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2014石景山高三数学一模理科

2014石景山高三数学一模理科
2014石景山高三数学一模理科

2014年石景山区高三统一测试

数学(理科)

本试卷共6页,满分为150分,考试时间为120分钟.请务必将答案答在答题卡上,在试卷上作答无效,考试结束后上交答题卡.

第Ⅰ卷(选择题 共40分)

一、选择题共8小题,每小题5分,共40分.在每小题给出的四个选项中,选出符合题

目要求的一项.

1.已知全集U =R ,集合{}

2

|20A x x x =-<,{}|10B x x =-≥,那么U A B = e( )

A .{}|01x x <<

B .{}|0x x <

C .{}|2x x >

D .{}|12x x <<

2.下列函数中,在(0)+∞,

内单调递减,并且是偶函数的是( ) A .2y x = B .1y x =+ C .lg ||y x =- D .2x y =

3.在25

1()x x

-的展开式中,x 的系数为( )

A .10

B .10-

C .20

D .20-

4.已知Rt △ABC 中,o 9054C AB BC ∠===,,,

以BC 为直径的圆交AB 于D ,则BD 的长为( ) A .4 B .95 C .125 D .16

5

A

C

B

5. 在平面直角坐标系xOy 中,抛物线22(0)x py p =>上纵坐标为1的点到焦点的距离

为3,则焦点到准线的距离为( ) A .2 B .8 C

D .4 6.右图是某个三棱锥的三视图,其中主视图是

等边三角形,左视图是直角三角形,俯视图是等腰直角三角形,则该三棱锥的体积是( )

A

B

3

C

D

7.阅读右面的程序框图,运行相应的程序, 输出的结果为( ) A .2- B .

12

C .1-

D .2

8.已知动点()P x y ,在椭圆22

:12516x y C +=上,F 为椭圆C 的右焦点,若点M 满足

||1MF = 且0MP MF ?=

,则||PM 的最小值为( )

主视图

左视图

俯视图

A

B .3

C .

12

5

D .1 第Ⅱ卷(非选择题 共110分)

二、填空题共6小题,每小题5分,共30分.

9.已知命题p :0x x e ?∈

10.在等比数列}{n a 中,14=2=16a a ,

,则数列}{

n a 的通项公式=n a _____________,设2log n n b a =,则数列}{

n b 的前n 项和=n S _____________.

11.已知圆C 的极坐标方程为=2ρ,以极点为原点,极轴为x 轴的正半轴建立平面直角

坐标系,则圆C 的直角坐标方程为_______________,若直线:30l kx y ++=与圆C 相切,则实数k 的值为_____________.

12.已知变量x y ,满足约束条件20170x y x x y -+≤??≥??+-≤?

,则x y 的取值范围是_________. 13.各大学在高考录取时采取专业志愿优先的录取原则.一考生从某大学所给的7个专

业中,选择3个作为自己的第一、二、三专业志愿,其中甲、乙两个专业不能同时兼报,则该考生有_____________种不同的填报专业志愿的方法(用数字作答). 14.若存在实常数k 和b ,使得函数()f x 和()g x 对其定义域上的任意实数x 分别满足:

()f x kx b ≥+和()g x kx b ≤+,则称直线:l y kx b =+为()f x 和()g x 的“隔离直

线”.已知函数2()1f x x =-和函数()2ln g x x =,那么函数()f x 和函数()g x 的隔离直线方程为_________.

三、解答题共6小题,共80分.解答应写出文字说明,演算步骤或证明过程. 15.(本小题满分13分)

在△ABC 中,角A B C ,,的对边分别为a b c ,,,且a b c <<

2sin b A =.

(Ⅰ)求角B 的大小; (Ⅱ)若2a =

,b =

c 边的长和△ABC 的面积.

16.(本小题满分13分)

经调查发现,人们长期食用含高浓度甲基汞的鱼类会引起汞中毒,其中罗非鱼体内汞含量比其它鱼偏高.现从一批数量很大的罗非鱼中随机地抽出15条作样本,经检测得各条鱼的汞含量的茎叶图(以小数点前的数字为茎,小数点后一位数字为叶)如下:

《中华人民共和国环境保护法》规定食品的汞含量不得超过1.0ppm .

