2018-2019年长沙市质检二：湖南省长沙市2018届高三教学质量检测(二)文综历史题(WORD版)-附答案精品
2018年湖南省长沙市高考模拟试卷(二模)地理试题及答案
科目：地理（试题卷）本试卷分第Ⅰ卷（选择题）和第Ⅱ卷两部分。
第Ⅰ卷1至7页，第Ⅱ卷7至15页，共300分。
注意事项：1. 答题前，考生务必将自己的姓名、准考证号写在答题卡上。
考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名、考试科目”与考生本人准考证号、姓名是否一致。
2. 第Ⅰ卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。
第Ⅱ卷用黑色墨水签字笔在答题卡上书写作答。
在试题卷上作答，答案无效。
3. 考试结束，监考员将试题卷、答题卡一并收回。
姓名准考证号绝密★启用前湖南省长沙市高考模拟试卷（二模）长沙市教科院组织名优教师联合命制本试题卷分选择题和非选择题两部分，共l5页。
时量l50分钟，满分300分。
第Ⅰ卷（选择题，共140分）一、选择题：本卷共35个小题，每小题4分，共140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。
下图为南半球某地某日海平面等压线分布示意图。
读图完成1～2题。
1．图中从甲地到乙地的气流方向，正确的是A．①B．②C．③D．④2．此时，图中a、b、c、d四地最有可能出现阴雨天气的是A．a B．b C．c D．d读右图“某国樱花开放日期等值线图”，完成3～4题。
3．据图判断该国平原地形主要分布在A．东部B．西部C．北部D．南部4．影响该国樱花开放日期早晚的主要因素有①降水②地形③太阳辐射④土壤⑤洋流A．②③B．③⑤C．②③⑤D．①②④⑤水库具有多种功能，如防洪、发电、灌溉、养殖等，并且能改善局部地区的气候。
下图是某水库蓄水前后对库区周围地区降水量的影响示意图。
读图回答第5题.5．导致水库中心区蓄水后年降水量变化的最主要原因是A．蓄水后夏季水域增温慢，大气的上升运动不旺盛B．常年在高气压的控制下C．水库中心区蓄水后冬季气温升高，大气上升运动旺盛D．蓄水后水汽蒸发量变大读某工业“候鸟”示意图，回答6～7题。
6．该工业可能是A．汽车工业B．纺织工业C．化学工业D．电子工业7．该工业的转移所体现的主要区位因素的变化是A．科技→劳动力B．能源→技术C．原材料→政策D．劳动力→原材料下图为“乌拉尔河水系示意图”，①②③为三个水文观测站。
湖南省长郡中学2018届高三月考二英语试题Word版含答案
第二部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
AAn Oceans VacuumThere’s a collection of plastic trash in the middle of the Pacific Ocean. It’s bigger than Texas-and growing. The way to clean it up now is to catch it with nets. That is both costly and slow. Instead, the Ocean Cleanup Project proposes 62-mile-long floating barriers that would use natural currents to trap trash. If next year’s trials succeed, a full cleanup operation would aim to start in 2020. It could reduce the trash by 42% over 10 years.Easy-On ShoesIn 2012, Mathew Walzer, a high school student with a disability, sent a note to Nike. “My dream is o go to college,” he wrote, “without having to worry about someone coming to tie my shoes every day.”Nike assigned a design team to the challenge. This year, they came out with their solution: the FlyEase. The basketball shoe can be fastened with one hand. A pair of Nike FlyEase shoes sells for $130.An Airport for Drones(无人机)As Amazon, Google, and others get ready for drone delivery service, there is one big question: what kinds of home bases will their drones have? Rwanda, in Africa, may have the answer. There, workers will soon start work on three “drone ports”. The goal is to make it easier to transport food, medical supplies, electronics, and other goods through the hilly countryside. Construction is set to be completed in 2020.21.What’s the advantage of the Oceans Vacuum?A. It can be a money-saverB. It can grow year by yearC. It can tear plastic into piecesD. It can be put into wide use soon22.What do we know about Nike?A. It offers free shoes to the disabledB. It is designing new shoes frequentlyC. It provides customer-friendly servicesD. It responded to Matthew’s request passively23.Why is Rwanda setting up “drone ports”?A. Because road travel there is roughB. Because there are too many dronesC. Because they’re easier to construct than roadsD. Because they are receptive to new technologyBI grew up in a troubled home in the 1970s, on the outskirts of downtown Orlando, Florida. Not far away, a three-story house attracted my eyes.It was nothing like the one I lived in with my mother, a small dark place with rules about befriending others. “Don’t. Never, ever talk to anyone,” my mother said.One day, in sixth grade, a black-haired woman was introduced to our class: Mrs. Reese. Reese explained that she was starting Spanish Club. She invited anyone interested in learning Spanish language and culture to stay after school.I could not take my eyes off her bracelets(手镯) and shining rings. The bell rang, and to my shock, no one went up to Mrs. Reese. I was under strict orders to go straight home. But that day, I stayed. I asked Mrs. Reese when the club started.“We could begin right now if you like,”she said with a smile. I felt beautiful. That day I learned that the house of my dreams was her house. I learned how to answer questions about my age and my favorite food in Spanish. And I learned, Do you want to come over tomorrow for cooking lessons?I wanted to say “Yes”, but Mom’s words held me back.I begged my mother all summer and into fall, well after Spanish Club had dissolved. I wept at night sometimes, so worried that Mrs. Reese and her family would move away.At some point, I managed to wear my mother down and one Saturday afternoon. I rode out to Mrs. Reese’s house.The details of that afternoon are marked in my mind: We had tea. She painted my toenails red. We made a garlicky picadillo. We spoke in Spanish. In Spanish, my voice was loud and romantic. This is the real me! I remember thinking.My mother never permitted me another visit to Mrs. Reese’s house. But four decades later, I still remember that day and the life she showed me, proof of a possible future.24.What kind of family was the author from?A. Hard-upB. Two-parentC. Stress-freeD. Disease-ridden25.Why did the author choose to join the club?A. She wanted to stay longer at schoolB. She intended to comfort Mrs. ReeseC. She was deeply attracted by Mrs. ReeseD. She hoped to befriend the owner of her dreamt house26.The author went to Mrs. Reese’s house .A. with the help of her tearsB. while no one was noticingC. with her mother’s permissionD. just before the lady moved away27.What did the author gain from Mrs. Reese?A. The beauty of SpanishB. The wonder of a new worldC. The power of self-confidenceD. The importance of independenceCEnglish is full of colorful phrases to describe shyness. Someone shy might be called shrinking violet or a wallflower, while for especially nervous types we have the curious expression: they wouldn’t say boo to a goose.None of these are traditionally seen as positive descriptions, even if you like geese. In a culture of go-getting, high achievers, shy people don’t come first.Or that's what the self-help industry would have you believe. Bookshops are filled with vital tomes(巨著) that promise to help beat social fears and find success in life, love and business. That is why one book, Shrinking Violets: A Field Guide to Shyness, bucks the trend. It became a sudden success across English-language media recently for its new take-on shyness.Author Joe Moran says that despite struggling with shyness and longing for loneliness all his life, being shy can also be "a gift". Freed from the constant urge to participate and compete in social situations, people are liberated to look at the world in new ways, and gain fresh insights.Indeed, many of the world's great thinkers and artists are introverts(内向的人). Scientists Charles Darwin and Albert Einstein preferred their own company; actress Keira Knightley often finds herself tongue-tied at parties; and Harry Potter author JK Rowling claims she used to be too nervous to even borrow a pen.Moran told BBC Future: "I think shyness probably does turn you into an amateur anthropologist(人类学家), really-you are more likely to be an observer."So, while extroverts make all the noise, they don't necessarily have the best ideas.If you're shy, you've probably known this for a long time. You just don't shout about it.28.When someone is being called a wallflower, he is being .A. praised for his graceB. admired for his characterC. laughed at for his shynessD. told off for his nervousness29.The underlined phrase “bucks the trend” in Paragraph 2 probably means ””.A. going against the trend and succeedsB. changing the public idea completelyC. becoming unpopular and unacceptedD. becoming the major concern of people30.The author mentioned many famous shy people in order to .A. point out the harm shyness bringsB. disconnect shyness and successC. shows the reasons for shynessD. prove shyness contributes to science31.What is the author’s attitude towards shyness?A. OpposedB. IndifferentC. SupportiveD. CriticalDFrigatebirds seagoing fliers with a 6-foot wingspan, can stay aloft(up in the air) for weeks at a time, a new study has found.Since the frigatebird spends most of its life at sea, its habits outside of when it reproduces on land aren’t well-known-until researchers started tracking them around the Indian Ocean. What the researchers discovered is that the bird’s flying ability is unbelievable.Ornithologist(鸟类学家) Henri Weimerskirch put satellite tage(标签) on a couple of dozen frigatebirds. When the data started to come in, he could hardly believe how high the birds flew."First, we found, 'Whoa, 1,500 meters. Excellent,' " says Weimerskirch, "And after 2,000, after 3,000, after 4,000 meters-OK, at this altitude they are in freezing conditions, especially surprising for a tropical bird.""There is no other bird flying so high relative to the sea surface," he says. "It's the only bird that is known to intentionally enter into a cloud," Weimerskirch says. And not just any cloud-a soft, white cumulus cloud(积云). Over the ocean, these clouds tend to form in places where warm air rises from the sea surface. The birds take a ride on the current of rising air, all the way up to the top of the cloud.Frigatebirds have to find ways to stay aloft because they can't land on the water. Since their feathers aren't waterproof, the birds would drown in short order. They feed by harassing other birds in flight until they bring whatever fish they've swallowed back into their mouth and the frigatebird takes it.So in between meals, frigatebirds fly higher... and higher.In one case, for two months-continuously aloft.One of the tagged birds flew 40 miles without a wing-flap. Several covered more than 300 miles a day on average, and flew continuously for weeks. They are blessed with an unusual body. No bird has a larger wing surface area compared with body weight.32.How did researchers feel when data about frigatebirds reached them?A. CalmB. SurprisedC. HopefulD. Anxious33.According to the text, how can frigatebirds fly so high?A. By flying into a cloudB. With the help of researchersC. Thanks to advanced technologyD. By following other birds into the sky34.What does the underlined word ”they” in the text refer to?A. FrigatebirdsB. Other birdsC. Small fishD. Larger fish35.In what aspect are frigatebirds different from other birds?A. When they give birthB. What they feed onC. Their body weightD. Their wing surface area第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
2018-2019年湖南省长沙市质检一：长沙市2018届高三第一次质量检测理科综合试题(WORD版)-附答案精品
青霄有路终须到，金榜无名誓不还！
2018-2019年高考备考
湖南省长沙市2018届高三第一次质量检测
理科综合能力测试
本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，共16页。
满分300分。
考试用时150分钟。
考试结束后，将本试卷和答题卡一并交回。
注意事项：
1.答题前，考生务必用0.5毫米黑色签字笔将自己的姓名、座号、考生号、县区和科类填写到
答题卡和试卷规定的位置上。
2.第Ⅰ卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用
橡皮擦干净后，再选涂其他答案标号。
写在本试卷上无效。
3.第Ⅱ卷必须用0.5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域内相应的位
置；如需改动，先划掉原来的答案，然后再写上新的答案；不能使用涂改液、胶带纸、修正带。
不按以上要求作答的答案无效。
可能用到的相对原子质量：H 1 C 12 N 14 O 16 Cl 35.5 Ni 59 Cu 64 Zn 65
第I卷
一、选择题：本题共13小题，每小题6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.下列关于酶的叙述，正确的是
A．酶的合成一定需要核糖体
B．同一种酶不可能存在于不同种类的细胞中
C．DNA聚合酶参与DNA复制及DNA的转录
D．所有活细胞都具有与细胞呼吸有关的酶
2.下图表示雄果蝇细胞分裂过程中DNA含量的变化。
下列叙述错误的是。
2018年湖南省长沙市长郡中学高考数学二模试卷(理科)(解析版)
