人教版2022-2021年中考一模數(shù)學(xué)試卷及答案

．．．．．．．．．．．．．．．．考生須知

中模數(shù)試1．卷分為試題和答題卡兩部分，所有試題均在答題卡作答．2．題前，在答題卡上考生務(wù)必將學(xué)校、班級、準(zhǔn)考證號、姓名填寫清楚．3．選擇題的所選選項(xiàng)填涂在答題卡上；作圖題用2B鉛．4．改時(shí)，用塑料橡皮擦干凈，不得使用涂改液．請保持卡面清潔，不要折疊．一、選擇題（本題共分，小題2）第題均有個(gè)選項(xiàng)，符合題意的選項(xiàng)只有一個(gè)．和日麗春光好一舞箏時(shí)箏我國人民非常喜愛的一項(xiàng)戶外娛樂活動(dòng)列風(fēng)箏剪紙作品中，不是軸對稱形的是A

B．

C．．．下面四幅圖中，用量角器測得AOB度是°的圖是O

B

A

90BA

B

B

90

A

O

AC．．如圖，數(shù)軸上每相鄰兩點(diǎn)距離表示1個(gè)位，點(diǎn)，B互為相反數(shù)，則點(diǎn)C表的數(shù)可能是A0B1C3D．．下圖可以折疊成的幾何體是A三棱柱B圓柱C．棱．錐

D．AC精品

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．中國有個(gè)名句“運(yùn)籌帷幄之中，決勝千里之外”．其中的“籌”原意是指《孫子算經(jīng)》中記載的“算籌”籌是古代用來行計(jì)算的工具，它是將幾寸長的小竹棍擺在平面上進(jìn)行運(yùn)算，算籌的擺放形式有縱橫兩種形式（如右圖）．當(dāng)表示一個(gè)多位數(shù)時(shí)，像阿拉伯計(jì)數(shù)一樣，把各個(gè)數(shù)位的數(shù)碼從左到右排列，但各位數(shù)碼的籌式需要縱橫相間：個(gè)位、百位、萬位數(shù)用縱式表示；十位，千位，十萬位數(shù)用橫式表示替以此類推．例如3306用算籌表示就是，則用籌可表示為AC.D.．一個(gè)正多邊形的每個(gè)內(nèi)角的度數(shù)都等于相鄰?fù)饨堑亩葦?shù)，則該正多邊形的邊數(shù)是A3B．C6D．．“龜兔賽跑”是同學(xué)們熟悉的寓言故事．如圖所示，表示了寓言中的龜、兔的路和時(shí)間的系（其中直線段表示烏龜，折線段表示兔子）．下列敘述正確的是A賽跑中，兔子共休息了分鐘B烏龜在這次比賽中的平均速度是0.1米分C．子比烏龜早到達(dá)終點(diǎn)分鐘D．龜上兔子用了分鐘中小學(xué)時(shí)期是學(xué)身心變化最為明顯的時(shí)期個(gè)時(shí)期孩子們的身高變化呈現(xiàn)一定的趨勢7~15歲間生子們會經(jīng)歷一個(gè)身高發(fā)育較迅速的階段們把這個(gè)年齡階段叫生速度峰值段明通過上網(wǎng)查閱2021年某市兒童體格發(fā)育調(diào)查表解市男女生7~15歲高平均值記錄情況制了如下統(tǒng)計(jì)圖，并得出以下結(jié)論：①歲之前齡女生的平均身高一般會略高于男生的平均身高；②10~12歲之間，女達(dá)到生長速度峰值段可能超過同齡男；③7~15歲期間生平均身高始終高于女生的平均身高；④歲生身高現(xiàn)生長速度峰值段男女生身高差距可能逐加大．以上結(jié)論正確的是A①③B．②③．②④D．④精品

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二、填空題（本題共分，小題2）．二次根式x意義，則x的值范圍是．10業(yè)部門要考察某種幼樹在一定條件下的移植成活率圖這種幼樹在移植過程中幼樹成活率的統(tǒng)計(jì)圖：估計(jì)該種幼樹在此條件下移植成活的概率為（結(jié)果精確到0.01m2n個(gè)=12如圖，測量小玻璃管口徑的量具上，AB的長為10毫米，被分為等份，如果小管口中正對著量具上20份處∥AB么小管口徑長是毫米．13已知：

，則代數(shù)式

的值是．14如圖AB是O的徑⊥弦CD點(diǎn)，若=10，CD=8，則=．432

B

1

A–3–2–1O–1–2–3D–4

23

15如圖，在平面直角坐標(biāo)系中可看作是經(jīng)若干次圖形的變化（平移、軸對稱、旋轉(zhuǎn)）得到的，寫出一種由得OCD過程精品

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．．．16下面是“作已知角的角平分線”的尺規(guī)作圖過程．已知：如圖1∠MONO

