\(x-\tan x=x-(x+\frac{1}{3} x^{3}+o(x^{3})) \sim-\frac{1}{3} x^{3}\) ,故 \(k=3\)

\(\lim _{x \to 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \to 0^{-}} \frac{x|x|}{x}=0\) ，\(\lim _{x \to 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \to 0^{+}} \frac{x \ln x}{x}=-\infty\) ，故 \(f(x)\) 不可导.

当 \(x>0\) 时, \(f(x)<0\) ;当 \(x<0\) 时, \(f(x)<0\) .故 \(f(x)\) 在 \(x=0\) 处取极大值.故选(B).

【解析】举反例：（A）\(u_{n}=\frac{n - 1}{n}\)（B）\(u_{n}=\frac{n - 1}{n}\)（C）\(u_{n}=-\frac{1}{n}\)

由 \(\frac{\partial Q}{\partial x}=\frac{\partial P}{\partial y}\) 得 \(\frac{\partial P}{\partial y}=\frac{1}{y^{2}}\) ，逐项验证选项，只有D满足（原题解析中“又上我有合 \(\frac{1}{x}\) ,我(焦,这()。”可能为排版错误，不影响结论）

\(\because A^{2}+A=2 E\) ,设 A 的特征值为 \(\lambda\) ，则 \(\lambda^{2}+\lambda=2\) ，即 \((\lambda+2)(\lambda-1)=0\) ，\(\therefore \lambda=-2\) 或 \(1\) 。

\(\because|A|=4\) ，故 A 的特征值为 \(\lambda_{1}=\lambda_{2}=-2\) , \(\lambda_{3}=1\) ，正惯性指数 \(p=1\) ，负惯性指数 \(q=2\) ，规范形为 \(y_{1}^{2}-y_{2}^{2}-y_{3}^{2}\) 。

(1) 由于三张平面无公共交点，方程组无解，故 \(r(A)

(2) 任意两张平面交线平行，说明系数矩阵中任意两行向量线性无关，故 \(r(A)=2\) ，增广矩阵秩 \(r(\bar{A})=3\) ，综上选A。

A选项 \(\Leftrightarrow P(AB)=0\) ，排除；

B选项 \(\Leftrightarrow A\) 、\(B\) 独立，排除；

C选项 \(\Leftrightarrow P(A)-P(AB)=P(B)-P(AB)\) ，即 \(P(A)=P(B)\) ，正确；

D选项 \(\Leftrightarrow 1=P(A)+P(B)\) ，排除。

X、Y独立，\(Z=X-Y \sim N(0, 2\sigma^{2})\) ，则

\[P(|X-Y|<1)=P(-1

【解析】\(\frac{\partial z}{\partial x} = f'(\sin y - \sin x)(-\cos x) + y\)

\(\frac{\partial z}{\partial y} = f'(\sin y - \sin x)\cos y + x\)

故

\[

\begin{aligned}

& \frac{1}{\cos x} \cdot \frac{\partial z}{\partial x} + \frac{1}{\cos y} \cdot \frac{\partial z}{\partial y} = -f'(\sin y - \sin x) + \frac{y}{\cos x} + f'(\sin y - \sin x) + \frac{x}{\cos y} \\

& = \frac{y}{\cos x} + \frac{x}{\cos y}

\end{aligned}

\]

【解析】\(y' = \frac{2 + y^{2}}{2y}\)

\[

\int \frac{2y}{2 + y^{2}} dy = \int 1 dx \quad \text{故} \quad \ln(2 + y^{2}) = x + C.

\]

由 \(y(0) = 1\) 得 \(C = \ln 3\)

则 \(\ln(2 + y^{2}) = x + \ln 3\)，故 \(e^{\ln(2 + y^{2})} = e^{x + \ln 3}\)

即 \(2 + y^{2} = 3e^{x}\)，故 \(y = \sqrt{3e^{x} - 2}\)

【解析】\(\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}(\sqrt{x})^{2n} = \cos \sqrt{x}\)

【解析】

\[

\begin{aligned}

& \iint_{\sum} \sqrt{4 - x^{2} - 4z^{2}} dxdy = \iint_{D_{xy}} |y| dxdy = 4 \iint_{D'} y dxdy = \\

& = 4 \int_{0}^{\frac{\pi}{2}} d\theta \int_{0}^{2} r^{2} \sin \theta dr = \frac{32}{3}

\end{aligned}

\]

