初中1年级数学

初中考试数学试卷一、选择题（本大题10个小题，每小题3分，共30分，每小题给出的四个选项中只有一个是正确的，请把正确选项的标号填在下面的选项栏内．）题号12345678910选项1．5−的相反数是（）A ．5−B ．5C ．15−D ．152．已知11x y =⎧⎨=−⎩是方程23x ay −=的一个解，则a 的值为（）A ．1B ．3C ．2−D ．1−-3．近年来，鄂尔多斯市政治、经济、文化等方面得到飞速发展，2008年全市经济总量突破1600亿元大关，居自治区首位．1600亿元用科学记数法可表示为（）A ．21.610×元B ．101.610×元C ．111.610×元D ．121.610×元4．为了美化城市，建设中的某小广场准备用边长相等的正方形和正八边形两种地砖镶嵌地面，在每一个顶点周围，正方形、正八边形地砖的块数分别是（）A ．1，2B ．2，1C ．2，3D ．3，25．下列运算正确的是（）A ．321a a −=B ．236()a a a −=i C ．22122a a −−=−D ．236()a a −=−6．下列事件中必然发生的事件是（）A ．一个不透明的袋子中有6个红球1个黑球，每次摸出一个球，然后放回搅匀．摸7次时一定会摸出一个黑球B ．任意一个五边形外角和等于540°C ．平移后的图形与原来图形的对应线段相等D ．在一个不等式的两边同时乘以一个数，结果仍是不等式7．如图是小丽学习时使用的圆锥形台灯灯罩的示意图，则围成这个灯罩的铁皮的面积是（不考虑缝隙等因素）A ．2600πcm B ．2525πcm C ．2300πcm D ．2150πcm 8．某闭合电路中，电源的电压为定值，电流()I A 与电阻()R Ω成反比例．如图所示的是该电路中电流I 与电阻R 之间的函数关系的图象，则用电阻R 表示电流I 的函数解析式为（）A ．2I R =B ．3I R =C ．5I R=D ．6I R=9．如图是小王早晨出门散步时，离家的距离s 与时间t 之间的函数图象．若用黑点表示小王家的位置，则小王散步行走的路线可能是（）10．如图，点A 和点B 相距60cm ，且关于直线l 对称，一只电动青蛙在距直线20cm ，距点A 为50cm 的点1P 处，按如下顺序循环跳跃：青蛙跳跃2009次后停下，此时它与直线l 相距（）A ．20cmB ．40cmC ．60cmD ．80cm 二、填空题（本大题8个小题，每小题3分，共24分）11．不等式组235321x x −<⎧⎨+−⎩≥的解集是．12．相交两圆的半径分别为6cm 和8cm ．请你写出一个符合条件的圆心距cm ．13．如图，刀柄外形是一个直角梯形（下底挖去一小半圆），刀片上、下是平行的，转动刀片时形成1∠，2∠，则12∠+∠=度．14．如图，将半径为4cm 的圆形纸片折叠后，圆弧恰好经过圆心O ，则折痕AB 的长是cm．从1P 点以A 点为对称中心跳至2P 点从2P 点以l 点为对称轴跳至3P 点从3P 点以B 点为对称中心跳至4P 点从4P 点以l 为对称轴跳至1P 点lA1P B 第10题图AOBAD BC第9题图ABCDt15．关于x 的一元二次方程2(1)210m x x −++=有两个不相等的实数根，那么m 的取值范围是．16．分式方程22510x x x x−=+−的解是．17．如图，正方形ABCD 的边长为1，以直线AB 为轴将正方形旋转一周，所得圆柱的主视图的周长是．18．如图，在平面直角坐标系中，直线334y x =−与x 轴、y 轴分别交于A B ，两点．现有半径为1的动圆位于原点处，以每秒1个单位的速度向右作平移运动，则经过秒，动圆与直线AB 相切．三、解答题（本大题8个小题，共66分．解答时要写出必要的文字说明、演算步骤或推证过程）19．（本小题满分9分）（1）计算：0(π 3.14)2cos60−−+°；（2）先化简，再求值．22112b a b a b a ab b⎛⎞−÷⎜⎟−+++⎝⎠，其中1a =+，1b =20．（本小题满分7分）为了提高农民抵御大病风险的能力，全国农村推行了新型农村合作医疗政策，农民只需每人每年交10元钱，就可以加入合作医疗．若农民患病住院治疗，出院后到新型农村合作医疗办公室按一定比例报销医疗费．小军与同学随机调查了他们镇的一些村民，根据收集到的数据绘制成了如图所示的统计图．根据以上信息，解答下列问题：（1）本次共调查了多少村民？被调查的村民中，有多少人参加合作医疗得到了报销款？（2）若该镇有村民10000人，请你计算有多少人参加了合作医疗？要使两年后参加合作医疗的人数增加到9680人，假设这两年的年增长率相同，求这个年增长率．x400100作医疗合作医疗类别参加合作医疗但没得到第20题图21．（本小题满分7分）甲、乙两商场同时开业，为了吸引顾客，都举办有奖酬宾活动，凡购物满100元，均可得到一次摸奖的机会．在一个纸盒里装有2个红球和2个白球，除颜色外，其他全部相同，摸奖者一次从中摸出两个球，根据球的颜色决定送礼金券的多少（如下表）．甲商场：乙商场：（1）请你用列表法（或画树状图）分别求出摸到两红、一红一白、两白的概率；（2）如果只考虑中奖因素，你将会选择去哪个商场购物？请说明理由．22．（本小题满分7分）花园小区有一朝向为正南方向的居民楼（如图），该居民楼的一楼是高4米的小区商场，商场以上是居民住房．在该楼的前面16米处要盖一栋高18米的办公楼．当冬季正午的阳光与水平线的夹角为35°时，问：（1）商场以上的居民住房采光是否有影响，为什么？（2）若要使商场采光不受影响，两楼应相距多少米？（结果保留一位小数）（参考数据：sin 350.57≈°，cos350.82≈°，tan 350.70≈°）23．（本小题满分8分）如图，已知AB 是O ⊙的直径，O ⊙过BC 的中点D ，且DE AC ⊥于点E ．（1）求证：DE 是O ⊙的切线；（2）若30C ∠=°，CE =O ⊙的半径．24．（本小题满分7分）某商场购进一种单价为40元的商品，如果以单价60元售出，那么每天可卖出300个．