文档介绍：该【2022～2023学年度上学期九年级数学期末学情分析样题(含答案) 】是由【1781111****】上传分享，文档一共【10】页，该文档可以免费在线阅读，需要了解更多关于【2022～2023学年度上学期九年级数学期末学情分析样题(含答案) 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..,、考试证号是否与本人相符合,再将自己的姓名、,请用橡皮擦干净后,,,并请加黑加粗,、选择题(本大题共6小题,每小题2分,,恰有一项是符合题目要求的,请将正确选项前的字母代号填涂在答题卡相应位置上).......1下列函数是二次函数的是21A.====,,8,8,10,1中,最后一个两位数的个位数字被墨迹覆盖,,l∥l∥l,则下列比例式成立的是123ABDEABDEABBEABADA.=B.=C.=D.=FyABADl13BElP2O-1O1xCFl3CD(第4题)(第5题)(第6题)=ax2+bx+c的图像,则不等式ax2+bx+c<<<-1或x><x<<0或x>,AB,CD分别是⊙O的内接正十边形和正五边形的边,AD,BC交于点P,则∠°°°°二、填空题(本大题共10小题,每小题2分,,请把答案直接填写在答题卡...:..上)72=2x的解是▲.=2y,则=▲.=3(x-1)2+2的图像的顶点坐标是▲.,AC>BC,若AB=2,则AC的长为▲.(结果保留根号),x是关于x的方程x2-kx-1=0的两个根,且x=-x,则k的值为▲.,母线长为6,则该圆锥侧面展开图的圆心角是▲°.,PB切⊙O于点B,PO交⊙O于点A,若PA=1,PB=2,则⊙O的半径为▲.AyBAPDPAOCBOxBC(第13题)(第14题)(第16题),在平面直角坐标系中,△AOB的边AO,AB的中点C,D的横坐标分别是1,4,则点B的横坐标是▲.,连接PA,PB,若AB=2,∠APB=30°,则点P到AB的最大距离为▲.,在ABC中,∠C=90°,BC=3,AC=4,动点P以每秒1个单位长度的速度从点A出发,沿着AB—C的路线运动,则以P为圆心,2为半径的⊙P与△ABC三边都有公共点的时间共▲、解答题(本大题共11小题,,解答时应写出文字说明、证明过.......程或演算步骤)17.(8分)解方程:(1)x2+2x-3=0;(2)(x-1)2=3x-.(8分)小明、小红两位同学邀请数学老师合影,3人随机站成一排.(1)数学老师站在中间的概率是▲;(2).(7分)甲、乙两名同学本学期五次某项测试的成绩(单位:分)(分)110甲100乙90:..1)甲、乙两名同学五次测试成绩的平均数分别是▲分、▲分;(2)利用方差判断这两名同学该项测试成绩的稳定性;(3)结合数据,请再写出一条与(1)(2).(8分)如图,⊙的弦AB,CD的延长线交于点P,连接AC,(1)求证△PAC∽△PDB;B(2)若PB=3,PD=4,AB=(第20题)21(8分)已知二次函数y=ax2+bx+c的图像经过(-1,0),(0,2),(1,0)三点.(1)求该二次函数的表达式;(2)当-1<x<2时,y的取值范围是▲.(3)将该函数的图像沿直线x=1翻折,.(7分)如图,在矩形ABCD中,点E,F分别在边BC,CD上,AE,:..BE)若,求证AE⊥BF;BCCFADAG(2)若E,F分别是BC,CD的中点,则的值为▲.EGFGBEC(第22题)23.(8分)如图,用长度均为12m的两根绳子分别围成矩形ABCD和扇形OEF,设AB的长为xm,半径OE为Rm,矩形和扇形的面积分别为Sm2,⌒(1)BC的长为▲m,EF的长为▲m;(用含x或R的代数式表示)(2)求S,S的最大值,(第23题)24.(8分),在以O为圆心的两个同心圆中,大圆的弦AB交小圆于C,D两点.(1)如图①,若大圆、小圆的半径分别为13和7,AB=24,则CD的长为▲.(2)如图②,大圆的另一条弦EF交小圆于G,H两点,若AB=EF,求证CD=①②(第24题)25.(8分)如图,在△ABC中,∠ACB=90°,点E,F分别在边AC,BC上,EF∥AB,以EF为直径的4CEOF:..与AB相切于点D,连接CD,DE,DF.(1)求证:①DEDF;②△ADE∽△DCF.(2)若CE=6,CF=8,则AB的长为▲.26.(8分)已知二次函数y=mx2+x-4m(m≠0).(1)求证:该二次函数图像与x轴总有两个公共点;(2)当m<0时,该二次函数图像顶点的纵坐标的最小值是▲.(3)若该二次函数图像的对称轴为直线x=n(n≠0),当-2<n<1时,结合图像,.(10分)5:..角平分线的有关联想就有很多……【】PAAC()如图①,是△PAB的角平分线,求证=.PBBC小明思路:关联“平行线、等腰三角形”,利用“三角形相似”.小红思路:关联“角平分线上的点到角的两边的距离相等”,利用“等面积法”.请根据小明或小红的思路,①【作图应用】PA(2)如图②,AB是⊙O的弦,在⊙O上作出点P,使得=:(1)用直尺和圆规作图;(2)保留作图的痕迹,②【深度思考】(3)如图③,PC是△PAB的角平分线,若AC=3,BC=1,则△PAB的面积最大值是▲.PACB③6:..2023学年度第一学期期末学情分析样题九年级数学试题参考答案及评分标准一、选择题(本大题共6小题,每小题2分,共12分)题号123456答案CABBDA二、填空题(本大题共10小题,每小题2分,共20分)7.=0,x=.(1,2)-+、解答题(本大题共11小题,共88分)17.(8分)(1)解:x2+2x-3=0x2+2x+1=3+1····················································································1分(x+1)2=4·····························································································2分x+1=±2·····························································································3分∴x=1,x=-3················································································4分12(2)解:(x-1)2-3(x-1)=0············································································5分(x-1)(x-1-3)=0···············································································6分∴x=1,x=4.