【解析】 \(\lim _{x \to a} \frac{\sin f(x)-\sin a}{x-a}=\frac{ 中值定理 }{ x \to a} \frac{\cos \xi[f(x)-a]}{x-a}=b \lim _{x \to a} \cos \xi, \xi\) 介于 \(a\) 与 \(f(x)\) 之间。 \(\lim _{x \to a} \cos \xi=\cos a\) 且 \(f(a)=a\) 故 \(\lim _{x \to a} \frac{\sin f(x)-\sin a}{x-a}=b \cos a\)

【解析】题目中 \(f(x)=\frac{e^{\frac{1}{x-1}} \ln |1+x|}{(e^{x}-1)(x-2)}\) 中

\[

\lim _{x \to 0} f(x)=\lim _{x \to 0} \frac{e^{\frac{1}{x-1}} \ln (1+x)}{(e^{x}-1)(x-2)}=\frac{e^{-1}}{-2} \lim _{x \to 0} \frac{x}{x}=\frac{1}{-2 e},

\]

\[

\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1^{+}} \frac{e^{\frac{1}{x-1}} \ln |1+x|}{(e^{x}-1)(x-2)}=\lim _{x \to 1^{-}} \frac{e^{\frac{1}{x-1}} \ln |1+1|}{(e-1)(1-2)}=\infty,

\]

\[

\lim _{x \to 2} f(x)=\lim _{x \to 2} \frac{e^{\frac{1}{x-1}} \ln |1+x|}{(e^{x}-1)(x-2)}=\lim _{x \to 2} \frac{e^{\frac{1}{2-1}} \ln |1+2|}{(e^{2}-1)(x-2)}=\infty,

\]

\[

\lim _{x \to -1} f(x)=\lim _{x \to -1} \frac{e^{\frac{1}{x-1}} \ln |1+x|}{(e^{x}-1)(x-2)}=\infty . 故选 (C)

\]

【答案】A.

【解析】对于(A)，由于\(F'(x)=\left[\int_{0}^{x}[\cos f(t)+f'(t)]dt\right]' = \cos f(x)+f'(x)\)，
\(F'(-x)=\cos [f(-x)]+f'(-x)\)
\(=\cos f(x)+f'(x)\)。

所以\(F'(x)\)为偶函数，则\(F(x)=\int_{0}^{x}[\cos f(t)+f'(t)]dt\)为奇函数，故(A)正确，(B)错误.

【解析】由于 \(\sum_{n=1}^{\infty} n a_{n}(x-2)^{n}\) 与 \(\sum_{n=1}^{\infty} a_{n}(x+1)^{n}\) 的收敛半径相同。

故由题设, \(\sum_{n=1}^{\infty} n a_{n}(x-2)^{n}\) 收敛区间为(-2,6),则收敛半径 \(R=4\) ,因此, \(\sum_{n=1}^{\infty} a_{n}(x+1)^{n}\) 的收敛区间为(-1-4,-1+4),即(-5,3)

【解析】因为 A 不可逆,所以 \(r(A)<4\) .又因为 \(A_{12} ≠0\) ,故 \(r(A^{*}) ≥1\) ,所以

\[

r(A) \geq n-1=3, r(A)=3, r\left(A^{*}\right)=1,|A|=0 .

\]

由于 \(A_{12} ≠0\) 所以 \(\alpha_{1}\) , \(\alpha_{3}\) , \(\alpha_{4}\) 线性无关. \(A^{*} x=0\) 的基础解系中有 \(4-1=3\) 个无关解向量, \(A^{*} A=|A| E=0\) ,所以 A 的列向量为 \(A^{*} x=0\) 的解.

因此 \(A^{*} x=0\) 的通解为 \(k_{1} \alpha_{1}+k_{2} \alpha_{3}+k_{3} \alpha_{4}\) ,

【解析】由于 \(\alpha_{1}\) , \(\alpha_{2}\) 线性无关,且为特征值1对应的特征向量,所以1至少为2重特征值.

因为 \(\alpha_{3}\) 为-1对应的特征向量, A 为3阶矩阵,所以 A 的特征值为1,1,-1. 矩阵 \(P^{-1} A P\) 中第二列为-1对应的特征向量,故 P 的第二列应为 \(\alpha_{3}\) 的线性组合,第一列、第三列为 \(\alpha_{1}\) , \(\alpha_{2}\) 的线性组合.故选 (D).