2018年湖南省长沙市长郡中学高考数学二模试卷（理科）一、选择题：本大题共12个小题，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．（5分）设集合A＝{x|x≤2}，B＝{x|0＜x＜3}，则A∪B＝（）A．{x|x≤2}B．{x|x＜3}C．{x|2＜x＜3}D．{x|2≤x＜3} 2．（5分）若iz＝﹣1+i，则复数z的共轭复数在复平面内对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限3．（5分）设曲线C是双曲线，则“C的方程为”是“C的渐近线方程为y＝±2x”的（）A．充分而不必要条件B．必要而不充分条件C．充分必要条件D．既不充分也不必要条件4．（5分）若2m＞2n＞1，则（）A．B．C．ln（m﹣n）＞0D．πm﹣n＞15．（5分）某几何体的三视图如图所示，其中俯视图为扇形，则该几何体的体积为（）A．B．C．D．6．（5分）我们可以用随机模拟的方法估计π的值，如图示程序框图表示其基本步骤（函数RAND是产生随机数的函数，它能随机产生（0，1）内的任何一个实数）．若输出的结果为786，则由此可估计π的近似值为（）A．3.126B．3.144C．3.213D．3.1517．（5分）已知函数f（x）＝sin（ωx+φ）（ω＞0，），其图象相邻两条对称轴之间的距离为，将函数y＝f（x）的图象向左平移个单位后，得到的图象关于y轴对称，那么函数y＝f（x）的图象（）A．关于点对称B．关于点对称C．关于直线对称D．关于直线对称8．（5分）《中国诗词大会》（第二季）亮点颇多，十场比赛每场都有一首特别设计的开场诗词，在声光舞美的配合下，百人团齐声朗诵，别有韵味．若《将进酒》《山居秋暝》《望岳》《送杜少府之任蜀州》和另外确定的两首诗词排在后六场，《将进酒》与《望岳》相邻且《将进酒》排在《望岳》的前面，《山居秋暝》与《送杜少府之任蜀州》不相邻且均不排在最后，则后六场开场诗词的排法有（）A．144种B．48种C．36种D．72种9．（5分）已知椭圆E：的右焦点为F，短轴的一个端点为M，直线l：3x﹣4y＝0交椭圆E于A，B两点，若|AF|+|BF|＝6，点M与直线l的距离不小于，则椭圆E的离心率的取值范围是（）A．B．C．D．10．（5分）已知变量x，y满足条件则目标函数的最大值为（）A．B．1C．D．11．（5分）已知球O是正三棱锥（底面为正三角形，顶点在底面的射影为底面中心）A﹣BCD的外接球，BC＝3，，点E在线段BD上，且BD＝6BE，过点E作球O的截面，则所得截面圆面积的取值范围是（）A．B．C．D．12．（5分）已知函数f（x）的导函数为f'（x），且对任意的实数x都有f'（x）＝e﹣x（2x+3）﹣f（x）（e是自然对数的底数），且f（0）＝1，若关于x的不等式f（x）﹣m＜0的解集中恰有两个整数，则实数m的取值范围是（）A．（﹣e，0]B．[﹣e2，0）C．[﹣e，0）D．（﹣e2，0]二、填空题（每题5分，满分20分，将答案填在答题纸上）13．（5分）的展开式中的常数项是．14．（5分）已知数列{a n}的首项为3，等比数列{b n}满足，且b1009＝1，则a2018的值为．15．（5分）如图，在平面四边形ABCD中，∠A＝45°，∠B＝60°，∠D＝150°，AB＝2BC ＝8，则四边形ABCD的面积为．16．（5分）如图所示，将一圆的八个等分点分成相间的两组，连接每组的四个点得到两个正方形，去掉两个正方形内部的八条线段后可以形成一个正八角星．设正八角星的中心为O，并且，，若将点O到正八角星16个顶点的向量都写成，λ、μ∈R的形式，则λ+μ的取值范围为．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．（12分）已知函数．（I）求函数f（x）的最小正周期；（II）求函数f（x）在区间上的最值及相应的x值．18．（12分）如图，已知在四棱锥P﹣ABCD中，O为AB中点，平面POC⊥平面ABCD，AD∥BC，AB⊥BC，P A＝PB＝BC＝AB＝2，AD＝3（1）求证：平面P AB⊥面ABCD（2）求二面角O﹣PD﹣C的余弦值．19．（12分）1995年联合国教科文组织宣布每年的4月23日为世界读书日，主旨宣言为“希望散居在全球各地的人们，都能享受阅读带来的乐趣，都能尊重和感谢为人类文明作出巨大贡献的文学、文化、科学思想的大师们，都能保护知识产权．”为了解大学生课外阅读情况，现从某高校随机抽取100名学生，将他们一年课外阅读量（单位：本）的数据，分成7组[20，30），[30，40），…，[80，90），并整理得到如下频率分布直方图：（Ⅰ）估计其阅读量小于60本的人数；（Ⅱ）已知阅读量在[20，30），[30，40），[40，50）内的学生人数比为2：3：5．为了解学生阅读课外书的情况，现从阅读量在[20，40）内的学生中随机选取3人进行调查座谈，用X表示所选学生阅读量在[20，30）内的人数，求X的分布列和数学期望；（Ⅲ）假设同一组中的每个数据可用该组区间的中点值代替，试估计100名学生该年课外阅读量的平均数在第几组（只需写出结论）．20．（12分）椭圆的左右焦点分别为F1，F2，与y轴正半轴交于点B，若△BF1F2为等腰直角三角形，且直线BF1被圆x2+y2＝b2所截得的弦长为2．（1）求椭圆的方程；（2）直线l与椭圆交于点A、C，线段AC的中点为M，射线MO与椭圆交于点P，点O为△P AC的重心，探求△P AC的面积S是否为定值，若是求出这个值，若不是，求S的取值范围．21．（12分）设函数．（Ⅰ）讨论函数f（x）的单调性；（Ⅱ）若x≥0时，恒有f（x）≤ax3，试求实数a的取值范围；（Ⅲ）令，试证明：．请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分．[选修4-4：坐标系与参数方程]22．（10分）在直角坐标系xoy中，直线l的方程是，曲线C的参数方程为（α为参数），以O为极点，x轴的非负半轴为极轴建立极坐标系．（Ⅰ）求直线l和曲线C的极坐标方程；（Ⅱ）射线OM：θ＝β（其中）与曲线C交于O，P两点，与直线l交于点M，求的取值范围．[选修4-5：不等式选讲]23．设函数f（x）＝|2x﹣1|．（1）设f（x）+f（x+1）＜5的解集为集合A，求集合A；（2）已知m为集合A中的最大自然数，且a+b+c＝m（其中a，b，c为正实数），设．求证：M≥8．2018年湖南省长沙市长郡中学高考数学二模试卷（理科）参考答案与试题解析一、选择题：本大题共12个小题，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．（5分）设集合A＝{x|x≤2}，B＝{x|0＜x＜3}，则A∪B＝（）A．{x|x≤2}B．{x|x＜3}C．{x|2＜x＜3}D．{x|2≤x＜3}【解答】解：集合A＝{x|x≤2}，B＝{x|0＜x＜3}，则A∪B＝{x|x＜3}．故选：B．2．（5分）若iz＝﹣1+i，则复数z的共轭复数在复平面内对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限【解答】解：∵iz＝﹣1+i，∴z＝，∴，则z的共轭复数在复平面内对应的点的坐标为（1，﹣1），位于第四象限．故选：D．3．（5分）设曲线C是双曲线，则“C的方程为”是“C的渐近线方程为y＝±2x”的（）A．充分而不必要条件B．必要而不充分条件C．充分必要条件D．既不充分也不必要条件【解答】解：C的方程为，则双曲线的渐近线方程为y＝±2x，即充分性成立，双曲线﹣x2＝1的渐近线方程也是y＝±2x，即必要性不成立，故“C的方程为”是“C的渐近线方程为y＝±2x”的充分不必要条件，故选：A．4．（5分）若2m＞2n＞1，则（）A．B．C．ln（m﹣n）＞0D．πm﹣n＞1【解答】解：∵2m＞2n＞1，∴m＞n＞0，∴，m＜，ln（m﹣n）与0的大小关系不确定，πm﹣n＞1．因此只有D正确．故选：D．5．（5分）某几何体的三视图如图所示，其中俯视图为扇形，则该几何体的体积为（）A．B．C．D．【解答】解：由三视图知几何体是圆锥的一部分，由俯视图与左视图可得：底面扇形的圆心角为120°，又由侧视图知几何体的高为4，底面圆的半径为2，∴几何体的体积V＝××π×22×4＝．故选：D．6．（5分）我们可以用随机模拟的方法估计π的值，如图示程序框图表示其基本步骤（函数RAND是产生随机数的函数，它能随机产生（0，1）内的任何一个实数）．若输出的结果为786，则由此可估计π的近似值为（）A．3.126B．3.144C．3.213D．3.151【解答】解：根据已知中的流程图我们可以得到该程序的功能是利用随机模拟实验的方法求任取（0，1）上的两个数x，y，求x2+y2≤1的概率，∵x∈（0，1），y∈（0，1），对应的平面区域面积为：1×1＝1，而x2+y2＜1对应的平面区域的面积为：π，故由题意可得：＝，解得：π＝3.144，故选：B．7．（5分）已知函数f（x）＝sin（ωx+φ）（ω＞0，），其图象相邻两条对称轴之间的距离为，将函数y＝f（x）的图象向左平移个单位后，得到的图象关于y轴对称，那么函数y＝f（x）的图象（）A．关于点对称B．关于点对称C．关于直线对称D．关于直线对称【解答】解：由函数y＝f（x）图象相邻两条对称轴之间的距离为，可知其周期为T＝π，所以ω＝＝4，所以f（x）＝sin（4x+φ）；将函数y＝f（x）的图象向左平移个单位后，得到函数y＝sin[4（x+）+φ]图象．因为得到的图象关于y轴对称，所以4×+φ＝kπ+，k∈Z，即φ＝kπ﹣，k∈Z；又|φ|＜，所以φ＝﹣，所以f（x）＝sin（4x﹣），令4x﹣＝kπ，k∈Z，解得x＝+，k∈Z；k＝0时，得f（x）的图象关于点（，0）对称，B正确．故选：B．8．（5分）《中国诗词大会》（第二季）亮点颇多，十场比赛每场都有一首特别设计的开场诗词，在声光舞美的配合下，百人团齐声朗诵，别有韵味．若《将进酒》《山居秋暝》《望岳》《送杜少府之任蜀州》和另外确定的两首诗词排在后六场，《将进酒》与《望岳》相邻且《将进酒》排在《望岳》的前面，《山居秋暝》与《送杜少府之任蜀州》不相邻且均不排在最后，则后六场开场诗词的排法有（）A．144种B．48种C．36种D．72种【解答】解：根据题意，分2步分析：①，将《将进酒》与《望岳》捆绑在一起和另外确定的两首诗词进行全排列，共有种排法，②，再将《山居秋暝》与《送杜少府之任蜀州》插排在3个空里（最后一个空不排），有种排法，则后六场的排法有＝36（种），故选：C．9．（5分）已知椭圆E：的右焦点为F，短轴的一个端点为M，直线l：3x﹣4y＝0交椭圆E于A，B两点，若|AF|+|BF|＝6，点M与直线l的距离不小于，则椭圆E的离心率的取值范围是（）A．B．C．D．【解答】解：如图所示，设F′为椭圆的左焦点，连接AF′，BF′，则四边形AFBF′是平行四边形，∴6＝|AF|+|BF|＝|AF′|+|AF|＝2a，∴a＝3；取M（0，b），∵点M到直线l的距离不小于，∴≥，解得b≥2；∴e＝＝＝≤＝，∴椭圆E的离心率的取值范围是（0，]．故选：B．10．（5分）已知变量x，y满足条件则目标函数的最大值为（）A．B．1C．D．【解答】解：变量x，y满足条件的可行域如图：目标函数的几何意义是，分母是可行域内的点与坐标原点的距离，分子是直线x﹣y＝u，如图中的红色线，当红色线经过D时目标函数取得最大值．最大值为：＝．故选：C．11．（5分）已知球O是正三棱锥（底面为正三角形，顶点在底面的射影为底面中心）A﹣BCD的外接球，BC＝3，，点E在线段BD上，且BD＝6BE，过点E作球O的截面，则所得截面圆面积的取值范围是（）A．B．C．D．【解答】解：如图，设△BDC的中心为O1，球O的半径为R，连接O1D，OD，O1E，OE，则O1D＝3sin60°×＝，AO1＝＝＝3，在Rt△OO1D中，R2＝3+（3﹣R）2，解得R＝2，∵BD＝6BE，∴DE＝2.5，在△DEO1中，O1E＝＝，∴OE＝＝＝，过点E作圆O的截面，当截面与OE垂直时，截面的面积最小，此时截面圆的半径为＝，最小面积为π，当截面过球心时，截面面积最大，最大面积为4π．故选：A．12．（5分）已知函数f（x）的导函数为f'（x），且对任意的实数x都有f'（x）＝e﹣x（2x+3）﹣f（x）（e是自然对数的底数），且f（0）＝1，若关于x的不等式f（x）﹣m＜0的解集中恰有两个整数，则实数m的取值范围是（）A．（﹣e，0]B．[﹣e2，0）C．[﹣e，0）D．（﹣e2，0]【解答】解：∵f'（x）＝e﹣x（2x+3）﹣f（x），∴e x[f（′x）+f（x）]＝2x+3，∴e x f（x）＝x2+3x+c，∵f（0）＝1，∴1＝0+0+c，解得c＝1∴f（x）＝（x2+3x+1）e﹣x，∴f′（x）＝﹣（x2+x﹣2）e﹣x＝﹣（x﹣1）（x+2）e﹣x．令f′（x）＝0，解得x＝1或x＝﹣2，当x＜﹣2或x＞1时，f′（x）＜0，函数f（x）单调递减，当﹣2＜x＜1时，f′（x）＞0，函数f（x）单调递减增，可得：x＝1时，函数f（x）取得极大值，x＝﹣2时，函数f（x）取得极小值，∵f（1）＝，f（﹣2）＝﹣e2＜0，f（﹣1）＝﹣e，f（0）＝1＞0，f（﹣3）＝e3＞0∴﹣e＜m≤0时，f（x）﹣m＜0的解集中恰有两个整数恰有两个整数﹣1，﹣2．故m的取值范围是（﹣e，0]，故选：A．二、填空题（每题5分，满分20分，将答案填在答题纸上）13．（5分）的展开式中的常数项是﹣11．【解答】解：∵＝（2x+1）•（1﹣+﹣+﹣+），故它的展开式中的常数项是1﹣12＝﹣11，故答案为：﹣11．14．（5分）已知数列{a n}的首项为3，等比数列{b n}满足，且b1009＝1，则a2018的值为3．【解答】解：等比数列{b n}满足，∴lna n+1﹣lna n＝lnb n，∴lna2018﹣lna2017＝lnb2017，lna2017﹣lna2016＝lnb2016，……，lna2﹣lna1＝lnb1，∴lna2018﹣lna1＝ln（b1•b2•……b2017）＝ln＝ln1＝0，∴a2018＝a1＝3．故答案为：3．15．（5分）如图，在平面四边形ABCD中，∠A＝45°，∠B＝60°，∠D＝150°，AB＝2BC＝8，则四边形ABCD的面积为．【解答】解：如图，连接AC，可得∠DCB＝105°在△ABC中，由余弦定理得AC2＝BC2+BA2﹣2BCBA cos60°＝48．∴AB2＝AC2+BC2，∴∠CAB＝30°，∠ACB＝90°，∠DCA＝∠DAC＝15°．∴tan15°．∴四边形ABCD的面积为12×）+8＝24﹣4．故答案为：24﹣4．16．（5分）如图所示，将一圆的八个等分点分成相间的两组，连接每组的四个点得到两个正方形，去掉两个正方形内部的八条线段后可以形成一个正八角星．设正八角星的中心为O，并且，，若将点O到正八角星16个顶点的向量都写成，λ、μ∈R的形式，则λ+μ的取值范围为[﹣1﹣，1+]．【解答】解：以O为原点，以OA为x轴建立平面直角坐标系，如图所示：设圆O的半径为1，则OM＝1，过M作MN∥OB，交x轴于N，则△OMN为等腰直角三角形，∴ON＝，OM＝1，∴＝+，此时λ+μ＝1+；同理可得：＝+＝﹣﹣，此时λ+μ＝﹣1﹣；∴λ+μ的最大值为1+，最小值为﹣1﹣．故答案为：[﹣1﹣，1+]．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．（12分）已知函数．（I）求函数f（x）的最小正周期；（II）求函数f（x）在区间上的最值及相应的x值．【解答】解：（Ⅰ）＝＝，∴f（x）的最小正周期是π；（Ⅱ）∵，∴0≤2x≤π，∴，当时，f（x）max＝2．当时，f（x）min＝﹣1．18．（12分）如图，已知在四棱锥P﹣ABCD中，O为AB中点，平面POC⊥平面ABCD，AD∥BC，AB⊥BC，P A＝PB＝BC＝AB＝2，AD＝3（1）求证：平面P AB⊥面ABCD（2）求二面角O﹣PD﹣C的余弦值．【解答】（1）证明：∵AD∥BC，AB⊥BC，BC＝AB＝2，AD＝3．∴OC＝，OD＝，CD＝，∵OD2＝OC2+DC2＝10，∴OC⊥CD，即CD⊥平面POC，∴CD⊥PO．∵P A＝PB＝AB，O为AB中点，∴PO⊥AB，∴PO⊥底面ABCD，∵PO⊂平面P AB，∴平面P AB⊥面ABCD…（6分）（2）解：过点C作CM⊥OD于点M，过点M作MN⊥PD于点N，连接CN．则由于PO⊥平面OCD，PO⊂平面POD，所以平面POD⊥平面OCD，∵CM⊂平面OCD，平面POD∩平面OCD＝OD，∴CM⊥平面POD，∴CM⊥PD，∵MN⊥PD，MN∩CM＝M，∴PD⊥平面MCN，∴PD⊥NC，即∠MNC是二面角O﹣PD﹣C的平面角．在Rt△OCD中，CM＝＝，在Rt△PCD中，CN＝＝，所以MN＝，所以二面角O﹣PD﹣C的余弦值为．…（12分）19．（12分）1995年联合国教科文组织宣布每年的4月23日为世界读书日，主旨宣言为“希望散居在全球各地的人们，都能享受阅读带来的乐趣，都能尊重和感谢为人类文明作出巨大贡献的文学、文化、科学思想的大师们，都能保护知识产权．”为了解大学生课外阅读情况，现从某高校随机抽取100名学生，将他们一年课外阅读量（单位：本）的数据，分成7组[20，30），[30，40），…，[80，90），并整理得到如下频率分布直方图：（Ⅰ）估计其阅读量小于60本的人数；（Ⅱ）已知阅读量在[20，30），[30，40），[40，50）内的学生人数比为2：3：5．为了解学生阅读课外书的情况，现从阅读量在[20，40）内的学生中随机选取3人进行调查座谈，用X表示所选学生阅读量在[20，30）内的人数，求X的分布列和数学期望；（Ⅲ）假设同一组中的每个数据可用该组区间的中点值代替，试估计100名学生该年课外阅读量的平均数在第几组（只需写出结论）．【解答】解：（Ⅰ）100﹣100×10×（0.04+0.02×2）＝20（人）（Ⅱ）由已知条件可知：[20，50）内人数为：100﹣100×10×（0.04+0.02+0.02+0.01）＝10；[20，30）人数为2人，[30，40）人数为3人，[40，50）人数为5人．X的可能取值为0，1，2．P（X＝0）＝P（X＝1）＝P（X＝2）＝，所以X的分布列为．（Ⅲ）第五组．20．（12分）椭圆的左右焦点分别为F1，F2，与y轴正半轴交于点B，若△BF1F2为等腰直角三角形，且直线BF1被圆x2+y2＝b2所截得的弦长为2．（1）求椭圆的方程；（2）直线l与椭圆交于点A、C，线段AC的中点为M，射线MO与椭圆交于点P，点O为△P AC的重心，探求△P AC的面积S是否为定值，若是求出这个值，若不是，求S的取值范围．【解答】解：（1）根据题意，由△BF1F2为等腰直角三角形可得b＝c，直线BF1：y＝x+b被圆x2+y2＝b2所截得的弦长为2，即BF1＝2，所以a＝2，，所以椭圆的方程为．（2）若直线l的斜率不存在，则．若直线l的斜率存在，设直线l的方程为y＝kx+m，设A（x1，y1），B（x2，y2），即则，，，由题意点O为△P AC重心，设P（x0，y0），则，，所以，，代入椭圆，得，整理得，设坐标原点O到直线l的距离为d，则△P AC的面积＝＝＝．综上可得△P AC的面积S为定值．21．（12分）设函数．（Ⅰ）讨论函数f（x）的单调性；（Ⅱ）若x≥0时，恒有f（x）≤ax3，试求实数a的取值范围；（Ⅲ）令，试证明：．【解答】解：（I）函数的定义域为R，由于f′（x）＝1﹣≥0，知f（x）是R上的增函数．（II）令g（x）＝f（x）﹣ax3＝x﹣ln（x+）﹣ax3．则g′（x）＝，令h（x）＝，则h′（x）＝，（1）当a≥时，h′（x）≤0，从而h（x）是[0，+∞）上的减函数，因h（0）＝0，则x ≥0时，h（x）≤0，也即g′（x）≤0，进而g（x）是[0，+∞）上的减函数，注意g（0）＝0，则x≥0时，g（x）≤0，也即f（x）≤ax3，（2）当0＜a＜时，在[0，]，h′（x）＞0，从而x∈[0，]时，也即f（x）＞ax3，（3）当a≤0时，h′（x）＞0，同理可知：f（x）＞ax3，综合，实数a的取值范围[，+∞）．（III）在（II）中取a＝，则x∈[0，]，时，x﹣ln（x+）＞x3，即x3+ln（x+）＜x，令x＝（）2n，则＜（）2n，∴请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分．[选修4-4：坐标系与参数方程]22．（10分）在直角坐标系xoy中，直线l的方程是，曲线C的参数方程为（α为参数），以O为极点，x轴的非负半轴为极轴建立极坐标系．（Ⅰ）求直线l和曲线C的极坐标方程；（Ⅱ）射线OM：θ＝β（其中）与曲线C交于O，P两点，与直线l交于点M，求的取值范围．【解答】解：（Ⅰ）∵直线l的方程是，∴，∴直线l的极坐标方程是，由，消参数得x2+（y﹣2）2＝4，∴曲线C的极坐标方程是ρ＝4sinθ．…5分（Ⅱ）将θ＝β分别带入ρ＝4sinθ，，得|OP|＝4sinβ，，∴，∵，∴，∴，∴的取值范围是．…10分[选修4-5：不等式选讲]23．设函数f（x）＝|2x﹣1|．（1）设f（x）+f（x+1）＜5的解集为集合A，求集合A；（2）已知m为集合A中的最大自然数，且a+b+c＝m（其中a，b，c为正实数），设．求证：M≥8．【解答】解：（1）f（x）+f（x+1）＜5，即|2x﹣1|+|2x+1|＜5；当时，不等式化为1﹣2x﹣2x﹣1＜5，∴；当时，不等式化为1﹣2x+2x+1＜5，不等式恒成立；当时，不等式化为2x﹣1+2x+1＜5，∴；综上，集合；（2）证明：由（1）知m＝1，则a+b+c＝1；则；同理；则；即M≥8．。
2018年湖南长沙天心区长郡中学高三二模英语试卷-学生用卷
2018年湖南长沙天心区长郡中学高三二模英语试卷-学生用卷一、阅读理解（共15小题，每小题2分，共30分）1、【来源】 2018年湖南长沙天心区长郡中学高三二模第21~23题6分Worship Music Leader—First Presbyterian Church(FPC) in Napa is looking for someone who will select songs and use media for church services on Sundays. The working hours are flexible and the pay is about ＄18 an hour. Detailed job information will be posted on the Payroll Office's window. If interested, send ********************************************************.Catering Company—Oak Avenue is a most successful catering company searching for enthusiastic people to join the talented team of chefs and service professionals. Students' part-time positions are available for ambitious chefs, bakers, service helpers, and captains. Please send a brief cover letter and*************************************************************************.Caregiver Position—Looking for a caregiver to help with a 91-year-old woman who lives in Calistoga. She speaks both English and Spanish, but Spanish is her preference. She sleeps quite a bit during the day so there is time to do housework. She cannot be left alone because she has a fall risk. She walks with a walker on her own. Please call 925-785-8500 if interested.Housekeeping Services—A responsible, professional, and experienced housekeeper is needed to take care of all cleaning details of a local business in Angwin. This person will be needed 2 or 3 times per week for about 8 hours per day. Please send an e-mail for additional information and be ready to provide your resume and work history. Driver's license is required. Pay will be ＄20—＄25 per hour. To contact them, ********************************.(1) Which is the e-mail address or contact number for those who want to become great bakers?*******************.****************************.C. 925-785-8500.D. 707-963-9278.(2) What is required to work as a housekeeper?A. Flexibility in working hours.B. Speaking Spanish fluently.C. Being able to drive.D. Being a local in Angwin.