数一2024

一、选择题:1 10 小题, 每小题 5 分, 共 50 分。下列每题给出的四个选项中, 只有一个选项是最符合题目要求的，请将所选项前的字母填在答题纸指定位置上。

(1)

已知函数 $\displaystyle f(x)=\int_0^x e^{\cos t} d t, g(x)=\int_0^{\sin x} e^{t^2} d t$, 则 ( ) (A) $f(x)$ 是奇函数, $g(x)$ 是偶函数 (B) $f(x)$ 是偶函数, $g(x)$ 是奇函数 (C) $f(x)$ 与 $g(x)$ 均为奇函数 (D) $f(x)$ 与 $g(x)$ 均为周期函数 李艳芳真题系列|考研数学一历年真题逐题精讲（2010-2024年）[更新至最新]-哔哩哔哩

• 读题:已知函数 $\displaystyle \underbrace{f(x)=\int_0^x e^{\cos t} d t, g(x)=\int_0^{\sin x} e^{t^2} d t}_{函数积分和求导后，奇偶性会发生改变}$, 则 ( )

• $\displaystyle \begin{aligned} e^{\cos x} \end{aligned}$ 是偶函数，$\displaystyle \begin{aligned} e^{\cos (-x)} = e^{\cos x} \end{aligned}$

  • $\displaystyle \begin{aligned} f(x) = \int_{0}^{x} e^{\cos t} dt \text{ 是奇函数} \end{aligned}$

• $\displaystyle \begin{aligned} e^{t^2} \end{aligned}$ 是偶函数

  • $\displaystyle \begin{aligned} h(x) = \int_{0}^{x} e^{t^2} dt \text{ 是奇函数} \end{aligned}$
  • $g(-x)=h(\sin (-x))=h(-\sin x)=-h(\sin x)=-g(x)$
    • $\displaystyle \begin{aligned} h(\sin x) = \int_{0}^{\sin x} e^{t^2} dt \text{ 是奇函数} \end{aligned}$

• 若 $f(x)$ 是连续函数， $\displaystyle \underbrace{F(x) 是 f(x) 的一个原函数}_{F(x)=\int_0^x f(t) d t}$，

  • 则当 $f(x)$ 是奇函数时， $F(x)$ 必是偶函数，
  • 当 $f(x)$ 是偶函数时，当且仅当 $F(0)=0$ 时， $F(x)$ 是奇函数.

• 小结：判定奇偶及周期性

  • (1) 用定义，如，证明
    • $\displaystyle \begin{aligned} f(x) = \int_x^{x+\frac{\pi}{2}} |\sin t| dt \end{aligned}$ 以 $\displaystyle \begin{aligned} \pi \end{aligned}$ 为周期
    • $\displaystyle \begin{aligned} F(x) = \int_{-a}^a |x-a| f(t) dt \end{aligned}$ 与 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 的奇偶性一致
  • (2) 用结论
    • 复合函数 $\displaystyle \begin{aligned} f[g(x)] \end{aligned}$，内 $\displaystyle \begin{aligned} (g(x)) \end{aligned}$ 偶则偶，内 $\displaystyle \begin{aligned} (g(x)) \end{aligned}$ 奇同外
      • 内部是偶函数，则复合函数也是偶函数，
      • 内部是奇函数，则奇偶性取决于外层函数
    • 求导改变奇偶性，但不改变周期性
    • $\displaystyle \begin{aligned} \int_0^x f(t) dt \end{aligned}$ 的奇偶性与 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 的计算奇偶性相反
    • $\displaystyle \begin{aligned} f(x) \end{aligned}$ 以 $\displaystyle \begin{aligned} T \end{aligned}$ 为周期，若 $\displaystyle \begin{aligned} \int_0^T f(t) dt = 0 \end{aligned}$，则 $\displaystyle \begin{aligned} \int_{x_0}^x f(t) dt \end{aligned}$ 也以 $\displaystyle \begin{aligned} T \end{aligned}$ 为周期

1999 年数一、二、三试题 设 $f(x)$ 是连续函数， $F(x)$ 是 $f(x)$ 的原函数，则（ ） （A） 当 $f(x)$ 是奇函数时， $F(x)$ 必是偶函数。 （B）当 $f(x)$ 是偶函数时， $F(x)$ 必是奇函数。 (C) 当 $f(x)$ 是周期函数时， $F(x)$ 必是周期函数。 (D) 当 $f(x)$ 是单调增函数时, $F(x)$ 必是单调增函数.

2005 年数一、二试题 设 $F(x)$ 是连续函数 $f(x)$ 的一个原函数， " $M \Leftrightarrow N$ "表示 " $M$ 的充分必要条件是 $N^{\prime \prime}$ ，则必有（） (A) $F(x)$ 是偶函数 $\Leftrightarrow f(x)$ 是奇函数. (B) $F(x)$ 是奇函数 $\Leftrightarrow f(x)$ 是偶函数. (C) $F(x)$ 是周期函数 $\Leftrightarrow f(x)$ 是周期函数. (D) $F(x)$ 是单调函数 $\Leftrightarrow f(x)$ 是单调函数.

2024 年数二试题 设函数 $\displaystyle f(x)=\int_0^{\sin x} \sin t^3 \mathrm{~d} t, g(x)=\int_0^x f(t) \mathrm{d} t$ ，则 $(\quad)$ (A) $f(x)$ 是奇函数， $g(x)$ 是奇函数. （B） $f(x)$ 是奇函数， $g(x)$ 是偶函数。 (C) $f(x)$ 是偶函数， $g(x)$ 是偶函数. (D) $f(x)$ 是偶函数， $g(x)$ 是奇函数.

(2)

设 $P=P(x, y, z), Q=Q(x, y, z)$ 均为连续函数, $\displaystyle \Sigma$ 为曲面 $Z=\sqrt{1-x^2-y^2}(x \leqslant 0, y \geqslant 0)$ 的上侧,则 $\displaystyle \iint_{\Sigma} P d y d z+Q d z d x=(\quad)$ (A) $\displaystyle \iint_{\Sigma}\left(\frac{x}{z} P+\frac{y}{z} Q\right) d x d y$ (B) $\displaystyle \iint_{\Sigma}\left(-\frac{x}{z} P+\frac{y}{z} Q\right) d x d y$ (C) $\displaystyle \iint_{\Sigma}\left(\frac{x}{z} P-\frac{y}{z} Q\right) d x d y$ (D) $\displaystyle \iint_{\Sigma}\left(-\frac{x}{z} P-\frac{y}{z} Q\right) d x d y$

• 读题
  设 $P=P(x, y, z), Q=Q(x, y, z)$ 均为连续函数, $\displaystyle \Sigma$ 为$\underbrace{曲面 Z=\sqrt{1-x^2-y^2}(x \leqslant 0, y \geqslant 0) 的上侧}_{画图+球面}$,则 $\displaystyle \underbrace{\iint_{\Sigma} P d y d z+Q d z d x=(\quad)}_{第二类曲面积分}$
• 解：$\displaystyle \begin{aligned} \iint\limits_{\Sigma} P dy dz + Q dx dz + R dx dy \xlongequal[]{转换投影法} \iint\limits_{\Sigma} \left[ P \cdot \left(-\frac{\partial z}{\partial x}\right) + Q \cdot \left(-\frac{\partial z}{\partial y}\right) + R \right] dx dy \end{aligned}$
  • 本题曲面 $\displaystyle \begin{aligned} \Sigma \end{aligned}$ 方程为 $\displaystyle \begin{aligned} z = \sqrt{1 - x^2 - y^2} \end{aligned}$，于是
    • $\displaystyle \begin{aligned} \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{1 - x^2 - y^2}} = \frac{-x}{z} \end{aligned}$
    • $\displaystyle \begin{aligned} \frac{\partial z}{\partial y} = \frac{-y}{z} \end{aligned}$
  • 按照上述公式，选 $\displaystyle \begin{aligned} (A) \end{aligned}$。

(3)

设幕级数 $\displaystyle \sum_{n=0}^{\infty} a_n x^n$ 的和函数为 $\ln (2+x)$, 则 $\displaystyle \sum_{n=0}^{\infty} n a_{2 n}$ ( ) (A) $-\frac{1}{6}$ (B) $-\frac{1}{3}$ (C) $\frac{1}{6}$ (D) $\frac{1}{3}$

• 读题:
  设$\displaystyle \underbrace{幂级数 \sum_{n=0}^{\infty} a_n x^n 的和函数为 \overbrace{\ln (2+x)}^{ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}}}_{已知和函数，反求幂级数的系数}$, 则 $\displaystyle \sum_{n=0}^{\infty} n a_{2 n}$ ( )

• $\displaystyle \ln (1+x)=x-\frac{x^2}{2}+\cdots+\frac{(-1)^{n-1} x^n}{n}+\cdots=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}(-1<x \leq 1)$

• 解：展开的关键在“变形”，由于
  $\displaystyle \begin{aligned} \ln(2+x) \xlongequal[]{恒等变形}\ln 2 \left(1 + \frac{x}{2}\right) = \ln 2 + \ln \left(1 + \frac{x}{2}\right) \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}]{幂级数展开} \ln 2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\left(\frac{x}{2}\right)^n}{n} \xlongequal[]{整理}\underbrace{\ln 2 + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n \cdot 2^n} x^n}_{\sum_{n=0}^{\infty} a_n x^n}\end{aligned}$
  • $\displaystyle \begin{aligned} = \sum_{n=0}^{\infty} a_n x^n \xlongequal[]{拆出1到∞} a_0 + \sum_{n=1}^{\infty} a_n x^n \end{aligned}$

• 所以 $\displaystyle \begin{aligned} a_0 = \ln 2 \end{aligned}$，$\displaystyle \begin{aligned} a_n = (-1)^{n-1} \frac{1}{n \cdot 2^n}, n = 1, 2, \cdots \end{aligned}$

  • $\displaystyle \begin{aligned} \xrightarrow[]{将n换成2n} \sum_{n=0}^{\infty} n a_{2n} \xlongequal[0项没用]{0×a_{2n} =0}\sum_{n=1}^{\infty} n a_{2n} = \sum_{n=1}^{\infty} n \cdot (-1)^{2n-1} \frac{1}{2n \cdot 2^{2n}} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{整理}-\sum_{n=1}^{\infty} \frac{1}{2^{2n+1}} = -\left(\frac{1}{2^{3}} + \frac{1}{2^{5}} + \frac{1}{2^{7}} + \cdots\right) \xlongequal[]{等比数列求和} -\frac{\frac{1}{2^{3}}}{1 - \frac{1}{2^{2}}} = -\frac{1}{6} \end{aligned}$

• 小结：如何找复杂函数 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 的幂级数展开式？

  • (1) 将 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 初等变形化成常用函数，然后套基本公式；
  • (2) 将 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 先求导化成常用函数，然后套基本公式；
    • $\arctan x$
  • (3) 将 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 先积分化成常用函数，然后套基本公式。
    • 数三2018#18

(4)

设函数 $f(x)$ 在区间上 $(-1,1)$ 有定义, 且 $\displaystyle \lim _{x \rightarrow 0} f(x)=0$, 则 ( ) (A) 当 $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x}=m$ 时, $f^{\prime}(0)=m$ (B) 当 $f^{\prime}(0)=m$ 时, $\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{x}=m$ (C) 当 $\displaystyle \lim _{x \rightarrow 0} f^{\prime}(x)=m$ 时, $f^{\prime}(0)=m$ (D) 当 $f^{\prime}(0)=m$ 时, $\displaystyle \lim _{x \rightarrow 0} f^{\prime}(x)=m$

解：

• A:$\displaystyle \begin{aligned} \lim_{x \to 0} f(x) = 0 \end{aligned}$，未必有 $\displaystyle \begin{aligned} f(0) = 0 \end{aligned}$
  • 故 $\displaystyle \begin{aligned} (A) \end{aligned}$ 选项 $\displaystyle \begin{aligned} \lim_{x \to 0} \frac{f(x)}{x} = m\xrightarrow[]{f(0)的值未知} \end{aligned}$未必有 $\displaystyle \begin{aligned} \lim_{x \to 0} \frac{f(x) - f(0)}{x} = m \end{aligned}$。
• B:当 $\displaystyle \begin{aligned} f'(0) = m \end{aligned}$ 时，必有 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 在 0 点连续，于是由 $\displaystyle \begin{aligned} \lim_{x \to 0} f(x) = 0 \end{aligned}$，必有 $\displaystyle \begin{aligned} f(0) = 0 \end{aligned}$，
  • 此时 $\displaystyle \begin{aligned} f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} \xlongequal[]{f(0) = 0}m \end{aligned}$
• 对 $\displaystyle \begin{aligned} (C) \end{aligned}$ 和 $\displaystyle \begin{aligned} (D) \end{aligned}$，当 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 在 0 点连续时，
  • 若 $\displaystyle \begin{aligned} \lim_{x \to 0} f'(x) = m \end{aligned}$（可以是 $\displaystyle \begin{aligned} \infty \end{aligned}$），此时有 $\displaystyle \begin{aligned} f'(0) = m \end{aligned}$（可以是 $\displaystyle \begin{aligned} \infty \end{aligned}$），
  • 但反之不成立，也就是说，若 $\displaystyle \begin{aligned} f'(0) = m \end{aligned}$，此时未必有 $\displaystyle \begin{aligned} \lim_{x \to 0} f'(x) = m \end{aligned}$。
    • 且还应知道，若 $\displaystyle \begin{aligned} \lim_{x \to 0} f'(x) \end{aligned}$ 不存在（非 $\displaystyle \begin{aligned} \infty \end{aligned}$），此时未必有 $\displaystyle \begin{aligned} f'(0) \end{aligned}$ 也不存在。
  • 以上所述便是“连续函数的导函数极限定理”，在 2009 年真题已证明过。

(5)

在空间直角坐标系 $O-x y z$ 中, 三张平面 $\pi_i: a_i x+b_i y+c_i z=d_i(i=1,2,3)$ 的位置关系如图所示,记 $\alpha_i=\left(a_i, b_i, c_i\right), \beta_i=\left(a_i, b_i, c_i, d_i\right)$ 若 $\displaystyle r\left(\begin{array}{l}\alpha_1 \\ \alpha_2 \\ \alpha_3\end{array}\right)=m, r\left(\begin{array}{l}\beta_1 \\ \beta_2 \\ \beta_3\end{array}\right)=n$, 则 $($, (A) $m=1, n=2$ (B) $m=n=2$ (C) $m=2, n=3$ (D) $m=n=3$