【解析】法1: A、B、C 中恰有一个事件发生的概率为 \(P(A \bar{B} \bar{C})+P(\bar{A} B \bar{C})+P(\bar{A} \bar{B} C)\)

\[

P(A \overline{B C})=P(A \overline{B})-P(A \overline{B} C)=P(A)-P(AB)-[P(AC)-P(ABC)]

\]

\[

P(\overline{A} B \overline{C})=P(\overline{A} B)-P(\overline{A} B C)=P(B)-P(AB)-[P(BC)-P(BCA)]

\]

\[

P(\overline{A B C})=P(\overline{B} C)-P(\overline{B} C A)=P(C)-P(CB)-[P(CA)-P(CAB)]

\]

因为 \(P(AB)=0\) ,而 \(ABC \subset AB\) ,则 \(0 ≤P(ABC) ≤P(AB)=0\) ,故 \(P(ABC)=0\)

将题干的已知代入以上三个式子中,得

\[

P(A \overline{B C})=\frac{1}{4}-\frac{1}{12}=\frac{2}{12}, P(\overline{A} B \overline{C})=\frac{1}{4}-\frac{1}{12}=\frac{2}{12}, P(\overline{A B C})=\frac{1}{4}-\frac{1}{12}-\frac{1}{12}=\frac{1}{12}

\]

故所求概率为 \(\frac{2}{12}+\frac{2}{12}+\frac{1}{12}=\frac{5}{12}\) ,故选(D)

法 2: \(p=P(A+B+C)-P(AB)-P(AC)-P(BC)+2 P(ABC)\)

\[

P(AB)=0 \Rightarrow P(ABC)=0

\]

\[

\begin{aligned}

& P(A+B+C)=P(A)+P(B)+P(C)-P(AB)-P(AC)-P(BC)+P(ABC) \\

& =\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-0-\frac{1}{12}-\frac{1}{12}+0=\frac{7}{12}

\end{aligned}

\]

\[

p=P(A+B+C)-P(AB)-P(AC)-P(BC)+2 P(ABC)=\frac{7}{12}-0-\frac{1}{12}-\frac{1}{12}+0=\frac{5}{12} . 故选(D)

\]

【解析】由已知, \(Cov(X, Y)=-\frac{1}{2} \cdot 1 \cdot \sqrt{4}=-1\)

\[

而只有 \(Cov(X, X+Y)=D(X)+Cov(X, Y)=0\) ,

\]

故与 X 相互独立的只有(A)(C); 又因为 \(D(X+Y)=D(X)+D(Y)+2 Cov(X, Y)=3\) ,所以 \(\frac{\sqrt{3}}{3}(X+Y) \sim N(0,1)\) 选(C)

【解析】\(\left.\frac{\partial z}{\partial x}\right|_{(0, \pi)} = \left.\frac{y + \cos(x + y)}{1 + [xy + \sin(x + y)]^{2}}\right|_{(0, \pi)} = \frac{\pi + \cos \pi}{1 + \sin^{2}\pi} = \pi - 1\)

\(\left.\frac{\partial z}{\partial y}\right|_{(0, \pi)} = \left.\frac{x + \cos(x + y)}{1 + [xy + \sin(x + y)]^{2}}\right|_{(0, \pi)} = \frac{\cos \pi}{1 + \sin^{2}\pi} = -1\)

\(\left.dz\right|_{(0, \pi)} = (\pi - 1)dx - dy\)

【解析】方程两边对 \(x\) 求导得，\(1 + y' + e^{2xy}(2y + 2xy') = 0\)，代入 \(x = 0\)，\(y = -1\) 得 \(y' = 1\)，故切线方程为 \(y = x - 1\)

【解析】由需求函数 \(Q(P) = \frac{800}{p + 3} - 2\)，可得 \(P = \frac{800}{Q + 2} - 3\)，利润函数 \(L(Q) = PQ - C(Q) = (\frac{800}{Q + 2} - 3)Q - (100 + 13Q) = -\frac{1600}{Q + 2} - 16Q + 700\)

