2021福州市第二次質檢數(shù)學文試題及答案

第題2021州市第二次質檢數(shù)學文試題及答案第題文科數(shù)學能力測試（卷刻120分鐘滿：分）注事：1．科考試分試題卷答題卷，考生須在答題卷上作答，答題前，請在答題卷的密封線內填寫學校班級、準考證號、姓名；2．本試卷分為第Ⅰ卷選擇題）和第Ⅱ卷（非選擇題）兩部分，全卷滿分150分考試時刻120分．參公：1．樣本數(shù)據(jù)

x,,x

的標準差

2．柱體體積式：，s其中x樣本平均數(shù)；

，

其中S為底面面，h高；3．錐體體積式：Sh，第卷選擇

其中S為底面面積，共60分）

為高一選題本大共小題，小5，60分在小題給四選項有只一選是正的把正選涂在題的應位上）1．函數(shù)y的義為A

B

．

D．2．若復滿足為數(shù)位），的虛部為A

2B5

．

D．

3．如圖某籃球聯(lián)賽中，甲、乙兩名運動員個場次得分的莖圖．設甲、乙兩人得分的平均數(shù)分別,x，中數(shù)分別為,m，甲乙甲乙Am甲乙甲乙

Bxm甲．x,mmD．xx,m甲乙甲乙甲乙甲乙x4．已知線x為雙曲線（a）一條漸近線則雙曲線的b離心率為A

32

B

52

．2D．5

．．5．執(zhí)行圖所示的程序框圖，輸出的有序實數(shù)對為A．．B

開xy．D．

是

否

輸

,y

6．已知線l與面平，則下列結錯的A直線l與平面沒公共點B存在通過直線l的面與平面平．直l與面內的意一條直線平行

y第題

結D．直線

l

上所有的點到平

的距離都相等7．已知函數(shù)(x)滿足：當,21af,bfcf(3)，a,b,的小關系為A

Bb

．

D．8．設變x,y滿足約束條件

yx,x2,則zx的值范疇2xyA

B

．

D．9．某四錐的三視圖如圖所示，該四棱錐的表面積為A

2B．14D．x0,10．函數(shù)f)，x

的零點個數(shù)為

4正視俯視

3側視第題圖A0B1C．D11．在ABC中G為ABC重心知3向GA與GB的夾角1則CA的小值是A

B6C9D12．已知函數(shù)有列三個結論①存在常數(shù)，對任的實數(shù)，恒有f②對任意給定的數(shù)M，都存在實數(shù)，使得f0

0

M；③直線y與數(shù)f且點許多多個．則所有正確結論序號是A①

B②

．③

D．②③

第卷（選題共分）二填空題本題共4小題，小題，共分把答案在題卡相位上13．已知集合★★★．14．已知函數(shù)(x)．在間一數(shù)，使得等式(x)成的概率為★★★．15．ABC的角,C所的分別是b,c．1052abcos105

的值為★★★．16．在各項均為正整數(shù)的單遞增數(shù)列

n

中，，a2

，且

a

kN

*則a的為★★．三解題本題小題共74分解承寫出字明、明程演算程)17（本小滿12分）已知函數(shù)f(x)3距離為．

cos

(0)的圖象與直線的相鄰兩個交點之間的（Ⅰ）求函數(shù)f)

的單調遞增區(qū)間（Ⅱ）若f

π，cos2的．318（本小滿12分）調查說明，中年的成就感與收入、學歷、職業(yè)中意度的指標有極強的相關性．將這三項的中意指標分別記為y，對它們進行量化示不中意表示差不多中意，示中意，再用綜合指標的評中年人的成就感等級：若w

