2020年江苏南京市联合体中考数学一模试卷

．．．．．．．．．．．．．．知识像烛光，能照亮一个人，也能照亮无数的人。培根．．．．．．．．．．．．．．数学注意事项：．本试卷共6页．全卷分120分．考试时间为120分．考生答题全部答在答题卡上，答在本试卷上无效．．请认真核对监考师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符，再将自己的姓名、准考证号用0.5毫黑色墨水签字笔填写在答题卡及本试卷上．．答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑．如需改动，请用橡皮擦干后，再选涂其他答案答非选择题必用毫米黑色墨水签字笔写在答题卡上的定位置其他位置答题一律无效．．作图必须用2B铅笔作答，并请加黑加粗，描写清楚．一、选择题（大题共6小题，每小题，共12分在每小题所给出的四个选项中，恰有项是符合题目要求的，请将正确项前的字母代号填涂在答题相应位置上）4的算术平方根是A2B－2CD±2．2019年江省粮食总产达吨居全国第四位．用科学记数法表示000是的结果是A4054×104．计算－a2Aa5

B．4B．－a

C．7C．a6

D．D．．已eq\o\ac(△,)ABC∽△DEFeq\o\ac(△,)eq\o\ac(△,)DEF面之比为．若BC，则的是A2B．C．4．．下列整数中，与7最接近是A1B．已一次函数y＝＋图像如图所示，则y＝－2kx－b的图像可能是

OOxOx

OA

B．

C．

D．

（6题）二、填空题（本大题共0小题，每小2分，共20分．不需写出解答程，请把答案直接填写在答题卡相应位置上）．使式子＋x－1意义的x取值范围是▲．．计算27－3

的结果是．/

⌒x．．．．．．．知识像烛光，能照亮一个人，也能照亮无数的人。培根⌒x．．．．．．．．分解因式(－－＋1的结果是．．已知关于x的方程＋mx－3一个根，则另一个根为，＝▲．11.若一组据，3，，，x的差比另一组数据，67，，9的方差小，可为举个满足条件的值）．如图，四边形ABCD是的接四边形，若⊙O半为4且C＝2∠，则BD长为▲．．如图，将正六边形ABCDEF点D逆针旋转得六边形′C′F′，则1＝

▲．

F′

F

O

′

1

′

′

D

′

DD

′

O

（第12题）

（第13题）

（第15）

（第16题）k14.反例数＝的像过点（2，ba＝－6则ab▲．15.如图，在eq\o\ac(△,Rt)ACB中，C＝°，BC4，＝5BD平∠AC于D＝．．如图，在平面直角坐标系中，点A的坐标(2，点的标是(，0)．作点关的对称点B，点B的标是（▲，三解答本大题共题共88分请在答题卡指定区域内作答解时应写出文字说明、证明过程或演算步骤）x－1x＋6＋9分计(2．x＋1x－＋3≥x＋，分解不等式组2＞－，

并解集在数轴上表示出来．－4－3－－013/

知识像烛光，能照亮一个人，也能照亮无数的人。培根分课外兴趣小组为了某段路上机动车的车速查了一段时间内若干辆车的车车速取整数，单位：千/时）并制成如图示的频数分布直方图．已知车速在41千米时到50千米时的车辆数占车辆总数的．（1在这段时间内他们抽查的车有▲辆；（2被抽查车辆的车速的中位数所在速度段（单位：千时）是（▲）AB．．50.5~60.5D．60.5~70.5（3补全频数分布直方图；（4如果全天超速（车速大于千米时）的车有辆，则当天的车流量约为多少辆？车辆数20161284

128533030.5（第19）

车速(千米/时)．（）甲、乙、丙医生志愿报名参加新冠肺炎救治工.（1随机抽取，则恰是甲的概率是▲；（2随机抽取，求甲在其中的概率．现有120台小两种型号的挖掘机同时工作型挖掘机每小时可挖掘土方360立米小型挖掘机每小时可挖掘土立方米小共挖掘土方704立米求小型号的挖掘机各多少台？/

112知识像烛光，能照亮一个人，也能照亮无数的人。培根11222.（8分）一辆货车从地发以每小时的度匀速驶往地一段时间后，一辆轿车从B地发沿同一条路匀速驶往A地车驶小后距B地与轿车相遇图中线段表示货车离B地距离y与车行驶的时间的系．（1）两之间的距离▲

