下载文档原格式
2019年四川省成都七中自主招生英语试卷及答案2019年四川省成都七中自主招生英语试卷第一部分阅读理解（共两节，满分40分）第一节（共5小题；每小题8分，满分40分）阅读下列短文，从每题所给的四个选项A．B．C和D中，选出最佳选项1．（8分）Brooke has just turned 12，with seven world records，is one of the best rock climbers in the world．Brooke comes from a climbing family．Both her parents are just past climbing champions．Her father stopped climbing some years ago，but her mother，Robyn，who won four world cup titles four years in a row，is still climbing．She runs a club for young climbers in Colorado，USA and coaches Brooke and her teammates．Sometimes it's difficult for mother and daughter to work together so closely，but they really respect and trust each other and Brooke says her mum is a great coach．Brooke says that her mother gives her a lot of good advice and it is a very important part of her climbing life．Robyn can be strict，but she is also passionate about climbing，and she passes this passion on to her students．Robyn says Brooke has very good strong wills and is very good at making herself want to practice．This helps her when she's facing the challenges of this difficult sport．She is also very hard﹣working．Success in rock climbing issomething you have to work at．To be a world﹣class athletes of any kind，you have to push yourself and train hard and that's what Brooke does，every day，at the club and at home．Brooke says that climbing is always there in their lives．They even have a climbing wall in their house！But climbing isn't only hard work．It's fun too．Brooke lovesclimbing and when she's on a high rock，she feels happy．Strangely，she says that when she looks down，she isn't scared．All she does is to think how cool it is to be small compared to the rock．（1）It's for Brooke and Robyn to work together so closely．A．difficultB．uneasyC．amazingD．annoying（2）T o be a successful climber，one has to．A．practice climbing at a very young ageB．have a great mother coach who is very strictC．face any difficult things bravely and keep on training hard D．keep pushing himself and train hard at any time and any place（3）What's the best title of this passage？A．An Excellent ClimberB．A Climbing FamilyC．Climbing ChampionsD．How to Be Successful Rock Climbers（4）Which of the sentences can the author agree with？A．Well begun is half done．B．No pains，no gains．C．He who climbs high falls heavily．D．Doing is better than saying．2．（8分）Papa，as a son of a dirt﹣poor farmer，left school early and went to work in a factory，for education was for the rich then．So，the world became his school．With great interest，he read everything he could lay his hands on，listened to the town elders and learned about the world beyond his tinyhometown．"There's so much to learn．"he'd say．"Though we' re born stupid，only the stupid remain that way．"He was determined that none of his children would be denied an education．