2015年电大《高等数学基础》期末模拟考试试题及参考答案资料小抄_第1页
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高等数学基础试题类型高等数学基础试题类型分为单项选择题、填空题、计算题和应用题。单项选择题的形式为四选一,即在每题的四个备选答案中选出一个正确答案;填空题只要求直接填写结果,不必写出计算过程和推理过程;计算题或应用题要求写出文字说明、演算步骤或推证过程。四种题型分数的百分比为单项选择题20,填空题20,计算题44,应用题16。期末考试采用闭卷笔试形式,卷面满分为100分,考试时间为90分钟。高等数学基础模拟题一、单项选择题1函数的图形关于(A)对称2EXYA坐标原点B轴C轴DYXY2在下列指定的变化过程中,(C)是无穷小量ABCD1SINX01SINX01LNE1X3设在可导,则(C)F0HFFH2LM0ABCDX2X0X20XF4若,则(B)CFFDFDLN1ABCDLNXXLCXFLCXF15下列积分计算正确的是(D)ABCD0DSI11DE0XD2SIN00DOS16设函数的定义域为,则函数的图形关于(A)对称XF,XFAB轴C轴D坐标原点YY7当时,变量(C)是无穷小量ABCDX1XSIN1EX28设,则(B)FEFFLIM0ABCD2E4E29(A)XFDABCD2FXFD121XFXFD210下列无穷限积分收敛的是(B)ABCD0DEX0EX1DX1二、填空题(每小题3分,共15分)1函数的定义域是241LNXY2,2若函数,在处连续,则012XKXFKE3曲线在处的切线斜率是33F,4函数的单调增加区间是YARCTN,5若,则XFSIDXFXSIN6函数的定义域是1L92|13,27函数的间断点是0SINXY0X8曲线在处的切线斜率是1F2,29函数的单调减少区间是2Y,110XDSINSINC三、计算题(每小题11分,共44分)1计算极限解1ILM2X21SINLM1SINL12XXX2设,求Y3ECOSYD解3DECOSDXXXL3DESIXXLNIN3计算不定积分XE21解由换元积分法得CUXED1ED21X14计算定积分E1LNX解由分部积分法得E1LNL1EX5计算极限解XX5SIN6LM000SI6SI6LLM55XX6设,求Y2Y解ISIN2LXXSINCO2LNX7计算不定积分D3CO解SISI3ISDXXXXN1ID3N1COS3C8计算定积分E1DL2X解EE112LND2LNDLXX221LN1|E5四、应用题(本题16分)1、某制罐厂要生产一种体积为V的有盖圆柱形容器,问容器的底半径与高各为多少时用料最省解设容器的底半径为,高为,则其表面积为RHHRS224V由,得唯一驻点,由实际问题可知,当时可使用料最省,此时,即当容器的底0S32R32VR34VH半径与高分别为与时,用料最省3342、圆柱体上底的中心到下底的边沿的距离为L,问当底半径与高分别为多少时,圆柱体的体积最大解如图所示,圆柱体高H,与底半径R满足22HRL圆柱体的体积公式为V将代入得22RL2LH求导得23HL令得,并由此解得,即当底半径为,高时,圆柱体得体积最大。0V3L6R63RL3HL典型例题例1计算极限321SINLM1XX解利用重要极限,及极限的运算法则得I013SINL2SNLI11XXXX1SIN3LIM1XXLILIM1X4例2计算极限27623解利用极限的运算法则得54LI2LI176LIM3323XXXXX例3设,求YSINY解利用导数的运算法则得XXXXY2333SINSINLLSINLX233SICOLLXX232SICOSLN1例4设,求YY解设,得2INUVLUSIN2XV利用复合函数求导法则,得XVXYYSIL2XVUSILUCO122TASINCO例5设是由方程确定的函数,求4ELYXDY解利用导数运算法则和复合函数求导法则,等式两端分别对求导得X左YX1LLN右YXXEEE444YX34E由此得整理得YYX3141X由微分定义得D4E例6计算X21解利用换元积分法得1DE1DE221XXXCUCX1E例7计算DLN解利用分部积分法得LND1LN1LL1XXXXN11XXDLCX1LN1例8求曲线上的点,使其到点的距离最短Y20,2A解曲线上的点到点的距离公式为XD与在同一点取到最小值,为计算方便求的最小值点,将代入得22DXY2令22XD令得可以验证是的最小值点,并由此解出,即曲线上的点和点011X2D2YXY22,1到点的距离最短,1,A高等数学基础第一次作业(一)单项选择题下列各函数对中,(C)中的两个函数相等A2XF,XGB2XF,XGC3LNF,LND1F,12设函数的定义域为,,则函数F的图形关于(C)对称A坐标原点BX轴CY轴DXY下列函数中为奇函数是(B)A1LN2YBXCOSC2XAD1LNXY下列函数中为基本初等函数是(C)AXBYC2YD0,1XY下列极限存计算不正确的是(D)A12LIMXB01LNIM0XXCSINLMXD01SINLMXX当0时,变量(C)是无穷小量ASINBCSID2LN若函数XF在点0满足(A),则XF在点0连续。