned\} \frac\{nf'(\xi)\}\{f(\xi)\}=\frac\{f'(1-\xi)\}\{f(1-\xi)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(重庆大学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) / 即要证 $\displaystyle \exists\ \xi\in (0,1),\mathrm\{ s.t.\}$
\begin\{aligned\} nf'(\xi)f(1-\xi)-f(\xi)f'(1-\xi)=&0,\\\\ f^\{n-1\}(\xi)\left\[nf'(\xi)f(1-\xi)-f(\xi)f'(1-\xi)\right\]=&0\\\\ \left.\left\[f^n(x)f(1-x)\right\]'\right|\_\{x=\xi\}=&0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle F(x)=f^n(x)f(1-x)$, 则 $\displaystyle F(0)=F(1)=0$. 而由 Rolle 定理知 $\displaystyle \exists\ \xi\in (0,1),\mathrm\{ s.t.\} F'(\xi)=0$. 结论得证.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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---
290、 (4)、 设 $\displaystyle y=\mathrm\{e\}^x\sin y+\mathrm\{e\}^y\sin x-7x$, 则 $\displaystyle y'=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学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) / 设 $\displaystyle F=\mathrm\{e\}^x\sin y+\mathrm\{e\}^y\sin x-7x-y$, 则
\begin\{aligned\} y'=-\frac\{F\_x\}\{F\_y\}=\frac\{7-\mathrm\{e\}^y\cos x-\mathrm\{e\}^x\sin y\}\{\mathrm\{e\}^x\cos y+\mathrm\{e\}^y\sin x-1\}. \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)/
---
291、 3、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上二阶可导, $\displaystyle f(a)=f(b)=0$. 证明:
\begin\{aligned\} \max\_\{x\in [a,b]\}|f(x)|\leq \frac\{1\}\{8\}(b-a)^2\max\_\{x\in [a,b]\}|f''(x)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京师范大学2023年数学分析考研试题) [微分法与不等式 ]
[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 $\displaystyle M=\max\_\{[a,b]\}|f''|$. 则由 Taylor 公式知
\begin\{aligned\} 0&=f(a)=f(x)+f'(x)(a-x)+\frac\{f''(\xi)\}\{2\}(a-x)^2,\qquad(I)\\\\ 0&=f(b)=f(x)+f'(x)(b-x)+\frac\{f''(\eta)\}\{2\}(b-x)^2,\qquad(II). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (b-x)\cdot(I)-(a-x)\cdot(II)$ 知
\begin\{aligned\} &(b-a)|f(x)|\leq\frac\{M\}\{2\}(b-x)(a-x)^2+\frac\{M\}\{2\}(x-a)(b-x)^2\\\\ =&\frac\{M\}\{2\}(x-a)(b-x)\left\[(x-a)+(b-x)\right\] \leq\frac\{(b-a)^3M\}\{8\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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](
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---
292、 4、 已知 $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上二阶可导, 且 $\displaystyle f''(x)\leq 0, \lim\_\{x\to 0\}\frac\{f(x)\}\{2x\}=2$. 证明: $\displaystyle f(x)\leq 4x$. (北京邮电大学2023年数学分析考研试题) [微分法与不等式 ]
[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(0)=&\lim\_\{x\to 0\}\frac\{f(x)\}\{2x\}\cdot 2x=2\cdot 0=0,\\\\ f'(0)=&\lim\_\{x\to 0\}\frac\{f(x)-f(0)\}\{x\}=4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle f''\leq 0$ 知 $\displaystyle f$ 是凹函数, 而
\begin\{aligned\} x\in\mathbb\{R\}\Rightarrow f(x)\leq f(0)+f'(0)x=4x. \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)/
---
293、 11、 已知 $\displaystyle 0 < a < b < 1$. 证明:
\begin\{aligned\} \arctan b-\arctan a < \frac\{b-a\}\{2ab\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京邮电大学2023年数学分析考研试题) [微分法与不等式 ]
[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 $\displaystyle f(x)=\arctan x+\arctan \frac\{1\}\{x\}, x > 0$, 则
\begin\{aligned\} f'(x)=0\Rightarrow f(x)=f(1)=\frac\{\pi\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故只要证
