解析：当\( x \neq 0 \)时，
\( f'(x)=\dfrac{\left(x\text{e}^{\frac{1}{x}}\right)'\left(1+\text{e}^{\frac{1}{x}}\right)-x\text{e}^{\frac{1}{x}}\left(1+\text{e}^{\frac{1}{x}}\right)'}{\left(1+\text{e}^{\frac{1}{x}}\right)^2}=\dfrac{\text{e}^{\frac{1}{x}}+\text{e}^{\frac{2}{x}}-\dfrac{1}{x}\text{e}^{\frac{1}{x}}}{\left(1+\text{e}^{\frac{1}{x}}\right)^2} \)，
\( \lim\limits_{x \to 0^-} f'(x) = \lim\limits_{x \to 0^-} \dfrac{\text{e}^{\frac{1}{x}}+\text{e}^{\frac{2}{x}}-\dfrac{1}{x}\text{e}^{\frac{1}{x}}}{\left(1+\text{e}^{\frac{1}{x}}\right)^2} \)。
又\( \lim\limits_{x \to 0^-} \dfrac{1}{x}\text{e}^{\frac{1}{x}} = \lim\limits_{x \to 0^-} \dfrac{\dfrac{1}{x}}{\text{e}^{-\frac{1}{x}}} = \lim\limits_{x \to 0^-} \dfrac{-\dfrac{1}{x^2}}{\text{e}^{-\frac{1}{x}} \cdot \left(-\dfrac{1}{x^2}\right)} = \lim\limits_{x \to 0^-} \dfrac{1}{\text{e}^{-\frac{1}{x}}} = 0 \)，故\( \lim\limits_{x \to 0^-} f'(x) = 0 \)。
同理，可求得
\( \lim\limits_{x \to 0^+} f'(x) = \lim\limits_{x \to 0^+} \dfrac{\text{e}^{\frac{1}{x}}+\text{e}^{\frac{2}{x}}-\dfrac{1}{x}\text{e}^{\frac{1}{x}}}{\left(1+\text{e}^{\frac{1}{x}}\right)^2} = \lim\limits_{x \to 0^+} \dfrac{\text{e}^{-\frac{1}{x}} + 1 - \dfrac{1}{x}\text{e}^{-\frac{1}{x}}}{\left(1+\text{e}^{-\frac{1}{x}}\right)^2} = 1 \)。
故\( x = 0 \)是\( f'(x) \)的跳跃间断点. 选项B正确.

答案 B

答案 B
$\lim\limits_{x \to 0} \frac{\|\boxed{\alpha} + x\boxed{\beta}\| - \|\boxed{\alpha}\|}{x} = \lim\limits_{x \to 0} \frac{\|\boxed{\alpha} + x\boxed{\beta}\|^2 - \|\boxed{\alpha}\|^2}{x(\|\boxed{\alpha} + x\boxed{\beta}\| + \|\boxed{\alpha}\|)}$
$= \lim\limits_{x \to 0} \frac{(\boxed{\alpha} + x\boxed{\beta}) \cdot (\boxed{\alpha} + x\boxed{\beta}) - \boxed{\alpha} \cdot \boxed{\alpha}}{x(\|\boxed{\alpha} + x\boxed{\beta}\| + \|\boxed{\alpha}\|)}$
$= \lim\limits_{x \to 0} \frac{2x\boxed{\alpha} \cdot \boxed{\beta} + x^2\|\boxed{\beta}\|^2}{2\|\boxed{\alpha}\|x} = \frac{\boxed{\alpha} \cdot \boxed{\beta}}{\|\boxed{\alpha}\|} + 0$
$= \|\boxed{\beta}\|\cos\langle \boxed{\alpha}, \boxed{\beta} \rangle = 2 \times \frac{1}{2} = 1$.
选项 B 正确.

