南昌大学高数试题及答案.pdf

该【南昌大学高数试题及答案 】是由【小屁孩】上传分享，文档一共【54】页，该文档可以免费在线阅读，需要了解更多关于【南昌大学高数试题及答案 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..南昌大学2006~2007学年第一学期期末考试试卷一、填空题(每空3分,共15分):(x)?x?2??lg(5?x)x?3的定义域为_______________.?e?x,x?0,(x)??ln(a?x),x?0,?则a为_____值时,f(x)在x=0处连续.(a>0)(x)在x=0可导,且f(0)=0,f(x)则lim??(x)?x,在[1,4]上使Lagrange(拉格朗日)中值定理成立的??(x)??sintdt,则dF(x)?、单项选择题(每题3分,共15分):?0是函数f(x)?xsin的().x(A)跳跃间断点.(B)可去间断点.(C)无穷间断点.(D)?e1?x2与直线x??1相交于点P,曲线过点P处的切线方程为().(A)2x?y?1?0.(B)2x?y?3?0.(C)2x?y?3?0.(D)2x?y?2?:..(x)在区间(a,b)内可导,x和x是区间(a,b)内任12意两点,且x?x,则至少存在一点?使().12(A)f(b)?f(a)?f'(?)(b?a),其中a???b.(B)f(b)?f(x)?f'(?)(b?x),其中x???(C)f(x)?f(x)?f'(?)(x?x),其中x???(D)f(x)?f(a)?f'(?)(x?a),其中a???(x)在(??,??)上连续,则d??f(x)dx?等于().??(A)f(x).(B)f(x)dx.(C)f(x)?C.(D)f'(x)??f(x)dx??f(x)dx??f'(x)dx存在,dxdx3则I?().(A)0.(B)f(x).(C)2f(x).(D)2f(x)?、计算下列极限(共2小题,每小题7分,共14分):1??01?(sinx)tanx.?x?22:..(共3小题,每小题7分,共21分):1??ln,求y''(0).1??y(x)由方程ln(x2?y)?x3y?sinx确定,求y'(0).?t2x??f(u)du,??其中f(u)具有二阶导数,且f(u)?0,2?y??f(t2)?,???(共2小题,每小题7分,共14分):1?1、(1?x8)2.?(2)?,f'(2)?0,及?f(x)dx?1,2012求?xf''(2x)dx.(7分)03:..?,试求其单增、单减区间,(1?x)2并求该函数的极值和拐点.(9分)(x)在[a,??)上连续,f''(x)在(a,??)内存在且大于f(x)?f(a)零,记F(x)?(x?a).证明:F(x)在(a,??)内x?a单调增加.(5分)南昌大学2006~2007学年第一学期期末考试试卷及答案一、填空题(每空3分,共15分):(x)?x?2??lg(5?x)x?3的定义域为(2?x?3与3?x?5;)?e?x,x?0,(x)??ln(a?x),x?0,?则a为(e)值时,f(x)在x=0处连续.(a>0)(x)在x=0可导,且f(0)=0,f(x)则lim?(f'(0))x?(x)?x,在[1,4]上使Lagrange(拉格朗日)中值定理成立的??(9/4)2x22一、(x)??sintdt,则dF(x)?(2sin(4x)dx)0二、单项选择题(每题3分,共15分):4:..?0是函数f(x)?xsin的(B).x(A)跳跃间断点.(B)可去间断点.(C)无穷间断点.(D)?e1?x2x??,曲线过点P处的切线方程为(C).(A)2x?y?1?0.(B)2x?y?3?0.(C)2x?y?3?0.(D)2x?y?2?(x)在区间(a,b)内可导,x和x是区间(a,b)内任12意两点,且x?x,则至少存在一点?使(C).12(A)f(b)?f(a)?f'(?)(b?a),其中a???b.(B)f(b)?f(x)?f'(?)(b?x),其中x???(C)f(x)?f(x)?f'(?)(x?x),其中x???(D)f(x)?f(a)?f'(?)(x?a),其中a???(x)在(??,??)上连续,则d??f(x)dx?等于(B).??(A)f(x).(B)f(x)dx.(C)f(x)?C.(D)f'(x)??f(x)dx??f(x)dx??f'(x)dx存在,dxdx3则I?(D).(A)0.(B)f(x).5:..(C)2f(x).(D)2f(x)?、计算下列极限(共2小题,每小题7分,共14分):1??01?cosx11??21解:x?0时,1?cosxx2,1?cosx2x2?x42221x41?cosx22?lim?lim?21?cosx1x?0x?