【答案】 (D).

【解】 当$x \to 0^+$时,$\int_{0}^{x}(e^{t^2} - 1)dt \sim \int_{0}^{x}t^2 dt = \frac{1}{3}x^3$;

$\int_{0}^{x}\ln(1 + \sqrt{t^3})dt \sim \int_{0}^{x}t^{\frac{3}{2}} dt = \frac{2}{5}x^{\frac{5}{2}}$;

$\int_{0}^{\sin x}\sin t^2 dt \sim \int_{0}^{x}t^2 dt = \frac{1}{3}x^3$;

$\int_{0}^{1 - \cos x}\sqrt{\sin t^3}dt \sim \int_{0}^{\frac{1}{2}x^2} t^{\frac{3}{2}} dt = \frac{\sqrt{2}}{20}x^5$,

应选(D).

【解析】当 \(f(x)\) 在 \(x=0\) 处可导,则 \(f(x)\) 在 \(x=0\) 处连续, \(f(0)=\lim _{x \to 0} f(x)=0\) ,

\[f'(0)=\lim _{x \to 0} \frac{f(x)-f(0)}{x-0}=\lim _{x \to 0} \frac{f(x)}{x}, 可得 \lim _{x \to 0} \frac{f(x)}{\sqrt{|x|}}=\lim _{x \to 0} \frac{f(x)}{x} \cdot \frac{x}{\sqrt{|x|}} .\]

\[

\lim _{x \to 0^{+}} \frac{f(x)}{x} \cdot \frac{x}{\sqrt{|x|}}=\lim _{x \to 0^{+}} \frac{f(x)}{x} \cdot \frac{x}{\sqrt{x}}=\lim _{x \to 0^{+}} \frac{f(x)}{x} \cdot \sqrt{x}=f'(0) \cdot 0=0,

\]

\[

\lim _{x \to 0^{-}} \frac{f(x)}{x} \cdot \frac{x}{\sqrt{|x|}}=\lim _{x \to 0^{-}} \frac{f(x)}{x} \cdot \frac{x}{\sqrt{-x}}=\lim _{x \to 0^{-}} \frac{f(x)}{x} \cdot(-\sqrt{-x})=f'(0) \cdot 0=0,

\]

即 \(\lim _{x \to 0} \frac{f(x)}{\sqrt{|x|}}=0\) ,故选(C)

【解析】因为\(f(x,y)\)在\((0,0)\)处可微，\(f(0,0)=0\)，所以 \(\lim\limits_{\substack{x \to 0 \\ y \to 0}} \frac{f(x,y) - f(0,0) - f_{x}^{\prime}(0,0) \cdot x - f_{y}^{\prime}(0,0) \cdot y}{\sqrt{x^{2} + y^{2}}} = 0\) 即 \(\lim\limits_{\substack{x \to 0 \\ y \to 0}} \frac{f(x,y) - f_{x}^{\prime}(0,0) \cdot x - f_{y}^{\prime}(0,0) \cdot y}{\sqrt{x^{2} + y^{2}}} = 0\)

因为\(\boldsymbol{n} \cdot (x, y, f(x,y)) = f_{x}^{\prime}(0,0)x + f_{y}^{\prime}(0,0)y - f(x,y)\)，

所以\(\lim\limits_{(x,y) \to (0,0)} \frac{|\boldsymbol{n} \cdot (x, y, f(x,y))|}{\sqrt{x^{2} + y^{2}}} = 0\)存在，故选\((\text{A})\) 。

【解析】由题知当\(\vert r\vert \lt R\)时，幂级数\(\sum_{n = 1}^{\infty}a_{n}x^{n}\)收敛，所以当\(\vert r\vert \lt R\)时，\(\sum_{n = 1}^{\infty}a_{n}r^{2n}\)收敛，因其逆否命题也成立，所以若当\(\sum_{n = 1}^{\infty}a_{n}r^{2n}\)发散时，\(\vert r\vert \geq R\)，故选 (A) 。

【解析】 \(A\) 经初等列变换化成 \(B\) ,所以存在可逆矩阵 \(P_{1}\) 使得 \(A P_{1}=B\) ,所以 \(A=B P_{1}^{-1}\) . 令 \(P=P_{1}^{-1}\) ,所以 \(A=B P\) .故选(B).

