2016年潍坊市学业水平考试数学试题(嵌入字体)(共6页).doc

精选优质文档-倾情为你奉上秘密启用前 试卷类型：A2016年潍坊市初中学业水平考试 数 学 试 题 2016.6注意事项：1本试题分第卷和第卷两部分第卷2页，为选择题，36分；第卷2页，为非选择题，84分；共120分考试时间为120分钟2答卷前务必将试题密封线内及答题卡上面的项目填涂清楚所有答案都必须涂、写在答题卡相应位置，答在本试卷上一律无效第卷 选择题（共36分）一、选择题（本题共12小题，在每个小题给出的四个选项中，只有一项是正确的，请把正确的选项选出来，每小题选对得3分，选错、不选或选出的答案超过一个均记0分）1. 计算：=（）A B C0 D82. 下列科学计算器的按键中，其上面标注的符号是轴对称图形但不是中心对称图形的是（） A B C D 3. 如图，几何体是由底面圆心在同一条直线上的三个圆柱构成的，其俯视图是（） A B C D 4. 近日，记者从潍坊市统计局获悉，2016年第一季度潍坊全市实现生产总值1256.77亿元，将1256.77亿用科学记数法可表示为（精确到百亿位）（）A B C D5. 实数a，b在数轴上对应点的位置如图所示，化简的结果是（）A BC D6. 关于的一元二次方程有两个相等的实数根，则锐角等于（）A15° B30° C45° D60°7. 木杆斜靠在墙壁上，当木杆的上端沿墙壁竖直下滑时，木杆的底端也随之沿着射线方向滑动下列图中用虚线画出木杆中点随之下落的路线，其中正确的是（） A B C D 8. 将下列多项式因式分解，结果中不含有因式的是（）A B C D9. 如图，在平面直角坐标系中，与轴相切于点，与轴分别交于点和点，则圆心到坐标原点的距离是（）A BC D10. 若关于的方程的解为正数，则的取值范围是（）A B且 C D且11. 如图，在中，以直角边为直径作交于点，则图中阴影部分的面积是（）A B C D12. 运行程序如图所示，规定：从“输入一个值”到“结果是否95”为一次程序操作，如果程序操作进行了三次才停止，那么的取值范围是（）数学试题（A） 第1页（共4页）数学试题（A） 第2页（共4页）A BCD第卷 （非选择题 共84分）说明：将第卷答案用0.5mm的黑色签字笔答在答题卡的相应位置上二、填空题（本大题共6小题，共18分，只要求填写最后结果，每小题填对得3分）13. 计算：14. 若与是同类项，则15. 超市决定招聘广告策划人员一名，某应聘者三项素质测试的成绩如表：测试项目创新能力综合知识语言表达测试成绩（分）将创新能力、综合知识和语言表达三项测试成绩按的比例计入总成绩，则该应聘者的总成绩是分16. 已知反比例函数的图象经过，则当时，自变量的取值范围是17. 已知，点是的平分线上的动点，点在边上，且，则点到点与到边的距离之和的最小值是18. 在平面直角坐标系中，直线：与轴交于点，如图所示依次作正方形、正方形、正方形，使得点、在直线上，点、在轴正半轴上，则点的坐标是三、解答题（本大题共7小题，共66分，解答要写出必要的文字说明、证明过程或演算步骤.）19（本小题满分6分）关于的方程有一个根是，求另一个根及的值20（本小题满分9分）今年5月，某大型商业集团随机抽取所属的家商业连锁店进行评估，将各连锁店按照评估成绩分成了四个等级，绘制了如图尚不完整的统计图表评估成绩（分）评定等级频数根据以上信息解答下列问题：（1）求的值；（2）在扇形统计图中，求等级所在扇形的圆心角的大小；（结果用度、分、秒表示）数学试题（A） 第4页（共4页）数学试题（A） 第3页（共4页）（3）从评估成绩不少于80分的连锁店中任选2家介绍营销经验，求其中至少有一家是等级的概率21（本小题满分8分）正方形内接于，如图所示，在劣弧上取一点，连接、，过点作交于点，连接，且与相交于点，求证：（1）四边形是矩形；（2）22（本小题满分9分）如图，直立于地面上的电线杆，在阳光下落在水平地面和坡面上的影子分别是，测得米，米，在处测得电线杆顶端的仰角为，试求电线杆的高度（结果保留根号）23（本小题满分10分）旅游公司在景区内配置了50辆观光车共游客租赁使用，假定每辆观光车一天内最多只能出租一次，且每辆车的日租金（元）是5的倍数发现每天的营运规律如下：当不超过100元时，观光车能全部租出；当超过100元时，每辆车的日租金每增加5元，租出去的观光车就会减少1辆已知所有观光车每天的管理费是1100元（1）优惠活动期间，为使观光车全部租出且每天的净收入为正，则每辆车的日租金至少应为多少元？（注：净收入 = 租车收入 - 管理费）（2）当每辆车的日租金为多少元时，每天的净收入最多？24（本小题满分12分）如图，在菱形中，过点作于点，于点（1）如图1，连接分别交于点，求证：； （2）如图2，将以点为旋转中心旋转，其两边分别与直线相交于点，连接，当的面积等于时，求旋转角的大小并指明旋转方向25（本小题满分12分）如图，已知抛物线经过的三个顶点，其中点，点，ACx轴，点时直线下方抛物线上的动点（1）求抛物线的解析式；（2）过点且与轴平行的直线与直线分别交于点，当四边形的面积最大时，求点的坐标；（3）当点为抛物线的顶点时，在直线上是否存在点，使得以为顶点的三角形与相似，若存在，求出点的坐标，若不存在，请说明理由2016年潍坊市初中学业水平考试 数学试题（A）参考答案及评分标准 2016.6一、选择题（本大题共12小题，每小题选对得3分，共36分）BDCBA BDCDB AC二、填空题（本大题共6小题，每小题填对得3分，共18分）13121431577.4161718三、解答题（本大题共7小题，共66分）19（本大题满分6分）解：设方程的另一个根是，由一元二次方程根与系数的关系，得： ····························································································································2分由，得·········································································································································3分代入，得解得·················································································································································5分所以，方程的另一个根是，的值是···························································································6分20（本大题满分9分）解：（1）由统计图表知，评定为等级的有家，占总评估连锁店数的，则···································································································································3分（2）由题意知等级的频数为·············································································4分则等级所在扇形的圆心角大小为········································································································································5分·······················································································································································6分（3）评估成绩不少于80分的为两个等级的连锁店等级有两家，分别用表示；等级有两家，分别用表示画树状图如下：·····································································································8分或列表如下：············································8分由树状图或列表可知，任选家共有种可能的情况，其中至少有一家是等级的情况有种所以，从评估成绩不少于分的连锁店中任选家，其中至少有一家是等级的概率························································································································································9分21（本大题满分8分）证明：（1）正方形内接于，············································································1分又··········································································································································3分四边形是矩形·····························································································