2019年9月湖南省高2020届高2017级炎德英才大联考长郡中学届高三月考试卷语文试题答题卡

长郡中学2020届高三月考试卷（二）英语试题注意事项：1.答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。

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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。

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.答案是C.1.What will the man do?A.Tie his shoes.B.Rise from a fall.C.Leave his shoes untied.2.How will the woman probably feel after the conversation?A.Glad.B.Worried.C.Disappointed.3.Who pushed the woman into the pool?A.Jane's brother.B.The man's brother.C.Amy's brother.4.Why will the man need to find another present?A.The store is closed.B.The store is too far away.C.The jacket is too expensive.5.Where are the speakers?A.At a concert.B.At Kate's house.C.At a restaurant.第二节（共15小题;每小题L 5分，满分22.5分）听下面5段对话或独白。

2019届长郡中学高三月考英语试卷（一）第一部分听力（共两节，满分30分）（略）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。

AChina has announced it's abolishing its one-child policy.What difference has it made, statistically speaking?400million births preventedThe one-child policy,officially in place since1979,has prevented400million births.Parents have faced fines and other punishments for having more children.The majori ty of the decrease in China's fertility(生育)rate happened in the1970s.It dropped from5.8children per woman in1970t02.7in1978.Despite the one-child policy the rate had only fallen t01.7by2013.21:28baby death rateSince the one-child policy was introduced,baby girls have become more likely to die than boys.In the1970s,according to the United Nations,60males per l,000live births died under the age of one.For girls the figure was53.In the1980s,after the one-child policy became official, the rate for both was36.By the1990s,26males per l,000live births died before the age of one-and33girls.The2000s saw21boys per l,000live births dying and28girls.1.16boys born for every girlSexually selective abortions have been considered as a major cause of China's unusual sexual imbalance.Gietel-Basten,associate professor in social policy at Oxford University,says the births of many girls are not registered if parents have broken the rule by having two children,adding officials often turn a blind eye.It's estimated there are now33million more men than women in China.4：2：1familiesWith the ageing of China's population and the continuation of the one-child policy,a“4: 2:1”home is the description given to households in which there are four grandparents cared for by two working age parents,who themselves have one child.By2050,it's predicted that a quarter of China's population will be65or older.The predicted decline in the number of people of working age is thought to have persuaded the government to drop the one-child policy.21.When was the baby death rate for both boys and girls equal?A.In the1970s.B.In the1980s.C.In the1990s.D.In the2000s.22.What makes the one-child policy abolished according to the passage?A.The decline of birth rate.B.The rise of baby death rate.C.The change of family structure.D.The decline of working age people.23.The passage sums up the one-child policy by.A.numberB.exampleC.contrastD.analysisBlt's a little hard to figure out the rules for sure,but the baby African buffalo(7K午)seems to have the upper hand.The two stand a couple feet apart,staring at each other.Behind them,their mothers look on with the kind of indifference(不关心)of mothers everywhere who see their kids playing a harmless game.The alarm went off about half an hour ago,the second time tonight.Here,one bell means “elephant”,two,“rhino”(犀牛),and three and four,1really can't remember for what,because when you're awoken by bells in the middle of the night,your first thought won't be“Animal!”. But as soon as I realized it was the rhino alarm,I was running for the stairs.Yet I'm the only one watching the animals stare-down.Either everyone else died of a heart attack w hen the bells went off,or l was the only one in the entire hotel who didn't turn the alarm switch to off before going to bed.The Ark,a hotel shaped like the biblical ark(圣经的方舟),lies in the highlands of Kenya's Aberdare National Park,about100kilometers north of Nairobi.At the Ark's prow(船首)are huge windows overlooking a waterhole.Earlier tonight,I'd watched a pack of wolves,buffalos and elephants.And now,at three or so in the morning,I'm down for rhino bells.The first time,2.5 hours ago,it was a single black rhino,which came down,got a drink,and left.I'd have been sorry for the lost sleep.I've already arranged with a guide to take me out at sunrise for the so-called“should_never_miss”bird-watching,where we'll get to see30species, but really,how many times in your life will you get to wake up and say,“Wow!Rhino!”