《高等数学(二)》全真押题试卷(二)

《高等数学(二)》全真押题试卷(二)
《高等数学(二)》全真押题试卷(二)

《高等数学(二)》全真押题试卷(二)

一、单项选择题(共10题,合计40分)

1

若事件A与B为互斥事件,且P(A)=0.3,P(A+B)=0.8,则P(B)等于().

A. 0.3

B. 0.4

C. 0.5

D. 0.6

[正确答案]C

本题分值:4分

试题解析:

本题考查的知识点是互斥事件的概念和加法公式.

2

A. 极大值1/2

B. 极大值-1/2

C. 极小值1/2

D. 极小值-1/2

[正确答案]D

本题分值:4分

试题解析:

本题主要考查极限的充分条件.

3

A. F(x)

B. -F(x)

C. 0

D. 2F(x)

[正确答案]B

本题分值:4分

试题解析:

用换元法将F(-x)与F(x)联系起来,再确定选项.

4

A. -2

B. -1

C. 0

D. 2

[正确答案]D

本题分值:4分

试题解析:

根据函数在一点导数定义的结构式可知

5

A.

B.

C.

D.

[正确答案]D

本题分值:4分

试题解析:

本题考查的知识点是复合函数的求导公式.

根据复合函数求导公式,可知D正确.

需要注意的是:选项A错误的原因是?是x的复合函数,所以必须通过对中间变量求导后才能对x求导.

6

当x→2时,下列函数中不是无穷小量的是().

A.

B.

C.

D.

[正确答案]C

本题分值:4分

试题解析:

7

设F(x)的一个原函数为xln(x+1),则下列等式成立的是().

A.

B.

C.

D.

[正确答案]A

本题分值:4分

试题解析:

本题考查的知识点是原函数的概念.

8

A. 0

B. 1/3

C. 1/2

D. 3

[正确答案]B

本题分值:4分

试题解析:

9

A. x=-2

B. x=-1

C. x=1

D. x=0

[正确答案]C

本题分值:4分

试题解析:

本题考查的知识点是函数间断点的求法.

如果函数?(x)在点x0处有下列三种情况之一,则点x0就是?(x)的一个间断点.

(1)在点x0处, ?(x)没有定义.

(2)在点x0处, ?(x)的极限不存在.

(3)

因此,本题的间断点为x=1,所以选C.

10若事件A发生必然导致事件B发生,则事件A和B的关系一定是( ).

A.

B.

C. 对立事件

D. 互不相容事件

[正确答案]A

本题分值:4分

试题解析:

本题考查的知识点是事件关系的概念.根据两个事件相互包含的定义,可知选项A正确.

二、填空题(共10题,合计40分)

11

[正确答案]应填2.

本题分值:4分

试题解析:

利用重要极限1求解.

12

[正确答案]

本题分值:4分

试题解析:

13

[正确答案]

应填π÷4.

本题分值:4分

试题解析:

14

[正确答案]应填0.

本题分值:4分

试题解析:

本题考查的知识点是函数在一点间断的概念.

15

[正确答案]应填1

本题分值:4分

试题解析:

16

[正确答案]

应填2In 2.本题考查的知识点是定积分的换元积分法.换元时,积分的上、下限一定要一起换.

本题分值:4分

试题解析:

17

已知P(A)=0.8,P(B\A)=0.5,则P(AB)=__________.

[正确答案]应填0.4

本题分值:4分

试题解析:

本题考查的知识点是乘法公式.

P(AB)=P(A)P(B\A)=0.8×0.5=0.4.

18

[正确答案]应填ln|x+1|-ln|x+2|+C

本题分值:4分

试题解析:

本题考查的知识点是有理分式的积分法.

简单有理函数的积分,经常将其写成一个整式与一个分式之和,或写成两个分式之和(如本题),再进行积分.

19

[正确答案]应填2xex2

本题分值:4分

试题解析:

20

二元函数?(x,y)=2+y2+xy+x+y的驻点是__________.[正确答案]应填x=-1/3,y=-1/3.

本题分值:4分

试题解析:

本题考查的知识点是多元函数驻点的概念和求法.

三、解答题(共8题,合计70分)

21

[正确答案]本题考查的知识点是重要极限Ⅱ.

