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专转本模拟试题与解析(二)

专转本模拟试题与解析(二)
专转本模拟试题与解析(二)

江苏省2013年普通高校“专转本”统一考试模拟试卷(二)解析

高等数学

注意事项:

1.考生务必将密封线内的各项填写清楚。

2.考生必须要钢笔或圆珠笔将答案直接写在试卷上,写在草稿纸上无效。

3.本试卷五大题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,共24分,在每小题给出的四个选项中,只有

一项是符合要求的,请把所选项前的字母填在题后的括号内)。 1、下列极限中,正确的是( ) A 、 sin 2lim

2x x x →∞= B 、arctan lim 1x x

x

→+∞=

C 、 224

lim

2

x x x →-=∞- D 、0lim 1x x x →+= 2、设2

1sin ,0(),0

x x f x x

ax b x ?>?

=??+≤?在0x =处可导,则( ) A 、1,0a b == B 、0,a b =为任意实数 C 、0,0b a == D 、1,a b =为任意实数 3、函数sin y x =在区间[]0,π上满足罗尔定理的ξ=( ) A 、0 B 、

4π C 、2

π

D 、π 4、设在[]b a ,上0)(>x f ,0)(<'x f ,0)(>''x f ,令dx x f y b a

?=)(1,

))((2a b b f y -=,[]()a b b f a f y -+=

)()(2

1

3,则有( ) A 、321y y y << B 、 312y y y << C 、213y y y << D 、132y y y <<

5、两个非零向量a 与b

垂直的充分必要条件是( )

A 、0=?b a

B 、 0

=?b a

C 、 0 =?a b

D 、0=?a a

6、下列级数发散的是( )

A 、1(1)ln(1)

n

n n ∞

=-+∑ B 、131n n n ∞

=-∑

C 、1

1(1)3n n n -∞

=-∑ D 、13

n n n ∞

=∑

二、填空题(本大题共6小题,每小题6分,共24分,请把正确答案的结果添在划线上)。

7、已知()g x 为有界函数,1

()sin ,0

(), 0

g x x f x x a x ?≠?=??=? 0x =处连续,则=a

8、已知0

(23)(23)

(2)1,lim

h f h f h f h

→--+'=则= ______

9、2x y xe -=,则()0y ''=

10、定积分1

4211ln 11x x x dx x --??

+- ?+??

?的值为 11、微分方程232x y y y xe '''-+=的特解y *

的形式应为____________________ 12、设22sin 2z x y x y =++,则(1,)2

dz π=_____________________________

三、计算题(本大题共8小题,每小题8分,共64分)。

13、考查1

1sin ()x x f x e x

-=的连续性,若有间断点,判别类型。

14、设函数)(x y y =由方程 1)cos(2-=-+e xy e y x 所确定,求

0|=x dx

dy

15、求不定积分6d (1)x

x x +?。

16、设1

sin ,0()20,0x x f x x x π

π?≤≤?=??<>?

或,求?=Φx dt t f x 0

)()(在),(+∞-∞内的表达式。

17、求过点)2,1,3(-A 且通过直线43:

521

x y z

L -+==的平面方程。

18、将函数)54ln()(-=x x f 展开为2-x 的幂级数,并指出其收敛域。

19、设(

)22,x y

z f x y e +=+,其中),(v u f 有二阶连续偏导数,求x z ??、y x z

???2。

20、计算积分y x

D

e

dxdy ??,其中2:,2D y x y x ==由所围成的区域。

四、证明题(每小题9分,共18分)

21、设()f x 在[0,1]上连续,且0()1f x ≤≤。证明:在[0,1]上至少存在一个ξ使()f ξξ=。

22、设)(x f 在0x =的邻域内有二阶连续导数,且0)0(=f ,???

