山东省青岛胶州市2020-2021学年高一下学期期中考试数学试题

—年度第二期期中业水平测高一数学试题本试共页22题全满150．试时120分钟注意项1．答前考务将己姓、生等写答卡试指位上并条形粘在题指位上2．回选题，出小答后用笔答卡对题的案号黑如需动用皮干后再涂他案号回非择时将案在题上。在试上效3．考结后请答卡交一、项择：大共小．小分共分在小给的个项，只有项符题要的1．

ππcos的值（）121236A．B．22

C．

64

D．

342四边BCD为形长4AB,ADBDa）A．

B．

C．

D．

3．已i虚单，列i相的（）A．C．

i11

B．D

(1i)(1i)i4

4已知

BC

3π)，为坐原下说正的）2A．

AB

B．

O,

三点线C．

,B

三点线

D

OBOC5已知

,B

是

的内，量

(sinA,sinBn(cos)

且

m

与

n共线则以ABC的状（）A．等三形

B．等腰角角

C．直三形

D．边角1

6．已复

m

m

m3)i

，

z3i

，其

i

为虚单，

，若

为纯数则列法确是）A．

B．复数

z

在复面对的在一限C．

2

D

|2z|7．如所，测山

MN

，选

和另座的山

C

为测观点从

点测

M

点的角C点仰CAB30C点得MCA60

以及，若高

2

米，山

等于）A．C．

300

米米

B．D

360

米米8设

tan124044,c1tan122

则

a,

大小系确是（）A．

c

B．

C．

D．

二、项择：大共小．小分共分在小给的个项，有多符题要．部对得5，对不的分有错得0分9关于组本据平数数率布方和差说法确）A．改其一数，均和位都发改B．频率布方中中数边右的方的积该等．若据频分直图单不称且左“尾均数于位D．本据方越，明本据离程越10．已平向

m(1,3)，2aa|

，则）A．

m

B．

m3

或

Ca与夹角大为

56

D

|a|2

222[80,90)222[80,90)．ABC，A,,C对分c

，

b13

．则下列法确是）A．

为锐三形

B．

面积

或

C．

AB

长度

6

D

ABC

外接的积

π12．下说正的（）A．在

ABC

中，

sinsinB

是

BCAC

的充条B．将函

x

的图向平

π

π个单长得函y)3

的图C．存实

x

，使等

32

成立D．在ABC中，sinAsinBsinC，ABC是角角三、空：大共小题每题分共分．13某任统5上步到位花时为8,12,9这组据标差．14．函ysinx3x在区[0,]的域．15．在中已知10，ABBC

．名高学的育试16．右为校成绩频分直图如要照层抽方抽取名学进分，则要取之的生数；估这000名学的育

频率组试平成为．

（小第一空分，第空分）

506070100

成绩3

四、答：大共小题共70分解应出字明证过或算骤17本题分分某蔬基准对有铺道施装能制泵统对棚菜行动制根滴，以大省力资，铺道达满最压态，泵动机，道以动续行注作最工压然水会新启不性的道系根最工压要置之力能对的泵统不品的菜于要的注度强不，以择备同能管系和泵统为分解地内有道统总情，随抽不蔬棚的干管进满测，这些道最工压据单：帕分

，将其按左右顺分编为一，二，

，第组下是据验据成的率布方，知a,b（）求的值

2ba

，

a,bR

．（）已最工压10以的道统需分配2台功水，第组第二共有条道求基需配的功水的．频率组b0.080.04

6

810

最小工作压（千帕）4

18本题分分已知a(cos2(1,sin数单，数z对应点．1

，

m

在平坐系，i

为虚（）|

；（）为曲|z|

(

z

为

z

的共复)上动，

Z

与

之间最距；（）

π

，求a上的影量．19本题分分已知数f(x2(x

π5)3sin(xxR)126

．（）将数

f(x)

化为

Asin(

形式求

f(x)

的最正期

和单递区；（）为的角f(

)恰为f(x)

的最值；π（）若tan()6

，求

f(

)

