## 张祖锦2023年数学专业真题分类70天之第11天
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231、 8、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上可导, 且满足 $\displaystyle f'\_+(a) < c < f'\_-(b)$. 证明: 存在 $\displaystyle \xi\in (a,b)$, 使得 $\displaystyle f'(\xi)=0$. (华东师范大学2023年数学分析考研试题) [微分 ]
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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. 设 $\displaystyle F(x)=f(x)-c$, 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)=c. \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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232、 3、 函数 $\displaystyle f(x)$ 在 $\displaystyle (a,+\infty)$ 上可导, 且 $\displaystyle f(a+0)=f(+\infty)$. 证明: 在 $\displaystyle (a,+\infty)$ 上至少存在一点 $\displaystyle c$, 使得 $\displaystyle f'(c)=0$. (华南理工大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/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](
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233、 2、 (40 分) 计算题. (1)、 $\displaystyle y=y(x)$ 由参数方程 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x=\frac\{t\}\{1+t^2\},\\\\ y=\frac\{t^2\}\{1+t^2\}\end\{array\}\right.$ 确定, 求 $\displaystyle \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}, \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}$. (华南师范大学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\} \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}&=\frac\{y'(t)\}\{x'(t)\}=\frac\{2t\}\{1-t^2\},\\\\ \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\{2(1+t^2)^3\}\{(1-t^2)^3\}. \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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234、 2、 (10 分) 设函数 $\displaystyle f$ 在 $\displaystyle \mathbb\{R\}$ 上二阶可导且 $\displaystyle a\in\mathbb\{R\}$. 若 $\displaystyle f(a) > 0, f'(a) < 0$, 且当 $\displaystyle x > a$ 时, 有 $\displaystyle f''(x) < 0$. 证明: 在 $\displaystyle (a,+\infty)$ 内, 方程 $\displaystyle f(x)=0$ 有唯一解. (华中师范大学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) / 由 $\displaystyle f'' < 0$ 知 $\displaystyle f'$ 严格递减, 而当 $\displaystyle x > a$ 时,
\begin\{aligned\} &f'(x) < f'(a) < 0\Rightarrow f\mbox\{严\}\searrow ,\\\\ &f(x)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f(a)+f'(\xi\_x)(x-a) < f(a)+f'(a)(x-a)\xrightarrow\{x\to+\infty\}-\infty. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由连续函数介值定理知 $\displaystyle \exists\ \xi\in (a,+\infty)$, 使得 $\displaystyle f(\xi)=0$. 又由 $\displaystyle f\mbox\{严\}\searrow $ 知 $\displaystyle \xi$ 是 $\displaystyle f(x)=0$ 在 $\displaystyle (a,+\infty)$ 上的唯一零点.跟锦数学微信公众号. [在线资料](
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235、 7、 (10 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, 且 $\displaystyle f(0)=0$. 证明: 对任意正整数 $\displaystyle k$, 至少存在一点 $\displaystyle \xi\in (0,1)$, 使得
\begin\{aligned\} \xi f'(\xi)+kf(\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\}
(吉林大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F(x)=(1-x)^kf(x)$, 则 $\displaystyle F(0)=F(1)=0$. 由 Rolle 定理知
\begin\{aligned\} \exists\ \xi\in (0,1),\mathrm\{ s.t.\} 0=F'(\xi) \Rightarrow \xi f'(\xi)+kf(\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\}
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236、 (0-11)、 设 $\displaystyle \mathrm\{e\}^y+2xy=\mathrm\{e\}$, 求 $\displaystyle \frac\{\mathrm\{ d\} y\}\{\mathrm\{ d\} x\}$. (吉林师范大学2023年(学科数学)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) / 等式两边关于 $\displaystyle x$ 求导知
