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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222212312312

132322+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,2

22123aaa>2123aaa...................................................................（6分）(2)由222123aaa>2123aaa,且22222212312333

loglogxxxxaaaaaa,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1co

s3C................................................................................................（6分）(2)由余弦定理得2222coscababC,即

2222223123838329bababaabbaabba,......................................（10分）所以,33,272ba

ba,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8CE

BD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25si

n5,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,11121n

aandna,............................................................................(4分)所以5711319+1

322=55aaaaaa......................................................................................................

...............(6分)(2)数列1212,,,,,,nbbbaaaaa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,............................................................................................(1

0分)4923231231323332122132nnnTnnnn............................................................(12

分)