## 张祖锦2023年数学专业真题分类70天之第14天
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300、 6、 证明下列问题. (1)、 $\displaystyle \forall\ x > 0, y\in\mathbb\{R\}$, 有
\begin\{aligned\} xy\leq x\ln x-x+\mathrm\{e\}^y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中科技大学2023年数学分析考研试题) [微分法与不等式 ]
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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) / 设 $\displaystyle f(x,y)=x\ln x-x+\mathrm\{e\}^y-xy$, 则
\begin\{aligned\} f\_x=\ln x-y, f\_y=\mathrm\{e\}^y-x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到 $\displaystyle f(x,\ln x)=f(\mathrm\{e\}^y,y)=0$, 我们知
\begin\{aligned\} y > \ln x\Rightarrow&\mathrm\{e\}^y > x\Rightarrow f\_y > 0\Rightarrow f(x,y) > f(x,\ln x)=0,\\\\ y < \ln x\Rightarrow&f\_x > 0\Rightarrow f(x,y) > f(\mathrm\{e\}^y,y)=0. \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 $\displaystyle f(x,y)\geq 0$.跟锦数学微信公众号. [在线资料](
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301、 2、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上二阶连续可微, $\displaystyle |f''(x)|\leq M$, 且 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 内可取到最大值. 证明: $\displaystyle |f'(0)|+|f'(1)|\leq M$. (厦门大学2023年数学分析考研试题) [微分法与不等式 ]
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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) / 由题设, $\displaystyle \exists\ c\in (0,1),\mathrm\{ s.t.\} f(c)=\max\_\{[0,1]\}f$. 由极值条件知 $\displaystyle f'(c)=0$. 于是
\begin\{aligned\} |f'(0)|&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} |f'(c)+f''(\xi)(0-c)|\leq Mc,\\\\ |f'(1)|&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} |f'(c)+f''(\eta)(1-c)|\leq M(1-c). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相加即得结论.跟锦数学微信公众号. [在线资料](
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302、 3、 (15 分) $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上二阶可导, $\displaystyle f'(a)=f'(b)=0$. 证明: 存在 $\displaystyle \xi\in (a,b)$, 使得
\begin\{aligned\} |f'(\xi)|\leq\frac\{4\}\{(b-a)^2\}|f(b)-f(a)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(山西大学2023年数学分析考研试题) [微分法与不等式 ]
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https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Taylor 定理,
\begin\{aligned\} \exists\ \eta\in \left(a,\frac\{a+b\}\{2\}\right),\mathrm\{ s.t.\} f\left(\frac\{a+b\}\{2\}\right)=&f(a)+f'(a)\frac\{b-a\}\{2\}+\frac\{f''(\eta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2\\\\ =&f(a)+\frac\{f''(\eta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2,\\\\ \exists\ \zeta\in \left(\frac\{a+b\}\{2\},b\right),\mathrm\{ s.t.\} f\left(\frac\{a+b\}\{2\}\right)=&f(b)+f'(b)\frac\{a-b\}\{2\}+\frac\{f''(\zeta)\}\{2\}\left(\frac\{a-b\}\{2\}\right)^2\\\\ =&f(b)+\frac\{f''(\zeta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相减有
\begin\{aligned\} |f(b)-f(a)| =&\left|\frac\{f''(\eta)-f''(\zeta)\}\{2\}\right|\left(\frac\{b-a\}\{2\}\right)^2 \leq \frac\{|f''(\eta)|+|f''(\zeta)|\}\{2\}\frac\{(b-a)^2\}\{4\}\\\\ \leq& \max\left\\{|f''(\eta)|,|f''(\zeta)|\right\\} \frac\{(b-a)^2\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle |f''(\eta)|\geq |f''(\zeta)|$, 则取 $\displaystyle \xi=\eta$ 即知结论成立; 若 $\displaystyle |f''(\eta) < |f''(\zeta)|$, 则取 $\displaystyle \xi=\zeta$ 即知结论成立.跟锦数学微信公众号. [在线资料](
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303、 4、 (15 分) 若 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上二阶可导, $\displaystyle f'(a)=f'(b)=0$, 试证: 存在 $\displaystyle \xi\in (a,b)$, 使得
