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）
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. -2
b>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
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