湖南省益阳市桃江县2019-2020学年高二下学期期末考试 数学试题
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这是一份湖南省益阳市桃江县2019-2020学年高二下学期期末考试 数学试题,共12页。
2019—2020学年度第二学期期末考试高 二 数 学 试 题 卷(时量:120分钟,满分:150分)一、单项选择题:本题共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.已知集合,,则=A. B. C. D.2.“”是“”的条件A. 充分不必要 B. 必要不充分 C.充要 D.既不充分又不必要3.已知双曲线的离心率为,则该双曲线的虚轴长为A.4 B. C. D.24.已知,则A. B. C. D.5.函数的图象可能是6.两个不同的小球要放到编号分别为1,2,3,4,5,6的盒子中,每个盒子中最多放入一个小球,则放入小球的盒子的编号不连续的概率为A. B. C. D. 7.已知函数是定义在上的奇函数,当时,,若实数满足,则的取值范围是A. B. C. D. 8.已知数列{an}是等比数列,Sn为其前n项和,若a1+a2+a3=4,a4+a5+a6=8,则S12等于A.40 B.60 C.32 D.509.已知菱形ABCD边长为4,,M为CD的中点,N为平面ABCD内一点,且满足AN = NM,则的值为A. B. 16 C. 14 D. 810.若将函数的图像向右平移个单位长度得到函数的图像,下列说法中正确的是A. 的图像关于直线对称 B. 上恰有两个零点C. 上单调递减 D. 上的值域为二、多项选择题:本题共2小题,每小题5分,共10分。在每小题给出的选项中,有多项符合题目要求。全部选对的得5分,部分选对的得3分,有选错的得0分。11.设公差不为0的等差数列的前n项和为,若,则下列各式的值为0的是A. B. C. D. 12.已知椭圆C:的左,右焦点分别为,,且,点在椭圆内部,点在椭圆上,则以下说法正确的是A. 的最小值为 B. 椭圆C的短轴长可能为2C. 椭圆C的离心率的取值范围为D. 若,则椭圆C的长轴长为三、填空题:本题共4小题,每小题5分,共20分;其中第16小题,第一空2分,第二空3分。13.已知平面向量,若,则 .14.已知,则的最小值为 .15.在正方体中,E为的中点,则异面直线与所成角的余弦值为 .16.已知抛物线C:的焦点为F,准线为,点P为准线上一点,且不在x轴上,直线交抛物线C于A,B两点,且,则 ;设坐标原点为O,则的面积为 .四、解答题:本题共6小题,共70分。解答应写出文字说明、证明过程或演算步骤。17.(本小题满分10分)已知函数. (1)求的定义域; (2)解关于的不等式. 18.(本小题满分12分) 已知的内角的对边分别为,且. (1)求角的大小; (2)已知,且,若的面积为,求b边的长以及外接圆的半径R. 19.(本小题满分12分)一饮料店制作了一款新饮料,为了进行合理定价先进行试销售,其单价x (元)与销量y(杯)的相关数据如下表:单价x (元)8.599.51010.5销量y(杯)120110907060 (1)已知单价x与销量y具有线性相关关系,求y关于x的线性回归方程; (2)若该款饮料每杯的成本为8元,试销售结束后,请利用(1)所求得的线性回归方程确定单价定为多少元时(单价保留到整数),销售利润最大?并求出利润的最大值.参考公式:线性回归方程的最小二乘法计算公式:,参考数据:20.(本小题满分12分)如图,直三棱柱中,,,,分别是棱,的中点.(1)证明:平面;(2)求二面角的余弦值. 21.(本小题满分12分)已知等比数列的前n项和为,,且成等差数列.(1)求数列的通项公式;(2)设,求数列的前项的和. 22.(本小题满分12分)已知椭圆的离心率为,且经过点.(1)求椭圆的方程;(2)若不过坐标原点的直线与椭圆相交于、两点,且满足,求面积取最大值时直线的方程.
2019—2020学年度第二学期期末考试高二数学参考答案一、单项选择题:本题共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.C 2.B 3.B 4.C 5.D 6.A 7.A 8.B 9.C 10.B二、多项选择题:本题共2小题,每小题5分,共10分。在每小题给出的选项中,有多项符合题目要求。全部选对的得5分,部分选对的得3分,有选错的得0分。11. BD 12. AD三、填空题:本题共4小题,每小题5分,共20分;其中第16小题,第一空2分,第二空3分。13. 14. 5 15. 16. 9;四、解答题:本题共6小题,共70分。解答应写出文字说明、证明过程或演算步骤。17.(本小题满分10分)【解析】(1)定义域为的解集∴当时,定义域为················································3分当时,定义域为··················································5分 (2)∵在定义域内,∴∴单调递增,结合定义域可知:的解集为················································10分注:直接给出函数f(x)单调性的给全分.18.(本小题满分12分)【解析】(1)由正弦定理以及得:·······················································2分∴,又sinA,∴,又sinB,∴·····················································6分(2),∴由,联立可得或∵,∴··················································8分根据余弦定理:∴·····················································10分由,即 综上:b边的长为,外接圆的半径R等于·························12分19.(本小题满分12分)【解析】(1)由表中数据可计算得∴y关于x的线性回归方程为(2)设定价为x元,则利润函数为,其中当时,y有最大值为148.所以单价定为10元时,销售利润最大,最大利润为148元.20.(本小题满分12分)【解析】(1)由题意知,平面,平面,.·························································2分又,,分别是棱,的中点,.······························································3分又平面,平面,,平面.··························································5分(2)不妨设,如图,以为原点,,,所在直线分别为,,轴,建立空间直角坐标系,则,,,,.····························································6分设平面的法向量为,则令,得,,.······························································8分因为y轴垂直平面,所以可取平面的法向量为,···························9分.又二面角显然为钝角所以二面角的余弦值为.············································12分【注】二面角的余弦值缺少负号扣2分21.(本小题满分12分)【解析】(1)设等比数列的公比为,由得, ∴,∴·················································2分由成等差数列得,即,∴····································3分数列的通项公式为·········································6分(2)当为偶数时,,当为奇数时,······································7分∴·······················································9分.·······················································12分22.(本小题满分12分)【解析】(1)∵,∴,,设椭圆C的方程为,将点A的坐标代入得:,∴.故椭圆的方程为···············································5分(2)依题意可知,直线的斜率存在,设其方程为,,由得,,,,,···························································6分,,,···························································7分则,即:且,······························································8分,····························································10分当且仅当,即时,等号成立.直线的方程为.···················································12分 【注】本题求出了直线方程而未求出最大值扣2分
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