2020湖北省重点高中联考协作体高三上学期期中考试数学（文）含答案

这是一份2020湖北省重点高中联考协作体高三上学期期中考试数学（文）含答案

www.ks5u.com2019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洁。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z＝i3(3－i)，则z＝A.1＋3i B.1－3i C.－1＋3i D.－1－3i2.已知集合A＝{－2，－1，0，1，}，B＝{x|x2－4≤0}，则A∩B＝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a，b满足a·b＝1，|b|＝2则(3a－2b)·b＝A.5 B.－5 C.6 D.65.函数y＝|x|＋1的图象与圆x2＋(y－1)2＝4所围成图形较小部分的面积是A. B. C. D.π6.已知方程表示焦点在x轴的双曲线，则m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e；(Il)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f(x)＝(x＋a)lnx，曲线y＝f(x)在点(1，f(1))处的切线与直线x＋2y＝0垂直。(I)求a的值；(Il)令，是否存在自然数n，使得方程f(x)＝g(x)在(n，n＋1)内存在唯一的根?如果存在，求出n，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xOy中，直线l的参数方程为(t为参数)，以原点为极点，x轴正半轴为极轴，建立极坐标系，⊙C的极坐标方程为ρ＝4sinθ。(I)写出⊙C的直角坐标方程；(II)P为直线l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x的不等式|x＋a|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：答案:,解析：由题意有: 故最小正周期为,最小值为.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AM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