【解析】令 \(L_{1}\) 的方程 \(\frac{x-a_{2}}{a_{1}}=\frac{y-b_{2}}{b_{1}}=\frac{z-c_{2}}{c_{1}}=t\) ,即有

\(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}a_2\\b_2\\c_2\end{pmatrix}+t\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}=\alpha_2 + t\alpha_1\)

\(L_{2}\) 方程为 \(\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}a_{3} \\ b_{3} \\ c_{3}\end{array}\right)+t\left(\begin{array}{l}a_{2} \\ b_{2} \\ c_{2}\end{array}\right)=\alpha_{3}+t \alpha_{2}\) ,由直线 \(L_{1}\) 与 \(L_{2}\) 相交得存在\(t\)使 \(\alpha_{2}+t \alpha_{1}=\alpha_{3}+t \alpha_{2}\) ,即 \(\alpha_{3}=t \alpha_{1}+(1-t) \alpha_{2}\) ,故\(\alpha_{3}\) 可由 \(\alpha_{1}\) , \(\alpha_{2}\) 线性表示,故应选(C).

【解析】\(P(A\overline{BC}) = P(A\overline{B\cup C}) = P(A) - P[A(B\cup C)]\)
\(= P(A) - P(AB + AC)\)
\(= P(A) + P(AB) - P(AC) + P(ABC)\)
\(=\frac{1}{4} - 0 - \frac{1}{12} + 0 = \frac{1}{6}\)，
\(P(C\overline{BA}) = P(C\overline{B\cup A}) = P(C) - P[C\cup (B\cup A)]\)
\(= P(C) + P(CB) - P(CA) + P(ABC)\)
\(=\frac{1}{4} - \frac{1}{12} - \frac{1}{12} + 0 = \frac{1}{12}\)，
\(P(A\overline{BC} + \overline{A}BC + \overline{A}\overline{B}C) = P(A\overline{BC}) + P(\overline{A}BC) + P(\overline{A}\overline{B}C)\)
\(=\frac{1}{6} + \frac{1}{6} + \frac{1}{12} = \frac{5}{12}\)。

故选 (D)。

【解析】由题意\(EX = \frac{1}{2}\)，\(DX = \frac{1}{4}\)，
\(E\left(\sum_{i = 1}^{100}X_{i}\right) = 100EX = 50\)，
\(D\left(\sum_{i = 1}^{100}X_{i}\right) = 100DX = 25\)。由中心极限定理\(\sum_{i = 1}^{100}X_{i} \sim N(50, 25)\)，所以
\(P\left\{\sum_{i = 1}^{100}X_{i} \leq 55\right\} = P\left\{\frac{\sum_{i = 1}^{100}X_{i} - 55}{5} \leq \frac{55 - 50}{5}\right\} = \varPhi(1)\)。故选\((B)\)。

【答案】\(-1\).

【解析】\(\lim\limits_{x \to 0} \left[ \frac{1}{\mathrm{e}^x - 1} - \frac{1}{\ln(1 + x)} \right]\)

\(=\lim\limits_{x \to 0} \frac{\ln(1 + x) - \mathrm{e}^x + 1}{(\mathrm{e}^x - 1)\ln(1 + x)}\)

\(=\lim\limits_{x \to 0} \frac{\ln(1 + x) - \mathrm{e}^x + 1}{x^2}\)

\(=\lim\limits_{x \to 0} \frac{\frac{1}{1 + x} - \mathrm{e}^x}{2x} = -1\).

