当前位置:文档之家› 2008年浙江省普通高校“专升本”联考《高等数学(一)》试卷及答案

2008年浙江省普通高校“专升本”联考《高等数学(一)》试卷及答案

2008年浙江省普通高校“专升本”联考《高等数学(一)》试卷及答案
2008年浙江省普通高校“专升本”联考《高等数学(一)》试卷及答案

2008年浙江省普通高校“专升本”联考《高等数学(一)》试卷及答案

考试说明:

1、考试时间为150分钟;

2、满分为150分;

3、答案请写在试卷纸上,用蓝色或黑色墨水的钢笔、圆珠笔答卷,否则无效;

4、密封线左边各项要求填写清楚完整。

一. 选择题(每个小题给出的选项中,只有一项符合要求:本题共有5个小题,每小题4分,共20分)

1.函数()()x x x f cos 12

+=是( ).

()A 奇函数 ()B 偶函数 ()C 有界函数 ()D 周期函数

2.设函数()x x f =,则函数在0=x 处是( ).

()A 可导但不连续 ()B 不连续且不可导

()C 连续且可导 ()D 连续但不可导 3.设函数()x f 在[]1,0上,

02

2

>dx

f d ,则成立( ).

()A ()()010

1

f f dx

df dx df x x ->>

== ()

B ()()0

1

10==>

->x x dx df f f dx

df

()

C ()()0

101==>

->x x dx

df f f dx

df ()D ()()10

01==>>

-x x dx

df dx

df f f

4.方程2

2y x z +=表示的二次曲面是( ).

()A 椭球面

()B 柱面 ()C 圆锥面

()D 抛物面

5.设()x f 在[]b a ,上连续,在()b a ,内可导,()()b f a f =, 则在()b a ,内,曲线()x f y =上平行于x 轴的切线( ).

()A 至少有一条 ()B 仅有一条

().C 不一定存在 ().D 不存在

二.填空题:(只须在横线上直接写出答案,不必写出计算过程,每小题4分,共40分)

1.计算_______

__________2

sin

1

lim

=

→x

x

x

2.设函数()x f 在1=x 可导, 且

()10

==x dx

x df ,则

()()

.

__________121lim

=

-+→x

f x f x .

3.设函数(),ln 2x x f =则().

________________________=

dx

x df

4.曲线x x x y --=2

33的拐点坐标._____________________

5.设x arctan 为()x f 的一个原函数,则()=x f ._____________________

6.

().

_________________________2=

?

x

dt t f dx

d

7.定积分()

.

________________________2

=

+?

π

dx x x

8.设函数()2

2

cos y

x z +=,则.

_________________________=

??x

z

9. 交换二次积分次序

().__________________________

,0

10

=?

?

x dy y x f dx

10. 设平面∏过点()1,0,1-且与平面0824=-+-z y x 平行,则平面∏的方程为

._____________________

三.计算题:(每小题6分,共60分) 1.计算x

e x

x 1lim 0

-→.

2.设函数()()x x g e x f x

cos ,==,且?

?

?

??=dx dg f y ,求dx dy .

3.计算不定积分()

?+.1x x dx

4.计算广义积分?+∞-0

dx xe

x

.

5.设函数()???<≥=0

,0,cos 4

x x x x x f ,求()?

-12

dx x f .

6. 设()x f 在[]1,0上连续,且满足()()?

+=10

2dt t f e x f x

,求()x f .

7.求微分方程

x

e dx

dy

dx

y

d =+22

的通解.

8.将函数()()x x x f +=1ln 2

展开成x 的幂级数.

9.设函数()y

x y x y x f +-=

,,求函数()y x f ,在2,0==y x 的全微分.

报考学校:______________________报考专业:______________________姓名: 准考证号: ------------------------------------------------------------------------------------------密封线---------------------------------------------------------------------------------------------------

10.计算二重积分,()??+D

dxdy y x 2

2

,其中1:22≤+y x D .

四.综合题:(本题共30分,其中第1题12分,第2题12分,第3题6分)

1.设平面图形由曲线x

e y =及直线0,==x e y 所 围成,

()1求此平面图形的面积;

()2求上述平面图形绕x 轴旋转一周而得到的

旋转体的体积.

2.求函数1323--=x x y 的单调区间、极值及曲线的凹凸区间.

3.求证:当0>x 时,e x x

?

+11.

《高等数学(一)答案

二..填空题:(每小题4分,共40分) 1.

2

1; 2. 2; 3.

x

1; 4. )3,1(-; 5.

2

11x

+;

6. ()x f -;

7.

3

3

2π; 8. ()2

2

sin 2y

x x +-; 9.()??1

1

,y

dx y x f dy ;

10. 224=+-z y x .

