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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】,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11211133262322adadaadad,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)=11125431

03309..........................................................................................................................

..(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sinsinsincosc

ossinABCBCBC,所以1sincossin3BCB,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,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)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=11122121122

nnnnn,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

)