【数学试卷+答案】2024福州初三一检

2023－2024学年第一学期福州市九年级数学适应性练习答案及评分标准评分说明：1．本解答给出了一种或几种解法供参考，如果学生的解法与本解答不同，可根据习题的主要考查内容比照评分参考制定相应的评分细则．2．对于计算题，当学生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示学生正确做到这一步应得的累加分数．4．只给整数分数．选择题和填空题不给中间分．一、选择题（本题共10小题，每小题4分，共40分）1．A2．B3．D4．B5．A6．C7．D8．C9．A10．C二、填空题（本题共6小题，每小题4分，共24分）11．（−2，1）12．>13．1∶2214．yx=+2（答案不唯一）15．14016．5三、解答题（本题共9小题，共86分）17．（本小题满分8分）解：a=1，b=−3，c=1．································································································1分2Δ=ba−4c··············································································································2分2=−−××(3)411=>50．··························································································4分方程有两个不相等的实数根2−±bba−4cx=······································································································5分2a−−±(3)5=···········································································································6分21×35±=，235+35−即x=，x=．·······················································································8分122218．（本小题满分8分）证明：∵OC=2OB，OD=2OA，OAOB1∴==．··································································································2分ODOC2A∵∠=AOB∠DOC，································································································3B分∴△AOB∽△DOC，································································································5O分ABOA1∴==，··································································································7分DCOD2DC∴CD=2AB．·······································································································8分19．（本小题满分8分）解：列表如下：x…−6−3−2−11236…y…−1−2−3−66321…····························································2分九年级数学试题答案平分参考第1页（共6页） yO1x····························································5分3当x<4时，y<0或y>．·························································································8分220．（本小题满分8分）解：记这3根细绳朝上的一端分别为A，B，C，其对应的朝下的一端分别为a，b，c．根据题意，可以列出如下表格：头ABC尾a（A，a）（B，a）（C，a）b（A，b）（B，b）（C，b）c（A，c）（B，c）（C，c）····························································4分由表（图）知，所有可能出现的结果共有9种，·······························································5分且这些结果出现的可能性相等．····················································································6分其中小亮选中的两端属于同一根细绳的结果有3种，·························································7分31∴小亮选中的两端属于同一根细绳的概率P==．························································8分9321．（本小题满分8分）解：（1）ABCO····························································3分如图，O为所求作的△ABC的外接圆圆心．·····························································4分（2）连接OA．∵AB==AC1，∠=°BAC120，∴∠=ABC∠=°ACB30．·····················································································5分∵ABA=B，ACA=C，A∴∠=AOB26∠=°ACB0，∠=AOC26∠=ABC0°，∴∠=BOC∠+AOB∠=°AOC120．······························································BC··········6分∵OA=OB，∴△AOB是等边三角形，O∴OA==AB1．·································································································7分根据题意，得半径OB，OC及BC所围成的图形是扇形，2120π1⋅⋅π∴S==，扇形OBC3603π即半径OB，OC及BC所围成的图形的面积是．····················································8分3九年级数学试题答案平分参考第2页（共6页） 22．（本小题满分10分）解：（1）①4；·················································································································1分②9；·················································································································2分③1；·················································································································3分2（2）第一步：将方程变形成xx+=514；·······································································4分5第二步：构造一个边长为x+的正方形（如图）；······················································6分2x25x25xS225第三步：求得右下角正方形面积S的值是；··························································7分4第四步：用两种方法表示图中大正方形的面积：55225()x+=+++xxxS，······································································8分2222将xx+=514代入，5821可得()x+=．··················································································9分24∵x>0，∴x=2．······························································································10分23．（本小题满分10分）（1）证明：连接OC．∵PA是⊙O的切线，AB是⊙O的直径，∴OA⊥AP，∴∠=°OAP90．·····························································································1分∵BCB=C，∴∠=BOC2∠BAC．······························································A·························2分∵∠=APC2∠BAC，∴∠=BOC∠APC．OP∵∠+BOC∠=°AOC180，∴∠+APC∠=°AOC180．·················································································3分BC∵在四边形OAPC中，∠+OAP∠+APC∠+AOC∠=°OCP360，∴∠=OCP90°，·····························································································4分∴OC⊥PC．