江苏省徐州市部分学校2021--2022学年九年级下学期第一次模拟考试数学试题(word版含答案)

这是一份江苏省徐州市部分学校2021--2022学年九年级下学期第一次模拟考试数学试题(word版含答案)，共12页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
2021--2022学年度第二学期一模检测九年级数学试题(全卷共140分    考试时间120分钟) 一、选择题(本大题共有8小题，每小题3分，共24分．在每小题给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前面的字母代号填涂在答题卡相应位置)1．2的相反数是    A.  2       B．2        C．        D  2．下列图形是轴对称图形的是            A                  B               C                D3．下列运算中，正确的是    ·‘   A.  ·=       B．      C.+=      D．÷=1  4．若代数式在实数范围内有意义，则x的取值范围是     A．X＞l              B．X≥1           C. x≤1          D．x≠15．“市长杯”足球赛中，七支参赛球队进球数如下(单位：个)：3、5、2、2、3、1、3，这组数据的中位数和众数分别是    A．1．5，3           B．2，2             C．3，3         D．2，36．数轴上在和之间的整数有   A．0个        B．1个       C.2个      D．3个7．已知正方形的对称中心在坐标原点，顶点A、B、C、D按逆时针依次排列，若A点的坐标为 (3，5)，则点B与点C的坐标分别为    A．(—3，5)，(—3，—5)    B．(—5，3)，(5，—3)    C．(—5，3)，(3，—5)      D．(—5，3)，(—3，—5)   九年级数学试题  第1页共6页   8.北京冬奥会跳台滑雪项目比赛其标准台高度是90m．运动员起跳后的飞行路线可以看作是抛物线的一部分，运动员起跳后的竖直高度y (单位：m)与水平距离x(单位：m)近似满 足函数关系=a+bx+c(a≠0)．下图记录了某运动员起跳后 的x与y的三组数据，根据上述函数模型和数据，可推断出该运动  员起跳后飞行到最高点时，水平距离为  A．10m            B．15m  C．20m            D．22．5m(第8题)二、填空题(本大题共有10小题，每小题3分，共30分．不需要写出解答过程，请将答案直接填写在答题卡相应位置) 9．4的平方根是_________· 10．分解因式：—4x+4=________· 11．新型冠状病毒呈球形或椭圆形有包膜，直径大约是80～160纳米，1纳米＝米．用科学计数法表示160纳米=__________米．  12．若分式 的值为0，则x的值为__________·  13．已知圆锥的底面半径为2cm，母线长为6cm，则圆锥的侧面积为__________c． 14．  《九章算术》原文如下：今有人共买物，人出八，盈三；人出七，不足四．问人数、物价各几何?译文：现有一些人共买一个物品，每人出8钱，还盈余3钱；每人出7钱，则还差4钱，问共多少人，物品价格多少钱?设共有x人，物品的价格是y钱，则可列方程组为____________· 15．如图，一艘轮船从位于灯塔C的北偏东60°方向，距离灯塔30海里的小岛A出发，沿正南方向航行一段时间后，到达位于灯塔C的南偏东45°方向上的B处，这时轮船B与小岛A的距离是________海里． 16．如图，AF是正五边形ABCDE的外接圆的切线，则∠CAF＝__________°．            (第15题)               (第16题)                (第17题) 九年级数学试题  第2页共6页17．如图，Rt△ABC中，∠ACB=90°，AC=BC＝4，点D是AB边上的一个动点，以DC为斜边作Rt△DCE，使∠CDE=30°，点E、A在CD的两侧，当点D从点A运动到点B时，点E的运动路程为___________·    18．已知反比例函数y＝的图像过点A(a-l,y)，B(a+1，)，若＞，则a的取值范围为__________·三、解答题(本大题共有10小题，共86分，请在答题卡制定区域内作答，解答时应写出文字  、说明、证明过程或演算步骤)    19．  (本题10分)计算：    ·      (1)┃－┃—+2sin60°＋      (2)   20．  (本题10分)     (1)解方程：  一6x一7＝0；       (2)解不等式组：   21．(本题7分)随着奥密克戎病毒的传播，部分地区采用了在线授课学习方式．某校计划为学生 提供以下四类在线学习方式：在线讲授、观看微课、在线答题和在线讨论．为了解学生需 求，该校随机对本校部分学生进行了“哪类在线学习方式最感兴趣”的调查，并根据调查结果绘制成如下两幅不完整的统计图．             ( 第21题)  根据图中信息，解答下列问题：(1)本次调查学生共________人，补全条形统计图：(2)扇形统计图中“观看微课”对应的扇形圆心角等于__________°；    (3)该校共有学生2600人，请你估计该校对“在线授课”最感兴趣的学生人数．九年级数学试题  第3页共6页 22．(本题8分)如图，在□ABCD中，DE⊥AB,点F在AB的延长线上，且CF⊥AB．求证：  (1)△ADE≌ABCF：  (2)四边形DEFC是矩形．    (第22题)  23．(本题7分)2022年徐州中考体育进行改革，男女考生各有七项可选，每位考生可以任选三项进行测试．某班对学生选项情况进行调查．随机抽取其中一组5名学生的报名情况如下表， 这5名学生分别标记为A，B，C，D，E，其中“√，’表示选报该项．     (1)5名学生中选项是1分钟跳绳、立定跳远、掷实心球的概率是__________：(2)每组随机抽取选项是"50米游泳”的两人进行测试，用画树状图的方法求该组中抽到的恰好是A、C的概率．  24．