解析
由\(\lim\limits_{\substack{x \to 0 \\ y \to 0}} f(x,y) = \lim\limits_{\substack{x \to 0 \\ y \to 0}} |xy| \ln (x^2 + y^2) = 0 = f(0,0)\)，知\( f(x,y) \)在点\((0,0)\)处连续，
\( f'_x(0,0) = \lim\limits_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = \lim\limits_{x \to 0} \frac{0 - 0}{x} = 0 \).
同理，\( f'_y(0,0) = 0 \).
\(\lim\limits_{\substack{x \to 0 \\ y \to 0}} \frac{f(x,y) - f(0,0) - [f'_x(0,0)x + f'_y(0,0)y]}{\rho}\)
\( = \lim\limits_{\substack{x \to 0 \\ y \to 0}} \frac{|xy| \ln (x^2 + y^2)}{\sqrt{x^2 + y^2}}, \ \rho = \sqrt{x^2 + y^2} \).
又\( 0 \leq \left| \frac{|xy| \ln (x^2 + y^2)}{\sqrt{x^2 + y^2}} \right| \leq \frac{1}{2\rho}(x^2 + y^2) |\ln (x^2 + y^2)| = \rho |\ln \rho| \)，
且\(\lim\limits_{\rho \to 0} \rho |\ln \rho| = 0\)，故\(\lim\limits_{\substack{x \to 0 \\ y \to 0}} \frac{|xy| \ln (x^2 + y^2)}{\sqrt{x^2 + y^2}} = 0\).
\( f(x,y) \)在点\((0,0)\)处可微. 选项D正确.

【注】$\vert xy\vert \leqslant \frac{1}{2}({x}^{2}+{y}^{2})$

答案 A
解析 依题设，曲面\( z = \sqrt{x^2 + y^2} \)的法向量为
\( \boxed{n} = (z'_x, z'_y, -1) = \left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}, -1 \right) = \left( \frac{x}{z}, \frac{y}{z}, -1 \right) \)，
其方向余弦为\( \cos\alpha = \frac{x}{\sqrt{2}z} \)，\( \cos\beta = \frac{y}{\sqrt{2}z} \)，\( \cos\gamma = \frac{-1}{\sqrt{2}} \)。
由转换公式\( \frac{\text{d}y\text{d}z}{\cos\alpha} = \frac{\text{d}z\text{d}x}{\cos\beta} = \frac{\text{d}x\text{d}y}{\cos\gamma} \)，有
\( \text{d}y\text{d}z = \frac{\cos\alpha}{\cos\gamma}\text{d}x\text{d}y = -\frac{x}{z}\text{d}x\text{d}y \)，\( \text{d}z\text{d}x = \frac{\cos\beta}{\cos\gamma}\text{d}x\text{d}y = -\frac{y}{z}\text{d}x\text{d}y \)。
故\( \iint_{\Sigma} P\text{d}y\text{d}z - Q\text{d}z\text{d}x = \iint_{\Sigma} \left( -\frac{x}{z}P + \frac{y}{z}Q \right) \text{d}x\text{d}y \)。选项 A 正确。

答案 A
解析 由\( A \)与\( B \)合同，知\( \boldsymbol{X^TAX} \)与\( \boldsymbol{X^TBX} \)有相同的正、负惯性指数.
又由\( |\lambda E - B| = \begin{vmatrix} \lambda & -1 & 0 \\ -1 & \lambda & 0 \\ 0 & 0 & \lambda - 2 \end{vmatrix} = (\lambda - 2)(\lambda^2 - 1) \)，得\( B \)的特征值为1,2,-1.\( \boldsymbol{X^TBX} \)的正惯性指数\( p = 2 \)，负惯性指数\( q = 1 \).
从而\( \boldsymbol{X^TAX} \)的规范形为\( y_1^2 + y_2^2 - y_3^2 \).
故选项A正确.

答案 C
解析 由$\boldsymbol{AX = 0}$有两个线性无关的解，知$\boldsymbol{AX = 0}$的基础解向量的个数$n - \mathrm{r}(\boldsymbol{A}) \geq 2$，即$\mathrm{r}(\boldsymbol{A}) \leq n - 2$. 故$\mathrm{r}(\boldsymbol{A}^*) = 0$，从而$\boldsymbol{A}^* = \boldsymbol{O}$，所以任意$n$维列向量均是$\boldsymbol{A^*X = 0}$的解. 因此，$\boldsymbol{AX = 0}$的解均是$\boldsymbol{A^*X = 0}$的解. 选项C正确，选项B不正确，选项A不正确.
对于选项D，由$\boldsymbol{AX = 0}$的基础解系至少包含两个解向量，知$\boldsymbol{AX = 0}$有无穷多个非零解，从而$\boldsymbol{AX = 0}$与$\boldsymbol{A^*X = 0}$的公共解有无穷多个非零解. 选项D不正确.

