2021福州市第二次质检数学文试题及答案

第题2021州市第二次质检数学文试题及答案第题文科数学能力测试（卷刻120分钟满：分）注事：1．科考试分试题卷答题卷，考生须在答题卷上作答，答题前，请在答题卷的密封线内填写学校班级、准考证号、姓名；2．本试卷分为第Ⅰ卷选择题）和第Ⅱ卷（非选择题）两部分，全卷满分150分考试时刻120分．参公：1．样本数据

x,,x

的标准差

2．柱体体积式：，s其中x样本平均数；

，

其中S为底面面，h高；3．锥体体积式：Sh，第卷选择

其中S为底面面积，共60分）

为高一选题本大共小题，小5，60分在小题给四选项有只一选是正的把正选涂在题的应位上）1．函数y的义为A

B

．

D．2．若复满足为数位），的虚部为A

2B5

．

D．

3．如图某篮球联赛中，甲、乙两名运动员个场次得分的茎图．设甲、乙两人得分的平均数分别,x，中数分别为,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．已知线x为双曲线（a）一条渐近线则双曲线的b离心率为A

32

B

52

．2D．5

．．5．执行图所示的程序框图，输出的有序实数对为A．．B

开xy．D．

是

否

输

,y

6．已知线l与面平，则下列结错的A直线l与平面没公共点B存在通过直线l的面与平面平．直l与面内的意一条直线平行

y第题

结D．直线

l

上所有的点到平

的距离都相等7．已知函数(x)满足：当,21af,bfcf(3)，a,b,的小关系为A

Bb

．

D．8．设变x,y满足约束条件

yx,x2,则zx的值范畴2xyA

B

．

D．9．某四锥的三视图如图所示，该四棱锥的表面积为A

2B．14D．x0,10．函数f)，x

的零点个数为

4正视俯视

3侧视第题图A0B1C．D11．在ABC中G为ABC重心知3向GA与GB的夹角1则CA的小值是A

B6C9D12．已知函数有列三个结论①存在常数，对任的实数，恒有f②对任意给定的数M，都存在实数，使得f0

0

M；③直线y与数f且点许多多个．则所有正确结论序号是A①

B②

．③

D．②③

第卷（选题共分）二填空题本题共4小题，小题，共分把答案在题卡相位上13．已知集合★★★．14．已知函数(x)．在间一数，使得等式(x)成的概率为★★★．15．ABC的角,C所的分别是b,c．1052abcos105

的值为★★★．16．在各项均为正整数的单递增数列

n

中，，a2

，且

a

kN

*则a的为★★．三解题本题小题共74分解承写出字明、明程演算程)17（本小满12分）已知函数f(x)3距离为．

cos

(0)的图象与直线的相邻两个交点之间的（Ⅰ）求函数f)

的单调递增区间（Ⅱ）若f

π，cos2的．318（本小满12分）调查说明，中年的成就感与收入、学历、职业中意度的指标有极强的相关性．将这三项的中意指标分别记为y，对它们进行量化示不中意表示差不多中意，示中意，再用综合指标的评中年人的成就感等级：若w

4则成就感为一级若

，成就感为二级；0w，成就感为三级．为了了解目前某群体中人的成就感情形，研究人员随采访了该群体的名年人，得到如下结果：人员编号z人员编号

A1

A

A

A

A

x,yz

（Ⅰ）若该群体200人，估量该群体中成就感等级为三级的人数是多少？（Ⅱ）从成就感级为一级的被采访者中随机抽两人，这两人的综合指标均的概率是多少？

19（本小满12分）如图在长方ABCDAC中，ABBCAA4，P为

1

P

C

1线段上点．

A

1

B

1（Ⅰ）求证：；（Ⅱ）当P线段D的点时，求三棱锥APBC的．

CA

B20（本小满12分）

第题图小辉是一位收藏好者，在第年初买了价值为元的收藏品M，于受到收藏品市场行情的碍，第2年第3年每初M的值为上年初的每年初M的价比上年初增加万.（Ⅰ）求第几年开始M价值超过原购买的价值；

；从第4年始，（Ⅱ）记

T

n

（

*

）表示收藏品M前年价的平均值，求

T

n

的最小值．21（本小满12分）已知函数fx)

e

x

,mR,

为自然对数的底．（Ⅰ）若x是f()的极值点，求的值；（Ⅱ）证明：当0时，ee．22（本小满14分）如图，已知椭圆

1）的心率e．点,A分为椭圆a2左焦点和右顶点且（Ⅰ）求椭圆方；

Q（Ⅱ）过点F作一条直线l交圆,Q两点，点关

F

O

A

xx轴的对点为Q

．若PF∥

，求证：

12

．

Q'第题图

年州市高中业班质量检测数测考答评则一选题本题有12个题每小5分，分分1A．A3C4．D5D6．7．．．10B11．12．二填题：大共小，小题分满分．13．

1415．216．55三解题本大共6小，共74分．17．本题紧考查三角函数的图象与性质（称性、周期性、单调性）、倍角的余弦公式等基础知，考查运算求解能力，考查数结合思想、化归与转化思想、函与方程思想．满分12分【解析】（Ⅰ）为f()(0,，因此

f()2sin()···········································································2分因此f()．max因为函数x)与线的邻两个交点之间距离，因此T·····························································································因此

