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福建省届三数学考冲刺适性模拟卷

5.知函yx切的斜为

R

上奇函,当

f

f(

2

处文科学一

A.

π

B.

D.

(福建省三毕业班复习教指导组泉市执笔整理)

6.执行图所示程序框图,输入

x

,输出

y本试卷共23题满分150分,共页考试用时120分钟注意事项

的值范是A.[

B

1[]41.题前考生将自的姓名、准考号填写在答卡上.2.生作时,答案在答题卡上.按照题号在题的答题区(黑色线框内

C

1[]4

D

作,超答题域书的答无效在草纸、题卷答题效.

7.知某圆的母与底所成角为,截的面积为3.择题案使铅填涂如需动,用橡皮擦净后,再选其它答案标;

4

,该圆的侧积为非择题案使0.5毫的色中性(签)笔或碳素书写字体整、迹清.4.持答卡卡清洁不折叠、不破.考试结束,将本试卷答题卡一并回.

A.

43

B

4

C

D

16一、选择:本大题共12小,每小题分,共60分,在每小题给的四个选项中,有一项是符合题要求的.

8.2020年的爆发之年,月中信通院发布了2020年4月国内手机市场运行分1.复数

z

i

z

在平面对应点位

析报告,报统计从2019年月到年4月这十个月国内手机市场总出货A.一象限

B第二象限

C第象限

D第四限

量与国内5G手机出货量同期手机出货量比重变化情况(简称市场占比得到下2.知集

0},B,则A

面两个统计图,A.

{1,1,2}

B

{}

D.

{1}3.椭圆

x2y9

的点和点分是双线E的点和焦点,的心率是A.

35413BCD.534.、乙丙三门组人员报名参加项志愿者活,已知甲、两部门各报2人丙部门报人若从5人随机抽取3人则人自不同门的概率为A.

13

B

232.D.3105第1页

二、填空:本大题共小题,每小题5分共20分.将答案填在答题卡的相应位置.13.已知向a

,实数t.114.角的点为坐标原,始与轴的非半轴合,边与位圆交点(,b2

,2)2

.15.若椭圆E:

x3

y

的左焦点F,坐标原点O

,定上的任意一,则则下列描述不正确的是

PO

2

PF|

2

的小值_____________.A.20204国内机出货量是十个中的大值B从2019年7月年2月国内机货量持稳增长

16.P

(,f))为函数f()sin(448

)(

上点,知向右平移个位2C相年个2019年半年的国内机市总出量相稳定

后落在C上D2019年月2020年1国内手机市场占比增长比年1月月的增

|

4,kN

*

}率

②在这的,得C上一点向平移后在上49.

ca

③在这的,得C

上点

(,())1212

向平移

56

后在

上A.

ca

cb

B2lnbln

.abb

D.

logab

f(x)

(,)54

单递减则

27410.

ABC

中角

A

的分线

D

,知

2AD

上四个论中所有确结的编.A.

B

3

C

D

三、解答:共70分.解答应写出文字说明,证明过程或演算步骤.第17~题为必考,每个.知函数

x0,fx)f(xx

()f(x)的有零点之和于

试题考生必须作答.第22、题为选考题考生根据要求作.A.

B

5

D

(一)必题:共60分12半径的与正体ABCDACD的个面均切1若DPC,P的轨对应的曲线长为6A.πB.πC.π.3333

为π

的面上动点

17分记S为正项数列和已知aaSnnnn();n(2)T为项等比数列和且bn

.n,,3

,求n的最值.第2页

②公司定店一按天算中销售不低58万天数应不少天结上图分析家分上个度是都有到这规定要求()果你是投资策者,你更意在哪家店资,请你根所学统计识,明你理由1912分如图,在六棱

中,底面ABCDEF

是边长为

的正六边,18分

2.某货公旗下甲、两家店.了调两家店的售

()

在棱上且PB平

,明:

为PE的中;情,现机抽了上年度家店20天日销额(位:元别到甲、两家分日销额的率分直方如下

()PB,点平面PCD

的离.2012分已函数

f)

axln

.()论

f()

的调性()

f)

