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

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:

·········.......................................................····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