2021南京联合体中考数学一模试卷及答案

【联合体数学】2021年一模考试试卷分析整体难度星级：★★★★优秀分数线：110良好分数线：102 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 号考点 内容 财务内部控制制度的内容财务内部控制制度的内容人员招聘与配置的内容项目成本控制的内容消防安全演练内容 难度1数与式有理数比大小★2数与式幂的运算★3数与式代数式无意义★4数与式实数与数轴★5图形的变换图形折叠+三角形倒角★6函数数形结合（利用函数图像判断方程根的情况）★★7数与式平方根与算数平方根★8数与式科学计数法★9数与式二次根式的化简★10方程与不等式根与系数的关系★11数与式乘法公式的简单应用★12统计与概率中位数★13图形的变换中心对称-求对称点的坐标★14方程与不等式二元一次方程★15圆切线的性质+圆周角定理★16圆几何综合（隐圆求弓形高度）★★17数与式分式的混合运算★18方程与不等式一元一次不等式组★19相似三角形+矩形矩形的性质+相似三角形的判定和性质★20统计与概率统计的应用（方差）★21统计与概率概率计算（树状图法或列表法）★22四边形平行四边形的性质和判定+菱形的判定★★23三角函数解直角三角形的应用（俯角仰角问题）★★24圆+四边形菱形的性质+圆周角定理+切线的证明★★一次函数的应用（行程问题）25函数★★☆数形结合（s-t图像中直线的斜率与速度的关系）二次函数与x轴交点、过定点问题；26函数★★☆分类讨论（根据不同参数值确定抛物线经过象限）27圆复杂尺规作图（作过定点并与定线相切的圆）★★★【联合体数学】2021联合体一模考试（试卷+ 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 ）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，请将正确选项的字母代号填涂在答题卡相应位置上）1．比－2小的数是（）A．－3B．－1C．0D．12．下列运算正确的是（）2A.a3a3a6B.a2a3a6C.a6a2a4D.a3a93．当x＝1时，下列式子没有意义的是（）xx－1xA．B．C．x－1D．x＋1xx－14．如图，数轴上两点M、N所对应的实数分别为m、n，则m－n的结果可能为（）A．4B．3C．2D．－0.3NM－3－1O－212(第4题)5．如图，在△ABC中，∠C＝90°，沿DE翻折使得A与B重合，若∠CBD＝26°，则∠ADE的度数是（）CDBAE(第5题)A．57°B．58°C．59°D．60°q6．关于x的方程px2＋p＝（p、q为常数且pq≠0）的根的情况，下列结论中正确的是（）xA．一个实数根B．二个实数根C．三个实数根D．无实数根2/16二、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的位置上．．．．．．．．．）7．5的平方根为，5的算术平方根为．8．华为正在研制厚度为0.000000005m的芯片．用科学记数法表示0.000000005是.9．计算3×（12－3）的结果是．210.若x－4x－7＝0的两个根为x1、x2，则x1＋x2－x1x2的值是．11．如图①，一个长为2a，宽为2b的长方形，用剪刀沿图中虚线（对称轴）剪开，把它分成四块全等的小长方形，然后按照图②那样拼成一个面积为49的大正方形，若中间小正方形的面积为1，则a＝、b＝．人数1514131210952①②012345分数（第11题）（第12题）12．光明中学全体学生参加社会实践活动，从中随机抽取50人的社会实践活动成绩制如图所示的条形统计图，则这50人的社会实践活动成绩的中位数是．13．若点A与点B（1，1）关于点C（－1，－1）对称，则点A的坐标是．14．笔记本4元/本，钢笔5元/支，某同学购买笔记本和钢笔恰好用去162元，那么最多购买钢笔支．15．如图，P是⊙O外一点，PB、PC是⊙O的两条切线，切点分别为B、C，若∠P为38°，点A在⊙O上（不与B、C重合），则∠BAC＝°．BADEOPlCBC（第15题）（第16题）16．如图，在边长为2cm的正方形ABCD中，直线l经过点D，作BE⊥l，垂足为E，连接AE．若AE＝BE，则△ABE的面积为cm2．三、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、证明过程或演算 步骤 新产品开发流程的步骤课题研究的五个步骤成本核算步骤微型课题研究步骤数控铣床操作步骤 ）a＋2a2－117．（7分）计算1＋÷．aa1x018．（7分）解不等式组x1，并写出不等式组的整数解．3219．（8分）如图，在矩形ABCD中，E是BC的中点，DF⊥AE，垂足为F．（1）求证△ADF∽△EAB；（2）若AB＝12，BC＝10，求DF的长．DCEFAB（第19题）20．（8分）某商场统计了A、B两种品牌洗衣机7个月的销售情况，结果如下：月份销量一月二月三月四月五月六月七月品牌A品牌16312924242420B品牌16202425262730（1）分别求这7个月A、B两种品牌洗衣机销量的方差；（2）由于库存不足，商场采购部欲从厂家采购A、B两种品牌洗衣机以满足市场需求．请你结合上述两种品牌洗衣机的销售情况，对商场采购部提出建议，并从两个不同角度说明理由.21．（8分）甲、乙、丙互相传球．假设他们相互之间传球是等可能的，并且由甲开始传球．（1）经过2次传球后，求球仍回到甲手中的概率；（2）经过3次传球后，球仍回到甲手中的概率为．22．（8分）如图，在□ABCD中，E、F分别为AD、BC的中点，点M、N在对角线AC上，且AM＝CN.（1）求证四边形EMFN是平行四边形；（2）若AB⊥AC，求证□EMFN是菱形．AEDMNBCF（第22题）23．（8分）为了测量悬停在空中A处的无人机的高度，小明在楼顶B处测得无人机的仰角为45°，小丽在地面C处测得A、B的仰角分别为56°、14°．楼高BD为20米，求此时无人机离地面的高度．（参考数据：tan14°≈0.25，tan56°≈1.50）A45°B56°14°CD（第23题）24．（8分）如图，在菱形ABCD中，E是CD上一点，且∠CAE＝∠B，⊙O经过点A、C、E．（1）求证AC＝AE；（2）求证AB与⊙O相切..ADEOBC（第24题）25．（8分）2020年江苏开通了多条省内高铁，其中一条可以从南京——镇江——扬州——淮安的高铁线路如图①所示，本线路高铁最高速度不超过每分钟5千米．