圖

N求作：射線，使它平分．作法：如圖2（1以點(diǎn)為心，任意長為半徑作弧，交OM于A，交于B；（2連結(jié)AB；（3分別以點(diǎn)，為心，大于AB的長為半徑作弧，兩弧相交于點(diǎn)；（4作射線OP．所以，射線即所求作的射線．請回答：該尺規(guī)作圖的依據(jù)是．三、解答題（本題共68分，第1722題，每小題5分，第23題7分，24題，第25題5分，第26題分，第27題7分，第28題）解答應(yīng)寫出文字說明、演算步驟證明過.17計(jì)算：3

18解不等式組

3

，并寫出它的所有整數(shù)解．19圖在△AB=AC點(diǎn)D是BC邊一點(diǎn)垂直平分CD交AC于，交于F，結(jié)，求證DE∥AB．AEB

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20關(guān)于x的元二次方程x（1求k的取值范圍；（2當(dāng)k為正整數(shù)時(shí)，求此時(shí)方程的根．

有兩個(gè)不相等的實(shí)數(shù)根．21如圖，在平面直角坐標(biāo)系中函數(shù)y

kx

的圖象與直線y+1交點(diǎn)A（1（1求，k的；（2連結(jié)，點(diǎn)P是函數(shù)y

kx

上一點(diǎn)，且滿足OP=OA直接寫出點(diǎn)的坐標(biāo)（點(diǎn)A除22如圖，eq\o\ac(□,)ABCD中分∠交AD點(diǎn)FAE⊥BF于，BC于，連接．（1求證：四邊形ABEF是形；（2連接若ABC=60，AB=4DF，求的．

AFDOBEC23為了解某區(qū)初二年級數(shù)學(xué)學(xué)科期末質(zhì)量監(jiān)控情況，進(jìn)行了抽樣調(diào)查，過程如下，請將有關(guān)問題補(bǔ)充完.收集數(shù)據(jù)隨機(jī)抽取甲乙兩所學(xué)校的20名生的數(shù)學(xué)成績進(jìn)行分：甲乙

91818490

89929388

77856667

86856988

71957691

31888796

97887768

93908297

72448559

91918888整理、描述數(shù)據(jù)按如下數(shù)據(jù)段整理、描述這兩組數(shù)據(jù)分段學(xué)校甲乙

30≤≤39≤≤49≤x≤≤≤6970≤≤≤x≤90≤≤00分析數(shù)據(jù)精品

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兩組數(shù)據(jù)的平均數(shù)、中位數(shù)、眾數(shù)、方差如下表：統(tǒng)計(jì)量學(xué)校甲乙

平均數(shù)

中位數(shù)

眾數(shù)m

方差經(jīng)統(tǒng)計(jì)，表格中m的是．得出結(jié)論若學(xué)校有名初二學(xué)生，估計(jì)這考試成績以上人數(shù)為以推斷出學(xué)校學(xué)生的數(shù)學(xué)水平較高.（少從兩個(gè)不同的角度說明推斷的合理性）24如圖，以為徑作⊙O，過點(diǎn)作⊙O的線，連結(jié)BC交O于D點(diǎn)E是BC邊中點(diǎn)，連結(jié)AE（1求證：∠=2；（2若=6，cos

，求DE的．DE25圖△C=60°厘米P從出發(fā)B→A以每秒米的速度勻速運(yùn)動(dòng)到點(diǎn)A設(shè)P的動(dòng)時(shí)間為x秒B兩點(diǎn)間的距離為y厘米．小新根據(jù)學(xué)習(xí)函數(shù)的經(jīng)驗(yàn)，對函數(shù)隨變量的化而變化的規(guī)律進(jìn)行了探究．下面是小新的探究過程，請補(bǔ)充完整：

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1212（1通過取點(diǎn)、畫圖、測量，得到了與y的組值，如下表：x（s）y（cm）

1.0

2.0

3.0

2.7

2.7

m

3.6經(jīng)測量的是（保留一位小數(shù)（）建立平面直角坐標(biāo)，描出表格中所有各對對應(yīng)值為坐標(biāo)的點(diǎn)，畫出該函數(shù)的圖象；（3結(jié)合畫出的函數(shù)圖象，解決問題：在曲線部分的最低點(diǎn)時(shí)，在ABC中出點(diǎn)P所在的位置．26在平面直角坐標(biāo)系中拋物線

ybx

的對稱軸為直線．（1求b的；（2y軸有一動(dòng)點(diǎn)（mP作直軸直線交拋物線于點(diǎn)xB（，中x．①當(dāng)x時(shí)，結(jié)合函數(shù)圖象，求出m的；②把直線PB下的函數(shù)圖象，沿直線向翻折，圖象的其余部分保持不變，得到一個(gè)新的圖象，圖象x5時(shí)y

