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静安区2012学年高三年级第一学期期末教学质量检测文理合并

静安区2012学年高三年级第一学期期末教学质量检测

数学试卷(文理科合并)

一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分.

1.已知函数)7

22sin(2

1)(π

+=ax x f 的最小正周期为π4,则正实数a = .

2.等比数列{}n a (*N n ∈)中,若16

12=

a ,2

15=

a ,则=12a .

3.两条直线0943:1=+-y x l 和03125:2=-+y x l 的夹角大小为 .

4.求和:n n n n n n C C C C 32793321++++ = .(*N n ∈)

5.(文)设x ,y 满足条件??

?≤+≤-≤-≤,

11,31y x y x 则点),(y x 构成的平面区域面积等于 .

6.(文)设y x ,满足约束条件??

?

??

??≤≤≤≤≤+≤+,40,30,

1223,

5y x y x y x 使目标函数y x z 56+=的值最大的点),(y x 坐标

是 . 7(文)设圆过双曲线

116

9

2

2

=-

y

x

右支的顶点和焦点,圆心在此双曲线上,则圆心到双曲线中

心的距离是 .

8.某旅游团要从8个风景点中选两个风景点作为当天上午的游览地,在甲和乙两个风景点中至少需选一个,不考虑游览顺序,共有 种游览选择.

9.(文)已知0

>++-x a ax 的解集是 .

10.(文)已知α、β为锐角,且2)2

tan

1)(2

tan

1(=++β

α

,则βαtan tan = .

11.(文)数列{}n a 的前n 项和为2

2n S n =(*N n ∈),对任意正整数n ,数列{}n b 的项都满足

等式022

121=+-++n n n n n a b a a a ,则n b = .

第12题

12.(文)机器人“海宝”在某圆形区域表演“按指令行走”.如图所示,“海宝”从圆心O 出发,先沿北偏西13

12arcsin

方向行走13米至点A 处,再沿正南方向行走14米至点B 处,最后沿正东

方向行走至点C 处,点B 、C 都在圆O 上.则在以圆心O 为坐标原点,正东方向为x 轴正方向,

正北方向为y 轴正方向的直角坐标系中圆O 的方程为

.

13(文)设P 是函数x

x y 2+

=(0>x )的图像上任意一点,

过点P 分别向直线x y =和y 轴作垂线,垂足分别为A 、B ,则PA

14.(文)设复数i a a z )sin 2()cos (θθ-++=(i 为虚数单位),若对任意实数θ,2≤z ,则实数a 的取值范围为 .

二、选择题(本大题满分20分)本大题共有4题,每题有且只有一个正确答案,考生应在答题纸的相应编号上,填上正确的答案,选对得5分,否则一律得零分.

15.(文)某地区的绿化面积每年平均比上一年增长10.4%,经过x 年,绿化面积与原绿化面积之

比为y ,则y=f(x)的图像大致为 ( )

(文)16.若复数021≠z z ,则2121z z z z =是12z z =成立的( )

(A) 充要条件 (B) 既不充分又不必要条件 (C) 充分不必要条件 (D) 必要不充分条件

17.(文)函数])3,1[(4

2)(2

∈+-=

x x

x x x f 的值域为 ( )

(A) ]3,2[ (B) ]5,2[ (C) ]3,3

7

[ (D) ]4,3

7

[

18.(文)已知向量a 和b 满足条件:a b ≠且0≠?b a .若对于任意实数t ,恒

有-≥-,则在a 、b 、b a +、b a -这四个向量中,一定具有垂直关系的两个向

量是( )

(A) a 与b a - (B) b 与b a - (C) a 与b a + (D)b 与b a +

三、解答题(本大题满分74分)本大题共5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤 .

19.(文)已知数列}{n a 的递推公式为??

?=∈≥+-=-.

2),2(,3231*1a N n n n a a n n

(1)令n a b n n -=,求证:数列}{n b 为等比数列;(2)求数列}{n a 的前 n 项和.

20. (文)已知c b a ,,分别为△ABC 三个内角A 、B 、

C 所对的边长,且c A b B a 53cos cos =-.

(1)求:

B

A tan tan 的值;

(2)若0

60=A ,5=c ,求a 、b .

21. 某仓库为了保持库内的湿度和温度,四周墙上均装有如图所示的自动通风设施.该设施的下部ABCD 是正方形,其中AB =2米;上部CDG 是等边三角形,固定点E 为AB 的中点.△EMN 是由电脑控制其形状变化的三角通风窗(阴影部分均不通风),MN 是可以沿设施边框上下滑动且始终保持和AB 平行的伸缩横杆.

