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2019哈三中二模理科数学题及答案

2019哈三中二模理科数学题及答案
2019哈三中二模理科数学题及答案

2019年哈尔滨市第三中学第二次高考模拟考试

理科数学

本试卷分为第I 卷(选择题)和第II 卷(非选择题)两部分,共24题,满分150分,考试时间120分钟。 注意事项

1.答题前,考生先将自己的姓名、准考证号码填写清楚;

2.选择题必须使用2B 铅笔填涂,非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整,字迹清楚。

3.请按照题号顺序在各题目的答题区域内作答,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。

4.保持卡面清洁,不得折叠,不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。

第I 卷(选择题,共60分)

一、选择题(本大题共12小题,每小题5分,共60分。在每小题给出的四个选项中,只有一项是符合题目要求的)

1. 集合{||1|2}A x x =-<,1

{|

39}3

x B x =<<,则A B = A .(1,2)

B .(1,2)-

C .(1,3)

D .(1,3)-

2.设S n 是公差为(0)d d ≠的无穷等差数列{}n a 的前n 项和,则“d < 0”是“数列{}n S 有最大项”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

3.ΔABC 中,(cos ,sin )m A A =,(cos ,sin )n B B =-,若1

2

m n ?=

,则角C 为 A .

3

π B .

23

π C .

6π D .

56

π 4.已知1

1e

a dx x =

?

,则61

()x ax

-展开式中的常数项为 A .20

B .-20

C .-15

D .15

5.正三棱柱ABC —A 1B 1C 1的所有棱长都为2,则异面直线AB 1与BC 1所成角的余弦值为

A .

12

B .

14

C .

23

D

6.已知函数()sin())(0,||)2

f x x x π

ωφωφωφ=++><

,其图象相邻的两条对称

轴方程为0x =与2

x π

=

,则

A .()f x 的最小正周期为2π,且在(0,)π上为单调递增函数

B .()f x 的最小正周期为2π,且在(0,)π上为单调递减函数

C .()f x 的最小正周期为π,且在(0,)2

π

上为单调递增函数 D .()f x 的最小正周期为π,且在(0,

)2

π

上为单调递减函数

7.一个几何体的三视图及尺寸如右图所示,则该几何体的 外接球半径为

A .

12

B

C .

174

D 8.过抛物线22(0)y px p =>的焦点F 的直线l 与抛物线在第一象限的交点为A ,直线l 与抛物线的准线的交点为B ,点A 在抛物线的准线上的摄影为C ,若AF FB =,

36BA BC ?=,则抛物线的方程为

A .26y x =

B .23y x =

C .212y x =

D .2y =

9.阅读右面的程序框图,输出结果s 的值为

A .

12

B C .

116

D .

18

10.在平行四边形ABCD 中,AE EB =,2CF FB =, 连接CE 、DF 相交于点M ,若AM AB AD λμ=+,则实数 λ与μ的乘积为

A .

14

B .

38

C .

34

D .

43

11.已知函数32()132

x mx m n x y +++=+的两个极值点分别为x 1,x 2,且1(0,1)x ∈,2(1,)x ∈+∞,记分别以m ,n 为横、纵坐标的点(,)P m n 表示的平面区域为D ,若函数log (4)(1)a y x a =+>的图象上存在区域D 内的点,则实数a 的取值范围为

A .(1,3]

B .(1,3)

C . (3,)+∞

D .[3,)+∞

12.设点P 在曲线x y e =上,点Q 在曲线1

1(0)y x x

=-

>上,则||PQ 的最小值为 A

(1)2

e - B

1)e -

C

2

D

第II 卷(非选择题,共90分)

二、填空题(本大题共4小题,每小题5分,共20分。将答案填在答题卡的相应位置上。) 13.若复数1z i =+,则

z

zi

=__________。 14.已知双曲线22

221(0,0)x y a b a b

-=>>的右焦点为F ,由F 向其渐近线引垂线,垂足为

P ,若线段PF 的中点在此双曲线上,则此双曲线的离线率为__________。

15.已知平面区域Ω

=0(,)y x y y ??≥?????≤????,直线l:2y mx m =+和曲线

C:y =有两个不同的交点,直线l 与曲线C 围城的平面区域为M ,向区域Ω内随机投一点A ,点A 落在区域M 内的概率为()P M ,若2

()[

,1]2P M ππ

-∈,则实数m 的取值范围是__________。 16.已知ΔABC 中,∠A,∠B,∠C 的对边分别为a,b,c ,若a = 1,2cosC + c = 2b ,则ΔABC 的周长的取值范围是__________。

三、解答题(本大题共6小题,共70分。解答应写出文字说明、证明过程或演算步骤) 17.(本小题满分12分)

已知正项数列满足24(1)n n S a =+。 (1)求数列{}n a 的通项公式; (2)设1

1

n n n b a a +=,求数列{}n b 的前n 项和T n 。

18.(本小题满分12分)

