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内蒙古巴彦淖尔一中2015-2016学年高一上学期10月月考(普通班)数学试卷

巴市一中2015-2016学年第一学期10月月考试题

高 一 数 学 试卷类型 A

出题人: 邬红

说明: 1.本试卷分第I 卷和第II 卷两部分,共120分。

2.将第I 卷选择题答案代号用2B 铅笔填在答题卡上。

第I 卷(选择题 共60分)

一、选择题(4分×15=60分)在每小题给出的四个选项中只有一项正确

1.已知全集{1,3,5,7,9}U =,集合{5,7}A =,{

}

a a A C U ,,12

=,则a 的值为( ) A .3 B .3- C. ±3 D.9±

2. 以下四个关系: φ}0{∈,∈0φ,{φ}}0{?,φ}0{,其中正确的个数是( )

A .1

B .2 C. 3

D.4

3.下列各组函数中,表示同一函数的是( ) A .x

x y y =

=,1 B .1,112

-=+?-=x y x x y C .33

,x y x y == D .2)(|,|x y x y ==

4.在映射中B A f →:,},|),{(R y x y x B A ∈==,且),(),(:y x y x y x f +-→,则A 中的元素)2,1(-对应集合B 中的元素为( ) A. )3,1(-- B.)3,1( C. )1,3(

D. )1,3(-

5.若集合{}

,1≥=x x A 且B B A = ,则集合B 可能是( )

A.{

}2,1 B .{}

1≤x x C. {}1,0,1- D.R 6. 如右图所示,I 为全集,M 、P 、S 为I 的子集,

则阴影部分所表示的集合为( )

A .(M ∩P)∪S

B .(M ∩P)∩S C. (M ∩P)∩(

C I S) D. (M ∩P)∪(C I S)

7.函数)31(2

≤≤-+=x x x y 的值域是( )

A .

B .

C .

D .[3

4

,12] 8.()1-=x x f 的图象是( ).

9.若()

()44

3

3

2,3ππ-=

-=

b a ,则=+b a (

A .1

B .5

C .1-

D .52-π 10.已知54)1(2

-+=-x x x f ,则=+)1(x f ( )

A .x x 62+

B .782++x x

C .322-+x x

D .1062-+x x

11.已知函数y f x =+()1定义域是[]-23,,则函数y f x =-()

21的定义域是( ) A .[]05

2

B .[]-14,

C .[]-55,

D .[]-37, 12.已知集合{}21≤≤-=x x A ,集合{}

a x x B ≤=,且φ=?B A ,则实数a 的取值范围是( )

A .{}2>a a

B .{}1-

C .{}1-≤a a

D .{}

21<≤-a a

13.设函数2

,0

(),0

x x f x x x -≤?=?

>?,若()4f a =,则实数a =( ) A .-4或-2 B .-2或2 C .-2或4 D .-4或2

14.已知偶函数)(x f 在区间]0,(-∞单调递减,则满足)3

1

()12(f x f <-的实数x 的取值范围是( )

A .)32,21[

B .)32,31[

C .)32,21(

D .)3

2,31( 15.若函数的则实数定义域为实数集a R ax ax y ,322+-=

取值范围是( )

A .]3,0(

B .)3,0(

C . )3,0[

D .]3,0[

第II 卷(非选择题 共60分)

二、填空题(5分×4=20分)将最后结果直接填在横线上. 16.已知()2

2

50,x x a a

a x R -+=>∈,则x x a a -+= .

17.若函数2

()(2)(1)3f x k x k x =-+-+是偶函数,则)(x f 的递减区间是 .

18.函数y =

的定义域为 19.已知()83

5

-++=cx bx ax x f ,且f (-2) = 10,则f (2) =________. 三、解答题(8分+10分+12分+10分=40分)

20. (1)计算:210232113(2)()(3)(1.5)488

-----++

(2)已知集合{}{}

101,84≤<=<≤=x x B x x A ,求B A C R ?)(.

21.已知定义域为R 的奇函数()x f ,当0x > 时, ()32

-=x x f .

(1)求函数()x f 在R 上的解析式; (2)解方程()x x f 2=.

22.已知函数()2m f x x x =-且()742

f =, (1)求m 的值;

(2)判断()f x 在()0,+∞上的单调性,并用定义证明. (3)求()f x 在上的值域

23.函数()222

+-=x x x f 在闭区间[]()R t t t ∈+1,上的最小值记为)(t g

(1)求)(t g 的函数表达式;

(2)作)(t g 的图像,并写出)(t g 的最小值.

巴市一中2015-2016学年第一学期10月月考试题

高 一 数 学(答案) 一、选择题(4分×15=60分)

1.C

2.A

3.C

4.D

5.A

6.C

7.B

8.B

9.A 10.B 11.A 12.B 13.D 14.D 15.D 二、填空题(5分×4=20分)

16. 23 . 17.()+∞,0 18.?

??

???≠≤

0,31x x x 且 19.__-26__ 三、解答题(8分+10分+12分+10分=40分)

20. (1

)计算:210232113(2)()(3)(1.5)488

-----++

(2)已知集合{}{}

101,84≤<=<≤=x x B x x A ,求B A C R ?)(. 【解析】(1)原式

1

222

3

927311482-

-??????

=--++- ? ? ?????

?

?1

2

2

32

3

334

11229

-????????--++-???? ? ?????

????????

3441112992

=

--++-=- (2) {}

10841)(≤≤<<=?x x x B A C R 或

21.已知定义域为R 的奇函数()x f ,当0x > 时, ()32

-=x x f .

(1)求函数()x f 在R 上的解析式; (2)解方程()x x f 2=.

【解析】(1)设0x <,则0x ->,

2222()()33,

()()()()3()3;

f x x x f x f x f x f x x f x x -=--=-∴-=-∴-=-∴=-+ 是奇函数 223,0,

()0,0,3,0.x x f x x x x ?->?

==??-

(2) 当0x =时,方程()x x f 2=即20x =,解之得0x =;

当0x >时,方程()x x f 2=即232x x -=,解之得3x =(1x =-舍去); 当0x <时,方程()x x f 2=即232x x -=,解之得3x =-(1x =舍去). 综上所述,方程()x x f 2=的解为0x =,或3x =,或3x =-. 22.已知函数()2m f x x x =-且()7

42

f =, (1)求m 的值;

(2)判断()f x 在()0,+∞上的单调性,并用定义证明. (3)求()f x 在上的值域

【解析】(1)因为()7

42

f =

,所以,所以1m =.

(2)()f x 在()0,+∞上为单调增函数

证明:设120x x >>,则()()()12121212122221f x f x x x x x x x x x ????

-=---=-+ ? ?????

因为120x x >>,所以120x x ->,12

2

10x x +>,所以()()12f x f x >, (3)

23.函数()222

+-=x x x f 在闭区间[]()R t t t ∈+1,上的最小值记为)(t g

(1)求)(t g 的函数表达式;

(2)作)(t g 的图像,并写出)(t g 的最小值.

【解析】(1)??

?

??≥+-<<≤+=1,2210,10,1)(22t t t t t t t g (2) 1

27442m -

=

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