• 解：由题意可知，$\displaystyle \begin{aligned} \pi_1, \pi_2, \pi_3 \end{aligned}$ 相交于一条直线，且不重合，
  • 于是方程组$\displaystyle \begin{aligned} \begin{cases}a_1 x + b_1 y + c_1 z = d_1, \\a_2 x + b_2 y + c_2 z = d_2, \\a_3 x + b_3 y + c_3 z = d_3\end{cases} \end{aligned}$有无穷多解。
  • 无穷多解要求①系数矩阵和增广矩阵的秩相等，②秩< n
  • 只有B选项符合要求
• $\displaystyle \begin{aligned} \pi_1, \pi_2, \pi_3 \end{aligned}$ 相交于一个点，则有唯一解
  • 唯一解要求①系数矩阵和增广矩阵的秩相等，②秩= n
  • 只有D选项符合要求

(6)

设向量 $\displaystyle \alpha_1=\left(\begin{array}{c}a \\ 1 \\ -1 \\ 1\end{array}\right), \alpha_2=\left(\begin{array}{l}1 \\ 1 \\ b \\ a\end{array}\right), \alpha_3=\left(\begin{array}{c}1 \\ a \\ -1 \\ 1\end{array}\right)$, 若 $\alpha_1, \alpha_2, \alpha_3$ 线性相关, 且其中任意两个向量均线性无关，则（ ） (A) $a=1, b \neq 1$ (B) $a=1, b=-1$ (C) $a \neq 2, b=2$ (D) $a=-2, b=2$

• 读题
  设向量 $\displaystyle \alpha_1=\left(\begin{array}{c}a \\ 1 \\ -1 \\ 1\end{array}\right), \alpha_2=\left(\begin{array}{l}1 \\ 1 \\ b \\ a\end{array}\right), \alpha_3=\left(\begin{array}{c}1 \\ a \\ -1 \\ 1\end{array}\right)$, 若 $\underbrace{\alpha_1, \alpha_2, \alpha_3 线性相关}_{r\left(\alpha_1, \alpha_2, \alpha_3\right) \leq 2 }$, 且其中$\underbrace{任意两个向量均线性无关}_{r\left(\alpha_1, \alpha_2, \alpha_3\right) \geqslant 2}$，则（ ）

• 解：显然 $\displaystyle \begin{aligned} a = 1 \end{aligned}$ 时，$\displaystyle \begin{aligned} \alpha_1 = \alpha_3 \end{aligned}$，这与 $\displaystyle \begin{aligned} \alpha_1, \alpha_2, \alpha_3 \end{aligned}$ 中任意两个向量都无关矛盾！

• 先算秩

  • $\alpha_1, \alpha_2, \alpha_3$ 线性相关，得到$\displaystyle \begin{aligned} & r\left(\alpha_1, \alpha_2, \alpha_3\right) \leq 2 \end{aligned}$
  • 其中任意两个向量均线性无关，得到$\displaystyle \begin{aligned} r\left(\alpha_1, \alpha_2, \alpha_3\right) \geqslant 2 \end{aligned}$
  • 从而$\displaystyle \begin{aligned} r\left(\alpha_1, \alpha_2, \alpha_3\right)=2 \end{aligned}$

• 初等行变化

  • $\displaystyle \left(\alpha_1, \alpha_2, \alpha_3\right)=\left(\begin{array}{ccc}a & 1 & 1 \\ 1 & b & a \\ -1 & b & -1 \\ 1 & a & 1\end{array}\right) \xrightarrow[]{行变化}\left(\begin{array}{ccc}1 & 1 & a \\ 1 & a & 1 \\ -1 & b & -1 \\ a & 1 & 1\end{array}\right)$
  • $\displaystyle \xrightarrow[]{第一行加到2，3，4行}\left(\begin{array}{ccc}1 & 1 & a \\ 0 & a-1 & 1-a \\ 0 & b+1 & a-1 \\ 0 & 1-a & 1-a^2\end{array}\right) \xrightarrow[]{第二行加到第四行}\left(\begin{array}{ccc}1 & 1 & a \\ 0 & a-1 & 1-a \\ 0 & b+1 & a-1 \\ 0 & 0 & 2-a-a^2\end{array}\right)$

• 求a的值

  • $\displaystyle \begin{aligned} & 2-a-a^2=0 \Rightarrow-(a+2)(a-1)=0 \end{aligned}$
  • $\displaystyle \begin{aligned} r\left(\alpha_1, \alpha_2, \alpha_3\right)=2 \xrightarrow[]{第四行全0} 2-a-a^2=0 \Rightarrow a=1 \text { 或 } a=-2 . \end{aligned}$

• $\displaystyle a=1时，\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & b+1 & 0 \\ c & 0 & 0\end{array}\right)$，$\alpha_1, \alpha_3$线性相关，和题意不合

• $\displaystyle a=-2时， 原式\xrightarrow[]{a=-2}\left(\begin{array}{ccc}1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & b+1 & -3 \\ 0 & 0 & 0\end{array}\right) \xrightarrow[]{第二行加到第三行}\left(\begin{array}{ccc}1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & b-2 & 0 \\ 0 & 0 & 0\end{array}\right)$

  • $r\left(\alpha_1, \alpha_2, \alpha_3\right)=2 \quad \Rightarrow b-2=0 \quad$ 则 $b=2$

(7)

设 $A$ 是秩为 2 的 3 阶矩阵, $\alpha$ 是满足 $A \alpha=0$ 的非零向量, 若对满足 $\beta^T \alpha=0$ 的 3 维向量 $\beta$ 均有 $A \beta=\beta$, 则（） (A) $A^3$ 的迹为 2 (B) $A^3$ 的迹为 5 (C) $A^2$ 的迹为 8 (D) $A^2$ 的迹为 9 解：

• 读题
  设 $A$ 是秩为 2 的 3 阶矩阵, $\underbrace{\alpha 是满足 A \alpha=0 的非零向量}_{一个特征值为0}$, 若对满足 $\beta^T \alpha=0$ 的 $\underbrace{3 维向量 \beta 均有 A \beta=\beta}_{另外两个特征值都是1}$, 则（）
• 得出特征值0有两个方法
  • $\displaystyle \begin{aligned} r(A) = 2 < 3 \end{aligned}$，于是 $\displaystyle \begin{aligned} |A| = 0 \end{aligned}$，于是 $\displaystyle \begin{aligned} \lambda_1 = 0 \end{aligned}$。
  • $\displaystyle \begin{aligned} A\alpha = 0 \end{aligned}$ ($\displaystyle \begin{aligned} a \neq 0 \end{aligned}$) → $\displaystyle \begin{aligned} a \end{aligned}$ 是 $\displaystyle \begin{aligned} A \end{aligned}$ 的属于0的一个特征向量。
• 求另外两个特征值
  • $\displaystyle \begin{aligned} \beta^T \alpha = 0 \end{aligned}$ → $\displaystyle \begin{aligned} \alpha^T \beta = 0 \end{aligned}$ ($\displaystyle \begin{aligned} \beta \end{aligned}$ 是 $\displaystyle \begin{aligned} \alpha^T x = 0 \end{aligned}$ 的解)
    • 因为只有一个方程，所以$\displaystyle \begin{aligned} \alpha^T x = 0 \end{aligned}$ 有非零解 $\displaystyle \begin{aligned} \beta_1, \beta_2 \end{aligned}$。
  • $\displaystyle \begin{aligned} A \beta_1 = \beta_1, A \beta_2 = \beta_2 \end{aligned}$，由特征向量的定义得出特征值
    • 据 $\displaystyle \begin{aligned} A\beta = \beta \end{aligned}$，得 $\displaystyle \begin{aligned} A\beta_1 = \beta_1 = 1 \cdot \beta_1 \end{aligned}$，且 $\displaystyle \begin{aligned} A\beta_2 = \beta_2 = 1 \cdot \beta_2 \end{aligned}$。
    • 则 $\displaystyle \begin{aligned} \beta_1, \beta_2 \end{aligned}$ 为特征值 $\displaystyle \begin{aligned} \lambda_2 = \lambda_3 = 1 \end{aligned}$ 的线性无关特征向量，
• $\displaystyle \begin{aligned} A \end{aligned}$ 有三个特征向量 $\displaystyle \begin{aligned} \beta_1, \beta_2, \alpha \end{aligned}$，特征值为 $\displaystyle \begin{aligned} 1, 1, 0 \end{aligned}$。
  • 所以 $\displaystyle \begin{aligned} \operatorname{tr}(A^n) = 0 + 1 + 1 = 2 \end{aligned}$

(8)

设随机变量 $X, Y$ 相互独立, 且 $X \sim N(0,2), Y \sim N(-2,2)$, 若 $P\{2 X+Y<a\}=P\{X>Y\}$,则 a=（） (A) $-2-\sqrt{10}$ (B) $-2+\sqrt{10}$ (C) $-2-\sqrt{6}$ (D) $-2+\sqrt{6}$

• 读题
  设随机变量 $X, Y$ 相互独立, 且 $\underbrace{X \sim N(0,2), Y \sim N(-2,2)}_{X,Y的期望E和方差D}$, 若 $\underbrace{\overbrace{P\{2 X+Y<a\}}^{正态分布的线性组合也服从正态}=P\{X>Y\}}_{用标准化求正态分布的概率}$,则 a=（）

• $2 X+Y, \quad Y-X$也服从正态分布
  分别由X和Y的期望方差来计算，这两个分布的期望和方差

  • $2 X+Y$
    • $\displaystyle \begin{aligned} E(2X+Y) \xlongequal[]{提出系数} 2EX + EY = 0+(-2)=-2\end{aligned}$
    • $D(2X+Y) \xlongequal[]{提出系数}4DX + DY = 4 \times 2 + 2 = 10$
    • 于是 $\displaystyle \begin{aligned} 2X+Y \sim N(-2, 10) \xrightarrow[]{标准化}\frac{2 X+Y+2}{\sqrt{10}} \sim N(0,1)\end{aligned}$
  • $X>Y$
    • $\displaystyle \begin{aligned} & E(Y-X)=E(Y)-E(X)=-2-0=-2 \end{aligned}$
    • $\displaystyle \begin{aligned} D(Y-X)=D(Y)+D(X)=2+2=4 \end{aligned}$
    • 同理可算得 $\displaystyle \begin{aligned} Y-X \sim N(-2, 4) \xrightarrow[]{标准化}\frac{Y-X+2}{2} \sim N(0,1)\end{aligned}$

• 标准化步骤$\displaystyle F(x)=P\{X \leq x\}=P\left\{\frac{X-\mu}{\sigma} \leq \frac{x-\mu}{\sigma}\right\}=\Phi\left(\frac{x-\mu}{\sigma}\right)$

  • $\displaystyle \begin{aligned} P\left\{ 2X+Y < a \right\} \xlongequal[]{标准化} P\left\{ \frac{2X+Y+2}{\sqrt{10}} < \frac{a+2}{\sqrt{10}} \right\} = \Phi\left( \frac{a+2}{\sqrt{10}} \right), \end{aligned}$
  • $\displaystyle \begin{aligned} P\{X > Y\} = P\{Y - X < 0\} \xlongequal[]{标准化}P\left\{ \frac{Y-X+2}{2} < \frac{2}{2} \right\} =\Phi(1)\end{aligned}$

• 由$P\{2 X+Y<a\}=P\{X>Y\}$，得$\displaystyle \begin{aligned} \frac{a+2}{\sqrt{10}} = 1 \end{aligned}$, 得 $\displaystyle \begin{aligned} a = \sqrt{10} - 2 \end{aligned}$

• ★正态分布的标准化: 若 $\displaystyle X \sim N\left(\mu, \sigma^{2}\right)$, 则 $\displaystyle Z=\frac{X-\mu}{\sigma} \sim N(0,1)$. 于是有

  • $X≤x$
    • $\displaystyle F(x)=P\{X \leq x\}=P\left\{\frac{X-\mu}{\sigma} \leq \frac{x-\mu}{\sigma}\right\}=\Phi\left(\frac{x-\mu}{\sigma}\right)$
      • 举例:$X$ 分布函数 $\left.\Phi(\frac{x-2}{4}\right) \Rightarrow X \sim N (2,4^2)$
      • 举例2:$\displaystyle \int_{-\infty}^{+\infty} φ\left(\frac{x-2}{4}\right) d x\underbrace{=4 \int_{-\infty}^{+\infty} \frac{1}{4} φ\left(\frac{x-2}{4}\right) d x}_{F_x(x) \sim N\left(2,4^2\right)}=4$
  • $a<X \leq b$
    • $\displaystyle P\{a<X \leq b\}\xlongequal[X减\mu除\sigma]{不等式三端同时标准化}P\left\{\frac{a-\mu}{\sigma}<\frac{X-\mu}{\sigma} \leq \frac{b-\mu}{\sigma}\right\}=\Phi\left(\frac{b-\mu}{\sigma}\right)-\Phi\left(\frac{a-\mu}{\sigma}\right)$
  • $|X| \leqslant a$
    • $\displaystyle P\{|X| \leqslant a\}=\Phi\left(\frac{a-\mu}{\sigma}\right)-\Phi\left(\frac{-a-\mu}{\sigma}\right) ;$

(9)

设随机变量 $X$ 的概率密度为 $\displaystyle f(x)=\left\{\begin{array}{l}2(1-x), 0<x<1 \\ 0, \text { 其他 }\end{array}\right.$, 在 $X=x(0<x<1)$ 的条件下, 随机变量 $Y$ 服从区间 $(x, 1)$ 上的均匀分布, 则 $\operatorname{Cov}(X, Y)=(\quad)$ (A) $-\frac{1}{36}$ (B) $-\frac{1}{72}$ (C) $\frac{1}{72}$ (D) $\frac{1}{36}$

• 读题
  设随机变量 $\displaystyle \underbrace{X 的概率密度为 f(x)=\left\{\begin{array}{l}2(1-x), 0<x<1 \\ 0, \text { 其他 }\end{array}\right.}_{用来求E(X)和E(X^2)和方差D(X)}$, 在 $X=x(0<x<1)$ 的条件下, 随机变量 $Y$ 服从区间 $(x, 1)$ 上的均匀分布, 则 $\underbrace{\operatorname{Cov}(X, Y)=}_{{Cov}(X, Y) = E(XY) - E(X)E(Y)}(\quad)$

