(完整word版)江苏省2012年专转本高数真题及答案

(完整word版)江苏省2012年专转本高数真题及答案
(完整word版)江苏省2012年专转本高数真题及答案

江苏省2012年普通高校“专转本”选拔考试

高等数学 试题卷(二年级)

注意事项:出卷人:江苏建筑大学-张源教授

1、考生务必将密封线内的各项目及第2页右下角的座位号填写清楚.

2、考生须用钢笔或圆珠笔将答案直接答在试卷上,答在草稿纸上无效.

3、本试卷共8页,五大题24小题,满分150分,考试时间120分钟. 一、 选择题(本大题共6小题,每小题4分,满分24分) 1、极限=+∞→)3sin 1sin

2(lim x

x

x x x ( )

A. 0

B. 2

C. 3

D. 5

2、设)

4(sin )2()(2--=

x x x

x x f ,则函数)(x f 的第一类间断点的个数为( )

A. 0

B. 1

C. 2

D. 3 3、设2

32

152)(x x x f -=,则函数)(x f ( )

A.只有一个最大值

B. 只有一个极小值

C.既有极大值又有极小值

D. 没有极值 4、设y

x z 3

)2ln(+

=在点)1,1(处的全微分为 ( ) A. dy dx 3- B. dy dx 3+ C. dy dx 321+ D. dy dx 32

1

- 5、二次积分dx y x f dy y

),(1

1

?

? 在极坐标系下可化为( )

A. ρθρθρθπ

θ

d f d )sin ,cos (40

sec 0

?

?

 

B.

ρρθρθρθπ

θ

d f d )sin ,cos (40

sec 0

?

?

 

C.

ρθρθρθπ

πθ

d f d )sin ,cos (2

4sec 0

??

D.

ρρθρθρθπ

πθ

d f d )sin ,cos (2

4

sec 0

??

 

6、下列级数中条件收敛的是( )

A. 12)1(1

+-∑∞

=n n

n n

B.

∑∞

=-1

)23()1(n n

n

C. ∑∞

=-12

)1(n n

n D. ∑∞

=-1

)1(n n

n 二、填空题(本大题共6小题,每小题4分,共24分)

7要使函数x

x x f 1

)21()(-=在点0=x 处连续,则需补充定义=)0(f _________.

8、设函数x

e x x x y 2221

2(+++=),则=)0()

7(y

____________.

9、设)0(>=x x y x

,则函数y 的微分=dy ___________.

10、设向量→→b a ,互相垂直,且,,23==→→b a ,则=+→

→b a 2___________.

11、设反常积分

2

1

=

?

+∞

-dx e a

x ,则常数=a __________. 12、幂级数n

n n

n x n )3(3)1(1

--∑∞

=的收敛域为____________. 三、计算题(本大题共8小题,每小题8分,共64分)

13、求极限)

1ln(2

cos 2lim 320x x x x x +-+→.

14、设函数)(x y y =由参数方程??

?

??+=-=t

t y t

t x ln 212所确定,求22,dx y d dx dy .

15、求不定积分?+dx x x 2cos 1

2.

16、计算定积分dx x x ?-2

1

1

21

17、已知平面∏通过)3,2,1(M 与x 轴,求通过)1,1,1(N 且与平面∏平行,又与x 轴垂直的直线方程.

18、设函数)(),(2

2

y x xy x f z ++=?,其中函数f 具有二阶连续偏导数,函数?具有二阶

连续导数,求y

x z

???2.

19、已知函数)(x f 的一个原函数为x

xe ,求微分方程)(44x f y y y =+'+''的通解.

20、计算二重积分??D

ydxdy ,其中D 是由曲线1-x y =,直线x y 2

1

=

及x 轴所围成的平面闭区域.

四、综合题(本大题共2小题,每小题10分,共20分)

21、在抛物线)0(2

>=x x y 上求一点P ,使该抛物线与其在点P 处的切线及x 轴所围成的平面图形的面积为3

2

,并求该平面图形绕x 轴旋转一周所形成的旋转体的体积.

22、已知定义在),(+∞-∞上的可导函数)(x f 满足方程3)(4)(31

-=-?

x dt t f x xf x

,试求:

(1)函数)(x f 的表达式; (2)函数)(x f 的单调区间与极值; (3)曲线)(x f y =的凹凸区间与拐点.

五、证明题(本大题共2小题,每小题9分,共18分)

23、证明:当10<

6

1arcsin x x x +>.

24、设??

???≠=?0)0(0)()(2

= x g x x dt

t g x f x ,其中函数)(x g 在),(+∞-∞上连续,且

3cos 1)(lim 0=-→x x g x 证明:函数)(x f 在0=x 处可导,且2

1)0(='f .

一.选择题 1-5 B C C A B D 二.填空题

7-12 2

-e 128 dx x x n

)ln 1(+ 5 2ln ]6,0(

三.计算题

13、求极限)

1ln(2

cos 2lim 320x x x x x +-+→.