\begin\{aligned\} |f''(\xi)|\geq \frac\{4\}\{(b-a)^2\}|f(b)-f(a)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(陕西师范大学2023年数学分析考研试题) [微分法与不等式 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 Taylor 定理,
\begin\{aligned\} \exists\ \eta\in \left(a,\frac\{a+b\}\{2\}\right),\mathrm\{ s.t.\} f\left(\frac\{a+b\}\{2\}\right)=&f(a)+f'(a)\frac\{b-a\}\{2\}+\frac\{f''(\eta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2\\\\ =&f(a)+\frac\{f''(\eta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2,\\\\ \exists\ \zeta\in \left(\frac\{a+b\}\{2\},b\right),\mathrm\{ s.t.\} f\left(\frac\{a+b\}\{2\}\right)=&f(b)+f'(b)\frac\{a-b\}\{2\}+\frac\{f''(\zeta)\}\{2\}\left(\frac\{a-b\}\{2\}\right)^2\\\\ =&f(b)+\frac\{f''(\zeta)\}\{2\}\left(\frac\{b-a\}\{2\}\right)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相减有
\begin\{aligned\} |f(b)-f(a)| =&\left|\frac\{f''(\eta)-f''(\zeta)\}\{2\}\right|\left(\frac\{b-a\}\{2\}\right)^2 \leq \frac\{|f''(\eta)|+|f''(\zeta)|\}\{2\}\frac\{(b-a)^2\}\{4\}\\\\ \leq& \max\left\\{|f''(\eta)|,|f''(\zeta)|\right\\} \frac\{(b-a)^2\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若 $\displaystyle |f''(\eta)|\geq |f''(\zeta)|$, 则取 $\displaystyle \xi=\eta$ 即知结论成立; 若 $\displaystyle |f''(\eta) < |f''(\zeta)|$, 则取 $\displaystyle \xi=\zeta$ 即知结论成立.跟锦数学微信公众号. [在线资料](
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304、 5、 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\}
(上海财经大学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](
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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](
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305、 7、 设 $\displaystyle f(x)$ 四阶可导, $\displaystyle f^\{(4)\}(x) > 0$, 且 $\displaystyle \lim\_\{x\to 0\}\frac\{f(x)\}\{x^3\}=1$. 证明: $\displaystyle f(x)\geq x^3$. (上海大学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) / $\displaystyle f(0)=\lim\_\{x\to 0\}f(x)=\lim\_\{x\to 0\}\frac\{f(x)\}\{x^3\}\cdot x^3=0$,
\begin\{aligned\} f'(0)=\lim\_\{x\to 0\}\frac\{f(x)\}\{x\}=\lim\_\{x\to 0\}\frac\{f(x)\}\{x^3\}\cdots x^2=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\} \lim\_\{x\to 0\}\frac\{f(x)\}\{x^2\}\left\\{\begin\{array\}\{llllllllllll\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{f'(x)\}\{2x\}=\frac\{1\}\{2\}f''(0),\\\\ =\lim\_\{x\to 0\}\frac\{f(x)\}\{x^3\}\cdot 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 $\displaystyle f''(0)=0$. 进一步,
\begin\{aligned\} 1=\lim\_\{x\to 0\}\frac\{f(x)\}\{x^3\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to 0\}\frac\{f'(x)\}\{3x^2\} \xlongequal\{\tiny\mbox\{L'Hospital\}\}\lim\_\{x\to 0\}\frac\{f''(x)\}\{6x\} =\frac\{1\}\{6\}f'''(0)\Rightarrow f'''(0)=6. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
对 $\displaystyle \forall\ x$, 由 Taylor 公式,
\begin\{aligned\} f(x)=\frac\{f'''(0)\}\{3!\}x^3+\frac\{f^\{(4)\}(\xi)\}\{4!\}x^4\geq x^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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---
306、 2、 (10 分) 设 $\displaystyle 0 < x < \pi$, 证明:
\begin\{aligned\} 0 < \frac\{1\}\{\ln \left(1+\frac\{\sin x\}\{x\}\right)\}-\frac\{x\}\{\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\}
(苏州大学2023年数学分析考研试题) [微分法与不等式 ]
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https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 我们先证明对数不等式: \begin\{aligned\} &\frac\{x\}\{1+x\}\leq \ln(1+x)\leq x\quad (x > -1),\\\\ \Leftrightarrow&0\leq\frac\{1\}\{\ln(1+x)\}-\frac\{1\}\{x\}\leq 1, \qquad(ex1.1.8: log inequality)\tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
等号当且仅当 $\displaystyle x=0$ 时成立. 用微分法.