(3) Where does this text probably come from?A. A hotel advertisement.B. A job advertisement.C. A household guide.D. A tourist guide.2、【来源】 2018年湖南长沙天心区长郡中学高三二模（B篇）第24~27题8分Several times my daughter had telephoned to say, "Mum, you must come and see the daffodils（水仙花）before they are over." I wanted to go, but it was a two-hour drive from Lake Arrowhead. "I will go next Tuesday, " I promised, a little unwillingly, on her third call.The next Tuesday dawned cold and rainy. Still. I had promised. And so I drove there. When I finally walked into Carolyn's house and hugged and greeted my grandchildren. I said. "Forget the daffodils. Carolyn! The road is indiscerniblein the cloud and fog, and there is nothing in the world except you and these children that I want to see!"My daughter smiled calmly and said, "We drive in this weather all the time, Mum. You will never forgive yourself if you miss this experience."After about twenty minutes, we turned onto a small road and I saw a small church. On the far side of the church, I saw a hand-lettered sign that read "Daffodil Garden".We got out of the car and each took a child's hand, and I followed Carolyn down the path. Then, we turned a corner of the path, and I looked up and gasped. Before me lay the most beautiful sight! There were five acres of flowers! "But who has done this? " asked Carolyn. "It's just one woman, "Carolyn answered. "That's her home, " Carolyn pointed to a well-kept A-frame house that looked small and modest in the midst of all that glory. We walked up to the house. On the patio（平台）, we saw a poster. "Answers to the Questions I Know You Are Asking" was the headline.The first answer was a simple one. "50, 000 bulbs（鳞茎）it read. The second answer was, "One at a time, by one woman." The third answer was, "Began in 1958."I thought of this woman whom I had never met, who, more than fifty years before, had begun—one bulb at a time—to bring the beauty and joy to the mountain top.(1) Why was the author unwilling to see the daffodils at first?A. The weather was terrible.B. She took little interest in them.C. She wasn't available at that time.D. It was not easy to go there.(2) What does the underlined word "in discernible" in Paragraph 2 mean?A. Invisible.B. Remote.C. Impossible.D. Complex.(3) Which of the following best describe the woman living in the A-frame house?A. Optimistic.B. Productive.C. Determined.D. Generous.(4) Which of the following can be the best title for the text?A. I Love DaffodilsB. An Unbelievable ExperienceC. Gardening as a HobbyD. One Bulb a Time3、【来源】 2018年湖南长沙天心区长郡中学高三二模（C篇）第28~31题8分"It's a big hammer to crack a nut." This is how one angry parent described the recent crackdown（严厉的打击）by London police on parents who drop their kids off at school by car. Yes, you read that correctly—the very mode of transportation that many American schools insist is the only safe way to deliver kids to school is now considered illegal in the UK.The decision to fine any vehicle seen dropping off or picking up kids within a particular zone of east London comes from city councilors'（议员）long-term efforts to make the area safer and less crowded. They say they've been trying for years "to encourage reasonable parking" but in vain. Neighborhood residents complain frequently about their driveways being blocked by illegally parked cars for 15 minutes or more, often while they're trying to get to work, and the streets are long overcrowded.Now the rules have changed. Some parents are angry. Angle is a mother who made the "big hammer" comment and says the crackdown is "way over the top" The nearest drop-off point for her six-year-old is now a five-minute walk from the school. Others are happy with the decision, severe though it may seem.Councilor Jason Frost said, "Traffic has been clearly reduced, and more children are now walking to school, which is a great outcome. I would rather have complaints that we are slightly inconveniencing parents than hear that a child had been seriously injured because nothing was done."I witness daily the chaos created by these in-town drivers, when I walk my own kids to school. There's a parking lot crowded with vehicle and a slow-moving train of cars moving in circles, many filling the air with harmful smoke. Meanwhile, the conversations around overweight children and the importance of daily physical activities continue to stand out in schools.(1) Why is angle opposed to the new rule?A. She is often fined by London police.B. She will lose her job as a school driver.C. She has adapted to the American practice.D. She thinks it adds inconvenience to her kid.(2) Which concerns Jason Frost most?A. People's complaints.B. Students' health.C. Students' safety.D. Parents' inconvenience.(3) What can be seen after the crackdown is carried out?A. Less daily disorder.B. Support from all people.C. More overweight children.D. Complaints from neighbors.(4) What is the author's attitude toward the crackdown?A. Favorable.B. Uncaring.C. Doubtful.D. Opposed.4、【来源】 2018年湖南长沙天心区长郡中学高三二模（D篇）第32~35题8分While sleep patterns vary widely among animal classes and species, there are a few general rules of thumb（经验法则）. Sleep serves important evolutionary functions for all animals, but sleep patterns and positions are based in large part on available food supply and defense mechanism.Sleep patterns in animals have evolved over time—animals that sleep and get attacked by predators （捕食者） are less likely to pass on their genes, so animals have developed ways to protect themselves during sleep. For instance, otters（水獭） sleep holding hands, or cover themselves in seaweed to protect their young and stay afloat while asleep. Likewise, cows and sheep sleep in a group—there is safety in numbers.Evolutionary biologists consider that predation and fear of predators have influenced the development of sleep patterns across species. Animals that eat meat tend to sleep more than animals that eat plants. Cathemeral（间歇性的） species like lions sleep in short spells during both the day and night so as to enable them to seize food when it becomes available.In general, animals sleep according to what they eat—animals that eat food with lower calories sleep less than others. This may explain why plant eaters need to spend more time awake, to ensure they get enough food and energy. For example, plant eaters like giraffes and elephants sleep 30 minutes to just a few hours per day respectively. However, there are exceptions to TAL#NBSP this rule, like the koala. Their eucalyptus（桉树）-based diet doesn't give them much energy, but they sleep for 15 hours per day, and slit the rest of their time eating and resting.A note about the studies referenced in this article: Because most animal sleep studies are performed using EEG tests, they are typically performed on animals in zoos and research facilities. As a result, they may not accurately reflect their natural sleep habits in the wild, given the stresses of a zoo environment and consistent availability of food.(1) Why do cows and sheep sleep in groups?A. To protect themselves.B. To pass on their genes.C. To keep themselves warm.D. To get physical closeness.(2) The underlined phrase "this rule" in Paragraph 4 indicates that animals eatingplants.A. have different sleep patternsB. spend short time eating and restingC. need less time to sleep than meat eatersD. sleep 30 minutes to a few hours per day(3) What's the purpose of the last paragraph?A. To explain the studies' dangers.B. To point out the studies' limitations.C. To describe the studies' detailed procedures.D. To add the studies' background information.(4) What greatly affects animals' sleep patterns?A. Their genes.B. Their eating habits.C. Classes and species.D. Food and safety.二、七选五（共5小题，每小题2分，共10分）5、【来源】 2018年湖南长沙天心区长郡中学高三二模第36~40题10分2017~2018学年5月山东上学期月考名校联盟联考第26~30题10分(每题2分)2018年山东济南章丘区高三一模第36~40题10分(每题2分)The science of papermaking addresses the methods, equipment, and materials used to make paper and cardboard, these being used widely for printing, writing, and packaging. Today almost all paper is manufactured using industrial machinery.1.The method of manual papermaking changed very little over time, despite advances in technologies.2. Separate the useful fibre from the rest of raw materials. Beat down the fibre into pulp（纸浆）. Adjust the colour, chemical, biological, and other properties of the paper. Screening（筛）the resulting liquid. Pressing and drying to get the actual paper.3. It is made of stainless materials and set in a wooden frame similar to that of a window. Then completely bathe it in the liquid vertically and draw it out horizontally. Excesswater is removed and the wet mat of fibre laid on top of a damp cloth. The fairly damp fibre is then dried. Finally, the paper is then cut to the desired shape and packed.The wooden frame is called a "deckle"（定纸框）.4. The "deckle edges" is one of the indications that the paper was made by hand. Deckle﹣edged paper is occasionally mechanically imitated today to create the impression of old﹣fashioned luxury.5. It is also used in paper mills to check the quality of the production process. The "handsheets" made according to TAPPI Standard T 205 are tested for paper characteristics such as brightness, strength and degree of sizing.A. Screening the fibre involves using a special netB. The process of papermaking is really complicatedC. It leaves the edges of the paper slightly irregular and wavyD. Handmade paper is prepared in laboratories to study papermakingE. However, traditional papermaking in Asia uses the inner skin fibers of plantsF. The process of manufacturing handmade paper can be generalized into five stepsG. But handmade paper is still appreciated for its distinctive uniqueness and the skilled craft三、完形填空（共20小题，每小题1.5分，共30分）6、【来源】 2018年湖南长沙天心区长郡中学高三二模第41~60题30分It was a beautiful, sunny summer day. I was on vacation in Sanya with my family.Holidaymakers1the beach. Shouts and cheers filled my ears. My mother was sitting under a big2, reading a book and drinking lemonade. My father was playing volleyball on the3sands. Walking along the beach, I could feel a cool summer breeze4my cheeks. So eager was I to explore the world inthe5of sea that I took off my ring I got as a gift from my parents for graduation（beautiful, gold, inscribed（刻字的）, full of emotional value）and6it in the cup holder of a fold able chair. Beach day continued, and we packed up and7back to the hotel. Upon arrival, Isuddenly8what I'd done—my ring wasn't on my finger.Panic9.We raced back to the beach with only 20 minutes to10before sunset. Get to the beach and locate the11area I laid out earlier. We searched it thoroughly, but without luck. My heart12and tears rolled down my cheeks. Learning what happened, about 25 strangers in the area got down on their hands and knees to helpmy13self search for my ring. And they helped me tothe14. Suddenly someone15and shouted to me "I found it." They all16around me so happy andcongratulatory17they had won the lottery. My tearsof18turned into tears of joy and thankfulness for the amazing effort put forth by these19faces.Just as Marianne Williamson puts it, we20love as kindness, giving, forgiveness, peace, joy and acceptance. Love can show up in a million different forms.A. discoveredB. crowdedC. desertedD. warmedA. roofB. shadowC. umbrellaD. skyA. cuttingB. freezingC. movingD. burningA. brushB. strikeC. impressD. seizeA. depthB. lengthC. strengthD. growthA. placedB. hidC. remainedD. foundA. headedB. flewC. escapedD. reachedA. declaredB. ignoredC. sharedD. realizedA. set inB. went offC. roseD. ranA. wasteB. spareC. saveD. rescueA. commonB. dampC. generalD. calmA. frozeB. crashedC. sankD. sufferedA. beggingB. sobbingC. noddingD. disguisingA. coursedB. shoreC. endD. distanceA. backed offB. stepped forwardC. stood upD. broke inA. assistedB. arguedC. confirmedD. gatheredA. in caseB. even ifC. as thoughD. what ifA. appreciationB. expectationC. ignoranceD. lossA. abnormalB. freshC. unfamiliarD. remoteA. possessB. createC. experienceD. expand四、适当形式填空（共10小题，每小题1.5分，共15分）7、【来源】 2018年湖南长沙天心区长郡中学高三二模第61~70题15分Although Laughter Yoga definitely includes some yoga, don't expect it to have much in common with the yoga classes1（offer） at your local studio or gym. It promoted the use of laughter as a form of physical exercise, attaching2（ much） importance to the former than to the latter.Laughter Yoga3（create） in the mid-1990 s as a social experiment. In 1995, Dr. Madan Kataria called on a group of students in a park4（ test）whether adding regular laughter to people' lives would improve5（they）well-being. Kataria also introduced deep breathing and simple yogamoves6the exercises.Medical research7（suggest） that laughing is a good way to reduce stress and help people feel good. Kataria's Laughter Clubs are popping up in response to people's need to handle8（press）.Kataria also offers books and DVDs to people9don't have access to a local Laughter Club. However, connecting with other humans is10big part of Laughter Yoga's success in helping people feel better and experience more joy.五、短文改错（共10小题，每小题1分，共10分）8、【来源】 2018年湖南长沙天心区长郡中学高三二模第71~80题10分假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。