亚当夏娃

• $\text{Cov}(X, Y) = E(XY) - E(X)E(Y)$
  • $\displaystyle \begin{aligned}E(x)=\int_{-\infty}^{+\infty} x f(x) d x=\int_0^1 x \cdot 2(1-x) d x\end{aligned}$
    • $\displaystyle \begin{aligned}=\int_0^1\left(2 x-2 x^2\right) d x=\left.\left(x^2-\frac{2 x^3}{3}\right)\right|_0 ^1=\frac{1}{3}\end{aligned}$
  • $\displaystyle \begin{aligned} E\left(X^2\right) =\int_{-\infty}^{+\infty} x^2 f(x) d x=2 \int_0^1 x^2 \cdot(1-x) d x \end{aligned}$
    • $\displaystyle \begin{aligned} =2 \int_0^1\left(x^2-x^3\right) d x=\left.2\left(\frac{x^3}{3}-\frac{x^4}{4}\right)\right|_0 ^1 \end{aligned}$
    • $\displaystyle \begin{aligned} =2 \cdot\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{1}{6} \end{aligned}$
  • $\displaystyle \begin{aligned} D(X)= E\left(X^2\right)－[E(X)]^2= \frac{1}{6} - \frac{1}{9} = \frac{1}{18} \end{aligned}$
• 在 $X=x(0<x<1)$ 的条件下, 随机变量 $Y$ 服从区间 $(x, 1)$ 上的均匀分布
  $\displaystyle \begin{aligned} Y|X \sim U(X, 1) \end{aligned}$
  • $\displaystyle \begin{aligned} E(Y|X) = \frac{1 + X}{2} \end{aligned}$, $\displaystyle \begin{aligned} D(Y|X) = \frac{(1 - X)^2}{12} \end{aligned}$
    • $EX=\frac{a+b}{2}$，$D(X)=\frac{(b-a)^2}{12}$
• $\displaystyle \begin{aligned} \text{Cov}(X, Y) = E(XY) - E(X)E(Y) = \frac{1}{4} - \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{36} \end{aligned}$
  • $\displaystyle \begin{aligned} E(Y) = E(E(Y|X)) = E \left( \frac{1 + X}{2} \right) = E\left(\frac{1}{2}\right)+\frac{1}{2} E(X)=\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{3} = \frac{2}{3} \end{aligned}$
  • $\displaystyle \begin{aligned} E(XY) = E(E(XY|X)) = E(X·E(Y|X) ) =E(X·\frac{1 + X}{2})\end{aligned}$
    • $\displaystyle \begin{aligned}= E \left( \frac{X + X^2}{2} \right) = \frac{1}{2} E(X)+\frac{1}{2} E\left(X^2\right)=\frac{1}{2}· \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{4}\end{aligned}$

常规法

• 由题意可知
  • $\displaystyle \begin{aligned} f(x) = \begin{cases} 2(1-x), & 0 < x < 1, \\ 0, & \text{其他}, \end{cases} \end{aligned}$
  • $\displaystyle \begin{aligned} f_{Y|X}(y \mid x) = \begin{cases} \frac{1}{1-x}, & x < y < 1, \\ 0, & \text{其他}. \end{cases} \end{aligned}$
  • 于是 $\displaystyle \begin{aligned} f(x, y) = f(x) \cdot f_{Y|X}(y \mid x) = \begin{cases} 2, & 0 < x < y < 1, \\ 0, & \text{其他}. \end{cases} \end{aligned}$
• $\text{Cov}(X, Y) = E(XY) - E(X)E(Y)$
  • $\displaystyle \begin{aligned} EX = \int_0^1 2x(1-x) dx = 2 \times \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{3}, \end{aligned}$
  • $\displaystyle \begin{aligned} EY = \int_0^1 dx \int_x^1 2y dy = \int_0^1 (1 - x^2) dx = \frac{2}{3}. \end{aligned}$
  • $\displaystyle \begin{aligned} E(XY) = \int_{0}^{1} dx \int_{x}^{1} 2xy dy = \int_{0}^{1} x(1-x^2) dx = \frac{1}{4}, \end{aligned}$
• $\displaystyle \begin{aligned} \text{于是Cov}(X, Y) = E(XY) - EX \cdot EY==\frac{1}{4}-\frac{2}{9}=\frac{1}{36} . \end{aligned}$

(10)

设随机变量 $X, Y$ 相互独立，且均服从参数为 $\lambda$ 的指数分布，令 $Z=|X-Y|$ ，则下列随机变量中与 $Z$ 同分布的是 (A) $\mathrm{X}+\mathrm{Y}$ (B) $\frac{X+Y}{2}$ (C) $2 X$ (D) $X$

• X的概率密度 $\displaystyle \begin{aligned} f(x) = \begin{cases} \lambda e^{-\lambda x}, & x > 0 \\ 0, & x \leq 0 \end{cases} \end{aligned}$
• Y的概率密度$\displaystyle \begin{aligned} f(y) = \begin{cases} \lambda e^{-\lambda y}, & y > 0 \\ 0, & y \leq 0 \end{cases} \end{aligned}$
• $\displaystyle \begin{aligned} X \end{aligned}$ 与 $\displaystyle \begin{aligned} Y \end{aligned}$ 的联合概率密度为 $\displaystyle \begin{aligned} f(x, y) = f_X(x) \cdot f_Y(y) = \begin{cases} \lambda^2 e^{-\lambda(x+y)}, & x > 0, y > 0, \\ 0, & \text{其他}. \end{cases} \end{aligned}$
• 
  
• 设 $\displaystyle \begin{aligned} Z \end{aligned}$ 的分布函数为 $\displaystyle \begin{aligned} F_Z(z) \end{aligned}$，则 $\displaystyle \begin{aligned} F_Z(z) = P\{Z \leqslant z\} = P\{\mid X-Y \mid \leqslant z\} \end{aligned}$。
  • ① 当 $\displaystyle \begin{aligned} z < 0 \end{aligned}$ 时$\displaystyle \begin{aligned} \xrightarrow[因此概率为零]{绝对值不可能小于一个负数}F_Z(z) = 0 \end{aligned}$；
  • ② 当 $\displaystyle \begin{aligned} z \geqslant 0 \end{aligned}$ 时，$\displaystyle \begin{aligned} F_Z(z) = P\left\{|X-Y| \leq z\right\} \xlongequal[]{去绝对值}P\left\{-z \leq X-Y \leq z\right\} \end{aligned}$
    • $=P\{x-z \leq y \leq x+z\}$
      
    • $\displaystyle \begin{aligned} = \iint_D \lambda^2 e^{-\lambda(x+y)} dx dy \xlongequal[]{关于y=x对称} 2 \int_0^{+\infty} \lambda e^{-\lambda y} dy \int_y^{y+z} \lambda e^{-\lambda x} dx \end{aligned}$
      • $\displaystyle \xlongequal[]{e^{-\lambda x} \rightarrow-\frac{1}{λ} e^{-\lambda x}}-\left.2 \lambda \int_0^{+\infty} e^{-\lambda y} \cdot e^{-\lambda x}\right|_{x=y} ^{x=y+z} d y$
      • $\displaystyle \begin{aligned} & =-2 \lambda \int_0^{+\infty} e^{-\lambda y} \cdot\left(e^{-\lambda y-\lambda z}-e^{-\lambda y}\right) d y\xlongequal[]{提出e^{-\lambda y}}-2 \lambda \int_0^{+\infty} e^{-2 \lambda y}\left(e^{-\lambda z}-1\right) d y \end{aligned}$
      • $\displaystyle \begin{aligned} \xlongequal[]{提出z}-2 \lambda\left(e^{-\lambda z}-1\right) \int_0^{+\infty} e^{-2\lambda y} d y\xlongequal[]{积分-2λe^{-2\lambda y} }\left.\left(e^{-\lambda z}-1\right) \cdot e^{-2 \lambda y}\right|_0 ^{+\infty} \end{aligned}$
      • $\displaystyle \begin{aligned} =\left(e^{-\lambda z}-1\right) \cdot(-1)=1-e^{-\lambda z} . \end{aligned}$
      • 
        
• $\displaystyle F_z(z)=\left\{\begin{array}{l}1-e^{-\lambda z}, \quad z≥0 \\ 0,z≤0\end{array} \Rightarrow f_z(z)=\left\{\begin{array}{cc}\lambda e^{-\lambda z}, & z>0 \\ 0, & 其他\end{array}\right.\right.$
• 从而选D

(11)

已知 $\displaystyle \lim _{x \rightarrow 0} \frac{\left(1+a x^2\right)^{\sin x}-1}{x^3}=6$, 则 $a=$ $\qquad$

• $\displaystyle \begin{aligned} \lim _{x \rightarrow 0} \frac{\left(1+a x^2\right)^{\sin x}-1}{x^3} & \xlongequal[]{\left(1+a x^2\right)^{\sin x}=e^{\sin x \ln \left(1+a x^2\right)}}\lim _{x \rightarrow 0} \frac{e^{\sin x \ln \left(1+a x^2\right)}-1}{x^3} \end{aligned}$
• $\displaystyle \begin{aligned} \xlongequal[]{e^n-1 \sim u}\lim _{x \rightarrow 0} \frac{\sin x \ln \left(1+a x^2\right)}{x^3} \end{aligned}$
• $\displaystyle \begin{aligned} \xlongequal[\ln (1+n) \sim u]{\sin u \sim u}\lim _{x \rightarrow 0} \frac{x \cdot a x^2}{x^3}=\lim _{x \rightarrow 0} \frac{a x^3}{x^3} \end{aligned}\begin{aligned} =a . \end{aligned}$

(12)

已知 $f(u, v)$ 具有二阶连续偏导数, 且 $\left.d f\right|_{(1,1)}=3 d u+4 d v$, 若 $y=f\left(\cos x, 1+x^2\right)$, 则 $\left.\frac{d^2 y}{d x^2}\right|_{x=0}=$

• $\displaystyle \begin{aligned} \left.d f\right|_{(1,1)}=f_1^{\prime}(1,1) d u+f_2^{\prime}(1,1) d v=3 d u+4 d v \end{aligned}$
  • $f_1^{\prime}(1,1)=3, f_1^{\prime}(1,1)=4$
• $y=f\left(\cos x, 1+x^2\right)$
  • 为了求二阶导所以求一阶导：
    $\displaystyle \begin{aligned} \frac{d y}{d x}=f_1^{\prime}\left(\cos x, 1+x^2\right) \cdot(-\sin x)+f_2^{\prime}\left(\cos x, 1+x^2\right) \cdot 2 x \end{aligned}$
  • 向题中条件靠近：
    $\displaystyle \begin{aligned} \frac{d^2 y}{d x^2}=\underbrace{f_{11}^{\prime \prime}\left(\cos x, 1+x^2\right) \cdot(-\sin x)^2+f_{12}^{\prime \prime}\left(\cos x, 1+x^2\right) \cdot(-\sin x) \cdot 2 x}_{=0}\end{aligned}$
    $\displaystyle \begin{aligned} +f_1^{\prime}\left(\cos x, 1+x^2\right) \cdot(-\cos x) \end{aligned}$
    $\displaystyle \begin{aligned} +\underbrace{f_{21}^{\prime \prime}\left(cos x, 1+x^2\right) \cdot 2 x \cdot(-\sin x)+f_{22}^{\prime \prime}\left(\cos x, 1+x^2\right) \cdot(2 x)^2}_{=0}\end{aligned}$
    $\displaystyle \begin{aligned} +f_2^{\prime}\left(\cos x \cdot 1+x^2\right) \cdot 2 \end{aligned}$
    • 求导乘法法则：右边不动，左边求导;左边不动，右边求导
    • $\xrightarrow[]{将x=0代入}$得$\displaystyle \begin{aligned} \left.\frac{d y}{d x^2}\right|_{x=0}=f_1^{\prime}(1,1) \cdot(-1)+f_2^{\prime}(1,1) \cdot 2=-3+4 x_2=5 \end{aligned}$

(13)

已知 $f(x)=1+x$, 若 $\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos n x, x \in[0, \pi]$, 则 $\displaystyle \lim _{n \rightarrow \infty} n^2 \sin a_{2 n-1}=$ $\qquad$

