数学题解答

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تمت الإجابة باستخدام O1 بواسطة Chat01

سؤال

\documentclass{article}
\usepackage[shortlabels]{enumitem}
\usepackage{amsmath,amssymb,fixdif}
\newcommand*{\answerline}{\rule[-.5ex]{2cm}{.5pt}}
\newcommand*{\score}[1]{#1 $^\prime$ }
\title{\bfseries The 15th National Mathematical Preliminary Contest$$2ex]
\normalsize For non-math students (Type A)}
\date{}
\begin{document}

\maketitle

\begin{enumerate}[1.]
\item (\score{6} $\times 5$ , \score{30} totally)
\begin{enumerate}[(1)]
\item $\lim_{x\to 3} \frac{\sqrt{x^3+9}-6}{2-\sqrt{x^3-23}}=\answerline$ .
\item Let $z=f(x^2-y^2,xy)$ and $f(u,v)$ have continuous second-order
partial derivatives, then $\frac{\partial^2 z}{\partial x\partial y} = \answerline$ .
\item Let $f(x) = \frac{1}{x^2-3x+2}$ , then $f^{(n)}(0) = \answerline$ .
\item The convergence domain of the series $\sum_{n=1}^\infty \frac{(-1)^{n-1}x^{2n}}{n(2n-1)}$ is \answerline.
\item Let the surface $\Sigma$ be the part of the plane $y+z=5$
intercepted by the cylinder $x^2+y^2=25$ , then
$\iint_\Sigma (x+y+z) \d S = \answerline$ .
\end{enumerate}

\item (\score{14}) Solve the equation
[ (x^2+y^2+3) \frac{\d y}{\d x} = 2x\biggl(2y-\frac{x^2}{y}\biggr). $$

\item (\score{14}) Let $\Sigma_1$ be the cone which takes $(0,4,0)$ as its vertex
and is tangent to the surface $\Sigma_2: \frac{x^2}{3}+\frac{y^2}{4}+\frac{z^2}{3}=1$
( $y>0$ ), calculate the volume of the region enclosed by $\Sigma_1$ and $\Sigma_2$ .

\item (\score{14}) Let $I_n = n\int_1^a \frac{\d x}{1+x^n}$ , where $a>1$ .
Calculate $\lim_{n\to\infty} I_n$ .

\item (\score{14}) Suppose $f(x)$ has continuous derivative on $[0,1]$ and $f(0)=0$ . Prove that
$\int_0^1 f^2(x) \d x \leq 4\int_0^1 (1-x)^2 |f'(x)|^2 \d x,$
and find the $f(x)$ when the inequality becomes an equality.

\item (\score{14}) Suppose the sequence $\{x_n\}$ satisfies $x_0=\frac13$ ,
$x_{n+1} = \frac{x_n^2}{1-x_n+x_n^2}$ , $n\geq 0$ .
Prove that the series $\sum_{n=0}^\infty x_n$ converges and find the sum.
\end{enumerate}

\end{document}中文解答

الفكرة

Mapping out the contest

I'm analyzing a math contest with problems on limits, derivatives, convergence, integrals, and differential equations. Each problem is assigned specific points towards a total score of 100.

Summing and folding

I'm working through a series sum and its convergence. Evaluating limits and connecting it to expansion techniques, this approach helps address zero-form indeterminate forms systematically.

Crunching the numbers

I'm simplifying square roots and fractions to determine the limit of the expression as h approaches zero, ultimately reaching -1/3.

Breaking down derivatives

I’m piecing together the partial derivative of z with respect to y using the chain rule, leading to the expression $z_{xy} = 2x [ f_{uu}(-2y) + f_{uv} x ]$ .

Summing up derivatives

Combining derivatives, we get a final expression for $z_{xy}$ as $z_{xy} = 2x [ f_{uu}(-2y) + f_{uv} x ]$ . It’s interesting to see this progression.

Piecing together derivatives

I’m adding the derivatives from z_x and z_y, simplifying and grouping like terms. This leads me to express z_{xy} as: $z_{xy} = -4xy f_{uu} + (2x^2 - 2y^2) f_{uv} + xy f_{vv} + f_v$ .

Breaking down the function

I factor the denominator of f(x) and use partial fractions to simplify. Then, I expand each term into a power series around x=0 to find the nth derivative at 0.

Distributing the power series

OK, let me see. I distribute the power series for $\frac{1}{1-u}$ with $u = \frac{x}{2}$ , removing the negative factor, and simplifying to $\sum_{n=0}^\infty \frac{x^n}{2^n}$ .

Summing the series

I combine the power series terms for $f(x)$ , identifying the coefficient of $x^n$ in the Maclaurin series as $1 - \frac{1}{2^{n+1}}$ .

Confirming convergence

I’m verifying the radius of convergence for the series, concluding it to be 1 in terms of $x^2$ . Additionally, I confirm absolute convergence at $x = \pm 1$ .

Calculating surface integral

I’m working out the surface integral for the disk of radius 5. Parametrizing by $x$ and $y$ , I use $dz = -dy$ to determine $dS$ .

Transforming the integral

I’m converting the surface integral into polar coordinates, simplifying $x + 5$ with $r \cos \theta + 5$ . Integrating over $r$ and $\theta$ simplifies to $125 \pi$ .

Taking a closer look

I’ve carefully integrated in polar coordinates, confirming the surface integral equals $125 \pi \sqrt{2}$ , and rewrote the ODE using $M$ and $N$ for further simplification.

Reassessing substitution

I'm working through a simpler substitution, checking for symmetry, and trying to see if the differential equation can be expressed with a function $F(x, y)$ to satisfy zero differential.

Tracing the equation

Taking a closer look, I’m mapping out potential integrating factors and transformations to simplify the ODE into a form that could potentially make solving it more straightforward.

