2021江苏省外国语学校高二下学期期中数学试题含答案

这是一份2021江苏省外国语学校高二下学期期中数学试题含答案，共10页。试卷主要包含了04，函数的大致图象可能是，已知函数，，若，则的最大值是，已知，，若，则，给定函数，下列结论正确的是等内容，欢迎下载使用。
2020~2021学年第二学期期中调研测试 高二数学 2021.04一、单项选择题：本大题共小题，每小题分，共计40分，每小题给出的四个选项中，只有一个是正确的，请把正确的选项填涂在答题卡的相应位置上。1.函数的单调递区间为（    ）A. B. C. D.2.用数字0，1，2，3可以组成无重复数字的四位偶数有（    ）A.12个 B.10个 C.20个 D.16个3.函数的大致图象可能是（    ）A.  B. C.  D. 4.在的展开式中，只有第5项的二项式系数最大，则展开式中x的系数为（    ）A. B. C. D.75.已知函数的图象在处的切线与函数的图象相切，则实数（    ）A. B. C. D.6.将编号为1，2，3，4，5，6，7的小球放入编号为1，2，3，4，5，6，7的七个盒子中，每盒放一球，若有且只有三个盒子的编号与放入的小球的编号相同，则不同的放法种数为（    ）A.315 B.640 C.840 D.50407.已知函数，，若，则的最大值是（    ）A. B. C. D.8.已知，，若，则（    ）A. B. C. D.二、多项选择题：本题共2小题，每小题5分，共10分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.给定函数.下列说法正确的有（    ）A.函数在区间上单调递减，在区间上单调递增B.函数的图象与x轴有两个交点C.当时，方程有两个不同的的解D.若方程只有一个解，则10.下列结论正确的是（    ）A.6本不同的书分给甲、乙、丙三人，每人两本，有种不同的分法；B.6本不同的书分给甲、乙、丙三人，其中一人1本，一人2本，一人3本，有种不同的分法；C.6本相同的书分给甲、乙、丙三人，每人至少一本，有10种不同的分法；D.6本不同的书分给甲、乙、丙三人，每人至少一本，有540种不同的分法.11.设，下列结论正确的是（    ）A.B.C.，，，…，中最大的是D.当时，除以2000的余数是112.已知函数，下述结论正确的是（    ）A.存在唯一极值点，且B.存在实数a，使得C.方程有且仅有两个实数根，且两根互为倒数D.当时，函数与的图象有两个交点三、填空题：本题共4小题，每小题5分，共20分.13.二项式的展开式中，常数项为___________.14.若函数在上单调递增，则实数a的取值范围是___________.15.如图，用五种不同的颜色涂在图中不同的区域内，要求每个区域只能涂一种颜色，且相邻（有公共边）区域涂的颜色不同，则不同的涂色方案一共有__________种.（用数字作答）.16.已知函数，若在上单调减函数，则实数a的最大值为_________.若，在上至少存在一点，使得成立，则实数a的最小值为___________.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.（本题满分10分）已知函数，的图象在点处的切线为.（1）求函数的解析式；（2）设，求证：；18.（本题满分12分）已知从的展开式的所有项中任取两项的组合数是21。（1）求展开式中所有二项式系数之和2）若的展开式中的常数项为，求a的值。19.按照下列要求，分别求有多少种不同的方法？（列式并用数字作答）（1）5个不同的小球放入4个不同的盒子，每个盒子至少放一个小球（2）6个不同的小球放入4个不同的盒子，每个盒子至少一个小球；（3）6个相同的小球放入4个不同的盒子，每个盒子至少一个小球；（4）6个不同的小球放入4个不同的盒子，恰有1个空盒.20.（本题满分12分）已知函数（）.（1）当时，求函数的单调区间；（2）是否存在实数a，使恒成立，若存在，求出实数a的取值范围；若不存在，说明理由.21.（本题满分12分）在杨辉三角形中，从第2行开始，除1以外，其它每一个数值是它上面的两个数值之和，该三角形数阵开头几行如图所示.第0行                  1第1行                1   1第2行              1   2   1第3行            1   3   3   1第4行          1   4   6   4   1第5行        1   5   10  10   5   1第6行      1   6   15  20  15   6   1（1）在杨辉三角形中是否存在某一行，使该行中三个相邻的数之比是？若存在，试求出是第几行；若不存在，请说明理由；（2）已知n，r为正整数，且，求证：任何四个相邻的组合数，，，不能构成等差数列.22.（本题满分12分）已知函数，（a，）.（1）若时，直线是曲线的一条切线，求b的值；（2）若，且在上恒成立，求a的取值范围；（3）令，且在区间上有零点，求的最小值.2020~2021学年第二学期期中调研测试 高二数学参考答案 2021.04一、单项选择题：本大题共小题，每小题分，共计40分，每小题给出的四个选项中，只有一个是正确的，请把正确的选项填涂在答题卡的相应位置上。1.B；  2.B；  3.A；  4.D；  5.B；  6.A；  7.A；  8.D二、多项选择题：本题共2小题，每小题5分，共10分.在每小题给出的四个选项中，有多项符合题目要求.全部选对的得5分，部分选对的得2分，有选错的得0分.9.AC；  10.ACD；  11.ABD；  12.ACD三、填空题：本题共4小题，每小题5分，共20分.13.；  14.；  15.180；  16.（1）  （2）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.（本题满分10分）【解析】（1），，····························································1分由已知得，···································································2分解得，故····································································4分（2），得.···································································6分当时，，单调递减；当时，，单调递增.····························································8分∴，从而，即.································································10分18.解析：的展开式共有项，由题意得，解得，·········································2分所以展开式中所有二项系数之和为.