湖北省武汉市重点中学5G联合体2025-2026学年高二下学期期末考试数学试题含答案

高二数学

—、选择题:本题共8小题,每小题5分,共40分.在每小题给出的选项中,只有一项

是符合题目要求的·

1.己知函数(f

A.B·-iC.1D·

2.已知等比数列{a,}的各项均为正数,q;a,l,a,a,4,则a,a,=()

A.B·4C·3D.2

A.20B.120C.60D.135

4.学校测试AI智能阅卷,启用了标注为甲,乙,丙的三款评卷系统·平台将随机调用

甲,乙,丙的概率依次为0.4,0.4,0.2.若甲,乙,丙批改一道数学题的正确率分别为

90%,80%,70%·现随机抽取一道题目,则该题目被正确批改的概率为()

A.0.81B.0.82C.0.83D.0.84

5.将标有5,5,2,3,4,6的六张数字卡片分成甲,乙,丙三组,要求每组都有奇

数数字卡片与偶数数字卡片,则不同的分法总数为()

A.12B.36C.24D.18

6.若fx()是定义在区间(-3,2)上的函数,其图象如图所示,设(x)的导函数为(),

则f(x-1)J"(x-1)>0的解集为()

A.(-2,-1)u(1,2)B.(1,0)u(1,2)

C.(-3,-2)u(0,1)D.(1,0)U(0,1)

2

7.设随机变量服从正态分布N2.,若P(5>4)=0.1,则函数

A.0.2B.0.3C.0.4D.0.5

8.己知定义在R上的增函数J(x)满足对wxeR,有f(-x)+f(4+x)=f(2),设

g(:)=J(x+1)+4x+4+"-e",若对wxe(0,),都有g(xe"+ax-+IX)s8恒成立,

则实数a的取值范围为()

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二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符

合题目要求,全部选对的得6分,部分选对的得部分分,有选错的得0分.

9.下列命题正确的是()

A.己知两个变量线性相关,若它们的相关性越强,则相关系数r越接近于1

B.残差平方和越小,回归模型的拟合效果越好

C.线性回归直线必然过样本中心点

2

D.已知px3.8410.05),根据分类变量x与y的成对样本数据,计算得到

X=24.1,依据a=0.05的独立性检验,变量x与y不独立,这个结论犯错误的概率不

超过0.05

10.已知函数s(x)=-46e2)",则下列结论正确的是(

A.函数f(x)有2个极值点

B.函数(x)无最小值

2

D.函数y=3f(x)+2f(t)-1有5个零点

11.杨辉为我国古代数学史上著述丰富的数学家,其传世著作包括《详解九章算法》,

《日用算法》及《杨辉算法》·公元1261年,他于《详解九章算法》中刊载了图1所

示数表,后世称之为杨辉三角,图2为该数表的数字呈现形式·杨辉三角的发现较欧洲

相关研究成果早约五百年,充分体现了我国古代数学所取得的卓越成就,亦足以令中华

民族引以为傲·据此材料,下列说法正确的是()

A.第8行所有数字的和等于256

B.第8行所有数字的平方和等于C16

C.记每一行的第kkN")个数组成的数列称为第k斜列,该三角形数阵前2026行

中第k斜列各项之和为c2027k

高二数学第2页共4页

填空题:本题共3小题,每小题5分,共5分.

一个公共点,则实数a的值为·

14.甲、乙、丙三人玩剪刀、石头、布"的游戏.游戏规则为:剪刀赢布,布赢石头,

石头赢剪刀·每一局游戏甲、乙、丙同时出cc剪刀、石头、布"中的一种手势,且相互独

立.在一局游戏中某人赢1个人得2分,赢2个人得5分,其他情况得0分.设一局游

戏后3人总得分为,则随机变量的数学期望E的值为·

四、解答题:本题共5小题,共77分.解答应写出文字说明,证明过程或演算步骤·

15.(本小题13分)

在AABC中,角A,B,C的对边分别为a,b,c,若bcosA+JbsinA-c-a=0.

