广西来宾市2020届高三4月教学质量诊断性联合考试数学（文）试题.pdf

高三·数学（文科） 第 1页（共 6页） 广西 2020 年 4 月份高三教学质量诊断性联合考试 数学（文科） 参考答案及评分标准 一、选择题（本大题共 12 小题，每小题 5 分，共 60 分） 1. A 【解答】由中不等式变形，得，解得，∴． ∵，∴．故选． 2. C 【解答】，所以虚部为．故选． 3. C 【解答】如果一次随机取出个球，用列举法可求得全是黄球的概率是，那么至少有个红球 的概率为．故选 C． 4. A 【解答】：根据指数函数的性质可知，对任意，总有成立，即为真命题； ：“”是“”的充分不必要条件，即为真命题，则为真命题．故选 A． 5. A 【解答】∵，∴，，∴，把代入，得．故选． 6. D 【解答】∵数列是各项不为的等差数列，由，得，，，∴，解得或（舍去）．则．又数列是 等比数列，则．故选． 7. A 【解答】∵，∴曲线上任意一点处的切线的斜率的取值范围是，∴切线的倾斜角的取值范 围是．故选． 8. A 【解答】根据程序框图知，该程序运行后是输出当时，令，解得；当时，，满足题意；当 时，令，解得，不满足题意．综上，若输出的，那么输入的为或 0． 9. B 【解答】如图，分别过点，在平面内作，，连接，，由题易证，均为直角三角形．设正方形 的边长为， 则 ， ，∴. 连接，由于是正方形的中心，∴点在上， ∴直线都在平面内，是相交直线．故选． 10. C 【解答】作出 表示的平面区域为如图所示的及其内部， 其中，，， 而的几何意义是可行域内的点到直线 的距离的倍，观察图形并结合计算可知， 点到直线的距离最大. 所以当时，取得最大值．故选． 11. C 【解答】由题意可得＝，设右焦点为， 由＝＝知，＝，＝，高三·数学（文科） 第 2页（共 6页） 所以＝． 由＝知，＝，即． 在中，由勾股定理，得． 由椭圆定义，得＝＝＝，从而＝，得＝， 于是 ＝＝，所以椭圆的方程为．故选． 12. B 【解答】， 易知在上为增函数，且的解为． ∴化为， ∴ 解得或．故选． 二、填空题（本大题共 4 小题，每小题 5 分，共 20 分） 13. （或） 【解答】∵，，，设向量与的夹角为，则，解得，∴．故答案为：． 14. 【解答】由，得， 两式相减，得，即，，∴是公比为的等比数列， 由，得， 所以，故答案为：. 15. 【解答】点，关于直线对称，可得直线为的垂直平分线， 的中点为，的斜率为，所以直线的斜率为， 可得直线的方程为，令，可得， 由题意可得，即有， 由，可得＝， 解得＝或（舍去）．故答案为：. 16. 【解答】由题意,知正四棱锥如图所示，则．高三·数学（文科） 第 3页（共 6页） 三、解答题（共 70 分） 17. 解：在 中，由正弦定理， 代入，得. ·································································· 1 分 又因为，得， ············································· 2 分 得， ···················································································· 3 分 . ···················································································· 4 分 由余弦定理，得 ···························································· 5 分 . ······························································· 6 分 由可得， ···························································· 7 分 由，得，即， ······················································ 8 分 则， ····································································· 9 分 ， ·····································································10 分 故 ········································································11 分 . ····················································································12 分 18. 解：由直方图，得. ···········3 分 ∴. ·····························································································6 分 由直方图可知，新生上学所需时间在[60,100]的频率为, ···········9 分高三·数学（文科） 第 4页（共 6页） ∴估计全校新生上学所需时间在[60,100]的概率为 0.12. ∴， ··················································································11 分 故估计 800 名新生中有 96 名学生可以申请住宿. ···············································12 分 19. 证明： 在中，因为， 所以为的中点. ················································· 2 分 又因为是的中点，所以. ························· 4 分 又平面，平面， 所以平面. ················································· 6 分 在中，因为，所以. ······· 7 分 又因为，所以. ·································· 8 分 又因为， 所以，即. ························· 9 分 又因为平面，平面， 所以平面. ··················································11 分 又因为平面， 所以平面平面. ·········································12 分 20. 解：若直线与的图象相切，设切点为， ， ································································································ 1 分 ∴ ········································································ 2 分 且， ········································································ 3 分 解得， ················································································· 4 分 . ················································································· 5 分 令， ······················································ 6 分 不等式对任意的 恒成立，等价于在上恒成立． ∴， ························································· 7 分 设． 若，则，在上单调递增，又，∴不成 立． ··············································································································· 8 分 若，的图象开口朝下，对称轴为． ①当，即时， ，高三·数学（文科） 第 5页（共 6页） ∴在上， ，∴，又，∴恒成立． ········· 9 分 ②当，即时， ， ∴在 上，存在，使 ， ∴时，，即； 时，，即． ······················································10 分 ∴，∴不满足恒成立． ·································11 分 综上可知，. ·······················································································12 分 21. 解：设，． 联立 消去，得， ········································································ 1 分 ∴. ·························································································· 2 分 显然直线过抛物线的焦点， ∴. ··········································································· 3 分 设线段的中点坐标为， 则， ······························································· 4 分 则以为直径的圆的方程为. ···································· 5 分 由得， 易得直线，直线， ····················· 6 分 联立 由①，得， 由②，同理可得， 由，得，得， ········· 7 分 联立 得，则，，∴， ，即. ······································· 8 分 ∴， ·········································· 9 分 点到直线的距离， ··················10 分 ∴. ··················· 11 分 显然，当时，的面积最小，最小值为. ········································ 12 分 22. 解：曲线的直角坐标方程为， 即， ················································································· 1 分 故曲线的极坐标方程为， ··················································· 2 分高三·数学（文科） 第 6页（共 6页） 即． ·························································································· 3 分 因为曲线的极坐标方程为， 即， ····································································· 4 分 故曲线的直角坐标方程为， 即． ····································································· 5 分 直线的极坐标方程化为直角坐标方程，得 ， ································· 6 分 由 得 或 ··················································· 7 分 则. ·············································································· 8 分 由 得 或 则. ··············· 9 分 故. ·······························································10 分 23. 解： ························································· 2 分 当时，结合，得； ······················································ 3 分 当时，结合，得； ···································· 4 分 当时，结合，得． 综上，不等式的解集为. ······································· 5 分 ··················································· 6 分. ······························································· 7 分 有解， 等价于， ············································· 8 分 ， ············································· 9 分 所以，解得. ·······························································10 分