上海市普陀区2019年初三二模数学试卷(含答案)

—1—普陀区2018学年第二学期初三质量调研数学试卷（时间：100分钟，满分：150分）考生注意：1．本试卷含三个大题，共25题．答题时，考生务必按答题要求在答题纸规定的位置上作答，在草稿纸、本试卷上答题一律无效．2．除第一、二大题外，其余各题如无特别说明，都必须在答题纸的相应位置上写出证明或计算的主要步骤．一、选择题：（本大题共6题，每题4分，满分24分）[下列各题的四个选项中，有且只有一个选项是正确的，选择正确项的代号并填涂在答题纸的相应位置上]1.下列计算中，正确的是·············································································（▲）（A）235()aa；（B）236aaa；（C）2236aaa；（D）2235aaa．2.如图1，直线1l//2l，如果130，250，那么3········································（▲）（A）20；（B）80；（C）90；（D）100．3.2011年，国际数学协会正式宣布，将每年的3月14日设为国际数学节，这与圆周率π有关．下列表述中，不正确的是·······································································（▲）（A）π=314.；（B）π是无理数；（C）半径为1cm的圆的面积等于πcm2；（D）圆周率是圆的周长与直径的比值．4.下列函数中，如果0x，y的值随x的值增大而增大，那么这个函数是·············（▲）（A）2yx；（B）2yx；（C）1yx；（D）21yx．5.如果一组数据3、4、5、6、x、8的众数是4，那么这组数据的中位数是············（▲）（A）4；（B）4.5；（C）5；（D）5.5．l1l2图1123—2—6.如图2，ABCD的对角线AC、BD交于点O，顺次联结ABCD各边中点得到的一个新的四边形，如果添加下列四个条件中的一个条件：①AC⊥BD；②△△ABOCBOCC；③DAOCBO；④DAOBAO，可以使这个新的四边形成为矩形，那么这样的条件个数是································································································（▲）（A）1个；（B）2个；（C）3个；（D）4个．二、填空题：（本大题共12题，每题4分，满分48分）7.分解因式：22aa＝▲．8.函数131yx的定义域是▲．9.不等式组21034xxx，≤的解集是▲．10.月球离地球近地点的距离为363300千米，数据363300用科学记数法表示是▲．11.如果2a、1b，那么代数式2ab的值等于▲．12.如果关于x的方程2320xxm有两个相等的实数根，那么m的值等于▲．13.抛物线225yaxax的对称轴是直线▲.14.张老师对本校参加体育兴趣小组的情况进行调查，图3-1和图3-2是收集数据后绘制的两幅不完整统计图．已知参加体育兴趣小组的学生共有80名，其中每名学生只参加一个兴趣小组．根据图中提供的信息，可知参加排球兴趣小组的人数占参加体育兴趣小组总人数的百分数是▲.15.如图4，传送带AB和地面BC所成斜坡的坡度为1:3，如果它把物体从地面送到离地面2米高的地方，那么物体所经过的路程是▲米.（结果保留根号）图2ABCDO图4BC2米A图3-1图3-2篮球45%足球排球排球足球篮球球24人数兴趣小组—3—16.如图5，AD、BE是△ABC的中线，交于点O，设OBa，ODb，那么向量AB用向量a、b表示是▲．17.如图6，一个大正方形被平均分成9个小正方形，其中有2个小正方形已经被涂上阴影，在剩余的7个白色小正方形中任选一个涂上阴影，使图中涂上阴影的三个小正方形组成轴对称图形，这个事件的概率是▲．18.如图7，AD是△ABC的中线，点E在边AB上，且DE⊥AD，将△BDE绕着点D旋转，使得点B与点C重合，点E落在点F处，联结AF交BC于点G，如果52AEBE，那么GFAB的值等于▲．三、解答题：（本大题共7题，满分78分）19．（本题满分10分）计算：312019212sin60227(1)2．20．（本题满分10分）解方程：242193xxx．E图5ABCDO图6图7ABCDE—4—21．（本题满分10分）如图8，已知点D、E分别在△ABC的边AB和AC上，DE//BC，13DEBC，△ADE的面积等于3．（1）求△ABC的面积；（2）如果9BC，且2cot3B，求AED的正切值．22．（本题满分10分）某工厂生产一种产品，当生产数量至少为20吨，但不超过60吨时，每吨的成本y（万元/吨）与生产数量x（吨）之间是一次函数关系，其图像如图9所示．（1）求出y关于x的函数解析式；（2）如果每吨的成本是4.8万元，求该产品的生产数量；（3）当生产这种产品的总成本是200万元时，求该产品的生产数量．ABCDE图80x（吨）图9y（万元/吨）5.646206028—5—23．（本题满分12分）已知：如图10，在四边形ABCD中，ADBC，点E在AD的延长线上，ACEBCD，ECEDEA2．（1）求证：四边形ABCD为梯形；（2）如果ECABEAAC，求证：ABEDBC2．24．（本题满分12分）在平面直角坐标系xOy中，直线243yxm(0)m与x轴、y轴分别交于点A、B如图11所示，点C在线段AB的延长线上，且2ABBC．（1）用含字母m的代数式表示点C的坐标；（2）抛物线21103yxbx经过点A、C，求此抛物线的表达式；（3）在第（2）题的条件下，位于第四象限的抛物线上，是否存在这样的点P：使2PABOBCSS△△，如果存在，求出点P的坐标，如果不存在，试说明理由．