Thus，Papa insisted that we learn at least one new thing each day．Though，as children，we thought this was crazy，it would never have occurred to us to deny Papa a request．And dinner time seemed perfect for sharing what we had learned．We would talk about the news of the day；no matter how insignificant，it was never taken lightly．Papa would listen carefully and was ready with some comment，always to the point．Then came the moment﹣the time to share the day's new learning．Papa，at the head of the table，would push back his chair and pour a glass of red wine，ready to listen．"Felice，"he'd say，"tell me what you learned today"．"I learned that the population of Nepal is .."Silence．Papa was thinking about what was said，as if the salvation of the world would depend upon it．"The population of Nepal．Hmm．Well…"he'd say．"Get the map，and let's see where Nepal is．"And the whole family went on a search for Nepal．This same experience was repeated until each family member had a turn．Dinner ended only after we hada clear understanding of at least half a dozen such facts．As children，we thought very little about these educational wonders．Our family，however，was growing together，sharing experiences and participating in one another's education．And by looking at us，listening to us，respecting our input，affirming our value，giving us a sense of dignity，Papa was unquestionably our mostinfluential teacher．Later during my training as a future teacher，I studied with some of the most famous educators．They were imparting what Papa had known all along﹣the value of continual learning．His technique has served me well all my life．Not a single day has been wasted，though I can never tell when knowing the population of Nepal might prove useful．（1）What do we know from the first paragraph？A．The author's father was born in a worker's family．B．Those born stupid could not change their life．C．The town elders wanted to learn about the world．D．The poor could hardly afford school education．（2）It can be learned from the passage that the author．A．enjoyed talking about newsB．knew very well about NepalC．felt regretted about those wasted daysD．appreciated his father's educational technique（3）What is the greatest value of"dinner time"to the author？A．Continual learning．B．Showing talents．C．Family get﹣together．D．Winning Papa's approval．（4）The author's father can be best described as．A．an educator expert at training future teachersB．a parent insistent on his children's educationC．a participant willing to share his knowledgeD．a teacher strict about everything his students did3．（8分）Some of the world's most famous musicians recently gathered in Paris and New Orleans to celebrate the first annual International Jazz Day．