山东省百校大联考2024-2025学年高三上学期12月学情诊断 数学答案

22绝密★启用前anan12(n1)12n1,22an1an22(n2)12n3,2025届高三上学期学情诊断22a3a22215,2024.12数学答案及评分标准22a2a12113,(n1)(32n1)将以上各式相加，得222，ana135(2n3)(2n1)n1一、选择题：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，只有一个选项是正确2将代入上式即得22，且当时也成立，所以22，的．请把正确的选项填涂在答题卡相应的位置上．a11annn1ann二、又因为数列为正项数列，所以．分anann(nN)·····································································61．A2．C3．B4．B5．D6．C7．A8．C由（）可得nn，令n，其前项和为，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（2）1bn(1)n3cn(1)n2nT2n部选对得6分，部分选对的得部分分，选对但不全的得部分分，有选错的得0分．则T2n1234(2n1)2nn，··············································································9分2n2n2n1．．．122n3(13)3(31)339AC10BCD11ACD又因为333，························································12分1322三、填空题：本题共3小题，每小题5分，共15分．32n13所以Sn．···························································································13分2n21π16．【解析】正弦定理+最值12．13．14．43，[2,6]29bcosC3bsinCsinBcosC3sinBsinC（1）根据正弦定理，1可化为1，1分ac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．sinAsinCsinBcosC3sinBsinCsinAsinC15．（1）【法一】构造常数列sinBcosC3sinBsinCsin(BC)sinC2222222222sinBcosC3sinBsinCsinBcosCsinCcosBsinC由an1an2n1=(n+1)n(nN)，a11，可得an1(n+1)anna110，sinC(3sinBcosB1)0.故数列22是恒为的常数列，所以22，分ann0ann·······································································5·············································································································································5分π1又因为数列为正项数列，所以分因为C(0,π)，所以sinC0，故有3sinBcosB10，进而有sin(B)，因为B(0,π)，anann(nN).·······································································662ππ5ππππ【法二】累加法所以B(,)，故有B，所以B．································································7分666663由题意得：n2且nN，有2ππ0ACπ32ππ（2）因为B，所以，进而有C．·················································8分3π620C2数学试卷答案第1页（共9页）{#{QQABZQYAogAoAAIAARhCEwFSCkGQkgCACagOBEAEoAABiRNABAA=}#}11ab1cos|cosn,n|1222由正弦定理可得，tantan122πsinA3sinCtantan()1221132π3113所以有sin(C)cosCsinC，221sinA32213cosCtan212tan1atan2sinCsinCsinC22sinCtan1π当且仅当2，即时等号成立，分tan2·························································14所以13313cosC313，分tan4SacsinBa()()·······································11ABC24422sinC422tanC3即平面AEFG与平面ABCD所成角的余弦值的最大值为．·········································15分ππ3因为，所以3，所以33133，分CtanCSABC()··························146238422tanC218．【解析】（1）显然yf(x)的图象经过(2,0)，当x0时，y2，所以f(x)的图象经过33的所有定点的坐标为(2,0)和(0,2)················································································1分所以ABC面积的取值范围是(,)．·················································································15分82xxx由题知f(x)eax(x2)(ea)(x1)e2a，····························································2分17．【解析】直线与平面所成的角，平面与平面所成的角，线面平行的判定定理和性质定理，若以(2,0)为切点，f(2)e22a，切线为y(e22a)(x2)；最值问题若以(0,2)为切点，f(0)2a1，切线为y(2a1)x2；············································4分（1）由BD//l，BD平面AEFG，l平面AEFG可得，BD//平面AEFG，····················2分（注：上述两条切线写出一条即可）分再由BD平面BB1D1D，平面BB1D1D平面AEFGEG，所以BD//EG，·······················4（）①当时，x恒成立，又因为BE//DG，所以四边形BDGE为平行四边形，所以BEDG．·······························5分2a0e2a0在正四棱柱中，均垂直于平面，所以直线、与平面ABCDA1B1C1D1BB1,DD1ABCDAEAG所以当x1时,f(x)0,f(x)在(,1)单调递减，ABCD所成的角分别为EAB,GAD，即EAB,GAD，····································6分当x1时，f(x)0，f(x)在(1,)单调递增；··················································5分又因为BEDG，ABAD，所以tantan，从而；········································7分②当a0时，由f(x)0，得x1或xln2a．············································6分（）y122以A为坐标原点，分别以AB,AD,AA1的方向为x轴、轴、z轴的正方向建立空间直角坐e标系．当ln(2a)1，即a时，f(x)0恒成立，则f(x)在R上单调递增，········································7分2e则E(1,0,tan)，G(0,1,tan)，所以AE(1,0,tan)，AG(0,1,tan)，····························9分当ln(2a)1时，即a时,当x1时，f(x)0,f(x)在(,1)单调递增;2AEn1xztan0当1xln(2a)时，f(x)0，f(x)在1,ln(2a)单调递减；设平面的法向量，则有，令，则有AEFGn1(x,y,z)z1AGn1yztan0当xln(2a)时，f(x)0，f(x)在ln(2a),单调递增；··················································9分xtan,ytan，所以，11分n1(tan,tan,1)·································································e当ln(2a)1时，即0a时，2平面的法向量，平面与平面所成角为，则ABCDn2(0,0,1)AEFGABCD当xln(2a)时，f(x)0，f(x)在，ln(2a)单调递增；数学试卷答案第2页（共9页）{#{QQABZQYAogAoAAIAARhCEwFSCkGQkgCACagOBEAEoAABiRNABAA=}#}11当ln(2a)x1时，f(x)0，f(x)在(ln(2a),1)单调递减；令g(x)x(lnxx1)，g(x)lnx2x，exex当x1时，f(x)0，f(x)在1,单调递增；·················································11分易知g(x)在(0,)上单调递增，x时，g(x)；x0时，g(x)，所以存在1综上所述：当a0时，f(x)在(,1)单调递减，在(1,)单调递增；t(0,)，使得g(t)lnt2t0，即etlnt2t即et(t)lntt，ete当0a时，f(x)在，ln(2a)单调递增，在(ln(2a),1)单调递减，在1,单调递增；xt12令G(x)ex，则G(t)G(lnt)，因为G(x)在R上单调递增，所以lntt，即e．te当a时，f(x)在R上单调递增；2当x(0,t)时，g(x)0，g(x)单调递减；当x(t,)时，g(x)0，g(x)单调递增．e当a时，f(x)在(,1)单调递增，在1,ln(2a)单调递减，在ln(2a),单调递增．···············12分1222所以g(x)g(t)tlnttt0，所以f(x2)(lnxx1)e0．················