山东省齐鲁教科研协作体19所2016届高三上学期第一次联考数学（理）试题 含答案

齐鲁名校教科研协作体山东省19所名校2016届高三第一次调研（新起点）联考数学（理科）试题命题学校：莱芜一中（侯伟华）审题学校：莱芜一中临沂一中邹城一中本试卷分第Ｉ卷和第II卷两部分,共4页.满分150分,考试用时120分钟.考试结束后,将本试卷和答题卡一并交回.注意事项：1.答题前,考生务必用毫米黑色签字笔将自己的姓名、座号、考生号、县区和科类填写在答题卡和试卷规定的位置上.2.第I卷每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动,用橡皮擦干净后,再选涂其它答案标号,答案写在试卷上无效.3.第II卷必须用毫米黑色签字笔作答,答案必须写在答题卡各题目指定区域内相应的位置,不能写在试卷上；如需改动,先划掉原来的答案,然后再写上新的答案；不能使用涂改液、胶带纸、修正带,不按以上要求作答的答案无效.4.填空题请直接填写答案,解答题应写出文字说明、证明过程或演算步骤.参考公式：锥体的体积公式：SKIPIF1<0,其中SKIPIF1<0是锥体的底面积,SKIPIF1<0是锥体的高.如果事件SKIPIF1<0互斥,那么SKIPIF1<0；如果事件A,B独立,那么SKIPIF1<0.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.SKIPIF1<0．（原创）若复数SKIPIF1<0,(SKIPIF1<0是虚数单位),则SKIPIF1<0的共轭复数是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】SKIPIF1<0,SKIPIF1<0.故选B.【考点】复数的概念、运算SKIPIF1<0．（原创）设全集为SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,则“SKIPIF1<0”是“SKIPIF1<0”的A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件【答案】C【解析】画出韦恩图可知若SKIPIF1<0,易得SKIPIF1<0；反之,若SKIPIF1<0,即集合SKIPIF1<0是集合SKIPIF1<0的子集,易得SKIPIF1<0.所以“SKIPIF1<0”是“SKIPIF1<0”的充要条件.故选C.【考点】集合间的基本关系、充分条件必要条件SKIPIF1<0．（原创）函数SKIPIF1<0的定义域为A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0SKIPIF1<0【答案】A【解析】由SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,即SKIPIF1<0,所以函数的定义域为SKIPIF1<0.故选A.【考点】函数的定义域、对数函数的图象与性质SKIPIF1<0．（改编）若实数SKIPIF1<0,SKIPIF1<0满足约束条件SKIPIF1<0,则SKIPIF1<0的取值范围是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【解析】画出可行域如图中阴影部分所示,易得在点SKIPIF1<0处,SKIPIF1<0取最小值SKIPIF1<0；在点SKIPIF1<0处,SKIPIF1<0取最大值SKIPIF1<0.所以SKIPIF1<0的取值范围是SKIPIF1<0.故选C.【考点】线性规划SKIPIF1<0．（原创）已知矩形SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】A【解析】解法一：以A为坐标原点,SKIPIF1<0为SKIPIF1<0轴,SKIPIF1<0为SKIPIF1<0轴建立平面直角坐标系,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0.故选A.解法二：记SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0.故选A.【考点】平面向量的数量积SKIPIF1<0．（原创）SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】B【解析】SKIPIF1<0SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0所以SKIPIF1<0,SKIPIF1<0SKIPIF1<0.故选B.【考点】二倍角公式、同角三角函数的基本关系式SKIPIF1<0．（原创）有下列4个命题：两个平面垂直,过一个平面内任意一点作交线的垂线,则此直线必垂直于另一平面;平面SKIPIF1<0内两条不平行的直线都平行于另一平面SKIPIF1<0,则SKIPIF1<0;SKIPIF1<0两条直线和一个平面所成的角相等,则这两条直线平行;直线SKIPIF1<0不平行于平面SKIPIF1<0,则平面SKIPIF1<0内不存在与直线SKIPIF1<0平行的直线.