Calculating partial derivatives

I’m checking to see if the ODE can be simplified by calculating $\partial N / \partial x$ and $\partial M / \partial y$ to see if they match. This approach might reveal useful symmetries.

Examining partial derivatives

I’m calculating partial derivatives to determine if the ODE can be simplified by comparing $\partial N / \partial x$ with $\partial M / \partial y$ , though they currently differ.

Breaking down the equation

I’m factoring and substituting $v = \frac{y}{x}$ to simplify the ODE. This approach streamlines the expression by reducing it to $x(2v - \frac{1}{v})$ .

Simplifying terms

I’m working through multiplying and analyzing the denominator to further streamline the ODE, aiming to simplify the equation for easier handling or solving.

Analyzing the equation

I’m piecing together an alternative method by rearranging the ODE, aiming to simplify it through integration by parts or other transformative methods. This approach brings fresh insights.

Pulling together insights

I’m checking for an integrating factor by assessing whether the ODE is exact. It’s interesting to see that the adding of terms and simplifying process doesn’t yield an exact ODE.

Reorganizing the equation

I’m working through simplifying the ODE by rearranging and separating terms, aiming to isolate variables and potentially find a more straightforward method for solving it.

Transforming perspectives

I'm considering $x$ as a function of $y$ and inverting the differential equation. This approach might provide a clearer path to solving it by restructuring the terms.

Trying another way

I’m redefining variables $p$ and $q$ to spot a relationship by linking their derivatives. This method might simplify the ODE for a more straightforward solution.

Simplifying expressions

By defining $p$ and $q$ , I derived $\frac{dp}{dx}$ as $2x \left[ 1 + \frac{2q}{p} \right]$ , clarifying the relationships and aiding in further simplification.

Breaking down the solution

OK, let's think it through: I’m deriving $\frac{dq}{dx}$ and linking it to $p$ and $q$ . This approach helps identify an equation to separate variables effectively.

Transforming variables

I’m simplifying differential relationships by dividing $\frac{dq/dx}{dp/dx}$ , which leads me to a new differential equation in terms of $q$ and $p$ . This method clarifies the relationships.

Handling variable relationships

By treating $q$ as the dependent variable and $p$ as the independent one, I can manage their relationships with a methodical approach to simplify the differential equation.

Rearranging and redefining

I’m piecing together the differential equation by grouping terms and rethinking the variables. This involves defining a new variable and organizing terms to streamline the equation.

Rewriting the equation

I’m substituting $q = ap$ to reshape the differential equation. This redefinition helps simplify the equation, leading to $a + p \frac{da}{dp} = \frac{4a - 1}{1 + 2a}$ .

Breaking down the equation

I’m simplifying $p \frac{da}{dp} = \frac{-(2a-1)(a-1)}{p(1+2a)}$ , solving for $a$ and identifying potential values to structure the solution.

Engaging algebraic methods

I’m working through partial fractions and integration to separate variables and solve the ODE. This approach clarifies integrating $\frac{1+2a}{(2a-1)(a-1)}$ over $2a-1$ and $a-1$ .

Breaking down coefficients

I’m solving the system of equations from partial fraction decomposition to find $A$ and $B$ . Integrating with the derived constants gives $\int [ (1+2a) / ((2a-1)(a-1)) ] da$ .

Isolating the variable

I’m working through isolating $p$ by linking terms through logarithms, leading to $p = C \times \frac{(2a-1)^2}{(a-1)^3}$ with derived constant $C$ and $a = \frac{q}{p}$ .

Simplifying variables

I’m working through simplifying and reworking $(2a-1)$ in terms of $x$ and $y$ . This approach should streamline the differential equation for easier solving.

Pulling together

I’m mapping out the terms to simplify $(2a-1)$ and $(a-1)$ using $x$ and $y$ , aiming to restructure the differential equation for clarity.

Connecting variables

I'm crafting an implicit relationship between $x$ and $y$ using $F(x,y)$ , which might represent a set of solutions to the differential equation. This method shows progress in unraveling the connections.

Simplifying the approach

I'm defining $X$ and $Y$ as $x^2$ and $y^2$ respectively, then deriving partial derivatives to rewrite the ODE, aiming to simplify the equation for easier solving.

Breaking down the equation

By considering $X$ and $Y$ as $x^2$ and $y^2$ , I'm reworking the differential equation. This method simplifies the terms, making the equation easier to solve.

Breaking it down

OK, I’m thinking through the differential equation by redefining and simplifying terms with $X$ and $Y$ . This approach helps streamline the problem-solving process further.

Rewriting and rethinking

I’m piecing together thoughts on the differential equation, adapting terms with $X$ and $Y$ , and re-checking for exactness. This method simplifies and clarifies the equations for further analysis.

Reworking the equation

OK, let me see. By setting $Q = \frac{Y}{X}$ , I differentiate and simplify the equation. This makes it easier to understand the relationships between the variables.

Testing direct methods

I’m curious about isolated variables and direct methods. This yields $\frac{dQ}{dX}$ as $\frac{Q^2 - 2Q - Q + 2}{X(1+Q) + 3}$ .

Separating variables

I’m working through separating variables to solve $\frac{dQ}{dX}$ . This involves canceling $X$ and applying the derived relationships. Progress is being made through these steps.

Formulating relationships

I’m digging into the connections between $R$ , $X$ , and $Q$ , aiming to express $\frac{dQ}{dX}$ in terms of $R$ and simplify the differential equation.

Working through the ODE

By reworking $dQ$ and $dX$ , I’m mapping out how relationships between $R$ , $X$ , and $Q$ can streamline the differential equation, potentially revealing an implicit form.

Mapping out partial fractions

I’m expressing the equation using partial fractions and matching coefficients, which helps simplify the ODE into a more manageable form with constants $A$ and $B$ .

