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2007年普通高等学校招生全国统一考试(浙江卷)数学(文科)第I卷(共50分)一、选择题:本大题共10小题,每小题5分,共50分在每小题给出的四个选项中,只有一项是符合题目要求的1设全集,则( )ABCD2已知,且,则( )ABCD3“”是“”的()充分而不必要条件必要而不充分条件充分必要条件既不充分也不必要条件4直线关于直线对称的直线方程是()第5题5要在边长为16米的正方形草坪上安装喷水龙头,使整个草坪都能喷洒到水假设每个喷水龙头的喷洒范围都是半径为6米的圆面,则需安装这种喷水龙头的个数最少是()65436展开式中的常数项是( )ABCD7若是两条异面直线外的任意一点,则( )A过点有且仅有一条直线与都平行B过点有且仅有一条直线与都垂直C过点有且仅有一条直线与都相交D过点有且仅有一条直线与都异面8甲、乙两人进行乒乓球比赛,比赛规则为“3局2胜”,即以先赢2局者为胜,根据经验,每局比赛中甲获胜的概率为,则本次比赛甲获胜的概率是( )ABCD9若非零向量满足,则() 10已知双曲线的左、右焦点分别为,是准线上一点,且,则双曲线的离心率是()第II卷(共100分)二、填空题:本大题共7小题,每小题4分,共28分11函数()的值域是 12若,则的值是 13某校有学生2000人,其中高三学生500人,为了解学生的身体素质情况,彩用按年级分层抽样的方法,从该校学生中抽取一个200人的样本,则样本中高三学生的人数为 14中的满足约束条件则的最小值是 15曲线在点处的切线方程是 16某书店有11种杂志,2元1本的8种,1元1本的3种小张用10元钱买杂志(每种至多买一本,10元钱刚好用完),则不同买法的种数是 (用数字作答)17已知点在二面角的棱上,点在内,且若对于内异于的任意一点,都有,则二面角的大小是三、解答题:本大题共5小题,共72分解答应写出文字说明,证明过程或演算步骤18(本题14分)已知的周长为,且(I)求边的长;(II)若的面积为,求角的度数(第20题)19(本题14分)已知数列中的相邻两项是关于的方程的两个根,且(I)求,及()(不必证明);(II)求数列的前项和20(本题14分)在如图所示的几何体中,平面,平面,且,是的中点(I)求证:;(II)求与平面所成的角的正切值(第21题)21(本题15分)如图,直线与椭圆交于两点,记的面积为(I)求在,的条件下,的最大值;(II)当,时,求直线的方程22(本题15分)已知(I)若,求方程的解;(II)若关于的方程在上有两个解,求的取值范围,并证明2007年普通高等学校招生全国统一考试(浙江卷)数学(文科)试题参考答案一、选择题:本题考查基本知识和基本运算每小题5分,满分50分12345678910二、填空题:本题考查基本知识和基本运算每小题4分,满分28分11121314151617三、解答题18本题主要考查利用正弦定理、余弦定理来确定三角形边、角关系等基础知识和基本运算能力满分14分解:(I)由题意及正弦定理,得,两式相减,得(II)由的面积,得,由余弦定理,得,所以19本题主要考查等差、等比数列的基本知识,考查运算及推理能力满分14分(I)解:方程的两个根为,当时,所以;当时,所以;当时,所以;当时,所以因为当时,所以(II)解:20本题主要考查空间线面关系、空间向量的概念与运算等基础知识,同时考查空间想象能力和推理运算能力满分14分方法一:(I)证明:因为,是的中点,所以又因为平面,所以(II)解:连结,设,则,在直角梯形中,是的中点,所以,因此因为平面,所以,因此平面,故是直线和平面所成的角在中,方法二:如图,以点为坐标原点,以,分别为轴和轴,过点作与平面垂直的直线为轴,建立直角坐标系,设,则,(I)证明:因为,所以,故(II)解:设向量与平面垂直,则,即,因为,所以,即,因为,与平面所成的角是与夹角的余角,所以21本题主要考查椭圆的几何性质、椭圆与直线的位置关系等基础知识,考查解析几何的基本思想方法和综合解题能力满分15分(I)解:设点的坐标为,点的坐标为由,解得所以当且仅当时,S取到最大值1()解:由得AB 又因为O到AB的距离所以代入并整理,得解得,代入式检验,0 故直线AB的方程是 或或或(22)本题主要考查函数的基本性质、方程与函数的关系等基础知识,以及综合运用所学知识、分类讨论等思想方法分析和解决问题的能力满分15分()解:(1)当k2时,当时,1或1时,方程化为2解得,因为,舍去,所以当时,11时,方程化为解得,由得当k2时,方程的解所以或 (II)解:不妨设0x1x22,因为所以在(0,1是单调函数,故0在(0,1上至多一个解,若1x1x22,则x1x20,故不符题意,因此0x11x22由得,所以;由得,所以;故当时,方程在(0,2)上有两个解因为0x11x22,所以,0消去k 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