UNESCO（United NationsEducational，Scientific and Cultural Organization）recently set April 30 as a day to raise awareness of jazz music，its significance，and itspotential as a unifying voice across cultures．Despite the celebrations，though，in the U．S．the jazz audience continues to shrink and grow older，and the music has failed to connect with younger generations．It's Jason Moran's job to help change that．As the Kennedy Center's artistic adviser for jazz，Moran hopes to widen the audience for jazz，make the music more accessible，and preserve its history and culture．"Jazz seems like it's not really a part of the American appetite．"Moran tells National Public Radio's reporter Neal Conan．"What I'm hoping to accomplish is that my generation and younger start to reconsider and understand that jazz is not black and white anymore．It's actually color，and it's actually digital．Moran says one of the problems with jazz today is that the entertainment aspect of the music has been lost．"The music can't be presented today the way it was in 1908 or 1958．It has to continue to move，because the way the world works is not the same，"says Moran．Last year，Moran worked on a project that arranged Fats Waller's music for a dance party，"just to kind of put it back in the mind that Waller is dance music as much as it is concert music，"says Moran．"For me，it's the re﹣contextualization．In music，where does the emotion lie？Are we，as humans，gaining any insight on how to talk about ourselves and how something as abstract as a Charlie Parker record gets us into a dialogue about our emotions and our thoughts？Sometimes we lose sight that the music has a wider context，"says Moran，"so I want to continue those dialogues．Those are the things I want tofoster．"（1）Why did UNESCO set April 30 as International Jazz Day？A．T o protect cultural diversity．B．To recognize the value of jazz．C．T o remember the birth of jazz．D．To encourage people to study music．（2）What does the underlined word"that"in paragraph 3 refer to？A．Jazz becoming more accessible．B．The jazz audience becoming larger．C．The production of jazz growing faster．D．Jazz being less popular with the young．（3）What can we infer about Moran's opinion on jazz？A．It will disappear gradually．。
2018-2019年最新四川成都市第七中学自主招生语文模拟精品试卷（第一套）（满分：100分考试时间：90分钟）一、语文基础知识（18分，每小题3分）1．下列词语中加点的字，读音全都正确的一组是（）A．连累(lěi) 角(juã)色河间相(xiàng) 冠冕(miǎn)堂皇B专横(hâng) 忖(cǔn)度涮(shuàn) 羊肉妄加揣(chuāi)测C．笑靥(yâ) 顷(qīng)刻汗涔(cãn)涔休戚(qì)相关D慨叹(kǎi) 俨(yǎn)然刽子手(kuàì) 刎(wěn)颈之交2、下列各项中字形全对的是（）A、橘子州偌大急躁光阴荏苒B、蒙敝犄角慰籍书生意气C、敷衍磕绊笔竿艰难跋涉D、翱翔斑斓屏蔽自怨自艾3、依次填入下列各句横线上的词语，最恰当．．的一项是（）⑴虽然他尽了最大的努力，还是没能住对方凌厉的攻势，痛失奖杯。
⑵那些见利忘义，损人利己的人，不仅为正人君子所，还很可能滑向犯罪的深渊。
⑶我认为，真正的阅读有灵魂的参与，它是一种个人化的精神行为。
A.遏制不耻必需B.遏止不耻必需C.遏制不齿必须D.遏止不齿必须4、下列句中加点的成语，使用恰当的一句是（）A、故宫博物院的珍宝馆里，陈列着各种奇珍异宝、古玩文物，令人应接不暇。
B、任何研究工作都必须从积累资料做起，如果不掌握第一手资料，研究工作只能是空中楼阁．．．．。
C、电影中几处看来是闲笔，实际上却是独树一帜之处。
D、这部精彩的电视剧播出时，几乎万人空巷，人们在家里守着荧屏，街上显得静悄悄的。
