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平谷区2022～2023学年度第一学期期末质量监控试卷初三数学2023年1月一、选择题（本题共16分，每小题2分）下面各题均有四个选项，其中只有一个．．是符合题意的.1．已知2x=3y（y0），下列比例式成立的是A

．xy=23B．xy=32C．x2=y3D．x3=2y2．如图，△ABC中，D、E分别为AB、AC边上的点，DE//BC，若AD=2BD，则DEBC的值为（）(A)12(B)32(C)21(D)233．将抛物线22yx=的图象先向左平移1个单位，再向下平移3个单位，得到的抛

物线的表达式是（A）22(1)-3yx=+(B)22(1)-3yx=−(C)22(1)3yx=++（D）22(1)3yx=−+4.如图，每个小正方形的边长为1，点A、B、C均在格点上，则sinB的值是（A）1（B）34（C）35（D）

455．如图，若点A是反比例函数2y=x0x（〉）的图象上一点，过点A作x轴的垂线交x轴于点B，点C是y轴上任意一点，则△ABC的面积为（A）1（B）2（C）3（D）46.如图，ABCDY中，点E为AD中点，若△AEO的面积为1，则△B

OC的面积为（A）2（B）3（C）4（D）87．“今有圆材，埋在壁中，不知大小，以锯锯之，深一寸，锯道长一尺，问径几何？”这是《九章算术》中的一个问题，用现代的语言表述为：如图，CD为⊙O的直径，弦AB

⊥CD于E，CE=1寸，弦AB=10寸，则⊙O的半径为多少寸（A）5（B）12（C）13（D）268.如果I表示汽车经撞击之后的损坏程度，经多次实验研究后知道，I与撞击时的速度v的平方之比是常数2，则I与v的函数关系为(A)正比例函数关系(

B)反比例函数关系(C)一次函数关系(D)二次函数关系二、填空题（本题共16分，每小题2分）9．函数y=x-3的自变量x的取值范围是．EOBDCAOEDACB10.扇形的圆心角为120°，半径为3，则扇形的弧长

为.11.如图，在Rt△ABC中，∠C=90°，如果cosA=23，AB=6，那么AC的长为．12．如图，在⊙O中，A，B，C是⊙O上三点，如果∠ACB=30º，弦AB=5，那么⊙O的半径长为_________________.13.已知二次函数y＝ax2＋bx＋c（a≠0）的部分图象,则关于x

的一元二次方程ax2＋bx＋c＝0的解为．14．如图，Rt△ABC中，∠BAC=90°，AD⊥BC于D，BD=1，CD=4，则AD的长为．15.青藏铁路是当今世界上海拔最高、线路最长的高原铁路，因路况、季节、天气等原因行车的平均速度在250-360（千

米/小时）之间变化，铁路运行全程所需要的时间（小时）与运行的平均速度（千米/小时）满足如图所示的函数关系，列车运行的平均速度最大和列车运行的平均速度最小时全程所用时间相差____________小时.16.张老师准备为书法兴趣小组的同学购买上课的用具，在文具商店看

到商店有A、B两种组合和C、D、E、F商品销售，组合及单件商品质量一样，若该小组共有12人，其中，笔和本每人各需要一份，砚台2人一方即可，墨汁n瓶（3n）。张老师共带了200元钱，请给出一个满足条件的购买方案__

_________（购买数量写前面商品代码写后面即可,例如：2A+3B+...）；n的最大值为__________.商品价格组合A（1支笔+1个本+1方砚台+1瓶墨汁）25元组合B（1支笔+1个本+1瓶墨汁）18元C：1支笔5元D：1个

本4元E：一方砚台10元F：一瓶墨汁12元OCAB三、解答题(本题共68分，第17、18、20、21、22、23、25题，每小题5分；第19、24题，每小题6分；第26-28题，每小题7分)解答应写出文字说

明、演算步骤或证明过程.17．计算：113272cos305−−+−+．18．已知：如图，在△ABC中，D为AB边的中点，连接CD，∠ACD＝∠B，AB=4，求AC的长.19．已知二次函数2-23yxx=−.（1）求该二次函数的顶点坐标；（2）求该二次函数图象与x轴、y轴的

交点；（3）在平面直角坐标系xOy中，画出二次函数2-23yxx=−的图象；（4）结合函数图象，直接写出当12x−时，y的取值范围．20.如图，已知劣弧AB，如何等分劣弧AB？下面给出两种作图方法，选择其中一种方法，利用直尺和圆规完成作图，并补全证明过程.方法一:①作射线

