2021届高三四川省雅安市高三第三次诊断考试数学（理）试题（pdf版）

$雅安市高中2018级第三次诊断性考试 数学（理科）参考答案及评分标准 一.选择题 1.A 2.C 3.D 4. B 5. C 6. C 7. C 8.B 9.C 10. A 11. D 12.A 二.填空题 13.3 14. 15. 3 16.②③④ 三.解答题 17.解：∵是与的等差中项 ∴= ∴ ∴........................................................3分 ∵∴∴ ∴ ....................6分 （2）由（1）得...........8分 ......10分 ∵数列为单调递增的数列，∴ ∴ . ...........12分 18.解：（1）由已知可得，岁及以下采用乘坐成雅高铁出行的有 人············..................................................1分 列联表如表: 40岁及以下 40岁上 合计 乘成雅高铁 40 10 50 不乘成雅高铁 20 30 50 合计 60 40 100 ·········.......................................................····4分 由列联表中的数据计算可得的观测值 ·································6分 由于，故有的把握认为“采用乘坐成雅高铁出行与年龄有关”.····························································7分 （2）采用分层抽样的方法，从“岁及以下”的人中抽取人， 从“岁以上”的人中抽取人，···································8分 的可能取值为: ···········································10分 故分布列如表: 数学期望.·············12分 19．解：（Ⅰ）证明：因为PA⊥底面ABCD，所以PA⊥CD， 因为∠PCD＝90，所以PC⊥CD， 所以CD⊥平面PAC， 所以CD⊥AC．······························································4分 （Ⅱ）因为底面ABCD是平行四边形，CD⊥AC，所以AB⊥AC．又PA⊥底面ABCD，所以AB，AC，AP两两垂直． 如图所示，以点A为原点，以为x轴正方向，以为单位长度，建立空间直角坐标系． 则B(1，0，0)，C(0，1，0)，P(0，0，1)，D(－1，1，0)． 设，则， 又∠DAE＝60°，则， 即，解得． ···································· 8分 则，， 所以. 因为，所以． 又，故二面角B-AE-D的余弦值为．·······················……12分 另解：同上，E(0，，)． 设是平面BAE的法向量，解得法向量， 设是平面DAE的法向量，解得法向量， ，由图可知，二面角的余弦值为. 20.解：（1），·············································1分 由题知，则 椭圆的标准方程··········································4分 （2）（i）若的斜率不存在，则 此时······································.5分 （ii）若的斜率存在，设，设的方程为：， ，·······················6分 由韦达定理得：·········································7分 ，·················································8分 ··············································11分 所以：为定值1.··············································12分 另解：（2）当直线AB的斜率为0时，，·········5分 当直线AB的斜率不为0时，设直线AB为：，设则： ，···································6分 ，·········································7分 则：，···········