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九年级数学试题第1页（共4页）人教版数学九年级上册期末模拟试卷一、选择题1．在平面直角坐标系中，点M（1，﹣2）与点N关于原点对称，则点N的坐标为A．（﹣2，1）B．（1，﹣2）C．（2，-1）D．（-1，2）2．用配方法解一元二次方程0122

xx，可将方程配方为A．212xB．012xC．212xD．012x3．下列事件中，属于随机事件的有①任意画一个三角形，其内角和为360°；②投一枚骰子得到的点数是奇数；③经过有交通信号灯的路口，遇到红灯；④从日历本上任选一天为星期天．A．

①②③B．②③④C．①③④D．①②④4．下列抛物线的顶点坐标为(4，－3)的是A．342xyB．342xyC．342xyD．342xy5．有n支球队参加篮球比赛，共比赛了15场，每两个

队之间只比赛一场，则下列方程中符合题意的是A．151nnB．151nnC．301nnD．301nn6．某小组在“用频率估计概率”的实验中，统计了某种结果出现的频率，绘制了如图所示的折线图，那么符合这一结果的实验最有可能的是A．袋子中

有1个红球和2个黄球，它们只有颜色上的区别，从中随机地取出一个球是黄球B．掷一个质地均匀的正六面体骰子，落地时面朝上的点数是6C．在“石头、剪刀、布”的游戏中，小宇随机出的是“剪刀”D．掷一枚质地均匀的硬币，落地时结果是“正面向上”7．如果一个正多边形的中心角为60°，那么这个正多边形的边数是

A．4B．5C．6D．78．已知点A（x1，y1），B（x2，y2）是反比例函数xy1的图象上的两点，若x1＜0＜x2，则下列结论正确的是A．y1＜0＜y2B．y2＜0＜y1C．y1＜y2＜0D．y2＜y1＜09．如图，AB为⊙O的直径，PD切⊙

O于点C，交AB的延长线于D，且CO=CD，则∠PCA=A．30°B．45°C．60°D．67.5°频率次数5000400030002000100000.250.200.150.100.05（第6题图）DCBOAP（

第9题图）九年级数学试题第2页（共4页）10．如图，在Rt△ABC和Rt△ABD中，∠ADB=∠ACB=90°，∠BAC=30°，AB=4，AD=22，连接DC，将Rt△ABC绕点B顺时针旋转一周，则线段DC长的取值范围是A．2≤DC≤4B．22≤DC≤4C．2

22≤DC≤22D．222≤DC≤222二、填空题11．如图，在平面直角坐标系xoy中，矩形OABC，OA=2，OC=1，写出一个函数0kxky，使它的图象与矩形OABC的边有两个公共点，这个函数的表达式可以为（答案不唯一）．

12．已知关于x的方程032axx有一个根为﹣2，a=．13．圆锥的底面半径为7cm，母线长为14cm，则该圆锥的侧面展开图的圆心角为°．14．设O为△ABC的内心，若∠A=48°，则∠BOC=

°．15．把球放在长方体纸盒内，球的一部分露出盒外，其截面如图所示，已知EF=CD=4cm，则球的半径为cm．16．抛物线cbxaxy2（a>0）过点（﹣1，0）和点（0，﹣3），且顶点在第四象限，则a的取值范围是．三、解答题17．解方程（每小题4分，共8分）（1）02

2xx（2）01232xx18．（8分）已知关于x的方程)0(03)3(2kxkkx．（1）求证：方程一定有两个实数根；（2）若方程的两个实数根都是整数，求正整数k的值．CABOyx（第11题图）CDAB（第10题图）CBEFAD（第15题图）九

年级数学试题第3页（共4页）19．（8分）有甲、乙两个不透明的布袋，甲袋中有3个完全相同的小球，分别标有数字0，1和2；乙袋中有3个完全相同的小球，分别标有数字1，2和3，小明从甲袋中随机取出1个小球，记录标有

的数字为x，再从乙袋中随机取出1个小球，记录标有的数字为y，这样确定了点M的坐标(x，y)．（1）写出点M所有可能的坐标；（2）求点M在直线3xy上的概率．20．（8分）如图，直线y=x+2与y轴交于点A，与反比例函数0kxky的图象交于点C，过

点C作CB⊥x轴于点B，AO=2BO，求反比例函数的解析式．21．（8分）如图，12×12的正方形网格中的每个小正方形的边长都是1，正方形的顶点叫做格点．矩形ABCD的四个顶点A，B，C，D都在格点上，将△ADC绕点A顺时针

方向旋转得到△AD′C′，点C与点C′为对应点．（1）在正方形网格中确定D′的位置，并画出△AD′C′；（2）若边AB交边C′D′于点E，求AE的长．C'ABDC（第21题图）CAOByx（第20题图）九年级数学试题第4页（共