根据销售经验，每降价1元，每天可多卖出20个．假设每个降价x （元），每天销售量y （个），球两红一红一白两白礼金券（元）5105球两红一红一白两白礼金券（元）10510居民楼第22题图B第23题图每天获得最大利润W （元）（1）求出y 与x 的函数关系式；（2）6000元是否为每天销售这种商品的最大利润？如果是，请说明理由；如果不是，请求出最大利润，此时这种商品的销售价应定为多少元？25．（本小题满分9分）在ABC △中，2AB AC ==，90A ∠=°，取一块含45°角的直角三角尺，将直角顶点放在斜边BC 边的中点O 处（如图1），绕O 点顺时针方向旋转，使90°角的两边与Rt ABC △的两边AB AC ，分别相交于点E F ，（如图2）．设BE x =，CF y =．（1）探究：在图2中，线段AE 与CF 之间有怎样的大小关系？试证明你的结论；（2）若将直角三角尺45°角的顶点放在斜边BC 边的中点O 处（如图3），绕O 点顺时针方向旋转，其他条件不变．①试写出y 与x 的函数解析式，以及x 的取值范围；②将三角尺绕O 点旋转（如图4）的过程中，OEF △是否能成为等腰三角形？若能，直接写出OEF △为等腰三角形时x 的值；若不能，请说明理由．26．（本小题满分12分）已知：12t t ，是方程22240t t +−=的两个实数根，且12t t <，抛物线223y x bx c =++的图象经过点12(0)(0)A t B t ，，，．（1）求这个抛物线的解析式；（2）设点()P x y ，是抛物线上一动点，且位于第三象限，四边形OPAQ 是以OA 为对角线的平行四边形，求OPAQ ▱的面积S 与x 之间的函数关系式，并写出自变量x 的取值范围；（3）在（2）的条件下，当OPAQ ▱的面积为24时，是否存在这样的点P ，使OPAQ▱为正方形？若存在，求出P AOCBC AOCBF EEF 图1图2图3图4第25题图2009年鄂尔多斯市初中毕业升学考试试卷数学试题参考答案及评分说明（一）阅卷评分说明1．正式阅卷前先进行试评，在试评中认真阅读参考答案，明确评分标准，不得随意或降低评分标准．试评的试卷必需在后期予以复查，防止阅卷前后期评分标准宽严不一致．2．评分方式为分步累计评分，解答过程的某一步骤发生笔误，只要不降低后继部分的难度，而后继部分再无新的错误，后继部分可评应得分数的50%；若是几个相对独立的得分点，其中一处错误不影响其他得分点的评分．3．最小记分单位为1分，不得将评分标准细化至1分以下（即不得计小数分）．4．解答题题头一律计该题的实际得分，不得用计负分的方式计分．对解题中的错误需用红笔标出，并继续评分，直至将解题过程评阅完毕，并在最后得分点处标上该题实际的得分．5．本参考答案只给出一至两种解法，凡有其他正确解法都应参照本评分说明分步确定得分，并同样实行分步累计评分．6．合理精简解题步骤者，其简化的解题过程不影响评分．（二）参考答案及评分标准一、选择题（本大题10个小题，每小题3分，共30分）题号12345678910选项BACADCCBDB二、填空题（本大题8个小题，每小题3分，共24分）11．14x −<≤12．（只要求写一个在2与14之间的数即可）；13．9014．15．2m <且1m ≠16．32x =18．73或173三、解答题（本大题8个小题，共66分）19．（本小题满分9分）（1）计算：0(π 3.14)2cos60−−+°；解：原式13122=−+×···························································3分（一处计算正确给1分）3=·······························································································································4分（2）先化简，再求值．22112b a b a b a ab b⎛⎞−÷⎜⎟−+++⎝⎠，其中1a =+，1b =解：原式2()()()a b a b ba b a b a b +−+=÷−++··············································································1分22()()()b a b a b a b b +=−+····························································································2分2()a b a b+=−····················································································································3分当1a =1b ===5分20．（本小题满分7分）解：（1）400100500+=（人）·······················································1分（无单位不扣分）400312×=%（人）·····································································································2分所以，本次共调查了500人，有12人参加合作医疗得到报销款．（2）参加合作医疗的百分率为40080500=%，·································································3分所以该镇参加合作医疗的村民有1000080×%=8000（人）············································4分设年增长率为x ，由题意：得28000(1)9680x +=························································6分解得10.1x =，2 2.1x =−（舍去），即年增长率为10%．