．20本题分分在ABC中a,bc

分别角

,BC的对，ccosB

，sin

23

．（）求

cos

；（）若

5，PCA延线一，接PB，BP，PBC的积5

，π，π21本题分分在

中，

M

为

所在面的点ACBAC

π4

，1MC，NANC3

．（）以

和作一基表NM，并|

；（）D为线MN上点设CDxABy(xR)，若线CD经ABC垂心求．22本题分分在平直坐系，知

2(,),(8m),Ct2

，

t,t

．（）若

t

，

为

x

轴上一点点

．（ⅰ当

三点线，点

的坐

；（ⅱ求

||

的最值（）若

sin

)且CA与CB的夹)2

，求

的取范．6

—年度第二期期中业水平测高一数学答案及评分标一、项择：大共小．小分共分1--8：BCDACAD二、项择：大共小．小分共分9．；

．AC

11．BD；

12．ABD三、空：大共小题每题分共分．13．

；

．

[3,2]

；

15．

152

；

16（）

40

2）

73

．四、答：大共小题共70分解应出字明证过或算骤17（本小题满分）解）根分列特，

(a0.080.08

·······················分因为a

23

，所0.18，a0.12··························································5（）第组第组频为

(0.120.08)0.4

········································分所以管总数

200

500

·································································7所以小作在以的道统

500(0.08120

条·················分所以基需配的功水的数18.（小题分分）

n

································分解）bsin

22sin

cos

······································································分所以

mb

··········································································2分所以z

3i)3i4i2i1

·············分所以z5

·················································································分（）z曲|zz|

··························································································分，4i)，因此线复面(2,

圆心半为

的圆·········································分7

a164a164故Z与之间距为1

(237

·····································7分所以与1（）因

之间最距为37···························································8分π，以b(1,)················································分2此时

b，a与的夹角余为

ab|b|

···································分与方向同单向为e

b4)||17

··········································分所以在上投向na|,)171719（本小题满分）

······································12分解）f)2(x

π5)3sin()(R)126

πcos(2x)πππ3)3sin(2x)x)6ππ2[sin(2)]66πππ2sin(2x)2sin(2x)63

·····················································3分所以

fx)

的最正

22

······························································分由

ππππ5kxπ，πxπ23212所以

fx)

的单递区[

π5π](kZ)12

···································分（）f

π)3因为

0

ππ5，所π33

······················································6分ππ所以32

······················································································7即

512

时，

f(

恰为

fx)

的最值······················································8分8

（）f

)3

4sin()tan()6πsin()2)2()6

··············································································································10分因为tan(

π)所f6

4)6tan2)6

85

·····················分20（本小题满分）解）由意

cosB

，根正定，得

sinCcosB

···········分所以

sin(B)sinBcosCCcosB即

sincosC

，即

sinB也即

B)

·····················································································2分因为

，所

B

，即

B

·············································分所cosA)B2sin

B)

49

·················分（）B，所b由余定知a

2

2

2

cos

2

12()9

···························分解得

b

···························································································分因cosBAC

19

································································分在中因所BCP所BCA

πBAC2

)BAC

·································分又因

459

···········································································分所以

PBC

14405BAC5=···················12分2921（本小题满分）解）由

13

，所

M

为线

BC

上靠

的三分······················1分由

，所

为线

AC

的中····················································2111NMNCCB(AC)2336

·············分9

所|NM|(ACAB所|NM|(ACAB因为ABcos

22

··································5分11221769363（）D为线MN上一，NDNM

···············6分则CDCNND

1111NMAC(AC)2231kk)3

·············································································分1111kAB)ACkABk)AC3211πkk)364

·····························································8分因CD直经的垂，以AB，CDAB

·····················9所以

1πk)k)2324解得

34

······························································································10分11所以CDAB)ABAC368因为

xABy

，所

13x,48

·················································分22.（小题分分）解）

P(x,0)tm，以B(4,2)

······························1分因为

1,2),A

(3,4)，A·········································2所以

x

，解x

52所以

三点线，

P

5的坐(,0)2

···············································分（ⅱ因

关于

x

轴的称为

···············································4分所以

||PBPA

|