黑龙江省哈尔滨市2013届高考数学第二次模拟考试试题理（扫描版）-12-\n-12-\n-12-\n-12-\n-12-\n-12-\n2013年哈尔滨市第三中学第二次高考模拟考试数学试卷（理工类）答案及评分标准一、选择题：题号123456789101112答案BABBBCCDCBBD二、填空题：13.14.15.16.三、解答题：17.(Ⅰ)整理得………………………………4分又得………………………………6分-12-\n(Ⅱ)由（1）知……………………………8分所以……………………………………12分18.解:(Ⅰ)第六组···························2分第七组···························4分估计人数为··························6分(Ⅱ)可能的取值为0,1,2,3.························7分所以的分布列0123·············10分=.·····················12分19.(Ⅰ)，分别为的中点，为矩形，·················2分-12-\n,又面,面,平面⊥平面·····················4分(Ⅱ),又，又,所以面,··················6分法一：建系为轴，为轴，为轴,，,平面法向量，平面法向量··········9分，可得.·············12分法二：连交于点,四边形为平行四边形，所以为的中点，连,则,面,,作于点，所以面,连,则,即为所求·············9分在中，,解得-12-\n·············12分20.(Ⅰ)由已知解得，，方程为·······3分(Ⅱ)设，则（1）当直线的斜率存在时，设方程为联立得：有①由以为直径的圆经过坐标原点O可得：·整理得：②将①式代入②式得：,···········6分又点到直线的距离··········8分-12-\n所以··········10分(2)当直线的斜率不存在时，设方程为（）联立椭圆方程得：代入得到即，综上：的面积是定值又的面积，所以二者相等.·······12分20.(Ⅰ)由原式，················1分令，可得在上递减，在上递增，所以即···············3分（Ⅱ），，时，函数在单调递增···············5分，，，，必有极值，在定义域上不单调··············8分-12-\n················9分（Ⅲ）由(I)知在(0,1)上单调递减∴时，即················10分而时，···············12分22.（I）∵，∴，又∵，∴，∴∽∴又∵，∴···5分（II），，是⊙的切线，，·······10分23.（Ⅰ）圆的极坐标方程为：·········5分（Ⅱ）圆心到直线距离为，圆半径为，所以弦长为···········10分24.（Ⅰ）的解集为：··········5分（Ⅱ）··········10分-12-
