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 130∠=︒250∠=︒3∠=20︒80︒90︒100︒314.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 （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 ）； （C ）5； （D ）．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△△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）l 12图1图2 AC DO7.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）16.如图5，AD 、BE 是△ABC 的中线，交于点O ，设OB a =u u u r r ，OD b =u u u r r ，那么向量AB u uu r用向量、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，图4米图3-1图3-2篮球45% 足球排球E 图5ACO图6图7ABCDE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 =--+．21．（本题满分10分），PC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）7.(2)a a+；8.13x≠；9.112x-<≤；10.53.63310⨯；1112.174；13．1x=；14．25%；15．；16．2a b+rr；17．57；18．1063．图12图13图1121．解：（1）∵DE BC ADE ABC 2△△ADE ABC S DE S BC ⎛⎫= ⎪⎝⎭13DE BC =19△△ADE ABC S S =3△ADE S =27△ABC S =A AH BC H 27△ABC S =1272BC AH ⋅⋅=9BC =6AH =AH BC 90AHB AHC ∠=∠=︒ABH 90AHB ∠=︒2cot 3B =23BH AH =4BH =5CH =ACH 90AHC ∠=︒6tan 5AH C HC ==DE BC AED C ∠=∠6tan 5AED ∠=AED ∠65y x y kx b =+(0)k ≠620,5.628.k b k b =+⎧⎨=+⎩1,207.k b ⎧=-⎪⎨⎪=⎩y x 1720y x =-+ 4.8y =14.8720x =-+44x =1(7)20020x x -+=140x =2100x =ACE BCD ∠=∠DCE BCA ∠=∠EC ED EA =⋅2ED ECEC EA=E ∠EDC ECA DCE CAE ∠=∠BCA CAE ∠=∠AD BC AD BC <AB CD ABCD EDC ECA EC CDEA AC=EC ABEA AC=AB DC =ABCD B DCB ∠=∠AD BC EDC DCB∠=∠EDC B ∠=∠ECD ACB ∠=∠EDC ABCED DCAB BC=AB ED BC =⋅2C CH OB H 243y x m =-+x y A B A ()6,0m B ()0,4m （2分）∴6OA m =，4OB m =. ∵CH ⊥OB ，∴CH OACH BH BCOA OB AB==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 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⎝⎭.·································································· （1分） 25．解：（1）在Rt △ABC 中，90ACB ∠=︒，4cos 5BAC ∠=，∴45AC AB =． ∵5AB =，∴4AC =． ··························································································· （1分） 由勾股定理得 3BC =． ···························································································· （1分） ∵OB OA x ==，∴4CO x =-． 在Rt △BCO 中，90C ∠=︒，由勾股定理得 2223(4)x x +-=． ············································································ （1分） 解得258x =． ············································································································· （1分） （2）过点O 、C 分别作OH ⊥AB 、CG ⊥AB ，垂足为点H 、G ．∵OH ⊥AB ，∴AH DH =． ·················································································· （1分） 同理 DG EG =． ∵4cos 5BAC ∠=，∴45AH x =． ∴85AD x =． ··············································································································· （1分） ∵CG ⊥AB ，∴90AGC ∠=︒． ∴90AGC ACB ∠=∠=︒．又∵CAB ∠是公共角，∴△AGC ∽△ACB ． ∴AG AC AC AB =．∴165AG =．∵AE y =，∴165GE y =-． ···················································································· （1分） ∴165DG y =-． ∴16168555y y y x -+-+=． ∴化简得 32855y x =-（2528≤x <）． ································································· （2分） （3）708x << ····················································································································· （2分）2x =． ·························································································································· （1分） 2548x <<． ················································································································· （2分）。