\(\begin{aligned}
&\text{【解析】}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{1}{t+\sqrt{t^2 + 1}}\left(1 + \frac{t}{\sqrt{t^2 + 1}}\right)}{\frac{t}{\sqrt{t^2 + 1}}}=\frac{1}{t},\\
&\frac{d^2y}{dx^2}=\frac{d\left(\frac{dy}{dx}\right)}{dx}=\frac{\frac{d\left(\frac{dy}{dx}\right)}{dt}}{\frac{dx}{dt}}=\frac{-\frac{1}{t^2}}{\frac{t}{\sqrt{t^2 + 1}}}=-\frac{\sqrt{t^2 + 1}}{t^3},\\
&\text{得}\left.\frac{d^2y}{dx^2}\right|_{t = 1}=-\sqrt{2}.
\end{aligned}\)

【解析】特征方程为\(\lambda^{2}+a \lambda+1=0\)，特征根为\(\lambda_{1}\)，\(\lambda_{2}\)，则\(\lambda_{1}+\lambda_{2}=-a\)，\(\lambda_{1} \cdot \lambda_{2}=1\)，特征根\(\lambda_{1}<0\)，\(\lambda_{2}<0\)。

\[
\begin{aligned}
\int_{0}^{+\infty} f(x)\,\mathrm{d}x &= -\int_{0}^{+\infty} \left[ f''(x)+af'(x) \right]\mathrm{d}x \\
&= -\left[ f'(x)+af(x) \right]_{0}^{+\infty} \\
&= n + am
\end{aligned}
\]

【解析】\[
\begin{aligned}
\frac{\partial f}{\partial y} &= e^{x(xy)^{2}} \cdot x = x e^{x^{3}y^{2}}, \\
\frac{\partial^{2} f}{\partial x \partial y} &= \frac{\partial\left(\frac{\partial f}{\partial y}\right)}{\partial x} = e^{x^{3}y^{2}} + 3x^{3}y^{2}e^{x^{3}y^{2}}, \\
\text{所以 }\left.\frac{\partial^{2} f}{\partial x \partial y}\right|_{(1,1)} &= e + 3e = 4e.
\end{aligned}
\]

【解析】\[
\begin{aligned}
&\left|\begin{array}{cccc}
a & 0 & -1 & 1 \\
0 & a & 1 & -1 \\
-1 & 1 & a & 0 \\
1 & -1 & 0 & a
\end{array}\right| = \left|\begin{array}{cccc}
a & 0 & -1 & 1 \\
0 & a & 1 & -1 \\
-1 & 1 & a & 0 \\
0 & 0 & a & a
\end{array}\right| \\
&= \left|\begin{array}{cccc}
0 & a & -1+a^{2} & 1 \\
0 & a & 1 & -1 \\
-1 & 1 & a & 0 \\
1 & 0 & a & a
\end{array}\right| = -\left|\begin{array}{ccc}
a & -1+a^{2} & 1 \\
a & 1 & -1 \\
0 & a & a
\end{array}\right| \\
&= -\left|\begin{array}{ccc}
a & a^{2}-2 & 1 \\
a & 2 & -1 \\
0 & 0 & a
\end{array}\right| = a^{4}-4a^{2}.
\end{aligned}
\]

【解析】\[
\begin{align*}
f(x) &= 
\begin{cases}
\frac{1}{\pi} & -\frac{\pi}{2} < x < \frac{\pi}{2}, \\
0 & \text{其他}.
\end{cases} \\
\\
\text{所以 } \operatorname{cov}(X, Y) &= E[XY] - E[X]E[Y] \\
&= E[X \sin X] - E[X]E[\sin X] \\
&= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \sin x \cdot \frac{1}{\pi} \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\pi} x \, dx \cdot \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\pi} \sin x \, dx \\
&= 2 \cdot \frac{1}{\pi} \int_{0}^{\frac{\pi}{2}} x \sin x \, dx - 0 \\
&= \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} (-x) \, d(\cos x) \\
&= \frac{2}{\pi} \left( -x \cos x \bigg|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} \cos x \, dx \right) \\
&= \frac{2}{\pi} \left( 0 + \sin x \bigg|_{0}^{\frac{\pi}{2}} \right) \\
&= \frac{2}{\pi}.
\end{align*}
\]