∵C为半径OC的外端点，∴PC为⊙O的切线．·······················································································5分（2）解：连接OP交AC于点D．∵PA，PC是⊙O的切线，PC==AB4，∴PA===PCAB4，PO平分∠APC，···································································6分∴OP⊥AC，ACA=2D．····················································································7分1在Rt△OAP中，∠=°OAP90，OA==AB2，2A22∴OP=+=OAPA25．·················································································8分11又SO=⋅=⋅AAPOPAD，······························································OP············9分△OAP22DOAAP⋅45∴AD==，OP5BC85∴AC=．··································································································10分5九年级数学试题答案平分参考第3页（共6页） 24．（本小题满分12分）−2a解：（1）根据题意得抛物线的对称轴为直线x=−=1．·······················································1分2a∵该抛物线与x轴交于A，B两点（点A在点B左侧），且AB=4，∴A（−1，0），B（3，0）．····················································································3分2将A（−1，0）代入ya=−+xa2xc，得aac++=20，即30ac+=．·····································································································4分（2）①该抛物线上到x轴距离为4的点的纵坐标可以为4或−4，22即ax−+=24axc或ax−+=24axc−．∵a<0，∴该抛物线开口向下．∵抛物线上到x轴距离为4的点恰有三个，2∴关于x的方程ax−+=24axc−有两个不相等的实数根，2关于x的方程ax−+=24axc有两个相等的实数根，即该抛物线的顶点纵坐标为4，············································································5分2∴设该抛物线的解析式为yax=−+(1)4．····························································6分2将A（−1，0）代入yax=−+(1)4，得440a+=，··································································································7分解得a=−1，2∴抛物线的解析式为yx=−−(1)4+．··································································8分②∵B（3，0），C（0，3），∴由待定系数法得直线BC的解析式为yx=−+3．记AP交BC于点G，分别过点B，C作AP的垂线，垂足分别为M，N，11∴∠=BMG∠=CNG90°，SA=⋅PBM，SA=⋅PCN．△APB△APC22∵△APC的面积为△APB面积的2倍，y11C∴APCN⋅=×⋅2APBM，22∴CN=2BM．································································································9分M∵∠=BGM∠CGN，GP∴△BGM∽△CGN，NBGBM1ABOHx∴==，CGCN2∴CG=2BG，∴点G在线段BC上，或者CB的延长线上．过点G作GH∥y轴，交x轴于点H，HOGC∴=．································································································10分BOBCHOGCGC2(ⅰ)当点G在线段BC上时，===，BOBCBG+CG3∴OH=2，即点G的横坐标为2，∴G（2，1），11∴由待定系数法得直线AG的解析式为yx=+．33112将yx=+代入yx=−−(1)4+，33258得xx−−=0，338解得x=−1，x=，123811∴P（，）．·························································································11分39九年级数学试题答案平分参考第4页（共6页） HOGCGC(ⅱ)当点G在CB的延长线上时，===2，BOBCCG−BG∴OH=6，即点G的横坐标为6，y∴G（6，−3），C33∴由待定系数法得直线AG的解析式为yx=−−．77332将yx=−−代入yx=−−(1)4+，77NOBH21724得xx−−=0，Ax7724解得x=−1，x=，12M7P2493∴P（，−）．749G8112493综上，点P的坐标为（，）或（，−）．················································12分3974925．（本小题满分14分）证明：（1）∵BC绕点B顺时针旋转α（01°<<α80°）得到BD，∴BDB=C，∠=CBDα，·················································································1分1∴∠=BCD∠=BDC90°−α．······························································A··············2分2∵01°<<α80°，1D∴09°<0°−α<90°．2E∵将DC绕点D逆时针旋转90°得到DE，∴∠=CDE90°，DCD=E，···············································································3分∵∠=°ACB90，∠=CDE90°，CB1∴∠=ACD∠−ACB∠=BCDα，21∠=BDE∠−CDE∠=BDCα，2即∠=ACD∠BDE．··························································································4分解：（2）连接AD．∵BC=BD，AC==BC4，A∴BD==AC4．在Rt△ABC中，∠=°ACB90，D22E∴AB=+=ACBC42．·················································································5分∵DEC=D，∠=ACD∠BDE，∴△ACD≌△BDE，·····························································································6分CB∴BE=AD．······································································································7分∵ADAB−=−BD424（当且仅当点D在线段AB上时，等号成立），························8分∴BE的最小值为424−．····················································································9分（3）延长ED至点F，使得DFD=E，连接AF，CF，AD．∵∠=CDE90°，∴CD垂直平分EF，∴CE=CF．·····································································································10分又∠=CED45°，A∴∠=CFD∠=°CED45，∴∠=°FCE90．F∵∠=°ACB90，D∴∠=FCA∠ECB．E∵CA=CB，∴△FCA≌△ECB，CB∴AFB=E，∠=FAC∠=°EBC45，······································································11分∴∠=FAB∠+FAC∠=°CAB90．九年级数学试题答案平分参考第5页（共6页） 在Rt△FAE中，D为EF中点，1∴ADE===FDEDF．2由（2）得△ACD≌△BDE，∴ADB=E，DCE=D∴DCA===DAFDF，∴△ADF是等边三角形，∠=DAC∠DCA，∴∠=DAF60°，································································································12分∴∠=DAC∠−DAF∠=CAF15°，∴∠=DCA∠=DAC15°，·····················································································13分1即α=°15，2∴α=°30．·······································································································14分九年级数学试题答案平分参考第6页（共6页）