第2期2版21.2.3因式分解法1.解：(1)因式分解，得(5x+6)(5x-6)=0.于是得5x+6=0,或5x-6=0,-56(2)移项，得2x(x+2)-5(x+2)=0.因式分解，得(2x-5)(x+2)=0.于是得2x-5=0,或x+2=0,5二、填空题31x1=2=-2.s+=，s=-2(3)移项，得(x+4)2-2(x+4)=0.7.-20228.=2,x39.10%因式分解，得(x+4)(x+4-2)=0.10=2k=-511-号-P=s-4=32-4x于是得x+4=0,或x+4-2=0,x1=-4,x2=-2,12.7-V2或7或7+V2=7,即-=tVT-2=422.x-1-2=0;x1=-1,x2=3三、解答题*21.2.4一元二次方程的根与系数的_1=s=V7.13.解.(1)移项，得3x(x-1)+(x-1)=0s t st关系因式分解，得(x-1)(3x+1)=0.:的值为V7或-V71.C于是得x-1=0,或3x+1=0,2.53.25x1=1,x2=-34解：设方程的另一根为x2,且x=4.(2)移项，得(x+3)2_(1-2x)2=0.根据根与系数的关系，得x+x2=4,因式分解，得(x+3+1-2x)(x+3xx2=1-m,1+2x)=0,即4+x2=4,4x2=1-m.即(-x+4)(3x+2)=0.解得x2=0,m=1.于是得-x+4=0,或3x+2=0所以m的值为1，另一个根为0.25.解：(1)根据题意，得△=(2m)2-x1=4,x2=-3·4(m2+m)≥0.14.解：设每个人转发x个好友解得m≤0.根据题意，得1+x+x2=157.(2)根据一元二次方程根与系数·解得x=12x=-13(不合题意，舍去)】的关系，得x1+x2=-2m,xx2=m2+m.答：每个人转发12个好友15.解：答案不唯一，如x2+x2=(x1+2)2-2x12=12,.(-2m)2-2(m2+m)=12,即m2-m-②利用因式分解法：x2-3x=0.因式分解，得x(x-3)=06=0.于是得x=0,或x-3=0,解得m=-2,m2=3(舍去).故m的值为-2.x1=0,x2=3.③利用配方法：x2-4x=4,21.3实际问题与一元二次方程配方，得x2-4x+4=8,第1课时(x-2)2=8.1B2解：(1)设该校这两年藏书的年·由此可得x-2=±2V/2,均增长率为x.x1=2+2V2,x2=2-2V2.根据题意，得5(1+x)2=9.816.解：(1)(36-3x).解得x=0.4=40%,x2=-2.4(不合题(2)根据题意，得x(36-3x)=96.意，舍去)解得x1=4,x2=8.答：该校这两年藏书的年均增·当x=4时，36-3x=36-3×4=24>22,长率为409%.不符合题意，舍去；（2)9.8×(1+40%)=13.72(万册).当x=8时，36-3x=36-3×8=12<22,答：预测到2023年年底该校的藏符合题意.书量是13.72万册.答：若围成的菜地面积为96方第2课时米，此时的宽AB为8米1.112.解：设每顶头盔应降价x元，则17解1)子分每顶头盔的销售利润为(68-x-40)元，(2).·一元二次方程2x2-3x-1=0均每周的销售量为(100+20x)顶.;的两个根分别为m,n,根据题意，得(68-x40)(100+20x)=31m+n=2mn=-24000.整理，得x2-23x+60=0..nmmitn_(mtn)2mnnmnmn解方程，得x1=3,x2=20,11.68-x≤58，3P-2x22113.x≥10.121.x=20.-2答：每顶头盔应降价20元(3).·实数s,t满足2s2-3s-1=3版10,22-3t-1=0,且s≠t,一、选择题.s,t是一元二次方程2x2-3x-1=01~6.ACADBA的两个实数根

