【全国甲卷】四川省凉山州2024届高中毕业班高三年级第三次诊断性检测(凉山三诊)(5.9-5.10)

凉山州第三次诊断性考试理科数学参考答案一，选择题（每题5分，共60分）：1-5：BDADB6-10：CCAAC11-12：CC12题解析：令xy0f(0)0,xy1f(1)0，①对；f(x)x2x2x1x2(0,),g(x)g(xy)g(x)g(y)，g(x2)g(x1)g(x1)g()xx1x1xf(2)xx21对；当时由知成立，当时，由g(x2)g(x1)g()0②x0①③x0x1x2x1②g(xn)g(x)g(xn1)g(xn)g(xn1)g(x)g(xn)g(x)(n1)g(x)f(xn)f(x)nn1ng(x)nnf(x)nxf(x)xx③正确.11由①得f(2)2f()f(1)0f(2)2由③得22f(xn)f(2n)nf(2i)nxn1f(x)2n1f(2)2i1f(2)2n12得④错.nni1ii1二，填空题（每题5分，共20分）【答案】114.【答案】315【答案】(1，1)16【答案】2e2etlnt16题解析：t为函数零点ett2lnt0tet2etlntt3lnt0tetetex(x2)exet(t2lnt)t3lntlntlnt.令g(x),g'(x)g(x)在(0,2)t2t2x2x3e21et1(2,)g(x)g(2)即.令h(x)xlnx,h'(x)lnx10min4et2e11etet2t3t3xh(x)在(,),h()h(t)tett32e2eeet2t2et1et三，解答题（共70分，17题10分，18-22每题12分）17(12分)解：根据扇形统计图易得选择物理类学生为1000(48%24%18%)900人，87其中男生900480人，女生900420，选择历史类100人，其中男生15152310040人，女生10060人55男生女生合计物理类480420900历史类4060100合计5204801000..............................................................................................................................................3分n(adbc)21000(4806042040)2250K26.4106.635(ab)(cd)(ac)(bd)90010052048039所以没有99%把握认为“该校学生选择物理类与性别有关”.................................6分(2)按照分层抽样选择物化生、物化政及物化地人数分别为8人，4人，3人，X0，1，2.C0C222C1C112C2C01312312312分P(X0)2,P(X1)2,P(X2)2..............10C1535C1535C1535所以X分布列如下：X012P2212135353522121E(X)0120.4...............................................................................12分353535解：取18(1)CC1中点Q，BF中点M，连接EM,MQ,QH.FMGQ1,FM//GQ四边形MQGF为平行四边形MQ//FG...①..................................................................3分又HQ//DC,DC//AB,AB//EMHQ//EM四边形EMQH为平行四边形EH//MQ...②由①②得EH//FGE,F,G,H四点共面，即点H在平面EFG中....................................................................................................................................6分以为坐标原点，为轴建系如图易得(2)DDA,DC,DD1x,y,z,E(2,0,1),P(0,0,3)，设平面与平面法向量分别为B1(2,2,4),G(0,2,3)PEGB1EGm(x,y,z),n(m,n,p)xz0x1mEP0EP(2,0,2),EG(2,2,2),EB1(0,2,3)xyz0y0mEG0令x1z1m(1,0,1)...................................................................................................8分2n3p0m1mEB10mnp0n3n（1,3,-2）...........................10分mEG0令n3p27设二面角PEGB平面角为则coscosm,n1147二面角PEGB的余弦值为......................................................................12分11413*19.解：（1）记数列{a}前n项和为S，则顶点B坐标为(Sa,a)(n2,nN)nnnn12n2n31在函数yx上aSa.....................................................................3分2nn12n321*321*anSn1an(n2,nN)an1Snan1(nN)42...①42...②31②-①得(aa)(aa)(aa)(n2,nN*)aa04n1nn1n2n1nn1n2aa(n2,nN*)............................................................................................6分n1n313213第一个等边三角形顶点B(a,a)代入yx得a，B(aa,a)代入12121132122224222yxaaa.故{a}是以a为首项为公差等差数列，23213n133222a(n1)n.............................................................................................8分n333（2）由（1）得1919111911111Tn(...)(1...)anan14n(n1)41223n(n1)4223nn1919n(1)4n14n4........................................................................................................12分yyy220.解:(1)设动点P(x,y),kk3x21(x1).................4分PA2PA1x1x1333(2)易知直线AB斜率不为0.设AB方程为xty2,且t(,).设A(x,y),3311B(x2,y2).xty2y2(3t21)y212ty90(3t210)x21312t9yy,yy，36(t21)0....................6分1213t21213t23(x1)3(x1)由题意易得kk3k2直线BA方程为y2(x1).....①BA1BA2BA11y2y2y直线方程为1分AA2y(x1)......②............................................................................8x11由①②得9x1yyyyyy212121213tx13(x1)(x1)3(ty1)(ty1)3(t2yyt(yy)1)9t212t2121212123(1)13t213t29113x点M横坐标为定值......12分3(9t212t213t2)22备注：非对称式处理方式比较多，此处只提供利用第三定义转化回避非对称式，整体代换，半配凑，硬解方式处理非对称式均给满分.21.解:f'(x)(2x1)(exm)..........................................................................................1分11当m0时f'(x)(2x1)(exm)0x函数f(x)在(,)单调递减，在221(,)单调递增；........................................................................................................2分21当m0时，f'(x)(2x1)(exm)0x或lnm21e1若lnm即0m时,f'(x)(2x1)(exm)0xlnm或x2e211函数f(x)在(,lnm)单调递增,(lnm,-)单调递减，(,)单调递增；......3分22e若m时，f'(x)(2x1)(exm)0函数f(x)在(,）单调递e增...............................................................................................4分1e1若lnm即m时，f'(x)(2x1)(exm)0x或xlnm.2e211函数f(x)在(,-)单调递增,(-,lnm)单调递减，(lnm,)单调递增；..........5分2211综上：当m0时，f(x)在(,)单调递减，在(,)单调递增；22e11当0m时,函数f(x)在(,lnm)单调递增,(lnm,-)单调递减，(,)e22单调递增；e当m时，f'(x)(2x1)(exm)0函数f(x)在(,）单调递增；ee11当m时，函数f(x)在(,-)单调递增,(-,lnm)单调递减，(lnm,)单调递e22增；........................................................................................................................................6分(2)由（1）知当m0时函数f(x)至多两个零点，不满足条件.e当m时，函数f(x)至多一个零点，不满足条件；ee11当0m时函数f(x)在(,lnm)单调递增,(lnm,-)单调递减，(,)e22单调递增,f(lnm)mlnm(1lnm)0，函数f(x)至多一个零点，不满足.........