现有甲、乙两车按以下方式营运，甲车从南京匀速行驶去淮安，在镇江和扬州两站都停靠5分钟；乙车从南京匀速行驶直达淮安，乙车比甲车晚出发20分钟．设甲车出发x分钟后行驶的路程为y1千米，图②中的折线O—A—B—C—D—E表示在整个行驶过程中y1与x的函数图像．（1）甲车速度为千米/分；（2）若乙车行驶1小时到达淮安，则乙车出发多久后与甲车相遇？（3）若乙车行驶的过程中不得与甲车在镇江站与扬州站的站台内相遇，并要在甲之前到达淮安，则乙车速度v乙的范围为.y/千米E270淮安240210120千米180150CD120A90B扬州6090千米30千米分钟60x/南京镇江O102030405060708090100110120①②26．（9分）已知二次函数y＝mx2－2(m＋1)x＋4（m为常数，且m≠0）．（1）求证：不论m为何值，该函数的图像与x轴总有公共点；（2）不论m为何值，该函数的图像都会经过两个定点，这两个定点的坐标分别为、；（3）该函数图像所经过的象限随m值的变化而变化，直接写出函数图像所经过的象限及对应的m的取值范围．27.（9分）如图，在△ABC中，∠ACB＝90°．用直尺与圆规分别作出满足下列条件的⊙O.（不写作法，保留作图痕迹）（1）在图①中，⊙O过点C且与AB相切；（作出一个即可）（2）在图②中，D为AB上一定点，⊙O过点C且与AB相切于点D；（3）在图③中，E为AC上一定点，⊙O过点C、E且与AB相切．AADBCBC①②AEBC③【联合体数学】2021年一模考试答案一、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，请将正确选项的字母代号填涂在答题卡相应位置上）题号123456答案ACDABAqq第6题解析:关于x的方程px2p的根的个数即为函数ypx2p与y的交点个数xx①当且时②当且时③当且时④当且时p0q0p0q0p0q0p0q0综上所述，两个函数始终有1个交点，故方程有1个实数根.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的位置上．．．．．．．．．）题号7891011答案5；551093114；3题号1213141516答案43,33071或10921第16题干中有“如图”，且没有提到动点，所以我们按照一个答案给的，与学校有差异，下面是两个答案的解析第16题解析：∵BE⊥l∴∠BED=90°∴E在以BD为直径的圆上，圆心为BD中点O，如图所示∵AE=BE∴E在AB的垂直平分线上∵OA=OB'∴EO所在直线为AB的垂直平分线，交⊙O于点E、E∴M为AB中点1∴OM=AD=12∵AB=AD=222∴BD=ABAD=22∴OE=2∴EM21，EM'2111'∴SABEM21或S'ABEM21ABE2ABE2三、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、证明过程或演算步骤）17．（本题7分）aa＋2a2－1解：原式＝（＋）÷··································································1分aaa2a＋2a＝··············································································3分aa2－12(a＋1)a＝····································································5分a(a＋1)(a－1)2＝··························································································7分a－118．（本题7分）解：解不等式①，得x≥1．·········································································2分解不等式②，得x＜5．·········································································4分∴原不等式组的解集为1≤x＜5．····························································6分∴原不等式组的整数解为：1、2、3、4．··················································7分19．（本题8分）（1）证明：∵四边形ABCD矩形，∴∠DAB＝∠B＝90°，···································································1分即∠EAB＋∠DAF＝90°．DC∵DF⊥AE，∴∠AFD＝90°．E∴∠ADF＋∠DAF＝90°，FAB∴∠ADF＝∠EAB．·······································································2分又∵∠DFA＝∠B＝90°，·································································3分∴△ADF∽△EAB．·······································································4分（2）∵在矩形ABCD中，E是BC的中点，BC＝10．∴AD＝BC＝10，BE＝5．·······························································5分在Rt△ABE中，AB＝12，BE＝5，由勾股定理得：AE＝13，··················6分∵△ADF∽△EAB，DFAD∴＝，·················································································7分ABAEDF10120∴＝，即DF＝．································································8分12131320．