(sinx)tanx.?x?2??tanx解:(1)令y?sinxlny?tanxlnsinx(2)limlny?limtanxlnsinx???x?x?221cosxlnsinxsinx?lim?lim?0cotx?csc2x??x?x?22(3)lim(sinx)tanx?limy?limelny?e0?1???x?x?x?(共3小题,每小题7分,共21分):1??ln,求y''(0).1?x21?x1解:y?ln?[ln(1?x)?ln(1?x2).1?x226:..1??12x?1?12x??y'?????.3分????2?1?x1?x2?2?1?x1?x2?1?12(1?x2)?4x2?11?x2y''????????.2(1?x)2(1?x2)22(1?x)2(1?x2)2??13于是y''(0)???1??.?y(x)由方程ln(x2?y)?x3y?sinx确定,求y'(0).解:方程两边对x求导,得1?2x?y'??3x2y?x3y'??y当x?0时,由原方程得y?1,代入上式得y'(0)??t2x??f(u)du,??其中f(u)具有二阶导数,且f(u)?0,2?y??f(t2)?,????4tf(t2)f'(t2),?f(t2).解:dtdt7:..dydy4tf(t2)f'(t2)dt????4tf'(t2).dxdxf(t2)dt?dy?d???dy??dx?d??d2ydx4f'(t2)?8t2f''(t2)??dt????.dx2dxdxf(t2)(共2小题,每小题7分,共14分):1?1、(1?x8)x7111?11?解:原式=?dx??dx8=??dx8????x8(1?x8)8888x81?x8x1?x??1?ln|x|?ln|1?x8|?.??cos2x11解:原式??xdx??xdx??xcos2xdx224x211??xsin2x?cos2x?(2)?,f'(2)?0,及?f(x)dx?1,2012求?xf''(2x)dx.(7分)0解:设t?2x,则1t2122?xf''(x)dx??f''(t)dt02048:..11?222??2??tf'(t)?2?tf'(t)dt??2?tdf(t)8?00?8?0?11?22???tf(t)??f(t)dt??(1?1)??00??,试求其单增、单减区间,(1?x)2并求该函数的极值和拐点.(9分)4x8x?4解:y'?,y''?.(1?x)3(1?x)41令y'?0,得x?0;令y''?0,得x??.2x(??,?1/2)?1/2(?1/2,0)0(0,1)(1,??)y'--0+-y''-0++++y拐点极小值故(0,1)为单增区间,(??,0)和(1,??)为单减区间;函数在x?0处取得极小值,极小值为0;点(?1/2,2/9)(x)在[a,??)上连续,f''(x)在(a,??)内存在且大于f(x)?f(a)零,记F(x)?(x?a).证明:F(x)在(a,??)内x?a单调增加.(5分)1?f(x)?f(a)?证明:F'(x)?f'(x)?.??x?a?x?a?由拉格朗日中值定理知存在??(a,x),使9:..f(x)?f(a)?f'(?).x?a1?F'(x)??f'(x)?f'(?)?.x?a由f''(x)?0可知f'(x)在(a,??)内单调增加,因此对任意x和?(a???x),有f'(x)?f'(?),从而F'(x)?0,故F(x)在(a,??):..南昌大学2009~2010学年第一学期期末考试试卷一、填空题(每空3分,共15分)?arcsin?ln?x?1?,则它的定义域为。3??x???e,则它在0,ln2上满足拉格朗日中值定理的结论中的??。?esin2x,则dy?。f?x??f?x?3?x?f??x?lim00?,则。0?x?x?0f?x??xe?。二、单项选择题(每小题3分,共15分)?x2,x?1,?f(x)?f?x?x??3,则在处的().x?,x?1?3(A)左导数存在,右导数存在(B)左导数存在,右导数不存在(C)左导数不存在,右导数存在(D)左导数不存在,右导数不存在1f?x??arctanx?,则是().x(A)可去间断点(B)无穷间断点(C)振荡间断点(D)跳跃间断点f?x??a,b?,则下列论断不正确的是()11:..b????(A)?fxdx是fx的一个原函数a??x????(B)在a,b上,?ftdt是fx的一个原函数a??b????(C)在a,b上,??ftdt是fx的一个原函数x????(D)fx在a,b上可积y?f?x?,则().y?1?1oxx??1f?x?(A)是的驻点,但不是极值点x??1f?x?(B)不是的极值点(C)x??1是f?x?的极小值点x??1f?x?