四川省资阳市安岳县2022—2023学年 九年级上学期期末学业质量检测 ·数学试题 (含答案)

这是一份四川省资阳市安岳县2022—2023学年 九年级上学期期末学业质量检测 ·数学试题 (含答案)，共11页。
安岳县2022—2023学年度上期期末学业质量检测九年级·数学全卷分为第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。满分150分，考试时间共120分钟。注意事项：1．答题前，考生务必将自己的姓名、座位号、报名号（考号）写在答题卡上，并将条形码贴在答题卡上对应的虚线框内。同时在答题卡背面第4页顶端用2B铅笔涂好自己的座位号。2．第Ⅰ卷每小题选出的答案不能答在试卷上，必须用2B铅笔在答题卡上把对应题目的答案标号徐黑，如需改动，用橡皮擦擦净后，再选涂其它答案。第Ⅱ卷必须用0.5mm黑色墨水签字笔书写在答题卡上的指定位置。不在指定区域作答的将无效。3．考试结束，监考人员只将答题卡收回。第Ⅰ卷（选择题 共40分）一、选择题（本大题共10个小题，每小题4分，共40分．在每小题给出的四个选项中，只有一个选项是符合题目要求的.）1．下列计算正确的是A．  B．  C．  B．2．下列事件为必然事件的是A．篮球运动员在罚球线上投篮一次，未投中B．在数轴上任取一点，则该点表示的数是有理数C．经过有交通信号灯的路口，遇到绿灯D．任意画一个四边形，其内角和为3603．估算：的值应在A．0和1之间  B．1和2之间  C．2和3之间  D．3和4之间4．在平面直角坐标系中，将点M(-2，3)向右平移1个单位，再向下平移2个单位，得到的点的坐标为A．(-1，1)   B．(-1，5)   C．(-3，1)   D．(-3，5)5．如图1，在△ABC中，D、E、F分别是AB、AC、BC的中点，若∠CFE=55°，则∠ADE的度数为A．65°    B．60°    C．55°    D．50°6．“读万卷书，行万里路.”某校为了丰富学生的阅历知识，坚持开展课外阅读活动，学生人均阅读量从七年级的每年100万字增加到九年级的每年121万字．设该校七至九年级人均阅读量年均增长率为x，则可列方程为A．100(1+x)2＝121       B．100(1+x%)2＝121C．100(1+2x)＝121       D．100+100(1+x)+100(1+x)2＝1217．如图2，在四边形ABCD中，AD‖BC，AC与BD相交于点O，若，则的值为A．    B．    C．    D．8．已知实数a在数轴上的位置如图3所示，则化简：的结果为A．2     B．-2    C．2a-6    D．-2a+69．如图4，在菱形ABCD中，∠ABC=60°，E是BC上一点，连结AE，将△ABE沿AE翻折，使点B落在点F处，连结BF、DF．若，则tan∠CDF的值为A．    B．    C．   D．10．如图5，直线l的解析式为，点M1(0，1)，M1N1⊥y轴交直线l于点N1；点M2为y轴上位于M1上方的一点，且M1M2=M1N1，M2N2⊥y轴交直线l于点N2；点M3为y轴上位于M2上方的一点，且M2M3=M2N2，M3N3⊥y轴交直线l于点N3，按此规律，线段N2022N2023的长为A．  B． C． D．  第Ⅱ卷（非选择题 共110分）注意事项： 1．请用0.5毫米的黑色签字笔在答题卡相应区域作答，超出答案区域的答案无效。2．试卷中横线及方框处是需要你在第Ⅱ卷答题卡上作答的内容或问题。请注意准确理解题意、明确题目要求，规范地表达、工整地书写解题过程或结果。二、填空题．（本大题6个小题，每小题4分，共24分）11．若代数式有意义，则x的取值范围为______________.12．如图6所示的电路中，若任意闭合一个开关，则灯泡L1发光的概率是_____________.13．若最简二次根式与是同类二次根式，则m =_____________.14．若，且2a+b-c=6，则a+b+c的值为_____________.15．如图7，在△ABC中，∠ACB=90°，CD为AB边上的中线，点G为△ABC的重心．若AC=12，BC=16，则DG的长为_____________.16．如图8-1，在四边形ABCD中，若△ABC∽△ACD，则称AC为四边形ABCD关于点A的“靓线”．如图8-2，在□ABCD中，AB=5，E为AD的中点，F为BA延长线上一点，连结BE、CE、EF，若BE为四边形BCEF关于点B的“靓线”，CE=6，则AF的长为_____________.            三、解答题（本大题共8个小题，共86分，解答应写出必要的文字说明、证明过程或演算步骤．）17．(本小题满分9分)计算：（1）  （2）   18．（本小题满分10分）先化简，再求值：，其中. 19．（本小题满分10分）如图9，已知：△ABC三个顶点的坐标分别为A(-2，-1），B（-5，-2），C（-1，-3）．（1）画出△ABC关于x轴对称的△A1B1C1；（2）以点O为位似中心，将△ABC放大为原来的2倍，得到△A2B2C2，请在网格中画出△A2B2C2，并写出点B2的坐标．   20．(本小题满分10分）为了更好落实“双减”政策，增强课后服务的时效性，我县一中学定于每周四下午进行兴趣课“走班制”，开设了5类兴趣课（每位学生均选其一）：A．音乐；B．体育；C．美术；D．信息技术；E．演讲．为了了解该校学生的参与情况，现随机抽取了部分学生进行调查，并将调查结果绘制成如图10所示的两幅不完整的统计图．根据图中信息，解答下列问题：（1）求此次调查的学生人数，并补全条形统计图；（2）求“C”类兴趣课所对应扇形的圆心角的度数；（3）若“E”类兴趣班中有2名男生和3名女生，从中随机抽取2名参加县级演讲比赛，请用列表或画树状图的方法，求恰好抽到1名男生和1名女生的概率． 