?Isn't that the definition of a pretty good night?Last week we went from Uganda into Kenya.Before I left home,1thought I'd be happy with ten elephants and five or six giraffes on the whole trip.I didn't dare to dream the rhino.24.The message that the hotel's bells sent was.A.a very wise way the author had never heard ofB.hard for the author to remember in detailC.a matter that all visitors had got used toD.like conditioned response training for animals25.What's the author's attitude toward the bell?A.It's bothering but two bells are welcome.B.It should be turned off whenever it rings.C.It lets visitors have a chance to see buffalos.D.It is a very bad way to wake the guests up.26.What can we learn from the passage?A.The author hates all the animals.B.There are lots of wild animals near the Ark.C.The author slept well during the whole trip.D.The author has no interest in thinos.27.What did the author see at about3:00a.m.?A.Two buffalos were fighting fiercely for food.B.A rhino and a buffalo were fighting for water.C.A rhino and a buffalo were staring at each other.D.A buffalo was teaching its baby to walk freely.CMany people trying to sell homes find that an increase in home prices has turned the market in their favor.But sellers can still get the short end of the deal if they aren't careful.Here are a few tips for you:Don't test your luck.Of course you think anyone who moves into your lovely home should be willing to pay top dollars,especially if you've recently invested in some improvements.But listing a home at a price that's too high above the market price could turn away some buyers.Buyers noticing that the home still hasn't sold may begin to assume there's something wrong with the house and use that as a reasonable excuse for offering a lower price.And if a home hasn't received any offers after two weeks,it might be time to reset the price.A price that's too low can bring about an undesired outcome.Listing your home at or slightly below the market price can have the effect of drawing in a large group of buyers and increase the chances that a home will receive multiple offers.But setting the price too low comes with several risks.One possibility is that buyers will get skeptical of the home that is listed for$15,000to $20,000less than similar homes in the area,especially if it's not properly marketed.Once again, people might assume there is something wrong with the home and may not bother to look at it.Spy on the competition.Going to other people's open houses can give you a better sense of how your home compares to others on the market.Check out the decoration in their kitchens,the size of their backyards and use the information to figure out where your home should fit in the range of the price.But don't set your pricing just on what you see elsewhere.28.What does the underlined part in Paragraph l probably mean?A.Be at a disadvantage.B.Get the upper hand.C.Have control over the situation.D.Be unable to fit in.29.What should home sellers do according to the passage?A.Price your house on the basis of its geographic location.B.Price your house slightly above the market price after decorating it.C.Change the price if no offer has been received within a week.D.Price your house at or slightly below the market price.30.If you set the price of your house too low,.A.buyers might think it not worthwhile to go to have a lookB.you are likely to come into conflict with the neighborhood sellersC.your house will be crowded with buyers within a couple of daysD.chances of your house being sold at a better price will be greatly increased31.What is the benefit of going to other people's open house?A.You can pick up some useful lessons on house selling.B.You can get your pricing mainly based on what you see elsewhere.C.You can work out how much money you should ask for your home.D.You can know how to make your house stand out against other houses.DWhat's your favorite type of music?Most people may prefer rock and roll,pop or jazz. These types are most publicized by television and the radio.Country music now would seem to mistakenly belong to the category of rock or pop.Publicity of this category started to drop only after the year2000;however,it did have its peak years just before falling down.What people may not know is that this type of music inspired(促成)rock and roll especially with one of the most important figures in music history-Elvis Presley-who is known as the6/King of Rock and Roll".Elvis used to be very famous at a radio program which broadcast country music back in the late1940s.Country music is influenced by jazz and blues.Rock music is also influenced by jazz and blues so that they may even seem similar.Nowadays,country music is closely related to pop music due to artists'preference and popular demand.Starting from the1980's it slowly divided itself into New and Old Country Music. Legends like Johnny Cash began to slip away because the new understanding of country music actually contains pop and his music just doesn't fit the category.Fortunately,country music is still heavily promoted through websites and reviews.Certain types of music nowadays are linked with different age groups.Classical for the people in the mid19th century,pop in the1920's,and then there's country music.All these types continue to grow and modernize while still appealing to certain age groups.More modern country music compositions tend to become different in their own ways which makes it even harder to identify