專升本證明題真題及答案

一、單項(xiàng)選擇題（共10題）1.函數(shù)\(y=\frac{1}{\ln(x-1)}\)的定義域是（）A.\(x>1\)B.\(x\neq2\)C.\(x>1\)且\(x\neq2\)D.\(x\geq1\)且\(x\neq2\)答案：C2.當(dāng)\(x\to0\)時(shí)，與\(x\)等價(jià)的無(wú)窮小是（）A.\(\sin2x\)B.\(1-\cosx\)C.\(\ln(1+x)\)D.\(e^{2x}-1\)答案：C3.設(shè)函數(shù)\(f(x)\)在點(diǎn)\(x_0\)處可導(dǎo)，且\(f^\prime(x_0)=2\)，則\(\lim\limits_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=（）\)A.\(1\)B.\(2\)C.\(0\)D.不存在答案：B4.函數(shù)\(y=x^3-3x\)的單調(diào)遞減區(qū)間是（）A.\((-\infty,-1)\)B.\((-1,1)\)C.\((1,+\infty)\)D.\((-\infty,+\infty)\)答案：B5.不定積分\(\int\frac{1}{x}dx=（）\)A.\(\ln|x|+C\)B.\(\lnx+C\)C.\(-\frac{1}{x^2}+C\)D.\(x^{-1}+C\)答案：A6.定積分\(\int_{0}^{1}x^2dx=（）\)A.\(\frac{1}{3}\)B.\(\frac{1}{2}\)C.\(1\)D.\(0\)答案：A7.設(shè)向量\(\vec{a}=(1,-2,3)\)，\(\vec=(2,k,6)\)，若\(\vec{a}\parallel\vec\)，則\(k=（）\)A.\(-4\)B.\(4\)C.\(-1\)D.\(1\)答案：A8.平面\(2x-y+3z-1=0\)的法向量為（）A.\((2,-1,3)\)B.\((-2,1,-3)\)C.\((2,1,3)\)D.\((2,-1,-3)\)答案：A9.級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{1}{n^2}\)是（）A.發(fā)散的B.條件收斂的C.絕對(duì)收斂的D.無(wú)法判斷答案：C10.微分方程\(y^\prime=2x\)的通解是（）A.\(y=x^2+C\)B.\(y=2x^2+C\)C.\(y=x^2+1\)D.\(y=2x^2+1\)答案：A二、多項(xiàng)選擇題（共10題）1.下列函數(shù)中，是偶函數(shù)的有（）A.\(y=x^2+1\)B.\(y=\cosx\)C.\(y=e^x\)D.\(y=\frac{1}{x^2}\)答案：ABD2.下列極限存在的有（）A.\(\lim\limits_{x\to0}\frac{\sinx}{x}\)B.\(\lim\limits_{x\to\infty}\frac{\sinx}{x}\)C.\(\lim\limits_{x\to0}\frac{1}{x}\)D.\(\lim\limits_{x\to\infty}\frac{1}{x}\)答案：ABD3.函數(shù)\(f(x)\)在點(diǎn)\(x_0\)處可導(dǎo)的充分必要條件是（）A.\(f(x)\)在點(diǎn)\(x_0\)處連續(xù)B.\(f(x)\)在點(diǎn)\(x_0\)處左導(dǎo)數(shù)和右導(dǎo)數(shù)都存在C.\(f(x)\)在點(diǎn)\(x_0\)處左導(dǎo)數(shù)和右導(dǎo)數(shù)都存在且相等D.\(f(x)\)在點(diǎn)\(x_0\)處的導(dǎo)數(shù)等于零答案：C4.下列函數(shù)中，在區(qū)間\([0,1]\)上滿足羅爾定理?xiàng)l件的有（）A.\(y=x^2-1\)B.\(y=x(1-x)\)C.\(y=x^3\)D.\(y=|x|\)答案：AB5.下列積分計(jì)算正確的有（）A.\(\int_{-1}^{1}x^3dx=0\)B.\(\int_{0}^{1}e^xdx=e-1\)C.\(\int_{0}^{\pi}\sinxdx=2\)D.\(\int_{1}^{e}\frac{1}{x}dx=1\)答案：ABCD6.設(shè)向量\(\vec{a}=(1,2,3)\)，\(\vec=(3,2,1)\)，則下列運(yùn)算正確的有（）A.\(\vec{a}+\vec=(4,4,4)\)B.\(\vec{a}-\vec=(-2,0,2)\)C.\(\vec{a}\cdot\vec=10\)D.\(\vec{a}\times\vec=(4,8,-4)\)答案：ABC7.下列平面方程中，過(guò)點(diǎn)\((1,0,0)\)的有（）A.\(x=1\)B.\(x+y+z=1\)C.\(y=0\)D.\(x-y+z=1\)答案：ABD8.下列級(jí)數(shù)中，收斂的有（）A.\(\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\)B.\(\sum_{n=1}^{\infty}\frac{1}{n^3}\)C.\(\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\)D.\(\sum_{n=1}^{\infty}(-1)^n\frac{1}{n}\)答案：ABD9.下列微分方程中，是一階線性微分方程的有（）A.\(y^\prime+y=x\)B.\(y^\prime+y^2=x\)C.\(y^\prime+xy=e^x\)D.\(y^{\prime\prime}+y=0\)答案：AC10.下列函數(shù)中，\(y=C_1e^x+C_2e^{-x}\)（\(C_1\)，\(C_2\)為任意常數(shù)）是其通解的微分方程有（）A.\(y^{\prime\prime}-y=0\)B.\(y^{\prime\prime}+y=0\)C.\(y^\prime-y=0\)D.