2023绵阳高三上学期第二次诊断性考试（1月）数学（文）PDF版含答案

这是一份2023绵阳高三上学期第二次诊断性考试（1月）数学（文）PDF版含答案，文件包含四川省绵阳市2022-2023学年高三上学期第二次诊断性考试1月数学文答案docx、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PDF版无答案pdf等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
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