2018年电大《财务管理》期末考试复习题及参考答案资料小抄_第1页
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小抄版高等数学基础归类复习考试小抄一、单项选择题11下列各函数对中,(C)中的两个函数相等A,B,2XFXG2XFXGC,D,3LNLN1121设函数的定义域为,则函数的图形关于(C)对称F,FA坐标原点B轴C轴DXYX设函数的定义域为,则函数的图形关于(D)对称XFAB轴C轴D坐标原点Y函数的图形关于(A)对称2EXA坐标原点B轴C轴DYXY1下列函数中为奇函数是(B)ABCD1LN2XYXYCOS2XA1LNX下列函数中为奇函数是(A)ABCD3XE1LNYYSI下列函数中为偶函数的是(D)ABCDXYSIN1X2XCOS1LN2X21下列极限存计算不正确的是(D)AB2LIMX01LNIM0XCD0SNS22当时,变量(C)是无穷小量ABCDXIX1X1IN2LX当时,变量(C)是无穷小量ABCD0SI1E2X当时,变量(D)是无穷小量ABCDXXLN下列变量中,是无穷小量的为(B)ABCD1SIN0XLN101XE24X31设在点X1处可导,则(D)XFHFFH12LIM0ABCD1112F设在可导,则(D)F0XFXFHLI00ABCDX20XF设在可导,则(D)F0FFHLI00ABCDFX20F设,则(A)ABCDXFEFF1LIM0E2E1432下列等式不成立的是(D)ABCDXXDCOSSINXDXD2LNXD下列等式中正确的是(B)ABARTN1221CDXXLXCOTTA41函数的单调增加区间是(D)42XFABCD,函数在区间内满足(A)5Y6A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单调上升函数在区间(5,5)内满足(A)62XA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单调上升函数在区间内满足(D)Y,2A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51若的一个原函数是,则(D)ABCDXFX1FXLN21X32若是的一个原函数,则下列等式成立的是(A)。FFABAFXDXAAFBFDXBA小抄版CDXFFAFBDXFBA52若,则(B)COSXFDABCDINCSCSINCXOS下列等式成立的是(D)ABDFXFXFFCDD(B)ABCDXFXD323XF32XF1XF1(D)ABCDFX22XFXFD12FXFD23若,则(B)CXFDABCDCX2CXFCXF1补充,无穷积分收敛的是XEFXDED12函数的图形关于Y轴对称。X10二、填空题函数的定义域是(3,)LN392XF函数的定义域是(2,3)(3,4XY4LN函数的定义域是(5,2)XF1若函数,则10,2XFF2若函数,在处连续,则E,1KXFK函数在处连续,则202SINXKXFK函数的间断点是X0,SIN1Y函数的间断点是X3。32X函数的间断点是X0EY13曲线在处的切线斜率是1/2F2,1曲线在处的切线斜率是1/4X曲线在(0,2)处的切线斜率是1F曲线在处的切线斜率是33,32曲线在处的切线方程是Y1切线斜率是0XFSIN1曲线YSINX在点0,0处的切线方程为YX切线斜率是14函数的单调减少区间是(,0)1L2函数的单调增加区间是(0,)EXF函数的单调减少区间是(,1)2函数的单调增加区间是(0,)F函数的单调减少区间是(0,)2XY51DEEX2XDSIN22SITANXCTAN若,则9SIN3XCF3SIF523035D21123DX0EXDX1LN下列积分计算正确的是(B)ABCDD1EX0D1XEX0D12X小抄版0D|1X三、计算题(一)、计算极限(1小题,11分)(1)利用极限的四则运算法则,主要是因式分解,消去零因子。(2)利用连续函数性质有定义,则极限0XFLIM00XFFX类型1利用重要极限,计算1SINLXKSNKXTANLI11求解X5SIN6LM056SIL5SI6L00XX12求解0TALI3XX3TANLM031TANLI1013求解IX类型2因式分解并利用重要极限,化简计算。1SLAX1SINLAA21求解1SINLM21XXSINLM21XX21I22解21SLX211SINLMSINL21XXX23解3SIN4MXLI3SI3SI43323X类型3因式分解并消去零因子,再计算极限31解4586LI24X4586LI24XX12LIXX32LI4X3231X3335M17X33解4LIM2X412LIM2LI4LI2XXXX其他,0SIN21LMSIN1L020XXXX21SINLMSINL00XX,546LI2X1LI2X54362LIX3LI2X(0807考题)计算解XSIN8TAL0XSIN8TAL048SITAL0X(0801考题)计算解X2LIM0X2LIM021LI0X(0707考题)1SIN31431SINL1X(二)求函数的导数和微分(1小题,11分)(1)利用导数的四则运算法则VUVU(2)利用导数基本公式和复合函数求导公式XLN1AAXEUEUXX2CSOTANICSSIXEXEXXXXSINCOSCIN2OCOSISIIN22XXXEEEUOSSSINI22XXEEEUSINSINCO2I22类型1加减法与乘法混合运算的求导,先加减求导,后乘法求导;括号求导最后计算。11XY3解322XEE1322XXE132XE12XYLNCOT解小抄版XXXXXYLN2CSLNLCSLNCOT22213设,求EXAY解XEXXEYXX1SCTA1TATLT2类型2加减法与复合函数混合运算的求导,先加减求导,后复合求导21,求解LNSI2YOLNSI2222,求ECOXY解222CSESICOSSINSIXXEXXX23,求,解X5LNYY5455LNL类型3乘积与复合函数混合运算的求导,先乘积求导,后复合求导,求。解EYXCOS2XEXEXSICO2CS22其他,求。