求一阶导可得 \(\frac{\partial f}{\partial x}=3 x^{2}-y\)，\(\frac{\partial f}{\partial y}=24 y^{2}-x\)。令 \(\left\{\begin{array}{l}\frac{\partial f}{\partial x}=0\\\frac{\partial f}{\partial y}=0\end{array}\right.\) 可得 \(\left\{\begin{array}{l}x=\frac{1}{6}\\y=\frac{1}{12}\end{array}\right.\)。求二阶导可得 \(\frac{\partial^{2} f}{\partial x^{2}}=6 x\)，\(\frac{\partial^{2} f}{\partial x \partial y}=-1\)，\(\frac{\partial^{2} f}{\partial y^{2}}=48 y\)。当 \(x=0\)，\(y=0\) 时，\(A=0\)，\(B=-1\)，\(C=0\)，\(AC-B^{2}<0\)，故不是极值。当 \(x=\frac{1}{6}\)，\(y=\frac{1}{12}\) 时，\(A=1\)，\(B=-1\)，\(C=4\)，\(AC-B^{2}>0\) 且 \(A=1>0\)，故 \((\frac{1}{6}, \frac{1}{12})\) 是极小值点。极小值 \(f(\frac{1}{6}, \frac{1}{12})=(\frac{1}{6})^{3}+8(\frac{1}{12})^{3}-\frac{1}{6} \times \frac{1}{12}=-\frac{1}{216}\)。

设 \(P=\frac{4 x-y}{4 x^{2}+y^{2}}\)，\(Q=\frac{x+y}{4 x^{2}+y^{2}}\)，则 \(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}=\frac{-4 x^{2}+y^{2}-8 x y}{(4 x^{2}+y^{2})^{2}}\)。取路径 \(L_{\varepsilon}: 4 x^{2}+y^{2}=\varepsilon^{2}\)，方向为顺时针方向。则 \(\int_{L} \frac{4 x-y}{4 x^{2}+y^{2}} d x+\frac{x+y}{4 x^{2}+y^{2}} d y=\int_{L+L_{\varepsilon}} \frac{4 x-y}{4 x^{2}+y^{2}} d x+\frac{x+y}{4 x^{2}+y^{2}} d y-\int_{L_{\varepsilon}} \frac{4 x-y}{4 x^{2}+y^{2}} d x+\frac{x+y}{4 x^{2}+y^{2}} d y=\iint_{D}(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}) d x d y+\frac{1}{\varepsilon^{2}} \int_{L_{\varepsilon}}(4 x-y) d x+(x+y) d y=0+\frac{1}{\varepsilon^{2}} \cdot 2 \cdot \frac{\pi \varepsilon^{2}}{2}=\pi\)。

【解析】\(R = \lim\limits_{n \to \infty} \left|\frac{a_n}{a_{n + 1}}\right| = \lim\limits_{n \to \infty} \left|\frac{n + 1}{n + 2}\right| = 1\)，故\(|x| \lt R = 1\)时，\(\sum\limits_{n = 1}^{\infty} a_n x^n\)收敛。

令\(S(x) = \sum\limits_{n = 1}^{\infty} a_n x^n\)，

则\(S'(x) = \sum\limits_{n = 1}^{\infty} na_n x^{n - 1} = \sum\limits_{n = 0}^{\infty} (n + 1)a_{n + 1} x^n = a_1 + \sum\limits_{n = 1}^{\infty} \left(n + \frac{1}{2}\right)a_n x^n\)