C. $\boldsymbol{AX = 0}$的解均是$\boldsymbol{A^*X = 0}$的解.
D. $\boldsymbol{AX = 0}$与$\boldsymbol{A^*X = 0}$没有非零公共解.

答案 A
解析 $\boldsymbol{A}^* = -\boldsymbol{A} - \boldsymbol{E}$两边左乘$\boldsymbol{A}$，得
$\boldsymbol{A}\boldsymbol{A}^* = -\boldsymbol{A}^2 - \boldsymbol{A}$。
由$|\boldsymbol{A}| = -2$，知$|\boldsymbol{A}|\boldsymbol{E} = -\boldsymbol{A}^2 - \boldsymbol{A}$，即$\boldsymbol{A}^2 + \boldsymbol{A} - 2\boldsymbol{E} = \boldsymbol{O}$。
设$\boldsymbol{A}$的任一特征值为$\lambda$，则$\lambda^2 + \lambda - 2 = 0$，解得$\boldsymbol{A}$的可能特征值为$1$，$-2$。
由$|\boldsymbol{A}| = -2$，知$\boldsymbol{A}$的特征值只能为$1$，$1$，$-2$。

故$\boldsymbol{A}^*$的特征值为$\frac{|\boldsymbol{A}|}{\lambda}$，即$-2$，$-2$，$1$。
$\boldsymbol{A}^* + a\boldsymbol{E}$的特征值为$-2 + a$，$-2 + a$，$1 + a$。
由$\boldsymbol{X}^{\mathrm{T}}(\boldsymbol{A}^* + a\boldsymbol{E})\boldsymbol{X} = 1$表示椭球面，知该二次型的正惯性指数为3，负惯性指数为0，
故$-2 + a > 0$，$-2 + a > 0$，$1 + a > 0$，解得$a > 2$。
选项A正确。

答案 B
解析 记 \( A = \{X \leq 1\} \)，\( B = \{Y \leq -1\} \)，由已知得 \( P(AB) = \frac{1}{4} \).
\( P\{X > 1, Y > -1\} = P(\overline{A}\overline{B}) = P(\overline{A \cup B}) = 1 - P(A \cup B) \)
\( = 1 - P(A) - P(B) + P(AB) \).
又 \( P(A) = P\{X \leq 1\} = \Phi\left(\frac{1}{\sigma}\right) \)，\( P(B) = P\{Y \leq -1\} = \Phi\left(-\frac{1}{\sigma}\right) = 1 - \Phi\left(\frac{1}{\sigma}\right) \)，故
\( P\{X > 1, Y > -1\} = 1 - \Phi\left(\frac{1}{\sigma}\right) - 1 + \Phi\left(\frac{1}{\sigma}\right) + \frac{1}{4} = \frac{1}{4} \).
选项 B 正确.

答案 B
解析 由已知，得 \( X_1, X_2, \cdots, X_n \) 相互独立，且都服从参数为 \( \lambda = \frac{1}{2} \) 的泊松分布。
故 \( EX_i = \frac{1}{2} \)，\( DX_i = \frac{1}{2} (i = 1, 2, \cdots, n) \)。
由独立同分布的中心极限定理知，对任意的 \( x \in \mathbf{R} \)，有
\[
\begin{align*}
\lim_{n \to \infty} P\left\{ \frac{\sum_{i=1}^{n} X_i - E\left( \sum_{i=1}^{n} X_i \right)}{\sqrt{D\left( \sum_{i=1}^{n} X_i \right)}} \leq x \right\} &= \lim_{n \to \infty} P\left\{ \frac{\sum_{i=1}^{n} X_i - \frac{n}{2}}{\sqrt{\frac{n}{2}}} \leq x \right\} \\
&= \lim_{n \to \infty} P\left\{ \frac{2\sum_{i=1}^{n} X_i - n}{2\sqrt{n}} \leq \frac{x}{\sqrt{2}} \right\} = \Phi(x)
\end{align*}
\]
取 \( \frac{x}{\sqrt{2}} = 1 \)，即 \( x = \sqrt{2} \)，则 \( \lim_{n \to \infty} P\left\{ \frac{2\sum_{i=1}^{n} X_i - n}{2\sqrt{n}} \leq 1 \right\} = \Phi(\sqrt{2}) \)。选项 B 正确。

答案 C
解析 由已知，得\(\overline{X}\sim N\left(\mu,\frac{\sigma^{2}}{n}\right)\)，且\(\overline{X}\)与\(S^{2}\)相互独立，故\(E\overline{X}=\mu\)，\(E(\overline{X}S^{2})=E\overline{X}\cdot E(S^{2})\)。

