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2017届上海市闵行区高三二模数学卷(含答案)

2017届上海市闵行区高三二模数学卷(含答案)
2017届上海市闵行区高三二模数学卷(含答案)

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闵行区2017届第二学期高三年级质量调研考试

数 学 试 卷

(满分150分,时间120分钟)

考生注意:

1.答卷前,考生务必在答题纸上将学校、班级、考生号、姓名等填写清楚.

2.请按照题号在答题纸各题答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效. 3.本试卷共有21道试题.

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 = .

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 ,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 '=

,O 是坐标原点,则PQ 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 =___.

x y

O

x y

O

x y

O

x

y

O

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 、分别是两条异面直线12l l 、的方向向量,向量a b 、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 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) 32

t =,s 的最小值为12π 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,

C

Q 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,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

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.

A B

C

M

B 1

C 1

A 1

(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

22

50x 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 --=++++ ,求1

lim

n n n

T T +→∞的值.

闵行区2016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.

29; 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

B 1

C 1

A 1

z

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 ===-

……………8分

设平面ABM 的一个法向量为(,,)n x y z =

由00n AB n AM ??=???=?? 得00x y z =??+=? 所以(0,1,1)n =-

……………………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?===

?? ……………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1

AM AM ⊥, 1,AB AC AB AA ⊥⊥ ,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,1122,25AM AB == 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

1sin1202ABC S x y ?=

?? y x ??=4

3 …………………………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 =+

…………………………8分

得2

22133AD AB AC ??=+ ???

2291

9494AC AC AB AB +?+=

…………………………10分 224411

7507501500()15009929=?+???-+?250000= ||500AD ∴= , …………………………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, 120cos 222AC AB AC AB BC ?-+=

22750150027501500cos120=+-?? 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( C ,即)3750,750(-C ,设),(00y x D ………8分 由2CD DB = ,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,22

440

4x 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

?=-=

22

23225

CM k

k k b k k b ∴=

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

22

54211b r k k -==

=++

()230,3k ∴=?,舍去

综上所述,直线l 的方程为:1,9x x == ……………………………10分 (3)(][)0,24,5r ∈ 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1111222

233x 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

()

132122111111

11221!n n n n n a a a a a a a a a a n n n ---????==???=

--- ………………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 --=++++ 得

11111

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

n T n n n n n =

+++++

?-?-?-?--? ……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 ??----?=

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

1012321

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

?-- ……16分 12!

n

n T n +?=

则12

lim

lim 0n n n n

T T n +→∞→∞== ……………………………18分

2018上海高三英语二模翻译汇总(含答案)

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2018年上海高三数学二模分类汇编

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