·················································4分（2）由（1）知，的展开式的通项为，····························································6分令，或2，解得或，····························································8分因为展开式中的常数项为，······················································10分解得.·······································································12分19.【解析】（1）；······························································3分（2）；或；··································································6分（3）；或；··································································9分（4）.或.···································································12分20.【解析】（1）函数的定义域为，，·················································1分21.【解析】（1）存在.·························································1分杨辉三角形的第n行由二项式系数，，1，2，…，n组成.若第n行中有三个相邻的数之比为3∶4∶5，则，，······································································3分即，，解得，.即第62行有三个相邻的数，，的比为3∶4∶5.········································5分（2）证明  若有n，r（），使得，，，成等差数列，则，，······································································6分即，···········································································7分所以，整理得，.两式相减得，·································································9分所以，，，成等差数.··························································10分由二项式系数的性质可知.·······················································11分这与等差数列的性质矛盾，从而要证明的结论成立.···································12分22.【解析】（1）当时，，，设切点，则在点A处的切线为，···························································2分化简得，因为是的一条切线，，，解得，；·································································3分（2）当时，令，则.····························································4分若，则当时，恒成立，在上单调递增，，即符合题意；·······························································5分若时，由，得，当时，，在上单调递减，，与已知在上恒成立矛盾，舍去.··················································6分综上，且.····································································7分（3），.若，则在区间上恒成立，在区间上单调递增，因为在区间上有零点，所以，解得.·······································································8分所以，当时，等号成立，此时.·························································9分若时，当时，，在上单调递减，当时，，在上单调递增.因为在区间上有零点，所以，所以，所以，································································10分令，，则，所以在（2）上单调递减.所以.若，则在区间上恒成立，在区间上单调递减.因为叫在区间上有零点，所以，解得.··········································································11分所以，当时，等号成立，此时；·······················································12分综上，的最小值是.当时，由，得，或，由，得，····································································2分故函数的单调递增区间为和，单调递减区间为，·······································4分当时，恒成立，故函数的单调递增区间为.············································5分（2）恒成立等价于恒成立，令，当时，即当时，，故在内不能恒成立，··············································6分当时，即当时，则，故在内不能恒成立，············································7分当时，即当时，，由解得，······················································8分当时，；当时，.······························································10分所以，解得.·································································11分综上，当时，在内恒成立，即恒成立，所以实数a的取值范围是.························································12分