(1)求角B;

2若b3C的值

16.(本小题15分)

如图,在四棱锥PABCD中,PD平面ABCD,底面ABCD为正方形,

(1)若N为PC的中点,求证:平面BND;

(2)若PDAD==3,求平面PAD与平面PBC夹角的余弦值.

高二数学第3页共4页

17.(本小题15分)

己知等差数列fa,}的前n项和为s,,且423,a3=7.

(1)求数列{a,}的通项公式;

(2)若b,=2n+,令cabrnr,求数列{c,}的前n项和T,.

18.(本小题17分)

2026年马年春晚《武BOT》节目中,宇树科技的人形机器人与塔沟武校的少年武者进

行了一场人机武术对抗赛·假设每局比赛中··少年武者获胜的

且每局胜负相互独立比赛采用k局k+胜制即先赢得k+局者获胜

(1)当K=l时,记结束比赛时的局数为X,求X的分布列和数学期望E(X);

(2)设在该赛制下机器人获胜的概率为pk).

①求p(1)和p(2)的值,并比较它们的大小,据此说明k=1和k=2哪种赛制对机器人更

有利;

②随着k的增大,机器人获胜的可能性如何变化?证明你的结论.

19.(本小题17分)

已知函数f(x)=2xlnx,

(1)求不等式0≤f(x)≤2e的解集;

(2)已知实数a<0,求g(x)=f(x)-4x+2a的零点个数;

高二数学第4页共4页

高二（下）期末考试参考答案

一．选择题

1234567891011

ADCBDBCBBCDADABD

二．填空题

13

12.13513.114.

3

三．解答题

15.解：（1）由bcosA+3bsinA-c-a=0及正弦定理得

sinBcosA+3sinBsinA-sinC-sinA=0·····················································1分

因为sinC=sin(π-A-B)=sin(A+B)=sinAcosB+cosAsinB，

所以3sinBsinA-sinAcosB-sinA=0·························································3分

由于sinA≠0，:3sinB-cosB-1=0···························································4分

所以sin·························································································5分

又0<B<π,故B···················································································6分

193

（2）由题得ΔABC的面积S=acsinB=，故ac=9①·································8分

24

而b2=a2+c2-2accosB，············································································9分

且b=2，故a2+c2=18②,··········································································11分

由①②得a=c=313分

16.（1）连接CA交BD于点M，连接MN.

因为ABCD为正方形，M为AC中点，·····························································2分

又因为N为PC的中点，

所以MN为ΔCPA中位线，MN//AP.································································4分

又因为MNC平面BND，AP丈平面BND，

所以AP//平面BND．····················································································6分

（2）（方法一）因为PD丄平面ABCD，AD，DCC平面ABCD，所以PD丄DA，PD丄DC，

在正方形ABCD中，DA丄DC，

所以以{,,}为正交基底建立空间直角坐标系O-xyz，······························7分

因为PD=AD=3，

所以C(0,3,0)，B(3,3,0)，P(0,0,3)，

所以=(3,3,-3),=(0,3,-3).·····································································8分

设平面PBC的一个法向量为=(x,y,z)，

所以·································································9分

解得x=0，取y=1，得z=1，所以=(0,1,1)，···············································11分

又平面PAD的一个法向量为=(0,1,0)，·························································13分

所以cos

2

所以平面PAD与平面PBC夹角的余弦值为．···············································15分

2

（方法二）因为PD丄平面ABCD，ADC平面ABCD，CDC平面ABCD，

所以PD丄AD,PD丄CD.············································································8分

因为底面ABCD为正方形，所以CD丄AD.