\begin\{aligned\} \mathrm\{e\}^y y'+2y+2xy'=0\Rightarrow y'=-\frac\{2y\}\{\mathrm\{e\}^y+2x\}. \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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237、 (0-12)、 设 $\displaystyle y=\mathrm\{e\}^\{x+1\}\sin x$, 求 $\displaystyle \frac\{\mathrm\{ d\}^2y\}\{\mathrm\{ d\} x^2\}$. (吉林师范大学2023年(学科数学)数学分析考研试题) [微分 ]
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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\} y'=&\mathrm\{e\}^\{x+1\}\sin x+\mathrm\{e\}^\{x+1\}\cos x,\\\\ y''=&(\mathrm\{e\}^\{x+1\}\sin x+\mathrm\{e\}^\{x+1\}\cos x)+(\mathrm\{e\}^\{x+1\}\cos x-\mathrm\{e\}^\{x+1\}\sin x) =2\mathrm\{e\}^\{x+1\}\cos x. \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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238、 (0-14)、 设 $\displaystyle g(x)=\int\_0^x (x-t)^2f(t)\mathrm\{ d\} t$, 且 $\displaystyle f(x)$ 为连续函数. 求 $\displaystyle g'(x)$. (吉林师范大学2023年(学科数学)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) / $\displaystyle g'(x)=\int\_0^x 2(x-t)f(t)\mathrm\{ d\} t=2\int\_0^x (x-t)f(t)\mathrm\{ d\} t$. 进一步,
\begin\{aligned\} g''(x)=2\int\_0^x f(t)\mathrm\{ d\} t, g'''(x)=2f(x). \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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239、 2、 证明题. (0-19)、 已知 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, 且 $\displaystyle f(1)=0$. 证明: 存在 $\displaystyle \xi\in (0,1)$, 使得
\begin\{aligned\} f'(\xi)=-\frac\{2\}\{\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\}
(吉林师范大学2023年(学科数学)数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F(x)=x^2f(x)$, 则 $\displaystyle F(0)=F(1)=0$. 由 Rolle 定理,
\begin\{aligned\} \exists\ \xi\in (0,1),\mathrm\{ s.t.\} 0=F'(\xi)\Rightarrow f'(\xi)=-\frac\{2\}\{\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\}
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240、 (0-21)、 确定 $\displaystyle y=\ln (x^2+1)$ 的凹凸区间和拐点. (吉林师范大学2023年(学科数学)数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle y'=\frac\{2x\}\{1+x^2\}, y''=\frac\{2(1-x^2)\}\{(1+x^2)^2\}$ 知 $\displaystyle f$ 在 $\displaystyle (-\infty,-1)$ 上凹, 在 $\displaystyle (-1,1)$ 上凸, 在 $\displaystyle (1,+\infty)$ 上凹. 于是 $\displaystyle y$ 的拐点为 $\displaystyle (\pm 1, \ln 2)$.跟锦数学微信公众号. [在线资料](
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241、 (4)、 设 $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}\mathrm\{e\}^\{-\frac\{1\}\{x^2\}\},&x\neq 0,\\\\ 0,&x=0.\end\{array\}\right.$ 求 $\displaystyle f^\{(4)\}(0)$. (暨南大学2023年数学分析考研试题) [微分 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (4-1)、 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 $\displaystyle =0$. (4-2)、 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\}
(4-3)、 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\}
(4-4)、 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\}
特别地, $\displaystyle f^\{(4)\}(0)=0$.跟锦数学微信公众号. [在线资料](
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242、 (9)、 设 $\displaystyle f(x)=\left(1-\frac\{1\}\{x\}\right)^x, x\in (-\infty,0)\cup (1,+\infty)$. 求 $\displaystyle f'(x)$. (暨南大学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) / $\displaystyle x$ 的取值范围保证了 $\displaystyle 1-\frac\{1\}\{x\} > 0$, 而
\begin\{aligned\} &\ln f(x)=x\ln \left(1-\frac\{1\}\{x\}\right)\Rightarrow \frac\{f'(x)\}\{f(x)\}=\ln \left(1-\frac\{1\}\{x\}\right)+x\frac\{\frac\{1\}\{x^2\}\}\{1-\frac\{1\}\{x\}\}\\\\ \Rightarrow&f'(x)=\left(1-\frac\{1\}\{x\}\right)^x\left\[\ln \left(1-\frac\{1\}\{x\}\right)+\frac\{1\}\{x-1\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
