第1.1节综合训练

第1.1节综合训练
第1.1节综合训练

第1.1节综合训练

刷能力

1.下列有关集合的写法正确的是()

A.B.C.D.

答案D

解析元素和集合是属于或不属于的关系,空集是没有元素的集合,所以D选项正确.

考点:元素和集合的关系.

2.设全集,集合

,则图中的阴影部分表示的集合为( )

A. B C D

答案B

解析由韦恩图可知阴影部分表示的集合为(C U A)∩B,根据集合的运算求解即可.解:全集U={1,2,3,4,5,6,7,8},集合A={1,2,3,5},B={2,4,6},由韦恩图可知阴影部分表示的集合为(C U A)∩B,∵C U A={4,6,7,8},∴(C U A)∩B={4,6}.故选B

3. 设集合,,则等于()。

A: B: C: D:

答案D

解析本题主要考查交集的运算,补集的运算。

由,

,故

4. 设a,,集合,则

答案1

解:集合,,,

计算得出,..因此,本题正确答案是:1.

5.设集合,,定义运算,则集合的子集的个数为()

A.3

B.4

C.8

D.16

答案C

解析因为,,,所以

,所以集合有3个元素,所以其子集的个数有8个.考点:集合子集的个数

6. 若U={n︱n是小于9的正整数},A={n∈U︱n是奇数},B={n∈U ︱n是3的倍数},则?U(A∪B)=.

解:∵U={n|n是小于9的正整数},∴U={1,2,3,4,5,6,7,8},则A={1,3,5,7},B={3,6,9},

所以A∪B={1,3,5,7,9},所以?U(A∪B)={2,4,8}

故答案为:{2,4,8}

7.设集合,或.

(1)若,求;

(2)若,求实数a的范围.

解:(1),集合,

或.或.

(2),可得:,计算得出.

8. 已知集合或,集合.

(1)若,求和;

(2)若,求实数a的取值范围.

解:(1),.

,或;

(2)由,得,

若,即,,满足;

当,即时,要使,

则或,计算得出.

使的a的取值范围是或.

9.

设,,

(1)若,求;

(2)如果,求实数的取值范围.

答案

解析

刷能力2

1. 设全集U={0,1,2,3,4},集合A={0,1,2,3},集合B={2,3,4},则( ?U A)∪( ?U B)=( )

A.{0}

B.{0,1}

C.{0,1,4}

D.{0,1,2,3,4}

解:由已知,得A∩B={2,3}.所以( ? U A)∪( ? U B)=? U (A∩B)={0,1,4}.故答案为:c

2.已知全集。,,则集合()。

A: B: C: D:

答案D

解析本题主要考查集合的基本运算。

,根据补集的定义,

3.已知集合,则满足条件

的集合C的个数为( )

A. 1

B. 2

C. 4

D. 8

答案D

解:,

因为,所以C中元素个数至少有1,2;至多为:0,1,2,3,4;

所以集合C的个数为子集的个数:.

4.定义,若,

,则

答案

解析

,;,;,,

,同理可求.

5.市场调查公司为了了解某小区居民在阅读报纸方面的取向,抽样调查了500户居民,调查的结果显示:订阅晨报的有334户,订阅晚报的有297户,其中两种都订的有150户,则两种都不订的有

户.

答案

19

解:绘制Venn图,由图可以知道,

,

因此,本题正确答案是:19.

6.已知集合A={x|},B={x|mx+1=0,m≠0}

(1)若m=1,求A∩B;

(2)若A∪B=A,求实数m的值组成的集合。

答案

(1)因为A={x|},所以A={-1,3}。当m=1,集合B满足x+1 =0,所以B={-1},所以A∩B={-1}。

(2)A={-1,3},A∪B=A,

∴B A,

①m=0时,B=,B A;

②m≠0时,由mx+1=0,得x=-,

∵B A,

∴-∈A,

∴-=-1或-=3,得m=1或,

∴满足题意的m的集合为{0,1,}.

7.已知集合A={x|2-a x2+a},B={x|x1或x4}。

(1)当a=3时,求A∩B;

(2)若A∩B=?,求实数a的取值范围。

答案

(1)当a=3时,A={x|-1x5} ,所以,A∩B={x|-1x1或4x5}

(2)当A∩B=?时,则有

①A≠?,有

解得0a<1;

②A=?,有2-a>2+a解得a<0综上所述,a的取值范围为(-∞,1)

8. 已知集合,,

(1)若B?A,求实数a的取值范围;(2)若,求实数a的取值范围.

解:(1)若B?A,则,计算得出;

(2),

(ⅰ) 当时,有;

(ⅱ)当时,有,

又,则有或,计算得出:或

或.综上可以知道:或.

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得113 a -<<- 三、解答题 1. 解:21616(2)0,21,m m m m ?=-+≥≥≤-或 222222min 1()21 2 11,()2 m m m αβαβαβαβ+=+-=--=-+= 当时 2. 解:(1)∵80 83,30x x x +≥?-≤≤? -≥? 得∴定义域为[]8,3- (2)∵222101011,110x x x x x x ?-≥? -≥=≠=-??-≠? 得且即∴定义域为{}1- (3)∵0 01110211 0101 1x x x x x x x x x x ?? ?? ?-≠?

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