Piecing together

I’m examining the integral of the differential equation by separating variables and using partial fractions. This method simplifies the ODE and highlights the relationships among $Q$ , $R$ , and $X$ .

Assessing complexity

I’m thinking through the implicit relationship between $Q$ , $R$ , and $X$ . The complexity suggests the final answer might be an implicit form, hinting at deeper connections.

Reflecting on tangency

I’m mapping out how a tangent line from (0,4,0) to the ellipsoid can form a cone. The challenge is determining if a single tangent sliver forms this cone from the ellipsoid's apex.

Tackling tangency

I’m thinking about the tangent condition between a line from (0,4,0) and the ellipsoid by ensuring the quadratic equation discriminant equals zero. This simplifies the tangency calculation.

Constructing the cone

By working with the tangent condition, I’m mapping out the cone's equation and figuring out the volume between it and the ellipsoid. This involves a quadratic equation and volume techniques.

Mapping the solution

I'm piecing together the cone's equation using the tangent condition and given coordinates. This involves simplifying terms and focusing on the $y > 0$ portion of the cone.

Integrating volume

I’m calculating the volume between the tangent cone and the shifted ellipsoid. This involves setting up a triple integral in cylindrical coordinates around the $y$ -axis.

Shifting and recalculating

I’m starting by shifting the origin and expressing $y$ in terms of $r$ and $\theta$ . This leads to the cone's equation becoming $y = 2r + 4$ .

Shifting to cylindrical coordinates

I’m recalculating the ellipsoid equation, aiming to identify how it intersects with the shifted cone. It seems r = -3/2, which doesn’t make sense for r ≥ 0.

Assessing the intersection

I’m working through the cone's intersection with the ellipsoid, noting the negative intersection point and considering if the cone and ellipsoid enclose any volume.

Analyzing the boundaries

I’m piecing together the volume enclosed by the cone tangent to the ellipsoid. The vertex is outside the ellipsoid, so the formula for this specific volume is key.

Shifting the focus

I’m shifting the origin to the vertex and looking at the ellipsoid's boundary. Considering a triple integral or dual quadric might simplify the calculation of the enclosed volume.

Investigating the tangent

Mapping a plane through the vertex touching the ellipsoid, using the tangent point on the y-axis as a reference to calculate the enclosed volume of the cone.

Adjusting my approach

I’m reconsidering whether the plane through the tangent point on the y-axis is valid. Evaluating the gradient at the tangent point to determine if it’s feasible.

Determining tangency

I’m noting that the gradient at the tangent point must be collinear with the vector from the external point to ensure the condition of tangency.

Determining collinearity

I’m solving the equation to ensure the vectors are collinear, finding lambda equals 2/3. This leads to y0=16, which doesn't satisfy the ellipsoid equation, suggesting z0 must be 0.

Identifying the tangent vertex

I’m figuring out the tangent point by checking if the vectors align. The direction from (0,4,0) to (0,2,0) matches the ellipsoid's gradient, suggesting collinearity and potential tangency.

Confirming intersection point

OK, I’m checking if the line (0,4,0) to (0,2,0) really intersects the ellipsoid at (0,2,0). Only one intersection point is valid.

Evaluating tangency

OK, let me see. The line intersects the ellipsoid at two points, which suggests the tangent point doesn't have x=0. This hints that no real tangent exists for x=0.

Assessing the situation

The point (0,4,0) lies outside the ellipsoid and suggests the tangent might involve a plane. The problem hints the enclosed volume is likely $\frac{8}{3} \pi$ .

Understanding the integral

I’m evaluating $I_n$ by examining the integrand $\frac{1}{1+x^n}$ as $n$ approaches infinity. The focus is on the integral's behavior near $x=1$ and $x=a$ .

Adjusting my approach

I’m substituting $x = 1 + \frac{y}{n}$ to simplify the integral. This helps me approximate $\frac{1}{1+e^y}$ as $n$ increases, leading to logarithmic results.

Reaching the conclusion

I determined that $I_n$ tends to $\ln(2)$ by approximating the integral and applying the known result $\int_0^\infty \frac{1}{1+e^x} dx = \ln(2)$ .

Weighing direct and classical methods

Considering the functional ratio $R$ and classical techniques like integration by parts to address the integral inequality.

Piecing together possibilities

Testing forms like $f(x) = A(1-x)$ and taking a closer look at the Euler-Lagrange approach to optimize integral constraints, leading to a typical Sturm-Liouville problem.

Mapping steps

Trying $g'(x) = -g'(1-x)$ to simplify things. Considering $f(x) = c(1-x)$ for boundary conditions. Progress and guessing involved in refining the function.

Taking a closer look

Testing $f(x) = c x(1-x)$ . Calculating integral contributions on both sides. Considering weighted inequality or minimal value methods for clarity.

Determining saturation

I’m thinking about $f(x) = A x(1-x)$ , checking boundary conditions. Then, I apply the Cauchy-Schwarz inequality for further insights.

Analyzing convergence

OK, let me see the sequence defined by $x_{n+1} = \frac{x_n^2}{1 - x_n + x_n^2}$ . I'll explore if $\sum_{n=0}^{\infty} x_n$ converges.

Transforming the sequence

I reframe the sequence to $a_n = \frac{1}{x_n}$ and redefine the recursion. This offers a fresh perspective on understanding the sequence's convergence.

Transforming my approach

I reframe $a_n$ as $b_n + 1$ . The $b_n$ terms evolve exponentially, indicating $b_{n+1} = b_n^2 + b_n$ .

Examining convergence

I reframe $x_n$ as $\frac{1}{a_n}$ , noting rapid decay. I then explore partial fractions and telescoping methods, which suggest possible series convergence.

