2023年12月5日发(作者：对口单招数学试卷难题)

准考证号： 姓名：

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2020-2021学年度第一学期福州市九年级期末质量抽测

数 学 试 题

本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，完卷时间120分钟，满分150分．

注意事项：

1．答题前，考生务必在试题卷、答题卡规定位置填写本人准考证号、姓名等信息．

考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致．

2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号．非选择题答案用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答，在试题卷上答题无效．

3．作图可先使用2B铅笔画出，确定后必须用0.5毫米黑色墨水签字笔描黑．

4．考试结束，考生必须将试题卷和答题卡一并交回．

第Ⅰ卷

一、选择题（本题共10小题，每小题4分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的）

1．下列图形中，是中心对称图形的是

A

B

2．下列事件中，是确定性事件的是

C

D

A．篮球队员在罚球线上投篮一次，未投中

B．经过有交通信号灯的路口，遇到绿灯

C．投掷一枚骰子（六个面分别刻有1到6的点数），向上一面的点数大于3

D．任意画一个三角形，其外角和是360

A．（3，1）

C．（3，1）

A．12

C．6

B．（1，3）

D．（1，3）

B．63

D．33

九年级数学 －

1

－ （共5页）

3．将点（3，1）绕原点顺时针旋转90°得到的点的坐标是

4．已知正六边形ABCDEF内接于⊙O，若⊙O的直径为2，则该正六边形的周长是 5．已知甲，乙两地相距s（单位：km），汽车从甲地匀速行驶到乙地，则汽车行驶的时间t（单位：h）关于行驶速度v（单位：km/h）的函数图象是

t/h t/h

t/h

t/h

O

v/(km/h)

O

v/(km/h)

2O

v/(km/h)

C

O

v/(km/h)

A

A．图象的开口向上

B．图象的对称轴为直线x1

C．函数有最小值

B D

6．已知二次函数yx2x3，下列叙述中正确的是

D．当x＞1时，函数值y随自变量x的增大而减小

A．m＜1

C．m＞1

B．m＞1且m≠0

D．m≥1且m≠0

A B

D

F

7．若关于x的方程mx22x10有两个不相等的实数根，则m的取值范围是

8．如图，AB∥CD∥EF，AF与BE相交于点G，若BG3，CG2，CE6，则EF的值是

ABG

C

E

A．6

5C．8

3

B．8

5D．4

9．某餐厅主营盒饭业务，每份盒饭的成本为12元．若每份盒饭的售价为16元，每天可卖出360份．市场调查反映：如调整价格，每涨价1元，每天要少卖出40份．若该餐厅想让每天盒饭业务的利润达到1680元，设每份盒饭涨价x元，则符合题意的方程是