them as country music while they,again,remain appealing only to certain age groups. Active promoters of country music videos,however,seem to appeal to a large number of age groups especially those active in Australia,for as long as country music isn't mixed with other types,it will eventually regain its charm as music for all ages.32.It can be learned from the first paragraph that country music____.A.is most publicized by television and the radioed to be liked by more people compared to jazzC.has become less popular since the year2000D.should be included in the category of rock or pop33.What is the relationship between rock music and country music?A.Both types influenced the formation of rock and roll.B.Presley turned pop music into country music.C.They are both influenced by jazz and blues.D.The use of the same instruments makes them sound similar.34.Some legends of country music began to slip away because____.A.their songs totally changed the style of country musicB.they began to work on pop music for more profitsC.there is no marketing space for country music anymoreD.they can't meet the popular demands on country music35.The author seems to believe that____.A.country music should keep its own style to attract more listenersB.it's necessary for country music to be mixed with other typesC.only a certain age group of people will like country music in the futuremon characteristics can be found in all modern music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。

2019-2020学年湖南省长沙市长郡中学高三（上）月考数学试卷（一）（9月份）一、选择题：本大题共12小题,每小题5分,共60分在每小题给出的四个选项中,只有一项是符合题目要求的.1. 已如集合P={x|x2−2x−3≥0}，Q={x|1<x<4}，则P∩Q=()A.(−1, 3)B.[3, 4)C.(−∞, −3)∪[4, +∞)D.(−∞, −1)∪(3, +∞)【答案】B【考点】一元二次不等式的解法交集及其运算【解析】根据不等式的性质求出集合P的等价条件，结合交集定义进行计算即可．【解答】解：P={x|x2−2x−3≥0}={x|x≥3或x≤−1}，∵Q={x|1<x<4}，∴P∩Q={x|3≤x<4}，故选B.2. 设复数z满足|z−i|+|z+i|＝4，z在复平面内对应的点为(x, y)，则（）A.x24−y23=1 B.x24+y23=1 C.y24−x23=1 D.y24+x23=1【答案】D【考点】轨迹方程【解析】设复数z对应的点为Z，由|z−i|+|z+i|＝4，知点Z到点A(0, 1)、点B(0, −1)的距离和大于|AB|，由此可得结论，求出方程即可．【解答】设复数z对应的点为Z，则|z−i|表示点Z到点A(0, 1)的距离，|z+i|表示点Z到点B(0, −1)的距离，又|AB|＝2，由|z−i|+|z+i|＝4，知点Z到点A、B的距离和大于|AB|，z的关键为椭圆，所以a＝2，c＝1，则b=√3，椭圆的焦点坐标就是AB，故z在复平面内对应的点的轨迹是：y 24+x23=1．3. 若0<x<y<1，0<a<1，则下列不等式正确的是（）A.log a x2<log a y3B.cos ax<cos ayC.a x<a yD.x a<y a【答案】 D【考点】不等式的基本性质 【解析】A ．由0<x <y <1，0<a <1，√x <√y <√y 3，即可判断出log a √x 与log a √y 3大小关系．B ．由0<x <y <1，0<a <1，可得0<ax <ay <1，即可得出cos (ax)与cos (ay)大小关系．C ．由0<x <y <1，0<a <1，利用指数函数的单调性即可判断出结论．D ．∵ 0<x <y <1，0<a <1，根据幂函数f(x)＝x a 的单调性即可得出大小关系． 【解答】A ．∵ 0<x <y <1，0<a <1，√x <√y <√y 3，因此log a √x >log a √y 3，因此不正确；B ．∵ 0<x <y <1，0<a <1，∴ 0<ax <ay <1，∴ cos (ax)>cos (ay)，因此不正确；C ．∵ 0<x <y <1，0<a <1，∴ a x >a y ，因此不正确；D ．∵ 0<x <y <1，0<a <1，根据幂函数f(x)＝x a 的单调性，可得x a <y a ．4. A 4纸是生活中最常用的纸规格．A 系列的纸张规格特色在于：①A 0、A 1、A 2…、A 5，所有尺寸的纸张长宽比都相同．②在A 系列纸中，前一个序号的纸张以两条长边中点连线为折线对折裁剪分开后，可以得到两张后面序号大小的纸，比如1张A0纸对裁后可以得到2张A1纸，1张A1纸对裁可以得到2张A2纸，依此类推．这是因为A 系列纸张的长宽比为√2:1这一特殊比例，所以具备这种特性．已知A0纸规格为84.1厘米×118.9厘米.118.9÷84.1≈1.41≈√2，那么A4纸的长度为（ ） A.14.8厘米 B.21.0厘米 C.29.7厘米 D.42.0厘米 【答案】 C【考点】根据实际问题选择函数类型 【解析】由已知可得A 4纸的长为(√2)4，计算得答案．【解答】由题意，A 0纸的长与宽分别为118.9厘米，84.1厘米， 则A 1纸的长为√2，A 2纸的长为118.9√2√2=(√2)2，A 3纸的长为(√2)2√2=(√2)3，A 4纸的长为(√2)3√2=(√2)4=29.7（厘米）．5. 函数f(x)＝x|x|−sin 2x 的大致图象是（ ）A. B.C. D.【答案】 B【考点】函数的图象变化 函数的图象 【解析】先判断函数的奇偶性和图象的对称性，利用当x ＝π时，f(x)的符号是否对应，利用排除法进行求解即可． 【解答】f(−x)＝−x|x|+sin 2x ＝−(x|x|−sin 2x)＝−f(x)， 则f(x)为奇函数，图象关于原点对称，排除C ，D 当x ＝π时，f(π)＝π2−sin 2π＝π2>0，排除A ，6. 中国是发现和研究勾股定理最古老的国家之一，古代数学家称直角二角形的较短的直角边为勾、另一直角边为股、斜边为弦，其三边长组成的一组数据称为勾股数，现从1∼15这15个数中随机抽取3个整数，则这三个数为勾股数的概率为（ ） A.1910 B.3910C.3455D.4455【答案】 D【考点】古典概型及其概率计算公式 【解析】从这15个数中随机选取3个整数，所有的基本事件个数n =C 153，利用列举法求出勾股数有4个，由此能求出这三个数为勾股数的概率． 【解答】从这15个数中随机选取3个整数，所有的基本事件个数n =C 153，其中，勾股数为：(3, 4, 5)，(6, 8, 10)，(9, 12, 15)，(5, 12, 13)，共4个， ∴ 这三个数为勾股数的概率为： p =4C 153=4455．7. 已知向量a →，b →满足|a →|=2,|b →|=√2，且a →⊥(a →+2b →)，则b →在a →方向上的投影为（ ） A.1 B.−√2 C.√2 D.−1【答案】 D【考点】平面向量数量积的性质及其运算 【解析】利用向量的垂直关系，推出a →⋅b →，然后求解b →在a →方向上的投影．【解答】向量a →，b →满足|a →|=2,|b →|=√2，且a →⊥(a →+2b →)， 可得a →2+2a →⋅b →=0， 可得a →⋅b →=−2， 则b →在a →方向上的投影为：a →⋅b →|a →|=−1．8. 已知函数f(x)=sin (x +π6)−m ，x ∈[0,7π3]有三个不同的零点x 1，x 2，x 3，且x 1<x 2<x 3，则x 1+2x 2+x 3的值为（ ） A.10π3B.4πC.11π3D.不能确定【答案】 A【考点】三角函数的最值 正弦函数的图象 【解析】令f(x)=sin (x +π6)−m =0，则sin (x +π6)=m ，由条件知函数y =sin (x +π6)与函数y ＝m 在[0,7π3]上有三个交点，然后根据函数的图象的对称性可得结果．【解答】令f(x)=sin (x +π6)−m =0，则sin (x +π6)=m ，∵ 函数y =sin (x +π6)的对称轴为x =π3+kπ，k ∈Z ，s∴ 当x ∈[0,7π3]时，x =π3或x =4π3．∵ 函数f(x)=sin (x +π6)−m ，x ∈[0,7π3]有三个不同的零点x 1，x 2，x 3，且x 1<x 2<x 3，∴ 函数y =sin (x +π6)与函数y ＝m 在[0,7π3]上有三个交点，∴ 由函数y =sin (x +π6)与函数y ＝m 在[0,7π3]上的图象知当y =sin (x +π6)与函数y ＝m 在[0,7π3]上有三个交点时，x 1+x 22=π3，x 2+x 32=4π3， ∴ x 1+2x 2+x 3=2π3+8π3=10π3．9. 若a ≠b ，数列a ，x 1，x 2，b 和数列a ，y 1，y 2，y 3，b 都是等差数列，则 x 2−x 1y 2−y 1=( )A.23B.34C.1D.43【答案】 D【考点】等差数列的性质 【解析】根据等差数列的性质可分别求x 2−x 1=13(b −a)，y 2−y 1=14(b −a)，即可求比值． 【解答】∵ a 、x 1、x 2、b 成等差数列 ∴ x 2−x 1=13(b −a)∵ a ，y 1，y 2，y 3，b 都是等差数列， ∴ y 2−y 1=14(b −a) ∴ x 2−x 1y 2−y 1=43．10. 已知椭圆C:x 2a 2+y 2b 2=1(a >b >0)的左右焦点分别为F 1、F 2，O 为坐标原点，A 为椭圆上一点，且AF 1→⋅AF 2→=0，直线AF 2交y 轴于点M ，若|F 1F 2|＝6|OM|，则△OMF 2与△AF 1F 2的面积之比为（ ） A.481 B.427C.25144D.518【答案】 D【考点】 椭圆的离心率 【解析】由题意画出图形，且得到tan ∠MF 2O =13，再由AF 1→⋅AF 2→=0，利用三角形相似可得|AF 1||AF 2|=13，设|AF 1|＝x ，则|AF 2|＝3x ，得2a ＝3x +x ＝4x ，又4c 2＝(3x)2+x 2＝10x 2，联立求得ca 的值，进一步得到△OMF 2与△AF 1F 2的面积之比． 