高等數(shù)學(xué)自考試題及答案

一、單項選擇題（每題2分，共10題）1.函數(shù)\(y=\sinx\)的周期是（）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)2.\(\lim\limits_{x\to0}\frac{\sinx}{x}\)的值為（）A.0B.1C.2D.不存在3.函數(shù)\(y=x^2\)在\(x=1\)處的導(dǎo)數(shù)是（）A.1B.2C.3D.44.\(\intxdx\)等于（）A.\(\frac{1}{2}x^2+C\)B.\(x^2+C\)C.\(\frac{1}{3}x^3+C\)D.\(x^3+C\)5.函數(shù)\(z=xy\)在點\((1,2)\)處的全微分\(dz\)是（）A.\(2dx+dy\)B.\(dx+2dy\)C.\(dx+dy\)D.\(2dx+2dy\)6.級數(shù)\(\sum_{n=1}^{\infty}\frac{1}{n}\)是（）A.收斂B.發(fā)散C.條件收斂D.絕對收斂7.曲線\(y=e^x\)與\(x=0\)，\(x=1\)及\(x\)軸圍成的面積為（）A.\(e-1\)B.\(e\)C.\(e+1\)D.\(1\)8.設(shè)\(f(x)\)是奇函數(shù)，則\(\int_{-a}^{a}f(x)dx\)的值為（）A.\(2\int_{0}^{a}f(x)dx\)B.0C.\(a\)D.\(2a\)9.方程\(x^2+y^2-z^2=0\)表示的曲面是（）A.球面B.柱面C.圓錐面D.拋物面10.函數(shù)\(f(x)=x^3-3x\)的駐點是（）A.\(x=1\)B.\(x=-1\)C.\(x=\pm1\)D.\(x=0\)二、多項選擇題（每題2分，共10題）1.下列函數(shù)中是偶函數(shù)的有（）A.\(y=x^2\)B.\(y=\cosx\)C.\(y=e^x\)D.\(y=|x|\)2.極限存在的條件有（）A.左極限存在B.右極限存在C.左右極限相等D.函數(shù)在該點有定義3.下列求導(dǎo)公式正確的有（）A.\((x^n)^\prime=nx^{n-1}\)B.\((\sinx)^\prime=\cosx\)C.\((\lnx)^\prime=\frac{1}{x}\)D.\((e^x)^\prime=e^x\)4.定積分的性質(zhì)有（）A.\(\int_{a}^kf(x)dx=k\int_{a}^f(x)dx\)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.\(\int_{a}^f(x)dx=\int_{a}^{c}f(x)dx+\int_{c}^f(x)dx\)5.下列級數(shù)中收斂的有（）A.\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)B.\(\sum_{n=1}^{\infty}\frac{1}{n!}\)C.\(\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\)D.\(\sum_{n=1}^{\infty}(-1)^n\frac{1}{n}\)6.多元函數(shù)\(z=f(x,y)\)在點\((x_0,y_0)\)可微的必要條件有（）A.\(z=f(x,y)\)在點\((x_0,y_0)\)連續(xù)B.\(z=f(x,y)\)在點\((x_0,y_0)\)偏導(dǎo)數(shù)存在C.\(z=f(x,y)\)在點\((x_0,y_0)\)偏導(dǎo)數(shù)連續(xù)D.\(z=f(x,y)\)在點\((x_0,y_0)\)全增量\(\Deltaz=A\Deltax+B\Deltay+o(\rho)\)7.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8.下列積分中，能用牛頓-萊布尼茨公式計算的有（）A.\(\int_{0}^{1}x^2dx\)B.\(\int_{-1}^{1}\frac{1}{x}dx\)C.\(\int_{0}^{2}\sqrt{4-x^2}dx\)D.\(\int_{1}^{e}\frac{1}{x}dx\)9.二元函數(shù)\(z=x^2+y^2\)的性質(zhì)正確的有（）A.圖形是旋轉(zhuǎn)拋物面B.有極小值C.在\((0,0)\)處偏導(dǎo)數(shù)為0D.是凸函數(shù)10.冪級數(shù)\(\sum_{n=0}^{\infty}a_nx^n\)的收斂區(qū)間求法涉及（）A.求收斂半徑B.討論端點處的斂散性C.比值審斂法D.根值審斂法三、判斷題（每題2分，共10題）1.函數(shù)\(y=\frac{1}{x}\)在定義域內(nèi)是單調(diào)遞減函數(shù)。（）2.若\(\lim\limits_{x\tox_0}f(x)\)存在，則\(f(x)\)在\(x_0\)處一定連續(xù)。（）3.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)\(y^\prime=\cosx\)，二階導(dǎo)數(shù)\(y^{\prime\prime}=-\sinx\)。（）4.定積分\(\int_{a}^f(x)dx\)的值與積分變量用什么字母表示無關(guān)。（）5.級數(shù)\(\sum_{n=1}^{\infty}u_n\)收斂，則\(\lim\limits_{n\to\infty}u_n=0\)。