，求的值圍．精品

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27在ABC，，CDBC于點(diǎn)C，交∠ABC平分線于點(diǎn)DAE平∠交BD于E，過點(diǎn)作EFAC于點(diǎn)F連接DF（1補(bǔ)全圖；（2如圖，當(dāng)∠°時(shí)，①求證BE=DE；②寫出判斷DFAB的置關(guān)系的思路（不用寫出證明過程（3如圖，當(dāng)∠BAC=時(shí)直接寫出，DF，的關(guān)系．AD

A

DEEB

圖

B

圖2

C28.在面角坐標(biāo)系xOy中M的標(biāo)為

x,y11

坐標(biāo)為

x,y2

xx1

2

，y1

2

，以MN為構(gòu)造菱形，若該菱形的兩條對角線分別平行于軸軸則稱該形為邊的“坐標(biāo)菱形”.（1已知點(diǎn)A2,0（0,2以為的“坐標(biāo)菱形”的最小內(nèi)角_______；（2若點(diǎn)（1,2在直線y上，以為的“坐標(biāo)菱形”為正形，求直線表式；（3O的徑為

，點(diǎn)P的標(biāo)為3,m)若在⊙O上存在一點(diǎn)Q，得以為的“坐標(biāo)菱形”為正方形，求的值范圍．精品

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數(shù)學(xué)試卷參答案及評分準(zhǔn)一、選擇題（本題共分，小題2）題答

B

A

C

A

C

B

D

C二、填空題（本題共分，小題2）．x；．；11

；．

；．；14．215不唯一沿軸下翻折軸左平移2個(gè)單位度得到16答案不唯一：到線段兩端點(diǎn)距離相等的點(diǎn)在線段的垂直平分線上；等腰三角形三線合一．三、解答題（本題共68分，第17-22題，每小題5分，第23題7分，第24題6分，第25題5分，第26題分，第27題7分，第28題）解答應(yīng)寫出文字說明、演算步驟證明過.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∴原不等式組的解集為

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

∴適合原不等式組的整數(shù)解為0,1,2·································································19證明：，∴∠B∠．································································································∵垂平分CD，∴．··································································································∴∠=C．····························································································∴∠=B．····························································································∴DF∥AB．·································································································

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AEBDFC20解）關(guān)于的一元二次方程有兩個(gè)不相等的實(shí)根．∴2································································1=8－>0.∴

·····································································································2（2∵k為正整數(shù)，∴k．······································································································3解方程

，得

0,x

．···························································21解）直線+1經(jīng)點(diǎn)A（1∴a=2．·····················································································∴（1,2∵函數(shù)y

kx

的圖象經(jīng)過點(diǎn)A（，∴k．·····················································································（2點(diǎn)P的標(biāo)2,1···················································22）明：∵平∠ABC，∴∠ABF∠CBF．·························································································1∵ABCD，∴AD．∴∠AFB∠CBF．∴∠ABF∠．∴AB=AF．∵⊥BF∴∠ABF∠∠CBF∠BEO°．∴∠=BEO．∴．∴AF=BE．∴四邊形ABEF是行四邊形．eq\o\ac(□,∴)是形2（2解：AD=BC，AF=BE∴DF=CE．∴BE=2．∵AB=4∴．∴CE=2．過點(diǎn)A作⊥BC于．3∵∠=60°，AB=BE，∴△ABE是等邊三角形．∴BG=GE=2．∴．∴四邊形是行四邊形．

B

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∴AGCF是形．∴AG=CF．在△中∠°，=4，∴AG=23．∴=23．523整理、描述數(shù)據(jù)分段學(xué)校

30≤≤39≤≤≤x≤≤x≤69≤≤79≤x≤90≤x≤甲乙

10

10

01

04

32

78

85分析數(shù)據(jù)經(jīng)統(tǒng)計(jì)，表格中m的是．·得出結(jié)論甲學(xué)校有400名初二學(xué)生，估計(jì)這次考試成績80分上人數(shù)為300··············4答不唯一，理由須支撐推斷結(jié).·································································724（）證明：∵AC⊙O的線，∴∠BAC=90°．···························································································1∵點(diǎn)E是BC的中點(diǎn)，∴AE=EC．∴∠=∠EAC，····························································································2∵∠=∠C+∠，∴∠=2C．···························································································（2解：連結(jié)AD∵AB為徑作O，∴∠ABD°．∵=，cosB，18∴=．·····································45

在eq\o\ac(△,Rt)ABC中AB=6，cos

，

∴BC．∵點(diǎn)E是BC的中點(diǎn)，∴．···············································5

C∴

75

．··········································25解）；·························································································（2如圖所示；···························································································精品

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1212APB（3如圖····································································································526解）拋物線

的對稱軸為直線，∴b=2．···································································（2①∴拋物線的表達(dá)式為

x

．∵A（x，（x，∴直線AB行x軸．∵，∴AB=3．∵對稱軸為x，∴=

．·············································∴當(dāng)x