上海市闵行区2017届高三4月质量调研考试(二模)数学试题(带答案)

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俯视图

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闵行区2016-2017学年第二学期高三年级质量调研考试

数 学 试 卷 2017.04

（满分150分，时间120分钟）

一、填空题（本大题共有12题，满分54分，第1～6题每题4分，第7～12题每题5分）考生应在答题纸上相应编号的空格内直接填写结果． 1. 方程()3log 212x +=的解是 ．

2. 已知集合{}

{}11,1,0,1,M x x N =+≤=-则M N =I ．

3. 若复数122,2z a i z i =+=+（i 是虚数单位），且12z z 为纯虚数，则实数a = ．

4. 直线2232x t y t

?=--??=+??（t 为参数）对应的普通方程是 ．

5. 若()

1(2),3n n n x x ax bx c n n -*+=++++∈≥N L ，且

4b c =，则a 的值为 ．

6. 某空间几何体的三视图如右图所示，则该几何体的侧面积

是 ．

7. 若函数()2()1x

f x x a =+-在区间[]0,1上有零点，则实

数

a 的取值范围是 ．

8. 在约束条件123x y ++-≤下，目标函数2z x y =+的

最

大值为 ．

9. 某学生在上学的路上要经过2个路口，假设在各路口是否遇到红灯是相互独立的，遇到红灯的概率都是

1

3

，则这名学生在上学的路上到第二个路口时第一次遇到红灯的概率是 ． 10. 已知椭圆()2

2

2101y x b b

+=<<，其左、右焦点分别为12F F 、，122F F c =．若此椭圆上存

在点P ，使P 到直线1

x c

=

的距离是1PF 与2PF 的等差中项，则b 的最大值为 ． 11. 已知定点(1,1)A ，动点P 在圆22

1x y +=上，点P 关于直线y x =的对称点为P '，向量

AQ OP '=u u u r u u u r

，O 是坐标原点，则PQ u u u r 的取值范围是 ．

12. 已知递增数列{}n a 共有2017项，且各项均不为零，20171a =，如果从{}n a 中任取两项,i j a a ，当i j <时，j i a a -仍是数列{}n a 中的项，则数列{}n a 的各项和2017S =___．

二、选择题（本大题共有4题，满分20分，每题5分）每题有且只有一个正确选项，考生应在答题纸的相应位置，将代表正确选项的小方格涂黑．

13. 设a b r r 、

分别是两条异面直线12l l 、的方向向量，向量a b r r 、的夹角的取值范围为A ，12l l 、所成的角的取值范围为B ，则“A α∈”是“B α∈”的 ( )

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

x y

O

①

②

③

④ x y

O

x y

O

x y

O

14. 将函数sin 12y x π??

=-

??

?

图像上的点,4P t π??

???

向左平移(0)s s >个单位，得到点P '，若P '位于函数sin 2y x =的图像上，则 ( )

(A) 12t =

，s 的最小值为6π (B) 32t =，s 的最小值为6π

(C) 12t =

，s 的最小值为12π (D) 32t =，s 的最小值为12

π

15. 某条公共汽车线路收支差额y 与乘客量x 的函数关系如下图所示（收支差额=车票收入-支出费用），由于目前本条线路亏损，公司有关人员提出了两条建议：建议(Ⅰ)不改变车票价格，减少支出费用；建议(Ⅱ)不改变支出费用，提高车票价格，下面给出的四个图形中，实线和虚线分别表示目前和建议后的函数关系，则 ( )

(A)

①反映了

建

议

（Ⅱ），③反映了建议

（Ⅰ）

(B) ①反映了建议（Ⅰ），③反映了建议（Ⅱ） (C) ②反映了建议（Ⅰ），④反映了建议（Ⅱ） (D) ④反映了建议（Ⅰ），②反映了建议（Ⅱ）

16. 设函数()y f x =的定义域是R ，对于以下四个命题： （1）若()y f x =是奇函数，则(())y f f x =也是奇函数； （2）若()y f x =是周期函数，则(())y f f x =也是周期函数； （3）若()y f x =是单调递减函数，则(())y f f x =也是单调递减函数； （4）若函数()y f x =存在反函数1

()y f x -=，且函数1()()y f x f x -=-有零点，

则函数()y f x x =-也有零点．

其中正确的命题共有 ( )