5、下列句子中，没有语病的一项是（）A 大学毕业选择工作那年，我瞒着父母和姑姑毅然去了西藏支援边疆教育。
B北京奥运会火炬接力的主题是‚和谐之旅‛，它向世界表达了中国人民对内致力于构建和谐社会，对外努力建设和平繁荣的美好世界。
C他不仅是社会的一员，同时还是宇宙的一员。
他是社会组织的公民，同时还是孟子所说的‚天民‛。
2019年四川省成都七中自主招生数学试卷副标题一、选择题（本大题共12小题，共60.0分）1. 若M =5x 2−12xy +10y 2−6x −4y +13(x 、y 为实数)，则M 的值一定是( )A. 非负数B. 负数C. 正数D. 零 2. 将一个棱长为m(m >2且m 为正整数)的正方体木块的表面染上红色，然后切成m 3个棱长为1的小正方体，发现只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的12倍，则m 等于( ) A. 16 B. 18 C. 26 D. 32 3. 已知6a 2−100a +7=0以及7b 2−100b +6=0，且ab ≠1，则ab 的值为( )A. 503B. 67C.1007D. 764. 若a =√3√2+√3+√5，b=2+√6−√10，则ab 的值为( )A. 12B. 14√2+√3√6+√105. 满足|ab|+|a −b|−1=0的整数对(a,b)共有( )A. 4个B. 5个C. 6个D. 7个6. 在凸四边形ABCD 中，E 为BC 边的中点，BD 与AE 相交于点O ，且BO =DO ，AO =2EO ，则S △ACD ：S △ABD 的值为( ) A. 2：5 B. 1：3 C. 2：3 D. 1：27. 从1到2019连续自然数的平方和12+22+32+⋯+20192的个位数字是( )A. 0B. 1C. 5D. 9 8. 已知x +y +z =0，且1x+1+1y+2+1z+3=0，则代数式(x +1)2+(y +2)2+(z +3)2的值为( ) A. 3 B. 14 C. 16 D. 369. 将一枚六个面编号分别为1、2、3、4、5、6的质地均匀的正方体骰子先后投掷两次，记第一次掷出的点数为a ，第二次掷出的点数为b ，则使关于x 、y 的方程组{ax +by =22x +y =3，只有正数解的概率为( ) A. 112B. 16C. 518D. 133610. 方程3a 2−8a −3b −1=0，当a 取遍0到5的所有实数值时，则满足方程的整数b 的个数是( ) A. 12个 B. 13个 C. 14个 D. 15个11. 若一个三角形的三边和为40，且各边长均为整数，则符合条件的三角形的个数为( ) A. 31个 B. 32个 C. 33个 D. 34个12. 若关于x 的方程x 2+ax +b −3=0有实根，则a 2+(b −4)2的最小值为( )A. 0B. 1C. 4D. 9二、填空题（本大题共7小题，共52.0分）13.已知x=3+√132，则代数式x4−3x3−3x+1的值为______．14.在正十边形的10个顶点中，任取4个顶点，那么以这4个顶点为顶点的梯形有______个．15.在Rt△ABC中，∠C=90°，AC=1，BC=2，D为AB中点，E为边BC上一点，将△ADE沿DE翻折得到△A′DE，使△A′DE与△BDE重叠部分的面积占△ABE面积的14，则BE的长为______．16.已知关于x的方程√x2−2x+1−√x2−4x+4+2√x2−6x+9=m恰好有两个实数解，则m的取值范围为______．17.如图，PA切⊙O于点A，PE交⊙O于点F、E，过点A作AB⊥PO于点D，交⊙O于点B，连接DF，若sin∠BAO=23，PE=5DF，则PFPE=______．18.如图，四边形ABCD中，AB=AD=5，BC=DC=12，∠B=∠D=90°.M和N分别是线段AD和线段BC上的点，且满足BN=DM，则线段MN的最小值为______．19.若−12<x<1，x1+x−2x2=a0+a1x+a2x2+a3x3…+a n x n，则a2+a3=______．三、解答题（本大题共2小题，共38.0分）20.已知二次函数y=x2+(a−7)x+6，反比例函数y=ax(1)当a=2时，求这两个函数图象的交点坐标；(2)若这两个函数的图象的交点不止一个，且交点横、纵坐标都是整数，求符合条件的正整数a的值；(3)若这两个函数的交点都在直线x=12的右侧，求a的取值范围．21.已知：四边形ABCD中，点E、F分别为边AD、AB上的点，连接BE、DF相交于点G，且满足∠ADF=∠ABE(1)如图1，若DE=BG=n，cos∠AEB=23，GE=3，求AE的长(用含n的代数式表示)；(2)如图2，若ABCD为矩形，G恰为BE中点，连接CG，AE=1，作点A关于BE，求DE的长．的对称点A′，A′到CG的距离为3√24答案和解析1.【答案】A【解析】解：M =5x 2−12xy +10y 2−6x −4y +13=4x 2−12xy +9y 2+y 2−4y +4+x 2−6x +9=(2x −3y)2+(y −2)2+(x −3)2≥0，故M 一定是非负数． 故选：A ．通过配方法配出平方根，从而判断M 值的大小．本题考查了配方法的应用，熟练配方法的应用是解答此题的关键． 2.【答案】C【解析】解：将一个棱长为m(m >2且m 为正整数)的正方体木块的表面染上红色，然后切成m 3个棱长为1的小正方体，则只有一个表面染有红色的小正方体的数量为6(m −2)2， 恰有两个表面染有红色的小正方体的数量12(m −2)，∵只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的12倍，∴6(m −2)2=12×12(m −2)， 解得m 1=26，m 2=2(舍去)， 故选：C ．只有一个表面染有红色的小正方体的数量为6(m −2)2，恰有两个表面染有红色的小正方体的数量12(m −2)，根据只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的12倍，即可得到m 的值． 本题主要考查了正方体，解决问题的关键是抓住表面涂色的正方体切割小正方体的特点：1面涂色的在面上，2面涂色的在棱长上，3面涂色的在顶点处，没有涂色的在内部，由此即可解决此类问题． 3.【答案】D【解析】解：∵7b 2−100b +6=0， ∴6×1b 2−100×1b+7=0，∵6a 2−100a +7=0，∴a 、1b 是方程6x 2−100x +7=0的两根， ∴由根与系数的关系可知：ab =76，故选：D ．