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高一数学第7周周练

高一数学第7周周练
高一数学第7周周练

运河中学高一数学周练试卷(10.20)

一、填空题:(本大题共14小题,每小题5分,共70分) 1

= 。

2.满足{}{}{}a b a A c b a A =??,,,且的集合A 的个数有 。

3.已知函数f (x)|x |=,

在①y =

②2y =,③

2x y x =,④x ,x 0 ;

y x,x 0 .>?=?-

4.如图,函数()f x 的图象是折线段ABC ,其中

A B C ,,的坐标分别为(04)(20)(,,,,,,则

))1((f f = 。

5.函数1

()3x f x a -=+的图象一定过定点P ,则P 点的坐标是___________。

6.设函数()()()x x f x x m x R ππ-=+∈是偶函数,则实数m =__ ___。

7.函数

x x x x f +-=

)2()(的定义域为 。

8. 已知集合{}{}0,10,A x x a B x ax =

-==-=且A

B B =,则实数a =______。

9.已知三个数0.70.80.7

6,0.7,0.8a b c ===,则三个数的大小关系为

____ (用“<”连接)。 10. 已知函数

3

()(1)5

x

f x a a =--的图象经过第二、三、四象限,则

实数a 的取值范围是 。

11.设奇函数()f x 在(0)+∞,

上为增函数,且(1)0f =,则不等式

[]()()0x f x f x -

-

<的解集为 。

12.设定义在R 上的函数()f x 满足()()213f x f x ?+=,若()12f =,

()99f = 。

13. 函数)(x f y =为偶函数且在[)+∞,0上是减函数,则

)4(2

x f -的单调递

增区间为 。 14.已知函数

2()(,)f x x ax b a b R =++∈的值域为[)0,+∞,若关

于x 的不等式

()f x c <的解集为(,6)m m +,则实数c = 。

二、解答题:(本大题共6小题,共70分.请在答题卷指定区域内作答,解答

应写出文字说明,证明过程或演算步骤)。 15.计算:(本题满分14分)

(1)0

121

3

2

)23()12()01.0()833(-+--+---;

(2)若23x

a =,求33x x

x

x

a a a a --++的值。

16.(本题满分14分)已知函数()12x x

f x -=+

(22x -<≤)。

(1)用分段函数的形式表示该函数;

(2)画出该函数的图象;

(3)写出该函数的值域、单调区间。

17.(本小题满分14分)已知函数

11

()(0,1)

12

x

f x a a

a

=->≠

+。

(1)求函数()

f x的定义域;(2)判断函数()

f x的奇偶性。

18.(本小题满分16分)设a为实数,函数

2

()||1

f x x x a

=+-+,x R

∈.

(1)若

()

f x是偶函数,试求a的值;

(2)在(1)的条件下,求

()

f x的最小值;

(3)王小平同学认为:无论a取何实数,函数()

f x都不可能是奇函数.

你同意他的观点吗?请说明理由。

19.(本小题满分16分)某民营企业生产A 、B 两种产品,根据市场调查和预测,

A 产品的利润与投资成正比,其关系如图1所示;

B 产品的利润与投资的算术平方根成正比,其关系如图2所示(利润与投资单位:万元)。 (1)分别将A 、B 两种产品的利润表示为投资的函数关系式;

(2)该企业已筹集到10万元资金,并全部投入A 、B 两种产品的生产,问:怎样分配这10万元投资,才能使企业获得最大利润,其最大利润为多少万元?

20、(本小题满分16分)设函数

)0(1

)(≠++

=a b x ax x f 的图像过点(0,-1)

且与直线y = -1有且只有一个公共点;设点P(x 0 ,y 0)是函数y =()f x 图象上任意一点,过点P 分别作直线y =x 和直线x =1的垂线,垂足分别是M ,N 。 (1)求()y f x =的解析式;

(2)证明:曲线()y f x =的图像是一个中心对称图形,并求其对称中心Q ; (3)证明:线段PM ,PN 长度的乘积PM · PN 为定值;并用点P 横坐标x 0 表

示四边形QMPN 的面积。

x y 0 4 2.5 图2

x 1.8 0 y 0.45 图1

一、填空题: 1,

4- 2,2 3,1 4, 0 5,(1,4) 6,-1 7,[]2,0

8,1或-1或0 9,b <c <a 10,5(,2)3

11,(1

0)(01)-,, 12,13

2 13,(]2.-∞-,[]2,0 14,9 二、解答题:

15. 解:(1)原式=2

449

-……………7分

(2)原式=73

……………14分

16.解:(1)

1(20)

()1(02)x x f x x --<

≤≤? ……………4分

(2)如图……………8分

(3)值域为[1,3)…………11分

单调减区间为(2,0]-

14分

17,解:(1)要使函数

()f x 有意义,需10x

a +≠ ,则x R ∈,所以()f x

的定义域是R ;……………4分

(2)()f x 的定义域是R ……………2分

对任意的

x R ∈, 因为

()()f x f x -+=11111212

x

x a a --+-++ ……………………10分

1

111011

x x x

a a a +-=-=++,所以()()f x f x -=-,……………12分 所以函数()f x 是奇函数。………………………14分

18. 解:(1)∵()f x 是偶函数,∴()()f x f x -=在R 上恒成立, 即2

2

()||1||1x x a x x a -+--+=+-+,

化简整理,得 0ax =在R 上恒成立, ……………3分 ∴0a =. ……………5分 (另解 :由()f x 是偶函数知,(1)(1)f f -=

22(1)|1|11|1|1a a -+--+=+-+ 整理得|1||1|a a +=-,解得 0a =

再证明

2

()||1f x x x =++是偶函数,所以 0a = ) (2)由(Ⅰ)知0a =,∴

2

()||1f x x x =++, ∵2

0x ≥,||0x ≥,∴()1f x ≥,当且仅当0x =时,()1f x =,…………8分

∴当0x =时,()f x 的最小值为1. ……………10分 (3)王小平同学的观点是正确的. ……………11分 若()f x 是奇函数,则()()f x f x -=-在R 上恒成立,

∴(0)(0)f f =-,∴(0)0f =, ……………14分 但无论a 取何实数,(0)||10f a =+>,

∴()f x 不可能是奇函数. ……………16分 19.解:(1)设投资为x 万元,

A 、

B 两产品获得的利润分别为f (x )、g (x )万元,

由题意,

112(),(),0;0)f x k x g x k k k x ==≠≥ ……………'3

又由图知f (1.8)=0.45 ,g(4)=2.5;解得

1215

,44k k =

=

1()(0);()0)4f x x x g x x =

≥=≥ ……………'8

(不写定义域扣2分)

(2)设对B 产品投资x 万元,则对A 产品投资(10-x )万元, 记企业获取的利润为y 万元,

1(10)0)4y x x =

-≥ ……………10分

t =,则2

x t =

,(0t ≤≤

21565()4216y t =--+

52t =也即254x =时,y 取最大值6516 ……………14分 答:对B 产品投资254万元,对A 产品投资15

4万元时,

可获最大利润65

16万元。(答1分,单位1分) ……………16分

20.解:(1) 函数)

0(1

)(≠++=a b x ax x f 的图像过点(0,-1)

1)0(-=∴f 得b =-1 所以

11

)(++

=x ax x f ,……………2分

()f x 的图像与直线y= -1有且只有一个公共点

11

1++

=-x ax 只有一解 即 []0)1(=-+a ax x 只有一解 ∴a =1

1

()1f x x x =+

-……………4分 (2)证明:已知函数1y x =,21y x =

都是奇函数. 所以函数

1

()g x x x =+

也是奇函数,其图像是以原点为中心的中心对称图形. 而1

()11

1f x x x =-++-

可知,函数()g x 的图像向右、向上各平移1个单位,即得到函数()f x 的图像,故函数

()f x 的图像是以点Q (11),为中心的中心对称图形.………… 9分

(3)证明: P 点00011x x x ??

+ ?

-?

?, 过P 作PA ⊥x 轴交直线y=1于A 点,交直线y =x 于点B , 则QA =PN =AB =x 0 -1 ,QB =

)1(20-x .

PA =y P -1=

11100-+

-x x , ∴PB =PA -AB =11

0-x ,

∴PM =BM =

)1(21220-=x PB . ∴PM ?PN =)1(210-x .( x 0 -1)=22

为定值。 ……………13分

连QP ;∵QM =QB +BM =

)1(20-x +)1(21

0-x ,

?=?=

?21

21PM QM S QMP

[)1(20-x +)1(210-x ].)1(21

0-x =2

0)1(4121-+x

又2121=?=?PA NP S QNP ( x 0 -1). (

11100

-+-x x )= 21)1(2120+-x ∴=QMPN S 21)1(2120+-x +20)1(4121-+x =2

0)1(21-x +20)1(41-x +1 。16分

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