【解析】：\(\alpha_{1}\)，\(\alpha_{2}\) 无关，\(A = (\alpha_{1}, \alpha_{2}, \alpha_{3})\)

\[

\therefore r(A) \geq 2

\]

\[

\because \alpha_{3} = -\alpha_{1} + 2\alpha_{2}

\]

\[

\therefore r(A) = 2

\]

故 \(Ax = 0\) 的基础解系中有 \(n - r(A) = 3 - 2 = 1\) 个解向量。

\[

\because \alpha_{1} - 2\alpha_{2} + \alpha_{3} = 0

\]

\(\therefore (1, -2, 1)^{T}\) 为 \(Ax = 0\) 的解，作为基础解系。

\(\therefore\) 通解为 \(x = k(1, -2, 1)^{T}\)，\(k \in R\)

【解析】\(f(x) = \begin{cases} \frac{x}{2}, & 0 < x < 2 \\ 0, & \text{其他} \end{cases}\)

\[

E(X) = \int_{-\infty}^{+\infty} x f(x) dx = \int_{0}^{2} \frac{x^{2}}{2} dx = \left. \frac{1}{6}x^{3} \right|_{0}^{2} = \frac{4}{3}

\]

\[

F(x) = \int_{-\infty}^{x} f(t) dt = \begin{cases} 0, & x < 0 \\ \int_{0}^{x} \frac{t}{2} dt, & 0 \leq x < 2 \\ 1, & x \geq 2 \end{cases}

\]

\[

\begin{aligned}

& P(F(X) > E(X) - 1) = P\left( F(X) > \frac{1}{3} \right) \\

& = P\left( \frac{x^{2}}{4} > \frac{1}{3} \right) = P\left( x > \frac{2}{\sqrt{3}} \right) = \int_{\frac{2}{\sqrt{3}}}^{2} \frac{x}{2} dx = \frac{2}{3}

\end{aligned}

\]

【解】(Ⅰ)$y'+xy = \mathrm{e}^{-\frac{x^2}{2}}$的通解为
$y = \left( \int \mathrm{e}^{-\frac{x^2}{2}} \cdot \mathrm{e}^{\int x \mathrm{d}x} \mathrm{d}x + C \right) \mathrm{e}^{-\int x \mathrm{d}x} = (x + C)\mathrm{e}^{-\frac{x^2}{2}},$
由$y(0) = 0$得$C = 0$，故$y = x\mathrm{e}^{-\frac{x^2}{2}}$.

(Ⅱ)$y' = (1 - x^2)\mathrm{e}^{-\frac{x^2}{2}}$，$y'' = (x^3 - 3x)\mathrm{e}^{-\frac{x^2}{2}} = x(x + \sqrt{3})(x - \sqrt{3})\mathrm{e}^{-\frac{x^2}{2}},$
令$y'' = 0$得$x = -\sqrt{3}$，$x = 0$，$x = \sqrt{3},$
当$x \in (-\infty, -\sqrt{3})$时，$y'' < 0$；当$x \in (-\sqrt{3}, 0)$时，$y'' > 0$；当$x \in (0, \sqrt{3})$时，$y'' < 0$；
当$x \in (\sqrt{3}, +\infty)$时，$y'' > 0$，
故$y = x\mathrm{e}^{-\frac{x^2}{2}}$的凸区间为$(-\infty, -\sqrt{3})$及$(0, \sqrt{3})$；凹区间为$(-\sqrt{3}, 0)$及$(\sqrt{3}, +\infty)$，
曲线$y = x\mathrm{e}^{-\frac{x^2}{2}}$的拐点为$(-\sqrt{3}, -\sqrt{3}\mathrm{e}^{-\frac{3}{2}})$，$(0, 0)$及$(\sqrt{3}, \sqrt{3}\mathrm{e}^{-\frac{3}{2}})$.