从某学校高三年级共1000名男生中随机抽取50人测量身高。据测量,被测学生身高全部介于155cm 到195cm 之间,将测量结果按如下方式分成八组,第一组[155,160),第二组[160,165),… ,第八组[190,195]。下图是按上述分组方法得到的频率分布直方图的一部

分、其中第六组、第七组、第八组人数依次构成等差数列。

(1)求第六组、第七组的频率,并估算高三年级全体男生身高在180cm 以上(含180cm )的人数;

(2)学校决定让这50人在运动会上组成一个高旗队,在这50人中要选身高在180cm 以上(含180cm )的三人作为队长,记X 为身高在[180,185)的人数,求X 的分布列和数学期望。

19.(本小题满分12分)

如图,在四棱锥P —ABCD 中,PA ⊥AD ,AB ∥CD ,CD ⊥AD ,AD = CD = 2AB = 2,E ,F 分别为PC ,CD 的中点,DE = EC 。

(1)求证:平面ABE ⊥平面BEF ;

(2)设PA = a ,若平面EBD 与平面ABCD 所成锐二面角[,]43

ππ

θ∈,求a 的取值范围。

20.(本小题满分12分)

已知椭圆2222:1(0)x y C a b a b +=>>过点,离心率12e =,若点00(,)M x y 在椭圆C 上,则点00

(

,)x y N a b

称为点M 的一个“椭点”,直线l 交椭圆C 于A 、B 两点,若点A 、

B 的“椭点”分别是P 、Q ,且以PQ 为直径的圆经过坐标原点O 。 (1)求椭圆

C 的方程;

(2)若椭圆C 的右顶点为D ,上顶点为E ,试探究ΔOAB 的面积与ΔODE 的面积的大小关系,并证明。

21.(本小题满分12分)

已知函数2()ln (0)f x ax x x x a =+->。

(1)若函数满足(1)2f =,且在定义域内2()2f x bx x ≥+恒成立,求实数b 的取值范围;

(2)若函数()f x 在定义域上是单调函数,求实数a 的取值范围; (3)当11x y e <<<时,试比较y x 与1ln 1ln y

x

++的大小。

选考题:请考生从第22、23、24三题中任选一题作答,如果多做,则按所做的第一题记分。 22.(本小题满分10分)

选修4-1:几何证明选讲

如图所示,已知PA 与⊙O 相切,A 为切点,过点P 的割线交圆于B 、C 两点,弦CD∥AP,AD 、BC 相交于点E ,F 为CE 上一点,且DE 2

= EF·EC。

(1)求证:CE ·EB = EF ·EP ;

(2)若CE:BE = 3:2,DE = 3,EF = 2,求PA 的长。

23.(本小题满分10分)

选修4-4:坐标系与参数方程

在极坐标系中,已知直线l 的极坐标方程为sin()14

π

ρθ+

=+C 的圆心是

)4

C π

(1)求圆C 的极坐标方程; (2)求直线l 被圆C 所截得的弦长。

24.(本小题满分10分)

选修4-5:不等式选讲

设函数()|21||3|f x x x =+--。 (1)解不等式()0f x >;

(2)已知关于x 的不等式3()a f x +<恒成立,求实数a 的取值范围。

2019年哈尔滨市第三中学第二次高考模拟考试

数学试卷(理工类)答案及评分标准

一、选择题:

13. 1- 14. 2 15. []1,0 16.

(]3,2

三、解答题:

17. (Ⅰ)整理得21=--n n a a ……………………………… 4分 又11=a 得12-=n a n ……………………………… 6分 (Ⅱ)由(1)知 )1

21121(21+--=

n n b n …………………………… 8分 所以1

2+=

n n

T n …………………………………… 12分 18. 解: (Ⅰ) 第六组

08.0=p ·························

··2分 第七组

06.0=p ·························

··4分

估计人数为

180 (6)

