武汉市常青联合体2023-2024学年度第二学期期中考试
高二数学试卷
考试时间：2024年4月18日 满分：150分
一、单选题：本题共8小题，每小题5分，共40分。在每小题给出的四个选项中只有一项符合题目要求。
1.设是可导函数，且，则( )
A．2 B． C． D．
2.下列求导数的运算中正确的是( )
A. B.
C. D.
3.已知函数,则的大致图象为( )
A. B. C. D.
4.根据马伯庸的小说《长安十二时辰》同名改编的电视剧中，靖安司通过长安城内的望楼传递信息。望楼传递信息的方式如下：如图所示，在九宫格中，每个小方格可以在白色和紫色(此处以阴影代表紫色)之间变换，从而一共可以有种不同的颜色组合，即代表种不同的信息。现要求最多出现个紫色格子，那么一共可以传递的不同的信息有( )
A. 种 B. 种 C. 种 D. 种
5.,则( )
A. B. C. D.
6.“”是“直线与曲线相切”的（ ）
A．充分不必要条件 B．必要不充分条件
C．充分必要条件 D．既不充分也不必要条件
7.已知,,,则的大小关系为( )
A. B. C. D.
8.若两曲线与存在公切线,则正实数a的取值范围是( )
A. B. C. D.
二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的四个选项中有多项符合要求，全部选对得6分，部分选对的得部分分。
9.下列说法中正确的是( )
A. 将个相同的小球放入个不同的盒子中,要求不出现空盒,共有种放法
B. 被除后的余数为
C.若,则
D. 10个朋友聚会,见面后每两个人握手一次,一共握手45次
10.已知函数在区间上单调递减,则实数可以是( )
A. 0 B. C. 1 D.
11.已知函数及其导函数的定义域为,若,函数和均为偶函数,则( )
A. 函数的图象关于点对称 B. 函数是周期为的周期函数
C. 函数的图象关于点对称 D.
三、填空题：本题共3小题，每小题5分，共15分。
12.在的展开式中,的系数是__________.
13.已知函数,若成立,则的最小值为__________.
14.若存在正数,使得不等式有解,则实数的取值范围是__________.
四、解答题：共77分，解答应写出文字说明、证明过程或演算步骤。
15.(13分)
(1)解关于x的不等式
(2)解关于n的方程
（15分）
在二项式的展开式中，第项和第项的系数比为
(1)求n的值及展开式中的常数项是第几项；
(2)展开式中系数最大的项是第几项。
（15分）
已知数列的前n项和为,且
(1)证明:数列是等差数列；
(2)已知,求数列的前n项和。
（17分）
已知函数
(1)求的极值；
(2)若对于任意,不等式恒成立,求实数的取值范围。
（17分）
已知函数
(1)讨论的单调性；
(2)若0时，f(x)的图象恒在x轴上方，求m的范围；
(3)若存在不相等的实数,使得,证明:。2024 学年度第二学期期中考试
数学试卷参考答案
一、单选题
1. B 2.D 3.C 4.C 5.B 6.C 7.D 8.B
二、选择题
9. ACD 10.ABD 11.ABD
三、填空题
12.9 13. 14.
四、解答题
15.(1) 由 ,得 ,········································································································································2 分
于是 ,
整理得 ,解得 ,················································································································5 分
所以 .······································································································································································································································6 分
(2)原方程变形为 ,即 ,显然 ,························9 分
因此 ,
化简整理,得 ,···········································································································································································12 分
而 ,解得 ,所以 .··························································································································································13 分
16. (1)二项式 展开式的通项公式为
( ).························2 分
因为第 项和第 项的系数比为 ,所以 ,
化简得 ,解得 ,················································································································································4 分
所以 ( ).
令 ,得 ,·············································································································································································6 分
1
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所以常数项为第 项.······························································································································································································7 分
(2)设展开式中系数最大的项是第 项,
则 ,·····························································11 分
解得 .·······················································································································································································································13 分
因为 ,所以 或 ,所以展开式中系数最大的项是第 项和第 项.·············15 分
17. (1)由 ,得 ,
令 ,则 ,解得 ;·······································································································································································1 分
当 时, ,
所以 ,所以 ,··········3 分
所以当 时, ,
有 ,········································6 分
又 满足上式,
所以 ,得 ,
所以数列 是首项为 0，公差为 1 的等差数列.·······················································································································8 分
(2)由(1)知 ,所以 ,····································································································9 分
所以 ,
故 ,
两式相减,得
2
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,
所以 .························································································································································································································15 分
18. (1)由函数 ,可得 ,························································································································2 分
令 ,即 ,解得 ;
令 ,即 ,解得 ,
所以函数 在区间 单调递减, 单调递增,··············································································4 分
当 时, 取得极小值,极小值为 ,无极大值.·········································································6 分
(2)由不等式 恒成立,即 恒成立,
即对于任意 ,不等式 恒成立,····················································································································8 分
设 ,可得 ,
令 ,即 ,解得 ;
令 ,即 ,解得 ,
所以 在 上单调递减;在 单调递增,···············································································12 分
所以,当 时,函数 取得极小值,同时也时最小值, ,·············································14 分
所以 ,即 ,所以实数 的取值范围为 .·························································17 分
19. (1)函数 的定义域为 , ,
当 时, ,所以 在 上单调递增;
当 时,由 得 ,所以 在 上单调递增;
由 得 ,所以 在 上单调递减;
故 时,所以 在 上单调递增;························································································································2 分
当 时, 在 上单调递增,在 上单调递减;·············································4 分
(2)由 f(x)的图象恒在 x 轴上方，可得 f(x) = x mlnx > 0
3
{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}
1
因为 x>0 且 m>0 ,不等式两边同时除以 mx ，可得 > ···········6 分
h(x)= 设 可得 h’(x)=1
2
令 h’(x)>0，解得 0令 h’(x)<0，解得 x>e
所以 h(x)在(0,e)上单调递增，在(e,+∞)上单调递减
所以当 x=e时 h(x) 1取得最大值，h(e)= ··············································································································8 分
1
所以 >h(x) 1 1max 即 > 所以 m 的范围是(0,e)······························································9 分
(3)证明: , ,
由(1)可知,当 时, 在 上是增函数,
故不存在不相等的实数 ,使得 ,所以 .
由 得 ,即 ,···············11 分
不妨设 ,则 ,则 ,
要证 ,只需证 ,
即证 ,只需证 ,····························································································································14 分
令 ,则只需证 ,即证 ,
令 ,则 ,
所以 在 上是增函数,所以 ,
从而 ,故 .·····················································································································································································17 分
4
{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}