（本题8分）－1解：（1）x＝（16＋31＋29＋24＋24＋24＋20）＝24········································1分A7－1x＝（16＋20＋24＋25＋26＋27＋30）＝24；·······························2分B71S2＝×[(16－24)2＋(31－24)2＋(29－24)2＋(24－24)2＋(24－24)2＋(24－24)2＋(20－24)2]＝22┈分A731130S2＝×[(16－24)2＋(20－24)2＋(24－24)2＋(25－24)2＋(26－24)2＋(27－24)2＋(30－24)2]＝┈分B774－－（）由，可知、两种品牌平均销量相当，由22，可知品牌销量的离散2xA＝xBABSB＜SAB程度较小，由表格可知，B品牌一月到七月的销量呈上升趋势，·····················7分故建议商场采购B品牌冰箱．·································································8分21．（本题8分）解：（1）所有可能出现的结果有：（乙，甲）、（乙，丙）、（丙，甲）、（丙，乙）共4种，它们出现的可能性相同．所有的结果中，满足“球仍回到甲手中”（记为事件A）的21结果有4种，所以P(A)＝＝．·····························································6分421（2）．··································································································8分422．（本题8分）证明：（1）∵四边形ABCD是平行四边形，∴AD∥BC，AD＝BC．····································································1分∴∠EAM＝∠FCN．∵E，F分别是AD，BC的中点，又AD＝BC，∴AE＝DE＝BF＝CF．∵AM＝CN，∴△AEM≌△CFN．········································································2分∴EM＝FN，∠AME＝∠CNF．∵∠AME＋∠EMN＝180°，∠CNF＋∠FNM＝180°．∴∠EMN＝∠FNM．∴EM∥FN．··················································································3分∴四边形EMFN是平行四边形．························································4分（2）如图，连接EF交AC于点O．·························································5分由（1）得AE∥BF，AE＝BF．∴四边形AEBF是平行四边形．·························································6分∴AB∥EF．∵AB⊥AC，∴∠BAC＝90°．∴∠COF＝∠BAC＝90°．∴EF⊥MN，··················································································7分∴平行四边形EMFN是菱形．···························································8分23．（本题8分）解：如图，过点A作AE⊥CD，垂足为E，过点B作BF⊥AE，垂足为F．设BF＝x，由题意可得，ED＝BF＝x，EF＝BD＝20．在Rt△BCD中，∠BCD＝14°，BD∵tan14°＝，···················································································1分CDBD20∴CD＝≈＝80．·····································································2分tan14°0.25在Rt△ABF中，∠ABF＝45°，AF∵tan45°＝，····················································································3分BFBF∴AF＝＝x．···············································································4分tan45°在Rt△ACE中，∠ACE＝56°，AE∵tan56°＝，···················································································5分CE∴AE＝CE·tan56°＝(CD－ED)·tan56°≈1.50(80－x)．····································6分∵AE＝AF＋EF，∴1.50(80－x)＝x＋20．··················································7分∴x≈40，即AE＝AF＋EF＝60．该无人机的高度为60m．·······································································8分24．（本题8分）证明：（1）∵菱形ABCD，∴DA＝DC，∠D＝∠B，AB∥CD，∴∠ACD＝∠CAD＝∠CAE＋∠DAE，····················································1分∵∠D＝∠B，∠CAE＝∠B，∴∠D＝∠CAE，···············································································2分又∵∠AEC＝∠D＋∠DAE，∴∠ACD＝∠AEC，············································································3分∴AC＝AE．