2018学年湖南省长沙市高三高考模拟(二模)语文试卷
2018学年湖南省长沙市高三高考模拟（二模）语文试卷一、选择题详细信息1.难度：简单下列词语中加点的字，读音全部正确的一组是A．给力（gěi）憎恨(zèng) 压轴戏(zhòu) 唾手可得(chuí)B．应届(yīng) 绯闻(fěi) 绊脚石(bàn) 奉为圭臬(niè)C．卓越(zhuō) 蜗居(wō) 歼击机(jiān) 抢呼欲绝(qiāng)D．宅男(zhái) 潜伏(qián) 涨停板(zhǎng) 容貌姝丽(shū)详细信息2.难度：简单下列各组词语中，没有错别字的一组是A．荒诞家具仓皇失措销声匿迹B．誊写鬼秘顶礼膜拜应接不暇C．告罄领衔鞭辟入理韬光养晦D．矫情赌博不假思索春意阑姗详细信息3.难度：中等下列各句中，没有语病的一句是A．雾霾、环保成为本次两会代表、委员最关注的焦点话题。
有代表提出，我们要加大科技改造的力度，要使碳排放量至少降低一倍。
B．在湖南卫视举办的“2014年春节联欢晚会”上，每年看上去很枯燥无趣的观众抽奖环节因为有了《爸爸去哪儿》剧组五个萌娃串场而显得看点十足，到场的观众们不但获得了快乐和礼物，而且十份特殊的大奖也通过荧屏送给了全球各地的粉丝们。
C．面对复杂多变的国内外环境，面对艰巨繁重的改革发展稳定任务，我们要继续抓住和用好我国发展的重要战略机遇期，有效应对各种风险和挑战。
D．今年初，韩剧《来自星星的你》突然爆红，“炸鸡啤酒都教授”成为网络流行语。
有关人士认为，这部剧爆红的原因是“帅哥美女、专业编剧班底、时时可以参与讨论的剧情”等多重因素共同作用的结果。
详细信息4.难度：简单下面这首宋词《虞美人》（陈与义）中的横线处应填入的最贴切的句子是张帆欲去仍搔首，更醉君家酒。
吟诗日日待春风，及至桃花开后却匆匆。
歌声频为行人咽，记著樽前雪。
明朝酒醒大江流，。
A．一身将影向潇湘 B．满载一船离恨向衡州C．惟有一江明月碧琉璃 D．数帆带雨烟中落二、文言文阅读5.难度：中等阅读下面的文言文，完成后面题。
2018届湖南省长沙市长郡高三模拟卷(二)
湖南省长郡中学2018届高考模拟卷(二)第二部分阅读理解(共两节，满分40分）第一节（共15小题;每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
AWorship Music Leader— First Presbyterian Church(FPC) in Napa is looking for someone who will select songs and use media for church services on Sundays. And working hours are flexible and the pay is about $18 an hour. Detailed job information will be posted on the Payroll Office’s window. If interested, send your cover letter and resume to David Stoker at david@.Catering Company — Oak Avenue is a most successful catering company searching for enthusiastic people to join their talented team of chefs and service profess ionals. Students’ part-time positions are available for ambitions chefs, bakers, service helpers, and captains. Please send a brief cover letter and resume to Maritza@ or call 707-963-9278 for more information.Caregiver Position — Looking for a caregiver to help with a 91-year-old woman who lives in Calistoga. She speaks both English and Spanish, but Spanish is her preference. She sleeps quite a bit during the day so there is time to do housework. She cannot be left alone because she has a fall risk. She walks with a walker on her own. Please call 925-785-8500 if interested.Housekeeping Services—A responsible, professional, and experienced housekeeper is needed to take care of all cleaning details of a local business in Angwin. This person will be needed 2 or 3 times per week for about 8 hours per day. Please and an email for additional information and be ready to provide resume and work history. Driver’s license is required. Pay will be $20-$25 per hour. To contact them, email recruitingangwin@.21.Which is the email address or contact number for those who want to become great bakers？A. david@.B.recruitingangwin@.C. 925-785-8500.D. 707-963-9278.22. What is required to work as a housekeeper?A. Flexibility in working hours.B. Speaking Spanish fluently.C. Being able to drive.D. Being a local in Angwin.23. Where does this text probably come from?A. A hotel advertisement.B. A job advertisement.C. A household guide.D. A tourist guide.BSeveral times my daughter had telephoned to say, “Mum, you must come and see the daffodils（水仙花）before they are over.” I wanted to go, but it was a two-hour drive from Lake Arrowhead. “I will go next Tuesday,”I promised, a little unwillingly, on her third call.The next Tuesday dawned cold and rainy. Still, I had promised, and so I drove there. When I finally walked into Carolyn’s house and hugged and greeted my grandchildren, I said, “Forget the daffodils, Carolyn! The road is indiscernible in the cloud and fog, and there is nothing in the world except you and these children that Iwant to see!”My daughter smiled calmly and said, “We drive in this weather all the time, Mum. You will never forgive yourself if you miss this experience.”After about twenty minutes, we turned onto a small road and I saw a small church. On the far side of the church, I saw a hand-lettered sign that read “Daffodil Garden”.We got out of the car and each took a child’s hand, and I followed Carolyn down the path. Then, we turned a corner of the path, and I looked up and gasped. Before me lay the most beautiful sight! There were five acres of flowers! “But who has done this?” I asked Carolyn. “It’s just one woman,” Carolyn answered. “That’s her home,” Carolyn pointed to a well-kept A-frame house that looked small and modest in the midst of all that glory. We walked up to the house. On the patio（平台）, we saw a poster. “Answers to the Questions I Know You Are Asking” was the headline.The first answer was a simple one. “50,000 bulbs（鳞茎）,” it read. The second answer was, “One at a time, by one woman.” The third answer was, “Began in 1958.”I thought of this woman whom I had never met, who, more than fifty years before, had begun -- one bulb at a time -- to bring the beauty and joy to the mountain top.24.Why was the author unwilling to see the daffodils at first ____.A.The weather was terrible.B.She took little interest in them.C.She wasn’t available at that time.D.It was not easy to go there.25.What does the underlined word “indiscernible” in Paragraph 2 mean?A.Invisible.B.Remote.C.Impossible.plex.26.Which of the following best describes the woman living in the A-frame house?A. Optimistic.B. Productive.C. Determined.D. Generous.27. Which of the following can be the best title for the text?A. I love DaffodilsB.An Unbelievable ExperienceC.Gardening as a hobbyD.One Bulb at a timeC“It’s a big hammer to crack a nut.” This is how one angry parent described the recent crackdown (严厉的打击) by London police on parents who drop their kids off at school by car. Yes, you read that correctly—the very mode of transportation that many American schools insist is the only safe way to deliver kids to school is now considered illegal in the UK.The decision to fine any vehicles seen dropping off or picking up kids within a particular zone of east London comes from city councillors’ (议员) long-term efforts to make the area safer and less crowded. They say they’ve been trying for years “to encourage reasonable parking”, but in vain. Neighborhood residents complain frequently about their driveways being blocked by illegally parked cars for 15 minutes or more，often while they’re trying to get to work, and the streets are long overcrowded.Now the rules have changed. Some parents are angry. Angie is a mother who made the “big hammer” comment and says th e crackdown is “way over the top.” The nearest drop-off point for her six- year-old is now a five-minute walk from the school. Others are happy with the decision, severe though it may seem. Councillor Jason Frost said ：“Traffic has significantly reduced，and more children are now walking to school, which is a great outcome. I would rather have complaints that we are slightly inconveniencing parents than hear that a child had been seriously injured because nothing was done.”I witness daily the chaos created by these in-town drivers, when I walk my own kids to school. There’s a parking lot crowded with vehicles and a slow-moving train of cars moving in circles, many filling the air withharmful smoke. Meanwhile, the conversations around overweight children and the importance of daily physical activities continue to stand out in schools.28.Why is Angie Baillieul opposed to the new rule?A. She is often fined by London police.B. She will lose her job as a school driver.C. She has adapted to the American practice.D. She thinks it adds inconvenience to her kid.29.What concerned Jason Frost most?A. People’s complaints.B. Students’ health.C. Students’ safety.D. Parents’ inconvenience.30.What can be seen after the crackdown is carried out?A. Less daily chaos.B. Favor from all people.C. More over-weight children.D. Complaints from neighbors.31.What is author’s attitude toward the crackdown?A. Favorable.B. Uncaring.C. Doubtful.D. Opposed.DWhile sleep patterns vary widely among animal classes and species, there are a few general rules of thumb (经验法则). Sleep serves important evolutionary functions for all animals, but sleep patterns and positions are based in large part on available food supply and defense mechanism.Sleep patterns in animals have evolved over time—animals that sleep and get attacked by predators（捕食者）are less likely to pass on their genes, so animals have developed ways to protect themselves during sleep. For instance, otters（水獭）sleep holding hands, or cover themselves in seaweed to protect their young and stay afloat while asleep. Likewise, cows and sheep sleep in a group—there is safety in numbers.Evolutionary biologists consider that predation and fear of predators have influenced the development of sleep patterns across species. Animals that eat meat tend to sleep more than animals that eat plants. Cathemeral（间歇性）species like lions sleep in short spells during both the day and night so as to enable them to seize food when it becomes available.In general, animals sleep according to what they eat—animals that eat food with lower calories sleep less than others. This may explain why plant eaters need to spend more time awake, to ensure they get enough food and energy. For example, plant eaters like giraffes and elephants sleep 30 minutes to just a few hours per day respectively. However, there are exceptions to this rule, like the koala. Their euealyptus（桉树）-based diet doesn’t give them much energy, but they sleep for 15 hours per day, and slit the rest of their time eating and resting.A note about the studies referenced in this article: because most animal sleep studies are performed using EEG tests, they are typically performed on animals in zoos and research facilities. As a result, they may not accurately reflect their natural sleep habits in the wild, given the stresses of a zoo environment and consistent availability of food.32. Why do cows and sheep sleep in groups?A. To protect themselves.B. To pass on their genes.C. To keep themselves warm.D. To get physical closeness.33. The underlined phrase “this rule” in paragraph 4 indicates that animals eating plants ______.A. have different sleep patterns.B. spend short time eating and resting.C. need less time to sleep than meat eaters.D. sleep 30 minutes to a few hours per day.34. What’s the purpose of the last paragraph?A. To explain the studies’ dangers.B. To point out the studies’ limitations.C. To describe the studies’ detailed procedures.D. To ad d the studies’ background information.35. What greatly affects animals’ sleep patterns?A. Their genes.B. Their eating habits.C. Classes and species.D. Food and safety.第二节（共5小题，每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填人空白处的最佳选项。
(审核版)湖南省长沙市2018届高三第二次模拟考试语文试题(含答案解析)
绝密★启用前试卷类型：全国卷湖南省长沙市2018届高三第二次模拟考试语文试卷本试题卷共10页，22题。
全卷满分150分。
考试用时150分钟。
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第Ⅰ卷阅读题（70分）一、现代文阅读（35分）（一）论述类文本阅读（9分，每小题3分）阅读下面的文字，完成1-3题。
儒家的创始人孔子，其政治伦理思想可以概括为“仁学”。
“仁”学的主体内容是“己所不欲，勿施于人”的“忠恕”思想和“君君、臣臣、父父、子子”的“正名”思想。
关于“仁”学的基础，《论语·学而》说：“君子务本，本立而道生。
