【文档说明】河南省八市重点高中2020-2021学年高二上学期11月联考数学（理）答案.pdf，共(4)页，182.294 KB，由小赞的店铺上传

转载请保留链接：https://www.doc5u.com/view-eb9394be688ffbbfb7adbf2cf55930fc.html

以下为本文档部分文字说明：

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)=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,3.4mtanABQ

B,.......................................................................................(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)12221nnnn

ba=11122121122nnnnn,1nnniSb=2135211222nnn,设2135211222nnnT,则2311352122222nnnT,两式相减得221111211122222nn

nnT=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)