北京大兴区2023-2024年九上期末数学试卷及答案.pdf

该【北京大兴区2023-2024年九上期末数学试卷及答案 】是由【小屁孩】上传分享，文档一共【12】页，该文档可以免费在线阅读，需要了解更多关于【北京大兴区2023-2024年九上期末数学试卷及答案 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..大兴区2023~,共三道大题,28道小题,满分100分,考试时间120分钟。、准考证号,并将条形码贴在指定区域。,在试卷上作答无效。,选择题、作图题用2B铅笔作答,其他试题用黑色字迹签字笔作答。,请将答题卡交回。一、选择题(共16分,每题2分)第1-8题均有四个选项,,共设置了“数字民航”“电动航空”“商业航天”“通航维修”,则他选中“电动航空”,是中心对称图形而不是轴对称图形的为 A. B. C. -3x-1=0的根的情况,=(x-2)2+=-==-=,将抛物线y=3x2先向右平移4个单位长度,再向上平移1个单位长度,=3(x+4)2-=3(x+4)2+=3(x-4)2-=3(x-4)2+1初三数学试卷第1页(共6页):..,则60°,菱形OABC的顶点A,B,C在☉O上,过点B作☉☉O的半径为2,,点A,B在☉O上,且点A,O,B不在同一条直线上,点P是☉O上一个动点(点P不与点A,B重合),在点P运动的过程中,有如下四个结论:①恰好存在一点P,使得∠PAB=90°;②若直线OP垂直于AB,则∠OAP=∠OBP;③∠,所有??所有??正确结论的序号是A.①②B.①③C.②③D.①②③二、填空题(共16分,每题2分)(a-3)x2-3x-4=0是关于x的一元二次方程,-3x+m=0有一个根为1,,若点(2,y),(4,y)在抛物线y=2(x-3)2-4上,12y(填“>”,“=”或“<”).,四边形ABCD内接于☉O,点E在AD的延长线上,°.若∠CDE=80°,则∠,△ABC的内切圆☉O与AB,BC,CA分别相切于点D,E,F,若AD=2,BC=6,则△(共6页):..(0,1)且当自变量x>0时,“琮琮”“宸宸”“莲莲”组合名为“江南忆”,出自唐朝诗人白居易的名句“江南忆,最忆是杭州”,它融合了杭州的历史人文、,,某商店吉祥物“江南忆”6月份的销售量为1200件,8月份的销售量为1452件,设吉祥物“江南忆”6月份到8月份销售量的月平均增长率为x,,在平面直角坐标系xOy中,二次函数y=ax2+bx+c(a<0)的图象经过点(0,1),(2,1).给出下面三个结论:①2a-b=0;②a+b+c>1;③关于x的一元二次方程ax2+bx+c-m=0(m<1),、解答题(共68分,第17-21题每题5分,第22题6分,第23题5分,第24-26题每题6分,第27-28题,每题7分)解答应写出文字说明、:x2+8x=-2x-1=0的一个根,求代数式(a-1)2+a(a-2)-x+2m-2=0有两个实数根.(1)求m的取值范围;(2)当m取最大整数值时,=x2+bx+c经过点(1,0),(0,-3).(1)求抛物线的解析式;(2),在△ABC中,∠C=45°,AB=2,☉O为△ABC的外接圆,求☉(共6页):..,、111枚银牌、71枚铜牌的优异成绩,,某校对八、九年级学生进行了杭州亚运会知识竞赛(测试满分为100分,得分x均为不小于80的整数),并从其中分别随机抽取了20名学生的测试成绩,整理、描述和分析如下(成绩得分用x表示,共分成四组:≤x<85;≤x<90;≤x<95;≤x≤100).:80,82,83,83,85,85,86,87,89,90,90,91,94,95,95,95,95,96,99,:90,90,91,92,92,93,93,、九年级抽取的学生竞赛成绩的平均数、中位数、众数如下::根据以上信息,解答下列问题:(1)写出表中m,n的值及九年级抽取的学生竞赛成绩在D组的人数;(2)若该校九年级共400人参加了此次知识竞赛活动,估计九年级竞赛成绩不低于90分的人数是;(3)为了进一步弘扬体育运动精神,学校决定组织学生开展亚运精神宣讲活动,准备从九年级抽取的竞赛成绩在D组的学生中,随机选取一名担任宣讲员,、乙是抽取的成绩在D组的两名学生,用画树状图或列表的方法,求甲、(共6页):..,函数y=kx+b(k≠0)的图象经过点A(-1,2)和B(1,4).(1)求该函数的解析式;1(2)当x>2时,对于x的每一个值,函数y=x+n的值小于函数y=kx+b(k≠0)的值且大2于5,,AB是☉O的直径,点C在☉O上,连接AC,BC,过点O作OD⊥BC于点D,过点C作直线CE交OD延长线于点E,使得∠E=∠B.(1)求证:CE为☉O的切线;(2)若DE=6,CE=35,,某公园一个圆形喷水池,,A处是喷头,喷出水流沿形状相同的曲线向各个方向落下,,,水流竖直高度的最高处位置C距离喷水池中心O的水平距离OD为1m.(1)求喷出水流的竖直高度y(m)与距离水池中心O的水平距离x(m)之间的关系式,并求水流最大竖直高度CD的长;(2)安装师傅调试时发现,喷头竖直上下移动时,抛物线形水流随之竖直上下移动(假设抛物线水流移动时,保持对称轴及形状不变),,(共6页):..,点(2,m)在抛物线y=ax2+bx+c(a>0)上,设抛物线的对称轴为x=t.(1)当m=c时,求t的值;(2)点(-1,y),(3,y)在抛物线上,若c<m,请比较y,y1212的大小,△ABC中,∠BAC=90°,AB=AC,点P为BA的延长线上一点,连接PC,以P为中心,将线段PC顺时针旋转90°得到线段PD,连接BD.(1)依题意补全图形;(2)求证:∠ACP=∠DPB;(3)用等式表示线段BC,BP,BD之间的数量关系,,在平面直角坐标系xOy中,已知点M(0,t),N(0,t+2),对于坐标平面内的一点P,给出如下定义:若∠MPN=30°,则称点P为线段MN的“亲近点”.