(Ⅰ)检查人员从这15条鱼中,随机抽出3条,求3条中恰有1条汞含量超标的概率; (Ⅱ)若从这批数量很大的鱼........中任选3条鱼,记ξ表示抽到的汞含量超标的鱼的条数.以此15条鱼的样本数据来估计...这批数量很大的鱼的总体数据,求ξ的分布列及数学期望E ξ.

1235567889 1

35567

罗非鱼的汞含量(ppm )

17.(本小题满分14分)

如图,正三棱柱111ABC A B C -的底面边长是2

D 是AC 的中点. (Ⅰ)求证:1B C ∥平面1A BD ; (Ⅱ)求二面角1A BD A --的大小;

(Ⅲ)在线段1AA 上是否存在一点E ,使得平面11B C E ⊥平面1A BD ,若存在, 求出AE 的长;若不存在,说明理由.

18.(本小题满分13分)

设函数2()ln ()f x x ax x a =+-∈R . (Ⅰ)若1a =,求函数()f x 的单调区间;

(Ⅱ)若函数()f x 在区间(01],

上是减函数,求实数a 的取值范围; (Ⅲ)过坐标原点O 作曲线)(x f y =的切线,证明:切点的横坐标为1.

A

1A

1B

1C

C

D

B

19.(本小题满分14分)

给定椭圆C :22

221(0)x y a b a b

+=>>,称圆心在原点O

是椭圆C 的“准圆”.若椭圆C

的一个焦点为0)F ,,

其短轴上的一个端点到F 的

(Ⅰ)求椭圆C 的方程和其“准圆”方程;

(Ⅱ)点P 是椭圆C 的“准圆”上的动点,过点P 作椭圆的切线12l l ,

交“准圆”于点M N ,.

(ⅰ)当点P 为“准圆”与y

求直线12l l ,

的方程并证明12l l ⊥; (ⅱ)求证:线段MN 的长为定值.

20.(本小题满分13分)

对于数列{}n a ,把1a 作为新数列{}n b 的第一项,把i a 或i a -(234i n = ,,,

,)作为新数列{}n b 的第i 项,数列{}n b 称为数列{}n a 的一个生成数列.例如,数列

12345,,,,的一个生成数列是12345--,,,,.

已知数列{}n b 为数列1

{

}()2

n n *∈N 的生成数列,n S 为数列{}n b 的前n 项和. (Ⅰ)写出3S 的所有可能值; (Ⅱ)若生成数列{}n b 满足311

(1)78

n n S =

-,求数列{}n b 的通项公式; (Ⅲ)证明:对于给定的n *∈N ,n S 的所有可能值组成的集合为

121

{|2}2

n n k x x k k *--=

∈≤N ,,.

2014年石景山区高三统一测试高三数学(理科)

参考答案

一、选择题:本大题共8个小题,每小题5分,共40分.

二、填空题:本大题共6个小题,每小题5分,共30分.两空的题目,第一空2分,

第二空3分. 三、解答题:本大题共6个小题,共80分.应写出文字说明,证明过程或演算步骤. 15.(本小题满分13分) 解:2sin b A =,

2sin sin A

B A =, ……………2分

因为0A π<<,所以sin 0A ≠, 所以sin B =

, …………… 4分 因为0B π<<,且a b c <<,所以60B =

. ……………6分 (Ⅱ)因为2a =,b =

所以由余弦定理得222

12222

c c =+-???

,即2

230c c --=, 解得3c =或1c =-(舍),

所以c 边的长为3. ……………10分

11=sin 2322ABC S ac B ?=??=

. ……………13分 16.(本小题满分13分)

解:(Ⅰ)记“15条鱼中任选3条恰好有1条鱼汞含量超标”为事件A ,则

1251031545()91

C C P A C ==,

∴15条鱼中任选3条恰好有1条鱼汞含量超标的概率为

45

91

. ……………4分 (Ⅱ)依题意可知,这批罗非鱼中汞含量超标的鱼的概率51

()153

P B =

=, ………5分 ξ可能取0,1,2,3. ……………6分 则3

0318(0)1327P C ξ??==-= ??? ,2

13114(1)1339P C ξ??==??-= ???,

223

112(2)1339P C ξ??