21．（本小题满分11分）第19届亚运会原定于2022年9月10日至25日在杭州举行，其吉祥物“琼琼”、“莲莲”、“宸宸”组成的“江南忆”毛绒玩具套件，已成为杭州店销人气款．某商场销售这种毛绒玩具，平均每天可售出50套，每套盈利60元．但由于受疫情影响，此届亚运会将延期至2023年举行，于是该商场决定采取降价措施，以尽快减少库存，经调查发现，每套毛绒玩具每降价1元，平均每天可多售出2套.（1）若每套毛绒玩具降价5元，则该商场平均每天可盈利多少元？（2）若该商场计划平均每天盈利3500元，则每套毛绒玩具应降价多少元？    22．（本小题满分11分）如图11，A、B两地是我国某海域一东西方向上的两个小岛．一天，一艘渔政船在C处巡逻时，测得小岛A在它的北偏东15°方向上，它沿西北方向航行海里后到达D处，测得小岛A在它的东北方向.（1）求D处与小岛A的距离；（2）若该渔政船在D处测得小岛B在它的北偏西53°方向上，求小岛A、B之间的距离.（参考数据：sin53°=，cos53°=，tan53°=）      23．（本小题满分12分）定义：已知x1，x2是关于x的一元二次方程ax2+bx+c=0(a≠0)的两个实数根，若x1<x2<0，且<4，则称这个方程为“限根方程”．如：一元二次方程x2+13x+30=0的两根为x1= -10，x2=-3，因-10<-3<0，，所以一元二次方程x2+13x+30=0为“限根方程”.请阅读以上材料，回答下列问题：（1）判断一元二次方程x2+9x+14=0是否为“限根方程”，并说明理由；（2）若关于x的一元二次方程2x2+(k+7)x+k2+3=0是“限根方程”，且两根x1、x2满足x1+x2+x1x2=-1，求k的值；（3）若关于x的一元二次方程x2+(1-m)x -m=0是“限根方程”，求m的取值范围.   24．（本小题满分13分）【情境再现】（1）如图12-1，在正方形ABCD中，点E、F分别在边AB、BC上，且DE⊥AF，求证：DE=AF.【迁移应用】（2）如图12-2，在矩形ABCD中，AD=kAB（k为常数），点E、F、G、H分别在矩形ABCD的边上，且EG⊥FH，求证：EG=kFH.【拓展延伸】（3）如图12-3，在四边形ABCD中，∠ABC=∠ADC=90°，∠BCD=60°，CD=4，点E、F分别在边AB、BC上，且CE⊥DF，，求AB的长.  2022—2023学年度第一学期期末义务教育九年级学情诊断数学学科参考答案及评分意见一、选择题（共10小题，每小题4分，共40分）题号12345678910答案BDCACABADC二、填空题（共6小题，每小题4分，共24分）11．x≥1        12．        13．314．18        15．       16．1.4三、解答题（共86分）17．解：（1）原式=····················································2分                   =····························································4分                   =-1···························································5分 （2）原式=·························································7分 =····························································8分 =····························································9分 18．解：原式= ························································3分      =··························································6分      =··························································8分 当时，原式=························································10分19．解：（1）如下图····················································3分（2）如下图························································8分B2(10，4) ··························································10分 20．解：（1）此次调查的学生人数：15÷30%=50（人）··························1分D：50-8-15-10-5=12（人），图略. ·······································3分（2）10÷50×360°=72°·················································5分（3）列表或树状图略. ················································8分恰好抽到1名男性和1名女性的概率为：.···································10分21．解：（1）(60-5)×(50+5×2)=3300（元）··································4分答：该商场平均每天可盈利3300元.