（）6.二元函數(shù)\(z=f(x,y)\)在點\((x_0,y_0)\)處偏導(dǎo)數(shù)存在，則在該點一定可微。（）7.函數(shù)\(y=x^3\)在\((-\infty,+\infty)\)上是凹函數(shù)。（）8.廣義積分\(\int_{1}^{+\infty}\frac{1}{x^2}dx\)是收斂的。（）9.向量\(\vec{a}=(1,2)\)與向量\(\vec=(2,4)\)平行。（）10.方程\(x^2+y^2=1\)表示的曲線是空間中的圓柱面。（）四、簡答題（每題5分，共4題）1.求函數(shù)\(y=\ln(1+x^2)\)的導(dǎo)數(shù)。答案：根據(jù)復(fù)合函數(shù)求導(dǎo)法則，令\(u=1+x^2\)，則\(y=\lnu\)。先對\(\lnu\)關(guān)于\(u\)求導(dǎo)得\(\frac{1}{u}\)，再對\(u\)關(guān)于\(x\)求導(dǎo)得\(2x\)，所以\(y^\prime=\frac{2x}{1+x^2}\)。2.計算定積分\(\int_{0}^{1}(x^2+1)dx\)。答案：根據(jù)定積分基本運算，\(\int_{0}^{1}(x^2+1)dx=\int_{0}^{1}x^2dx+\int_{0}^{1}1dx\)。由積分公式\(\intx^2dx=\frac{1}{3}x^3+C\)，\(\int1dx=x+C\)，可得\([\frac{1}{3}x^3+x]_{0}^{1}=\frac{1}{3}+1-0=\frac{4}{3}\)。3.求函數(shù)\(z=x^3-3xy^2\)的偏導(dǎo)數(shù)\(\frac{\partialz}{\partialx}\)和\(\frac{\partialz}{\partialy}\)。答案：求\(\frac{\partialz}{\partialx}\)時，把\(y\)看作常數(shù)，得\(\frac{\partialz}{\partialx}=3x^2-3y^2\)；求\(\frac{\partialz}{\partialy}\)時，把\(x\)看作常數(shù)，得\(\frac{\partialz}{\partialy}=-6xy\)。4.簡述判斷級數(shù)\(\sum_{n=1}^{\infty}u_n\)收斂的比值審斂法。答案：設(shè)\(\lim\limits_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\rho\)。當(dāng)\(\rho\lt1\)時，級數(shù)\(\sum_{n=1}^{\infty}u_n\)絕對收斂，從而收斂；當(dāng)\(\rho\gt1\)（或\(\rho=+\infty\)）時，級數(shù)\(\sum_{n=1}^{\infty}u_n\)發(fā)散；當(dāng)\(\rho=1\)時，比值審斂法失效。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=x^3-3x\)的單調(diào)性與極值。答案：先求導(dǎo)\(y^\prime=3x^2-3=3(x+1)(x-1)\)。令\(y^\prime=0\)，得駐點\(x=\pm1\)。當(dāng)\(x\lt-1\)或\(x\gt1\)時，\(y^\prime\gt0\)，函數(shù)單調(diào)遞增；當(dāng)\(-1\ltx\lt1\)時，\(y^\prime\lt0\)，函數(shù)單調(diào)遞減。所以\(x=-1\)取極大值\(2\)，\(x=1\)取極小值\(-2\)。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\((1,-2)\)。再求二階偏導(dǎo)數(shù)，\(A=\frac{\partial^2z}{\partialx^2}=2\)，\(B=\frac{\partial^2z}{\partialx\partialy}=0\)，\(C=\frac{\partial^2z}{\partialy^2}=2\)。\(AC-B^2=4\gt0\)且\(A\gt0\)，所以在\((1,-2)\)處取極小值\(-5\)。3.討論廣義積分\(\int_{0}^{+\infty}e^{-x}dx\)的斂散性。答案：\(\int_{0}^{+\infty}e^{-x}dx=\lim\limits_{b\to+\infty}\int_{0}^e^{-x}dx\)，由\(\inte^{-x}dx=-e^{-x}+C\)，則\(\lim\limits_{b\to+\infty}[-e^{-x}]_{0}^=\lim\limits_{b\to+\infty}(1-e^{-b})=1\)，所以該廣義積分收斂。4.討論冪級數(shù)\(\sum_{n=0}^{\infty}nx^n\)的收斂區(qū)間。答案：先求收斂半徑\(R\)，\(\lim\limits_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim\limits_{n\to\infty}\frac{n+1}{n}=1\)，所以\(R=1\)。當(dāng)\(x=1\)時，級數(shù)\(\sum_{n=0}^{\infty}n\)發(fā)散；當(dāng)\(x=-1\)時，級數(shù)\(\sum_{n=0}