1 (每小题6分,共48分)
(1) 求
解答:
(2)
求
解答:
(3) 求
解答:
(4) 求
解答:
(5) 方程
解答: 在方程
(6) 求曲线
解答: 由分析可知,曲线
(7) 计算二重积分
解答:
(8) 判别级数
解答: 方法一:由于
方法二:
注记: 由于
2 (16分) 求函数
解答: 由题意可知
1) 当
2) 当
3) 当
4) 当
5) 当x =
0
时,此时f‘_{+}(0) = \lim\limits_{x \to
0^{+}}\cfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^{+}}\cfrac{xe^{(x - 1)}
- 0}{x - 0} = \lim\limits_{x \to 0^{+}}e^{x - 1} = e^{-1},
综上可知,所求f(x) 的导函数为f‘(x) = \left\{ \begin{array}{ll} (1 - x)e^{1 - x}, & x > 1 \\ \textrm{不存在}, & x = 1 \\ (x + 1)e^{x - 1}, & 0 < x < 1 \\ \textrm{不存在}, & x = 0 \\ -(x + 1)e^{x - 1}, & x < 0 \end{array} \right. ; 1\,^{\circ} , x > 1, f‘(x) < 0, 0 < x < 1, f‘(x) > 0 \Longrightarrow f(x) 在x = 1 处取得极大值f(1) = 1; 2\,^{\circ} , 0 < x < 1, f‘(x) > 0, -1 < x < 0, f‘(x) < 0 \Longrightarrow f(x) 在x = 0 处取得极小值f(0) = 0; 3\,^{\circ} , -1 < x < 0, f‘(x) < 0, x < -1, f‘(x) > 0 \Longrightarrow f(x) 在x = -1 处取得极大值f(-1) = e^{-2};
综上可知,f(x) 的极大值为1和e^{-2} , 极小值为0.
3 (10分) 设f(x) 在[0,1] 上有一阶连续导数,且f(0) = f(1) = 0, 记M = \max\limits_{0 \le x \le 1}|f‘(x)|, 求证:|\int_{0}^{1}f(x)\rd x|\le \cfrac{1}{4}M.
证明:
方法一: \begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| & \le &
|\int_{0}^{\cfrac{1}{2}}f(x)\rd x| + |\int_{\cfrac{1}{2}}^{1}f(x)\rd x|\\ &
= & |\int_{0}^{\cfrac{1}{2}}[f(x) - f(0)]\rd x| +
|\int_{\cfrac{1}{2}}^{1}[f(x) -f(1)]\rd x| \\ & = &
|\int_{0}^{\cfrac{1}{2}}f‘(\xi)(x-0)\rd x| +
|\int_{\cfrac{1}{2}}^{1}f‘(\eta)(x-1)\rd x| \\ & \le &
\int_{0}^{\cfrac{1}{2}}|f‘(\xi)(x-0)|\rd x +
\int_{\cfrac{1}{2}}^{1}|f‘(\eta)(x-1)|\rd x \\ & \le &
M\int_{0}^{\cfrac{1}{2}}x \rd x + M\int_{\cfrac{1}{2}}^{1}(1 - x)\rd x \\ &
= & M\cfrac{x^2}{2}|_{0}^{\cfrac{1}{2}} + M(x -
\cfrac{x^2}{2})|_{\cfrac{1}{2}}^{1} = \cfrac{1}{4}M, (\textrm{其中}\xi \in
(0,\cfrac{1}{2}), \eta \in (\cfrac{1}{2},1)). \end{eqnarray*}
方法二: \begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| &
\stackrel{t = x - \cfrac{1}{2}}{=} & |\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}f(t
+ \cfrac{1}{2})\rd t| \stackrel{\textrm{ 分步积分}}{=} |t
f(t+\cfrac{1}{2})|_{\cfrac{1}{2}}^{\cfrac{1}{2}} -
\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f‘(t + \cfrac{1}{2})\rd t|\\ & = &
|\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f‘(t + \cfrac{1}{2})\rd t| \le
\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t f‘(t + \cfrac{1}{2})|\rd t\\ & \le
& M\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t|\rd t =
2M\int_{0}^{\cfrac{1}{2}}|t|\rd t = 2M\int_{0}^{\cfrac{1}{2}}t \rd t \\ & =
& 2M\cdot \cfrac{t^2}{2}|_{0}^{\cfrac{1}{2}} =
\cfrac{1}{4}M.\end{eqnarray*}
4 (18分) 设函数f(x,y) = \left\{ \begin{array}{ll} x - y + \cfrac{(xy)^2}{(x^2 + y^2)^{3/2}}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0)\end{array} \right. , 证明:
(1) f(x,y) 在原点处连续;
(2) f(x,y) 在原点的偏导数f_x(0,0) 和f_y(0,0) 存在;
(3) f(x,y) 在原点不可微.
