2012厦门市中考数学试题及答案

数学试题第1页共16页2012年厦门市初中毕业及高中阶段各类学校招生考试数学（试卷满分：150分考试时间：120分钟）准考证号姓名座位号注意事项：1．全卷三大题，26小题，试卷共4页，另有答题卡．2．答案一律写在答题卡上，否则不能得分．3．可直接用2B铅笔画图．一、选择题（本大题有7小题，每小题3分，共21分.每小题都有四个选项，其中有且只有一个选项正确）1．－2的相反数是A．2B．－2C．±2D．－122．下列事件中，是必然事件的是A.抛掷1枚硬币，掷得的结果是正面朝上B.抛掷1枚硬币，掷得的结果是反面朝上C.抛掷1枚硬币，掷得的结果不是正面朝上就是反面朝上D．抛掷2枚硬币，掷得的结果是1个正面朝上与1个反面朝上3．图1是一个立体图形的三视图，则这个立体图形是A．圆锥B．球C．圆柱D．三棱锥4．某种彩票的中奖机会是1%，下列说法正确的是A．买1张这种彩票一定不会中奖B．买1张这种彩票一定会中奖C．买100张这种彩票一定会中奖D．当购买彩票的数量很大时，中奖的频率稳定在1%5．若二次根式x－1有意义，则x的取值范围是A．x＞1B．x≥1C．x＜1D．x≤16．如图2，在菱形ABCD中，AC、BD是对角线，若∠BAC＝50°，则∠ABC等于A．40°B．50°C．80°D．100°CB图2DA图1俯视图左视图正视图数学试题第2页共16页7．已知两个变量x和y，它们之间的3组对应值如下表所示.x－101y－113则y与x之间的函数关系式可能是A．y＝xB．y＝2x＋1C．y＝x2＋x＋1D．y＝3x二、填空题（本大题有10小题，每小题4分，共40分）8．计算：3a－2a＝．9．已知∠A＝40°，则∠A的余角的度数是．10．计算：m3÷m2＝.11．在分别写有整数1到10的10张卡片中，随机抽取1张卡片，则该卡片上的数字恰好是奇数的概率是．12．如图3，在等腰梯形ABCD中，AD∥BC，对角线AC与BD相交于点O，若OB＝3，则OC＝．13．“x与y的和大于1”用不等式表示为.14．如图4，点D是等边△ABC内一点，如果△ABD绕点A逆时针旋转后能与△ACE重合，那么旋转了度.15．五边形的内角和的度数是．16．已知a＋b＝2，ab＝－1，则3a＋ab＋3b＝；a2＋b2＝.17．如图5，已知∠ABC＝90°，AB＝πr，BC＝πr2，半径为r的⊙O从点A出发，沿A→B→C方向滚动到点C时停止.请你根据题意，在图5上画出圆心．．O运动路径的示意图；圆心O运动的路程是.三、解答题（本大题有9小题，共89分）18．（本题满分18分）（1）计算：4÷(－2)＋(－1)2×40；（2）画出函数y＝－x＋1的图象；（3）已知：如图6，点B、F、C、E在一条直线上，∠A＝∠D，AC＝DF，且AC∥DF.求证：△ABC≌△DEF.图6ABCDFE图4ABCDE图3ABDCO→图5ABCO数学试题第3页共16页19．（本题满分7分）解方程组：3x＋y＝4，2x－y＝1.20．（本题满分7分）已知：如图7，在△ABC中，∠C＝90°，点D、E分别在边AB、AC上，DE∥BC，DE＝3，BC＝9.（1）求ADAB的值；（2）若BD＝10，求sin∠A的值.21.（本题满分7分）已知A组数据如下：0，1，－2，－1，0，－1，3.（1）求A组数据的平均数；（2）从A组数据中选取5个数据，记这5个数据为B组数据.要求B组数据满足两个条件：①它的平均数与A组数据的平均数相等；②它的方差比A组数据的方差大.你选取的B组数据是，请说明理由.【注：A组数据的方差的计算式是SA2＝17[(x1－—x)2＋(x2－—x)2＋(x3－—x)2＋(x4－—x)2＋(x5－—x)2＋(x6－—x)2＋(x7－—x)2]】22．（本题满分9分）工厂加工某种零件，经测试，单独加工完成这种零件，甲车床需用x小时，乙车床需用(x2－1)小时，丙车床需用(2x－2)小时.（1）单独加工完成这种零件，若甲车床所用的时间是丙车床的23，求乙车床单独加工完成这种零件所需的时间；（2）加工这种零件，乙车床的工作效率与丙车床的工作效率能否相同？请说明理由.23．（本题满分9分）已知：如图8，⊙O是△ABC的外接圆，AB为⊙O的直径，弦CD交AB于E，∠BCD＝∠BAC.（1）求证：AC＝AD；（2）过点C作直线CF，交AB的延长线于点F，若∠BCF＝30°,则结论“CF一定是⊙O的切线”是否正确？若正确，请证明；若不正确，请举反例.图7ABCDE图8FBCEDOA数学试题第4页共16页24．（本题满分10分）如图9，在平面直角坐标系中，已知点A(2，3)、B(6，3)，连结AB.如果点P在直线y＝x－1上，且点P到直线AB的距离小于1，那么称点P是线段AB的“邻近点”．（1）判断点Ｃ(72，52)是否是线段AB的“邻近点”，并说明理由；（2）若点Q(m，n)是线段AB的“邻近点”，求m的取值范围．25．（本题满分10分）已知□ABCD，对角线AC与BD相交于点O，点P在边AD上，过点P分别作PE⊥AC、PF⊥BD，垂足分别为E、F，PE＝PF．（1）如图10，若PE＝3，EO＝1，求∠EPF的度数；（2）若点P是AD的中点，点F是DO的中点，BF＝BC＋32－4，求BC的长．26．（本题满分12分）已知点A(1，c)和点B(3，d)是直线y＝k1x＋b与双曲线y＝k2x（k2＞0）的交点．（1）过点A作AM⊥x轴，垂足为M，连结BM．若AM＝BM，求点B的坐标；（2）设点P在线段AB上，过点P作PE⊥x轴，垂足为E，并交双曲线y＝k2x（k2＞0）于点N．当PNNE取最大值时，若PN＝12，求此时双曲线的解析式．EF图10ABCDOPxyB42642O图9A数学试题第5页共16页2012年厦门市初中毕业及高中阶段各类学校招生考试数学参考答案及评分标准说明：1．解答只列出试题的一种或几种解法．如果考生的解法与所列解法不同，可参照解答中评分标准相应评分；2．