??=='≠=0

,0)0(0,)

()(x f x x x f x g ;

证明:)(x g 在0x =处可导。

五、综合题(每小题10分,共20分) 23、分析函数x

xe y -=的单调性,凹凸性,极值,拐点及渐近线。

24、在曲线ln y x =上(),1e 点处作切线l ,

(1)求由曲线切线、曲线本身及x 轴所围的图形面积。 (2)求上述所围图形绕x 轴旋转所得立体的体积。

江苏省2013年普通高校“专转本”统一考试模拟试卷解析(二)

高等数学

一、选择题(本大题共6小题,每小题4分,共24分,在每小题给出的四个选项中,只有一项是符合要求的,请把所选项前的字母填在题后的括号内)。 1、下列极限中,正确的是( ) A 、 sin 2lim

2x x x →∞= B 、arctan lim 1x x

x

→+∞=

C 、 224

lim

2

x x x →-=∞- D 、0lim 1x x x →+= 解析:求极限时,先判断极限类型,若是00或∞

型可以直接使用罗比达法则,其余类型可以转化为

00或∞

型。罗比达法则求极限的好处主要有两方面,一是通过求导降阶,二是通过求导将难求极限的极限形式转变为容易求极限的形式。不过,在求极限时应灵活使用多种方法,特别是无穷小量或是无穷大量阶的比较,无穷小量与有界变量的乘积还是无穷小量等性质。使用等价无穷小或是等价无穷大的目的是将函数转换为幂的形式,方便判别阶数。

0lim ln 00lim 1x x x

x x x e e →+

→+

===,故本题答案选D

2、设2

1sin ,0

(),0

x x f x x

ax b x ?>?=??+≤?在0x =处可导,则( ) A 、1,0a b == B 、0,a b =为任意实数 C 、0,0b a == D 、1,a b =为任意实数

解析:分段函数在分段点处的极限、连续性与可导性,若分段点的左右两侧的表达式互不相

同,则必须使用定义左右分别讨论。本题只需按照导数定义讨论即可。

00()(0)(0)lim

lim x x f x f ax

f a

x x -→-→--'===

2001sin ()(0)(0)lim lim x x x b

f x f x f x x

+→+→+--'== ,因为左、右导数应相等,易知0,0b a ==; 故本题答案选C

3、函数sin y x =在区间[]0,π上满足罗尔定理的ξ=( ) A 、0 B 、4π C 、2

π

D 、π

解析:熟记罗尔定理、拉格朗日定理的条件与结论及其几何解释,本题答案选C

4、设在[]b a ,上0)(>x f ,0)(<'x f ,0)(>''x f ,令dx x f y b a

?

=

)(1,

))((2a b b f y -=,[]()a b b f a f y -+=

)()(2

1

3,则有( ) A 、321y y y << B 、 312y y y << C 、213y y y << D 、132y y y <<

解析:本题利用导数考查曲线的形态,定积分的几何意义——曲边梯形的面积(代数和)。比较图形面积即可知本题答案选B

5、两个非零向量a 与b

垂直的充分必要条件是( )

A 、0=?b a

B 、 0

=?b a

C 、 0

=?a b D 、0=?a a

解析:本题考查向量平行与垂直的充要条件,本题答案选A 6、下列级数发散的是( )

A 、1

(1)ln(1)n

n n ∞

=-+∑ B 、131n n n ∞

=-∑

C 、1

1(1)3n n n -∞

=-∑ D 、13

n n n ∞

=∑ 解析:该题考察级数的收敛性质、级数收敛的必要条件,(交错)P -级数等。故本题答案选B ,因为该选项破坏级数收敛的必要条件。 记住

11

:p

n n

=∑当1p >时收敛,1p ≤时发散。 交错1(1):n

p n n

=-∑当1p >时绝对收敛,01p <≤时条件收敛,0p ≤时发散。

二、填空题(本大题共6小题,每小题6分,共24分,请把正确答案的结果添在划线上)。

7、已知()g x 为有界函数,1

()sin ,0

(), 0

g x x f x x a x ?≠?=??=? 0x =处连续,则=a

解析:因为0

1lim ()lim ()sin 0(0)x x f x g x f x

→→===,又(0)f a =,故0a =。

8、已知0

(23)(23)

(2)1,lim

h f h f h f h

→--+'=则= ______

解析:该题考察导数定义

0000

()()()lim

h f x h f x f x h →+-'=或0000()()

()lim h f x h f x f x h

→+-'=; 式子当中的h 应当理解为中间变量,看成文字。该题答案6

-

9、2x y xe -=,则()0y ''= 解析:()222x

x y e

x e --'=+-,()212x y e x -'=-

()()22122x x y e x e --''=-+--,()244x y e x -''=-+,(0)4y ''=-。

10、定积分1

4211ln 11x x x dx x --??

