广西柳州市2021届高三下学期第三次模拟考试理科数学试题（PDF版，含答案）

柳州市 2021 届高三第三次模拟考试 理科数学参考答案及评分标准 一、选择题：(每小题 5 分, 满分 60 分) 1 2 3 4 5 6 7 8 9 10 11 12 C B C B A B A D B A D C 二、填空题：(本大题共 4小题，每小题 5 分，共 20 分) 13． 10 2 14．96 15. 9 1 4 n  16.①② 三、解答题：（本大题共 6小题，共 70 分） 17．(12 分)解：（1）因为 2sina A ，由正弦定理得 2 sin sin sin b c a B C A    .·························1 分 因为 ( )cos 2 cos 2 2 a b aC B   ，所以 2 2 22sin 2sin 2 ab aB C   .······································3 分 所以 2 2 22( ) 2( ) 2 2 2 b c ab a   ，即 2 2 2a b c ab   .································································ 4 分 所以 2 2 2 1cos 2 2 a b cC ab     .··························································································5 分 因为 0 πC  ，所以 π 3 C  .························································································ 6 分 （2）由（1）知 2sinb B . 所以 2 2 2 24sin 4sina b A B   1 cos2 1 cos24( ) 2 2 A B    ············································································ 7 分 4πcos( 2 )cos2 34[1 ] 2 2 AA     1 3cos2 sin 2 2 24(1 ) 2 A A   πcos(2 ) 34(1 ) 2 A    π4 2cos(2 ) 3 A   ······················································································ 9 分 因为 π π0 ,0 2 2 A B    ，所以 π0 , 2 2π π0 . 3 2 A A          所以 π π 6 2 A  .·····································10 分 所以 π 11 cos(2 ) 3 2 A     .··························································································11 分 所以 2 25 6a b   ，即 2 2a b 的取值范围是 (5,6] .···························································12 分 18．（12 分）（1）证明：在直角梯形 AEFB中， AE EF ，且直角梯形 1 1DEFC 是通过直角梯形 AEFB 以直线 EF 为轴旋转而得， 所以 1D E EF .所以 BF EF , 1C F EF .··························································· 2 分 所以 EF 平面 1BC F .··························································································3 分 所以 平面 1 1C D EF 平面 1BC F .············································································· 4 分 （2）由(1)可知BF EF , 1C F EF . 因为 二面角 1C EF B  为 3  ，所以 1 3 C FB   ··