\(y^\prime+y=0\)答案：A三、判斷題（共10題）1.函數(shù)\(y=\sqrt{x-1}+\frac{1}{x-2}\)的定義域是\([1,2)\cup(2,+\infty)\)。（√）2.若\(\lim\limits_{x\tox_0}f(x)\)存在，則\(f(x)\)在點(diǎn)\(x_0\)處一定連續(xù)。（×）3.函數(shù)\(y=x^3\)在\(x=0\)處的導(dǎo)數(shù)為\(0\)，所以\(x=0\)是函數(shù)\(y=x^3\)的極值點(diǎn)。（×）4.不定積分\(\intf^\prime(x)dx=f(x)\)。（×）5.定積分\(\int_{a}^f(x)dx\)的值與積分變量\(x\)的選取無(wú)關(guān)。（√）6.若向量\(\vec{a}\)與向量\(\vec\)垂直，則\(\vec{a}\cdot\vec=0\)。（√）7.平面\(Ax+By+Cz+D=0\)（\(A\)，\(B\)，\(C\)不全為\(0\)）的法向量為\((A,B,C)\)。（√）8.級(jí)數(shù)\(\sum_{n=1}^{\infty}a_n\)收斂，則\(\lim\limits_{n\to\infty}a_n=0\)。（√）9.微分方程\(y^\prime=f(x)g(y)\)是可分離變量的微分方程。（√）10.二階常系數(shù)齊次線性微分方程\(y^{\prime\prime}+py^\prime+qy=0\)（\(p\)，\(q\)為常數(shù)）的特征方程為\(r^2+pr+q=0\)。（√）四、簡(jiǎn)答題（共4題）1.求函數(shù)\(y=x^3-3x^2+1\)的極值。對(duì)函數(shù)求導(dǎo)得\(y^\prime=3x^2-6x\)，令\(y^\prime=0\)，即\(3x^2-6x=0\)，\(3x(x-2)=0\)，解得\(x=0\)或\(x=2\)。當(dāng)\(x<0\)時(shí)，\(y^\prime>0\)，函數(shù)遞增；當(dāng)\(0<x<2\)時(shí)，\(y^\prime<0\)，函數(shù)遞減；當(dāng)\(x>2\)時(shí)，\(y^\prime>0\)，函數(shù)遞增。所以\(x=0\)時(shí)取極大值\(y(0)=1\)；\(x=2\)時(shí)取極小值\(y(2)=-3\)。2.計(jì)算定積分\(\int_{0}^{1}(x^2+e^x)dx\)。根據(jù)定積分的運(yùn)算法則，\(\int_{0}^{1}(x^2+e^x)dx=\int_{0}^{1}x^2dx+\int_{0}^{1}e^xdx\)。\(\int_{0}^{1}x^2dx=\left[\frac{1}{3}x^3\right]_{0}^{1}=\frac{1}{3}\)，\(\int_{0}^{1}e^xdx=[e^x]_{0}^{1}=e-1\)。所以\(\int_{0}^{1}(x^2+e^x)dx=\frac{1}{3}+e-1=e-\frac{2}{3}\)。3.求過(guò)點(diǎn)\((1,2,3)\)且與平面\(2x-3y+z-1=0\)平行的平面方程。已知平面\(2x-3y+z-1=0\)的法向量為\((2,-3,1)\)，所求平面與之平行，法向量相同。設(shè)所求平面方程為\(2x-3y+z+D=0\)，將點(diǎn)\((1,2,3)\)代入得\(2\times1-3\times2+3+D=0\)，解得\(D=1\)，所以平面方程為\(2x-3y+z+1=0\)。4.求級(jí)數(shù)\(\sum_{n=1}^{\infty}\frac{x^n}{n}\)的收斂半徑和收斂區(qū)間。根據(jù)冪級(jí)數(shù)收斂半徑公式\(R=\lim\limits_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\)，這里\(a_n=\frac{1}{n}\)，\(a_{n+1}=\frac{1}{n+1}\)，則\(R=\lim\limits_{n\to\infty}\frac{\frac{1}{n}}{\frac{1}{n+1}}=1\)。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\(\sum_{n=1}^{\infty}(-1)^n\frac{1}{n}\)，收斂。所以收斂區(qū)間為\([-1,1)\)。五、討論題（共4題）1.討論函數(shù)\(f(x)=\begin{cases}x^2+1,&x\leq0\\2x+1,&x>0\end{cases}\)在\(x=0\)處的連續(xù)性與可導(dǎo)性。首先求\(\lim\limits_{x\to0^{-}}f(x)=\lim\limits_{x\to0^{-}}(x^2+1)=1\)，\(\lim\limits_{x\to0^{+}}f(x)=\lim\limits_{x\to0^{+}}(2x+1)=1\)，且\(f(0)=0^2+1=1\)，所以函數(shù)在\(x=0\)處連續(xù)。再求左導(dǎo)數(shù)\(f_{-}^\prime(0)=\lim\limits_{x\to0^{-}}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0^{-}}\frac{x^2+1-1}{x}=0\)，右導(dǎo)數(shù)\(f_{+}^\prime(0)=\lim\limits_{x\to0^{+}}\frac{f(x)-f(0)}{x-0}=\lim\limits_{x\to0^{+}}\frac{2x+1-1}{x}=2\)，左右導(dǎo)數(shù)不相等，所以函數(shù)在\(x=0\)處不可導(dǎo)。2.討論積分\(\int_{1}^{+\infty}\frac{1}{x^p}dx\)的斂散性。當(dāng)\(p=1\)時(shí)，\(\int_{1}^{+\infty}\frac{1}{x}dx=\lim\limits_{b\to+\infty}\int_{1