\(= 1 + \sum\limits_{n = 1}^{\infty} na_n x^n + \frac{1}{2}\sum\limits_{n = 1}^{\infty} a_n x^n = 1 + xS'(x) + \frac{1}{2}S(x)\)，

则\(S'(x) - \frac{1}{2(1 - x)}S(x) = \frac{1}{1 - x}\)，故\(S(x) = e^{\int \frac{1}{2(1 - x)}dx} \left[\int e^{\int \frac{-1}{2(1 - x)}dx} \frac{1}{1 - x}dx + C\right] = \frac{C}{\sqrt{1 - x}} - 2\)，

由\(S(0) = C - 2 = 0\)，故\(C = 2\)，则\(S(x) = \frac{2}{\sqrt{1 - x}} - 2\)。

由题意 \(F(x, y, z)=z-\sqrt{x^{2}+y^{2}}\)，则 \(F_{x}'=-\frac{x}{\sqrt{x^{2}+y^{2}}}\)，\(F_{y}'=-\frac{y}{\sqrt{x^{2}+y^{2}}}\)，\(F_{z}'=1\)。\(I=\iint_{\sum}\left[[x f(x y)+2 x-y] \frac{F_{x}'}{F_{z}'}+[y f(x y)+2 y+x] \frac{F_{y}'}{F_{z}'}+z f(x y)+z\right] d x d y=\iint_{\sum}\left[[x f(x y)+2 x-y]\left(-\frac{x}{\sqrt{x^{2}+y^{2}}}\right)+[y f(x y)+2 y+x]\left(-\frac{y}{\sqrt{x^{2}+y^{2}}}\right)+z f(x y)+z\right] d x d y=-\iint_{\sum} \sqrt{x^{2}+y^{2}} d x d y=\iint_{D_{x y}} \sqrt{x^{2}+y^{2}} d x d y=\int_{0}^{2 \pi} d \theta \int_{1}^{2} r^{2} d r=\frac{14}{3} \pi\)。

(I) 设 \(x=x_{0}\) 时 \(|f(x_{0})|=M\)，由拉格朗日定理得：\(\exists \xi_{1} \in(0, x_{0})\) 使得 \(|f'(\xi_{1})|=|\frac{f(x_{0})-f(0)}{x_{0}-0}|=\frac{|f(x_{0})|}{x_{0}}=\frac{M}{x_{0}}\)；\(\exists \xi_{2} \in(x_{0}, 2)\) 使得 \(|f'(\xi_{2})|=|\frac{f(x_{0})-f(2)}{x_{0}-2}|=\frac{|f(x_{0})|}{2-x_{0}}=\frac{M}{2-x_{0}}\)。若 \(x_{0} \in(0,1]\)，取 \(\xi=\xi_{1}\)，有 \(|f'(\xi)|=\frac{M}{x_{0}} \geq M\)；若 \(x_{0} \in(1,2)\)，取 \(\xi=\xi_{2}\)，有 \(|f'(\xi)|=\frac{M}{2-x_{0}} \geq M\)。

(II) \(M=|f(x_{0})|=|f(x_{0})-f(0)|=|\int_{0}^{x_{0}} f'(x) d x| \leq \int_{0}^{x_{0}}|f'(x)| d x \leq \int_{0}^{x_{0}} M d x = M x_{0}\)；\(M=|f(x_{0})|=|f(2)-f(x_{0})|=|\int_{x_{0}}^{2} f'(x) d x| \leq \int_{x_{0}}^{2}|f'(x)| d x \leq \int_{x_{0}}^{2} M d x = M(2-x_{0})\)，故 \(\begin{cases}M(1-x_{0}) \leq 0 \\ M(1-x_{0}) \geq 0\end{cases}\) 同时成立。若 \(x_{0} \neq 1\)，显然 \(M=0\)；若 \(x_{0}=1\)，\(M \leq \int_{0}^{1}|f'(x)| d x \leq M\) 且 \(M \leq \int_{1}^{2}|f'(x)| d x \leq M\)，当且仅当区间 \((0,1)\) 和 \((1,2)\) 上 \(|f'(x)| \equiv M\)，若 \(M \neq 0\)，则 \(f'(x) \equiv \pm M\) 与 \(f(0)=f(2)=0\) 矛盾，或 \(f'(x)\) 在 \(x=1\) 处不可导，与已知矛盾，故 \(M=0\)。

