專科高數(shù)下試題及答案

單項(xiàng)選擇題（每題2分，共10題）1.函數(shù)\(z=\ln(x+y)\)的定義域是（）A.\(x+y\gt0\)B.\(x+y\geq0\)C.\(x\gt0\)且\(y\gt0\)D.\(x\geq0\)且\(y\geq0\)2.設(shè)\(z=x^2y\)，則\(\frac{\partialz}{\partialx}\)等于（）A.\(2xy\)B.\(x^2\)C.\(y\)D.\(2x\)3.交換積分次序后，\(\int_{0}^{1}dx\int_{x}^{1}f(x,y)dy\)等于（）A.\(\int_{0}^{1}dy\int_{y}^{1}f(x,y)dx\)B.\(\int_{0}^{1}dy\int_{0}^{y}f(x,y)dx\)C.\(\int_{1}^{0}dy\int_{y}^{1}f(x,y)dx\)D.\(\int_{1}^{0}dy\int_{0}^{y}f(x,y)dx\)4.級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{1}{n^p}\)當(dāng)（）時(shí)收斂。A.\(p\gt1\)B.\(p\lt1\)C.\(p\geq1\)D.\(p\leq1\)5.向量\(\vec{a}=(1,-2,3)\)與\(\vec=(-2,4,-6)\)的關(guān)系是（）A.垂直B.平行C.相交D.異面6.直線\(\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\)與平面\(x+2y+3z-6=0\)的位置關(guān)系是（）A.平行B.垂直C.相交但不垂直D.直線在平面內(nèi)7.設(shè)\(u=xyz\)，則\(du\)在點(diǎn)\((1,1,1)\)處的值為（）A.\(dx+dy+dz\)B.\(xdx+ydy+zdz\)C.\(dx-dy-dz\)D.\(dx+dy-dz\)8.函數(shù)\(f(x,y)=x^2+y^2\)在點(diǎn)\((0,0)\)處（）A.有極大值B.有極小值C.無(wú)極值D.不是駐點(diǎn)9.曲線\(x=t\)，\(y=t^2\)，\(z=t^3\)在點(diǎn)\((1,1,1)\)處的切線方程為（）A.\(\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{3}\)B.\(\frac{x-1}{1}=\frac{y-1}{1}=\frac{z-1}{1}\)C.\(\frac{x-1}{3}=\frac{y-1}{2}=\frac{z-1}{1}\)D.\(\frac{x-1}{1}=\frac{y-1}{3}=\frac{z-1}{2}\)10.設(shè)\(D\)是由\(x=0\)，\(y=0\)，\(x+y=1\)圍成的區(qū)域，則\(\iint_{D}dxdy\)的值為（）A.\(\frac{1}{2}\)B.\(1\)C.\(\frac{1}{3}\)D.\(\frac{1}{4}\)答案：1.A2.A3.A4.A5.B6.B7.A8.B9.A10.A多項(xiàng)選擇題（每題2分，共10題）1.下列哪些是多元函數(shù)的求導(dǎo)法則（）A.鏈?zhǔn)椒▌tB.乘積法則C.商法則D.復(fù)合函數(shù)求導(dǎo)法則2.下列級(jí)數(shù)中，收斂的有（）A.\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)B.\(\sum_{n=1}^{\infty}\frac{1}{n}\)C.\(\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\)D.\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)3.關(guān)于向量的運(yùn)算，正確的有（）A.\(\vec{a}\cdot\vec=\vec\cdot\vec{a}\)B.\(\vec{a}\times\vec=-\vec\times\vec{a}\)C.\((\vec{a}+\vec)\cdot\vec{c}=\vec{a}\cdot\vec{c}+\vec\cdot\vec{c}\)D.\((\vec{a}\times\vec)\times\vec{c}=\vec{a}\times(\vec\times\vec{c})\)4.以下哪些屬于二次曲面（）A.球面B.橢球面C.拋物面D.圓錐面5.函數(shù)\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微的充分條件有（）A.\(f_x(x,y)\)和\(f_y(x,y)\)在\((x_0,y_0)\)處連續(xù)B.\(f(x,y)\)在\((x_0,y_0)\)處的偏導(dǎo)數(shù)存在C.\(\lim\limits_{\Deltax\to0,\Deltay\to0}\frac{\Deltaz-f_x(x_0,y_0)\Deltax-f_y(x_0,y_0)\Deltay}{\sqrt{(\Deltax)^2+(\Deltay)^2}}=0\)D.\(f(x,y)\)在\((x_0,y_0)\)處連續(xù)6.交換二重積分\(\int_{0}^{2}dx\int_{x}^{2}f(x,y)dy\)積分次序后可能為（）A.\(\int_{0}^{2}dy\int_{0}^{y}f(x,y)dx\)B.\(\int_{0}^{x}dy\int_{y}^{2}f(x,y)dx\)C.\(\int_{x}^{2}dy\int_{0}^{y}f(x,y)dx\)D.\(\int_{0}^{2}dy\int_{0}^{2}f(x,y)dx\)7.下列關(guān)于冪級(jí)數(shù)\(\sum_{n=0}^{\infty}a_n(x-x_0)^n\)的說(shuō)法正確的是（）A.存在收斂半徑\(R\)B.在收斂區(qū)間內(nèi)絕對(duì)收斂C.在收斂區(qū)間端點(diǎn)處可能收斂也可能發(fā)散D.收斂半徑\(R\)可以通過(guò)公式\(R=\lim\limits_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)（當(dāng)極限存在時(shí)）計(jì)算8.對(duì)于平面\(Ax+By+Cz+D=0\)，說(shuō)法正確的是（）A.