243、 (3)、 设 $\displaystyle f(x)$ 连续, 且 $\displaystyle f'(0)$ 存在, 若
\begin\{aligned\} \forall\ x,y\in\mathbb\{R\}, f(x+y)=\frac\{f(x)+f(y)\}\{1-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\}
证明: $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上可微. (暨南大学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) / 令 $\displaystyle y=0$ 知
\begin\{aligned\} f(x)=\frac\{f(x)+f(0)\}\{1-f(x)f(0)\}\Rightarrow -f^2(x)f(0)=f(0)\Rightarrow 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\}
对 $\displaystyle \forall\ x\in\mathbb\{R\}$,
\begin\{aligned\} &\lim\_\{h\to 0\}\frac\{f(x+h)-f(x)\}\{h\} =\lim\_\{h\to 0\}\frac\{\frac\{f(x)+f(h)\}\{1-f(x)f(h)\}-f(x)\}\{h\}\\\\ =&\lim\_\{h\to 0\}\frac\{1+f^2(x)\}\{1-f(x)f(h)\}\frac\{f(h)\}\{h\} =[1+f^2(x)]f'(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f$ 可导.跟锦数学微信公众号. [在线资料](
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---
244、 4、 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上可导, $\displaystyle \lim\_\{x\to+\infty\}f'(x)=0$. 证明: $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=0$. (南昌大学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 $\displaystyle \lim\_\{x\to+\infty\}f'(x)=0$ 知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\geq X, |f'(x)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(X)\}\{x\}=0$ 知
\begin\{aligned\} \exists\ X' > X,\mathrm\{ s.t.\} \forall\ x\geq X', \left|\frac\{f(X)\}\{x\}\right| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle x\geq X'$ 时,
\begin\{aligned\} \left|\frac\{f(x)\}\{x\}\right|\leq&\left|\frac\{f(X)\}\{x\}\right|+\frac\{\left|f(x)-f(X)\right|\}\{x\}\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\}&\left|\frac\{f(X)\}\{x\}\right|+\frac\{|f'(\xi\_x)|(x-X)\}\{x\} < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=0$. 一言以蔽之, 这就是函数极限的 L'Hospital 法则.跟锦数学微信公众号. [在线资料](
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---
245、 5、 (15 分) $\displaystyle f(x)$ 在 $\displaystyle (-1,1)$ 上 $\displaystyle 2k+2$ 阶连续可微, $\displaystyle k\in \mathbb\{N\}$ 满足
\begin\{aligned\} f^\{(i)\}(0)=0, 0\leq i\leq 2k\mbox\{且\} f^\{(2k+1)\}(0)\neq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求证: $\displaystyle f(x)$ 在 $\displaystyle x=0$ 附近可逆. (南京大学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 $\displaystyle f^\{(2k+1)\}(0) > 0$. 由 Taylor 公式,
\begin\{aligned\} f'(x)=&\sum\_\{k=0\}^\{2k-1\}\frac\{(f')^\{(k)\}(0)\}\{k!\}+\frac\{(f')^\{(2k)\}(0)\}\{(2k)!\}x^\{2k\}+o(x^\{2k\})\\\\ =&\left\[\frac\{f^\{(2k+1)\}(0)\}\{(2k)!\}+o(1)\right\]x^\{2k\}. \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\} \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x\in \mathring\{U\}(0;\delta), 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 $\displaystyle f$ 在 $\displaystyle U(0;\delta)$ 上严增, 而是可逆的.跟锦数学微信公众号. [在线资料](
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---
246、 6、 (15 分) $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上二阶可导,
\begin\{aligned\} f(0)=f(1)=0, \min\_\{x\in [0,1]\}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\}
求证: 存在 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f''(\xi)\geq 8$. (南京大学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 $\displaystyle f(0)=f(1)=0$ 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)、 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 $\displaystyle f''(\eta)=\frac\{2\}\{\xi^2\}, f''(\zeta)=\frac\{2\}\{(1-\xi)^2\}.$ (3)、 当 $\displaystyle 0 < \xi\leq \frac\{1\}\{2\}$ 时, $\displaystyle f''(\eta)=\frac\{2\}\{\eta^2\}\geq 8$. (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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---
