山西省三重教育联盟2023-2024学年高三上学期九月联考数学答案

＿、12345678DACCBABD9101112ACABDACDBCD＿＿、＿填空攫213.8;14.34;15.(16,625];16.y=——(x—4);3四、···············17.-·.----一�.-./!分32·:BE(0,兀），:.sinB*0,..smA=Slll(A——)＋—,..............·2分321.fj.fj:.sinA=—sinA——cosA+—,···············3分222'3'2兀冗4兀·:AE(0,兀），:.A+-E(—,-),..............·4分333冗2兀冗:.A+—=—，解得A=..............·5分333冗3./3A=—，且MBC的面积为，(2)·:32.1../33./3..-bcsmA=—be=,前牟1叶bc=6,..............·6分242..16=b2+c2-2bccosA=b2+c2-be=b2+c2-6,:.b2+c2=22...............·7分'''·:D为BC的中点，:.2AD=AB+AC,..............·8分－一2一2---2两端同时平方，得4AD=AB+2AB-AC+AC,－一222...............:.4AD=c+2bccosA+b=28,9分···············解得m了=7,:.AD的长为打．10分参考答案（数学）第1页共6页{#{QQABAQCAogAgAABAAQhCAQFwCEAQkBGAAIoOBBAEMAIAARNABCA=}#}18.,...............·2分10a1+45d=145,27——2—..............·3分:.2a1+=29,2a广29a1+27=0(或9d29d+6=0),a1—27a1=,I2或\···············4分d=32d=—,92239a=—n+,n918l..............·6分=n-n-2丸3=2bn2n-5'31n-(2)由题e=(3n—2)·(—)l,...............7分n31121n-1:.兀=lx1+4x—+7x(—)+···+(3n—2)·(—)，...............8分3331112131-11n:.—T=Ix—+4x(—)+7x(—)+···+(3n—5)·(—)n+(3n—2)·(—),...............9分3n33333211131:.—T=1+3x(—+(千＋（—）+···+(丿尸）—(3n—2)·(-t···············10分3n3333311(—)n1551=1+3·33—(3n—2)·(—)n=——(3n+—).(—)n···············11分1l—32233153n51曰:.Tn=——(—+—)·(-)···············12分424319.(1)证明：？四边形BDEF为矩形，:.BF_l_BD,···············1分又·:AC_l_BF,AC与BD相交，..............·3分:.BF_l_平面ABCD;···············4分(2)解：过A作BF的平行线AH,则AH_l_平面ABCD,以A为原点建立空间直角坐标系A—xyz,如图所示．zFycX参考答案（数学）第2页共6页{#{QQABAQCAogAgAABAAQhCAQFwCEAQkBGAAIoOBBAEMAIAARNABCA=}#}不妨设BF=h(h>0),..............·6分则A(0,0,0),B(4,0,0),C(4,4,0),D(0,4,0),G(2,2,h),••:.BG=(—2,2,h),BC=(0,4,0),一设平面CBG的个法向量为n=(x,y,z),—2x+2y+hz=0,-...............7分4y=O,•••.....•......8分·:AC..lBF,AC..lBD,BF门BD=B,BF,BD已平面BGD,一•••.....•......9分:.ACl_平面BGD,平面BGD的个法向量为飞三=(4,4,0),一，°ln·ACI4hI...............分由题意得cos60=＝—,解得h=2,111�11兀五言]二21此时—S16—...............分V=3战CD·h=3.1220.(1)解：根据题意，得列联表如下：比较关注不太关注总计男240160400女15050200..............·2分总计390210600零假设为H。：性别与对新能源汽车关注度无关联，..............·3分X2由表中的数据计算得：2600(24050—160xl50)..............·4分X=�13.187>6.635,400X200X390X2102：．依据小概率值a=O.OI的x独立性检验，推断H。不成立，即认为性别..............