2007级高数下试题及答案

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1、1南昌大学南昌大学南昌大学南昌大学 202020200 0 0 07 7 7 7202020200 0 0 08 8 8 8 学年第二学期期末考试试卷学年第二学期期末考试试卷学年第二学期期末考试试卷学年第二学期期末考试试卷一、一、一、一、 填空题填空题填空题填空题( ( ( (每空每空每空每空 3 3 3 3 分，共分，共分，共分，共 15151515 分分分分) ) ) )1. 1. 1. 1. 设设32 ,2,32 ,2,32 ,2,32 ,2,aijkbijkaijkbijkaijkbijkaijkbijk=+=+=+=+则则( 2 ) (3 )( 2 ) (3 )( 2 ) (3 )(

2、 2 ) (3 )abababab=_._._._.2. 2. 2. 2. 函数函数2222222222222222ln(25)(4)ln(25)(4)ln(25)(4)ln(25)(4)zxyxyzxyxyzxyxyzxyxy=+=+=+=+的的定义域是定义域是_._._._.3. 3. 3. 3. 设函数设函数(cossin )(cossin )(cossin )(cossin )x x x xzeyxyzeyxyzeyxyzeyxy=+=+=+=+, , , , 则则1 1 1 10 0 0 0x x x xy y y ydzdzdzdz=_. . . .4. 4. 4. 4. 交换累次

3、积分的次序交换累次积分的次序( , )2 2 2 22 2 2 21111111101010101y y y yy y y ydyf x y dxdyf x y dxdyf x y dxdyf x y dx=_. . . .5. 5. 5. 5. 微分方程微分方程2 2 2 2 y y y yy y y yx x x x= = = =的通解为的通解为_._._._.二、二、二、二、 单项选择题单项选择题单项选择题单项选择题 ( ( ( (每小题每小题每小题每小题 3 3 3 3 分分分分, , , ,共共共共 15151515 分分分分) ) ) )1.1. 过点过点(3,0, 1)(3,0,

4、 1)(3,0, 1)(3,0, 1) 且与平面且与平面375120375120375120375120xyzxyzxyzxyz+=+=+=+=平行的平面方程是平行的平面方程是( ( ( (). ). ). ).(A)(A)(A)(A)3540354035403540xzxzxzxz=. . . .(B)(B)(B)(B) 37540375403754037540xyzxyzxyzxyz+=+=+=+= . . . .(C)(C)(C)(C) 350350350350xyzxyzxyzxyz+=+=+=+=(D)(D)(D)(D)75120751207512075120xyzxyzxyzxyz

5、+=+=+=+= . . . .2 2设设2 2 2 2u u u uz z z zv v v v= = = =, , , , 而而2 ,22 ,22 ,22 ,2uxy vyxuxy vyxuxy vyxuxy vyx=+=+=+=+, , , , 则则z z z zx x x x=( ( ( (). ). ). ).(A)(A)(A)(A)()()()2 2 2 22232232232232 2 2 2xy xyxy xyxy xyxy xyyxyxyxyx+. . . .(B)(B)(B)(B)()222222222 2 2 2xyxyxyxyyxyxyxyx+. . . .(C)(C)

6、(C)(C)()()2232232232232 2 2 2xy xyxy xyxy xyxy xyyxyxyxyx+. . . .(D)(D)(D)(D)()()2 2 2 22 2 2 2222222222 2 2 2xyxyxyxyyxyxyxyx+. . . .3 3 设可微函数设可微函数( , )( , )( , )( , )f x yf x yf x yf x y在点在点00000000(,)(,)(,)(,)xyxyxyxy取得极小值取得极小值, , , ,则下列结论正确的是则下列结论正确的是 ( ( ( (). ). ). ).(A)(A)(A)(A)0 0 0 0(, )(,

7、)(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数大于零处的导数大于零. . . .2(B)(B)(B)(B)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数等于零处的导数等于零. . . .(C)(C)(C)(C)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数小于零处的导数小于零. . . . . . . .(D)(D)(D)(D)0 0 0 0(, )(, )(,

