山东省济南市钢城区2022-2023学年九年级上学期期中数学试题

这是一份山东省济南市钢城区2022-2023学年九年级上学期期中数学试题，共10页。
济南市钢城区2022-2023学年度上学期期中考试初四数学试题本试题分选择题部分和非选择题部分，共6页，满分为150分，考试用时120分钟。答题前，谓考生务必将自己的姓名、学校、班级、准考证号写在答题卡规定位置，将条形码粘贴在规定位置。答题时，选择题部分每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题部分，用0.5mm黑色签字笔在答题卡上题号所提示的答题区域作答，直接在试题上作答无效；解答题作图需用黑色签字笔。考试结束后，将本试题和答题卡一并交回。第Ⅰ卷（选择题40分）一、选择题（本题共1小题，每小题选对得4分，选错、不选或选出的答案超过一个均记零分，共40分）1．已知是锐角，，则等于（    ）A．30° B．60° C．45° D．90°2．已知反比例函数，则它的图象经过点（    ）A． B． C． D．3．将二次函数的图象向左平移2个单位，则平移后的二次函数表达式为（    ）A． B． C． D．4．已知在中，，，，则AC等于（    ）A．6 B．16 C．12 D．45．若双曲线，经过点，，则与的大小关系为（    ）A． B． C． D．无法比䢂与的大小6．已知二次函数，下面结论正确的是（    ）A．图象的开口向下  B．最小值是3C．图象的对称轴是直线 D．当时，y随x的增大而增大7．双曲线与在第一象限内的图象如图所示，作一条平行于y轴的直线分别交双曲线于A、B两点，连接OA、OB，则的面积为（    ）A．1 B．2 C．3 D．48．一次函数与二次函数在同一平面直角坐标系中的大致图象可能是（    ）A．    B．C．    D．9．2002年世界数学家大会（ICM）会徽的主体图案是由四个全等的直角三角形组成的四边形．若，，则CD的长为（    ）A．  B．C．  D．10．已知二次函数（）在时有最小值，则m等于（    ）A．5 B．或 C．5或 D．或第Ⅱ卷（非选择题110分）二、填空题（本大题共6小题，只要求填写最后结果，每小题填对得4分，共24分）11．若反比例函数的图象在二、四象限，则m的取值范围是________．12．如图，已知的三个顶点均在格点上，则________．13．已知抛物线的顶点在x轴上，那么c的值是________．14．同一坐标系中，反比例函数和正比例函数图象交于、，则________．15．如图是抛物线型拱桥，当拱顶离水面2 m时，水面宽4 m．若水面再上升1.5 m，则水面的宽度为________m．16．如图，将矩形对折，使点A点与D重合，点B与C重合，折痕为EF；展开后再次折叠，使点A与点D重合于EF上的点P处，折痕分别为BM、CN，若，，则________．三、解答题（本大题共10小题，共86分，解答要写出必要的文字说明、证明过程或推演步骤）17．（本题满分6分）计算：18．（本题满分6分）求：二次函数的顶点坐标和对称轴．19．（本题满分6分）为检测某品牌一次性注射器的质量，将注射器里充满一定量的气体，当温度不变时，注射器里的气体的压强p(kPa)是气体体积V(ml)的反比例函数，其图象如图所示．（1）求这一函数的表达式：（2）当气体体积为时，求气体压强的值．20．（本题满分8分）某天，北海舰队在中国南海例行训练，位于A处的济南舰突然发现北偏西30°方向上的C处有一可疑舰艇．济南舰马上通知位于正东方向200海里B处的西安舰，西安舰测得C处位于其北偏西60°方向上，请问此时两舰距C处的距离分别是多少？（结果保留整数，，）21．（本题满分8分）如图，一次函数与反比例函数的图象交于点、，与x轴交于点D，与y轴交于点C．（1）求m、n的值；（2）观察函数图象，直接写出不等式的解集：（3）连接AO，BO，求的面积．22．（本题满分8分）如图，已知抛物线与x轴交于，，与y轴交于点C．（1）求c、t的值；（2）若点P是抛物线第一象限内的一个动点，且满足，求点P坐标．23．（本题满分10分）如图，某数学研究小组测量山体AC的高度，在点B处测得山体A的仰角为45°，沿BC方向前行20 m至点D处，斜坡DE的坡度为1：2，在观景台E处测得山顶A的仰角为58°，且点E到水平地面BC的垂直距离EF为10 m．点B，D，C在一条直线上，AB，AE，AC在同一竖直平面内．（1）求斜坡DE的水平宽度DF的长；（2）求山体AC的高度．（结果精确到1 m．参考数据，，，）24．（本题满分10分）某景区超市销售一种纪念品，这种商品的成本价为15元/件，要求销售单价不低于成本，每件最高利润不高于10元．市场调查发现，当销售单价定为18元/件时，每天可销售42件，销售单价每涨1元，销售量减少1个．（1）求该纪念品每天销售量y（件）与销售单价x（元/件）之间的函数关系式，并直接写出自变量x的取值范围；（2）若每天的销售利润为w（元），求销售单价为多少元时，每天获得的利润最大？最大利润是多少元？25．（本满分12分）如图，已知一次函数与反比例的图象相交于点，与x轴相交于点B．（1）求k的值以及点B的坐标；（2）以AB为边作菱形，使点C在x轴正半轴上，点D在第一象限，求点D的坐标；（3）在y轴上是否存在点P，使的值最小？若存在，请求出的最小值，若不存在，请说明理由．26．（本题满分12分）如图，抛物线过点，，与y轴交于点C．在x轴上有一动点，过点P作x轴的垂线l分别交抛物线和直线BC于F、G．（1）求抛物线的解折式；（2）求线段FG长度的最大值，并求此时点F的坐标；（3）当时，H是直线l上的点且在第一象限内，若是以BH为直角边的直角三角形，求点H的坐标． 济南市钢城区2022-2023学年度上学期期中考试初四数学试题参考答案及评分标准一、选择题（本大题共10小题，每小题4分，共40分．）题号12345678910答案BCADBBABAC二、填空题（本大题共6小题，每小题4分，共24分．）11．    12．    13．4    14．    15．2    16．三、解答题（本大题共10小题，共86分．解答要写出必要的文字说明、证明过程或推演步骤．）17．解：原式······································································4分······································································6分18．解：∵·················································································4分∴顶点坐标········································································5分对称轴：直线······································································6分（本题方法不限，用配方法或公式法求解均可，评分标准相同）19．