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泸州市初2026届学业水平模拟试题一
数 学
全卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共4页．全卷满分150分．考试时间共120分钟．
第Ⅰ卷（选择题 共48分）
一、选择题（本大题共12个小题，每小题4分，共48分．在每小题给出的四个选项中，只有一项是符合题目要求的）
1．某地区某日最高气温是零上，记作，最低气温是零下，应该记作
A． B． C． D．
2．新时代十年来，我国建成世界上规模最大的社会保障体系，其中基本医疗保险的参保人数由亿增加到亿，参保率稳定在．将数据亿用科学记数法表示为
A． B． C． D．
3．如图所示的几何体，其主视图是
A. B. C. D.
4．将一副三角尺（厚度不计）按如图所示摆放，使有刻度的两条边互相平行，则 图中的度数为
A． B． C． D．
5．下列运算正确的是
A． B．
C． D．
6．如图，在平面直角坐标系中，的顶点坐标分别是，，，以原点为位似中心，在第三象限画与位似，若与的相似比为，则点的对应点的坐标为
A． B． C． D．
7．如图，四边形是平行四边形，给出下列四个条件：①；②；③；④平分．若添加其中一个条件，不能使四边形是菱形的为
A．① B．② C．③ D．④
8．如图，直线与双曲线交于点和点，则不等式的解集是
A． B． C．或 D．或
9.如图，已知四边形是的内接四边形，为延长线上一点，，则
等于
A. B．
C． D．
10．如图，用半径为，圆心角为的扇形纸板，做一个圆锥形的生日帽，在不考
虑接缝的情况下，这个圆锥形生日帽的底面圆的周长是
A. B．
C． D．
11．如图，在中，，以为边作，，点D与点A在的两侧，则的最大值为
A． B．
C．5 D．8
12．已知关于x的二次函数的最小值为k，若，，则k的取值范围是
A． B．
C． D．
第Ⅱ卷（非选择题 共102分）
注意事项：用0.5毫米黑色墨迹签字笔在答题卡上对应题号位置作答，在试卷上作答无效．
二、填空题（本大题共5个小题，每小题4分，共20分）
13．已知一组数据：3，4，5，5，6，则这组数据的众数是___________．
14．分解因式________．
15．如图，的边和的边共线（即点A，B，C，D在同一条直线上），且，．若只需添加一个条件就能使得，则这个条件可以是________________．
16．若是一元二次方程的两个实数根，则________．
17．如图，抛物线与x轴交于点A，B，与y轴交于点C，点在抛物线上，点E在直线上，若，则点E的坐标是____________．