···············································7分21．（本小题满分8分）（1）树状图：··································································································2分（每两个树状图1分）也可以用列表法：红1红2白1白2红1（红1，红2）（红1，白1）（红1，白2）红2（红2，红1）（红2，白1）（红2，白2）白1（白1，红1）（白1，红2）（白1，白2）白2（白2，红1）（白2，红2）（白2，白1）P ∴（两红）21126==；P （一红一白）82123==；P （两白）21126==······································································································5分（一处正确得1分）（2）P ∵（甲市场得10元）P =（一红一白）23=P ∴（乙市场得10元）P =（两红）P +（两白）111663=+=····································6分因此，购物去乙市场，因乙市场得10元奖金的概率大．················································7分22．（本小题满分7分）解：（1）如图，光线交CD 于点E ，过点E 作EF BD ∥交AB 于点F ．·····················1分设DE x =米，则(18)AF x =−米在Rt AFE △中，35AEF ∠=∵°，tan 35AFEF ∴=°·····································2分180.7016x −=， 6.8x =··························3分6.84>∵，∴居民住房的采光有影响．·····················4分居民楼第22题图红1212红2112白1122白2121（2）如图，在Rt ABD △中tan ABADB BD ∠=···································5分18tan 35BD =°，1825.70.70BD =≈········6分答：两楼相距25.7米．····························7分23．（本小题满分7分）（1）证法一：连接OD ．·······················1分∵点D 为BC 的中点，点O 为AB 的中点．OD ∴为ABC △的中位线，OD AC ∴∥·························2分DEC ODE∴∠=∠DE AC ⊥∵，90DEC ∴∠=°，90ODE ∴∠=°，DE OD ∴⊥···········································3分DE ∴是O ⊙的切线································4分证法二：连接OD AD ，．·····················1分AB ∵为直径，90BDA ∴∠=°，90CDA ∠=°．30C ∠=∵°，60CAD ∴∠=°．DE AC ⊥∵，90AED ∴∠=°．30ADE ∴∠=°．·········································································································2分∵点D 为BC 的中点，AD BC ⊥，60BAD CAD ∴∠=∠=°OA OD =∵，60ODA OAD ∴∠=∠=°，···································································3分90ODE ∴∠=°，DE OD ∴⊥DE ∴是O ⊙的切线．···································································································4分（2）解法一：连接AD ，AB ∵为直径，90BDA ∴∠=°．DE AC ⊥∵，90CED ∴∠=°在Rt CED △中，cos CEC CD ∠=cos30CD=°，10CD =····························································································5分∵点D 为BC 的中点，10BD CD ∴==AC AB ∴=，30B C ∴∠=∠=°·················································································6分在Rt ABD △中．cos DB B AB ∠=，10cos 30AB ∠=°，3AB =···························7分O ∴⊙的半径为．·······························································································8分解法二：连接AD ，过O 点作OF BD ⊥，垂足为F ．·················································5分AB ∵为直径，90BDA ∴∠=°，D ∵是BC 的中点，BD CD ∴=，AC AB ∴=，30B C ∴∠=∠=°．