【解答】由|F 1F 2|＝2c ＝6|OM|，得|OM|=c3，故tan ∠MF 2O =13，由AF 1→⋅AF 2→=0，得∠F 1AF 2＝90∘，故|AF 1||AF 2|=13，设|AF 1|＝x ，则|AF 2|＝3x ，∴ 2a ＝3x +x ＝4x ， 又4c 2＝(3x)2+x 2＝10x 2，∴ e =ca =√104． ∴ S △OMF 2S △AF 1F 2=4c 29a 2=518．11. 已知偶函数f(x)满足f(4+x)=f(4−x)，且当x ∈(0, 4]时，f(x)=ln 2x x，关于x的不等式f 2(x)+af(x)>0在[−200, 200]上有且只有300个整数解，则实数a 的取值范围是（ ） A.(−ln 2,−13ln 6)B.(−ln 2,−13ln 6]C.(−13ln 6,−3ln 24) D.(−13ln 6,−3ln 24]【答案】 D【考点】函数的零点与方程根的关系 【解析】 此题暂无解析 【解答】解：∵ 偶函数f(x)满足f(4+x)=f(4−x)， ∴ f(x +4)=f(4−x)=f(x −4)，∴ f(x)的周期为8，且f(x)的图象关于直线x =4对称． 由于f(x)在[−200, 200]上含有50个周期， 且f(x)在每个周期内都是轴对称图形，∴ 关于x 的不等式f 2(x)+af(x)>0在(0, 4]上有3个正整数解． 又当x ∈(0, 4]时，f′(x)=1−ln 2x x 2，∴ f(x)在(0, e2)上单调递增，在[e2,4]上单调递减．∵ f(1)=ln 2，f(2)>f(3)>f(4)=ln 84=34ln 2>0，∴ 当x =k(k =1, 2, 3, 4)时，f(x)>0，∴ 当a ≥0时，f 2(x)+af(x)>0在(0, 4]上有4个整数解，不符合题意， ∴ a <0，由f 2(x)+af(x)>0可得f(x)<0或f(x)>−a ， 显然f(x)<0在(0, 4]上无正整数解，故而f(x)>−a 在(0, 4]上有3个整数解，分别为1，2，3， ∴ −a ≥f(4)=34ln 2，−a <f(3)=ln 63，−a <f(1)=ln 2，∴ −ln 63<a ≤−34ln 2．故选D .12. 已知SC 是球O 的直径，A ，B 是球O 球面上的两点，且CA =CB =1,AB =√3，若三棱锥S −ABC 的体积为1，则球O 的表面积为（ ） A.4π B.13π C.16π D.52π 【答案】 D【考点】球的体积和表面积 【解析】推导出∠SAC ＝∠SBC ＝90∘，SA ＝SB ＝AB =√3，S △ABC =12×√3×12=√34，V =13×√34×ℎ=1，求出ℎ＝4√3，α＝30∘，R 2＝r 2+ℎ2＝(12sin 30)2+(4√32)2＝13，从而能求出球O 的表面积．【解答】∵ SC 是球O 的直径，A ，B 是球O 球面上的两点， 且CA =CB =1,AB =√3，∴ ∠SAC ＝∠SBC ＝90∘，S △ABC =12×√3×12=√34， V =13×√34×ℎ=1，解得ℎ＝4√3，sin α=121=12，∴ α＝30∘，R 2＝r 2+ℎ2＝(12sin 30)2+(4√32)2＝13， 球O 的表面积S ＝4πR 2＝52π．二、填空题：本大题共4小题，每小题5分，共20分若直线y ＝kx(k ≠0)是曲线f(x)＝2x 3−x 2的一条切线，则k ＝________−18 ． 【答案】−18【考点】利用导数研究曲线上某点切线方程 【解析】设切点为(x 0, kx 0)，求出函数的导数，可得切线的斜率，由已知切线的方程，可得含x 0的方程组，解得x 0，即可得到所求k 的值． 【解答】f(x)＝2x 3−x 2的导数为f′(x)＝6x 2−2x ， 设切点为(x 0, kx 0)， 由切线y ＝kx ，可得：{6x 02−2x 0=k,2x 03−x 02=kx 0,， 将①代入②得2x 03−x 02=6x 03−2x 02，即4x 03=x 02，∴ x 0＝0或x 0=14，∴ k ＝0（舍去）或k =−18．已知等比数列{a n }的前n 项和为S n ，且S 3＝7a 1，则{a n }的公比q 的值为________． 【答案】 2或−3 【考点】等比数列的通项公式 等比数列的前n 项和 【解析】设等比数列的公比为q ，由S 3＝7a 1，可得a 1+a 1q +a 1q 2=7a 1，解出即可． 【解答】设等比数列的公比为q ，∵ S 3＝7a 1，∴ a 1+a 1q +a 1q 2=7a 1，化为q 2+q −6＝0，解得q ＝2或−3．为了庆祝六一儿童节，某食品厂制作了3种不同的精美卡片，每袋食品随机装入一张卡片，集齐3种卡片可获奖，现购买该种食品5袋，能获奖的概率为________． 【答案】5081【考点】排列、组合及简单计数问题 等可能事件等可能事件的概率 【解析】利用对立事件，不能获奖的概率，即可得到结论 【解答】因为5袋食品中放入的卡片所有的可能的情况有35种，而不能获奖表明此五袋中所放的卡片类型不超过两种，故所有的情况有C 32⋅25−3种（此处减有是因为五袋中所抽取的卡片全是相同的情况每一种都重复记了一次，故减3）． 所以小明获奖的概率是P ＝1−C 32⋅25−335=5081．已知P 为双曲线C:x 2a 2−y 2b 2=1(a >0, b >0)右支上一点，直线l 是双曲线C 的一条渐近线，P 在l 上的射影为Q ，F 1是双曲线的左焦点，若|PF 1|+|PQ|的最小值为3a ，则双曲线C 的离心率为________√2 ． 【答案】√2【考点】双曲线的离心率 【解析】P 在双曲线的右支上，所以PF 1−PF 2＝2a ，F 2是双曲线的右焦点，|PF 2|+|PQ|＝2a +|PF 2|+|PQ|≥2a +|QF 2|结合已知条件推出PQ ⊥l ，通过√a 2+b 2=a ，求解即可．【解答】因为P 在双曲线的右支上，所以PF 1−PF 2＝2a ，F 2是双曲线的右焦点，|PF 2|+|PQ|＝2a +|PF 2|+|PQ|≥2a +|QF 2|当且仅当P 在线段QF 2上时取等号， 因为|PF 1|+|PQ|的最小值为3a ， 所以|QF 2|＝a ，不妨设l 为：y =ba x ，由P 在l 上的射影为Q ，PQ ⊥l ， 所以22=a ，又c 2＝a 2+b 2，可得e 2＝2，所以e =√2．三、解答题本大题共70分.解答应写出文字说明、证明过程或演算步骤．第17～21题为必考题，每个试题考生都必须作答.第22、23题为选考题，考生根据要求作答.（一）必考题：共60分如图，D 是直角△ABC 斜边BC 上一点，AC =√3DC ． (Ⅰ)若∠BAD ＝60∘，求∠ADC 的大小； (Ⅱ)若BD ＝2DC ，且AB =√6，求AD 的长．【答案】（本题满分为1（1）∵ ∠BAD ＝60∘，∠BAC ＝90∘， ∴ ∠DAC ＝30∘，…1分 在△ADC 中，由正弦定理可得：DC sin ∠DAC=AC sin ∠ADC，…2分∴ sin ∠ADC =ACDC sin ∠DAC =√32，…3分 ∴ ∠ADC ＝120∘，或60∘，…4分又∠BAD ＝60∘，∴ ∠ADC ＝120∘...6分 （2）∵ BD ＝2DC ， ∴ BC ＝3DC ，在△ABC 中，由勾股定理可得：BC 2＝AB 2+AC 2，可得：9DC 2＝6+3DC 2， ∴ DC ＝1，BD ＝2，AC =√3，…8分 令∠ADB ＝θ，由余弦定理：在△ADB 中，AB 2＝AD 2+BD 2−2AD ⋅BD ⋅cos θ，…9分在△ADC 中，AC 2＝AD 2+CD 2−2AD ⋅CD ⋅cos (π−θ)，…10分可得：{6=AD 2+4−4AD cos θ3=AD 2+1+2AD cos θ，∴ 解得：AD 2＝2，可得：AD =√2⋯12分 【考点】 正弦定理 【解析】(Ⅰ)由已知可求∠DAC ＝30∘，在△ADC 中，由正弦定理可得sin ∠ADC =√32，即可解得∠ADC ＝120∘．(Ⅱ)由已知在△ABC 中，由勾股定理可得DC ＝1，BD ＝2，AC =√3，令∠ADB ＝θ，由余弦定理{6=AD 2+4−4AD cos θ3=AD 2+1+2AD cos θ，即可解得AD 的值．【解答】（本题满分为1（1）∵ ∠BAD ＝60∘，∠BAC ＝90∘， ∴ ∠DAC ＝30∘，…1分在△ADC 中，由正弦定理可得：DCsin ∠DAC =ACsin ∠ADC ，…2分 ∴ sin ∠ADC =ACDC sin ∠DAC =√32，…3分 ∴ ∠ADC ＝120∘，或60∘，…4分又∠BAD ＝60∘，∴ ∠ADC ＝120∘...6分 （2）∵ BD ＝2DC ， ∴ BC ＝3DC ，在△ABC 中，由勾股定理可得：BC 2＝AB 2+AC 2，可得：9DC 2＝6+3DC 2， ∴ DC ＝1，BD ＝2，AC =√3，…8分 令∠ADB ＝θ，由余弦定理：在△ADB 中，AB 2＝AD 2+BD 2−2AD ⋅BD ⋅cos θ，…9分在△ADC 中，AC 2＝AD 2+CD 2−2AD ⋅CD ⋅cos (π−θ)，…10分可得：{6=AD 2+4−4AD cos θ3=AD 2+1+2AD cos θ，∴ 解得：AD 2＝2，可得：AD =√2⋯12分如图，四面体ABCD 中，△ABC 是正三角形，△ACD 是直角三角形，∠ABD ＝∠CBD ，AB ＝BD ．（1）证明：平面ACD ⊥平面ABC ；（2）过AC 的平面交BD 于点E ，若平面AEC 把四面体ABCD 分成体积相等的两部分，求二面角D −AE −C 的余弦值． 【答案】证明：如图所示，取AC 的中点O ，连接BO ，OD ． ∵ △ABC 是等边三角形，∴ OB ⊥AC ．△ABD 与△CBD 中，AB ＝BD ＝BC ，∠ABD ＝∠CBD ， ∴ △ABD ≅△CBD ，∴ AD ＝CD ． ∵ △ACD 是直角三角形，∴ AC 是斜边，∴ ∠ADC ＝90∘． ∴ DO =12AC ．∴ DO 2+BO 2＝AB 2＝BD 2． ∴ ∠BOD ＝90∘． ∴ OB ⊥OD ．又DO ∩AC ＝O ，∴ OB ⊥平面ACD ． 又OB ⊂平面ABC ，∴ 平面ACD ⊥平面ABC ．设点D ，B 到平面ACE 的距离分别为ℎD ，ℎE ．则ℎD ℎE=DEBE ．∵ 平面AEC 把四面体ABCD 分成体积相等的两部分， ∴13S △ACE ⋅ℎD 13S △ACE ⋅ℎE=ℎD ℎE=DEBE =1．∴ 点E 是BD 的中点．建立如图所示的空间直角坐标系．不妨取AB ＝2．则O(0, 0, 0)，A(1, 0, 0)，C(−1, 0, 0)，D(0, 0, 1)，B(0, √3, 0)，E(0,√32,12)． AD →=(−1, 0, 1)，AE →=(−1,√32,12)，AC→=(−2, 0, 0)．设平面ADE 的法向量为m →=(x, y, z)，则{m →⋅AD →=0m →⋅AE →=0，即{−x +z =0−x +√32y +12z =0，取m →=(3,√3,3)．同理可得：平面ACE 的法向量为n →=(0, 1, −√3)． ∴ cos <m →,n →>=m →⋅n→|m →||n →|=−2√3√21×2=−√77． ∴ 二面角D −AE −C 的余弦值为√77．【考点】平面与平面垂直二面角的平面角及求法 【解析】（1）如图所示，取AC 的中点O ，连接BO ，OD ．△ABC 是等边三角形，可得OB ⊥AC ．由已知可得：△ABD ≅△CBD ，AD ＝CD ．