（2）设每套毛绒玩具应降价x元，由题意得(60-x)(50+2x)=3500··················································8分解之，得：x1=10，x2=25··············································10分∵该商场是为了尽快减少库存，∴x=25.答：该商场每套毛绒玩具应降价25元. ····································11分22．解：（1）由题意，得∠ADC=90°，∠ACD=60°······························1分∴tan60°=··························································3分∵，∴····························································4分答：D处与小岛A的距离为海里.（2）过点D作DE⊥AB于点E.由题意得：∠ADE=45°，∠BDE=53°······································5分∵，∴DE=AE=30····················································7分在Rt△BDE中，tan53°==···············································9分∴BE=×30=40······················································10分∴AB=40+30=70（海里）··············································11分答：小岛A、B之间的距离为70海里．23．解：（1）(x+2)(x+7)=0，∴x1=-7，x2=-2··································1分∵，∴此方程为“限根方程”. ··········································2分（2）由根与系数的关系，得x1+x2=，x1x2=································3分∵x1+x2+x1x2=-1，∴+=-1，∴k=2或-1····································4分①当k=2时，x1=，x2=-1，∴<4，∴k=2符合题意·····························5分②当k=-1时，x1=-2，x2=-1，∴=2，∴k=-1不合题意，舍去．∴k的值为2 ··········6分（3）解此方程得：x= -1或m···········································7分∵此方程为“限根方程”，∴△>0，且m<0，即(1-m)2+4m>0，∴(m+1)2 > 0，∴m<0且m≠-1······················································8分①当-1<m<0时，x1=-1，x2=m，∵<4，∴，∴·······························10分②当m<-1时，x1= m，x2= -1，∵<4，∴，∴-4<m<-3综上所述，m的取值范围为或-4<m<-3. ···································12分24．（1）证明：∵四边形ABCD为正方形，∴AD =AB，∠EAD=∠B=90°··············1分又∵DE⊥AF，∴∠ADE+∠DAF=90°=∠BAF+∠DAF，∴∠BAF=∠ADE··············2分在△ADE和△BAF中， ，∴△ADE≌△DAF．∴DE=AF··························3分（2）证明：过点D作DM∥EG交AB于点M，过点A作AN∥FH交BC于点N.易证：四边形EGDM为平行四边形，∴EG∥DM且EG=DM，同理AN∥FH且AN=FH·································································4分由（1）同理可得∠ADM=∠BAN，∴△ADM ∽△BAN··························5分∴·······························································6分∴，∴EG=kFH······················································7分（3）解：过点D作DG∥BC交BA的延长线于点G，过点C作CH⊥DG于点H.易证：四边形BCHG为矩形，由（2）同理可得：····························8分∵∠BCD=60°，∴∠DCH=30°，∵CD=4，∴DH=2，CH=························9分∴GH==···························································10分∴DG=····························································11分又∵∠ADC=90°，∴∠ADG=∠DCH=30°，∴AG=1·····························12分∴AB=BG-AG=······················································13分