高二数学第1页共4页参考答案12.513.1214.20224.【详解】奇函数图像关于原点对称，所以在关于原点对称区域内单调性相同，函数()yfx是奇函数且在区间5,1上单调递增，则函数()yfx在区间1,5上单调递增，由于区间为半开半闭区间，所以存在最大值.故选B.5.【详解】设圆锥底面半径为r，则高也为r,2122322613rrSVrrr,所以3r,母线长为26r.故选D6.【详解】由题可先将1号球分给甲，再将情况进行分类：第一类，甲只有一个球，则另外三个球分给乙和丙两位同学，有C32A22=6种方法，第二类，甲有两个球，则需要将3个球分发给3位同学的方法数有33A6种，所以共有6+6=12种分法.故选B.7.【详解】随机变量X的密度曲线关于2x对称，03PXPX43PXPX4434PXPXPX1313412PX，所以13412PX.故选A8.【详解】设1a，2a，3a，4a平均数为a，因为4222114iisaa，得225a，由0a，故5a.则样本数据141a，241a，341a，441a的平均数为4121a.故选A11.【详解】由题意可知1231()()()3PAPAPA,14()5PBA,11()5PCA,22()5PBA,A正确，B错误；3()1PBA，11223311()()()()()()()15PBPAPBAPAPBAPAPBA,C正确；1114()435()11()1115PABPABPB,D正确；故选ACD.14.【详解】21fx为奇函数，所以fx关于点2,1对称，21fx为偶函数，所以fx关于直线1x对称，故fx的周期为4，241ff，132ff，202512ff所以20251()506((1)(2)(3)(4))(2025)506422022ififffff，故答案为2022.15.(本小题满分13分)（1）由525S可得35a，·····························································································2分由253412aaaa可得2d，····················································································4分故21nan.·················································································································6分（2）由（1）得2nSn,································································································8分题号12345678答案CDDBDBAA题号91011答案ABDBCACD{#{QQABTQYg4gi4wBbACY4LUQU4CwoQsJMjLcokwQASKAQKCRFABKA=}#}高二数学第2页共4页故2111111()4141(21)(21)22121nnbSnnnnn··············································11分则111111(1)2335212121nnTnnn.··························································13分16.(本小题满分15分)解：(1)221afxxax,113fa，所以2a·············································2分232lnfxxxx，14f，切点为1,4··································································4分切线方程为431yx，即310xy········································································6分(2)2lnxxxmx，0x，minln1xmxx······...