【解】（Ⅰ）\(\text{grad}\ z = \{2ax, 2by\}\)，\(\text{grad}\ z\big|_{(3,4)} = \{6a, 8b\}\)，
因为梯度的方向即为方向导数最大的方向，
所以有\(\frac{6a}{-3} = \frac{8b}{-4}\)，即\(a = b\)，
再由\(\sqrt{36a^2 + 64b^2} = 10\)得\(a = b = -1\)．
（Ⅱ）曲面\(\Sigma: z = 2 - x^2 - y^2\)，\((x, y) \in D_{xy}\)，其中\(D_{xy} = \{(x, y) \mid x^2 + y^2 \leq 2\}\)，
则曲面\(\Sigma\)的面积为
\[
\begin{align*}
S &= \iint_{D_{xy}} \sqrt{1 + z_x'^2 + z_y'^2} \, dxdy = \iint_{D_{xy}} \sqrt{1 + 4x^2 + 4y^2} \, dxdy \\
&= 2\pi \int_{0}^{\sqrt{2}} r \sqrt{1 + 4r^2} \, dr = \frac{\pi}{4} \int_{0}^{\sqrt{2}} (1 + 4r^2)^{\frac{1}{2}} d(1 + 4r^2) \\
&= \frac{\pi}{4} \times \frac{2}{3}(1 + 4r^2)^{\frac{3}{2}} \bigg|_{0}^{\sqrt{2}} = \frac{\pi}{6}(27 - 1) = \frac{13}{3}\pi.
\end{align*}
\]

【解】 所求的面积为
$A = \int_{0}^{+\infty} \text{e}^{-x}|\sin x| \text{d}x$
$= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (-1)^k \int_{k\pi}^{(k+1)\pi} \text{e}^{-x}\sin x \text{d}x$
$= \lim\limits_{n \to \infty} \sum_{k=0}^{n} (-1)^k \left[ -\frac{1}{2}\text{e}^{-x}(\sin x + \cos x) \right] \bigg|_{k\pi}^{(k+1)\pi}$
$= \frac{1}{2}\lim\limits_{n \to \infty} \sum_{k=0}^{n} (-1)^{k+1} \left[ \text{e}^{-(k+1)\pi}(-1)^{k+1} - \text{e}^{-k\pi}(-1)^k \right]$
$= \frac{1}{2}\lim\limits_{n \to \infty} \sum_{k=0}^{n} \left[ \text{e}^{-(k+1)\pi} + \text{e}^{-k\pi} \right] = \frac{1}{2}\lim\limits_{n \to \infty} \left[ 1 + 2\sum_{k=1}^{n} \text{e}^{-k\pi} + \text{e}^{-(n+1)\pi} \right]$
$= \frac{1}{2}\left( 1 + 2\sum_{k=1}^{\infty} \text{e}^{-k\pi} \right) = \frac{1}{2}\left( 1 + \frac{2\text{e}^{-\pi}}{1 - \text{e}^{-\pi}} \right) = \frac{1}{2}\left( 1 + \frac{2}{\text{e}^{\pi} - 1} \right) = \frac{1}{2} + \frac{1}{\text{e}^{\pi} - 1}$.

（Ⅰ）【证明】 因为当\( 0 \leq x \leq 1 \)时，\( x^{n + 1} \sqrt{1 - x^{2}} \leq x^{n} \sqrt{1 - x^{2}} \)，
所以\( \int_{0}^{1} x^{n + 1} \sqrt{1 - x^{2}} \mathrm{d}x < \int_{0}^{1} x^{n} \sqrt{1 - x^{2}} \mathrm{d}x \)，即\( a_{n + 1} < a_{n} \)，故\(\{ a_{n} \}\)单调递减．
\( a_{n} \stackrel{x = \sin t}{=} \int_{0}^{\frac{\pi}{2}} \sin^{n} t \cdot \cos^{2} t \mathrm{d}t = \int_{0}^{\frac{\pi}{2}} (\sin^{n} t - \sin^{n + 2} t) \mathrm{d}t \)
\( = \int_{0}^{\frac{\pi}{2}} \sin^{n} t \mathrm{d}t - \int_{0}^{\frac{\pi}{2}} \sin^{n + 2} t \mathrm{d}t = I_{n} - \frac{n + 1}{n + 2} I_{n} = \frac{1}{n + 2} I_{n} \)，
\( a_{n - 2} \stackrel{x = \sin t}{=} \int_{0}^{\frac{\pi}{2}} \sin^{n - 2} t \cdot \cos^{2} t \mathrm{d}t = \int_{0}^{\frac{\pi}{2}} (\sin^{n - 2} t - \sin^{n} t) \mathrm{d}t \)
\( = \int_{0}^{\frac{\pi}{2}} \sin^{n - 2} t \mathrm{d}t - \int_{0}^{\frac{\pi}{2}} \sin^{n} t \mathrm{d}t = I_{n - 2} - I_{n} \)，
因为\( I_{n} = \frac{n - 1}{n} I_{n - 2} \)，所以\( I_{n - 2} = \frac{n}{n - 1} I_{n} \)，
于是\( a_{n - 2} = \frac{n}{n - 1} I_{n} - I_{n} = \frac{1}{n - 1} I_{n} \)，故\( a_{n} = \frac{n - 1}{n + 2} a_{n - 2} (n = 2, 3, \cdots) \)．

