2021北京大兴区初三上期末数学试卷及答案

2020~2021学年度第一学期期末检测初三数学参考答案及评分标准一、选择题（本题共分，每小题3题号答案12345678CABDBCDB二、填空题（本题共分，每小题3分）9.m0；10.=；（2，-4）；15.①②③；12.2；4yx；13.；14.答案不唯一，例如：16.m<5.3三、解答题（本题共17-21题，每小题5分，第题6分，第23-25题,每小题7分）解答应写出文字说明、演算步骤或证明过程.217.解:314分2-1-原式2222-1-3122-35分yxc2bc4c3∴.2分b解得.c-2x．（2）令y=0,yx2∴3分2x0.∴xx3解得:.12∴抛物线与x5分初三数学参考答案及评分标准第1页（共7页）19.解:（1）补全的图形如图所示：··························································2分（2）60.·········································································3分一条弧所对的圆周角等于它所对的圆心角的一半······················5分20.解:作DEAC，垂足为E，ED在△CED中，＝∵∠C=30º,CD=∴DE＝，CD····················································1分··················································2分∵＝CE，CD3CE∴.220∴CE＝103.∵∠是△的外角，∠ADB=75º,∠C=，∴∠CAD＝°.∴在△中，∠EAD＝ED1，AE∴AE＝10.·······················································3分·······························4分∴AC＝+CE＝10+103.AB∴在△ABC中，∠＝∴AB＝3.，AC答:这棵树的高度是(5+53)米.·····················5分初三数学参考答案及评分标准第2页（共7页）ykxk0)中得k=1.·····················1分21.1）将点A1，2m将点A（12）代入ymm2.0得=·····················2分x（2）①当点P在点A下方时,过点A作AGx轴，交直线H，∵平行于x轴，∴∽∆ACB22SAPQAHAG14APAC∴SACBAH1∴AG2∵点A（1，∴点P纵坐标为1.m2，2y.x∴P点坐标为（2，）.·····················4分②P在点A上方时，过点A作AGx轴，交直线H.∵平行于x轴，∴∽∆ACB.22SAHAG14APACAPQ∴SACBAH12∴AG∵点A（1，∴P点纵坐标为3.22代入yx得，x323∴P点坐标为,3.·····················5分2∴P点坐标为（2，1，3）.3初三数学参考答案及评分标准第3页（共7页）22.(1)证明：连接O∵OC=，∴∠＝∠∵四边形是矩形，∴∠BD=DCB=90°.又∵∠DAFBAC，∴∠AFD=∠···········································1分∵∠ACBACD＝90°，∴∠AFD＋∠＝90°.∴∠AFO＝90°.∴⊥AF于∴直线AF与⊙O相切．·········································2分(2)解：2∵∠DAF＝，∠＝∠BAC，222∴∠BAC＝∵∠B°，.∴∠BAC＝BC＝22.AB∵AB＝4，∴BC＝22.·········································3分ACAB2BC226.∴又∵四边形ABCD是矩形，∴BC=AD=222又∠D＝90°，∠DAF＝，22∴＝AD·tan∠DAF＝22×＝2.·······························4分∴＝23.设⊙O的半径为r在△中，AFO＝90°.初三数学参考答案及评分标准第4页（共7页）222∴OA＝OF＋AF.22即(26－r)＝r＋12.········································5分········································6分6解得r＝.26∴⊙O的半径为.223.解：b（1）∵1，2a∴ba.························································1分（2）把=-2ayax++a得y=-2ax++1.：配方得：=ax-1）∴顶点M（1，1）.···················································2分（3）①1个.···················································3分由①得，a12+1.②时，区域W内有1个整点.（Ⅰ）当抛物线过（1，0）时，区域W3个整点.将（-1，0代入y=-2ax++1,1得a.····················································4分41结合图象可得1a.······························5分4（Ⅱ）（0，-）时，区域W内恰有3个整点.将（0，-2代入y=-2ax++1a3.,得1综上所述，a的值范围是-1a或a3···············7分4初三数学参考答案及评分标准第5页（共7页）24.（1）AE=BE.·························································1分·························································2分·························································3分（2）依题意补全图形①AE=BE.如图，作EM⊥于M.∵∠DBC=∠ABC+∠=60°+∠ABD，∠EBM=∠+∠=∠ABD，∴∠DBC=∠EBM.在△DBC中，EBMC∴△DBC≌△EBM.∴BC=BM.在△ABC中∠C=90°，BAC=30°，1∴BC.21∴BMAB.2∴EM垂直平分AB.∴AE=BE.∴AE=BD.······································································5分1②CD2AB2AE2.······················································7分4初三数学参考答案及评分标准第6页（共7页）25.解：（1）①，0)E(1，5)··················································2分3②（Ⅰ）当点（4，）在直线y=kx+3上时，k，k=-42（Ⅱ）当点（3，1）在直线y=kx+3上时，3+3=1，=-（Ⅲ）当点（2，2）在直线y=kx+3上时，2+3=2，=-312（Ⅰ）（Ⅱ）（Ⅲ）结合图象可得312·····