································4分（2）正方形内接于的度数是··········································································································································5分又·······················································································································6分··············································································································································7分又在矩形中，···············································································································································8分22（本大题满分9分）解：延长交的延长线于点，过点作于点在中，···········································································································1分则········································2分······································3分在中，···························································4分解得·········································································································································6分在中·············································································7分解得即电线杆的高度是米···············································································································9分23（本大题满分10分）解：（1）由题意知，若观光车能全部租出，则由·······································································································································2分解得··················································································································································3分又因为是的倍数，所以，每辆车的日租金至少应为元····································································································4分（2）设每天的净收入为元当时，········································································································6分因为随的增大而增大，所以当时，的最大值为································································7分当时，数学试题（A）答案 第2页（共4页）数学试题（A）答案 第1页（共4页）···················································································9分当时，的最大值是，因为所以，每辆车的日租金为元时，每天的净收入最多是元····················································10分24（本大题满分12分）（1）证明：连接，设交于在菱形中，为等边三角形·····························································································································1分为中点······················································································3分同理是线段的三等分点···········································································································································4分（2）解：，又··········································································································································5分当顺时针旋转时，由旋转的性质知，···············································································································································7分是等边三角形则由，又，解得················································································9分所以，当顺时针旋转时，的面积是同理可得，当逆时针旋转时，的面积也是································································10分综上所述，当以点为旋转中心顺时针或逆时针旋转时，的面积是····12分25（本大题满分12分）解：（1）把，的坐标代入得 ·················································································································2分解得数学试题（A）答案 第4页（共4页）数学试题（A）答案 第3页（共4页）所以，抛物线的解析式是·····························································································3分（2）轴，由，解得··················································································································································4分设直线的解析式是由 解得则直线的解析式是············································································································5分设点的坐标为，则点的坐标为则四边形···············································7分又则当时，四边形面积的最大值是此时点的坐标是··················································································································8分（3）由，得顶点的坐标是此时则在中，同理可求···································································································································9分在直线上满足条件的点，如图或可求，当时，设，由得，解得·················································10分当时，设设，由得，解得····················································11分综上，满足条件的点有两个，坐标分别是或······································································································12分说明：本参考答案每题只给出了一种解题方法，其他正确方法应参考本标准给出相应分数专心-专注-专业