普陀区2019学年度第一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．已知35x y =，那么下列等式中，不一定正确的是（ ▲ ） （A ）5=3x y ； （B ）+8x y =； （C ）+85x y y =； （D ）35x x y y +=+． 2．下列二次函数中，如果函数图像的对称轴是y 轴，那么这个函数是（ ▲ ）（A ）22y x x =+； （B ）221y x x =++； （C ）22y x =+； （D ）2(1)y x =-． 3．已知在Rt △ABC 中，90C ∠=︒，1sin 3A =，那么下列说法中正确的是（ ▲ ） （A ）1cos 3B =； （B ）1cot 3A =； （C）tan A =； （D）cot B =．4．下列说法中，正确的是（ ▲ ）（A ）如果，a 是非零向量，那么0ka =； （B ）如果e 是单位向量，那么1e =； （C ）如果b a =，那么b a =或b a =-；（D ）已知非零向量a ，如果向量5b a =-，那么a ∥b ．0k =5．如果二次函数()2y x m n =-+的图像如图1所示，那么一次函数y mx n =+的图像经过（ ▲ ） （A ）第一、二、三象限； （B ）第一、三、四象限； （C ）第一、二、四象限； （D ）第二、三、四象限．6．如图2，在Rt △中，90ACB ∠=︒，CD AB ⊥，垂足为点D ，如果32ADC CDB C C =△△，9AD =，那么BC 的长是（ ▲ ）（A ）4； （B ）6； （C ）213； （D ）310．二、填空题：（本大题共12题，每题4分，满分48分） 7．化简：12()()2a b a b →→→→+--= ▲ ． 8．抛物线2(2)y a x =-在对称轴左侧的部分是上升的，那么a 的取值范围是 ▲ ． 9．已知函数2()321f x x x =--，如果2x =，那么()f x = ▲ ．10．如果抛物线22y ax ax c =++与x 轴的一个交点的坐标是(1,0)，那么与x 轴的另一个交点的坐标是 ▲ ．11．将二次函数222y x x =-+的图像向下平移m (0)m >个单位后，它的顶点恰好落在x轴上，那么m 的值等于 ▲ ．12．已知在Rt △ABC 中，90C ∠=︒，1cot 3B =，2BC =，那么AC = ▲ ． 13．如图3，△ABC 的中线AD 、CE 交于点G ，点F 在边AC 上，GF //BC ，那么GFBC的值是 ▲ ．14．如图4，在△ABC 与△AED 中，AB BCAE ED=，要使△ABC 与△AED 相似，还需添加 一个条件，这个条件可以是 ▲ ．（只需填一个条件）ABC 图3ABCDEG F图2AD CB图5ABCD 图4ABCEDxyO图115． 如图5，在Rt △中，90C ∠=︒，AD 是三角形的角平分线，如果35AB =，25AC =，那么点D 到直线AB 的距离等于 ▲ ．16．如图6，斜坡AB 长为100米，坡角30ABC ∠=︒，现因“改小坡度”工程的需要，将斜坡AB 改造成坡度1:5i =的斜坡BD （、、C 三点在地面的同一条垂线上），那么由点到点下降了 ▲ 米．（结果保留根号）17．如图7，在四边形ABCD 中，90ABC ∠=︒，对角线AC 、BD 交于点O ，AO CO =，CD BD ⊥，如果3CD =，5BC =，那么AB = ▲ ．18．如图8，在Rt △ABC 中，90C ∠=︒，5AC =，5sin 13B =，点P 为边BC 上一点，3PC =， 将△ABC 绕点P 旋转得到△A B C '''（点A 、B 、C 分别与点A '、B '、C '对应），使B C ''//AB ，边A C ''与边AB 交于点G ，那么A G '的长等于 ▲ ． 三、解答题：（本大题共7题，满分78分）19．（本题满分10分）计算：222sin 60cos60tan 604cos45︒-︒︒-︒．20.（本题满分10分）如图9，在△ABC 中，点D 、E 、F 分别在边AB 、AC 、BC 上，DE //BC ，EF //AB ，:1:3AD AB =．（1）当5DE =时，求FC 的长；（2）设AD a =，CF b =，那么FE = ▲ ，EA = ▲ （用向量a 、b 表示）．ABC A D A D ABCDE F图9图8ABC图7ADC BOAD B图6C如图10，在△ABC 中，点P 、D 分别在边BC 、AC 上，PA AB ⊥，垂足为点A ，DP BC ⊥，垂足为点P ，AP BPPD CD=． （1）求证：APD C ∠=∠；（2）如果3AB =，2DC =，求AP 的长．22．（本题满分10分）函数m y x =与函数xy k=（m 、k 为不等于零的常数）的图像有一个公共点()3,2A k -，其中正比例函数y 的值随x 的值增大而减小，求这两个函数的解析式．23．（本题满分12分）已知：如图11，四边形ABCD 的对角线AC 、BD 相交于点O ，AOD BOC S S =△△． （1）求证：OACOOB DO =； （2）设△OAB 的面积为S ，k ABCD=，求证：2(1)ABCD S k S =+四边形．CDBAO图11图10CDBAP在平面直角坐标系中（如图12），已知抛物线28()3y ax a x c =+++(0)a ≠经过点A ()3,2--，与y 轴交于点B ()0,2-，抛物线的顶点为点C ，对称轴与x 轴交于点D ．（1）求抛物线的表达式及点C 的坐标；（2）点E 是x 轴正半轴上的一点，如果AED BCD ∠=∠，求点E 的坐标；（3）在（2）的条件下，点P 是位于y 轴左侧抛物线上的一点，如果△PAE 是以AE 为直角边的直角三角形，求点P 的坐标．xOy 图12O11如图13，在梯形ABCD 中，AD //BC ，90C ∠=︒，2AD =，5BC =，3DC =，点E 在边BC 上，tan 3AEC ∠=．点M 是射线DC 上一个动点（不与点D 、C 重合），联结BM 交射线AE 于点N ，设DM x =，AN y =． （1）求BE 的长；（2）当动点M 在线段DC 上时，试求y 与x 之间的函数解析式，并写出函数的定义域； （3）当动点M 运动时，直线BM 与直线AE 的夹角等于45︒，请直接写出这时线段DM 的长．备用图ABCD ENM图13AB CDE普陀区2019学年度第一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．(B)； 2．(C)； 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212⨯-= ··································································· （4分）31-=······················································································· （3分）3=+． ······················································································ （3分）20．