考研数学三提升试题及答案

考研数学三提升试题及答案考研数学三提升试题及答案一、选择题题目1设函数$f(x)$在$x=0$处连续，且$\lim_{x\to0}\frac{f(x)}{x^2}=1$，则以下结论正确的是（）A.$f(0)=0$且$f^\prime(0)=0$B.$f(0)=0$且$f^\prime(0)$不存在C.$f(0)=1$且$f^\prime(0)=0$D.$f(0)=1$且$f^\prime(0)$不存在**答案**：A**解析**：因为$\lim_{x\to0}\frac{f(x)}{x^2}=1$，当$x\to0$时，分母$x^2\to0$，根据极限存在的条件，分子$f(x)\to0$。又因为$f(x)$在$x=0$处连续，所以$\lim_{x\to0}f(x)=f(0)$，则$f(0)=0$。由导数定义$f^\prime(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x}$，而$\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\left(\frac{f(x)}{x^2}\cdotx\right)$。已知$\lim_{x\to0}\frac{f(x)}{x^2}=1$，$\lim_{x\to0}x=0$，根据极限乘法法则$\lim_{x\to0}\left(\frac{f(x)}{x^2}\cdotx\right)=1\times0=0$，所以$f^\prime(0)=0$。题目2设$z=f(x^2-y^2,e^{xy})$，其中$f$具有二阶连续偏导数，则$\frac{\partial^2z}{\partialx\partialy}=$（）A.$-2xe^{xy}f^\prime_1+2ye^{xy}f^\prime_2+e^{2xy}(x^2+y^2)f^{\prime\prime}_{12}$B.$-2xe^{xy}f^\prime_1+2ye^{xy}f^\prime_2+e^{2xy}(x^2-y^2)f^{\prime\prime}_{12}$C.$-2xe^{xy}f^\prime_1-2ye^{xy}f^\prime_2+e^{2xy}(x^2+y^2)f^{\prime\prime}_{12}$D.$-2xe^{xy}f^\prime_1-2ye^{xy}f^\prime_2+e^{2xy}(x^2-y^2)f^{\prime\prime}_{12}$**答案**：D**解析**：令$u=x^2-y^2$，$v=e^{xy}$，则$z=f(u,v)$。先对$x$求偏导数：$\frac{\partialz}{\partialx}=f^\prime_1\cdot\frac{\partialu}{\partialx}+f^\prime_2\cdot\frac{\partialv}{\partialx}=f^\prime_1\cdot2x+f^\prime_2\cdotye^{xy}$再对$y$求偏导数：$\frac{\partial^2z}{\partialx\partialy}=\frac{\partial}{\partialy}(2xf^\prime_1+ye^{xy}f^\prime_2)$$=2x\frac{\partialf^\prime_1}{\partialy}+e^{xy}f^\prime_2+ye^{xy}\frac{\partialf^\prime_2}{\partialy}$其中$\frac{\partialf^\prime_1}{\partialy}=f^{\prime\prime}_{11}\cdot\frac{\partialu}{\partialy}+f^{\prime\prime}_{12}\cdot\frac{\partialv}{\partialy}=f^{\prime\prime}_{11}\cdot(-2y)+f^{\prime\prime}_{12}\cdotxe^{xy}$$\frac{\partialf^\prime_2}{\partialy}=f^{\prime\prime}_{21}\cdot\frac{\partialu}{\partialy}+f^{\prime\prime}_{22}\cdot\frac{\partialv}{\partialy}=f^{\prime\prime}_{21}\cdot(-2y)+f^{\prime\prime}_{22}\cdotxe^{xy}$因为$f$具有二阶连续偏导数，所以$f^{\prime\prime}_{12}=f^{\prime\prime}_{21}$。$\frac{\partial^2z}{\partialx\partialy}=2x(-2yf^{\prime\prime}_{11}+xe^{xy}f^{\prime\prime}_{12})+e^{xy}f^\prime_2+ye^{xy}(-2yf^{\prime\prime}_{21}+xe^{xy}f^{\prime\prime}_{22})$由于我们只关心含$f^{\prime\prime}_{12}$等二阶偏导数的项，化简可得：$\frac{\partial^2z}{\partialx\partialy}=-2xe^{xy}f^\prime_1-2ye^{xy}f^\prime_2+e^{2xy}(x^2-y^2)f^{\prime\prime}_{12}$二、填空题题目1设$\int_{0}^{1}(3x^2+ax)dx=3$，则$a=$______。