19南京联合体一模数学(含答案)

19南京联合体一模数学(含答案)-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIANG注意事项：2019年初中毕业生学业考试模拟卷数学1.本试卷共6页．全卷满分120分．考试时间为120分钟．考Th答题全部答在答题卡上，答在本试卷上无效．2.请认真核对监考教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合，再将自己的姓名、考试证号用0.5毫米黑色墨水签字笔填写在答题卡及本试卷上．3.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，请用橡皮擦干净后，再选涂其他答案．答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定位置，在其他位置答题一律无效．4.作图必须用2B铅笔作答，并请加黑加粗，描写清楚．一、选择题（本大题共6小题，每小题2分，共12分．在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答．题．卡．相．应．位．置．上）1.9的值等于A．3B．－3C．±3D．±32.下列运算结果正确的是A．a÷a＝aB．（a）＝aC．（ab）＝abD．aa＝a3．已知a为整数，且满足5＜a＜10，则a的值为A．4B．3C．214..已知反比例函数y＝的图像经过点（1，3），若x＜－1，则y的取值范围为A．y＞－3B．y＜3C．－3＜y＜0D．0＜y＜35.如图，将△ABC绕点A旋转任意角度得到△AB'C'，连接BB'、CC'，则BB'：CC'等于A．AB：ACB．BC：ACC．AB：BCD．AC：ABB´ADFC´CAB（第5题）(第6题)BEC6.如图，在边长为4的正方形ABCD中，点E、F分别是BC、CD上的动点，且EF＝4，G是EF的中点，下列结论正确的是A．AG⊥EFB．AG长度的最小值是42－2C．BE＋DF＝4D．△EFC面积的最大值是2ABAEDBCF二、填空题（本大题共10小题，每小题2分，共20分．不需写出解答过程，请把答案直接填写在答．题．卡．相．应．位．置．上）7.在－3、4、－2、5四个数中，任意两个数之积的最小值为▲．8．2018年江苏省实现GDP约92500亿元．用科学记数法表示92500是▲．．若式子在实数范围内有意义，则x的取值范围是▲．10．计算＋×的结果是▲．11．已知关于x的方程x2＋mx－2＝0的两个根为x、x，若xx－xx＝6，则m＝▲．12．点（m，y），（m＋1，y）都在函数y＝kx＋b的图像上，若y－y＝3，则k＝▲．13.某校九年级(1)班40名同学期末考试成绩统计表如下．成绩x（单位：分）60≤x＜7070≤x＜8080≤x＜9090≤x≤100人数414166能为70；④成绩的极差可能为40．其中所有正确结论的序号是▲．14.如图，将边长为2的正六边形ABCDEF绕顶点A顺时针旋转60°，则旋转后所得图形与正六边形ABCDEF重叠部分的面积为▲．EDFC（第14题）（第15题）（第16题）15.如图，在矩形ABCD中，AB＝4，BC＝6，E为AD的中点，△CED的外接圆与BE交于点F，则BF的长度为▲．OAB916.如图，AB是⊙O的弦，若⊙O的半径长为6，AB＝62，在⊙O上取一点C，使得AC＝82，则弦BC的长度为▲．三、解答题（本大题共11小题，共88分．请在答．题．卡．指．定．区．域．内作答，解答时应写出文字说明、 证明 住所证明下载场所使用证明下载诊断证明下载住所证明下载爱问住所证明下载爱问 过程或演算步骤）17．（7分）计算m＋2＋÷．18．（7分）解不等式组x＋2＜5，并把不等式组的解集在数轴上表示出来．－4－3－2－10123419．（7分）某区对参加2019年中考的3000名初中毕业生进行了一次视力抽样调查，绘制出如下频数分布表和频数分布直方图．某区2019年初中毕业生视力抽样频数分布表某区2019年初中毕业生视力抽样频数分布直方图频数/人数60力的值为▲；（3）若视力在4.9以上（含4.9）均为正常，根据以上信息估计全区初中毕业生中视力正常的学生有多少人？20．（8分）在课外活动时间，小明、小华、小丽做“互相传球”游戏（球从一人随机传给另一人），球从一人传到另一人就记为一次传球．现从小明开始传球．（1）经过三次传球后，求球仍传到小明处的概率；（2）经过四次传球后，下列说法：①球仍传到小明处的可能性最大；②球传到小华处的可能性最大；③球传到小华和小丽处的可能性一样大．其中所有正确结论的序号是（▲）A．①③B．②③C．①②③21．(7分)如图，在△ABC中，D是BC的中点，DE⊥AB，DF⊥AC，垂足分别是点E、F，BE＝CF．求证AD是△ABC的角平分线．AEFBDC22．（6分）【阅读材料】南京市地铁公司规定：自2019年3月31日起，普通成人持储值卡乘坐地铁出行，每个自然月内，达到规定消费累计金额后的乘次，享受相应的折扣优惠（见下图）.地铁出行消费累计金额月底清零，次月重新累计．比如：李老师二月份无储值卡消费260元，若采用新规持储值卡消费，则需付费150×0.95＋50×0.9＋60×0.8＝235.5元．【解决问题】甲、乙两个成人二月份无储值卡乘坐地铁消费金额合计300元（甲消费金额超过150元，但不超过200元）．若两人采用新规持储值卡消费，则共需付费283.5元．求甲、乙二月份乘坐地铁的消费金额各是多少元？16°37°45°23．（9分）甲、乙两艘快艇同时从A港口沿直线驶往B港口，甲快艇在整．个．航．行．的．过．程．中速度v海里/小时与航行时间t小时的函数关系如图①所示（图中的空心圈表示不含这一点），乙快艇一直保持匀速航行，两快艇同时到达B港口．（1）A、B两港口之间的距离为▲海里；（2）若甲快艇离B港口的距离为s海里，乙快艇离B港口的距离为s海里，请在图②中分别画出s、s与t之间的函数图像．（3）在整个行驶过程中，航行多少小时时两快艇相距5海里？（4）s/海里165150135120105907560453015O123②t/小时24．（8分）如图，有两座建筑物AB与CD，从A上有一点E，点E到B的距离为24米，从E°、45°．求建筑物CD的高度．（参考数据：DAv/(海里/小时)6030O13t/小时1BEC25．（9分）已知二次函数y＝mx－2mx（m为常数，且m≠0）．（1）求证：不论m为何值，该函数的图像与x轴有两个公共点．（2）将该函数的图像向左平移2个单位．①平移后函数图像所对应的函数关系式为▲；②若原函数图像顶点为A，平移后的函数图像顶点为B，△OAB为直角三角形（O为原点），求m的值．GH26．（10分）如图，在☐ABCD中，连接AC，⊙O是△ABC的外接圆，⊙O交AD于点E．（1）求证CE＝CD；（2）若∠ACB＝∠DCE．②求证CD与⊙O相切；②若⊙O的半径为5，BC长为45，则AE=▲．AEDOB27．（10分）如图①，在☐ABCD中，点E、F分别在AD、BC上，且AE＝CF，连接AF、BE交于点G，连接CE、DF交于点H．