又因为DCC平面PCD，PCC平面PCD，DC∩PC=C,

所以AD丄平面PCD.···················································································10分

在平面PAD过P作lAD，有l丄平面PCD,

PDC平面PCD,PCC平面PCD,所以PD丄l,PC丄l,LDPC为平面PAD与平面PBC夹角

或其补角，································································································12分

因为底面ABCD为正方形，所以AD=CD=3，又PD=3，PD丄CD，故LDPC即为所求

又LDPC···························································································14分

平面PAD与平面PBC夹角的余弦值为·····················································15分

（）设等差数列的首项为，，则前项和为

17.1{an}a1公差为dnSn=nad.

所以

即，··············································································3分

a1+3d

解得a1=3，d=2，·····················································································4分

所以an=3+(n-1)×2=2n+1.

因此数列{an}的通项公式为an=2n+1·····························································6分

n+1

（2）cn=anbn=(2n+1).2······································································7分

234n+1

Tn=3×2+5×2+7×2++(2n+1)×2，

345n+2

2Tn=3×2+5×2+7×2++(2n+1)×2···············································9分

34n+1n

所以Tn-2Tn=12+2×2+2×2++2×2-(2n+1)×2··························11分

×2n+2=2n+2-4，···································13分

n+2

即__Tn=(1_2n)2_4··············································································14分

n+2

所以Tn=(2n_1)2+4···········································································15分

18.（1）当k=1时，赛制为三局两胜制，故X的可能取值为2,3，

P(X

P=C················································································2分

所以X的分布列为：

X23

P5/94/9

E···············································································4分

（2）①因为每局比赛中，机器人获胜的概率为p，

由题可知P(1)为3局2胜制时，机器人获胜的概率，机器人获胜的情形有两种：2:0或2:1，

21222(|)

所以P(1)=p+C2p(1_p)=p(3_2p)=（）×3_2l=，··························6分

(丿

P(2)为5局3胜制时，机器人获胜的概率，机器人获胜的情形有三种：3:0或3:1或3:2，

32323232

P(2)=p+C3p(1_p)+C4p(1_p)=p(6p_15p+10)

··········································8分

所以P(2)>P(1)，

所以k=2时，5局3胜制对机器人更有利···························································9分

②随着k的增大，机器人获胜的可能性越来越大.

证明如下：

2k+1_i

由①可知，P=（用全场打满的方法算）·························10分

i

下面讨论2k+3局与前2k+1局的递推关系：（用先打2k+1局，再打2局的方法）

（i)若前2k+1局中机器人恰好赢了k局，则后两场机器人都要赢才能获胜，

kkk+12kk+2k+1

其概率为C2k+1p(1_p).p，即C2k+1p(1_p)···········································11分

（ii)若前2k+1局中机器人恰好赢了k+1局，则后两场机器人至少要赢一场才能获胜，

k+1k+k+2k

其获胜概率为Cp即C2k1p(1_p)(2_p)··················12分

（iii）若前2k+1局中机器人至少赢了k+2局，则后两场机器人无论输赢都获胜，

i2k+1_i

其获胜概率为k+1p································································13分

i

21

kk+2k+1k+k+2kii2k+1_i

：P(k+1)=C2k+1p(1_p)+C2k1p(1_p)(2_p)+C2k+1p(1_p)

i=k+2

kk+2k+1k+k+2kk+k+1k

：P(k+1)_P(k)=C2k+1p(1_p)+C2k1p(1_p)(2_p)_C2k1p(1_p)

kk+1k+1

=C2k+1p(1_p)(2p_1)，·········································································15分

kk+1k+1

p=,2×_1>0：C2k+1p(1_p)(2p_1)>0，即P(k+1)>P(k).················17分

19.解（1）已知f(x)=2xlnx，对其求导可得f'(x)=2(lnx+1)，令f,(x)=0，解得x

当x变化时，f,(x)，f(x)的变化情况如下表：

(|1

x0,l

(e

f,(x)-0+

f(x)单调递减极小值单调递增

·······················3分

当x，f(x)<0，又f(1)=0，f(e)=2e，

则不等式0≤f(x)≤2e的解集为····················································5分

（2）由题意g(x)=2xlnx_4x+2a的定义域为(0,+∞)，且g,(x)=2(lnx_1