·6分与对新能源汽车关注度有差异，此推断犯错误的概率不大于0.01;(2)根据(1)可知男女比例为2:1,故9人中女性的人数为3人，男性为6人，:.X的可能取值为0,I,2,3.···············7分3sc2-c11sC63P(X=0)=二＝—，P(X=1)==—c:21c:28'c1.c23633C1P(X=2)==—,P(X=3)=一二，c:14c:84:.x的分布列如下：X。12351531...............11分2128148451531分E(X)=Ox—+Ix—+2x—+3x—=1................1221281484参考答案（数学）第3页共6页{#{QQABAQCAogAgAABAAQhCAQFwCEAQkBGAAIoOBBAEMAIAARNABCA=}#}分—3l.)（或因为X服从超几何布，所以E(X)=3x9=(1)21.解：设M(xi,y1),N(x2,y2),G(x。,y。)，则X1*还，x。*o,2222..31+a1=3a．�xy,2,__2__2')__22:.3(x广—矿）+a(y/—卢）=0'···············1分Y1-Y2.Yi+Y2Y1-Y2.2y。=Yi-Y2.Y。=.3整理得：=kkOG=-2'..............·2分x1-x2x1+x2x1-x22x。x1-x2x0a•2—33•;koG=—，即koG·k=——,••a=4,···············3分4k4=：．椭圆C的方程为王－＋上-1;..............·4分43(2)由题P(—2,0),Q(2,0),设直线MN的方程为y=kx+m,···············5分y=kx+m2—.°.—8km.4m12...............7分Xl+X2=2,X1X2='24k+34k+3—1•:丸·k2=—,4．．Y1.Y2_(kx1+m)(从+m)X1+2X2+2(x1+2)(x2+2)22_kX1X2+km(xl+X2)+mX1X2+2(x1+x2)+42—22k24m12—8km222+m4k+34k+32—4m12—16km22+44k+34k+3l-42—23m12k，＝＝224m—16km+16k解得m=2k(舍去）或m=-k,..............·9分22—8k.4k12..X1+X2='2X1X2=2．4k+34k+3——·:QM=(x12,Yi),QN=(x22,y2)'——:.QM·QN=(x12)(x22)+y心——2——=(x12)(x22)+k(x1l)(x21)参考答案（数学）第4页共6页{#{QQABAQCAogAgAABAAQhCAQFwCEAQkBGAAIoOBBAEMAIAARNABCA=}#}2—22=(k+l)X1X2(k+2)(x1+x2)+k+42222(4k—12)(k+1)8k(k+2)2=—+k+44k2+34k2+3222222(4k—12)(k+1)—8k(k+2)+(k+4)(4k+3)4k2+32—k5<0'···············11分4k2+3:．点Q在以MN为直径的圆内．...............12分x22.前牟：0)·;f(x)=ae(I+lnx),aexxlnx+x+I:.f'(x)=aex(I+lnx)+=aex·...............·1分XX令t(x)=xlnx+x+l,则t'(x)=lnx+2................2分1·:t'(x)单调递增且t'(下)=0'...............3分e1:．当店(0,飞－）时，t'(x)<0,t(x)单调递减，e1当XE(—+oo)时，t'(x)>0,t(x)单调递增，..............·4分e2'11:.t(x)�t()=1—>0'···············5分了e了e又·:a>0,扩>0,x>O,:.f'(x)>0,即f(x)在定义域(0,+oo)上单调递增；•••.....•......6分Inx+II—Inx设rp(x)=,则叭—)=0'且例(x)=..............·7分(2)XeX:.当XE(O,l)时，叭x)>0,rp(x)单调递增，当XE(l,+oo)时，砑(x)<0,rp(x)单调递减，11且当XE(0,—)时，叭x)<0,当XE仁，+oo)时，叭x)>0................8分eex2由题g(x)—f(x)>0,即ae(I+Inx)rp(x)对任意的XE(O,—)恒成立．e参考答案（数学）第5页共6页{#{QQABAQCAogAgAABAAQhCAQFwCEAQkBGAAIoOBBAEMAIAARNABCA=}#}x1XI:•ae>X对任意XE(O,-)恒成立，即a>—对任意XE(0,—)恒成立．...............10分eXexI1—X设G(x)=,xE(0,-),则G'(x)=>0,eXeeX1:.G(x)在(0,—)上单调递增，e11:.G(X)