+- ?+??

?的值为 解析:该题考察奇偶函数的定积分在对称区间上的积分性质以及定积分的几何意义。

0,()()2(),()a

a

a

f x f x dx f x dx f x -??

=????

?为奇函数为偶函数 1

4211142111ln 111ln 1012

x x x dx x x x dx x dx x π----??+- ?+??

-=+-=++???

这里因为函数4

1()ln 1x f x x x

-=+是奇函数,故积分为零,积分1211x dx --?表示上半单位圆

的面积。

11、微分方程232x

y y y xe '''-+=的特解y *

的形式应为_____________________ 解析:解微分方程首先要判别类型,该方程是二阶常系数线性非齐次方程。 (1)齐次方程0y py qy '''++=,其中,p q 为常数。

求解步骤:1)特征方程

02=++q p λλ,求根21,λλ。

2)21,λλ 互异实根,x x

e c e c y 2121λλ+=,

21λλ=,x x xe c e c y 2

1

21λλ+=;

)0(2,1≠±=ββαλi ,12(cos sin )x y e c x c x αββ=+。

(2)非齐次方程()y py qy f x '''++=,通解为其所对应的齐次方程通解加上本身特解y *。 第一种:()()x

m f x e P x α=,其中()m P x 表示m 次多项式。

解结构:y =齐次方程通解y +特解y *。

特解y *形式设定如下: (1)识别,m α;

(2)考查α作为特征根的重数个数k ; (3)特解可设为()()k x

m y x x e Q x α*=,

其中()m Q x 表示m 次多项式。0,1,2,k ααα??

=???

不是特征根;是单根;是二重根;

第二种:()()()()()()cos sin x

m

n

f x e

P x x P x x αββ=+,

其中()m P x ,()n P x 表示,m n 次多项式。 解结构:y =齐次方程通解y +特解y *。

特解y *形式设定如下: (1)识别,,,m n αβ;

(2)计算i μαβ=+,k μ=和特征根12,λλ相等个数,()max ,l m n =。 (3)特解可设为()()()()()()?cos sin k x

l

l

y x x e

Q x x Q

x x αββ*=+,

其中()(),l l Q x Q x 为l 次多项式。

其中01i k i αβαβ+?=?+?,不是特征根;,是特征根;

故本题答案为 2()x

x Ax B e +,其中,A B 待定系数。

12、设22

sin 2z x y x y =++,则(1,)2

dz π=_____________________________

解析:该题考察多元函数的全微分

若(,)z f x y =可微,则(,)(,)x y dz f x y dx f x y dy ''=+,

00(,)

0000(,)(,)x y x y dz

f x y dx f x y dy ''=+

本题中,(2sin 2)(22cos 2)dz x y dx y x y dy =+++ (1,)2

2(2)dz dx dy ππ=+-

三、计算题(本大题共8小题,每小题8分,共64分)。

13、考查1

1sin ()x x f x e x

-=的连续性,若有间断点,判别类型。 解析:函数()f x 在0x 处连续的定义为0

0lim ()()x x

f x f x →=。实际上包含三个条件 (1) 函数()f x 在0x 处必须有定义; (2) 函数()f x 在0x 处的极限存在;

(3) 函数()f x 在0x 处的极限值必须等于函数值;

当上述三个条件不全满足时的点即为函数()f x 的间断点。而初等函数在定义区间之内均是连续的,所以,没有定义的点一定是间断点,分段函数的分段点是可能的间断点。 根据点0x 处的极限情况来加以分类:

????

???

∞?

???

??