其中正确命题的个数是A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】D【解析】①错误,举反例：以图1中的长方体为例，平面SKIPIF1<0平面SKIPIF1<0,交线为SKIPIF1<0,点SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0,但SKIPIF1<0不垂直于平面SKIPIF1<0;②正确,依据平面与平面平行的判定定理;③错误,举反例：图2的圆锥中,SKIPIF1<0,SKIPIF1<0与底面所在平面所成的角相等,但SKIPIF1<0与SKIPIF1<0不平行;=4\*GB3④错误,举反例：当直线SKIPIF1<0平面SKIPIF1<0时,平面SKIPIF1<0内存在与直线SKIPIF1<0平行的直线.故选D.【考点】直线与平面、平面与平面位置关系SKIPIF1<0．（改编）如图，该程序框图的算法思路源于我国古代数学专著《九章算术》中的“更相减损术”,执行此程序框图,若输入的SKIPIF1<0,SKIPIF1<0分别为SKIPIF1<0,SKIPIF1<0,则输出的SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0开始输入m、nm=nm>n输出mm=m-nn=n-m结束否是否是开始输入m、nm=nm>n输出mm=m-nn=n-m结束否是否是第8题图【答案】C【解析】第一次执行,输入SKIPIF1<0,SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0；第二次执行,因为SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0；第三次执行,因为SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0；第四次执行,因为SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,此时SKIPIF1<0.故选C.【考点】程序框图SKIPIF1<0．（原创）定义在SKIPIF1<0上的函数SKIPIF1<0满足SKIPIF1<0,且当SKIPIF1<0时,SKIPIF1<0,则SKIPIF1<0A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<0【答案】C【解析】由SKIPIF1<0知，SKIPIF1<0.故选C.【考点】函数的性质SKIPIF1<0（改编）设函数SKIPIF1<0的零点为SKIPIF1<0,函数SKIPIF1<0的零点为SKIPIF1<0,则A.SKIPIF1<0,SKIPIF1<0B.SKIPIF1<0,SKIPIF1<0C.SKIPIF1<0,SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0【答案】A【解析】解法一：因为函数SKIPIF1<0在SKIPIF1<0上单调递增,且SKIPIF1<0，SKIPIF1<0，由零点存在性定理知SKIPIF1<0；因为函数SKIPIF1<0在SKIPIF1<0上单调递增,SKIPIF1<0,SKIPIF1<0,由零点存在性定理知SKIPIF1<0.因为函数SKIPIF1<0在SKIPIF1<0上单调递增,且SKIPIF1<0,所以SKIPIF1<0；因为函数SKIPIF1<0在SKIPIF1<0上单调递增,且SKIPIF1<0,所以SKIPIF1<0.故选A.解法二：由SKIPIF1<0SKIPIF1<0得SKIPIF1<0；由SKIPIF1<0SKIPIF1<0得SKIPIF1<0.记SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0,记SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0.函数SKIPIF1<0的零点SKIPIF1<0即函数SKIPIF1<0与SKIPIF1<0交点的横坐标；函数SKIPIF1<0的零点SKIPIF1<0即函数SKIPIF1<0与SKIPIF1<0交点的横坐标.如图在同一平面直角坐标系中分别作出函数SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0的图象,易看出SKIPIF1<0即SKIPIF1<0；SKIPIF1<0即SKIPIF1<0.故选A.【考点】函数的零点二.填空题：本大题共5小题,每小题5分,共25分,答案须填在答题卡题中横线上.SKIPIF1<0（改编）在等差数列SKIPIF1<0中,已知SKIPIF1<0,则SKIPIF1<0.【答案】SKIPIF1<0【解析】解法一：SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0SKIPIF1<0.解法二：利用等差数列的性质得SKIPIF1<0.【考点】等差数列通项公式、性质SKIPIF1<0（原创）由曲线SKIPIF1<0与SKIPIF1<0围成的封闭图形的面积是________.【答案】SKIPIF1<0【解析】如图在同一平面直角坐标系内作出SKIPIF1<0与SKIPIF1<0的图象,则封闭图形的面积SKIPIF1<0.【考点】幂函数的图象、定积分SKIPIF1<0（原创）在SKIPIF1<0的展开式中,SKIPIF1<0的系数为________（用数字作答）.【答案】SKIPIF1<0【解析】SKIPIF1<0的展开式的通项是SKIPIF1<0,所以在SKIPIF1<0的展开式中,含SKIPIF1<0的项为SKIPIF1<0,所以SKIPIF1<0的系数为SKIPIF1<0.