······················································································4分（2）连接OA、OC，∵OA＝OC，∠AOC＝2∠AEC，1∴∠OAC＝∠OCA＝(180°－2∠AEC)＝90°－∠AEC，·······························5分2∵AB∥CD，∴∠ACD＝∠BAC，又∵∠ACD＝∠AEC，∴∠BAC＝∠AEC，············································································6分∴∠BAC＋∠OAC＝90°，···································································7分又∵点A在⊙O上，∴AB与⊙O相切．·············································································8分25．（本题8分）解：（1）3；································································································2分（2）法一：由题可知：y乙＝4.5（x－20），···················································3分y＝90+3（x－35），······························································4分BC当y乙＝y时，4.5（x－20）＝90+3（x－35）·····························5分BC解得：x＝5050－20＝30．所以，乙车出发30分钟后与甲相遇.··········································6分法二：设乙车出发t分钟后与甲相遇.由题可知：4.5t＝3（t+15）.··············5分解得：x＝30所以，乙车出发30分钟后与甲相遇..··········································6分271530（3）v或v5······························································8分8乙47乙解析：①甲、乙两车在镇江站之前相遇，则恰好到镇江站时速度最小（取不到，下同）90v乙93020由题意得：v5，所以不成立乙②甲、乙两车在镇江站和扬州站间相遇，则恰好离开镇江站时速度最大，恰好到达扬州站时速度最小1509030v乙v乙655203520，730因为所以v乙5，v乙57③甲、乙两车在扬州站和淮安站间相遇，则恰好离开扬州站时速度最大，恰好到达淮安站时速度最小2701502715v乙v乙100206020，84综上所述：271530v乙或v乙584726．（本题9分）解：（1）法一：令y＝0，即mx2－2(m＋1)x＋4＝0··········································1分b2－4ac＝[－2(m＋1)]2－4m×4························································2分＝4m2－8m＋4＝4(m－1)2∵m120∴b24ac4m120所以方程总有实数根．∴该函数的图像与x轴总有公共点∴该函数的图像与x轴总有公共点；······································3分法二：令y＝0，即mx2－2(m＋1)x＋4＝0··················································1分2∵m≠0，解得x＝2，x＝，··························································2分12m所以方程总有实数根．∴该函数的图像与x轴总有公共点；··················································3分（2）(0，4)、(2，0)·······································································5分解析：ymx22mx14mx22mx24xmx22x2x4当x22x0即x0时，y4或x2时，y0故不论m为何值，始终过定点(0，4)、(2，0)（3）m＜0时，函数图像过一、二、三、四象限；m＝1时，函数图像过一、二象限；0＜m＜1或m＞1时，函数图像过一、二、四象限．···························9分2解析：由（1）知，函数与x轴交点坐标(2，0)，（，0）m由（2）知，函数图像必过(0，4)、(2，0)两点分为两种情况2①m0时，抛物线开口向下，与x轴的另外一个交点（，0）在x轴负半轴，m此时抛物线过一、二、三、四象限2②m0时，抛物线开口向上，与x轴的另外一个交点（，0）在x轴正半轴m若与x轴只有一个交点，即两个交点重合，m1时，此时抛物线过一、二象限若与x轴有两个交点，即0m1或m1时，抛物线过一、二、四象限.27．（本题9分）（1）如图，⊙O即为所求．（答案不唯一，以下两种解法供参考）·······································3分AAMOBOCBC作法1：作BAC的角平分线，交BC于点O，以O为圆心，OC为半径画圆，即为所求.作法2：过C点作AB的垂线，垂足为M，作CM的中垂线，交CM于点O，以O为圆心，OC为半径画圆，即为所求.（2）如图，⊙O即为所求．································································A························6分DCBO作法：连接CD，作CD的中垂线，过D点作AB的垂线，交CD中垂线于点O，以O为圆心，OD为半径画圆，即为所求.（3）如图，⊙O即为所求．························································································9分右图痕迹为左图痕迹的一部分，两个图中的A、E、C相同，在AB上截取AN＝AM．AANEMEOBCC作法：右图：以AC为直径画圆，过E作AC的垂线，交该圆于M，连接AM左图：作CE中垂线，在AB上截取AN＝AM，过N作AB的垂线，交CE中垂线于点O，以O为圆心，OC为半径画圆，即为所求.