相等:可去间断点

左右极限均存在:第一类不相等:跳跃间断点若有一个为:无穷间断点左右极限至少有一个不存在:第二类均不为无穷,函数不停振荡:振荡间断点根据以上分析,本题间断点为1=x ,0

11

10

10

(10)l i m ()s i n (1)l i m 0

x x x f f x e -→-→--===,

1

1

10

10

(10)l i m ()

s i n (1)l i m x x x f f x e -→+→++=

==+∞。

即1=x 为第Ⅱ类无穷间断点。

又 1

lim ()x f x e -→=,0x =为可去间断点。

故1

1sin ()x x f x e x

-=的连续区间为()()(),00,11,-∞??+∞。 14、设函数)(x y y

=由方程 1)cos(2-=-+e xy e y x 所确定,求

0|=x dx

dy

。 解析:隐函数的导数是常考的一个内容,它的本质实际上是复合函数的导数问题。一般隐函数很难甚至不可能显化。其求导方法是方程(等式)两边对x 求导数,将y 看成x 的函数(中间变量)。

方程1)cos(2-=-+e xy e

y

x 两边对x 求导,得

))(sin()2(2='++'++y x y xy y e y x

)

sin()sin(222xy x e xy y e dx dy y x y x ++-=++ 将0=x 代入原方程得1=y ,于是得到2|0-==x dx

dy

15、求不定积分

6d (1)x

x x +?。

解析:该题使用凑微分法,当次数超过4次的有理分式,一般不用将其分解成简单分式之和来积分,原因有二个,首先分解较困难,其次,待定系数较多,不太容易确定。

???+=+=+)1(d 61)1(d )1(d 666

6656x x x x x x x x x x

=?+-6

6

6d )111(61x x x =C x x ++1

ln 6166

16、设1

sin ,0()20,

0x x f x x x π

π?≤≤?=??<>?或,求?=Φx dt t f x 0

)()(在),(+∞-∞内的表达式。

解析:变上限积分的计算,首先弄清楚变量t 与x 的关系, t 为积分变量,对t 积分时,将x

当作常量,注意分段函数的积分要分区间考虑。

0)(,00

==Φ

dt x x

11

0,()sin (1cos )22

x x x tdt x π≤≤Φ==-? 0

1

,()sin 012x x x tdt dt πππ>Φ=+=?

?

综上

0,

01()(1cos ),

021,

x x x x x ππ

Φ=-≤≤??>??

另外,变上限积分的求导公式,对于很多同学可能会觉得不容易记牢。(几乎每年必考) 在记忆时不彷考虑牛顿莱布尼兹公式辅助记忆

()

()()

()

()()

[()][()]b x b x a x a x f t dt F t F b x F a x ==-?

()()

(())([()][()])(())()(())()

b x a x f t dt F b x F a x f b x b x f a x a x ''''=-=-?

2

1224224()(sin )sin(1)(1)sin ()sin(1)2sin x

x x t dt x x x x x x x ?-''''==---=---?

17、求过点)2,1,3(-A 且通过直线43:

521

x y z

L -+==的平面方程。 解析:求平面方程,基本方法是使用点法式。求出平面上的一个定点和法向量n 。 平面上的定点)2,1,3(-A 已知,又直线1

2354:

z

y x L =+=-过点{}0,3,4-=B ,其方向向量法向量{}5,2,1s =,{}2,4,1-=AB ;故

{}5

2

18,9,22142

i

j k

n s AB =?==---

平面的方程为0)2(22)1(9)3(8=+----z y x 即为0592298=---z y x 。

18、将函数)54ln()(-=x x f 展开为2-x 的幂级数,并指出其收敛域。

解析:将函数()f x 展开为x 或0()x x -的幂级数,并指出其收敛域的方法,前面已祥述,这里不再赘述。

)]2(3

4

1ln[3ln ]3)2(4ln[)(-++=+-=x x x f

??? ??≤

5,)2(34)1(3ln 1x n x n

n

19、设(

)22,x y

z f x y e

+=+,其中),(v u f 有二阶连续偏导数,求x z ??、y x z

???2。

解析:该题型是几乎每年必考。需要认真掌握。

第一步:变量z y x ,,的关系网络图

12x y z x y

?→????→??

其中1,2分别表示22,x y

x y e

++

第二步:寻找与x 对应的路径()∨,计算的过程可以总结为“路中用乘,路间用加”

2121x y u

f f e x

+?''=?+??, 22211

1221222211

1222222222()2x y x y x y x y z

f y f e f y f e x y y f f e y e f ++++?''''''''=?+??+?+????''''''=?+?++?

