高等数学B2－期末总复习

一、选择题

1.\begin{aligned} \lim_{\substack{ x\to0 \\ y\to0}}\frac{3xy}{x^2+y^2}=(\ {\color{red}{ D }} \ ) \end{aligned}

A.\begin{aligned} \frac32 \end{aligned}

B.0

C.\begin{aligned} \frac65 \end{aligned}

D.不存在

2.对函数\begin{aligned} f(x,y)=xy \end{aligned}，点（0，0）（ B ）

A.不是驻点

B.是驻点却非极值点

C.是极大值点

D.是极小值点

3.若方程\begin{aligned} y''+py'+qy=0 \end{aligned}的系数满足1-p+q=0，则该方程有特解（ C ）

A.\begin{aligned} y=x \end{aligned}

B.\begin{aligned} y={\rm e}^x \end{aligned}

C.\begin{aligned} \color{red}{y={\rm e}^{-x}} \end{aligned}

D.y=\sin x

4.微分方程\begin{aligned} y''-y'=x^2 \end{aligned}的特解形式为（ D ）

A.\begin{aligned} y=Ax^2 \end{aligned}

B.\begin{aligned} y=Ax^2+Bx+C \end{aligned}

C.\begin{aligned} Ax^3 \end{aligned}

D.\begin{aligned} \color{red}y=x(Ax^2+Bx+C) \end{aligned}

5.若微分方程\begin{aligned} y'+p(x)y=q(x) \end{aligned}有两个不同的特解\begin{aligned} y_1=y_1(x) \end{aligned}，\begin{aligned} y_2=y_2(x) \end{aligned}，C 为任意常数，则该方程的通解为（ B ）.

A.\begin{aligned} y=C(y_1-y_2) \end{aligned}

B.\begin{aligned} \color{red}y=y_1+C(y_1-y_2) \end{aligned}

C.\begin{aligned} y=C(y_1+y_2) \end{aligned}

D.\begin{aligned} y=y_1+C(y_1+y_2) \end{aligned}

6.把二重积分\begin{aligned} \iint_D{\rm e}^{-x^2-y^2}{\rm d}x{\rm d}y \end{aligned}在极坐标系中化为二次积分，其中D由x^2+y^2\leq 1所围成（ D ）.

A.\begin{aligned} \int_0^{2\pi}{\rm d}\theta \int_0^1{\rm e}^{-r^2}{\rm d}r \end{aligned}

B.\begin{aligned} 4\int_0^\frac{\pi}{2}{\rm d}\theta \int_0^1{\rm e}^{-r^2}{\rm d}r \end{aligned}

C.\begin{aligned} 2\int_0^\frac{\pi}2 {\rm d}\theta \int_0^1r{\rm e}^{-r^2}{\rm d}r \end{aligned}

D.\begin{aligned} \color{red}\int_0^{2\pi}{\rm d}\theta\int_0^1r{\rm e}^{-r^2}{\rm d}r \end{aligned}

7.若幂级数\begin{aligned} \sum_{n=0}^\infty a_n{x}^n \end{aligned}的收敛半径是3，则幂级数\begin{aligned} \sum_{n=0}^\infty a_nx^{2n} \end{aligned}的收敛半径是（ A ）.

A.\begin{aligned} \color{red}\sqrt 3 \end{aligned}

B.3

C.9

D.\begin{aligned} \frac 13 \end{aligned}

8.已知\begin{aligned} \lim_{n \to \infty}u_n=a \end{aligned}，则\begin{aligned} \sum_{n=1}^\infty(u_n-u_{n+1}) \end{aligned}（ C ）.

A.收敛于0

B.收敛于a

C.收敛于u₁-a

D.发散

9.设级数\begin{aligned} \sum_{n=0}^\infty a_n(x-1)^n \end{aligned}的收敛半径是1，则级数在x=3点（ A ）.

A.发散

B.条件收敛

C.绝对收敛

D.不能确定收敛性

二、填空题

1.极限\begin{aligned} \lim_{\substack{x\to1\\y\to0}}{\frac{\arctan(x+y)}{\sqrt[3]{x^3+y}}}=\color{red}\frac{\pi}{4} \end{aligned}.

2.设\begin{aligned} z=x^4+y^4-4x^2y^2 \end{aligned}，则\begin{aligned} \frac{\partial^2z}{\partial x^2}= \color{red}12x^2-8y^2 \end{aligned}.