ALI0FB在点的某个邻域内有定义CM0FXDLIMLI00XFFXX(二)填空题函数1LN392XF的定义域是|3已知函数,则XFX2XXX21LIM若函数0,1XKF,在处连续,则KE函数,SINXY的间断点是X(二)计算题设函数0,EXXF求1,2F解2F,0F,1FE求函数1LGXY的定义域解21LGXY有意义,要求20X解得102X或,则定义域为1|02X或在半径为R的半圆内内接一梯形,梯形的一个底边与半圆的直径重合,另一底边的两个端点在半圆上,试将梯形的面积表示成其高的函数解DAROHEBC设梯形ABCD即为题中要求的梯形,设高为H,即OEH,下底CD2R直角三角形AOE中,利用勾股定理得22AEORH则上底2AERH故22HSRHRA求XXSIN3LM0解000SISIN3LILLM22XXX132求1SINL21X解11LMLILISNISNXXX求X3TANLI0解00SI1SIN31LLM3COCOSXXXA求XSIN1LM20解222200011LLILIMISIN1SINXXXXX02LIIX求XX31LIM解1433111LIMLILIMLI333XXXXXE求4586LI24XX解422MLILIM113XXX设函数1,1,2XXF讨论F的连续性,并写出其连续区间解分别对分段点,处讨论连续性(1)11LIMLI10XXF所以1LILIXXFF,即FX在1处不连续(2)2211LILIMXXFF所以11LILIXXFFF即FX在1处连续由(1)(2)得在除点1外均连续故F的连续区间为,高等数学基础第四次作业(一)单项选择题若XF的一个原函数是X1,则F(D)ALNB2CD32X下列等式成立的是(D)ADXFFBDXFFCDFFDDXFFX若COS,则(B)AINBCCCSINDCOSXFXD32(B)A3FB32FC1XFD31XF若CXFFD,则XFD1(B)AB2CCF2DCXF1由区间,BA上的两条光滑曲线XFY和G以及两条直线A和B所围成的平面区域的面积是(C)AAXGFDBBAFGDCBAXGFDDAXGFD(二)填空题函数F的不定积分是XF若函数XF与G是同一函数的原函数,则XF与G之间有关系式。CXGF常数DE22TANCT若XXF3OS,则XF3COS9X35D21SI3若无穷积分1XP收敛,则0P(三)计算题CXDXX1SINCOSDCOS2EE2ECXDXXLNLN1LCXXD2SIN41O2OS2CS2SIE11E1LN3LN3LDL3EXX414222021002DEXXLNDL12E1XEEEXDX121E2L(四)证明题证明若XF在,A上可积并为奇函数,则0DAXF证AAATFTDTFDXFT令0AAAXDF证毕证明若X在,上可积并为偶函数,则AAXFXF0D2D证AAAXFXFXF00DDAAXFTFTT0,是偶函数则令证毕AAAAXFFFF0000D2D证明XXXDD证AAAAAFXFFFF0000XXX00证毕高等数学基础第二次作业(一)单项选择题设0F且极限XFLIM0存在,则XFLI0(C)ABFCDCVX设XF在0可导,则HFFH2LI00(D)A2BXCXD0XF设XFE,则FF1LIM0(A)AB2CEDE4设91XXF,则0F(D)A9BCD9下列结论中正确的是(C)A若XF在点0有极限,则在点0可导B若XF在点0连续,则在点0X可导C若在点可导,则在点X有极限D若在点有极限,则在点连续(二)填空题设函数0,01SIN2XXF,则F0设XXFE5E2,则FDLX5LN2曲线1在,处的切线斜率是1K曲线XFSIN在4处的切线方程是42XY设Y2,则YLN2XX设L,则1(三)计算题求下列函数的导数YXYE3XXEY2123LNCOT2LNCSXLXY2L3SY42COS32LSINXXXSINL2Y2SINL1IYL4XXLCO43X3SI2XXY223LNSICSYLNTAEEEX1OSTAN求下列函数的导数Y21X212XX3COSLNY323TANCOSINXYX87X81Y3Y231321XYXECOS2SINXE22XYNYISINCOSSI1XN2SI5X2IN5LXXY2SINEXEY2SINSI22XX22LXXEXXYEEXXXEEYXN在下列方程中,是由方程确定的函数,求Y2COSEXYY2INYEX2COSINL1COSSILNIYX2SIN2ICO2YXYXYXYSIN2COS2222COSINXYXL1Y12ELNXYYEXYXSI12XXYEYINCOYXCOS2IN3XY223YEXY5LNLYXY2LN15YX求下列函数的微分DCSOTDXXSICO22XYSINLDY2INLS11ARCDXXDXXDY22221131两边对数得1LNL3LNXY13XY13XYESI2DEDXEDX2SISI23TANXX2CC33求下列函数的二阶导数YLXN1XY1SIYCOXCOS2SINXYARCTN2121X3X3LN2Y223LN23LN4XXY(四)证明题设XF是可导的奇函数,试证F是偶函数证因为FX是奇函数所以X两边导数得1FFFXF所以F是偶函数。