第十五届全国大学生数学竞赛初赛试题(非数学专业类A卷)

题目速览

1. 求极限：
   $$\underset{x\to 3}{\lim }\frac{\sqrt{x^3+9}-6}{2-\sqrt{x^3-23}}=.$$

2. 设函数 $z=f(x^2-y^2,xy)$, 且 $f(u,v)$ 具有连续的二阶偏导数, 则
   $$\frac{{\partial }^{2}z}{\partial x\partial y}=.$$

3. 设 $n$ 为正整数, $f(x)=\frac{1}{x^2-3x+2}$, 则:
   $$f^{(n)}(0)=.$$

4. 幂级数
   $$\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n(2n-1)}{x}^{2n}$$
   的收敛域为 $.$

5. 设曲面 $\Sigma$ 是平面 $y+z=5$ 被柱面 $x^2+y^2=25$ 所截得的部分, 则曲面积分
   $$\underset{\Sigma }{\iint }(x+y+z)\mathrm{d}S=$$

6.$$(x^2+y^2+3)\frac{\mathrm{d}y}{\mathrm{d}x}=2x\left(2y-\frac{x^2}{y}\right).$$

7. 设 ${\Sigma }_1$ 是以 $(0,4,0)$ 为顶点且与曲面 ${\Sigma }_2$: $\frac{x^2}{3}+\frac{y^2}{4}+\frac{z^2}{3}=1(y>0)$ 相切的圆锥面, 求曲面 ${\Sigma }_1$ 与 ${\Sigma }_2$ 所围成的空间区域的体积.

8. 设 $a>1$, 求极限
   $$\underset{n\to \infty }{\lim }n{\int }_1^a\frac{\mathrm{d}x}{1+x^n}.$$

9. 设 $f(x)$ 在 $[0,1]$ 上有连续的导数, 且 $f(0)=0$. 求证:
   $${\int }_0^1{f}^2(x)\mathrm{d}x⩽4{\int }_0^1(1-x{)}^2{\left[f^{\prime }(x)\right]}^2\mathrm{d}x,$$
   并求使得上式成为等式的 $f(x)$.

10. 设数列{xn}\{x_n\}{xn​} 满足 $x_0=\frac{1}{3},x_{n+1}=\frac{x_n^2}{1-x_n+x_n^2},n⩾0$. 证明无穷级数
    $$\sum _{n=0}^{\infty }{x}_n$$
    收敛, 并求其和.

逐题详解

求极限：

$$\underset{x\to 3}{\lim }\frac{\sqrt{x^3+9}-6}{2-\sqrt{x^3-23}}=.$$

解

$$\begin{array}{rl}\underset{x\to 3}{\lim }\frac{\sqrt{x^3+9}-6}{2-\sqrt{x^3-23}}& =\underset{x\to 3}{\lim }\frac{\left(\sqrt{x^3+9}-6\right)\left(\sqrt{x^3+9}+6\right)}{\left(2-\sqrt{x^3-23}\right)\left(2+\sqrt{x^3-23}\right)}\cdot \underset{x\to 3}{\lim }\frac{2+\sqrt{x^3-23}}{\sqrt{x^3+9}+6}\\ & =\underset{x\to 3}{\lim }\frac{x^3+9-36}{4-x^3+23}\cdot \underset{x\to 3}{\lim }\frac{2+\sqrt{x^3-23}}{\sqrt{x^3+9}+6}\\ & =-\frac{1}{3}\end{array}$$

设函数 $z=f(x^2-y^2,xy)$, 且 $f(u,v)$ 具有连续的二阶偏导数, 则

$$\frac{{\partial }^{2}z}{\partial x\partial y}=.$$

解

$$\begin{array}{rl}\frac{\partial z}{\partial x}& =f_u\cdot 2x+f_v\cdot y\\ \frac{{\partial }^{2}z}{\partial x\partial y}& =2x(f_{uu}\cdot (-2y)+f_{uv}\cdot x)+y(f_{uv}\cdot (-2y)+f_{vv}\cdot x)+f_v\\ & =-4xy{f}_{uu}+2(x^2-y^2)f_{uv}+xy{f}_{vv}+f_v\end{array}$$

