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上海市虹口区2017届高三4月期中教学质量监控(二模)数学试题(带答案)

上海市虹口区2017届高三4月期中教学质量监控(二模)数学试题(带答案)
上海市虹口区2017届高三4月期中教学质量监控(二模)数学试题(带答案)

虹口区2016-2017学年度第二学期期中教学质量监控测试

高三数学 试卷

(时间120分钟,满分150分) 2017.4

一、填空题(1~6题每小题4分,7~12题每小题5分,本大题满分54分)

1、集合{}1,2,3,4A =,{}(1)(5)0B x x x =--<错误!未找到引用源。,错误!未找到引用源。

则A B ?= .

2、复数21i

z i

-=

+所对应的点在复平面内位于第 象限. 3、已知首项为1公差为2的等差数列{}n a ,其前n 项和为n S ,则2

()lim n n n a S →∞= .

4、若方程组23

22ax y x ay +=??+=?

无解,则实数a = .

5、若7)(a x +的二项展开式中,含6x 项的系数为7,则实数=a .

6、已知双曲线2

2

21(0)y x a a

-=>,它的渐近线方程是2y x =±,则a 的值为 .

7、在ABC ?中,三边长分别为2a =,3b =,4c =,则

sin 2sin A

B

= ___________. 8、在平面直角坐标系中,已知点(2,2)P -,对于任意不全为零的实数a 、b ,直线

:(1)(2)0l a x b y -++=,若点P 到直线l 的距离为d ,则d 的取值范围是 . 9、函数2

1()(2)1

x

x f x x x ?≤?=?->??,如果方程()f x b =有四个不同的实数解1x 、2x 、3x 、4x ,则

1234x x x x +++= .

10、三条侧棱两两垂直的正三棱锥,其俯视图如图所示,主视图的边界是底边长为2的等腰三角形,则主视图的面积等于 .

11、在直角ABC ?中,2

A π

∠=

,1AB =,2AC =,M 是ABC ?内一点,且1

2

AM =

,若AM AB AC

λμ=+

,则2λμ+的最大值 . 12、无穷数列{}n a 的前n 项和为n S ,若对任意的正整数n 都有{}12310,,,,n S k k k k ∈ ,则10a 的可能取值最多..

有 个. 二、选择题(每小题5分,满分20分)

13、已知a ,b ,c 都是实数,则“a ,b ,c 成等比数列”是“2b a c =?的( )

.A 充分不必要条件 .B 必要不充分条件 .C 充要条件 .D 既不充分也不必要条件

14、1l 、2l 是空间两条直线,α是平面,以下结论正确的是( ).

.A 如果1l ∥α,2l ∥α,则一定有1l ∥2l . .B 如果12l l ⊥,2l α⊥,则一定有1l α⊥.

.C 如果12l l ⊥,2l α⊥,则一定有1l ∥α. .D 如果1l α⊥,2l ∥α,则一定有12l l ⊥.

15、已知函数()2

x x

e e

f x --=,1x 、2x 、3x R ∈,且120x x +>,230x x +>,310x x +>,则

123()()()f x f x f x ++的值( )

.A 一定等于零. .B 一定大于零. .C 一定小于零. .D 正负都有可能.

16、已知点(,)M a b 与点(0,

1)N -在直线3450x y -+=的两侧,给出以下结论:

①3450a b -+>;②当0a >时,a b +有最小值,无最大值;③221a b +>; ④当0a >且1a ≠时,

1

1b a +-的取值范围是93(,)(,)44

-∞-+∞ . 正确的个数是( )

.A 1 .B 2 .C 3 .D 4

F E

D

C

B

A

C 1

B 1

A 1

三、解答题(本大题满分76分)

17、(本题满分14分.第(1)小题7分,第(2)小题7分.)

如图111ABC A B C -是直三棱柱,底面ABC ?是等腰直角三角形,且4AB AC ==,直三棱柱的高等于4,线段11B C 的中点为D ,线段BC 的中点为E ,线段1CC 的中点为F .

(1)求异面直线AD 、EF 所成角的大小; (2)求三棱锥D AEF -的体积.

18、(本题满分14分.第(1)小题7分,第(2)小题7分.)

已知定义在(,

)2

2

π

π

-

上的函数()f x 是奇函数,且当(0,

)2

x π

∈时,tan ()tan 1

x

f x x =

+.

(1)求()f x 在区间(,

)2

2

π

π

-

上的解析式;

(2)当实数m 为何值时,关于x 的方程()f x m =在(,

)2

2

π

π

-

有解.

19、(本题满分14分.第(1)小题6分,第(2)小题8分.)

已知数列{}n a 是首项等于

1

16

且公比不为1的等比数列,n S 是它的前n 项和,满足325416

S S =-

. (1)求数列{}n a 的通项公式;

(2)设log n a n b a =(0a >且1)a ≠,求数列{}n b 的前n 项和n T 的最值.

20、(本题满分16分.第(1)小题3分,第(2)小题5分,第(3)小题8分.)

已知椭圆:C 22

221(0)x y a b a b

+=>>,定义椭圆C 上的点00(,)M x y 的“伴随点”为00(,)x y N a b .

