人教版2022-2021年中考一模数学试卷及答案

．．．．．．．．．．．．．．．．考生须知

中模数试1．卷分为试题和答题卡两部分，所有试题均在答题卡作答．2．题前，在答题卡上考生务必将学校、班级、准考证号、姓名填写清楚．3．选择题的所选选项填涂在答题卡上；作图题用2B铅．4．改时，用塑料橡皮擦干净，不得使用涂改液．请保持卡面清洁，不要折叠．一、选择题（本题共分，小题2）第题均有个选项，符合题意的选项只有一个．和日丽春光好一舞筝时筝我国人民非常喜爱的一项户外娱乐活动列风筝剪纸作品中，不是轴对称形的是A

B．

C．．．下面四幅图中，用量角器测得AOB度是°的图是O

B

A

90BA

B

B

90

A

O

AC．．如图，数轴上每相邻两点距离表示1个位，点，B互为相反数，则点C表的数可能是A0B1C3D．．下图可以折叠成的几何体是A三棱柱B圆柱C．棱．锥

D．AC精品

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．中国有个名句“运筹帷幄之中，决胜千里之外”．其中的“筹”原意是指《孙子算经》中记载的“算筹”筹是古代用来行计算的工具，它是将几寸长的小竹棍摆在平面上进行运算，算筹的摆放形式有纵横两种形式（如右图）．当表示一个多位数时，像阿拉伯计数一样，把各个数位的数码从左到右排列，但各位数码的筹式需要纵横相间：个位、百位、万位数用纵式表示；十位，千位，十万位数用横式表示替以此类推．例如3306用算筹表示就是，则用筹可表示为AC.D.．一个正多边形的每个内角的度数都等于相邻外角的度数，则该正多边形的边数是A3B．C6D．．“龟兔赛跑”是同学们熟悉的寓言故事．如图所示，表示了寓言中的龟、兔的路和时间的系（其中直线段表示乌龟，折线段表示兔子）．下列叙述正确的是A赛跑中，兔子共休息了分钟B乌龟在这次比赛中的平均速度是0.1米分C．子比乌龟早到达终点分钟D．龟上兔子用了分钟中小学时期是学身心变化最为明显的时期个时期孩子们的身高变化呈现一定的趋势7~15岁间生子们会经历一个身高发育较迅速的阶段们把这个年龄阶段叫生速度峰值段明通过上网查阅2021年某市儿童体格发育调查表解市男女生7~15岁高平均值记录情况制了如下统计图，并得出以下结论：①岁之前龄女生的平均身高一般会略高于男生的平均身高；②10~12岁之间，女达到生长速度峰值段可能超过同龄男；③7~15岁期间生平均身高始终高于女生的平均身高；④岁生身高现生长速度峰值段男女生身高差距可能逐加大．以上结论正确的是A①③B．②③．②④D．④精品

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二、填空题（本题共分，小题2）．二次根式x意义，则x的值范围是．10业部门要考察某种幼树在一定条件下的移植成活率图这种幼树在移植过程中幼树成活率的统计图：估计该种幼树在此条件下移植成活的概率为（结果精确到0.01m2n个=12如图，测量小玻璃管口径的量具上，AB的长为10毫米，被分为等份，如果小管口中正对着量具上20份处∥AB么小管口径长是毫米．13已知：

，则代数式

的值是．14如图AB是O的径⊥弦CD点，若=10，CD=8，则=．432

B

1

A–3–2–1O–1–2–3D–4

23

15如图，在平面直角坐标系中可看作是经若干次图形的变化（平移、轴对称、旋转）得到的，写出一种由得OCD过程精品

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．．．16下面是“作已知角的角平分线”的尺规作图过程．已知：如图1∠MONO

图

N求作：射线，使它平分．作法：如图2（1以点为心，任意长为半径作弧，交OM于A，交于B；（2连结AB；（3分别以点，为心，大于AB的长为半径作弧，两弧相交于点；（4作射线OP．所以，射线即所求作的射线．请回答：该尺规作图的依据是．三、解答题（本题共68分，第1722题，每小题5分，第23题7分，24题，第25题5分，第26题分，第27题7分，第28题）解答应写出文字说明、演算步骤证明过.17计算：3

18解不等式组

3

，并写出它的所有整数解．19图在△AB=AC点D是BC边一点垂直平分CD交AC于，交于F，结，求证DE∥AB．AEB

C精品

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20关于x的元二次方程x（1求k的取值范围；（2当k为正整数时，求此时方程的根．

有两个不相等的实数根．21如图，在平面直角坐标系中函数y

kx

的图象与直线y+1交点A（1（1求，k的；（2连结，点P是函数y

kx

上一点，且满足OP=OA直接写出点的坐标（点A除22如图，eq\o\ac(□,)ABCD中分∠交AD点FAE⊥BF于，BC于，连接．（1求证：四边形ABEF是形；（2连接若ABC=60，AB=4DF，求的．

AFDOBEC23为了解某区初二年级数学学科期末质量监控情况，进行了抽样调查，过程如下，请将有关问题补充完.收集数据随机抽取甲乙两所学校的20名生的数学成绩进行分：甲乙

91818490

89929388

77856667

86856988

71957691

31888796

97887768

93908297

72448559

91918888整理、描述数据按如下数据段整理、描述这两组数据分段学校甲乙

30≤≤39≤≤49≤x≤≤≤6970≤≤≤x≤90≤≤00分析数据精品

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两组数据的平均数、中位数、众数、方差如下表：统计量学校甲乙