??==?-= ?

?

??

??,3

3311(3)327

P C ξ??=== ???.…………10分 其分布列如下:

……………12分

所以842101231279927

E ξ=?

+?+?+?=. ……………13分 17.(本小题满分14分)

(Ⅰ)证明:连结1AB 交1A B 于M ,连结1B C DM ,

, 因为三棱柱111ABC A B C -是正三棱柱, 所以四边形11AA B B 是矩形, 所以M 为1A B 的中点. 因为D 是AC 的中点,

M

A

1A

1B

1C

B

C

D

所以MD 是三角形1AB C 的中位线, ……………2分 所以MD ∥1B C . ……………3分

因为MD ?平面1A BD ,1B C ?平面1A BD ,

所以1B C ∥平面1A BD . ……………4分 (Ⅱ)解:作CO AB ⊥于O ,所以CO ⊥平面11ABB A ,

所以在正三棱柱111ABC A B C -中如图建立空间直角坐标系O xyz -.

因为2AB =

,1AA =D 是

AC 的中点. 所以(100)A ,,,(100)B -,,

,(00C

,1(10)A , ……………5分 所以1(022D ,,,3(022

BD = ,,,

1(20)BA =

设()n x y z =

,,是平面1A BD 的法向量, 所以100n BD n BA ??=???=?? ,,即302

20x z x ?=???+=?

,,

令x =2y =,3z =,

所以(23)n =

是平面1A BD 的一个法向量.

……………6分 由题意可知1(00)AA =

是平面ABD 的一个法向量, ……………7分 所以11

cos 2

n AA <>== ,.

……………8分 所以二面角1A BD A --的大小为

3

π

. ……………9分 x

(Ⅲ)设(10)E x ,,

,则1(1C E x =-

,11(10C B

,=-

设平面11B C E 的法向量1111()n x y z ,,=, 所以111100n C E n C B ,,??=???=??

即11111)00x x y x ,

?-++=??

-=??

令1z =13x =

,1y =,

1(3n =, ……………12分

又10n n ?=

,即0-=

,解得x =, 所以存在点E ,使得平面11B C E ⊥平面1A BD

且AE =

. ……………14分 18.(本小题满分13分)

解: (Ⅰ)1a =时, 2()ln (0)f x x ax x

x =+->,

1(21)(1)

()21x x f x x x x

-+'∴=+-= , ……………1分

11(0)()0()()022

x f x x f x ''∈<∈+∞>,,,,,,

()f x 的减区间为1(0)2,,增区间1

()2+∞,. ……………3分

(Ⅱ)1

()2f x x a x

'=+-

()f x 在区间(01],上是减函数, ()0f x '∴≤对任意(01]x ∈,恒成立,

即1

20x a x

+-

≤对任意(01]x ∈,

恒成立, ……………5分 1

2a x x ∴≤

-对任意(01]x ∈,

恒成立, 令1

()2g x x x

=-,

min ()a g x ∴≤, ……………7分

易知()g x 在(01],

单调递减,min ()(1)1g x g ∴==-. 1a ∴≤-. ……………8分

(Ⅲ)设切点为(())M t f t ,

,1

()2f x x a x

'=+-, 切线的斜率12k t a t

=+-,又切线过原点()

f t k t

=

, ()2221

2ln 211ln 0f t t a t at t t at t t t t

=+-+-=+-∴-+=,即:, 存在性:1t =满足方程21ln 0t t -+=,

所以,1t =是方程21ln 0t t -+=的根. ……………11分 再证唯一性:设()2

1ln t t t ?=-+,()1

'20t t t

?=+>,

()t ?在(0,)+∞单调递增,且()1=0?,

所以方程21ln 0t t -+=有唯一解.

综上,切点的横坐标为1. ……………13分

19.(本小题满分14分) 解:

(Ⅰ)1c a b ==∴= ,

∴椭圆方程为2

213

x y +=, ………………2分

准圆方程为224x y +=. ………………3分

(Ⅱ)(ⅰ)因为准圆224x y +=与y 轴正半轴的交点为(02)P ,

, 设过点(02)P ,

且与椭圆相切的直线为2y kx =+, 所以由22

213

y kx x y =+???+=??,

,得22(13)1290k x kx +++=.

