湖北省普通高中联考协作体2020届高三上学期期中考试 数学（文） 12页

‎2019年秋季湖北省重点高中联考协作体期中考试 高三数学文科试卷 考试时间：2019年11月12日下午15:00 —17:00 试卷满分：150分 ‎★祝考试顺利★‎ 注意事项：‎ ‎1.答卷前，考生务必将自己的学校、考号、班级、姓名等填写在答题卡上。‎ ‎2.选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号，答在试题卷、草稿纸上无效。‎ ‎3.填空题和解答题的作答：用0.5毫米黑色签字笔直接答在答题卡上对应的答题区域内，答在试题卷、草稿纸上无效。‎ ‎4.考生必须保持答题卡的整?吉.考试结束后，将试题卷和答题卡一并交回。‎ 第I 卷 选择题(共60分）‎ 一、选择题（本大题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的。)‎ ‎1.已知复数，则 A. 1+3i B. 1-3i C.-1 + 3i D.-1-3i ‎2.已知集合 A= {-2，-1,0,1，},B={}，则 ‎ A. {-1，0，1，2} B. { 0,1,2} C. { -1,0,1} D. {-2，-1，0，1，2}‎ ‎3.产品质检实验室有5件样品，其中只有2件检测过某成分含量.若从这5件样品中随机取出3 件，则恰有2件检测过该成分含量的概率为 A. B. C. D. ‎ ‎4.已知向量满足，则 A.5 B.-5 C.6 D.6‎ ‎5.函数的图象与圆所围成图形较小部分的面积是 ‎ A. B. C. D. ‎ ‎6.已知方程表示焦点在轴的双曲线，则的取值范围是 ‎ A. -2b>0)，点O为坐标原点，点M满足，OM所在直线的斜率为. ‎ ‎(I)试求椭圆的离心率；‎ ‎(II)设点C的坐标为（0，-b），N为线段AC的中点，证明MN⊥AB.‎ ‎21.(本小题满分12分）‎ ‎ 已知函数，曲线在点(1，)处的切线与直线垂直.‎ ‎(I)求a的值.‎ ‎(II)令，是否存在自然数,使得方程在 内存在唯一的根？ 如果存在，求出，如果不存在，请说明理由.‎ ‎(二）选做题：共10分.请考生在第22,23题中任选一题作答，如果多做，则按所做的第一题计分.‎ ‎22.(本小题满分10分)选修4 —4:坐标系与参数方程 ‎ 在直角坐标系中，直线的参数方程为为参数），以原点为极点，轴正半轴为极轴，建立极坐标系，⊙C的极坐标方程为.‎ ‎(I)写出⊙C的直角坐标方程；‎ ‎(II)P为直线(上的一动点，当P到圆心C的距离最小时，求P的直角坐标.‎ ‎23.(本小题满分10分)选修4 — 5 :不等式选讲 ‎ 已知关于的不等式的解集为{}.‎ ‎(1)求实数的值；‎ ‎(2)求的最大值.‎ ‎ 2019秋高三文数参考答案及评分细则 一、选择题：本大题共12小题，每小题5分，共60分.‎ ‎1.答案：C，注意是求z的共轭复数.‎ ‎2.答案：D，集合，故 ‎3.答案：B，列举后容易知道，基本事件总数有10种，恰有2件检测过该成分含量的事件共有3种，所以所求概率为 ‎4.答案：B，‎ ‎5.答案：D，如下图，所围成的图形的面积，‎ ‎6.答案：B，易知即 ‎7.答案：C，①l不变，有l∥α且l⊥βα⊥β；②m不变，有m∥α且α⊥βm⊥β；③n不变，有α∥β且n⊥αn⊥β；分析知①，③正确.‎ ‎8.答案：A，‎ ‎9.答案：C，由对数运算的性质知，，所以l=n，又为增函数，时，，所以m>l,所以有 ‎10答案：C，（1）错，（2）（3）（4）对 ‎11.答案：A，假设C为钝角，则，，显然充分性不成立，又由可知，即，此时有，即A为钝角或B为钝角，从而△ABC为钝角三角形，必要性成立 ‎12.答案：D，由知，令，则所以有，即的图像关于直线对称.当时，；当时，。作出的图像可知，当时，有两个零点.‎ 二、填空题（本大题共4小题，每小题5分，共20分）‎ ‎13.答案：‎ ‎14答案：‎ 5. 答案:,‎ 解析：由题意有: ‎ 故最小正周期为,最小值为.‎ 16. 答案：‎ 解析：设每天生产A药品x吨，B药品y吨，利润，则有作出可行域知，z在点处取得最大值.‎ 三、解答题：本大题共6小题，共70分。解答应写出必要的文字说明。证明过程或者演算步骤。第17-21题为必考题，每个试题考生必须作答，第22,23题为选考题，考生根据要求作答。 （一）必考题：共60分 ‎17.解：（I）由直方图可知，用户所用流量在区间内的频率依次是0.1,0.15,0.2,0.25,0.15，········································································································3分 所以该月所用流量不超过3GB的用户占85%，所用流量不超过2GB的用户占45%，故k至少定为3；‎ ‎·····································································································································6分 ‎（II）由所用流量的频率分布图及题意，用户该月的人均流量费用估计为：‎ ‎2×1×0.1+2×1.5×0.15+2×2×0.2+2×2.5×0.25+3×2×0.15+(3×2+0.5×4)×0.05+(3×2+1×4)×0.05+(3×2+1.5×4)×0.05=5.1元······················································································12分 ‎18.