\begin\{aligned\} &f(x)\equiv \ln (1+x)-x\\\\ \Rightarrow& f'(x)=\frac\{1\}\{1+x\}-1 =-\frac\{x\}\{1+x\}\left\\{\begin\{array\}\{llllllllllll\} > 0,&-1 < x < 0\\\\ < 0,&x > 0\end\{array\}\right.\\\\ \Rightarrow& \forall\ x > -1, x\neq 0, f(x) < 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\}
\begin\{aligned\} &g(x)=x-(1+x)\ln (1+x)\\\\ \Rightarrow& g'(x)=-\ln(1+x)\left\\{\begin\{array\}\{llllllllllll\} > 0,&-1 < x < 0\\\\ < 0,&x > 0\end\{array\}\right.\\\\ \Rightarrow& \forall\ x > -1, x\neq 0, g(x) < g(0)=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle h(x)=\frac\{\sin x\}\{x\}, 0 < x < \pi$, 则
\begin\{aligned\} h'(x)=\frac\{x\cos x-\sin x\}\{x^2\}=\frac\{\cos x-\frac\{1\}\{x\}\int\_0^x \cos t\mathrm\{ d\} t\}\{x^3\} < 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\} 0 < x < \pi\Rightarrow& \pi=h(\pi) < h(x)=\frac\{\sin x\}\{x\} < \lim\_\{t\to 0^+\}h(t)=1\\\\ \overset\{\tiny\mbox\{第1步\}\}\{\Longrightarrow\}&0 < \frac\{1\}\{\ln \left(1+\frac\{\sin x\}\{x\}\right)\}-\frac\{x\}\{\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\}
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307、 1、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上可导, 且 $\displaystyle f(0)=0$. (1)、 若 $\displaystyle \lim\_\{x\to+\infty\}\frac\{f(x)\}\{x\}=0$, 求证: $\displaystyle \varliminf\_\{x\to+\infty\}|f'(x)|=0$; (2)、 若存在 $\displaystyle M > 0$, 使得
\begin\{aligned\} \forall\ x\in [0,+\infty), |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\}
求证: $\displaystyle f(x)\equiv 0, x\in [0,+\infty)$. (同济大学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)、
\begin\{aligned\} 0\leq&\varliminf\_\{x\to+\infty\}|f'(x)|\leq |f'(\xi\_x)| \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} \lim\_\{x\to+\infty\}\left|\frac\{f(2x)-f(x)\}\{2x-x\}\right|\\\\ =&\lim\_\{x\to+\infty\}\left\[2\frac\{f(2x)\}\{2x\}-\frac\{f(x)\}\{x\}\right\] \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} 2\cdot 0-0=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故有结论. (2)、 (2-1)、 设 $\displaystyle |f|$ 在 $\displaystyle \left\[0,\frac\{1\}\{2M\}\right\]$ 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\[0,\frac\{1\}\{2M\}\right\]$. (2-2)、 设 $\displaystyle |f|$ 在 $\displaystyle \left\[\frac\{1\}\{2M\},\frac\{1\}\{M\}\right\]$ 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\[\frac\{1\}\{2M\},\frac\{1\}\{M\}\right\]$. (2-3)、 如此一直做下去, 我们得到
\begin\{aligned\} f(x)\equiv 0, x\in \left\[\frac\{n\}\{2M\},\frac\{n+1\}\{2M\}\right\]\left(\forall\ n\in\mathbb\{N\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle f\equiv 0$.跟锦数学微信公众号. [在线资料](