··················································································8分1218.(8分)1(1).·······································································································2分3(2)解:3人随机站成一排,共有6种等可能性结果:(小明,小红,老师)、(小明,老师、小红)、(小红,小明,老师)、(小红,老师,小明)、(老师,小明,小红)、(老师,小红,小明).其中满足“小2明与数学老师相邻而站”(记为事件A)的结果有4种,所以P(A)=.·······················8分319.(7分)(1)80,80·································································································2分(2)方差分别是:(80-80)2+(90-80)2+(80-80)2+(70-80)2+(80-80)2s2==40分2····································3分甲5(60-80)2+(70-80)2+(90-80)2+(80-80)2+(100-80)2s2==160分2·································4分乙5由s2<s2可知,甲同学的成绩更加稳定.···························································5分甲乙(3)甲同学的成绩在70,80,90间上下波动,而乙的成绩从60分到100分,呈现上升趋势,越来越好,进步明显.·································································································7分20.(8分)(1)证明:∵四边形ABDC是⊙O的内接四边形,∴∠A+∠BDC=180°.································································A············1分∵∠BDC+∠PDB=180°,∴∠A=∠PDB.······················································B·······························2分又∠P=∠P,P∴△PAC∽△PDB.······························································O···················4分(2)∵△PAC∽△PDB,DPAPC∴=,···························································································5分PDPBC10PC∴=,(第20题)437:..15=.··························································································7分27∴CD=PC-PD=.·············································································8分221.(8分)(1)根据题意,可得图像顶点坐标为(0,2),设二次函数的表达式为y=ax2+2.·······1分将(1,0)代入,求得a=-2,······································································2分∴y=-2x2+2.·······················································································4分(可直接解三元一次方程组得一般式y=-2x2+2,或交点式y=-2(x+1)(x-1).)(2)-6<y≤2;······························································································6分(-6,2和不等号正确各1分)(3)y=-2(x-2)2+2.······················································································8分(或y=-2x2+8x-6,y=-2(x-1)(x-3).)22.(7分)ABBE(1)∵在矩形ABCD中,∠ABC=∠BCF=90°,.BCCF∴△ABE∽△BCF.·················································································2分AD∴∠BAE=∠CBF.·················································································3分∵∠ABG+∠CBF=90°.F∴∠ABG+∠BAE=90°,··········································································4分∴∠AGB=90°.G即AE⊥BF.···························································································5分(2)4.·······································································B·········E·············C·············7分23.(8分)(第22题)(1)6-x,12-2R.