【考点】二项式定理SKIPIF1<0（原创）SKIPIF1<0,SKIPIF1<0,SKIPIF1<0分别是SKIPIF1<0的三边,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,则SKIPIF1<0的面积是________.【答案】SKIPIF1<0【解析】SKIPIF1<0,因为SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0的面积SKIPIF1<0.【考点】余弦定理、三角形的面积公式SKIPIF1<0（改编）观察下列等式SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0照此规律,SKIPIF1<0SKIPIF1<0.【答案】SKIPIF1<0【解析】观察规律可知,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0【考点】归纳推理三．解答题：本大题共6小题,共75分,解答应写出文字说明、证明过程或演算步骤.SKIPIF1<0（原创）（本小题满分12分）已知函数SKIPIF1<0（Ⅰ）求SKIPIF1<0的单调递减区间;（Ⅱ）将函数SKIPIF1<0的图象向左平移SKIPIF1<0个单位,再将得到的图象上各点的横坐标伸长到原来的SKIPIF1<0倍,纵坐标不变,得到函数SKIPIF1<0的图象,求函数SKIPIF1<0在SKIPIF1<0上的值域.【解析】解法一：（Ⅰ）SKIPIF1<0SKIPIF1<0SKIPIF1<0由SKIPIF1<0,SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0的单调递减区间为SKIPIF1<0,SKIPIF1<0.（Ⅱ）将SKIPIF1<0的图象向左平移SKIPIF1<0个单位,得到SKIPIF1<0SKIPIF1<0,再将SKIPIF1<0图象上各点的横坐标伸长到原来的SKIPIF1<0倍,纵坐标不变,得到SKIPIF1<0.SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0函数SKIPIF1<0在SKIPIF1<0上的值域为SKIPIF1<0.解法二：SKIPIF1<0SKIPIF1<0SKIPIF1<0下同解法一.【考点】两角和差公式、三角函数的图象和性质、诱导公式SKIPIF1<0（改编）（本小题满分12分）如图,三棱柱SKIPIF1<0中,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0、SKIPIF1<0分别是线段SKIPIF1<0、SKIPIF1<0的中点.（Ⅰ）求证：SKIPIF1<0∥平面SKIPIF1<0;（Ⅱ）求平面SKIPIF1<0与平面SKIPIF1<0所成锐二面角的余弦值.【解析】解法一：（Ⅰ）连接SKIPIF1<0,交SKIPIF1<0于SKIPIF1<0,连接SKIPIF1<0,SKIPIF1<0.SKIPIF1<0四边形SKIPIF1<0为平行四边形,SKIPIF1<0SKIPIF1<0为线段SKIPIF1<0的中点.SKIPIF1<0SKIPIF1<0为SKIPIF1<0的中点,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0为SKIPIF1<0的中点,且SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0四边形SKIPIF1<0为平行四边形,SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥平面SKIPIF1<0.（Ⅱ）SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥SKIPIF1<0,SKIPIF1<0SKIPIF1<0平面SKIPIF1<0.又SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0.又SKIPIF1<0SKIPIF1<0,以SKIPIF1<0为坐标原点,SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分别为SKIPIF1<0轴、SKIPIF1<0轴、SKIPIF1<0轴建立空间直角坐标系如图所示.则SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0设平面SKIPIF1<0的一个法向量为SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0SKIPIF1<0令SKIPIF1<0,得SKIPIF1<0SKIPIF1<0SKIPIF1<0.又平面SKIPIF1<0的一个法向量SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0平面SKIPIF1<0与平面SKIPIF1<0所成锐二面角的余弦值为SKIPIF1<0.