解：（1）设···································································1分将代入上式，得····································································2分∴···············································································3分∴···············································································4分（2）当时，，即当气体体积为时，气体压强为·············································6分20．解：过点C作的延长线于点D，如图由题意可得：，，∴．即，∴（海里）．······································································4分在中，（海里）．···································································6分在中，（海里）．···································································8分故A处的济南舰距C处距离200海里，B处的西安舰距C处距离346海里．21.解：（1），（过程略）····························································2分（2）或··········································································4分（3）将、代入得：∴∴将代入得  ∴即·····································································6分∴···············································································8分22．解：（1）将代入得，∴···························································2分令，解得：，，∴即·································································4分（2）设点P的纵坐标为n，其中，∵∴∴∵，∴，即，∴．·············································································6分令，解得，（舍）故．·······························································8分23．解：（1）∵斜坡DE的坡度，，∴，∴．即斜坡DE的水平宽度DF长为20米．··············································2分（2）过点E作于点H，则四边形为矩形，∴，·············································································3分设，在中，  ∴·········································································5分∴，·············································································6分在中，∴即，解得···································································8分∴···············································································9分∴．.即山体AC的高度为90米．························································10分24．解：（1）．···································································3分其中，自变量x的取值范围是：·························································4分（2）············································································7分∵∴抛物线开口向下，∴当即时，w随x的增大而增大··························································8分∵∴当时，即当销售单价为25元时，每天获得利润最大，最大利润是350元．······························10分25．解：（1）将代入得，····························································1分将代入得．········································································2分∵一次函数与x轴交于点B∴令，解得，∴．···································································4分（2）∵，，∴，···································································6分∵四边形是菱形，∴，∵点C在x轴正半轴上，点D在第一象限，∴················································8分（3）作点关于y轴对称点，连接交y轴于点P，连接PB，此时值最小，且最小值为．·················10分∵，∴．即的最小值为35．···························································12分26．解：（1）解：（1）把、代入，得到·············································································2分解得，∴抛物线的解析式为····························································3分（2）∵，∴，由，可求直线BC：．·····················································4分设，则，∴···············································································6分∴当时，，此时····································································8分（3）设，其中∵，∴，，，①当时，有，即，解得······························································10分②当时，有，即，解得（舍），．综上所述，点H的坐标为或．·························································12分