(1)当t=0时,①在点A(23,0),B(3,2),C(-23,2),D(-1,-3)中,线段MN的“亲近点”的是;②点P在直线y=1上,若点P为线段MN的“亲近点”,则点P的坐标为;(2)若直线y=-3x-3上总存在线段MN的“亲近点”,(共6页):..大兴区2023~2024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(共16分,每题2分)题号12345678答案CAABDDCA二、填空题(共16分,每题2分)题号9101**********答案不唯一,如:答案a?32=80161200(1+x)2=1452②③y=x2+1三、解答题(共68分,第17-21题每题5分,第22题6分,第23题5分,第24-26题每题6分,第27-28题,每题7分)解答应写出文字说明、:x2+8x=+8x+16=9+16.···································································1分(x+4)2=25.………………………………………………………………2分x+4=±5.·············································································3分解得x=1,x=-9.································································:(a?1)2+a(a?2)=a2?2a+1+a2?2a·····························································2分=2a2?4a+1········································································3分∵a是方程x2?2x?1=0的一个根,∴a2?2a?1=0,∴a2?2a=1.·······································································4分=2(a2-2a)+1∴原式=2?1+1=3··············································································5分初三数学参考答案及评分标准第1页(共6页):..:(1)∵方程有两个实数根,???0·················································································1分∵Δ=(-1)2-4×1×(2m-2)=1?8m+8=9?8m?9?8m?09?m?················································································2分89(2)m?,m为最大整数,8?m=1.···············································································3分∴x2﹣x=:x=0,x=1.································································:(1)∵抛物线y=x2+bx+c经过点(1,0),(0,-3),?1+b+c=0∴?.··········································································2分?c=?3?b=2解得?.?c=-3∴y=x2+2x-3.·····································································3分(2)y=x2+2x-3.=(x+1)2-4∴顶点坐标为(-1,-4).···························································5分初三数学参考答案及评分标准第2页(共6页):..:连接OA,OB,············································1分∵∠C=45°,∴∠AOB=2∠C=90°.··········································2分在Rt△AOB中,∵OA2+OB2=AB2,AB=2,OA=OB,∴2OA2=4.························································4分∴OA2=2.∴OA=2(舍负).∴⊙O的半径是2.···········································:(1)m=95,n=,九年级抽取的学生竞赛成绩在D组的人数为4人;····3分(2)240.·····················································································4分(3)设D组的另外两名同学为丙,,所有可能出现的结果共12种,,21所以P==.································································6分(甲乙同时被选上):(1)把A(-1,2)和B(1,4)代入y=kx+b(k≠0)中,??k+b=2,?………………………………………………………………1分?k+b=4.?k=1,解得:?………………………………………………………………2分?b==x+3.