• 正弦级数与余弦级数 ：设 $f(x)$ 是周期为 $2 l$ 的周期函数. 记 $C=\left\{x \left\lvert\, f(x)=\frac{1}{2}\left[f\left(x^{-}\right)+f\left(x^{+}\right)\right]\right.\right\}$
  • 当 $f(x)$ 为奇函数时, $f(x)$ 的傅里叶级数是只含正弦项的正弦级数
    • $\displaystyle f(x)=\sum_{n=1}^{\infty} b_n \sin \frac{n \pi x}{l} \quad(x \in C)$
    • 其中 $\displaystyle b_n=\frac{2}{l} \int_0^l f(x) \sin \frac{n \pi x}{l} \mathrm{~d} x \quad(n=1,2, \cdots)$.
  • 当 $f(x)$ 为偶函数时, $f(x)$ 的傅里叶级数是只含常数项和余弦项的余弦级数
    • $\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos \frac{n \pi x}{l} \quad(x \in C),$
    • 其中 $\displaystyle a_n=\frac{2}{l} \int_0^l f(x) \cos \frac{n \pi x}{l} \mathrm{~d} x \quad(n=0,1,2, \cdots)$.
• $\displaystyle \xrightarrow[]{将l=\pi}a_n=\frac{2}{\pi} \cdot \int_0^\pi(x+1) \cdot \frac{\cos n \pi x}{\pi} d x$
• $\displaystyle \begin{aligned} a_n & =\frac{2}{\pi} \int_0^\pi(x+1) \cos n x d x\xlongequal[]{凑微分}\frac{1}{4} \cdot \frac{2}{\pi} \int_0^\pi(x+1) d(\sin n x) \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{分布}\underbrace{\left.\frac{2}{n \pi} \sin n x\right|_0 ^\pi}_{0-0=0}+\frac{2}{n \pi} \int_0^\pi x d(\sin n x)\xlongequal[]{分布}\frac{2}{n \pi}\left(\underbrace{\left.x \sin n x\right|_0 ^\pi}_{0-0=0}-\int_0^\pi \sin n x d x\right) \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[-\sin n x \rightarrow \frac{1}{n} \cos n x]{牛莱}\left.\frac{2}{n^2 \pi} \cos n x\right|_0 ^\pi\xlongequal[]{\cos n \pi=(-1)^n}\frac{2}{n^2 \pi}\left[(-1)^n-1\right] \end{aligned}$
• $a_{2 n-1}=\frac{2}{(2 n-1)^2 \pi} \cdot(-2)=-\frac{4}{(2 n-1)^2 \pi}$
  • $\displaystyle \begin{aligned} \lim _{n \rightarrow \infty} n^2 \sin a_{2n-1} & =\lim _{n \rightarrow \infty} n^2 \cdot \sin \left(-\frac{4}{(2 n+1)^2 \pi}\right)\xlongequal[]{\sin n \sim n}-\lim _{n \rightarrow \infty} \frac{4 n^2}{(2 n-1)^2 \pi} \end{aligned}$
• $\displaystyle \begin{aligned} =-\lim _{n \rightarrow \infty} \frac{4 n^2}{\left(4 n^2-4 n+1\right) \pi}=-\frac{1}{\pi} . \end{aligned}$
• 相似题目
  • 1991 年数一试题
    • 将函数 $f(x)=2+|x|(-1 \leq x \leq 1)$ 展开成以 2 为周期的傅里叶级数, 并由此求级数 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ 的和.
  • 1993 年数一试题
    • 设函数 $f(x)=\pi x+x^2(-\pi<x<\pi)$ 的傅里叶级数展开式为 $\displaystyle \frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos n x+b_n \sin n x\right)$, 则其中系数 $b_3$ 的值为 .
  • 1995 年数一试题
    • 将函数 $f(x)=x-1(0 \leq x \leq 2)$ 展开成周期为 4 的余弦级数.
  • 2003 年数一试题
    • 设 $\displaystyle x^2=\sum_{n=0}^{\infty} a_n \cos n x(-\pi \leq x \leq \pi)$, 则 $a_2=$ $\qquad$ .
  • 2008 年数一试题
    • 将函数 $f(x)=1-x^2(0 \leq x \leq \pi)$ 展开成余弦级数, 并求 $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}$ 的和.
  • 2023 年数一试题
    • 设 $f(x)$ 是周期为 2 的周期函数，且 $f(x)=1-x, x \in[0,1]$. 若 $\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos n \pi x$, 则 $\displaystyle \sum_{n=1}^{\infty} a_{2 n}=$ $\qquad$
  • 狄利克雷定理

(14)

微分方程 $y^{\prime}=\frac{1}{(x+y)^2}$ 满足条件 $y(1)=0$ 的解为 $\qquad$ .

• $\displaystyle \begin{aligned} 令u=x+y, \quad \frac{d u}{d x}=1+\frac{d y}{d x} 则原结程y^{\prime}=\frac{1}{(x+y)^2}化为 \frac{d y}{d x}-1=\frac{1}{u^2} \text { 。 } \end{aligned}$
• $\displaystyle \xrightarrow[]{移项}\begin{aligned}& \frac{d u}{d x}=\frac{1+u^2}{u^2} \xrightarrow[]{分离变量} \frac{u^2}{1+u^2} d u=d x \end{aligned}$
• $\displaystyle \begin{aligned} \xrightarrow[]{作差}\left(1-\frac{1}{1+u^2}\right) d u=d x \end{aligned}$
• $\displaystyle \begin{aligned} \xRightarrow{积分} u-\arctan u=x+C \xrightarrow[]{x=1，y=0， u=1 } 1-\frac{\pi}{4}=1+C \Rightarrow C=-\frac{\pi}{4} \end{aligned}$
  • $\displaystyle \begin{aligned}\xrightarrow[u=x+y]{C=-\frac{\pi}{4}} x+y-\arctan (x+y)=x-\frac{\pi}{4} \Rightarrow y-\arctan (x+y)=-\frac{\pi}{4} .\end{aligned}$

(15)

已知矩阵 $\displaystyle A=\left(\begin{array}{cc}a+1 & a \\ a & a\end{array}\right)$, 对于任意的 $\alpha=\binom{x_1}{x_2}, \beta=\binom{y_1}{y_2}$, 都有 $\left(\alpha^T A \beta\right)^2 \leq \alpha^T A \alpha \cdot \beta^T A \beta$,则 $a$ 的取值范围是 $\qquad$

• $\displaystyle \begin{aligned} \alpha^T = (x_1, x_2), \quad \beta^T = (y_1, y_2) \end{aligned}$
• $\displaystyle \begin{aligned} (\alpha^T A \beta)^2 \leq \alpha^T A \alpha \cdot \beta^T A \beta \xrightarrow[]{分别计算三者的行列式}\end{aligned}$
  • $\displaystyle \begin{aligned} \alpha^T A \beta \xlongequal[]{代入题中给出的行列式} (x_1, x_2) \begin{pmatrix}a+1 & a \\a & a\end{pmatrix}\begin{pmatrix}y_1 \\y_2\end{pmatrix} \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{拆开，方便计算}(x_1, x_2) \left[\begin{pmatrix}a & a \\a & a\end{pmatrix}+\begin{pmatrix}1 & 0 \\0 & 0\end{pmatrix}\right]\begin{pmatrix}y_1 \\y_2\end{pmatrix} \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{A的行乘B的列} [a x_1 + a x_2, \, a x_1 + a x_2] \begin{pmatrix}y_1 \\y_2\end{pmatrix}+ (x_1, 0) \begin{pmatrix}y_1 \\y_2\end{pmatrix} \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{A的行×B的列}a(x_1 + x_2) \cdot (y_1 + y_2) + x_1 y_1 \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{对称}\alpha^T A \alpha = (x_1, x_2) \begin{pmatrix}a+1 & a \\a & a\end{pmatrix}\begin{pmatrix}x_1 \\x_2\end{pmatrix}= a(x_1 + x_2)(x_1 + x_2) + x_1 x_1 \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{对称}\beta^T A \beta = (y_1, y_2) \begin{pmatrix}a+1 & a \\a & a\end{pmatrix}\begin{pmatrix}y_1 \\y_2\end{pmatrix}= a(y_1 + y_2)(y_1 + y_2) + y_1 y_1 \end{aligned}$
• $\displaystyle \begin{aligned} (\alpha^T A \beta)^2 \leq \alpha^T A \alpha \cdot \beta^T A \beta \xrightarrow[]{分别计算等号左右两个部分}\end{aligned}$
  • $\displaystyle \begin{aligned} \alpha^T A \alpha \cdot \beta^T A \beta = \left[a(x_1 + x_2)^2 + x_1^2 \right]\left[a(y_1 + y_2)^2 + y_1^2 \right] \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{乘法分配} a(x_1 + x_2)^2 y_1^2 + a(x_1 + x_2)^2 \cdot a(y_1 + y_2)^2 + x_1^2 y_1^2 + x_1^2 \cdot a(y_1 + y_2)^2 \end{aligned}$
  • $\displaystyle \begin{aligned} (\alpha^T A \beta)^2 = \left[a(x_1 + x_2) \cdot (y_1 + y_2) + x_1 y_1 \right]^2 \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{完全平方}a^2 (x_1 + x_2)^2 (y_1 + y_2)^2 + x_1^2 y_1^2 + 2 a x_1 y_1 (x_1 + x_2)(y_1 + y_2) \end{aligned}$
• $\displaystyle \begin{aligned} (\alpha^T A \beta)^2 \leq \alpha^T A \alpha \cdot \beta^T A \beta \end{aligned}$
  • $\displaystyle \begin{aligned} \underline{a^2 (x_1 + x_2)^2 (y_1 + y_2)^2} + \underline{x_1^2 y_1^2 }+ 2 a x_1 y_1 (x_1 + x_2)(y_1 + y_2) \end{aligned}$
    $\displaystyle \begin{aligned} \leq a(x_1 + x_2)^2 y_1^2 + \underline{a(x_1 + x_2)^2 \cdot a(y_1 + y_2)^2} + \underline{x_1^2 y_1^2} + x_1^2 \cdot a(y_1 + y_2)^2 \end{aligned}$
  • $\displaystyle \begin{aligned} a(x_1 + x_2)^2 y_1^2 + a x_1^2 (y_1 + y_2)^2 - 2 a x_1 y_1 (x_1 + x_2)(y_1 + y_2) \geq 0 \end{aligned}$
    • $\displaystyle \begin{aligned} \xrightarrow[]{完全平方}a \underbrace{\left[(x_1 + x_2) y_1 - x_1 (y_1 + y_2)\right]^2}_{平方为正数}\geq 0 \end{aligned}$
    • $\displaystyle \begin{aligned} a \geq 0 \end{aligned}$

(16)

设随机试验每次成功的概率为 $p$, 现进行 3 次独立重复试验, 在至少成功 1 次的条件下, 3 次试验全部成功的概率为 $\frac{4}{13}$, 则 $p=$ $\qquad$ .

• $C_n^k=\frac{n!}{k!(n-k)!}$，比如：$C_5^3=\frac{5!}{3!\cdot 2!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \cdot(2 \times 1)}=10$
• 二项分布$\displaystyle \begin{aligned} & X \sim B(n, p) ， p\{x=k\}=C_n^k p^k(1-p)^{n-k}\end{aligned}$
• $\displaystyle \begin{aligned} P\{x=3 \mid x \geqslant 1\}=\frac{P\{x=3, x \geqslant 1\}}{P\{x \geqslant 1 y}=\frac{P\{x=3\}}{P\{x \geqslant 1 \}}=\frac{C_3^3 p^3}{1-P\{x=0\}} \end{aligned}$
  • $\displaystyle \begin{aligned} =\frac{p^3}{1-(1-p)^3}=\frac{4}{13} \end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[]{交叉相乘} 13 p^3=4-4(1-p)^3=4-4\left(1-3 p+3 p^2-p^3\right) \end{aligned}$
    • $\xRightarrow{移项} 9 p^3+12 p^2-12 p=0 \xRightarrow{提取p} p\left(3 p^2+4 p-4\right)=0$
      • $\displaystyle \begin{aligned} \xrightarrow[]{因式分解}p(p+2)(3 p-2)=0 \Rightarrow p=\frac{2}{3} . \end{aligned}$

(17)

(本题满分 10 分) 设 $D=\left\{(x, y) \mid \sqrt{1-y^2} \leq x \leq 1,-1 \leq y \leq 1\right\}$, 求 $\iint_D \frac{x}{\sqrt{x^2+y^2}} d x d y$.

• 
  

按极坐标算(2倍夜雨)

• $\displaystyle \begin{aligned} & \iint_D=2 \iint_{D_1}=2\left(\iint_{D_3}+\iint_{D_4}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} =2\left[\iint_{D_3} \cos \theta r d r d \theta+\iint_{D_4} \cos \theta r d r d \theta\right] \end{aligned}$
  • $\displaystyle \begin{aligned} =2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \int_1^{\frac{1}{\sin \theta}} \cos \theta \cdot r d r+2 \int_0^{\frac{\pi}{4}} d \theta \int_1^{\frac{1}{\cos \theta}} \cos \theta \cdot r d r \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{对r积分}\left.2 \cdot \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta \cdot \frac{r^2}{2}\right|_1 ^{\frac{1}{\sin \theta}} d \theta+\left.2 \cdot \int_0^{\frac{\pi}{4}} \cos \theta \cdot \frac{r^2}{2}\right|_1 ^{\frac{1}{\cos \theta}} d \theta \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{牛莱}\underbrace{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta \cdot\left(\frac{1}{\sin ^2 \theta}-1\right) d \theta}_{\cos \theta d \theta=d(\sin \theta)}+\int_0^{\frac{\pi}{4}} \underbrace{\left(\frac{1}{\cos ^2 \theta}-1\right) \cos \theta}_{\sec \theta-\cos \theta} d \theta \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{对\theta积分}\left.\left(-\frac{1}{\sin \theta}-\sin \theta\right)\right|_{\frac{\pi}{4}} ^{\frac{\pi}{2}}+\left.(\ln |\sec \theta+\tan \theta|-\sin \theta)\right|_0 ^{\frac{\pi}{4}} \end{aligned}$
    • $\displaystyle \begin{aligned} =\left[(-1-1)-\left(-\sqrt{2}-\frac{\sqrt{2}}{2}\right)+\left[\ln (\sqrt{2}+1)-\frac{\sqrt{2}}{2}\right]-[\ln (1+0)-0]\right. \end{aligned}$
    • $\displaystyle \begin{aligned} =-2+\sqrt{2}+\frac{\sqrt{2}}{2}+\ln (\sqrt{2}+1)-\frac{\sqrt{2}}{2}-0 \end{aligned}$
    • $\displaystyle \begin{aligned} =\ln (\sqrt{2}+1)-2+\sqrt{2} \end{aligned}$

按极坐标算(2倍李艳芳)

• D 关于 x 轴对称，$D_1$关于$y=x$ 轴对称
  
• $\displaystyle \begin{aligned}\iint_D \frac{x}{\sqrt{x^2+y^2}} dxdy =\begin{aligned}\iint_{D_0} \frac{x}{\sqrt{x^2+y^2}} d x d y+\underbrace{\iint_{D_1|D_n} \frac{x}{\sqrt{x^2+y^2}} d x d y}_{\iint_{D_0} \frac{y}{\sqrt{x^2+y^2}} d x d y}=\iint_{D_0} \frac{x+y}{\sqrt{x^2+y^2}} d x d y\end{aligned}\end{aligned}$
• $\displaystyle \begin{aligned} \iint_{D_0} \frac{x+y}{\sqrt{x^2+y^2}} dxdy = \iint_{D_0} \frac{r(\cos\theta+\sin\theta)}{r} \cdot r drd\theta \xlongequal[]{消去r}\iint_{D_0} (\cos\theta+\sin\theta) \cdot r drd\theta \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{填入积分区域}\int_0^{\frac{\pi}{4}} (\cos\theta+\sin\theta) d\theta\underbrace{\int_1^{\sec\theta} r dr}_{\frac{\sec^2\theta-1}{2}} \xlongequal[]{\int r d r=\frac{r^2}{2}} \int_0^{\frac{\pi}{4}} (\cos\theta+\sin\theta) \cdot \frac{\sec^2\theta-1}{2} d\theta \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{提出系数} \frac{1}{2} \int_0^{\frac{\pi}{4}} (\cos\theta+\sin\theta)(\sec^2\theta-1) d\theta \end{aligned}$
  • $\displaystyle \begin{aligned}\xlongequal[]{去括号} \frac{1}{2} \int_0^{\frac{\pi}{4}} (\sec\theta-\cos\theta+\underbrace{\frac{\sin\theta}{\cos^2\theta}}_{-\sin \theta=(\cos \theta)^{\prime}}-\sin\theta) d\theta\end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\int \sec u d u=\ln |\operatorname{sec} u+\tan u|+C}\frac{1}{2} \left[ \ln(\sec\theta+\tan\theta) - \sin\theta + \frac{1}{\cos\theta} + \cos\theta \right]_0^{\frac{\pi}{4}} \end{aligned}$
  • $\displaystyle \begin{aligned} = \frac{1}{2} \left[ \ln(\sqrt{2}+1) + \sqrt{2} - 2 \right] \end{aligned}$