247、 3、 设 $\displaystyle f(x),g(x),h(x)$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 内可导. 证明: 存在 $\displaystyle \xi\in (a,b)$, 使得
\begin\{aligned\} \left|\begin\{array\}\{cccccccccc\}f(a)&g(a)&h(a)\\\\ f(b)&g(b)&h(b)\\\\ f'(\xi)&g'(\xi)&h'(\xi)\end\{array\}\right|=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
并由此推出柯西中值定理. (南京航空航天大学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 $\displaystyle F(x)=\left|\begin\{array\}\{cccccccccc\}f(a)&g(a)&h(a)\\\\ f(b)&g(b)&h(b)\\\\ f(x)&g(x)&h(x)\end\{array\}\right|$, 则 $\displaystyle F(x)$ 在 $\displaystyle [a,b]$ 上连续, 在 $\displaystyle (a,b)$ 内可导, 且由行列式的性质知 $\displaystyle F(a)=F(b)=0$. 据 Rolle 定理,
\begin\{aligned\} \exists\ \xi\in (a,b),\mathrm\{ s.t.\} 0=F'(\xi)=\left|\begin\{array\}\{cccccccccc\}f(a)&g(a)&h(a)\\\\ f(b)&g(b)&h(b)\\\\ f'(\xi)&g'(\xi)&h'(\xi)\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\}
令 $\displaystyle h(x)=1$, 则
\begin\{aligned\} 0=\left|\begin\{array\}\{cccccccccc\}f(a)&g(a)&1\\\\ f(b)&g(b)&1\\\\ f'(\xi)&g'(\xi)&0\end\{array\}\right| =\left|\begin\{array\}\{cccccccccc\}f(a)&g(a)&1\\\\ f(b)-f(a)&g(b)-g(a)&0\\\\ f'(\xi)&g'(\xi)&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\}
若 $\displaystyle g'(x)\neq 0, \forall\ x\in (a,b)$, 则上式化简为 $\displaystyle \frac\{f'(\xi)\}\{g'(\xi)\} =\frac\{f(b)-f(a)\}\{g(b)-g(a)\}$. 此即 Cauchy 中指定理.跟锦数学微信公众号. [在线资料](
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---
248、 5、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,2]$ 上连续, 在 $\displaystyle (0,2)$ 上可导, 且 $\displaystyle f(0)=f(2)=0$. 记 $\displaystyle M=\max\_\{x\in [0,1]\}|f(x)|$. (1)、 证明: 存在 $\displaystyle \xi\in (0,2)$, 使得 $\displaystyle |f'(\xi)|\geq M$; (2)、 若对任意的 $\displaystyle x\in (0,2)$, 有 $\displaystyle |f'(x)|\leq M$, 证明: $\displaystyle f(x)\equiv 0, x\in [0,2]$. (南京师范大学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 $\displaystyle M=0$, 则 $\displaystyle f\equiv 0$, 则任取 $\displaystyle (0,2)$ 的点 $\displaystyle \xi$ 都满足 $\displaystyle f'(\xi)=0=M$. 若 $\displaystyle M > 0$, 则通过 $\displaystyle f\to -f$ 可不妨设 $\displaystyle M=\max\_\{x\in [0,1]\}f(x) > 0$ 于 $\displaystyle (0,2)$ 内某 $\displaystyle c$ 处取得. 由极值条件知 $\displaystyle f'(c)=0$. (1-1)、 当 $\displaystyle 0 < c\leq 1$ 时,
\begin\{aligned\} \exists\ 0 < \xi < 1,\mathrm\{ s.t.\} f'(\xi)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{f(c)-f(0)\}\{c-0\} =\frac\{M\}\{c\}\geq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1-2)、 当 $\displaystyle 1 < c < 2$ 时,
\begin\{aligned\} \exists\ 1 < \xi < 2,\mathrm\{ s.t.\} f'(\xi)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \frac\{f(2)-f(c)\}\{2-c\} =\frac\{-M\}\{2-c\} < -M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故总有 $\displaystyle \xi\in (1,2),\mathrm\{ s.t.\} |f'(c)|\geq M$. (2)、 若 $\displaystyle |f'|\leq M$, 往用反证法证明 $\displaystyle f\equiv 0$. 若不然, 则由第 1 步的结果知 $\displaystyle c=1$, 即 $\displaystyle f$ 在 $\displaystyle x=1$ 处取得最大值 $\displaystyle M$, 且