有个零,求

的值范.()经计得到店日售额的平均数

,差为

33.87

.①计乙日销额的均数同一中的据用组区的中值作表第3页

Ox()当Ox

时求不式

f(

的集;21分已点F

,线l:y

,线l

垂l

于,线段PF的直平线交l

(2)

3,3],(≤,求a的值范围.()求点Q的轨的程;()点H

作的条切线切分别为

记HAB的接圆G,论a

取何,试断以HG为直的圆否恒定点?若是,出该定点坐;若不是,说理由(二)选题:共10分.请考生在第22、题中任选一题答,如果多做则按所做的第一计分.22修44:坐系与参数方10分在面直坐标xOy中坐标原点为极点,轴半轴极轴立极标系线C的坐标程为

sin,曲线C绕点O顺针旋转得曲线.4

2020届高三学考前刺适应性模拟卷()求曲C的坐标程和角坐方程()过点P于A,B两,的小值23修45:不式选讲分)

文科数学试题答案及分参考评分说明.解答给出了种或几种解供参考,如考生的解法本解不同可根试题主要查内比照分标制定应的分细..计算题,当生的解答在一步出现错时,如果后部分解答改变题的容和度,视影的程决定继部的给,但得超该部正确答应给分数的一;如后继分的答有严重错误就不给分3.答右所注数,示考生正确做这一步应得累加分数.4.给整分数选择和填空题不给间分.已函数

f()x

.第4页

f222bf222b一、单项择题:本题共12小题,每小题分,共60分.在每小题给的四个选项中只有一

上的奇函数,可导的奇函数导数是偶函数,所以项是符合目要求的.1..B3.A4.5.A6.B

f

,导数几何义可所求线的率为,

故A7.C.B9.C.A11C12.D21析依题意zi,其在复面内对应的位于第四象,故D.i2析依题意AN,A{1,2},选B

2,6析由程序图可知,函0.1x时y[],选B.4

,制图如下示,合图可知当3

x229

的焦点分为,

5,0

顶分别为

7图是锥的截面意得SAB60是边三形,圆锥面圆径为r,则r,

SABx2y设曲线E:a

,有a5,c

33,其离率e5

1所rrr故C.

,以

r4

,以圆侧面为选A.

8析】因为年月内手机市场出货量和国手机市场占比为4析设甲部的两人为

,2

,部门两人

,2

,部门一人,

十月中最大,所国内手机出量最大,故A正;从随机取人则所有基本件为

{AA,B}1

{A,AB}2

{C}2

从2019年7月到2020年2月国内5G手的市场占比持稳定增长受国内手机{AB,},{C}{C},{,},{A,},{A,}1111

出量影2月国内手机的货量比1月所下,故B错;由图知相比年个2019年半年的内手机市场出货量相对定,{C}2

,10种

故正;年12月2020年1月国内手机市占比增长为3人自不同部门含的基本事为{,,C}

{,C}

{,C}

17.8%26.3%0.48年1月2月的增率为

前者{,C}

,种

大故正;故B则3人来不同门的率为5析】方法一由函

42.故选D.105上奇函可得当

x

f

c9析通过a))(1),构造函数f(x,根其在(0,调ax递,可f()f(b),故A错;以

因与大不能确,故B错误;f

f

,导数几何义可所求

aa)abb

,所a

b

b

,C确线斜率,选A.方二:x,f

x,所f

,为

令c,cc0,D错.故选ab10析】解法一依题sinsin,sinsinADC

,正弦第5页

233434解3cos233434解3cos|PO

△ABD

CDAC①ACD,

在角

△OO,可求得OP1

63

,圆O的长为

23

故选D.将②得

:ACCD

,设

x

x

二、填空:本大题共小题,每小题5分共20分.将答案填在答题卡的相应位置.又为

cos

③由余定理知,

13析解法一由已知

得△ABD中ADB

AD

2BD2AB2AD

2

④△中

,得

t

.ADC

2CD2AC2

2

解二:,构以为边直角三形,又5,,9理,得

,联③④,可得

x

2

,A.