按直角坐标算(2倍)

• D 关于 x 轴对称，$\displaystyle \begin{aligned} \frac{x}{\sqrt{x^2+y^2}} \end{aligned}$ 关于 y 是偶函数。
  • 记 $\displaystyle \begin{aligned} D_1 \end{aligned}$ 为 D 位于 x 轴上方的部分，则
    $\displaystyle \begin{aligned} \iint_D \frac{x}{\sqrt{x^2+y^2}} dxdy = 2 \iint_{D_1} \frac{x}{\sqrt{x^2+y^2}} dxdy \end{aligned}$
• 记 $\displaystyle \begin{aligned} D_1 = \{(x, y) \mid \sqrt{1-y^2} \leq x \leq 1, 0 \leq y \leq 1\} \end{aligned}$，则
• $\displaystyle \begin{aligned} \iint_D \frac{x}{\sqrt{x^2 + y^2}} dxdy \xlongequal[]{对称性}2 \iint_{D_1} \frac{x}{\sqrt{x^2 + y^2}} dxdy \xlongequal[]{代入上下限}2 \int_0^1 dy \int_{\sqrt{1-y^2}}^1 \frac{x}{\sqrt{x^2 + y^2}} dx \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{统一}\int_0^1 dy \int_{\sqrt{1-y^2}}^1 \frac{1}{\sqrt{x^2 + y^2}} d(x^2 + y^2) = \int_0^1 2\sqrt{x^2 + y^2} \Big|_{x=\sqrt{1-y^2}}^{x=1} dy \end{aligned}$
    • 或$\displaystyle \begin{aligned} \xrightarrow[]{\frac{\partial\left(\sqrt{x^2+y^2}\right)}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}} \left.\int_0^1 \sqrt{x^2+y^2}\right|_{x=\sqrt{1-y^2}} ^{x=1} d y\end{aligned}$
  • $\displaystyle \begin{aligned} = 2 \int_0^1 (\sqrt{1+y^2} - 1) dy = 2 \int_0^1 \sqrt{1+y^2} dy - 2. \end{aligned}$
    • 方法1，$\displaystyle \begin{aligned} \int_{0}^{1} \sqrt{1+y^{2}} dy \xlongequal[]{分部积分} y\sqrt{1+y^{2}} \Big|_0^1 - \int_0^1 y \cdot \frac{y}{\sqrt{1+y^2}} dy \end{aligned}$
      • $\displaystyle \begin{aligned} \xlongequal[]{+1-1}\sqrt{2} - \int_0^1 \frac{y^2+1-1}{\sqrt{1+y^2}} dy \end{aligned}$
      • $\displaystyle \begin{aligned} \xlongequal[]{拆开变成两个} \sqrt{2} - \int_0^1 \sqrt{1+y^2} dy + \int_0^1 \frac{1}{\sqrt{1+y^2}} dy \end{aligned}$
      • $\displaystyle \begin{aligned} \xlongequal[]{绕回来了} \sqrt{2} - \int_0^1 \sqrt{1+y^2} dy + \ln\left(y+\sqrt{1+y^2}\right) \Big|_0^1. \end{aligned}$
      • 故 $\displaystyle \begin{aligned}\xrightarrow[]{相加除以二} \int_0^1 \sqrt{1+y^2} dy = \frac{\sqrt{2} + \ln(1+\sqrt{2})}{2}. \end{aligned}$
• 求$\displaystyle \int_0^1 \sqrt{1+y^2} d y-1$
  • $\displaystyle \begin{aligned} \xlongequal[1+\tan ^2 \alpha=\sec ^2 \alpha ]{y=\tan u，(\tan x)' = \sec^2 x，\tan \frac{\pi}{4}=1 } \int_0^{\frac{\pi}{4}} \sec u \cdot \sec ^2 u d u-1\end{aligned}$
  • $\displaystyle =\int_0^{\frac{\pi}{4}} \sec ^3 u d u-1$
    • $\displaystyle \begin{aligned} \int_0^{\frac{\pi}{4}} \sec^3 u \, du = \int_0^{\frac{\pi}{4}} \sec u \cdot d(\tan u) = \sec u \cdot \tan u \Big|_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} \tan u \cdot \sec u \cdot \tan u \, du \end{aligned}$
      • $\displaystyle \begin{aligned} = \sqrt{2} - \int_0^{\frac{\pi}{4}} \tan^2 u \sec u \, du \xlongequal[]{\tan ^2 u+1=\sec ^2 u} \sqrt{2} - \int_0^{\frac{\pi}{4}} (\sec^2 u - 1) \sec u \, du \end{aligned}$
      • $\displaystyle \begin{aligned} = \sqrt{2} - \int_0^{\frac{\pi}{4}} \sec^3 u \, du + \int_0^{\frac{\pi}{4}} \sec u \, du \xrightarrow[]{\int \sec u d u=\ln |\operatorname{sec} u+\tan u|+C}\end{aligned}$
    • $\displaystyle \begin{aligned}2 \int_0^{\frac{\pi}{4}} \sec ^3 u d u = \sqrt{2} + \ln|\sec u + \tan u| \Big|_0^{\frac{\pi}{4}} = \sqrt{2} + \ln(\sqrt{2} + 1) \end{aligned}$
  • $=\frac{1}{2}[\ln (\sqrt{2}+1)+\sqrt{2}]-1$

(18)

(本题满分 12 分) 设 $f(x, y)=x^3+y^3-(x+y)^2+3$, 已知 $z=f(x, y)$ 在 $(1,1,1)$ 处的切平面方程为 $T, T$ 与坐标平面所围成的有界区域在 $x o y$ 面上的投影为 $D$ 。
(1)求 $T$ 的方程。
(2)求 $f(x, y)$ 在 $D$ 上的最大值和最小值。
求二元函数在有界闭区域上的最值

第二问(李艳芳版)

• 图示
  解题过程：

• (II) 记 $\displaystyle \begin{aligned} F(x, y) = z - f(x, y) \end{aligned}$ 由题意 $\displaystyle \begin{aligned} z = f(x, y) \end{aligned}$ 即 $\displaystyle \begin{aligned} F(x, y) = 0 \end{aligned}$。

  • $\displaystyle \begin{aligned} F_x(x_0, y_0) = -f_x(x, y)= -3x^2 + 2(xy)\end{aligned}$
  • $\displaystyle \begin{aligned} F_y(x_0, y_0) = -f_y(x, y)= -3y^2 + 2(xy) \end{aligned}$
  • $\displaystyle \begin{aligned} F_z(x_0, y_0) = 1 \end{aligned}$

• 代入 $\displaystyle \begin{aligned} x = 1, y = 1，z=1 \end{aligned}$ 得 $\displaystyle \begin{aligned} L \end{aligned}$ 处的法向量为 $\displaystyle \begin{aligned} (1, 1, 1) \end{aligned}$。

• T的点法式方程：

  • $\displaystyle \begin{aligned} F(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0 \end{aligned}$，
  • 即 $\displaystyle \begin{aligned} 1 \cdot (x - 1) + 1 \cdot (y - 1) + 1 \cdot (z - 7) = 0 \end{aligned}$
    • $\displaystyle \begin{aligned} \Rightarrow x + y + z = 3 \end{aligned}$

• 图示
  

• (i) 先求 $\displaystyle \begin{aligned} D \end{aligned}$ 内部的驻点。

  • $\displaystyle \begin{aligned}\left\{\begin{array}{l} f_x(x, y) = 3x^2 - 2(xy) = 0 \\ f_y(x, y) = 3y^2 - 2(xy) = 0 \end{array}\right. \end{aligned}$，得$\displaystyle \begin{aligned} \Rightarrow 3x^2 - 3y^2 = 0 ，从而(x + y)(x - y) = 0 \end{aligned}$
    • 若 $\displaystyle \begin{aligned} x = -y \end{aligned}$，则 $\displaystyle \begin{aligned} (x, y) \end{aligned}$ 不在 $\displaystyle \begin{aligned} D \end{aligned}$ 内部
    • 故 $\displaystyle \begin{aligned} x =y \end{aligned}$，代入 $\displaystyle \begin{aligned} 3x^2 - 2(x + y) = 0 \end{aligned}$ 得 $\displaystyle \begin{aligned} 3x^2 - 4x = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \Rightarrow x = 0 \end{aligned}$ 或 $\displaystyle \begin{aligned} x = \frac{4}{3} \end{aligned}$
      • $\displaystyle \begin{aligned} \Rightarrow \end{aligned}$ 区域内部的一驻点为 $\displaystyle \begin{aligned} \left(\frac{4}{3}, \frac{4}{3}\right) \end{aligned}$。
    • $f\left(\frac{4}{3}, \frac{4}{3}\right)=\left(\frac{4}{3}\right)^3+\left(\frac{4}{8}\right)^3-\left(\frac{8}{3}\right)^2+3=\frac{128}{27}-\frac{64}{9}+3=3-\frac{64}{27}=\frac{17}{27}$

• 计算三个边界的值
  

  • $\displaystyle \begin{aligned} 当y = 0 \end{aligned}$，$\displaystyle \begin{aligned} 0 \leq x \leq 3 \end{aligned}$ 时， $\displaystyle \begin{aligned} f(x, 0) = x^3 - x^2 + 3 = \varphi(x) \end{aligned}$
    • $\displaystyle \begin{aligned} \varphi'(x) = 3x^2 - 2x = x(3x - 2) \end{aligned}$
      • $\displaystyle \begin{aligned} x = \frac{2}{3} \end{aligned}$ 是 $\displaystyle \begin{aligned} \varphi(x) \end{aligned}$ 在 $\displaystyle \begin{aligned} (0, 3) \end{aligned}$ 内的驻点。
      • $\displaystyle \begin{aligned} \varphi\left(\frac{2}{3}\right) = \left(\frac{2}{3}\right)^3 - \left(\frac{2}{3}\right)^2 + 3 = \frac{8}{27} - \frac{12}{27} + 3 = \frac{77}{27} \end{aligned}$
      • $\varphi(0)=3$
      • $\varphi(3)=21$
  • $x=0,0 \leqslant y \leqslant 3$，最值情况与上述类似，是对称的
  • 当$\displaystyle \begin{aligned} & x+y=3,0 \leq x \leq 3, \end{aligned}$
    • $\displaystyle \begin{aligned} f(x, y)=x^3+(3-x)^3-9+3=x^3+(3-x)^3-6=h(x) \end{aligned}$
      • $\displaystyle \begin{aligned} h^{\prime}(x)=3 x^2-3(3-x)^2 \xrightarrow[]{-3 \cdot\left(9-6 x+x^2)\right.} 18 x-27=9(2 x-3) \end{aligned}$
        • $\displaystyle \begin{aligned} x=\frac{3}{2} \text { 是 } h(x) \text { 在 }(0,3) \text { 内的驻点 } \end{aligned}$
      • $\displaystyle \begin{aligned} h\left(\frac{3}{2}\right)=\left(\frac{3}{2}\right)^3+\left(3-\frac{3}{2}\right)^3-6=\frac{27}{8}+\frac{27}{8}-6=\frac{3}{4} \end{aligned}$
    • 两个端点 x=0，x=3
      • $h(0)=3$
      • $h(3)=27-6=21$

• 比较可得

  • $f(3,0)=f(0,3)=21$
  • $f\left(\frac{4}{3}, \frac{4}{3}\right)=\frac{17}{27}$ 是 $f(x, y)$ 是区域 D上的最小值

高昆仑版

• 因为 $\displaystyle \begin{aligned} f(x, y) = x^3 + y^3 - (x + y)^2 + 3 \end{aligned}$，所以

  • $\displaystyle \begin{aligned} \frac{\partial f}{\partial x} = 3x^2 - 2(x + y), \quad \frac{\partial f}{\partial y} = 3y^2 - 2(x + y) \end{aligned}$
    • $\displaystyle \begin{aligned} \frac{\partial f(1, 1)}{\partial x} = -1, \quad \frac{\partial f(1, 1)}{\partial y} = -1 \end{aligned}$
      • 法向量 $\displaystyle \begin{aligned} \vec{n} = (-1, -1, -1) \end{aligned}$。
  • 切平面 $\displaystyle \begin{aligned} T \end{aligned}$ 的方程为 $\displaystyle \begin{aligned} -1 \cdot (x - 1) - 1 \cdot (y - 1) - 1 \cdot (z - 1) = 0 \end{aligned}$。
    • 即 $\displaystyle \begin{aligned} x + y + z = 3 \end{aligned}$。
    • 图示
      

• (2) 由 (1) 知，切平面 $\displaystyle \begin{aligned} T \end{aligned}$ 与坐标面所围有界区域在 $\displaystyle \begin{aligned} xOy \end{aligned}$ 平面上的投影
  $\displaystyle \begin{aligned} D = \{(x, y) \mid x \geq 0, y \geq 0, x + y \leq 3\} \end{aligned}$

• 令$\displaystyle \begin{aligned} \begin{cases}\frac{\partial f}{\partial x} = 0, \\\frac{\partial f}{\partial y} = 0,\end{cases} \end{aligned}$即$\displaystyle \begin{aligned} \begin{cases}3x^2 - 2(x + y) = 0, \\3y^2 - 2(x + y) = 0,\end{cases} \end{aligned}$

  • 解得 $\displaystyle \begin{aligned} D \end{aligned}$ 内部的可疑点$\displaystyle \begin{aligned} \begin{cases}x = \frac{4}{3} \\y = \frac{4}{3}\end{cases} \end{aligned}$

• 求边界上的点

  • 当y=0时
    • 当 $\displaystyle \begin{aligned} (x, y) \in \{(x, y) \mid 0 \leq x \leq 3, y = 0\} \end{aligned}$ 时，$\displaystyle \begin{aligned} f(x, 0) = x^3 - x^2 + 3 \end{aligned}$, $\displaystyle \begin{aligned} x \in [0, 3] \end{aligned}$。
    • 令$\displaystyle \begin{aligned} \frac{d}{dx}[f(x, 0)] = 3x^2 - 2x = 0, \end{aligned}$
      • 得驻点 $\displaystyle \begin{aligned} x = \frac{2}{3} \end{aligned}$。另有可疑点（端点）$\displaystyle \begin{aligned} x = 0 \end{aligned}$ 及 $\displaystyle \begin{aligned} x = 3 \end{aligned}$。
  • 当y=0时
    • 当 $\displaystyle \begin{aligned} (x, y) \in \{(x, y) \mid 0 \leq y \leq 3, x = 0\} \end{aligned}$ 时，由对称性，直接得驻点 $\displaystyle \begin{aligned} y = \frac{2}{3} \end{aligned}$。另有可疑点（端点）$\displaystyle \begin{aligned} y = 0 \end{aligned}$ 及 $\displaystyle \begin{aligned} y = 3 \end{aligned}$。
  • $x + y = 3$
    • 当 $\displaystyle \begin{aligned} (x, y) \in \{(x, y) \mid 0 \leq x \leq 3, x + y = 3\} \end{aligned}$ 时，
    • $\displaystyle \begin{aligned} f(x, 3-x) = x^3 + (3-x)^3 - 6, \quad x \in [0, 3] \end{aligned}$
      • 令 $\frac{\mathrm{d}}{\mathrm{d} x}[f(x, 3-x)]=3 x^2-3(3-x)^2=0$, 得 $x=\frac{3}{2}$.