解法二：（Ⅰ）取SKIPIF1<0中点SKIPIF1<0,连接SKIPIF1<0,SKIPIF1<0.SKIPIF1<0SKIPIF1<0为线段SKIPIF1<0的中点,又四边形SKIPIF1<0为平行四边形.SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.SKIPIF1<0SKIPIF1<0,SKIPIF1<0分别是SKIPIF1<0,SKIPIF1<0的中点,SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.SKIPIF1<0SKIPIF1<0,SKIPIF1<0、SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0∥平面SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.（Ⅱ）同解法一.【考点】直线与平面平行的判定、二面角及空间向量SKIPIF1<0（原创）（本小题满分12分）已知数列SKIPIF1<0是递增的等比数列,SKIPIF1<0,SKIPIF1<0.（Ⅰ）求数列SKIPIF1<0的通项公式;（Ⅱ）若SKIPIF1<0,求数列SKIPIF1<0的前SKIPIF1<0项和SKIPIF1<0.【解析】解法一：（Ⅰ）由SKIPIF1<0即SKIPIF1<0,消SKIPIF1<0得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,SKIPIF1<0SKIPIF1<0或SKIPIF1<0,SKIPIF1<0SKIPIF1<0是递增数列,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0（Ⅱ）SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.解法二：因为SKIPIF1<0是等比数列,SKIPIF1<0,所以SKIPIF1<0.又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0是方程SKIPIF1<0的两根,SKIPIF1<0SKIPIF1<0或SKIPIF1<0.SKIPIF1<0SKIPIF1<0是递增数列,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0.下同解法一.【考点】等比数列的通项公式、错位相减求和SKIPIF1<0（原创）（本小题满分12分）中秋节吃月饼是我国的传统习俗.设有两种月饼礼盒,甲礼盒中装有2个五仁月饼,2个豆沙月饼,2个莲蓉月饼；乙礼盒中装有3个五仁月饼,3个豆沙月饼.这12个月饼外观完全相同,从中随机选取4个.（Ⅰ）设事件SKIPIF1<0为“选取的4个月饼中恰有2个五仁月饼,且这2个五仁月饼选自同一个礼盒”,求事件SKIPIF1<0发生的概率;（Ⅱ）设SKIPIF1<0为选取的4个月饼中豆沙月饼的个数,求随机变量SKIPIF1<0的分布列和数学期望.【解析】（Ⅰ）由已知有SKIPIF1<0,所以事件SKIPIF1<0发生的概率为SKIPIF1<0.（Ⅱ）SKIPIF1<0的所有可能取值为SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所以随机变量SKIPIF1<0的分布列为SKIPIF1<001234SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0随机变量SKIPIF1<0的数学期望为SKIPIF1<0【考点】排列组合、古典概型、随机变量的分布列及数学期望SKIPIF1<0（改编）（本小题满分13分）已知一工厂生产某种产品的年固定成本为100万元,每生产1千件需另投入27万元.设该工厂一年内生产这种产品SKIPIF1<0千件并全部销售完,每千件的销售收入为SKIPIF1<0万元,且SKIPIF1<0（Ⅰ）写出年利润SKIPIF1<0（万元）关于年产量SKIPIF1<0（千件）的函数关系式;（Ⅱ）年产量为多少千件时,该工厂在这种产品的生产中所获得的年利润最大?(注：年利润SKIPIF1<0年销售收入SKIPIF1<0年总成本)【解析】（Ⅰ）SKIPIF1<0SKIPIF1<0（Ⅱ）当SKIPIF1<0时,SKIPIF1<0,令SKIPIF1<0得SKIPIF1<0SKIPIF1<0（SKIPIF1<0舍去）且当SKIPIF1<0时,SKIPIF1<0；当SKIPIF1<0时,SKIPIF1<0.所以当SKIPIF1<0时,SKIPIF1<0SKIPIF1<0.当SKIPIF1<0时,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0当且仅当SKIPIF1<0即SKIPIF1<0SKIPIF1<0时取等号.所以当SKIPIF1<0时,SKIPIF1<0SKIPIF1<0.因为SKIPIF1<0,所以当SKIPIF1<0时,SKIPIF1<0SKIPIF1<0.答：年产量为9千件时,该工厂在这种产品的生产中所获得的年利润最大. SKIPIF1<0【考点】函数应用、分段函数、基本不等式、导数SKIPIF1<0（原创）（本小题满分14分）设函数SKIPIF1<0,其中SKIPIF1<0.（Ⅰ）SKIPIF1<0时,求曲线SKIPIF1<0在点SKIPIF1<0处的切线方程;（Ⅱ）讨论函数SKIPIF1<0的单调性;（Ⅲ）当SKIPIF1<0时，证明对SKIPIF1<0,都有SKIPIF1<0.