[]山东省枣庄市2017届高三上学期期末质量检测数学（理）试题（扫描版）

2017 届第一学期期末考试 高三数学（理科）参考答案及评分标准 2017.1 一、选择题：本大题共 10 小题，每小题 5 分，共 50 分. ADAD BCAD CC 二、填空题：本大题共 5 小题，每小题 5 分，共 25 分． 11. 12. 63 6 7 13. π [ π 8 k  ， 5π π ] 8 k  ， kZ 14. 10 15. 2 5 三、解答题：本大题共 6 小题，共 75 分．解答应写出文字说明、证明过程或演算步骤． 16. 解：（1）由角 , ,A B C 的度数成等差数列，得 2B A C  . 又 ，所以πA B C   π 3 B  . ············································································ 2 分 由正弦定理，得3 4c a ，即 3 4 c a  .··································································· 3 分 由余弦定理，得 ，即2 2 2 2 cosb a c ac B   2 2 3 3 13 ( ) 2 4 4 c c c 1 2 c      .··· 5 分 解得 . ·············································································································· 6 分 4c  （2）解法一：由正弦定理，得 13 2 13 sin sin sin 3 3 2 a c b A C B     . 所以 2 13 sin 3 a A ， 2 13 sin . 3 c  C ··································································· 8 分 所以 2 13 (sin sin ) 3 a c A C   2 13 [sin sin( )] 3 A A B   2 13 π [sin sin( )] 33 A A   2 13 3 3 ( sin cos ) 2 23 A A  π 2 13 sin( ). 6 A  ·····································10 分 由 2π 0 3 A  ，得 π π 5π 6 6 A   6 . 高三数学(理科)答案第 1 页 共 11 页 所以当 π π 6 2 A  ，即 π 3 A  时， max( ) 2 1a c  3 . ··········································12 分 解法二：（2）由余弦定理，得 ， 2 2 2 2 cosb a c ac B   即 2 2 2( 13) 2 cos60a c ac    2 2a c ac   2( ) 3a c ac   ························ 8 分 由 2 a c ac  ，得 2( ) 4 a c ac  . 因此 2 2 2 3( )13 ( ) 3 ( ) 4 a c a c ac a c       2( ) 4 a c  . ··································10 分 所以 2 13a c  .当且仅当 13a c  时，等号成立. 所以 max( ) 2 1a c  3 n 1a . . ·························································································12 分 17.解：（1）当 时，由 ，得 1n  2 3 2 6n na a S   2 1 13 2 6a a   ，即 2 1 13 2 0a a   又 ， ，解得1 (0a  2) 1 1.a  ··············································································· 1 分 由 ，可知 . 2 3 2 6n