△ACD 是直角三角形，可得AC 是斜边，∠ADC ＝90∘．可得DO =12AC ．利用DO 2+BO 2＝AB 2＝BD 2．可得OB ⊥OD ．利用线面面面垂直的判定与性质定理即可证明．（2）设点D ，B 到平面ACE 的距离分别为ℎD ，ℎE ．则ℎD ℎE=DEBE ．根据平面AEC 把四面体ABCD 分成体积相等的两部分，可得13S △ACE ⋅ℎD 13S △ACE ⋅ℎE=ℎD ℎE=DEBE =1，即点E 是BD 的中点．建立如图所示的空间直角坐标系．不妨取AB ＝2．利用法向量的夹角公式即可得出． 【解答】证明：如图所示，取AC 的中点O ，连接BO ，OD ． ∵ △ABC 是等边三角形，∴ OB ⊥AC ．△ABD 与△CBD 中，AB ＝BD ＝BC ，∠ABD ＝∠CBD ， ∴ △ABD ≅△CBD ，∴ AD ＝CD ． ∵ △ACD 是直角三角形，∴ AC 是斜边，∴ ∠ADC ＝90∘． ∴ DO =12AC ．∴ DO 2+BO 2＝AB 2＝BD 2． ∴ ∠BOD ＝90∘． ∴ OB ⊥OD ．又DO ∩AC ＝O ，∴ OB ⊥平面ACD ． 又OB ⊂平面ABC ，∴ 平面ACD ⊥平面ABC ．设点D ，B 到平面ACE 的距离分别为ℎD ，ℎE ．则ℎD ℎE=DEBE ．∵ 平面AEC 把四面体ABCD 分成体积相等的两部分， ∴13S △ACE ⋅ℎD 13S △ACE ⋅ℎE=ℎD ℎE=DEBE =1．∴ 点E 是BD 的中点．建立如图所示的空间直角坐标系．不妨取AB ＝2．则O(0, 0, 0)，A(1, 0, 0)，C(−1, 0, 0)，D(0, 0, 1)，B(0, √3, 0)，E(0,√32,12)． AD →=(−1, 0, 1)，AE →=(−1,√32,12)，AC →=(−2, 0, 0)．设平面ADE 的法向量为m →=(x, y, z)，则{m →⋅AD →=0m →⋅AE →=0 ，即{−x +z =0−x +√32y +12z =0，取m →=(3,√3,3)．同理可得：平面ACE 的法向量为n →=(0, 1, −√3)． ∴ cos <m →,n →>=m →⋅n→|m →||n →|=√3√21×2=−√77． ∴ 二面角D −AE −C 的余弦值为√77．在平面直角坐标系xOy中，已知抛物线C:y2=4x的焦点为F，过F的直线l交抛物线C 于A，B两点．(1)求线段AF的中点M的轨迹方程；(2)已知△AOB的面积是△BOF面积的3倍，求直线l的方程．【答案】解：(1)根据题意：抛物线的焦点为F(1, 0)，设M(x, y)，则A(2x−1, 2y)，把A(2x−1, 2y)代入y2=4x可得：4y2=8x−4，即y2=2x−1．(2)设直线l的方程为x=my+1，代入y2=4x可得y2−4my−4=0，设A(x1, y1)，B(x2, y2)，则y1y2=−4，①若A在第一象限，B在第四象限，则y1>0，y2<0，则S△AOB=12⋅OF⋅(y1−y2)，S△BOF=12⋅OF⋅(−y2)，∵S△AOB=3S△BOF，∴y1−y2=−3y2，∴y1=−2y2，又y1y2=−4，∴y1=2√2，y2=−√2．故x1=2，x2=12，把A(2, 2√2)代入x=my+1可得m=2√2=√24，∴直线l的方程为x−√24y−1=0，即4x−√2y−4=0．②若A在第四象限，B在第一象限，则y1<0，y2>0，S△AOB=12⋅OF⋅(y2−y1)，S△BOF=12⋅OF⋅y2，∵S△AOB=3S△BOF，∴y2−y1=3y2，∴y1=−2y2，又y1y2=−4，∴y1=−2√2，y2=√2．故x1=2，x2=12，把A(2, −2√2)代入x=my+1可得m=2√2=−√24，∴直线l的方程为x+√24y−1=0，即4x+√2y−4=0．综上，直线l的方程为：4x−√2y−4=0或4x+√2y−4=0．【考点】与抛物线有关的中点弦及弦长问题三角形的面积公式圆锥曲线的轨迹问题直线的斜率【解析】（1）设M(x, y)，表示出A点坐标，代入抛物线方程化简即可；（2）设A(x1, y1)，B(x2, y2)，直线l的方程为x＝my+1，联立方程组可得则y1y2＝−4，三角形的面积比得出y1＝−2y2，讨论A，B所在象限得出A的坐标，进而可得出直线l的方程．【解答】解：(1)根据题意：抛物线的焦点为F(1, 0)，设M(x, y)，则A(2x−1, 2y)，把A(2x−1, 2y)代入y2=4x可得：4y2=8x−4，即y2=2x−1．(2)设直线l的方程为x=my+1，代入y2=4x可得y2−4my−4=0，设A(x1, y1)，B(x2, y2)，则y1y2=−4，①若A在第一象限，B在第四象限，则y1>0，y2<0，则S△AOB=12⋅OF⋅(y1−y2)，S△BOF=12⋅OF⋅(−y2)，∵S△AOB=3S△BOF，∴y1−y2=−3y2，∴y1=−2y2，又y1y2=−4，∴y1=2√2，y2=−√2．故x1=2，x2=12，把A(2, 2√2)代入x=my+1可得m=2√2=√24，∴直线l的方程为x−√24y−1=0，即4x−√2y−4=0．②若A在第四象限，B在第一象限，则y1<0，y2>0，S△AOB=12⋅OF⋅(y2−y1)，S△BOF=12⋅OF⋅y2，∵S△AOB=3S△BOF，∴y2−y1=3y2，∴y1=−2y2，又y1y2=−4，∴y1=−2√2，y2=√2．故x1=2，x2=12，把A(2, −2√2)代入x=my+1可得m=2√2=−√24，∴直线l的方程为x+√24y−1=0，即4x+√2y−4=0．综上，直线l的方程为：4x−√2y−4=0或4x+√2y−4=0．已知函数f(x)=x2e x（1）求函数f(x)的单调区间；（2）若函数g(x)＝f2(x)−kf(x)+1恰有四个零点，求实数k的取值范围．【答案】函数f(x)=x 2e x ，可得f′(x)=2x−x2e x，令f′(x)>0得，得2<x或x<0，故函数f(x)的单调增区间为(0, 2)单调减区间为(−∞, 0)或(2, +∞)．令t＝f(x)因为关于t的方程至多有两个实根，①当△<0时g(x)显然无零点，此时不满足题意；②当△＝0时t2−kt+1＝0有且只有一个实根，结合函数f(x)的图象，可得此时至多2个零点，也不满足题意．③当△>0时即k>2或k<−2，此时关于t的方程t2−kt+1＝0有两个不等实根t1、t2设t1<t2且t1+t2＝k，t1t2＝1，若要g(x)有四个零点则0<t1<f(x)极大值<t2而f(x)=f(2)=4e2，所以(4e2)2−k⋅4e2+1<0，解得k>4e +e24又4e+e24>2，故k>4e2+e24．【考点】利用导数研究函数的极值利用导数研究函数的单调性【解析】（1）求出函数f(x)=x 2e x 的导函数f′(x)=2x−x2e x，通过导函数的符号，求解函数的单调区间．（2）令t＝f(x)因为关于t的方程至多有两个实根，通过①当△<0时，②当△＝0时，③当△>0时，判断方程的解的情况，要g(x)有四个零点则0<t1<f(x)极大值<t2，然后区间极大值列出不等式推出k的范围．【解答】函数f(x)=x 2e x ，可得f′(x)=2x−x2e x，令f′(x)>0得，得2<x或x<0，故函数f(x)的单调增区间为(0, 2)单调减区间为(−∞, 0)或(2, +∞)．令t＝f(x)因为关于t的方程至多有两个实根，①当△<0时g(x)显然无零点，此时不满足题意；②当△＝0时t2−kt+1＝0有且只有一个实根，结合函数f(x)的图象，可得此时至多2个零点，也不满足题意．③当△>0时即k>2或k<−2，此时关于t的方程t2−kt+1＝0有两个不等实根t1、t2设t1<t2且t1+t2＝k，t1t2＝1，若要g(x)有四个零点则0<t1<f(x)极大值<t2而f(x)=f(2)=4e2，所以(4e2)2−k⋅4e2+1<0，解得k>4e2+e24又4e2+e24>2，故k>4e2+e24．为了让幼儿园大班的小朋友尝试以客体区分左手和右手，左肩和右肩，在游戏中提高细致戏察和辨别能力，同时能大胆地表达自己的想法，体验与同伴游戏的快乐，某位教师设计了一个名为【肩手左右】的游戏，方案如下：游戏准备：选取甲、乙两位小朋友面朝同一方向并排坐下进行游戏．教师站在两位小朋友面前出示游戏卡片．游戏卡片为两张白色纸板，一张纸板正反两面都打印有相同的”左“字，另一张纸板正反两面打印有相同的“右”字．游戏进行：一轮游戏（一轮游戏包含多次游戏直至决出胜者）开始后，教师站在参加游戏的甲、乙两位小朋友面前出示游戏卡片并大声报出出示的卡片上的“左”或者“右”字．两位小朋友如果听到“左”的指令，或者看到教师出示写有“左”字的卡片就应当将左手放至右肩上并大声喊出“停!”．小朋友如果听到“右”的指令，或者看到教师出示写有“右”字的卡片就应当将右手放至左肩上并大声喊出“停!”．最先完成指令动作的小朋友喊出“停!”时，两位小朋友都应当停止动作，教师根据两位小朋友的动作完成情况进行评分，至此游戏完成一次．游戏评价：为了方便描述问题，约定：对于每次游戏，若甲小朋友正确完成了指令动作且乙小朋友未完成则甲得1分，乙得−1分；若乙小朋友正确完成了指令动作且甲小朋友未完成则甲得−1分，乙得1分；若甲，乙两位小朋友都正确完成或都未正确完成指令动作，则两位小朋友均得0分．当两位小朋友中的一位比另外一位小朋友的分数多8分时，就停止本轮游戏，并判定得分高的小朋友获胜．现假设“甲小朋友能正确完成一次游戏中的指令动作的概率为α，乙小朋友能正确完成一次游戏中的指令动作的概率为β”，一次游戏中甲小朋友的得分记为X．（1）求X的分布列；（2）若甲小朋友、乙小朋友在一轮游戏开始时都赋予4分，p i(i＝0, 1,…,8)表示“甲小朋友的当前累计得分为i时，本轮游戏甲小朋友最终获胜”的概率，则P0＝1，p8＝1，p i ＝ap i−1+bp i+cp i+1(i＝1, 2,…,7)，其中a＝P(X＝−1)，b＝P(X＝0)，c＝P(X＝1)．假设α＝0.5，β＝0.6．(i)证明：{p i+1−p i)(i＝0, 1, 2,…,7)为等比数列；(ii)求p4，并根据p4的值说明这种游戏方案是否能够充分验证“甲小朋友能正确完成一次游戏中的指令动作的概率为0.5，乙小朋友能正确完成一次游戏中的指令动作的率为0.6”的假设．【答案】由题意知X所有可能的取值为−1，0，1，P(X＝−1)＝(1−α)β，P(X＝0)＝αβ+(1−α)(1−β)，P(X＝1)＝α(1−β)，∴X的分布列为：∴a＝0.5×0.6＝0.3，b＝0.5×0.6+0.5×0.4＝0.5，c＝0.5×0.4＝0.2，∵p i＝3p i−1+2p i+1，(i＝1, 2,…,7)，即p i＝0.3p i−1+0.5p i+0.2p i+1，(i＝1, 2,…,7)，整理，得5p i＝3p i−1+2p i+1，(i＝1, 2,…,7)，∴p i+1−p i=3(p i−p i−1)，(i＝1, 2,…,7)，2∴{p i+1−p i}(i＝0, 1, 2,…,7)是以p1−p0为首项，3为公比的等比数列．2(ii)由(i)知p i+1−p i ＝(p 1−p 0)⋅(32)i ＝p 1⋅(32)i ，∴ p 8−p 7＝p 1⋅(32)7，p 7−p 6=p 1⋅(32)6，…，p 1−p 0=p 1⋅(32)0， 累加求和得： p 8−p 0＝p 1•[(32)0+(32)1+...+(32)7]=1−(32)81−32p 1=(32)8−112p 1＝1，∴ p 1=12(32)8−1，∴ p 4＝p 4−p 0＝p 1•[(32)0+(32)1+(32)2+(32)3]=1−(32)41−32p 1=(32)4−132−1×32−1(32)8−1=1(32)4+1=1697≈0.16．