图10ABCDE图11xyOAB11—6—25．（本题满分14分）如图12，在Rt△ABC中，90ACB，5AB，4cos5BAC，点O是边AC上一个动点（不与A、C重合），以点O为圆心，AO为半径作⊙O，⊙O与射线AB交于点D；以点C为圆心，CD为半径作⊙C，设OAx．（1）如图13，当点D与点B重合时，求x的值；（2）当点D在线段AB上，如果⊙C与AB的另一个交点E在线段AD上时，设AEy，试求y与x之间的函数解析式，并写出x的取值范围；（3）在点O的运动的过程中，如果⊙C与线段AB只有一个公共点，请直接写出x的取值范围．备用图BAC图12ABCOD图13AB（D）CO—7—普陀区2018学年第二学期初三质量调研数学试卷参考答案及评分说明一、选择题：（本大题共6题，每题4分，满分24分）1．(C)；2．(B)；3．(A)；4．(D)；5．(B)；6．(C).二、填空题：（本大题共12题，每题4分，满分48分）三、解答题（本大题共7题，其中第19---22题每题10分，第23、24题每题12分，第25题14分，满分78分）19．解：原式=322338(1)2·······································································（6分）=233381····················································································（2分）=235.···································································································（2分）20．解：去分母得，242(3)(9)xxx.···································································（3分）整理得，2230xx.····················································································（3分）解得1x，3x．·······················································································（2分）经检验，3x是增根，舍去．········································································（1分）所以，原方程的解是1x．···············································································（1分）21．解：（1）∵DE//BC，7.(2)aa；8.13x；9.112x≤；10.53.63310；11．5；12.174；13．1x；14．25%；15．210；16．2ab；17．57；18．1063．—8—∴△ADE∽△ABC．·······························································································（1分）∴2△△ADEABCSDESBC．···································································································（1分）又∵13DEBC，∴19△△ADEABCSS．··············································································（1分）∵3△ADES，∴27△ABCS．···············································································（1分）（2）过点A作AH⊥BC，H为垂足．·········································································（1分）∵27△ABCS，∴1272BCAH．∵9BC，∴6AH．····························································································（1分）∵AH⊥BC，∴90AHBAHC．在Rt△ABH中，90AHB，2cot3B，∴23BHAH．∴4BH．··················································································································（1分）∴5CH．··················································································································（1分）在Rt△ACH中，90AHC，∴6tan5AHCHC．··········································（1