又\(\frac{(n-1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n-1)\)，所以
\(E(S^{2})=E\left[\frac{\sigma^{2}}{n-1}\cdot\frac{(n-1)S^{2}}{\sigma^{2}}\right]=\frac{\sigma^{2}}{n-1}E\left[\chi^{2}(n-1)\right]=\frac{\sigma^{2}}{n-1}(n-1)=\sigma^{2}\)。
故\(E(\overline{X}S^{2})=\mu\sigma^{2}\)。选项C正确。
【注】 也可以直接应用结论：\(E(S^{2})=\sigma^{2}\)。

答案 4
解析
\( y = \int_{0}^{t} \sin(t - s)ds \stackrel{t - s = u}{=} \int_{0}^{t} \sin u du \),
\( \frac{dy}{dx} = \frac{y_{t}'}{x_{t}'} = \frac{\sin t}{2e^{-t^{2}}} = \frac{1}{2}e^{t^{2}} \sin t, \frac{dy}{dx}\big|_{t=0} = 0 \),
\( \frac{d^{2}y}{dx^{2}} = \frac{(\frac{1}{2}e^{t^{2}} \sin t)'}{x_{t}'} = \frac{1}{4}e^{2t^{2}}(2t\sin t + \cos t) \),
所以\( \frac{d^{2}y}{dx^{2}}\big|_{t=0} = \frac{1}{4} \)，故曲率半径为\( R = \frac{(1 + y'^{2})^{\frac{3}{2}}}{|y''|} = 4 \)

答案 $\sqrt{3}\pi + \ln 3$
解析 $\frac{6x}{x^3 + 1} = \frac{-2}{x + 1} + \frac{2x + 2}{x^2 - x + 1} = \frac{-2}{x + 1} + \frac{2x - 1}{x^2 - x + 1} + \frac{3}{\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$，
$\int \frac{6x}{x^3 + 1}dx = \int \frac{-2}{x + 1}dx + \int \frac{2x - 1}{x^2 - x + 1}dx + \int \frac{3}{\left(x - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}dx$
$= \ln \frac{x^2 - x + 1}{(x + 1)^2} + 2\sqrt{3}\arctan \frac{2x - 1}{\sqrt{3}} + c$．
故 $\int_{\frac{1}{2}}^{+\infty} \frac{6x}{x^3 + 1}dx = \left[\ln \frac{x^2 - x + 1}{(x + 1)^2} + 2\sqrt{3}\arctan \frac{2x - 1}{\sqrt{3}}\right] \bigg|_{\frac{1}{2}}^{+\infty}$
$= \sqrt{3}\pi + \ln 3$．

答案 $\frac{2}{3}$

解析 由已知，有

$\frac{\partial u}{\partial x}=\frac{2x}{x^{2}+y^{2}+z^{2}},\frac{\partial u}{\partial y}=\frac{2y}{x^{2}+y^{2}+z^{2}},\frac{\partial u}{\partial z}=\frac{2z}{x^{2}+y^{2}+z^{2}},$

$\text{grad}\ u=\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)\stackrel{\text{记}}{=}(P,Q,R)$，则

$\frac{\partial P}{\partial x}=\frac{2(-x^{2}+y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{2}},\frac{\partial Q}{\partial y}=\frac{2(x^{2}+z^{2}-y^{2})}{(x^{2}+y^{2}+z^{2})^{2}},\frac{\partial R}{\partial z}=\frac{2(x^{2}+y^{2}-z^{2})}{(x^{2}+y^{2}+z^{2})^{2}}.$

所以$\text{div}(\text{grad}\ u)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}=\frac{2}{x^{2}+y^{2}+z^{2}}$，于是$\text{div}(\text{grad}\ u)\big|_{(1,1,1)}=\frac{2}{3}.$13.

答案 \( \frac{3}{2}\pi \)

解析 记\( x^2 + y^2 = R^2 \)所围区域为\( D \)，则由格林公式，有
\[
\begin{align*}
I(R) &= \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dxdy = 3\iint_D (1 - x^2 - y^2) dxdy \\
&= 3\int_0^{2\pi} d\theta \int_0^R (1 - r^2)r dr = 6\pi \left( \frac{1}{2}r^2 - \frac{1}{4}r^4 \right) \bigg|_0^R = 3\pi R^2 \left( 1 - \frac{R^2}{2} \right).
\end{align*}
\]
令\( I'(R) = 6\pi R(1 - R^2) = 0 \)，得\( R = 1 \)。
当\( R > 1 \)时，\( I'(R) < 0 \)；当\( 0 < R < 1 \)时，\( I'(R) > 0 \)。
故\( R = 1 \)为\( I(R) \)的极大值点，也是最大值点，最大值为\( I(1) = \frac{3}{2}\pi \)。