因为直线2y kx =+与椭圆相切,

所以2214449(13)0k k ?=-?+=,解得1k =±, ………………6分

所以12l l ,

方程为22y x y x =+=-+,. ………………7分 121l l k k ?=- ,12l l ∴⊥. ………………8分

(ⅱ)①当直线12l l ,

中有一条斜率不存在时,不妨设直线1l 斜率不存在, 则1l

:x = 当1l

:x =

1l

与准圆交于点1)1)-, 此时2l 为1y =(或1y =-),显然直线12l l ,

垂直; 同理可证当1l

:x =12l l ,

垂直. ………………10分 ②当12l l ,

斜率存在时,设点00()P x y ,,其中22

004x y +=. 设经过点00()P x y ,

与椭圆相切的直线为00()y t x x y =-+, 所以由002

2

()13

y t x x y x y =-+??

?+=??,

, 得 222

0000(13)6()3()30t x t y tx x y tx ++-+--=.

由0?=化简整理得 222

0000(3)210x t x y t y -++-=, 因为2

2

004x y +=,所以有2

2

2

0000(3)2(3)0x t x y t x -++-=.

设12l l ,

的斜率分别为12t t ,,因为12l l ,与椭圆相切, 所以12t t ,

满足上述方程2

2

2

0000(3)2(3)0x t x y t x -++-=, 所以121t t ?=-,即12l l ,

垂直. ………………12分

综合①②知:因为12l l ,

经过点00(,)P x y ,又分别交其准圆于点M N ,,且12l l , 垂直.

所以线段MN 为准圆224x y +=的直径, ||4MN =,

所以线段MN 的长为定值. ………………14分

20.(本小题满分13分) 解:(Ⅰ)由已知,112b =

,1||(,2)2n n b n n *

=∈≥N , ∴231148

b b =±=±,, 由于1117111511131111,2488248824882488

++=+-=-+=--=,,,

∴3S 可能值为1357

8888,,,. ……………3分

(Ⅱ)∵311

(1)78

n n S =-,

当1n =时,1233111

(1)788

a a a S ++==-=,

当2n ≥时,32313333111111

(1)(1)78788

n n n n n n n n a a a S S ----++=-=---=,

323131

8

n n n n a a a --∴++=,*n ∈N , ……………5分

∵{}n b 是1()2n n *

??∈????

N 的生成数列,

∴323212n n b --=±

;31311

2n n b --=±

;3312

n n

b =±

; ∴323133231311111(421)()22288

n n n n n n n n b b b n *

----++=±±±=±±±=∈N ,

在以上各种组合中, 当且仅当32313421()888

n n n n n n b b b n *

--=

=-=-∈N ,,时,才成立. ∴132213 2.2n

n n

n k b k n k *?=-??=∈??-≠-??N ,,(),. ……………8分

(Ⅲ)231111

2222

n n S =

±±±± 共有12n -种情形. 23231111111122222222n n n S ----≤≤++++ ,即121

22

n n n n S -≤≤, 又12322212

n n n n n

S ---±±±±= ,分子必是奇数, 满足条件121

222

n n n n

x -≤≤的奇数x 共有12n -个. ……………10分 设数列{}n a 与数列{}n b 为两个生成数列,数列{}n a 的前n 项和为n S ,数列{}n b 的前n 项和为n T ,从第二项开始比较两个数列,设第一个不相等的项为第k 项.

由于1

||||2

k k k a b ==

,不妨设00k

k a b ><,, 则11()()n n k k n k k n S T a a a b b b ++-=+++-+++

12111122()2222k k k n

++≤?

-?+++

1

111122()02222k k n n -=?

-?-=>, 所以,只有当数列{}n a 与数列{}n b 的前n 项完全相同时,才有n n S T =.…12分 ∴231111

2222

n n S =

±±±± 共有12n -种情形,其值各不相同. ∴n S 可能值必恰为13521

2222

n n n n n - ,,,,,共12n -个.

即n S 所有可能值集合为1

21{|2}2

n n k x x k k *--=

∈≤N ,,. ……………13分 【注:若有其它解法,请酌情给分】

丰台区2017年度高三一模英语试题

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