20、计算积分

y

x

D

e dxdy ??,其中2

:,2D y x y x ==由所围成的区域。 解析:二重积分问题是很多“专转本”同学的难点。首先要理解二重积分的几何意义,特别是对称型简化积分计算。 首先要画出积分区域,然后根据被积函数的特点与区域的形状选择适当的坐标以及适当的积分顺序。一般当被积函数形如22()f x y +,区域形状为圆形、圆环、扇形(环)等,往往使用极坐标计算;否则,往往用直角坐标计算。

2220

d d d d y

y x

x

x

x

D

e

x y x e y =????

2

20

()d x x e e x =-?

21e =-

四、证明题(每小题9分,共18分)

21、设()f x 在[0,1]上连续,且0()1f x ≤≤。证明:在[0,1]上至少存在一个ξ使()f ξξ=。 解析:遇到至少存在一点ξ的问题,通常使用介质定理,零点定理。还有罗尔定理和拉格朗日中值定理。这种问题一般逆向思维构造辅助函数。 令)()(x f x x F -=

则)(x F 在]1,0[上连续,且0)0()0(≤-=f F 0)1(1)1(≥-=f F

0)0(=F 或0)1(=F 成立,那么就相应地有0=ξ或1

否则可假设(0)0F <0)1(>F ,则由闭区间上的连续函数零点定理可知, 在)1,0(上存在一点ξ,使0)(=ξF ,即()f ξξ=; 综上所述,得到题设结论。

22、设)(x f 在0x =的邻域内有二阶连续导数,且0)0(=f ,???

??=='≠=0

,0)0(0,)

()(x f x x x f x g ;

证明:)(x g 在0x =处可导。

解析:分段函数在分段点处的极限、连续性与可导性,若分段点的左右两侧的表达式互不相

同,则必须使用定义左右分别讨论。本题只需按照导数定义讨论即可。

02

0000()(0)(0)lim

()

(0)

()(0)lim lim

()(0)()(0)lim lim 222

h h h h h g h g g h

f h f f h f h

h h h f h f f h f h →→→→→-'='-'-==''''''-=== 故)(x g 在0x =处可导,且(0)

(0)2f g '''=。

五、综合题(每小题10分,共20分)

23、分析函数x xe y -=的单调性,凹凸性,极值,拐点及渐近线。 解析:该题考察导数的综合应用,一般列表处理。 (1)定义域为R x ∈,

渐近线:因01lim lim

lim ===+∞→+∞→-+∞

→x

x x x x

x e e x xe

0y =,即x 轴为水平渐近线 (2)(1)x

y x e -'=- 1(1)(1)(2)x x x y e

x e x e ---''=-+--=-,由0y '=得1x =,由0y ''=得2x =

(3)列表分析

x )1,(-∞

1

)2,1(

2

),2(+∞

y '

y '' +

-

--

-+

y ↑

极大值

()1

1y e

-=

拐点

()222y e -=

(4)x xe y -=在)1,(-∞上单调上升向上凸,)2,1(上单调下降,向上凸,),2(+∞上单调下降,向上凸,(1,1-e )为极大值点,(2,22-e )为拐点。

24、在曲线ln y x =上(),1e 点处作切线l ,

(1)求由曲线切线、曲线本身及x 轴所围的图形面积。 (2)求上述所围图形绕x 轴旋转所得立体的体积。

解析:该类题型是定积分应用中常考的题型,但是近两年在该知识点常出综合题。结合切线(法线)、微分方程,极限等知识点出题。 (1) e e y k x y 1)(,1='==

' ,故切线为)(11e x e y -=- 或 e

x

y = 121

0122

y y e e

S e ey dy y e ????=-=-+=

-??????

?, (2) 22

01

()(ln )e

e x V dx x dx e ππ=-??3201211ln 2ln 3e e e x x x x x dx e x ππ??=--???

?? 12ln 3e e e xdx πππ=

-+?222(1)3e e e πππ=-+--)3

1(2e

-=π。

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