(1)设MN 与AB 之间的距离为x 米,试将△EMN 的面积S (平方米)表示成关于x 的函数; (2)求△EMN 的面积S (平方米)的最大值.

22. 已知椭圆

12

22

2=+

b

y a

x 的两个焦点为)0,(1c F -、)0,(2c F ,2c 是2a 与2

b 的等差中项,其

中a 、b 、c 都是正数,过点),0(b A -和)0,(a B 的直线与原点的距离为2

3.

(1)求椭圆的方程;

(2)(文)过点A 作直线交椭圆于另一点M ,求AM 长度的最大值;

(3)已知定点)0,1(-E ,直线t kx y +=与椭圆交于C 、D 相异两点.证明:对任意的0>t ,都存在实数k ,使得以线段CD 为直径的圆过E 点.

A

B

C

(文21题)

23已知x x f 2

1log

)(=,当点),(y x M 在)(x f y =的图像上运动时,点),2(ny x N -在函数

)(x g y n =的图像上运动(*N n ∈). (1)求)(x g y n =的表达式;

(2)若方程)2()(21a x g x g +-=有实根,求实数a 的取值范围; (3)设)

(2

)(x g n n x H =,函数)()()(11x g x H x F +=(b x a ≤≤<0)的值域为

]2

2

log

,2

2

[log

4

2

5

2

++a b ,求实数a ,b 的值.

高三年级 文理科数学试卷答案及评分标准

1.4

1=

a ; 2.64; 3.(理)65

33arccos

;(文)14-n

4.(理)

3

16;(文)同理3 5.(理)13;(文)2 6.(理)12-?n n ;(文))3,2(

7.(理)(2)(3)(4);(文)3

16 8.(理)4;(文)同理5 9.(理)

9

1093;(文))2,2(

a

10.(文理)1; 11.(理)2252

2=+y x ;(文)1

41422

-+=n n b n ; 12.(理)x y 162=;

(文)同理11

13.(理)),3[+∞-;(文)-1 14.(理)

6

π

;(文)]5

5,

5

5[-

.

15.(文理)D ; 16.(理)C ;(文)D ; 17.(文理)A ;18.(文理)B

19(理)解:(1)

①如图1所示,当MN 在矩形区域滑动, 即0<x ≤1时,

△EMN 的面积S =x ??221

=x ; ··························· 1分

②如图2所示,当MN 在三角形区域滑动, 即1<x <31+时,

如图,连接EG ,交CD 于点F ,交MN 于点H , ∵ E 为AB 中点,

∴ F 为CD 中点,GF ⊥CD ,且FG =3. 又∵ MN ∥CD , ∴ △MNG ∽△DCG . ∴

GF

GH DC

MN =

,即M N

=

. ························ 4分

故△EMN 的面积S

=12

x ?

=x x )3

31(3

32+

+-;

············································· 6分

综合可得:

C

图2

N 图1

(

)(

2011113

3x x S x x x ??=??

?-+++ ?? ????,<≤.<< ···························································7分

(2)①当MN 在矩形区域滑动时,x S =,所以有10≤

·······································8分

②当MN 在三角形区域滑动时,S =x x )3

31(3

32

+

+-

.

因而,当2

3

1+=x (米)时,S 得到最大值,最大值S =33

2

1

+

(平方米).

13

321>+

∴ S 有最大值,最大值为3

32

1+

平方米. ··································································· 12分

(文)解:(1)11113))1((3333323----=--=+-=-+-=-=n n n n n n b n a n a n n a n a b ,

2≥n

又1111=-=a b ,所以0≠n b (*N n ∈),

)2(31

≥=-n b b n n

所以,数列}{n b 是以1为首项3为公比的等比数列. ··················································6分

(2)1

3-=n n b ,n b a n n +=······················································································8分

所以数列}{n a 的前 n 项和)21()(21n b b b S n n +++++++= =

2

1

32

-++n n n .