平均数

中位数

众数m

方差经统计，表格中m的是．得出结论若学校有名初二学生，估计这考试成绩以上人数为以推断出学校学生的数学水平较高.（少从两个不同的角度说明推断的合理性）24如图，以为径作⊙O，过点作⊙O的线，连结BC交O于D点E是BC边中点，连结AE（1求证：∠=2；（2若=6，cos

，求DE的．DE25图△C=60°厘米P从出发B→A以每秒米的速度匀速运动到点A设P的动时间为x秒B两点间的距离为y厘米．小新根据学习函数的经验，对函数随变量的化而变化的规律进行了探究．下面是小新的探究过程，请补充完整：

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1212（1通过取点、画图、测量，得到了与y的组值，如下表：x（s）y（cm）

1.0

2.0

3.0

2.7

2.7

m

3.6经测量的是（保留一位小数（）建立平面直角坐标，描出表格中所有各对对应值为坐标的点，画出该函数的图象；（3结合画出的函数图象，解决问题：在曲线部分的最低点时，在ABC中出点P所在的位置．26在平面直角坐标系中抛物线

ybx

的对称轴为直线．（1求b的；（2y轴有一动点（mP作直轴直线交抛物线于点xB（，中x．①当x时，结合函数图象，求出m的；②把直线PB下的函数图象，沿直线向翻折，图象的其余部分保持不变，得到一个新的图象，图象x5时y

，求的值围．精品

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27在ABC，，CDBC于点C，交∠ABC平分线于点DAE平∠交BD于E，过点作EFAC于点F连接DF（1补全图；（2如图，当∠°时，①求证BE=DE；②写出判断DFAB的置关系的思路（不用写出证明过程（3如图，当∠BAC=时直接写出，DF，的关系．AD

A

DEEB

图

B

图2

C28.在面角坐标系xOy中M的标为

x,y11

坐标为

x,y2

xx1

2

，y1

2

，以MN为构造菱形，若该菱形的两条对角线分别平行于轴轴则称该形为边的“坐标菱形”.（1已知点A2,0（0,2以为的“坐标菱形”的最小内角_______；（2若点（1,2在直线y上，以为的“坐标菱形”为正形，求直线表式；（3O的径为

，点P的标为3,m)若在⊙O上存在一点Q，得以为的“坐标菱形”为正方形，求的值范围．精品

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数学试卷参答案及评分准一、选择题（本题共分，小题2）题答

B

A

C

A

C

B

D

C二、填空题（本题共分，小题2）．x；．；11

；．

；．；14．215不唯一沿轴下翻折轴左平移2个单位度得到16答案不唯一：到线段两端点距离相等的点在线段的垂直平分线上；等腰三角形三线合一．三、解答题（本题共68分，第17-22题，每小题5分，第23题7分，第24题6分，第25题5分，第26题分，第27题7分，第28题）解答应写出文字说明、演算步骤证明过.17解：3=33

60··············································································································································································①18解：x3解不等式①，得x≤2．····················································································1解不等式②，得x＞-1．··················································································3∴原不等式组的解集为

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

∴适合原不等式组的整数解为0,1,2·································································19证明：，∴∠B∠．································································································∵垂平分CD，∴．··································································································∴∠=C．····························································································∴∠=B．····························································································∴DF∥AB．·································································································

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AEBDFC20解）关于的一元二次方程有两个不相等的实根．∴2································································1=8－>0.∴

·····································································································2（2∵k为正整数，∴k．······································································································3解方程

，得

0,x

．···························································21解）直线+1经点A（1∴a=2．·····················································································∴（1,2∵函数y

kx

的图象经过点A（，∴k．·····················································································（2点P的标2,1···················································22）明：∵平∠ABC，∴∠ABF∠CBF．·························································································1∵ABCD，∴AD．∴∠AFB∠CBF．∴∠ABF∠．∴AB=AF．∵⊥BF∴∠ABF∠∠CBF∠BEO°．∴∠=BEO．∴．∴AF=BE．∴四边形ABEF是行四边形．eq\o\ac(□,∴)是形2（2解：AD=BC，AF=BE∴DF=CE．∴BE=2．∵AB=4∴．∴CE=2．过点A作⊥BC于．3∵∠=60°，AB=BE，∴△ABE是等边三角形．∴BG=GE=2．∴．∴四边形是行四边形．

B

AFDOGEC精品

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∴AGCF是形．∴AG=CF．在△中∠°，=4，∴AG=23．∴=23．523整理、描述数据分段学校

30≤≤39≤≤≤x≤≤x≤69≤≤79≤x≤90≤x≤甲乙

10

10

01

04

32

78

85分析数据经统计，表格中m的是．·得出结论甲学校有400名初二学生，估计这次考试成绩80分上人数为300··············4答不唯一，理由须支撑推断结.·································································724（）证明：∵AC⊙O的线，∴∠BAC=90°．···························································································1∵点E是BC的中点，∴AE=EC．∴∠=∠EAC，····························································································2∵∠=∠C+∠，∴∠=2C．···························································································（2解：连结AD∵AB为径作O，∴∠ABD°．∵=，cosB，18∴=．·····································45

在eq\o\ac(△,Rt)ABC中AB=6，cos

，

∴BC．∵点E是BC的中点，∴．···············································5

C∴

75

．··········································25解）；·························································································（2如图所示；···························································································精品

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1212APB（3如图····································································································526解）抛物线

的对称轴为直线，∴b=2．···································································（2①∴抛物线的表达式为

x

．∵A（x，（x，∴直线AB行x轴．∵，∴AB=3．∵对称轴为x，∴=

．·············································∴当x