2021雅安高三下学期5月第三次诊断考试数学（理）试题PDF版含答案

这是一份2021雅安高三下学期5月第三次诊断考试数学（理）试题PDF版含答案，文件包含2021届高三四川省雅安市高三第三次诊断考试数学理答案docx、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pdf等2份试卷配套教学资源，其中试卷共15页， 欢迎下载使用。
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401050不乘成雅高铁203050合计6040100·········.......................................................····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：，··············································8分，··················································11分所以：为定值1.·················································12分21题：解：（1）由题意，，可得a＝（x>0），·············1分转化为函数T(x)＝与直线y＝a在(0，＋∞)上有两个不同交点，········2分T′(x)＝(x>0)，故当x∈(0,1)时，T′(x)>0；当x∈(1，＋∞)时，T′(x)<0，故T(x)在(0,1)上单调递增，在(1，＋∞)上单调递减，·······················4分所以T(x)max＝T(1)＝1.又T＝0，故当x∈时，T(x)<0，当x∈时，T(x)>0. ·······5分可得a∈(0,1)．························································6分另解：则：，·········1分①当时，恒成立，不满足题意；······3分②当时，单调递减，则，当····························5分综上：·······················································6分(2)证明：　h′(x)＝－a，   因为x1，x2是ln x－ax＋1＝0的两个根，故ln x1－ax1＋1＝0，ln x2－ax2＋1＝0⇒a＝，·················..·8分要证h′(x1x2)<1－a，只需证x1x2>1，即证ln x1＋ln x2>0，即证(ax1－1)＋(ax2－1)>0，即只需证明 a>成立，即证>.·························9分不妨设0<x1<x2，故ln <＝.(*)····························10分令t＝∈(0,1)，φ(t)＝ln t－，······································11分φ′(t)＝－＝>0，则h(t)在(0,1)上单调递增，则φ(t)< φ(1)＝0，故（*）式成立，即要证不等式得证．······································12分22. 解 ：（1）直线l的直角坐标方程为：························2分曲线C的极坐标方程为：，即，化为直角坐标方程：.将曲线C上所有点的横坐标缩短为原来的一半，纵坐标不变，得到曲线：.  ··············································5分（2）直线的极坐标方程为，展开可得：.可得直角坐标方程：.可得参数方程：（为参数）.  ····························7分代入曲线的直角坐标方程可得：.解得，.∴.·······································10分23.解：（1）或解出或无解,  所以，原不等式的解集为[0，1]····················5分另解：（1）当时，等价于，则∴或无解综上，原不等式的解集为[0，1]···········································5分 （2）当时，，因为，所以恒成立，即恒成立，所以满足的解集为；而，当时，，当时，，作出的图像如上图所示， 要使的解集为，则需或，解得或；综上可得：a的取值范围是. ·····································10 分