·················································3分(2)n=4·······················································································5分初三数学参考答案及评分标准第3页(共6页):..24.(1)证明:连接OC.∵OB=OC,∴∠B=∠OCB.∵∠E=∠B,∴∠E=∠OCB.·······························································1分∵OD⊥BC,∴∠E+∠DCE=90°.∴∠OCB+∠DCE=90°.∴∠OCE=90°.即OC⊥CE.∴CE是⊙O的切线.···························································2分(2)∵OD⊥BC,∴∠CDE=90°.在Rt△CDE中,DE=6,CE=35,∴CD=CE2?DE2=3.…………………………..........................………3分∵OE⊥BC,∴BC=2CD=6.∴DE=BC.………………………………………………………………4分∵AB是直径,∴∠ACB=90°.∴∠CDE=∠△ABC与△CED中,??B=?E,??BC=DE,???ACB=?CDE.∴△ABC≌△CED.……………………………………….………5分∴AC=CD=3.∵O是AB的中点,D是BC的中点,13∴OD=AC=.····································································6分22初三数学参考答案及评分标准第4页(共6页):..:(1)由题意,A点坐标为(0,),B点坐标为(,0).…………………………1分设抛物线的解析式为y=a(x-1)2+k(a≠0)…………….………………….…2分∵抛物线经过点A,点B.?=a+k,?∴?0=a(?1)2+k.?????a=?1,解得:??k=.∴y=-(x-1)2+(0≤x≤).……………………………….……………3分∴x=1时,y=.∴.………………………………..………4分(2)························································································:(1)∵点(2,m)在y=ax2+bx+c(a?0)上,∴m=4a+2b+∵m=c,∴4a+2b=0.∴b=-?2a∴t=?=?=1.…………..………………………………………2分2a2a(2)∵点(2,m)在抛物线yax2bxc(a0)上,∴m=4a+2b+c.∵c<m,∴m-c>0.∴m-c=4a+2b>0.∴2a+b>0.············································································3分∵点(-1,y),(3,y)在抛物线yax2bxc(a0)上,12∴y=a-b+c,y=9a+3b+c,12∴y-y=(9a+3b+c)-(a-b+c)=8a+4b=4(2a+b).································4分21∵2a+b>0,∴4(2a+b)>0,∴y-y>∴y>y………………………………………………………………….(共6页):..27.(1)解:补全图形如图所示;·························································1分(2)证明:∵∠BAC=90°,∴∠ACP+∠APC=90°.∵以P为中心,将线段PC顺时针旋转90°得到线段PD,∴∠DPC=90°.∴∠APC+∠BPD=90°.∴∠ACP=∠DPB.····························································3分(3)线段BC,BP,BD之间的数量关系是2BP=BD+BC.………………4分证明:过点P作PE⊥PB交BC的延长线于点E.∵PE⊥PB,∴∠BPE=90°.∵∠DPC=90°,∴∠1+∠BPC=∠2+∠BPC=90°.∴∠1=∠2.·······································································5分∵AB=AC,∠BAC=90°,∴∠ABC=∠ACB=45°.∵∠BPE=90°,∴∠PBE=∠PEB=45°.∴PB=PE.········································································6分在△PBD与△PEC中,?PB=PE,???1=?2,??PD=PC.∴△PBD≌△PEC.∴BD=EC.∵BE=BP2+PE2=BP2+BP2=2BP.∴2BP=BD+BC.·····························································:(1)①A,C;········································································2分()()②?3?2,1,3+2,1;······················································5分(2)-11≤t≤3.············································································7分初三数学参考答案及评分标准第6页(共6页)