(完整)安徽普高专升本统考《高等数学》试题答案解析

(完整)安徽普高专升本统考《高等数学》试题答案解析
(完整)安徽普高专升本统考《高等数学》试题答案解析

安徽省2012年普通高等学校专升本招生考试

高等数学

注意事项:

1.试卷共8页,请用签字笔答题,答案按要求写在指定的位置。 2.答题前将密封线内的项目填写完整。

一、选择题(下列每小题的选项中,只有一项是符合题意的,请将表示该选项的字母填在题后的括号内。共10小题,每小题3分,共30分)

1.若函数???

??>+≤=0,sin 0,3)(x a x

x x e x f x 在0=x 在处连续,则=a ( C )

A. 0

B. 1

C. 2

D. 3

解:由)0()00()00(f f f =-=+得231=?=+a a ,故选C. 2.当0→x 时,与函数2

)(x x f =是等价无穷小的是( A ) A. )1ln(2

x + B. x sin C.

x tan D. x cos 1-

解:由11ln(lim 1ln()(lim )

22

0)20=+=+→→x x x x f x x ,故选A.

3.设)(x f y =可导,则'-)]([x e f =( D )

A. )(x

e

f -' B. )(x e f -'- C. )(x x e f e --' D. )(x x e f e --'-

解:)()()()]([x

x x x x

e f e e e f e f -----'-='?'='

,故选D. 4.设

x 1是)(x f

的一个原函数,则?=dx x f x )(3

( B ) A.

C x +2

2

1 B. C x +-221 C. C x +331 D. C x x +ln 414

解:因x 1是)(x f

的一个原函数,所以211)(x x x f -='

???

??=,所以

C x xdx dx x f x +-=-=??23

2

1)(故选B. 5.下列级数中收敛的是( C )

A. ∑∞

=-1

374n n

n n B. ∑

=-1

2

31

n n C.

∑∞

=13

2

n n

n D. ∑∞

=1

21sin

n n

解:因121

)1(lim 212

2)1(lim 33313

<=+=+∞→+∞→n n n n n n n n ,所以∑∞=132n n n 收敛,

故选C. 6.交换?

???+=

10

2121

1

21),(),(y y

y

dx y x f dy dx y x f dy I 的积分次序,则下列各项正确的

是( B ) A. ??

1

22

),(x

x dy y x f dx B.

??

10

22

),(x x dy y x f dy C.

?

?2

1

22

),(x x dy y x f dx D.

?

?

2

1

22

),(x x dy y x f dx

解:由题意画出积分区域如图:故选B.

7.设向量21,αα是非齐次线性方程组AX =b 的两个解,则下列向量中仍为该方程组解的是( D ) A.

21αα+ B. 21αα- C. 212αα+ D. 212αα-

解:因,2)(2121b b b A A A =+=+=+αααα同理得

,0)(21=-ααA ,3)2(21b A =+αα,)2(21b A =-αα

故选D.

8.已知向量)2,5,4,0(),0,,0,2(),1,1,2,1(321--==-=αααk 线性相关,则=k ( D ) A. -2 B. 2 C. -3 D. 3

解:????? ??+--+--→????? ??---+--→????? ?

?---=????? ??03002240112125402240112

12540002112132

1k k k k ααα 由于123,,ααα线性相关,所以123(,,)2r ααα≤,因此3=k

9.设B A ,为事件,且,2.0)(,4.0)(,6.0)(===AB P B P A P 则=)(B A P ( A ) A.0.2 B. 0. 4 C. 0.6 D. 0.8

解: 2.0)]()()([1)(1)()(=-+-=+-=+=AB P B P A P B A P B A P B A P 10.有两个口袋,甲袋中有3个白球和1个黑球,乙袋中有1个白球和3个黑球.现从甲袋中任取一个球放入乙袋,再从乙袋中任取一个球,则取出白球的概率是( B ) A.

163 B. 207 C. 41 D. 2

1 解: 由全概率公式得20

7

51415243=

?+?=

p

二、填空题(本题共10小题,每小题3分,共30分,把答案填在题中横线上。) 11.设函数2161

31arcsin

x

x y --

-=,则函数的定义域为)4,2[-. 解:424442016,131

12<≤-??

??<<-≤≤-?>-≤-≤-x x x x x .

12.设曲线22-+=x x y 在点M 处的切线斜率为3,则点M 的坐标是)0,1(. 解:12+='x y ,由1312=?=+='x x y ,从而0=y ,故填)0,1(.

13.设函数x x y arctan =,则=

''y 2

2)1(2

x +.

解:2

1arctan x

x x y ++=',2222222)1(2

)1(2111x x x x x y +=+-+++=''. 14.

=+?dx x

x 2012)1(ln

C x ++2013)1(ln 2013

. 解:

C x x d x dx x x ++=++=+??

2013

)1(ln )1(ln )1(ln )1(ln 20132012

2012.

15.=?∞++-dx xe x 01= e . 解:

e dx xe e dx xe x x ==?

?

+∞-∞++-0

1.

16.幂级数∑∞

=-15

)2(n n

n

n x 的收敛域为)7,3[-.

解:由152215lim 5)2(15)2(lim )()(lim 11

1<-=-+=-+-=∞→++∞→+∞→x x n n n

x n x x u x u n n n n n n n

n n .

得73<<-x 级数收敛,

当3-=x 时,级数为∑∞

=-1

)1(n n

n 收敛; 当7=x 时,级数为∑∞

=11n n 发散;

故收敛域为)7,3[-.

17.设A 是n 阶矩阵,E 是n 阶单位矩阵,且,032=--E A A 则=--1

)2(E A E A +.

解:)()2())(2(031

2

E A E A E E A E A E A A +=-?=+-?=---

18.设????