• 综上，$\displaystyle \begin{aligned} f(x, y) \end{aligned}$ 在 $\displaystyle \begin{aligned} D \end{aligned}$ 上的可能最值点是 $\displaystyle \begin{aligned} (0,0), \left(\frac{4}{3}, \frac{4}{3}\right), \left(\frac{2}{3}, 0\right), \left(0, \frac{2}{3}\right), (3,0), (0,3), \left(\frac{3}{2}, \frac{3}{2}\right) \end{aligned}$。

  • $\displaystyle \begin{aligned} f(0,0) = 3\end{aligned}$
  • $\displaystyle \begin{aligned}f\left(\frac{4}{3}, \frac{4}{3}\right) = \frac{17}{27}\end{aligned}$
  • $\displaystyle \begin{aligned}f\left(\frac{2}{3}, 0\right) = f\left(0, \frac{2}{3}\right) = \frac{77}{27}\end{aligned}$
  • $\displaystyle \begin{aligned} f(3,0) = f(0,3) = 21\end{aligned}$
  • $\displaystyle \begin{aligned}f\left(\frac{3}{2}, \frac{3}{2}\right) = \frac{3}{4}\end{aligned}$

• 所以 $\displaystyle \begin{aligned} f(x, y) \end{aligned}$ 在 $\displaystyle \begin{aligned} D \end{aligned}$ 上的最大值为 21，最小值为 $\displaystyle \begin{aligned} \frac{17}{27} \end{aligned}$。

(19)

(本题满分 12 分)

设函数 $f(x)$ 具有 2 阶导数，且 $f^{\prime}(0)=f^{\prime}(1),\left|f^{\prime \prime}(x)\right| \leq 1$ 。证明
(1)当 $x \in(0,1)$ 时, $\underbrace{|f(x)-f(0)(1-x)-f(1) x|}_{\varphi(x)}\leq \underbrace{\frac{x(1-x)}{2}}_{一眼看出积分是\frac{1}{12}}$;
(2) $\displaystyle \left|\int_0^1 f(x) d x-\frac{f(0)+f(1)}{2}\right| \leq \frac{1}{12}$

• $\displaystyle \begin{aligned} \left|\varphi(x)\right| \leq \frac{x(1-x)}{2} \end{aligned}$

• $\displaystyle \begin{aligned} &\int_0^1 \varphi(x) \, dx = \int_0^1 f(x) \, dx - f(0) \underbrace{\int_0^1 (1-x)}_{\frac{1}2}\, dx - f(1) \underbrace{\int_0^1 x \, dx}_{\frac{1}2}\end{aligned}$

  • $\displaystyle \begin{aligned} = \int_0^1 f(x) \, dx - \frac{f(0) + f(1)}{2}\end{aligned}$对应于第二问

• 分别写出 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 在 $\displaystyle \begin{aligned} x = 0 \end{aligned}$ 与 $\displaystyle \begin{aligned} x = 1 \end{aligned}$ 处的一阶泰勒公式

  • $\displaystyle \begin{aligned} f(x) = f(0) + f'(0)x + \frac{f''(\xi_1)}{2}x^2, \quad \xi_1 \in (0, x) \xrightarrow[]{×2}\end{aligned}$
  • $\displaystyle \begin{aligned} f(x) = f(1) + f'(1)(x-1) + \frac{f''(\xi_2)}{2}(x-1)^2, \quad \xi_2 \in (x, 1) \xrightarrow[]{×(x-1)}\end{aligned}$

• (2) 式 $\times 2$ - (1) 式 $×(x-1)$ ，结合 $f^{\prime}(0)=f^{\prime}(1)$ 可得

  • $\displaystyle \begin{aligned} f(x) & =\underbrace{x f(1)-(x-1) f(0)}_{f(0)和f(1)}+\frac{f^{\prime \prime}\left(\xi_2\right)}{2}(x-1)^2 \cdot x-\frac{f^{\prime \prime}\left(\xi_1\right)}{2} \cdot x^2(x-1) \end{aligned}$
    • $\displaystyle \begin{aligned} =x f(1)+(1-x) f(0)+\underbrace{\frac{x(x-1)}{2}}_{提出公因式}\left[f^{\prime \prime}\left(\xi_2\right) \cdot(x-1)-f^{\prime \prime}\left(\xi_1\right) x\right] \end{aligned}$
  • 移项并整理，当 $x \in(0,1)$ 时
    $|f(x)-f(0)(1-x)-f(1) x|=\left|\frac{x(x-1)}{2}\left[f^{\prime \prime}\left(\xi_2\right)(x-1)-f^{\prime \prime}\left(\xi_1\right) x\right]\right|$
    • $\xlongequal[]{将x从绝对值中拿出}\frac{x(1-x)}{2}\left|f^{\prime \prime}\left(\xi_2\right)(x-1)-f^{\prime \prime}\left(\xi_1\right) x\right|$
    • $\displaystyle \begin{aligned} \xrightarrow[|a \pm b| \leqslant|a|+|b|]{放缩} \leqslant \frac{x(1-x)}{2}\left[\left|f^{\prime \prime}\left(\xi_2\right)(x-1)\right|+\left|f^{\prime \prime}\left(\xi_1\right) x\right|\right] \end{aligned}$
    • $\displaystyle \begin{aligned} =\frac{x(1-x)}{2}\underbrace{\left[\left|f^{\prime \prime}\left(\xi_2\right)\right| \cdot(1-x)+\left|f^{\prime \prime}\left(\xi_1\right)\right| x\right]}_{继续从绝对值提出x}\end{aligned}$
    • $\displaystyle \begin{aligned} \xrightarrow[]{\left|f^{\prime \prime}(x)\right| \leqslant 1}\leq \frac{x(1-x)}{2}[(1-x)+x]\xlongequal[]{(1-x)+x=1}\frac{x(1-x)}{2} \end{aligned}$

• 第二问也可以拿分

• 记$\displaystyle \begin{aligned} \varphi(x)=f(x)-f(0)(1-x)-f(1) x,\end{aligned}$由第一问可知:$|\varphi(x)| \leq \frac{x(1-x)}{2}$

  • $\displaystyle \begin{aligned} \left|\int_0^1 \varphi(x) d x\right|\xlongequal[]{三项分别积分}\int_0^1 f(x) d x-f(0) \int_0^1(1-x) d x-f(1) \int_0^1 x d x \end{aligned}$
    • 
      
  • $\displaystyle \begin{aligned} =\left|\int_0^1 f(x) d x-\frac{f(0)+f(1)}{2}\right| \end{aligned}$

• 由

  • 积分的绝对值小于等于绝对值的积分$\displaystyle \begin{aligned} \left|\int_0^1 \varphi(x) d x\right| \leq \int_0^1 \mid \varphi(x)\mid d x \end{aligned}$
  • 以及 $\displaystyle \begin{aligned}\left| \varphi(x) \right|\leq \frac{x(1-x)}{2} \text {, 得 } \int_0^1|\varphi(x)| d x \leq \int_0^1 \frac{x(1-x)}{2} d x\end{aligned}$

• $\displaystyle \begin{aligned}\left|\int_0^1 \varphi(x) d x\right| = \left|\int_0^1 f(x) d x-\frac{f(0)+f(1)}{2}\right| \leq \int_0^1 | \varphi(x)| d x \leq \int_0^1 \frac{x(1-x)}{2} d x \end{aligned}$

  • $\displaystyle \begin{aligned} =\frac{1}{2} \int_0^1\left(x-x^2\right) d x \end{aligned}$
  • $\displaystyle \begin{aligned} =\left.\frac{1}{2}\left(\frac{x^2}{2}-\frac{x^3}{3}\right)\right|_0 ^1=\frac{1}{12} \end{aligned}$

(20)

(本题满分 12 分) 已知有向曲线 $L$ 为球面 $x^2+y^2+z^2=2 x$ 与平面 $2 x-z-1=0$ 的交线, 从 $z$轴正向往 $z$ 轴负向看为逆时针方向, 计算曲线积分 $\oint_L\left(6 x y z-y z^2\right) d x+2 x^2 z d y+x y z d z$

夜雨版

• $\displaystyle \begin{aligned} &\text { 取上侧方向，那么 } \int_{\Gamma}\left(6 x y z-y z^2\right) \mathrm{d} x+2 x^2 z \mathrm{~d} y+x y z \mathrm{~d} z \end{aligned}$

  • $\displaystyle \begin{aligned} & =\iint_{\Sigma}\left|\begin{array}{ccc}\mathrm{d} y \mathrm{~d} z & \mathrm{~d} z \mathrm{~d} x & \mathrm{~d} x \mathrm{~d} y \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\6 x y z-y z^2 & 2 x^2 z & x y z\end{array}\right| \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{按第一行展开}\iint_{\Sigma}\left(x z-2 x^2\right) \mathrm{d} y \mathrm{~d} z+(6 x y-3 y z) \mathrm{d} z \mathrm{~d} x+\left(z^2-2 x z\right) \mathrm{d} x \mathrm{~d} y \end{aligned}$
    • 斯托克斯把曲线转换为平面，用高斯直接等于0
    • 高斯只适用于曲面，不适用平面
    • 选择将面投影到xOy
  • $\displaystyle \xlongequal[\iint_{\Sigma}P \cdot\left(-z_x\right)+Q \cdot\left(-z_y\right)+R \cdot 1d x d y]{z=2x-1, \quad z_x'=2, \quad z_y'=0}\iint_{\sum}\left(xz-2x^3\cdot(-2)+(6xy-3yz)\cdot(0)+\left(z^2-2xz\right)\cdot1\right) dxdy$
  • $\displaystyle \xlongequal[]{整理}\iint_{\sum}\left(4x^2-2xz\right)+\left(z^2-2xz\right) dxdy$
  • $\displaystyle \xlongequal[]{整理}\iint_{\sum}(2x-z)^2 dxdy$
  • $\displaystyle \xlongequal[]{2x+z=1}\iint_{\sum} 1^2 dxdy$

• $\displaystyle \begin{aligned} 将 z = 2x - 1 代入 x^2 + y^2 + z^2 = 2x \end{aligned}$，得$\displaystyle \begin{aligned} x^2 + y^2 +\underbrace{(2x - 1)^2}_{展开}- 2x = 0, \end{aligned}$

  • $\displaystyle \begin{aligned} 5x^2 - 6x + y^2 + 1 = 0. \end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[椭圆标准形式]{配方}5\left(x^2 - \frac{6}{5}x\right) + y^2 + 1 = 0\end{aligned}$
    • $\displaystyle \begin{aligned} \underbrace{5\left(x - \frac{3}{5}\right)^2 + y^2 = \frac{4}{5}}_{x^2-\frac{6}{5} x+\frac{9}{5}和椭圆} \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[\frac{\left(x-x_0\right)^2}{a^2}+\frac{\left(y-y_0\right)^2}{b^2}=1]{同乘\frac{5}{4}}\frac{(x - \frac{3}{5})^2}{\frac{4}{25}} + \frac{y^2}{\frac{4}{5}} = 1 \end{aligned}$
    • 行如:椭圆面积：$\displaystyle \begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{aligned}$、面积 $\displaystyle \begin{aligned} \pi ab \end{aligned}$
    • $\displaystyle \begin{aligned} D \end{aligned}$ 是中心 $\displaystyle \begin{aligned} \left(\frac{3}{5}, 0\right) \end{aligned}$、长半轴长 $\displaystyle \begin{aligned} \sqrt{\frac{4}{5}} \end{aligned}$，短半轴长 $\displaystyle \begin{aligned} \sqrt{\frac{4}{25}} \end{aligned}$
    • 则$\displaystyle \begin{aligned} D \end{aligned}$ 的面积为 $\displaystyle \begin{aligned} \pi \sqrt{\frac{4}{25}} \sqrt{\frac{4}{5}} = \frac{4\pi}{5\sqrt{5}} = \frac{4\sqrt{5}}{25}\pi \end{aligned}$

