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东城区2011-2012二模试题理定稿 (修复的)

东城区2011-2012二模试题理定稿 (修复的)
东城区2011-2012二模试题理定稿 (修复的)

北京市东城区2011-2012学年度第二学期高三综合练习(二)

数学 (理科)

本试卷分第Ⅰ卷和第Ⅱ卷两部分,第Ⅰ卷1至2页,第Ⅱ卷3至5页,共150分。考试时长120分钟。考生务必将答案答在答题卡上,在试卷上作答无效。考试结束后,将本试卷和答题卡一并交回。

第Ⅰ卷(选择题 共40分)

一、本大题共8小题,每小题5分,共40分。在每小题列出的四个选项中,选出符合题目要求的一项。 (1)下列命题中,真命题是

(A )x ?∈R ,210x --< (B )0x ?∈R ,2001x x +=- (C )21,04

x x x ?∈-+>R (D )2

000,220x x x ?∈++

(2)将容量为n 的样本中的数据分成6组,若第一组至第六组数据的频率之比为2:3:4:6:4:1,且前三

组数据的频数之和等于27,则n 的值为

(A )70 (B )60 (C )50 (D )40

(3)4

1(2)x x

-

的展开式中的常数项为

(A )24- (B )6- (C )6 (D )24

(4)若一个三棱柱的底面是正三角形,其正(主)视图如图所示,则它的体积为

(A ) (B )2

(C )

(D )4

(5)若向量a ,b 满足1=a

,=

b ()⊥a a +b ,则a 与b 的夹角为

(A )2

π (B )

23

π (C )

34

π (D )

56

π

(6)已知m 和n 是两条不同的直线,α和β是两个不重合的平面,那么下面给出的条件中一定能推出

m ⊥β 的是

(A )⊥αβ,且m ?α (B )m ∥n ,且n ⊥β (C )⊥αβ,且m ∥α (D )m ⊥n ,且n ∥β

(7)若m 是2和8的等比中项,则圆锥曲线2

2

1y

x m

+

=的离心率为

(A

)2

(B

(C

)2

或2

(D

2

(8)定义:()00>>=y ,x y )y ,x (F x ,已知数列{}n a 满足:()()

n ,F ,n F a n 22=

()n *

∈N ,若对任意正整数

n ,都有k n a a ≥()k *

∈N 成立,则k a 的值为

(A )

12

(B )2 (C )

89

(D )

98

第Ⅱ卷(共110分)

二、填空题:本大题共6小题,每小题5分,共30分。

(11)在平面直角坐标系xOy 中,将点A 绕原点O 逆时针旋转 90到点B ,那么点B 的坐标为____, 若直线O B 的倾斜角为α,则sin 2α的值为 .

(12) 如图,直线P C 与 O 相切于点C ,割线P A B 经过圆心O , 弦C D ⊥A B 于点E ,4P C =,8P B =,则C E = .

(13) 已知函数sin 1

()1

x x f x x -+=

+()x ∈R 的最大值为M ,最小值为m ,则M m +的值为__.

(14) 已知点(,)A a b 与点(1,0)B 在直线34100x y -+=的两侧,给出下列说法: ①34

100a b -+>;

②当0a >时,a b +有最小值,无最大值; 2>;

④当0a >且1a ≠,0b >时,

1

b a -的取值范围为53(,)(

,)2

4

-∞-

+∞ .

其中,所有正确说法的序号是 .

三、解答题:本大题共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)(本小题共13分)

已知函数()sin()f x A x =+ω?(其中∈R x ,0A >,ππ0,2

2

ω?>-<<

)的部分

图象如图所示.

(Ⅰ)求函数()f x 的解析式;

(Ⅱ)已知在函数()f x 的图象上的三点,,M N P 的横坐标分别为1,1,5-,求sin M N P ∠的值.

(16)(本小题共13分)

某公园设有自行车租车点, 租车的收费标准是每小时2元(不足1小时的部分按1小时计算).甲、 乙两人各租一辆自行车,若甲、乙不超过一小时还车的概率分别为21

41,;一小时以上且不超过两小时还

车的概率分别为4

1

21,;两人租车时间都不会超过三小时.

(Ⅰ)求甲、乙两人所付租车费用相同的概率;

(Ⅱ)设甲、乙两人所付的租车费用之和为随机变量ξ,求ξ的分布列与数学期望ξE .

(17)(本小题共13分)

如图,矩形A M N D 所在的平面与直角梯形M B C N 所在的平面互相垂直,M B ∥N C ,M N M B ⊥, 且M C C B ⊥,2B C =,4M B =,3D N =. (Ⅰ)求证://A B 平面D N C ;

(Ⅱ)求二面角D B C N --的余弦值.

(18)(本小题共14分)

已知抛物线C :2

4x y =,M 为直线:l 1y =-上任意一点,过点M 作抛物线C 的两条切线

,M A M B ,切点分别为A ,B .