(I) 记 \(A=(\begin{array}{cc}1 & -2 \\ -2 & 4\end{array})\)，\(B=(\begin{array}{cc}a & 2 \\ 2 & b\end{array})\)，则 \(Q^{T} A Q=B\)，\(A\) 与 \(B\) 相似，有相同特征值。\(|\lambda E-A|=\lambda^{2}-5 \lambda=0\)，特征值为 \(\lambda_{1}=0\)，\(\lambda_{2}=5\)。\(|\lambda E-B|=\lambda^{2}-(a+b) \lambda+ab-4=0\)，故 \(a+b=5\)，\(ab=4\)，又 \(a \geq b\)，所以 \(a=4\)，\(b=1\)。

(II) 当 \(\lambda_{1}=0\) 时，\((0 E-A) x=0\) 的基础解系为 \(\alpha_{1}=(2,1)^{T}\)；当 \(\lambda_{2}=5\) 时，基础解系为 \(\alpha_{2}=(1,-2)^{T}\)。当 \(\lambda_{1}=0\) 时，\((0 E-B) x=0\) 的基础解系为 \(\beta_{1}=(1,-2)^{T}\)；当 \(\lambda_{2}=5\) 时，基础解系为 \(\beta_{2}=(2,1)^{T}\)。令 \(P_{1}=(\begin{array}{cc}2 & 1 \\ 1 & -2\end{array})\)，\(P_{2}=(\begin{array}{cc}1 & 2 \\ -2 & 1\end{array})\)，则 \(P_{1}^{-1} A P_{1}=P_{2}^{-1} B P_{2}=(\begin{array}{ll}0 & \\ & 5\end{array})\)，所以 \(B=P_{2} P_{1}^{-1} A P_{1} P_{2}^{-1}\)，计算得 \(P_{1} P_{2}^{-1}=\frac{1}{5}(\begin{array}{cc}4 & -3 \\ -3 & -4\end{array})\)，故正交矩阵 \(Q=\frac{1}{5}(\begin{array}{cc}4 & -3 \\ -3 & -4\end{array})\)。

(I) 法1：假设 \(P\) 不可逆，则 \(\alpha\) 与 \(A \alpha\) 线性相关，存在不全为0的实数 \(k_{1}\)，\(k_{2}\)，使得 \(k_{1} \alpha+k_{2} A \alpha=0\)。若 \(k_{2}=0\)，则 \(k_{1}=0\)，矛盾，故 \(k_{2} \neq 0\)，从而 \(A \alpha=-\frac{k_{1}}{k_{2}} \alpha\)，与 \(\alpha\) 不是特征向量矛盾，所以 \(P\) 可逆。

法2：设 \(A \alpha=\beta\)，则 \(\beta \neq k \alpha\)，即 \(\alpha\)，\(\beta\) 线性无关，故 \(P=(\alpha, \beta)\) 可逆。

(II) 由 \(A^{2} \alpha+A \alpha-6 \alpha=0\)，得 \(A P=A(\alpha, A \alpha)=(A \alpha, A^{2} \alpha)=(A \alpha, 6 \alpha-A \alpha)=(\alpha, A \alpha)(\begin{array}{cc}0 & 6 \\ 1 & -1\end{array})=P(\begin{array}{cc}0 & 6 \\ 1 & -1\end{array})\)，故 \(P^{-1} A P=(\begin{array}{cc}0 & 6 \\ 1 & -1\end{array})\)。令 \(B=(\begin{array}{cc}0 & 6 \\ 1 & -1\end{array})\)，\(|\lambda E-B|=\lambda^{2}+\lambda-6=0\)，特征值为 \(\lambda_{1}=2\)，\(\lambda_{2}=-3\)，\(A\) 与 \(B\) 相似，故 \(A\) 的特征值为2和-3，可相似对角化。