·································································································································· 14分

20(理)解:(1)∵a 、b 、c 成等比数列,∴b 2

=ac ························································1分 则cos B =

ac

b

c

a

22

2

2

-+=

ac ac

c

a

22

2

-+ ·················································································3分

而a 2

+c 2

≥2ac ∴cos B =

ac

ac

c

a

22

2

-+≥2

12=ac

ac ,等号当且仅当a =c 时取得,即

2

1≤cos B <1,得到

3

(2)cos2x -4sin(

2

4x π+)sin(

24

x π-

)=cos2x -4sin(2

4

x π+

)cos(2

4

x π+

)

=2cos x 2

-2cos x -1=2(cos x -

2

1)2-2

3 ·················································································· 11分

∵x =B

2

1≤cos x <1

∴2(cos x -

2

1)2-

2

3≥-

23

则由题意有:-m <-2

3即m >

2

3··················································································· 14分

(说明:这样分离变量1cos 2cos 22cos cos 22++-=->x x x x m 参照评分) (文)解:(1)由正弦定理

C

c B

b A

a sin sin sin =

=

得C A B B A sin 5

3cos sin cos sin =

-,2分

又B A B A B A C sin cos cos sin )sin(sin +=+=,所以A B B A cos sin 5

8cos sin 5

2=

,5分

可得

4cos sin cos sin tan tan ==

A

B B A B

A . ·

··················································································7分 (2)若060=A ,则2

3s i n =

A ,2

1cos =

A ,3tan =A ,得4

3t a n =

B ,可得

19

194cos =

B ,19

193sin ?=

B . ··········································································· 10分

38

1935sin cos cos sin )sin(sin ?=

+=+=B A B A B A C ,

由正弦定理

C

c B

b A

a sin sin sin ==得

19sin sin =?=

A C

c a ,2sin sin =?=

B C

c

b ························································ 14分

21(理)解:(1)当1=n 时,3

12

1a a =,由01≠a 得11=a .····································1分 当2=n 时,3

2221)1(a a +=+,由02≠a 得22=a 或12-=a .当3=n 时,3

33

22

321)1(a a a a ++=++,若22=a 得33=a 或23-=a ;若12-=a 得13=a ; 5分

综上讨论,满足条件的数列有三个:

1,2,3或1,2,-2或1,-1,1. ·············································································6分 (2)令n n a a a S +++= 21,则3

3

23

12

n n a a a S +++= (*N n ∈).

从而3

13

3

23

12

1)(++++++=+n n n n a a a a a S . ·······················································7分 两式相减,结合01≠+n a ,得12

12++-=n n n a a S . ······················································8分 当1=n 时,由(1)知11=a ;当2≥n 时,)(221--=n n n S S a =)()(2

12

1n n n n a a a a ---++,即0)1)((11=--+++n n n n a a a a ,所以n n a a -=+1或11+=+n n a a . ·························· 12分 又11=a ,20122013-=a ,所以无穷数列{}n a 的前2012项组成首项和公差均为1的等差数列,从第2013项开始组成首项为-2012,公比为-1的等比数列.故

???>-?≤≤=)2012()

1(2012)20121(n n n

a n

n . ··········································································· 14分 (说明:本题用余弦定理,或者正弦定理余弦定理共同使用也可解得,请参照评分)

(文)解:(1)

①如图1所示,当MN 在正方形区域滑动, 即0<x ≤2时,

△EMN 的面积S =x ??221

=x ; ··························· 2分

②如图2所示,当MN 在三角形区域滑动, 即2<x <32+

时,

如图,连接EG ,交CD 于点F ,交MN 于点H , ∵ E 为AB 中点,

∴ F 为CD 中点,GF ⊥CD ,且FG =3. 又∵ MN ∥CD , ∴ △MNG ∽△DCG . ∴

GF

GH DC

MN =,即3

)

23(2x MN -+=

. ················· 5分

故△EMN 的面积S =

x x ?-+?3

)

23(221 =x x )3

321(3

32++-; ··········································· 7分

综合可得: ??

?

??+<<++-≤<=322,)33

21(332

0,2x x x x x S ····························································8分 说明:讨论的分段点x=2写在下半段也可. (2)①当MN 在正方形区域滑动时,x S =,所以有20≤

②当MN 在三角形区域滑动时,S =x x )3

321(3

32

+

+-.

因而,当22

31<+

=x (米),S 在)32,2(+上递减,无最大值,20<

所以当2=x 时,S 有最大值,最大值为2平方米. ···················································· 14分 22.解:(1)在椭圆中,由已知得2

2

22

2

2

b a b a

c +=

-= ···········································1分

E

N

图1

E

B

C

图2

过点),0(b A -和)0,(a B 的直线方程为

1=-+

b

y a

x ,即0=--ab ay bx ,该直线与原点的距离

为2

3,由点到直线的距离公式得:

2

32

2

=

+b

a a

b ···················································3分

解得:1,32

2==b a ;所以椭圆方程为

11

3

2

2

=+

y

x

·

···················································4分 (2)(理))0,2(1-F ,直线11A F 的方程为22+=x y ,611=

