2022年1993考研数二真题及解析 .pdf

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1、1993 年全国硕士研究生入学统一考试数学二试题一、填空题(本题共 5 小题,每小题 3 分,满分 15 分.把答案填在题中横线上.)(1)0limlnxxx.(2)函数()yy x由方程222sin()0 xxyexy所确定,则dydx.(3)设11()(2)(0)xF xdt xt,则函数()F x的单调减少区间是.(4)tancosxdxx.(5)已知曲线()yf x过点1(0,)2,且其上任一点(,)x y处的切线斜率为2ln(1)xx,则()f x.二、选择题(本题共 5 小题,每小题3 分,满分 15 分.每小题给出的四个选项中,只有一项符合题目要求,把所选项前的字母填在题后的括号

2、内.)(1)当0 x时,变量211sinxx是 ()(A)无穷小 (B)无穷大(C)有界的,但不是无穷小 (D)有界的,但不是无穷大(2)设2|1|,1,()1 2,1,xxf xxx则在点1x处函数()f x ()(A)不连续 (B)连续,但不可导(C)可导,但导数不连续 (D)可导,且导数连续(3)已知2,01,()1,12,xxf xx设1()()xF xf t dt(02)x,则()F x为 ()(A)31,013,12xxxx (B)311,0133,12xxxx(C)31,0131,12xxxx (D)311,01331,12xxxx(4)设常数0k,函数()lnxf xxke在(

3、0,)内零点个数为 ()(A)3 (B)2 (C)1 (D)0(5)若()()f xfx,在(0,)内()0,()0fxfx,则()f x在(,0)内 ()(A)()0,()0fxfx (B)()0,()0fxfx(C)()0,()0fxfx (D)()0,()0fxfx三、(本题共 5 小题,每小题 5 分,满分 25 分.)(1)设2sin()yf x,其中f具有二阶导数,求22d ydx.(2)求2lim(100)xxxx.(3)求401cos2xdxx.(4)求30(1)xdxx.(5)求微分方程2(1)(2cos)0 xdyxyx dx满足初始条件01xy的特解.四、(本题满分9 分

4、)设二阶常系数线性微分方程xyyye的一个特解为2(1)xxyex e,试确定常数,并求该方程的通解.五、(本题满分9 分)设平面图形A由222xyx与yx所确定,求图形A绕直线2x旋转一周所得旋转体的体积.六、(本题满分9 分)作半径为r的球的外切正圆锥,问此圆锥的高h为何值时,其体积V最小,并求出该最小值.七、(本题满分6 分)设0 x,常数ae,证明()aaxaxa.八、(本题满分6 分)设()fx在0,a上连续,且(0)0f,证明：20()2aMaf x dx,其中0max|()|x aMfx.文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:C

5、H5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9

6、R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:C

7、H5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9

8、R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:C

9、H5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9

10、R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:C

11、H5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C6文档编码:CH5Q6B9V9Q4 HE4S4U6L9R8 ZB3U6K6F2C61993 年全国硕士研究生入学统一考试数学二试题解析一、填空题(本题共 5 小题,每小题 3 分,满分 15 分.)(1)【答案】0【解析】这是个0型未定式,可将其等价变换成型,从而利用洛必达法则进行求解.000021lnlimlnlimlimlim011xxxxxxxxxxx洛.(2)【答案】222222 cos()2 cos()2xyexxyyxyxy【解析】这是一个由复合函数和隐函数所确定的函数,将方程222sin()0 xxyexy两边对x

12、求导,得222cos()(22)20 xxyxyyeyxyy,化简得222222 cos()2 cos()2xyexxyyyxyxy.【相关知识点】复合函数求导法则：如果()ug x在点x可导,而()yf x在点()ug x可导,则复合函数()yfg x在点x可导,且其导数为()()dyfug xdx或dydy dudxdudx.(3)【答案】104x【解析】由连续可导函数的导数与0的关系判别函数的单调性.文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J

13、3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I

14、2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J

15、3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I

16、2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J

17、3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I

18、2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J

19、3 HI5Y9K1F5K2 ZH2B5I2A6O4将函数11()(2),xF xdtt两边对x求导,得1()2Fxx.若函数()F x严格单调减少,则1()20Fxx,即12x.所以函数()F x单调减少区间为104x.【相关知识点】函数的单调性：设函数()yfx在,a b上连续,在(,)a b内可导.(1)如果在(,)a b内()0fx,那么函数()yfx在,a b上单调增加；(2)如果在(,)a b内()0fx,那么函数()yfx在,a b上单调减少.(4)【答案】1/22cosxC【解析】32tansinsincoscoscoscosxxdxdxxxdxxxx3122coscos2cos