**答案**：$0$**解析**：先计算定积分$\int_{0}^{1}(3x^2+ax)dx=\left(x^3+\frac{1}{2}ax^2\right)\big|_{0}^{1}$$=1^3+\frac{1}{2}a\times1^2-0^3-\frac{1}{2}a\times0^2=1+\frac{1}{2}a$已知$\int_{0}^{1}(3x^2+ax)dx=3$，则$1+\frac{1}{2}a=3$，解得$a=4-2=0$。题目2设$A$为三阶矩阵，$|A|=2$，则$|2A^{-1}|=$______。**答案**：$4$**解析**：根据矩阵的性质，若$A$为$n$阶矩阵，$k$为常数，则$|kA|=k^n|A|$，$|A^{-1}|=\frac{1}{|A|}$。已知$A$为三阶矩阵，$n=3$，$|A|=2$，则$|2A^{-1}|=2^3|A^{-1}|$$=8\times\frac{1}{|A|}=8\times\frac{1}{2}=4$三、解答题题目1求函数$f(x,y)=x^2+2y^2-x^2y$的极值。**答案**：先求驻点：对$f(x,y)$分别求关于$x$和$y$的偏导数：$f^\prime_x=2x-2xy=2x(1-y)$$f^\prime_y=4y-x^2$令$\begin{cases}f^\prime_x=0\\f^\prime_y=0\end{cases}$，即$\begin{cases}2x(1-y)=0\\4y-x^2=0\end{cases}$当$x=0$时，代入$4y-x^2=0$得$y=0$；当$y=1$时，代入$4y-x^2=0$得$x^2=4$，即$x=\pm2$。所以驻点为$(0,0)$，$(2,1)$，$(-2,1)$。再求二阶偏导数：$f^{\prime\prime}_{xx}=2-2y$$f^{\prime\prime}_{xy}=-2x$$f^{\prime\prime}_{yy}=4$然后根据判别式$A=f^{\prime\prime}_{xx}$，$B=f^{\prime\prime}_{xy}$，$C=f^{\prime\prime}_{yy}$，$\Delta=B^2-AC$判断驻点性质：对于驻点$(0,0)$：$A=f^{\prime\prime}_{xx}(0,0)=2$，$B=f^{\prime\prime}_{xy}(0,0)=0$，$C=f^{\prime\prime}_{yy}(0,0)=4$$\Delta=B^2-AC=0-2\times4=-8\lt0$，且$A=2\gt0$，所以$f(0,0)=0$是极小值。对于驻点$(2,1)$：$A=f^{\prime\prime}_{xx}(2,1)=0$，$B=f^{\prime\prime}_{xy}(2,1)=-4$，$C=f^{\prime\prime}_{yy}(2,1)=4$$\Delta=B^2-AC=(-4)^2-0\times4=16\gt0$，所以$(2,1)$不是极值点。对于驻点$(-2,1)$：$A=f^{\prime\prime}_{xx}(-2,1)=0$，$B=f^{\prime\prime}_{xy}(-2,1)=4$，$C=f^{\prime\prime}_{yy}(-2,1)=4$$\Delta=B^2-AC=4^2-0\times4=16\gt0$，所以$(-2,1)$不是极值点。综上，函数$f(x,y)$的极小值为$f(0,0)=0$，无极大值。题目2计算二重积分$\iint_{D}(x+y)dxdy$，其中$D$是由直线$y=x$，$y=0$，$x=1$所围成的区域。**答案**：先确定积分区域$D$，$D$可以表示为$0\leqx\leq1$，$0\leqy\leqx$。则$\iint_{D}(x+y)dxdy=\int_{0}^{1}dx\int_{0}^{x}(x+y)dy$先对$y$积分：$\int_{0}^{x}(x+y)dy=\int_{0}^{x}xdy+\int_{0}^{x}ydy$$=x\cdoty\big|_{0}^{x}+\frac{1}{2}y^2\big|_{0}^{x}=x^2+\frac{1}{2}x^2=\frac{3}{2}x^2$再对$x$积分：$\int_{0}^{1}\frac{3}{2}x^2dx=\frac{3}{2}\times\frac{1}{3}x^3\big|_{0}^{1}=\frac{1}{2}$题目3设$y=y(x)$由方程$e^y+xy=e$所确定，求$y^{\prime\prime}(0)$。**答案**：先对$e^y+xy=e$两边关于$x$求导：$e^yy^\prime+y+xy^\prime=0$当$x=0$时，代入原方程$e^y+0\timesy=e$，得$e^y=e$，解得$y=1$。将$x=0$，$y=1$代入$e^yy^\prime+y+xy^\prime=0$得：$e\cdoty^\prime(0)+1+0\timesy^\prime(0)=0$，解得$y^\prime(0)=-\frac{1}{e}$。对$e^yy^\prime+y+xy^\prime=0$两边再关于$x$求导：$e^y(y^\prime)^2+e^yy^{\prime\prime}+y^\prime+y^\prime+xy^{\prime\prime}=0$将$x=0$，$y=1$，$y^\prime(0)