高等數(shù)學(xué)a卷試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是（）A.\(\cosx\)B.\(-\cosx\)C.\(\sinx\)D.\(-\sinx\)2.\(\lim\limits_{x\to0}\frac{\sinx}{x}=\)（）A.0B.1C.\(\infty\)D.不存在3.曲線\(y=x^2\)在點(diǎn)\((1,1)\)處的切線斜率為（）A.1B.2C.3D.44.不定積分\(\intx^2dx=\)（）A.\(\frac{1}{3}x^3+C\)B.\(\frac{1}{2}x^3+C\)C.\(x^3+C\)D.\(\frac{1}{4}x^3+C\)5.定積分\(\int_{0}^{1}xdx=\)（）A.\(\frac{1}{2}\)B.1C.\(\frac{1}{3}\)D.\(\frac{1}{4}\)6.函數(shù)\(f(x)=x^3-3x\)的駐點(diǎn)為（）A.\(x=1\)B.\(x=-1\)C.\(x=\pm1\)D.\(x=0\)7.若\(f(x)\)的一個(gè)原函數(shù)是\(x^2\)，則\(f(x)=\)（）A.\(2x\)B.\(x^2\)C.\(\frac{1}{2}x^2\)D.\(x\)8.向量\(\vec{a}=(1,2)\)，\(\vec=(2,-1)\)，則\(\vec{a}\cdot\vec=\)（）A.0B.1C.2D.39.二元函數(shù)\(z=x^2+y^2\)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)10.級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{1}{n}\)是（）A.收斂的B.發(fā)散的C.條件收斂D.絕對(duì)收斂二、多項(xiàng)選擇題（每題2分，共10題）1.下列函數(shù)中，在\(x=0\)處連續(xù)的有（）A.\(y=|x|\)B.\(y=\frac{1}{x}\)C.\(y=x^2\)D.\(y=\sinx\)2.以下哪些是求導(dǎo)公式（）A.\((x^n)^\prime=nx^{n-1}\)B.\((\sinx)^\prime=\cosx\)C.\((\lnx)^\prime=\frac{1}{x}\)D.\((e^x)^\prime=e^x\)3.下列積分計(jì)算正確的有（）A.\(\int1dx=x+C\)B.\(\int\cosxdx=\sinx+C\)C.\(\inte^xdx=e^x+C\)D.\(\int\frac{1}{x}dx=\ln|x|+C\)4.函數(shù)\(f(x)\)在\([a,b]\)上滿足羅爾定理的條件是（）A.在\([a,b]\)上連續(xù)B.在\((a,b)\)內(nèi)可導(dǎo)C.\(f(a)=f(b)\)D.\(f(x)\)為多項(xiàng)式函數(shù)5.向量\(\vec{a}=(1,1,1)\)，\(\vec=(1,-1,0)\)，則（）A.\(\vec{a}+\vec=(2,0,1)\)B.\(\vec{a}-\vec=(0,2,1)\)C.\(\vec{a}\cdot\vec=0\)D.\(|\vec{a}|=\sqrt{3}\)6.對(duì)于二元函數(shù)\(z=f(x,y)\)，下列說(shuō)法正確的是（）A.偏導(dǎo)數(shù)\(\frac{\partialz}{\partialx}\)是將\(y\)看作常數(shù)對(duì)\(x\)求導(dǎo)B.全微分\(dz=\frac{\partialz}{\partialx}dx+\frac{\partialz}{\partialy}dy\)C.駐點(diǎn)處一定取得極值D.可微一定連續(xù)7.下列級(jí)數(shù)中，收斂的有（）A.\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)B.\(\sum_{n=1}^{\infty}\frac{1}{n!}\)C.\(\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\)D.\(\sum_{n=1}^{\infty}n\)8.曲線\(y=f(x)\)的漸近線類型有（）A.水平漸近線B.垂直漸近線C.斜漸近線D.拋物線漸近線9.下列屬于基本初等函數(shù)的有（）A.冪函數(shù)B.指數(shù)函數(shù)C.對(duì)數(shù)函數(shù)D.三角函數(shù)10.關(guān)于定積分的性質(zhì)，正確的有（）A.\(\int_{a}^kf(x)dx=k\int_{a}^f(x)dx\)（\(k\)為常數(shù)）B.\(\int_{a}^[f(x)+g(x)]dx=\int_{a}^f(x)dx+\int_{a}^g(x)dx\)C.\(\int_{a}^f(x)dx=-\int_^{a}f(x)dx\)D.若\(f(x)\geq0\)在\([a,b]\)上成立，則\(\int_{a}^f(x)dx\geq0\)三、判斷題（每題2分，共10題）1.函數(shù)\(y=\frac{1}{x-1}\)在\(x=1\)處間斷。（）2.若\(f^\prime(x_0)=0\)，則\(x_0\)一定是\(f(x)\)的極值點(diǎn)。