\(L'(Q) = \frac{1600}{(Q + 2)^{2}} - 16 = 0\)，解得 \(Q = 8\)

\(L''(Q) = \frac{-3200}{(Q + 2)^{3}}|_{Q = 8} = -\frac{16}{5} < 0\)，当 \(Q = 8\) 时，\(L(Q)\) 最大

【解析】\(V_{y} = \int_{0}^{1}2\pi x(\frac{1}{1 + x^{2}} - \frac{x}{2})dx = 2\pi \int_{0}^{1}\frac{x}{1 + x^{2}}dx - \pi \int_{0}^{1}x^{2}dx\)

\(= \left.\pi \ln(1 + x^{2})\right|_{0}^{1} - \left.\frac{\pi}{3}x^{3}\right|_{0}^{1} = \pi \ln 2 - \frac{\pi}{3}\)

【答案】$a^{4}-4a^{2}.$

【解析】$\begin{vmatrix}a&0&-1&1\\0&a&1&-1\\-1&1&a&0\\1&-1&0&a\end{vmatrix}=a\begin{vmatrix}a&1&0\\1&a&a\\-1&0&a\end{vmatrix}-\begin{vmatrix}0&a&0\\-1&1&a\\1&-1&a\end{vmatrix}$

$=-a\begin{vmatrix}1&a^{2}\\a&2a\end{vmatrix}-2a\begin{vmatrix}0&a\\-1&1\end{vmatrix}$

$=-a(2a - a^{3}) - 2a^{2}$

$=a^{4}-4a^{2}.$

【解析】\(Y\) 表示 \(X\) 被 \(3\) 除的余数，取值为 \(0, 1, 2\)

\(P\{Y = 0\} = \sum_{k=1}^{\infty}P\{X = 3k\} = \sum_{k=1}^{\infty}\frac{1}{2^{3k}} = \frac{\frac{1}{8}}{1 - \frac{1}{8}} = \frac{1}{7}\)

\(P\{Y = 1\} = \sum_{k=0}^{\infty}P\{X = 3k + 1\} = \sum_{k=0}^{\infty}\frac{1}{2^{3k + 1}} = \frac{\frac{1}{2}}{1 - \frac{1}{8}} = \frac{4}{7}\)

\(P\{Y = 2\} = \sum_{k=0}^{\infty}P\{X = 3k + 2\} = \sum_{k=0}^{\infty}\frac{1}{2^{3k + 2}} = \frac{\frac{1}{4}}{1 - \frac{1}{8}} = \frac{2}{7}\)

\(E(Y) = 0 \times \frac{1}{7} + 1 \times \frac{4}{7} + 2 \times \frac{2}{7} = \frac{8}{7}\)

【解析】由等价无穷小的定义可知

\[
\begin{align*}
&\lim_{n \to \infty} \frac{\left(1 + \frac{1}{n}\right)^{n}-e}{\frac{b}{n^{a}}}=\lim_{n \to \infty} \frac{e^{n\ln\left(1 + \frac{1}{n}\right)}-e}{\frac{b}{n^{a}}}=\lim_{n \to \infty} \frac{e\left(e^{n\ln\left(1 + \frac{1}{n}\right)-1}-1\right)}{\frac{b}{n^{a}}}=\lim_{n \to \infty} \frac{e\left(n\ln\left(1 + \frac{1}{n}\right)-1\right)}{\frac{b}{n^{a}}}=1\\
&\lim_{t \to 0^{+}} \frac{e\left(\frac{\ln(1 + t)}{t}-1\right)}{bt^{a}}=\lim_{t \to 0^{+}} \frac{e(\ln(1 + t)-t)}{bt^{a + 1}}=\lim_{t \to 0^{+}} \frac{e\left(-\frac{1}{2}t^{2}\right)}{bt^{a + 1}} = 1
\end{align*}
\]

则若使该极限存在必有\(a + 1 = 2\)，即\(a = 1\)。可得\(b = -\frac{1}{2}e\)。

【解析】由\(\begin{cases} \dfrac{\partial f}{\partial x} = 3x^{2}-y = 0 \\ \dfrac{\partial f}{\partial y}=24y^{2}-x = 0 \end{cases}\)计算可得

