專升本：2018年《高數(shù)》真題及答案

2018年成人高等學(xué)校專升本招生全國統(tǒng)一考試高等數(shù)學(xué)（一）一、選擇題：每小題4分，共40分，在每小題給出的四個選項(xiàng)中，只有一項(xiàng)是符合題目要求。x1.lim（）x0cosxA.eB.2C.1D.02.設(shè)y1cosx，則dy=（）A.1sinxdxB.1sinxdx，則fx（C.5D.5ln5xC.sinxdxD.sinxdx3.若函數(shù)fx5x）A.5x1B.x5x1x14.dx（）2x11A.ln2xCB.ln2xCC.2CD.2C2x2x5.f2xdx（）A.1f2xCB.f2xCC.2f2xCD.1fxC2216.若fx為連續(xù)的奇函數(shù)，則fxdx1A.0B.2C.2f1D.2f17.若二元函數(shù)zxA.2xy32y2y3x2y，則z（x）B.xy32yC.2xy3D.xy38.方程x2y22z0表示的二次曲面是（）A.柱面B.球面C.旋轉(zhuǎn)拋物面D.橢球面9.已知區(qū)域Dx,y1x1,1y1，則xdxdy（）DA.0B.1C.2D.410.微分方程yy1的通解為（）A.y2xCB.1y2xCC.y2CxD.2y2xC2二、填空題:11～20小題,每小題4分,共40分11.曲線yx3x4的拐點(diǎn)為___________6x32112.lim13xx___________x013.若函數(shù)fxxarctanx，則fx___________14.若ye2x，則dy___________15.2x3dx___________16.1xxdx___________52117.0sinxdx___________218.1n___________3n019.0exdx___________20.若二元函數(shù)zxy22，則z___________xy2三、解答題:21～28題,共70分.解答應(yīng)寫出推理、演算步驟3sinx21.設(shè)函數(shù)fxx,x<0，在x0處連續(xù)，求a3xa,x03x32x1222.求limx1sinx1223.設(shè)函數(shù)fx2xln3x2，求f0xsin3tdt24.求limx00x225.求xcosxdx26.求函數(shù)fx1x1x5的極值3232127.求微方程yy2lnx的通解x28.設(shè)區(qū)域Dx,yxy9,y0，計算xydxdy2222D2018年成人高等學(xué)校專升本招生全國統(tǒng)一考試高等數(shù)學(xué)（一）試題答案解析1.【答案】Dlimx00【解析】limx0cosxlimcosxx01x02.【答案】D【解析】y1cosxsinx，故dysinxdx3.【答案】C5ln5x【解析】fx5x4.【答案】B1【解析】dxln2xC2x5.【答案】A【解析】f2xdx12f2xd2x1f2xC26.【答案】A1【解析】因?yàn)閒x為連續(xù)的奇函數(shù)，故fxdx017.【答案】C【解析】zx2y3x2y，故z2xy3x8.【答案】C【解析】x2y22z0可化為xyz，故表示的是旋轉(zhuǎn)拋物面22229.【答案】A111【解析】xdxdyxdxdy2xdx0111D10.【答案】B【解析】原方程分離變量得ydydx，兩邊同時積分得1yxC，故22方程的通解為1y2xC211.【答案】（2，-6）【解析】y3x212x3，y6x12，令y0，則x2,y6，故拐點(diǎn)為（2，-6）12.【答案】e3【解析】lim13xxlim13x13x3e31x0x0x213.【答案】1x21x2【解析】fxxarctanx，則fx121x1x214.【答案】2e2x【解析】ye2x2e2x，則dy2e2xdxdx15.【答案】x23xC【解析】2x3dxx23xC16.【答案】23【解析】1xxdx16x1x252631113317.【答案】2【解析】0sinxdx2sindxx2cosx022222018.【答案】32111【解析】1lim3n3113lim232nx03n11x0n0319.【答案】1【解析】0exdxex1020.【答案】4xy【解析】zx2y2,z2xyx2,z4xyxy23sinx321.【答案】limfxlimx0x0xlimfxlim3xaax0x0且f0a因?yàn)閒x在x0處連續(xù)所以limfxlimfxf0x0x0a322.【答案】lim3x32x21lim3x32x21sinx1x122x1x1lim3x2x1x1x1x1x13x2x1limx1x15233x29fx2fx23.【答案】3x22故f0941cos3txx0sin3tdtlim324.【答案】limx00x2x2x0131cos3xlimx0x292x2x21lim3x03225.【答案】xcosxdxxsinxxsinxcosxC26.【答案】fxxx，令fx0，得x10，x21，sinxdx2當(dāng)x<0或x>1時，fx>0，此時fx為單調(diào)增加函數(shù)當(dāng)0<x<1時，fx<0，此時fx為單調(diào)減少函數(shù)故當(dāng)x0時，fx取極大值，極大值f0529當(dāng)x1時，，fx取極小值，極小值f1627.【答案】這是個一階線性非齊次微分方程1Px，Qx2lnxx11故通解