（）3.\(\int_{a}^f(x)dx\)的值與積分變量用什么字母表示無(wú)關(guān)。（）4.兩個(gè)向量平行，則它們的對(duì)應(yīng)坐標(biāo)成比例。（）5.二元函數(shù)\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的偏導(dǎo)數(shù)存在，則函數(shù)在該點(diǎn)一定可微。（）6.級(jí)數(shù)\(\sum_{n=1}^{\infty}a_n\)收斂，則\(\lim\limits_{n\to\infty}a_n=0\)。（）7.函數(shù)\(y=x^3\)在\((-\infty,+\infty)\)上是單調(diào)遞增的。（）8.若\(f(x)\)為奇函數(shù)，則\(\int_{-a}^{a}f(x)dx=0\)。（）9.曲線\(y=x^2\)沒(méi)有漸近線。（）10.不定積分\(\intf(x)dx\)表示\(f(x)\)的所有原函數(shù)。（）四、簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(y=x^3-3x^2+5\)的單調(diào)區(qū)間。答案：對(duì)\(y\)求導(dǎo)得\(y^\prime=3x^2-6x=3x(x-2)\)。令\(y^\prime>0\)，解得\(x<0\)或\(x>2\)，此為單調(diào)遞增區(qū)間；令\(y^\prime<0\)，解得\(0<x<2\)，此為單調(diào)遞減區(qū)間。2.計(jì)算定積分\(\int_{0}^{1}(x^2+1)dx\)。答案：由定積分運(yùn)算法則，\(\int_{0}^{1}(x^2+1)dx=\int_{0}^{1}x^2dx+\int_{0}^{1}1dx\)。\(\int_{0}^{1}x^2dx=[\frac{1}{3}x^3]_0^1=\frac{1}{3}\)，\(\int_{0}^{1}1dx=[x]_0^1=1\)，所以結(jié)果為\(\frac{1}{3}+1=\frac{4}{3}\)。3.求函數(shù)\(z=x^2y+y^2\)關(guān)于\(x\)和\(y\)的偏導(dǎo)數(shù)。答案：把\(y\)看作常數(shù)對(duì)\(x\)求導(dǎo)，\(\frac{\partialz}{\partialx}=2xy\)；把\(x\)看作常數(shù)對(duì)\(y\)求導(dǎo)，\(\frac{\partialz}{\partialy}=x^2+2y\)。4.簡(jiǎn)述判斷級(jí)數(shù)\(\sum_{n=1}^{\infty}a_n\)收斂的比值判別法。答案：計(jì)算\(\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\rho\)。當(dāng)\(\rho<1\)時(shí)，級(jí)數(shù)絕對(duì)收斂；當(dāng)\(\rho>1\)時(shí)，級(jí)數(shù)發(fā)散；當(dāng)\(\rho=1\)時(shí)，比值判別法失效。五、討論題（每題5分，共4題）1.討論函數(shù)\(f(x)=\begin{cases}x+1,&x\geq0\\x-1,&x<0\end{cases}\)在\(x=0\)處的連續(xù)性與可導(dǎo)性。答案：\(\lim\limits_{x\to0^+}f(x)=1\)，\(\lim\limits_{x\to0^-}f(x)=-1\)，左右極限不相等，所以\(f(x)\)在\(x=0\)處不連續(xù)，不連續(xù)則不可導(dǎo)。2.討論二元函數(shù)\(z=x^2+y^2-2x+4y\)的極值情況。答案：先求偏導(dǎo)數(shù)\(\frac{\partialz}{\partialx}=2x-2\)，\(\frac{\partialz}{\partialy}=2y+4\)。令偏導(dǎo)數(shù)為\(0\)，得駐點(diǎn)\((1,-2)\)。再求二階偏導(dǎo)數(shù)，判斷\(A=2\)，\(B=0\)，\(C=2\)，\(AC-B^2=4>0\)且\(A>0\)，所以在\((1,-2)\)處有極小值\(z(1,-2)=-5\)。3.討論冪級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{x^n}{n}\)的收斂區(qū)間。答案：用比值判別法，\(\lim\limits_{n\to\infty}\left|\frac{\frac{x^{n+1}}{n+1}}{\frac{x^n}{n}}\right|=|x|\)。當(dāng)\(|x|<1\)時(shí)收斂，當(dāng)\(|x|>1\)時(shí)發(fā)散。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\([-1,1)\)。4.討論函數(shù)\(y=\frac{1}{x^2-1}\)的漸近線情況。答案：垂直漸近線：令\(x^2-1=0\)，得\(x=\pm1\)，即\(x=\pm1\)是垂直漸近線；水平漸近線：\(\lim\limits_{x\to\infty}\frac{1}{x^2-1}=0\)，所以\(y=0