(1)求椭圆C 上的点M 的“伴随点”N 的轨迹方程; (2)如果椭圆C 上的点3(1,

)2的“伴随点”为1

3

(,)22b

,对于椭圆C 上的任意点M 及它的“伴随点”N ,求OM ON

的取值范围;

(3)当2a =,3b =时,直线l 交椭圆C 于A ,B 两点,若点A ,B 的“伴随点”分别是P ,

Q ,且以PQ 为直径的圆经过坐标原点O ,求OAB ?的面积.

21、(本题满分18分.第(1)小题3分,第(2)小题6分,第(3)小题9分.)

对于定义域为R 的函数()y f x =,部分x 与y 的对应关系如下表:

x

2-

1-

0 1 2 3 4 5 y

2

3 2

1-

2

(1)求{[(0)]}f f f ;

(2)数列{}n x 满足12x =,且对任意n N *

∈,点1(,

)n n x x +都在函数()y f x =的图像上,求

124n x x x +++ ;

(3)若()sin()y f x A x b ω?==++,其中0A >,0ωπ<<,0?π<<,03b <<,求此函数的解析式,并求(1)(2)(3)f f f n +++ (n N *

∈).

F

E

D

C

B

A

C 1

B 1

A 1

虹口区2016-2017学年度第二学期高三年级数学学科

期中教学质量监控测试题答案

一、填空题(1~6题每小题4分,7~12题每小题5分,本大题满分54分) 1、{2,3,4}; 2、四; 3、4; 4、2±; 5、1; 6、2 ;

7、

76; 8、[0,5]; 9、4; 10、63

; 11、22; 12、91; 二、选择题(每小题5分,满分20分)

13、A ; 14、D ; 15、B ; 16、B ; 三、解答题(本大题满分76分)

17、(14分)解:(1)以A 为坐标原点,AB 、AC 、1AA 分别为x 轴和y 轴建立直角坐标系.

依题意有D (2,2,4),A (0,0,0),E (2,2,0),F (0,4,2)

所以(2,2,4),(2,2,2)AD EF ==-

.……………………3分

设异面直线AD 、EF 所成角为角,

||cos ||||

AD EF AD EF α?==?

|448|4416444-++=++++ 23 所以2

arccos 3α=, 所以异面直线AD 、EF 所成角的大小为2

arccos

3

…………7分 (2) 线段11B C 的中点为D ,线段BC 的中点为E ,由4AB AC ==,高14A A =,得

42BC =,∴22AE =,42DEF S = ………………3分

由E 为线段BC 的中点,且AC AB =,BC AE ⊥∴,由⊥1BB 面ABC ,1BB AE ⊥∴, 得⊥AE 面C C BB 11,

1116

4222333

D AEF A DEF DEF V V S A

E --==?=??=

∴三棱锥D AEF -的体积为16

3

体积单位.……………………7分

18、(14分)解:(1)设02

x π

-

<<,则02

x π

<-<

()f x 是奇函数,则有tan()tan ()()tan()11tan x x

f x f x x x

-=--=-

=-+-…………4分

∴t a n

0t a n 12

()00t a n 0

1t a n 2x x x f x x x x x

ππ?<

??-<<-? ………………7分 (2)设02

x π

<<

,令tan t x =,则0t >,而tan 1

()1tan 111x t y f x x t t

==

==-+++.

11t +>,得1011t <

<+,从而10111t <-<+,∴()y f x =在02

x π

<<的取值范围是01y <<.…………………………11分

又设02

x π

-

<<,则02

x π

<-<

,由此函数是奇函数得()()f x f x =--,0()1f x <-<,从而

1()0f x -<<.………………13分

综上所述,()y f x =的值域为(1,1)-,所以m 的取值范围是(1,1)-.…………14分

19、(14分)解:(1)325416S S =- , 1q ≠,3211(1)(1)5

41116

a q a q q q --∴=?---.……2分

整理得2320q q -+=,解得2q =或1q =(舍去).………………4分

1512n n n a a q --∴=?=.………………6分

(2)log (5)log 2n a n a b a n ==-.………………8分

1)当1a >时,有log 20,a > 数列{}n b 是以log 2a 为公差的等差数列,此数列是首项为负的递增的等差数列.

由0n b ≤,得5n ≤.所以min 45()10log 2n a T T T ===-.n T 的没有最大值.………11分

2)当01a <<时,有log 20a <,数列{}n b 是以log 2a 为公差的等差数列,此数列是首项为正的递减的等差数列.

0n b ≥,得5n ≤,max 45()10log 2n a T T T ===-.n T 的没有最小值.…………14分

20、(16分)解:(1)解.设N (,x y )由题意 0

0x x a

y y b

?

=??

?

?=??

