广西来宾市2020届高三4月教学质量诊断性联合考试 数学（理）试题（PDF版）

广西来宾市2020届高三4月教学质量诊断性联合考试 数学（理）试题（PDF版）

高三·数学（理科） 第 1页（共 4页） 广西 2020 年 4 月份高三教学质量诊断性联合考试 数学(理科) 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】∵, ∴. 2.C 【解答】∵,∴. 3.B 【解答】从随机数表第 6 行的第 9 列和第 10 列数字开始从左到右依次选取两个数字, 位于 01 至 50 中间(含端点),选出的四个数依次为 41,48,28,19,则选出的第 4 个个体的编号 为 19. 4.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. 5.C 【解答】∵,∴. 6.A 【解答】由双曲线 1,得,∴,∴双曲线的离心率. 7.C 【解答】.令,解得或或或,观察各选项中的图形,可知只有选项 C 符合题意. 8.A 【解答】①若,且,表示两个不同的点,则由平面的基本性质的公理 1,可得,故正确. ②若若,且,表示两个不同的点,分两种情况:若,表示两个不同的平面,则由平面的基本性质的 公理 2,可得;若与表示相同的平面,则与重合,故不正确. ③若,则不能判定是否在平面上,故不正确. ④若,,,,,,分两种情况:若,,不共线,由平面的基本性质的公理 3,可得 与重合;若,,共线,则不能 判定与重合,故不正确.所以其中正确的有 1 个. 9.B 【解答】二项式的展开式中,第项为.令,解得,此时为;令,解得,此时.∴展开式中含的项的 系数是. 10.D 【解答】由,得,由题意,得,∴,当且仅当,即时取等号,此时. 11.B 【解答】①∵集合表示直线上的点构成的点集合,集合表示圆心为,半径为 3 的圆上的 点构成的点集合,由圆心到直线的距离,知有两个交点,故①错误;②当时,显然定义域不是,当 时,分母恒不为 0,∴,解得,故②正确;③的定义域为且,∴可化简为.∵,∴是奇函数,故③错误; ④令,则,∴,∵,当或或或或时,,故④正确. 12.C 【解答】∵,∴,则,即,即,即.∵,∴,解得,此时,即. 二､填空题(本大题共 4 小题,每小题 5 分,共 20 分) 13. 【解答】∵共线,∴,∴,∴. 14. 【解答】∵在数列中,,∴数列是首项为 1,公差为 2 的等差数列,为前 n 项和,∴.∵,∴,解得 或(舍去). 15. 【解答】如图,设椭圆的左焦点为.由椭圆定义得, 即,∵为线段的中点,为线段的中点,∴, 代入,得,解得,∴,∴的离心率为. 16. 【解答】由题意,知正四棱锥如图所示,则. 高三·数学（理科） 第 2页（共 4页） 三､解答题(共 70 分) 17.解:(1)由直方图,得. ············3 分 ∴. ·····················································································6 分 (2)由直方图可知,新生上学所需时间在[60,100]的频率为, ······8 分 ∴估计全校新生上学所需时间在[60,100]的概率为 0.12. ································9 分 ∴(名), ·······································································11 分 故估计 800 名新生中有 96 名学生可以申请住宿. ·········································12 分 18.解:(1)∵,且, ∴, ·······························································2 分 则有, ∴, ·····································································4 分 ∵B 为三角形的内角, ∴,∴. ·····································································5 分 ∵C 为三角形的内角,∴. ································································6 分 (2)∵∴ ······················································8 分 又∵, ∴, ································································10 分 ∴,当且仅当时取等号. 故的最小值为 12. ············································································12 分 19.解:(1)证明:在中,, 由余弦定理,得,∴, ∴,即. ····································································2 分 ∵, 故⊥平面. ···············································································3 分 ∵平面, ∴. ························································································4 分 又∵, ∴⊥平面. ·················································································6 分 (2)由题意,知两两垂直,.如图,以为坐标原点,所在的直线分别为轴,轴,轴建立空间直角坐标 系,则,,,, ,,. ·······································7 分 高三·数学（理科） 第 3页（共 4页） 设,由, 得, 得, ∴ ·················································································9 分 设平面 MBD 的法向量为, 则⊥⊥, ∴即 ·····························································10 分 令,得平面的一个法向量为. 设直线与平面所成的角为 则. ···············································12 分 20.解:(1)由已知可得的定义域为. ∵, ∴, ············································································1 分 ∴, ····························································································2 分 ∴. ··········································································3 分 令,得,令,得, ···········································4 分 ∴的单调递增区间为,单调递减区间为. ·······························5 分 (2)不等式可化为. ·········6 分 令, 则, ····················································7 分 令,则的对称轴为. ①当,即时,易知在上单调递减, ∴, ·········································································8 分 若,则,∴, ∴在上单调递减,,不合题意. ······························9 分 若,则,∴必存在,使得时,,∴, ∴在上单调递增,恒成立,符合题意. ······················10 分 ②当,即时,易知必存在,使得在上单调递增, ∴,∴, ∴在上单调递增,恒成立,符合题意. ······················11 分 综上所述,的取值范围是. ···························································12 分 21.解:(1)由题意,设过点的直线的斜率为,则,设,. ∵,∴, ∴①. ·······························································2 分 联立直线与抛物线方程,得整理,得 则②. ································································3 分 由①②,得, ∴ ,解得, ·············································································4 分 ∴直线的方程为. ·································································5 分 (2)证明:根据(1),联立直线与抛物线的方程,得整理,得 则. ···································································7 分 ∵,,∴. 高三·数学（理科） 第 4页（共 4页） ∴,. ···········································8 分 ∵ . ··································································································10 分 ∴∥. ·······················································································12 分 22.解:(1)∵曲线的参数方程为(为参数), 消去参数,得, ································································2 分 由得, ··························································3 分 ∴曲线的极坐标方程为. ························································5 分 (2)∵, ∴直线的方程为. ······························································7 分 联立解得或 ········································9 分 ∴交点的极坐标为和. ························································10 分 23.解:(1) ···································1 分 ∵, ∴或或 ·······················3 分 ∴或或, ∴或, ··················································································4 分 ∴不等式的解集为或. ························································5 分 (2)证明:由(1)知, ∴, ∴, ····················································································7 分 ∴, ···9 分 当且仅当时取等号, ∴. ·········································································10 分