答案 \(\begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{3} \\ 0 & 0 & \frac{1}{3} \end{bmatrix}\)

解析 由已知，有\((\beta_1,\beta_2,\beta_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}\stackrel{记}{=}(\alpha_1,\alpha_2,\alpha_3)P\)，故\((\alpha_1,\alpha_2,\alpha_3)=(\beta_1,\beta_2,\beta_3)P^{-1}\)，\(P^{-1}=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{bmatrix}^{-1}=\begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} & -\frac{1}{3} \\ 0 & 0 & \frac{1}{3} \end{bmatrix}\)。

答案 \(\frac{1}{n^2} + \frac{1}{n^3}\)
解析 \( E(T^2) = E\left[ \left( \frac{\overline{X} - S}{n} \right)^2 \right] = \frac{1}{n^2}E\left[ (\overline{X} - S)^2 \right] \)
\( = \frac{1}{n^2}\left[ E(\overline{X}^2 - 2\overline{X}S + S^2) \right] = \frac{1}{n^2}\left[ E(\overline{X}^2) - 2E(\overline{X}S) + E(S^2) \right] \)

依题设，知\( E\overline{X} = EX = 0 \)，\( D\overline{X} = \frac{1}{n} \)，\( E(S^2) = DX = 1 \)。
又\( \overline{X} \)与\( S^2 \)相互独立，故\( \overline{X} \)与\( S \)相互独立，所以
\( E(\overline{X}^2) = D\overline{X} + (E\overline{X})^2 = \frac{1}{n} + 0 = \frac{1}{n} \)，
\( E(\overline{X}S) = E\overline{X} \cdot ES = 0 \)。
故\( E(T^2) = \frac{1}{n^2}\left( \frac{1}{n} - 0 + 1 \right) = \frac{1}{n^2} + \frac{1}{n^3} \)。

### 解析
由$\lim\limits_{t\to 0}\frac{f[(x+t)^2]-f(x^2+t)}{(2x-1)t}=1-2\ln x$，有
$\lim\limits_{t\to 0}\frac{f[(x+t)^2]-f(x^2+t)}{t}=(2x-1)(1-2\ln x)$。

而
$\lim\limits_{t\to 0}\frac{f[(x+t)^2]-f(x^2+t)}{t}$
$=\lim\limits_{t\to 0}\frac{f[(x+t)^2]-f(x^2)}{t}-\lim\limits_{t\to 0}\frac{f(x^2+t)-f(x^2)}{t}$
$=[f(x^2)]'-f'(x^2)=f'(x^2)\cdot 2x - f'(x^2)=f'(x^2)(2x-1)$。

故$f'(x^2)(2x-1)=(2x-1)(1-2\ln x)$，从而$f'(x^2)=1-\ln x^2$，于是$f'(x)=1-\ln x$，
$f(x)=\int (1-\ln x)dx = x - \int \ln x dx = 2x - x\ln x + c$。

由$f(1)=2$，得$c=0$，故$f(x)=2x - x\ln x$。

令$f'(x)=1-\ln x=0$，得$x=e$，又$f''(e)=-\frac{1}{e}<0$，知$f(e)=e$为$f(x)$的极大值。

【注】$\lim\limits_{t\to 0}\frac{f[(x+t)^2]-f(x^2)}{t}=\lim\limits_{t\to 0}\frac{f(x^2+2xt+t^2)-f(x^2)}{2xt+t^2}\cdot \frac{2xt+t^2}{t}=f'(x^2)\cdot 2x$。

解析 （Ⅰ）由
$\mathrm{d}f(x,y) = (3x^2 - 2ax)\mathrm{d}x + (3y^2 - 2ay)\mathrm{d}y$
$= \mathrm{d}(x^3 - ax^2) + \mathrm{d}(y^3 - ay^2)$
$= \mathrm{d}(x^3 + y^3 - ax^2 - ay^2 + c)$，
知$f(x,y) = x^3 + y^3 - ax^2 - ay^2 + c$. 由$f(0,0) = 0$，得$c = 0$.
故$f(x,y) = x^3 + y^3 - ax^2 - ay^2$.