则00

x ax y by =??=?,又22

00221(0)x y a b a b +=>>

∴22

22()()1(0)ax by a b a b

+=>>,从而得221x y +=……………………3分

(2)由

11

2a

=,得2a =.又221914a b +=,得3b =.…………5分

点00(,)M x y 在椭圆上,22

00143

x y +=,2200334y x =-,且2

004x ≤≤,

∴22

2

000000023(,)(,

)32

2433

x

y x y OM ON x y x -=?=+=+

由于23

04

->,OM ON 的取值范围是3,2????……8分 (3) 设1122(,),(,)A x y B x y ,则1122,,,2233x y x y P Q ????

? ?

????

; 1)当直线l 的斜率存在时,设方程为y kx m =+, 由2214

3y kx m x y =+??

?+=??

得222

(34)84(3)0k x kmx m +++-=; 有22122

212248(34)08344(3)34k m km x x k m x x k ?

??=+->?

-?+=?+?

?-=

?+?

① ……10分 由以PQ 为直径的圆经过坐标原点O 可得: 1212340x x y y +=; 整理得:221212(34)4()40k x x mk x x m ++++= ②

将①式代入②式得: 2

2

342k m +=,………………………… 12分

048,0,043222>=?>∴>+m m k

又点O 到直线y kx m =+的距离2

1m d k

=

+

22

2

2

2222

212

23414334143433411m m

k k m k

k m k k x x k AB ?+=+?+=+-++=-+= 所以1

32

OAB S AB d ?==……………………14分

2) 当直线l 的斜率不存在时,设方程为(22)x m m =-<<

联立椭圆方程得22

3(4)4

m y -=;代入1212340x x y y +=得22

3(4)3404m m --?

=,解得22m =,从而2

32

y =,

3212121=-==?y y m d AB S OAB 综上:OAB ?的面积是定值,3……………………16分

21、(18分)解:(1) {[(0)]}((3))(1)2f f f f f f ==-= ……………………3分 (2) 11212,()()(2)0,n n x x f x x f x f +==∴===

32()3,x f x == 43()1,x f x ==-54()2x f x ==51x x ∴=,周期为

4 , 所以

12n x x x +++ =

4n .……………………9分 (3)由题意得 (1)2(1)

(1)2(2)(0)3(3)(2)0(4)

f f f f -=??=?

?=??=? 由(1)(2)sin()sin()sin cos 0ω?ω?ω?-∴+=-+∴=

又 0ωπ<

π

?∴=…………11分

从而有2

3cos 32cos 23(2cos 1)30cos20

A b A A A b b A A A A b ωωωω+=?

+-=??

+=?=-???-+-=??+=? 22242230 2.1A A A A A b ∴-+-+=∴== 1c o s 2ω

= 0ωπ<<3

πω∴= ()2cos 13

f x x π

∴=+…………………………13分

此函数的最小正周期为6, (6)(0)3f f ==

(1)(2)(3)4)+(5)(6)6f f f f f f ++++= (…………14分

1)当2n k =()k N *

∈时.

(1)(2)(3)(1)(2)(6)f f f n f f f k +++=+++

[(1)(2)(6)]63k f f f k n =+++== .……………………16分

2)当21n k =-()k N *

∈时.

(1)(2)(3)(1)(2)(6)(62)(61)(6)f f f n f f f k f k f k f k +++=+++-----

[(1)(2)(6)]56532k f f f k n =+++-=-=- .………………18分

2017届上海市徐汇区高三英语二模卷(含听力文本和答案)

2016学年第二学期徐汇区高三模拟考英语试卷2017.4 I. Listening Comprehension Section A Short Conversations Directions: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard. 1. A. He knows who is knocking. B. He is eager to know who it is. C. He doesn’t want to open the door. D. He is ready to open the door. 2. A. By plane. B. By bus. C. By taxi. D. By train. 3. A. $100. B. $200. C. $300. D. $400. 4. A. She went to cinema. B. She went to an exhibition. C. She stayed at home. D. She stayed with her classmates. 5. A. In a doctor’s office. B. In a professor’s office. C. In an operating room. D. In an emergency ward. 6. A. The man paid the tuition for learning physics. B. The man got a lot of money for his hard work. C. His hard work was not rewarding at all. D. His work before the test led to a good result. 7. A. A furnished house. B. A recent book. C. A further study. D. A new record. 8. A. They will go swimming. B. They will climb mountains. C. They will buy some clothes. D. They will forecast the weather conditions. 9. A. He has another lecture to attend. B. He has no interest in the lecture. C. He’s attended the same lecture given by Professor Wilson before. D. He might miss the lecture, if the woman didn’t remind him. 10.A. She fully agrees with the man. B. They are uncertain about the weather. C. She disagrees with the man. D. She thought the man was always late. Section B Directions: In Section B, you will hear several longer conversation(s) and short passage(s), and you will be asked several questions on each of the conversation(s) and the passage(s). The conversation(s) and the passage(s) will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard. Questions 11 through 13 are based on the following passage. 11. A. People are encouraged to be a craftsman. B. Learning woodworking is not as hard as you think. C. Learning woodworking will help you know more people. D. Taking a class in woodworking will be very helpful. 12. A. Because I am a talent in this art and want to share it with others. B. Because I am interested in it and want to show it to others. C. Because I wonder how to pick materials and how to do it well. 1 / 16

2017年上海虹口区初三二模语文试题

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