法向量為\(\vec{n}=(A,B,C)\)B.當(dāng)\(D=0\)時(shí)，平面過(guò)原點(diǎn)C.平面與\(x\)軸的交點(diǎn)為\((-\frac{D}{A},0,0)\)（\(A\neq0\)）D.兩平面\(A_1x+B_1y+C_1z+D_1=0\)與\(A_2x+B_2y+C_2z+D_2=0\)垂直的充要條件是\(A_1A_2+B_1B_2+C_1C_2=0\)9.設(shè)\(z=f(u,v)\)，\(u=x+y\)，\(v=x-y\)，則下列正確的是（）A.\(\frac{\partialz}{\partialx}=f_u+f_v\)B.\(\frac{\partialz}{\partialy}=f_u-f_v\)C.\(\frac{\partial^2z}{\partialx^2}=f_{uu}+2f_{uv}+f_{vv}\)D.\(\frac{\partial^2z}{\partialy^2}=f_{uu}-2f_{uv}+f_{vv}\)10.下列哪些是計(jì)算二重積分\(\iint_{D}f(x,y)dxdy\)常用的方法（）A.直角坐標(biāo)法B.極坐標(biāo)法C.柱坐標(biāo)法D.球坐標(biāo)法答案：1.ABD2.ACD3.ABC4.ABCD5.AC6.A7.ABCD8.ABCD9.ABCD10.AB判斷題（每題2分，共10題）1.函數(shù)\(z=\sqrt{x-y}\)的定義域?yàn)閈(x\geqy\)。（）2.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處偏導(dǎo)數(shù)存在，則\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處一定連續(xù)。（）3.級(jí)數(shù)\(\sum_{n=1}^{\infty}(-1)^n\)收斂。（）4.向量\(\vec{a}=(1,0,0)\)與\(\vec=(0,1,0)\)的數(shù)量積為\(0\)，則\(\vec{a}\)與\(\vec\)垂直。（）5.平面\(2x+3y-z+1=0\)的法向量為\((2,3,-1)\)。（）6.函數(shù)\(z=x^2+y^2\)的駐點(diǎn)為\((0,0)\)。（）7.交換積分次序時(shí)，積分上下限可能會(huì)發(fā)生變化。（）8.冪級(jí)數(shù)\(\sum_{n=0}^{\infty}x^n\)的收斂半徑為\(1\)。（）9.若\(f(x,y)\)在區(qū)域\(D\)上連續(xù)，則\(\iint_{D}f(x,y)dxdy\)一定存在。（）10.曲線\(x=\cost\)，\(y=\sint\)，\(z=t\)是螺旋線。（）答案：1.√2.×3.×4.√5.√6.√7.√8.√9.√10.√簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(z=x^3+y^3-3xy\)的駐點(diǎn)。答案：分別求偏導(dǎo)數(shù)，\(z_x=3x^2-3y\)，\(z_y=3y^2-3x\)。令\(z_x=0\)，\(z_y=0\)，即\(\begin{cases}3x^2-3y=0\\3y^2-3x=0\end{cases}\)，解得駐點(diǎn)為\((0,0)\)和\((1,1)\)。2.計(jì)算\(\int_{0}^{1}dx\int_{x}^{1}e^{y^2}dy\)。答案：交換積分次序，原積分\(=\int_{0}^{1}dy\int_{0}^{y}e^{y^2}dx\)。先對(duì)\(x\)積分得\(\int_{0}^{1}ye^{y^2}dy\)，令\(u=y^2\)，\(du=2ydy\)，則積分變?yōu)閈(\frac{1}{2}\int_{0}^{1}e^{u}du=\frac{1}{2}(e-1)\)。3.求向量\(\vec{a}=(2,-3,1)\)與\(\vec=(1,-1,3)\)的夾角余弦值。答案：根據(jù)向量夾角余弦公式\(\cos\theta=\frac{\vec{a}\cdot\vec}{\vert\vec{a}\vert\vert\vec\vert}\)。\(\vec{a}\cdot\vec=2\times1+(-3)\times(-1)+1\times3=8\)，\(\vert\vec{a}\vert=\sqrt{2^2+(-3)^2+1^2}=\sqrt{14}\)，\(\vert\vec\vert=\sqrt{1^2+(-1)^2+3^2}=\sqrt{11}\)，所以\(\cos\theta=\frac{8}{\sqrt{14\times11}}=\frac{8}{\sqrt{154}}\)。4.求冪級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{x^n}{n}\)的收斂半徑和收斂區(qū)間。答案：由\(R=\lim\limits_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)，這里\(a_n=\frac{1}{n}\)，\(a_{n+1}=\frac{1}{n+1}\)，則\(R=1\)。當(dāng)\(x=1\)時(shí)，級(jí)數(shù)為\(\sum_{n=1}^{\infty}\frac{1}{n}\)發(fā)散；當(dāng)\(x=-1\)時(shí)，級(jí)數(shù)為\(\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\)收斂，所以收斂區(qū)間為\([-1,1)\)。討論題（每題5分，共4題）1.討論多元函數(shù)連續(xù)、偏導(dǎo)數(shù)存在、可微之間的關(guān)系。答案：可微能推出偏導(dǎo)數(shù)存在且函數(shù)連續(xù)；偏導(dǎo)數(shù)存在不一定連續(xù)，也不一定可微；函數(shù)連續(xù)不能推出偏導(dǎo)數(shù)存在，也不能推出可微。例如\(z=\sqrt{x^2+y^2}\)在\((0,0)\)連續(xù)但偏導(dǎo)數(shù)不存在；\(z=\begin{cases}\frac{xy}{x^2+y^2},(x,y)\neq(0,0)\\0,(x,y)=(0,0)\end{cases}\)偏導(dǎo)數(shù)存在但不連續(xù)。2.討論級(jí)數(shù)