（Ⅱ）【解】 因为\(\{ a_{n} \}\)单调递减，所以\( a_{n} = \frac{n - 1}{n + 2} a_{n - 2} > \frac{n - 1}{n + 2} a_{n - 1} \)，
从而有\( \frac{n - 1}{n + 2} < \frac{a_{n}}{a_{n - 1}} < 1 \)，由夹逼定理得\( \lim\limits_{n \to \infty} \frac{a_{n}}{a_{n - 1}} = 1 \)．

【解】 设Ω的形心坐标为\( (\bar{x},\bar{y},\bar{z}) \)，
由对称性得\( \bar{x} = 0 \)，且\( \bar{y} = \frac{\iiint_{\Omega} y \, dxdy dz}{\iiint_{\Omega} dxdy dz} \)，\( \bar{z} = \frac{\iiint_{\Omega} z \, dxdy dz}{\iiint_{\Omega} dxdy dz} \)，
\( \iiint_{\Omega} dxdy dz = \int_{0}^{1} dz \iint_{x^2 + (y - z)^2 \leq (1 - z)^2} dxdy = \pi \int_{0}^{1} (1 - z)^2 dz = \frac{\pi}{3}(z - 1)^3 \big|_{0}^{1} = \frac{\pi}{3} \)；
\( \iiint_{\Omega} y \, dxdy dz = \int_{0}^{1} dz \iint_{x^2 + (y - z)^2 \leq (1 - z)^2} y \, dxdy \)，
由\( \iint_{x^2 + (y - z)^2 \leq (1 - z)^2} y \, dxdy \stackrel{y = u + z}{=} \iint_{x^2 + u^2 \leq (1 - z)^2} (u + z) \, dxdu \)
\( = \iint_{x^2 + u^2 \leq (1 - z)^2} z \, dxdu = \pi z (1 - z)^2 \)得
\( \iiint_{\Omega} y \, dxdy dz = \pi \int_{0}^{1} z (1 - z)^2 dz = \frac{\pi}{12} \)；
\( \iiint_{\Omega} z \, dxdy dz = \int_{0}^{1} z dz \iint_{x^2 + (y - z)^2 \leq (1 - z)^2} dxdy = \pi \int_{0}^{1} z (1 - z)^2 dz = \frac{\pi}{12} \)，
故Ω的形心坐标为\( (0, \frac{1}{4}, \frac{1}{4}) \)。

（Ⅰ）【解】 由题意得$b\alpha_1 + c\alpha_2 + \alpha_3 = \beta$，即$\begin{cases} b + c + 1 = 1, \\ 2b + 3c + a = 1, \\ b + 2c + 3 = 1, \end{cases}$
解得$a=3$，$b=2$，$c=-2$。

（Ⅱ）【证明】 因为$|\alpha_2,\alpha_3,\beta| = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 3 & 1 \\ 2 & 3 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 0 & 0 & -2 \\ 0 & 1 & -1 \end{vmatrix} = 2 \neq 0$，所以$\alpha_2,\alpha_3,\beta$线性无关，故$\alpha_2,\alpha_3,\beta$为$\mathbb{R}^3$的一个基。
设由$\alpha_2,\alpha_3,\beta$到$\alpha_1,\alpha_2,\alpha_3$的过渡矩阵为$Q$，即$(\alpha_1,\alpha_2,\alpha_3) = (\alpha_2,\alpha_3,\beta)Q$，于是$Q = (\alpha_2,\alpha_3,\beta)^{-1}(\alpha_1,\alpha_2,\alpha_3)$，
由$\begin{pmatrix} 1 & 1 & 1 & \vdots & 1 & 0 & 0 \\ 3 & 3 & 1 & \vdots & 0 & 1 & 0 \\ 2 & 3 & 1 & \vdots & 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 & \vdots & 1 & 0 & 0 \\ 0 & 0 & -2 & \vdots & -3 & 1 & 0 \\ 0 & 1 & -1 & \vdots & -2 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 1 & \vdots & 1 & 0 & 0 \\ 0 & 1 & -1 & \vdots & -2 & 0 & 1 \\ 0 & 0 & 1 & \vdots & \frac{3}{2} & -\frac{1}{2} & 0 \end{pmatrix}$
$\to \begin{pmatrix} 1 & 1 & 0 & \vdots & -\frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 1 & 0 & \vdots & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 0 & 1 & \vdots & \frac{3}{2} & -\frac{1}{2} & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & \vdots & 0 & 1 & -1 \\ 0 & 1 & 0 & \vdots & -\frac{1}{2} & -\frac{1}{2} & 1 \\ 0 & 0 & 1 & \vdots & \frac{3}{2} & -\frac{1}{2} & 0 \end{pmatrix}$得
$(\alpha_2,\alpha_3,\beta)^{-1} = \begin{pmatrix} 0 & 1 & -1 \\ -\frac{1}{2} & -\frac{1}{2} & 1 \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{pmatrix}$，则
$Q = \begin{pmatrix} 0 & 1 & -1 \\ -\frac{1}{2} & -\frac{1}{2} & 1 \\ \frac{3}{2} & -\frac{1}{2} & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 0 \\ -\frac{1}{2} & 0 & 1 \\ \frac{1}{2} & 1 & 0 \end{pmatrix}$。