（1）求证四边形EGFH为平行四边形．（2）提出问题：AEDBFC1在AD、BC边上是否存在点E、F，使得四边形EGFH为矩形小明从特殊到一般探究了以下问题．【特殊化】如图②，若∠ABC=90°，AB=2，BC=6．在AD、BC边上是否存在点E、F，使得四边形EGFH为矩形？若存在，求出此时AE的长度；若不存在，说明理由．ADBC②【一般化】如图③，若∠ABC=60°，AB=m，BC=n．在AD、BC边上是否存在点E、F使得四边形EGFH为矩形？指出点E、F存在（或不存在）的可能情况，写出此时m、n满足的条件，并直接写出存在时AE的长度．（用含m、n的代数式表示）BC③2019年初中毕业生学业考试模拟测试数学试卷参考答案及评分 标准 excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载 说明：本评分标准每题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，参照本评分标准的精神给分．一、选择题（本大题共6小题，每小题2分，共12分）题号123456答案ADBCAB二、填空题（本大题共10小题，每小题2分，共20分）7．－15．8．9.25×10．9．x≠1．10．33．11．－4．12．－3．13．①④．14．23．15．3.6．16．8＋22或8—22.．三、解答题（本大题共11小题，共88分）17．（本题7分）m－4＋3解：原式＝÷m－2m1································································3分2(m－2)(m1)(m－1)＝m－2·2(m－2)m1··············································································6分＝2m－2··································································································7分18．（本题7分）解：解不等式①，得x＜3．························2分解不等式②，得x＞－3．·······················4分∴原不等式组的解集为－3＜x＜3．···················6分－4－3－2－11234···············································································································7分19．（本题7分）解：（1）a=50，b=0.05；········································································2分（2）补图略；····························4分（3）0.3×3000=900．·························7分20．（本题8分）解：（1）用a，b，c分别表示小明，小华，小丽，所有可能出现的结果有：（b，a，c）、（b，a，b）、（b，c，a）、（b，c，b）、（c，a，b）、（c，a，c）、（c，b，a）、（c，b，c）共8种，它们出现的可能性相同．所有的结果中，满足“球仍传到小明处”（记为事件A）的结果有2种，所以P(A)==（2）A····································································································8分21．（本题7分）证明：∵DE⊥AB，DF⊥AC，∴∠DEB＝∠DFC＝90°．············································································1分∵D是BC的中点，∴BD＝DC．·····························································································2分在Rt△EBD和Rt△FCD中，BE＝CF，BD＝DC，∴Rt△EBD≌Rt△FCD，···········································································4分∴ED＝FD．·····························································································5分∵DE⊥AB，DF⊥AC，∴AD是△ABC的角平分线．········································································7分22.（本题6分）解：设甲二月份乘坐地铁消费的金额是x元，乙二月份乘坐地铁消费的金额是y元．根据x＋y＝300，题意列方程组得····································4分150×0.95＋0.9(x－150)＋0.95y＝283.5．x＝180，解得y＝120．答：甲二月份乘坐地铁消费的金额是180元，乙二月份乘坐地铁消费的金额是120元．···············································································································6分23．（本题9分）（1）150·································································································2分（2）如图·······························4分s/海里165150135120105907560453015O123t/小时（3）当0≤t≤1时，s所对应的函数关系式为s＝–30t＋150；·········5分当1＜t≤3时，s所对应的函数关系式为s＝–60t＋180；···········6分当0≤t≤3时，s所对应的函数关系式为s＝–50t＋150；···········7分当0≤t≤1时，(–30t＋150)–(–50t＋150)＝5；解得t＝0.25小时；·······8分当1≤t≤3时，(–60t＋180)–(–50t＋150)＝5；解得t＝2.5小时；当航行0.25小时或2.5小时时，两快艇相距5海里．·············9分24．（本题8分）解：如图，过点A作AF⊥CD，垂足为F．设CD＝xm．在Rt△ECD中，∠DEC＝45°，∵tan45°＝CD，·····························1分CE∴CE＝CD＝x．