；（2求y与之的函数关系式；（3若两车同时到达各自目的地，在同一坐系中画出轿车离B地距离与车行驶时间x的数图像，用字说明该图像与x轴点所表示的实际意义．y∕160O3（第22题）

x∕h23.（8分）图①，在四边形中∠＝C＝90°，AB＝，求证：四边形是矩形；（2如图②，若四边形满∠A＝∠C＞90°＝，求证：四边形是行四边形DD

B

（图①）

（图②）/

知识像烛光，能照亮一个人，也能照亮无数的人。培根24.（8分如图，位于南偏西37°方向港口C位于A偏东35°方向，B位于西方向轮船甲从A出发沿正南向行驶40海里到点D，此时轮乙从出发沿正东向行驶海里至处，E位于D南偏西45°方向这时处距离港口C有多？（参考据：tan37°，）北37°D45

东E

（第24题）25.（）如图①，在矩形ABCD中，AB，BC，点E是BC边一点连接、，eq\o\ac(△,)的外接⊙O，交AD于，交于点，接FG.（1求eq\o\ac(△,)AFG∽△；（2当的长为时eq\o\ac(△,)为腰三角形；（3如图②，若＝，求证ABO相F

D

F

DGG

O

O

E

（图①）

（图②）26.（分）已知二次函数y＝x22＋m＋m－（是常数（1求证：不论为值，该函数的图像的顶点都在函数＝－1的像．（2若该函数的图像与函数y＝x＋的像有两个交点，则b的值范围为（▲）（

AbB．b＞

C．＞－

D．＞－2（3该函数图像与坐标轴交点的个数随的值变化而变化，直接写出交点个数及对应的m的值范围．/

．．知识像烛光，能照亮一个人，也能照亮无数的人。培根．．27.（分）【概念认识】在同一个圆中两条互相垂直且相等的弦定义等垂”两弦所在直线交点为等垂弦的分割点．如图①AB是O的AB＝CD，⊥，垂足为，AB是等垂弦为等垂弦AB、的割．

O

OD（图①）（图②）

D【数学理解】（1如图②，AB是⊙的，⊥OAOD⊥OB，分别交O点、，连接．求证：AB、CD是的垂弦．（在⊙中⊙O的半径为，E为垂弦AB、的割点，

BE1＝．AB的长度．AE3【题决（3）、⊙O的条弦，＝AB，且⊥AB，垂足为F．①在图③中，利用直尺和圆规作弦CD保留作图痕迹，不写作法②若⊙O的径为r，＝（m为数足与O的置关系随m的变化而化，直接写出点F与⊙O的置关系及对应点的值范围．BO（图③）/

知识像烛光，能照亮一个人，也能照亮无数的人。培根南市2020初毕生模试数试参答及分准说明本分标准每题给出了一或几种解法供参考果考生的解法与本解答不同参照本评分标准的精神给分．（错扣！该的了出要分明笔出1次面回扣；笔一直改来扣；面错，面法确算没的后面低一的）一、选择题（本大题共6小，每小题2分共12分题号答案

35ACDBC二、填空题（本大题共10题，每小题2分共分．，．

11．x≥．

．3．

．a－．（有，1）（案唯，＜＜6即．π．．

．．

．．

．．

三、解答题（本大题共题，共88分）题6）x＋2x－＋1)(－1)解：原式＝（－·························································x＋1x＋1(x＋3)2（分1分除乘1分分、母式解各1分x＋3(＋1)(x－1)＝·······························································5（加分x＋1(＋2＝

x－1x＋3

············································································6分（约分题6）解：解不等式①，得x≤1········································································解不等式②，得x＞－．·····································································4分∴原不等式组的解集为－2≤1····························································5分－－－2－014··············································································································分题8）解）40················································································（2）································································································4分（3图（方1分，数分）························································（4200÷＝（过、果分）·················································题8）解）．··························································································分（2）所有可能出现的结有乙）共，它们出现的可能性相同．所有的结果中，满足“选中甲事件）结/