(I) \((X_{1}, Y)\) 的分布函数为:

\[
\begin{aligned}
F(x, y) &= P\left\{X_{1} \leq x, Y \leq y\right\} \\
&= P\left\{X_{3}=0, X_{1} \leq x, Y \leq y\right\} + P\left\{X_{3}=1, X_{1} \leq x, Y \leq y\right\} \\
&= P\left\{X_{3}=0, X_{1} \leq x, X_{2} \leq y\right\} + P\left\{X_{3}=1, X_{1} \leq x, X_{1} \leq y\right\} \\
&= \frac{1}{2} P\left\{X_{1} \leq x\right\} P\left\{X_{2} \leq y\right\} + \frac{1}{2} P\left\{X_{1} \leq x, X_{1} \leq y\right\}
\end{aligned}
\]

(1) \(x \leq y\) 时, \(F(x, y)=\frac{1}{2} P\left\{X_{1} \leq x\right\} P\left\{X_{2} \leq y\right\}+\frac{1}{2} P\left\{X_{1} \leq x\right\}=\frac{1}{2} \Phi(x) \Phi(y)+\frac{1}{2} \Phi(x)\)

(2) \(x>y\) 时, \(F(x, y)=\frac{1}{2} P\left\{X_{1} \leq x\right\} P\left\{X_{2} \leq y\right\}+\frac{1}{2} P\left\{X_{1} \leq y\right\}=\frac{1}{2} \Phi(x) \Phi(y)+\frac{1}{2} \Phi(y)\)

综上所述: \(F(x, y)=\left\{\begin{array}{l}\frac{1}{2} \Phi(x) \Phi(y)+\frac{1}{2} \Phi(x), x \leq y \\ \frac{1}{2} \Phi(x) \Phi(y)+\frac{1}{2} \Phi(y), x>y \end{array}\right.\)

(II) 

\begin{aligned}
F_{Y}(y) & = P\{Y \leq y\} = P\left\{X_{3}=0, Y \leq y\right\} + P\left\{X_{3}=1, Y \leq y\right\} \\
& = P\left\{X_{3}=0, X_{3} X_{1} + (1-X_{3}) X_{2} \leq y\right\} + P\left\{X_{3}=1, X_{3} X_{1} + (1-X_{3}) X_{2} \leq y\right\} \\
& = P\left\{X_{3}=0, X_{2} \leq y\right\} + P\left\{X_{3}=1, X_{1} \leq y\right\} = \frac{1}{2} P\left\{X_{2} \leq y\right\} + \frac{1}{2} P\left\{X_{1} \leq y\right\} = \Phi(y),
\end{aligned}

\(\therefore Y \sim N(0,1)\)

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

\begin{aligned}
P\{T > s + t \mid 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(\frac{s}{\theta}\right)^{m}}} \\
&= e^{\frac{s^{m} - (s + t)^{m}}{\theta^{m}}} 。
\end{aligned}

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

\begin{aligned}
\ln L(\theta) &= -\frac{1}{\theta^{m}} \sum_{i=1}^{n} t_{i}^{m} + \ln m^{n} \\
&\quad + \ln \left(\prod_{i=1}^{n} t_{i}^{m-1}\right) - n m \ln \theta \\
\end{aligned}

\begin{aligned}
\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}}
\end{aligned}

所以 \(\theta\) 的最大似然估计值为 \(\hat{\theta}=\sqrt[m]{\frac{1}{n} \sum_{i=1}^{n} t_{i}^{m}}\)。