A F ,当椭圆上的点P 到直

线11A F 距离最大时,△11PA F 面积取得最大值·······························································6分 设与直线11A F 平行的直线方程为d x y +=

2,将其代入椭圆方程

11

3

2

2

=+

y

x

得:

0122372

2

=-++d

x d

x ,0=?,即03

283

2882

2

=+

-

d

d

,解得72

=d

当7-=d 时,椭圆上的点P 到直线11A F 距离最大为

3

7

2+

,此时△11PA F 面积为

2

14

223

7

26

2

1+

=

+

··························································································9分

(文)设),(y x M ,则)1(322y x -=,422)1(2

222

++-=++=y y y x AM

,其中1

1≤≤-y ·····································································································································6分 当2

1=

y 时,2

AM

取得最大值

2

9,所以AM 长度的最大值为

2

23····························9分

(3)将t kx y +=代入椭圆方程,得0336)31(2

2

2

=-+++t ktx x k ,由直线与椭圆有两个交

点,所以0)1)(31(12)6(222>-+-=?t k kt ,解得3

12

2

->t k

································ 11分

设),(11y x C 、),(22y x D ,则2

21316k

kt x x +-

=+,2

2

2131)1(3k

t x x +-=

?,因为以CD 为直径的圆

过E 点,所以0=?ED EC ,即0)1)(1(2121=+++y y x x , ····································· 13分

而))((2121t kx t kx y y ++==221212)(t x x tk x x k +++,所以 01316)

1(31)1(3)

1(2

2

2

2

2

=++++-+-+t k

kt tk k

t k

,解得t

t k 3122

-=

······························ 14分

如果3

12

2

->

t k

对任意的0>t 都成立,则存在k ,使得以线段CD 为直径的圆过E 点.

09)1(3

1)312(

2

2

222

2

2

>+-=

--

-t

t

t t t

t ,即3

12

2

->

t k

.所以,对任意的0>t ,都存在k ,

使得以线段CD 为直径的圆过E 点. ··········································································· 16分 23(理)解:(1)x y =

,3

x y =是对等函数; ·

························································4分 (2)研究对数函数x y a log =,其定义域为),0(+∞,所以x x a

a

log

log

=,又0log

≥x a

所以当且仅当0log

≥x a

时)()(x f x f =成立.所以对数函数x y a

log

=在其定义域),0(+∞内

不是对等函数. ·············································································································6分 当10<

,此时x y a

log

=是对等函数;

当1>a 时,若),1[+∞∈x ,则0log

≥x a

,此时x y a

log

=是对等函数;

总之,当10<

log =是对等函数;当1>a 时,在)

,1[+∞及其任意非空子集内x y a

log

=是对等函数. ····························································· 10分

(3)对任意D x ∈,讨论)(x f 与)(x f -的关系. 1)若D 不关于原点对称,如x y =

虽是对等函数,但不是奇函数或偶函数; ··········· 11分

2)若{}0=D ,则0)0()0(≥=f f .当0)0(=f 时,)(x f 既是奇函数又是偶函数;当0)0(>f 时,)(x f 是偶函数. ·································································································· 13分 3)以下均在D 关于原点对称的假设下讨论. 当0>x 时,0)()()(≥==x f x f x f ;

当0x 时,

0<-x ,令t x =-,则t x -=,且0

综上讨论,当0x 时,0<-x ,令t x =-,则t x -=,且0

综上讨论,若当0

??-==)2(),(x g ny x f y n 得x n x nf x g n 2

1log

)()2(==-,

所以)2(log )(2

1+=x n x g n ,

(2->x ). ·················································································································4分 (2))(log 2)2(log

2

12

1a x x +=+,即a x x +=+2(02>+x ) ··························6分

2++

-=x x a ,令02>+=

x t ,所以4

922

++-=t t a ,当4

7-

=x 时,4

9=

a .即实

数a 的取值范围是]4

9

,(-∞··························································································· 10分

(3)因为n

x n n x x H )

2(12

)()

2(log

2

1+=

=+,所以)2(log

2

1)(2

1+++=

x x x F .

)(x F 在),2(+∞-上是减函数.·

··················································································· 12分 所以??

?

???

?

+=+=2

2

log )(2

2log )(5

2

42

b b F a a F 即???

????

+=++++=+++2

2

log

)2(log

2

12

2

log )2(log

215

2

2

14

2

2

1

b b b a a a ,所以??

?==3

,2b a ·········· 16分

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