(本题8分)为更好支撑徐州城市功能区发展，提升公共交通服务水平，完善城市综合交通运输体系，国家发展改革委原则同意徐州市城市轨道交通第二期建设规划．为使工程提前 年完成，需将工作效率提高 原计划完成第二期建设需要多少年?    九年级数学试题  第4页共6页 25．(本题8分)如图，已知一次函数y=-2x+8的图像与坐标轴交于A，B两点，并与反比例    函数y=的图像只有一个公共点C．    (1)点C的坐标是__________(2)点M为线段BC的中点，将点C和点M向左平移m(m＞0)个单位，平移后的对应点都落在反比例函数y＝  (k≠0)的图像上时，求众的值．         (第25题)26．(本题8分)如图，AC与⊙O交于点C，点B在⊙O上，OA＝6，AC＝4，  OB=2，  BC∥OA．  (1)求证：AC是⊙O的切线．  (2)求四边形AOBC的面积．  (第26题) 27．  (本题10分)已知线段AC．  (1)用无刻度的直尺和圆规作Rt△ABC，∠ACB=90°，∠BAC=60°．(不写作法，保留作图痕迹)  (2)在(1)的Rt△ABC中，若AC=-2，点D在线段CB上以每秒1个单位的速度从点C出发运动到点B停止，过点D作AC的平行线，交AB于点E．以DE为边向运动的相反方向作等边△DEF，设点D的运动时间为t(秒)．  ①求当点F在AC上时，t的值：  ②在整个运动过程中，是否存在这样的时刻t，使得以C、D、F为顶点的三角形为等腰三角形?若存在，请求出t的值；若不存在，请说明理由．  九年级数学试题  第5页共6页       (第27题) 28．(本题10分)如图，以AB为直径的⊙D与抛物线y＝abx＋c交于点A、B、C，与y    轴交于点E，点A、C、F的坐标分别是(-3，0)、  (0，－3)，过点B作y轴的垂线垂足为F(0，－4)．    (1)求线段CE的长；    (2)求抛物线的函数表达式：    (3)抛物线对称轴上是否存在点P，使⊙P与直线AB和x轴都相切?若存在，求出点P的坐标；若不存在，说明理由．         (第28题)        九年级数学试题  第6页共6页 2021—2022学年度第二学期一模检测九年级数学（答案及评分标准）一、选择题(每题3分，共24分)题号12345678答案ACABCCDB 二、选择题 (每题3分，共30分)9.  ；       10. ；       11. ；  12 .2；      13. 24π；14. ；   15. ；    16.72°；       17. ；   18. 三、解答题 (共86分)19.（1）原式=-1++2·························································4分=2+1. ··························································5分   （2）原式=······························································3分            = - 1··························································5分20.(1)解：··································································2分          x - 7＝0或x + 1＝0························································3分          x1 =7 ，x2 = -1  ·························································5分(2) 解不等式①得，，······················································2分解不等式①得，························································4分所以不等式组的解集是··················································5分21. (1)  120；条形图高12（图略）···················································2分(2)  72°·······························································4分(3) ···································································6分答：对“在线讲授”最感兴趣的学生人数是780人································7分22. 解：（1）∵四边形ABCD是平行四边形，∴，AD∥BC,·····························1分∴ ,······································································2分∵DE⊥AB, CF⊥AB,∴········································································3分∴△ADE≌△BCF····························································4分（2）∵，∴DE∥CF,·························································5分∵四边形ABCD是平行四边形，∴DC∥AB,··········································6分∴四边形DEFC是平行四边形,···················································7分∵∴四边形DEFC是矩形·······················································8分23. 