12006年河南省普通高等学校选拔优秀专科生进入本科阶段学习考试《高等数学》试卷题号一二三四五六总分核分人分数一、单项选择题(每小题2分,共计60分)在每小题的四个备选答案中选出一个正确答案,并将其代码写在题干后面的括号内。不选、错选或多选者,该题无分.1.已知函数)12(xf的定义域为]1,0[,则)(xf的定义域为()A.]1,21[B.]1,1[C.]1,0[D.]2,1[解:Bxx112110.2.函数)1ln(2xxy)(x是()A.奇函数B.偶函数C.非奇非偶函数D.既奇又偶函数解:01ln)1ln()1ln()()(22xxxxxfxfA.3.当0x时,xxsin2是x的()A.高阶无穷小B.低阶无穷小C.同阶非等价无穷小D.等价无穷小解:1sinlim20xxxxC.4.极限nnnnsin32lim()A.B.2C.3D.5解:Bnnnnnnn2]sin32[limsin32lim.5.设函数0,10,1)(2xaxxexfax,在0x处连续,则常数a()A.0B.1C.2D.3解:Baaaaexexfaxxaxxx1122lim1lim)(lim20200.6.设函数)(xf在点1x处可导,则xxfxfx)1()21(lim0()A.)1(fB.)1(2fC.)1(3fD.-)1(f解:xxfffxfxxfxfxx)1()1()1()21(lim)1()21(lim00Cfxfxfxfxfxx)1(3)1()1(lim2)1()21(lim2007.若曲线12xy上点M处的切线与直线14xy平行,则点M的坐标()A.(2,5)B.(-2,5)C.(1,2)D.(-1,2)解:Ayxxxy5,2422000.8.设202cossintyduuxt,则dxdy()A.2tB.t2C.-2tD.t2得分评卷人2解:Dttttdxdy2sinsin222.9.设2(ln)2(nxxyn,为正整数),则)(ny()A.xnxln)(B.x1C.1)!2()1(nnxnD.0解:Bxyxyxxynnn1ln1ln)()1()2(.10.曲线233222xxxxy()A.有一条水平渐近线,一条垂直渐近线B.有一条水平渐近线,两条垂直渐近线C.有两条水平渐近线,一条垂直渐近线,D.有两条水平渐近线,两条垂直渐近线解:Ayyyxxxxxxxxyxxx2122lim,4lim,1lim)2)(1()3)(1(2332.11.下列函数在给定的区间上满足罗尔定理的条件是()A.]2,0[|,1|xyB.]2,0[,)1(132xyC.]2,1[,232xxyD.]1,0[,arcsinxxy解:由罗尔中值定理条件:连续、可导及端点的函数值相等C.12.函数xey在区间),(内()A.单调递增且图像是凹的曲线B.单调递增且图像是凸的曲线C.单调递减且图像是凹的曲线D.单调递减且图像是凸的曲线解:Ceyeyxx0,0.13.若CxFdxxf)()(,则dxefexx)(()A.CeFexx)(B.CeFx)(C.CeFexx)(D.CeFx)(解:DCeFedefdxefexxxxx)()()()(.14.设)(xf为可导函数,且xexf)12(,则)(xf()A.Cex1221B.Cex)1(212C.Cex1221D.Cex)1(212解:BCexfexfexfxxx)1(21)1(212)()()12(.15.导数batdtdxdarcsin()A.xarcsinB.0C.abarcsinarcsinD.211x解:baxdxarcsin是常数,所以Bxdxdxdba0arcsin.16.下列广义积分收敛的是()A.1dxexB.11dxxC.1241dxxD.1cosxdx解:Cxdxx)21arctan4(412arctan4141112.17.设区域D由)(),(,),(,xgyxfyabbxax所围成,则区域D的面积为()A.badxxgxf)]()([B.badxxgxf)]()([3C.badxxfxg)]()([D.badxxgxf|)()(|解:由定积分的几何意义可得D的面积为badxxgxf|)()(|D.18.若直线32311znyx与平面01343zyx平行,则常数n()A.2B.3C.4D.5解:Bnnn30943}3,43{}3,,1{.19.设yxyxyxfarcsin)1(),(,则偏导数)1,(xfx为()A.2B.1C.-1D.-2解:Bxfxxfx1)1,()1,(.20.设方程02xyzez确定了函数),(yxfz,则xz=()A.)12(zxzB.)12(zxzC.)12(zxyD.)12(zxy解:令xyeFyzFxyzezyxFzzxz222,),,(Azxzxyxyzyzxyeyzxzz)12(222.21.设函数xyyxz2,则11yxdz()A.dydx2B.dydx2C.dydx2D.dydx2解:222xydxxdydyxxydxdzAdydxdxdydydxdzyx2211.22.