【解析】（Ⅰ）SKIPIF1<0时,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.又SKIPIF1<0SKIPIF1<0曲线SKIPIF1<0在点SKIPIF1<0处的切线方程为SKIPIF1<0.（Ⅱ）SKIPIF1<0的定义域为SKIPIF1<0SKIPIF1<0 令SKIPIF1<0得SKIPIF1<0或SKIPIF1<0 当SKIPIF1<0即SKIPIF1<0时,当SKIPIF1<0时,SKIPIF1<0；当SKIPIF1<0时,SKIPIF1<0.当SKIPIF1<0即SKIPIF1<0时当SKIPIF1<0时SKIPIF1<0；当SKIPIF1<0时SKIPIF1<0；当SKIPIF1<0时SKIPIF1<0.当SKIPIF1<0即SKIPIF1<0时SKIPIF1<0.当SKIPIF1<0即SKIPIF1<0时当SKIPIF1<0时SKIPIF1<0；当SKIPIF1<0时SKIPIF1<0；当SKIPIF1<0时SKIPIF1<0.综上所述：当SKIPIF1<0时,SKIPIF1<0的增区间为SKIPIF1<0,减区间为SKIPIF1<0；当SKIPIF1<0时,SKIPIF1<0的增区间为SKIPIF1<0和SKIPIF1<0；减区间为SKIPIF1<0；当SKIPIF1<0时,SKIPIF1<0的增区间为SKIPIF1<0,无减区间；当SKIPIF1<0时,SKIPIF1<0的增区间为SKIPIF1<0和SKIPIF1<0,减区间为SKIPIF1<0.（Ⅲ）证法一：①当SKIPIF1<0时,由（Ⅱ）知：SKIPIF1<0在SKIPIF1<0上单调递增,在SKIPIF1<0上单调递减,在SKIPIF1<0上单调递增,所以SKIPIF1<0.SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0记SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0在SKIPIF1<0上单调递增.SKIPIF1<0当SKIPIF1<0时,SKIPIF1<0即SKIPIF1<0成立.又SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.所以SKIPIF1<0.SKIPIF1<0当SKIPIF1<0时,SKIPIF1<0时SKIPIF1<0.②当SKIPIF1<0时,SKIPIF1<0在SKIPIF1<0上单调递增,SKIPIF1<0SKIPIF1<0.=3\*GB3③当SKIPIF1<0时,由（Ⅱ）知,SKIPIF1<0在SKIPIF1<0上单调递增,在SKIPIF1<0上单调递减,在SKIPIF1<0上单调递增.故SKIPIF1<0在SKIPIF1<0上只有一个极大值SKIPIF1<0,所以当SKIPIF1<0时,SKIPIF1<0.SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0当SKIPIF1<0时,SKIPIF1<0时SKIPIF1<0.综①②=3\*GB3③知：当SKIPIF1<0时，对SKIPIF1<0,都有SKIPIF1<0.注：判断当SKIPIF1<0时,SKIPIF1<0,也可用如下两种方法：方法一：SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.所以SKIPIF1<0.方法二：SKIPIF1<0SKIPIF1<0令SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0.（Ⅲ）证法二：SKIPIF1<0SKIPIF1<0.记SKIPIF1<0,先证SKIPIF1<0,SKIPIF1<0.记SKIPIF1<0,SKIPIF1<0,令SKIPIF1<0得SKIPIF1<0.SKIPIF1<0SKIPIF1<0时,SKIPIF1<0;SKIPIF1<0时,SKIPIF1<0.SKIPIF1<0SKIPIF1<0即SKIPIF1<0.SKIPIF1<0SKIPIF1<0在SKIPIF1<0上单调递减,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0.故证SKIPIF1<0.（Ⅲ）证法三：SKIPIF1<0SKIPIF1<0同证法二得SKIPIF1<0即SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0.故证SKIPIF1<0.【考点】导数的几何意义、用导数研究函数的单调性、恒成立问题、分类讨论的思想方法齐鲁名校教科研协作体山东省19所名校2016届高三第一次调研（新起点）联考数学试题（理科）评分标准选择题：本大题共10小题,每小题5分,共50分.二.填空题：本大题共5小题,每小题5分,共25分.11.1412.SKIPIF1<013.12014.SKIPIF1<015.SKIPIF1<0三．解答题：本大题共6小题,共75分.SKIPIF1<0（本小题满分12分）【解析】解法一：（Ⅰ）SKIPIF1<0SKIPIF1<0SKIPIF1<0………………….4分由SKIPIF1<0,SKIPIF1<0,………………….5分得SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0的单调递减区间为SKIPIF1<0,SKIPIF1<0.