6,点A铜仁市2023年7月高二年级质量监测试卷AB方A.√3数学7.已知本试卷共4页，22题。全卷满分150分。考试用时120分钟。A.0注意事项：1,答题前，考生务必将自己的姓名、准考证号填写在本试卷相应的位置。8.已为2.答案全部填写在答题卡上，写在本试卷上无效。值秀3.考试结束后，将本试卷和答题卡一并交回。A.第I部分选择题（共60分）一选择题：本大题共8个小题，每小题5分，共40分.在每小题给出的四个选项中，只1c.有一项是符合题目要求的：才21,0、-11,已知集合A=(-2,-1,0,1,2),B={xx2<3),则A∩BC)二、进凡{-2，-1,0,1,2}是A{0,1,2,3〉9.下C-1,0,1P.(-2,2)(A)2.复数z=i(2一i)对应的点在复面内的24第一象限B.第二象限C.第三象限乙D.第四象限3.函数f(x)的图象与函数y=2x的图象关于x轴对称，则f(2)(D)A.-2B-号D/44.音程由两个音组成，是和声的最小单位.有的音听起来和谐而有的则不和谐，这和C.4音与音之间的波形（正弦型）有关.比如，d到高音d0)可以构成纯八度音程，听感上十分和谐，这是因为两者波形的周期比为2：1两个声波在1个(2个)周期后就立即重合，并有规律的进行下去.再此如T(o)到5(so)可以构成纯五度音程，两者周期比从3：2两个声波在？个(3个)周期后就立即重合，听感上也很和谐也就是说，两个音波形的周期比例越简单，听感越和谐.已知在一个调性中〔(d0)的波形符合函数f(t)=An80t(A为振幅，t为时间)，在音与音之间振帽相同的情况下，与1(do构成纯度音程的（高音do)、纯五度音程的5(0的波形函数分(A)Af()-Asin 360mt;f(t)-Asin 270xB.f(t)=Asin90πt;f(t)=Asin120πtC.f(t)=Asin 360xt;f()=Asin 120xD.f(t)=Asin90πt;f(t)-Asin270πt5已知双线写兰-士的渐近线方程为y-±》则n的位为AD.-3 m Asimw tB.-9W=0元A.9c高数学试卷共4页第页口。之

第八章面解析几何3.(2022福建福州第八中学高三翔末)已知椭圆C:。4.(2023河南郑州三模)已知点M为直线11：x=-1上的动点，N(1,0),过M作直线l,的垂线l2,l2交MW1(e>60)的离心率为亭且经过M,夏y的中垂线于点P,记点P的轨迹为C2,经(1)求曲线C的方程；过定点T(1,0)且斜率不为0的直线l交C于E,F两(2)设过点(5,0)的直线与曲线C交于A,B两点，在点，A,B分别为椭圆C的左、右两顶点x轴上求一定点Q(Q异于点N,且异于点(5,0)，使(1)求椭圆C的方程；点N到直线QA和QB的距离相等(2)设直线AE与BF的斜率分别为k1,k,求的值：(3)设直线AE与BF的交点为P,求证：点P在一条定直线上0·375·

答案与详解所以能得3分的概率是号号，A正确：对于B,乙P1

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度，得到y=2sin2x+2π+2的图像，所以g(x)=2sin2x+2π+2.…8分33令g(x)=0,得x=kπ5πkeZ,…10分1因为g(x)[0,m]上有且仅有5个零点，所以53π≤m<年65π1212。…12分19.（12分)131己知函数f(x)=三x+二(a-1)x2+ax。3若f)在x=处取得极值，求∫)的单调递减躯(2)若f(x)在区间(0,2)上存在极小值且不存在极大值，求实数a的取值范围.【解析】f'(x)=x2+(a-1)x+a.…1分(1)因为f(x)在x=-处取得极值，3所以八3=0,o-10+a=0,解利e=即3…3分所以r=号〔+》-2。令f(x)<0,故-30-号a0…11分综上，实数a的取值范围是（号，0…12分20.(12分)

19.(★★)（本小题满分12分）就小★★已知函数f(x)=xlnx,g(x)=-x2十a.x-3(a∈R).20.(k★)((1)求f(x)在点(e,f(e)处的切线方程；已知函数Q)若函(2)若对于任意的x∈[，e],都有2f()≥g(x成立，求实数a的取(2)若a值范围定器得