三.计算题(每小题6分,共60分) 1.解法一.由洛必达法则,得到1

lim

1lim

x

x x

x e

x

e →→=- …………..4分

1=. …………6分

解法二.令t e x

=-1, 则 ()t x +=1ln ……….. 2分

于是, ()

11ln lim

1lim

=+=-→→t t x

e t x

x . …………6分

2.解.

x dx dg

sin -=, ()x

e x

f dx d

g f y sin sin -=-=??

? ??= …………3分 故 x e

dx

dy x

cos sin --=. ………..6分

3. 解法一.令t x =,,则2

t x =, ………..2分

()

()

?

??

+=+=+=

+.arctan 2121212

2

C t t

dt t t tdt

x x dx ……….5分

C x +=arctan 2. ……….6分

解法二.

()

()

?

?

=+

=+2

1)(21x x d x x dx ……….4分

C x +=arctan 2. ……….6分

4.解.

??

+∞

-∞+-+∞

-+

-=0

dx e

xe

dx xe

x

x

x

……….3分

10

=-=+∞-x

e

. ………..6分

5.解.

()()()?????+

=

+=

---1

2

4

1

2

1

2

cos

xdx dx x

dx

x f dx x f dx

x f ……….3分

1sin 5

32sin 5

110

2

5+=

+=

-x

x

. ……….6分

6.解. 设()A dx x f =?1

0,两边对已给等式关于x 从0到1积分,得到

()()????+-=+=+=

1

10

1

1

1

2122dx x f e A e

Adx dx e

dx

x f x

x

……….4分

从而解得()e dx x f -=?11

.. ………..5分

代入原式得()()e e x f x

-+=12. ……….6分

7.解.特征方程为02

=+k k ,得到特征根1,021-==k k , ………..1分 故对应的齐次方程的通解为x

e

c c y -+=21, ………..3分

由观察法,可知非齐次方程的特解是x

e y 2

1=*

, ………..5分

因而,所求方程的通解为 x

x

e e c c y 2

121+

+=-,其中21,c c 是任意常数. ……….6分

8.解.因为()()

)11(1

1432

1ln 1

4

3

2

≤<-++-++-

+

-

=++x n x

x

x

x

x x n n

, ….3分

所以()2

2

1ln x x x =+()

)1

14

32(1

4

3

2

++-++-

+

-

+n x

x

x

x

x n n

=()

)11(1

14

3

2

3

6

543

≤<-++-++-

+

-+x n x

x

x

x

x n n

. ……..6分

9解.

()()

222,2y x x y x y x y y f y x y y x y x x x

f +-=???? ??+-??=??+=???? ??+-??=

??, ……….2分 从而

()

()

0,

12,02,0=??=??y

f x

f , ……….4分

所以()()()

()

dx dy y

f dx x

f y x df =??+

??=2,02,02,0,. ………6分

10.解.采用极坐标变换,令θθsin ,cos r y r x == ,πθ20,10<≤≤

()????=

+1

3

20

2

2

dr r

d dxdy

y

x

D

πθ ……….4分

2

π

=

. ……..6分

四.综合题:(每小题10分,共30分) 1.解法一(1).()?-=

1

dx e e S x

……….4分

(

)

1110

=+-=-=e e e

ex x

. ………..6分

(2).()?-=1

22

dx e

e

V x

π

………..9分

()()

12121212

221

22

+=??

????--=?

?

? ??-=e e e e x e x πππ ………..12分

解法二.(1)?-=1

dx e e S x

……….3分

110

=-=x

e

e . ………..6分

(2).?-=1

22

dx e e V x

ππ ……….9分

()

12

2

2

10

22

+=

-

=e

e

e x

π

π

π. …………12分

2.解.定义域为),(+∞-∞,

()23632

-=-=x x x x dx

dy ,令

0=dx

dy ,得到 2,021==x x (驻点), …….2分

(),162

2

-=x dx

y d 由

02

2

=dx

y d ,得到13=x , …….3分

……..8分 故 )0,(-∞),2(+∞为单调增加区间,(0,2)为单调减少区间; ……….10分 极大值为-1,极小值为-5, ……..11分

)1,(-∞为凸区间,),1(+∞为凹区间 ………12分

3.证明. 令()()],ln )1[ln(11ln x x x x x x F -+=??

?

?

?

+

= ()(),11ln 1ln 111

ln 1ln +--+=??

? ??-++-+=x x x x x x x x dx dF

……….2分 利用中值定理,()ξ

1

ln 1ln =

-+x x ,其中1+<

所以

01

11

>+-

=

x dx

dF ξ

,因此,当0>x 时,()x F 是单调增加的, ………5分

而e x x

x =??? ?

?

++∞→11lim ,

所以当0>x 时,e x x

?

+11. ………..6分

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