2021福州市第二次質(zhì)檢數(shù)學(xué)文試題及答案

第題2021州市第二次質(zhì)檢數(shù)學(xué)文試題及答案第題文科數(shù)學(xué)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(nèi)填寫學(xué)校班級、準(zhǔn)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(biāo)準(zhǔn)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(yīng)位上）1．函數(shù)y的義為A

B

．

D．2．若復(fù)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(shè)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?qū)崝?shù)對為A．．B

開xy．D．

是

否

輸

,y

6．已知線l與面平，則下列結(jié)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(nèi)的意一條直線平行

y第題

結(jié)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(dāng),21af,bfcf(3)，a,b,的小關(guān)系為A

Bb

．

D．8．設(shè)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(cè)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(jié)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(jié)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(diào)遞增區(qū)間（Ⅱ）若f

π，cos2的．318（本小滿12分）調(diào)查說明，中年的成就感與收入、學(xué)歷、職業(yè)中意度的指標(biāo)有極強的相關(guān)性．將這三項的中意指標(biāo)分別記為y，對它們進(jìn)行量化示不中意表示差不多中意，示中意，再用綜合指標(biāo)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(jié)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(biāo)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(dāng)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(dāng)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(guān)

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(zhì)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(zhì)（稱性、周期性、單調(diào)性）、倍角的余弦公式等基礎(chǔ)知，考查運算求解能力，考查數(shù)結(jié)合思想、化歸與轉(zhuǎn)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(diào)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(chǔ)知識，考查數(shù)據(jù)處理能力、運算求解能力、應(yīng)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(biāo)，可得下表：

9C11AB212人員編號AAAA9C11AB212綜合指標(biāo)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(biāo)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(jié)果為：A15種··················································其中綜合指標(biāo)w有,，共3名事件發(fā)的所有可能結(jié)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(guān)系及幾何體的體積等基礎(chǔ)知識，考查空間想象能論證能力求解能力數(shù)形結(jié)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(shè)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(chǔ)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(zhuǎn)化思想、分類與整合思．滿分分．解：（Ⅰ）設(shè)第年的價值為a，依題意，當(dāng)3時數(shù)列20，公為

12

的等比數(shù)列，因此

．故10，a，此aa．··············································································································當(dāng)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(shè)S表示前年初M的值的和，則Tnn由（Ⅰ）知，當(dāng)13時

40

，T

40n

①；··············································································································當(dāng)n4時由于S35，

2

nn，

2

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

322n

11當(dāng)且僅當(dāng)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(dǎo)不式等基礎(chǔ)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(jié)合想、化歸與轉(zhuǎn)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(gòu)造函數(shù)h))，········································································因此

(1x)

(

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

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

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(chǔ)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(jié)合思想、化歸與轉(zhuǎn)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(biāo)軸不垂直設(shè)Q與Q關(guān)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(biāo)軸不垂直．設(shè)l程為y）yy因為點與點軸對,因

14分又因為橢圓關(guān)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(biāo)軸不垂直．設(shè)l方為y）y因為點與點軸對,因

3k3又因為橢圓關(guān)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(shè)=）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(biāo)軸不垂直．設(shè)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(shè)FP=），因此x·································································8分因