数学答案第1页（共4页）保密★启用前毕节市2025届高三年级第一次诊断性考试数学参考答案及评分建议一、单项选择题：本题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项符合题目要求.题号12345678答案CBCDADAC二、多项选择题：本题共3小题，每小题6分，共18分.全部选对的得6分，部分选对的得部分分，有选错的得0分.(备注：第9题选对一个得2分，第10，11题选对一个得3分)91011ABCADAC三、填空题：本题共3小题，每小题5分，共15分．12.)3,1(13.2114.12,1(第一空2分，第二空3分)四、解答题：本题共5小题，共77分.15.解：（Ⅰ）nm，0cos32sin2AbBa································（2分）又结合正弦定理可得：0cossin3sinsinABBA······························（4分）0sinB，0cos3sinAA，3tanA·······························（5分）),0(A，3A·······································································（6分）（Ⅱ）由（Ⅰ）可知3A，3Asinbc21SABC··································（8分）3bc433sinbc21，4bc····················································（10分）由42cos222222bcbcbcbccbAbccba（当且仅当2cb时取等）······························································（12分）2a，即a的最小值为2································································（13分）{#{QQABQYwAggCoAAJAABhCQQGyCgOQkACAAYgOAAAQoAAACQFABAA=}#}数学答案第2页（共4页）16.解：（Ⅰ） ]1)5([)(2xaaxexfx∴]4)53([)(2axaaxexfx·····················································（2分） 直线l的斜率为e4,···································································（3分）由题意知eeaf4)95()1(，解得1a·····································（5分）∴)14()(2xxexfx∴ef2)1(，即)2,1(eP····························································（6分）∴曲线)(xfy在点)2,1(eP处的切线方程为)1(42xeey即024eyex········································································（7分）（Ⅱ）由（Ⅰ）知）（32)(2xxexfx由0)(xf得1x或3x由0)(xf得31x·································································（10分）∴)(xf的单调递增区间为），（1,），（3，)(xf的单调递减区间为），（31·························································（13分）∴1x时,)(xf有极大值，极大值为ef6)1(································（14分）3x时,)(xf有极小值，极小值为32)3(ef···········（15分）17.（Ⅰ）证明:连接BD交AC于点F,连接EF··················（2分） 底面ABCD是菱形∴F是BD的中点,又E是PD的中点∴PBEF//···············...