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308、 (2)、 已知 $\displaystyle f(x)$ 在 $\displaystyle [0,2]$ 上二阶可导, 且 $\displaystyle \max\_\{0\leq x\leq 2\} \left\\{|f(x)|,|f''(x)|\right\\}\leq 1$. 证明: 对 $\displaystyle \forall\ x\in [0,2]$, $\displaystyle |f'(x)|\leq 2$. (武汉理工大学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)&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} f(x)+f'(x)(0-x)+\frac\{f''(\xi\_x)\}\{2\}(0-x)^2,\\\\ f(2)&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} f(x)+f'(x)(2-x)+\frac\{f''(\eta\_x)\}\{2\}(2-x)^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相减得
\begin\{aligned\} 2|f'(x)|\leq 2+\frac\{1\}\{2\}x^2+\frac\{1\}\{2\}(2-x)^2 \leq 2+\frac\{1\}\{2\}\cdot 4=4. \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](
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---
309、 5、 (15 分) 证明: 当 $\displaystyle x\geq 0$ 且 $\displaystyle y\geq 0$ 时, 有 $\displaystyle \frac\{x^2+y^2\}\{4\}\leq \mathrm\{e\}^\{x+y-2\}$. (西北大学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\} &4\mathrm\{e\}^\{x+y-2\}=4\mathrm\{e\}^\{x-1\}\mathrm\{e\}^\{y-1\}\\\\ \geq& 4\left\[1+(x-1)+\frac\{(x-1)^2\}\{2\}\right\]\left\[1+(y-1)+\frac\{(y-1)^2\}\{2\}\right\]\\\\ \geq& 4\frac\{x^2+1\}\{2\}\frac\{y^2+1\}\{2\}\geq x^2+y^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](
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---
310、 8、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上三阶可导, 且 $\displaystyle f(-1)=0, f(1)=1, f'(0)=0$. 证明:
\begin\{aligned\} \sup\_\{x\in [-1,1]\}f'''(x)\geq 3. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西南大学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) / 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\}\leq \frac\{1\}\{3\}\sup\_\{x\in [-1,1]\}f'''(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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---
311、 5、 (8 分) 设 $\displaystyle f$ 为 $\displaystyle (0,+\infty)$ 上的二阶可微函数, 且 $\displaystyle |f(x)|\leq M\_0, |f''(x)|\leq M\_2$. 证明:
\begin\{aligned\} |f'(x)|\leq 2\sqrt\{M\_0M\_2\}, x\in (0,+\infty). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(长安大学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) / 若 $\displaystyle M\_2=0$, 则 $\displaystyle f$ 是线性函数, 而 $\displaystyle M\_0 < \infty$ 蕴含 $\displaystyle f$ 为常值函数, 于是 $\displaystyle M\_1=0$. 若 $\displaystyle M\_2 > 0$, 则由 Taylor 定理知
\begin\{aligned\} f(x+h)=f(x)+f'(x)h+\frac\{f''(\xi)\}\{2\}h^2, \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\} \left|f'(x)\right|=&\left|\frac\{f(x+h)-f(x)\}\{h\}-\frac\{f''(\xi)\}\{2\}h\right|\leq \frac\{2M\_0\}\{h\}+\frac\{M\_2\}\{2\}h. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle h=2\sqrt\{\frac\{M\_0\}\{M\_2\}\}$, 则 $\displaystyle |f'(x)|\leq 2\sqrt\{M\_0M\_2\}$.跟锦数学微信公众号. [在线资料](
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---
312、 (2)、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上二阶可导, 且 $\displaystyle f(a)=f(b)=0$. 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\}