本题分值:8分

试题解析:

对于重要极限Ⅱ:

22

[正确答案]

本题考查的知识点是二元隐函数全微分的求法.

利用公式法求导的关键是需构造辅助函数

然后将等式两边分别对x(或y或z)求导.读者一定要注意:对x求导时,y,z均视为常数,而对y或z求导时,另外两个变量同样也视为常数.也即用公式法时,辅助函数F(x,y,z)中的三个变量均视为自变量.

求全微分的第三种解法是直接对等式两边求微分,最后解出出,这种方法也十分简捷有效,建议考生能熟练掌握.

解法1等式两边对x求导得

解法3

本题分值:10分

试题解析:

23

[正确答案]

本题的关键是求出切线与坐标轴的交点.

本题分值:10分

试题解析: 24

已知袋中装有8个球,其中5个白球,3个黄球.一次取3个球,以X 表示所取的3个球中黄球的个数.

(1)求随机变量X 的分布列; (2)求数学期望E(X). [正确答案]

本题考查的知识点是随机变量X 的概率分布的求法.

【解析】 本题的关键是要分析出随机变量X 的取值以及算出取这些值时的概率.

因为一次取3个球,3个球中黄球的个数可能是0个,1个,2个,3个,即随机变量X 的取值为X=0,X=1,X=2,X=3.取这些值的概率用古典概型的概率公式计算即可. 解

(1)

所以随机变量X 的分布列为

>

注意:如果计算出的分布列中的概率之和不等于1,即不满足分布列的规范性,则必错无疑,考生可自行检查.

本题分值:9分 试题解析: 25

[正确答案]

本题考查的知识点是定积分的计算方法.

【解析】 本题既可用分部积分法计算,也可用换元积分法计算.此处只给出分部积分法,

有兴趣的读者可以尝试使用换元积分法计算.

本题分值:8分

试题解析:

26

设函数y=αx3+bx+c在点x=1处取得极小值-1,且点(0,1)为该函数曲线的拐点,试求常数a,b,c.

[正确答案]

本题考查的知识点是可导函数在某一点取得极小值的必要条件以及拐点的概念.

联立①②③,可解得α=1,b=-3,c=1.

本题分值:9分

试题解析:

27

求由曲线y=2-x2,),=2x-1及x≥0围成的平面图形的面积S以及此平面图形绕X轴旋转一周所得旋转体的体积Vx.

[正确答案]

本题考查的知识点有平面图形面积的计算及旋转体体积的计算.

本题的难点是根据所给的已知曲线画出封闭的平面图形,然后再求其面积S.求面积的关键是确定对x积分还是对Y积分.

确定平面图形的最简单方法是:题中给的曲线是三条,则该平面图形的边界也必须是三条,多一条或少一条都不是题中所要求的.

确定对x积分还是对y积分的一般原则是:尽可能用一个定积分而不是几个定积分之和来表示.本题如改为对y积分,则有

计算量显然比对x积分的计算量要大,所以选择积分变量的次序是能否快而准地求出积分的关键.

在求旋转体的体积时,一定要注意题目中的旋转轴是戈轴还是y轴.

由于本题在x轴下面的图形绕x轴旋转成的体积与x轴上面的图形绕x轴旋转的旋转体的体积重合了,所以只要计算x轴上面的图形绕戈轴旋转的旋转体体积即可.如果将旋转体的体积写成

上面的这种错误是考生比较容易出现的,所以审题时一定要注意.

由已知曲线画出平面图形为如图2—1—2所示的阴影区域.

本题分值:8分

试题解析:

28甲、乙二人单独译出某密码的概率分别为0.6和0.8,求此密码被破译的概率.

[正确答案]

本题考查的知识点是事件相互独立的概念和概率的加法公式.

本题的关键是密码被破译这一事件是指密码被甲破译或被乙破译,如果理解成甲破译密码且乙破译密码就错了!另外要注意:甲、乙二人破译密码是相互独立的.

解设A=“甲破译密码”,B=“乙破译密码”,C=“密码被破译”,则C=A+B,所以P(C)=P(A+B)=P(A)+P(B)-P(AB)=P(A)+P(B)-P(A)P(B)=0 6+0.8-0. 6×0.8=0. 92

本题分值:8分

试题解析:

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