解答: (1) \begin{eqnarray*}\mbox{原极限}
& \stackrel{x=\rho\cos\theta,y=\rho\sin\theta}{=} & \lim\limits_{\rho
\to 0}\left[\rho\cos\theta - \rho\sin\theta +
\cfrac{(\rho\cos\theta\rho\sin\theta)^2}{\rho^3}\right] \\ & = &
\lim\limits_{\rho \to 0}\rho\left[\cos\theta - \sin\theta +
(\cos\theta\sin\theta)^2\right] \\ & = & 0 =
f(0,0),\end{eqnarray*}
(2) f_x(0,0) = \lim\limits_{x \to 0}\cfrac{f(x,0) -
f(0,0)}{x - 0} = \lim\limits_{x \to 0}\cfrac{x}{x} = 1,
(3) \begin{eqnarray*}&
& \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{f(\Delta x,\Delta y) -
f(0,0) - f_x(0,0)\Delta x - f_y(0,0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta
y)^2}}\\ & = & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{\Delta
x - \Delta y + \cfrac{(\Delta x \Delta y)^2}{((\Delta x)^2 + (\Delta
y)^2)^{3/2}} - \Delta x + \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} \\ &
= & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{(\Delta x\Delta
y)^2}{((\Delta x)^2 + (\Delta y)^2)^2}\\ & \stackrel{\Delta y=k \Delta x}{=}
& \lim\limits_{\Delta x \to 0}\cfrac{(k(\Delta x)^2)^2}{((\Delta x)^2 +
k^2(\Delta x)^2)^2} \\ & = & \cfrac{k^2}{(1 +
k^2)^2}(\textrm{随着}k\textrm{的值的变化而变化}),\end{eqnarray*}
5 (16分) 求曲面z = xy -1 上与原点最近的点的坐标.
解答: 首先构造拉格朗日函数F(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda(xy - z -1) , 于是有\left\{ \begin{array}{l} F_x = 2x + \lambda y = 0\\ F_y = 2y + \lambda x = 0\\ F_z = 2z - \lambda = 0\\ F_{\lambda} = xy - z - 1 =0\end{array} \right. \Longrightarrow \left\{ \begin{array}{l} x = 0\\ y = 0\\ z = -1\\ \lambda = -2,\end{array} \right. 由于(0,0,-1) 是此问题的唯一驻点 (稳定点) ,而此问题一定有最小值,故(0,0,-1) 为所求点.
6 (16分) 设\vec{F} = \cfrac{y\vec{i} - x\vec{j}}{x^2 + y^2}, 曲线L 由圆x^2 + y^2 = 1 和椭圆\cfrac{x^2}{4} + y^2 = 1 组成,方向均为逆时针方向,求\int_{L}\vec{F}\rd \vec{s}.
解答: 方法一:记圆x^2 +
y^2 = 1
为曲线L_1
, 椭圆\cfrac{x^2}{4} + y^2 = 1
为曲线L_2
, 于是L = L_1 + L_2
,又设圆x^2 + y^2 = a, (0 < a < 1, a \to
0)
为曲线L_3
, 方向为逆时针方向,于是L = (L_1 - L_3) + (L_2 - L_3) + 2
L_3
,再记P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) =
\cfrac{y}{x^2 + y^2}
,于是在(L_1 - L_3) + (L_2 - L_3)
上, \cfrac{\partial P}{\partial x} =
\cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 +
y^2)^2}\textrm{且连续},
由此可知 \begin{eqnarray*} \int_{L}\vec{F}\rd \vec{s}
&=& \int_{(L_1 - L_3) + (L_2 - L_3) + 2 L_3}\vec{F}\rd \vec{s}\\
&=& \int_{L_1 - L_3}\vec{F}\rd \vec{s} + \int_{L_2 - L_3}\vec{F}\rd
\vec{s} + \int_{2 L_3}\vec{F}\rd \vec{s}\\ &\equiv& I_1 + I_2 + I_3.
\end{eqnarray*}
方法二:记圆x^2 + y^2
= 1
为曲线L_1
, 椭圆\cfrac{x^2}{4} + y^2 = 1
为曲线L_2
, 于是L = L_1 + L_2
,再记P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) =
\cfrac{y}{x^2 + y^2}
,于是在L_2 - L_1
上, \cfrac{\partial P}{\partial x} =
\cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 +
y^2)^2}\textrm{且连续},
由此可知 \int_{L}\vec{F}\rd \vec{s} = \int_{(L_2 -
L_1) + 2 L_1}\vec{F}\rd \vec{s} = \int_{L_2 - L_1}\vec{F}\rd \vec{s} + \int_{2
L_1}\vec{F}\rd \vec{s} = I_1 + I_2,
7 (16分) 求函数项级数\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n} 的和函数,并讨论在x \in (-\infty,+\infty) 上的一致收敛性.