、26.(10分)问题提出】(I)如I图D,在R△MBc中，∠B=0,点D为nG上一点，且CD=2AC,过点D作DE上AC于点E,若AB=4,则DE的长为一：【问题探究】(2)如图②，在RI△ABC中，LABC=90°，点D是BC边.上一点，连接AD,过点D作DE⊥AD交ACD于点E,过点B作B1AC于点F,交AD于点G,试判断△ABG与△DCE是否相似，并说明理由：【问题解决】(3)如图③，RI△ABC是一块菜园而示意图∠ABC=90°，BC=2AB,AD是BC边上的中线，BF⊥AC于点F,交AD于点G,DE⊥AD交AC于点E,经测量，DE=50米，现欲沿DG修一条灌溉水渠，请你求出灌溉水渠的长度DG.图①图②图③(第26题图)九年级数学期中调研H-6-(共6页)

46l令7%☐100：24人辽宁实验高三上(.试题+答案.pdf郑3火1共)少(2)求数列{an}前n项的和S,18.在aMBC中，角A、B、C的对边分别为a、b、c,且cosC=a-csinBb(1)求角B的大小：(2)若a=3,c=√2，求sinC的值.19.某职称考试有A,B两门课程，每年每门课程均分别有一次考试机会，若某门课程上一年通过，则下一年不再参加该科考试，只要在连续两年内两门课程均通过就能获得该职称某考生准备今年两门课程全部参加考试，预测每门课程今年通过的概率均为；：若两门均没有通过，则明年每门课程通过的概率均为号：若只有一门没过，则明年这门课程道过的概率为}(1)求该考生两年内可获得该职称的概率；(2)设该考生两年内参加考试的次数为随机变量X,求X的分布列与数学期望20.直三棱柱ABC-AB,C,中，底面△ABC是等腰直角三角形，∠ACB=90°，AC=BC=2,D,E分别为CC,AB的中点且E在面ABD上的射影是△ABD的重心G(1)求证：DE/面ABC:(2)求二面角B-AD-E的面角的余弦值，21.已知函数f(x)=e(ax-2)+x+2.(1)若曲线f(x)在点(1，∫()处的切线与直线y=x行，求该切线方程；(2)当x≥0时，f(x)≥0恒成立，求a的取值范围22.已知函数f(x)=e-a,x∈(0，小，f'(x)为其导函数.函数f(x)在其定义域(0，刂内有零点x(1)求实数a的取值范围：(2)设函数g(x)=f"(x)(m-x)-f(m),求证：对任意的m∈(0，]且m≠x,g(m)g()<0.第4页/共5页3)求证：k≤1-.兰心④也宣目录打开方式转存下载云打印√IO