(A)1个 (B) 2个 (C) 3个 (D) 4个

三、解答题（本大题共有5题，满分76分）解答下列各题必须在答题纸的相应位置写出必要的步骤．

17. （本题满分14分，本题共有2个小题，第1小题满分6分，

A B

C

M

B 1

C 1

A 1

A

B

C

P

Q D

第2小题满分8分）

直三棱柱111C B A ABC -中，底面ABC 为等腰直角三角形， AC AB ⊥，2==AC AB ，41=AA ,

M 是侧棱1CC 上一点，设h MC =． （1）若C A BM 1⊥，求h 的值；

（2）若2h =，求直线1BA 与平面ABM 所成的角．

18. （本题满分14分，本题共有2个小题，第1小题满分6分，第2小题满分8分）

设函数()2x

f x =，函数()

g x 的图像与函数()f x 的图像关于y 轴对称． （1）若()4()3f x g x =+，求x 的值；

（2）若存在[]0,4x ∈，使不等式(+)(2)3f a x g x --≥成立，求实数a 的取值范围．

19. （本题满分14分，本题共有2个小题，第1小题满分6分，第2小题满分8分）

如图所示，PAQ ∠是某海湾旅游区的一角，其中ο

120=∠PAQ ，为了营造更加优美的旅游环境，旅游区管委会决定在直线海岸AP 和AQ 上分别修建观光长廊AB 和AC ，其中AB 是宽长廊，造价是800元/米，AC 是窄长廊，造价是400元/米，两段长廊的总造价为120万元，同时在线段BC 上靠近点B 的三等分点D 处建一个观光平台，并建水上直线通道AD （平台大小忽略不计），水上通道的造价是1000元/米．

(1) 若规划在三角形ABC 区域内开发水上游乐项目，要求ABC △的面积最大，那么AB 和AC 的长度

分别为多少米？

(2) 在(1)的条件下，建直线通道AD 还需要多少

钱？

20. (本题满分16分，本题共有3个小题，第1小题

满分4分，第2小题满分6分，第3小题满分6分)

设直线l 与抛物线24y x =相交于不同两点A B 、，与圆()()2

2250x y r r -+=>相切于点M ，且M 为线段AB 的中点．

(1) 若AOB △是正三角形（O 为坐标原点），求此三角形的边长； (2) 若4r =，求直线l 的方程；

(3) 试对()0,r ∈+∞进行讨论，请你写出符合条件的直线l 的条数（只需直接写出结果）．

21. (本题满分18分，本题共有3个小题，第1小题满分4分，第2小题满分8分，第3小题满分6分)

已知()y f x =是R 上的奇函数，(1)1f -=-，且对任意(),0x ∈-∞，()11x f x f x x ??= ?-??

都成立．

(1) 求12f ??-

???、13f ??

- ???

的值；

(2) 设1

()()n a f n n

*

=∈N ，求数列{}n a 的递推公式和通项公式；

(3) 记121321n n n n n T a a a a a a a a --=++++L ，求1

lim n n n

T T +→∞的值．

闵行区2016-2017学年第二学期高三年级质量调研考试

数学试卷参考答案与评分标准

一. 填空题 1．4x =； 2．{1,0}-； 3．1； 4．10x y +-=； 5．16； 6．410π； 7．1,12??

-????

； 8．9； 9．

2

9

； 10．32； 11．2,6????； 12．1009；

二. 选择题 13．C ； 14．A ； 15．B ； 16．B ． 三. 解答题

17．[解]（1）以A 为坐标原点，以射线AB 、AC 、1AA 分别为x 、y 、z 轴建立空间直角坐标系，

如图所示，

则)0,0,2(B ,)4,0,0(1A ，)0,2,0(C ，),2,0(h M ……………………2分

)

,2,2(h BM -=，

)4,2,0(1-=C A ……………………4分

由C A BM 1⊥得01=?C A BM ，即0422=-?h

解得1=h ． ……………………6分 (2) 解法一：此时(0,2,2)M

()()()12,0,0,0,2,2,2,0,4AB AM BA ===-u u u r u u u u r u u u r

……………8分

设平面ABM 的一个法向量为(,,)n x y z =r

由00n AB n AM ??=???=??r u u u r r u u u u r

得00x y z =??+=?

A B

C M

B 1

C 1

A 1

x

y

z

所以(0,1,1)n =-r

……………………10分 设直线1BA 与平面ABM 所成的角为θ

则1

1

410

sin 5220n BA n BA θ?==

=??r u u u r r u u u r ……………12分 所以10

sin

5

arc θ= 所以直线1BA 与平面ABM 所成的角为10

sin 5

arc ………………14分 解法二：联结1A M ，则1A M AM ⊥，

1,AB AC AB AA ⊥⊥Q ，AB ∴⊥平面11AAC C …………………8分 1AB A M ∴⊥

1A M ∴⊥平面ABM

所以1A BM ∠是直线1BA 与平面ABM 所成的角； ……………………10分 在1A BM Rt △中，1122,25A M A B == 所以1112210

sin 5

25A M A BM A B ∠=

==

……………………12分 所以110

arcsin

5

A BM ∠= 所以直线1BA 与平面ABM 所成的角为10

sin

5

arc ………………14分 18．[解]（1）由()4()3f x g x =+得2423x

x

-=?+ ……………………2分

223240x x ?-?-=

所以21x =-（舍）或24x

=， ……………………4分 所以2x = ……………………6分 （2）由()(2)3f a x g x +--≥得22

23a x

x +-≥ ……………………8分

2223a x x +≥+2232a x x -?≥+? ……………………10分

而232

23x

x

-+?≥，当且仅当[]4232,log 30,4x x x -=?=∈即时取等号…12分

所以223a

≥，所以21

1log 32

a ≥+

．………………………………14分 19．[解]（1）设AB 长为x 米，AC 长为y 米，依题意得8004001200000x y +=， 即23000x y +=， ………………………………2分

1

sin1202ABC S x y ?=??o y x ??=43 …………………………4分 y x ??=2832

2283??

