湖北省武汉市常青联合体2023-2024学年高二下学期期中考试数学试卷（pdf版，含答案）

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f(x)是可导函数，且im
f0+3Ax)f=2,则f(四)=(）
△x
A.2
B.
C.-1
D.-2
2.下列求导数的运算中正确的是(
A.(log
B.(sinx+cos'=cosx-sin
元
C.(3)y=x3-
D.[n(2x-l)'=
2x-1
3.已知函数f=c2r-
2,则f(x)的大致图象为(
x-1
B
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2种不同的颜色组合，即代表512种不同的信息。现要求最多出现2个紫
色格子，那么一共可以传递的不同的信息有()
高二数学试卷第1页共4页
A.36种
B.45种
C.46种
D.84种
5.
(1-3x)3=a+a,x+ax2+ax3+a4x4+a,x3
则
aa +aa as =
A.31
B.1023
C.1024
D.32
6.“b=-1”是“直线y=3x+b与曲线y=x(nx+2)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a=
巨，b=,c-n5,则ab,c的大小关系为()
4
e2,c=
3
A.aC.bD.a8.若两曲线y=lnx-1与y=ax2存在公切线，则正实数a的取值范围是()
A.(0,2e]
B.e3,+o)
D.[2e,to∞)
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4个相同的小球放入3个不同的盒子中，要求不出现空盒，共有3种放法
B.482022-3被7除后的余数为2
C.
若
(x+1)4+(x-103=4+ax+ax2+4x3+ax4+ax
则
4w+a2+1=-8
高二数学试卷第2页共4页
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f(x)=lnx-(x-a(a∈R)在区间[1，oo)上单调递减，则实数G
可以是()
A.0
B.V2-1
C.1
D.月
11.已知函数f(x)及其导函数f(x)的定义域为R,若f'(2)=8,函数
f(2x+1)和f”(x+2)均为偶函数，则()
A.函数（x的图象关于点(1,0)对称
B.函数(x)是周期为4的周期
函数
C.函数f(x)的图象关于点(3，O)对称D.
f=8
三、填空题：本题共3小题，每小题5分，共15分。
12.在(x-1)(x-2)(x-3)(x+4)(x+5)(x+6)的展开式中，x5的系数是
13.已知函数f(x)=xg(x)=e,若f(m=g(n)成立，则m(n+1)的最小值
为
14.若存在正数x,使得不等式e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的不等式戌<642
(2)解关于n的方程C+C=19
5
高二数学试卷第3页共4页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
{#{QQABBQQAgggoQIJAABhCUQHwCAGQkBACCKoGxBAEIAIBiANABAA=}#}
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=}#}