(D)????,n222n?n?1n?n?2n?n?n则limx?()nn??11(A)0(B)1(C)(D)23三、计算题(共7小题,每小题7分,共49分)esin3x??1?cosx?x??cos2x?lntanx,求y?.??ex?x2?ax??2,求a,?012:..?y?x?是由方程ey?xy?e所确定的隐函数,求y??0?.???ex?1?e3x且f?0??1,求f?x?.?.sin2x?5cos2x?0,x?1,?3??(x)??,计算?fx?,x?10??x2四、解答题(共2小题,每小题8分,共16分)?t?2?x???0?3y?asint??x??0,1???31??且fx??x??fxdx,1?x201??求?、证明题(5分)??0f?x??x,x???设,若在上连续,00?x,x???f??x??A在内可导且lim,000x?x?0f??x??A证明:.?013:..南昌大学2009~2010学年第一学期期末考试试卷及答案一、填空题(每空3分,共15分)xy?arcsin?ln?x?1??1,3?,则它的定义域为3??x???e,则它在0,ln2上满足拉格朗日中值定理的结论中的???lnln2。?esin2x,则dy?dx。sin2xf??x?,则0f?x??f?x?3?x?lim00??3f??x?。?x0?x?0???x???xe的凹区间为2,??.二、单项选择题(每小题3分,共15分)?x2,x?1,?f(x)?f?x?x??3,则在处的(B).x?,x?1?3(A)左导数存在,右导数存在(B)左导数存在,右导数不存在(C)左导数不存在,右导数存在(D)左导数不存在,右导数不存在1f?x??arctanx?,则是(D).x(A)可去间断点(B)无穷间断点14:..(C)振荡间断点(D)跳跃间断点f?x??a,b?,则下列论断不正确的是(A)b????(A)?fxdx是fx的一个原函数a??x????(B)在a,b上,?ftdt是fx的一个原函数a??b????(C)在a,b上,??ftdt是fx的一个原函数x????(D)fx在a,b上可积y?f?x?,则(C).y?1?1oxx??1f?x?(A)是的驻点,但不是极值点(B)x??1不是f?x?的极值点x??1f?x?(C)是的极小值点x??1f?x?(D)????,nn2?n?1n2?n?2n2?n?n则limx?(C)nn??11(A)0(B)1(C)(D)23三、计算题(共7小题,每小题7分,共49分)esin3x??1?cosx?x?015:..sin3x解:原式=lim?21x?02x??cos2x?lntanx,求y?.y???2sin?2x??lntanx?cos2x?cotx?sec2x解:??2sin?2x??lntanx?2cot2x??ex?x2?ax??2,求a,?0??ex?x2?ax?b解:lim?2xx?0???lim?ex?x2?ax?b??0??x?0?1?b?0即b?1??ex?x2?ax?b??又lim?limex?2x?a?1?a?2xx?0x?0?a??1y?y?x?ey?xy?ey??0?,:方程两边同时对x求导,有:eyy??y?xy??0yy???ey?x当x?0时,从原方程得y?:y'(0)??.e16:..??f?ex?1?e3xf?0??1f?x?,?exf??t??1?t3则:??1f?t??1?t3dt?t?t4?C?4f?0??1C?1又所以11f?t??t?t4?1f?x??x?x4??.sin2x?5cos2x11??解:原式?dx??dcotx??2sin2x1?5cot2x1?5cotx11??1?????d5cotx??arctan5cotx?C5??251?5cotx?0,x?1,?3??(x)??,计算?fx?,x?10??x2解:令t?x?1,则dt?dx由于x:0?3t:?1?23??2????fx?1dx??ftdt?0?11??2????ftdt??ftdt??1117:..112??0dt??dt??11t2211???t21四、解答题(共2小题,每小题8分,共16分)?t?2?x??3acosusinudu1、求由参数方程?0?3y?asint??costdt解:????tantdxdx?3acos2t?sintdt?dy?d???dy??dx?d??d2ydx?sec2x1??dt????dx2dxdx?3acos2tsint3acos4tsintdt????2、设fx在0,1上连续1??31??且fx??x??fxdx,1?x201??求???解:令A??fxdx,018:..1f?x???Ax31?x2?1?13?A???Axdx???0?1?x2?1?A??1?arctanx?x4??A???4?440??1??