8、 )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数不存在处的导数不存在. . . .4 4设设L L L L为取正向的圆周为取正向的圆周222222224 4 4 4xyxyxyxy+=+=+=+=, , , , 则曲线积分则曲线积分22222222()()()()()()()()L L L Lxy dxxydyxy dxxydyxy dxxydyxy dxxydy+ 之值为之值为 ( ( ( (). ). ). ).(A)(A)(A)(A)0 0 0 0. . . .(B)(B)(B)(B)4 4 4 4 . . . .(C)(C)(C)(C

9、)4 4 4 4. . . .(D)(D)(D)(D) . . . .5 5函数函数( )cos( )cos( )cos( )cosf xxf xxf xxf xx= = = =关于关于x x x x的幂级数展开式为的幂级数展开式为 ( ( ( (). ). ). ).(A)(A)(A)(A)2422422422421( 1)( 11)1( 1)( 11)1( 1)( 11)1( 1)( 11)nnnnnnnnxxxxxxxxxxxxxxxx+ + + + + + + + (B)(B)(B)(B)2422422422421( 11)1( 11)1( 11)1( 11)n n n nxxxxxx

10、xxxxxxxxxx+ + + + . . . .(C)(C)(C)(C)2 2 2 21( 11)1( 11)1( 11)1( 11)n n n nxxxxxxxxxxxxxxxx+ + + + . . . .(D)(D)(D)(D)2422422422421( 1)()1( 1)()1( 1)()1( 1)()2!4!(2 )!2!4!(2 )!2!4!(2 )!2!4!(2 )!n n n nn n n nxxxxxxxxxxxxx x x xn n n n+ + + + + + + + +=+=+=到点到点(0,0)(0,0)(0,0)(0,0)O O O O的弧段的弧段. . . .

11、32 2、利用高斯公式计算曲面积分利用高斯公式计算曲面积分xdydzydzdxzdxdyxdydzydzdxzdxdyxdydzydzdxzdxdyxdydzydzdxzdxdy+, , , ,其中其中为上半球面为上半球面222222222222zRxyzRxyzRxyzRxy=的上侧。的上侧。五、五、五、五、解下列各题解下列各题解下列各题解下列各题( ( ( (共共共共 2 2 2 2 小题小题小题小题, , , , 每小题每小题每小题每小题 8 8 8 8 分分分分, , , ,共共共共 16161616 分分分分):):):):1 1、判定正项级数、判定正项级数1 1 1 1! ! !

12、!n n n nn n n nn n n nn n n n = = = = 的敛散性的敛散性2 2、设幂级数、设幂级数1 1 1 11 1 1 14 4 4 4n n n nn n n nn n n nx x x xn n n n = = = = . .(1).(1). 求收敛半径与收敛区间求收敛半径与收敛区间 ; ;(2).(2). 求和函数求和函数. .六、六、六、六、计算题（共计算题（共计算题（共计算题（共 2 2 2 2 小题小题小题小题. . . . 每小题每小题每小题每小题 8 8 8 8 分分分分, , , , 共共共共 16161616 分分分分）: : : :1 1、求微分方

13、程、求微分方程2 2 2 2109109109109x x x xyyyeyyyeyyyeyyye+=的通解的通解. .2 2、( (应用题应用题) ) 计算由平面计算由平面0 0 0 0z z z z= = = =和旋转抛物面和旋转抛物面222222221 1 1 1zxyzxyzxyzxy=所围成的立体的体积所围成的立体的体积. .七、七、七、七、(6(6(6(6 分分分分) ) ) )已知连续可微函数已知连续可微函数( )( )( )( )f xf xf xf x满足满足1 1 1 1(0)(0)(0)(0)2 2 2 2f f f f= = = = , ,且能使曲线积分且能使曲线积分(

