江苏省太仓市2012年初中毕业暨升学考试模拟试卷(数学)

数学一、选择题1．下面四个由−2和3组成的算式中，运算值最小的是（）A．−2−3B．−23C．3−2D．(−3)22．一个正方形的面积为28，则它的边长应在（）A．3到4之间B．4到5之间C．5到6之间D．6到7之间3．一组数据4，5，6，7，7，8的中位数和众数分别是（）A．7，7B．7，6.5C．5.5，7D．6.5，74．如图，在平面直角坐标系中，菱形OACB的顶点O在原点，点C的坐标为（4，0），点B的纵坐标是−1，则顶点A坐标是（）A．（2，−1）B．（1，−2）C．（1，2）D．（2，1）5．如图，△ABC是边长为2的等边三角形，将△ABC沿射线BC向右平移得到△DCE，连接AD、BD，下列结论错误．．的是（）A．AD∥BCB．AC⊥BDC．四边形ABCD面积为43D．四边形ABED是等腰梯形6．不等式组11224(1)xxx≤的解集是（）A．−2＜x≤3B．−2＜x＜3C．2＜x≤3D．−2≤x＜37．关于x的两个方程220xx与122xxa有一个解相同，则a的值为（）A．−2B．−3C．−4D．−58．如图，△ABC是边长为6的等边三角形，AD=2，AE∥BC，直线BD交AE于点E，则BE的长为（）A．37B．43C．33D．59．已知P是⊙O内一点，⊙O的半径为15，P点到圆心O的距离为9，则通过P点且长度是整数的弦的条数是（）A．5B．7C．10D．1210．在△ABC中，∠ABC=30°，AB边长为10，AC边的长度可以在3、5、7、11中取值，满足这些条件的互不全等的三角形的个数是（）A．3B．4C．5D．6二、填空题11．分解因式：x3−4x=．12．去年，太仓全市实现全口径财政收入226.5亿元，同比增长25.8%．则226.5亿元用科学记数法可表示为元．OABCyxABCED13．函数2xyx中，自变量x的取值范围是．14．现有四条线段，长度依次是2，3，4，5，从中任选三条，能组成三角形的概率是．15．已知抛物线的顶点坐标为（2，9），且它在x轴上截得的线段长为6，则该抛物线的解析式为．16．如图是函数y=3−|x−2|的图象，则这个函数的最大值是．17．若一个圆锥的侧面积是它底面积的2倍，则这个圆锥的侧面展开图的圆心角是．18．如图，已知直线334yx交x轴、y轴于点A、B，⊙P的圆心从原点出发以每秒1个单位的速度向x轴正方向移动，移动时间为t(s)，半径为2t，则t=s时⊙P与直线AB相切．三、解答题19．计算：202012o113π1cos602．20．解方程组22,3210.xyxy21．先化简22214244xxxxxxxx，再从−2，0，1，2中选择一个合适．．的数代入，求出这个代数式的值．ABxOyP22．(本题共6分)如图，已知四边形ABCD的对角线AC、BD相交于点O，△ABC≌△BAD．(1)求证：OA=OB；(2)若∠CAB=35，求∠CDB的度数．23．(本题共6分)太仓人杰地灵，为了了解学生对家乡历史文化名人的知晓情况，某校对部分学生进行了随机抽样调查，并将调查结果绘制成如图所示统计图的一部分．根据统计图中的信息，回答下列问题：(1)本次抽样调查的样本容量是_；(2)在扇形统计图中，“了解很少”所在扇形的圆心角是度；(3)若全校共有学生1300人，那么该校约有多少名学生“基本了解”太仓的历史文化名人？24．(本题共6分)我们在配平化学方程式时，对于某些简单的方程式可以用观察法配平，对于某些复杂的方程式，还可以尝试运用方程的思想和比例的方法．例如方程式：322NHONO+HO催化剂，可以设NH3的系数为1，其余三项系数分别为x、y、z，即：3221NHONO+HOxyz催化剂，依据反应前后各元素守恒，得：1,32,2yzxyz，解之得四项系数之比为1:54:1:32，扩大4倍得整数比为4:5:4:6，即配平结果为：3224NH5O4NO+6HO催化剂．请运用上述方法，配平化学方程式：223Al+MnOAlOMn高温．ABCDO不了解10%很了解基本了解了解很少不了解了解很少基本了解很了解了解程度51015202530人数/人255525．(本题共6分)智能手机如果安装了一款测量软件“SmartMeasure”后，就可以测量物高、宽度和面积等．如图，打开软件后将手机摄像头的屏幕准星对准脚部按键，再对准头部按键，即可测量出人体的高度．其数学原理如图②所示，测量者AB与被测量者CD都垂直于地面BC．(1)若手机显示AC=1m，AD=1.8m，∠CAD=60，求此时CD的高．（结果保留根号）(2)对于一般情况，试探索手机设定的测量高度的公式：设AC=a，AD=b，∠CAD=α，即用a、b、α来表示CD．（提示：sin2α+cos2α=1）26．(本题共8分)如图，已知一次函数y1=k1x+6与反比例函数22kyx（x0）的图象交于点A、B，且A、B两点的横坐标分别为2和4．(1)k1=，k2=；(2)求点A、B、O所构成的三角形的面积；(3)对于x0，试探索y1与y2的大小关系（直接写出结果）．ABOxy图①ABCD图②27．如图，已知矩形ABCD中，AB=10，AD=4，点E为CD边上的一个动点，连结AE、BE，以AE为直径作圆，交AB于点F，过点F作FH⊥BE于H，直线FH交⊙O于点G．(1)求证：⊙O必经过点D；(2)若点E运动到CD的中点，试证明：此时FH为⊙O的切线；(3)当点E运动到某处时，AE∥FH，求此时GF的长．28．如图，将□OABC放置在平面直角坐标系xOy内，已知AB边所在直线的解析为：y=−x+4．(1)点C的坐标是（，）；(2)若将□OABC绕点O逆时针旋转90得OBDE，BD交OC于点P，求△OBP的面积；(3)在(2)的情形下，若再将四边形OBDE沿y轴正方向平移，设平移的距离为x（0≤x≤8），与□OABC重叠部分面积为S，试写出S关于x的函数关系式，并求出S的最大值．