112018年成人高等学校专升本招生全国统一考试高等数学(一)一、选择题:每小题4分,共40分,在每小题给出的四个选项中,只有一项是符合题目要求。1.limx()x0cosxA.eB.2C.1D.02.设y1cosx,则dy=()A.1sinxdxB.1sinxdxC.sinxdxD.sinxdx3.若函数fx5x,则fx()A.5x1B.x5x1C.5xln5D.5x4.1dx()2xA.ln2xCB.ln2xCC.2CD.2C5.f2xdx()2x2xA.1f2xC2B.f2xCC.2f2xCD.1fxC26.若fx为连续的奇函数,则11fxdxA.0B.2C.2f1D.2f17.若二元函数zx2y3x2y,则z()xA.2xy32yB.xy32yC.2xy3D.xy38.方程x2y22z0表示的二次曲面是()A.柱面B.球面C.旋转抛物面D.椭球面9.已知区域Dx,y1x1,1y1,则xdxdy()DA.0B.1C.2D.41110.微分方程yy1的通解为()A.y2xCB.1y2xC2C.y2CxD.2y2xC二、填空题:11~20小题,每小题4分,共40分11.曲线yx36x23x4的拐点为12.lim13xxx013.若函数fxxarctanx,则fx14.若ye2x,则dy15.2x3dx16.1x5x2dxx17.0sin2dx18.1n03n19.exdx020.若二元函数zx2y2,则2zxy三、解答题:21~28题,共70分.解答应写出推理、演算步骤3sinx,x<021.设函数fxx,在x0处连续,求a3xa,x0x22.求limx13x32x21sinx2123.设函数fx2xln3x2,求f024.求lim0sin3tdtx0x225.求xcosxdx26.求函数fx1x31x25的极值3227.求微方程y1y2lnx的通解x28.设区域Dx,yx2y29,y0,计算x2y2dxdyD2018年成人高等学校专升本招生全国统一考试高等数学(一)试题答案解析1.【答案】D【解析】limxlim0x00x0cosx2.【答案】Dlimcosx1x0【解析】y1cosxsinx,故dysinxdx3.【答案】C【解析】fx5x5xln54.【答案】B【解析】1dxln2xC2x5.【答案】A【解析】f2xdx1f2xd2x1f2xC226.【答案】A【解析】因为fx为连续的奇函数,故11fxdx07.【答案】C【解析】zx2y3x2y,故z2xy3x8.【答案】C【解析】x2y22z0可化为x2y2z,故表示的是旋转抛物面229.【答案】A111【解析】xdxdyxdx1dy2xdx011D10.【答案】B【解析】原方程分离变量得ydydx,两边同时积分得1y2xC,故2方程的通解为1y2xC211.【答案】(2,-6)【解析】y3x212x3,y6x12,令y0,则x2,y6,故拐点为(2,-6)12.【答案】e31133【解析】lim13xxlim13x3xex013.【答案】x21x2x0【解析】,则1x2fxxarctanxfx11x21x214.【答案】2e2xdx【解析】ye2x2e2x,则dy2e2xdx15.【答案】x23xC【解析】2x3dxx23xC16.【答案】23【解析】1x5x2dx1x61x3121613317.【答案】2xxxx【解析】0sin2dx20sin2d22cos20218.【答案】3211113n313【解析】limlim1n03nx0113x023n2,x19.【答案】1xx【解析】edxe10020.【答案】4xy【解析】zx2y2,zx2xy22zxy4xy21.【答案】limfxlim3sinx3limx0x0fxlim3xaax0x0x且f0a因为fx在x0处连续所以limx0fxlimx0fxf0a322.【答案】3x32x213x32x21limx1sinx21limx1x21limx1limx13x2x1x1x1x13x2x1x15223.【答案】fx233x2fx9故f0943x2sin3tdt1cos3tx24.【答案】lim0x0x2lim3x0x22011cos3xlim3x0x29x21lim23...