…………………6分（=2\*ROMANⅡ）将SKIPIF1<0的图象向左平移SKIPIF1<0个单位,得到SKIPIF1<0SKIPIF1<0,………………….7分再将SKIPIF1<0图象上各点的横坐标伸长到原来的SKIPIF1<0倍,纵坐标不变,得到SKIPIF1<0,………………….8分SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.………………….9分SKIPIF1<0SKIPIF1<0.………………….11分SKIPIF1<0SKIPIF1<0.SKIPIF1<0函数SKIPIF1<0在SKIPIF1<0上的值域为SKIPIF1<0.………………….12分解法二：（Ⅰ）SKIPIF1<0SKIPIF1<0SKIPIF1<0………………….4分下同解法一.SKIPIF1<0（本小题满分12分）【解析】解法一：（Ⅰ）连接SKIPIF1<0,交SKIPIF1<0于SKIPIF1<0,连接SKIPIF1<0、SKIPIF1<0,SKIPIF1<0四边形SKIPIF1<0为平行四边形,SKIPIF1<0SKIPIF1<0为线段SKIPIF1<0的中点.SKIPIF1<0SKIPIF1<0为SKIPIF1<0的中点,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.………………….1分SKIPIF1<0为SKIPIF1<0的中点,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.SKIPIF1<0SKIPIF1<0SKIPIF1<0…………………2分SKIPIF1<0四边形SKIPIF1<0为平行四边形.SKIPIF1<0∥SKIPIF1<0.……………….3分SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0.SKIPIF1<0∥平面SKIPIF1<0.………………….5分（Ⅱ）SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥SKIPIF1<0,SKIPIF1<0SKIPIF1<0平面SKIPIF1<0.又SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0.又SKIPIF1<0SKIPIF1<0,以SKIPIF1<0为坐标原点,SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分别为SKIPIF1<0轴、SKIPIF1<0轴、SKIPIF1<0轴建立空间直角坐标系.………………….6分则SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0,SKIPIF1<0.………………….7分设平面SKIPIF1<0的一个法向量为SKIPIF1<0SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0SKIPIF1<0令SKIPIF1<0,得SKIPIF1<0SKIPIF1<0SKIPIF1<0.………………….9分又平面SKIPIF1<0的一个法向量SKIPIF1<0………………….10分SKIPIF1<0SKIPIF1<0………………….11分SKIPIF1<0平面SKIPIF1<0与平面SKIPIF1<0所成锐二面角的余弦值为SKIPIF1<0.………………….12分解法二：（Ⅰ）取SKIPIF1<0中点SKIPIF1<0,连接SKIPIF1<0,SKIPIF1<0.SKIPIF1<0SKIPIF1<0为线段SKIPIF1<0的中点,又四边形SKIPIF1<0为平行四边形.SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.………………….2分SKIPIF1<0SKIPIF1<0,SKIPIF1<0分别是SKIPIF1<0,SKIPIF1<0的中点,SKIPIF1<0SKIPIF1<0∥SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0SKIPIF1<0∥平面SKIPIF1<0.………………….4分SKIPIF1<0SKIPIF1<0,SKIPIF1<0、SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0平面SKIPIF1<0∥平面SKIPIF1<0.SKIPIF1<0SKIPIF1<0平面SKIPIF1<0,SKIPIF1<0∥平面SKIPIF1<0.………………….5分（Ⅱ）同解法一.SKIPIF1<0（本小题满分12分）【解析】解法一：（Ⅰ）由SKIPIF1<0即SKIPIF1<0…………………2分消SKIPIF1<0得SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,SKIPIF1<0SKIPIF1<0或SKIPIF1<0………………….4分SKIPIF1<0SKIPIF1<0是递增数列,SKIPIF1<0SKIPIF1<0…….5分SKIPIF1<0SKIPIF1<0.…….6分（Ⅱ）SKIPIF1<0…….7分SKIPIF1<0SKIPIF1<0………………….8分SKIPIF1<0SKIPIF1<0………………….9分SKIPIF1<0