p 4表示甲小朋友当前累计得分为4分时，本轮游戏最终甲获胜的概率， 由计算结果可以看出，假设一次游戏中甲小朋友完成指令动作的概率为0.5， 乙小朋友完成指令动作的概率为0.6，本轮游戏甲小朋友获胜的概率p 4≈0.16，这种情况发生的概率非常小，说明这种游戏方案不能够充分验证-次游戏中甲小朋友完成指令动作的概率为0.5， 乙小朋友完成指令动作的概率为0.6的假设． 【考点】 数列的应用离散型随机变量及其分布列 【解析】（1）由题意知X 所有可能的取值为−1，0，1，分别求出相应的概率，由此能求出X 的分布列．(2)(i)由α＝0.5，β＝0.6，求出a ，b ，c ，推导出5p i ＝3p i−1+2p i+1，(i ＝1, 2,…,7)，从而p i+1−p i =32(p i −p i−1)，(i ＝1, 2,…,7)，由此能证明{p i+1−p i }(i ＝0, 1, 2,…,7)是以p 1−p 0为首项，32为公比的等比数列．(ii)推导出p i+1−p i ＝(p 1−p 0)⋅(32)i ＝p 1⋅(32)i ，从而p 8−p 7＝p 1⋅(32)7，p 7−p 6=p 1⋅(32)6，…，p 1−p 0=p 1⋅(32)0，累加求和得p 8−p 0=(32)8−112p 1＝1，求出p 1=12(32)8−1，由此能求出p 4=1697≈0.16．由此得到这种游戏方案不能够充分验证-次游戏中甲小朋友完成指令动作的概率为0.5，乙小朋友完成指令动作的概率为0.6的假设． 【解答】由题意知X 所有可能的取值为−1，0，1， P(X ＝−1)＝(1−α)β，P(X ＝0)＝αβ+(1−α)(1−β)， P(X ＝1)＝α(1−β)， ∴ X 的分布列为：∴ a ＝0.5×0.6＝0.3，b ＝0.5×0.6+0.5×0.4＝0.5，c ＝0.5×0.4＝0.2， ∵ p i ＝3p i−1+2p i+1，(i ＝1, 2,…,7)，即p i ＝0.3p i−1+0.5p i +0.2p i+1，(i ＝1, 2,…,7)， 整理，得5p i ＝3p i−1+2p i+1，(i ＝1, 2,…,7)， ∴ p i+1−p i =32(p i −p i−1)，(i ＝1, 2,…,7)，∴ {p i+1−p i }(i ＝0, 1, 2,…,7)是以p 1−p 0为首项，32为公比的等比数列． (ii)由(i)知p i+1−p i ＝(p 1−p 0)⋅(32)i ＝p 1⋅(32)i ，∴ p 8−p 7＝p 1⋅(32)7，p 7−p 6=p 1⋅(32)6，…，p 1−p 0=p 1⋅(32)0， 累加求和得： p 8−p 0＝p 1•[(32)0+(32)1+...+(32)7]=1−(32)81−32p 1=(32)8−112p 1＝1，∴ p 1=12(32)8−1，∴ p 4＝p 4−p 0＝p 1•[(32)0+(32)1+(32)2+(32)3]=1−(32)41−32p 1=(32)4−132−1×32−1(32)8−1=1(32)4+1=1697≈0.16．p 4表示甲小朋友当前累计得分为4分时，本轮游戏最终甲获胜的概率， 由计算结果可以看出，假设一次游戏中甲小朋友完成指令动作的概率为0.5， 乙小朋友完成指令动作的概率为0.6，本轮游戏甲小朋友获胜的概率p 4≈0.16，这种情况发生的概率非常小，说明这种游戏方案不能够充分验证-次游戏中甲小朋友完成指令动作的概率为0.5， 乙小朋友完成指令动作的概率为0.6的假设．（二）选考题：共10分.请考生在22、23题中任选一题作答，如果多做，则按所做的第一题计分.[选修4-4：坐标系与参数方程]在平面直角坐标系xOy 中，曲线C 1的参数方程为{x =2+√32t ，y =1+12t ， （t 为参数）．以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，曲线C 2的极坐标方程为ρ=4cos θ． (1)求曲线C 1的普通方程和C 2的直角坐标方程；(2)已知点P(2, 1)，曲线C 1与C 2的交点为A ，B ，求|PA|+|PB|的值． 【答案】解：(1)由{x =2+√32t ，y =1+12t ，消去t 得：x −2=√3(y −1)， 整理得C 1的普通方程为：x −√3y −2+√3=0； 在ρ=4cos θ两边同乘以ρ得：ρ2=4ρcos θ，由ρ2=x 2+y 2，x =ρcos θ得C 2的直角坐标方程为：x 2+y 2=4x ， 即(x −2)2+y 2=4；(2)将C 1的参数方程{x =2+√32t ，y =1+12t ，代入x 2+y 2=4x 整理得：t 2+t −3=0． 设A ，B 对应的参数分别为t 1，t 2，则 t 1+t 2=−1，t 1t 2=−3， 由(1)知C 2是圆心为(2, 0)，半径为2的圆． 检验知点P(2, 1)在该圆内， ∴ t 1，t 2异号，由参数的几何意义知|PA|+|PB|=|t 1−t 2|=√(t 1+t 2)2−4t 1t 2=√13． 【考点】圆的极坐标方程与直角坐标方程的互化 参数方程与普通方程的互化 直线和圆的方程的应用 【解析】(Ⅰ)直接把直线参数方程中的参数t 消去，可得直线的普通方程；把ρ＝4cos θ两边同乘以ρ，结合极坐标与直角坐标的换算关系式可得曲线C 2的直角坐标方程；(Ⅱ)将C 1的参数方程代入x 2+y 2＝4x ，整理得关于t 的一元二次方程，再由根与系数的关系及参数t 的几何意义求解． 【解答】（1）由{x =2+√32ty =1+12t，消去t 得：x −2=√3(y −1)， 整理得C 1的普通方程为：x −√3y −2+√3=0； 在ρ＝4cos θ两边同乘以ρ得：ρ2＝4ρcos θ，由ρ2＝x 2+y 2，x ＝ρcos θ得C 2的直角坐标方程为：x 2+y 2＝4x ， 即(x −2)2+y 2＝4；（2）将C 1的参数方程{x =2+√32ty =1+12t ，代入x 2+y 2＝4x 整理得：t 2+t −3＝0． 设A ，B 对应的参数分别为t 1，t 2，则 t 1+t 2＝−1，t 1t 2＝−3， 由(Ⅰ)知C 2是圆心为(2, 0)，半径为2的圆． 检验知点P(2, 1)在该圆内， ∴ t 1，t 2异号，由参数的几何意义知|PA|+|PB|=|t 1−t 2|=√(t 1+t 2)2−4t 1t 2=√13． (2)将C 1的参数方程{x =2+√32t ，y =1+12t ， 代入x 2+y 2=4x 整理得：t 2+t −3=0． 设A ，B 对应的参数分别为t 1，t 2，则 t 1+t 2=−1，t 1t 2=−3， 由(1)知C 2是圆心为(2, 0)，半径为2的圆． 检验知点P(2, 1)在该圆内， ∴ t 1，t 2异号，由参数的几何意义知|PA|+|PB|=|t 1−t 2|=√(t 1+t 2)2−4t 1t 2=√13． [选修4-5：不等式选讲]已知函数f(x)＝|2x +2|−|ax −2|（1）当a ＝1时，求不等式f(x)≥2x −1的解集；（2）若存在x ∈(1, 3)，使不等式f(x)>2x 成立，求a 的取值范围．【答案】当a ＝1时，f(x)＝|2x +2|−|x −2|={−x −4,x <−13x,−1≤x ≤2x +4,x >2，当x <−1时，由−x −4≥2x −1解得x <−1；当−1≤x ≤2时，由3x ≥2x −1解得x ≥−1，∴ −1≤x ≤2； 当x >2时，由x +4≥2x −1解得x ≤5，∴ 2<x ≤5； 综上可得，原不等式的解集为{x|x ≤5}．∵ x ∈(1, 3)，∴ f(x)>2x 等价于|ax −2|<2，即−2<ax −2<2， 即等价于0<a <4x，∴ 由题设可得，存在x ∈(1, 3)使0<a <4x成立，又由x ∈(1, 3)，可知4x<4，∴ a 的取值范围为(0, 4)．【考点】绝对值不等式的解法与证明 【解析】（1）利用分段函数求不等式的解集，注意要分情况讨论； （2）将恒成立问题转化为求最值问题，即可求出a 的范围． 【解答】当a ＝1时，f(x)＝|2x +2|−|x −2|={−x −4,x <−13x,−1≤x ≤2x +4,x >2，当x <−1时，由−x −4≥2x −1解得x <−1；当−1≤x ≤2时，由3x ≥2x −1解得x ≥−1，∴ −1≤x ≤2； 当x >2时，由x +4≥2x −1解得x ≤5，∴ 2<x ≤5； 综上可得，原不等式的解集为{x|x ≤5}．∵ x ∈(1, 3)，∴ f(x)>2x 等价于|ax −2|<2，即−2<ax −2<2， 即等价于0<a <4x ，∴ 由题设可得，存在x ∈(1, 3)使0<a <4x 成立， 又由x ∈(1, 3)，可知4x <4，∴ a 的取值范围为(0, 4)．。

2019-2020学年长沙市长郡中学高三英语月考试卷及答案解析第一部分阅读（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、D四个选项中选出最佳选项AElectric Shocks Can Be FatalGovernment statistics recently showed that in theUK, more than 3,000 people a year experience electric shocks in their homes. A smaller number of people are killed after they touch the power lines outside their homes. Electric shocks can cause a person's heart or breath to stop and are potentially fatal. It is essential for people to learn basic techniques to deal with such emergencies.What to do?● If you are the first person to reach someone who has an electric shock, don't touch him or her!● If the victim is still holding the appliance that gives him or her the shock (e.g. a hair dryer), unplug it or turn off the power at its source.Under no circumstances can you try to move the appliance with your hands!● Ifyou can't turn off the power, use a piece of wood, like a broom handle or a chair, to separate the victim from the appliance or the power source. You may even be able to do this with a folded newspaper.● The victim must remain lying down. If he or she isunconscious, the victim should be placed on his or her side. But he or she should not be moved if there is a possibility of neck or spine injuries unless it is ly necessary.