解：（1）∵DE //BC ，EF //AB ，∵DE BF =． ···················································································· （1分） ∵5DE =，∵5BF =． ········································································ （1分） ∵DE //BC ，∵AD DEAB BC =． ··················································································· （1分） ∵13AD AB =，∵513BC =． ······································································ （1分） 7． 2a b →→+； 8． 2a <； 9． 7； 10．30-(,) ； 11．1； 12．6；13． 13； 14．B E ∠=∠（AB ACAE AD=等）； 15．2 ； 16．50-； 17．154； 18．2013．解得 15BC =， ················································································· （1分） 10FC =． ························································································ （1分） （2）FE =2a -，EA =12a b -+． ························································ （2分+2分）21．解：（1）∵PA AB ⊥，DP PC ⊥，∵90BAP CPD ∠=∠=︒． ···································································· （1分） 在Rt △ABP 与Rt △PCD 中，AP BPPD CD=， ∵Rt △ABP ∵Rt △PCD ． ···································································· （1分） ∵APB PDC ∠=∠． ············································································ （1分） ∵DPB APB APD ∠=∠+∠，DPB PDC C ∠=∠+∠，得APD C ∠=∠． ··············································································· （2分） （2）∵Rt △ABP ∵Rt △PCD ． ∵B C ∠=∠．∵AB AC =． ···················································································· （1分） ∵3AB =，2DC =，∵1AD =． ··························································· （1分） ∵APD C ∠=∠，PAD CAP ∠=∠，∵△APD ∵△ACP ． ·········································································· （1分） ∵AD APAP AC=． ················································································· （1分）得AP = ···················································································· （1分）22．解：由点A ()3,2k -在函数xy k=的图像上，可得 32k k-=． ················································································ （1分） 整理，得2230k k --=． ··································································· （1分） 解得 13k =，21k =-． ····································································· （2分） ∵正比例函数y 的值随x 的值增大而减小，∵1k =-． ························································································ （2分）得 y x =-，点A ()3,3-． ································································· （2分） 由点A ()3,3-在函数my x=的图像上，可得 9m =-． ··················································································· （1分） ∵9y x=-． ······················································································ （1分） 两个函数的解析式分别为y x =-，9y x=-．23．