·······················································6分B 第23题图居民楼第22题图在Rt CED △中，cos CE C CD ∠=，cos30CD=°，10CD =．10DB CD ∴==，5BF ∴=················································7分在Rt BFO △中，cos BFB OB ∠=5cos30OB =°，OB =··················8分即O ⊙的半径为3．（此题解法较多，只要正确，可参考以上评分标准给分）24．（本小题满分7分）解：（1）30020y x =+································································································2分（2）(6040)(30020)Wx x =−−+(20)(30020)x x =−+2201006000x x =−++252061252x ⎛⎞=−−+⎜⎟⎝⎠·································································································4分其中，020x ≤≤·········································································································5分当52x =时，W 有最大值，最大值是6125．60006125<∵，∵6000不是最大利润，······································································6分60 2.557.5−=∴销售价应定为57.5元．······························································································7分25．（本小题满分9分）（1）答：线段AE 与CF 之间有相等关系．··································································1分证明：连接AO ．如图2································································································2分AB AC =∵，点O 为BC 的中点，90BAC ∠=°，C OC图1图2图3图4B第23题图90AOC ∴∠=°，45EAO C ∠=∠=°，AO OC=90EOF ∠=∵°，90EOA AOF ∠+∠=°，90COF AOF ∠+∠=°，EOA FOC ∴∠=∠EOA FOC ∴△≌△，AE CF ∴=···············································································3分（还可证BOE AOF △≌△，得BE AF =，由AB AC =，可得AE CF =）（2）解：①连接AO ．如图4AB AC =∵，90BAC ∠=°，45C B ∴∠=∠=°，135BEO EOB ∴∠+∠=°，45EOF ∠=∵°，135FOC EOB ∴∠+∠=°，FOC BEO ∴∠=∠，BEO COF ∴△∽△，BE OB OC CF∴=···········································4分在Rt ABC △中，BC ===O 为BC的中点，BO OC ∴==BE x =∵，CF y =，2xy =，2y x ∴=···········································5分取值范围是：12x ≤≤·································································································6分②OEF △能构成等腰三角形当1x =时，OE EA =（或OE EF =）；2x =时，OA OF =（或EF OF =）；x =OE OF =，OEF △能构成等腰三角形．····9分（每写出一个x 的值给1分）26．（本小题满分12分）（1）解：22240t t ++=，(6)(4)0t t +−=，16t =−，24t =··································1分12t t <∵，(60)A ∴−，，(04)B ，···············2分∵抛物线223y x bx c =++经过A B ，两点．42460c b c =⎧∴⎨−+=⎩解得1434b c ⎧=⎪⎨⎪=⎩2214433y x x ∴=++······························4分（2）∵点()P x y ，在抛物线上，位于第三象限，0y ∴<，即0y −>．又12262APO S S OA y OA y y ==××==i i ∵△6S y ∴=−····················································································································6分2222147644(76)425332x x x x x ⎛⎞⎛⎞=−++=−++=−++⎜⎟⎜⎟⎝⎠⎝⎠········································7分x 第26题图。