4页）22．（10分）在矩形ABCD中，AB=8，BC=6，将矩形按图示方式进行分割，其中正方形AEFG与正方形JKCI全等，矩形GHID与矩形EBKL全等．（1）当矩形LJHF的面积为43时，求AG的长；（

2）当AG为何值时，矩形LJHF的面积最大．23．（10分）如图，点A，C，D，B在以O点为圆心，OA长为半径的圆弧上，AC=CD=DB，AB交OC于点E．求证：AE=CD．LHIKJFEDBCAG（第22题图）1）OABCDE（

第23题图）九年级数学试题第5页（共4页）24．（12分）如图，在等边△BCD中，DF⊥BC于点F，点A为直线DF上一动点，以B为旋转中心，把BA顺时针方向旋转60°至BE，连接EC．（1）当点A在线段DF的延长线上时，①

求证：DA=CE；②判断∠DEC和∠EDC的数量关系，并说明理由；（2）当∠DEC=45°时，连接AC，求∠BAC的度数．25．（14分）如图，在平面直角坐标系xoy中，二次函数cbxaxy2（0a）的图象经过A（0，4），B（2，0），C（-2，0）三点．

（1）求二次函数的解析式；（2）在x轴上有一点D（-4，0），将二次函数图象沿DA方向平移，使图象再次经过点B．①求平移后图象顶点E的坐标；②求图象A，B两点间的部分扫过的面积．FDBCAOExy（第25题图）EDFBCA（第24题图）九年级数学试题第6页（共4页）参

考答案一、选择题（本大题共10小题，每小题4分，共40分）1．D；2．A；3．B；4．C；5．C；6．B；7．C；8．B；9．D；10．D．二、填空题（本大题共6小题，每小题4分，共24分）11．如：xy1（答案不唯一，0＜k＜2的

任何一个数）；12．2；13．180；14．114；15．2.5；16．0＜a＜3.三、解答题（本大题共9小题，共86分）17.（每小题4分，共8分）(1)解：0)2(xx„„„„„„„„„„„„„„„„„„„„„„„2分∴2,021xx.„

„„„„„„„„„„„„„„„„„„„4分（2）解：1,2,3cba∴161-34-22）（∴64232162x„„„„„„„„„„„„„„„„2分∴1,3121xx.„„„„„„„„„„„„„„„„„„„4分18．（8分）（1）证明：

9634)3(22kkkk0)32k（，„„„„„„„„„„„„„„„„„„„„2分∴方程一定有两个实数根.„„„„„„„„„„„„„„„„3分（2）解：3,3,ckbka，22)3(34)3kk

k（，kkkkkkx2)3(32)3()3(2，kxx3,121，„„„„„„„„„„„„„„„„„„6分∵方程的两个实数根都是整数，∴正整数31或k．„„„„„„„„„„„„„„„„„„„8分19．（8分）解：（

1）九年级数学试题第7页（共4页）方法一：列表：yx1230(0,1)(0,2)(0,3)1(1,1)(1,2)(1,3)2(2,1)(2,2)(2,3)从表格中可知，点M坐标总共有九种可能情况：（0,1），（0,2），（0,3），（1,1），（1,2），（

1,3），（2,1），（2,2），（2,3）．„„„„„„„„„„„„„„„„„„„„„„„3分方法二：从树形图中可知，点M坐标总共有九种可能情况：（0,1），（0,2），（0,3），（1,1），（1,2），（1,3），

（2,1），（2,2），（2,3）．„„„„„„„„„„„„„„„„„„„„„„„3分（2）当x=0时，y=-0+3=3，当x=1时，y=-1+3=2，当x=2时，y=-2+3=1，„„„„„„„„„„„„„„„„„„„„6分由（1）可得点M坐标总

共有九种可能情况，点M落在直线y=-x+3上（记为事件A）有3种情况．∴P(A)3193．„„„„„„„„„„„„„„„„8分20．（8分）解：当x=0时，y=2，∴A(0，2)，„„„„„„„„„„„„„2分∴AO=2，∵AO=2BO，∴BO=1，„„„„„„„„„„„

„„„„„„„4分当x=1时，y=1+2=3，∴C(1，3)，„„„„„„„„„„„„„„„„„6分把C(1，3)代入xky，解得：3kxy3:反比例函数的解析式为„„„„„„„„„„„„„„„„„„„8分21．（8分）解：（1）准确画出图形；„„„„„„„„„„„„„„„„

„„„3分102321321321甲袋：乙袋：ED'C'ABDC（第21题答题图）九年级数学试题第8页（共4页）（2）∵将△ADC绕点A顺时针方向旋转得到△AD′C′，点C与点C′为对应点，∴△ADC≌△AD′C′，∴AC