A．(16x12)(36040x)1680

B．(x12)(36040x)1680

C．(x12)[36040(x16)]1680

D．(16x12)[36040(x16)]1680

10．已知抛物线y(xx1)(xx2)1（x1＜x2），抛物线与x轴交于（m，0），（n，0）两点（m＜n），则m，n，x1，x2的大小关系是

九年级数学 －

2

－ （共5页）

A．x1＜m＜n＜x2

C．m＜x1＜n＜x2

B．m＜x1＜x2＜n

D．x1＜m＜x2＜n 第Ⅱ卷

注意事项：

1．用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答，在试题卷上作答，答案无效．

2．作图可先用2B铅笔画出，确定后必须用0.5毫米黑色墨水签字笔描黑．

二、填空题（本题共6小题，每小题4分，共24分）

11．若⊙O的半径为2，则270°的圆心角所对的弧长是 ．

12．若x2是关于x的方程x2x2m0的一个解，则m的值是 ．

13．已知反比例函数y4，当3＜x＜1时，y的取值范围x是 ．

14．如图，将一块等腰直角三角尺的锐角顶点P放在以AB为直径的半圆O上，∠P的两边分别交半圆O于B，Q两点，若AB2，则BQ的长是 ．

15．《易经》是中华民族聪明智慧的结晶．如图是《易经》中的一种卦图，每一卦由三根线组成（线形为““”），如正北方向的卦为“卦，这一卦中恰有1根“是

．

16．如图，在平行四边形ABCD中，AB23，BC6，A

ADC120，点E，F分别在边AD，AB上运动，且F

B C

E D

P

A O B

Q

”或”的概率”．从图中任选一”和2根“满足BF3DE，连接BE，CF，则CF3BE的最小值是 ．

三、解答题（本题共9小题，共86分．解答应写出文字说明、证明过程或演算步骤）

17．（本小题满分8分）

解方程：x22x10．

18．（本小题满分8分）

如图，AB是⊙O的直径，C为半圆O上一点，直线l经过点C，过点A作AD⊥l于点D，连接AC，当AC平分∠DAB时，求证：直线l是⊙O的切线．

D

C

l

A O B

九年级数学 －

3

－ （共5页）

19．（本小题满分8分）

一名男生推铅球，铅球行进高度y（单位：m）与水平距离x（单位：m）之间的函数关系是y1(x4)23．如图，A，B是该函数图象上的两点．

12（1）画出该函数的大致图象；

（2）请判断铅球推出的距离能否达到11 m，并说明理由．

y/m

3

A

-3 -2 -1

O

1 2 3 4 5 6 7 8 9 10 11

x/m

B

20．（本小题满分8分）

为发展学生多元能力，某校九年级开设A，B，C，D四门校本选修课程，要求九年级每个学生必须选报且只能选报其中一门．图1，图2是九年（1）班学生A，B，C，D四门校本选修课程选课情况的不完整统计图．请根据图中信息，解答下列问题．

频数（人）

16

16

12

8

4

0 A B C D

课程

12

D

C

图1 图2

A

10%

B

40%

（1）求九年（1）班学生的总人数及该班选报A课程的学生人数；

（2）在统计的信息中，我们发现九年（1）班的甲同学和乙同学选报了A课程，若从该班选报A课程的同学中随机抽取2名进行选修学习效果的测评，求甲，乙同时被抽中的概率．

21．（本小题满分8分）

如图，点D是等边三角形ABC内一点，连接DA，DC，将△DAC绕点A顺时针旋转60，点D的对应点为E．

（1）画出旋转后的图形；

（2）当C，D，E三点共线时，求∠BEC的度数．

22．（本小题满分10分）

如图，一次函数yxb的图象与y轴正半轴交于点C，与反比例函数yk的图象交于A，B两点，若OC2，点Bx的纵坐标为3．

（1）求反比例函数的解析式；

（2）求△AOB的面积．

九年级数学 －

4

－ （共5页）

A

D

B

y

C

O

B

x

C

A 23．（本小题满分10分）

如图，作△ADC，使得点B，D在AC异侧，且ADCD，ABAC，ADCBAC，

E是BC延长线上一点，连接AE交CD于点F．

（1）求证：△ABC∽△DAC；

（2）若AB22CFAD，试判断△ACF的形状，并说明理由．

A D

F

B C E

24．（本小题满分12分）

如图，四边形ABCD内接于⊙O，BAD90，ABAD，点E是AB上一点，连接DE交AB于点F，连接AE，BE．

（1）若AD52，BE6，求DE的长；

（2）若CEDE，且DE8，CD9.6，求AF的值．

BFE

A

F

B

O

D

C

25．（本小题满分14分）

如图，A，B分别为x轴正半轴，y轴正半轴上的点，已知点B的坐标是（0，6），2BAO45．过A，B两点的抛物线y1xbxc与x轴的另一个交点落在线段2OA上．该抛物线与直线ykxm（k＞0）在第一象限交于C，D两点，且点C的横坐标为1．

（1）求该抛物线的解析式；

（2）若直线CD与线段AB的交点记为E，当BE1时，求点D的坐标；

AE2（3）P是x轴上一点，连接PC，PD，当CPD90时，若满足条件的点P有两个，且这两点间的距离为1，求直线CD的解析式．

y

B

O

九年级数学 －

5

－ （共5页）

A x

2020-2021学年度第一学期福州市九年级期末质量抽测

数学试题答案及评分标准

评分说明：

1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制定相应的评分细则．

2．对于计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．

3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．

4．只给整数分数．选择题和填空题不给中间分．

一、选择题（共10小题，每小题4分，满分40分；在每小题给出的四个选项中，只有一项是符合题目要求的，请在答题卡的相应位置填涂）

1．B

6．D

二、填空题（共6小题，每小题4分，满分24分，请在答题卡的相应位置作答）

11．3π

14．2

12．3

13．4＜y＜4

316．66

2．D

7．B

3．B

8．C

4．C

9．A

5．B

10．A

15．3

8

三、解答题（共9小题，满分86分，请在答题卡的相应位置作答）

17．（本小题满分8分）

解法一：a1，b2，c1， ··················································································b4ac