\(\begin{cases} x = 0 \\ y = 0 \end{cases}\)或\(\begin{cases} x=\dfrac{1}{6} \\ y = \dfrac{1}{12} \end{cases}\)

\(\dfrac{\partial^{2} f}{\partial x^{2}} = 6x\)，\(\dfrac{\partial^{2} f}{\partial x\partial y}=- 1\)，\(\dfrac{\partial^{2} f}{\partial y^{2}} = 48y\)

当\(x = 0,y = 0\)时，\(A = 0,B=-1,C = 0\)，则\(AC - B^{2}=-1\lt0\)，所以\((0,0)\)点不是极值点；

当\(x=\dfrac{1}{6},y = \dfrac{1}{12}\)时，\(A = 1,B=-1,C = 4\)，则\(AC - B^{2}=3\gt0\)且\(A\gt0\)，所以\(\left(\dfrac{1}{6},\dfrac{1}{12}\right)\)为极小值点，极小值\(f\left(\dfrac{1}{6},\dfrac{1}{12}\right)=-\dfrac{1}{216}\) 

【解析】（I）\(y'' + 2y' + 5y = 0\)，由特征方程知，\(r^{2} + 2r + 5 = 0\)，可得
\(r_{1} = -1 + 2i\)，\(r_{2} = -1 - 2i\)，
则方程的通解
\(y = e^{-x}(C_{1}\cos2x + C_{2}\sin2x)\)。
又因\(f(0) = 1\)，\(f'(0) = -1\)，所以\(C_{1} = 1\)，\(C_{2} = 0\)，因此\(f(x) = e^{-x}\cos2x\)。
（II）
\(a_{n} = \int_{n\pi}^{+\infty} e^{-x}\cos2xdx = \frac{1}{5e^{n\pi}}\)，
设
\(I = \int_{n\pi}^{+\infty} e^{-x}\cos2xdx\stackrel{分部积分}{=\!=\!=}\frac{1}{2}\int_{n\pi}^{+\infty} e^{-x}d\sin2x\)
\(=\frac{1}{2}\left[ e^{-x}\sin2x\big|_{n\pi}^{+\infty} - \int_{n\pi}^{+\infty} \sin2xde^{-x} \right]\)
\(= -\frac{1}{4}\int_{n\pi}^{+\infty} e^{-x}d\cos2x\)
\(= -\frac{1}{4}\left[ e^{-x}\cos2x\big|_{n\pi}^{+\infty} + \int_{n\pi}^{+\infty} e^{-x}\cos2xdx \right]\)，
所以
\(I = \frac{1}{4}e^{-x}\cos2x\big|_{x = n\pi} - \frac{1}{4}I\)，
即
\(\frac{5}{4}I = \frac{1}{4}e^{-x}\cos2x\big|_{x = n\pi}\)，
故\(I = \frac{1}{5e^{n\pi}}\)，\(\sum_{n = 1}^{\infty} a_{n} = \frac{1}{5}\sum_{n = 1}^{\infty} e^{-n\pi} = \frac{1}{5}\sum_{n = 1}^{\infty}(e^{-\pi})^{n} = \frac{1}{5(e^{\pi} - 1)}\)。

【解析】令\(\iint_{D} f(x, y)dxdy = A\)，则\(f(x, y) = y\sqrt{1 - x^{2}} + Ax\)
即\(\iint_{D} f(x, y)dxdy = \iint_{D}(y\sqrt{1 - x^{2}} + Ax)dxdy = \iint_{D} y\sqrt{1 - x^{2}}dxdy\)
\(= 2\int_{0}^{1} \sqrt{1 - x^{2}}dx\int_{0}^{\sqrt{1 - x^{2}}} ydy = \int_{0}^{1}(1 - x^{2})^{\frac{3}{2}}dx = \frac{3\pi}{16}\)，故\(A = \frac{3\pi}{16}\)
\(\iint_{D} xf(x, y)dxdy = \iint_{D}\left[ xy\sqrt{1 - x^{2}} + Ax^{2} \right]dxdy = A\iint_{D} x^{2}dxdy\)
\(= A\int_{0}^{\pi} \cos^{2}\theta d\theta\int_{0}^{1} r^{3}dr = \frac{A}{2}\int_{0}^{\frac{\pi}{2}} \cos^{2}\theta d\theta = \frac{A}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi^{2}}{128}\) 