=AC′，AD′＝AD=5，CD′＝CD=10，∠AD′C′＝∠ADC＝90°，∠AC′D′＝∠ACD，∵AB∥CD，∴∠BAC＝∠ACD，∵AB⊥CC′，AC=AC′，∴∠BAC＝∠C′AB，∴∠AC′D′＝∠C′AB，∴C′

E＝AE．„„„„„„„„„„„„„„„„„„„5分222RECBEBCBECt中，在，xAEABBExAE-10-,则设，222)-105xx（，„„„„„„„„„„„„„„„„„„„„„„„„„„7分425:x解得.425的长为答：AE

„„„„„„„„„„„„„„„„„„„„„„„„„„8分22．（10分）解：（1）正方形AEFG和正方形JKCI全等，矩形GHID和矩形EBKL全等，设AG=x，DG=6-x，BE=8-x，FL=x-(6-x)=2x-6，LJ=8-2x，方法1：LJFL

SLIHF矩形，∴43)28)(62(xx„„„„„„„„„„„„„„„„„„„„„„„„2分∴415,41321xx，AG=413或AG=415．„„„„„„„„„„„„„„„4分方法2：AEFGDGH

IABCDLIHFSSSS正方形矩形矩形矩形22)6)(8(2248432xxx，„„„„„„„„„„„„„„„„„„„2分∴415,41321xx，AG=413或AG=415．„„„„„„„„„„„„„„„4分（2）设矩形LJH

F的面积为S，)28)(62(xxS„„„„„„„„„„„„„„„„„„„„„„„„„6分九年级数学试题第9页（共4页）482842xx1)27(42x„„„„„„„„„„„„„„„„„„„„„„„„„8分04a，S有最大值，当AG=27时

，矩形LJHF的面积最大．„„„„„„„„„„„„„„„10分23．（10分）证明：方法一：连接OC，OD，∵AC=CD=DB，∴DBCDAC弧弧弧，∴BODCODAOC，„„„„„„„„„„„„„„„„„„„„2分∴AOCCODDOBCODCOB22，∵

CAECOB2，∴CAEAOC，„„„„„„„„„„„„„„„4分OCOAAOC中，在，2-902180AOCAOC-ACO，„„„„5分ACECAEAECACE--18

0中，在)290(180AOCAOC2-90AOC，„„„„„„„„„„„„„„„„„„„„„„„„„„6分AECACE，„„„„„„„„„„„„„„„„„„„„„„„„7分AEAC，„„„„„„„„„„„„„„„

„„„„„„„„„„„8分CDAC，CDAE．„„„„„„„„„„„„„„„„„„„„„10分方法二：连接OC，OD，∵AC=CD=DB，∴DBCDAC弧弧弧，∴BODCODAOC，„„„„„„„„„„„„„

„„„„„„„2分∴AOCCODDOBCODCOB22，∵CAECOB2，∴CAEAOC，„„„„„„„„„„„„„„„4分∵∠CAO=∠CAE+∠EAO，∠AEC=∠AOC

+∠EAO，∴∠CAO=∠AEC，„„„„„„„„„„„„„„„„„„„„„„„„„6分OCOAAOC中，在，∴∠ACO=∠CAO，∴∠ACO=∠AEC，AEAC，„„„„„„„„„„„„„„„„„„8分OABCDE（第23题答题图

）九年级数学试题第10页（共4页）CDAC，CDAE„„„„„„„„„„„„„„„„„„„„„„10分方法三：连接AD，OC，OD，∵AC=DB，∴弧AC=弧BD，∴∠ADC=∠DAB，„„„„„„„„„„„„„„„„„„„„„„„„„2分∴CD∥AB

，∴∠AEC=∠DCO，„„„„„„„„„„„„„„„„„„„„„„„„„4分∵AC=CD，AO=DO，∴CO⊥AD，∴∠ACO=∠DCO，„„„„„„„„„„„„„„„„„„„„„„„„„6分∴∠ACO=∠AEC，∴AC=AE，

„„„„„„„„„„„„„„„„„„„„8分∵AC=CD，∴AE=CD．„„„„„„„„„„„„„„„„„„„„„„„10分24．（12分）（1）①证明：∵把BA顺时针方向旋转60°至BE，∴AB

EBEBA，60°，„„„„„„„„„„„„1分在等边△BCD中，BCDB，60DBCFBAFBADBCDBA60，FBACBE60，CBEDBA，„„„„„„„„„„„„„„„„2分∴△BAD≌△BEC，∴DA=CE；„„„„„„„„„„„„