22=(2)41(1)8＞0． ···········································································方程有两个不等的实数根

bb24ac ·························································································x2a九年级数学 －

6

－ （共5页） （2）8=12， ·················································································2即x11+2，x212． ··············································································2解法二：x22x1， ·································································································2····························································································x2x12， ·(x1)2， ·································································································x12， ································································································x2+1，

即x121，x221． ···········································································

18．（本小题满分8分）

证明：连接OC．

∵AC平分∠DAB，

∴DACCAB． ··························································································∵OAOC，

D

·∴OCACAB， ·························································································C

·l

·∴OCADAC， ·························································································∴OC∥AD， ····································································································∴ADCOCD180． ··················································································A O B

∵AD⊥l，

∴ADC90， ······························································································∴OCD90， ······························································································∴OC⊥CD．

∵点C为半径OC的外端点，

∴直线l是⊙O的切线． ·····················································································

19．（本小题满分8分）

解：（1）

y/m

3

A

-3 -2 -1

O

1 2 3 4 5 6 7 8 9 10 11

x/m

B

··················································································································该函数的大致图象如图所示．

（2）铅球推出的距离不能达到11 m． ·······································································2理由如下：当x10时，y1(104)30， ··············································12∴该男生此次推球最远距离为10 m， ··················································而10＜11， ···················································································九年级数学 －

7

－ （共5页） ∴铅球推出的距离不能达到11 m．

20．（本小题满分8分）

······················································解：（1）九年（1）班学生的总人数是1640%40， ·······················································该班选报A课程的学生人数是4010%4． ·（2）由（1）得，九年（1）班选报A课程的人数是4，将甲，乙以外的两人记为丙，丁．

根据题意，可以列出如下表格：

第一个人

甲 乙 丙 丁

第二个人

甲 （乙，甲） （丙，甲） （丁，甲）

乙 （甲，乙） （丙，乙） （丁，乙）

丙 （甲，丙） （乙，丙） （丁，丙）

丁 （甲，丁） （乙，丁） （丙，丁）

··················································································································由表可知，所有可能出现的结果共有12种，且这些结果出现的可能性相等． ··················································································································其中他们“甲，乙同时被抽中”的结果有2种．

∴P（甲，乙同时被抽中）＝2＝1． ······························································612∴甲，乙同时被抽中的概率是1．

6

21．（本小题满分8分）

解：（1）

A

E

D

B C

··················································································································如图，△EAB是所求作的△DAC绕点A顺时针旋转60后得到的三角形． ··················································································································（2）连接DE．

∵△DAC绕点A顺时针旋转60后得到△EAB，

A

∴EADBAC60，△EAB≌△DAC， ·························································∴EBADCA． ························································································E

∵C，D，E三点共线，

F

·∴EFBAFC． ·······················································································D

∵三角形的内角和为180，

九年级数学 －

8

－ （共5页）

B C ∴BECBAC60．·················································································

22．（本小题满分10分）

解：（1）∵点C在y轴正半轴，OC2，

∴b2， ·····································································································∴一次函数解析式为yx2． ········································································将y3代入yx2，得x1，

∴B（1，3）． ································································································将点B（1，3）代入yk，

x得k3， ·····································································································1∴k3，

∴反比例函数的解析式为y3． ······································································x（2）将y0代入yx2，得x2，

∴点D的坐标是（0，2），

∴OD2． ·································································································y

·将yx2代入y3，得x23，

xxB

C

解得x11，x23． ·····················································································D

当x3时，y321，

O

x

A

∴点A的坐标是（3，1），

∴点A到x轴的距离是1．···············································································∵点B的纵坐标为3，

∴点B到x轴的距离是3，···············································································∴S△AOBS△AODS△BOD1211234． ·················································22

23．（本小题满分10分）

（1）证明：∵ABAC，ADCD，

∴ABAC． ··························································································DADC∵BACADC，

∴△ABC∽△DAC． ···················································································（2）解：△ACF是直角三角形． ··················································································理由如下：由（1）得△ABC∽△DAC，

∴ACBACD，ABBC． ······················································DAAC∵ABAC，

∴AB2ADBC．

∵AB22CFAD，

A D

ADBC2CFAD，

即BC2CF． ··············································································取BC中点G，连接AG，

F

九年级数学 －

9

－ （共5页）

B G C E ∴BC2CG，

∴CGCF． ···············································································∵ABAC，

∴AG⊥BC，

∴AGC90．

∵ACAC，

∴△AGC≌△AFC， ·······································································∴AGCAFC，

∴AFC90， ···········································································∴△ACF是直角三角形．