14、 )( )( )( )( )( )( )( )x x x xL L L Lef x ydxf x dyef x ydxf x dyef x ydxf x dyef x ydxf x dy + 与路径无关与路径无关, , 求求( )( )( )( )f xf xf xf x. .4南昌大学南昌大学南昌大学南昌大学 202020200 0 0 07 7 7 7202020200 0 0 08 8 8 8 学年第二学期期末考试试卷及答案学年第二学期期末考试试卷及答案学年第二学期期末考试试卷及答案学年第二学期期末考试试卷及答案一、一、一、一、 填空题填空题填空题填空题( ( ( (每空每空每空每空 3

15、 3 3 3 分，共分，共分，共分，共 15151515 分分分分) ) ) )1. 1. 1. 1.设设32 ,2,32 ,2,32 ,2,32 ,2,aijkbijkaijkbijkaijkbijkaijkbijk=+=+=+=+则则( 2 ) (3 )( 2 ) (3 )( 2 ) (3 )( 2 ) (3 )abababab=18181818 . . . .2. 2. 2. 2. 函数函数2222222222222222ln(25)(4)ln(25)(4)ln(25)(4)ln(25)(4)zxyxyzxyxyzxyxyzxyxy=+=+=+=+的的定义域是定义域是 22222222(

16、 , ) 425( , ) 425( , ) 425( , ) 425x yxyx yxyx yxyx yxy+. . . .3. 3. 3. 3. 设函数设函数(cossin )(cossin )(cossin )(cossin )x x x xzeyxyzeyxyzeyxyzeyxy=+=+=+=+, , , , 则则1 1 1 10 0 0 0x x x xy y y ydzdzdzdz=()()()()e dxdye dxdye dxdye dxdy+ + + +. . . .4. 4. 4. 4. 交换累次积分的次序：交换累次积分的次序：( , )2 2 2 22 2 2 21111

17、111101010101y y y yy y y ydyf x y dxdyf x y dxdyf x y dxdyf x y dx=2 2 2 21111111110101010( , )( , )( , )( , )x x x xdxf x y dydxf x y dydxf x y dydxf x y dy . . . .5. 5. 5. 5. 微分方程微分方程2 2 2 2 y y y yy y y yx x x x= = = =的通解为：的通解为：11111111C C C CxxxxxxxxyCeyeyCeyeyCeyeyCeye+=或或. . . .二、二、二、二、 单项选择题单

18、项选择题单项选择题单项选择题 ( ( ( (每小题每小题每小题每小题 3 3 3 3 分分分分, , , ,共共共共 15151515 分分分分) ) ) )1.1. 过点过点(3,0, 1)(3,0, 1)(3,0, 1)(3,0, 1) 且与平面且与平面375120375120375120375120xyzxyzxyzxyz+=+=+=+=平行的平面方程是平行的平面方程是( ( ( (B B B B). ). ). ).(A)(A)(A)(A)3540354035403540xzxzxzxz=. . . .(B)(B)(B)(B) 37540375403754037540xyzxyzxyz

19、xyz+=+=+=+= . . . .(C)(C)(C)(C) 350350350350xyzxyzxyzxyz+=+=+=+=(D)(D)(D)(D)75120751207512075120xyzxyzxyzxyz+=+=+=+= . . . .2 2设设2 2 2 2u u u uz z z zv v v v= = = =, , , , 而而2 ,22 ,22 ,22 ,2uxy vyxuxy vyxuxy vyxuxy vyx=+=+=+=+, , , , 则则z z z zx x x x=( ( ( (A A A A). ). ). ).(A)(A)(A)(A)()()()2 2 2

20、22232232232232 2 2 2xy xyxy xyxy xyxy xyyxyxyxyx+. . . .(B)(B)(B)(B)()222222222 2 2 2xyxyxyxyyxyxyxyx+. . . .5(C)(C)(C)(C)()()2232232232232 2 2 2xy xyxy xyxy xyxy xyyxyxyxyx+. . . .(D)(D)(D)(D)()()2 2 2 22 2 2 2222222222 2 2 2xyxyxyxyyxyxyxyx+. . . .3 3 设可微函数设可微函数( , )( , )( , )( , )f x yf x yf x yf

