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福建省南安市某重点中学2012-2013学年高二下学期期末数学理试卷

福建省南安市某重点中学2012-2013学年高二下学期期末数学理试卷
福建省南安市某重点中学2012-2013学年高二下学期期末数学理试卷

南安市重点中学2012~2013学年度下学期期末考

高二(理)数学科试卷

本试卷考试内容为:坐标系与参数方程,集合,常用逻辑用语,函数、导数及其应用。分第I 卷(选择题)和第II 卷,共4页,满分150分,考试时间120分钟。

注意事项:

1.答题前,考生务必先将自己的姓名、准考证号填写在答题纸上。

2.考生作答时,请将答案答在答题纸上,在本试卷上答题无效。按照题号在各题的答题区域内作答,超出答题区域书写的答案无效。

3.答案使用0.5毫米的黑色中性(签字)笔或碳素笔书写,字体工整、笔迹清楚(英语科选择题答案使用2B 铅笔填涂,如需改动,用橡皮擦干净后,再选涂其它答案标号)。

4.保持答题纸纸面清洁,不破损。考试结束后,将本试卷自行保存,答题纸交回。

第I 卷(选择题 , 共60分)

一.选择题:本大题共12小题,每小题5,共60分。

1、设全集(){}{}

,1,02,-<=<+==x x B x x x A R U 则右图中阴影部分表示的集合为( )

A 、{}02<<-x x

B 、{}

12-<<-x x C 、{}

0>x x D 、{}

1-

2、下列函数中与y =

为同一函数的是( )

A 、x y =

B 、{

,(0)

,(0)

x x y x x >=

-< C 、||y x = D 、 log a x y a =

3、参数方程?

???

?

x =1+cos θy =-2+sin θ(θ为参数)所表示的图形是( )

A 、直线

B 、射线

C 、圆

D 、半圆

4、当10<

33

log 3<< B 、 x x x

33

log 3<< C 、3

33log x x x

<< D 、x

x x 3log 3

3<<

5、方程)(x f y =的图象如图所示,那么函数)2(x f y -=的图象是( )

6、下列有关命题的说法正确的是()

A、命题“若0

xy=,则0

x=”的否命题为:“若0

xy=,则0

x≠”

B、“若0

=

+y

x,则x,y互为相反数”的逆命题为真命题

C、命题“R

?x,使得2

210

x-<”的否定是:“R

?x,均有2

210

x-<”

D、命题“若cos cos

x y

=,则x y

=”的逆否命题为真命题

7、函数x

x

f x3

2

)

(+

=的零点所在的一个区间是( )

A、(-2,-1)

B、(-1,0)

C、(0,1)

D、(1,2)

8、若关于x的方程0

1

22=

-

-x

ax在区间(0,1)内恰有一解,则a的取值范围是()

A、(1,+∞)

B、(-∞,-1)

C、(-1,1)

D、(0,1)

9、)

(x

f是R上周期为3的奇函数,若1

)1(<

f, 1

)2(2-

+

=a

a

f,则a的取值范围是()

A、a<0.5且a≠1

B、-1

C、a<-1或a>0

D、-1

10、对于每一个实数x,)

(x

f是4

2+

-

=x

y和x

y3

=这两个函数中较小者,则)

(x

f的最大值是()

A、3

B、4

C、0

D、-4

11、

,(>1)

()=

(4-)+2,(1)

2

x

a x

f x a

x x

?

?

?

??

是R上的单调递增函数,则实数a的取值范围为()

A、(1,+∞)

B、[4,8]

C、[4,8)

D、(1,8)

12、已知函数)

(x

f

y=是定义在R上的减函数,函数)1

(-

=x

f

y的图象关于点)0,1(对称. 若对任意的R

y

x∈

,,不等式0

)1

2

(

)1

(2

2≤

-

+

-

+

-

+x

x

f

y

x

f恒成立,2

2y

x+的最小值是()

A、0

B、

5

5

C、

5

5

2

D、3

第II卷(非选择题,共90分)

二.填空题:本大题共4小题,每小题4,共16分。 13、已知函数()2log ,0

,3,0

x

x x f x x >?=?≤?则=??

?

???)41(f f ____ ____. 14

1

1

)x dx -=?

15、若曲线的极坐标方程为=2sin 4cos ,ρθθ+,以极点为原点,极轴为x 轴正半轴建立直角坐标系,则该曲线的直角坐标方程为 。

16、函数()f x 的定义域为A ,若1212,()()x x A f x f x ∈=且时总有12x x =,则称()f x 为单函数,例如:函数()21()f x x x R =+∈是单函数。 给出下列命题:

①函数2

()()f x x x R =∈是单函数; ②指数函数)(2)(R x x f x

∈=是单函数;

③若()f x 为单函数,121212,,()()x x A x x f x f x ∈≠≠且则; ④在定义域上具有单调性的函数一定是单函数。

其中的真命题是 。(写出所有的真命题的序号)

三.本大题共6小题,共74分。

17、(12分)在直角坐标系中,以原点为极点,x 轴的正半轴为极轴建坐标系,已知曲线

)0(cos sin :2>=a a C θθρ,已知过点)4,2(--P 的直线l 的参数方程为:???