故A?即?fxdx?505五、证明题(5分)??0f?x??x,x???设,若在上连续,00?x,x???limf??x??A在内可导且,00x?x?0f??x??A证明:.?0证明:f?x??f?a?f??x??lim?证法一:?0x?ax?x?0f??x??lim?limf??x??A1x?x?x?x?00证法二:f?x??f?a?f??x??lim因为?0x?ax?x?0f?x??x,x???又在上连续,00?x,x???在内可导,所以由拉格朗日中值定理可知,00?x??x,x???f?x??f?x??f?????x?x?,有,0000其中x???x0所以19:..f?x??f?a?f?????x?x?f??x??lim?lim0?limf?????0x?ax?ax?x?x?x?x?x?000x?x???x?limf??x??A当时,且,00x?x?0limf????limf?????A所以=,x?x???x?00f??x??A即.?020:..南昌大学2008~2009学年第一学期期末考试试卷三、填空题(每空3分,共15分)(x)的定义域是[0,1],则函数f(x?a)(a?0)的定义域是。?。3nn???f(ex),f(x)为可导函数,则dy?,单调增加区间为。(1,3)是曲线y?ax3?bx2的拐点,则a?,b?。四、单项选择题(每小题3分,共15分)()。?cosx,x?0,(A)f(x)??sinx,x?0.?(B)f(x)?lnx?cosx.?1,x?0,?(C)f(x)?x???0,x?0.?x?1,x?0,?(D)f(x)??0,x?0,?x?1,x?0.??arctanx在横坐标为1的点处的切线方程是()。21:..?11(A)y??(x?1);(B)y?(x?1);422?(C)y??x?1;(D)y?x?1。[?1,1]上满足拉格朗日中值定理条件的函数是()。1(A);(B)lnx;x?211(C)?2x?1?;(D)arctan。?x3?4x2?3x?4的凹区间是().44(A)[,??);(B)(??,];33(C)[?2,0];(D)没有凹区间。?y(x)是可微函数且由方程yt20?edt??costdt?0所确定,则y'(x)?()。0x(A)ey2cosx;(B)ey2;(C)?e?y2cosx;(D)e?y2cosx。三、求下列极限(共2小题,每小题8分,共16分)x?1?2x?3?。??x???2x?1?x2t2?。xex2x?0四、求下列导数(共2小题,每小题7分,共14分)22:..1、设y?3x?3x,求y'(x)。?x?ln(1?t2),d2y2、设?求。y?t?arctant,dx2?五、求下列不定积分(共2小题,每小题7分,共14分)1、?tan3xsecxdx。lnx?2、dx。x2六、计算题(共2小题,每小题7分,共14分)?1、计算定积分?cos2x?cos4xdx。02、(应用题)某车间靠墙壁要盖一间长方形小屋,现有存砖只够砌20m长的墙壁,问应围成怎样的长方形才能使这间小屋的面积最大?七、解下列各题(共2小题,第1小题7分,第2小题5分,共12分)?1x3sin,x?0,?1、讨论函数f(x)??x??0,x?0在x?0处的连续性与可导性。2、设函数f(x)在[0,2]上连续,在(0,2)内可导,且f(0)?f(1)?2,f(2)?:必存在??(0,2),使f'(?)?0。23:..南昌大学2008~2009学年第一学期期末考试试卷及答案三、填空题(每空3分,共15分)(x)的定义域是[0,1],则函数f(x?a)(a?0)的定义域是??a,1?a?。?2。3nn???f(ex),f(x)为可导函数,则dy?f'(ex)exdx。?ax3?bx2的拐点,(1,3)39则a??,b?。22四、单项选择题(每小题3分,共15分)1、下列函数在其定义域内连续的是(B)。?cosx,x?0,(A)f(x)??(B)f(x)?lnx?,x?0.??1?x?1,x?0,,x?0,??(C)f(x)?x(D)f(x)?0,x?0,????0,x??1,x?0.??2、曲线y?arctanx在横坐标为1的点处的切线方程是(A)?11(A)y??(x?1);(B)y?(x?1);422?(C)y??x?1;(D)y?x?1。424:..3、在区间[?1,1]上满足拉格朗日中值定理条件的函数是(A)。1(A);(B)lnx;x?211(C)?2x?1?;(D)arctan。3x4、曲线y?x3?4x2?3x?4的凹区间是(A).44(A)[,??);(B)(??,];33(C)[?2,0];(D)没有凹区间。