● It is essential to maintain the victim's body heat, so make sure you cover him or her with a blanket before you do anything else. If the victim is not breathing, apply mouth-to-mouth resuscitation (人工呼吸). Keep the victim's head low until professional help arrives.● If the electric shock has been caused by an external power line, the dangers to the victim and to anybody providing first aid are much greater.1. What kind of passage is it?A. An advertisement.B. A horror story.C. A news report.D. First aid emergency advice.2. The underlined sentence implies that ________.A. you should move the applianceB. you should pick up the appliance and turn off the electricityC. it is very dangerous to touch the appliance with your handsD. it is unnecessary to unplug the appliance with your hands3. When a person has got an electric shock, you should ________.A. separate the victim from the appliance and let him sit upB. keep the victim warm and help him or her breathe againC. move the victim onto his or her side if he or she has got neck injuriesD. keep the victim's head high until professional help arrivesBLight pollution is a significant but overlooked driver of the rapid decline of insect populations, according to the most comprehensive review of the scientific evidence to date.Artificial light at night can affect every aspect of insects' lives, the researchers said. "We strongly believe artificial light at night — in combination with habitat loss, chemical pollution.invasive (入侵的) species, and climate change — is driving insect declines, " the scientists concluded after assessing more than 150 studies.Insect population collapses have been reported around the world, and the first global scientific review published in February,said widespread declines threatened to cause a "catastrophic collapse of nature's ecosystems".There are thought to be millions of insect species, most still unknown to science, and about half are active at night. Those active in the day may also be disturbed by light at night when they are at rest.The most familiar impact of light pollution is moths (飞蛾) flapping around a bulb, mistaking it for the moon. Some insects use the polarisation of light to find the water they need to breed, as light waves line up after reflecting from a smooth surface. But artificial light can scupper (使泡汤) this. Insects areimportant prey (猎物) for many species, but light pollution can tip the balance in favour of the predator if it traps insects around lights. Such increases in predation risk were likely to cause the rapid extinction of affected species, the researchers said.The researchers said most human-caused threats to insects have analogues in nature, such as climate change and invasive species. But light pollution is particularly hard for insects to deal with.However, unlike other drivers of decline, light pollution is ly easy to prevent. Simply turning off lights that are not needed is the most obvious action, he said, while making lights motion-activated also cuts light pollution. Shading lights so only the area needed is lit up is important. It is the same with avoiding blue-white lights, which interfere with daily rhythms. LED lights also offer hope as they can be easily tuned to avoid harmful colours and flicker rates.4. What is discussed in the passage?A. Causes of declining insect populations.B. Consequences of insect population collapses.C. Light pollution: the key bringer of insect declines.D. Insect declines: the driver of the collapsed ecosystem.5. What is the 5th paragraph mainly about?A. How light travels in space.B. How light helps insects find food.C. How the food chain is interrelated.D. How light pollution affects insects.6. What does the underlined word"analogues"in Paragraph 6probably mean?A. Selective things.B. Similar things.C. Variations.D. Limitations.7. What is the purpose of the last paragraph?A. To offer solutions.B. To give examples.C. To make comparisons.D. To present arguments.CIt is a question people have been asking for ages. Is there a way to turn back the aging process?For centuries, people have been looking for a “fountain of youth”. The idea is that if you find a magical fountain, and drink from its waters, you will not age.Researchers in New York did not find an actual fountain of youth, but they may have found a way to turn back the aging process. It appears the answer may be hidden right between your eyes, in an area called the hypothalamus (下丘脑). The hypothalamus is part of your brain. It controls important activities within the body.Researchers at New York’s Albert Einstein College of Medicine found that hypothalamus neural (神经的) stem cells also influence how fast aging takes place in the body.What are stem cells(干细胞)? They are simple cells that can develop into specialized cells, like blood or skin cells. Stem cells can also repair damaged tissues and organs.Dongsheng Cai is a professor at the Albert Einstein College of Medicine. He was the lead researcher in a study on aging in mice. He and his team reported their findings in the journal Nature, Cai explained when hypothalamus function is in decline, particularly the loss of hypothalamus