函数2033222yxxyz在定义域上内()A.有极大值,无极小值B.无极大值,有极小值C.有极大值,有极小值D.无极大值,无极小值解:,6)0,0(),(062,06222xzyxyxyzxyxz2,6222yxzyz是极大值A.23设D为圆周由012222yxyx围成的闭区域,则Ddxdy()A.B.2C.4D.16解:有二重积分的几何意义知:Ddxdy区域D的面积为.24.交换二次积分axadyyxfdx000(),(,常数)的积分次序后可化为()A.aydxyxfdy00),(B.aaydxyxfdy0),(C.aadxyxfdy00),(D.ayadxyxfdy0),(解:积分区域},0|),{(}0,0|),{(axyayyxxyaxyxDB.25.若二重积分20sin20)sin,cos(),(rdrrrfddxdyyxfD,则积分区域D为4()A.xyx222B.222yxC.yyx222D.220yyx解:在极坐标下积分区域可表示为:}sin20,20|),{(rrD,在直角坐标系下边界方程为yyx222,积分区域为右半圆域D26.设L为直线1yx上从点)0,1(A到)1,0(B的直线段,则Ldydxyx)(()A.2B.1C.-1D.-2解:L:,1xyxxx从1变到0,012)(DdxdxdydxyxL.27.下列级数中,绝对收敛的是()A.1sinnnB.1sin)1(nnnC.12sin)1(nnnD.1cosnn解:22sinnn12sinnn收敛C.28.设幂级数nnnnaxa(0为常数,2,1,0n),在点2x处收敛,则0)1(nnna()A.绝对收敛B.条件收敛C.发散D.敛散性不确定解:0nnnxa在2x收敛,则在1x绝对收敛,即级数0)1(nnna绝对收敛A.29.微分方程0sincoscossinydxxydyx的通解为()A.CyxcossinB.CyxsincosC.CyxsinsinD.Cyxcoscos解:dxxxdyyyydxxydyxsincossincos0sincoscossinCCyxCxyxxdyydsinsinlnsinlnsinlnsinsinsinsin.30.微分方程xxeyyy2的特解用特定系数法可设为()A.xebaxxy)(B.xebaxxy)(2C.xebaxy)(D.xaxey解:-1不是微分方程的特征根,x为一次多项式,可设xebaxy)(C.二、填空题(每小题2分,共30分)得分评卷人531.设函数,1||,01||,1)(xxxf则)(sinxf_________.解:1)(sin1|sin|xfx.32.xxxx231lim22=_____________.解:)31(1lim)31)(2()2(lim231lim2222xxxxxxxxxxxx123341.33.设函数xy2arctan,则dy__________.解:dxxdy2412.34.设函数bxaxxxf23)(在1x处取得极小值-2,则常数ba和分别为___________.解:bababaxxxf12,02323)(25,4ba.35.曲线12323xxxy的拐点为__________.解:)1,1(),(0662632yxxyxxy.36.设函数)(),(xgxf均可微,且同为某函数的原函数,有1)1(,3)1(gf则)()(xgxf_________.解:2)1()1()()(gfCCxgxf2)()(xgxf.37.dxxx)sin(32_________.解:3202sin)sin(3023232dxxxdxdxxdxxx.38.设函数0,0,)(2xxxexfx,则20)1(dxxf__________.解:201110012132)()1(edxedxxdttfdxxfxtx.39.向量}1,1,2{}2,1,1{ba与向量的夹角为__________.解:3,21663||||,cosbabababa.40.曲线022zxyL:绕x轴旋转一周所形成的旋转曲面方程为_________.解:把xy22中的2y换成22yz,即得所求曲面方程xyz222.41.设函数yxxyzsin2,则yxz2_________.解:yxyxzsin2yxyxzcos212.42.设区域}11,10|),{(yxyxD,则________)(2Ddxdyxy.解:Ddxxdyxydxdxdyxy102101122322)()(.43.函数2)(xexf在00x处展开的幂级数是________________.解:0!nnxnxe0022),(,!1)1(!)()(2nnnnnxxxnnxexf.44.幂级数0112)1()1(nnnnnx的和函数为_________.6解:0111011)21ln()2()1(1)2