根据根与系数的关系即可求出答案． 本题考查根与系数的关系，解题的关键是熟练运用根与系数的关系，本题属于基础题型． 4.【答案】B【解析】解：a =√3√2+√3+√5√2+√3−√5√2+√3−√5=√3(√2+√3−√5)2√6=√2(√2+√3−√5)4=b4．∴ab =14． 故选：B ． 将a 乘以√2+√3−√5√2+√3−√5可化简为关于b 的式子，从而得到a 和b 的关系，继而能得出ab 的值．本题考查二次根式的乘除法，有一定难度，关键是在分母有理化时要观察b的形式．5.【答案】C【解析】解：∵|ab|+|a−b|=1，∴0≤|ab|≤1，0≤|a−b|≤1，∵a，b是整数，∴|ab|=0，|a−b|=1或|a−b|=0，|ab|=1①当|ab|=0，|a−b|=1时，Ⅰ、当a=0时，b=±1，∴整数对(a,b)为(0,1)或(0,−1)，Ⅱ、当b=0时，a=±1，∴整数对(a,b)为(1,0)或(−1,0)，②当|a−b|=0，|ab|=1时，∴a=b，∴a2=b2=1，∴a=1，b=1或a=−1，b=−1，∴整数对(a,b)为(1,1)或(−1,−1)，即：满足|ab|+|a−b|=1的所有整数对(a,b)为(0,1)或(0,−1)或(1,0)或(−1,0)或(1,1)或(−1,−1)．∴满足|ab|+|a−b|−1=0的整数对(a,b)共有6个．故选：C．先判断出|ab|=0，|a−b|=1或|a−b|=0，|ab|=1，再借助a，b是整数即可得出结论．此题考查了绝对值，以及数对，分类讨论的思想，确定出|ab|=0，|a−b|=1或|a−b|= 0，|ab|=1是解题的关键．6.【答案】D【解析】解：如图，过点B作BF//AD交AE延长线于F，连接OC，∵BF//AD∴∠F=∠DAO∵BO=DO，∠BOF=∠DOA∴△FOB≌△AOD(AAS)∴FO=AO∵AO=2EO∴FO=2EO∴EO=EF，∵E为BC边的中点∴BE=CE∵∠BEF=∠CEO∴△BEF≌△CEO(SAS)∴∠BFE=∠COE∴BF//OCAD//OC∴S△ACD=S△AOD，∵BD=2OD∴S△ABD=2S△AOD，∴S△ABD=2S△ACD∴S△ACD：S△ABD=1：2；故选：D ．过点B 作BF//AD 交AE 延长线于F ，连接OC ，先证明△FOB≌△AOD ，再证明△BEF≌△CEO ，可得AD//OC ，可得S △ACD =S △AOD ，由S △ABD =2S △AOD ，可得S △ACD ：S △ABD =1：2；本题考查了全等三角形判定和性质，三角形面积，平行线间的距离等知识点，有一定的难度，解题关键是作平行线构造全等三角形． 7.【答案】A【解析】解：以2为指数的幂的末位数字是1，4，9，6，5，6，9，4，1，0依次循环的，∵2019÷10=201…9，(1+4+9+6+5+6+9+4+1+0)×201+(1+4+9+6+5+6+9+4+1) =45×201+45 =9045+45 =9090，∴12+22+32+42+⋯+20192的个位数字是0． 故选：A ．由题中可以看出，故个位的数字是以10为周期变化的，用2019÷10，计算一下看看有多少个周期即可．此题主要考查了找规律，要求学生通过观察，分析、归纳发现其中的规律，并应用发现的规律解决问题．解决本题的关键是找到以2为指数的末位数字的循环规律． 8.【答案】D【解析】解：∵x +y +z =0，且1x+1+1y+2+1z+3=0，[(x +1)2+(y +2)2+(z +3)2][12+12+12]≥[(1×(x +1)+1×(y +2)+1×(z +3)]2=(x +y +z +6)2(x +1)2+(y +2)2+(z +3)2≥36∴(x +1)2+(y +2)2+(z +3)2的值为36． 故选：D ．根据已知条件可得x 、y 、z 的值即可求解．本题考查了分式的加减法，解决本题的关键是合理分析已知条件． 9.【答案】B【解析】解：①当a −2b =0时，方程组无解；②当a −2b ≠0时，方程组的解为由a 、b 的实际意义为1，2，3，4，5，6可得． 易知a ，b 都为大于0的整数，则两式联合求解可得x =3b−22b−a ，y =4−3a2b−a ， ∵使x 、y 都大于0则有x =3b−22b−a >0，y =4−3a2b−a >0， ∴解得a <43，b >23或者a >43，b <23，∵a ，b 都为1到6的整数，∴可知当a 为1时b 只能是1，2，3，4，5，6；或者a 为2，3，4，5，6时b 无解， 这两种情况的总出现可能有6种； (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)，又掷两次骰子出现的基本事件共6×6=36种情况，故所求概率为=636=16， 故选：B ．首先分两种情况：①当a −2b =0时，方程组无解；②当a −2b ≠0时，方程组的解为由a 、b 的实际意义为1，2，3，4，5，6可得．把方程组两式联合求解可得x =3b−22b−a ，y =4−3a2b−a ，再由x 、y 都大于0可得x =3b−22b−a >0，y =4−3a 2b−a>0，求出a 、b 的范围，列举出a ，b 所有的可能结果，然后求出有正数解时，所有的可能，进而求出概率．此题主要考查了列表法求概率，以及二元一次方程的解法，题目综合性较强． 10.【答案】B【解析】解：∵3a 2−8a −3b −1=0， ∴b =a 2−83a −13=(a −43)2−259，∵0≤a ≤5， ∴−43≤a −43≤113， ∴0≤(a −43)2≤1219， ∴−259≤(a −43)2−259≤969，即−259≤b ≤969，∴整数b =−2，−1，0，1，…，10，共13个，故选：B ．首先将方程3a 2−8a −3b −1=0进行变形，变成用含a 的代数式表示b ，然后把含a 的代数式配方，再根据a 的取值求出b 的取值范围，由于是求b 的整数的个数，所以再找b 的取值范围内的整数解即可．此题主要考查了利用配方法求一元二次方程的整数根，做此题的关键是用含a 的代数式表示b ，然后根据a 的取值求b 的取值，综合性较强，难度不大． 11.