3.交换积分\begin{aligned} \int_0^2{\rm d}y\int_{y^2}^{2y}f(x,y){\rm d}x \end{aligned}的积分次序得：\begin{aligned} \color{red}\int_0^4{\rm d}x\int_{\frac x2}^{\sqrt x}f(x,y){\rm d}y \end{aligned}.

4.比较二重积分\begin{aligned} I_1=\iint_D(x+y)^2{\rm d}\sigma \end{aligned}与\begin{aligned} I_2=\iint_D(x+y)^3{\rm d}\sigma \end{aligned}的大小，其中D是由圆周\begin{aligned} (x-2)^2+(y-1)^2=2 \end{aligned}所围成，则\begin{aligned} \color{red}I_1\leq I_2. \end{aligned}

5.设\begin{aligned} D=\{ (x,y)||x|\leq 3,|y|\leq 2 \} \end{aligned}，则\begin{aligned} \iint_Dx(1+y){\rm d}\sigma= \end{aligned}0 .

6.若级数\begin{aligned} \sum_{n=1}^{\infty}u_n=S \end{aligned}，则级数\begin{aligned} \sum_{n=1}^{\infty}(u_n+u_{n+1})= \color{red} 2S-u_1 \end{aligned}.

7.幂级数\begin{aligned} \sum_{n=1}^\infty \frac{x^n}{\sqrt n} \end{aligned}的收敛区间是(-1,1) .

8.微分方程\begin{aligned} \frac{{\rm d}y}{x}+\frac{{\rm d}x}{y}=0 \end{aligned}的通解为\begin{aligned} \color{red} x^2+y^2=C \end{aligned}.

9.微分方程\begin{aligned} y''-6y'+10y=e^{3x} \end{aligned}的一个特解是\begin{aligned} \color{red}{\rm y}_1=e^{3x} \end{aligned}.

10.若某个二阶常系数线性微分方程的通解为\begin{aligned} y=C_1{\rm e}^x+C_2 \end{aligned}，其中\begin{aligned} C_1 \end{aligned}，\begin{aligned} C_2 \end{aligned}为独立的任意常数，则该方程为\begin{aligned} \color{red}y''-y'=0 \end{aligned}.

三、计算题

1.求微分方程\begin{aligned} y''+2y'=3x-1 \end{aligned}的通解.

解：特征方程为\begin{aligned} r^2+2r=0 \end{aligned}，特征根为\begin{aligned} r_1=-2 \end{aligned}，\begin{aligned} r_2=0 \end{aligned}.

对应的齐次方程的通解为\begin{aligned} Y=C_1+C_2e^{-2x} \end{aligned}.

\begin{aligned} P_n(x)e^{\alpha x}=3x-1 \end{aligned}，\begin{aligned} \alpha=0 \end{aligned}为单特征根.

设方程的一个特解为\begin{aligned} \bar {\rm y}=x(ax+b) \end{aligned}，则\begin{aligned} \bar{\rm y}'=2ax+b \end{aligned}，\begin{aligned} \bar{\rm y}''=2a \end{aligned}.

代入方程得\begin{aligned} 2a+4ax+2b=3x-1 \end{aligned}，比较系数得\begin{aligned} 4a=3 \end{aligned}，\begin{aligned} 2a+2b=-1 \end{aligned}，解得\begin{aligned} a=\frac34 \end{aligned}，\begin{aligned} b=-\frac54 \end{aligned}.

所求方程的一个特解为\begin{aligned} \bar{\rm y}=\frac34x^2-\frac54x \end{aligned}.

所求的通解为\begin{aligned} y=C_1+C_2e^{-2x}+\frac34x^2-\frac54x \end{aligned}.

四、解答题

1.某厂要建造一个容积为32立方米的长方体无盖蓄水池，底面和四个侧面的单位造价相同，为100元/立方米，应如何选择蓄水池的尺寸，方可使它的造价最少？

解：设蓄水池的长、宽和高分别为x,y,z，则容积为\begin{aligned} V=xyz \end{aligned}，因此\begin{aligned} z=\frac V{xy} =\frac{32}{xy} \end{aligned}，

解方程组\begin{cases} g_x'=100(y-\frac{64}{x^2})=0\\ g_y'=100(x-\frac{64}{y^2})=0 \end{cases}得x=y=4，因此，\begin{aligned} z=\frac{32}{4\times 4}=2 \end{aligned}，

根据问题实际意义可知，最小值在定义域内存在，因此，所求的唯一驻点（4，4）就是最小值点，即：

当长和宽均为4，高为2时，蓄水池的造价最少，为f(4,4,2)=4800.

高等数学B2期末习题PDF版：https://pan.yunxge.cn/gsb2qmzfx.pdf

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