得

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

f())6令2k

x

k

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

,．··································································因此函数x)

的单调递增区间[k

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

，为f6

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

2·····························································11分67．············································································分918．本小题要紧考查概率、计等基础知识，考查数据处理能力、运算求解能力、应用意识，考查必定与然思想．满分12分【解析）运算10名被采访者综合指标，可得下表：

9C11AB212人员编号AAAA9C11AB212综合指标46243··············································································································1由上表可知：成感为三级（即0）只有A一位，频率为．··········3分101用样本的频率估总体的频率估该群体中成就感等级为三级的数有20．10··············································································································（Ⅱ事件为“从成就感等级是一级的被采访者中随机抽取两人们的合指标w均为4（）可知成就感是一级的（）：,AA,AAA，6位从中随机抽取两人，有可能的结果为：A15种··················································其中综合指标w有,，共3名事件发的所有可能结果为A种，··························分因此P()

31．················································································12分1519小题要紧考查直线与直线直线与平面的位置关系及几何体的体积等基础知识，考查空间想象能论证能力求解能力数形结合思想与化思想分分．证明：（Ⅰ）连BD．因为ABCDABC是方体，且AB2，因此四边形ABCD是正形，·······································································因此ACBD．························································································2分因为在长方体BCD中，BB平面，AC面，因此．···························································分A11因为BD面BBD平DD且BDB，因此面D．················································分因为BP平BBD，D因此ACBP·······························································6（Ⅱ）点P到面ABC的离AA，1ABC的积····························································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的面积S217．··10分设三棱锥APBC的为，

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

．817即三棱锥的高···································································分1720本小题要紧考查数等数列等比数不等式等基础知识考查运算求解能力、抽象概括能力、用意识，考查函数与方程思想化归与转化思想、分类与整合思．满分分．解：（Ⅰ）设第年的价值为a，依题意，当3时数列20，公为

12

的等比数列，因此

．故10，a，此aa．··············································································································当n4时数列为首项，公差为的差数列，又a，此·········································································令a，

274

，又因为n

*因n．··········································4分因此，第年初M开的价值超过原购买的价值．······································（Ⅱ）设S表示前年初M的值的和，则Tnn由（Ⅰ）知，当13时

40

，T

40n

①；··············································································································当n4时由于S35，

2

nn，

2

n32n．································································nn35当n，由①得，，T，此················10分当n4时由②知，2nn

322n

11当且仅当2n

32n

，即时等号成立．即．···································································11分由于T，故在第4年T的值最小，其最小值为．································分21本小题要紧考查函导不式等基础知考查推理论证能力抽象括能力、运算求解能力，查函数与方程思想、数形结合想、化归与转化思想．满分分．

4x4解：（Ⅰ）因为f()，因f

e

x)(e

，·············由x是f()的极值点，得

m)

，············································解得m，····························································································现在f(，检验，x是f(x)的值．e因此所求的实数m的为．·······································································4分（Ⅱ）证明：取m，(x)

e

，现在f

e)

．····················6分构造函数h))，········································································因此

(1x)

(

在恒，因此h(x)单调递减，···································································8分因此(x)(0),···················································································9分故

在成立，说明f(x)在单调递减．···············10分e因此当

b时，，因为因此0,e，e因此

(e

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

e

成．········································································12分22．本小题要紧考查直线、圆等基础知识，考查推理论证能力、运算求解能力，考查函数与方程思想、形结合思想、化归与转化思想分类与整合思想．满分14分．【解析】（Ⅰ）椭圆半焦距为c，

c1e,23,

··································································2分解得c，······················································································3分因此················································································4分因此椭圆方程为4

．································································（Ⅱ方一依题意得，PQ与标轴不垂直设Q与Q关于x对,因此Q）论可知，因为PFAQ直线FQ与线相，故，············1解得．····························································································235又因为点x,y在椭圆，此y，或y

34

5．························3y由椭圆对称性，妨取y5，直线的率4x2

．

4881x814881x8181因此直线方为

52

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

10分由

x,x12,

3得点P坐为,

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

11分因此PF

8164

，················

12分

16

13分1因此PFAQ··················································································2方法二：依题意，PQ与坐标轴不垂直．设l程为y）yy因为点与点轴对,因

14分又因为椭圆关于x轴对，因此点Q

也在椭圆上·······································由

312,

消去得．因此x

k4x3

．·················································因为PFAQ

，因此直线

的方程为由消y312,

k

x

．因为直线AQ

交椭圆于

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：依题意得PQ坐标轴不垂直．设l方为y）y因为点与点轴对,因

3k3又因为椭圆关于轴对称，因此3k3

也在椭圆上·······································x由消x得3y312,

．6k因此y．········································································3k因为PFAQ直线为x,由消x得，3y312,

．因为直线AQ

交椭圆于

x因此

k12，即3k3k

．·····························································设=）则y，此y．分6因此y，得，············································13分3k11因此FPPF22

．···························································14分方法四：依题意得PQ坐标轴不垂直．设l方为y）y因为点Q与轴称因Q·············································6分因为,F,Q三共线，因此yy因此．·····································································7分因为PFAQ可设FP=），因此x·································································8分因