???

?

+-=+-=t y t x 2

2422

2, 直线l 与曲线C 分别交于M,N 两点.

(Ⅰ)写出曲线C 和直线l 的普通方程; (Ⅱ)若2=a ,求线段|MN |的长度.

18、(12分)已知0>a ,设命题p :函数x

a y =在R 上单调递增;

命题q :不等式0222

>+-ax ax 对任意x ∈R 恒成立.若p 且q 为假,p 或q 为真,求a 的取值范围.

19、( 12分)已知函数??

?

??<-=>-=.0 ,21,0 ,2,0 ,4)(2x x x x x x f

(Ⅰ)求)]2([-f f 的值;

(Ⅱ)求)1(2

+a f (a R ∈)的值;

(Ⅲ)当34<≤-x 时,求函数)(x f 的值域. 20、(12分)某种商品每件进价9元,售价20元,每天可卖出69件.若售价降低,销售量可以

增加,且售价降低)110(≤≤x x 元时,每天多卖出的件数与2x x +成正比.已知商品售价降低3 元时,一天可多卖出36件.

(Ⅰ)试将该商品一天的销售利润表示成x 的函数;(Ⅱ)该商品售价为多少元时一天的销售利润最大?

21、(12分)已知函数()42

x x

n g x -=是奇函数,()()4log 41x

f x mx =++是偶函数. (Ⅰ)求m n +的值; (Ⅱ)设()()1

,2

h x f x x =

+若()()4log 21g x h a >+????对任意1x ≥恒成立,求实数a 的取值范围.

22、(14分)已知函数()ln 1a

f x x x

=

+-(a 是常数). (Ⅰ)若2x =是函数()f x 的极值点,求曲线()y f x =在点

()1,(1)f 处的切线方程; (Ⅱ)当1a =时,方程()f x m =在∈x 21,e e ??

????

上有两解,求实数m 的取值范围;

(Ⅲ)求证: 1

ln 1n n n

>-1(>n ,且)*N n ∈.

南安市重点中学2012~2013学年度下学期期末考—参考答案

高二(理)数学科试卷

一.选择题:(60分)

二.填空题:(16分) 13、

91 14、2

π 15、0242

2=--+y x y x 16、②③④ 三.本大题共6小题,共74分。

17、解:(Ⅰ) ax y =2

,2-=x y ……………..6分

(Ⅱ)直线的参数方程为???

?

??

?

+-=+-=t y t

x 2

2422

2(t 为参数),

代入x y 22=, 得到0402102

=+-t t , ………………8分 则有21021=+t t ,4021=t t . ………………10分 因为|MN|=|21t t -|,所以404)()(212212212

=-+=-=t t t t t t MN

.

解得|MN|=102 . ………………12分 18、解:由命题p ,得a >1,对于命题q ,

因x ∈R ,0222

>+-ax ax 恒成立,

又因a >0,所以Δ=2a 2

-8a <0,

即0

当p 真q 假时 ,?

????

a >1,

a ≤0或a ≥4.所以a ≥4 8分

当p 假q 真时,?

??

??

a ≤1,

0

综上可知,a 的取值范围为(0,1]∪[4,+∞) 12分

19、解:(Ⅰ) []2(2)5=45=21f

f f -=

--() (3分)

(Ⅱ)2

2

2

4

2

(1)4(1)23f a a a a +=-+=--+ (6分) (Ⅲ)①当40x -≤

<时,

∵()12f x x =- ∴1()9f x <≤

②当0x =时,(0)2f = ③当03x <

<时,∵2()4f x x =-

∴ 5()4f x -<<

故当43x -≤

<时,函数()f x 的值域是(]5,9-

20、解:(1)由题意可设,每天多卖出的件数为2

()k x x +,∴2

36(33)k =+,∴3k =

又每件商品的利润为)920(x --元,每天卖出的商品件数为)(3692

x x ++

∴该商品一天的销售利润为)110(75936303)](369)[11()(23

2

≤≤+-+-=++-=x x x x x x x x f

(2)由)6)(23(336609)(2

'

---=-+-=x x x x x f 令'()0f x =可得2

3

x =或6=x

当x 变化时,'()f x 、()f x 的变化情况如下表:

∴当商品售价为14元时,一天销售利润最大,最大值为975元

21、

()()min

3

12

g x g ==……………9分

由题意得到

3

224132210

a a a ?+

?

?-<

?+>??

……………11分

……………12分

22、(14分)已知函数()ln 1a

f x x x

=

+-(a 是常数). (Ⅰ)若2x =是函数()f x 的极值点,求曲线()y f x =在点()1,(1)f 处的切线方程; (Ⅱ)当1a =时,方程()f x m =在∈x 21,e e ??

????

