湖南省益阳市2024-2025学年高三期末考试数学答案

益阳市2024年下学期普通高中期末质量检测高三数学（参考答案）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，只有一项是符合题目要求的．题号12345678答案CCBAABDB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．全部选对的得6分，部分选对的得部分分，有选错的得0分．题号91011答案BCDACBD三、填空题：本题共3小题，每小题5分，共15分．51112．3i13．14．[-,]4e4e4四、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．15．(本小题满分13分)x2y221解：(Ⅰ)点P(2,1)代入方程C:1(a0,b0)得1①a2b2a2b2且2b2c②，a2b2c2③···································································3分由①②③可解得：a24,b22，x2y2所以椭圆C:1··········································································5分422(Ⅱ)设直线l的方程：yxm，点A(x,y)、B(x,y).21122x2y2直线方程代入椭圆C:1得2x222mx2m240，42由8m28(2m24)32(4m2)0，得2m2，2分x1x22mx1x2m2·································································8则弦长222，AB1k(x1x2)4x1x23(4m)2点P(2,1)到直线l：yxm的距离2221m2m2dm2·······················································10分2331()2221121ABP面积SABd3(4m2)m2m2(4m2)2········12分2232当且仅当m24m2即m2时等号成立，2故ABP面积取得最大值2时直线l的方程为yx2························13分2高三数学参考答案第1页共5页{#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}16．(本小题满分15分)b2cosA2cosC【详解】(Ⅰ)法一：由可得bsinC2cosAsinC2sinAcosC,sinAsinCbsinC2sin(AC)2sinB····································································3分由正弦定理可得：bc2b·········································································5分所以c2·······························································································7分b2cosA2cosCb2cosA2cosC法二：由可得·····································2分sinAsinCacb2c2a2a2b2c2b由余弦定理得：bcab···········································4分acb2c2a2a2b2c2化简得：bc，即bc2b，所以c2··················7分bb(Ⅱ)法一：tanC2tanB，则sinCcosB2sinBcosC，所以3sinCcosB2sinBcosC2sinCcosB2sinBC2sinA·····················9分由正弦定理可得：3ccosB2a，又因为c2，所以3cosBa························11分133所以VABC面积为：SacsinB3cosBsinBsin2B···························13分222π3当sin2B1即B时取等.所以VABC面积的最大值为·····························15分4212sinAsinB2sinAsinB法二：SabsinC2sinCsin(AB)2sinAsinB2tanAtanB················································9分sinAcosBcosAsinBtanAtanBtanBtanC3tanB又tanAtan(BC)································.12分tanBtanC12tan2B13tan2B226tanB6tanB3则S2tanB1,当B时取等··········15分3tanB2tanB2tanB24tanB242tan2B1法三：tanC2tanB，易知B,C都为锐角如图，作边BC上的高AD，则有BD2AD2AB24,tanC2tanBAD2AD,即BD2CD·····································································11分CDBD11333BD2AD23SBCADBDADBDADABC2224422当且仅当BDAD,即B时取等··························································15分4法四：tanC2tanB，则sinCcosB2sinBcosC，ccosB2bcosC,a2c2b2a2b2c2余弦定理可得，即a23b23c212······················10分2aa由余弦定理可得：a2b2c22bccosAb244bcosA,高三数学参考答案第2页共5页{#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}则有12a23b24b244bcosA,22b2化简可得bbcosA2，即cosA····················································12分b1b22所以SbcsinAbsinAb1()2b45b242b53当b2时，S············································································15分2max2法五：同法四可得a23b23c212···························································10分如图过点C作底边AB的高CD，不妨设AD1d,BD1d,CDh,则有b2(1d)2h2,a2(1d)2h2,则a23b23(1d)23h2(1d)2h212，19整理可得h2(d)2,······································································13分24931所以h2，即h，当且仅当d时取等，42213则有Sch·················································································15分2217．(本小题满分15分)解：(Ⅰ)设ACBDO，连接PO，过点A作PO的垂线，垂足即为点A在平面PBD内的射影H.·····························································································2分下面证明:ABAD,CDBC，点A,C在线段BD的中垂线上，即有ACBDPA平面ABCD,BD平面ABCD，PABD又PAAC=A，PA,AC平面PAC，BD平面PAC·························4分又BD平面PBD，平面PBD平面PAC又平面PBD平面PACAO，AHPO，AH平面PACAH平面PBD，故点H为点A在平面PBD内的射影·······························6分(Ⅱ)由(Ⅰ)可知，可建立如图空间直角坐标系O-xyz，ABAD2,，又BAD120，BCD60，PBPD且PBPD，易知OA1，OB3，PB2OB6,3在RtPAB中，PAPB2AB22，在RtPAO中，PO3，OH312则O(0,0,0),A(0,1,0),B(3,0,0),H(0,,)······································9分33设平面HAB的法向量为nx,y,z，22AB(3,1,0)，AH(0,,)333xy0HnAB0则，22nAH0yz033不妨取x1,则n(1,3,6)平面DAB的法向量m0,0,1································································12分高三数学参考答案第3页共5页{#{QQABKYQEggiAABBAARgCAQEACECQkBAACYgOBEAYIAIASBFABAA=}#}mn6二面角HABD的平面角为，|cos||cosm,n|··········14分mn1010又(0,)，sin1cos2510二面角HABD的正弦值为···························································15分518．(本小题满分17分)解：(Ⅰ)记事件A为抽到一件合格品，事件B为抽到另一件为不合格品，C1×C11608020分P(AB)=2=········································································2C100495C2-C247610020分P(A)=2=········································································4C100495P(AB)16040P(B|A)===······························································5分P(A)4761191(Ⅱ)(i)由题：若X~B(100,)，则EX50,DX25·······································7分21又PXkCk()100PX100k,100211所以P0X25P(0X25或75X100)PX5025···········.9分22251由切比雪夫不等式可知，PX5025252251所