AFBCHEDOGABCDEOxyP29．如图，已知点A(−3，5)在抛物线y=12x2+c的图象上，点P从抛物线的顶点Q出发，沿y轴以每秒1个单位的速度向正方向运动，连结AP并延长，交抛物线于点B，分别过点A、B作x轴的垂线，垂足为C、D，连结AQ、BQ．(1)求抛物线的解析式；(2)当A、Q、B三点构成以AQ为直角边的直角三角形时，求点P离开点Q多少时间？(3)试探索当AP、AC、BP、BD与一个平行四边形的四条边对应相等（即这四条线段能构成平行四边形）时，点P离开点Q的时刻．ACDOBPyxQ2012年太仓市初中毕业暨升学考试模拟试卷数学参考答案及评分标准一、选择题（每小题3分，共30分）题号12345678910答案BCDDCADADB二、选择题（每小题3分，共24分）11．x(x+2)(x−2)12．2.265101013．x≥−2且x≠014．3415．y=−(x+1)(x−5)16．317．18018．2411或24三、解答题（共10大题，共76分）19．（共4小题，每小题4分，共16分）解：原式=14+1+12=512········································································4’+1’20．（共2小题，每小题4分，共8分）解：①2得：4x+2y=4③····································································1’②+③得：7x=14················································································2’∴x=2·····························································································3’把x=2代入①得：y=−2······································································4’∴原方程组的解为：2,2.xy······························································5’21．(本题共6分)解：原式=221(2)42xxxxxxx·························································2’=2(2)(2)(1)(2)4xxxxxxxx······························································3’=24(2)4xxxxx···············································································4’=212x······················································································5’取x=1代入得，原式=−1·····································································6’22．（本题6分）（1）证明：∵△ABC≌△BAD，∴∠BAC=∠ABD．····································1’∴OA=OB．···················································································2’（2）解：∵△ABC≌△BAD，∴AC=BD．·················································3’∵OA=OB，∴OC=OD，∴∠OCD=∠ODC．········································4’∵∠OAB+∠OBA=2∠CAB=70，∴∠OCD+∠ODC=70．·······················5’∴∠CDB=35．··············································································6’23．（本题共6分）(1)50··································································································2’(2)180·································································································4’(3)解：由题意得，“很了解”占10%，故“基本了解”占30%．···················5’∴“基本了解”的学生有：130030%=390(人)········································6’24．（本题共6分）解：设Al的系数为1，其余三项分别为x，y，z即：2231Al+MnOAlOMnxyz高温·················································1’由题意得：12,,23yxzxy···