四川省绵阳市2022-2023学年高三上学期第二次诊断性考试（1月）数学理答案

绵阳市高中2020级第二次诊断性考试

理科数学参考答案及评分意见

一、选择题：本大题共12小题，每小题5分，共60分．

    DACBD   ABCAD   CA

二、填空题：本大题共4小题，每小题5分，共20分．

13．9 14．     15． 5          16．[1，3)

三、解答题：本大题共6小题，共70分．

17．解：（1）由，

可得，·····················································2分

又，·······················································4分

∴，5分

且，可得；··················································6分

（2）由，可得，················································8分

由余弦定理.··················································9分

∵，

平方可得，····················································10分

即，所以．·················································12分

18．解：（1）因为0.92<0.99，根据统计学相关知识，越大，意味着残差平方和越小，那么拟合效果越好，因此选择非线性回归方程②进行拟合更加符合问题实际．              4分

（2）令，则先求出线性回归方程：，·································5分

∵，·······················································7分

=374，·····················································9分

∴，······················································10分

由，得，···················································11分

即，

∴所求非线性回归方程为：．····································12分

19．解：设的公比为q，则，所以，

所以，·····················································2分

因为的各项都为正，所以取，·····································3分

所以．·····················································4分

若选①：由，得，·············································5分

两式相减得：，整理得，········································6分

因为，所以是公比为2，首项为1的等比数列，·························7分

∴，·······················································8分

∴，·······················································9分

∵在R上为增函数，···········································10分

∴数列单调递增，没有最大值，··································11分

∴不存在，使得对任意的恒成立．································12分

若选择②：因为，且，

∴为等比数列，··············································6分

公比，·····················································7分

∴，·······················································8分

所以．····················································10分

当且仅当时取得最大值，·······································11分

∴存在，使得对任意的恒成立．··································12分

若选择③：因为，所以，········································5分

∴是以1为公差的等差数列，又，··································6分

∴，所以，··················································8分

设，则，···················································9分

∴当时，，所以，

当时，，所以，

当时，，所以，

则，······················································11分

∴存在，使得对任意的恒成立．··································12分

20．解：（1）设，，

直线BC的方程为：()，··········································1分

联立，消x整理得：，···········································2分

∴，·······················································3分

从而：

∴为定值.···················································5分

（2）直线AB的方程为：，·········································6分

令，得到，··················································7分

同理：．····················································8分

从而

····················································9分

又，

，························································10分

所以，·····················································11分

因为：，所以，

即线段MN长度的取值范围为．···································12分

21．解：（1）由a=1时，，··········································1分

由解得：x>0或x<−1；由解得：−1< x<0.······························3分

故f(x)在区间上单调递增，在区间(−1，0)上单调递减.····················4分

∴f(x)的极大值是f(−1)=，极小值是f(0)=0；····························5分

（2）时，，且，

，恒有等价于．

i)若时，，故，所以f(x)在区间[0，2]上单调递增，故

，解得：，··················································6分

ii) 若时，，故，所以f(x)在区间[0，2]上单调递减，故

，解得：，··················································7分

iii) 若时，由解得：，故f(x)在区间上单调递增；························8分

由解得：，故f(x)在区间上单调递减.

∴，或f(0) ．

又f(2)−f(0)=，················································9分

①当时， f(2)−f(0)，故，

解得：，又，故此时．·········································10分

②当时， f(2)−f(0)<0，故，

令，则，又，

故>0，即h(a)在区间上单调递增，

又，则恒成立．··············································11分

综上：．···················································12分

22．解：（1）①当B在线段AO上时，由|OA|‧|OB|=4，则B（2，）或（2，）；

②当B不在线段AO上时，设B（ρ，θ），且满足|OA|‧|OB|=4，

∴A，····················································1分

又∵A在曲线l上，则，········································3分

∴，·····················································4分

又∵，即.

综上所述，曲线C的极坐标方程为：

，或．···················································5分

（2）①若曲线C为：，此时P，Q重合，不符合题意；

②若曲线C为：，设l1：，

又l1与曲线C交于点P，联立

得：，·····················································6分

又l1与曲线l交于点Q，联立

得：，·····················································7分

又∵M是P，Q的中点，

，·························································8分

令，则，

又∵，则，且，

∴，且在上是增函数，··········································9分

∴，且当时，即时等号成立．

∴的最大值为．··············································10分

23．解：（1）由≤3的解集为[n，1]，可知，1是方程=3的根，

∴=3+|m+1|=3，则m=−1，········································1分

∴=|2x+1|+|x−1|，

①当x≤时，=−3x≤3，即x≥−1，解得：−1≤x≤，························2分

②当时，=x+2≤3，解得：，······································3分

③当x≥1时，=3x≤3，解得：x=1．··································4分

综上所述：的解集为[−1，1]，所以m=−1，n=−1．······················5分

（2）由（1）可知m=−1，则．········································6分

令，，则，，

又a，b 均为正数，则（），

由基本不等式得，，···········································7分

∴，当且仅当x=y=1时，等号成立．

所以有，当且仅当x=y=1时，等号成立．······························8分

又

(当且仅当x=y时，等号成立)．···················9分

∴成立，(当且仅当，时等号成立)．································10分