2024届湖南省岳阳市高三一模数学参考答案

岳阳市2024届高三教学质量监测(一)数学参考答案及评分标准一、单项选择题：本题共8小题，每小题5分，共40分．1．B2．C3．C4．D5．A6．A7．B8．D二、多项选择题：本题共4小题，每小题5分，共20分．9．ABC10．BD11．ACD12．ABD三、填空题：本题共4个小题，每小题5分，共20分．7713．（1，2）(满足k(1,2),k0均可)14．(,)(,)222225115．xy16．2(2分)；21，(3分)42四、解答题：本题共6小题，共70分．解答应写出文字说明，证明过程或演算步骤．nn1．由题意，∵≥17(1)an1(n2)SnSn1nn1nn1分anSnSn11n(n1).......................................................(2)SnSn1SnSn1nn1SnnSn1（n1）SnSn1n数列为常数列分{Snn}........................................................................................................(4)Snn1，分(2)S1a11SnnS111........................................................................(5)SnS12Sn(n1)Snn(Snn)(Sn1)0又，，分an0Sn0Snn..................................................................................................(7)1111........................................................................................................(8分)Sn1Sn(n1)nn1n1111111S1S2S2S3S3S4Sn1Sn1223(n1)n111111111....................................................................................(10分)223(n1)nn1{#{QQABJQCEogCoQBAAAAhCAw3oCACQkAEACAoOxAAAMAIAiQFABAA=}#}18．(1)由题意，sinBsinAsinAatanB2ctanAatanAa2ca0...............................................(1分)cosBcosAcosAab2caa20cosBcosAcosAb2casinB2sinCsinA又a0，0，0..................................(3分)cosBcosAcosAcosBcosAcosAsinBcosA2sinCcosBsinAcosB0,即sin（AB）2sinCcosB01即sinC2sinCcosB0又sinC0,cosB,B为三角形内角2B............................................................................................................................................(5分)313(2)SacsinBac3，ac4......................................................................(7分)ABC2411SSSa1sinc1sin3，ac43.......................(9分)ABCABDCBD2626（ac）2（43）212................................................................................................................(10分)ac4ccc整理得()210()10，解得526............................................................................(12分)aaa．取的中点连接，则又平面19(1)BCQ,FQ,EQFQ//C1C,A1A//C1C,FQ//A1A,A1AABCFQ平面ABC........................................................................(2分)分FQBC,B1C1//BC,EFB1C1,BCEF.........................................................................(3)FQEFF，BC平面EFQ,又EQ平面EFQBCEQ..............................(4分)EQ为ABC的中位线，EQ//AC，BCAC............................................................(5分)(2)过Q作QMCE于M，连接FM，由(1)知FQ平面ABCFQEC,FQQMQ，EC平面FMQ，ECFMFQ二面角FECB的平面角为FMQ,tanFMQ22QM2又QMCQsin，FQ2....................................(7分)42又两两垂直，以为坐标原点如图建系CA,CB,CC1C，E(1,1,0),F(0,1,2),C(0,0,0),A1(2,0,2)EF(1,0,2)分CE(1,1,0),CA1(2,0,2)...............................................................(8)设平面的法向量为A1ECn（x,y,z）nCE0xy0,得，令x1，y1,z1，n（1,1,1）...............................(10分)xz0nCA1015设直线EF与平面AEC夹角为，sincosEF,n...............................(11分)1510直线EF与平面AEC夹角的余弦值为..................................................................................(12分)152{#{QQABJQCEogCoQBAAAAhCAw3oCACQkAEACAoOxAAAMAIAiQFABAA=}#}20．教师学生合计分文学类103040.........................................................................................................(1)理工类101020合计204060(1)提出零假设H0：老师与学生的借阅情况不存在差异分60(10101030)2............................................................................(3)23.752.706x204020400.1根据小概率值的独立性检验，推断不成立，即认为判断老师与学生的借阅情况是存在差异，此0.1H0推断犯错误的概率不大于0.1....................................................................................................................(4分)201(2)设借阅理工类书籍的概率为p，则p............................................................................(5分)603设随机抽取m人中借阅理工类书籍的人数为随机变量X1212C5()5()m5C6()6()m6P(X5)P(X6)m33m33则，.....................................................(8分)P(X5)P(X4)1212C5()5()m5C4()4()m4m33m3321m!2m!1C（5）C（6）m3m35!(m5)!36!(m6)!3，5142m!1m!2C（m）C（m）335!(m5)!34!(m4)!31211(m5）363得解得14m17.............................................................................................(11分)11215m43又mNm可取14，15，16，17.......................................................................................................(12分)21．(1)由题意，其准线为x1，N点坐标为(1，0)...................................................................(1分)y24x不妨设直线的方程为，设直线的方程为，联立NAyk1(x1)NByk2(x1)yk1(x1)得2222，由题知424，得2分k1x(2k14)xk104k116k1164k10k11....................(3)42k21，同理，故直线的方程为分xA21xB1ABx1.........................................................(4)2k1(2)由题可知，若存在实数，则0，AHAEAM1111AHAEAM又H、E、M在同一条直线上1，2故只需证明M为H、E的中点即可.....................................................................................................(5分)设直线为，设点，点NDxty1C(x1,y1)D(x2,y2)y24x联立，得2，分y4ty40y1y24t,y1y24...........................................................(6)xty13{#{QQABJQCEogCoQB