OA、OB；②作∠AOB的平分线OD，与弧AB交于点C；点C即为所求作.证明：∵OC平分∠AOB∴∠AOC=∠BOC∴（）(填推理的依据).方法二:①连接AB；②作线段AB的垂直平分线EF，直线EF与弧AB交于点C；点C即为所求作.证明：

∵EF垂直平分弦AB∴直线EF经过圆心O，∴（）(填推理的依据).21.某班同学们来到操场，想利用所学知识测量旗杆的高度.方法如下：如图，线段AB表示旗杆，已知A，C，D三点在一条直线上，首先用1.5米高的测角仪在点C处测得旗杆顶端B的仰角为65°，在点D处测得旗杆顶端

B的仰角为45°，其中，线段CE和DF均表示测角仪，然后测量出CD的距离为5.5米，连接EF并延长交AB于点G，.根据这些数据，请计算旗杆AB的长约为多少米.（1.265tan,4.065cos,9.065sin）22.已知：一次函数)0(2

−=kkxy，与反比例函数)0,0(=xmxmy交与点A（2,4）（1）求一次函数和反比例函数的表达式；（2）已知点P（0，n）（n>0）过点P作垂直于y轴的直线，与反比例函数交于点B，与一次函数交于点C，横、纵坐标都是整数的点叫做整点．若线段BC、AC与反

比例函数图象上AB之间的部分围成的图象中（不含边界）恰有3个整点，直接写出n的取值范围.23.如图，在Rt△ABC中，∠ACB=90°，AD平分∠BAC交BC边于点D，DE⊥AB于点E，若BD=5，,54cos=B求AC的长.24.如图，已知锐角∠

ABC，以AB为直径画⊙O，交BC于点M，BD平分∠ABC与⊙O交于点D，过点D作DE⊥BC于点E．（1）求证：DE是⊙O的切线；（2）连接OE交BD于点F，若∠ABC=60°，AB=4，求DF长．25.某景观公园内人工湖里有一组小型喷泉,水柱从垂直于湖面的水枪喷出，若设距水枪水平距离为

x米时水柱距离湖面高度为y米，y与x近似的满足函数关系2y=ax-h+ka0（）（〈）.现测量出x与y的几组数据如下,x(米)01234…y(米)1.753.03.754．03.75…请解决以下问题:(1)求出满足条件的函数关系式；(2)身高1.75米的小明

与水柱在同一平面中，设他到水枪的水平距离为m米（m≠0），画出图象，若小明被水枪淋到,结合图象求m的取值范围.26.在平面直角坐标系xOy中，抛物线2(0)yaxbxa=+，设抛物线的对称轴为.xt=（1）当抛物线过点（

-2，0）时，求t的值；（2）若点(2,)(1,n)m−和在抛物线上，若,mn且0,amn求t的取值范围.27.如图，△ABC中，D为AC边中点，E为BC延长线上一点，连接ED并延长，使DF=ED，连接BF.（1）依题意补全图形；（2）连接BD，若222ABBFCE=+，猜想BD与

DE的数量关系，并证明。28.如图，平面直角坐标系中，矩形ABCD，其中A（1,0）、B（4,0）、C（4,2）、D（1,2）.定义如下：若点P关于直线l的对称点P’在矩形ABCD的边上，则称点P为矩形ABCD关于直线l的“关联点”.（1）已知点P1(-1,2)、点P2（

-2,1）、点P3（-4,1）、点P4（-3,-1）中是矩形ABCD关于y轴的关联点的是；（2）Me的圆心M（27−,1）半径为23，若Me至少存在一个点是矩形ABCD关于直线x=t的关联点，求t的取值范围

；（2）Me的圆心M（m,1）（m<0)半径为r，若存在t值使Me上恰好存在四个点是矩形ABCD关于直线x=t的关联点，写出r的取值范围，并写出当r取最小值时t的取值范围(用含m的式子表示）。平谷区2022-2023年期末试卷评分标准初三数

学2023年1月一、选择题（本题共16分，每小题2分）二、填空题（本题共16分，每小题2分）题号910111213141516答案3x2451,321=−=xx22.2答案不唯一例如5A+1E+7C+7D4A+2E+8

C+8D3A+3E+9C+9D+1F等;5三、解答题(本题共68分，第17、18、20、21、22、23、25题，每小题5分；第19、24题，每小题6分；第26-28题，每小题7分)解答应写出文字说明

、演算步骤或证明过程.17.解：113272cos305−−+−+．3=35-3322++··············································································

·····4=5-3·······································································································

·518.解：∵D为AB中点，AB=4∴AD=2································································································