.t8Ge8uEG8v猛6.It's time for us to have a d iscussion(讨论)about our school sports meeting送弟评报社7.They were free yesterday and planned(计划)to have a trip to the mountains.EARNING ENGLISH8.It was a hot summer and many peoplecouldn't stand(忍受)the hot weather.大报第6期9.Eric watches the news_(新闻节目)every day八年级安徽课时讲练because he can learn a lot from it.Unit 5l0.The students had an educational(有教育意义的)lesson on the farm last weekend..Section A 1a-2d要点归纳Ⅱ.根据汉语意思完成句子，每空一词。☆mind v.介意；对（某事）烦恼1.你认为《学英语》这份报纸怎么样？其后可接名词或动词-ing形式，常用于疑问句What do you think of the newspaper或否定句中。Learning English?☆plan n.&v.打算；计划2.我们需要查清楚是谁写的这封信。·n.make a plan(for sth.)(为某事)做计划·v.plan to do sth.计划/打算做某事We need to find out who wrote this letter.3.我妹妹梦想着有一天能驾驶飞机。☆expectv.预料：期待My sister dreams of flying a plane one day.·expect sth.from sb.期望从某人那里得4.萨莉认为她能从访谈节目中学到很多东西。到某物Sally thinks that she can learn a lot from·expect(sb./sth.)to do sth.期待（某人/某talk shows.物)做某事5.没人知道是怎么回事。大家都很担心。No one knows what's going on.Everyone isworried.同m.根据2d对话内容，完成下面的短文。每空一词。根据汉语提示完成短文，每空词数不限。Grace and Sarah are having a 1.discussion步Jack and his family are 1.making a plan forabout TV shows.Grace can't 2.stand game(为..做计划their weekend.They2.shows or sports shows.She loves soap operas.She练plan to go camping(计划去野营)by a river..likes to follow the story and see what 3.happensThey will stay there for two days.They don't 3.mind living(介意住在)in the tent for a night.next.Sarah doesn't mind soap operas.HerThey4.expect to have a wonderful trip(期待favorite TV shows are the 4.news and talk有一个愉快的旅行).shows.They may not be very 5.exciting,but shecan learn a lot from them.She hopes to be a TVreporter one day.基【,根据首字母及汉语提示完成句子。Section A 3a-3cL.Ihope(希望)to get a new bike on my14th动词不定式础birthday.、语法练2.In fact,.Ididn't expect(预料)the man to be a训big movie star.请将例句中的句子补充完整，并完善笔记。3.It is very noisy outside and we wonder whate.g.1.The children decided to ride(ride)bikesto the beach.练is happening(发生).4.Would you m ind_(介意)waiting outside for2.It was a rainy day so they chose to stay(stay)a minute,please?at home.5.When I'm free,I like staying at home andNote:动词不定式可以跟在动词后面watching sitcoms_(情景喜剧).作宾语。

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C.若a+b=1,则m,n有最小值6+4√2D.若m=>42,则2a12.设函数f(x)=Asin(ox+p)(A,o>0,0≤p<2π)，如图是函数f(x)及其导函数f'(x)的部分图像，则()A.A=B.p=5元6C.f(x)与y轴交点坐标为3V30,2D.f(:与f'(x)的所有交点中横坐标绝对值的最小值为V6三、填空题（本题共4小题，每题5分，共20分）13.己知sin14.已知数列{an}满足a,=1,a2=2，an+2-an=(-1)+2,则数列{an}的前30项和为·15.已知x∈-22ππ则不等式emx-cosx-tanx≥0的解集为16.在△ABC中，角A,B,C所对的边分别为a,b,c,点O为△ABC外接圆的圆心，若a=√3，且c+2W3cosC=2b,AO=mAB+nAC,则m+n的最大值为四、解答题（本题共6小题，共70分)7,0分)设△4C的内角4，B,C的边分别为a,h,c,巴知c=3,且sim/C-cosC三清(1)求角C的大小：(2)若向量m=(1,sinA)与n=(2,sinB)共线，求△ABC的周长.

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This made a good home for bats,and soon the bridge was the home of thousands of bats.38 Now,they have come to val- ue their winged neighbors.The bats are a tourist attraction,and they eat lots of bugs every night. There are also structures built with the aim of bringing wild- life into the city.The Beijing Olympic Forest Park is a good ex- ample.The park used native plants and created open,natural spaces for wildlife.The result is a zone in Beijing with over 160 species of birds.In many ways,the park is the opposite of a zoo. 39 If we learn to share our space,we can become better neigh bors to the wildlife around us.40 Our own future will be en- dangered too.

【答案】 One possible version: Dear Peter, I have some good news to tell you that a bookstore nearby our school was opened.So I can't wait to invite you to visit it with me tomorrow.How about meeting at 10 am at the school gate?Is it convenient for you? The newly-opened bookstore is really unique.Not only can we purchase books but also we can get e-books for free.There is also a coffee shop inside,where we can read books joyfully. More information about the discount of the books will be sent to your email box later. Looking forward to your early reply. Yours sincerely, Li Hua

C.The bright future of Al in song writing. D.The wide application of algorithms in Al. 14.What can we infer from the last paragraph? A.Al is mainly used in writing lyrics. B.AI can take the place of instruments. C.AI will likely change how art is made. D.Al will be the major tool for art creation.

60.Yes,(I agree).Perseverance is important /crucial to suc- cess. Or:It depends.Perseverance may keep us going in a wrong direction when we remain stubborn.