【答案】C【解析】解：根据题意得三角形的三边都小于20， 设最小的两边为x ≤y ≤19，x +y >20 当x =2时，y =19， 当x =3时，y =18， 当x =4时，y =17，18， 当x =5时，y =16，17， 当x =6时，y =15，16，17， 当x =7时，y =14，15，16， 当x =8时，y =13，14，15，16， 当x =9时，y =12，13，14，15，当x =10时，y =11，12，13，14，15， 当x =11时，y =11，12，13，14， 当x =12时，y =12，13，14， 当x =13时，y =13，符合条件的三角形的个数为1+1+2+2+3+3+4+4+5+4+3+1=33， 故选：C ．首首先根据三角形的两边之和大于第三边以及三边和为40长，得到三角形的三边都必须小于20；再结合三角形的两边之差小于第三边进行分析出所有符合条件的整数．本题考查了三角形三边关系，关键是列出约束条件．12.【答案】B【解析】解：由x2+ax+b−3=0知b关于a的函数解析式为b+ax+x2−3=0，∵a2+(b−4)2的最小值可看做点(a,b)到(0,4)距离的最小值，则两点的距离d=2√12+x2=2√x2+1=√x2+1≥1，∴点(a,b)到(0,4)距离的最小值为1，即a2+(b−4)2的最小值为1，故选：B．由x2+ax+b−3=0知b关于a的函数解析式为b+ax+x2−3=0，而a2+(b−4)2的最小值可看做点(a,b)到(0,4)距离的最小值，再根据点到直线的距离公式求解可得．本题主要考查两点间的距离公式，熟练掌握公式的定义是解题关键．13.【答案】2【解析】解：当x=3+√132时，原式=x4−3x3−3x+1=(x2)2−3x(x2+1)+1=[(3+√132)2]2−3×3+√132[(3+√132)2+1]+1=(11+3√132)2−3×3+√132×13+3√132+1=119+33√132−117+33√132+1=1+1=2．故答案为：2．将原式适当变形，再代入进行计算便可．本题主要考查了求整式的值，二次根式的计算，适当进行整式的变形，可以减小计算的难度．14.【答案】60【解析】解：设正十边形为A1A2 (10)以A1A2为底边的梯形有A1A2A3A10、A1A2A4A9、A1A2A5A8共3个．同理分别以A2A3、A3A4、A4A5、…、A9A10、A10A1为底边的梯形各有3个，这样，合计有30个梯形．以A1A3为底边的梯形有A1A3A4A10、A1A3A5A9共2个．同理分别以A2A4、A3A5、A4A6、…、A9A1、A10A2为底边的梯形各有2个，这样，合计有20个梯形．以A1A4为底边的梯形只有A1A4A5A101个．同理分别以A2A5、A3A6、A4A7、…、A9A2、A10A3为底边的梯形各有1个，这样，合计有10个梯形，则以4个顶点为顶点的梯形有：30+20+10=60(个)，故答案为：60．分以A1A2为底边、A1A3为底边、A1A4为底边，根据梯形的概念、正多边形的性质解答．本题考查的是梯形的概念、正多边形的性质，灵活运用分情况讨论思想是解题的关键．15.【答案】√52【解析】解：如图，连接AA′，延长ED交AA′于点M∵∠C=90°，AC=1，BC=2，∴AB=√AC2+BC2=√5∵D为AB中点，∴AD=DB=√5 2∵将△ADE沿DE翻折得到△A′DE，∴AD=A′D，AE=A′E∴ED垂直平分AA′∴EM⊥AA′，∵AD=DB=AA′=√5 2∴△ABA′是直角三角形∴∠AA′B=90°，即AA′⊥A′B∴ME//A′B∴∠MEF=∠FA′B，∵△A′DE与△BDE重叠部分的面积占△ABE面积的14，∴S△DEF=14S△AEB，∴DF=14AB=12DB∴DF=FB，且∠MEF=∠FA′B，∠A′FB=∠EFD ∴△A′FB≌△EFD(AAS)∴EF=A′F，且DF=FB，∠EFB=∠A′FD∴△BFE≌△DFA′(SAS)∴AD=BE=√5 2故答案为：√52连接AA′，延长ED交AA′于点M，由勾股定理可求AB=√5，可得AD=DB=√52，由折叠的性质可得AD=A′D=DB，AE=A′E，可得AA′⊥A′B，EM⊥AA′，由题意可得DF= BF，由“AAS”可证△A′FB≌△EFD，可得EF=A′F，由“SAS”可得△BFE≌△DFA′，即可求BE的长．本题考查了翻折变换，勾股定理，直角三角形的判定和性质，全等三角形的判定和性质，证明△A′FB≌△EFD是本题的关键．16.【答案】1≤m<3或m>3【解析】解：原方程变形为：|x−1|−|x−2|+2|x−3|=m，①当x≥3时，x−1−(x−2)+2(x−3)=m，x=m+52≥3，∴m=2x−5，此时m≥1；②当2≤x<3时，x−1−(x−2)+2(3−x)=m，x=7−m 2∴m=7−2x，此时1<m≤3；③当1≤x<2时，x−1−(2−x)+2(3−x)=m，∴m=3(不符合题意)；④当x<1时，1−x−(2−x)+2(3−x)=m，∴m=5−2x，此时m>3．恰好有两个实数解，所以1≤m<3或m>3，故答案为1≤m<3或m>3．解无理方程关键是要去掉根号，将其转化为整式方程．解无理方程的基本思想是把无理方程转化为有理方程来解，在变形时要注意根据方程的结构特征选择解题方法．本题主要考查无理方程，解题的关键是掌握二次根式的性质、绝对值的性质等知识点．17.【答案】310【解析】解：连接OE，如图，∵AB⊥PO，∴∠ADO=90°，在Rt△ADO中，sin∠DAO=ODOA =23，设OD=2x，OA=3x，∵PA切⊙O于点A，∴OA⊥PA，∴∠APO=∠OAD，在Rt△APO中，sin∠APO=OAOP =23，∴OP=32×3x=92x，∵∠APD=∠OPA，∴Rt△PAD∽Rt△POA，∴PD：PA=PA：PO，即PA2=PD⋅PO，∵PA切⊙O于点A，PE交⊙O于点F、∴PA2=PF⋅PE，∴PD⋅PO=PF⋅PE，即PF：PO=PD：PE，而∠DPF=∠EPO，∴△PDF∽△PEO，∴DFOE =PFPO，∴PF=92x3x⋅DF=32DF，而PE=5DF，∴PFPE =32DF5DF=310．故答案为310．连接OE，如图，利用正切的定义得到sin∠DAO=ODOA =23，则可设OD=2x，OA=3x，再根据切线的性质得OA⊥PA，所以∠APO=∠OAD，利用正弦的定义得到OP=92x，证明Rt△PAD∽Rt△POA，利用相似比得到PA2=PD⋅PO，而PA2=PF⋅PE，所以PD⋅PO=PF⋅PE，则可判断△PDF∽△PEO，利用相似比得到PF=32DF，然后利用PE=5DF可得到PFPE的值．