人教版必修一课时跟踪检测(一) 集 合

2016高考理科数学一轮复习 课时跟踪检测(一) 集 合

一、选择题

1．(2015·广州测试)已知集合A ＝???

x ???

??x ∈Z ，且32－x ∈Z ，则集合A 中的元素个数为( )

A ．2

B ．3

C ．4

D ．5 2．(2014·江西高考)设全集为R ，集合A ＝{x |x 2－9＜0}，B ＝{x |－1＜x ≤5}，则A ∩(?R B )＝( )

A ．(－3,0)

B ．(－3，－1)

C ．(－3，－1]

D ．(－3,3)

3．已知集合A ＝{x |y ＝1－x 2}，B ＝{x |x ＝m 2，m ∈A }，则( )

A ．A B

B ．B A

C ．A ?B

D ．B ?A

4．设函数f (x )＝lg(1－x 2)，集合A ＝{x |y ＝f (x )}，B ＝{y |y ＝f (x )}，则

图中阴影部分表示的集合为( )

A ．[－1,0]

B ．(－1,0)

C ．(－∞，－1)∪[0,1)

D ．(－∞，－1]∪(0,1)

5．(2015·西安一模)设集合A ＝{(x ，y )|x ＋y ＝1}，B ＝{(x ，y )|x －y ＝3}，则满足M ?(A ∩B )的集合M 的个数是( )

A ．0

B ．1

C ．2

D ．3

6．在整数集Z 中，被5除所得余数为k 的所有整数组成一个“类”，记为[k ]，即[k ]＝{5n ＋k |n ∈Z }，k ＝0,1,2,3,4.给出如下四个结论：

①2 014∈[4]；②－3∈[3]；③Z ＝[0]∪[1]∪[2]∪[3]∪[4]；④“整数a ，b 属于同一‘类’”的充要条件是“a －b ∈[0]”．

其中，正确结论的个数是( )

A ．1

B ．2

C ．3

D ．4

二、填空题

7．已知A ＝{0，m,2}，B ＝{x |x 3－4x ＝0}，若A ＝B ，则m ＝________.

8．(2014·重庆高考)设全集U ＝{n ∈N |1≤n ≤10}，A ＝{1,2,3,5,8}，B ＝{1,3,5,7,9}，则(?U A )∩B ＝________.

9．(2015·昆明二模)若集合A ＝{x |x 2－9x ＜0，x ∈N *}，B ＝??????y ??

4y ∈N *，y ∈N *，则A ∩B 中元素的个数为________．

10．(2015·南充调研)已知集合A ＝{x |4≤2x ≤16}，B ＝[a ，b ]，若A ?B ，则实数a －b 的取值范围是________．

三、解答题

11．已知集合A ＝{－4,2a －1，a 2}，B ＝{a －5,1－a,9}，分别求适合下列条件的a 的值．

(1)9∈(A ∩B )；

(2){9}＝A ∩B .

12．(2015·福州月考)已知集合A ＝{x |1＜x ＜3}，集合B ＝{x |2m ＜x ＜1－m }．

(1)当m ＝－1时，求A ∪B ；

(2)若A ?B ，求实数m 的取值范围；

(3)若A ∩B ＝?，求实数m 的取值范围．

答案

1．选C ∵32－x

∈Z ，∴2－x 的取值有－3，－1,1,3，又∵x ∈Z ，∴x 值分别为5,3,1，－1，

故集合A 中的元素个数为4，故选C.

2．选C 由题意知，A ＝{x |x 2－9＜0}＝{x |－3＜x ＜3}，∵B ＝{x |－1＜x ≤5}，∴?R B ＝{x |x ≤－1或x ＞5}．

∴A ∩(?R B )＝{x |－3＜x ＜3}∩{x |x ≤－1或x ＞5}＝{x |－3＜x ≤－1}．

3．选B 由题意知A ＝{x |y ＝1－x 2}，∴A ＝{x |－1≤x ≤1}，∴B ＝{x |x ＝m 2，m ∈A }＝{x |0≤x ≤1}，∴B A ，故选B.