M:Yes,please,Taylor.Where are you going?The Jungle Cafe? W:No.I'm going outside.There's a coffee truck in the parking lot, M:OK.Can I have a cappuccino,please? W:Sure.Do you want regular or non-fat milk? M:Regular with sugar,please.I'm not on a diet. W:OK.But Linda wants non-fat milk in her coffee. (Text 7) M:Hi,Rachel,what are you doing tomorrow? W:I'm going on a mountain hike.Why do you ask? M:I wanted you to take care of my dog while I'm at school. W:You go to school on Saturdays? M:Well,I have to take a subject test.If I pass with a good

If you are lucky granted an interview for volunteers,I can meet you at the airport and put you up with in my house.Also,I will provide you with some other help you need.Wished you good luck!

from that,such factors as lack of active exercise and bad living habit also contribute to obesity.

30.What can replace the underlined phrase in the last paragraph? A.With more time. B.With more assistance. C.More accurately. D.More passionately. 31.What can be the best title of the text? A.Open monitoring meditation helps remove errors B.Open monitoring meditation helps detect mistakes C.Open monitoring meditation helps change your perform- ance D.Open monitoring meditation helps improve your mental health

was about to go back to my house.I got into a taxi and told the driver my destination.To my surprise,the driver made an apology,says he didn't know the way.What come a taxi driver didn't know the way?I was a little of angry.At the moment,an old couple walked up and wanted to share the car.Their destina- tion was the same as me,so he let them in.The two greeted us but the grandpa told the driver the way.From their conversation I knew the driver has been a taxi driver just for three days.He

2022一2023学年度高三一轮复周测卷（十）5.在△ABC中，内角A,B满足A十B<,则下列结论一定正确的是班级理数·三角函数的图象与性质、A.sin A0,p<)的最小正周期为3.函数y=si(2x-)在区间[-x上的大致图象是元，其最小值为一2，且满足f(x)=-f(2-x)小，则p=6A士小核的验图其丸雷面个贵的（行，）米（国&±8m8禄线面单0商()农（©）c或写4.函数y=sim(-2x)的单调递增区间是A[-kx-君-kx+]∈z刀D-8或-38.已知曲线f(r)=sin(or十)与g(x)=cos(r十p:)有公共点(0,2)(u>0B-2x-君，-2kx+3]k∈Z)0<9<20<9<)若f(x)与g(x)在区间(-2,2)上的单调性相反，则。C[2x+,2kr+g]k∈z的最大值为D()A号C.1D理数·周测卷（十）第1页（共4页）理数·周测卷（十）第2页（共4页）