5t,解得t.解二:作AE,因,以ECED,

114析】已知得,sin(+.2由平分定理知,AB:AC:CDx,则ED,x.22在△AED中①

,设

15析】法一由已知,得,设P(,y)则OP||22

,在

AE2AB2

x

43

42x联①②得已可作出数

2故,选A.f

222法:设,F的象的点的坐标别为,

所PO

PF|

3cos

2

等5,选.

2

2

4

2

12析】题意

P

的迹为面

1

的面对的圆

O

t

PF

4t

6

,依意,计算,

OO1

33

D的点为1

64

2min

6424.16162解

三:

线

理,

得第6页

22|PO|2PF2PM

2PMPM

19(,)542

单递减成立所④错误故选③.设3cossin

,0

三、解答:共70分.解答应写出文字说明,证明过程或演算步骤.第17~题为必考,每个试题考生必须作答.第22、题为选考题考生根据要求作.则

3cos

sin

t

(一)必题:共60分17分26cos

332t2t22

48

32

32

记S为正项数列nn();n

项,已

n

2

n

n

.t所

|POPF|2PM

3PM4

52

(2)T为项等比数列n最值.

n

项,且

283

T

n

的【题意】本主要查递数列等差列、比数通项和等础知识;考查算求解推理证等本能;考分类整合化归转化本思;取向数学运算逻辑理核素养解析1)

n

S11

,得

.····················································分当n时,a

n

2

n

n

①可得

n

2

a

n

S

n

②·····················2分16

析由已知得,图象C的周为

2

(kZ)

,一条称轴

142

①—②得

n

2

n

2

an

n

.··························································分故

k

3k4

,以①误;

整得

n

n

n

n

a,以n

n

25分存,,所以②正确4π因图象一条对轴为x,则(,())2

π关于x的对称点为2

所2·····································································分n()题,设为,4n12(

1112

,f())12

,存在,得C

上点

(,(12

12

))

向平移

11512

又q,所以q,·············································································分仍上,所③正;11因时,(x),)4

单递减,),)42542

11时,f(4

n

411

n

2

n

································································分所

T

,第7页

5

,得n,所n的小值为5.··································分

能;或与必的统概率本思;取数据析、学运核心素养.18分

解法一)估计乙店日销额平数为20250.9乙

.························4分某货公旗下甲、两家店.了调两家店的售情,现机抽查了上个度两店天日销额(位:元别到甲、两家店日售额频率布直

②销售超过58万的天占比不少于甲销售不低万概率约为

901360

,··········································6分图下:

58)0.007520乙销售不低万概率约为

,······································分58)200.0100.325

,两均大

14

,店均达到一规的要.···········································分()经计得到店日售额的平均数49,差为.

()案唯一但需合数据与统计率相关知识以说理,方给分.答一:店日售额均值高于店,计算乙店差为,甲店销售况比店要定,以我甲店答二:店日售额均值高于店,频率布直图可,甲的销售额差明低于店,甲店售情比乙要稳,所我选店;答三:然甲日销额平值略于乙,但店日售额万-100万出的概比甲高,我认乙店有潜,所我选店.·········分解法二)同解一.·························································································4分901②销售超过58万的天占比不少于;·········································6分360由店的率分直方可知若甲日销额不于x万时的概不低①计乙日销额的均数同一中的据用组区的中值作表②公司定,店一(按360计算中日售额低于58万的天数应少于90天结合上图,析两分店个年是否有达这一定的求?

14则

,60

0.25580.03

,···························8分()如果是投决策,你更愿意在家店投资,你根据所学统计知识,明你的理

由店的率分直方可知乙店销售不低万元的率约由

0.00520

14

,店均达到一规的要.·············10分【题意】本主要查平数、差、方图础知;考数据理、运算求解本

()同解一.··························································································12分第8页

在六边

ABCDEF

BC4

,以

AO3

2

.19分

又为

2,以PO如图,六棱锥

中,底面

是边长为

的正六边形,

,以

2

2

2

BO

.·································分7.

BO

,

ABCDEF

,()点

在棱PE上且PB平

,明:

为的点;

所PO面.·········································································8分

ABCDEF

,以平

ABCDEF

,()若,点到面PCD

的离.