stem cells, and this protection against the agingdevelopment is lost. it eventually leads to aging.Using this information, the researchers began trying to activate, or energize, the hypothalamus laboratory mice. They did this by injecting the animals with stem cells, Later, the researchers examined tissues and tested for changes in behavior. They looked for changes in the strength and coordination (协调) of the animals muscles. They also studied the social behavior and cognitive ability of the mice. The researchers say the results show that the treatment slowed aging in the animals, Cai says injecting middle-aged mice with stem cells from younger mice helped the older animals live longer.But these results were just from studying mice in a laboratory. If the mice can live longer, does that mean people could have longer lives? The next step is to see if the anti-aging effects also work in.8. In Paragraph 2 a “fountain of youth” is mentioned to ________.A. introduce the main topicB. show a hidden secret.C. describe scientists researchD. recommend a way to stay young9. Aging takes place in the body when _______.A. stem cells develop into specialized cellsB. there are important activities within the bodyC. hypothalamus neural stem cells fail to protect against agingD. the hypothalamus fails to repair damaged tissues and organs10. What do we know about the researchers at Albert Einstein College of Medicine from the text?A. They did experiments to see how stem cells work.B. They studied mice to find their connection with humans.C. They have found a possible way to slow the aging progress.D. They have found no changes in mice s behavior during the experiment11. What will the researchers probably do next?A. They will help some animals live longer.B. They will announce the fountain of youth doesn’t existC. They will develop products to help people live a longer life immediatelyD. They will do research to see if what they have found in mice will apply to humans.DEver wondered if dogs can learn new words? Yes, say researchers as they have found that talented dogs may have the ability to grasp new words after hearing them only four times.While previous evidence seems to show that most dogs do not learn words, unless eventually very well trained, a few individuals have shown some extraordinary abilities, according to a study published in the journal Scientific Reports.“We wanted to know under which conditions the gifted dogs may learn novel words” said researcher xuekw Claudia Fugazza from theEötvösLorándUniversityinHungary. For the study, the team involved two gifted dogs, Whisky and Vicky Nina. The team exposed the dogs to the new words in two different conditions.In the exclusion-based task, presented with seven known toys and one new toy, the dogs were able to select the new toy when presented with a new name. Researchers say this proves that dogs can choose by exclusion when faced with a new word, they selected the only toy which did not have a known name.However, this was not the way they would learn the name of the toy. In fact, when they were presented with one more equally new name to test their ability to recognize the toy by its name, the dogs got totally confused and failed.The other condition, the social one, where the dogs played with their owners who pronounced the name of the toy while playing with the dog, proved to be the successful way to learn the name of the toy, even after hearing it only 4 times. “The rapid learning that we observed seems to equal children’s ability to learn many new words at a fast rate around the age of 18 months,” Fugazza says. “But we do not know whether the learning mechanisms(机制) behind this learning are the same for humans and dogs. ”To test whether most dogs would learn words this way, 20 other dogs were tested in the same condition, but none of them showed any evidence of learning the toy names, confirming that the abilityto learn words rapidly in the absence of formal training is very rare and is only present in a few gifted dogs.12. What was the purpose of the study published in Scientific Reports?A. To better train dogs’ ability to learn new words.B. To further confirm previous evidence about dogs.C. To prove extraordinary memory abilities of gifted dogs.D. To explore favorable conditions for gifted dogs’ new-word learning.13. How did the dogs react when exposed to two new names in the first condition?A. Slow to understand.B. Quick to learn.C. At a loss.D. In a panic.14. What was found about dogs’ new-word learning in the social condition?A. Learning through playing applied to most dogs.B. The social condition helped dogs learn new words.CDogs’ new-word learning turned out to be less effective.D. Dogs shared similar learning mechanisms with children.15. Which of the following is the best title for the text?A. Gifted Dogs Can Learn New Words Rapidly.B. Dogs Identify Newly-named Toys by Exclusion.C. Dogs Can Acquire Vocabulary through Tons of Training.D. Gifted Dogs Have Similar Learning Abilities to Humans.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。