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高二理数参考答案第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（理科）1~5CDBCA6~10CBBAD11~12DC13．【正确答案】1,1,214．【正确答案】33415．【正确答案】20,1316．【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,12,22,1,因为22mam是pq为真命题的必要条件,所以1222mm,解得01m,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1121113326

2322adadaadad,解得122,611dda舍去,所以1162124naandnn....................................

....................................................(6分)(2)1125125111423423nnnnnnnnbaannnn,............

.......................................(8分)高二理数参考答案第2页共4页所以100111111111114344556101102102103T

.........................(10分)=1112543103309..........................

..................................................................................................(12分)19．【解析】解：不等式2440xaxa,即

24400xaxaa,即40xaxa................................................................

............................................................（4分）当02a时4aa,40xaxa444axaxxaaaa或,................................

.（6分）当2a时24020xaxxa,该不等式解集为；.......................................（8分）当2a时4aa时40xaxa444xaxaaxaaa

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1sincossin3

BCB,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33,272baba,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1.6CDm,π,+4CBDECA,从点C向AB作垂线,垂足为E,1t

an5CDBD,所以8CEBDm,..................................................................................................(3分)11+π1+tan35

tan+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6.8t

anABPB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)由112nnnaa得122nnnaa,两式相除得22nnaa,所以212,nna

a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1222

2,2,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311216622nnnn

T,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)