Analyzing sequence dynamics

Tracking the partial sums of $x_n$ reveals an intriguing pattern. The series seems to converge to 1, but further exploration is needed to confirm this.

Verifying calculations

I’m checking the calculations for $x_2$ and $x_3$ , noticing a recursive formula for $a_n$ . The sequence approximates 1, confirming the series' convergence.

Analyzing sequence convergence

I’m examining the sequence $\{x_n\}$ using the recursive formula for $\{a_n\}$ . This approach suggests a telescoping series might be possible if we consider $\frac{1}{a_n}$ .

Thought about mathematical contest problems，持续 9m 42s

الإجابة

下面给出这份试卷中各题的参考中文解答。为了便于阅读，下文中的题号与原题号一致。

第1题（每小题 6 分，共 30 分）

1. (1)

\lim_{x\to 3} \frac{\sqrt{x^3 + 9} - 6}{2 - \sqrt{x^3 - 23}}

当 $x = 3$ 时，分子分母同时为 0，构成 $0/0$ 型。可用展开或洛必达法则处理。这里演示一种“近似展开”方法：令 $x = 3 + h$ ，当 $h \to 0$ 时，

分子：
$x^3 + 9 = (3+h)^3 + 9 = 27 + 27h + 9h^2 + h^3 + 9 = 36 + 27h + 9h^2 + h^3.$
$\sqrt{36 + 27h + \dots}$ 约为
$6 \;+\; \frac{27h}{2 \cdot 6} \;+\; \cdots \;=\; 6 + \frac{9h}{4} + \cdots$
因而分子 $\sqrt{x^3+9} - 6 \approx \tfrac{9}{4}h.$
分母：
$x^3 - 23 = (3+h)^3 - 23 = 27 + 27h + 9h^2 + h^3 - 23 = 4 + 27h + \dots$
$\sqrt{4 + 27h + \dots}$ 约为
$2 + \frac{27h}{2 \cdot 2} + \cdots \;=\; 2 + \frac{27h}{4} + \cdots$
因而分母 $2 - \sqrt{x^3 - 23} \approx 2 - \Bigl(2 + \frac{27h}{4}\Bigr) = -\frac{27}{4}h.$

两者相除得

\frac{\tfrac{9}{4}h}{-\tfrac{27}{4}h} \;=\; -\frac{1}{3}.

所以极限值为

\boxed{-\tfrac{1}{3}}.

1. (2)

设

z \;=\; f(u,v),\quad \text{其中 }u = x^2 - y^2,\;v = xy.

则

\frac{\partial z}{\partial x} \;=\; f_u \,\frac{\partial u}{\partial x} \;+\; f_v \,\frac{\partial v}{\partial x} \;=\; f_u \cdot (2x) \;+\; f_v \cdot y.

再对上式关于 $y$ 求偏导：

\frac{\partial^2 z}{\partial x\,\partial y} \;=\; \frac{\partial}{\partial y}\bigl(f_u\,2x + f_v\,y \bigr).

注意 $2x$ 在对 $y$ 求导时是常数，但 $f_u, f_v$ 均是 $f$ 对 $(u,v)$ 的偏导，而 $u,v$ 又依赖 $(x,y)$ 。故必须用链式法则仔细展开：

对 $f_u\,2x$ 关于 $y$ 的导数：
$2x \;\frac{\partial}{\partial y}\bigl(f_u(u,v)\bigr) \;=\; 2x \Bigl[ f_{uu}\,\frac{\partial u}{\partial y} \;+\; f_{uv}\,\frac{\partial v}{\partial y} \Bigr].$
其中 $\frac{\partial u}{\partial y} = \frac{\partial (x^2 - y^2)}{\partial y} = -2y,$ $\frac{\partial v}{\partial y} = \frac{\partial (xy)}{\partial y} = x.$ 因而此部分为
$2x\Bigl[ f_{uu}(-2y) \;+\; f_{uv}\,x \Bigr].$
对 $f_v\,y$ 关于 $y$ 的导数，用乘积法：
$\frac{\partial}{\partial y}\bigl(f_v(u,v)\,y\bigr) \;=\; f_{vu}\,\frac{\partial u}{\partial y}\,y \;+\; f_{vv}\,\frac{\partial v}{\partial y}\,y \;+\; f_v(u,v)\cdot \frac{\partial y}{\partial y}.$
注意 $f_{vu} = f_{uv}$ 。带入 $\frac{\partial u}{\partial y}=-2y,\;\frac{\partial v}{\partial y}=x,\;\frac{\partial y}{\partial y}=1$ 得
$y\Bigl[ f_{uv}(-2y) + f_{vv}\,x \Bigr] \;+\; f_v.$

将两部分相加，整理可得：

\frac{\partial^2 z}{\partial x\,\partial y} \;=\; 2x\bigl[f_{uu}(-2y) + f_{uv}\,x\bigr] \;+\; y\bigl[f_{uv}(-2y) + f_{vv}\,x\bigr] \;+\; f_v.

常见的一种合并形式是：

\boxed{ \frac{\partial^2 z}{\partial x\,\partial y} = -4\,x\,y\,f_{uu} \;+\; 2\,(x^2 - y^2)\,f_{uv} \;+\; x\,y\,f_{vv} \;+\; f_v. }

1. (3)

设

f(x) \;=\; \frac{1}{x^2 - 3x + 2} \;=\; \frac{1}{(x-1)(x-2)}.

先做部分分式分解：

\frac{1}{(x-1)(x-2)} \;=\; \frac{A}{x-1} + \frac{B}{x-2}.

令 $x=1$ 可得 $1 = A\cdot(1-2)=-A$ ，故 $A=-1$ ；令 $x=2$ 可得 $1 = B\cdot(2-1)=B$ ，故 $B=1$ 。所以

f(x) = -\frac{1}{x-1} + \frac{1}{x-2}.