【解析】(I)设\(x = x_0\)时\(\vert f(x_0)\vert = M\)，由拉格朗日定理

\(\exists\xi_1\in(0,x_0)\)使得\(\vert f'(\xi_1)\vert=\left\vert\frac{f(x_0)-f(0)}{x_0 - 0}\right\vert=\frac{\vert f(x_0)\vert}{x_0}=\frac{M}{x_0}\)

\(\exists\xi_2\in(x_0,2)\)使得\(\vert f'(\xi_2)\vert=\left\vert\frac{f(x_0)-f(2)}{x_0 - 2}\right\vert=\frac{\vert f(x_0)\vert}{2 - x_0}=\frac{M}{2 - x_0}\)

若\(x_0\in(0,1]\)则取\(\xi = \xi_1\)，有\(\vert f'(\xi)\vert=\vert f'(\xi_1)\vert=\frac{M}{x_0}\geq M\)

若\(x_0\in(1,2)\)则取\(\xi = \xi_2\)，有\(\vert f'(\xi)\vert=\vert f'(\xi_2)\vert=\frac{M}{2 - x_0}\geq M\)

(II) \(M = \vert f(x_0)\vert=\vert f(x_0)-f(0)\vert=\left\vert\int_{0}^{x_0}f'(x)dx\right\vert\leq\int_{0}^{x_0}\vert f'(x)\vert dx\leq\int_{0}^{x_0}Mdx\leq Mx_0\)

\(M = \vert f(x_0)\vert=\vert f(2)-f(x_0)\vert=\left\vert\int_{x_0}^{2}f'(x)dx\right\vert\leq\int_{x_0}^{2}\vert f'(x)\vert dx\leq\int_{x_0}^{2}Mdx\leq M(2 - x_0)\)

故\(\begin{cases}M(1 - x_0)\leq0 \\ M(1 - x_0)\geq0 \end{cases}\)同时成立

若\(x_0\neq1\)则显然\(M = 0\)

若\(x_0 = 1\)则\(M\leq\int_{0}^{1}\vert f'(x)\vert dx\leq\int_{0}^{1}Mdx = M\)且\(M\leq\int_{1}^{2}\vert f'(x)\vert dx\leq\int_{1}^{2}Mdx\leq M\)

当且仅当区间\((0,1)\)和\((1,2)\)上\(\vert f'(x)\vert = M\)成立，若假设\(M\neq0\)

要么\(f'(x)\equiv\pm M\)与\(f(0)=f(2)=0\)矛盾

要么\(f'(x)=\begin{cases}\pm M & 0\lt x\lt1 \\ \mp M & 1\lt x\lt2 \end{cases}\)，\(f(x)\)在\(x = 1\)处不可导，与已知矛盾

故\(M = 0\) 

【解析】(I)记 \( A = \begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix} \)，\( B = \begin{pmatrix} a & 2 \\ 2 & b \end{pmatrix} \)，则 \( Q^TAQ = B \)，其中 \( Q \) 是正交矩阵。\( A \) 与 \( B \) 相似，所以 \( A \) 与 \( B \) 有相同的特征值。\( |\lambda E - A| = \begin{vmatrix} \lambda - 1 & 2 \\ 2 & \lambda - 4 \end{vmatrix} = \lambda^2 - 5\lambda = 0 \)，得到 \( A \) 的特征值为 \( \lambda_1 = 0 \)，\( \lambda_2 = 5 \)。

\( |\lambda E - B| = \begin{vmatrix} \lambda - a & -2 \\ -2 & \lambda - b \end{vmatrix} = \lambda^2 - (a + b)\lambda + ab - 4 = 0 \)。由 \( B \) 的特征值也为 \( \lambda_1 = 0 \)，\( \lambda_2 = 5 \)，得 \( ab = 4 \)，\( a + b = 5 \)。因为 \( a\geq b \)，所以 \( a = 4 \)，\( b = 1 \)。

(II)当 \( \lambda_1 = 0 \) 时，\( (0E - A)x = 0 \) 的基础解系为 \( \alpha_1 = (2, 1)^T \)；