本题考查了相似三角形的判定与性质：在判定两个三角形相似时，应注意利用图形中已有的公共角、公共边等隐含条件，以充分发挥基本图形的作用，寻找相似三角形的一般方法是通过作平行线构造相似三角形，灵活运用相似三角形的性质表示线段之间的关系；也考查了切线的性质和切割线定理．18.【答案】60√213【解析】解：连接BD交AC于H，作∠ABC的平分线BP，交AC于P，连接PD，作PE⊥BC于E，连接PM、PN，如图所示：则PN≥PE，在△ABC和△ADC中，{AB=AD BC=DC AC=AC，∴△ABC≌△ADC(SSS)，∴∠BAP=∠DAP，在△ABP和△ADP中，{AB=AD∠BAP=∠DAP AP=AP，∴△ABP≌△ADP(SAS)，∴∠ABP=∠ADP=12∠ABC=45°，BP=DP，∵∠ABP=∠NBP=12∠ABC=45°，∴∠NBP=∠MDP，在△NBP和△MDP中，{BN=DM∠NBP=∠MDP BP=DP，∴△NBP≌△MDP(SAS)，∴PM=PN，∠BPN=∠DPM，∴∠BPD=∠MPN，∵BP=DP，PM=PN，∴∠BDP=∠DBP=∠MNP=∠NMP，∴△PMN∽△PBD，∴MNBD =PNBP≥PEPB，∵sin∠NBP=PEPB =sin45°=√22，∴MNBD ≥√22，∴MN≥√22BD，在△ABH和△ADH中，{AB=AD∠BAH=∠DAH AH=AH，∴△ABH≌△ADH(SAS)，∴BH=DH，∠BHA=∠DHA=90°，AC=√AB2+BC2=√52+122=13，S△ABC=12AB⋅BC=12BH⋅AC，∴BH=AB⋅BCAC =5×1213=6013，∴BD=2BH=12013，∴MN≥√22×12013=60√213，∴线段MN的最小值为60√213，故答案为：60√213．连接BD交AC于H，作∠ABC的平分线BP，交AC于P，连接PD，作PE⊥BC于E，连接PM、PN，则PN≥PE，证明△ABC≌△ADC(SSS)，得出∠BAP=∠DAP，证明△ABP≌△ADP(SAS)，得出∠ABP=∠ADP=12∠ABC=45°，BP=DP，易证∠NBP=∠MDP，证明△NBP≌△MDP(SAS)，得出PM=PN，∠BPN=∠DPM，推出∠BPD=∠MPN，证出∠BDP=∠DBP=∠MNP=∠NMP，得出△PMN∽△PBD，则MNBD =PNBP≥PEPB，由sin∠NBP=PEPB =sin45°=√22，推出MNBD≥√22，即MN≥√22BD，证明△ABH≌△ADH(SAS)，得出BH=DH，∠BHA=∠DHA=90°，AC=√AB2+BC2=13，由S△ABC=1 2AB⋅BC=12BH⋅AC，求出BH=6013，得出BD=2BH=12013，即可得出结果．本题考查了相似三角形的判定与性质、全等三角形的判定与性质、勾股定理、三角函数等知识；本题综合性强，证明三角形相似和三角形全等是解题的关键． 19.【答案】2【解析】解：x =(1+x −2x 2)(a 0+a 1x +a 2x 2+a 3x 3…+a n x n )， 当x =0时，a 0=0，∴1=(1+x −2x 2)(a 1+a 2x +a 3x 2…+a n x n−1)， 当x =0时，a 1=1，a 1+a 2=0，a 2+a 3−2a 1=0， ∴a 2=−1，a 3=3， ∴a 3+a 2=2， 故答案为2．先去分母，第一次赋值x =0求出a 0=0，再化简式子为1=(1+x −2x 2)(a 1+a 2x +a 3x 2…+a n x n−1)，第二次赋值x =0，求出a 1=1，再由等式的性质得到a 1+a 2=0，a 2+a 3−2a 1=0即可求解．本题考查数字的变化规律；能够通过所给例子，找到式子的规律，给式子恰当的赋值运算是解题的关键．20.【答案】解：(1)联立y =x 2+(a −7)x +6，y =ax 并整理得：x 3+(a −7)x 2+6x −a =0…①，a =2时，上式为：(x −1)(x 2−4x +2)=0， 解得：x =1或2+√2或2−√2，故函数交点坐标为：(1,2)或(2+√2,2−√2)或(2+√2,2−√2)； (2)①式中含有(x −1)的因式，即：(x −1)[x 2+(a −6)x +a]=0， 故其中一个根：x =1，a 为正整数，x 2+(a −6)x +a =0方程有一个到两个的根， △=(a −6)2−4a ≥0，交点横、纵坐标都是整数，则△一定是完全平方数(设为k)， 即(a −6)2−4a =k 2(k 为非负整数)， 整理得：(a −8)2−k 2=28，即：(a −8+k)(a −8−k)=28=4×7=2×14=1×28， 而a −8+k ≥a −8−k ，当a −8+k =7，a −8−k =4时，解得：a =13.5(舍去)； 当a −8+k =14，a −8−k =2时，解得：a =16； 当a −8+k =28，a −8−k =1时，a =23.5(舍去)； 故a =16；(3)两个函数的交点都在直线x =12的右侧，只会出现如下图所示的情况，两个函数三个交点在x =12的右侧，其中一个交点横坐标为x =1在x =12的右侧， 故只需要确定x 2+(a −6)x +a =0根的情况，只要左侧的根在x =12右侧即可， 解上述方程得：x =6−a±√a 2−16a+362，即6−a−√a2−16a+362>12，解得：a >116．故：a 的取值范围为：a >116．【解析】(1)联立y =x 2+(a −7)x +6，y =ax 并整理得：x 3+(a −7)x 2+6x −a =0，a =2时，上式为：(x −1)(x 2−4x +2)=0，即可求解；(2)(x −1)[x 2+(a −6)x +a]=0，故其中一个根：x =1，a 为正整数，x 2+(a −6)x +a =0方程有一个到两个的根，△=(a −6)2−4a ≥0，交点横、纵坐标都是整数，则△一定是完全平方数(设为k)，即(a −6)2−4a =k 2(k 为非负整数)，讨论确定a 的值； (3)两个函数的交点都在直线x =12的右侧，两个函数三个交点在x =12的右侧，其中一个交点横坐标为x =1在x =12的右侧，即6−a−√a2−16a+362>12，即可求解．本题考查的是二次函数与反比例函数的交点问题、根的判别式、整数的性质，涉及面较广，难度较大．21.