【解】(Ⅰ)因为\( A \sim B \)，所以\( \text{tr} A = \text{tr} B \)，即\( x - 4 = y + 1 \)，或\( y = x - 5 \)，
再由\( |A| = |B| \)得\( -2(-2x + 4) = -2y \)，即\( y = -2x + 4 \)，
解得\( x = 3 \)，\( y = -2 \).

(Ⅱ)\( A = \begin{pmatrix} -2 & -2 & 1 \\ 2 & 3 & -2 \\ 0 & 0 & -2 \end{pmatrix} \)，\( B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix} \)，
显然矩阵\( A,B \)的特征值为\( \lambda_1 = -2 \)，\( \lambda_2 = -1 \)，\( \lambda_3 = 2 \)，
由\( 2E + A \to \begin{pmatrix} 0 & -2 & 1 \\ 2 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & \frac{1}{4} \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0 \end{pmatrix} \)得\( A \)的属于特征值\( \lambda_1 = -2 \)的特征向量为
\( \alpha_1 = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix} \)；

由\( E + A \to \begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \)得\( A \)的属于特征值\( \lambda_2 = -1 \)的特征向量为
\( \alpha_2 = \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix} \)；

由\( 2E - A \to \begin{pmatrix} 2 & 1 & -2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & \frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \)得\( A \)的属于特征值\( \lambda_3 = 2 \)的特征向量为
\( \alpha_3 = \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix} \)，

令\( P_1 = \begin{pmatrix} -1 & -2 & -1 \\ 2 & 1 & 2 \\ 4 & 0 & 0 \end{pmatrix} \)，则\( P_1^{-1}AP_1 = \begin{pmatrix} -2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \)；

由\( 2E + B \to \begin{pmatrix} 4 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \)得\( B \)的属于特征值\( \lambda_1 = -2 \)的特征向量为\( \beta_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \)；

由\( E + B \to \begin{pmatrix} 3 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & \frac{1}{3} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \)得\( B \)的属于特征值\( \lambda_2 = -1 \)的特征向量为
\( \beta_2 = \begin{pmatrix} -1 \\ 3 \\ 0 \end{pmatrix} \)；

由\( 2E - B \to \begin{pmatrix} 0 & -1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{pmatrix} \to \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \)得\( B \)的属于特征值\( \lambda_3 = 2 \)的特征向量为\( \beta_3 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \)，

令\( P_2 = \begin{pmatrix} 0 & -1 & 1 \\ 0 & 3 & 0 \\ 1 & 0 & 0 \end{pmatrix} \)，则\( P_2^{-1}BP_2 = \begin{pmatrix} -2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \)

由\( P_1^{-1}AP_1 = P_2^{-1}BP_2 \)得\( (P_1P_2^{-1})^{-1}A(P_1P_2^{-1}) = B \)，

故\( P = P_1P_2^{-1} = \begin{pmatrix} -1 & -1 & -1 \\ 2 & 1 & 2 \\ 0 & 0 & 4 \end{pmatrix} \).