参考答案及解析151靶向提升(5分)所以an+1一an=1(n≥1)，所以数列{an}为等差数列.(x+In z),(4分)易知-e(x+1)<0,(2)解：由(1)可知an=a1十n-1=n.(5分)设Aa)=x+1n,则N(o)=1十>0，在数列{b,〉中，在a+1之前的所有项数和为十所以h(x)在区间(0，十∞)内单调递增，(7分)(1+2+3+…+)=k+3)k≤90，2又h(）=。-1<0,h1)=1>0,故存在唯-实解得k≤12，所以当=12时，在数列(bn}中，413之前的所有项数∈(，1)使得h()=+ln=0,数恰好为90，(7分)(8分)所以T0=(a1+a2+ag.+a12)+(1X21+2X所以当x∈(0，x)时，g(x)>0;当x∈（x,十o∞)2+3×23+.+12×22)=1+12)X12+(1×2时，g'(x)<0.21+2×22+3×23+…+12X212)=78+(1×21+所以g(x)在区间(0，xo)内单调递增，在区间(，2×22+3×23+…+12×212).(9分十∞)内单调递减，(9分)令Q2=1×2+2×22+3×23+…+12×212,故g(x)≤g(xo)=e(xo-x6-xol1nxo),则2Q2=1×22+2X23+3×24+.+12X213,又xo+lnxo=0,即lnx0=-xo,xo=eo,所以-Q2=2+22+23+24+…十212-12×213=(11分)-11×213-2,所以g(xo)=e。[x0-x6-x0（-x0)]=xoe=所以Q2=11×213+2=90114,e·eo=1,所以T90=78+Q2=11×213+80=90192.所以g(x)≤1，即xef(x)≤1成立.(12分)(12分)20.(1)解：函数f(x)的定义域为(0，十∞)，(1分21,解：1)由题意得F(多，0），过点A作AQ⊥x轴，f(x)=-a-1=-ax+1(2分)垂足为点Qxx当a≥0时，f(x)<0,f(x)单调递减，无极值点；所以cos∠AFQ=cos(π-∠OFA)=-cos∠OFA=当a<0时，令了)=0，得=日，当xe(0,号，在R△AFQ中，eos∠APQ=FO4=2AF日)时，f(x)<0:当xe(-,+)时号f(x)>0.会解得2所以f(x)在区间(0，一是)内单调递减，在区间所以C的方程为y2=4x.(4分)(2)由题知M(-1,-4),设P(-1,m),且-4≤(一是，十)内单调递增。m≤4.(5分)所以当工=一】时，f(x)有极小值，极小值点为x依题意得切线PH,PG的斜率存在，设H(,M),G(2,2),PH的方程为y=k(x-)十1,无极大值点(4分)(k≠0)，(2)证明：当a=1时，xef(x)=xe(1-x-lnx)联立y=4x,e'(x-x2-xIn x),y=k(x-1)+,得ky2-4y+4(y一k)=0,设g(x)=e(x-x2-xlnx),△=16-16k(y1-kx1)=0g(x)=e(-x2-x-xIn x-In x)=-e'(x+1).整理得-4k十4=0，解得=2(7分)·18.

4πX22=16元，故D错误，故选C形PF,QFz面积等于8.12.A因f(x)=ae+(ar+2)c-1=(a.x+a+15.1设了()=g)=>0,期e所以sin2分2)-1,因为f(-1)-是-1<0所以函/2x-T=1,所以x1=lnt,x,-1所以sin轻于数f(x)不可能在R上单调递增.所以若函以x-2+2h>0,今0-所以立数(x)为R上单调函数，则必是单调递减函数，即了(x)≤0恒成立.由f(0)=a十1P+名n>0.所以0=-因为c≤0、可得a≤一1.令g(x)=f(x)=(ax十a+2)e-1,则g'(x)=(ax+2a+2)e.由+1)-D,当0<1<1时，k(0<0,所t因为0g(x)>0,可得x<-(8+2),由g'(x)<0()在(0,1)上单调递减，当>1时，0>06分a所以h(D在(0,1)上单调递增，所以=1时，(2)词可得>-(2+子，8在（-0，-2(取得最小值，最小值为0)=之+合-0AC-因为名上单调造增在(-2一号十止单洞二1.n(m+3)(2分)在A递减故g)=g(-2-二)=-ae91626n+g(3分)由题意可知，a=2,当n≥2时，a,-a,1-1,要g(x)≤0，即e≤-1，令t=得：n十1，则a=a1+(a2-a)+(a-a)+…则G1,即2+2a,令y解(a,-am-:)=2+3+4十…+(n+1)=nn+3211n2t+2,y无=2<0，所以y-n当n=1时，a:=2也成立，所以a,=nn3》22t卡2递减，所以当t=1时，ym=0,故所以bm=（一1)”·n(2n+5)(n于2)a,=(-1)”·中g(x)m≤0恒成立，选A.,8113.5因为f=a2±2a型，所以f在x)=n(2n十5)2(n+2)m(n+3=（-1):,2(2n+5)(1+2)(n+3】=(-1)”·(2千2广n千3，所以24=(-号2“2因为)为2-1奇函数，故f(-x)=一f八x),即a·25+(2a-是+导+号+（号-号+中品1)=a+(2a-1)·2,整理得到(a一1)(2r-1)2-2广22n=0,所以a=1.3-25X2n+3=2m十336n+914.8因为P,Q为C上关于坐标原点对称的两17.解：(1)设∠DAC=a,点，且PQ|=FF,,所以四边形PEQF,为则∠ADC=R一a一音6=5r-a,矩形，不妨设点P在第一象限，PF:=m,因为AC=√3DC,PF2=1,则m-n=43,m2十2=4c2=64,由正弦定理知ACDC所以m一2mn十n=48,解得mn=8,即四边in∠ADCsin aS3名师原创模拟·数学理科答案第3页（共32页）