21、 x y在点在点00000000(,)(,)(,)(,)xyxyxyxy取得极小值取得极小值, , , ,则下列结论正确的是则下列结论正确的是 ( ( ( (B B B B). ). ). ).(A)(A)(A)(A)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数大于零处的导数大于零. . . .(B)(B)(B)(B)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数等于零处的导数等于零. . . .(C)(C)(C)(C

22、)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数小于零处的导数小于零. . . . . . . .(D)(D)(D)(D)0 0 0 0(, )(, )(, )(, )f xyf xyf xyf xy在在0 0 0 0yyyyyyyy= = = =处的导数不存在处的导数不存在. . . .4 4设设L L L L为取正向的圆周为取正向的圆周222222224 4 4 4xyxyxyxy+=+=+=+=, , , , 则曲线积分则曲线积分22222222()()()()()()()()L L L Lxy dx

23、xydyxy dxxydyxy dxxydyxy dxxydy+ 之值为之值为 ( ( ( (A A A A). ). ). ).(A)(A)(A)(A)0 0 0 0. . . .(B)(B)(B)(B)4 4 4 4 . . . .(C)(C)(C)(C)4 4 4 4. . . .(D)(D)(D)(D) . . . .5 5函数函数( )cos( )cos( )cos( )cosf xxf xxf xxf xx= = = =关于关于x x x x的幂级数展开式为的幂级数展开式为 ( ( ( (D D D D). ). ). ).(A)(A)(A)(A)2422422422421( 1)

24、( 11)1( 1)( 11)1( 1)( 11)1( 1)( 11)nnnnnnnnxxxxxxxxxxxxxxxx+ + + + + + + + (B)(B)(B)(B)2422422422421( 11)1( 11)1( 11)1( 11)n n n nxxxxxxxxxxxxxxxx+ + + + . . . .(C)(C)(C)(C)2 2 2 21( 11)1( 11)1( 11)1( 11)n n n nxxxxxxxxxxxxxxxx+ + + + . . . .(D)(D)(D)(D)2422422422421( 1)()1( 1)()1( 1)()1( 1)()2!4!(2

25、 )!2!4!(2 )!2!4!(2 )!2!4!(2 )!n n n nn n n nxxxxxxxxxxxxx x x xn n n n+ + + + + + + + +=+=+=到点到点(0,0)(0,0)(0,0)(0,0)O O O O的弧段的弧段. . . .7解解: : : :2222222222222 ,.2 ,.2 ,.2 ,.yyyyyyyyPxeyQx eyPxeyQx eyPxeyQx eyPxeyQx ey=222222222,22.2,22.2,22.2,22.yyyyyyyyQPQPQPQPxexexexexexexexexyxyxyxy=2. 2. 2. 2.Q

26、PQPQPQPxyxyxyxy=连接连接 OAOAOAOA 构成闭路构成闭路 OABO,OABO,OABO,OABO, 其围成区域为其围成区域为 D.D.D.D.沿沿2 2 2 21 1 1 10 0 0 01 1 1 1:0,:0,:0,:0,2 2 2 2a a a aOAyIxdxaOAyIxdxaOAyIxdxaOAyIxdxa= . . . .1 1 1 1L L L LD D D DQPQPQPQPIdIIdIIdIIdIxyxyxyxy =1 1 1 12 2 2 2D D D DdIdIdIdI = 2 2 2 22 2 2 22 2 2 2111111112(2).2(2).

