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专升本试题及解答(四川理工2017)

专升本试题及解答(四川理工2017)
专升本试题及解答(四川理工2017)

2017年四川理工学院专升本《高等数学》考试题(理工类)

一、单项选择题(每题3分,共15分)

1、当0→x 时,下列选项中是x 的高阶无穷小的是( C )

(A )x 2sin (B )11--x (C )1cos -x (D ))51ln(x + 【知识点】无穷小的比较。

解析:021lim 1cos lim 200=-

=-→→x

x

x x x x ,由定义知,1cos -x 是x 的高阶无穷小。 2、已知c x F dx x f +=?)()(,则=+?dx x

f )12

(( D )

(A )C x F +)(2 (B )C x F +)2( (C )C x F ++)12( (D )C x

F ++)12

(2

【知识点】第一类换元积分法(凑微分法)。 解析:

C x

F x d x f dx x f ++=++=+??

)12

(2)12()12(2)12(。 3、可设方程x

xe y y y 396-=+'+''特解的待定系数形式为( B )

(A )x

e

b ax 3)(-+ (B )x

e

b ax x 32

)(-+ (C )x

axe

3- (D )x

e

3-

【知识点】二阶非齐次方程的特解形式)(*x Q e x y n x

k λ=。

解析:特征方程0962

=++r r ,321-==r r (重根),3-=λ 故,特解形式可设为:x

e

b ax x y 32

)(*-+=。

4、下列级数中,条件收敛的是( C ) (A )

n n n )32()

1(1

1

∑∞

=-- (B )∑∞=--11)1(n n n (C )12)1(11+-∑∞=-n n n n (D )311

51)1(n

n n ∑∞

=-- 【知识点】条件收敛的概念。 解析:对级数

1

2)1(1

1

+-∑∞

=-n n

n n : ∑∑∞

=∞

=+=1112n n n n n u ,021

12lim ≠=+∞→n n n ,由级数收敛的必要条件知,级数∑∞

=1n n u 发散; 由交错级数的审敛法知,

12)

1(1

1

+-∑∞

=-n n

n n 收敛,即∑∞

=1

n n u 收敛, 故,级数

1

2)1(1

1

+-∑∞

=-n n

n n 条件收敛。

5、设21,αα是非齐次线性方程组b AX =的解,β是对应齐次方程组的一个解,则b AX =一定有一个解是( D )

(A )21αα+ (B )21αα- (C )21ααβ++ (D )βαα-+213

2

31 【知识点】方程组解的定义。

解析:由题设知,0,,21===βααA b A b A ,而

b b b A A A A =-+=-+=-+03

2

313231)3231(2121βααβαα, 即,向量βαα-+213231满足方程组b AX =,故βαα-+213

2

31为方程组的一个解。

二、填空题:(每题3分,共15分) 1、已知3)31(lim -∞

→=+

e n

nk

n ,则=k 。

【1-】 【知识点】重要极限。

解析:3333)

31(lim )31(lim -?∞→∞→==+=+e e n

n k k

n

n nk n ,即1-=k 。 2、设???>≤=1

,ln 1,)(x x x e x f x ,则=?e dx x f 0

)( 。【e 】

【知识点】定积分的区间可加性。 解析:

e e dx x dx e dx x

f e

x e

=+-=+=???

1)1(ln )(1

10

3、曲线2

1x x

y -=

的渐近线条数为 ;【3】

【知识点】渐近线的定义。 解析:01lim

2=-∞→x x

x ,即0=y 为曲线的水平渐近线;

∞=-→211lim x x x ,∞=--→2

11lim x x

x ,即1±=x 为曲线的垂直渐近线。

4、幂级数∑∞

=-1

)3(n n n x 的收敛域为 。【)4,2[】

【知识点】幂级数的收敛域。(考虑端点) 解析:由13lim

1

<-=+∞→x u u n

n n 得:42<

当2=x 时,级数∑∞

=-1

)1(n n

n 为收敛的交错级数,

当4=x 时,级数

∑∞

=1

1

n n 为发散的调和级数,故收敛域为)4,2[。

5、若矩阵X 满足???? ??--=???? ?

?123431

52X ,则=X 。【???

?

??-1042】 【知识点】逆矩阵及矩阵的乘法。 解析:由???

?

??--=????

??12343152X 得: ???? ??--???? ??=-123431521

X ???

?

??-=???? ??--???? ??--=104212342153

三、求解下列各题(每小题6分,共60分)

1、确定b a ,的值,使函数???≤+>=0

,)sin(0

,)(x b ax x e x f x 在0=x 处可导。【1==b a 】

【知识点】可导、连续的定义。

解析:由连续知:])[sin(lim )(lim 00b ax e x x

x +=-

→+

→,即1=b ;

由可导知:x

x a x e x x x ??=?--→??+→?)

sin(lim 1lim 00,即1=a 。

2、求曲线2

x y =与直线2+=x y 所围成平面图形的面积,并求此图形绕x 轴旋转一周所成旋转体的的体积。

【知识点】定积分的应用(求面积和体积)。 解析:(图略)29)2(2

1

2=

-+=?

