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()a a =； （B ）236a a a ⋅=； （C ）2236a a a ⋅=； （D ）2235a a a +=．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 π cm 2； （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 ）2y x =-； （B ）2y x=； （C ）1y x =-+； （D ）21y x =-．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．l 1l 2图11236.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△△ABO CBO C C =；③DAO CBO ∠=∠；④DAO BAO ∠=∠，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：22a a +＝ ▲ ． 8.函数131y x =-的定义域是 ▲ ． 9.不等式组21034x x x -<⎧⎨-⎩，≤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 =-，的值等于 ▲ ．12.如果关于x 的方程2320x x m -+-=有两个相等的实数根，那么m 的值等于 ▲ ． 13.抛物线225y ax ax =-+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2A BCDO 图4C米图3-1图3-2篮球45% 足球排球16.如图5，AD 、BE 是△ABC 的中线，交于点O ，设OB a =，OD b =，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52AE BE =，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：312019212sin 60227(1)2-⎛⎫︒-+--- ⎪⎝⎭．20．（本题满分10分）解方程：242193x x x =--+．E 图5AB CDO 图6图7ABCDE如图8，已知点D 、E 分别在△ABC 的边AB 和AC 上，DE //BC ，13DE BC =，△ADE 的面积等于3．（1）求△ABC 的面积； （2）如果9BC =，且2cot 3B =，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．A BCDE图8（吨）图9已知：如图10，在四边形ABCD 中，AD BC <，点E 在AD 的延长线上， ACE BCD ∠=∠，EC ED EA =⋅2． （1）求证：四边形ABCD 为梯形； （2）如果EC ABEA AC=，求证：AB ED BC =⋅2．24．（本题满分12分）在平面直角坐标系xOy 中，直线243y x m =-+(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2AB BC =． （1）用含字母m 的代数式表示点C 的坐标；（2）抛物线21103y x bx =-++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2PAB OBC S S =△△，如果存在，求出点P 的坐标，如果不存在，试说明理由．图10A BCD ExyO AB11如图12，在Rt △ABC 中，90ACB ∠=︒，5AB =，4cos 5BAC ∠=，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OA x =． （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AE y =，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12AB C OD图13AB （D ）C O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=228(1)-+-- ······································································· （6分）=281++ ···················································································· （2分）=5. ··································································································· （2分）20．解：去分母得，242(3)(9)x x x =---. ··································································· （3分）整理得，2230x x +-=. ···················································································· （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)a a +； 8. 13x ≠； 9. 112x -<≤； 10. 53.63310⨯； 1112.174； 13．1x =； 14．25%； 15． 16．2a b +；17．57； 18．1063．∴△ADE ∽△ABC ． ······························································································· （1分）∴2△△ADE ABC S DE S BC ⎛⎫= ⎪⎝⎭． ··································································································· （1分） 又∵13DE BC =，∴19△△ADE ABC S S =． ·············································································· （1分） ∵3△ADE S =，∴27△ABC S =． ··············································································· （1分） （2）过点A 作AH ⊥BC ，H 为垂足． ········································································· （1分）∵27△ABC S =，∴1272BC AH ⋅⋅=．∵9BC =，∴6AH =． ···························································································· （1分） ∵AH ⊥BC ，∴90AHB AHC ∠=∠=︒． 在Rt △ABH 中，90AHB ∠=︒，2cot 3B =，∴23BH AH =． ∴4BH =．·················································································································· （1分） ∴5CH =． ·················································································································· （1分） 在Rt △ACH 中，90AHC ∠=︒，∴6tan 5AH C HC ==．·········································· （1分） ∵DE //BC ，∴AED C ∠=∠．∴6tan 5AED ∠=． ······································································································ （1分） 即AED ∠的正切值65．22．解：（1）设y 关于x 的函数解析式为y kx b =+(0)k ≠， ···················································· （1分）由题意，得620,5.628.k b k b =+⎧⎨=+⎩·························································································· （2分）解得 1,207.k b ⎧=-⎪⎨⎪=⎩ ········································································································· （1分） ∴y 关于x 的函数解析式为1720y x =-+． ···························································· （1分）（2）将 4.8y =代入解析式，得14.8720x =-+． ························································· （1分）解得 44x =． ·············································································································· （1分）所以，该产品的生产数量是44吨． （3）由题意，得1(7)20020x x -+=． ············································································· （1分） 解得 140x =，2100x =（不符合题意，舍去）． ················································ （2分） 所以，该产品的生产数量是40吨． 23．证明：（1）∵ ACE BCD ∠=∠，∴DCE BCA ∠=∠． ······················································· （1分）∵EC ED EA =⋅2，∴ED ECEC EA=． ······································································· （1分） 又∵E ∠是公共角，∴△EDC ∽△ECA ． ····························································· （1分） ∴DCE CAE ∠=∠． ································································································· （1分） ∴BCA CAE ∠=∠．∴AD ∥BC ． ············································································································· （1分） ∵AD BC <，∴AB 与CD 不平行．∴四边形ABCD 是梯形． ··························································································· （1分） （2）∵△EDC ∽△ECA ．∴EC CDEA AC =． ∵EC AB EA AC=，∴AB DC =． ············································································· （1分） ∴四边形ABCD 是等腰梯形． ··············································································· （1分） ∴B DCB ∠=∠．··································································································· （1分） ∵AD ∥BC ．∴EDC DCB ∠=∠． ∴EDC B ∠=∠．∵ECD ACB ∠=∠，∴△EDC ∽△ABC ． ····················································· （1分） ∴ED DCAB BC=． ········································································································· （1分） ∴AB ED BC =⋅2． ····························································································· （1分） 24．解：（1） 过点C 作CH ⊥OB ，垂足为点H ．∵直线243y x m =-+与x 轴、y 轴分别相交于点A 、B ，∴点A 的坐标是()6,0m ，点B 的坐标是()0,4m . ··················································· （2分）∴6OA m =，4OB m =. ∵CH ⊥OB ，∴CH //OA ． ∴CH BH BCOA OB AB==． ·································································································· （1分） ∵2AB BC =，∴3CH m =，2BH m =.∴点C 的坐标是()3,6m m -.······················································································· （1分）（2） ∵抛物线21103y x bx =-++经过点A 、点C ，可得 221(6)6100,31(3)3106.3m m b m m b m ⎧-⨯+⋅+=⎪⎪⎨⎪-⨯--⋅+=⎪⎩ ··································································· （2分）∵0m >，解得 1,13m b =⎧⎪⎨=⎪⎩. ························································································ （1分） ∴抛物线的表达式是2111033y x x =-++. ···························································· （1分）（3）过点P 分别作PQ ⊥OA 、垂足为点Q ．设点P 的坐标为211(,10)33n n n -++．可得OQ n =，2111033PQ n n =--．∵2PAB OBC S S =△△，2AB BC =．∴△PAB 与△OBC 等高，∴OP //AB . ·································································· （1分） ∴BAO POQ ∠=∠.∴tan tan BAO POQ ∠=∠.∴211102333n n n --=. ································································································· （1分）解得1n =，2n =（舍去）． ····················································· （1分） ∴点P的坐标是3323⎛- ⎝⎭.································································· （1分） 25．解：（1）在Rt △ABC 中，90ACB ∠=︒，4cos 5BAC ∠=，∴45AC AB =． ∵5AB =，∴4AC =． ··························································································· （1分）。