XY解2CLNOXYX2CSINL2XX0807设,求解SIEY2SININCOSXEYX0801设,求解2X222XXXEE0707设,求解SINCOSININSI0701设,求解XYECOLYXXY1L(三)积分计算(2小题,共22分)凑微分类型11D2XX计算解XCOS2CXD1SINCOSCOS20707计算解XD1SIN2CXX1OSDSIN1SI20701计算解E2ED121XE凑微分类型2X计算解XDCOSCXDXXSIN2COSDCOS0807计算解INOIN2IN0801计算解XEDCEXDEXEX凑微分类型3,DLN1LN1A计算解XDLNCXDUX|LLL计算解E12E1E1LN2DL2X5L1EX5定积分计算题,分部积分法类型1CXAXDAXAXDAXDAAA12111LNLNLNLN计算解,E124L241LNLN2LXD21EXXE0LLNE1E小抄版计算解,E12DLNX2ACXXDX1LN1LNL2E2NE1E1计算解,DXE1LN2ACXXDX4LN2LLDE1N44LLN21EXXDE0807EL94219LN32ND3233E1EX0707EE12XLLX33类型2CEAXDAXA21XXEDE2101041042XX11EXXXDEDE21010241304222EXX(0801考题)E10X类型3CAXADAXAXDSIN1COCOSCSSIN2XXSIN1IN1ICO20SINXD02SICOS20X20COSXD120COSINSI20XXCXDSIN421COIN20SXD402I1COSS20XX2200011COIN|IN|4D四、应用题(1题,16分)类型1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,圆柱体高与底半径满足HR22RH圆柱体的体积公式为LV2求导并令032L得,并由此解出LR6即当底半径,高时,圆柱体的体积最大LR36H类型2已知体积或容积,求表面积最小时的尺寸。21(0801考题)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,则其容积RH22,RHR表面积为RS2,由得,此时。24RV0S32V34VRH由实际问题可知,当底半径与高时可使用料最省。3R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解本题的解法和结果与21完全相同。生产一种体积为V的无盖圆柱形容器,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,则无盖圆柱形容器表面积为RHL小抄版,令,得RVRHS2202RVS,VR,3由实际问题可知,当底半径与高时可使用料最省。3RRH22欲做一个底为正方形,容积为32立方米的长方体开口容器,怎样做法用料最省(0707考题)解设底边的边长为,高为,用材料为,由已知,XHY322VHX2XH表面积,XVY422令,得,此时2042XV63,4X2由实际问题可知,是函数的极小值点,所以当,时用料最省。H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本题的解法与22同,只需把V625代入即可。类型3求求曲线上的点,使其到点的距离最短KXY20,AA曲线上的点到点的距离平方为,AAXL2,0KX231在抛物线上求一点,使其与轴上的点的距离最短XY420,3解设所求点P(X,Y),则满足,点P到点A的距离之平方为Y4322令,解得是唯一驻点,易知是函数的极小值点,0L1X1X当时,或,所以满足条件的有两个点(1,2)和(1,2)132求曲线上的点,使其到点的距离最短XY2,A解曲线上的点到点A(2,0)的距离之平方为X令,得,由此,L122XYY即曲线上的点(1,)和(1,)到点A(2,0)的距离最短。XY208074求曲线上的点,使其到点A(0,2)的距离最短。解曲线上的点到点A(0,2)的距离公式为2XY222YYXD与在同一点取到最大值,为计算方便求的最大值点,2D31Y令得,并由此解出,0223Y6X即曲线上的点()和点()到点A(0,2)的距离最短X,6,2AGANEMPLOYMENTTRIBUNALCLAIEMLOYMENTTRIBUNALSSORTOUTDISAGREEMENTSBETWEENEMPLOYERSANDEMPLOYEESYOUMAYNEEDTOMAKEACLAIMTOANEMPLOYMENTTRIBUNALIFYOUDONTAGREEWITHTHEDISCIPLINARYACTIONYOUREMPLOYERHASTAKENAGAINSTYOUYOUREMPLOYERDISMISSESYOUANDYOUTHINKTHATYOUHAVEBEENDISMISSEDUNFAIRLYFORMOREINFORMU,TAKEADVICEFROMONEOFTHEORGANISATIONSLISTEDUNDERFURTHERHELPEMPLOYMENTTRIBUNALSARELESSFORMALTHANSOMEOTHERCOURTS,BUTITISSTILLALEGALPROCESSANDYOUWILLNEEDTOGIVEEVIDENCEUNDERANOATHORAFFIRMATIONMOSTPEOPLEFINDMAKINGACLAIMTOANEMPLOYMENTTRIBUNALCHALLENGINGIFYOUARETHINKINGABOUTMAKINGACLAIMTOANEMPLOYMENTTRIBUNAL,YOUSHOULDGETHELPSTRAIGHTAWAYFROMONEOFTHEORGANISATIONSLISTEDUNDERFURTHERHELPATIONABOUTDISMISSALANDUNFAIRDISMISSAL,SEEDISMISSALYOUCANMAKEACLAIMTOANEMPLOYMENTTRIBUNAL,EVENIFYOUHAVENTAPPEALEDAGAINSTTHEDISCIPLINARYACTIONYOUREMPLOYERHASTAKENAGAINSTYOUHOWEVER,IFYOUWINYOU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