又为

BE

ABCDEF

,面

PAC

ABCDEFAC

,【题意本主要查线平行线面垂多体的体

BE

,因为

CD

,以

,积点面等基知识查间想象运求解推理证等本能;考转化化归

又为

,以

,得

S

△PCD

.·······················10分形合等本思;取数学算、观想、

E

到面

的离,

E-PCD

P-

S

···················分逻推理核心养.

13

PCD

13

,得hCDE

47

.···········································分解析1)设

CF

BER

,正六形ABCDEF

中易知为BE中点.······················)因PB平面

,面PBE,面

PBE

QR

已函数

f)aln

.所∥QR.·························································································3分因为中,以

为的点.·····················································分

f()()论的调性f)有个零,求()

的值范.【题意】本主要查函单调、零基础识;查运求解、推理论基本能力;考数形合、类与合等本思;取数学算、辑推等核素养.解法一)f

1x

(x.·······································1分()

BEO连结PO.

①a时,ax,以f

f

上减.············分在六边ABCDEF,易BE,CO.又为,所POAC.······························································分第9页

②时由f

1,f可x,a

22x222x2000x所fx在0,上递减在上增.·······································4分

,··········7分()当时由(1)可,减,不可能有个零点.··········分(a2)1②时fln,minaa

令H,H分x而H时

1,g,以g在0,上增,而a

···························10分

······························································································7分

G

xG

x

若f

x当时,g

min

,而f.·························8分

两零点则y与G所的值围是············12分当0时min1在上取x,f2),ae

解法三)解法.····························································································分()题价于程axaxx有两解,即a.x令kx1所fx在

上1个点···························································10分

则f价y····························6分在上,a因ln,

lnx因,由可得0,可所x2e·············································································································分y,过点所fx在上个点.综上所,的取值围为

.············分

当yPy解法二1)解法.·····························································································4分xx()方程axx等于,以x有个零等价xxx于有两解,······································································分xxlnx令Gx,x第10页

lnxyxlnx1则,去和,得00,xx20020即ln.········································10分0

HB2x即22112,HA28222112k,HA128x214x2令,然pHB2x即22112,HA28222112k,HA128x214x2

代上面子,

ax24

2

于x,时切点.···················································11分所当以a的值范围是

同可得的垂线程是

ax4

2

,·····························分·····························································································································12分

联方程得圆坐标

32G()22

·························································8分21分

以HG为径的的方为

axyy

···················分已点Fl:y,直l()求点Q的迹C的程;

垂l

于,线段PF的直平线交l

化整理得

5aax222(过点

作的条切线切分别为

记的外圆为

不论取

ax2

2

y

2

·······················································分何,试断以HG为直的圆否恒过定点?是,求出该点坐标;若是,请说理由【题意】本主要查曲的方、垂平分的性等基知识考查运算求解力;

由的任性,

22

2

2

,体数形合思;取逻辑理、学运和直想象核心养.解)依意,FQPQ,··············································································1分

,得,············································11分yyy所以为径的圆过定点·····················································分假Q点的坐为

,x

y

y

,···········································3分

解二)解法·····························································································分化,得xy所点Q的轨C的程是.··································分()法一假设A(,x),B(,),H1抛线方化成x,求导得,··············································分212x2,中垂HA的斜是HA中坐标是(12x28x2HA的中垂方程11·····································6分14又,x8,xx21第11页

()假设(x),B(,x),H抛线方化成,求导y,··············································分2x2,垂线的斜率是HA中点坐标(),x2xxHA中垂方程1···································6分14又,即8,11ax2代上面子,y1

30000002223000000222同可得的中垂线方是

ax224x

22

,··························7分

的坐标程为

sin,将曲线C绕O顺针旋得曲线C.4联方程得圆坐标G(,1.··························································分22由称性知,点存且必轴,为点M

()曲线的坐标程和直角坐标程;()点P于A两,求PA的小.【题意】本主要查极标方、参方程直角标方、参数几何意基础知识;3aGMa)2

,HMy

········································分

考推理证、算求等基能力考查形结、化与转等基思想;则32aGMy22y221y0·························

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