• 第二类曲面积分的计算方法三：合一投影法（这个方法用起来和上一个方法差不多）

  • 若积分曲面 $\displaystyle \Sigma$ 由方程 $z=z(x, y)$ 给出， $\displaystyle \Sigma$ 在 $x O y$ 坐标面的投影为 $D_{x y}$, 函数 $z(x, y)$ 在 $D_{x y}$ 上有一阶连续偏导数, $f(x, y, z)$ 在 $\displaystyle \Sigma$ 上连续,
  • 则有 $\displaystyle \iint_{\Sigma} P d y d z+Q d x d z+R d x d y=\iint_{\Sigma}\left[P \cdot\left(-z_x\right)+Q \cdot\left(-z_y\right)+R \cdot 1\right] d x d y$

高昆仑版

• 
  
• 记曲线 $\displaystyle \begin{aligned} L \end{aligned}$ 在平面 $\displaystyle \begin{aligned} 2x - z - 1 = 0 \end{aligned}$ 上所围部分为 $\displaystyle \begin{aligned} \Sigma \end{aligned}$。
  根据右手法则，取上侧。于是平面 $\displaystyle \begin{aligned} \Sigma \end{aligned}$ 的法向量 $\displaystyle \begin{aligned} \vec{n} = -\{2, 0, -1\} = \{-2, 0, 1\} \end{aligned}$。
• 由斯托克斯公式:空间线转空间面
  $\displaystyle I=\iint_{\Sigma}\left|\begin{array}{ccc}\cos \alpha & \cos \beta & \cos \gamma \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R\end{array}\right| d S\xlongequal[]{\cos \alpha = \frac{-2}{\sqrt{5}}, \cos \beta = 0, \cos \gamma = \frac{1}{\sqrt{5}}}\iint_{\Sigma}\left|\begin{array}{ccc}\frac{-2}{\sqrt{5}} & 0 & \frac{1}{\sqrt{5}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 6 x y z-y z^2 & 2 x^2 z & x y z\end{array}\right| d S$
  • $\underbrace{\iint\left[\frac{-2}{\sqrt{5}}\left(x z-2 x^2\right)-0+\frac{1}{\sqrt{5}}\left(\underbrace{4 x z-6 x z}_{-2xz}+z^2\right)\right]}_{\frac{1}{\sqrt{5}}\left(-2 x z-2 x^2-2 x z+z^2\right)} \mathrm{d} S$
  • $\displaystyle =\iint_{\Sigma} \frac{1}{\sqrt{5}}(2 x-z)^2 d S\xlongequal[一投影到xOy面:ds=\sqrt{1+z_x^{\prime 2}+z_y^{\prime 2}}]{二代:2x - z =1 }\iint_{Dxy} \frac{1}{\sqrt{5}} \cdot 1^2 \cdot \underbrace{\sqrt{1+z_x^{\prime 2}+z_y^{\prime 2}}}_{\sqrt{1+2^2+1^2}=\sqrt{5}}d x d y$
  • $\displaystyle \begin{aligned} = \iint_{D_{xy}} \frac{1}{\sqrt{5}} \cdot 1^2 \cdot \sqrt{5} \, dx \, dy = \iint_{D_{xy}} 1 \, dx \, dy = A_{D_{xy}}. \end{aligned}$
• 求投影椭圆的面积，由$\displaystyle \begin{aligned} \left\{\begin{array}{l}x^2 + y^2 + z^2 = 2x \\2x - z - 1 = 0\end{array}\right. \end{aligned}$
  • 消去 $\displaystyle \begin{aligned} z \end{aligned}$，得 $\displaystyle \begin{aligned} D_{xy} \end{aligned}$：$\displaystyle \begin{aligned} x^2 + y^2 + (2x - 1)^2 \leq 2x \end{aligned}$。
  • 即 $\displaystyle \begin{aligned} D_{xy} \end{aligned}$：$\displaystyle \begin{aligned} 5x^2 + y^2 - 6x + 1 \leq 0 \end{aligned}$。，配方$\displaystyle \begin{aligned} D_{xy} \end{aligned}$：$\displaystyle \begin{aligned} \frac{\left(x - \frac{3}{5}\right)^2}{\frac{4}{25}} + \frac{y^2}{\frac{4}{5}} \leq 1 \end{aligned}$。
  • $\displaystyle \begin{aligned} \Rightarrow A_{D_{xy}} \xlongequal[]{椭圆面积=\pi ab} \pi \cdot \frac{2}{5} \cdot \frac{2}{\sqrt{5}} = \frac{4\pi}{5\sqrt{5}}. \end{aligned}$

李艳芳版

• 
  

• 曲线L的方程是:$\displaystyle \left\{\begin{array}{l}x^2+y^2+z^2=2 x \\ 2 x-z-1=0\end{array}\right.$，曲线L是封闭曲线L

  • L 围成平面 $\displaystyle \begin{aligned} 2x - z - 1 = 0 \end{aligned}$ 上的有界部分为 $\displaystyle \begin{aligned} \Sigma \end{aligned}$、取上侧。记所求曲线积分为 $\displaystyle \begin{aligned} I \end{aligned}$。

• L 的方向与 $\displaystyle \begin{aligned} \Sigma \end{aligned}$ 的方向符合右手法则(从上往下看)，则由斯托克斯公式

  • $\displaystyle \begin{aligned} I = \oint_L (6xyz - yz^2) \, dx + 2x^2z \, dy + xyz \, dz \end{aligned}$
    • $\displaystyle \begin{aligned} = \iint_{\Sigma} \left| \begin{array}{ccc}dy \, dz & dz \, dx & dx \, dy \\\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\6xyz - yz^2 & 2x^2z & xyz \\\end{array} \right| \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{将第一行展开} \iint_{\Sigma} (xz - 2x^2) \, dy \, dz - (yz - 2xy + 2yz) \, dz \, dx + (z^2 - 2xz) \, dx \, dy \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{合并整理} \iint_{\Sigma} (xz - 2x^2) \, dy \, dz + (6xy - 3yz) \, dz \, dx + (z^2 - 2xz) \, dx \, dy \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{提取公因式}\iint_{\Sigma} (-x)(2x - z) \, dy \, dz + 3y(2x - z) \, dz \, dx + (-z)(2x - z) \, dx \, dy \end{aligned}$
    • $\displaystyle \begin{aligned} \stackrel{代入2x-z=1}{\xlongequal[]{}} \iint_{\Sigma} (-x) \, dy \, dz + 3y \, dz \, dx + (-z) \, dx \, dy \end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[]{投影垂直于ZoX面}\iint_z(-x) d y d z+(-z) d x d y.\end{aligned}$

• 利用两类曲面积分之间的联系得

  • $\displaystyle \begin{aligned} \iint_{\Sigma} (-x) \, dy \, dz + (-z) \, dx \, dy \quad \text{转化为关于} \, x, y \, \text{的曲面积分} \end{aligned}$
  • 令:F(x,y,z) $\displaystyle =z-2 x+1 \quad \sum: F(x, y, z)=0$
    • $\displaystyle \begin{aligned} n \end{aligned}$ 可取为 $\displaystyle \begin{aligned} (F_x', F_y', F_z') = (-2, 0, 1) \end{aligned}$
    • $\displaystyle \begin{aligned} \cos \alpha = \frac{-2}{\sqrt{5}}, \quad \cos \beta = 0, \quad \cos \gamma = \frac{1}{\sqrt{5}} \end{aligned}$
    • $\displaystyle \begin{aligned} dy \, dz = \frac{\cos \alpha}{\cos \gamma} \, dx \, dy = -2 \, dx \, dy \end{aligned}$

• 于是$\displaystyle \begin{aligned} &\iint_{\Sigma} (-x) \, dy \, dz + (-z) \, dx \, dy \end{aligned}$

  • $\displaystyle \begin{aligned} \xrightarrow[]{dy \, dz = \frac{\cos \alpha}{\cos \gamma} \, dx \, dy = -2 \, dx \, dy }\iint_{\Sigma} [(-2)(-x) + (-z)] \, dx \, dy = \iint_{\Sigma} (2x - z) \, dx \, dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow{2x - z = 1} \iint_{\Sigma} \, dx \, dy = \iint_{D} \, dx \, dy\end{aligned}$

• $\displaystyle \begin{aligned} 将 z = 2x - 1 代入 x^2 + y^2 + z^2 = 2x \end{aligned}$，得$\displaystyle \begin{aligned} x^2 + y^2 + (2x - 1)^2 - 2x = 0, \end{aligned}$

  • $\displaystyle \begin{aligned} 5x^2 - 6x + y^2 + 1 = 0. \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{配方}5\left(x^2 - \frac{6}{5}x\right) + y^2 + 1 = 0 \end{aligned}$
  • $\displaystyle \begin{aligned} \underbrace{5\left(x - \frac{3}{5}\right)^2 + y^2 = \frac{4}{5}}_{椭圆} \end{aligned}$
  • $\displaystyle \begin{aligned} \frac{(x - \frac{3}{5})^2}{\frac{4}{25}} + \frac{y^2}{\frac{4}{5}} = 1 \end{aligned}$
    • 行如:椭圆面积：$\displaystyle \begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{aligned}$、面积 $\displaystyle \begin{aligned} \pi ab \end{aligned}$
    • $\displaystyle \begin{aligned} D \end{aligned}$ 是中心 $\displaystyle \begin{aligned} \left(\frac{3}{5}, 0\right) \end{aligned}$、长半轴长 $\displaystyle \begin{aligned} \sqrt{\frac{4}{5}} \end{aligned}$，短半轴长 $\displaystyle \begin{aligned} \sqrt{\frac{4}{25}} \end{aligned}$
    • 则$\displaystyle \begin{aligned} D \end{aligned}$ 的面积为 $\displaystyle \begin{aligned} \pi \sqrt{\frac{4}{25}} \sqrt{\frac{4}{5}} = \frac{4\pi}{5\sqrt{5}} = \frac{4\sqrt{5}}{25}\pi \end{aligned}$

• 1997 年数一试题

  • 计算曲线积分$\oint_C(z-y) \mathrm{d} x+(x-z) \mathrm{d} y+(x-y) \mathrm{d} z,$
    其中 $C$ 是曲线 $\displaystyle \left\{\begin{array}{l}x^2+y^2=1, \\ x \rightarrow y+z=2,\end{array}\right.$ 从 $z$ 轴正向往 $z$ 轴负向看， $C$ 的方向是顺时针的.

• 2001 年数一试题

  • 计算 $I=\oint_L\left(y^2-z^2\right) \mathrm{d} x+\left(2 z^2-x^2\right) \mathrm{d} y+\left(3 x^2-y^2\right) \mathrm{d} z$ ，其中 $L$ 是平面 $x+y+z=2$ 与柱面 $|x|+|y|=1$ 的交线，从 $z$ 轴正向看去， $L$ 为逆时针方向.

• 2015 年数一试题

  • 已知曲线 $L$ 的方程为 $\displaystyle \left\{\begin{array}{l}z=\sqrt{2-x^2-y^2}, \\ z=x,\end{array}\right.$ 起点为 $A(0, \sqrt{2}, 0)$, 终点为 $B(0,-\sqrt{2}, 0)$, 计算曲线积分 $\displaystyle I=\int_L(y+z) \mathrm{d} x+\left(z^2-x^2+y\right) \mathrm{d} y+x^2 y^2 \mathrm{~d} z$.

• 2022 年数一试题

  • 已知 $\displaystyle \Sigma$ 为曲面 $4 x^2+y^2+z^2=1(x \geq 0, y \geq 0, z \geq 0)$ 的上侧, $L$ 为 $\displaystyle \Sigma$ 的边界曲线，其正向与 $\displaystyle \Sigma$ 的正法向量满足右手法则，计算曲线积分$\displaystyle I=\int_L\left(y z^2-\cos z\right) \mathrm{d} x+2 x z^2 \mathrm{~d} y+(2 x y z+x \sin z) \mathrm{d} z .$

(21)

(本题满分 12 分)

已知数列 $\left\{x_n\right\},\left\{y_n\right\},\left\{z_n\right\}$ 满足 $x_0=-1, y_0=0, z_0=2$, 且 $\displaystyle \left\{\begin{array}{l}x_n=-2 x_{n-1}+2 z_{n-1} \\ y_n=-2 y_{n-1}-2 z_{n-1} \\ z_n=-6 x_{n-1}-3 y_{n-1}+3 z_{n-1}\end{array}\right.$, 记 $\displaystyle \alpha_n=\left(\begin{array}{l}x_n \\ y_n \\ z_n\end{array}\right)$,写出满足 $\alpha_n=A \alpha_{n-1}$ 的矩阵 $A$, 并求 $A^n$ 及 $x_n, y_n, z_n(n=1,2, \ldots)$.

