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高二理数参考答案第1页共4页高二数学参考答案(理科)1~5CDBCA6~10CBBAD11~12DC13.【正确答案】,11,214.【正确答案】33415.【正确答案】20,1316.【正确答案】④17.
【解析】解:(1)若命题p是真命题,则2240a,所以1a;若命题q是真命题,由22222314xyyxy及210y得24x,所以22,22xa.若pq为真命题,则p是真命题或q是真命题,所以实数a的取值范围是,
12,2,2...............................................................................(5分)(2)若pq为真命题,由(1)得实数a的取值范围,1
2,22,1,因为22mam是pq为真命题的必要条件,所以1222mm,解得01m,即实数m的取值范围是)0,1(......................
.................................(10分)18.【解析】解:(1)由342Sa及134,2,2aaa成等比数列得342143222Saaaa,即112111332
62322adadaadad,解得122,611dda舍去,所以1162124naandnn...............................
.........................................................(6分)(2)1125125111423423nnnnnnn
nbaannnn,...................................................(8分)高二理数参考答案第2页共4页所以100111111111114344
556101102102103T.........................(10分)=1112543103309
...........................................................................................................................
.(12分)19.【解析】解:不等式2440xaxa,即24400xaxaa,即40xaxa..........................................
..................................................................................(4分)当02a时4aa,40xaxa444axaxxaaaa
或,.................................(6分)当2a时24020xaxxa,该不等式解集为;.......................................(8分)当2a时4aa
时40xaxa444xaxaaxaaa或.............(10分)综上可得02a时原不等式解集为44,,aaaa,2a时原不等式解集为,2a时原不等式解集为44,,aaaa
.............................................................................................................
......(12分)20.【解析】解:(1)由正弦定理及1cos3acBb得1sinsincossin3ACBB,........................(2分)因为sinsinsincoscossinABCBCBC,所以1sincossin
3BCB,因为0,π,sin0BB,所以1cos3C......................................................................................
.....(6分)(2)由余弦定理得2222coscababC,即:222222)(31)2(38)(38)(329bababaabbaabba,....................................(10
分)所以33,272baba,当332ab时取等号.所以ab的最大值为33..................................................................................
.................................(12分)高二理数参考答案第3页共4页21.【解析】(1)如图所示,甲的眼睛到地面距离1.6CDm,π,+4CBDECA,从点C向A
B作垂线,垂足为E,1tan5CDBD,所以8CEBDm,..........................................................................
........................(3分)11+π1+tan35tan+141tan215,π3tan+81242AECE(m),所以121.613.6ABAEEB(m),即灯的顶端A到
地面的距离AB为13.6m................................................................................................
.(7分)(2)由25sin5,可得tan2,所以6.8tanABPBm,3.4mtanABQB,..............................................................................
.........(9分)因为60PBQ,所以2216.83.426.83.43.433.41.72PQ5.8m.所以,PQ两点之间的距离约为5.8m...............................
.........................................................................(12分)22.【解析】(1)由112nnnaa得122nnnaa,两式相除得22nnaa,所以
212,nnaa都是公比为2的等比数列,由422aa及243aa得21a,又01221aa,所以11a,高二理数参考答案第4页共4页所以n为奇数时11122122nnnaa,n为偶数时2122222nnnaa,所
以12222,2,nnnnan为奇数为偶数........................................................................................................................
....(6分)(2)12221nnnnba=11122121122nnnnn,1nnniSb=2135211222nnn,设2135211222nnnT,则2311352122222nnnT
,两式相减得221111211122222nnnnT=11121211212nnn=2121322nnn,所以311216622nnnnT,6nnSnTn,因为
12362nnnT所以12362nnnSn所以125612nnnSn所以121102nnnnSS所以nS单调递增所以12nSS所以成立所以26nSn............................................
........................................(12分)