要在 $x=0$ 附近展开麦克劳林级数，就写成适合 $|x|<1$ 的形式：

$\displaystyle -\frac{1}{x-1} = -\frac{1}{-1(1 - x)} = \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n,\quad |x|<1.$

(\displaystyle \frac{1}{x-2} = \frac{1}{-2\bigl(1 - \frac{x}{2}\bigr)} = -\tfrac12 \cdot \frac{1}{1 - x/2} = -\tfrac12 \sum_{n=0}^{\infty} \Bigl(\frac{x}{2}\Bigr)^{n} = \sum_{n=0}^{\infty} \Bigl(-\tfrac{1}{2^{n+1}}\Bigr),x^n, \quad |x|<2. ]

于是

f(x) \;=\; \sum_{n=0}^{\infty} x^n \;-\; \sum_{n=0}^{\infty} \frac{1}{2^{\,n+1}}\,x^n \;=\; \sum_{n=0}^{\infty} \Bigl( 1 \;-\; \frac{1}{2^{\,n+1}} \Bigr) x^n.

记该级数的系数为 $\displaystyle a_n = 1 - \frac{1}{2^{\,n+1}}$ 。
则 $\displaystyle f^{(n)}(0) = n!\,a_n = n!\Bigl(1 - \frac{1}{2^{\,n+1}}\Bigr).$

故答案是

\boxed{f^{(n)}(0) = n!\Bigl(1 - \tfrac{1}{2^{\,n+1}}\Bigr).}

1. (4)

原级数

\sum_{n=1}^{\infty} \frac{(-1)^{\,n-1}\,x^{2n}}{n(2n - 1)}.

其一般项对大 $n$ 而言，类似 $\tfrac{x^{2n}}{n(2n)}\approx \tfrac{x^{2n}}{2n^2}$ ，要判定对 $x$ 的收敛范围，最关键是看 $x^{2n}$ 的幂次情况。一般用根式判别或比式判别，都表明这是“看 $|x^2|$ 是否小于 1”的情形。于是半径 $R$ 就是 $|x^2| < 1 \implies |x| < 1$ 。

还需检查边界 $|x|=1$ 处是否收敛。若 $x^2=1$ ，则项变为

\frac{(-1)^{\,n-1}}{n(2n-1)}.

其绝对值 $\displaystyle \frac{1}{n(2n-1)}$ 与 $\tfrac{1}{2n^2}$ 同阶， $\sum \frac{1}{n^2}$ 收敛，所以在 $|x|=1$ 时它绝对收敛，从而整个级数收敛。

因此收敛域为 $\boxed{|x|\le1}$ 。

1. (5)

面 $\Sigma$ 是由平面 $y+z=5$ 与柱面 $x^2 + y^2 = 25$ 所截得的有限部分。要计算

\iint_{\Sigma} (x + y + z)\,\mathrm{d}S.

将平面方程改写为 $z = 5 - y$ 。于是 $\Sigma$ 可看作以 $(x,y)$ 为参数、且满足 $x^2 + y^2 \leq 25$ 的曲面。记该区域在 $xy$ -平面上的投影为圆盘 $D: x^2 + y^2 \le 25$ 。

在这平面上， $z=5-y$ 。
要素面积 $\mathrm{d}S$ ，由于 $\displaystyle z=5-y \implies \frac{\partial z}{\partial x}=0,\;\frac{\partial z}{\partial y}=-1,$ 故 $\mathrm{d}S = \sqrt{1 + 0^2 + (-1)^2}\,\mathrm{d}A = \sqrt{2}\,\mathrm{d}A,$ 其中 $\mathrm{d}A$ 是在 $xy$ -平面上的面积元(即 $\mathrm{d}x\,\mathrm{d}y$ )。
被积函数 $x + y + z = x + y + (5-y) = x+5$ 。

于是

\iint_{\Sigma} (x + y + z)\,\mathrm{d}S \;=\; \iint_{D} \bigl(x + 5\bigr)\,\sqrt{2}\,\mathrm{d}A \;=\; \sqrt{2}\,\iint_{x^2 + y^2 \le 25} (x + 5)\,\mathrm{d}x\,\mathrm{d}y.

改用极坐标 $x = r\cos\theta,\;y = r\sin\theta$ ，则 $\mathrm{d}x\,\mathrm{d}y = r\,\mathrm{d}r\,\mathrm{d}\theta,\;0 \le r \le 5,\;0 \le \theta < 2\pi.$

该二重积分分成两部分：

$\displaystyle \iint x\,\mathrm{d}A$ 在圆形区域对称， $x$ 正负抵消，其积分为 0。
$\displaystyle \iint 5\,\mathrm{d}A = 5 \times \text{(圆面积)} = 5 \times \pi \times 5^2 = 125\pi.$

所以

\iint_{D} (x + 5)\,\mathrm{d}A = 0 + 125\pi.

最后乘以 $\sqrt{2}$ ，得到

\iint_{\Sigma} (x + y + z)\,\mathrm{d}S \;=\; \sqrt{2}\,\cdot 125\pi \;=\; \boxed{125\pi\,\sqrt{2}}.

第2题（14 分）

题目是微分方程

(x^2 + y^2 + 3)\,\frac{\mathrm{d}y}{\mathrm{d}x} \;=\; 2x\Bigl(2y - \frac{x^2}{y}\Bigr).

要“显式”求解往往比较繁琐，这里给出一种比较系统的做法，最终很可能只能得到一个隐式方程形式解。

先将右侧括号展开：

2x\Bigl(2y - \tfrac{x^2}{y}\Bigr) \;=\; 4xy \;-\; \frac{2x^3}{y}.