（Ⅱ）由$\begin{cases} f'_x = 3x^2 - 2ax = 0, \\ f'_y = 3y^2 - 2ay = 0, \end{cases}$得$(0,0)$，$(0,\frac{2}{3}a)$，$(\frac{2}{3}a,0)$，$(\frac{2}{3}a,\frac{2}{3}a)$，且$f_{xx}'' = 6x - 2a$，$f_{xy}''=0,f_{yy}''=6y - 2a$.
对于点$(0,0)$，有$A = -2a < 0$，$B = 0$，$C = -2a$，$AC - B^2 = 4a^2 > 0$，故$f(0,0) = 0$为极大值.
对于点$(0,\frac{2}{3}a)$，有$A = -2a$，$B = 0$，$C = 2a$，$AC - B^2 = -4a^2 < 0$，故$f(x,y)$在点$(0,\frac{2}{3}a)$处不取得极值.
同理，在$(\frac{2}{3}a,0)$处，$f(x,y)$不取得极值.
对于点$(\frac{2}{3}a,\frac{2}{3}a)$，$A = 2a > 0$，$B = 0$，$C = 2a$，$AC - B^2 = 4a^2 > 0$，故$f(\frac{2}{3}a,\frac{2}{3}a) = -\frac{4}{27} \cdot 2a^3 = -8$为极小值，即$a^3 = 27$，得$a = 3$.

解析 （Ⅰ）因\( \int_{0}^{x} 4t f(x^2) f(x^2 - t^2) dt = 2f(x^2) \int_{0}^{x} 2t f(x^2 - t^2) dt \)，
又因\( \int_{0}^{x} 2t f(x^2 - t^2) dt \stackrel{x^2 - t^2 = s}{=} \int_{0}^{x^2} f(s) ds = \int_{0}^{x^2} f(t) dt \)，
故原等式化为\( 2f(x^2) \int_{0}^{x^2} f(t) dt = x^6 \)，记\( x^2 = u \)，上式可化为
\[ 2f(u) \int_{0}^{u} f(t) dt = u^3 \] ①
令\( F(u) = \int_{0}^{u} f(t) dt \)，则①式为
\[ 2F'(u)F(u) = u^3 \]
上式两边同时积分，得\( F^2(u) = \frac{1}{4} u^4 + C \)。
由\( F(0) = 0 \)，得\( C = 0 \)，故\( F(u) = \frac{1}{2} u^2 \)（因\( F(u) \geq 0 \)），即\( \int_{0}^{u} f(t) dt = \frac{1}{2} u^2 \)。
上式两边同时对\( u \)求导，得\( f(u) = u \)，即
\[ f(x) = x (x \geq 0) \]

（Ⅱ）由（Ⅰ），知\( \sum_{n=1}^{\infty} 2^{1-n} f(n^2) = \sum_{n=1}^{\infty} \frac{n^2}{2^{n-1}} \)，
构造幂级数\( S(x) = \sum_{n=1}^{\infty} n^2 x^{n-1} \)，可求得其收敛半径为1，且当\( x = \pm 1 \)时，级数发散，故其收敛域为\( (-1,1) \)。

\( S(x) = \sum_{n=1}^{\infty} n^2 x^{n-1} = \sum_{n=1}^{\infty} (n x^n)' = \left( \sum_{n=1}^{\infty} n x^n \right)' = \left( x \sum_{n=1}^{\infty} n x^{n-1} \right)' \)

记\( S_1(x) = \sum_{n=1}^{\infty} n x^{n-1} \)，则\( S_1(x) = \sum_{n=1}^{\infty} (x^n)' = \left( \sum_{n=1}^{\infty} x^n \right)' = \left( \frac{x}{1 - x} \right)' = \frac{1}{(1 - x)^2} \)，
故\( S(x) = [x S_1(x)]' = \left[ x \cdot \frac{1}{(1 - x)^2} \right]' = \frac{1 + x}{(1 - x)^3} \)，\(| x | < 1\)。

令\( x = \frac{1}{2} \)，则\( \sum_{n=1}^{\infty} \frac{n^2}{2^{n-1}} = S\left( \frac{1}{2} \right) = \frac{1 + \frac{1}{2}}{\left( 1 - \frac{1}{2} \right)^3} = 12 \)。

