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高二文数参考答案第1页共4页高二数学参考答案(文科)1~5BCBCA6~10CBBBD11~12BD13.【答案】0,514.【答案】33415.【答案】,1515,16.【答案】④17.【解析
】(1)由2343,23aSa成等比数列得111333263adadad,......................................(2分)解得133,22da,..................
.......................................................................................................................(4分)所以
133311222naandnn..............................................................................................(6分)(2)
因为1nnnba,所以100123499100Taaaaaa=5075d.................................................................................
......................................................................(10分)18.【解析】:(1)若命题p是真命题,则2240a,所
以1a;..................................................(2分)若命题q是真命题,由22222314xyyxy及210y得24x,所以22,22xa....
.......................................................................................................................(4分)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...........................................................................
.....(6分)(2)若pq为真命题,由(1)得实数a的取值范围,12,22,1,..........................................(8分)因为22mam是pq为真命题的必要条件,
所以1222mm,..........................................................................................
................................................(10分)解得01m,即实数m的取值范围是)0,1(.....................................................
.................................(12分)19.【解析】:(1)因为222212312312132322+2aaaaaaaaaaaa高二文数参考答案第2页共4页=2222111111122221aaqaaq
aqaqaqqq..........................................................................(3分)因为数列na是等比数列,2110,0aa,又22130,1024qqqq
,所以221210aqqq,222123aaa>2123aaa...................................................................(6分)(2)由222123aaa
>2123aaa,且22222212312333loglogxxxxaaaaaa,所以1302xx,......................................................
..............................................................................(9分)由230xx得03xx或,由231x
x得31331322x,所以原不等式的解集为313313,0322,............................................................
.................(12分)20.【解析】:(1)由正弦定理及1cos3acBb得1sinsincossin3ACBB,..................................(2分
)因为sinsinsincoscossinABCBCBC,所以1sincossin3BCB,因为0,π,sin0BB,所以1cos3C.....................................................
...........................................(6分)(2)由余弦定理得2222coscababC,即2222223123838329bababaabbaabba,...............
.......................(10分)所以,33,272baba,当332ab时取等号.所以ab的最大值为33.................................
.......................................................................................(12分)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AB
作垂线,垂足为E,高二文数参考答案第3页共4页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)设等差数列na的公差为d,由939SS,得11936933adad,所以1120daa,11121naandna,...........
.................................................................(4分)所以5711319+1322=55aaaaaa.......................
..............................................................................................(6分)(2)数列1212,,,,,,nb
bbaaaaa成等比数列,其公比213aqa,由nba是该等比数列的第2n项得113nnbaa,...................................................................................(7分
)高二文数参考答案第4页共4页又111121nbnnaabdaab,所以111121=3nnaaba,1312nnb,.............................
...............................................................(10分)4923231231323332122132nnnTnnnn...................
.........................................(12分)