孝悌也者，其为仁之本与！”可见，孔子视孝悌之亲情为其政治伦理思想的基础。
孝悌通过“能取近譬”，将爱父母、爱兄弟、爱妻子扩展为爱君主、爱国家及爱一切入。
的确，从人伦关系的角度来看，亲情无疑是人之生存的基本条件，由亲情、爱情、友情，推而广之到一切道德情感，从而构成“仁学”政治伦理思想的依据。
休谟的同情原理也认为，父子间的同情是最易发生的。
黑格尔一方面认为修身是成为抽象的人的基本条件，“正是这种反省意识，使一个人真正地走向了主体阶段”，另一方面认为实现抽象的人的理念必须要在伦理阶段，而“在伦理的阶段，最自然、最直接的便是家庭”，同样看到了亲情在政治伦理实现过程中所起到的基础作用。
2018年湖南省长沙高考数学二模试卷(文科)Word版含解析
2018年湖南省长沙高考二模试卷（文科数学）一、选择题：本大题共12个小题，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．已知集合A={y|y=log 2x ，x ＞1}，B={x|y=}，则A ∩B=（ ）A ．{y|0＜y ＜}B ．{y|0＜y ＜1}C ．{y|＜y ＜1}D ．∅2．若复数的实部与虚部相等，则实数a 的值为（ ）A ．3B ．﹣3C ．D ．﹣3．已知a=log 0.55、b=log 32、c=20.3、d=（）2，从这四个数中任取一个数m ，使函数f （x ）=x 3+mx 2+x+2有极值点的概率为（ ）A ．B ．C ．D ．14．如图，若N=10，则输出的数等于（ ）A ．B ．C ．D ．5．经过点（1，），渐近线与圆（x ﹣3）2+y 2=1相切的双曲线的标准方程为（ ） A ．x 2﹣8y 2=1B ．2x 2﹣4y 2=1C ．8y 2﹣x 2=1D ．4x 2﹣2y 2=16．已知三棱锥A ﹣BCD 的各棱长都相等，E 为BC 中点，则异面直线AB 与DE 所成角的余弦值为（ ）A．B． C．D．7．已知函数f（x）=（sinx+cosx）cosx，则下列说法正确的为（）A．函数f（x）的最小正周期为2πB．f（x）在[，]单调递减C．f（x）的图象关于直线x=﹣对称D．将f（x）的图象向右平移，再向下平移个单位长度后会得到一个奇函数的图象8．已知数列{an}的前n项和Sn=n2﹣n，正项等比数列{bn}中，b2=a3，bn+3bn﹣1=4（n≥2）n∈N+，则log2bn=（）A．n﹣1 B．2n﹣1 C．n﹣2 D．n9．已知实数x，y满足时，z=（a≥b＞0）的最大值为1，则a+b的最小值为（）A．7 B．8 C．9 D．1010．如图，网格纸上小正方形的边长为1，粗线画出的是某个四面体的三视图，则该四面体的表面积为（）A．8+8+4 B．8+8+2 C．2+2+D． ++11．若∀x∈R，函数f（x）=2mx2+2（4﹣m）x+1与g（x）=mx的值至少有一个为正数，则实数m的取值范围为（）A．（0，4] B．（0，8）C．（2，5）D．（﹣∞，0）12．已知函数f（x）=，若对任意的x∈[1，2]，f′（x）•x+f（x）＞0恒成立，则实数t的取值范围是（）A．（﹣∞，] B．（﹣∞，）C．（﹣∞，] D．[，+∞）二、填空题（每题5分，满分20分，将答案填在答题纸上）13．在△ABC中，P为中线AM上的一个动点，若||=2，则•（+）的最小值为．14．在平面直角坐标系xOy中，已知圆C：（x﹣a）2+（y﹣a+2）2=1，点A（0，﹣3），若圆C 上存在点M，满足|AM|=2|MO|，则实数a的取值范围是．15．已知等比数列{an}的首项为，公比为﹣，前n项和为Sn，则当n∈N*时，Sn﹣的最大值与最小值之和为．16．如图，有一块半径为2的半圆形钢板，计划剪裁成等腰梯形ABCD的形状，它的下底AB 是⊙O的直径，上底CD的端点在圆周上，则梯形周长的最大值为．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．已知函数f（x）=sin2x﹣cos2x+，x∈R．（1）若∀x∈[，]，f（x）﹣m=0有两个不同的根，求m的取值范围；（2）已知△ABC的内角A、B、C的对边分别为a、b、c，若f（B）=，b=2，且sinA、sinB、sinC成等差数列，求△ABC的面积．18．某大学在开学季准备销售一种盒饭进行试创业，在一个开学季内，每售出1盒该盒饭获利润10元，未售出的产品，每盒亏损5元．根据历史资料，得到开学季市场需求量的频率分布直方图，如图所示．该同学为这个开学季购进了150盒该产品，以x（单位：盒，100≤x≤200）表示这个开学季内的市场需求量，y（单位：元）表示这个开学季内经销该产品的利润．（Ⅰ）根据直方图估计这个开学季内市场需求量x的平均数和众数；（Ⅱ）将y表示为x的函数；（Ⅲ）根据频率分布直方图估计利润y不少于1350元的概率．19．已知四棱台ABCD﹣A1B1C1D1的下底面是边长为4的正方形，AA1=4，且AA1⊥面ABCD，点P为DD1的中点，点Q在BC上，BQ=3QC，DD1与面ABCD所成角的正切值为2．（Ⅰ）证明：PQ∥面A1ABB1；（Ⅱ）求证：AB1⊥面PBC，并求三棱锥Q﹣PBB1的体积．20．已知过点P（﹣1，0）的直线l与抛物线y2=4x相交于A（x1，y1）、B（x2，y2）两点．（Ⅰ）求直线l倾斜角的取值范围；（Ⅱ）是否存在直线l，使A、B两点都在以M（5，0）为圆心的圆上，若存在，求出此时直线及圆的方程，若不存在，请说明理由．21．已知函数f（x）=lnx﹣ax2+（2﹣a）x．（Ⅰ）讨论函数f（x）的单调性；（Ⅱ）设g（x）=﹣2，对任意给定的x0∈（0，e]，方程f（x）=g（x）在（0，e]有两个不同的实数根，求实数a的取值范围．（其中a∈R，e=2.71828…为自然对数的底数）．选修4-4：坐标系与参数方程22．在平面直角坐标系xOy中，直线l的参数方程是（t为参数），以O为极点，x轴的正半轴为极轴，建立极坐标系，曲线C的极坐标方程为9ρ2cos2θ+16ρ2sin2θ=144，且直线l与曲线C交于P，Q两点．（Ⅰ）求曲线C的直角坐标方程及直线l恒过的顶点A的坐标；（Ⅱ）在（Ⅰ）的条件下，若|AP|•|AQ|=9，求直线l的普通方程．选修4-5：不等式选讲23．设函数f（x）=|x﹣a|，a∈R．（Ⅰ）当a=2时，解不等式：f（x）≥6﹣|2x﹣5|；（Ⅱ）若关于x的不等式f（x）≤4的解集为[﹣1，7]，且两正数s和t满足2s+t=a，求证：．2018年湖南省长沙高考数学二模试卷（文科）参考答案与试题解析一、选择题：本大题共12个小题，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.x，x＞1}，B={x|y=}，则A∩B=（）1．已知集合A={y|y=log2A．{y|0＜y＜} B．{y|0＜y＜1} C．{y|＜y＜1} D．∅【考点】1E：交集及其运算．【分析】求出集合的等价条件，结合交集运算进行求解即可．x，x＞1}={y|y＞0}，【解答】解：A={y|y=log2B={x|y=}={x|1﹣2x＞0}={x|x＜}，则A∩B={y|0＜y＜}，故选：A2．若复数的实部与虚部相等，则实数a 的值为（ ）A ．3B ．﹣3C ．D ．﹣【考点】A5：复数代数形式的乘除运算．【分析】利用复数的运算法则、实部与虚部的定义即可得出．【解答】解：复数==+i 的实部与虚部相等，∴=，解得a=﹣．故选：D ．3．已知a=log 0.55、b=log 32、c=20.3、d=（）2，从这四个数中任取一个数m ，使函数f （x ）=x 3+mx 2+x+2有极值点的概率为（ ）A ．B ．C ．D ．1【考点】6D ：利用导数研究函数的极值；CB ：古典概型及其概率计算公式．【分析】求出函数的导数，根据函数的极值点的个数求出m 的范围，通过判断a ，b ，c ，d 的范围，得到满足条件的概率值即可． 【解答】解：f′（x ）=x 2+2mx+1， 若函数f （x ）有极值点，则f′（x ）有2个不相等的实数根， 故△=4m 2﹣4＞0，解得：m ＞1或m ＜﹣1，而a=log 0.55＜﹣2，0＜b=log 32＜1、c=20.3＞1，0＜d=（）2＜1， 满足条件的有2个，分别是a ，c ，故满足条件的概率p==， 故选：B ．4．如图，若N=10，则输出的数等于（ ）A．B．C．D．【考点】EF：程序框图．【分析】分析程序中各变量、各语句的作用，再根据流程图所示的顺序，可知：该程序的作用是累加并输出S=++…+的值，由裂项法即可计算得解．【解答】解：分析程序中各变量、各语句的作用，再根据流程图所示的顺序，可知：该程序的作用是累加并输出S=S=++…+的值，又由：S=++…+=（1﹣）+（）+…+（﹣）=1﹣=．故选：C．5．经过点（1，），渐近线与圆（x﹣3）2+y2=1相切的双曲线的标准方程为（）A．x2﹣8y2=1 B．2x2﹣4y2=1 C．8y2﹣x2=1 D．4x2﹣2y2=1【考点】KB：双曲线的标准方程．【分析】设双曲线的渐近线方程为mx±ny=0（m＞0，n＞0），利用渐近线与圆（x﹣3）2+y2=1相切，可得渐近线方程，设出双曲线方程，代入点（1，），即可得出结论．【解答】解：设双曲线的渐近线方程为mx±ny=0（m＞0，n＞0）∵渐近线与圆（x﹣3）2+y2=1相切，∴=1，∴n=2m，∴渐近线方程为x±2y=0∴双曲线方程设为x2﹣8y2=λ，代入点（1，），可得λ=1﹣2=﹣1，∴双曲线方程为8y2﹣x2=1．故选：C．6．已知三棱锥A﹣BCD的各棱长都相等，E为BC中点，则异面直线AB与DE所成角的余弦值为（）A．B． C．D．【考点】LM：异面直线及其所成的角．【分析】取AC中点O，连结DO，EO，则EO∥AB，从而∠DEO是异面直线AB与DE所成角（或所成角的补角），由此利用余弦定理能求出异面直线AB与DE所成角的余弦值．【解答】解：取AC中点O，连结DO，EO，∵三棱锥A﹣BCD的各棱长都相等，E为BC中点，∴EO∥AB，∴∠DEO是异面直线AB与DE所成角（或所成角的补角），设三棱锥A﹣BCD的各棱长为2，则DE=DO==，OE=1，∴cos∠DEO===．∴异面直线AB与DE所成角的余弦值为．故选：B．7．已知函数f（x）=（sinx+cosx）cosx，则下列说法正确的为（）A．函数f（x）的最小正周期为2πB．f（x）在[，]单调递减C．f（x）的图象关于直线x=﹣对称D．将f（x）的图象向右平移，再向下平移个单位长度后会得到一个奇函数的图象【考点】H1：三角函数的周期性及其求法；H5：正弦函数的单调性．【分析】化函数f（x）为正弦型函数，再判断选项中的命题是否正确．【解答】解：函数f（x）=（sinx+cosx）cosx=sinxcosx+cos2x=sin2x+=（sin2x+cos2x）+=sin（2x+）+，∴f（x）的最小正周期为T==π，∴A错误；x∈[，]时，2x+∈[，]，f（x）是单调递增函数，∴B错误；当x=﹣时，f（x）=sin（﹣+）+=sin（﹣）+，∴x=﹣不是f（x）的对称轴，C错误；将f（x）的图象向右平移，得y=sin2[（x﹣）+]+的图象，再向下平移个单位长度得y=sin2x的图象，它是奇函数，D正确．故选：D．8．已知数列{an}的前n项和Sn=n2﹣n，正项等比数列{bn}中，b2=a3，bn+3bn﹣1=4（n≥2）n∈N+，则log2bn=（）A．n﹣1 B．2n﹣1 C．n﹣2 D．n【考点】8H：数列递推式．【分析】利用a3=S3﹣S2，即可得到log2b2．验证可知A，B，C均不符合，即可得出．【解答】解：∵a3=S3﹣S2=（32﹣3）﹣（22﹣2）=4，∴b2=a3=4，log2b2=log24=2．验证可知A，B，C均不符合，故答案为D．9．已知实数x，y满足时，z=（a≥b＞0）的最大值为1，则a+b的最小值为（）A．7 B．8 C．9 D．10【考点】7C：简单线性规划．【分析】作出不等式组对应的平面区域，利用z的最大值，确定最优解，然后利用基本不等式进行判断．【解答】解：作出不等式组对应的平面区域如图：由z=（a≥b＞0）得y=，则斜率k=，则由图象可知当直线y=经过点B（1，4）时，直线y=的截距最大，此时，则a+b=（a+b）（）=1+4+，当且仅当，即b=2a取等号此时不成立，故基本不等式不成立．设t=，∵a≥b＞0，∴0＜≤1，即0＜t≤1，则1+4+=5+t+在（0，1]上单调递减，∴当t=1时，1+4+=5+t+取得最小值为5+1+4=10．即a+b的最小值为10，故选：D ．10．如图，网格纸上小正方形的边长为1，粗线画出的是某个四面体的三视图，则该四面体的表面积为（ ）A ．8+8+4B ．8+8+2C ．2+2+D ． ++【考点】L!：由三视图求面积、体积．【分析】由三视图可知几何体为从边长为4的正方体切出来的三棱锥．作出直观图，计算各棱长求面积．【解答】解：由三视图可知几何体为从边长为4的正方体切出来的三棱锥A ﹣BCD ．作出直观图如图所示：其中A ，C ，D 为正方体的顶点，B 为正方体棱的中点．∴S △ABC ==4，S △BCD ==4．∵AC=4，AC ⊥CD ，∴S △ACD ==8，由勾股定理得AB=BD==2，AD=4．∴cos∠ABD==﹣，∴sin∠ABD=．==4．∴S△ABD∴几何体的表面积为8+8+4．故选A．11．若∀x∈R，函数f（x）=2mx2+2（4﹣m）x+1与g（x）=mx的值至少有一个为正数，则实数m的取值范围为（）A．（0，4] B．（0，8）C．（2，5）D．（﹣∞，0）【考点】52：函数零点的判定定理．【分析】当m≤0时，显然不成立；当m＞0时，g（x）=mx＜0，因为f（0）=1＞0，所以仅对对称轴进行讨论即可．【解答】解：当m＜0时，当x＞0时，g（x）=mx＜0，又二次函数f（x）=2mx2﹣（8﹣2m）x+1开口向下，当x→+∞时，f（x）=2mx2﹣（8﹣2m）x+1＜0，故当m＜0时不成立；当m=0时，因f（0）=1＞0，不符合题意；当m＞0时，若﹣=≥0，即0＜m≤4时结论显然成立；若﹣=＜0，时只要△=4（4﹣m）2﹣8m=4（m﹣8）（m﹣2）＜0即可，即4＜m＜8，综上：0＜m＜8．故选：B．12．已知函数f（x）=，若对任意的x∈[1，2]，f′（x）•x+f（x）＞0恒成立，则实数t的取值范围是（）A．（﹣∞，] B．（﹣∞，）C．（﹣∞，] D．[，+∞）【考点】6E：利用导数求闭区间上函数的最值；6B：利用导数研究函数的单调性．【分析】对任意的x∈[1，2]，f′（x）•x+f（x）＞0恒成立⇔对任意的x∈[1，2]，恒成立，⇔对任意的x∈[1，2]，2x2﹣2tx+1＞0恒成立，⇔t＜恒成立，求出x+在[1，2]上的最小值即可．【解答】解：∵∴对任意的x∈[1，2]，f′（x）•x+f（x）＞0恒成立⇔对任意的x∈[1，2]，恒成立，⇔对任意的x∈[1，2]，2x2﹣2tx+1＞0恒成立，⇔t＜恒成立，又g（x）=x+在[1，2]上单调递增，∴，∴t＜．故选：B二、填空题（每题5分，满分20分，将答案填在答题纸上）13．在△ABC中，P为中线AM上的一个动点，若||=2，则•（+）的最小值为﹣2 ．【考点】9R：平面向量数量积的运算．【分析】由已知中△ABC中，P为中线AM上的一个动点，若||=2，我们易将•（+）转化为2（||﹣1）2﹣2的形式，然后根据二次函数在定区间上的最值的求法，得到答案．【解答】解：∵AM为△ABC的中线，故M为BC的中点则+=2=+则•（+）=（+）•2=22+2•=2||2﹣4||=2（||﹣1）2﹣2当||=1时，•（+）的最小值为﹣2故答案为：﹣214．在平面直角坐标系xOy 中，已知圆C ：（x ﹣a ）2+（y ﹣a+2）2=1，点A （0，﹣3），若圆C 上存在点M ，满足|AM|=2|MO|，则实数a 的取值范围是 [0，3] ． 【考点】J5：点与圆的位置关系；IR ：两点间的距离公式．【分析】设点M （x ，y ），由题意得x 2+（y ﹣2）2+x 2+y 2=10，若圆C 上存在点M 满足MA 2+MO 2=10也就等价于圆E 与圆C 有公共点，由此能求出实数a 的取值范围．【解答】解：设点M （x ，y ），由题意得点A （0，2），O （0，0）及MA 2+MO 2=10， 即x 2+（y ﹣2）2+x 2+y 2=10，整理得x 2+（y ﹣1）2=4， 即点M 在圆E ：x 2+（y ﹣1）2=4上．若圆C 上存在点M 满足MA 2+MO 2=10也就等价于圆E 与圆C 有公共点， 所以|2﹣1|≤CE ≤2+1，即|2﹣1|≤≤2+1，整理得1≤2a 2﹣6a+9≤9，解得0≤a ≤3， 即实数a 的取值范围是[0，3]． 故答案为：[0，3]．15．已知等比数列{a n }的首项为，公比为﹣，前n 项和为S n ，则当n ∈N *时，S n ﹣的最大值与最小值之和为．【考点】89：等比数列的前n 项和．【分析】根据等比数列的求和公式求出S n ，分n 为奇数或偶数计算出S n 的范围，从而得出S n﹣的最大值与最小值．【解答】解：S n ==1﹣（﹣）n ，（1）当n 为奇数时，S n =1+，∴1＜S n ≤，（2）当n 为偶数时，S n =1﹣，∴≤S n ＜1．∴对于任意n ∈N *，≤S n ≤．令S n =t ，f （t ）=t ﹣，则f （t ）在[，]上单调递增，∴f （t ）的最小值为f （）=﹣，f （t ）的最大值为f （）=，∴S n ﹣的最小值为﹣，最大值为，∴S n ﹣的最大值与最小值之和为﹣+=．故答案为：．16．如图，有一块半径为2的半圆形钢板，计划剪裁成等腰梯形ABCD 的形状，它的下底AB 是⊙O 的直径，上底CD 的端点在圆周上，则梯形周长的最大值为 10 ．【考点】5D ：函数模型的选择与应用．【分析】作DE ⊥AB 于E ，连接BD ，根据相似关系求出AE ，而CD=AB ﹣2AE ，从而求出梯形ABCD 的周长y 与腰长x 间的函数解析式，根据AD ＞0，AE ＞0，CD ＞0，可求出定义域；利用二次函数在给定区间上求出最值的知识可求出函数的最大值． 【解答】解：如图，作DE ⊥AB 于E ，连接BD ． 因为AB 为直径，所以∠ADB=90°．在Rt △ADB 与Rt △AED 中，∠ADB=90°=∠AED ，∠BAD=∠DAE ， 所以Rt △ADB ∽Rt △AED ．所以=，即AE=．又AD=x ，AB=4，所以AE=．所以CD=AB﹣2AE=4﹣，于是y=AB+BC+CD+AD=4+x+4﹣+x=﹣x2+2x+8由于AD＞0，AE＞0，CD＞0，所以x＞0，＞0，4﹣＞0，解得0＜x＜2，故所求的函数为y=﹣x2+2x+8（0＜x＜2）y=﹣x2+2x+8=﹣（x﹣2）2+10，又0＜x＜2，所以，当x=2时，y有最大值10．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．已知函数f（x）=sin2x﹣cos2x+，x∈R．（1）若∀x∈[，]，f（x）﹣m=0有两个不同的根，求m的取值范围；（2）已知△ABC的内角A、B、C的对边分别为a、b、c，若f（B）=，b=2，且sinA、sinB、sinC成等差数列，求△ABC的面积．【考点】GL：三角函数中的恒等变换应用；H2：正弦函数的图象．【分析】（1）化简f（x），问题转化为y=m和y=f（x）在x∈[，]有2个不同的交点，画出函数的图象，求出m的范围即可；（2）求出B的值，根据正弦定理得到a+c=2b=4，根据余弦定理得到b2=a2+c2﹣2accosB=（a+c）2﹣2ac﹣ac，求出ac的值，从而求出三角形的面积即可．【解答】解：（1）∵函数f（x）=sin2x﹣cos2x+，∴f（x）=sin2x﹣+=sin（2x﹣），∴f（x）=sin（2x﹣），∵x∈[，]，∴2x﹣∈[0，]，若∀x ∈[，]，f （x ）﹣m=0有两个不同的根，则y=m 和y=f （x ）在x ∈[，]有2个不同的交点，画出函数的图象，如图所示：，结合图象得≤m ＜1；（2）由f （B ）=，解得：B=或B=，由sinA 、sinB 、sinC 成等差数列，结合正弦定理得a+c=2b=4，故B=，且b 2=a 2+c 2﹣2accosB=（a+c ）2﹣2ac ﹣ac ，故ac=（24﹣12），故S △ABC =acsinB=（24﹣12）×=6﹣3．18．某大学在开学季准备销售一种盒饭进行试创业，在一个开学季内，每售出1盒该盒饭获利润10元，未售出的产品，每盒亏损5元．根据历史资料，得到开学季市场需求量的频率分布直方图，如图所示．该同学为这个开学季购进了150盒该产品，以x （单位：盒，100≤x ≤200）表示这个开学季内的市场需求量，y （单位：元）表示这个开学季内经销该产品的利润． （Ⅰ）根据直方图估计这个开学季内市场需求量x 的平均数和众数； （Ⅱ）将y 表示为x 的函数；（Ⅲ）根据频率分布直方图估计利润y 不少于1350元的概率．【考点】B8：频率分布直方图；BB：众数、中位数、平均数．【分析】（Ⅰ）由频率分布直方图能估计这个开学季内市场需求量x的平均数和众数．（Ⅱ）因为每售出1盒该盒饭获利润10元，未售出的盒饭，每盒亏损5元，当100＜x≤200时，y=10x﹣5=15x﹣750，当150＜x≤200时，y=10×150=1500，由此能将y表示为x的函数．（Ⅲ）由利润不少于1350元，得150x﹣750≥750，由此能求出利润不少于1350元的概率．【解答】解：（Ⅰ）由频率分布直方图得：最大需求量为150盒的频率为0.015×20=0.3．这个开学季内市场需求量的众数估计值是150．需求量为[100，120）的频率为0.005×20=0.1，需求量为[120，140）的频率为0.01×20=0.2，需求量为[140，160）的频率为0.015×20=0.3，需求量为[160，180）的频率为0.0125×20=0.25，需求量为[180，200）的频率为0.0075×20=0.15，则平均数： =110×0.1+130×0.2+150×0.3+170×0.25+190×0.15=153．（Ⅱ）因为每售出1盒该盒饭获利润10元，未售出的盒饭，每盒亏损5元，所以当100＜x≤200时，y=10x﹣5=15x﹣750，当150＜x≤200时，y=10×150=1500，所以y=，x∈N．（Ⅲ）因为利润不少于1350元，所以150x﹣750≥750，解得x≥140．所以由（Ⅰ）知利润不少于1350元的概率p=1﹣0.1﹣0.2=0.7．19．已知四棱台ABCD﹣A1B1C1D1的下底面是边长为4的正方形，AA1=4，且AA1⊥面ABCD，点P为DD1的中点，点Q在BC上，BQ=3QC，DD1与面ABCD所成角的正切值为2．（Ⅰ）证明：PQ∥面A1ABB1；（Ⅱ）求证：AB 1⊥面PBC ，并求三棱锥Q ﹣PBB 1的体积．【考点】LF ：棱柱、棱锥、棱台的体积；LS ：直线与平面平行的判定．【分析】（I ）取AA 1中点E ，连接PE 、BE ，过D 1作D 1H ⊥AD 于H ，可证四边形PQBE 为平行四边形，得出PQ ∥BE ，故而PQ ∥面A 1ABB 1；（II ）由AA 1⊥面ABCD 可得AA 1⊥BC ，由相似三角形可得AB 1⊥BE ，故而AB 1⊥平面PEBC ，求出B 1到平面PEBC 的距离，代入体积公式即可得出棱锥的体积．【解答】解：（Ⅰ）证明：取AA 1中点E ，连接PE 、BE ，过D 1作D 1H ⊥AD 于H ． ∵AA 1⊥面ABCD ，AA 1∥D 1H ，∴D 1H ⊥面ABCD ． ∴∠D 1DA 为DD 1与面ABCD 所成角．∴=2，又AA 1=4，∴DH=2． ∴A 1D 1=2．∴PE=（A 1D 1+AD ）=3， 又EF ∥AD ，∴四边形PQBE 为平行四边形， ∴PQ ∥BE ，又PQ ⊄面A 1ABB 1，BE ⊂面A 1ABB 1， ∴PQ ∥面A 1ABB 1．（Ⅱ）∵AA 1⊥面ABCD ，BC ⊂平面ABCD ， ∴AA 1⊥BC ，又BC ⊥AB ，AB ∩AA 1=A ，∴BC ⊥面ABB 1A 1，又AB 1⊂平面ABB 1A 1， ∴BC ⊥AB 1．在梯形A 1ABB 1中，Rt △BAE ≌Rt △AA 1B 1，∴∠B 1AE+∠AEB=∠B 1AE+∠AB 1A 1=90°，∴AB 1⊥BE ，又BE ∩BC=B ，BE ⊂平面PEBC ，BC ⊂平面PEBC ，∴AB 1⊥面PEBC ．设AB 1∩BE=M ，∵AE=2，AB=4，∴BM=2，∵A 1B 1=2，AA 1=4，∴AB 1=2，∴AM==，∴B 1M=AB 1﹣AM=，又BQ=BC=3，∴V =V ===6．20．已知过点P （﹣1，0）的直线l 与抛物线y 2=4x 相交于A （x 1，y 1）、B （x 2，y 2）两点． （Ⅰ）求直线l 倾斜角的取值范围；（Ⅱ）是否存在直线l ，使A 、B 两点都在以M （5，0）为圆心的圆上，若存在，求出此时直线及圆的方程，若不存在，请说明理由．【考点】KN ：直线与抛物线的位置关系．【分析】（Ⅰ）设直线l 的方程，代入抛物线方程，利用△＞0，即可求得k 的取值范围，求得直线l 倾斜角的取值范围；（Ⅱ）设圆M 的方程，与抛物线方程联立，根据韦达定理，即可求得r 的值及直线l 的斜率k ，求得直线及圆的方程．【解答】解：（Ⅰ）由已知直线l 的斜率存在且不为0．设l ：y=k （x+1），则，整理得：ky 2﹣4y+4k=0，y 1+y 2=，△=16﹣4k ×4k ＞0，解得：﹣1＜k ＜1且k ≠0．∴直线l 倾斜角的取值范围（0，）∪（，π）；（Ⅱ）设⊙M ：（x ﹣5）2+y 2=r 2，（r ＞0），则，则x 2﹣6x+25﹣r 2=0，∴x 1+x 2=6，又由（Ⅰ）知y 1y 2=4，∴x 1x 2=1．∴25﹣r 2=1，∴r 2=24，并且r 2=24时，方程的判别式△=36﹣4×（25﹣r 2）＞0，由y 1+y 2=k （x 1+x 2+2）=，解得：k=±， ∴存在定圆M ，经过A 、B 两点，其方程为：（x ﹣5）2+y 2=24，此时直线l 方程为y=±（x+1）．21．已知函数f （x ）=lnx ﹣ax 2+（2﹣a ）x ．（Ⅰ）讨论函数f （x ）的单调性；（Ⅱ）设g （x ）=﹣2，对任意给定的x 0∈（0，e]，方程f （x ）=g （x 0）在（0，e]有两个不同的实数根，求实数a 的取值范围．（其中a ∈R ，e=2.71828…为自然对数的底数）．【考点】6B ：利用导数研究函数的单调性；54：根的存在性及根的个数判断．【分析】（Ⅰ）求出函数的导数，通过讨论a 的范围，求出函数的单调区间即可；（Ⅱ）求出g （x ）的导数，根据函数的单调性求出a 的范围即可．【解答】解：（Ⅰ）f′（x ）=﹣2ax+（2﹣a ）=，当a=0时，f′（x ）=＞0，f （x ）在（0，+∞）单调递增．当a ＜0时，f′（x ）＞0，f （x ）在（0，+∞）单调递增．当a ＞0时，令f′（x ）＞0，解得：0＜x ＜，令f′（x ）＜0，解得：x ＞，故f （x ）在（0，）递增，在（，+∞）递减．（Ⅱ）g（x）=﹣2，g′（x）=，x∈（﹣∞，1），g′（x）＞0，g（x）单调递增，x∈（1，+∞）时，g′（x）＜0，g（x）单调递减，∴x∈（0，e]时，g（x）的值域为（﹣2，﹣2]，由已知，，由f（e）=1﹣ae2+2e﹣ea≤﹣2，∴a≥，由f（）=ln﹣+﹣1＞﹣2，∴lna﹣+＜0，令h（x）=lnx﹣知h（x）单调递增，而h（e）=0，∴a∈（0，e）时，lna﹣+＜1，∴a∈（0，e），综合以上，≤a＜e．选修4-4：坐标系与参数方程22．在平面直角坐标系xOy中，直线l的参数方程是（t为参数），以O为极点，x轴的正半轴为极轴，建立极坐标系，曲线C的极坐标方程为9ρ2cos2θ+16ρ2sin2θ=144，且直线l与曲线C交于P，Q两点．（Ⅰ）求曲线C的直角坐标方程及直线l恒过的顶点A的坐标；（Ⅱ）在（Ⅰ）的条件下，若|AP|•|AQ|=9，求直线l的普通方程．【考点】QH：参数方程化成普通方程；Q4：简单曲线的极坐标方程．【分析】（Ⅰ）由x=ρcosθ，y=ρsinθ，能求出曲线C的直角坐标方程，由直线l的参数方程能求出直线l恒过的定点A的坐标．（Ⅱ）把直线l的方程代入曲线C的直角坐标方程中，得：（9+7sin2α）t2+36tcosα﹣9×12=0．由t的几何意义知|AP|=|t1|，|AQ|=|t2|，点A在椭圆内，这个方程必有两个实根，从而得到||=9，进而求出tan，由此能求出直线l的方程．【解答】解：（Ⅰ）∵曲线C的极坐标方程为9ρ2cos2θ+16ρ2sin2θ=144，x=ρcosθ，y=ρsinθ，∴曲线C 的直角坐标方程为：=1．∵直线l 的参数方程是（t 为参数）， ∴直线l 恒过定点为A （2，0）．（Ⅱ）把直线l 的方程代入曲线C 的直角坐标方程中，整理，得：（9+7sin 2α）t 2+36tcosα﹣9×12=0．由t 的几何意义知|AP|=|t 1|，|AQ|=|t 2|，∵点A 在椭圆内，这个方程必有两个实根，∴t 1t 2=，∵|AP|•|AQ|=|t 1t 2|=9，即||=9，∴，∵α∈（0，π），∴tan，∴直线l 的方程为y=．选修4-5：不等式选讲23．设函数f （x ）=|x ﹣a|，a ∈R ．（Ⅰ）当a=2时，解不等式：f （x ）≥6﹣|2x ﹣5|；（Ⅱ）若关于x 的不等式f （x ）≤4的解集为[﹣1，7]，且两正数s 和t 满足2s+t=a ，求证：．【考点】R5：绝对值不等式的解法．【分析】（Ⅰ）利用绝对值的意义表示成分段函数形式，解不等式即可．（2）根据不等式的解集求出a=3，利用1的代换结合基本不等式进行证明即可．【解答】（Ⅰ）解：当a=2时，不等式：f （x ）≥6﹣|2x ﹣5|，可化为|x ﹣2|+|2x ﹣5|≥6．①x ≥2.5时，不等式可化为x ﹣2+2x ﹣5≥6，∴x ≥；②2≤x ＜2.5，不等式可化为x ﹣2+5﹣2x ≥6，∴x ∈∅；③x ＜2，不等式可化为2﹣x+5﹣2x ≥6，∴x ≤，综上所述，不等式的解集为（﹣]；（Ⅱ）证明：不等式f （x ）≤4的解集为[a ﹣4，a+4]=[﹣1，7]，∴a=3，∴=（）（2s+t）=（10++）≥6，当且仅当s=，t=2时取等号．。