\begin\{aligned\} 0\leq x\leq 1\Rightarrow&|f(x)|=|f(x)-f(0)|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\xi\_x)|x\leq Mx\\\\ \Rightarrow& \frac\{f(x)-f(1)\}\{x-1\}=\frac\{f(1)-f(x)\}\{1-x\}\geq\frac\{M-Mx\}\{1-x\}=M, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} 1 < x < 2\Rightarrow&|f(x)|=|f(x)-f(2)|\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} |f'(\eta\_x)(2-x)|\leq M(2-x)\\\\ \Rightarrow&\frac\{f(x)-f(1)\}\{x-1\}\leq \frac\{M(2-x)-M\}\{x-1\}=-M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由上述第 1 式知 $\displaystyle f'\_-(1)\geq M$. 由上述第 2 式知 $\displaystyle f'\_+(1)\leq -M$. 但 $\displaystyle f'(1)$ 存在, 而得 $\displaystyle -M\geq f'(1)\geq M\Rightarrow M=0$. 这与已设矛盾. 故有结论.跟锦数学微信公众号. [在线资料](
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---
249、 7、 (15 分) 令
\begin\{aligned\} \varphi(x)=\left\\{\begin\{array\}\{llllllllllll\}\sin x, &x\geq 0,\\\\ \cos x,&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\}
证明: 不存在 $\displaystyle (-\infty,+\infty)$ 上的可微函数 $\displaystyle f(x)$ 使得对任意实数 $\displaystyle x$ 都有
\begin\{aligned\} f\left(f(x)\right)-|x|f'(x)=\varphi(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学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. 若存在满足题设的 $\displaystyle f$, 则 (1)、 令 $\displaystyle x > 0$, 题中等式两边除以 $\displaystyle x$ 后令 $\displaystyle x\to 0^+$ 得
\begin\{aligned\} \lim\_\{x\to 0^+\}\frac\{f\left(f(x)\right)\}\{x\}-f'(0)=\lim\_\{x\to 0^+\}\frac\{\sin x\}\{x\}=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\}
故
\begin\{aligned\} &f\left(f(0)\right)=\lim\_\{x\to 0^+\}\frac\{f\left(f(x)\right)\}\{x\}\cdot x\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} 0\\\\ \stackrel\{(I)\}\{\Rightarrow\}&f'\left(f(0)\right)\cdot f'(0)-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)、 令 $\displaystyle x < 0$, 题中等式两边除以 $\displaystyle x$ 后令 $\displaystyle x\to 0^-$ 得
\begin\{aligned\} f'\left(f(0)\right)\cdot f'(0)-f'(0)=\lim\_\{x\to 0^-\}\frac\{\cos x\}\{x\}=-\infty. \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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---
250、 3、 求 $\displaystyle y=\frac\{(x+1)^2\}\{(x-1)^3\}$ 与 $\displaystyle y=\sqrt[3]\{6x^2-x^3\}$ 的所有渐近线 (包括水平, 垂直和斜渐近线). (上海大学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)、 对 $\displaystyle y=\frac\{(x+1)^2\}\{(x-1)^3\}$, $\displaystyle x=1$ 是垂直渐近线. 由 $\displaystyle \lim\_\{x\to\infty\}y=0$ 知 $\displaystyle y$ 有水平渐近线, 没有斜渐近线. (2)、 对 $\displaystyle y=\sqrt[3]\{6x^2-x^3\}$, 由
\begin\{aligned\} \lim\_\{x\to\infty\}\frac\{y\}\{x\}=\lim\_\{x\to\infty\}\sqrt[3]\{\frac\{6\}\{x\}-1\}=-1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} &\lim\_\{x\to\infty\}[y-(-1)x]=\lim\_\{x\to\infty\}\left(\sqrt[3]\{6x^2-x^3\}+x\right)\\\\ =&\lim\_\{x\to\infty\}\frac\{6x^2\}\{\sqrt[3]\{(6x^2-x^3)^2\}-\sqrt[3]\{6x^2-x^3\}\cdot x+x^2\}\\\\ =&\lim\_\{x\to\infty\}\frac\{6\}\{\sqrt[3]\{\left(\frac\{6\}\{x\}-1\right)^2\}-\sqrt[3]\{\frac\{6\}\{x\}-1\}+1\}=2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle y$ 只有斜渐近线 $\displaystyle y=-x+2$.跟锦数学微信公众号. [在线资料](
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251、 12、 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, 在 $\displaystyle (0,1)$ 内可导, 且 $\displaystyle \forall\ x\in (0,1)$, 有
\begin\{aligned\} \left|xf'(x)-f(x)+f(0)\right|\leq\frac\{Mx\}\{\ln^2x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \lim\_\{x\to 0^+\}f'(x)$ 存在. (首都师范大学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 $\displaystyle F(x)=\frac\{f(x)-f(0)\}\{x\}$, 则
\begin\{aligned\} |F'(x)=\frac\{|xf'(x)-f(x)+f(0)|\}\{x^2\}\leq \frac\{M\}\{x\ln^2x\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再设