··························································································2分(2)S=x(6-x)=-(x-3)2+9,··········································································4分1∵-1<0,∴当x=3时,S有最大值9.·························································5分11S=(12-2R)R=-(R-3)2+9,·····································································6分22∵-1<0,∴当R=3时,S有最大值9.·························································7分2∴S的最大值=S的最大值.·········································································8分1224.(8分)(1)46;······································································································2分(2)证明:过O作OM⊥AB,作ON⊥EF,垂足分别为M、N1111∴DM=CD,HN=GH,AM=AB,EN=EF,3分2222又AB=EF,∴AM=EN.…………………………………………...4分连接OA、OE、OD、OH,在Rt△OAM和Rt△OEN中,FH?OA=OE,?NAM=ENG…………………………………...O∴Rt△OAM≌Rt△∴OM=ON.…………………………………………………...6分ACMDB∴在Rt△ODM和Rt△OHN中,?OD=OH,??OM=ON(第24题)∴Rt△ODM≌Rt△OHN.…………………………………………………………………...7分∴DM=HN.∴CD=GH.……………………………………………………………………………...8分25.(8分)(1)①证明::..⊙与AB边相切于点D,∴OD⊥AB.······························································································1分∴∠ODB=90°.∵EF∥AB,∴∠EOD=∠ODB=90°.∴∠FOD=∠EOD=90°.··········································································2分∴DEDF.······························································································3分⌒⌒⌒⌒②证明:∵CF=CF,DF=DF,∴∠CEF=∠CDF,∠DEF=∠DCF.····························································4分∵EF∥AB,∴∠CEF=∠A,∠DEF=∠ADE.∴∠CDF=∠A,∠DCF=∠ADE.································································5分∴△ADE∽△DCF.····················································································6分245(2).·······································································································8分1226.(8分)(1)根据题意,当y=0时,mx2+x-4m=0,∵b2-4ac=1-4m(-4m)=1+16m2>0,·················································1分∴方程mx2+x-4m=0有两个不相等的实数根.·········································2分∴该二次函数图像与x轴总有两个公共点.················································3分(2)2.·······································································································5分1111(3)m>或m<-.(,-,不等号正确各1分)············································8分4242D27.(9分)(1)方法一:过点B作BD∥∴=,∠1=∠D,∠2=∠3.·······························································2分PDBC12∵PC平分∠APB,∴∠1=∠∴∠D=∠3.①∴PD=PB.·······························································································3分PAAC∴=.································································································4分PBBC方法二:分别过点P,A作PD⊥AB,CE⊥PA,CF⊥PB,垂足为D,E,F.∵PC平分∠APB,∴CE=∵S=PA·CE=AC·PD,S=PB·CF=BC·PD,△PAC22△PBC22EFPA·CEAC·PDPAAC∴=.∴=.PB·CFBC·PDPBBCACDB①(2)点P,P即为所求;(作出一点3分,两点4分)·············································8分129:..1P1OOBBACACPP22方法一方法二PP11OOABACBODCPP22方法三方法四3)3.········································································································10分10