·····················································································2分tan45°＝，在Rt△ABE中，∠AEB＝37°，∵tan37°3分∴AB＝BEtan37°≈0.75×24＝18···································································4分∴FC＝AB＝18∴DF＝DC－FC＝x－18在Rt△AFD中，∠DAF＝16°，∵tan16°＝DF，·····························5分AF∴AF＝≈∴BC＝AF＝···································································································6分又∵BC＝BE＋EC7分解得x＝36答：建筑物CD的高度为36米．·····················8分25．（本题9分）（1）证明：当y＝0时，mx－2mx＝0，···························································1分解得x＝0，x＝2．····················································································2分∴函数图像与x轴的交点坐标为（0，0），（2，0）．即不论m为何值时，函数的图像与x轴有两个公共点．···········3分（2）①y＝mx＋2mx或y＝m(x＋1)－m·············································································5分②A(1，－m)，B(－1，－m)，······················7分则OA＝1＋m，OB＝1＋m，AB＝4，∴在Rt△OAB中，OA＋OB＝AB，即1＋m＋1＋m＝4····································8分∴m＝±1····························································································9分26．（本题10分）（1）证明：∵四边形ABCD是平行四边形，∴∠D＝∠B．·····························1分∵⊙O是四边形ABCE的外接圆，∴∠B＋∠AEC＝180°．·············································································2分∵∠DEC＋∠AEC＝180°，∴∠B＝∠DEC．·····················································································3分∴∠D＝∠DEC．∴CE＝CD．·····························································································4分（2）证明：连接CO并延长交⊙O于点F，∵在△ABC和△DCE中，∠B＝∠D，∠ACB＝∠DCE．∴∠DEC＝∠BAC················································································································5分又∵∠DEC＝∠D∴∠B＝∠BAC，即AC＝BC·······························································································6分∴CF平分∠ACB∴∠BAF＝∠BCF＝∠ACF又∵∠BAF＋∠BAC＝90°，AB∥CD，∴∠BAC＝∠ACD⑥图5图6∴∠ACF＋∠ACD＝∠DCF＝90°，即CD⊥CF，···························································································7分∵点C在⊙O上∴CD与⊙O相切·····················································································8分（3）45·······························································································10分527．（本题10分）（1）∵四边形ABCD是平行四边形；∴AD＝BC，AD∥BC．∵AE＝CF；∴ED＝BF．···························································································1分∵AE＝CF，AE∥CF；∴四边形AECF是平行四边形．···································································2分∴AF∥EC．∵ED＝BF，ED∥BF；∴四边形EDFB是平行四边形．··································································3分∴BE∥DF．∵AF∥EC，BE∥DF，∴四边形EGFH是平行四边形．··································································4分（2）如图1，以BC为直径作⊙O，⊙O与AD有两个不同公共点，即为所求点E，5分由题意易证△BAE∽△EDC，∴ABED＝AECD，26－AE＝AE2，AE＝3±5．·············6分BCB图1图2图3图4（3）以BC为直径作⊙O，⊙O的半径是n，2①如图2，当0＜n＜3m时，⊙O与AD无公共点，没有符合条件的点E；···7分②如图3当n＝3m时，⊙O与AD有1个公共点，即为所求的点E，AE＝(－1)m(也可写为AE＝(n-m)或AE＝(n)；8分③如图4当3m＜n＜2m时，⊙O与AD有2个公共点，即为所求的点E，④·········································································································AE＝(n－m－)或AE＝(n－m)；9分④如图5，图6当n≥2m时，符合条件的点E有1个，AE＝(n－m)．10分