··················1∵∠ACD=∠B，∠A=∠A∴ADCACB:··········································································

········································3∴ADACACAB=·····················································································4∴24

ACAC=∴22AC=···················································································519.（1）322−−=xxy题号12345678答案

BDACACCD22113xx=−+−−2(1)4x=−−∴顶点坐标为（1，-4）······································1（2）x0,y=-3y0-3).=令∴与轴的交点为（，···························

························2（3）2120,x-2x-3=0,x=x=-x)-1,0.=令y解得3，1∴与轴的交点为（3，0和（）··4（4）画出图象····································

·······················································5（5）04−y·······································

··················································620.方法一:作图正确..........................................................................

...........................................2证明：∵OC平分∠AOB∴∠AOC=∠BOC∴弧AC=弧BC··················································

·············3（在同圆或等圆中，如果两个圆心角，那么它所对的弧相等，所对的弦也相等）(填推理的依据).····································································

··············································5方法二:作图正确····························································································2证

明：∵EF垂直平分弦AB∴直线EF经过圆心O，∴弧AC=弧BC··························································································

························3（垂径定理）.·····················································································

·····························521.解：由题意，∠BGF=90°∠BEG=65°，∠BFG=45°，EF=CD=5.5米AG=EC=FD=1.5米·····························

·······································1在Rt△BGE中，∵∠BGF=90°，∠BEG=65°21tan.BG∠BEG=EG··········································

··························2设EG=x，则BG=2.1x在Rt△BGF中，∵∠BGF=90°，∠BFG=45°∴BG=FG············································

······································3∴2.1x=x+5.5解得，5x=·············································

························4∴2.1x=10.5AB=10.5+1.5=12米·····························································5∴旗杆高约为12米.22.解：（1）∵反比例函数

)0,0(=xmxmy过点A（2,4）∴m=8xy8=：反比例函数的解析式为·········································1∵一次函数)0(2−=kkxy过点A（2,4）∴k=32-x3=y一次函数的解析式为：······

··································2（2）8721nn或·······································································523.解：∵D

E⊥AB∴∠BED=90°5,54cos==BDB∴BE=4，DE=3····································································2∵AD平分∠BAC，

∠ACB=90°，DE⊥AB∴DC=DE=3····································································3∴BC=BD+DC=88,54cos==BCB∴AB=10··············

···········································4由勾股，AC=6·················································524．（1）解：连结OD．∵BD平分∠ABC∴∠DCO=90°·········

······································1∴∠ABD=∠CBD，∵OD=OB∴∠ODB=∠ABD．∴∠ODB=∠CBD．············································2∴OD∥BC.∵DE⊥BC∴∠ODE=

∠DEB=90°∴DE是⊙O的切线·········································3（2）连接AD∵∠ABC=60°BD平分∠ABC，∴∠ABD=∠CBD=30°∵AB是直径∴∠ADB=90°∵AB=4∴AD=2，23BD=∵∠CBD=30°，

∠DEB=90°∴3DE=，BE=3··································4∵OD∥BCDFOBFE:∴ODDF=BEFB设DF=x2x=323x−∴435=DF·····

················625.解：（1）由表格可知抛物线的顶点坐标为（3,4）设抛物线的解析式为2340ya(x)(a)=−+...........................1∵抛物线过点（1,3）代入得，4a+4=3........................

...............................................................214∴a=-................................................................

..............................3∴21344y(x)=−−+(2）67m.................................................

....................................................526.（1）对称轴x=-1···················································

·····································2（2）若a>0,当m>n>0时，如图211tttt+−−此时，1122t-··················

·······································································4当0>m>n时，没有满足条件的抛物线.若a<0,则有m>0>

n时，如图此时，1t−..........................................................................................

..................................61122t-或1t−..................................................................

..........................................727.(1)补全图形................................................

.....................................................1(2)结论：BD=DE........................................

.....................................2证明：连接AF.∵D为AC中点∴AD=DC.....................................................................3∵DF=DE,AD=DC，∠

ADF=∠EDC∴△ADF≌△CDE.∴AF=EC，∠AFD=∠DEC..........................................4∴AF∥CE222ABBFCE=+∴∠AFB=90°...............

.............................................5∵FA∥BE∴∠FBE=90°.......................................................

..............6∵FD=DEDEEFBD==21.........................................................728．解：（1）P1，P3；··

··················································································2（2）12−t·················

··········································································4(3)222)4(1rr=−+35;=r解得35r1rr且的范围为：满

足条件的........................524232221mtmmtm++++或·······································································7