?

??-=10010

1110A ,记1

-A 表示A 的逆矩阵, *A 表示A 的伴随矩阵,则 =-*

1)(A ????

?

??----100101110.

19.设型随机变量),8,1(~N X 且),()(c X P c X P ≥=<则c = 1.

解:由正态分布的对称性得1==μc .

20.设型随机变量X 在区间]4,2[上服从均匀分布,则方差=

)(X D 3

1

.

解:直接由均匀分布得3

1

12)24()(2=-=X D .

三、计算题:本大题共8小题,其中第21-27题每题7分,第28题11分,共60分。

21.计算极限x x

x x 20tan sin lim

-→.

解:原式= 2

0sin lim x x

x x -→

=x

x

x 2cos 1lim

0-→

=2

sin lim 0x

x →=0.

22.求由方程xy y x =确定的隐函数的导数dx

dy

. 解:两边取对数得y x y x ln ln ln +=, 两边求导得y y

x y y x y '+='+

11ln , 从而)

1()

ln 1(--=

x x y x y dx dy .

23.计算定积分

?

-22

2

2

1

1dx x x

解:令t x sec =,则,tan sec tdt t dx =当2=x 时, 4

π=

t ;当2=x 时, 3

π=

t .

所以原式= ?3

4

2

tan sec tan sec π

πdt t t t t = ?34

cos π

πtdt = =|34

sin π

πt =

)23(2

1

-.

24.求微分方程02=--'x e y y 的通解.

解:原方程可整理为x

e y y =-'2

这是一阶线性微分方程,其中x

e x Q x P =-=)(,2)(. 所以原方程的通解为

??

????+??=?-C dx e x Q e y dx x P dx x P )()()( )(22C dx e e e dx

x dx

+??

=?-.

)(2C dx e e x x +=?-)(2C e e x x +-=-

x x Ce e 2+-=

25.计算二重积分

??D

yd x

σ2

,其中D 是由直线222===xy x y x 和、所围成的区

域.

解:区域D 如图阴影部分所示.

??D

yd x σ2

??=x

x

y y x x 22221

d d

?=

2

12222d 21|y y x x

x

?-=2

1

4)d 44(21x x |215

)252(x x -=5

210=.

26.设矩阵????

?

??---=320031

10

1

A ,,231?????

??=B 且满足X B A B AX +=+2

,求矩阵X .

解:由X B A B AX +=+2可得B E A E A B E A X E A ))(()()(2+-=-=-

因024

2

041

100||≠-=---=-E A ,所以E A -可逆, 因此B E A X )(+=????? ???????

?

?---=23122

0021

102????

?

??-=250

27.设行列式132

1

31213

2

113211)(++++=

x x x x x D ,求)(x D 在0=x 处的导数.

解:132731273

21732171321

31213

2113211)(+++++++=

++++=

x x x x x x x x x x x x D

2

11

10

1110

10001)

7(1

32

1

31213

2

113211)

7(--+=++++=x x x x x x x x

)23)(7()2)(1)(7(22+-+=--+=x x x x x x x x .

故)32)(7()23)(72()(22-+++-+='x x x x x x x D . 从而14)0(='D .

28.已知离散型随机变量X 的密度函数为?????

????≥<≤<≤<=.2,1,21,21,10,,0,0)(x

x x a x x F 且数学期望

3

4

)(=

X E .

图5-7

求: (1) a 的值; (2) X 的分布列;(3)方差D (X ).

解:(1) 由分布函数的性质知,随机变量X 的可能取值为0、1、2,且

21)2(,21)1(,)0(==-=

===X P a X P a X P 因3

4

23212)21(10)(=-=?+-?+?=a a a X E

所以61

=a .

(2) 由(1)即得

(3) 3

223160)(2

222=?+?+?=X E ,

四、证明题与

应用题:本大题共3小题,每小题10分,共30分。 29.设)(2y

x f xy u =,其中)(t f 可微,u y

z

y x z x

3:=??+??证明. 证明:因为

y

y x f xy y x f y x u 1

)()(22?'+=?? ),()(2

y

x f xy y x

f y '+=

???

?

??-?'+=??22)()(2y x y x f xy y x xyf y u )()(22y

x

f x y x xyf '-=,

故)()(2)()(2222y

x f y x y x f xy y x f y x y x f xy y u y x u x

'-+'+=??+?? )(32

y

x

f xy =u 3=. ????(9分)

30.设D 是由曲线e x x y ==,ln 及x 轴所围成的的平面区域

求: (1) 平面区域D 的面积S ; (2) D 绕y 轴旋转一周所成的旋转体的体积V . 解:区域D 如图阴影部分所示。曲线x y ln =与x 轴及 e x =的交点坐标分别为)1,(),0,1(e (1)平面区域D 的面积

1)ln (d ln |1

1

=-==?e

e

x x x x x S .

(2)D 绕y 轴旋转一周所成的旋转体的体积V

).

1(2

2

d d )(121

221

22

1

22|+=

-

==-=-??=??e e e y e e y

e e V y y y π

π

πππππ

31.证明不等式:当e b a >>时,

)71828.2(ln ln ≈<

a a

b a b . 证明: 设),(,ln )(+∞∈=e x x x x f ,则),(,0ln 1)(+∞∈>+='e x x x f , 所以),(ln )(+∞∈=e x x x x f 在上单调递增,从而当当e b a >>时,有

)()(b f a f >,即b b a a ln ln >,即

b

a

a b

+∞∈<-='e x x x

x g , 所以),(ln )(+∞∈=e x x

x

x g 在上单调递减,从而当当e b a >>时,有 )()(b f a f <,即b b a a ln ln <,从而a b

a b ln ln <.

综上所述:当e b a >>时,有b

a a

b a b <

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