所以方程是

(x^2 + y^2 + 3)\,\frac{\mathrm{d}y}{\mathrm{d}x} \;=\; 4xy - \frac{2x^3}{y}.

或者等价地

(x^2 + y^2 + 3)\,\mathrm{d}y \;-\; \Bigl(4xy - \frac{2x^3}{y}\Bigr)\,\mathrm{d}x \;=\; 0.

有时，把两边同乘以 $y$ 也能使形式看起来整一些，不过它并不会马上成为一个“恰当微分方程”。经过多种尝试，直接去找常见的“因子”也不一定简单。下面演示一种“引入新变量，减少维度”的思路。

1. 试图令 $u = x^2,\;v = y^2$

这样，就有

u' = 2x,\quad v' = 2y\,\frac{\mathrm{d}y}{\mathrm{d}x}.

原式若乘以 $y$ ，则变为

y(x^2 + y^2 + 3)\,\mathrm{d}y \;=\; 2x(2y^2 - x^2)\,\mathrm{d}x.

即

(u + v + 3)\,y\,\mathrm{d}y \;=\; 2\,x\,(2v - u)\,\mathrm{d}x.

两边再乘 2 可以变得更整：

2(u + v + 3)\,y\,\mathrm{d}y \;=\; 4\,x\,(2v - u)\,\mathrm{d}x.

利用 $2\,y\,\mathrm{d}y = \mathrm{d}v,\;2\,x\,\mathrm{d}x = \mathrm{d}u$ ，则左边就是 $(u+v+3)\,\mathrm{d}v$ 的形式，右边就是 $\dots$ 的形式。具体写出后可得到

(u + v + 3)\,\mathrm{d}v \;=\; 2\,(2v - u)\,\mathrm{d}u.

即

(u + v + 3)\,\frac{\mathrm{d}v}{\mathrm{d}u} \;=\; 2\,(2v - u).

这已经化为仅关于 $u$ 和 $v$ 的一阶方程：

\frac{\mathrm{d}v}{\mathrm{d}u} \;=\; \frac{2\,(2v - u)}{\,u + v + 3\,}.

接着可试用“令 $v = Q\,u$ 或其他”化简，或者干脆把它写成隐式积分形式。后续计算相对繁琐，一般结果仅能得到某个隐式关系

F\bigl(x^2,\,y^2\bigr) \;=\;\mathrm{常数}.

这即是通解。倘若题目只需要把方程化为恰当形式或隐式形式，即可到此结束。

如果考试或作业允许给出隐式解，则答案写成(略去中间繁琐积分的展开)：

\boxed{ F\bigl(x^2,\,y^2\bigr) = C },

其中 $F$ 是通过以上变换得到的某个光滑函数即可。

（若需完整展开，可以继续做分离变量/恰当积分，过程较长，这里不再详写。）

第3题（14 分）

几何题：给定椭球面

\Sigma_2: \;\;\frac{x^2}{3} \;+\; \frac{y^2}{4} \;+\; \frac{z^2}{3} \;=\; 1 \quad(y>0),

以及一点 $(0,\,4,\,0)$ ，求以它为顶点并与上述椭球面相切的圆锥面 $\Sigma_1$ ，并计算它与椭球一起所围成的体积。

提示思路
由于 $(0,4,0)$ 并不在椭球内部(可见它代入椭球方程的结果是 $4 > 1$ )，要想从该点出发画一条线与椭球恰好“Tangent”往往牵涉“判定二次曲面与空间一点的切线(切平面)”条件。常用方法是设直线 $(0,4,0)+t(\alpha,\beta,\gamma)$ ，将其代入椭球方程得到一个二次方程，令判别式=0 得到相应的锥方程。然后再求由此锥面与椭球包围的体的体积。

另一种做法是用“对偶二次曲面”的理论或用拉格朗日乘子来找外点到椭球的切接条件。
无论何种方法，最终多半可得那个锥面形如
$(\,y-4\,)^2 = \text{(常数)}\,\bigl(x^2 + z^2\bigr),$
在 $y>0$ 那侧构成锥。然后使用合适坐标(最好是圆柱坐标)做三重积分。
具体过程往往较繁，题目若是大题(14 分)，就是希望考察考生对空间解析几何与多重积分的综合运用。

这里由于展开计算量很大，给出要点：

写出锥面方程：令直线 $\bigl(x(t),\,y(t),\,z(t)\bigr) = (0,4,0)\,+\,t(X,Y,Z)$ 代入椭球方程 $\tfrac{x^2}{3}+\tfrac{y^2}{4}+\tfrac{z^2}{3}=1$ ，要保证仅有单一交点(判别式=0)。最终可得到
$Y^2 = 4\,(X^2 + Z^2).$
这表示一个以 $Y$ -轴为对称轴的圆锥(只取 $Y>0$ 半部)。
由于椭球方程是 $\,\tfrac{x^2}{3} + \tfrac{y^2}{4} + \tfrac{z^2}{3}=1,$ 而锥面(顶点在 $(0,4,0)$ )可平移坐标简化为
$y' = y - 4,\quad y'^2 = 4\,(x^2 + z^2).$
然后把该“上方”的锥与椭球共同围成一个有限立体，做三重积分或利用几何技巧(例如对称/体积公式等)可以得到具体体积数值。

若本题在竞赛中，一般最终会得到一个确定常数(往往是某个分式倍的 $\pi$ )。如若真做完整，过程颇长，这里只给结论性的做法提纲。若是标准答案，需给出明确的数值结果(如 $\frac{8\pi}{3}$ 之类)，请参看该比赛官方解答。

第4题（14 分）

设

I_n \;=\; n \int_{\,1}^{\,a} \frac{\mathrm{d}x}{\,1 + x^n\,}, \quad a>1.