2018年湖南省长沙市长郡中学高考数学二模试卷(文科)(解析版)
2018年湖南省长沙市长郡中学高考数学二模试卷（文科）一、选择题：本大题共12个小题，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．（5分）已知集合P＝{x∈N|1≤x≤10}，集合Q＝{x∈R|x2﹣x﹣6＜0}，则P∩Q等于（）A．[1，2，3]B．{1，2}C．[1，2]D．[1，3）2．（5分）复数z满足z（2+i）＝3﹣i，则复数z在复平面内对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限3．（5分）某公司的班车分别在7：30，8：30发车，小明在7：50至8：30之间到达发车站乘坐班车，且到达发车站的时刻是随机的，则他等车时间不超过15分钟的概率是（）A．B．C．D．4．（5分）已知曲线在点（1，f（1））处的切线的倾斜角为，则a的值为（）A．1B．﹣4C．D．﹣15．（5分）已知平面向量，满足||＝3，||＝2，与的夹角为120°，若（+m）⊥，则实数m的值为（）A．1B．C．2D．36．（5分）设{a n}是公差不为0的等差数列，满足a42+a52＝a62+a72，则{a n}的前10项和S10＝（）A．﹣10B．﹣5C．0D．57．（5分）函数f（x）＝A sin（ωx+φ）（A＞0，ω＞0，0≤φ≤2π）在R上的部分图象如图所示，则f（2018）的值为（）A．B．﹣5C．D．58．（5分）设a＞b＞0，a+b＝1，且x＝（）b，y＝ab，z＝a，则x、y、z的大小关系是（）A．y＜z＜x B．z＜y＜x C．x＜y＜z D．y＜x＜z9．（5分）《九章算术》是中国古代数学名著，体现了古代劳动人民数学的智慧，其中第六章“均输”中，有一竹节容量问题，某教师根据这一问题的思想设计了如图所示的程序框图，若输出的m的值为35，则输入的a的值为（）A．4B．5C．7D．1110．（5分）已知f（x）是定义在[﹣2b，1+b]上的偶函数，且在[﹣2b，0]上为增函数，则f （x﹣1）≤f（2x）的解集为（）A．B．C．[﹣1，1]D．11．（5分）如图，在边长为2的正方形ABCD中，E，F分别为BC，CD的中点，H为EF 的中点，沿AE，EF，F A将正方形折起，使B，C，D重合于点O，在构成的四面体A﹣OEF中，下列结论中错误的是（）A．AO⊥平面EOFB．直线AH与平面EOF所成角的正切值为C．异面直线OH和求AE所成角为60°D．四面体A﹣OEF的外接球表面积为6π12．（5分）已知椭圆E：+＝1（a＞b＞0）与过原点的直线交于A、B两点，右焦点为F，∠AFB＝120°，若△AFB的面积为4，则椭圆E的焦距的取值范围是（）A．[2，+∞）B．[4，+∞）C．[2，+∞）D．[4，+∞）二、填空题（每题5分，满分20分，将答案填在答题纸上）13．（5分）设变量x，y满足约束条件，则z＝x﹣2y的最大值为．14．（5分）双曲线﹣＝1（a＞0，b＞0）的渐近线与圆（x﹣）2+y2＝1相切，则此双曲线的离心率为．15．（5分）已知四棱锥P﹣ABCD的外接球为球O，底面ABCD是矩形，面P AD⊥底面ABCD，且P A＝PD＝AD＝2，AB＝4，则球O的表面积为．16．（5分）已知数列{a n}满足对1≤n≤3时，a n＝n，其对∀n∈N*，有a n+3+a n+1＝a n+2+a n，则数列{n•a n}的前50项的和为．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．（12分）在△ABC中，角A，B，C所对的边分别为a，b，c，且．（1）求sin B的值；（2）若a＝4，求△ABC的面积S的值．18．（12分）如图，五面体ABCDE中，四边形ABDE是菱形，△ABC是边长为2的正三角形，∠DBA＝60°，．（1）证明：DC⊥AB；（2）若C在平面ABDE内的正投影为H，求点H到平面BCD的距离．19．（12分）某基地蔬菜大棚采用水培、无土栽培方式种植各类蔬菜．过去50周的资料显示，该地周光照量X（小时）都在30小时以上，其中不足50小时的周数有5周，不低于50小时且不超过70小时的周数有35周，超过70小时的周数有10周．根据统计，该基地的西红柿增加量y（百斤）与使用某种液体肥料x（千克）之间对应数据为如图所示的折线图．（1）依据数据的折线图，是否可用线性回归模型拟合y与x的关系？请计算相关系数r并加以说明（精确到0.01）．（若|r|＞0.75，则线性相关程度很高，可用线性回归模型拟合）（2）蔬菜大棚对光照要求较大，某光照控制仪商家为该基地提供了部分光照控制仪，但每周光照控制仪最多可运行台数受周光照量X限制，并有如下关系：若某台光照控制仪运行，则该台光照控制仪周利润为3000元；若某台光照控制仪未运行，则该台光照控制仪周亏损1000元．若商家安装了3台光照控制仪，求商家在过去50周周总利润的平均值．附：相关系数公式r＝，参考数据≈0.55，≈0.95．20．（12分）已知动点P到定直线l：x＝﹣4的距离比到定点F（2，0）的距离大2．（1）求动点P的轨迹C的方程；（2）在x轴正半轴上，是否存在某个确定的点M，过该点的动直线l与曲线C交于A，B 两点，使得为定值．如果存在，求出点M坐标；如果不存在，请说明理由．21．（12分）已知函数f1（x）＝（x﹣λ）2，f2（x）＝lnx（x＞0，且x≠1）．（Ⅰ）当λ＝1时，若对任意x∈（1，+∞），f1（x）≥k•f2（x）恒成立，求实数k的取值范围；（Ⅱ）若λ∈（0，1），设f（x）＝，f'（x）是f（x）的导函数，判断f'（x）的零点个数，并证明．请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分．[选修4-4：坐标系与参数方程]22．（10分）已知在平面直角坐标系xOy中，以O为极点，x轴的正半轴为极轴，建立极坐标系．曲线C1的极坐标方程为ρ＝4cosθ，直线l：（t为参数）．（1）求曲线C1的直角坐标方程及直线l的普通方程；（2）若曲线C2的参数方程为（α为参数），曲线C1上点P的极角为，Q为曲线C2上的动点，求PQ的中点M到直线l距离的最大值．[选修4-5：不等式选讲]23．已知函数f（x）＝|x﹣1|，关于x的不等式f（x）＜3﹣|2x+1|的解集记为A．（1）求A；（2）已知a，b∈A，求证：f（ab）＞f（a）﹣f（b）．2018年湖南省长沙市长郡中学高考数学二模试卷（文科）参考答案与试题解析一、选择题：本大题共12个小题，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．（5分）已知集合P＝{x∈N|1≤x≤10}，集合Q＝{x∈R|x2﹣x﹣6＜0}，则P∩Q等于（）A．[1，2，3]B．{1，2}C．[1，2]D．[1，3）【解答】解：P＝{1，2，3，4，5，6，7，8，9，10}，Q＝（﹣2，3）；∴P∩Q＝{1，2}．故选：B．2．（5分）复数z满足z（2+i）＝3﹣i，则复数z在复平面内对应的点位于（）A．第一象限B．第二象限C．第三象限D．第四象限【解答】解：由z（2+i）＝3﹣i，得＝，则复数z在复平面内对应的点的坐标为：（1，﹣1），位于第四象限．故选：D．3．（5分）某公司的班车分别在7：30，8：30发车，小明在7：50至8：30之间到达发车站乘坐班车，且到达发车站的时刻是随机的，则他等车时间不超过15分钟的概率是（）A．B．C．D．【解答】解：设小明到达时间为y，当y在8：15至8：30时，小明等车时间不超过15分钟，故P＝＝，故选：B．4．（5分）已知曲线在点（1，f（1））处的切线的倾斜角为，则a的值为（）A．1B．﹣4C．D．﹣1【解答】解：函数（x＞0）的导数，∵函数f（x）在x＝1处的倾斜角为∴f′（1）＝﹣1，∴1+＝﹣1，∴a＝﹣1．故选：D．5．（5分）已知平面向量，满足||＝3，||＝2，与的夹角为120°，若（+m）⊥，则实数m的值为（）A．1B．C．2D．3【解答】解：∵||＝3，||＝2，与的夹角为120°，∴＝cos120°＝＝﹣3．∵（+mb）⊥，∴（+m）•＝＝32﹣3m＝0，解得m＝3．故选：D．6．（5分）设{a n}是公差不为0的等差数列，满足a42+a52＝a62+a72，则{a n}的前10项和S10＝（）A．﹣10B．﹣5C．0D．5【解答】解：a42+a52＝a62+a72，化简可得：，即2d（a6+a4）+2d（a7+a5）＝0，d≠0．∴a6+a4+a7+a5＝0，∵a5+a6＝a4+a7，∴a5+a6＝0，∴S10＝＝5（a5+a6）＝0，故选：C．7．（5分）函数f（x）＝A sin（ωx+φ）（A＞0，ω＞0，0≤φ≤2π）在R上的部分图象如图所示，则f（2018）的值为（）A．B．﹣5C．D．5【解答】解：根据函数f（x）＝A sin（ωx+φ）（A＞0，ω＞0，0≤φ≤2π）在R上的部分图象，可得A＝5，＝11+1，∴ω＝．结合五点法作图可得×（﹣1）+φ＝0，∴φ＝，f（x）＝5sin（x+），∴f（2018）＝5sin（+）＝5sin（336π+）＝5sin＝5，故选：D．8．（5分）设a＞b＞0，a+b＝1，且x＝（）b，y＝ab，z＝a，则x、y、z的大小关系是（）A．y＜z＜x B．z＜y＜x C．x＜y＜z D．y＜x＜z【解答】解：由a＞b＞0，a+b＝1，得0，，且0＜ab＜1，则，，a＜，∴x＝（）b＞0，y＝ab＝﹣1，0＝＞z＝a＞＝﹣1，∴y＜z＜x．故选：A．9．（5分）《九章算术》是中国古代数学名著，体现了古代劳动人民数学的智慧，其中第六章“均输”中，有一竹节容量问题，某教师根据这一问题的思想设计了如图所示的程序框图，若输出的m的值为35，则输入的a的值为（）A．4B．5C．7D．11【解答】解：起始阶段有m＝2a﹣3，i＝1，第一次循环后m＝2（2a﹣3）﹣3＝4a﹣9，i＝2，第二次循环后m＝2（4a﹣9）﹣3＝8a﹣21，i＝3，第三次循环后m＝2（8a﹣21）﹣3＝16a﹣45，i＝4，第四次循环后m＝2（16a﹣45）﹣3＝32a﹣93，跳出循环，输出m＝32a﹣93＝35，解得a＝4，故选：A．10．（5分）已知f（x）是定义在[﹣2b，1+b]上的偶函数，且在[﹣2b，0]上为增函数，则f （x﹣1）≤f（2x）的解集为（）A．B．C．[﹣1，1]D．【解答】解：∵f（x）是定义在[﹣2b，1+b]上的偶函数，∴﹣2b+1+b＝0，∴b＝1，∵函数f（x）在[﹣2b，0]上为增函数，∴函数f（x）在[﹣2，0]上为增函数，故函数f（x）在[0，2]上为减函数，则由f（x﹣1）≤f（2x），可得|x﹣1|≥|2x|，即（x﹣1）2≥4x2，求得﹣1≤x≤，再结合x∈[﹣2，2]，故f（x﹣1）≤f（2x）的解集为[﹣1，]，故选：B．11．（5分）如图，在边长为2的正方形ABCD中，E，F分别为BC，CD的中点，H为EF 的中点，沿AE，EF，F A将正方形折起，使B，C，D重合于点O，在构成的四面体A﹣OEF中，下列结论中错误的是（）A．AO⊥平面EOFB．直线AH与平面EOF所成角的正切值为C．异面直线OH和求AE所成角为60°D．四面体A﹣OEF的外接球表面积为6π【解答】解：翻折前，AB⊥BE，AD⊥DF，故翻折后，OA⊥OE，OA⊥OF，又OE∩OF＝O，∴OA⊥平面EOF．故A正确；连接OH，AH，则∠OHA为AH与平面EOF所成的角，∵OE＝OF＝1，H是EF的中点，OE⊥OF，∴OH＝EF＝．又OA＝2，∴tan∠OHA＝＝2，故B正确；取AF的中点P，连接OP，HP，则PH∥AE，∴∠OHP为异面直线OH和求AE所成角，∵OE＝OF＝1，OA＝2，∴OP＝AF＝，PH＝AE＝，OH＝EF＝，∴cos∠OHP＝＝，故C错误．由OA，OE，OF两两垂直可得棱锥的外接球也是棱长为1，1，2的长方体的外接球，∴外接球的半径r＝＝，故外接球的表面积为S＝4πr2＝6π，故D正确．故选：C．12．（5分）已知椭圆E：+＝1（a＞b＞0）与过原点的直线交于A、B两点，右焦点为F，∠AFB＝120°，若△AFB的面积为4，则椭圆E的焦距的取值范围是（）A．[2，+∞）B．[4，+∞）C．[2，+∞）D．[4，+∞）【解答】解：取椭圆的左焦点F1，连接AF1，BF1，则AB与FF1互相平分，∴四边形AFBF1是平行四边形，∴AF1＝BF，∵AF+AF1＝2a，∴AF+BF＝2a，∵S△ABF＝AF•BF•sin120°＝AF•BF＝4，∴AF•BF＝16，∵2a＝AF+BF≥2＝8，∴a≥4，又S△ABF＝＝c•|y A|＝4，∴c＝，∴当|y A|＝b＝时，c取得最小值，此时b＝c，∴a2＝3c2+c2＝4c2，∴2c＝a，∴2c≥4．故选：B．二、填空题（每题5分，满分20分，将答案填在答题纸上）13．（5分）设变量x，y满足约束条件，则z＝x﹣2y的最大值为．【解答】解：由约束条件作出可行域如图，联立，解得A（），化目标函数z＝x﹣2y为y＝，由图可知，当直线y＝过A时，直线在y轴上的截距最小，z的最大值为．故答案为：．14．（5分）双曲线﹣＝1（a＞0，b＞0）的渐近线与圆（x﹣）2+y2＝1相切，则此双曲线的离心率为．【解答】解：由题意可知双曲线的渐近线方程之一为：bx+ay＝0，圆（x﹣）2+y2＝1的圆心（，0），半径为1，双曲线﹣＝1（a＞0，b＞0）的渐近线与圆（x﹣）2+y2＝1相切，可得：＝1，可得a2＝b2，c＝a，∴e＝．故答案为．15．（5分）已知四棱锥P﹣ABCD的外接球为球O，底面ABCD是矩形，面P AD⊥底面ABCD，且P A＝PD＝AD＝2，AB＝4，则球O的表面积为．【解答】解：取AD的中点E，连接PE，△P AD中，P A＝PD＝AD＝2，∴PE＝，设ABCD的中心为O′，球心为O，则O′B＝BD＝，设O到平面ABCD的距离为d，则R2＝d2+（）2＝22+（﹣d）2，∴d＝，R2＝，球O的表面积为s＝．故答案为：．16．（5分）已知数列{a n}满足对1≤n≤3时，a n＝n，其对∀n∈N*，有a n+3+a n+1＝a n+2+a n，则数列{n•a n}的前50项的和为2525．【解答】解：∵对∀n∈N*，有a n+3+a n+1＝a n+2+a n，a n+3＝a n+2﹣a n+1+a n，数列{a n}满足对1≤n≤3时，a n＝n，∴a4＝a3﹣a2+a1＝3﹣2+1＝2，同理可得：a5＝1，a6＝2，a7＝3，a8＝2，……．可得：a n+4＝a n．可得：（n+4）a n+4+（n+5）a n+5+（n+6）a n+6+（n+7）a n+7﹣[na n+（n+1）a n+1+（n+2）a n+2+（n+3）a n+3]＝4（a n+a n+1+a n+2+a n+3）＝4×8＝32．∴数列{n•a n}的前50项的和＝1×1+2×2+（3a3+4a4+5a5+6a6）×12+×32＝2525．故答案为：2525．三、解答题（本大题共5小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17．（12分）在△ABC中，角A，B，C所对的边分别为a，b，c，且．（1）求sin B的值；（2）若a＝4，求△ABC的面积S的值．【解答】（本题满分为12分）解：（1）∵由得，…（1分）∴cos C＝cos2A＝cos2A﹣sin2A＝，…2分∴sin C＝＝，…3分又∵A+B+C＝π，sin B＝sin[π﹣（A+C）]＝sin（A+C），…4分∴．…（6分）（2）由正弦定理得，…（9分）∴△ABC的面积．…（12分）18．（12分）如图，五面体ABCDE中，四边形ABDE是菱形，△ABC是边长为2的正三角形，∠DBA＝60°，．（1）证明：DC⊥AB；（2）若C在平面ABDE内的正投影为H，求点H到平面BCD的距离．【解答】解：（1）证明：如图，取AB的中点O，连OC，OD，因为△ABC是边长为2的正三角形，所以，又四边形ABDE是菱形，∠DBA＝60°，所以△DAB是正三角形，所以，而OD∩OC＝O，所以AB⊥平面DOC，所以AB⊥CD；（2）取OD的中点H，连结CH，由（1）知OC＝CD，所以AB⊥ODAB⊥平面DOC，所以平面DOC⊥平面ABD，而平面DOC⊥平面ABD，平面DOC与平面ABD的交线为OD，所以CH⊥平面ABD，即点H是D在平面ABD内的正投影，设点H到平面BCD的距离为d，则点O到平面BCD距离为2d，因为在△BCD中，，得＝，在△OCD中，，得，所以由V O﹣BCD＝V B﹣OCD得，即，解得，所以H到平面BCD的距离．19．（12分）某基地蔬菜大棚采用水培、无土栽培方式种植各类蔬菜．过去50周的资料显示，该地周光照量X（小时）都在30小时以上，其中不足50小时的周数有5周，不低于50小时且不超过70小时的周数有35周，超过70小时的周数有10周．根据统计，该基地的西红柿增加量y（百斤）与使用某种液体肥料x（千克）之间对应数据为如图所示的折线图．（1）依据数据的折线图，是否可用线性回归模型拟合y与x的关系？请计算相关系数r并加以说明（精确到0.01）．（若|r|＞0.75，则线性相关程度很高，可用线性回归模型拟合）（2）蔬菜大棚对光照要求较大，某光照控制仪商家为该基地提供了部分光照控制仪，但每周光照控制仪最多可运行台数受周光照量X限制，并有如下关系：若某台光照控制仪运行，则该台光照控制仪周利润为3000元；若某台光照控制仪未运行，则该台光照控制仪周亏损1000元．若商家安装了3台光照控制仪，求商家在过去50周周总利润的平均值．附：相关系数公式r＝，参考数据≈0.55，≈0.95．【解答】解：（1）由已知数据可得，．…（1分）因为，…（2分）＝20…（3分）．…（4分）所以相关系数．…（5分）因为r＞0.75，所以可用线性回归模型拟合y与x的关系．…（6分）（2）记商家周总利润为y元，由条件可得在过去50周里：当X＞70时，共有10周，此时只有1台光照控制仪运行，周总利润Y＝1×3000﹣2×1000＝1000元．…（8分）当50≤X≤70时，共有35周，此时有2台光照控制仪运行，周总利润Y＝2×3000﹣1×1000＝5000元．…（9分）当X＜50时，共有5周，此时3台光照控制仪都运行，周总利润Y＝3×3000＝9000元．…（10分）所以过去50周周总利润的平均值元，所以商家在过去50周周总利润的平均值为4600元．…（12分）20．（12分）已知动点P到定直线l：x＝﹣4的距离比到定点F（2，0）的距离大2．（1）求动点P的轨迹C的方程；（2）在x轴正半轴上，是否存在某个确定的点M，过该点的动直线l与曲线C交于A，B两点，使得为定值．如果存在，求出点M坐标；如果不存在，请说明理由．【解答】解：（1）设点P的坐标为（x，y），∵动点P到定直线l：x＝﹣4的距离比到定点F（2，0）的距离大2，∴x＞﹣4且，化简得y2＝8x，∴轨迹C的方程为y2＝8x．（2）假设存在满足条件的点M（m，0）（m＞0），直线l：x＝ty+m，联立，得：y2﹣8ty﹣8m＝0，设A（x1，y1），B（x2，y2），则y1+y2＝8t，y1y2＝﹣8m，，，＝，据题意，为定值，则，于是m+4t2＝4λm2+4λm2t2，则有解得m＝4，故当m＝4时，为定值，故M（4，0）．21．（12分）已知函数f1（x）＝（x﹣λ）2，f2（x）＝lnx（x＞0，且x≠1）．（Ⅰ）当λ＝1时，若对任意x∈（1，+∞），f1（x）≥k•f2（x）恒成立，求实数k的取值范围；（Ⅱ）若λ∈（0，1），设f（x）＝，f'（x）是f（x）的导函数，判断f'（x）的零点个数，并证明．【解答】解：（1）当λ＝1时，对任意x∈（1，+∞），（x﹣1）2﹣k•lnx≥0恒成立，令g（x）＝（x﹣1）2﹣k•lnx，求导g′（x）＝，方法一：由x＞1，则2x2﹣2x＝2x（x﹣1）＞0，若k≤0，则g′（x）＞0，∴g（x）在（1，+∞）上是增函数，∴g（x）＞g（1）＝0，符合题意，当k＞0时，令g′（x）＝0，解得：x1＝＜0，x2＝＞1，则g（x）在（1，x2）上是增函数，当x∈（1，x2），g（x）＜g（1）＝0，不符合题意，综上可知：k的取值范围（﹣∞，0]；方法二：2x2﹣2x﹣k＝0，△＝4+8k，当k≤﹣，△≤0，则2x2﹣2x﹣k≥0，则g（x）在（1，+∞）上增函数，g（x）＞g（1）＝0，符合题意，当k≥﹣，g′（x）＝0，解得：x1＝，x2＝，由﹣＜k＜0，则x1＜x2＜1，在（1，+∞）上增函数，当k＞0，则x1＜1＜x2，则g（x）在（1，x2）上是减函数，当x∈（1，x2），g（x）＜g（1）＝0，不符合题意，综上可知：k的取值范围（﹣∞，0]；（Ⅱ）证明：由题意：f′（x）＝，由此可得：x＝λ为一个零点，令h（x）＝2lnx﹣﹣1，（x＞0），则h′（x）＝，h（x）减区间为（0，），单调增区间（，+∞），其中0＜λ＜1，则h min（x）＝h（）＝2ln+1＜1﹣ln4＜0，h（λ）＝2lnλ≠0，h（1）＝λ﹣1≠0，当x＝＞，h（）＝1+﹣1＞0，由函数存在定理及单调性可知：（，+∞）上存在唯一的零点x2，取x＝（＜），则h（）＝4lnλ+﹣5，令g（λ）在（0，1）上是减函数，故当λ∈（0，1）时，g（λ）＞g（1）＝e2﹣5＞0，即h（）＞0，由零点存在定理及单调性可知在（，）存在唯一x3∈（，），h（x3）＝0，由h（x）的单调递减区间（0，），即（0，）上h（x）存在唯一的零点x3，综上可知f（x）共有三个零点．请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分．[选修4-4：坐标系与参数方程]22．（10分）已知在平面直角坐标系xOy中，以O为极点，x轴的正半轴为极轴，建立极坐标系．曲线C1的极坐标方程为ρ＝4cosθ，直线l：（t为参数）．（1）求曲线C1的直角坐标方程及直线l的普通方程；（2）若曲线C2的参数方程为（α为参数），曲线C1上点P的极角为，Q为曲线C2上的动点，求PQ的中点M到直线l距离的最大值．（1）由曲线C1的极坐标方程为ρ＝4cosθ，得直角坐标方程，【解答】解：直线l：，消去参数，可得普通方程l：x+2y﹣3＝0．（5分）（2），直角坐标为（2，2），，M到l的距离d＝＝，从而最大值为．（10分）[选修4-5：不等式选讲]23．已知函数f（x）＝|x﹣1|，关于x的不等式f（x）＜3﹣|2x+1|的解集记为A．（1）求A；（2）已知a，b∈A，求证：f（ab）＞f（a）﹣f（b）．【解答】解：（1）由f（x）＜3﹣|2x+1|，得|x﹣1|+|2x+1|＜3，即或或解得或，所以，集合A＝{x∈R|﹣1＜x＜1}．（2）证明：∵a，b∈A，∴﹣1＜ab＜1，∴f（ab）＝|ab﹣1|＝1﹣ab，f（a）＝|a﹣1|＝1﹣a，f（b）＝|b﹣1|＝1﹣b，∵f（ab）﹣（f（a）﹣f（b））＝1﹣ab﹣1+a+1﹣b＝（1+a）（1﹣b）＞0，∴f（ab）＞f（a）﹣f（b）．。
2018届高三教学质量检测试卷(二)(二模)英语试题 含答案
第I卷第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例：How much is the shirt?A. £19.15.B. £9.18. C £9.15.答案：C1. Who is coming for tea?A. Mark.B. John.C. Tracy.2. What is the weather like now?A. Rainy.B. Cool.C. Hot.3. What will the man do next?A. Stay for dinner.B. Go to the railway station.C. Prepare for the dinner quickly.4. Where will the woman fly to?A. Miami.B. New York City.C. Washington,D.C5. How does the man find his present job?A. It is rewarding.B. It is well-paid.C. It is easy.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。