(郑州大学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](
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---
313、 4、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,a]$ 上二阶可导, $\displaystyle |f''(x)|\leq M$, $\displaystyle f(x)$ 在 $\displaystyle (0,a)$ 上取得最大值. 求证:
\begin\{aligned\} |f'(0)|+|f'(a)|\leq Ma. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中南大学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 $\displaystyle \xi\in (0,a),\mathrm\{ s.t.\} f(\xi)=\max\_\{(0,a)\}f$, 则 $\displaystyle f'(\xi)=0$. 据 Lagrange 中值定理,
\begin\{aligned\} &\left\\{\begin\{array\}\{llllllllllll\} f'(0)-f'(\xi)=(f')'(\eta)(0-\xi)=-f''(\eta)\xi \Rightarrow |f'(0)|\leq M\xi\\\\ f'(a)-f'(\xi)=(f')'(\zeta)(a-\xi)=f''(\zeta)(a-\xi) \Rightarrow |f'(a)|\leq M(a-\xi) \end\{array\}\right.\\\\ \Rightarrow& |f'(0)|+|f'(a)|\leq M\xi+M(a-\xi)=Ma. \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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---
314、 (2)、 计算不定积分 $\displaystyle \int \mathrm\{e\}^\{\sin x\}\frac\{x\cos^3x-\sin x\}\{\cos^2x\}\mathrm\{ d\} x$. (河海大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\int x\mathrm\{ d\} \mathrm\{e\}^\{\sin x\}-\int \mathrm\{e\}^\{\sin x\}\frac\{\sin x\}\{\cos^2x\}\mathrm\{ d\} x\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} x\mathrm\{e\}^\{\sin x\}-\int \mathrm\{e\}^\{\sin x\}\mathrm\{ d\} x-\int \mathrm\{e\}^\{\sin x\}\frac\{\sin x\}\{\cos^2x\}\mathrm\{ d\} x\\\\ &=x\mathrm\{e\}^\{\sin x\}-\int \frac\{\mathrm\{e\}^\{\sin x\}\cos x\cdot \cos x-\mathrm\{e\}^\{\sin x\}(-\sin x)\}\{\cos^2x\}\mathrm\{ d\} x\\\\ &=x\mathrm\{e\}^\{\sin x\}-\int\frac\{(\mathrm\{e\}^\{\sin x\})'\cos x-\mathrm\{e\}^\{\sin x\}(\cos x)'\}\{\cos^2x\}\mathrm\{ d\} x\\\\ &=x\mathrm\{e\}^\{\sin x\}-\frac\{\mathrm\{e\}^\{\sin x\}\}\{\cos x\}+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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---
315、 (0-13)、 求 $\displaystyle \int\frac\{1+\ln x\}\{(x\ln x)^2\}\mathrm\{ d\} x$. (吉林师范大学2023年(学科数学)数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\int \frac\{1\}\{x^2\ln^2x\}\mathrm\{ d\} x+\int \frac\{1\}\{x^2\ln x\}\mathrm\{ d\} x\\\\ =&-\int \frac\{1\}\{x\}\mathrm\{ d\} \frac\{1\}\{\ln x\}+\int \frac\{1\}\{x^2\ln x\}\mathrm\{ d\} x\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&-\left\[\frac\{1\}\{x\}\cdot \frac\{1\}\{\ln x\} -\int \left(-\frac\{1\}\{x^2\}\frac\{1\}\{\ln x\}\right)\mathrm\{ d\} x\right\]+\int \frac\{1\}\{x^2\ln x\}\mathrm\{ d\} x\\\\ =&-\frac\{1\}\{x\ln x\}+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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---