4．选D 因为A ＝{x |y ＝f (x )}＝{x |1－x 2＞0}＝{x |－1＜x ＜1}，则u ＝1－x 2∈(0,1]， 所以B ＝{y |y ＝f (x )}＝{y |y ≤0}，

A ∪

B ＝(－∞，1)，A ∩B ＝(－1,0]，

故图中阴影部分表示的集合为(－∞，－1]∪(0,1)，选D.

5．选C 由题中集合可知，集合A 表示直线x ＋y ＝1上的点，集合B 表示直线x －y ＝3上的点，联立????

? x ＋y ＝1，x －y ＝3可得A ∩B ＝{(2，－1)}，M 为A ∩B 的子集，可知M 可能为{(2，

－1)}，?，所以满足M ?(A ∩B )的集合M 的个数是2，故选C.

6．选C 因为2 014＝402×5＋4，又因为[4]＝{5n ＋4|n ∈Z }，所以2 014∈[4]，故①正确；因为－3＝5×(－1)＋2，所以－3∈[2]，故②不正确；因为所有的整数Z 除以5可得的余数为0,1,2,3,4，所以③正确；若a ，b 属于同一‘类’，则有a ＝5n 1＋k ，b ＝5n 2＋k ，所以a －b ＝5(n 1－n 2)∈[0]，反过来，如果a －b ∈[0]，也可以得到a ，b 属于同一“类”，故④正确．故有3个结论正确．

7．解析：由题知B ＝{0，－2,2}，A ＝{0，m,2}，若A ＝B ，则m ＝－2.

答案：－2

8．解析：由题意，得U ＝{1,2,3,4,5,6,7,8,9,10}，故?U A ＝{4,6,7,9,10}，所以(?U A )∩B ＝{7,9}．

答案：{7,9}

9．解析：解不等式x 2－9x ＜0可得0＜x ＜9，所以A ＝{x |0＜x ＜9，x ∈N *}＝

{1,2,3,4,5,6,7,8}，又4y

∈N *，y ∈N * ，所以y 可以为1,2,4，所以B ＝{1,2,4}，所以A ∩B ＝B ，A ∩B 中元素的个数为3.

答案：3

10．解析：集合A ＝{x |4≤2x ≤16}＝{x |22≤2x ≤24}＝{x |2≤x ≤4}＝[2,4]，因为A ?B ，所以a ≤2，b ≥4，所以a －b ≤2－4＝－2，即实数a －b 的取值范围是(－∞，－2]．

答案：(－∞，－2]

11．解：(1)∵9∈(A ∩B )，∴2a －1＝9或a 2＝9， ∴a ＝5或a ＝3或a ＝－3.

当a ＝5时，A ＝{－4,9,25}，B ＝{0，－4,9}；

当a ＝3时，a －5＝1－a ＝－2，不满足集合元素的互异性； 当a ＝－3时，A ＝{－4，－7,9}，B ＝{－8,4,9}， 所以a ＝5或a ＝－3.

(2)由(1)可知，当a ＝5时，A ∩B ＝{－4,9}，不合题意， 当a ＝－3时，A ∩B ＝{9}．

所以a ＝－3.

12．解：(1)当m ＝－1时，B ＝{x |－2

则A ∪B ＝{x |－2

(2)由A ?B 知????? 1－m ＞2m ，2m ≤1，

1－m ≥3，解得m ≤－2，

即实数m 的取值范围为(－∞，－2]．

(3)由A ∩B ＝?，得

①若2m ≥1－m ，即m ≥13

时，B ＝?，符合题意； ②若2m ＜1－m ，即m ＜13时，需????? m ＜13，1－m ≤1或????? m ＜13，2m ≥3，

得0≤m ＜13或?，即0≤m ＜13

. 综上知m ≥0，即实数m 的取值范围为[0，＋∞)．

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