解析 （Ⅰ）设$f(x_0)=m$，由$f(0)=f(1)=0$，$m<0$，知$x_0\in(0,1)$，且$f'(x_0)=0$．
由拉格朗日中值定理，知存在$\xi\in(0,x_0)$，使得
$$f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0}=\frac{m}{x_0}<\frac{m}{n}<f'(x_0).$$
对$f'(x)$在$[\xi,x_0]$上应用介值定理，存在$x_n\in(\xi,x_0)\subset(0,1)$，使得$f'(x_n)=\frac{m}{n}$，
即$nf'(x)=m$在$x\in(0,1)$内有实根$x_n$．又$f''(x)>0$，知$f'(x)$在$(0,1)$内严格单调递增．故$x_n$是唯一的．
（Ⅱ）由$f'(x)$严格单调递增，知
$$f'(x_n)=\frac{m}{n}<\frac{m}{n+1}=f'(x_{n+1}),$$
故$x_n<x_{n+1}<x_0<1$，$\{x_n\}$单调递增有上界，由单调有界准则，知$\lim\limits_{n\to\infty}x_n$存在．
记$\lim\limits_{n\to\infty}x_n=A$，则$0<A\leq x_0$，由$f'(x)$在$(0,1)$内连续，知
$$f'(A)=f'\left(\lim\limits_{n\to\infty}x_n\right)=\lim\limits_{n\to\infty}f'(x_n)=\lim\limits_{n\to\infty}\frac{m}{n}=0.$$
由$f'(x)$严格单调递增，知$A=x_0$．
故$\lim\limits_{n\to\infty}f(x_n)=f\left(\lim\limits_{n\to\infty}x_n\right)=f(x_0)=m$．

解析 （Ⅰ）由$\boldsymbol{A}^{\mathrm{T}}\boldsymbol{X}=\boldsymbol{0}$的解均是$\boldsymbol{\beta}^{\mathrm{T}}\boldsymbol{X}=\boldsymbol{0}$的解，知$\boldsymbol{A}^{\mathrm{T}}\boldsymbol{X}=\boldsymbol{0}$与$\begin{cases}\boldsymbol{A}^{\mathrm{T}}\boldsymbol{X}=\boldsymbol{0}\\\boldsymbol{\beta}^{\mathrm{T}}\boldsymbol{X}=\boldsymbol{0}\end{cases}$同解，
故$\mathrm{r}(\boldsymbol{A}^{\mathrm{T}})=\mathrm{r}\begin{pmatrix}\boldsymbol{A}^{\mathrm{T}}\\\boldsymbol{\beta}^{\mathrm{T}}\end{pmatrix}$，即$\mathrm{r}(\boldsymbol{A})=\mathrm{r}(\boldsymbol{A},\boldsymbol{\beta})$。对$(\boldsymbol{A},\boldsymbol{\beta})$作初等行变换，有
$(\boldsymbol{A},\boldsymbol{\beta})=\begin{bmatrix}a&1&1&b\\0&a - 1&0&1\\1&1&a&1\end{bmatrix}\to\begin{bmatrix}1&1&a&1\\0&a - 1&0&1\\a&1&1&b\end{bmatrix}$
$\to\begin{bmatrix}1&1&a&1\\0&a - 1&0&1\\0&1 - a&1 - a^2&b - a\end{bmatrix}\to\begin{bmatrix}1&1&a&1\\0&a - 1&0&1\\0&0&1 - a^2&b - a + 1\end{bmatrix}$。

当$a=1$时，$\mathrm{r}(\boldsymbol{A})=1$，$\mathrm{r}(\boldsymbol{A},\boldsymbol{\beta})=2$，故$a=1$舍去。
当$a=-1$时，$(\boldsymbol{A},\boldsymbol{\beta})\to\begin{bmatrix}1&1&-1&1\\0&-2&0&1\\0&0&0&b + 2\end{bmatrix}$，故$b=-2$，有$\mathrm{r}(\boldsymbol{A})=\mathrm{r}(\boldsymbol{A},\boldsymbol{\beta})=2$。
综上所述，$a=-1$，$b=-2$。

（Ⅱ）由（Ⅰ）知，$\boldsymbol{A}=\begin{bmatrix}-1&1&1\\0&-2&0\\1&1&-1\end{bmatrix}$，则由
$|\lambda\boldsymbol{E}-\boldsymbol{A}|=\begin{vmatrix}\lambda + 1&-1&-1\\0&\lambda + 2&0\\-1&-1&\lambda + 1\end{vmatrix}=\lambda(\lambda + 2)^2=0$，
得$\boldsymbol{A}$的特征值为$\lambda_1=0$，$\lambda_2=\lambda_3=-2$。
由$0\boldsymbol{E}-\boldsymbol{A}=\begin{bmatrix}1&-1&-1\\0&2&0\\-1&-1&1\end{bmatrix}\to\begin{bmatrix}1&0&-1\\0&1&0\\0&0&0\end{bmatrix}$，得$\boldsymbol{\alpha}_1=(1,0,1)^{\mathrm{T}}$。
由$-2\boldsymbol{E}-\boldsymbol{A}=\begin{bmatrix}-1&-1&-1\\0&0&0\\-1&-1&-1\end{bmatrix}\to\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$，得$\boldsymbol{\alpha}_2=(-1,1,0)^{\mathrm{T}}$，$\boldsymbol{\alpha}_3=(-1,0,1)^{\mathrm{T}}$。
令$\boldsymbol{P}=(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3)=\begin{bmatrix}1&-1&-1\\0&1&0\\1&0&1\end{bmatrix}$，则$\boldsymbol{P}^{-1}\boldsymbol{AP}=\begin{bmatrix}0&0&0\\0&-2&0\\0&0&-2\end{bmatrix}=\boldsymbol{\Lambda}$。