316、 (5)、 求不定积分 $\displaystyle \int \frac\{\arctan x\}\{x^2\}\mathrm\{ d\} x$. (暨南大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\int \arctan x\mathrm\{ d\}\frac\{-1\}\{x\} =-\frac\{\arctan x\}\{x\}-\int\frac\{1\}\{1+x^2\}\cdot\frac\{-1\}\{x\}\mathrm\{ d\} x\\\\ =&-\frac\{\arctan x\}\{x\}+\int \left(\frac\{1\}\{x\}-\frac\{x\}\{1+x^2\}\right)\mathrm\{ d\} x\\\\ =&-\frac\{\arctan x\}\{x\}+\ln x-\frac\{1\}\{2\}\ln (1+x^2)+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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---
317、 (2)、 求不定积分 $\displaystyle \int x\cos^2x\mathrm\{ d\} x$. (南京师范大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{2\}\int x(1+\cos 2x)\mathrm\{ d\} x =\frac\{x^2\}\{4\}+\frac\{1\}\{2\}\int x\cos 2x\mathrm\{ d\} x\\\\ =&\frac\{x^2\}\{4\}+\frac\{1\}\{4\}\int x\mathrm\{ d\} \sin 2x \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{x^2\}\{4\}+\frac\{1\}\{4\}\left\[x\sin 2x-\int \sin 2x\mathrm\{ d\} x\right\]\\\\ =&\frac\{x^2\}\{4\}+\frac\{1\}\{4\}x\sin 2x+\frac\{1\}\{8\}\cos 2x+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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---
318、 (2)、 求不定积分 $\displaystyle \int x\mathrm\{e\}^\{2x\}\mathrm\{ d\} x$. (陕西师范大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\frac\{1\}\{2\}\int x\mathrm\{ d\} \mathrm\{e\}^\{2x\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{1\}\{2\}\left(x\mathrm\{e\}^\{2x\}-\int \mathrm\{e\}^\{2x\}\mathrm\{ d\} x\right)\\\\ =&\frac\{1\}\{2\}x\mathrm\{e\}^\{2x\}-\frac\{1\}\{4\}\mathrm\{e\}^\{2x\}+C=\left(\frac\{x\}\{2\}-\frac\{1\}\{4\}\right)\mathrm\{e\}^\{2x\}+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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---
319、 2、 定义函数 $\displaystyle f(x)=\lim\_\{n\to\infty\}\frac\{x^2\mathrm\{e\}^\{n(x-1)\}+ax+b\}\{\mathrm\{e\}^\{n(x-1)\}+1\}$, 其中 $\displaystyle x\in\mathbb\{R\}$, $\displaystyle a,b$ 是常数. (1)、 讨论 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上不定积分的存在性. (2)、 当 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上不定积分存在, 求 $\displaystyle \int f(x)\mathrm\{ d\} x$. (首都师范大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}ax+b,&x < 1,\\\\ \frac\{x^2+ax+b\}\{2\}=\frac\{1+a+b\}\{2\},&x=1,\\\\ \lim\_\{n\to\infty\}\frac\{x^2+(a+b)\mathrm\{e\}^\{n(1-x)\}\}\{1+\mathrm\{e\}^\{n(1-x)\}\}=x^2,&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\}
若 $\displaystyle f$ 的不定积分存在 $\displaystyle \Leftrightarrow \exists\ F,\mathrm\{ s.t.\} F'=f$, 由导函数极限定理知 $\displaystyle f$ 至多有第二类间断点. 从而 $\displaystyle f$ 连续:
\begin\{aligned\} a+b=f(1-0)=f(1)=\frac\{1+a+b\}\{2\}=f(1+0)=1\Leftrightarrow b=1-a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此时, $\displaystyle f(x)=\left\\{\begin\{array\}\{llllllllllll\}a(x-1)+1,&x < 1,\\\\ 1,&x=1,\\\\ x^2,&x > 1.\end\{array\}\right.$ 求不定积分即知
