2023—2024学年第一学期期末检测高三数学参考答案2024.011．B2．A3．B4．C5．D6．A7．A8．C9．BC10．AD11．AC12．ABD1613．814．315．(0,]16．27ab17.【答案】(1)在△ABC中，由正弦定理得：，sinsABin又因为ab4，所以sinABBAC4sin4sin()4sin()，13又因为C，所以sin4sin()4(sincos)2sin23AAAAAAcos，3322所以sin23cosAA，·················································································3分因为A(0,)，所以sin0A，所以cos0A，sinA所以tan23A．·············································································5分cosA(2)方法一：在△ABC中，由余弦定理得：cababC2222cos，又c1，ab4，C，所以11624cosbbbb22，331解得b2，······························································································8分131133所以SabCbbbsin432．·············································10分△ABC22213sinA12方法二：由(1)知23，又sincos122AA，解得sin2A.cosA13ac在△ABC中，由余弦定理得，sinsinACcA22sin16所以a2，················································································8分sin132C11a333所以SabsinCaa2.···············································10分△ABC2242161318.【答案】(1)因为aabnnn13，babnnn13，所以abababnnnnnn11444()，······························································2分又a13，b11，所以ab1140，所以abnn各项均不为0，························3分ab所以nn114是常数，abnn所以数列abnn是等比数列．·······································································5分n(2)由(1)知，abnn4.①··········································································6分方法一：因为an13anbn，babnnn13，所以an11bn2an2bn2(anbn)，······························································8分又a13，b11，所以ab1120，所以abnn各项均不为0，ab所以nn112是常数，abnn第1页（共5页）{#{QQABQQCAogCgAAAAAAgCEwVKCEEQkAGAAAoOAFAIIAAASBNABAA=}#}所以数列abnn是首项为2，公比为2的等比数列，n所以abnn2.②1①+②：24a2nn，所以a(42nn).·······················································12分nn2nn方法二：因为annnab13，abnn4，所以aann124，·····························8分aa所以nn12n1，22nn1aaa1所以n2时，n1112222122211…nnn，2222n211nn所以ann22(2)≥，1又n1时，上式也成立，所以a(42nn)．··················································12分n219．【答案】(1)方法一：连结PM，MB，BD.因为△PAD为等边三角形，M是AD的中点，所以PMAD.又因为平面PAD平面ABCD，平面PAD平面AD，PM平面PAD，所以PM平面ABCD.···················································································2分因为MB、BC平面，所以PMMB，PMBC.在Rt△PMB中，PM3，PB6，所以MBPBPM223，在△MAB中，MA1，AB2，所以MAMBAB222，所以AMB，则MBAD.·······································4分2又ADBC∥，所以BCMB，又因为BCPM，PMMBM，PMMB、平面PBM，所以BC平面PBM，又MN平面PBM，所以BCMN.·································6分zzPPNNMDCMDCAAQyxByxB(方法一图)(方法二图)方法二：连结PM，因为△PAD为等边三角形，M是AD的中点，所以PMAD.又因为平面PAD平面ABCD，平面PAD平面AD，PM平面PAD，所以PM平面.···················································································2分如图，在平面内，作MQMA，分别以MA,,MQMP为x,,yz轴，建立如图所示的空间直角坐标系，则A(1,0,0)，P(0,0,3).设C(a,b,0)（b0），则B(a2,b,0).因为AB2，所以(ab1)224.①因为PC10，所以ab22310.②····························································4分由①②，解得：a2，b3（舍负）.所以C(2,3,0)，B(0,3,0)，3333因为N为PB的中点，所以N(0,,)，所以BC(2,0,0)，MN(0,,)，2222所以BCMN0，所以BCMN．·································································6分第2页（共5页）{#{QQABQQCAogCgAAAAAAgCEwVKCEEQkAGAAAoOAFAIIAAASBNABAA=}#}(2)由(1)可知，PM平面ABCD，又MA、MB平面ABCD，所以PMMA，PMMB，又ADMB，所以以M点为坐标原点，MA、MB、MP所在直线分别为x轴、y轴、z轴建立如图所示的空间直角坐标系.则A(1,0,0)，B(0,3,0)，P(0,0,3)，M(0,0,0).33因为MPMB3，N为PB的中点，所以MNPB，N(0,,)，22由(1)知MNBC，又PBBCB，PB、BC平面PBC，33所以MN平面PBC，所以MN(0)，，为平面PBC的一个法向量.··················8分22nAB0,设nxyz(,,)为平面PAB的一个法向量，则nAP0.xy30,因为AB(1,3,0)，AP(1,0,3)，所以xz30,取y1，则x3，z1，则n(3,1,1)为平面PAB的一个法向量.·····················10分330311MNn10所以cos,MNn22，·············11分MNn3350()()(3)1122222222由图可知二面角APBC的平面角为钝角，10所以二面角的余弦值为.··························································12分520．【答案】(1)由题可知XB~(10000,0.25%)，则EX()100000.002525，········································································2分记该公司今年这一款保险产品利润为变量Y，则YX2005，所以E()(2005)2005()75YEXEX万元.················································4分(2)因为XBnp~(,)，当n较大且p较小时，EX()25，则DX()25.由于n较大，XN~(,)2，其中EX()25，2DX()25，·······················6分若该公司今年这一款保险产品利润YX2005(50,100)，则X(20,30)，P(200YXPXPX5(50,100))(2030)()0.683；··············9分若该公司今年这一款保险产品利润YX20050，则X40，10.997P(200YXPXP50)(40)(3X)0.0015.······················11分2答：(1)EX()25，该公司今年这一款保险产品利润的期望为75万元；(2)①该公司今年这一款保险产品利润为50~100万元的概率为0.683；②亏损的概率为0.0015.·················································································12分xy2221．【答案】(1)因为双曲线E:1的渐近线方程为bxay0，左焦点Fc(,0)，ab22c6,a23所以则b3，又abc222，所以aa223，所以a26，bc23,ba22xy22故双曲线E的标准方程为1.································································4分63第3页（共5页）{#{QQABQQCAogCgAAAAAAgCEwVKCEEQkAGAAAoOAFAIIAAASBNABAA=}#}(2)由题设可知l11y:k(x3)，l22yk:(x3).设A(x,y)11，Bx(,y)22，2ykx1(3),12k则由得(12)121860kxkxk2222，所以xx1，22111122xy2612k16k263kk2又是的中点，所以1，11，MABxM2ykM1(3)2212k11212kk1163kk2则11M(,)22.1212kk1163kk2同理22分N(,)22.··················································································612