2024年河南省五市高三第一次联考（数学）参考答案一、选择:题号12345678答案CCCBDABA二、选择:题号9101112答案ABDBCDABBD三、填空:23913.114.6015.101216.13四、解答题:17.(Ⅰ)由条件b2a2ac，根据正弦定理可得sin2BAACsin2sinsin...........................................1分即cos2ABACcos22sinsin.......................................2分所以2sin(ABABAC)sin()2sinsin..............................3分也即sin(BAA)sin...............................................4分从而证得BA2...................................................5分(Ⅱ)若ABC为锐角三角形，根据(Ⅰ)BA2BAA2222则CABA322得A......................................................6分64sin(CABABABAA)sinsin()sinsin4sin2式子sinAAAsinsin数学参考答案第1页共7页{#{QQABZYQAggiAAoBAAAgCAQnwCEIQkAACACoOgBAEMAAACRNABAA=}#}sin(3AAAA)sin(3)2cos3A.....................................................................8分sinA32由3AAcos3(,0)242sin(CAB)sin因此2cos3A(2,0)即为所求.........................................10分sinAa18.(Ⅰ)根据条件则f'(x)2x(x0)...............................1分x当a0时，f'(x)0在定义域(0,)内恒成立，因此f()x在(0,)递减；...............................2分2a当a0时，由f'(x)0，解得0x；22af'(x)0，解得x.................................3分2因此：当a0时，f()x的单调减区间为(0,)，无增区间；2a2aa0时，f()x的单调减区间为(,)，增区间为(0,)；..........4分222a注：区间端点x处可以是闭的2(Ⅱ)若f()x有两个零点，有(Ⅰ)可知a02a且f()()xf22a2a2a则必有f()aln()2a0.............................6分222a即ln1022解得a........................................................8分e11又因f()0,f(4a)aln4a16a2aa(ln4a16a1).......................9分ee28114t即g()lntt41(tt4a)g()'t4ett数学参考答案第2页共7页{#{QQABZYQAggiAAoBAAAgCAQnwCEIQkAACACoOgBAEMAAACRNABAA=}#}11可得g(t)g()ln110，448也即得g(t)0在t(,)恒成立....................................................................10分e12a2a从而可得f()x在(,),(,4a)区间上各有一个零点...............11分e222综上所述，若f()x有两个零点实数a的范围为(,)..................12分e(Ⅰ)记等差数列的公差为，则由条件19.anda3a4a11a12da13da110d3(a15d)3a684得.........................................................2分a628从而da7a633285.................................................3分ana6(n6)d5n2即为所求...................,..........................4分(Ⅱ)对任意，由k2k，即k2k..................5分nN5an555n2522整理得5k1n52k1.........................................6分55故5k11n52k1，...............................................8分从而得2k1k1...............................................9分bk55则的前项和为bkk5(25k1)1(5k1)T..........................................10分k251511(52k165k1)(kN).......................................12分2420.(Ⅰ)取BC中点为E,由条件则OE为梯形ABCD的中位线，数学参考答案第3页共7页{#{QQABZYQAggiAAoBAAAgCAQnwCEIQkAACACoOgBAEMAAACRNABAA=}#}则OEBC.............................................................................................................1分又PBPC,则PEBC且PEOEE,根据线面垂直的判定定理可得BC面POE.............2分得BCPO.......................................................3分又由PAPD,则POAD,AD,BC为梯形的两腰，则AD与BC相交即可得PO面ABCD，.............................................5分又OC面ABCD,进而得COPO...................................6分1(Ⅱ)取CD的中点为Q,由ABCD1,CDA60,2则AQCD,ADCD2QD2,因此ACD为等边三角形,COAD由(Ⅰ)知PO面ABCD，OPOAOC..............................7分如图，分别以OC,,OAOP分别为x,,yz轴正方向，建立空间直角坐标系由CDDAPAPD2,CDA60,则OPOC3,33ABPD(0,1,0),(,,0),C(3,0,0),(0,0,3),(0,1,0)22123又由DM2MPM(0,,)3333得PC(3,0,3),BC(,,0)22423AC(3,1,0),AM(0,,).....................................................................8分33设平面的一个法向量为PCBn1(,,)abc3a3c0nPC0由133n1BC0aa022取，得，得分a3b1,c3n1(3,1,3).......................................................9同理可得平面的一个法向量为分ACMn2(1,3,2)............................................10数学参考答案第4页共7页{#{QQABZYQAggiAAoBAAAgCAQnwCEIQkAACACoOgBAEMAAACRNABAA=}#}记平面PCM与平面ACM所成的角的大小为，nn311323则cos|12|.......................................................11分|n1||n2|7842........................................................................................................12分721.(Ⅰ)由条件，每次抢题+答题，甲得1分的概率为131111P甲..............................................1分252220119每次抢答题乙得1分的概率为PP乙1甲1...................2分2020若第二题答完比赛结束，则前两次答题甲得2分或者乙得2分，因此第二题发完119101比赛结束事件发生的概率P（）（）22.................................................4分2020200(Ⅱ)根据题意，竞赛结束时抢答题目的总数X的所有可能取值为2,4,6,8，，，2n(nN)...........................................5分记pnP(X2n)101由(Ⅰ)知，当X2时，pP(X2)，120099且pp()nN.............................................7分n1200n则X的分布列可表示为：X246……2n……Pp1p2p3……pn……...............................................8分∴EX2p14p26p38p42npn101999999994p6p8p2np100200120022003200n110199(4p6p8p2np(2n2)p)100200123n1n10199[(246ppp2np)2(pppp)]100200123n123n10199(EX2)...........................................11分100200400解得EX....................................................12分101数学参考答案第5页共7页{#{QQABZYQAggiAAoBAAAgCAQnwCEIQkAACACoOgBAEMAAACRNABAA=}#}22.(Ⅰ)根据条件则2a6,a3,当点D位于短轴顶点时，DFF面积的最大，且(S)bc22，12DF1F2max也即a2b2c29,bc22........................................2分解得b22,c1或b1,c221又e，因此a3,