\begin\{aligned\} \int f(x)\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}\frac\{a\}\{2\}(x-1)^2+x+C\_1,&x < 1,\\\\ \frac\{x^3\}\{3\}+C\_2,&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\}
由不定积分的连续性知
\begin\{aligned\} C\_1+1=\frac\{1\}\{3\}+C\_2\Rightarrow \int f(x)\mathrm\{ d\} x=\left\\{\begin\{array\}\{llllllllllll\}\frac\{a\}\{2\}(x-1)^2+x+C,&x < 1,\\\\ \frac\{x^3+2\}\{3\}+C,&x\geq 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\}
跟锦数学微信公众号. [在线资料](
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320、 2、 计算下列积分 (每小题 9 分, 共 36 分). (1)、 $\displaystyle \int x\ln \frac\{1+x\}\{1-x\}\mathrm\{ d\} x$. (四川大学2023年数学分析考研试题) [不定积分 ]
[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 $\displaystyle f(x)=\ln\frac\{1+x\}\{1-x\}$, 则 $\displaystyle f'(x)=\frac\{2\}\{1-x^2\}$, 而
\begin\{aligned\} \mbox\{原式\}=&\int f(x)\mathrm\{ d\} \frac\{x^2\}\{2\} =\frac\{x^2\}\{2\}f(x)-\int \frac\{2\}\{1-x^2\}\frac\{x^2\}\{2\}\mathrm\{ d\} x\\\\ =&\frac\{x^2\}\{2\}f(x)-\int \left(-1+\frac\{1\}\{1-x^2\}\right)\mathrm\{ d\} x =\frac\{x^2\}\{2\}f(x)+x-\frac\{1\}\{2\}f(x)+C\\\\ =&\frac\{x^2-1\}\{2\}\ln \frac\{1+x\}\{1-x\}+x+C. \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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321、 (2)、 $\displaystyle \int\frac\{\mathrm\{ d\} x\}\{2+\tan^2x\}$. (天津大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\frac\{1\}\{2\}\int \frac\{\mathrm\{ d\} x\}\{1+\left(\frac\{\tan x\}\{\sqrt\{2\}\}\right)^2\} \stackrel\{\frac\{\tan x\}\{\sqrt\{2\}\}=t\}\{=\}\frac\{1\}\{2\} \int\frac\{1\}\{1+t^2\}\cdot\frac\{1\}\{1+2t^2\}\mathrm\{ d\} t\\\\ &=\frac\{1\}\{\sqrt\{2\}\}\int\left(-\frac\{1\}\{1+t^2\}+\frac\{2\}\{1+2t^2\}\right)\mathrm\{ d\} t\\\\ &=\frac\{1\}\{\sqrt\{2\}\}\left\[-\arctan t+\sqrt\{2\}\int \frac\{\mathrm\{ d\} \left(\sqrt\{2\}t\right)\}\{1+(\sqrt\{2\}t)^2\}\mathrm\{ d\} t\right\]\\\\ &=\frac\{1\}\{\sqrt\{2\}\} \left\[-\arctan t+\sqrt\{2\}\arctan \left(\sqrt\{2\}t\right)\right\]+C\\\\ &=x-\frac\{1\}\{\sqrt\{2\}\}\arctan\left(\frac\{\tan x\}\{\sqrt\{2\}\}\right)+C . \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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322、 (4)、 $\displaystyle \int \arctan \sqrt\{x\}\mathrm\{ d\} x=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [不定积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&=\int \arctan t\cdot 2t\mathrm\{ d\} t\left(\sqrt\{x\}=t\right)\\\\ &=\int \arctan t\mathrm\{ d\} (t^2)\\\\ &=t^2\cdot \arctan t -\int\frac\{t^2\}\{1+t^2\}\mathrm\{ d\} t\left(\mbox\{分部积分\}\right)\\\\ &=t^2\cdot \arctan t -\int \left(1-\frac\{1\}\{1+t^2\}\right)\mathrm\{ d\} t\\\\ &=(1+t^2)\cdot \arctan t -t+C\\\\ &=(1+x)\cdot \arctan \sqrt\{x\} -\sqrt\{x\}+C. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 08:48:21