(Ⅰ)当M 的坐标为(0,1)-时,求过,,M A B 三点的圆的方程; (Ⅱ)证明:以A B 为直径的圆恒过点M .

(19)(本小题共13分)

已知函数11()()ln f x a x x a x

=+

+

-(1a >)

. (Ⅰ)试讨论()f x 在区间(0,1)上的单调性;

(Ⅱ)当[)3,a ∈+∞时,曲线()y f x =上总存在相异两点11(,())P x f x ,22(,())Q x f x ,使得曲线

()y f x =在点P ,Q 处的切线互相平行,求证:1265

x x +>

.

(20)(本小题共14分)

对于数列{}n a (1,2,,)n m = ,令k b 为1a ,2a , ,k a 中的最大值,称数列{}n b 为{}n a 的“创新数列”.例如数列2,1,3,7,5的创新数列为2,2,3,7,7.

定义数列{}n c :123,,,,m c c c c 是自然数1,2,3, ,(3)m m >的一个排列. (Ⅰ)当5m =时,写出创新数列为3,4, 4,5,5的所有数列{}n c ;

(Ⅱ)是否存在数列{}n c ,使它的创新数列为等差数列?若存在,求出所有的数列{}n c ,若不存在,请说

明理由.

北京市东城区2011-2012学年度高三综合练习(二)

数学参考答案及评分标准 (理科)

一、选择题(本大题共8小题,每小题5分,共40分)

(1)A (2)B (3)D (4)A (5)C (6)B (7)D (8)C 二、填空题(本大题共6小题,每小题5分,共30分)

(9)1- (10)(1,0) 2 (11))3,1(-

2

-

(12)125

(13)2 (14)③④

注:两个空的填空题第一个空填对得3分,第二个空填对得2分.

三、解答题(本大题共6小题,共80分) (15)(共13分)

解:(Ⅰ)由图可知,1A =,最小正周期428T =?=.

由2π

8T =

,得4

π=

ω. ……………3分

又π(1)sin()14

f ?=+= ,且ππ22

?-

<<

所以

ππ42

+=

?, 即4

π=? . ………………5分

所以π()sin(

)sin

(1)44

4

f x x x =+

=+π

π

. ………………6分 (Ⅱ)因为(1)0,(1)1,f f -==π(5)sin

(51)1,4

f =+=-

所以(1,0),(1,1),(5,1)M N P --. …………………7分 所以MN PN MP =

=

=

.

由余弦定理得3cos 5

M N P ∠=

=-. ……………11分

因为[)0,M NP ∠∈π, 所以4sin 5

M N P ∠=

. ……………13分

(其它解法酌情给分) (16)(共13分)

解:(Ⅰ)甲、乙两人所付费用相同即为2,4,6元. ……………2分

都付2元的概率为1111428P =?=; 都付4元的概率为2111248P =?=;

都付6元的概率为31114416

P =

?=;

故所付费用相同的概率为12311158816

16

P P P P =++=

++=. ……………6分

(Ⅱ)依题意,ξ的可能取值为4,6,8,10,12. ……………8分

1(4)8P ξ==; 11

115(6)442216P ξ==

?+?=; 1111115(8)4

4242416

P ξ==

?+?+?=; 11113(10)4

4

2

416

P ξ==?+?=

111(12)4416P ξ==?=.

故ξ的分布列为

……………11分

所求数学期望1553

11546810128

16

16

16

16

2

E ξ=?

+?+?+?

+?=

. ……………13分

(17)(共13分)

(Ⅰ)证明:因为M B //N C ,M B ?平面D N C ,N C ?平面D N C ,

所以M B //平面D N C . ……………2分 因为A M N D 为矩形,

所以M A //D N .

又M A ?平面D N C ,D N ?平面D N C , 所以M A //平面D N C . ……………4分 又MA MB M = ,且M A ,M B ?平面AM B , 所以平面AM B //平面D N C . ……………5分 又A B ?平面AM B ,

所以//A B 平面D N C . ……………6分

(Ⅱ)解:由已知平面AM N D ⊥平面M B C N ,且平面AMND 平面M B C N M N =,D N M N ⊥,

所以D N ⊥

平面M B C N ,又M

N N C

⊥,故以点N 为坐标原点,建立空间直角坐标系N xyz -

.

……………7分

由已知得30M C M C N =∠=

,易得M N =

,3N C =.

则(0,0,3)D ,(0,3,0)C

,4,0)B .

(0,3,

3)D C =-

,0)C B =

. ……………8分

设平面D B C 的法向量1(,,)x y z =n ,

则110,0.

D C C B ??=???=??

n n

即330,0.

y z y -=??+=令1x =-

,则y =

,z =

所以1(=-n . ……………10分 又2n (0,0,1)=是平面N BC 的一个法向量,

所以122112

cos ,7

?=

=

=

n n n n n n .