三、本大题共2个小题，每小题8分，共16分
18． 计算：；
19.先化简，再求值：，其中．
四、本大题共3个小题，每小题10分，共30分．
20．为了解全校学生对篮球、足球、乒乓球、羽毛球四项球类运动的喜爱情况，在全校随机抽取了名学生进行问卷调查，每名学生只选择一项球类运动填写问卷．将调查结果绘制成如下统计图，请你根据图中所提供的信息解答下列问题．
(1)求______，并补全条形统计图．
(2)若该校共有1200名学生，请估计喜欢乒乓球运动的学生有多少名？
(3)学校羽毛球队计划从甲、乙、丙、丁四名同学中挑选两名同学加入球队．请用画树状图或列表的方法计算恰好选中甲、乙两名同学的概率．
21．学校通过劳动教育促进学生树德、增智、强体、育美全面发展，计划组织八年级学生到“开心”农场开展劳动实践活动．到达农场后分组进行劳动，若每位老师带38名学生，则还剩6名学生没老师带；若每位老师带40名学生，则有一位老师少带6名学生．劳动实践结束后，学校在租车总费用2300元的限额内，租用汽车送师生返校，每辆车上至少要有1名老师．现有甲、乙两种大型客车，它们的载客量和租金如下表所示
甲型客车 乙型客车
载客量/（人/辆） 45 30
租金/（元/辆） 400 280
(1)参加本次实践活动的老师和学生各有多少名？
(2)租车返校时，既要保证所有师生都有车坐，又要保证每辆车上至少有1名老师，则共需租车多少辆？
(3)学校共有几种租车方案？最少租车费用是多少？
22．为了增强学生体质、锤炼学生意志，某校组织一次定向越野拉练活动．如图，A点为出发点，途中设置两个检查点，分别为点和点，行进路线为．点在点的南偏东方向处，点在A点的北偏东方向，行进路线和所在直线的夹角为．
(1)求行进路线和所在直线的夹角的度数；
(2)求检查点和之间的距离（结果保留根号）．
五、本大题共3个小题，每小题12分，共36分．
23．如图，在平面直角坐标系中，直线与双曲线相交于点，点在轴正半轴上，点，连接，四边形为菱形．
(1)求和的值；
(2)设点是直线上一动点，且，求点的坐标．
24．如图，内接于，直径交于点G，过点D作射线，使得，延长交过点B的切线于点E，连接．
(1)求证：是的切线；
(2)若．
①求的长；②求的半径．
25．在平面直角坐标系中，已知抛物线与x轴交于点和点B，与y轴交于点．
(1)求这条抛物线的函数解析式；
(2)P是抛物线上一动点（不与点A，B，C重合），作轴，垂足为D，连接．
①如图，若点P在第三象限，且，求点P的坐标；
②直线交直线于点E，当点E关于直线的对称点落在y轴上时，请求出四边形的周长．泸州市初2026届学业水平模拟试题一
数学参考答案
一．选择题
题号 1 2 3 4 5 6 7 8 9 10
答案 A C A B D B B B A D
题号 11 12
答案 D B
二．填空题
13．5 14． 15．（答案不唯一）． 16． 17．和
三．解答题
18．解：（1）原式=················································5分
·····················································7分
；·····························································8分
19.原式，···························································2分 ···············································4分
···································································6分
当时，原式．······················································8分
20．（1）解： （名，···············································2分
喜欢乒乓球的人数；（名，·······································3分
补全统计图：
··················································4分
故答案为：200；
（2）解：（名，······················································6分
答：估计喜欢乒乓球运动的学生有312名；
（3）解：画树状图得：
············································8分
一共有12种等可能出现的结果，符合条件的结果有2种，·······················9分
恰好选中甲、乙两名同学的概率为．····································10分
21．（1）解：设参加本次实践活动的老师有x名，·································1分
，解得：，················································2分
∴，····················································3分
答：参加本次实践活动的老师有6名，学生有234名；
（2）解：∵每辆车上至少有1名老师，参加本次实践活动的老师有6名，
∴汽车总数不超过6辆，
∵要保证所有师生都有车坐，
∴汽车总数不少于（辆），则汽车总数最少为6辆，······················5分
∴共需租车6辆，
（3）解：设租用甲客车a辆，则租用乙客车辆，
，······················································6分
解得：，·····························································7分
∵a为整数，
∴或，方案一：租用甲客车4辆，则租用乙客车2辆；
方案二：租用甲客车5辆，则租用乙客车1辆；······································8分
设租车费用为y元，
，
∵，∴y随a的增大而增大，··················································9分
∴当时，y最小，，
综上：学校共有两套租车方案，最少租车费用是2160元．································10分
22．（1）解：如图，根据题意得，，，
，
．····································2分
在中，，
．·························································4分
答：行进路线和所在直线的夹角为．
（2）过点A作，垂足为．
，
，
．,·······················································6分
在中，
，·······························································7分
．,······································8分
在中，，
，······························································9分
．······················································10分
答：检查点和之间的距离为．
23．（1）解：∵四边形为菱形，∴与关于即轴对称，··················1分
∵点，∴点的坐标为．·················································2分
∵点在双曲线上，
∴；···································································3分
∵点在直线上，
∴，解得；······························································4分
（2）解：如图，连接，设直线与轴的交点为，则．
∵四边形为菱形，，，
∴根据菱形对称性，，
∴，，
∴．················································5分
∵，∴．
点是直线上一动点，设点的坐标为，则···················7分
①当时，，解得
∴；·········································································9分
②当时：，解得
∴．············································································11分
综上，点的坐标为或．·····················································12分
24．（1）证明：连接，
∵，，
∴，·····························································1分
设，则，
∵，
∴，·············································2分
∴，··········································3分
∵是圆的半径，
∴是的切线；·······························································4分
（2）解：①连接，
∵，∴，
∵是切线，∴，
∵，，
∴，
∵，
~······································································6分
∴，
∴，
∴；·············································································7分
②∵，
∵，
∵，
∴，，
∴，
在中，，··········································9分
∵，，
∴，······································································10分
∴，
∴，
∴，··········································································11分
∴
∴的半径为．···································································12 分
25．解：（1）∵抛物线与x轴交于点，与y轴交于点，
∴把，代入得，
，解得，，·····················································2分
∴抛物线的函数解析式为；···········································3分
（2）①设，过点作于点，如图，
∴
∵
∴
∵轴，
∴
又
∴四边形是矩形，
∴
∴
∵
∴
∴（不合题意，舍去）
∴
∴；·································································5分
②设，
对于，当时，解得，·················6分
∴∵
由勾股定理得，
当点在第三象限时，如图，过点作轴于点，

则四边形是矩形，
∵点与点关于对称，
∴
∵轴，
∴
∴
∴
∴
∴四边形是平行四边形，
∴四边形是菱形，
∵
∴
∴
∴
∴·······························································7分
设直线的解析式为，
把代入得，，解得，，
∴直线的解析式为，·············································8分
∴，
∴,
又且
∴
解得，（舍去）······················································9分
∴
∴四边形的周长；······························································10分
当点在第二象限时，如图，

同理可得：
解得，（舍去）∴····························································11分
∴四边形的周长；······························································12分
综上，四边形的周长为或．

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