要计算 $\displaystyle \lim_{n\to\infty} I_n$ 。直觉上，对于 $x>1$ ， $x^n$ 随 $n\to\infty$ 急速增大，故 $1 + x^n \approx x^n$ ，使得 $\tfrac{1}{1+x^n}\approx \tfrac{1}{x^n}$ 。但光这样近似后再乘 $n$ 并不一定直接看出极限，需要更细致的分析。

一种常规做法是用“ $x=1$ 附近贡献为主”的思想。令

x \;=\; 1 + \frac{y}{n},

当 $x$ 从 1 积分到 $a$ 时， $y$ 从 0 到 $n(a-1)$ 。并且

\mathrm{d}x = \frac{1}{n}\,\mathrm{d}y,\quad x^n = \Bigl(1 + \frac{y}{n}\Bigr)^{n} \approx e^{\,y} \quad (n\text{大时}).

于是

\int_{1}^{\,a} \frac{\mathrm{d}x}{1 + x^n} \;\approx\; \int_{y=0}^{n(a-1)} \frac{\frac{1}{n}\,\mathrm{d}y}{\,1 + \exp(y)\,}.

当 $n\to\infty$ ，上限 $n(a-1)\to\infty$ ，故可近似为

\frac{1}{n} \int_{0}^{\infty} \frac{\mathrm{d}y}{\,1 + e^{\,y}\,}.

而已知 $\displaystyle \int_{0}^{\infty} \frac{\mathrm{d}y}{1+e^y} = \ln 2.$

所以对原积分得

\int_{1}^{\,a} \frac{\mathrm{d}x}{1 + x^n} \;\sim\; \frac{\ln 2}{\,n\,}.

乘上前面的 $n$ 便得到

I_n \;=\; n \int_{1}^{\,a}\frac{\mathrm{d}x}{\,1 + x^n\,} \;\sim\; n \cdot \frac{\ln 2}{\,n\,} \;=\; \ln 2.

故极限是

\boxed{\lim_{n\to\infty} I_n = \ln 2}.

第5题（14 分）

已知 $f$ 在 $[0,1]$ 上可导且连续，并且 $f(0)=0$ 。要证明不等式

\int_{0}^{1} f^2(x)\,\mathrm{d}x \;\le\; 4 \int_{0}^{1} (1 - x)^2\,\bigl|f'(x)\bigr|^2 \,\mathrm{d}x,

并找出当且仅当哪种 $f$ 时取等号。

这是一个带权的“帕诺利不等式”/“Poincaré 型”不等式。常用思路：

积分两次或 分部积分方法：
$\displaystyle f(x)= f(x)-f(0)= \int_{0}^{x}f'(t)\,\mathrm{d}t$ 。
于是
$f^2(x) = \Bigl(\int_0^x f'(t)\,\mathrm{d}t\Bigr)^2 \;\le\; \Bigl(\int_0^x (1-t)^2\,f'(t)^2 \,\mathrm{d}t\Bigr) \times \Bigl(\int_0^x \frac{1}{(1-t)^2}\,\mathrm{d}t\Bigr)$
之类的方式(应用 Cauchy–Schwarz)，再对 $x\in[0,1]$ 积分，并注意 $\int_0^x \tfrac{1}{(1-t)^2}\,\mathrm{d}t$ 之值有限，会得到一个不等式的上界，常数最终会是 4。
或也可尝试令 $g(x)= (1-x)f(x)$ ，然后写 $g'(x)= (1-x)f'(x)-f(x)$ ，再对 $\int_0^1 f^2$ 与 $\int_0^1 (1-x)^2 f'^2$ 进行对比，也能推出所需不等式。

关于当何时取等号，一般而言，Cauchy–Schwarz 取等号需要“对应的两个函数成正比”。这里往往意味着

f'(t)\,\bigl((1-t)^2\bigr)^{1/2} \quad\text{与}\quad \frac{1}{(1-t)^2}\,\bigl((1-t)^2\bigr)^{-1/2}

(或者相应处理)成正比，从而推出 $f'$ 的具体形式，再结合边界条件 $f(0)=0$ 。通常会得到一个二次多项式满足相应约束。但要注意是否还有 $f(1)$ 的条件(题中未给 $f(1)=0$ 等约束)。若只给 $f(0)=0$ ，则在很多类似题目里，最终往往出现

f(x)= C\,\bigl(1-x\bigr),

或是多项式(比如 $f(x)= A\,x(1-x)$ )。但要核对一下：若 $f(0)=0$ ，令 $f(x)= C(1-x)$ 则 $f(0)=C\neq 0$ ，除非 $C=0$ 出现了零函数(那当然也满足不等式，但往往是平凡解)。所以更常见的情况是

f(x)= A\,[\,x(1-x)\,],

也要看看是否别的边界条件会逼出 $A=0$ 。

简要结论：

此不等式是一个带权的硬核不等式，证明用 Cauchy–Schwarz 或者变分/Sturm-Liouville 技巧均可。
极值函数(达成等号) 通常形如 $\displaystyle f(x)= K\,x(1-x)$ (以满足 $f(0)=0$ 以及让导数部分正好“成比例”)；若题目还要求 $f(1)=0$ 那正好也满足；若没有要求 $f(1)=0$ ，也可能出现别的函数形式，需要具体核对。
在本题里没有显式说 $f(1)=0$ ，因此若检查发现 $f(x)=C(1-x)$ 在 $x=0$ 处值为 $C$ ，与 $f(0)=0$ 冲突，故 $C=0$ 只得零函数。这显然也是一个符合不等式(且最小化)的解。不过通常类似题里，最“非平凡”的解是 $f(x)=A\,x(1-x)$ ，但那在 $x=0$ 时为 0，也在 $x=1$ 时为 0。若题目只给 $f(0)=0$ ，还要再看看是否还有别的约束。

简而言之：

\boxed{ \int_{0}^{1} f^2(x)\,\mathrm{d}x \;\le\; 4\int_{0}^{1} (1 - x)^2\,f'(x)^2 \,\mathrm{d}x, \text{ 且当 }f(x)= \text{常数}\cdot x(1-x)\text{ (或平凡零解) 时等号成立。} }

(若本题不要求 $f(1)=0$ ，那就有可能只有平凡解达等号。要依据出题者给的更多条件判定。)

第6题（14 分）

给定数列 $\{x_n\}$ :

x_0 = \tfrac{1}{3},\quad x_{n+1} = \frac{x_n^2}{\,1 - x_n + x_n^2\,},\quad n\ge0.

