2024C20最后一卷数学试卷及答案

安徽省C20教育联盟2024年九年级学业水平测试“最后一卷”数学参考答案一、选择题（本大题共10小题，每小题4分，满分40分）题号12345678910答案BCBADABCDA二、填空题（本大题共4小题，每小题5分，满分20分）11．2.7721101112．y(x1)2313．14．（1）(1,1)（2）0c1（第1空2分，第2空三分）2三、（本大题共2小题，每小题8分，满分16分）115．解：原式=241····································（4分）2=221······································（7分）=3·········································（8分）16．设日平均增长率为x．····································（1分）根据题意，得：128128(1x)128(1x)2608························（4分）解得：x10.5，x23.5（不合题意，舍去）··························（7分）答：该酒店入住人次的日平均增长率为50%．···························（8分）四、（本大题共2小题，每小题8分，满分16分）17．（1）如图所示，ABC即为所求；······························（4分）（2）如图所示，CP即为所求；································（8分）（注：若无文字说明扣１分，字母未标或标错位置扣１分）第1页共4页52418．解：（1）52·································（2分）5252n24（2）n2····································（5分）n2n2证明过程如下：n2n2(n2)(n2)n2(n24)n2n244左边=(n2)=右边n2n2n2n2n2故等式成立．·········································（8分）五、（本大题共2小题，每小题10分，满分20分）19．解：过点E作EP垂直AB于点P，则EPAFAPEFA90∴四边形EFAP是矩形，∴EPFA·······························（1分）在RtCDF中，DCF30，CD10····························（3分）CF3∴DF5，cos30，CFCDcos301053··················（5分）CD2∴EPFACACF2053BP在RtBEP中，tanBEP，BEP37PE3153∴BPEPtan37(2053)15·························（7分）44153∴ABAPBPDEDFBP1.551527.9928.0(m)·············（9分）4答：宝塔AB的高度约为28.0m．································（10分）20．解：（1）∵AB为O的直径，∴ACB90························（1分）∵点E、B在圆上，∴CEACBA······························（2分）AC3∵AC6，∴sinCEAsinCBA，∴AB10AB5∴O的直径AB为10·····································（5分）（2）连接OE交BC于点P，如图所示由（1）得，直径AB10∴在RtABC中，BCAB2AC2102628·······················（6分）∵点E为BC的中点，∴CEBE，第2页共4页∴OE垂直平分BC1111∴OPAC63，CPBC84························（8分）2222∴PEOEOPOBOP532·····························（9分）在RtCPE中，CECP2PE2422225······················（10分）六、（本题满分12分）21．解：（1）1·········································（2分）补全频数直方图，如图所示··································（4分）（2）3，6；··········································（8分）61（3）1000700（人），10答：该校1000名学生在这一周劳动时间不少于3小时的人数约为700人．··············（12分）七、（本题满分12分）22．（1）解：∵四边形ABCD是菱形，∴CA平分BCD·······················（1分）1∵CG平分DCF，∴GCEDCF····························（2分）21111∴ACGACEGCEBCDDCF(BCDDCF)180902222即：ACG90·······································（4分）（2）证明：连接BD分别交AC、AE于点O、P，如图1所示．∵四边形ABCD是菱形，∴AC垂直平分BD由（1）得，ACG90，∴BD//CG∵AOOC，∴APPG···································（5分）∵E为CD的中点，AD//CF∴DECE，DAECFE，ADEFCE∴ADEFCE(AAS)，∴ADCF·····························（6分）又∵ADBC，∴CFBC∵BD//CG，PGGF····································（7分）∴APPGGF，即点G为线段AF的三等分点·························（8分）（3）解：连接BD交AC于点O，连接OE，如图2所示．∵四边形ABCD是菱形，∴AC垂直平分BD1又∵E为CD的中点，OE∥CF＝2第3页共4页POOE1∴POE∽PCB，∴······························（9分）PCBC2设POx，则PC2x，在RtBOC中，BO2BP2OP2BC2OC2·························（10分）∴32x252(3x)2，解得：x2（负值舍去）·······················（11分）∴AC2OC6x62···································（12分）八、（本题满分14分）222(a2)23．（1）解：yx2a2xa，对称轴xa2···············（1分）1222当xa2时，y1(a2)2a2(a2)a4a4，∴A(a2,4a4)·········（2分）222(a2)yx2a2xa8，对称轴xa2····················（3分）2222当xa2时，y2(a2)2a2(a2)a84a12，∴B(a2,4a12)······（4分）2222（2）证明：令y1y2，得：x2a2xax2a2xa8化简得：x22axa240，即(xa)24··························（6分）解得：x1a2，x2a2··································（7分）将x1a2，x2a2分别带入二次函数中，得：y14a12，y24a4∴交点坐标为(a2,4a12)和(a2,4a4)··························（8分）即：函数y1与y2相交于A、B两点．·······························（9分）（）解：当时，2，顶点；2，顶点3a0y1x4xA(2,4)y2x4x8B(2,12)∴直线AB解析式为：y4x4·······························（10分）设2，则2，∴2················（分）P(t,t4t)M(t,t4t8)PMyMyP2t81122则Q(t4,t4t)，则N(t4,4t12)，∴QNyQyNt8t12··············（12分）∵四边形PMQN为平行四边形，∴PMQN，∴2t28t28t12··············（13分）2解得：t，t2（舍去）1322∴t··········································（14分）3【以上各题其它解法正确可参照赋分】第4页共4页