• $\displaystyle \begin{aligned} &\alpha_n=A \alpha_{n-1} \quad \text {正好对应 }\left(\begin{array}{l}x_n \\y_n \\z_n\end{array}\right)=A\left(\begin{array}{l}x_{n-1} \\y_{n-1} \\z_{n-1}\end{array}\right) \end{aligned}$
• $\displaystyle \begin{aligned} \text { 由题中方程 } \left\{\begin{array}{l}x_n=-2 x_{n-1}+2 z_{n-1} \\ y_n=-2 y_{n-1}-2 z_{n-1} \\ z_n=-6 x_{n-1}-3 y_{n-1}+3 z_{n-1}\end{array}\right.\end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{提取系数矩阵}\left(\begin{array}{l}x_n \\y_n \\z_n\end{array}\right) =\underbrace{\left(\begin{array}{ccc}-2 & 0 & 2 \\0 & -2 & -2 \\-6 & -3 & 3\end{array}\right)}_{A}\left(\begin{array}{l}x_{n-1} \\y_{n-1} \\z_{n-1}\end{array}\right) \end{aligned}$
    • 得$\displaystyle \begin{aligned} A=\left(\begin{array}{ccc}-2 & 0 & 2 \\0 & -2 & -2 \\-6 & -3 & 3\end{array}\right) \end{aligned}$
• 求$A^n$，要通过$A^n=P \Lambda^n P^{-1}$求，所以要求矩阵$A$对应的特征向量和特征值
  • 特征值组成对角矩阵
  • 特征向量组成可逆矩阵
  • 行初等变换求可逆矩阵的逆矩阵
• 求特征值:$\displaystyle \begin{aligned} A-\lambda E=\left|\begin{array}{ccc}-2-\lambda & 0 & 2 \\0 & -2-\lambda & -2 \\-6 & -3 & 3-\lambda\end{array}\right| \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow{r_1+r_2} \left|\begin{array}{ccc}-2-\lambda & -2-\lambda & 0 \\0 & -2-\lambda & -2 \\-6 & -3 & 3-\lambda\end{array}\right| \xrightarrow{c_2-c_1}\left|\begin{array}{ccc}-2-\lambda & 0 & 0 \\0 & -2-\lambda & -2 \\-6 & 3 & 3-\lambda\end{array}\right| \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{按第一行展开}(-2-\lambda)[(-2-\lambda)(3-\lambda)-(-6)] \end{aligned}$
  • $\displaystyle \begin{aligned} =(-2-\lambda)\left[\lambda^2-\lambda\underbrace{-6+6}_{0}\right] \end{aligned}$
  • $\displaystyle \begin{aligned} =(-2-\lambda) \cdot \lambda(\lambda-1)\end{aligned}$
    • 解得，$\displaystyle \begin{aligned} &\lambda_1=0, \lambda_2=1, \lambda_3=-2 \end{aligned}$
• 求每个特征值对应的特征向量:$|A-λ \cdot E| x=0$
  • $\displaystyle \begin{aligned}\lambda=0, 特征多项式|A-0 \cdot E| x=0 \end{aligned}$
    • $\displaystyle \begin{aligned} \left|\begin{array}{ccc}-2 & 0 & 2 \\0 & -2 & -2 \\-6 & -3 & 3\end{array}\right| \xrightarrow{r_3-3 r_1}\left|\begin{array}{ccc}-2 & 0 & 2 \\0 & -2 & -2 \\0 & -3 & -3\end{array}\right| \xrightarrow[]{2，3两行成比例}\left|\begin{array}{ccc}-1 & 0 & 1 \\0 & 1 & 1 \\0 & 0 & 0\end{array}\right| \end{aligned}$
    • $\displaystyle \begin{aligned} 令x_3=1, \text { 则 } x=k_1\left(\begin{array}{c}1 \\-1 \\1\end{array}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} \lambda=1，特征多项式|A-E| x=0 \end{aligned}$
    • $\displaystyle \begin{aligned} \left|\begin{array}{ccc}-3 & 0 & 2 \\0 & -3 & -2 \\-6 & -3 & 2\end{array}\right| \xrightarrow{r_3-2 r_1}\left|\begin{array}{ccc}-3 & 0 & 2 \\0 & -3 & -2 \\0 & -3 & -2\end{array}\right| \rightarrow\left|\begin{array}{lll}3 & 0 & -2 \\0 & 3 & 2 \\0 & 0 & 0\end{array}\right| \end{aligned}$
    • $\displaystyle 设x_3=3 ，则 \lambda=k_2\left(\begin{array}{c}2 \\-2 \\3\end{array}\right)$
  • $\displaystyle \begin{aligned} \lambda=-2，特征多项式\left|\begin{array}{ccc}A+2 E\end{array}\right| x=0 \end{aligned}$
    • $\displaystyle \left|\begin{array}{ccc}0 & 0 & 2 \\0 & 0 & -2 \\-6 & -3 & 5\end{array}\right| \longrightarrow\left|\begin{array}{lll}6 & 3 & 0 \\0 & 0 & 1 \\0 & 0 & 0\end{array}\right|$
    • $\displaystyle \begin{aligned} 由x_3=0, 设x_2=2，则 x=k_3\left(\begin{array}{c}-1 \\2 \\0\end{array}\right) \end{aligned}$
• 将特征向量拼接成可逆矩阵
  $\displaystyle \begin{aligned} & P=\left(\begin{array}{ccc}1 & 2 & -1 \\-1 & -2 & 2 \\1 & 3 & 0\end{array}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} P^{-1} A P=\Lambda \end{aligned}$
  • $\displaystyle \begin{aligned} A=P \wedge P^{-1}, \quad \Lambda=\left(\begin{array}{lll}0 & & \\& 1 & \\& & -2\end{array}\right) \end{aligned}$
• 由对角矩阵和可逆矩阵求n阶矩阵
  $\displaystyle \begin{aligned} A^n=P \wedge^n P^{-1} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{代入三个矩阵}\left(\begin{array}{ccc}1 & 2 & -1 \\-1 & -2 & 2 \\1 & 3 & 0\end{array}\right)\left(\begin{array}{lll}0 & & \\& 1 & \\& & -2\end{array}\right)\left(\begin{array}{ccc}1 & 2 & -1 \\-1 & -2 & 2 \\1 & 3 & 0\end{array}\right)^{-1} \end{aligned}$
    • 通过化行最简矩阵求可逆矩阵$\displaystyle \begin{aligned}\left(\begin{array}{ccc}1 & 2 & -1 \\-1 & -2 & 2 \\1 & 3 & 0\end{array}\right)^{-1}\end{aligned}$
      • $\displaystyle \begin{aligned} \left(\begin{array}{ccc:ccc}1 & 2 & -1 & 1 & 0 & 0 \\-1 & -2 & 2 & 0 & 1 & 0 \\1 & 3 & 0 & 0 & 0 & 1\end{array}\right) \xrightarrow{r_2+r_1, r_3-r_1}\left(\begin{array}{ccc:ccc}1 & 2 & -1 & 1 & 0 & 0 \\0 & 0 & 1 & 1 & 1 & 0 \\0 & 1 & 1 & -1 & 0 & 1\end{array}\right) \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow{r_3-r_2, r_2 \leftrightarrow r_3}\left(\begin{array}{ccc:ccc}1 & 2 & -1 & 1 & 0 & 0 \\0 & 1 & 0 & -2 & -1 & 1 \\0 & 0 & 1 & 1 & 1 & 0\end{array}\right) \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow{r_1-2 r_2, r_1+r_3}\left(\begin{array}{ccc:ccc}1 & 0 & 0 & 6 & 3 & -2 \\0 & 1 & 0 & -2 & -1 & 1 \\0 & 0 & 1 & 1 & 1 & 0\end{array}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} A^n=\left(\begin{array}{ccc}1 & 2 & -1 \\-1 & -2 & 2 \\1 & 3 & 0\end{array}\right)\left(\begin{array}{lll}0 & & \\& 1 & \\& & (-2)^n\end{array}\right)\left(\begin{array}{ccc}6 & 3 & -2 \\-2 & -1 & 1 \\1 & 1 & 0\end{array}\right) \end{aligned}$
    • $\displaystyle \begin{aligned} \xlongequal[]{A的行乘B的列}\left(\begin{array}{ccc}0 & 2 & -(-2)^n \\0 & -2 & 2(-2)^n \\0 & 3 & 0\end{array}\right)\left(\begin{array}{ccc}6 & 3 & -2 \\-2 & -1 & 1 \\1 & 1 & 0\end{array}\right) \end{aligned}$
    • $\displaystyle \begin{aligned} &\xlongequal[]{A的行乘B的列}\left(\begin{array}{ccc}-4-(-2)^n & -2-(-2)^n & 2 \\4+2 \cdot(-2)^n & 2+2(-2)^n & -2 \\-6 & -3 & 3\end{array}\right) \end{aligned}$
• 由递推关系求$\alpha_n$
  $\displaystyle \begin{aligned} \alpha_n=A \alpha_{n-1}=A^2 \alpha_{n-2}=A^3 \alpha_{n-3}=A^n \alpha_0 \end{aligned}$
  • $\displaystyle \begin{aligned} \underbrace{\left(\begin{array}{l}x_n \\y_n \\z_n\end{array}\right)=A^n\left(\begin{array}{l}x_0 \\y_0 \\z_0\end{array}\right)=A^n\left(\begin{array}{c}-1 \\0 \\2\end{array}\right)}_{ \alpha_{n}=A^n \alpha_0 }\end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{代入A^n}\left(\begin{array}{c}-4+(-2)^n-4 \\-4-2(-2)^n-4 \\6+6\end{array}\right) \end{aligned}$

2000 年数一试题

某试验性生产线每年一月份进行熟练工与非熟练工的人数统计，然后将 $\frac{1}{6}$ 熟练工支援其他生产部门，其缺额由招收新的非熟练工补齐。新、老非熟练工经过培训及实践至年终考核有 $\frac{2}{5}$ 成为熟练工。设第 $n$ 年一月份统计的熟练工和非熟练工所占百分比分别为 $x_n$ 和 $y_n$, 记成向量 $\binom{x_n}{y_n}$.

(1) 求 $\binom{x_{n+1}}{y_{n+1}}$ 与 $\binom{x_n}{y_n}$ 的关系式并写成矩阵形式: $\binom{x_{n+1}}{y_{n+1}}=\boldsymbol{A}\binom{x_n}{y_n}$;

(2) 验证 $\eta_1=\binom{4}{1}, \eta_2=\binom{-1}{1}$ 是 $\boldsymbol{A}$ 的两个线性无关的特征向量, 并求出相应的特征值；

(3) 当 $\binom{x_1}{y_1}=\binom{\frac{1}{2}}{\frac{1}{2}}$ 时, 求 $\binom{x_{n+1}}{y_{n+1}}$.

1991 年数一试题

设3阶矩阵 $\boldsymbol{A}$ 的特征值为 $\lambda_1=1, \lambda_2=2, \lambda_3=3$, 对应的特征向量依次为 $\displaystyle \xi_1=\left(\begin{array}{l}1 \\ 1 \\ 1\end{array}\right), \quad \xi_2=\left(\begin{array}{l}1 \\ 2 \\ 4\end{array}\right), \xi_3=\left(\begin{array}{l}1 \\ 3 \\ 9\end{array}\right)$, 又向量 $\displaystyle \boldsymbol{\beta}=\left(\begin{array}{l}1 \\ 1 \\ 3\end{array}\right)$.

(1) 将 $\boldsymbol{\beta}$ 用 $\xi_1, \xi_2, \xi_3$ 线性表出; (2) 求 $\boldsymbol{A}^n \boldsymbol{\beta}$ ( $n$ 为自然数).

(22)

(本题满分 12 分)

设总体 $X$ 服从 $[0, \theta]$ 上的均匀分布, 其中 $\theta \in(0,+\infty)$ 为未知参数, $X_1, X_2 \ldots \ldots . X_n$ 是来自总体 $X$ 的简单随机样本, 记 $X_{(n)}=\max \left\{X_1, X_2 \ldots . . X_n\right\}, T_c=c X_{(n)}$.
(1)求 $c$, 使得 $T_c$ 为 $\theta$ 的无偏估计量;
(2)记 $h(c)=E\left(T_c-\theta\right)^2$, 求 $c$ 使得 $h(c)$ 最小.

• 恒等变形:$\displaystyle \begin{aligned} E\left(T_c\right)=\theta \xrightarrow{T_c=c X_{(n)}} E\left(c X_{(n)}\right)=cE\left(X_{(n)}\right)=\theta \end{aligned}$

  • 求期望
    • 先求分布函数
      $\displaystyle \begin{aligned} F_{X_{(n)}}(x)=P\left\{X_{(n)} \leq x\right\} \xrightarrow{X_{(n)}=\max \left\{X_1, X_2, \cdots, X_n\right\}} P\left\{\max \left\{X_1, x_2, \cdots, X_n\right\} \leq x\right\} \end{aligned}$
      • $\displaystyle \begin{aligned} =P\left\{X_1 \leqslant x\right\} \cdots P\left\{X_n \leqslant x\right\} \end{aligned}$
      • $\displaystyle \begin{aligned} = \begin{cases}1, & x>\theta \\0, & x<0 \\\left(\frac{x-0}{\theta-0}\right)^n=\frac{x^n}{\theta^n}, & 0 \leqslant x \leqslant \theta\end{cases} \end{aligned}$
    • 分布函数求导得概率密度
      $\displaystyle \begin{aligned} f_{X_{(n)}}(x)=\left\{\begin{array}{cc}\frac{n x^{n-1}}{\theta^n}, & 0 \leq x \leq \theta \\0, & \text { 其他 }\end{array}\right. \end{aligned}$
    • 何处求概率，何处算积分
      $\displaystyle \begin{aligned} E(X(n))=\int_0^\theta x \cdot \frac{n \cdot x^{n-1}}{\theta^n} d x=\frac{n}{\theta^n} \cdot \int_0^\theta x^n d x=\left.\frac{n}{\theta^n} \cdot \frac{x^{n+1}}{n+1}\right|_0 ^\theta \end{aligned}$
      • $\displaystyle \begin{aligned} =\frac{n}{\theta^n} \cdot \frac{1}{n+1} \cdot \theta^{n+1}=\frac{n \theta}{n+1} \end{aligned}$
  • 代入原式，得$\displaystyle \begin{aligned} E\left(T_c\right)=cE\left(X_{(n)}\right)=\theta\end{aligned}$
    • $\displaystyle \begin{aligned} c \cdot \frac{n \theta}{n+1}=\theta \text { ，得 } c=\frac{n+1}{n} \end{aligned}$

• 恒等变形，得$\displaystyle \begin{aligned} & h(c)=E\left[\left(T_c-\theta\right)^2\right] \xrightarrow{T_c=c X(n)} E\left[\left(c X_{(n)}-\theta\right)^2\right] \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[]{完全平方}E\left[c^2 X_{(n)}^2+\theta^2-2 \cdot c X(n) \cdot \theta\right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{提出系数}c^2 E\left(X_{(n)}^2\right)+\theta^2-2 c \cdot \theta \cdot E(X(n)) \end{aligned}$
    • 定义法求:
      $\displaystyle \begin{aligned} E\left(X^2\right)=\int_0^\theta x^2 \cdot \frac{n x^{n-1}}{\theta^n} d x=\frac{n}{\theta^n} \cdot \int_0^\theta X^{n+1} d x=\left.\frac{n}{\theta^n} \cdot \frac{x^{n+2}}{n+2}\right|_0 ^\theta=\frac{n}{n+2} \cdot \theta^2 \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[E(X)=\frac{n}{n+1} \theta]{E\left(X^2\right)=\frac{n}{n+2} \theta^2}c^2 \cdot \frac{n}{n+2} \theta^2+\theta^2-2 c \cdot \theta \cdot \frac{n}{n+1} \theta \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{提出\theta^2}\theta^2\left(\frac{n}{n+2} \cdot c^2-1-2 c \cdot \frac{n}{n+1}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[二元函数求最小值]{x=-\frac{b}{2 a} }-\frac{-2 \frac{n}{n+1}}{2 \cdot \frac{n}{n+2}}=\frac{n+2}{n+1} \end{aligned}$