\begin\{aligned\} &\left(\frac\{\pi\}\{2\}-\arctan a\right)-\left(\frac\{\pi\}\{2\}-\arctan b\right) < \frac\{1\}\{2\}\left(\frac\{1\}\{a\}-\frac\{1\}\{b\}\right), 0 < a < b < 1\\\\ \Leftrightarrow&\arctan \frac\{1\}\{a\}-\arctan \frac\{1\}\{b\} < \frac\{1\}\{2\}\left(\frac\{1\}\{a\}-\frac\{1\}\{b\}\right), 0 < a < b < 1\\\\ \Leftrightarrow&\arctan x-\arctan y < \frac\{x-y\}\{2\}, 1 < y < x\\\\ \Leftarrow&\frac\{\arctan x-\arctan y\}\{x-y\}\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{1\}\{1+\xi\_x^2\} < \frac\{1\}\{1+y^2\} < \frac\{1\}\{2\}, 1 < y < 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)/
---
294、 9、 设函数 $\displaystyle f: [0,1]\to \mathbb\{R\}$ 是连续的且在 $\displaystyle (0,1)$ 上可微, 若 $\displaystyle f$ 满足: (1)、 $\displaystyle f(0)=0$; (2)、 存在常数 $\displaystyle M > 0$ 使得 $\displaystyle |f'(x)|\leq M|f(x)|$ 对任意 $\displaystyle x\in (0,1)$ 成立. 证明: 在 $\displaystyle [0,1]$ 上 $\displaystyle f(x)\equiv 0$. (东北师范大学2023年数学分析考研试题) [微分法与不等式 ]
[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)、 设 $\displaystyle |f|$ 在 $\displaystyle \left\[0,\min\left\\{\frac\{1\}\{2M\},1\right\\}\right\]$ 上的最大值在 $\displaystyle x\_0$ 处取得, 则
\begin\{aligned\} |f(x\_0)|&=|f(x\_0)-f(0)| =|f'(\xi)x| \leq M|f(\xi)| \frac\{1\}\{2M\}\leq \frac\{1\}\{2\}|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\}
这表明 $\displaystyle f(x\_0)=0$. 故 $\displaystyle f(x)\equiv 0, x\in \left\[0,\min\left\\{\frac\{1\}\{2M\},1\right\\}\right\]$. 若 $\displaystyle \frac\{1\}\{2M\}\geq 1$, 则结论已证. (2)、 若不然, 设 $\displaystyle |f|$ 在 $\displaystyle \left\[\frac\{1\}\{2M\},\min\left\\{\frac\{1\}\{M\},1\right\\}\right\]$ 上的最大值在 $\displaystyle x\_1$ 处取得, 则
\begin\{aligned\} |f(x\_1)|&=\left|f(x\_1)-f\left(\frac\{1\}\{2M\}\right)\right| =\left|f'(\eta)\left(x\_1-\frac\{1\}\{2M\}\right)\right|\\\\ & \leq M|f(\eta)|\cdot \frac\{1\}\{2M\} \leq \frac\{1\}\{2\}|f(x\_1)|. \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 $\displaystyle f(x\_1)=0$. 故 $\displaystyle f(x)\equiv 0, x\in \left\[\frac\{1\}\{2M\},\min\left\\{\frac\{1\}\{M\},1\right\\}\right\]$. (3)、 如此一直做下去, 经过有限步后知 $\displaystyle f\equiv 0$.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](
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---
295、 3、 设 $\displaystyle f$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 上有二阶连续导数,
\begin\{aligned\} f(0)=f(1)=1, f''(x) < 8. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 对任意的 $\displaystyle x\in [0,1]$, 有 $\displaystyle f(x) > 0$. (复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [微分法与不等式 ]
[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 $\displaystyle \min\_\{[0,1]\}f=m$ 在 $\displaystyle c\in [0,1]$ 处取得. (1)、 若 $\displaystyle c=0$ 或 $\displaystyle 1$, 则
\begin\{aligned\} \forall\ x\in [0,1], f(x)\geq m=f(0)=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 若 $\displaystyle 0 < c < 1$, 则由极值条件知 $\displaystyle f'(c)=0$. 再由 Taylor 定理知
\begin\{aligned\} 1&=f(0)=f(c)+\frac\{f''(\xi)\}\{2\}c^2 < m+4c^2,\\\\ 1&=f(1)=f(c)+\frac\{f''(\eta)\}\{2\}(1-c)^2 < m+4(1-c)^2. \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\} &m > 1-4c^2, m > 1-4(1-c)^2\\\\ \Rightarrow&m > \max\left\\{1-4c^2,1-4(1-c)^2\right\\} =1-4\min\left\\{c^2,(1-c)^2\right\\}\\\\ &=1-4\min\left\\{c,1-c\right\\}^2 \geq 1-4\left(\frac\{1\}\{2\}\right)^2=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此事, 亦有
