江苏通州高级中学 高届三模拟试卷 -数学.doc

你的首选资源互助社区通州高级中学高三数学模拟试卷开始结束是否输出(第5题)一、填空题：1若复数满足（i是虚数单位），则= 2已知命题：“，”，请写出命题的否定： 3已知平面向量a=(-1,1),b=(x-3,1)，且ab，则 4从标有数字1到4的四张卡片中任取2张，则积为偶数的概率为 5右图是一程序框图，则其输出结果为 6射击运动员甲、乙两人在6次射击中取得的成绩分别为：第1次第2次第3次第4次第5次第6次甲8环9环x环10环6环7环乙7环9环7环8环y环9环若甲、乙两人的平均成绩都是8环，则方差较小的运动员是 7已知双曲线的顶点与焦点分别是椭圆的焦点与顶点，若双曲线的两条渐近线与椭圆的交点构成的四边形恰为正方形，则椭圆的离心率为 8已知圆：，过圆外一点作圆的切线（为切点），当点在直线上运动时，则四边形PAOB的面积的最小值为 9动点在不等式组表示的平面区域内部及其边界上运动，则的取值范围是 10已知是实数且若，那么=_，此时=_.12 48 16 32（第12题）11在ABC中有如下结论：“若点M为ABC的重心，则”，设a，b，c分别为ABC的内角A，B，C的对边，点M为ABC的重心.如果，则内角A的大小为 12将首项为1，公比为2的等比数列的各项排列如右表，其中第行第个数表示为，例如若，则 13记数列的前n项和为Sn，若是公差为d的等差数列，则为等差数列时d的值为 14已知函数,若，且，则的最小值为 二、解答题： 15.已知函数，是的导函数.（I）求及函数y=的最小正周期；（）当时，求函数的值域.16.如图1所示，在边长为12的正方形中，点在线段上，且，作/，分别交、于点、，作/，分别交、于点、，将该正方形沿，折叠，使得与重合，构成如图2所示的三棱柱http（1）求证：平面；（2）求四棱锥的体积；17. 上海某玩具厂生产万套世博会吉祥物海宝所需成本费用为元,且,而每套售出价格为元,其中 ,问:该玩具厂生产多少套吉祥物时,使得每套成本费用最低？ 若产出的吉祥物能全部售出,问产量多大时,厂家所获利润最大?18.圆锥曲线上任意两点连成的线段称为弦.若圆锥曲线上的一条弦垂直于其对称轴，我们将该弦称之为曲线的垂轴弦.已知点、是圆锥曲线C上不与顶点重合的任意两点，是垂直于轴的一条垂轴弦，直线分别交轴于点和点.（）试用的代数式分别表示和；yEPNMxOF（第18题）（）已知“若点是圆C：上的任意一点（），是垂直于轴的垂轴弦，直线分别交轴于点和点，则”. 类比这一结论，我们猜想：“若曲线C的方程为（如图），则也是与点M、N、位置无关的定值”，请你对该猜想给出证明.19如图，在直角坐标系中，有一组底边长为的等腰直角三角形，底边依次放置在轴上（相邻顶点重合），点的坐标为，。（）若在同一条直线上，求证数列是等比数列；（）若是正整数，依次在函数的图象上，且前三个等腰直角三角形面积之和不大于，求数列的通项公式。20已知：二次函数，其中，且函数在处取得极值.（I）求所满足的关系；（II）若直线：与函数在上的图象恒有公共点，求的最小值；（III）试判断是否存在，使得对任意的，不等式恒成立？如果存在，请求出符合条件的的所有值；如果不存在，说明理由.参考答案一、填空题：本大题共14小题，每小题5分，共70分1 2 , 3 4 4 5 6乙 78 9 10 11 12122 131或 14二、解答题：本大题共6小题，共90分.解答应写出文字说明、证明过程或演算步骤15.解：（I）， 2分 =，所以y=的最小正周期为T2. 5分（），9分， 12分函数的值域为. 14分16. （1）证明：在正方形中，因为， 所以三棱柱的底面三角形的边因为，所以，所以因为四边形为正方形，所以，而，所以平面（2）解：因为平面，所以为四棱锥的高因为四边形为直角梯形，且，所以梯形的面积为所以四棱锥的体积17. 解：（1）3分（当且仅当时，取等号）生产100万套时，每套成本费用最低.6分（2）由题设，利润，9分当，即时，当产量为万套时，利润最大12分当时，函数在上是增函数，当产量为200万套时，14分18.（）解：因为是垂直于轴的一条垂轴弦，所以，则 2分 令则 4分 同理可得：， 6分（）证明：由（）可知：， 8分 在椭圆C：上， . 10分 则（定值）.是与和点位置无关的定值. 15分19解：（）点的坐标依次为， 则，若共线；则，即，即，所以数列是等比数列。 （）依题意，两式作差，则有：， 又，故， 即数列是公差为的等差数列；此数列的前三项依次为，由，可得，故，或，或。数列的通项公式是，或，或。 由知，时，不合题意；时，不合题意；时，； 所以，数列的通项公式是。 20解：（I） 由已知得，代入， 2分（II）由题意得：方程在时总有解， ，即， 当时， 在时单调递减， 4分 当时，由，同理可得， 当时，由（当且仅当时，取“=”）得， 当时，同理可得. 6分 要使得直线：与函数在上的图像总有交点，实数应取、（），（）三者中的最大值，（），又（）， 的最小值为. 10分（III） ， 当时 ，由得，时，函数单调递增， 时成立. 13分 当且时，类似地可由单调性证得，又，成立，当时，等价于 且.由上可知，此时不成立. 综上，存在符合条件的，其所有值的集合为. 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