故所求二面角D B C N --

7

……………13分

(18)(共14分)

(Ⅰ)解:当M 的坐标为(0,1)-时,设过M 点的切线方程为1y kx =-,

由24,1,

x y y kx ?=?=-?消y 得2440x kx -+=. (1) 令2

(4)440k ?=-?=,解得1k =±.

代入方程(1),解得(2,1),(2,1)A B -. ……………3分 设圆心P 的坐标为(0,)a ,由PM PB =,得12a +=,解得1a =.

故过,,M A B 三点的圆的方程为2

2

(1)4x y +-=. ……………5分

(Ⅱ)证明:设0(,1)M x -,由已知得24

x

y =

,12

y x '=

,设切点分别为2

1

1(,

)4

x A x ,2

22(,

)4

x B x ,

所以12

M A x k =

,22

M B x k =

切线M A 的方程为2

1

11()4

2

x x y x x -

=

-即2

11112

4

y x x x =

-

切线M B 的方程为2

222()4

2

x x y x x -

=

-即2

221124y x x x =

-

. ……………7分 又因为切线M A 过点0(,1)M x -,所以得2

0111112

4

x x x -=-. ① 又因为切线M B 也过点0(,1)M x -,所以得20221112

4

x x x -=-

. ②

所以1x ,2x 是方程2

011124

x x x -=

-

的两实根,

由韦达定理得1202,x x x +=124x x =-. ……………9分

因为2110(,1)4x M A x x =-+ ,22

20(,1)4

x M B x x =-+ ,

所以2212

1020()()(1)(1)44

x x M A M B x x x x ?=--+++

22

2

22

12120120121()()1164x x x x x x x x x x =-+++

+

++

2

2

2

21212012012121

()()2116

4x x x x x x x x x x x x ??=-+++

+

+-+?

?. 将1202,x x x +=124x x =-代入,得0MA MB ?=

. ……………13分

所以以A B 为直径的圆恒过点M . ……………14分

(19)(共13分)

(Ⅰ)解:由已知0x >,2

2

2

2

1

11

()1

()()

1()1a x a x x a x a a

a f x x x

x

x

+

-+

+--

'=

--=-

=-

. ………2分

由()0f x '=,得11x a =,2x a =. ………4分

因为1a >,所以101a

<<,且1a a

>

所以在区间1

(0,

)a 上,()0f x '<;在区间1

(

,1)a 上,()0f x '>.

故()f x 在1(0,)a 上单调递减,在1

(,1)a

上单调递增. ……………6分

(Ⅱ)证明:由题意可得,当[)3,a ∈+∞时,12()()f x f x ''=(12,0x x >,且12x x ≠).

2

2

11

22

1

1

1111

a a a a x x x x +

+

--=-- ,

所以121

2

12

111x x a a

x x x x ++

=

+

=

,[)3,a ∈+∞. ……………8分

因为12,0x x >,且12x x ≠,所以2

12

12()2

x x x x +<恒成立,

所以

2

12

1214()x x x x >

+,又120x x +>,

所以1212

1x x a a x x ++=12

4x x >

+,整理得1241x x a a

+>

+

. ……………11分

令()g a 41a a =

+

,因为[)3,a ∈+∞,所以()g a 在[)3,+∞上单调递减,

所以()g a =

41a a

+

在[)3,+∞上的最大值为6(3)5

g =

所以1265

x x +>

. ……………13分

(20)(共14分)

解:(Ⅰ)由题意,创新数列为3,4, 4,5,5的所有数列{}n c 有两个,即数列3,4,1,5,2; 数列3,4,2,5,1. ……………4分

(Ⅱ)存在数列{}n c ,使它的创新数列为等差数列. 数列{}n c 的创新数列为{}n e (1,2,3,,)n m = , 因为m e 是12,,,m c c c 中的最大值, 所以m e m =.

由题意知,k e 为12,,,k c c c 中最大值,1k e +为121,,,,k k c c c c + 中的最大值, 所以k e 1k e +≤,且{}1,2,,k e m ∈ .

若{}n e 为等差数列,设其公差为d , 则1k k d e e +=-0≥且d ∈N ,

当0d =时,{}n e 为常数列,又m e m =, 所以数列{}n e 为m ,m , ,m .

此时数列{}n c 是首项为m 的任意一个符合条件的数列; ……………8分

当1d =时,因为m e m =,所以数列{}n e 为1,2,3, ,m .

此时数列{}n c 为1,2,3, ,m ; ……………10分

当2d ≥时,因为111(1)(1)222m e e m d e m m e =+-≥+-?=-+ ,

又3m >,10e > ,所以m e m >,这与m e m =矛盾,所以此时{}n e 不存在,即不存在{}

n c 使得它的创新数列为公差2d ≥的等差数列. ……………13分 综上,当数列{}n c 为以m 为首项的任意一个符合条件的数列或{}n c 为数列1,2,3, ,m

时,它的创新数列为等差数列. ……………14分

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