\begin\{aligned\} \forall\ x\in [0,1], f(x)\geq m > 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](
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---
296、 5、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,2]$ 上满足
\begin\{aligned\} |f(x)|\leq 1, |f''(x)|\leq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 在 $\displaystyle [0,2]$ 上, $\displaystyle |f'(x)|\leq 2$; 并举例说明存在 $\displaystyle f(x)$ 满足题目条件且存在 $\displaystyle c\in [0,2]$, 使得 $\displaystyle |f'(c)|\leq 2$. (合肥工业大学2023年数学分析考研试题) [微分法与不等式 ]
[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 Taylor 公式,
\begin\{aligned\} f(0)&=f(x)+f'(x)(0-x)+\frac\{f''(\xi\_x)\}\{2\}(0-x)^2,\\\\ f(2)&=f(x)+f'(x)(2-x)+\frac\{f''(\eta\_x)\}\{2\}(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\}
相减即得
\begin\{aligned\} &2|f'(x)|\leq 2+\frac\{1\}\{2\} [x^2+(2-x)^2]\leq 4\\\\ \Rightarrow&|f'(x)|\leq 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
另外, $\displaystyle f(x)=\frac\{x^2\}\{2\}-1$ 满足题设, 且 $\displaystyle |f'(2)|=2$.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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---
297、 4、 (15 分) 设 $\displaystyle a > 1, n\in\mathbb\{Z\}\_+$, 证明:
\begin\{aligned\} \frac\{a^\frac\{1\}\{n+1\}\}\{(n+1)^2\} < \frac\{a^\frac\{1\}\{n\}-a^\frac\{1\}\{n+1\}\}\{\ln a\} < \frac\{a^\frac\{1\}\{n\}\}\{n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(河南大学2023年数学分析考研试题) [微分法与不等式 ]
[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 $\displaystyle f(x)=a^x$, 则
\begin\{aligned\} a^\frac\{1\}\{n\}-a^\frac\{1\}\{n+1\} =&f\left(\frac\{1\}\{n\}\right)-f\left(\frac\{1\}\{n+1\}\right) \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f'(\xi\_n)\left(\frac\{1\}\{n\}-\frac\{1\}\{n+1\}\right)\\\\ =&\frac\{a^\{\xi\_n\}\ln a\}\{n(n+1)\} \in \left(\frac\{a^\frac\{1\}\{n+1\}\}\{(n+1)^2\},\frac\{a^\frac\{1\}\{n\}\}\{n^2\}\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)/
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298、 12、 设 $\displaystyle x,y\in\mathbb\{R\}$, 试证:
\begin\{aligned\} |\arctan x-\arctan y|\leq |x-y|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(黑龙江大学2023年数学分析考研试题) [微分法与不等式 ]
[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\} |\arctan x-\arctan y| \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{1\}\{1+\xi^2\}|x-y|\leq |x-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)/
---
299、 3、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 内可导, $\displaystyle f(a)=0$, 且 $\displaystyle \forall\ x\in (a,b), f(x) > 0$. 证明: 不存在常数 $\displaystyle M > 0$, 使得
\begin\{aligned\} \left|\frac\{f'(x)\}\{f(x)\}\right|\leq M, x\in (a,b). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(湖南大学2023年数学分析考研试题) [微分法与不等式 ]
[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
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ x\in(a,b), |f'(x)|\leq M|f(x)|, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} &-Mf(x)\leq f'(x)\leq Mf(x)\Rightarrow [\mathrm\{e\}^\{-Mx\}f(x)]'\leq 0, [\mathrm\{e\}^\{Mx\}f(x)]'\geq 0\\\\ \Rightarrow&\mathrm\{e\}^\{-Mx\}f(x)\leq \mathrm\{e\}^\{-Ma\}f(a)=0, \mathrm\{e\}^\{Mx\}f(x)\geq \mathrm\{e\}^\{Ma\}f(a)=0 \Rightarrow f(x)\equiv 0. \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](
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发表于 2023-3-5 08:47:48