设 $n$ 为正整数, $f(x)=\frac{1}{x^2-3x+2}$, 则 :
$$f^{(n)}(0)=.$$

解

将$f(x)=\frac{1}{x^2-3x+2}$变为：$(x^2-3x+2)f(x)=1$两边同时求导：

$$\begin{array}{rl}{\left((x^2-3x+2)f(x)\right)}^{(n)}& =0\\ \Rightarrow 2f^{(n)}-3n{f}^{(n-1)}+n(n-1)f^{(n-2)}& =0\end{array}$$
取$b_n=\frac{f^{(n)}}{n!}$得到方程：

$$2b_n-3b_{n-1}+b_{n-2}=0$$
同时有初值条件：$b_0=\frac{1}{2}$与$b_1=\frac{3}{4}$
解差分方程得到：
$$b_n=1-\frac{1}{2^{n+1}}$$
故：
$$f^n(0)=n!\left(1-\frac{1}{2^{n+1}}\right)$$

幂级数
$$\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n(2n-1)}{x}^{2n}$$
的收敛域为 $.$

解

$$\begin{array}{rl}L& =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{(-1{)}^n}{(n+1)(2n+1)}\cdot \frac{n(2n-1)}{(-1{)}^{n-1}}\right|\\ & =\underset{n\to \infty }{\lim }\frac{n(2n-1)}{(n+1)(2n+1)}=1.\end{array}$$
故，$x\in (-1,1)$，取$x=\pm 1$.得到数项级数：

$$\sum _{n=0}^{+\infty }\frac{(-1{)}^n}{n(2n-1)}$$

显然绝对收敛。

因此：$$x\in [-1,1]$$

设曲面 $\Sigma$ 是平面 $y+z=5$ 被柱面 $x^2+y^2=25$ 所截得的部分, 则曲面积分
$$\underset{\Sigma }{\iint }(x+y+z)\mathrm{d}S=$$

解
$$\begin{array}{rl}{\iint }_{\Sigma }(x+y+z)\mathrm{d}S& =\sqrt{2}{\iint }_{{\Sigma }_{xOy}:x^2+y^2\le 25}(x+5)\mathrm{d}x\mathrm{d}y\\ & =5\sqrt{2}{\iint }_{{\Sigma }_{xOy}:x^2+y^2\le 25}\mathrm{d}x\mathrm{d}y\\ & =125\sqrt{2}\pi \end{array}$$

解微分方程:
$$(x^2+y^2+3)\frac{\mathrm{d}y}{\mathrm{d}x}=2x\left(2y-\frac{x^2}{y}\right).$$

解

做换元：$\{\begin{array}{l}z=\frac{y^2+1}{x^2+2}\\ P=x^2+2\end{array}$

$$\begin{array}{rl}(x^2+y^2+3)\frac{\mathrm{d}y}{\mathrm{d}x}& =2x\left(2y-\frac{x^2}{y}\right)\\ \Rightarrow P\frac{\mathrm{d}z}{\mathrm{d}P}& =-\frac{z^2-3z+2}{z+1}\\ \Rightarrow \frac{(z-2{)}^3}{(z-1{)}^2}& =-\frac{C}{P}\\ \Rightarrow (2x^2-y+2+3{)}^3& =C(x^2-y^2+1{)}^2\end{array}$$

设 ${\Sigma }_1$ 是以 $(0,4,0)$ 为顶点且与曲面 ${\Sigma }_2$: $\frac{x^2}{3}+\frac{y^2}{4}+\frac{z^2}{3}=1(y>0)$ 相切的圆锥面, 求曲面 ${\Sigma }_1$ 与 ${\Sigma }_2$ 所围成的空间区域的体积.

解
容易得到圆锥面方程：${\Sigma }_1:4(x^2+z^2)=(y-4{)}^2$
$$\begin{array}{rl}V& =\underset{x^2+z^2\le \frac{9}{4}}{\iint }\left[(4-2\sqrt{x^2+z^2})-2\sqrt{1-\frac{1}{3}(x^2+z^2)}\right]\mathrm{d}x\mathrm{d}z\\ & =4\pi \cdot \frac{9}{4}-2{\int }_0^{2\pi }{\int }_0^{\frac{3}{2}}{r}^2\mathrm{d}r-2{\int }_0^{2\pi }{\int }_0^{\frac{3}{2}}\sqrt{1-\frac{r^2}{3}}r\mathrm{d}r\\ & =9\pi -4\pi {\int }_0^{\frac{3}{2}}{r}^2\mathrm{d}r-4\pi {\int }_0^{\frac{3}{2}}\sqrt{1-\frac{r^2}{3}}r\mathrm{d}r\\ & =9\pi -\frac{9\pi }{2}-\frac{7\pi }{2}\\ & =\pi \end{array}$$