\begin\{aligned\} G(x)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{1\}\{\ln x\},&0 < x\leq 1,\\\\ 0,&x=0,\end\{array\}\right.\quad H(x)=\left\\{\begin\{array\}\{llllllllllll\}\frac\{Mx\}\{\ln^2 x\},&0 < x\leq 1,\\\\ 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\}
则 $\displaystyle G,H$ 在 $\displaystyle x=0$ 处连续, $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ 0 < x < y < \delta$,
\begin\{aligned\} |G(x)-G(y)| < \frac\{\varepsilon\}\{3M\}, |H(x)|=|H(x)-H(0)| < \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\}
从而
\begin\{aligned\} &|F(x)-F(y)|=\left|\frac\{F(x)-F(y)\}\{G(x)-G(y)\}\right|\cdot |G(x)-G(y)|\\\\ \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Cauchy\}\}& \left|\frac\{F'(\xi)\}\{G'(\xi)\}\right|\cdot |G(x)-G(y)| =\left|\frac\{F'(\xi)\}\{-\frac\{1\}\{\xi\ln^2\xi\}\}\right||G(x)-G(y)| \leq M\cdot\frac\{\varepsilon\}\{3M\}=\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\}
再由题设,
\begin\{aligned\} |f'(x)-f'(y)| \leq& |f'(x)-F(x)|+|F(x)-F(y)|+|F(y)-f'(y)|\\\\ \leq&\frac\{Mx\}\{\ln^2x\}+\frac\{\varepsilon\}\{3\}+\frac\{My\}\{\ln^2y\} < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
有函数极限的 Cauchy 收敛准则知 $\displaystyle f'(0+0)$ 存在. 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)/
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252、 8、 (本题 15 分) 设 $\displaystyle f(x),g(x)$ 在 $\displaystyle (a,b)$ 上连续可导, 且
\begin\{aligned\} f'(x)g(x)-g'(x)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\}
证明: $\displaystyle f(x)$ 的相邻两个零点之间必有 $\displaystyle g(x)$ 的零点. (四川大学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)、 先用反证法证明 $\displaystyle f,g$ 不可能有相同的零点. 若不然, 设 $\displaystyle f,g$ 有公共零点 $\displaystyle x\_0$, 则
\begin\{aligned\} F(x\_0)=f'(x\_0)g(x\_0)-g'(x\_0)f(x\_0)=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. (2)、 用反证法证明题目. 设 $\displaystyle x\_0 < x\_1$ 为 $\displaystyle f$ 的相邻零点, 而 $\displaystyle g(x)\neq 0$, $\displaystyle x\in (x\_0,x\_1)$. 又由 (1) 知
\begin\{aligned\} g(x)\neq 0,\quad x\in [x\_0,x\_1]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle G(x)=\frac\{f(x)\}\{g(x)\}$, 则
\begin\{aligned\} G'(x)=\frac\{F(x)\}\{g^2(x)\} > 0,\quad x\in [x\_0,x\_1]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle G$ 在 $\displaystyle [x\_0,x\_1]$ 上严格递增, $\displaystyle G(x\_0) < G(x\_1)$. 这与
\begin\{aligned\} 0=\frac\{f(x\_0)\}\{g(x\_0)\}=G(x\_0)=G(x\_1)=\frac\{f(x\_1)\}\{g(x\_1)\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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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253、 9、 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: (1)、 存在 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f(\xi)=2022-2023\xi$; (2)、 存在 $\displaystyle \eta,\zeta\in (0,1), \eta\neq \zeta$, 使得
\begin\{aligned\} [1+f'(\eta)][2021+f'(\zeta)]=4044. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(太原理工大学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) / (1)、 设 $\displaystyle F(x)=f(x)-2022+2023x$, 则 $\displaystyle F(0)=-2022, F(1)=2$. 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)=2022-2023\xi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 Lagrange 中值定理, $\displaystyle \exists\ 0 < \eta < \xi < \zeta < 1,\mathrm\{ s.t.\}$
\begin\{aligned\} f'(\eta)=\frac\{f(\xi)-f(0)\}\{\xi-0\}=\frac\{f(\xi)\}\{\xi\}, f'(\zeta)=\frac\{f(1)-f(\xi)\}\{1-\xi\}=\frac\{1-f(\xi)\}\{1-\xi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle f(\xi)=2022-2023\xi$ 代入, 经过简单计算即知 $\displaystyle [1+f'(\eta)][2021+f'(\zeta)]=4044$.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 08:46:17