? ??+≤y x =28125032

m 当且仅当y x =2，即750,1500x y ==时等号成立，

所以当ABC △的面积最大时，AB 和AC 的长度分别为750米和1500米……6分 （2）在(1)的条件下，因为750,1500AB m AC m ==．

由2133

AD AB AC =+u u u r u u u r u u u r

…………………………8分

得2

22133AD AB AC ??=+ ???

u u u r u u u r u u u r

2291

9494AC AC AB AB +?+=

…………………………10分 224411

7507501500()15009929=?+???-+?250000= ||500AD ∴=u u u r

， …………………………12分

1000500500000?=元

所以，建水上通道AD 还需要50万元． …………………………14分 解法二：在ABC ?中，ο120cos 222AC AB AC AB BC ?-+=

22750150027501500cos120=+-??o 7750= ………8分

在ABD ?中，AC

AB AC BC AB B ?-+=2cos 2

22

7

75075021500)7750(750222??-+=

77

2= …………………………10分 在ABD ?中，B BD AB BD AB AD cos 222?-+=

7

7

2)7250(7502)7250(75022?

??-+==500 …………12分 1000500500000?=元

所以，建水上通道AD 还需要50万元． …………………………14分

解法三：以A 为原点，以AB 为x 轴建立平面直角坐标系，则)0,0(A ，)0,750(B

)120sin 1500,120cos 1500(οοC ,即)3750,750(-C ，设),(00y x D ………8分

由2CD DB =u u u r u u u r ，求得?????==3

25025000y x ， 所以()

250,2503D …………10分

所以，2

2)03250()0250(||-+-=AD 500=……………………12分

1000500500000?=元

所以，建水上通道AD 还需要50万元． …………………………14分

20．[解] (1)设AOB △的边长为a ，则A 的坐标为31

(

,)22

a a ±………2分 所以2

134,22a a ??

±=? ???

所以83a =

此三角形的边长为83． ……………………………4分 (2)设直线:l x ky b =+

当0k =时，1,9x x ==符合题意 ……………………………6分

当0k ≠时，224404x ky b

y ky b y x =+??--=?=?

…………………8分

222121216()0,4,42(2,2)k b y y k x x k b M k b k ?=+>+=+=+?+

1

1,AB CM AB k k k k

?=-=Q 22

23225

CM k

k k b k k b ∴=

=-?=-+- 22216()16(3)003k b k k ∴?=+=->?<<

22

54211b r k k -==

=++Q

()230,3k ∴=?，舍去

综上所述，直线l 的方程为：1,9x x == ……………………………10分 (3)(][)0,24,5r ∈U 时，共2条；……………………………12分

()2,4r ∈时，共4条； ……………………………14分 [)5,r ∈+∞时，共1条． ……………………………16分

21．[解]（1）对等式()11x f x f x x ??=

?-??

， 令11(1)12x f f ??

=-?-=-=-

???

所以112f ??

-

=- ???

……………………………2分 令1111222233x f f f ??

??

??=-

?-=-=- ? ? ???

??

??

, 所以11

32

f ??-=- ???

……………………………4分 （2）取1x n =-

，可得111()()1f f n n n =--+，………………6分 即111()()1f f n n n

=+，

所以11()n n a a n n *

+=∈N

1(1)(1)1,a f f ==--=

所以数列{}n a 的递推公式为111

1,()n n a a a n n

*+==∈N ……………………………8分 故

()

13212211111111221!n n n n n a a a a a a a a a a n n n ---????==???=---L ………………10分 所以数列{}n a 的通项公式为1

(1)!

n a n =

-． …………………12分

（3）由（2）1

(1)!

n a n =

-代入121321n n n n n T a a a a a a a a --=++++L 得

11111

0!(1)!1!(2)!2!(3)!3!(3)!(1)!0!

n T n n n n n =

+++++?-?-?-?--?L ……14分

1(1)!(1)!(1)!(1)!

11(1)!1!(2)!2!(3)!3!(3)!(2)!1!n n n n n T n n n n n ??----?=

++++++??-?-?-?--???

L 1012321

11111112(1)!(1)!n n n n n n n n n n T C C C C C C n n ---------???=++++++=?

?--L ……16分 12!

n

n T n +?=

则12

lim

lim 0n n n n

T T 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

北京市海淀区2017届高三英语二模试题答案(最新整理)

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