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·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.20B.120C.60D.1354.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.0.845.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.186.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(),A.(-2,-1)u(1,2)B.(1,0)u(1,2)C.(-3,-2)u(0,1)D.(1,0)U(0,1)A.0.2B.0.3C.0.4D.0.5则实数a的取值范围为()高二数学第1页共4页二、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,有多项符高二数学第2页共4页合题目要求,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1B.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D.已知px23.8410.05),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0.0510.已知函数s(x)=-4A.函数f(x)有2个极值点B.函数(x)无最小值2211.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256B.第8行所有数字的平方和等于C16C.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3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;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4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(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.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;一．选择题123456789ADCBDBCBBCDADABD二．填空题3三．解答题15.解1）由bcosA+3bsinA-c-a=0及正弦定理得因为sinC=sin(π-A-B)=sin(A+B)=sinAcosB+cosAsinB，由于sinA≠0，:3sinB-cosB-1=0···························································4分所以sin·························································································5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，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所以以{,,}为正交基底建立空间直角坐标系O-xyz，······························7分所以=(3,3,-3),=(0,3,-3).···········································设平面PBC的一个法向量为=(x,y,z)，所以cos所以平面PAD与平面PBC夹角的余弦值为2．···············································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,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17.（1）设等差数列{an}的首项为a1，公差为d，则前n项和为Sn=nad.所以+3d，··············································································3分解得a1=3，d=2，·····················································································4分因此数列{an}的通项公式为an=2n+1·····························································6分n+1······································································7分T234n+1，345n+2·34n (1_2n)2n+2_4··············································································14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(XP=C················································································2分所以X的分布列为：X23P5/94/9（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，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，+Cp3(1_p)+Cp3(1_p)2=p3(6p2_15p+10)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：由①可知，P=i2k+1_i（用全场打满的方法算）·························10分下面讨论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，（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，（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，k+1pk+2(1_p)k+1+C1pk+2(1_p)k(2_p)+21Ck+1pi(1_p)2k+1_i：P(k+1)_P(k)=Ck+1pk+2(1_p)k+1+C1pk+2(1_p)k(2_p)_C1pk+1(1_p)kk+1pk+1(1_p)k+1(2p_1)，········