证明：（1）过点A 作AH ⊥BD ，垂足为点H . ···················································· （1分）∵S △AOD =AH DO ⋅⋅21, S △AOB =AH OB ⋅⋅21, ∴OB DO AH OB AHDO S S AOBAOD=⋅⋅⋅⋅=∆∆2121. ····························································· （2分）同理，BOC AOB S COS OA∆∆=． ·········································································· （1分） ∵AOD BOC S S =△△， ∵DO COOB OA=． ················································································ （1分） （2）∵OACOOB DO =，AOB COD ∠=∠， ∵△OCD ∵△OAB ． ······································································· （1分） ∵CD DO COk AB BO AO===． ··································································· （1分） 22k AB CD S S OAB OCD =⎪⎭⎫ ⎝⎛=∆∆． ·································································· （1分） ∵△OAB 的面积为S ，∴S k S OCD ⋅=∆2. ············································· （1分） 又∵k OBDOS S OAB AOD ==∆∆，∵S k S AOD ⋅=∆． ············································· （1分） 同理，S k S BOC ⋅=∆. ······································································ （1分）∴AOB BOC COD DOA ABCD S S S S S =+++△△△△四边形S k S k S k S ⋅+⋅+⋅+=2 S k k ⋅++=)12(2S k 2)1(+=．································································· （1分）24．解：（1）由抛物线28()3y ax a x c =+++经过点A ()3,2--和点B ()0,2-，得2,893() 2.3c a a c =-⎧⎪⎨-++=-⎪⎩ 解得4,32.a c ⎧=⎪⎨⎪=-⎩ ··············································· （2分） ∵抛物线的表达式是24423y x x =+-． ·············································· （1分） 点C 的坐标是3(,5)2--． ··································································· （1分） （2）联结AB 交CD 于点F ，过点A 作AH OD ⊥，H 为垂足．∵A ()3,2--，B ()0,2-，∵3AB =． 由对称性可得 32BF =． ····································································· （1分） ∵5CD =，∵3CF =．在Rt △BCF 中，1tan 2BF BCF CF ∠==． ················································· （1分） 在Rt △AEH 中，tan AHAEH EH∠=，∵AED BCD ∠=∠， ∵12AH EH =．∵4EH =． ····································································· （1分） ∵3OH =，∵1OE =．∵点E 的坐标是()1,0． ······································································· （1分） （3）∵△PAE 是以AE 为直角边的直角三角形， ∵90PAE ∠=︒或90PEA ∠=︒．设点P 点的坐标为24(,42)3m m m +-． ①当90PAE ∠=︒时，点P 只能在AE 的下方． 过点P 作PG AH ⊥，G 为垂足．∵3PG m =+，2443AG m m =--． ∵GAE AHE AEH ∠=∠+∠，GAE PAE PAG ∠=∠+∠，∵PAG AEH ∠=∠．∵tan tan PAG AEH ∠=∠． ∵PG AH AG EH =．∵2314243m m m +=--． ···················································· （1分） 解得3m =-，32m =-． ∵3m =-不合题意舍去，∵32m =-． ∵点P 的坐标是3(,5)2--． ································································ （1分） ②当90PEA ∠=︒时．同理可得点P的坐标是913(42-+． ··································· （2分）25．解：（1）过点A 作AH BC ⊥，H 为垂足．∵AH BC ⊥，∴90AHE ∠=︒．∵90C ∠=︒，∴AHE C ∠=∠．∴AH //DC ．∵AD //BC ，3DC =∴3AH DC ==． ······························································· （1分） 同理可得2HC AD ==． ··························································································· （1分） 在Rt △AEH 中，90AHE ∠=︒，tan 3AEH ∠=，∴3AH HE=． ∴1EH =． ················································································································ （1分） ∵5BC =，∴2BE =． ····························································································· （1分）（2）延长BM 、AD 交于点G ． ············································································· （1分） ∵DG //BC ，∴DG DM BC MC=． 由DM x =，3DC =，5BC =， 得53DG x x =-，解得53x DG x=-． ·········································································· （1分） ∴633x AG x+=-． ········································································································· （1分） ∵AG //BC ，∴AN AG BN BE =． 在Rt △AEH 中，90AHE ∠=︒，1EH =，3AH =，。