由直线y=3-2xln2是曲线y=f(x)的一条切线，切点在y轴上，得f'(0)=-2ln2,f(0)=3,即厂2alh2=-21n2,得a=1,b=1.年(4分)2a+b=3,(2证明：由得)=1+（合），f0=2fe-是因为f(1)=2<5,f(1)·f(2)=3<5,所以当n=1或n=2时，f(1)·f(2)·…·f(n)<5成立.(5分)当n≥3时，设M=f(3)·f(4)·…·f(n),则lnM=lnf(3)+lnf(4)+…+lnf(n)=[1+(合)]+[1+(2)]++[+（合分）]游当渠。道中(6分)设g(x)=ln(x+l)-x,延前的)衣，（河的关大示年立则g)=1=千因为当x∈(0，十∞)时，g'(x)<0,所以g(x)在区间[0，十∞)上单调递减.又因为g(0)=0,所以当x>0时，g(x)<0,m即ln(x+l)0x①当a≤0时，f'(x)>0,f(x)在区间(0，十∞)上单调递增；②当a>0时，令fr)=0,则x=日当0<<时，fx)>0:当>时，f'x)0,所以fx)在区同(0，日)上单调递增，在区间（合+上单调递成。(2)由(1)可知，当a≤0时，f(x)单调递增，所以f(x)至多有一个零点，舍去；与，当a>0,x>0时，由x趋近于0，f(x)趋近于一∞，x趋近于十∞，f(x)趋近于-∞，可知要使fx)有两个零点，只禽f(日)=-lna-1>0,从而0

二、填空题：本题共4小题，每小题5分，共20分。带题目为能力提升题，分值不计入总分。13.已知a,b∈R,(a+bi)2=3+4i(i是虚数单位)，则a2+b2=,ab=【答案】52【解析】由题意可得a2-b2+2abi=3+4i,则a2-b2=3解得4，则a2十b2=5,ab=2.ab=2,1b2=1,14.某篮球队6名主力队员在最近三场比赛中投进的三分球个数如下表所示：队员i123456三分球个数a2a304a6如图是统计该6名队员在最近三场比赛中投进的三分球总数的程序框图，则图中判断框应填,输出的s=开始/输入41,2，…，6=0,=0=计1s=s+ai否输出s结束【答案】i≤6a1十a2十…十a6【解析】因为是统计该6名队员在最近三场比赛中投进的三分球总数的程序框图，所以图中判断框应填i≤6，输出的s=a1十a2十…十a6,15.设函数f)-千2c>0),观察」xxfi(z)-f(x)-2f:(z)=f(F (x))-3f(x)=f(f:(z))=78'f.(x)=f(f(z))=x15x+16…根据以上事实，归纳推理可得：当n∈N且n≥2时，fm(x)=f(fm-1(x)=【答案】(2-1)x+2【解析】观察知：四个等式等号右边的分母为x十2,3x十4,7x+8,15x十16，即(2-1)x+2,(4-1)x+4,(8-1)x+8,(16-1)x+16,所以归纳出fn(x)的分母为(2”-1)x+2",故当n∈N且n≥2时，fn(x)=f(f.-1(x)=(2-1)x+2116.分形几何学不仅让人们感悟到数学与艺术审美的统一，而且还有其第1行深刻的科学方法论意义按照如图①所示的分形规律可得如图②所第2行示的一个树形图，记图②中第n行白圈的个数为am,黑圈的个数为…第3行bn,则a4=;数列{an十bn}的通项公式为②【答案】14am十bn=3"-1①2,a2=2=32-1+1【解析】对于白圈，由图可得a1=1=3十1，2,a3=5=33-1+12，所以a4=31+1=14,2则a,=3”-1+12对于黑周，61=0,6：=1,6，=46，=13，，所以6.=0-1=32，所以a,十6，=3m-1-13-1+1+3-1-1=3-1.22·77·