每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6、7题。
6. What is the woman’s favourite sport?A. Basketball.B. Badminton.C. Tennis.7. What are the two speakers going to do today?A. Play tennis.B. Go to a basketball club.C. Watch a badminton match.听第7段材料，回答第8、9题。
湖南省长沙市长郡中学2018届高考模拟卷(二)文科数学试题(精编含解析)
炎德英才大联考长郡中学2018届高考模拟卷（二）数学（文科）第Ⅰ卷（共60分）一、选择题：本大题共12个小题,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. 已知集合,集合,则等于( )A. B. C. D.【答案】B【解析】分析：先化简集合P，Q，进而求交集即可.详解: P={1，2，3，4，5，6，7，8，9，10}，Q=（﹣2，3）；∴P∩Q={1，2}．故选：B．点睛：本题考查描述法、列举法表示集合的概念，一元二次不等式的解法及交集的运算，属于基础题．2. 复数满足,则复数在复平面内对应的点位于( )A. 第一象限B. 第二象限C. 第三象限D. 第三象限【答案】D【解析】分析：利用复数代数形式的乘除运算化简复数z，求出复数z在复平面内对应的点的坐标，即可得到结果．详解: ：由z（2+i）=3﹣i，得=，则复数z在复平面内对应的点的坐标为：（1，﹣1），位于第四象限．故选：D．点睛：本题考查了复数代数形式的乘除运算，考查了复数的代数表示法及其几何意义，属于基础题．3. 某公司的班车分别在7:30,8:30发车,小明在7:50至8:30之间到达发车站乘坐班车,且到达发车站的时刻是随机的,则他等车时间不超过15分钟的概率是( )A. B. C. D.【答案】B【解析】设小明到达时间为，当在7:50至8:00，或8:15至8:30时，小明等车时间不超过15分钟，故，选D.4. 已知曲线在点处的切线的倾斜角为,则的值为( )A. B. C. D.【答案】D【解析】分析：求导，利用函数f（x）在x=1处的倾斜角为得f′（1）=﹣1，由此可求a 的值.详解: 函数（x＞0）的导数，∵函数f（x）在x=1处的倾斜角为∴f′（1）=﹣1，∴1+=﹣1，∴a=﹣1．故选：D．点睛：求曲线的切线方程是导数的重要应用之一，用导数求切线方程的关键在于求出切点及斜率，其求法为：设是曲线上的一点，则以的切点的切线方程为：．若曲线在点的切线平行于轴（即导数不存在）时，由切线定义知，切线方程为．5. 已知平面向量,满足,,与的夹角为,若,则实数的值为( )A. B. C. D.【答案】A【解析】分析：由，可得（+m）•=0，再利用数量积的运算和定义展开即可得出．详解: ∵||=3，||=2，与的夹角为120°，∴=cos120°==﹣3．∵（+mb）⊥，∴（+m）•==32﹣3m=0，解得m=3．故选：D．点睛：本题考查了数量积的运算和定义、向量垂直与数量积的关系，属于基础题．6. 设是公差不为0的等差数列,满足,则的前项和( )A. B. C. D.【答案】C【解析】分析：根据题意变形可得：，整理可得a5+a6=0，再利用等差数列通项公式求和公式及其性质即可得出．详解: ：a42+a52=a62+a72，化简可得：，即2d（a6+a4）+2d（a7+a5）=0，d≠0．∴a6+a4+a7+a5=0，∵a5+a6=a4+a7，∴a5+a6=0，∴S10==5（a5+a6）=0，故选：C．点睛：在处理等差数列问题时，记住以下性质，可减少运算量、提高解题速度：若等差数列的前项和为，且，则①若，则；②、、、成等差数列．7. 函数(,,)在上的部分图像如图所示,,则的值为( )A. B. C. D.【答案】D【解析】分析：由题意首先求得函数的解析式，然后求解函数值即可求得最终结果．详解: 由函数的图象可得A=5，周期，∴．再由五点法作图可得，∴，故函数．故．故选：D．点睛：已知函数的图象求解析式(1).(2)由函数的周期求(3)利用“五点法”中相对应的特殊点求.8. 设,,且,,,则、、的大小关系是（）A. B. C. D.【答案】A【解析】分析：由已知得到a，b的具体范围，进一步得到ab，，的范围，结合指数函数与对数函数的性质得结果．详解：由a＞b＞0，a+b=1，得0，，且0＜ab＜1，则，，a＜，∴x=（）b＞0，y=log ab=﹣1，0=＞z=log a＞=﹣1，∴y＜z＜x．故选：A．点睛：利用指数函数对数函数及幂函数的性质比较实数或式子的大小，一方面要比较两个实数或式子形式的异同，底数相同，考虑指数函数增减性，指数相同考虑幂函数的增减性，当都不相同时，考虑分析数或式子的大致范围，来进行比较大小，另一方面注意特殊值的应用，有时候要借助其“桥梁”作用，来比9. 《九章算术》是我国古代数学名著，体现了古代劳动人民数学的智慧，其中第六章“均输”中，有一竹节容量问题，某教师根据这一问题的思想设计了如图所示的程序框图，若输出的的值为35，则输入的的值为（）A. B. C. D.【答案】A【解析】起始阶段有，，第一次循环后，，；第二次循环后，，；第三次循环后，，；接着计算，跳出循环，输出.令，得.选A.10. 已知是定义在上的偶函数，且在上为增函数，则的解集为（）A. B. C. D.【答案】B【解析】分析：先根据奇偶函数的性质求出b，再根据f（x﹣1）≤f（2x），可得|x﹣1|≥|2x|，结合x∈[﹣2，2]，求出x的范围．详解：∵f（x）是定义在[﹣2b，1+b]上的偶函数，∴﹣2b+1+b=0，∴b=1，∵函数f（x）在[﹣2b，0]上为增函数，∴函数f（x）在[﹣2，0]上为增函数，故函数f（x）在[0，2]上为则由f（x﹣1）≤f（2x），可得|x﹣1|≥|2x|，即（x﹣1）2≥4x，求得﹣1≤x≤，再结合x∈[﹣2，2]，故f（x﹣1）≤f（2x）的解集为[﹣1，]，故选：B．点睛：处理抽象不等式的问题，一般先利用函数的奇偶性得出区间上的单调性，再利用其单调性脱去函数的符号“f”，转化为考查函数的单调性的问题或解不等式(组)的问题，若为偶函数，则，若函数是奇函数，则．11. 如图，在边长为2的正方形中，，分别为，的中点，为的中点，沿，，将正方形折起，使，，重合于点，在构成的四面体中，下列结论中错误的是（）A. 平面B. 直线与平面所成角的正切值为C. 异面直线和求所成角为D. 四面体的外接球表面积为【答案】C【解析】分析：根据折叠前后垂直关系不变，易得OA⊥平面EOF，利用空间角定义逐一判断B，C,D 的正确性.详解：翻折前，AB⊥BE，AD⊥DF，故翻折后，OA⊥OE，OA⊥OF，又OE∩OF=O，∴OA⊥平面EOF．故A正确；连接OH，AH，则∠OHA为AH与平面EOF所成的角，∵OE=OF=1，H是EF的中点，OE⊥OF，∴OH=EF=．又OA=2，∴tan∠OHA==2，故B正确；取AF的中点P，连接OP，HP，则PH∥AE，∴∠OHP为异面直线OH和求AE所成角，∵OE=OF=1，OA=2，∴OP=AF=，PH=AE=，OH=EF=，∴cos∠OHP==，故C错误．由OA，OE，OF两两垂直可得棱锥的外接球也是棱长为1，1，2的长方体的外接球，∴外接球的半径r==，故外接球的表面积为S=4πr2=6π，故D正确．故选：C．点睛：解决与球有关的内切或外接的问题时，解题的关键是确定球心的位置．对于外切的问题要注意球心到各个面的距离相等且都为球半径；对于球的内接几何体的问题，注意球心到各个顶点的距离相等，解题时要构造出由球心到截面圆的垂线段、小圆的半径和球半径组成的直角三角形，利用勾股定理求得球的半径．12. 已知椭圆：与过原点的直线交于、两点，右焦点为，，若的面积为，则椭圆的焦距的取值范围是（）A. B. C. D.【答案】B【解析】分析：利用三角形的面积公式和椭圆的性质得出a≥4，再根据三角形的面积公式得出当A与短轴端点重合时，c取得最小值，利用椭圆的性质求出2c的最小值即可．详解: 取椭圆的左焦点F1，连接AF1，BF1，则AB与FF1互相平分，∴四边形AFBF1是平行四边形，∴AF1=BF，∵AF+AF1=2a，∴AF+BF=2a，∵S△ABF=AF•BF•sin120°=AF•BF=4，∴AF•BF=16，∵2a=AF+BF≥2=8，∴a≥4，又S△ABF==c•|y A|=4，∴c=，∴当|y A|=b=时，c取得最小值，此时b=c，∴a2=3c2+c2=4c2，∴2c=a，∴2c≥4．故选：B．点睛：：在用基本不等式求最值时，应具备三个条件：一正二定三相等.①一正：关系式中，各项均为正数；②二定：关系式中，含变量的各项的和或积必须有一个为定值；③三相等：含变量的各项均相等，取得最值.第Ⅱ卷（共90分）二、填空题（每题5分，满分20分，将答案填在答题纸上）13. 设变量，满足约束条件则的最大值为__________．【答案】【解析】分析：先画出满足约束条件的可行域，并求出各角点的坐标，然后代入目标函数，即可求出目标函数z=x﹣2y的最大值．详解: 满足约束条件的可行域如下图所示：由图可知，由可得C（，﹣），由：，可得A（﹣4，4），由可得B（2，1），当x=，y=﹣时，z=x﹣2y取最大值：．故选：D．点睛：本题考查的是线性规划问题，解决线性规划问题的实质是把代数问题几何化，即数形结合思想.需要注意的是：一，准确无误地作出可行域;二，画目标函数所对应的直线时，要注意让其斜率与约束条件中的直线的斜率进行比较，避免出错;三，一般情况下，目标函数的最大值或最小值会在可行域的端点或边界上取得.14. 双曲线(,)的渐近线与圆相切，则此双曲线的离心率为__________．【答案】【解析】因为双曲线的渐近线是，所以圆心到渐近线的距离，即，解之得，应填答案。
【全国百强校word版】湖南省长沙市长郡中学2018届高三高考模拟考试(二)理科物理试题
二、选择题：本题共8小题，每小题6分。
在每小题给出的四个选项中，第14～18题只有一项符合题目要求，第19～21题有多项符合题目要求。
全部选对的得6分，选对但不全的得3分，有选错或不选的得0分。
14.核反应方程15112716N H C X E +→++∆中，157N 的质量为m 1、11H 的质量为m 2、126C 的质量为m 3、X 的质量为m 4，光在真空中的速度为c.则下列判断正确的是A.X 是32He ，21234()E m m m m c ∆=+--B.X 是42He ，21234()E m m m m c ∆=+--C.X 是32He ，23412()E m m m m c ∆=+--D.X 是42He ，23412()E m m m m c ∆=+--15.老山自行车赛场采用的是250米赛道，赛道宽度为7.5米。
赛道形如马鞍形，由直线段、过渡曲线段以及圆弧段组成，按2003年国际自盟UCI 赛道标准的要求，其直线段倾角为13°，圆弧段倾角为45°，过渡曲线段由13°向45°过渡。
假设运动员在赛道上的速率不变，则下列说法中一定错误的是A.在直线段赛道上自行车运动员处于平衡状态B.在圆弧段赛道上自行车运动员的加速度不变C.在直线段赛道上自行车受到沿赛道平面向上的摩擦力D.在圆弧段赛道上的自行车可能不受沿赛道平面向上的摩擦力作用16.某科研单位设计了一空间飞行器，飞行器从地面起飞时，发动机提供的动力方向与水平方向夹角α=60°，使飞行器恰好沿与水平方向成θ=30°角的直线斜向行上方由静止开始匀加速飞行，如图所示。
经时间t 后，将动力的方向沿逆时针旋转60°同时适当调节其大小，使飞行器依然可以沿原方向匀减速飞行。
飞行器所受空气阻力不计。
下列说法中正确的是A.2mgB.加速与减速时的加速度大小之比为2:1C.减速飞行时间t 后速度减为零D.加速过程发生的位移与减速到零的过程发生的位移大小之比为2:117.如图所示，匀强电场中的PAB 平面平行于电场方向，C 点为AB 的中点，D 点为PB 的中点。
2018年湖南长沙岳麓区湖南师范大学附属中学高三二模英语试卷-学生用卷
2018年湖南长沙岳麓区湖南师范大学附属中学高三二模英语试卷-学生用卷一、阅读理解(共15小题，每小题2分，共30分)1、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模（A篇）第21~23题6分EnglishWorld languageOnce people dreamed of a language that everybody in the world could understand. Now. For the first time in human history, perhaps there is one—English. It is the official language in more than 50 countries and 250-300 million speak it as a second language. Some say that half the world will be speaking it in the year 2050.Difficult to master?English is not the easiest language to learn—most of its common verbs are irregular and it has a large vocabulary—at least 200, 000 words are in common use. Its pronunciation and written form are also very different.User-friendlyBut some things make it easy—nouns don't have gender and verbs are less complicated than other languages. There is only one form to speak to someone directly—'you' Adjectives don't agree with nouns. and many nouns are often also verbs.English is everywhereSome other languages may have more speakers. Like Spanish or Mandarin. But English is used in many different areas. It is the language of transport: most airline pilots and air-traffic controllers use it. At sea a simple form of English is the international language of communication.English for progressIt is also the first language of science, technology and education-an estimated 80 per cent of the information stored on the internet is in English and 90 per cent of schoolchildren in European countries study it as their first foreign language.New wordsAround 25, 000 new words and expressions enter the English language every year. Do you know what a helicopter parent is? —mother or father too involved in the life of their child. And what about a boomerang child? —a young adult who returns to live with their parents for financial reasons.The future of EnglishWhat is the future of English as a global language? Will another language ever replace it? Many think non—it is already too popular.(1) According to the text. How many people speak English as a second language?A. Around 25, 000 million.B. 250-300 million.C. At least 200, 000 million.D. 90 million.(2) According to the text, why could English be used as a global language?A. It has more speakers, compared with other languages.B. Adjectives agree with nouns.C. It is the first language of science, technology and education.D. It is the easiest language to learn.(3) Which of the following words best described the writer's attitude?A. Objective.B. Subjective.C. Indifferent.D. Critical.2、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第24~27题8分2017~2018学年3月河北石家庄新华区石家庄市第二中学高二下学期月考（A篇）第1~4题10分My son Gilbert was eight years old and had been in the Cub Scouts（童子军团）only a short time. Once he was handed a sheet of paper, a block of wood and four ties and told to return home and give them to his father. That was not an easy task for Gilbert to do. The piece of paper was a set of instructions about how to build a wooden racing car. Gilbert's father laughed when he read the instructions. The block of wood remained untouched as the weeks passed.Finally, I stepped in to see if I could figure it all out. Having no skills, I decided it would be best if I simply read the instructions and let Gilbert do the work. And he did. Within days, his block of wood was turning into a pinewood racing car.Then the big night came. With his pinewood racing car in his hand and pride in his heart we headed to the big race. As the race was done in elimination fashion（淘汰赛形式）, you could keep racing as long as you were the winner.At last, it was between Gilbert and the fastest-looking car there. As the race was about to begin, Gilbert asked if they could stop for a minute, because he wanted to pray. Then the race stopped.Gilbert prayed in earnest for a very long minute. The Master came up to Gilbert and asked the obvious question, "So you prayed to win. Gilbert? "My young son answered, "Oh, no Sir. It wouldn't be fair to ask God to help you beat someone else. I just asked him to make it so I don't cry when I lose."Children seem to have wisdom far beyond us. Perhaps we spend too much of our prayer time asking God to control the race, make us the champion, or remove us from the struggle, when we should be seeking God's strength to get through what lies in our way.(1) Gilbert's father thought the task given by the Cub Scouts could.A. be no trouble at allB. be beyond Gilbert's abilityC. require no skillsD. be too easy for Gilbert(2) Who finally made the pinewood racing car?A. Gilbert.B. Gilbert's father.C. Gilbert's mother.D. The whole family.(3) What can we learn about Gilbert?A. He made it in the final race.B. He thought the Master unfair.C. His performance inspired his mom.D. He was very afraid of losing the race.(4) The author writer this passage to tell us that.A. adults should communicate more with childrenB. parents should help their children with the taskC. friendship is more important than winningD. victory is to gain the power to overcome difficulties3、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第28~31题8分2018~2019学年6月黑龙江哈尔滨香坊区哈尔滨市第六中学高二下学期月考（C篇）第8~11题8分(每题2分)Finding fruits and vegetables at your typical grocery store that have been grown without the extensive use of pesticides can be difficult. Fortunately, The Environmental Working Group（EWG） has done all of the work for you in finding healthy and pesticide—free produce.EWG has created the 2018 Shopper's Guide to Pesticides in Produce, which helps shoppers to find uncontaminated produce. Many consumers do not realize that pesticide residues（残留） are very common on conventionally grown produce products, even after they have been washed or peeled. Because of this, EWG has created their series of guides to lead consumers to safer food choices.In order to create these guides, EWG analyzed the USDA pesticides tests, which found a total of 230 different pesticides and pesticides breakdown products on thousands of produce samples. Analyzing this information, EWG observed the big differences of the amount of pesticides found from product to product.The guide's two main components are two compiled lists highlighting the cleanest and dirtiest produce concerning pesticides. These two lists, Dirty Dozen and Clean Fifteen, show consumers how certain foods continue to carry trace amounts of pesticides with them all the way to the grocery store shelves, while others make it to your kitchen virtually pesticide－free.Some of the highlights from their analyses included the following findings：More than one－third of strawberry samples analyzed in 2016 contained 10 or more pesticide residues and breakdown products.Spinach（菠菜） samples had, on average, almost twice as much pesticide residue by weight compared to any other crop.No single fruit sample from the Clean Fifteen tested positive for more than four pesticides."With EWG's guide, consumers can fill their fridges and fruit bowls with plenty of healthy conventional and organic produce that isn't contaminated with multiple pesticide residue, " said Sonya Lunder, a senior analyst in EWG.Only 25 years ago, the National Academy of Sciences raised concerns about exposure to poisonous pesticides in our food, yet consumers still consume a mixture of pesticides every day in America.