-dx x x A ;ππ5

62])2[(214

2=-+=?-dx x x V 。

3、设

x

x

sin 是)(x f 的一个原函数,求?'dx x f x )(。 【知识点】分部积分法。 解析:

c x

x

x x c x x x x x dx x f x xf x xdf dx x f x +-=+-'=-=='???sin 2cos sin )sin (

)()()()(。 4、求曲线3

2

,,:t z t y t x L ===上的点,使该点处的切线平行于平面133=++z y x 。 【知识点】空间曲线的切线。

解析:2

3,2,1t z t y x ='='=',设对应点的的参数为t ,则切向量}3,2,1{2

t t s =; 平面的法向量}1,3,3{=n ,由条件得:0=?s n ,即03632

=++t t ,即1-=t , 故,所求点的坐标为)1,1,1(-。

5、方程023

=+-y xz z 确定函数),(y x z z =,求dz 。 【知识点】隐函数的全微分(微分的形式不变性)。 解析:两边求微分:0232

=+--dy xdz zdx dz z ,即x

z dy

zdx dz --=2

32。 6、将函数x

x f 1

)(=

展开成)3(+x 的幂级数。 【知识点】幂级数的展开式。

解析:3

3113

1

)3(311+-

-=+--

=x x x ,由

∑∞

==-011

n n x x (1

=+∞=-+=+-=0

1

03)3()33(31)(n n n

n n x x x f (06<<-x )。 7、求可导函数)(x y ,使其满足方程20

)(2)(x dt t y x y x

=+?

【知识点】变上限函数、微分方程。

解析:两边求导:x y y 22=+',由通解公式得:

2

1

)21(]2[222222-+=+-=+??=---?x Ce C e xe e C dx xe e y x x x x dx dx ;

由0)0(=y 得21=C ,故,2

1

212-+=-x e y x 。

8、计算dx y y e dy x y y e I L

x x )cos ()22sin (2

--++-=?,其中L 是上半圆x

y x 222=+从点)0,2(A 到点)0,0(B 的一段弧。 【知识点】曲线积分。 解析:

22sin +=??y e x Q

x ,12sin -=??y e y P x ,由格林公式得: πσσ23

3)(

==??-??=+

?????

?

d d y P x Q D

D

BA

L

, 而

22

1e dx e x BA

-=-=??,故12

3

2-+=

?e L

π。 9、计算??

+=

D

dxdy y x I )(2

2,其中D 由4)2(22=+-y x 所围成的闭区域。 【知识点】二重积分(极坐标系)。 解析:(图略)θπ

θπ

cos 40;2

2

:≤≤≤

≤-

R D ,于是

128]4cos 8

1

2cos 2183[128cos 12820204

cos 40

3

22

-=++===??

?-π

πθ

π

πθθθθθd dr r d I 。

10、当a 为何值时,线性方程组???????=+++=-++=+++=+++a

x x x x x x x x x x x x x x x x 432143214

3214321710535105363132有解,在有解的情况下,求其

通解。

【知识点】非齐次线性方程组的通解。

解析:???????

?

?-=a A 7105354105131631

13

21

1???????

?

?----→324204784

02242

013211a ??

?

?

?

?

?

??---→

50000

03000

22420

13

2

11

a

要使方程组有解,必须05=-a ,即5=a ,此时方程组有无穷多解。 取3x 为自由变量,令13=x 得齐次方程组的基础解系:T

)0,1,2,0(-=α; 令03=x 得非齐次方程组的一个特解:T

)0,0,1,0(=β,故通解为αβk X +=。

四、证明题(每小题5分,共10分)

1、已知向量组321,,ααα线性无关,且211ααβ+=,232ααβ+=,313ααβ+=,证明:向量组321,,βββ线性无关。 【知识点】向量组线性无关的定义。

证明:)()()(313232211332211ααααααβββ+++++=++x x x x x x =332221131)()()(αααx x x x x x +++++

因321,,ααα线性无关,所以???

??===??????=+=+=+0

000003

21322131x x x x x x x x x ,

由定义知,向量组321,,βββ线性无关。

2、设函数)(x f 在]1,0[上连续,在)1,0(内可导,且0)0(=f ,证明:在)1,0(内至少存在一点ξ,使得ξ

ξξ-=

'1)(3)(f f 。 【知识点】罗尔定理。

证明:令)()1()(3

x f x x F -=,)()1()()1(3)(3

2

x f x x f x x F '-+--=';

显然)(x F 在]1,0[上连续,在)1,0(内可导,且0)1()0(==F F ,

由罗尔定理,)1,0(∈?ξ使0)(='ξF ,即0)()1()()1(33

2

='-+--ξξξξf f 因)1,0(∈ξ,所以01≠-ξ,于是,0)()1()(3='-+-ξξξf f , 即,ξ

ξξ-=

'1)

(3)(f f 。

山东省2017年普通高等教育专升本统一考试英语试题及答案

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