27、2(2).2(2).2224222422242224aaaaaaaaa a a a=2 2、利用高斯公式计算曲面积分利用高斯公式计算曲面积分xdydzydzdxzdxdyxdydzydzdxzdxdyxdydzydzdxzdxdyxdydzydzdxzdxdy+, , , ,其中其中为上半球面为上半球面222222222222zRxyzRxyzRxyzRxy=的上侧。的上侧。0 0 0 0A A A A（a a a a， 0 0 0 0）B B B BD D D Dx x x xy y y y8解解: : : :记记1 1 1 1 为平面为平面0 0 0 0z z z z= = = = 的下侧

28、的下侧. . . .1,1,1.1,1,1.1,1,1.1,1,1.PQRPQRPQRPQRxyzxyzxyzxyz=由高斯公式有由高斯公式有原式原式11111111+=30303030 = 3 3 3 32.2.2.2.R R R R = = = =五、五、五、五、解下列各题解下列各题解下列各题解下列各题( ( ( (共共共共 2 2 2 2 小题小题小题小题, , , , 每小题每小题每小题每小题 8 8 8 8 分分分分, , , ,共共共共 16161616 分分分分):):):):1 1、判定正项级数、判定正项级数1 1 1 1! ! ! !n n n nn n n nn n n n

29、n n n n = = = = 的敛散性的敛散性解解: :1 1 1 11 1 1 1(1)!(1)!(1)!(1)!limlimlimlimlimlimlimlim! ! ! !(1)(1)(1)(1)n n n nn n n nn n n nnnnnnnnnn n n nu u u unnnnnnnnununununn n n n+ + + + + + + + + +=+ + + +limlimlimlim1 1 1 1n n n nn n n nn n n nn n n n= = = =+ + + +11111111lim1.lim1.lim1.lim1.1 1 1 11 1 1 1n

30、n n nn n n ne e e en n n n=+ + + +所以原级数收敛所以原级数收敛. .2 2、设幂级数、设幂级数1 1 1 11 1 1 14 4 4 4n n n nn n n nn n n nx x x xn n n n = = = = . .(1).(1). 求收敛半径与收敛区间求收敛半径与收敛区间 ; ;(2).(2). 求和函数求和函数. .9解解: : (1).(1).1 1 1 11 1 1 1lim4.lim4.lim4.lim4.4 4 4 4n n n nn n n nn n n na a a aR R R Ra a a a + + + +=当当1 1 1

31、14 4 4 4x x x x= = = =时时, ,1 1 1 11 1 1 14 4 4 4n n n nn n n n = = = = 发散发散; ;当当1 1 1 14 4 4 4x x x x= = = = 时时, ,1 1 1 11 1 1 1( 1)( 1)( 1)( 1)4 4 4 4n n n nn n n nn n n n = = = = 收敛收敛. .故收敛区间为故收敛区间为 1/4,1/4). 1/4,1/4). 1/4,1/4). 1/4,1/4). (2).(2). 设设 ( )( )( )( )S xS xS xS x= = = =1 1 1 11 1 1 14

32、4 4 4n n n nn n n nn n n nx x x xn n n n = = = = . .111111111111111111111 1 1 1( )4(4 ).( )4(4 ).( )4(4 ).( )4(4 ).14141414nnnnnnnnnnnnnnnnnnnnSxxxSxxxSxxxSxxxx x x x= 0000000011111111( )ln(14 ).( )ln(14 ).( )ln(14 ).( )ln(14 ).144144144144xxxxxxxxSx dxdxxSx dxdxxSx dxdxxSx dxdxxx x x x= = = = 即即1 1

33、 1 1( )ln(14 ).( )ln(14 ).( )ln(14 ).( )ln(14 ).4 4 4 4S xxS xxS xxS xx= = = = 1/4,1/4). 1/4,1/4). 1/4,1/4). 1/4,1/4). ( (0)0).( (0)0).( (0)0).( (0)0).S S S S= = = =六、六、六、六、计算题（共计算题（共计算题（共计算题（共 2 2 2 2 小题小题小题小题. . . . 每小题每小题每小题每小题 8 8 8 8 分分分分, , , , 共共共共 16161616 分分分分）: : : :1 1、求微分方程、求微分方程2 2 2 21

34、09109109109x x x xyyyeyyyeyyyeyyye+=的通解的通解. .解解: :2 2 2 2121212121090.9,1.1090.9,1.1090.9,1.1090.9,1.rrrrrrrrrrrrrrrr+=+=+=+=9 9 9 912121212. . . .xxxxxxxxYC eC eYC eC eYC eC eYC eC e=+=+=+=+2 2 2 2 = = = =不是特征根不是特征根, , 所以设所以设2 2 2 2*.*.*.*.x x x xyAeyAeyAeyAe= = = =代入原方程得代入原方程得: :2 2 2 211111111.*.