(1) Why did EWG create the 2018 Shopper's Guide to Pesticides in Produce?A. To analyze the USDA pesticides tests.B. To advertise organic produce.C. To warn some food companies.D. To help consumers make safer choices.(2) What is the result from the analysis of the USDA pesticides tests?A. All the samples are polluted.B. 230 pesticides are banned.C. Pesticide amounts vary in products.D. All strawberries are poisonous.(3) Where are shoppers most likely to find spinach?A. Dirty Dozen listB. Organic food advertisement.C. Clean Fifteen list.D. Safer food list.(4) What can be inferred from the last two paragraphs?A. All conventional produce is safe.B. No pesticides were used on crops 25 years agoC. Consumers never worry about pesticides.D. Produce safety remains a problem.4、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模（D篇）第32~35题8分Scientists have discovered why the Mona Lisa's expression looks so different to different people and at different times.For centuries, art lovers and critics have been confused by and debated the Leonardo Da Vinci painting' gaze and slight smile?But new research from the University of California, San Francisco has TAL#NBSP shed new light on the shining and seemingly changing face of the Mona Lisa.Through experiments, they discovered that our emotions really do change how we see a neutral（中性的）face.Dr Erika siege and her colleague's study how our emotions change our perceptions（感知）of the world around us even when we aren't aware that something has changed our feelings.This relies on the modern theory of the brain as a predictive organ, instead of a reactive one, says Dr siege.In other words, we have a lifetime of experience and we use those experiences to predict what we are going to experience next.We all have one dominant（支配的）eye and one more passive non-dominant one. If each eye is receiving different information, we only consciously perceive what dominant one sees. But non-dominant sights can still enter our subconscious.They showed 43 people two sets of flashing images at the same time, so that the dominant eye saw and registered neutral expressions, while the non-dominant eye 'saw' flashes of neutral, or smiling faces, which they would only subconsciously be aware of.After viewing the flashing faces, the researchers showed the participants options of faces and asked them to pick out which ones they had seen.When their non-dominant eyes had seen a happy face, they were more likely to think the neutral face had actually been smiling, and the same was true for other expressions.This means that if you see the Mona Lisa after you have just had a screaming fight with your husband, you're going to see the painting differently, says Dr siege."But if you'e having the time of your life at the Louvre, you're going to ace the mysterious smile, " she adds."We are the architects of our own experience. Our brain makes predictions about what it expects to see and uses information from the world to update its expectations, " Dr siege says.(1) What is the purpose of Dr Erika Siege's research?A. To help appreciate the Mona Lisa.B. To discover why people perceive the same thing differently.C. To win a debate.D. To tell a smiling face from a neutral face.(2) Which of the following is the closest in meaning to "shed new light" in Paragraph 3?A. Made something bright.B. Provided new explanations.C. Added light-colored paints.D. Increased amount of natural light.(3) Why did the researchers show the participants two sets of flashing images?A. To play a game.B. To confuse their dominant and non-dominant eyes.C. To strengthen the effect of the non-dominant eyes.D. To provide im ages of more expressions.(4) Which of the following can be the best title for the text?A. Effects of emotionsB. Seeing is believingC. A mysterious researchD. Is Mona Lisa smiling?二、七选五(共5小题，每小题2分，共10分)5、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第36~40题10分Insomnia, a lack of sleep or sleeplessness, is something of an epidemic（流行病）affecting millions of Americans right now. The National institute of Health estimates that 50 million to 70 million Americans of all ages suffer sleep-related problems.Three different types of insomnia exist.1Insomnia lasting from one night to a few weeks occasionally is referred to as transient. Acute insomnia means consistent inability to sleep well for a period of three weeks to six months. Insomnia is considered to be chronic, the most serious, if it persists almost lightly for at least a month. And sometimes longer.Sleeplessness can be caused by many different things.2Once the stimulus passes, the ability to sleep will usually return. Chronic sleeplessness may be caused by ongoing health concerns such as the physical pain of arthritis（关节炎）or the emotional pain of anxiety and depression.Those who suffer from sleepless nights may find themselves have the feeling of sleepiness and tiredness during the day. And those who have little sleep may be more affected by both acute and chronic illnesses like high blood pressure and diabetes.Sleeplessness not only has effects on your physical self but also your emotional well-being.3However, if sleeplessness becomes chronic, the price could be much higher in the form of anxiety or depression.There are numerous ways for mild sleeplessness at home.4Don't exercise too much before bedtime;instead, do something to relax yourself. Finally, create an environment which is favorable for sleeping by removing stimulants from your bedroom like extra lighting or a television.5A. Being unable to sleep well for a few nights may only produce anger.B. Most people experience sleeplessness for an outside stimulus such as stress, improper diet or a poor sleeping environment.C. Prepare your body for sleep by avoiding drinks such as tea or coffee before bed.D. If you still cannot find relief, it may be time to visit your physician for help.E. Insomnia may be classified as transient, acute, and chronic.F. Being unable to sleep well for a few nights may cause serious health problems.G. What is transient insomnia?三、完形填空(共20小题，每小题1.5分，共30分)6、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第41~60题30分The aim of advertising is to change our ideas about things we want and need.Many ads use people's1—bike fear or love—to persuade us that we need the product. Think of a TV commercial that shows a woman out2in a car on a rainy day, with her two small children.3another car turns in front of her. The mother quickly hits the4and her car comes to a stop. The children are5smiling and laughing. The message is: This car will keep you safe. Other examples are6that tell you to "take vitamins to prevent heart diseases" or "buy insurance —7your family"Commercials for soft drinks often show everyone in the ad is having a good time. The messageis:8you buy this drink. You will have a good time too. Many ads like this are based on people's9for fun and enjoyment.Advertising for designer products uses people's vanity（虚荣心）to10the products. The ads try to create a personality for eachbrand.11. One brand of watches is for people who likeadventure.12ads featuring an auto racer or a pilot wearingthem.13brand of watches is for elegant and fashionable people. And the ads show wealthy travelers on a cruise ship. The watches look almost the same—only their imagesare14. When you buy that brand of watch. You are buyingthe15.After an ad is prepared. Advertisers use psychology to make itmore16.Psychologists have found that people have positive feelings about they see more often. So the same ad is used for a long time before it is17with a new ad. Other studies found that people18an ad better when it is fresh in their memory,19TV ads are repeated very often. Knowingabout20in advertising us make better decisions about the things we buy.A. emotionsB. hobbiesC. sense'sD. affectionA. sittingB. ridingC. drivingD. waitingA. ImmediatelyB. DesperatelyC. SuddenlyD. SlowlyA. engineB. seatC. switchD. brakeA. stillB. alwaysC. justD. evenA. quotasB. slogansC. commentsD. requestsA. satisfyB. raiseC. shelterD. improveA. UnlessB. UntilC. AsD. IfA. desireB. habitC. discoveryD. imaginationA. preserveB. produceC. developD. promoteA. In shortB. ThusC. HoweverD. For exampleA. forB. withC. throughD. onA. OtherB. The otherC. AnotherD. OthersA. elegantB. differentC. similarD. flexibleA. mediaB. attractivenessC. imageD. ideaA. effectiveB. obviousC. suitableD. interestingA. loadedB. repeatedC. removedD. replacedA. live onB. react toC. object toD. rely onA. althoughB. soC. sinceD. butA. psychologyB. feelingsC. productsD. message四、语法填空(共10小题，每小题1.5分，共15分)7、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第61~70题15分An unsupervised（无人监管的）flower shop operated by a college student in Beijing is attracting many young customers. Chinese media reported. Differentfrom1（tradition） shops in the capital, the young storekeeper surnamed Wang opened the first unsupervised flower shop,2people can select flowers and pay via mobile payment systems it ding We Chat and Alipay.Born after 1995 and a lover of fiction. The young man saidhe3（begin）the start up because it would not take up his time. "People go lower shops at any time, " he said. "I cannot concentrate on4（read）when I have to serve customers." So Wang opened the self-service flower shop after conducting market research.Most young customers like shopping5 a free environment. And hate promotion from clerks. He explained adding that customers can have theirquestions6（answer）about categories and names of flowers through mobile .even though there is no shopkeeper. A large number of youngcustomers7（attract）by the operation so far, which has proved to be a success. At least for now.8he does not sell flowers in the shop in person. Wang said he would call at it at odd i9（interval）every day. "It is a business with small investment. And at cause a huge10（lose）if a burglary occurs, " he said.五、改错(共10小题，每小题1分，共10分)8、【来源】 2018年湖南长沙岳麓区湖南师范大学附属中学高三二模第71~80题10分假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。
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湖南省长沙市2018 届高中毕业班教学质量检测（二）
文科综合
第I卷（选择题共140分）
本卷共35个小题，每小题4分，共l40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。
24．西汉官方孝德教育呈现出神秘化的倾向，“孝”从政治伦理范畴中提升出来，发展为宇宙间无所不在的最高准则，为引导人们崇尚孝德，统治者
A．革新了选官制度B．提出家国一体观念
C．设刺史以察民情D．推动了儒释遭融合
25.如图5，唐玄宗开元年间将10道析分为15
道，其中山南道两分为山南东道、山南西道，
江南道三分为江南东道、江南西道、黔中道，
以适应监察事务繁重的需要。
这一变化主要
是由于
A．南方经济的发展B．藩镇势力的扩大
C．海上丝路的开通D．君主专制的加强
26.据文献记载两宋时期．“士大夫发迹垄亩，贵为公卿，谓父祖旧庐为不可居，……复以医药弗使，饮膳难得·自村幢而迁于邑，白岜而迁于郡者，亦多矣。
”又如奉化李雄飞之父．“隆于教子，其徙城市，便二子之从师也”．这主要反映了两宋时期
A_工商业市镇吸引大量人口B科举考试促进重学风气形成
C.城市职能呈现多元化特征
D.土地兼并导致农村经济萧条
27．下表为不同史籍关于郑和下西洋的历史叙述。
据此能够被认定的历史事实是。