20、xdxxC.(5)【答案】222111(1)ln(1)222xxx【解析】这是微分方程的简单应用.由题知2ln(1)dyxxdx,分离变量得2ln(1)dyxxdx,两边对x积分有2221ln(1)ln(1)(1)2yxxdxxd x.由分部积分法得2222221112ln(1)(1)(1)ln(1)(1)2221xxd xxxxdxx222221(1)ln(1)211(1)ln(1).22xxxdxxxxC因为曲线()yf x过点1(0,)2,故12C,所以所求曲线为222111(1)ln(1)222yxxx.二、选择题(本题共 5 小题,每小题 3 分,满分 15 分.)(1)【答案】(D

21、)文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI

22、5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O

23、4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI

24、5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O

25、4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI

26、5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O

27、4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4【解析】因为当0 x时,1sinx是振荡函数,所以可用反证法.若取11kxk,则221111sin()sin0kkkkxx,211(2)2kxk,则22222111sin(2),(1,2,)2kkkkxx.因此,当k时,有10kx及20kx,但变量211sinxx或等于0 或趋于,这表明当0 x时它是无界的,但不是无穷大量,即(D)选项正确.(2)【答案】

28、(A)【解析】利用函数连续定义判定,即如果函数在0 x处连续,则有000lim()lim()()xxxxf xf xf x.由题可知221111|1|1lim()limlimlim(1)211xxxxxxf xxxx,221111|1|1lim()limlimlim(1)211xxxxxxf xxxx.因()f x在1x处左右极限不相等,故在1x处不连续,因此选(A).(3)【答案】(D)【解析】这是分段函数求定积分.当01x时,01xt,故2()f tt,所以23311111()()(1)33xxxF xf t dtt dttx.当12x时,12,tx故()1f t,所以111()()11x

29、xxF xf t dtdttx.应选(D).(4)【答案】(B)【解析】判定函数()f x零点的个数等价于判定函数()yf x与x的交点个数.对函数()lnxfxxke两边对x求导,得11()fxxe.令()0fx,解得唯一驻点xe,文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K

30、1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编

31、码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K

32、1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编

33、码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K

34、1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编

35、码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4即()0,0;(),()0,;(),fxxef xfxexf x严格单调增加严格单调减少所以xe是极大值点,也是最大值点,最大

36、值为()ln0ef eekke.又因为00lim()lim(ln)lim()lim(ln)xxxxxf xxkexf xxke,由连续函数的介值定理知在(0,)e与(,)e各有且仅有一个零点(不相同).故函数()lnxfxxke在(0,)内零点个数为2,选项(B)正确.(5)【答案】(C)【解析】方法一：由几何图形判断.由()(),f xfx知()f x为奇函数,图形关于原点对称；在(0,)内()0,()0,()fxfxf x图形单调增加且向上凹,根据图可以看出()f x在(,0)内增加而凸,()0,()0fxfx,选(C).方法二：用代数法证明.对恒等式()()f xfx两边求导,得()()

37、,()()fxfxfxfx.当(,0)x时,有(0,)x,所以()()0,()()0fxfxfxfx,故应选(C).三、(本题共 5 小题,每小题 5 分,满分 25 分.)(1)【解析】222sin()cos()()2yf xf xfxx,22cos()()2yf xfxx2222cos()()2cos()()2f xfxxf xfxx22cos()()(2)f xfxx2222222sin()()(2)cos()()(2)f xfxxf xfxx文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B

38、5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E

39、4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B

40、5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E

41、4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B

42、5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E

43、4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B

44、5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O422cos()()2f xfx.【相关知识点】复合函数求导法则：如果()ug x在点x可导,而()yf x在点()ug x可导,则复合函数()yfg x在点x可导,且其导数为()()dyfug xdx或dydy dudxdudx.(2)【解析】应先化简再求函数的极限,2222(100)(100)lim(100)lim100 xxxxxxxxxxxx22100100limlim11001001xxxxxxx.因为0 x,所以22100100100limlim5011 11 10011001xxxxx.(

45、3)【解析】先进行恒等变形,再利用基本积分公式和分部积分法求解.2444000sec1tan1cos222xxxdxdxxdxx4440001111sintantan(0)22242cosxxxxdxdxx4400111cosln(cos)82cos82dxxx1121ln(cos)ln(cos0)lnln 282482284.(4)【解析】用极限法求广义积分.2333000(1)1(1)(1)(1)(1)(1)xxdxdxxxdxxx12200121(1)(1)lim22(1)bbxxxx221111lim02(1)222bbb.(5)【解析】所给方程是一阶线性非齐次微分方程,其标准形式是文

46、档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y

47、9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文

48、档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y

49、9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文

50、档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y9K1F5K2 ZH2B5I2A6O4文档编码:CM5K1K7E4J3 HI5Y