**解析 （Ⅰ）** 先求$X$的概率密度.
记$X$的分布函数为$F_X(x)$，则当$x \leq 0$时，$F_X(x) = 0$；当$x > 0$时，有
$$F_X(x) = P\{\min(X_1,X_2) \leq x\} = 1 - P\{X_1 > x, X_2 > x\}$$
$$= 1 - P\{X_1 > x\}P\{X_2 > x\} = 1 - \mathrm{e}^{-\lambda x} \cdot \mathrm{e}^{-\lambda x} = 1 - \mathrm{e}^{-2\lambda x}.$$
故$X$的概率密度为
$$f_X(x) = F_X'(x) = \begin{cases} 2\lambda \mathrm{e}^{-2\lambda x}, & x > 0, \\ 0, & x \leq 0. \end{cases}$$
$Z$的分布函数为
$$F_Z(z) = P\{XY \leq z\} = P\{XY \leq z, Y = -1\} + P\{XY \leq z, Y = 1\}$$
$$= P\{X \geq -z, Y = -1\} + P\{X \leq z, Y = 1\}$$
$$= P\{X \geq -z\} \cdot P\{Y = -1\} + P\{X \leq z\} \cdot P\{Y = 1\}$$
$$= \frac{1}{3}[1 - F_X(-z)] + \frac{2}{3}F_X(z)$$

故$f_Z(z) = F_Z'(z) = \frac{1}{3}f_X(-z) + \frac{2}{3}f_X(z)$.
当$z < 0$，即$-z > 0$时，有$f_X(-z) = 2\lambda \mathrm{e}^{2\lambda z}$，$f_X(z) = 0$.
当$z > 0$，即$-z < 0$时，有$f_X(z) = 2\lambda \mathrm{e}^{-2\lambda z}$，$f_X(-z) = 0$.
当$z = 0$时，$f_Z(z) = 0$，故
$$f_Z(z) = \begin{cases} \frac{2}{3}\lambda \mathrm{e}^{2\lambda z}, & z < 0, \\ 0, & z = 0, \\ \frac{4}{3}\lambda \mathrm{e}^{-2\lambda z}, & z > 0. \end{cases}$$

**（Ⅱ）** 先求$T = \max(X, X_3)$的概率密度，由$X = \min(X_1,X_2)$与$X_3$相互独立，有
$$F_T(t) = P\{X \leq t\} \cdot P\{X_3 \leq t\} = (1 - \mathrm{e}^{-2\lambda t})(1 - \mathrm{e}^{-\lambda t}), t > 0.$$
故$f_T(t) = F_T'(t) = \begin{cases} \lambda \mathrm{e}^{-\lambda t} + 2\lambda \mathrm{e}^{-2\lambda t} - 3\lambda \mathrm{e}^{-3\lambda t}, & t > 0, \\ 0, & t \leq 0, \end{cases}$
$$ET = \int_0^{+\infty} t(\lambda \mathrm{e}^{-\lambda t} + 2\lambda \mathrm{e}^{-2\lambda t} - 3\lambda \mathrm{e}^{-3\lambda t}) \mathrm{d}t = \frac{1}{\lambda} + \frac{1}{2\lambda} - \frac{1}{3\lambda} = \frac{7}{6\lambda}.$$
于是$\lambda$的矩估计量为$\hat{\lambda} = \frac{7}{6\bar{T}}(\bar{T} = \frac{1}{n}\sum_{i=1}^n T_i)$.

【注】 第（Ⅱ）问中的计算用到了$\Gamma$函数，$\int_0^{+\infty} x^{\alpha - 1}\mathrm{e}^{-x} \mathrm{d}x = \Gamma(\alpha), \alpha > 0$.
特别地，$\Gamma(n + 1) = n!$（$n$为正整数）.