高等数学基础第三次作业(一)单项选择题若函数XF满足条件(D),则存在,BA,使得ABFFA在,BA内连续B在,内可导C在,内连续且可导D在,BA内连续,在,内可导函数142XF的单调增加区间是(D)A,B,C,2D,2函数52Y在区间6内满足(A)A先单调下降再单调上升B单调下降C先单调上升再单调下降D单调上升函数XF满足0F的点,一定是XF的(C)A间断点B极值点C驻点D拐点设F在,BA内有连续的二阶导数,,0BA,若XF满足(C),则XF在0取到极小值A00XFXB0XFC,FD,0F设在内有连续的二阶导数,且X,则XF在此区间内是(A)A单调减少且是凸的B单调减少且是凹的C单调增加且是凸的D单调增加且是凹的(二)填空题设XF在,BA内可导,,0BAX,且当0X时F,当0时F,则0X是F的极小值点若函数F在点0可导,且0是F的极值点,则0XF0函数1LN2XY的单调减少区间是,函数EF的单调增加区间是若函数在,BA内恒有XF,则XF在,BA上的最大值是AF函数352XF的拐点是X0(三)计算题求函数21Y的单调区间和极值令XX5,2驻点列表极大值27F极小值05求函数3YX在区间,0内的极值点,并求最大值和最小值令。XXY驻点1026F最大值21最小值试确定函数DCXBAY3中的DCBA,,使函数图形过点4,2和点10,,且2X是驻点,1X是拐点解BAC2604824163DCBA求曲线XY上的点,使其到点0,2A的距离最短解上的点是设P,,D为P到A点的距离,则XD22101XX令。AXY的距离最短到点上点0,12圆柱体上底的中心到下底的边沿的距离为L,问当底半径与高分别为多少时,圆柱体的体积最大设园柱体半径为R,高为H,则体积LHV22LHLH。33032令。L时其体积最大当,33一体积为V的圆柱体,问底半径与高各为多少时表面积最小设园柱体半径为R,高为H,则体积222RVSH表面积33204。令34VH答当2R3VH时表面积最大。欲做一个底为正方形,容积为625立方米的长方体开口容器,怎样做法用料最省解设底连长为X,高为H。则225656XHXX2,22,55,Y极大极小Y上升27下降0上升2130FF侧面积为XHXS25042令510532答当底连长为5米,高为25米时用料最省。(四)证明题当X时,证明不等式LNX证由中值定理得011LLX。XX时当0L1LN当0时,证明不等式EXEF设00FXFX单调上升且时当时当证毕即1,F请您务必删除一下内容,O_O万分谢谢2015年中央电大期末复习考试小抄大全,电大期末考试必备小抄,电大考试必过小抄INDONESIAHASEMERGEDASATOPVACATIONDESTINATIONFORCHINESETHISSUMMER,ALONGWITHTHAILANDSPHUKETISLANDANDTHEMALDIVESARCHIPELAGOENCOURAGEDBYGROWTHINCHINESETRAVELERSTOINDONESIA,THEGOVERNMENTINJAKARTAHASRECENTLYRELAXEDITSVISAPOLICYSINCEJUNE10,CHINESETOURISTSCANENTERINDONESIATHROUGHNINEAPPOINTEDLOCATIONS,INCLUDINGTHESOEKARNOHATTAINTERNATIONALAIRPORTINJAKARTA,THENGURAHRAIINTERNATIONALAIRPORTINBALIANDTHEKUALANAMUINTERNATIONALAIRPORTINMEDAN,BYJUSTGETTINGTHEIRPASSPORTSSTAMPEDONARRIVALOFFICIALLY,ITISCALLEDTHE“FREEVISA“SCHEMETHENEWPOLICYALSOMEANSCHINESETOURISTSCANSAVE35ONVISAFEES,WHICHWILLLIKELYMAKETRIPSTOINDONESIAMOREALLURINGWHILESUCHANARRANGEMENTALLOWSCHINESEVISITORSTOSTAYUPTO30DAYSFORTHEPURPOSEOFTRAVELING,THOSESEEKINGEXTENSIONSWILLNEEDTOGETPAIDVISASFROMTHATCOUNTRYSMISSIONSINBEIJING,SHANGHAI,GUANGZHOUORHONGKONG,ACCORDINGTOANOFFICIALATTHEINDONESIAEMBASSYINBEIJINGSANTODARMOSUMARTO,HEADOFTHEEMBASSYSINFORMATION,SOCIALANDCULTURALSECTION,SAYSTHENEWPOLICYISAIMEDATSTRENGTHENINGPEOPLETOPEOPLECONTACTSBETWEENTHETWOCOUNTRIESDAIYU,MARKETINGDIRECTOROFCTRIP,AMAJORCHINESEONLINETRAVELAGENCY,SAYS“WEVESEENANEARLY50PERCENTGROWTHOVERTHEPREVIOUSMONTHINTHENUMBEROFCHINESETOURISTSTOBALISINCETHEPOLICYWASANNOUNCED“DURINGHISTRIPTOCHINAINMARCH,INDONESIANPRESIDENTJOKOWIDODOPROPOSEDTHATHISCOUNTRYANDCHINAAIMTOINCREASETWOWAYVISITSTOAMAXIMUMOF10MILLIONPEOPLEINTHENEXTFEWYEARSTHEINDONESIANGOVERNMENTHASSETTHEIRSIGHTSONGREETING2MILLIONCHINESETOURISTSBYTHEENDOFTHISYEAR,ACCORDINGTOTHEOFFICIAL“THEINDONESIANGOVERNMENTHOPESTHATINDONESIANNATIONALSWOULDALSOBEGIVENPREFERENTIALTREATMENTWHILEVISITINGCHINA,“SAYSDARMOSUMARTOIN2013,THENUMBEROFCHINESETOURISTSWHOVISITEDINDONESIASTOODAROUND807,000LASTYEAR,ITINCREASEDTO959,000,HEADDSBALIREMAINSTHEMOSTPOPULARSITEAMONGCHINESETOURISTSINFEBRUARY,MORETHAN92,200CHINESETOURISTSMADETRIPSTOBALI,RANKINGFIRSTAMONGOVERSEASVISITORS,THEBALITIMESREPORTEDLASTYEAR,THEISLANDWITNESSEDMORETHAN586,000ARRIVALSFROMCHINA,ANINCREASEOF51PERCENTOVER2013“BALIISNTJUSTFORSIGHTSEEING,BUTALSOFORWEDDINGS,HONEYMOONS,GOLFANDMANYOUTDOORACTIVITIES,“SAYSFANWENQING,AMARKETINGEXECUTIVEATGARUDA,INDONESIASNATIONALAIRLINERGARUDAOFFERSTHREENONSTOPFLIGHTSBETWEENBEIJINGANDBALI,ANDFOUREACHWEEKBETWEENBEIJINGANDJAKARTATHEREAREDAILYFLIGHTSFROMGUANGZHOU,INGUANGDONGPROVINCE,ANDSHANGHAITOJAKARTAASWELLITTAKESFROMSIXTOEIGHTHOURSONANONSTOPFLIGHTTOREACHINDONESIAFROMCHINAINDONESIAISHOMETOMORETHAN17,000ISLANDSTOURISTSCANTOURYOGYAKARTATOSEEPRAMBANAMANDBOROBUDUR,TWOHISTORICALANDRELIGIOUSSITES,ANDEXPLORETHELOCALHANDICRAFTMARKETSTHEBASEOFAVOLCANO,BANDUNG,WHICHISALSOKNOWNASTHE“PARISOFJAVA“,ISIDEALFORADVENTURESEEKERSMANYECOTOURISMDESTINATIONS,SUCHASTOBALAKEINSUMATRA,TANJUNGPUTINGNATIONALPARKINKALIMANTANANDBUNAKENINMANADO,AREPLACESTHEEMBASSYOFFICIALRECOMMENDSFORTOURISTSINDONESIABOASTSWORLDCLASSSURFINGFACILITIESALONGTHESHORESOFSUMATRADOWNTONUSATENGGARAISLANDSRECENTLY,LOMBOK,ASISTERISLANDEASTOFBALIHASGOTTENSOMUCHATTENTIONFROMCHINESETRAVELERSTHATSOMETRAVELAGENCIESAREOFFERINGTWINPACKAGESFORBALIANDLOMBOKTHEBESTTIMETOVISITBALIISFROMMAYTOOCTOBER,WHENITDOESNTRAINMUCHANDTHEWEATHERISCOOLATPRESSTIME,AWEEKLONGPACKAGEFROMSHANGHAIORGUANGZHOUTOBALIWASALITTLEMORETHAN3,000YUAN480ONCTRIPSWEBSITECHINESETOURISTSHAVETAKENASHINETORAFTINGINUBUD,SIGHTSEEINGINTANAHLOT,ANDSURFINGANDDIVINGINNUSALEMBONGAN,SAYSDAICIVETCOFFEE,ESSENTIALOILANDWOODCARVINGSAREAMONGTHEMOSTPOPULARPRODUCTSFORCHINESETOURISTSININDONESIA,SHESAYSINDONESIANFOODISALSOSOMETHINGTHATVISITORSSHOULDNTMISSLOCALCUISINESFEATUREMANYSPICES,INCLUDINGPEPPERSANDCLOVES,ANDCOCONUTPULPFRIEDRICE,“DIRTYDUCK“ANDROASTSUCKLINGPIGARERECOMMENDEDCHILDRENSILLUSTRATIONBOOKSFIRSTAPPEAREDINTHE1600SINEUROPETHEIRPASSAGETOCHINA,HOWEVER,TOOKMUCHLONGER,ROUGHLY300YEARSINCHINA,CHILDRENSILLUSTRATIONBOOKSHAVEAHISTORYOFJUST100YEARSANDTHEIRDEVELOPMENTCANBEDIVIDEDINTOFOURSTAGESFROMEARLY1900STOMID1900S,ALARGENUMBEROFBOOKSANDMAGAZINESWITHILLUSTRATIONSAPPEAREDWITHTHEFASTDEVELOPMENTOFTHEMODERNPUBLISHINGINDUSTRYTHEILLUSTRATIONARTSATTHATTIMEWEREAMIXEDGENREOFCHINESETRADITIONALPAINTINGSANDTHEWESTERNSTYLE,ANDMOSTOFTHESTORIESORIGINATEFROMTHEFUNDAMENTALCHANGESTHATCHINAEXPERIENCEDAFTERTHECOLLAPSEOFQINGDYNASTY16441911,THEWARLORDSTRUGGLESANDTHEWARAGAINSTTHEJAPANESEINVASIONTHESECONDSTAGEWASFROMMID1900STOLATE1970S,WHENTHE“CULTURALREVOLUTION“196676ENDEDTHEFASTDEVELOPMENTOFSOMESPECIALPUBLISHINGHOUSESFORCHILDRENGAVEBIRTHTOALARGENUMBEROFILLUSTRATIONSTORYBOOKS,WHICHMADEAGOODUSEOFFOLKARTTECHNIQUES,SUCHASWOODBOARDCARVING,FRESCOES,ANDINKANDWASHPAINTING,ETCMANYOFTHEPOPULARSTORIESAREFROMCHINESEHISTORYANDWARSAFTER1900STHETHIRDSTAGEISFROMEARLY1980STOLATE1990S,AGOLDENPERIODFORTHEDEVELOPMENTOFCHILDRENSILLUSTRATIONBOOKSTHEREFORMANDOPENINGUPBROUGHTINNEWCONCEPTSANDIDEASFROMTHEWESTT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