广西壮族自治区2025年4月高三毕业班诊断学考试数学试卷及答案（广西三模）

数学参考答案一、单选题:log2a3+log2a11=log2a3a11=4,选C.取第五位数据2.21，选B.4.解：∵l的方向向量为=(2,1)，则l斜率为，因为直线与l垂直，所以斜率为－2又∵过A点，所以直线为2x＋y－1＝0.选D.5.解：以AB,AD,AA1为基底，则DEQ\*jc3\*hps23\o\al(\s\up2(–),C)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),C)EQ\*jc3\*hps23\o\al(\s\up2(–),1)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),A)EQ\*jc3\*hps23\o\al(\s\up2(–),1)2-4x-3>0开口向上有解，:f(x)=f(2-x)→f(x)关于x=1对称.:f(x)关于x=1对称.→f(x)在x<1为减函数二、多选题通项公式为Tk+1=CEQ\*jc3\*hps16\o\al(\s\up4(k),6)26k(1)kx62k,所以6-2k=0,k=3则第四项为常数项，B错二项式系数最大项为中间项第四项，所以为CEQ\*jc3\*hps16\o\al(\s\up5(3),6)=20，C对选ACD.10.解：递推公式取倒得构造新数列得得到为公比为3，首项为3的等比数列→→则A对，C错.选ABD.22三、填空题)13.解：正四面体内半径最大的球为内切球兀R2=2兀N*kk-1-k(1-μ)k-1=k[μk-1-(1-μ)k-1]所以即则恒成立，只需2λ2-tλ+2大于的最大值。可得bk+1<bk,则bk的最大值为b1=1λλ|(λ2,λλ|(λ2,四、解答题:b2=(a+c)2-ac:b2=又据余弦定理b2=a2+c2-2accosB得，········································································2分:B=2EQ\*jc3\*hps22\o\al(\s\up6(兀),3)·······························································3分2-ac·····························································4分分:CΔABC≤3+2解得a=c=2······················································································11分:BD丄AC,又AD=:据勾股定理得BD=1······································································13分分(2)由题可知X=0,20,40,60·········································································5分P······································································6分分甲总得分的分布列：X0204060P1 4············································································10分分所以甲获胜概率更大···············································································15分:E、F、N三点分别为PB、AO、PA的中点:在平面PAO中，NF//PO,又:NF丈平面POC，PO平面POC:NF//平面POC 2分同理，EN//平面POC·····································································4分又:NF∩EN=N，NF平面ENF，EN平面ENF，所以平面ENF//平面POC:EF平面EFN:EF//平面POC··························································6分建立以BG,BC,BQ分别为x,y,z的空间直角坐标系.A,C,P假设在PC上存在点M使得OM丄AB,设M(x,y,z)EQ\*jc3\*hps15\o\al(\s\up4(→),CM)EQ\*jc3\*hps15\o\al(\s\up4(→),CP)→Mλ,2-3λ,2λEQ\*jc3\*hps17\o\al(\s\up0(→),BA)EQ\*jc3\*hps17\o\al(\s\up4(→),OM)λ-1,-3λ,2λ)5EQ\*jc3\*hps15\o\al(\s\up4(→),BA)EQ\*jc3\*hps15\o\al(\s\up4(→),OM)5EQ\*jc3\*hps18\o\al(\s\up12(→),BM)EQ\*jc3\*hps18\o\al(\s\up12(→),n)设直线BM与平面PAO所成角为θ,则分可得椭圆方程：1···································································3分（2）解:(ⅰ)设直线PF的方程为：x=my+,点Q(x1,y1),R(x2,y2)，则x2(ⅱ)①当直线PF斜率为0时，不妨设R(-2,0),Q(2,0),P(t,0)②当直线PF斜率不为0时，设P(x0,y0)，由(ⅰ)得-2··························14分所以点P在定直线l:x=上，MF平行直线l，点P到直线MF的距离 综上可知，△PFM的面积为·······························································17分·················································································2分f,,(x)<0；所以f(x)为(0，+∞)的凸函数··············································3分所以f在为“凹函数”···················6分2n时，等号成立:h(x)最小值为····································