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）数学文答案

绵阳市高中2020级第二次诊断性考试

文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．

    DDCAA   BCDBA   CA

二、填空题：本大题共4小题，每小题5分，共20分．

13． 14．     15．        16．[1，3)

三、解答题：本大题共6小题，共70分．

17．解：（1）由，及正弦定理

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

∵·························································4分

∴，·······················································6分

即，且，可得；···············································8分

（2）由，可得，················································10分

由余弦定理.·················································12分

18．解：（1）由题意知，2=+，①······································1分

当n=1时，2=+，则；············································2分

当时，2=+，②··················································3分

①②相减可得，2an =−+−，······································4分

∴an+= −，则an-=1，

∴数列是以为首项，1为公差的等差数列，····························5分

所以，an = n(n∈N ∗).···········································6分

（2），·····················································7分

设，则，····················································8分

∴当时，，所以，·············································9分

当时，，所以，·············································10分

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

则，

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

19．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，

即，······················································11分

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

20．解：（1）设，，

直线BC的方程为：，其中， ······································1分

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

所以：，，··················································3分

从而

所以：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）解：(1) a=2时，，

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

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

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

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

（2），且，·····················································6分

①当时，，，

故f(x)在区间[1，2]上单调递增，所以， ·······························7分

②当时，，，

故f(x)在区间[1，2]上单调递减，

所以，显然在区间上单调递增，

故<0．·······················································9分

③当时，由解得：；由解得：．

故f(x)在区间上单调递增，在区间上单调递减．

此时，则，

故在区间上单调递增，故h(a)<h(1)=0. ································11分

综上：，且h(a)的最大值是0.·······································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