yt205、函数y?y(x)是可微函数且由方程?edt??costdt?00x所确定,则y'(x)?(D)。(A)ey2cosx;(B)ey2;(C)?e?y2cosx;(D)e?y2cosx。三、求下列极限(共2小题,每小题8分,共16分)x?1?2x?3?。??x???2x?1?x?1?2?解:原式?lim1???x???2x?1?2?x?1??2x?1?2x?1?2?2???lim1?????2x?1??x?????e25:..x2t2?。x?0xex2解:由洛必达法则有x2ex2原式?limx?0ex2?2x2ex2x2?lim?01?2x2x?0四、求下列导数(共2小题,每小题7分,共14分)1、设y?3x?3x,求y'(x)。2?2?1???1?解:y'?x?3x3?1?x3???33???x?ln(1?t2),d2y2、设?求。y?t?arctant,dx2?dy1t2dx2t解:?1??,?.dt1?t21?t2dt1?t2dydydtt???.dxdx2dt?dy?d???dx?1?.dt2?dy?d???dy?dx1??d??d2ydx1?t2??dt2?????.dx2dxdx2t4tdt1?t226:..五、求下列不定积分(共2小题,每小题7分,共14分)1、?tan3xsecxdx。解:原式??tan2xd(secx)??(sec2x?1)d(secx)1?sec3x?secx?C3lnx?。x21?解:原式??lnxdxlnx1????dxxx2lnx1????Cxx六、计算题(共2小题,每小题7分,共14分)?1、计算定积分?cos2x?cos4xdx。0??解:原式??cos2x(1?cos2x)dx??cos2xsin2xdx00???cosxsinxdx0????2cosxsinxdx??cosxsinxdx?02????2sinxd(sinx)??sinxd(sinx)?0227:..?0121?sin2x?sin2x??022.(应用题)某车间靠墙壁要盖一间长方形小屋,现有存砖只够砌20m长的墙壁,问应围成怎样的长方形才能使这间小屋的面积最大?解:,长为y,则小屋的面积为S??y?20,即y?20??x(20?2x)?20x??(0,10).S'?20?'?0,得驻点x?''??4?0知x?5为极大值点,又驻点唯一,,长为10m时,、解下列各题(共2小题,第1小题7分,第2小题5分,共12分)?1x3sin,x?0,?(x)??x??0,x?0在x?0处的连续性与可导性。28:..1解:limf(x)?limx3sin?0x?0x?0x而f(0)?0,?f(x)在x??0f(x)?f(0)x又f'(0)?lim?limx?0x?0x?0x?01?limx2sin??0x故f(x)在x?(x)在[0,2]上连续,在(0,2)内可导,且f(0)?f(1)?2,f(2)?:必存在??(0,2),使f'(?)?0。证明:f(x)在[0,2]上连续,?f(x)在[0,1]上连续,且在[0,1]上必有最大值M和最小值m,于是m?f(0)?M,m?f(1)??f(0)?f(1)?2Mf(0)?f(1)m??,至少存在一点c?[0,1],使f(0)?f(1)f(c)??(c)?f(2)?1,且f(x)在[c,2]上连续,在(c,2)内可导,由罗尔定理知,必存在??(c,2)?(0,2),使f'(?)?:..南昌大学2010~2011学年第一学期期末考试试卷五、填空题(每空3分,共15分)y?ex2f???x???1?x??x??0??x??,且,则。?????2.?2x2011?sinxdx?。??21???dx?。2??2xlnxlimn?ln?n?1??lnn??。??n???x3?2x?xx,则dy?。六、单项选择题(每小题3分,共15分)f?x?g?x?,dx????则?fxgtdt?().dxaf?x?g?x?f??x?g??x?(A)(B)f??x?g?x??f?x?g??x?(C)??????x??(D)fxgx?f?x?gtdtaf?x??ex?3x?2x?0f?x?,当时,是比的()(A)高阶无穷小(B)低阶无穷小(C)等价无穷小(D)非等价的同阶无穷小f?x??a,b??a,b??,则在上至少有一点,使得()f?????0f????0(A)(B)30:..b???????fxdxfb?faf????af?????(C)(D)b?ab?a?12?lnx?x?1??e?1?f?x??,?,在内()??1?e???11?x?3????x(A)不满足拉格朗日定理条件;9e?3(B)满足拉格朗日定理条件且??;5e(C)满足拉格朗日定理条件,但?无法求出;(D)不满足拉格朗日定理条件,9e?3但有??满足中值定理的结论。5