4則成就感為一級若

，成就感為二級；0w，成就感為三級．為了了解目前某群體中人的成就感情形，研究人員隨采訪了該群體的名年人，得到如下結果：人員編號z人員編號

A1

A

A

A

A

x,yz

（Ⅰ）若該群體200人，估量該群體中成就感等級為三級的人數(shù)是多少？（Ⅱ）從成就感級為一級的被采訪者中隨機抽兩人，這兩人的綜合指標均的概率是多少？

19（本小滿12分）如圖在長方ABCDAC中，ABBCAA4，P為

1

P

C

1線段上點．

A

1

B

1（Ⅰ）求證：；（Ⅱ）當P線段D的點時，求三棱錐APBC的．

CA

B20（本小滿12分）

第題圖小輝是一位收藏好者，在第年初買了價值為元的收藏品M，于受到收藏品市場行情的礙，第2年第3年每初M的值為上年初的每年初M的價比上年初增加萬.（Ⅰ）求第幾年開始M價值超過原購買的價值；

；從第4年始，（Ⅱ）記

T

n

（

*

）表示收藏品M前年價的平均值，求

T

n

的最小值．21（本小滿12分）已知函數(shù)fx)

e

x

,mR,

為自然對數(shù)的底．（Ⅰ）若x是f()的極值點，求的值；（Ⅱ）證明：當0時，ee．22（本小滿14分）如圖，已知橢圓

1）的心率e．點,A分為橢圓a2左焦點和右頂點且（Ⅰ）求橢圓方；

Q（Ⅱ）過點F作一條直線l交圓,Q兩點，點關

F

O

A

xx軸的對點為Q

．若PF∥

，求證：

12

．

Q'第題圖

年州市高中業(yè)班質量檢測數(shù)測考答評則一選題本題有12個題每小5分，分分1A．A3C4．D5D6．7．．．10B11．12．二填題：大共小，小題分滿分．13．

1415．216．55三解題本大共6小，共74分．17．本題緊考查三角函數(shù)的圖象與性質（稱性、周期性、單調性）、倍角的余弦公式等基礎知，考查運算求解能力，考查數(shù)結合思想、化歸與轉化思想、函與方程思想．滿分12分【解析】（Ⅰ）為f()(0,，因此

f()2sin()···········································································2分因此f()．max因為函數(shù)x)與線的鄰兩個交點之間距離，因此T·····························································································因此

得

···········································································4分因此

f())6令2k

x

k

,Z，···························································5分解得

,．··································································因此函數(shù)x)

的單調遞增區(qū)間[k

,],kZ．·································7（Ⅱ）由（Ⅰ），f

，為f6

，因此．·················································································63因此cos2································································10分6

2·····························································11分67．············································································分918．本小題要緊考查概率、計等基礎知識，考查數(shù)據(jù)處理能力、運算求解能力、應用意識，考查必定與然思想．滿分12分【解析）運算10名被采訪者綜合指標，可得下表：

9C11AB212人員編號AAAA9C11AB212綜合指標46243··············································································································1由上表可知：成感為三級（即0）只有A一位，頻率為．··········3分101用樣本的頻率估總體的頻率估該群體中成就感等級為三級的數(shù)有20．10··············································································································（Ⅱ事件為“從成就感等級是一級的被采訪者中隨機抽取兩人們的合指標w均為4（）可知成就感是一級的（）：,AA,AAA，6位從中隨機抽取兩人，有可能的結果為：A15種··················································其中綜合指標w有,，共3名事件發(fā)的所有可能結果為A種，··························分因此P()

31．················································································12分1519小題要緊考查直線與直線直線與平面的位置關系及幾何體的體積等基礎知識，考查空間想象能論證能力求解能力數(shù)形結合思想與化思想分分．證明：（Ⅰ）連BD．因為ABCDABC是方體，且AB2，因此四邊形ABCD是正形，·······································································因此ACBD．························································································2分因為在長方體BCD中，BB平面，AC面，因此．···························································分A11因為BD面BBD平DD且BDB，因此面D．················································分因為BP平BBD，D因此ACBP·······························································6（Ⅱ）點P到面ABC的離AA，1ABC的積····························································7分118因此V4=．························································8分33在eq\o\ac(△,Rt)P中BB4,BP2,因此BP，··········································9分1同理又BC=2因此PBC的面積S217．··10分設三棱錐APBC的為，