解：（1）；·······························································2分（2）树状图（略）；.························································5分共有6种等可能情况，其中是A、C的有2种情况，则抽到恰好是A、C的概率是···················································7分24.解：设原计划完成第二期建设需要x年.············································1分根据题意，得：， ···························································4分解得：x＝5································································6分经检验x＝5是原方程的解······················································7分答：设原计划完成第二期建设需要5年.············································8分25. 解：点坐标为·······························································2分（2）∵一次函数的图象与坐标轴交于，两点，∴点······································································3分∵点为线段的中点，∴点······································································5分∴点和点平移后的对应点坐标分别为，···········································6分∴,   ∴····································································8分26. (1)证明：连接OC∵点C在圆上，OB=2，∴OC=OB=2··············································1分∵OA＝6，AC＝，∴,∴·······································································3分∴,∴AC是⊙O的切线．······················································4分                                                       (2)过点O作BC的垂线垂足为D∵BC∥OA, ∴∵, ∴△ODC∽△OCA,∴∴,······················································6分∴S=·····································································8分27.（1）作图(略)······························································2分（2）①解：过点F作DE的垂线，垂足为H∵AC=，∴BC=6,∵DC=t，∴BD=6 –t, ·························································1分∵DE∥AC，△DEF是等边三角形，∴，在RT△DFC中，,····························································2分在RT△DBE中，,····························································3分∵2FC=DE，∴，∴t=2 .·······················································4分②当CD=CF时,  过点C作DF的垂线，垂足为M ，在RT△DMC中，∴DM=2DF=2DE，∴，∴t=·····················································6分   当CD=DF时   ，∴·······································································8分当CD=DF时过点F作DC的垂线，垂足为N ，∴在RT△DFN中，∵DF=DE，∴，∴t=3. ··························································10分      28. 解：（1）过点D作OF的垂线，垂足为H∵BF⊥y轴，∴BF∥DH∥AO, ∴，············································1分∵OF=4，∴OH=2，∵OC=3，∴ CH=OC-OH=1∵DH⊥EC，∴CE=2CH=2·····················································2分（2）连接AC、BC∵OA=OC，，∴∵AB是⊙D的直径，∴，∴，∴BF=CF=FO -CO=1,∴点B的坐标是（-1，-4）·····················································4分将A (－3，0)、B (－1，－4)、   C (0，－3)代入得               ． ∴ ∴y=x2+2x-3．·····························································6分（3）∵设存在点P，∴BP=m+4，过点P作x轴的垂线,垂足为H，,∵AH=2，BH=4, ∴AB=,∵⊙P与直线AB和x轴都相切,∴,即∴，∴存在P，.····························································10分