当 \( \lambda_2 = 5 \) 时，\( (5E - A)x = 0 \) 的基础解系为 \( \alpha_2 = (1, -2)^T \)。

当 \( \lambda_1 = 0 \) 时，\( (0E - B)x = 0 \) 的基础解系为 \( \beta_1 = (1, -2)^T \)；

当 \( \lambda_2 = 5 \) 时，\( (5E - B)x = 0 \) 的基础解系为 \( \beta_2 = (2, 1)^T \)。

令 \( P_1 = \begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix} \)，\( P_2 = \begin{pmatrix} 1 & 2 \\ -2 & 1 \end{pmatrix} \)，则 \( P_1^{-1}AP_1 = \begin{pmatrix} 0 \\ & 5 \end{pmatrix} \)，\( P_2^{-1}BP_2 = \begin{pmatrix} 0 \\ & 5 \end{pmatrix} \)。

所以 \( P_1^{-1}AP_1 = P_2^{-1}BP_2 \)，所以 \( B = P_2P_1^{-1}APP_2^{-1} \)。

\( P_1P_2^{-1} = \begin{pmatrix} 2 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ -2 & 1 \end{pmatrix}^{-1} = \frac{1}{5}\begin{pmatrix} 4 & -3 \\ -3 & -4 \end{pmatrix} \)。

因为 \( \frac{1}{5}\begin{pmatrix} 4 & -3 \\ -3 & -4 \end{pmatrix} \) 是正交矩阵，所以 \( Q = \frac{1}{5}\begin{pmatrix} 4 & -3 \\ -3 & -4 \end{pmatrix} \)。 

(I)(解法一)证明:假设 P 不是可逆矩阵,则 α 与 \(A \alpha\) 线性相关.根据线性相关的定义,存在不全为0的实 数 \(k_{1}, k_{2}\) ,使得成立 \(k_{1} \alpha+k_{2} A \alpha=0\) .若 \(k_{2}=0\) ,则有 \(k_{1} \alpha=0\) . α 是非零向量,则 \(k_{1}=0\) ,与 \(k_{1}\) \(,k2\) 不全为0矛盾,所以 \(k_{2} ≠0\) 所以有 \(A \alpha=-\frac{k_{1}}{k_{2}} \alpha\) ,与 a 不是 A 的特征向赋子后,所以程设不成,放 P 是可道矩碎。

(解法二)设 \(A \alpha=\beta\) ,则 \(\beta ≠k \alpha\) ,即 α β 线性无关,得 \(P=(\alpha, \beta)=(\alpha, A \alpha)\) 可逆矩阵.

(II)因为 \(A^{2} \alpha+A \alpha-6 \alpha=0\) ,所以 \(A P=A(\alpha, A \alpha)=(A \alpha, A^{2} \alpha)=(A \alpha, 6 \alpha-A \alpha)=(\alpha, A \alpha)[\begin{array}{rr}0 & 6 1 & -1\end{array}]\) 又因为 P 是可 逮明间,所以 \(P^{-1} A P=[\begin{array}{ll}0 & 6 1 & -1\end{array}]\) 记 \(B=[\begin{array}{ll}0 & 6 1 & -1\end{array}]\) 令 \(|\lambda E-B|=|\begin{array}{cc}\lambda & -6 -1 & \lambda+1\end{array}|=\lambda^{2}+\lambda-6=0\) 解到 B 的转证慎为 \(\lambda_{1}=2\) \(\lambda_{2}=-3\) A 与 B 相队则 A 的狗证留也为 \(\lambda_{1}=2\) , \(\lambda_{2}=-3\) ,所以 A 可相似对角化.

\[P\left\{Z_{1}=1\right\}=P\{X-Y>0\}=\iint_{x>y} f(x, y) d x d y=\frac{\frac{1}{8} \pi × 1^{2}}{\frac{1}{2} \pi × 1^{2}}=\frac{1}{4}\]

\[P\left\{Z_{1}=0\right\}=1-\frac{1}{4}=\frac{3}{4}\]

\[P\left\{Z_{2}=0\right\}=P\{X+Y \leq 0\}=\iint_{y \leq-x} f(x, y) d x d y=\frac{\frac{1}{8} \pi × 1^{2}}{\frac{1}{2} \pi × 1^{2}}=\frac{1}{4}\]

\[P\left\{Z_{2}=1\right\}=1-\frac{1}{4}=\frac{3}{4}\]

\[又 P\left\{Z_{1}=1, Z_{2}=0\right\}=0\]