解：（I）设等差数列的公差为d，因为，所以 又，所以d=2，即，································ ··································3分 设正项等比数列的公比为q，因为即，由，知，所以·················································································································6分 （II）······················································································8分 设，则 ‎····································································································································12分 8. 解：（I）证明:如图,由直三棱柱知，··············································2分 又M为BC的中点知AM⊥BC,又,所以·······································4分 又AMÌ平面AMN,所以平面AMN⊥平面B1BCC1·······································································6分 ‎(II)如图：设AB的中点为D,连接A1D,CD.因为△ABC是正三角形,所以CD⊥AB.由直三棱柱知CD⊥AA1.所以CD⊥平面A1ABB1,所以∠CA1D为直线A1C与平面A1ABB1所成的角.即∠CA1D=30°,···8分 所以A1C=2CD=2×=,所以A1D=6,在Rt△AA1D中，AA1=，NC=·······························································································10分 三棱锥的体积即为三棱锥的体积,所以V=···································································12分 ‎20.解：（I）由，，知,··································2分 由kOM=知 ‎······································································································4分 所以,,所以e=.·································································6分 ‎(II)证明:由N是AC的中点知,点N,所以，···············································8分 又，所以·····················································10分 由（I）知，即，所以=0，‎ 即MN⊥AB. ····················································································································12分 ‎21.解:(I)易知切线的斜率为2,即,又,所以a=1; ················· ·············4分 ‎ (II)设,‎ 当时,.又所以存在,使得.······················································································································6分 又··························································································8分 所以当时,‎ ‎,‎ 当[2,)时,即时,为增函数,所以时,方程在 内存在唯一的根. ···············································································································12分 ‎（二）选考题：共10分.请考生在第22,23题中任选一题作答，如果多做，则按所做的第一题计分.‎ ‎22.解: （I）由知所以所以⊙C的直角坐标方程为··········································································································5分 (II)由(I)知⊙C的标准方程为,即圆心,设P点坐标为，则，所以当t=0时，|PC|有最小值，此时P点坐标为(6,0).·····························································································································10分 ‎23.解：（I）由知，所以即；······························5分 ‎（II）依题意知:‎ ‎····························································8分 当且仅当即时等号成立，‎ 所以所求式子的最大值为.··························································································10分 版权所有：高考资源网(www.ks5u.com)‎