原式=3

0304202sin lim 4sin 22lim 2cos 2lim x x

x x x x x x x x x x -=-=-+→→→

12

1621lim 6cos 1lim 22

020==-=→→x x

x x x x

14、设函数)(x y y =由参数方程??

???

+=-

=t

t y t

t x ln 212所确定,求22,dx y d dx dy . 原式=t t t t dt dx dt dy dx dy 211222=++==12112)()(222

22+=+===t t t

dt dx dt dx dy

d dx dx dy d dx y d

15、求不定积分?+dx x x 2cos 1

2.

原式=

???+-+=+=+)12(tan tan )12(tan )12(cos 1

22x xd x x x d x dx x x

C x x x xdx x x +++=-+=?

cos ln 2tan )12(tan 2tan )12(

16、计算定积分

dx x x ?-2

1

1

21

. 原式=令t x =-12,则原式=

613arctan 211

22

13123

1

2π==+=+??

t dt t dt t t t

17、已知平面∏通过)3,2,1(M 与x 轴,求通过)1,1,1(N 且与平面∏平行,又与x 轴垂直的直线方程.

解:平面∏的法向量)2,3,0(-=?=→→→i OM n ,直线方向向量为)3,2,0(--=?=→

→→i n S , 直线方程:3

1

2101--=

--=-z y x

18、设函数)(),(2

2

y x xy x f z ++=?,其中函数f 具有二阶连续偏导数,函数?具有二阶

连续导数,求y

x z

???2.

解:x y f f x

z

221?'+?'+'=???

?''??+''+'+?''=???y x f xy f x f y x z 22222122

19、已知函数)(x f 的一个原函数为x

xe ,求微分方程)(44x f y y y =+'+''的通解.

解:x

x

e x xe x

f )1()()(+='=,先求044=+'+''y y y 的通解,特征方程:0442

=++r r , 221-=、r ,齐次方程的通解为x e

x C C Y 221)(-+=.令特解为x e B Ax y )(+=*, 代入原方程得:1969+=++x B A Ax ,有待定系数法得:

???=+=19619B A A ,解得??

???

==27191B A ,所以通解为x x e x e x C C Y )27191()(22

1+++=-

20、计算二重积分??D

ydxdy ,其中D 是由曲线1-x y =,直线x y 2

1

=

及x 轴所围成的平面闭区域. 原式=?

?

+=

1

21

212

1

y y

dx ydy .

四.综合题

21、在抛物线)0(2

>=x x y 上求一点P ,使该抛物线与其在点P 处的切线及x 轴所围成的平面图形的面积为

3

2

,并求该平面图形绕x 轴旋转一周所形成的旋转体的体积. 解:设P 点)0)(,(02

00>x x x ,则02x k =切,切线:)(2,002

0x x x x y -=- 即x x x y 02

02,=+,由题意

3

2

)2(2

00

02

0?

=-+x dy y x x y ,得20=x ,)4,2(P πππ15

16

)44(2

1

22

4=

--=??x d x x d x V x

22、已知定义在),(+∞-∞上的可导函数)(x f 满足方程3)(4)(31

-=-?

x dt t f x xf x

,试求:

(1)函数)(x f 的表达式; (2)函数)(x f 的单调区间与极值; (3)曲线)(x f y =的凹凸区间与拐点. 解:(1)已知3)(4

)(31

-=-?

x dt t f x xf x

两边同时对x 求导得:23)(4)()(x x f x f x x f =-'+

即:x y x

y 33

=-

',则323cx x y +-=由题意得:2)1(-=f ,1=c ,则323)(x x x f +-= (2)2,0,063)(212

===-='x x x x x f 列表讨论得在),2()0,(+∞?-∞单调递增,在

)2,0(单调递减。极大值0)0(=f ,极小值4)2(-=f

(3)1,066)(==-=''x x x f

列表讨论得在)1,(-∞凹,在),1(+∞凸。拐点)2,1(-

五、证明题

23、证明:当10<

6

1arcsin x x x +>. 解:令0)0(,61arcsin )(3

=-

-=f x x x x f ,0)0(,21111)(22

='---=

'f x x x f 0)1)

1(1(

)

1()(3

23

2>--=--=

''x x x x x x f ,在10<

0)0()(='>'f x f ,所以在10<f x f ,得证。

24、设??

???≠=?0)0(0)()(2

= x g x x

dt

t g x f x ,其中函数)(x g 在),(+∞-∞上连续,且3cos 1)(lim 0=-→x x g x 证明:函数)(x f 在0=x 处可导,且2

1)0(='f .

解:因为3cos 1)(lim

0=-→x x g x ,即32

1)

(lim 20=→x

x g x 所以有23)(lim 20=→x x g x

又因为)(x g 在),(+∞-∞上连续,所以0)(lim )0(0

==→x g g x ,则

)0(2

1

23313)(lim

)(lim

)

0()(lim

2

3

2

f x x

g x dt t g x

g x dt t g x x

x x

x '==?===-→→→??

江苏省2015年专转本高等数学真题

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