1225知识像烛光，能照亮一个人，也能照亮无数的人。培根1225果有种，所以(A)＝＝．······································································分（解所可出的果（甲乙甲，（乙，）共3种它出的能相．有结中满“中（为件）的果种，以P(＝（举列或状过正3；子分正各1分；果1分无过仅正确果得分结没约不分结正但没列所结或有明可性1分；树图不结扣1分；表用勾扣1分列法举不但中有确果只有个的只1）题7）（设没方不分设且面思才设1分列程解问，个程2分解：设大型挖掘机台则小型挖掘(120－x)．·········································根据题意得［360＋200(120)］＝704000·····································5分解得x＝，120＝答：大型挖掘机台，小型挖掘机50台·············································722.（题8分解）；………（2y＝400x＋400；………………分（过程，结果分结果的2个子都对）（3如图，线段y即所求的像线失扣1分…………………6分货车行驶的时间为÷80＝5h可求出y的数表达式：y＝120－200该图像与x轴点标为（0分13

y∕400160

它表示的实际意义：货车从A地发时后，轿车从B地出发．……………………8

O

353（第22题）

x∕h题8）（1证明：如图①，连接，∵∠A＝∠＝90°在eq\o\ac(△,Rt)和eq\o\ac(△,Rt)中＝，＝，∴eq\o\ac(△,Rt)≌eq\o\ac(△,Rt)（···································································2分∴，∴四边形是行四边形，··································································3分∵∠A＝90°，∴四边形是形．············································································分D

A

D

（图①）

B

（图②）

F/

知识像烛光，能照亮一个人，也能照亮无数的人。培根（2如图②，分别过点B、作⊥点E，⊥BC于，····················分∵∠BAD∠BCD，∴∠＝∠，在△ABE△CDF中∠AEB＝∠＝，∠＝∠，AB＝CD∴△≌△AAS······································································6分∴BEDF，＝CF，由（）可得四边形EBFD是形，····························································7分∴＝BF，∴＝BC∵ABCD，＝BC∴四边形是行四边形·································································题8）解：如图，延长交于点，⊥.设EF＝x里．在eq\o\ac(△,Rt)DEF中∠＝，∵∠EDF＝，………分DFx∴tan45＝，DF，…………DF在eq\o\ac(△,Rt)中，∠DFE＝90°，∵∠＝，………分∴＝AFtan37°…………4分∴20≈0.75(40＋)，…

37°40D45°x

北

东∴x＝，………分∴AF＝ADDF＝80．

20E

x

F在eq\o\ac(△,Rt)中∠AFC＝，CF∵tan∠CAF＝，

（第24题）CF∵＝，························································································∴＝≈800.70∴＝EF＋＝40＋＝96······································································答：E处离港口约海里．题9）（1证明：∵四边形FGED是O的接四边形，∴∠FGE∠ADE＝180．·······································································1分∵∠AGF∠FGE＝180，∴∠AGF∠ADE．··················································································2分又∠GAFDAE，∴AFG∽△；···················································································3分（25、、9分·····························································6分（3证明：图②，过O作OHAB于H，反向延长OH交CD于点I∴∠AHI＝90°，/

F

DGH

O

I

O知识像烛光，能照亮一个人，也能照亮无数的人。培根O在矩形ABCD中∠BAD∠＝，∴∠AHI＝∠BAD＝∠ADC＝，∴四边形为形，∴HI＝＝，OID90°，即OICD∴DI＝CD3∵BE1，＝，∴＝，∵∠＝90，∴DE为直径OD为径，在eq\o\ac(△,Rt)DEC中，由勾股定理得DE＝10∴OD5在eq\o\ac(△,Rt)DIO中由勾股定理得∴IOOD2

－DI

＝4…………分∴OHHI－OI＝－4，……………8分∴是⊙O的径，又OH，∴AB与O相．···················································································题10分（1证明：∵y＝x

－2＋m＋m＝(x－)2＋m1··································································1分∴该函数的图像的顶点坐标为，m－························································2分将x＝代入＝－1得＝－1··································································∴不论为何值，该函数的图像的顶点都在数＝－1的像上．························（2）．································································································（3①当＞时，该函数图像与坐标轴交点的个数为；······························－15－1－②当m，＝，m时，该函数图像与坐标轴交点的个数为2；8分－15－－5－15－＋5③当m，＜m＜，＜＜1时该函数图像与标轴交点222的个数为3······················································································10题10分（1如图①，连接BC∵⊥O、⊥，∴∠＝∠BOD90°，

∴∠AOB，∴ABCD············································································