设 $a>1$, 求极限
$$\underset{n\to \infty }{\lim }n{\int }_1^a\frac{\mathrm{d}x}{1+x^n}.$$

解

Changing $x$to $1/x$gives

$$I_n:=n{\int }_1^a\frac{dx}{1+x^n}=n{\int }_b^1\frac{x^{n-2}}{1+x^n}dx,$$
where $b=\frac{1}{a}\in (0,1)$. Now, use integration by parts with $\frac{1}{x}=u$ and $\frac{n{x}^{n-1}}{1+x^n}dx=dv$. Then $du=-\frac{dx}{x^2}$ and $v=\ln (1+x^n)$ and so
$$I_n=\ln 2-\frac{\ln (1+b^n)}{b}+{\int }_b^1\frac{\ln (1+x^n)}{x^2}dx.$$
Since $b\in (0,1)$, we have $\underset{n\to \infty }{\lim }{b}^n=0$ and so
$$\underset{n\to \infty }{\lim }{I}_n=\ln 2+\underset{n\to \infty }{\lim }{\int }_b^1\frac{\ln (1+x^n)}{x^2}dx.\text{\hspace{1em}(*)}$$
Since $\ln(1 + t) \leq t, \ t > -1 $, we have $0\le \frac{\ln (1+x^n)}{x^2}\le x^{n-2}$ and hence $0\le {\int }_b^1\frac{\ln (1+x^n)}{x^2}dx\le \frac{1-b^{n-1}}{n-1}$ implying that $\underset{n\to \infty }{\lim }{\int }_b^1\frac{\ln (1+x^n)}{x^2}dx=0$, by the squeeze theorem. Thus, by (*), $\lim\limits_{n \to \infty} I_n = \ln 2 $.

Now, as an exercise, try to prove or disprove this claim: for any $b\in (-1,1)$:
$$\underset{n\to \infty }{\lim }n{\int }_b^1\frac{dx}{1+x^n}=\ln 2.$$

设 $f(x)$ 在 $[0,1]$ 上有连续的导数, 且 $f(0)=0$. 求证:
$${\int }_0^1{f}^2(x)\mathrm{d}x⩽4{\int }_0^1(1-x{)}^2{\left[f^{\prime }(x)\right]}^2\mathrm{d}x,$$
并求使得上式成为等式的 $f(x)$.

解

$$\begin{array}{rl}{\int }_0^1{f}^2(x)\mathrm{d}x& ={\int }_0^1(1-x)f^{\prime }(x)\cdot f(x)\mathrm{d}x\\ & ⩽{\left({\int }_0^1(1-x{)}^2(f^{\prime }(x){)}^2\mathrm{d}x\right)}^{\frac{1}{2}}{\left({\int }_0^1{f}^2(x)\mathrm{d}x\right)}^{\frac{1}{2}}\\ & ⩽4{\int }_0^1(1-x{)}^2[f^{\prime }(x){]}^2\mathrm{d}x\end{array}$$
当且仅当$cf(x)=(1-x)f^{\prime }(x)$时，可以取等,此时：
$$f(x)=0$$

设数列{xn}\{x_n\}{xn​} 满足 $x_0=\frac{1}{3},x_{n+1}=\frac{x_n^2}{1-x_n+x_n^2},n⩾0$. 证明无穷级数
$$\sum _{n=0}^{\infty }{x}_n$$
收敛, 并求其和.

解
$$\begin{array}{rl}{x}_{n+1}& =\frac{x_n^2}{1-x_n+x_n^2}\\ \Rightarrow x_n& =\frac{x_n}{1-x_n}-\frac{x_{n+1}}{1-x_{n+1}}\end{array}$$

那么：

$$\begin{array}{rl}\sum _{n=0}^{\infty }{x}_n& =\sum _{n=0}^{\infty }\left(\frac{x_n}{1-x_n}-\frac{x_{n+1}}{1-x_{n+1}}\right)\\ & =\frac{x_0}{1-x_0}=\frac{1}{2}\end{array}$$