解答: 记f_n(x) = \cfrac{x^2}{(1 + x^2)^n} , 函数项级数\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n} 的前n 项部分和函数为S_n(x) , 和函数为S(x) , 于是有
(1) 当x =
0
时,此时S_n(x) = 0
, 从而S(x) = \lim\limits_{n \to \infty}S_n(x) =
\lim\limits_{n \to \infty}0 = 0;
(2) 当x \ne
0
时,此时S_n(x) = \cfrac{\cfrac{x^2}{1 + x^2}\left[1 -
(\cfrac{1}{1 + x^2})^n\right]}{1 - \cfrac{1}{1 + x^2}},
综上可知, S(x) =
\left\{ \begin{array}{ll} 1, & x \ne 0 \\ 0, & x = 0 \end{array} \right.
;
8 (10分) 研究级数\sqrt{2} + \sqrt{2 - \sqrt{2}} + \sqrt{2 - \sqrt{2 + \sqrt{2}}} + \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}} + \cdots 的敛散性.
解答: 方法一:设a_1 = \sqrt{2}, a_2 = \sqrt{2 - \sqrt{2}}, a_3 = \sqrt{2 - \sqrt{2 + \sqrt{2}}}, \cdots, 从而可得
a_1 = \sqrt{2} =
2\sin\cfrac{\pi}{4} = 2\cos\cfrac{\pi}{4}, a_2 = \sqrt{2 - \sqrt{2}} = \sqrt{2 -
2\cos\cfrac{\pi}{4}} = 2\sin\cfrac{\pi}{8},
(1) 当n = 1 时,此时a_1=2\sin\cfrac{\pi}{4} 显然成立;
(2) 假设n = k 时,a_k 成立,即a_k = 2\sin\cfrac{\pi}{2^{k + 1}} , 下面证明当n = k + 1 时, a_{k+1} = \sqrt{2 - \sqrt{2 + 2 - a_k^2}} = \sqrt{2 - 2\cos\cfrac{\pi}{k+1}} = 2\sin\cfrac{\pi}{2^{k+2}}, 可知当n = k + 1 时也成立.
于是可得a_n = 2\sin\cfrac{\pi}{2^{n + 1}}(n=1,2,\cdots) , 而显然可得a_n \le 2\cfrac{\pi}{2^{n + 1}} = \cfrac{\pi}{2^{n}} , 而级数\sum\limits_{n=1}^{\infty}\cfrac{\pi}{2^{n}} 收敛,由正项级数的比较原则可知,所求原级数收敛.
方法二:设a_n = \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}, (n 个根号) ,满足a_{n + 1} = \sqrt{2 + a_n},
现在用数学归纳法来证明数列\{a_n\} 是有界的.
显然,a_1 = \sqrt{2} \in (0,2);
假设n = k 时,0 < a_k < 2,
则当n = k +
1
时,0 < a_{k + 1} = \sqrt{2 + a_k} < \sqrt{2 +
2} = 2,
所以0 < a_n < 2 (n = 1,2,\cdots),
数列\{a_n\}
有界的. 由于\cfrac{a_{n + 1}}{a_n} = \cfrac{\sqrt{2 +
a_n}}{a_n} = \sqrt{\cfrac{2}{a_n^2} + \cfrac{1}{a_n}} >1,
由单调有界原理,数列\{a_n\}
有极限,记为a
.由于a_{n + 1} = \sqrt{2 + a_n},
a = \sqrt{2 +
a}
, \Longrightarrow a = 2
, 即\lim\limits_{n \to \infty}a_n = 2.
从而 \begin{eqnarray*}\lim\limits_{n \to
\infty}\cfrac{\sqrt{2 - a_{n+1}}}{\sqrt{2 - a_n}} & = & \lim\limits_{n
\to \infty}\sqrt{\cfrac{2 - a_{n+1}}{2 - a_n}} = \lim\limits_{n \to
\infty}\sqrt{\cfrac{2 -\sqrt{2 + a_n}}{2 - a_n}} \\ & = & \lim\limits_{n
\to \infty}\sqrt{\cfrac{(2 -\sqrt{2 + a_n})(2 +\sqrt{2 + a_n})}{(2 - a_n)(2
+\sqrt{2 + a_n})}} \\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{2 -
a_n}{(2 - a_n)(2 +\sqrt{2 + a_n})}}\\ & = & \lim\limits_{n \to
\infty}\sqrt{\cfrac{1}{2 +\sqrt{2 + a_n}}} = \cfrac{1}{2} <
1.\end{eqnarray*}
[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答,布布扣,bubuko.com
[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答
原文地址:http://www.cnblogs.com/zhangzujin/p/3738112.html