2009年鄂尔多斯市初中毕业升学考试

2009年鄂尔多斯市初中毕业升学考试数 学注意事项：1．本试题满分120分，考试用时120分钟 2．答题前将密封线内的项目填写清楚3．考试结束后将试卷按页码顺序排好，全部上交一、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．）1．5-的相反数是（ ） A ．5-B ．52．已知11x y =⎧⎨=-⎩是方程23x -A ．1B ．3C ．2-31600亿元大关，居自治区首位．1600A ．21.610⨯元B ．1.6⨯1210⨯元4A ．1，2B ．2，15．下列运算正确的是（ ） A ．B ．236()a a a -=C ．236()a a -=-6 ）A 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 ．2525πcmC ．2300πcmD ．2150πcm第7题图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=D ．6I =9101PA 111213，则14A1P 第10题图AOB15．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．（1（2． 20．10元钱，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．A D BC21．（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（222．（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：sin350.57≈°，cos350.82≈°，tan350.70≈°）23．O 过BC 的中点D ， 且（1（2O 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），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，此第22题图B第23题图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 点坐标；若不存在，说明理由．图12009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．345步累计评分．611．2与14之间的数即可）；13 15．2m <且1m ≠ 16．32x = 1866分） 19（10(π 3.14)2cos 60-+°； 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 =时，原式==·····························································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） P ∴（两红）126==；P （一红一白）123==；P （两白）126== ·············································································································· 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． ·········································红1红2白1 白2红2红1白1 白2白1红1红2 白2白2红1红2 白122．（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． ····（2）如图，在Rt ABD △中tan ABADB BD∠= ·····················18tan 35BD =°，180.70BD =≈答：两楼相距25.7米． ············23．（本小题满分7分）（1）证法一：连接OD ． ······ 点D 为BC 的中点，点O 为OD ∴为ABC △的中位线，OD AC ∴∥··························· 2DEC ODE ∴∠=∠DE AC ⊥ ，90DEC ∴∠=°DE OD ∴⊥ ·····························DE ∴是O 的切线 ·················证法二：连接OD AD ，． ······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 第23题图点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3． ········································································································8分 解法二：连接AD ，过O 点作OF BD ⊥，垂足为F ． ·····················································5分AB 为直径，∴∠D 是BC 的中点，BD CD ∴=，AC ∴在Rt CED △中，cos 10DB CD ∴==，5BF ∴= ··················在Rt BFO △中，cos 5cos30OB =°，OB =即O 的半径为324．（本小题满分7解：（1）30020y =+（2）(60W x =--(20)(30020)x x =-+220100x x =-++=-···································································································4分·································································································5分 当x 6125．············································································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）A FE（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 ∠=°， 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 ∴∠=∠，···············4分 在Rt ABC △中，BC =BO OC ∴==BE x = ，CF y =，············5分 取值范围是：12x ≤≤············6分 ②OEF △当1x =时，OE EA =x =OE OF =1分）26．（本小题满分12（1(6)(4)0t t +-=，16t =-，24t =······································1分1t 4) ··············· 2分 A B ，两点． c ⎧∴⎨⎩1434c ⎪=⎩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 ==⨯⨯== △6Sy ∴=- ······························································································································6分 x第26题图中小学教育资源站 2222147644(76)425332x x x x x ⎛⎫⎛⎫=-++=-++=-++ ⎪ ⎪⎝⎭⎝⎭ ···········································7分 令0y =时，22144033x x ++=，解得16x =-，21x =-． 抛物线与x 轴的交点坐标为(60)-，，(10)-，． x ∴的取值范围为61x -<<- ······························································································8分 （3）当24S =时，得27244252x ⎛⎫=-++ ⎪⎝⎭，解得：13x =-，24x =- ·······················9分 代入解析式得：14y =-，24y =-． ∴点P 的坐标为(34)--，，(44)--， 当点P 为(34)--，时，满足PO PA =，此时，平行四边形OPAQ 是菱形．当点P 为(44)--，时，不满足PO PA =，此时，平行四边形OPAQ 不是菱形． ········· 10分 而要使平行四边形OPAQ 此时，点P 的坐标为(33)--，，而(33)--，不在抛物线223y x =OPAQ 为正方形． 12分。