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高二文数参考答案第1页共4页2020—2021学年上学期全国百强名校“领军考试”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01m,即实数m的取值范围是)0,1(......................................................................................(12分)19.【解析】：(1)因为222212

312312132322+2aaaaaaaaaaaa高二文数参考答案第2页共4页=2222111111122221aaqaaqaqaqaqqq...........................................................

...............（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22222212312333loglogxxxxaaaa

aa,所以1302xx,...............................................................................

.....................................................（9分）由230xx得03xx或,由231xx得31331322x,所以原不等式的解集为3133

13,0322，.............................................................................（12分）20.【解析】：(1)由正弦定理及1cos3acBb得1

sinsincossin3ACBB,..................................（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,33,272baba,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1.6CDm,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8CEBDm,..............................................................................

.......................(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6.8tanABPB

m,3.4mtanABQB,.......................................................................................(9分)因为60PBQ,所以2216.

83.426.83.43.433.41.72PQ5.8m.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)4923231231323332

122132nnnTnnnn............................................................(12分)