24．（本小题满分12分）

解：（1）在Rt△ABD中，BAD90，ABAD，AD52，

22∴BDABAD10． ··············································································∵BD是直径，

∴BED90． ····························································································在Rt△BED中，BE6，

22∴DEBDBE8． ···············································································（2）连接EO并延长交CD于点I，连接OC，EC．

过点A作AG⊥AE交DE于点G，作AH⊥DE于点H，

∴EAGAHGAHE90．

∵CEDE，

∴CEDE，CDEDBE．

∵OCOD，CD=9.6

∴EI垂直平分CD， ·······················································································∴DI1CD4.8，EID90．

2在Rt△DEI中，DE8，

22∴EIDEDI6.4． ··············································································∵BEDDIE90，

∴△BDE∽△DEI，

∴BDEBDE8，

DEIDEI6.4∴BD10，EB6． ····················································································A

·E

∵BAD90，

F

∴BADBAGEAGBAG，

H

G

即EABGAD．

B

D

∵ABAD，

O

∴ABD45．

∵ADAD，

C

∴AEDABD45，

∴AGE45AED，

∴AEAG， ································································································∴△ABE≌△ADG，点H为EG中点， ·······························································九年级数学 －10

－ （共5页）

I ∴DGBE6，

∴EGDEDG2，

∴AH1EG1． ·························································································2∵AHEBED90，AFHBFE，

∴△AHF∽△BEF，

∴AFAH1． ·························································································BFBE6

25．（本小题满分14分）

解：（1）∵B（0，6），

∴OB6． ···································································································∵BAO45，AOB90，

∴ABO45BAO，

∴OAOB6． ····························································································∵点A在x轴正半轴，

∴A（6，0）．

186bc0，将A（6，0），B（0，6）代入y1x2bxc，得

2c6．b4，解得 ·································································································c6．∴该抛物线的解析式为y1x24x6． ····························································2（2）过点E作EF⊥x轴于点F，

y

∴AFE90AOB，

∴EF∥BO，AEF45ABOOAB，

∴OFBE1，OAOB6，

B

AFAE2E

∴OF1OA2，

3∴点F的横坐标为2，FA4，

O F A

x

∴点E的横坐标为2，EF4，

∴点E的坐标是（2，4）． ···············································································2将x1代入y1x4x6，得y2.5，∴点C的坐标是（1，2.5）． ·····················2由C（1，2.5），E（2，4）得直线CD的解析式为y3x1， ································222将y3x1代入y1x4x6，得1x4x63x1，

2222解得x11，x210． ·····················································································∵点C的横坐标为1，

∴点D的横坐标为10，

将x10代入y3x1，得y16，

2∴点D的坐标是（10，16）． ············································································（3）由（2）得C（1，2.5）．

y

11

九年级数学 － － （共5页）

B

D 设D（xD，yD），P（t，0）．

根据题意可知，点D在点C的上方，

点P是以CD为直径的圆与x轴的交点，

∴1＜t＜xD．

分别过点C，D作x轴的垂线，垂足为点H，I，

∴CHIDIH90，

∴HCPHPC90．

∵CPD90，

∴IPDHPC90，

∴IPDHCP，

∴△HCP∽△IPD，

∴CHHP，即2.5t1，

xDtyDPIID∴(t1)(xDt)2.5yD…①． ···········································································将点C（1，2.5）代入ykxm中，得2.5km，

∴m2.5k， ·····························································································∴直线CD的解析式为yk(x1)2.5．

将yk(x1)2.5代入y1x24x6，

22整理可得x(2k8)x2k70，

解得xC1，xD2k7，

∴D（2k7，2k26k2.5）． ········································································将点D（2k7，2k26k2.5）代入①，

整理可得t2(2k8)t5k217k530，

42224k32k6420k68k5316k36k11．

∵满足条件的点P有两个，

不妨设满足条件的两个点P的横坐标分别为t1，t2，且t1＜t2．

2由求根公式可得t1t22k8，t1t25k17k53． ···········································4依题意得t2t11，

22∴(t2t1)(t1t2)4t1t21，

22∴(2k8)4(5k17k53)1，

42即8k18k50，

解得k11，k25＜0（舍去），

42当k1时，16k228k11＞0，满足题意，

4∴直线CD的解析式为y1x9． ···································································44

九年级数学 －12

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