3183因為V，此，·····················································分331717因此h，解得h317

．817即三棱錐的高···································································分1720本小題要緊考查數(shù)等數(shù)列等比數(shù)不等式等基礎知識考查運算求解能力、抽象概括能力、用意識，考查函數(shù)與方程思想化歸與轉化思想、分類與整合思．滿分分．解：（Ⅰ）設第年的價值為a，依題意，當3時數(shù)列20，公為

12

的等比數(shù)列，因此

．故10，a，此aa．··············································································································當n4時數(shù)列為首項，公差為的差數(shù)列，又a，此·········································································令a，

274

，又因為n

*因n．··········································4分因此，第年初M開的價值超過原購買的價值．······································（Ⅱ）設S表示前年初M的值的和，則Tnn由（Ⅰ）知，當13時

40

，T

40n

①；··············································································································當n4時由于S35，

2

nn，

2

n32n．································································nn35當n，由①得，，T，此················10分當n4時由②知，2nn

322n

11當且僅當2n

32n

，即時等號成立．即．···································································11分由于T，故在第4年T的值最小，其最小值為．································分21本小題要緊考查函導不式等基礎知考查推理論證能力抽象括能力、運算求解能力，查函數(shù)與方程思想、數(shù)形結合想、化歸與轉化思想．滿分分．

4x4解：（Ⅰ）因為f()，因f

e

x)(e

，·············由x是f()的極值點，得

m)

，············································解得m，····························································································現(xiàn)在f(，檢驗，x是f(x)的值．e因此所求的實數(shù)m的為．·······································································4分（Ⅱ）證明：取m，(x)

e

，現(xiàn)在f

e)

．····················6分構造函數(shù)h))，········································································因此

(1x)

(

在恒，因此h(x)單調遞減，···································································8分因此(x)(0),···················································································9分故

在成立，說明f(x)在單調遞減．···············10分e因此當

b時，，因為因此0,e，e因此

(e

，···········································································11分因此b

e

成．········································································12分22．本小題要緊考查直線、圓等基礎知識，考查推理論證能力、運算求解能力，考查函數(shù)與方程思想、形結合思想、化歸與轉化思想分類與整合思想．滿分14分．【解析】（Ⅰ）橢圓半焦距為c，

c1e,23,

··································································2分解得c，······················································································3分因此················································································4分因此橢圓方程為4

．································································（Ⅱ方一依題意得，PQ與標軸不垂直設Q與Q關于x對,因此Q）論可知，因為PFAQ直線FQ與線相，故，············1解得．····························································································235又因為點x,y在橢圓，此y，或y

34

5．························3y由橢圓對稱性，妨取y5，直線的率4x2

．

4881x814881x8181因此直線方為

52

····························································

10分由

x,x12,

3得點P坐為,

．·············································

11分因此PF

8164

，················

12分

16

13分1因此PFAQ··················································································2方法二：依題意，PQ與坐標軸不垂直．設l程為y）yy因為點與點軸對,因

14分又因為橢圓關于x軸對，因此點Q

也在橢圓上·······································由

312,

消去得．因此x

k4x3

．·················································因為PFAQ

，因此直線

的方程為由消y312,

k

x

．因為直線AQ

交橢圓于

x8k因此，故．························································9分8884k因此xx,x3k

，解得

57x．44因此

k．·············································································11分3k2因此PF分64

分161因此PFAQ··················································································分2方法三：依題意得PQ坐標軸不垂直．設l方為y）y因為點與點軸對,因

3k3又因為橢圓關于軸對稱，因此3k3

也在橢圓上·······································x由消x得3y312,

．6k因此y．········································································3k因為PFAQ直線為x,由消x得，3y312,

．因為直線AQ

交橢圓于

x因此

k12，即3k3k

．·····························································設=）則y，此y．分6因此y，得，············································13分3k11因此FPPF22

．···························································14分方法四：依題意得PQ坐標軸不垂直．設l方為y）y因為點Q與軸稱因Q·············································6分因為,F,Q三共線，因此yy因此．·····································································7分因為PFAQ可設FP=），因此x·································································8分因