故联合概率分布:

\begin{array}{c|cc|c} \diagbox{Z_1}{Z_2} & 0 & 1 & \\ \hline 0 & \frac{1}{4} & \frac{1}{2} & \frac{3}{4} \\ 1 & 0 & \frac{1}{4} & \frac{1}{4} \\ \hline & \frac{1}{4} & \frac{3}{4} & \\ \end{array}

\[E\left(Z_{1}\right)=\frac{1}{4}, E\left(Z_{2}\right)=\frac{3}{4}, E\left(Z_{1}^{2}\right)=\frac{1}{4}, E\left(Z_{2}^{2}\right)=\frac{3}{4}, E\left(Z_{1} Z_{2}\right)=\frac{1}{4}\]

\[D\left(Z_{1}\right)=E\left(Z_{1}^{2}\right)-E^{2}\left(Z_{1}\right)=\frac{1}{4}-\left(\frac{1}{4}\right)^{2}=\frac{3}{16},\]

\[D\left(Z_{2}\right)=E\left(Z_{2}^{2}\right)-E^{2}\left(Z_{2}\right)=\frac{3}{4}-\left(\frac{3}{4}\right)^{2}=\frac{3}{16}\]

\[Cov\left(Z_{1}, Z_{2}\right)=E\left(Z_{1} Z_{2}\right)-E\left(Z_{1}\right) E\left(Z_{2}\right)=\frac{1}{16}\]

\[\rho_{Z_{1} Z_{2}}=\frac{Cov\left(Z_{1}, Z_{2}\right)}{\sqrt{D\left(Z_{1}\right)} \sqrt{D\left(Z_{2}\right)}}=\frac{\frac{1}{16}}{3}=\frac{1}{3}\]

(I) \(P\{T>t\}=1-P\{T \leq t\}=1-F(t)=e^{-\left(\frac{t}{\theta}\right)^{m}}\)

\[P\{T>s+t | T>s\}=\frac{P\{T>s+t, T>s\}}{P\{T>s\}}=\frac{P\{T>s+t\}}{P\{T>s\}}=\frac{e^{-\left(\frac{s+t}{\theta}\right)^{m}}}{e^{-\left(s^{t}\right)^{m}}}=e^{\frac{s^{m}-(s+t)^{m}}{\theta^{m}}}\]

(II) \(f(t)=F'(t)=\left\{\begin{array}{l}e^{-\left(\frac{t}{\theta}\right)^{m}} \frac{m t^{m-1}}{\theta^{m}}, t>0 \\ 0, 其他 \end{array}\right.\)

似然函数:

\[L(\theta)=\prod_{i=1}^{n} e^{-\left(\frac{t_{i}}{\theta}\right)^{n}} \frac{m t_{i}^{m-1}}{\theta^{m}}=e^{-\frac{1}{\theta^{n}} \sum_{i=1}^{n} t_{i}^{m-1}} m^{n}\left(\prod_{i=1}^{n} t_{i}^{m-1}\right) \theta^{-n m},\left(t_{i}>0, i=1,2, \cdots, n\right)\]

\[ln L(\theta)=-\frac{1}{\theta^{m}} \sum_{i=1}^{n} t_{i}^{m}+ln m^{n}+ln \left(\prod_{i=1}^{n} t_{i}^{m-1}\right)-n m ln \theta\]

\[\frac{d ln L(\theta)}{d \theta}=\frac{m}{\theta^{m+1}} \sum_{i=1}^{n} t_{i}^{m}-\frac{n m}{\theta}=0 \Rightarrow \theta=\sqrt[m]{\frac{1}{n} \sum_{i=1}^{n} t_{i}^{m}}\]

所以 θ 的是大口格治计最为: \(\theta=\sqrt[m]{\frac{1}{n} \sum_{i=1}^{n} t_{i}^{m}}\)