山東省青島膠州市2020-2021學年高一下學期期中考試數(shù)學試題

—年度第二期期中業(yè)水平測高一數(shù)學試題本試共頁22題全滿150．試時120分鐘注意項1．答前考務將己姓、生等寫答卡試指位上并條形粘在題指位上2．回選題，出小答后用筆答卡對題的案號黑如需動用皮干后再涂他案號回非擇時將案在題上。在試上效3．考結后請答卡交一、項擇：大共?。》止卜衷谛〗o的個項，只有項符題要的1．

ππcos的值（）121236A．B．22

C．

64

D．

342四邊BCD為形長4AB,ADBDa）A．

B．

C．

D．

3．已i虛單，列i相的（）A．C．

i11

B．D

(1i)(1i)i4

4已知

BC

3π)，為坐原下說正的）2A．

AB

B．

O,

三點線C．

,B

三點線

D

OBOC5已知

,B

是

的內(nèi)，量

(sinA,sinBn(cos)

且

m

與

n共線則以ABC的狀（）A．等三形

B．等腰角角

C．直三形

D．邊角1

6．已復

m

m

m3)i

，

z3i

，其

i

為虛單，

，若

為純數(shù)則列法確是）A．

B．復數(shù)

z

在復面對的在一限C．

2

D

|2z|7．如所，測山

MN

，選

和另座的山

C

為測觀點從

點測

M

點的角C點仰CAB30C點得MCA60

以及，若高

2

米，山

等于）A．C．

300

米米

B．D

360

米米8設

tan124044,c1tan122

則

a,

大小系確是（）A．

c

B．

C．

D．

二、項擇：大共?。》止卜衷谛〗o的個項，有多符題要．部對得5，對不的分有錯得0分9關于組本據(jù)平數(shù)數(shù)率布方和差說法確）A．改其一數(shù)，均和位都發(fā)改B．頻率布方中中數(shù)邊右的方的積該等．若據(jù)頻分直圖單不稱且左“尾均數(shù)于位D．本據(jù)方越，明本據(jù)離程越10．已平向

m(1,3)，2aa|

，則）A．

m

B．

m3

或

Ca與夾角大為

56

D

|a|2

222[80,90)222[80,90)．ABC，A,,C對分c

，

b13

．則下列法確是）A．

為銳三形

B．

面積

或

C．

AB

長度

6

D

ABC

外接的積

π12．下說正的（）A．在

ABC

中，

sinsinB

是

BCAC

的充條B．將函

x

的圖向平

π

π個單長得函y)3

的圖C．存實

x

，使等

32

成立D．在ABC中，sinAsinBsinC，ABC是角角三、空：大共小題每題分共分．13某任統(tǒng)5上步到位花時為8,12,9這組據(jù)標差．14．函ysinx3x在區(qū)[0,]的域．15．在中已知10，ABBC

．名高學的育試16．右為校成績頻分直圖如要照層抽方抽取名學進分，則要取之的生數(shù)；估這000名學的育

頻率組試平成為．

（小第一空分，第空分）

506070100

成績3

四、答：大共小題共70分解應出字明證過或算驟17本題分分某蔬基準對有鋪道施裝能制泵統(tǒng)對棚菜行動制根滴，以大省力資，鋪道達滿最壓態(tài)，泵動機，道以動續(xù)行注作最工壓然水會新啟不性的道系根最工壓要置之力能對的泵統(tǒng)不品的菜于要的注度強不，以擇備同能管系和泵統(tǒng)為分解地內(nèi)有道統(tǒng)總情，隨抽不蔬棚的干管進滿測，這些道最工壓據(jù)單：帕分

，將其按左右順分編為一，二，

，第組下是據(jù)驗據(jù)成的率布方，知a,b（）求的值

2ba

，

a,bR

．（）已最工壓10以的道統(tǒng)需分配2臺功水，第組第二共有條道求基需配的功水的．頻率組b0.080.04

6

810

最小工作壓（千帕）4

18本題分分已知a(cos2(1,sin數(shù)單，數(shù)z對應點．1

，

m

在平坐系，i

為虛（）|

；（）為曲|z|

(

z

為

z

的共復)上動，

Z

與

之間最距；（）

π

，求a上的影量．19本題分分已知數(shù)f(x2(x

π5)3sin(xxR)126

．（）將數(shù)

f(x)

化為

Asin(

形式求

f(x)

的最正期

和單遞區(qū)；（）為的角f(

)恰為f(x)

的最值；π（）若tan()6

，求

f(

)

．20本題分分在ABC中a,bc

分別角

,BC的對，ccosB

，sin

23

．（）求

cos

；（）若

5，PCA延線一，接PB，BP，PBC的積5

，π，π21本題分分在

中，

M

為

所在面的點ACBAC

π4

，1MC，NANC3

．（）以

和作一基表NM，并|

；（）D為線MN上點設CDxABy(xR)，若線CD經(jīng)ABC垂心求．22本題分分在平直坐系，知

2(,),(8m),Ct2

，

t,t

．（）若

t

，

為

x

軸上一點點

．（ⅰ當

三點線，點

的坐

；（ⅱ求

||

的最值（）若

sin

)且CA與CB的夾)2

，求

的取范．6

—年度第二期期中業(yè)水平測高一數(shù)學答案及評分標一、項擇：大共?。》止卜?--8：BCDACAD二、項擇：大共?。》止卜?．；

．AC

11．BD；

12．ABD三、空：大共小題每題分共分．13．

；

．

[3,2]

；

15．

152

；

16（）

40

2）

73

．四、答：大共小題共70分解應出字明證過或算驟17（本小題滿分）解）根分列特，

(a0.080.08

·······················分因為a

23

，所0.18，a0.12··························································5（）第組第組頻為

(0.120.08)0.4

········································分所以管總數(shù)

200

500

·································································7所以小作在以的道統(tǒng)

500(0.08120

條·················分所以基需配的功水的數(shù)18.（小題分分）

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)，因此線復面(2,

圓心半為

的圓·········································分7

a164a164故Z與之間距為1

(237

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

之間最距為37···························································8分π，以b(1,)················································分2此時

b，a與的夾角余為

ab|b|

···································分與方向同單向為e

b4)||17

··········································分所以在上投向na|,)171719（本小題滿分）

······································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)

的單遞區(qū)[

π5π](kZ)12

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

π)3因為

0

ππ5，所π33

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

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

512

時，

f(

恰為

fx)

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

（）f

)3

4sin()tan()6πsin()2)2()6

··············································································································10分因為tan(

π)所f6

4)6tan2)6

85

·····················分20（本小題滿分）解）由意

cosB

，根正定，得

sinCcosB

···········分所以

sin(B)sinBcosCCcosB即

sincosC

，即

sinB也即

B)

·····················································································2分因為

，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（本小題滿分）解）由

13

，所

M

為線

BC

上靠

的三分······················1分由

，所

為線

AC

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

·············分9

所|NM|(ACAB所|NM|(ACAB因為ABcos

22

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

···············6分則CDCNND

1111NMAC(AC)2231kk)3

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

·····························································8分因CD直經(jīng)的垂，以AB，CDAB

·····················9所以

1πk)k)2324解得

34

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

xABy

，所

13x,48

·················································分22.（小題分分）解）

P(x,0)tm，以B(4,2)

······························1分因為

1,2),A

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

x

，解x

52所以

三點線，

P

5的坐(,0)2

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

關于

x

軸的稱為

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

||PBPA

|