(Ⅱ) X 可能的取值为0,1, 2,

3. ························7分

425

)0(3935===C C x P 4220)1(3

92514===C C C x P 4215)2(391524===C C C x P 422

)3(39

3

4===C C x P 所以X 的分布列

·············10分

)(X E =

3

4

. ····················· 12分

19.(Ⅰ) ,//CD AB ,AD CD ⊥22===AB CD AD ,F 分别为CD 的中点,

ABFD ∴为矩形,

BF AB ⊥ ················· 2分

EF DC EC DE ⊥∴=, ,又EF AB CD AB ⊥∴,// ⊥∴=AE E EF BF , 面BEF ,?AE 面ABE ,

∴平面ABE ⊥平面

BEF ····················· 4分

(Ⅱ) EF DC EC DE ⊥∴=, ,又EF PD //,PD AB CD AB ⊥∴,// 又PD AB ⊥,所以⊥AB 面

PAD ,PA AB ⊥ ··················6分

法一:建系AB 为x 轴,AD 为y 轴,AP 为z 轴,

)0,2,0(),0,0,1(D B ),0,0(a P ,)0,2,2(C ,)2

,1,1(a

E

平面BCD 法向量1(0,0,1)n =,平面EBD 法向量

)2,,2(2-=a a n ··········9分

]22,21[4

52

cos 2∈+=

a θ,可得]5

15

2,552[

∈a . ·············12分 法二:连AC 交BF 于点K ,四边形ABCF 为平行四边形,所以K 为AC 的中点,连EK ,

则PA EK //,⊥EK 面ABCD ,EK BD ⊥, 作BD KH ⊥于H 点,所以⊥BD 面EKH ,

连EH ,则EH BD ⊥,EHK ∠即为所

求 ············· 9分

在EHK Rt ?中,5

15221=?=HK ,]3,1[255

12tan ∈==a a

θ

解得

]5

15

2,552[

∈a ·············12 分

20. (Ⅰ)由已知???

????=+==+211433

2

2222a c c b a b a 解得42=a ,32=b ,方程为

13

42

2=+y x ·······3 分 (Ⅱ) 设),(),,(2211y x B y x A ,则)3

,2(),3,2(2211y

x Q y x P (1)当直线l 的斜率存在时,设方程为m kx y +=

?????=++=13422

y x m kx y 联立得:0)3(48)43(222=-+++m kmx x k 有???

?

?

????

+-=

+-=+>-+=?22212

212243)3(44380)43(48k m x x k km x x m k ① 由以PQ 为直径的圆经过坐标原点O 可得:0432121=+y y x x · 整理得:04)(4)43(221212=++++m x x km x x k ②

将①式代入②式得:22243m k =+, ··········· 6 分

048,0,043222>=?>∴>+m m k

又点O 到直线m kx y +=的距离2

1k

m d +=

2

2

2

2

2222

212

23414334143433411m m

k k m k

k m k k x x k AB ?+=+?+=+-++=-+=

··········

8 分

所以

3232212

2

===?m

m d AB S OAB ·········· 10 分 (2) 当直线l 的斜率不存在时,设方程为m x =(22<<-m )

联立椭圆方程得:4

)4(322

m y -=

代入0432121=+y y x x 得到04)4(3322

=--m m 即552±

=m ,5

152±=y 32

1

2121=-==?y y m d AB S OAB

综上:OAB ?的面积是定值3

又ODE ?的面积3322

1

=??=,所以二者相

等. ·······12分 21. (Ⅰ) 由原式

b x

x x ≥--

?ln 11, ················ 1分

令x

x

x x g ln 11)(--

=,可得)(x g 在(]1,0上递减, 在[)+∞,1上递增,所以0)1()(min ==g x g

0≤b ···············3分

(Ⅱ))

0(,ln 2)(>-='x x ax x f x x a x f ln 2,0)(≥

≥'得令,

x x x h ln )(=

设,

时当e x =e x h 1

)(max =

e

a 21

∴当时,函数)(x f 在),0(+∞单调递增 ···············5分

e a 210<

<若,x a x g x x ax x g 12)(),0(,ln 2)('

-=>-=

a x x g 21

,0)('=

=,0)(),,21(,0)(),21,0(//>+∞∈<∈x g a x x g a x

a

x 21

=

∴时取得极小值即最小值 时而当e a 210<< 021

ln 1)21(<-=a

a g ,

必有根0)(/=x f ,)(x f 必有极值,在定义域上不单调··············8分

e a 21

···········

·····9分 (Ⅲ)由(I)知x

x

x g ln 11)(+-

=在(0,1)上单调递减 ∴

11

<<

时,)()(y g x g >即y

y

x x ln 1ln 1+<+ ················ 10分 而

11

<<

时,0ln 1,0ln 1>+∴<<-x x x

y x y ln 1ln 1++<

∴ ··············· 12分

22.(I )∵EC EF DE ?=2

,∴C EDF ∠=∠,

又∵C P ∠=∠,∴P EDF ∠=∠,∴EDF ?∽PAE ?

∴EP EF ED EA ?=?又∵EB CE ED EA ?=?,∴EP EF EB CE ?=?···5分 (II )3=BE ,29=

CE ,4

15=BP PA 是⊙O 的切线,PC PB PA ?=2

,4

3

15=PA ·······10分

23.(Ⅰ)圆C 的极坐标方程为:

)4

sin(22π

θρ+= (5)

(Ⅱ)圆心到直线距离为1,圆半径为2,所以弦长为2 ··········· 10分 24.(Ⅰ)0)(>x f 的解集为:

),3

2

()4,(+∞?--∞ ·········· 5分

(Ⅱ)

2

13

-

哈三中2018届高中三年级二模英语试题(卷)(有答案)0001

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