【解】(Ⅰ)因为$X\sim E(1)$，所以X的分布函数为$F(x)=\begin{cases}1-e^{-x},\ &x\geq0,\\0,\ &x<0.\end{cases}$
$F_Z(z)=P\{XY\leq z\}=P\{Y=-1\}P\{XY\leq z\mid Y=-1\}+P\{Y=1\}P\{XY\leq z\mid Y=1\}$
$=pP\{-X\leq z\}+(1-p)P\{X\leq z\}=pP\{X\geq -z\}+(1-p)P\{X\leq z\}$
$=p[1-P\{X\leq -z\}]+(1-p)P\{X\leq z\}=p[1-F(-z)]+(1-p)F(z),$
当$z<0$时，$F_Z(z)=pe^z$；
当$z\geq0$时，$F_Z(z)=p+(1-p)(1-e^{-z})$，
故
$f_Z(z)=\begin{cases}pe^z,\ &z<0,\\(1-p)e^{-z},\ &z\geq0.\end{cases}$

(Ⅱ)$\text{Cov}(X,Z)=\text{Cov}(X,XY)=E(X^2Y)-E(X)\cdot E(XY)$
$=E(X^2)E(Y)-[E(X)]^2E(Y)=D(X)\cdot E(Y),$
因为$X\sim E(1)$，所以$E(X)=1$，$D(X)=1$，
又因为$Y\sim\begin{pmatrix}-1&1\\p&1-p\end{pmatrix}$，所以$E(Y)=(-1)p+(1-p)=1-2p$，
X与Z不相关的充分必要条件是$\text{Cov}(X,Z)=0$，
故当$p=\frac{1}{2}$时，X与Z不相关.

(Ⅲ)设$F(x,y)$为$(X,Z)$的联合分布函数，
$F(1,1)=P\{X\leq1,Z\leq1\}=P\{X\leq1,XY\leq1\}$
$=P\{Y=-1\}P\{X\leq1,XY\leq1\mid Y=-1\}+P\{Y=1\}P\{X\leq1,XY\leq1\mid Y=1\}$
$=\frac{1}{2}P\{X\leq1,-X\leq1\}+\frac{1}{2}P\{X\leq1\}=\frac{1}{2}P\{-1\leq X\leq1\}+\frac{1}{2}P\{X\leq1\}$
$=P\{X\leq1\}=F(1)=1-\frac{1}{e},$
$F_X(1)=P\{X\leq1\}=1-\frac{1}{e},$
$F_Z(1)=P\{XY\leq1\}=P\{Y=-1\}P\{XY\leq1\mid Y=-1\}+P\{Y=1\}P\{XY\leq1\mid Y=1\}$
$=\frac{1}{2}P\{-X\leq1\}+\frac{1}{2}P\{X\leq1\}=\frac{1}{2}P\{X\geq-1\}+\frac{1}{2}P\{X\leq1\}$
$=\frac{1}{2}+\frac{1}{2}(1-\frac{1}{e})=1-\frac{1}{2e},$
因为$F(1,1)\neq F_X(1)\cdot F_Z(1)$，所以X与Z不相互独立.

【解】（Ⅰ）由归一性得
\[
\begin{align*}
1&=\int_{\mu}^{+\infty}\frac{A}{\sigma}\mathrm{e}^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}\mathrm{d}x = A\int_{\mu}^{+\infty}\mathrm{e}^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}\mathrm{d}\left(\frac{x-\mu}{\sigma}\right)=A\int_{0}^{+\infty}\mathrm{e}^{-\frac{x^{2}}{2}}\mathrm{d}x\\
&=\sqrt{2\pi}A\int_{0}^{+\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^{2}}{2}}\mathrm{d}x = \frac{\sqrt{2\pi}}{2}A\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^{2}}{2}}\mathrm{d}x = \frac{\sqrt{2\pi}}{2}A,
\end{align*}
\]
解得\(A = \sqrt{\frac{2}{\pi}}\)。

（Ⅱ）\(L(\sigma^{2})=\frac{A^{n}}{(\sigma^{2})^{\frac{n}{2}}}\mathrm{e}^{-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}(x_{i}-\mu)^{2}}\)，
\[
\ln L(\sigma^{2})=n\ln A - \frac{n}{2}\ln\sigma^{2} - \frac{1}{2\sigma^{2}}\sum_{i=1}^{n}(x_{i}-\mu)^{2},
\]
由\(\frac{\mathrm{d}}{\mathrm{d}\sigma^{2}}\ln L(\sigma^{2})=-\frac{n}{2}\cdot\frac{1}{\sigma^{2}} + \frac{1}{2\sigma^{4}}\sum_{i=1}^{n}(x_{i}-\mu)^{2}=0\)得
\(\sigma^{2}\)的最大似然估计值为\(\hat{\sigma}^{2}=\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\mu)^{2}\)，
故\(\sigma^{2}\)的最大似然估计量为\(\hat{\sigma}^{2}=\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\mu)^{2}\)。