广西柳州市2021届高三下学期第三次模拟考试文科数学试题（PDF版）

柳州市 2021 届高三第三次模拟考试 文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 A D B A D D B A B 二、填空题：(本大题共 4小题，每小题 5 分，共 20 分) 13. �㔨 � 14.5 � 15.� ��� � 16. �� � � �� � �त� �� �� 三、解答题: （本大题共 6 小题，共 70 分） 17.解：（1）在 ABC 中，由正弦定理得 sinsin a CA c  ，···················································1 分 所以条件可转化为 2 sin cos tan 3a C a C C c    ，·······················································3 分 0 C   3 C   ·······················································································4 分 （2）由余弦定理得 2 212 a b ab ab    ，····························································· 5 分 当且仅当 2 3a b  时等号成立，所以 ab的最大值为12，········································· 7 分 故三角形面积 1 sin 2 ab C的最大值为3 3，································································ 8 分 设 AB边上的高为 h，则三角形面积 1 3 3 3 3 2 ch h h    ，····························· 10 分 从而 AB边上高的取值范围为 (0,3]．··································································· 12 分 18．（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 1DE 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）∵直角梯形 AEFB绕 EF转到 D1EFC1 ∴DE1⊥EF·············································································································································· 5 分 ∵AE⊥EF且 AE∩D1E=E··························································································6 分 ∴EF⊥平面 AD1E·································································································· 7 分 且∠AED1为二面角 C1-EF-B的平面角∴∠���� � � � ·······················································8 分 又∵AE=D1E ∴△AED1为等边三角形········································································ 9 分 ∴������ � � � ��� � � � � � �