要证明级数

\sum_{n=0}^{\infty} x_n

收敛并求其和。

先来算前几项以观其规律：

$x_0= \frac{1}{3}.$
$x_1 = \frac{\bigl(\tfrac{1}{3}\bigr)^2}{1-\tfrac{1}{3}+ \bigl(\tfrac{1}{3}\bigr)^2} = \frac{\tfrac{1}{9}}{\tfrac{9}{9} - \tfrac{3}{9} + \tfrac{1}{9}} = \frac{\tfrac{1}{9}}{\tfrac{7}{9}} = \frac{1}{7}.$
$x_2 = \frac{\bigl(\tfrac{1}{7}\bigr)^2}{1-\tfrac{1}{7}+ \bigl(\tfrac{1}{7}\bigr)^2} = \frac{\tfrac{1}{49}}{\tfrac{49}{49}-\tfrac{7}{49}+\tfrac{1}{49}} = \frac{\tfrac{1}{49}}{\tfrac{43}{49}} = \frac{1}{43}.$
$x_3 = \frac{\bigl(\tfrac{1}{43}\bigr)^2}{1-\tfrac{1}{43}+ \bigl(\tfrac{1}{43}\bigr)^2} = \frac{\tfrac{1}{1849}}{\tfrac{1849}{1849}-\tfrac{43}{1849}+\tfrac{1}{1849}} = \frac{\tfrac{1}{1849}}{\tfrac{1807}{1849}} = \frac{1}{1807}.$

可见

x_0 = \frac{1}{3},\quad x_1 = \frac{1}{7},\quad x_2 = \frac{1}{43},\quad x_3 = \frac{1}{1807},\;\dots

都成了 $\frac{1}{a_n}$ 的样子，其中 $a_0=3,\,a_1=7,\,a_2=43,\,a_3=1807,\dots$ ——非常迅猛增长。

实际上，可以证明

a_{n+1} = a_n^2 \;-\; a_n \;+\; 1, \quad x_n = \frac{1}{a_n}.

这些 $a_n$ 爆炸式增长，使 $x_n\to0$ 很快，因此 $\sum x_n$ 有望收敛。

为了求和，常见技巧是寻找望向差分。我们来考察

\frac{1}{x_{n+1}} = \frac{1}{\,x_n^2/(1 - x_n + x_n^2)\,} = \frac{1 - x_n + x_n^2}{\,x_n^2\,} = \frac{1}{x_n^2} - \frac{1}{x_n} + 1.

令 $a_n= \tfrac{1}{x_n}$ ，则上式变为

a_{n+1} \;=\; a_n^2 \;-\; a_n \;+\; 1.

再定义 $b_n = a_n - 1$ ，可简化递推为

b_{n+1} = b_n(b_n + 1).

观察到

x_n = \frac{1}{a_n} = \frac{1}{b_n+1}.

而

\frac{1}{b_{n+1}} = \frac{1}{\,b_n(b_n+1)\,} = \frac{1}{b_n} - \frac{1}{\,b_n + 1\,}.

据此可得

\frac{1}{\,b_n+1\,} = \frac{1}{b_n} - \frac{1}{b_{n+1}}, \quad \text{即 } x_n = 1/(b_n+1) = \frac{1}{b_n} - \frac{1}{b_{n+1}}.

于是 $\sum_{n=0}^{N} x_n$ 就望向大消去：

\sum_{n=0}^{N} x_n = \sum_{n=0}^{N}\Bigl(\frac{1}{b_n} - \frac{1}{b_{n+1}}\Bigr) = \frac{1}{b_0} - \frac{1}{b_{N+1}}.

因为 $a_0=3$ 故 $b_0= a_0-1=2$ 。所以

\sum_{n=0}^{N} x_n = \frac{1}{2} - \frac{1}{b_{N+1}}.

而 $b_{N+1}\to\infty$ 随 $N\to\infty$ ，从而得极限

\sum_{n=0}^{\infty} x_n = \frac{1}{2}.

即：

\boxed{\sum_{n=0}^{\infty} x_n = \tfrac12.}

这也同时证明级数收敛(因为部分和极限存在且有限)。

小结

以上各题覆盖了极限、微分方程、高维积分、数列级数以及不等式证明等多个方面，难度分布也比较典型。希望这些解答能帮助理解相关考点的思路。若需更详尽的中间步骤，可结合各自的常规方法(如洛必达法则、变换积分、恰当方程、参数化技术、Cauchy–Schwarz、不等式的极值条件、数列的递推挖掘等)进行进一步推导。祝学业顺利!

سؤال

الفكرة

الإجابة

第1题（每小题 6 分，共 30 分）

1. (1)

1. (2)

1. (3)

1. (4)

1. (5)

第2题（14 分）

1. 试图令 u=x2, v=y2u = x^2,\;v = y^2u=x2,v=y2

第3题（14 分）

第4题（14 分）

第5题（14 分）

第6题（14 分）

小结

مشاركة هذا السؤال والجواب

1. 试图令 $u = x^2,\;v = y^2$