2023上海崇明區(qū)初三二模數(shù)學(xué)試題及答案

崇明區(qū)2022學(xué)年第二學(xué)期教學(xué)質(zhì)量調(diào)研測試卷九年級數(shù)學(xué)評分標(biāo)準(zhǔn)（測試時間：100分鐘，滿分：150分）一、選擇題（本大題共6題，每題4分，滿分24分）．B2．A3．C4．D5．B6．C二、填空題（本大題共12題，每題4分，滿分48分）．2；8．3；．1x3；．x2；1．m1；．增大；13．．；．135；31337．45；16．a(chǎn)b；；18．.32三、解答題（本大題共7題，滿分78分）．（本題滿分10121()2082221412(222………（841222225………（2．（本題滿分10解：由②得xy0或x2y0（22xy，2xy，原方程組可化為或（2分+2xyx2yx，x，12··················································（2分+2解得原方程的解是y；y.12．（本題滿分10分，第(1)小題5分，第(2)小題5解：（1）聯(lián)結(jié)OC交ABH...........................................................(1OC過圓心，點C是的中點,AD=43HA=DH=23，OC1∴∠CHBCHA=90°九年級數(shù)學(xué)共6頁第1頁∴22AC2又∵21．∴3（1OAxOCxOHx3，⊥AB∴∠OHA=90°，xx22∴22AO2，∴··················································（17x，（1277∴，即：⊙O的半徑長為222）過點D作⊥BC，垂足為MRt233（12133233sinB（123RtBDM中，BDsinB（12點C是的中點∴CDCA21（13∴DMCD27...........................................................................................(12114sinBCD22.（本題滿分10分，每小題各5k0..........................................................(11）設(shè)所求函數(shù)解析式為＝＋b（20kb80，）............................................................(1由題意得：50kbk，b..................................................(2解得：y2x∴所求的yx的函數(shù)解析式為．..................................................(1210024002x12030..................................................(22）由題意得：x整理得，x2170x42000解得：1140,x230..................................................(1九年級數(shù)學(xué)共6頁第2頁經(jīng)檢驗，1140,x230均為原方程的解，但當(dāng)x1401y160，不符合題意舍去.........................................................(123012060..................................................(1∴x30∴答：所選購的甲、乙消毒劑的數(shù)量分別為30瓶、60．（本題滿分12分，每小題各6證明：(1)∵四邊形ABCD是平行四邊形∥CD,ADBC................................................(2AGBGGFBG∴,................................................(2GMGDAGGDAGGF∴,................................................(1GMAGGMGF................................................(1∵∥∴AG2BAGBAGBCG................................................(1M∴∵∴BCG,MMGCCGF,∴FCG∽CMG................................................(1CGGF∴,GMCG∴CG2GMGF................................................(1AG2GMGF∴CGAG................................................(1∵四邊形ABCD是平行四邊形∴AEEC∴ACBD................................................(1∴平行四邊形ABCD是菱形................................................(1．（本題滿分12分，每小題各4解：（1）由題意得A（50）（，5................................................(1把A5，）B05）代入yx2bxc255b0b6解得：................................................(2c5yx∴c526x5D（，-4）................................................(12M作y軸的平行線交直線ABH九年級數(shù)學(xué)共6頁第3頁7M(a,-)yx6x52把47a26a5439a,a解得：12227M(a,-)在拋物線對稱軸左側(cè)的圖像上4337∴a，M(,-)................................................(123247237H(,)................................................(1xyx5y把∴∴222................................................(14M落在△74214m∴................................................(1P(x,xQ(x,x6x23由題意可得∥，∵．∴四邊形PEDQ為平行四邊形或等腰梯形．PE∥DQ時，四邊形PQDE為平行四邊形，此時P在對稱軸左側(cè)，則=DEE2)，ED6由題意x5x26x5x5x................................................(12∴x25x6xx3舍）解得：12∴Q(2,................................................(1與DQ不平行時，四邊形為等腰梯形，此時P在對稱軸右側(cè)，PQPE=DQ∴∠AED=∠QDE由題意可得：∠OBA=AED=QDE=45°；作QN⊥DE,垂足為N，可得△QND是等腰直角三角形，九年級數(shù)學(xué)共6頁第4頁6x則x3；Nx2x3x26x5(.............................................(1分)xx3舍）解得：12∴Q(4,................................................(1．（本題滿分14分，第(1)小題4分，第(2)小題5分，第(3)小題5C90ACFBDC90................................................(11）∵CFBEC90∴∵∴∴∴C90DBCACFBDC90................................................(1DBC................................................(1BCCDBC23∴tanACFtanDBC................................................(1F作FH⊥AC,垂足為H................................................(1AHFACB90∴∴AFAHFHAH∴;BFCHBCAC∵BCAC6∴AH2FH................................................(1AF2FH∴BFCHtanACFtanDBCFHCDx................................................(1CHBCAF2FH323∴yx（0x6................................................(2BFCH(3)∵DCGACB90∴△DGC與△相似有兩種情況................................................(1GA時①當(dāng)ADFCDG∴FAD∽CGD九年級數(shù)學(xué)共6頁第5頁DCGACFAFD90∴∵DBC,EDCCDB∴ECD∽CDBCDBD∴∴CD2DEBDEDCDFDDEBD2同理：∴CDFDxBCFD3x353∵sinA∴∴x................................................(1ABAD356x22AF∴yx5151................................................(13BFGABC時CDGG90CDGADF②當(dāng)AABC90∵∴A∴AFFDF作FHAC,垂足為H6x