„„„„„„„3分②判断：∠DEC+∠EDC=90°．„„„„„„„„„„4分DCDB，BCDA，3021BDCBDA，∵△BAD≌△BEC，∴∠BCE=∠BDA=30°，„„„„„„„„„„„„„„„„„„„„„„„5分在等边△BCD中，∠BCD=

60°，∴∠ACE=∠BCE+∠BCD＝90°，∴∠DEC+∠EDC=90°．„„„„„„„„6分（2）分三种情况考虑：①当点A在线段DF的延长线上时（如图1），由（1）可得，为直角三角形DCE，90DCE，EDFB

CA（第24题答题图1）九年级数学试题第11页（共4页）459045DECEDCDEC时，当，DECEDC，CECD，由（1）得DA=CE，∴CD=DA，CDBDDBC中，在等边，CDDABD60BDC，

BCDA，3021BDCCDABDA，„„„„„„„„„„„„„„„„„7分DADBBDA中，在，752-180BDABAD，DCDADAC中，在，752-180ADCDAC，1507

575DACBADBAC．„„„„„„„„„„„„„8分②当点A在线段DF上时（如图2），BEBAB至顺时针方向旋转为旋转中心，把以60，60ABEBEBA，，60DBCBCBDBDC，中，在等边，ABEDBC，ABC

ABEABCDBC--，EBCDBA即，DBA≌CBE，CEDA，„„„„„„„„„„9分90RDFCDFCt中，在，DF＜DC，∵DA＜DF，DA=CE，∴CE＜DC，由②可知为直角三角形DCE，∴∠DEC≠45°．„„„„„„„„„„„10分③当点A

在线段FD的延长线上时（如图3），同第②种情况可得DBA≌CBE，ECBADBCEDA，，60BCDBDCBDC中，在等边，BCDA，EDFBCA（第24题答题图3）EDFBCA（第24题答题图2）九年级数学试

题第12页（共4页）3021BDCCDFBDF，150180BDFADB，150ADBECB，90BCDECBDCE，45904

5DECEDCDEC时，当，DECEDC，CECD，∴AD=CD=BD，„„„„„„„„„„„„„„„„„11分∵150ADCADB，152-180ADBBAD，152-180

CDACAD，30CADBADBAC，.30150或的度数为综上所述，BAC„„„„„„„12分25．（14分）（1）得）代入（）（）（把cbxaxyCBA20,2-,0,2,4,0，

0240244cbacbac，„„„„„„„„„„2分401:cba解得，42xy．„„„„„„„„„„„„4分（2）设直线DA得解析式为y=kx+d(k≠0)，把A（0，4）

，D（-4，0）代入得，044dkd，41:dk解得，∴y=x+4，„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分FDBCAOExy（第25题答题图）GIKHL九年级数学试题第13页（共4页

）设E（m，m+4），平移后的抛物线的解析式为：4)(2mmxy．把B（2，0）代入得：04)-2-2mm（不符合题意，舍去），解得(0521mm，∴E（5，9）．„„„„„„„„„„„„„„„„„„„„„„„„„„8分（3）如图，连接AB，过点B作BL∥AD交平移后的抛

物线于点G，连结EG，∴四边形ABGE的面积就是图象A，B两点间的部分扫过的面积．„„„„10分过点G作GK⊥x轴于点K，过点E作EI⊥y轴于点I，直线EI，GK交于点H．方法一：由点A（0，4）平移至点E（5，9），可知点B先向右平移5个单位，再向上平移5个单位至点G．∵B（2，0），∴点G（7

，5），„„„„„„„„„„„„„„„„„„„12分∴GK=5，OB=2，OK=7，∴BK=OK-OB=7-2=5，∵A（0,4），E（5,9），∴AI=9-4=5，EI=5，∴EH=7-5=2，HG=9-5=4，∴GBKEHGAEIAOBIOKHABGHS-

S-S-S-SS矩形四边形3025-8-635521-4221-5521-4221-97答：图象A，B两点间的部分扫过的面积为30.„„„„„„„„„„„14分方法二：bxyBL的解析式为设直线，02:0,2bB）代入得

（把点，2b，2xy，9)5(22xyxy联立，02:11yx解得，5722yx，∴点G（7，5），„„„„„„„„„„„„„„„„„„„„„„„„„12分∴

GK=5，OB=2，OK=7，∴BK=OK-OB=7-2=5，∵A（0,4），E（5,9），∴AI=9-4=5，EI=5，∴EH=7-5=2，HG=9-5=4，∴GBKEHGAEIAOBIOKHABGH

S-S-S-S-SS矩形四边形九年级数学试题第14页（共4页）3025-8-635521-4221-5521-4221-97答：图象A，B两点间的部分扫过的面积为30.„„„„„„„„„„„14分