评阅试卷，要坚持每题评阅到底，不能因考生解答中出现错误而中断对本题的评阅．如果考生的解答在某一步出现错误，影响后续部分而未改变本题的内容和难度，视影响的程度决定后继部分的给分，但原则上不超过后续部分应得分数的一半；3．解答题评分时，给分或扣分均以1分为基本单位．一、选择题（本大题共7小题，每小题3分，共21分）题号1234567选项ACADBCB二、填空题（本大题共10小题，每题4分，共40分）8.a.9.50°.10.m.11.12.12.3.13.x＋y＞1.14.60.15.540°.16.5；6.17.；2πr.三、解答题（本大题共9小题，共89分）18．（本题满分18分）（1）解：4÷(－2)＋(－1)2×40＝－2＋1×1···································································4分＝－2＋1·······································································5分＝－1.·········································································6分（2）解：正确画出坐标系······························································8分正确写出两点坐标··························································10分画出直线······································································12分（3）证明：∵AC∥DF，……13分∴∠ACB＝∠DFE.……15分又∵∠A＝∠D，……16分AC＝DF，……17分∴△ABC≌△EDF.……18分19．（本题满分7分）ABCDFE数学试题第6页共16页解1：3x＋y＝4，①2x－y＝1.②①＋②，得····································································1分5x＝5，·········································································2分x＝1.··········································································4分将x＝1代入①，得3＋y＝4，······································································5分y＝1．··········································································6分∴x＝1，y＝1.······································································7分解2：由①得y＝4－3x．③·······································1分将③代入②，得2x－(4－3x)＝1．···························································2分得x＝1.·······································································4分将x＝1代入③，得y＝4－3×1····································································5分＝1．··········································································6分∴x＝1，y＝1.······································································7分20．（本题满分7分）（1）解：∵DE∥BC，∴△ADE∽△ABC.……1分∴ADAB＝DEBC.……2分∴ADAB＝13.……3分（2）解1：∵ADAB＝13，BD＝10，∴ADAD＋10＝13·······························································4分∴AD＝5·····································································5分经检验，符合题意.∴AB＝15.在Rt△ABC中，·····························································6分sin∠A＝BCAB＝35.····························································7分解2：∵ADAB＝13，BD＝10，ABCDEG数学试题第7页共16页∴ADAD＋10＝13·······························································4分∴AD＝5·····································································5分经检验，符合题意.∵DE∥BC，∠C＝90°∴∠AED＝90°在Rt△AED中，····································