上有两解,求实数m 的取值范围;

(Ⅲ)求证: 1

ln 1n n n

>-1(>n ,且)*N n ∈. 解:(Ⅰ) 2

()x a f x x

-'=.0)2('=f ,得2=a ,1)1(=f ,1)1('

-=f

切线方程为02=-+y x . ………….4分

(Ⅱ)当1=a 时,21

1()lg 1,()x f x x f x x

x -'=

+-=

,其中21,x e e ??

∈????

, 当??

?

???∈1,1e x 时,0)(<'x f ;(2

1,x e ?∈?时,0)(>'x f ,

∴1=x 是)(x f 在21,e e ??

????

上唯一的极小值点,∴ []0)1()(min ==f x f

2222111()2,()lg 11,f e f e e e e e =-=+-=+2211()210f f e e e e ??

-=---< ???

综上,所求实数m 的取值范围为{|02}m m e <≤-. …………8分 (Ⅲ)若1=a 时,由(2)知x x

x

x f ln 1)(+-=在[)+∞,1上为增函数, 当1>n 时,令1-=

n n

x ,则1>x ,故0)1()(=>f x f , 即01ln 11ln 1

111>-+-=-+---

=??

? ??-n n n n n n n n n n n f ,∴1ln 1n n n >-.…….14分

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3.用准确简明的语句压缩下面这个语段。(不超过40个字)(3分) 眼前就物候来说,不但已不是春天,而且也已经不是夏天:眼前是 西风劲吹、落叶辞树的深秋天气。“悲哉秋之为气也”,眼前是古代诗人高呼“悲哉”的时候。然而在这秋之声大合唱中,在我们燕园图书馆的 草坪上,在黄叶丛中,在红树枝下,我看到的却是阳春艳景,姹紫嫣 红,这些男女大孩子一下子变成了巨大的花朵,一霎时,开满了校园。连黄叶树顶上也似乎开出了碗口大的山茶花和木棉花,红红的一片, 把碧空都映得通红。至于那些“红于二月花”的霜叶,真的变成了红艳 艳的鲜花。整个燕园变成了一座花山,一片花海。 4.甲(选物理考生做)下面是校长在一次高三学生大会上讲话的一部 分,假如你是听众之一,请根据校长讲话的内容回答后面的问题。(6分) 离高考的日子越来越近了,老师们的辅导也越来越勤了,同学们 的学习也越来越紧了,大家都在做最后的冲刺。据反映大多数同学每 晚伏案读书做练习要到12点钟,这种精神是可贵的。但也有老师反映相当一部分同学课堂上精神不振,昏昏欲睡,课堂学习效率大打折扣。 校长肯定的是: 校长否定的是: 校长的目的是: 乙(选历史考生做)“空城计”是《三国演义》中一个有名的故事,请 写出该故事中蜀魏双方的主帅并概述主要情节,不超过100个字。(6分)

高二英语下学期期末考试试题(含答案)

第二学期期末考试 高二级 英语试卷 第一部分 阅读理解(共两节,满分40分) 第一节:(共15小题;每小题2分,满分30分) 阅读下列短文,从每题所给的四个选项(A 、B 、C 和D )中,选出最佳选项,并在答题卡上将该项 涂黑。 A Shakespeare’s Birthplace and Exhibition of Shakespeare’s World Welcome to the world-famous house where William Shakespeare was born in l564 and where he grew up. The property (房产) remained in the ownership of Shakespeare’s family until 1806.The House has welcomed visitors traveling from all over the world for over 250 years. ◆Enter through the Visitors’ Centre and see the highly-praised exhibition Shakespeare’s World, a lively and full introduction to the life and work of Shakespeare. ◆Stand in the rooms where Shakespeare grew up. ◆Discover examples of furniture and needlework from Shakespeare’s period. ◆Enjoy the traditional (传统的) English garden, planted with trees and flower mentioned in the poet’s works. ◆The Birthplace is within easy walking distance of all the car parks shown on the map; nearest is Windsor Street(3 minutes’ Walk) ◆The Ho use may present difficulties but the Visitors’ Centre, its exhibition, and the garden are accessible to wheelchair users. ◆The Shakespeare Coffee House(opposite the Birthplace) 1. How much is the admission for a family of two gr own-ups and two children? A.£9.80. B.£12.00. C.£14.20. D.£16.40. 2. Where is the nearest parking place to Shakespeare’s Birthplace? A. Behind the exhibition hall. B. Opposite the Visito rs’ Centre. C. At Windsor Street. D. Near the Coffee House. 3. A wheelchair user may need help to enter . A. the House B. the garden C. the Visitors’ Cent re D. the exhibition hall B OPENING TIMES: 20 Mar. to 19 Oct. Mon. to Sat. : 9:00 a. m. to 5:00 p. m. Sun: 9:30 a. m. to 5:00 p. m. 20 Oct. to 19 Mar Mon. to Sat : 9:00 a. m. to 4:00 p. m. Sun. : 10:00 a. m. to 4:00 p.m. ADMISSION: Adult: £4.90 Child: £2.20 Family: £12.00 (2 adults+ up to 3 children)

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