35、*.*.*.77777777x x x xAyeAyeAyeAye= = = = = = = = 故原方程的通解为故原方程的通解为: :92929292121212121 1 1 1. . . .7 7 7 7xxxxxxxxxxxxyC eC eeyC eC eeyC eC eeyC eC ee=+=+=+=+102 2、( (应用题应用题) ) 计算由平面计算由平面0 0 0 0z z z z= = = =和旋转抛物面和旋转抛物面222222221 1 1 1zxyzxyzxyzxy=所围成的立体的体积所围成的立体的体积. .解法一解法一: :D D D DVzdVzdVzdVzd = =

36、 = = 22222222(1)(1)(1)(1)D D D Dxydxdyxydxdyxydxdyxydxdy= 212121212 2 2 200000000(1)(1)(1)(1)drrdrdrrdrdrrdrdrrdr =1 1 1 1242424240 0 0 0111111112.2.2.2.242242242242rrrrrrrr =解法二解法二: :VdvVdvVdvVdv = = = = 2 2 2 2211211211211000000000000r r r rdrdrdzdrdrdzdrdrdzdrdrdz = = = =212121212 2 2 200000000(1

37、)(1)(1)(1)drrdrdrrdrdrrdrdrrdr =1 1 1 1242424240 0 0 0111111112.2.2.2.242242242242rrrrrrrr =七、七、七、七、(6(6(6(6 分分分分) ) ) )已知连续可微函数已知连续可微函数( )( )( )( )f xf xf xf x满足满足1 1 1 1(0)(0)(0)(0)2 2 2 2f f f f= = = = , ,且能使曲线积分且能使曲线积分( )( )( )( )( )( )( )( )x x x xL L L Lef x ydxf x dyef x ydxf x dyef x ydxf x

38、dyef x ydxf x dy + 与路径无关与路径无关, , 求求( )( )( )( )f xf xf xf x. .解解: :( ),( ).( ),( ).( ),( ).( ),( ).x x x xPef xyQf xPef xyQf xPef xyQf xPef xyQf x =+= =+= =+= =+= ( ),( ).( ),( ).( ),( ).( ),( ).x x x xPQPQPQPQef xfxef xfxef xfxef xfxyxyxyxyx =+= =+= =+= =+= 因为曲线积分与路径无关因为曲线积分与路径无关, , 所以所以QPQPQPQPxyx

39、yxyxy= = = =. .于是得：于是得：( )( ).( )( ).( )( ).( )( ).x x x xfxef xfxef xfxef xfxef x =+=+=+=+11即：即：( )( ).( )( ).( )( ).( )( ).x x x xfxf xefxf xefxf xefxf xe += += += += ( )()( )()( )()( )()dxxdxdxxdxdxxdxdxxdxf xeeedxCf xeeedxCf xeeedxCf xeeedxC=+=+=+=+ ()()()()xxxxxxxxxxxxeee dxCeee dxCeee dxCeee dxC=+=+=+=+ ( 1)().( 1)().( 1)().( 1)().xxxxxxxxedxCeCxedxCeCxedxCeCxedxCeCx=+=+=+=+= 由由1 1 1 1(0)(0)(0)(0)2 2 2 2f f f f= = = = , , 得得1 1 1 1. . . .2 2 2 2C C C C= = = = 1 1 1 1( )().( )().( )().( )().2 2 2 2x x x xf xexf xexf xexf xex = += += += +