【答案】解：(1)作GH ⊥AD 于H ，AI ⊥BE 于I ， ∵GE =3，cos∠AEB =23，∴EH =2，HG =√5，设AE =3x ，则EI =2x ，AI =√5x ，∴GI =3−2x ，BI =BG +GI =n +3−2x ， ∴DH =DE +EH =n +2， ∵∠ADF =∠ABE ，∴∠DHG =∠AIB =90°， ∴△GHD∽△AIB ， ∴DH BI=HG AI，∴n+2n+3−2x =√5√5x ， 解得：x =n+3n+4， ∴AE =3x =3n+9n+4；(2)如图2，连接AA′交BE 于M ，连接按个，作A′N ⊥CG 于N ，∵四边形ABCD 为矩形，G 恰为BE 中点，∴CG =DG ，∴∠GCD =∠GDC ，∴∠BCG =∠ADG =∠ABE =90°−∠CBG ， ∴∠BCG +∠CBG =90°， ∴CG ⊥BE ，∵AA′⊥BE ，A′N ⊥CG ， ∴四边形MA′NG 是矩形， ∴GM =A′N =3√24，设ME =x ，则AG =BG =GE =x +34√2， ∴AM 2=AG 2−GM 2=AE 2−EM 2=(x +3√24)2−(34√2)2=1−x 2， 解得：x =√24，∴BG =GE =ME +GM =√2， ∴BE =2√2，∵∠ABE =∠BCG ， ∴△GCB∽△ABE ， ∴BC BE =BG AE，∴2√2=√21， 解得：BC =4，∴AD =BC =4， ∴DE =AD −AE =4−1=3．【解析】(1)作GH ⊥AD 于H ，AI ⊥BE 于I ，根据已知条件得到EH =2，HG =√2，设AE =3x ，则EI =2x ，AI =√5x ，得到GI =3−2x ，BI =BG +GI =n +3−2x ，根据相似三角形的性质得到AE =3x =3n+9n+4；(2)如图2，连接AA′交BE 于M ，连接按个，作A′N ⊥CG 于N ，根据矩形的性质得到CG =DG ，求得∠GCD =∠GDC ，推出四边形MA′NG 是矩形，得到GM =A′N =3√24，设ME =x ，则AG =BG =GE =x +34√2，根据勾股定理列方程得到BG =GE =ME +GM =√2，求得BE =2√2，根据相似三角形的性质即可得到结论．本题考查了矩形的性质，相似三角形的判定和性质，轴对称的性质，勾股定理，正确的作出辅助线是解题的关键．。
自主招生考试数学试卷一、选择题（共10小题，每小题6分，满分60分）1．（6分）有一个角为60°的菱形，边长为2，其内切圆面积为（）A．B．C．D．2．（6分）若方程组的解为（a，b，c），则a+b+c=（）A．1 B．0 C．﹣1 D．23．（6分）圆O1与圆O2半径分别为4和1，圆心距为2，作圆O2的切线，被圆O1所截得的最短弦长为（）A．﹣1 B．8 C．2D．24．（6分）如下图，梯形ABCD中，AD∥BC，AC与BD交于O，记△AOD、△ABO、△BOC的面积分别为S1、S2、S3，则S1+S3与2S2的大小关系为（）A．无法确定B．S1+S3＜2S2C．S1+S3=2S2D．S1+S3＞2S25．（6分）关于x的分式方程2k﹣4+仅有一个实数根，则实数k的取值共有（）A．1个B．2个C．3个D．4个6．（6分）两本不同的语文书、两本不同的数学书和一本英语书排放在书架上，若同类书不相邻，英语书不放在最左边，则排法的种数为（）A．32 B．36 C．40 D．447．（6分）若a=，则的值的整数部分为（）A．1 B．2 C．3 D．48．（6分）在圆内接四边形ABCD中，∠BAD、∠ADC的角平分线交于点E，过E作直线MN平行于BC，与AB、CD交于M、N，则总有MN=（）A．BM+DN B．AM+CN C．BM+CN D．AM+DN9．（6分）由若干个边长为1的小正方形组成一个空间几何体（小正方形可以悬空），其三视图如图，则这样的小正方体至少应有（）A．8个B．10个C．12个D．14个10．（6分）正方体ABCD的边长为1，点E在边AB上，BE=，BF=，动点P从E出发沿直线向F运动，每当碰到正方形的边时反弹，反弹时反射角等于入射角，而当碰到正方形顶点时沿入射路径反弹，当点P第一次返回E时，P所经过的路程为（）A．B．C．2D．二、填空题（共8小题，每小题6分，满分48分）11．（6分）对任意实数k，直线y=kx+（2k+1）恒过一定点，该定点的坐标是．12．（6分）如图，圆锥母线长为2，底面半径为，∠AOB=135°，经圆锥的侧面从A到B的最短距离为．13．（6分）设（3x﹣2）6=a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6，那么a1+a2+a3+a4+a5+a6=．14．（6分）如图，向正五边形ABCDE区域内均匀掷点，落在五边形FGHJK区域内的概率为．15．（6分）函数y=kx﹣1与y=x2的图象交于两点（x1，y1）（x2，y2），若+=18，则k=．16．（6分）在△ABC中，∠C=90°，D、E分别是BC、CA上的点，且BD=AC，AE=CD，BE、AD相交于点P，则∠BPD=．17．（6分）函数y=2+的最大值为．18．（6分）若x≥y≥z，则（2x+1）（2y+1）（2z+1）=13xyz的正整数解（x，y，z）为．三、解答题（共2小题，满分42分）19．（22分）正方形ABCD边长为2，与函数x=（x＞0）的图象交于E、F两点，其中E位于线段CD上，正方形ABCD可向右平移，初始位置如图所示，此时，△DEF的面积为．正方形ABCD在向右平移过程中，位于线段EF上方部分的面积记为S，设C点坐标为（t，0）（1）求k的值；（2）试写出S与t的函数关系式及自变量t的取值范围；（3）若S=2，求t的值；（4）正方形ABCD在向右平移过程中，是否存在某些位置，沿线段EF折叠，使得D点恰好落在BC边上？若存在，确定这些位置对应t的值得大致范围（误差不超过0.1）；若不存在，说明理由．20．（20分）（1）求函数y=|x﹣1|+|x﹣3|的最小值及对应自变量x的取值；（2）求函数y=|x﹣1|+|x﹣2|+|x﹣3|的最小值及对应自变量x的取值；（3）求函数y=|x﹣1|+|x﹣2|+…+|x﹣n|的最小值及对应自变量x的取值；（4）求函数y=|x﹣1|+|2x﹣1|+…+|8x﹣1|+|9x﹣1|的最小值及对应自变量x 的取值．2017年四川省成都七中自主招生考试数学试卷参考答案与试题解析一、选择题（共10小题，每小题6分，满分60分）1．（6分）有一个角为60°的菱形，边长为2，其内切圆面积为（）A．B．C．D．【解答】解：过A作AE⊥BC，如图所示：∵菱形ABCD的边长为2，∠ABC═60°，∴∠BAE=30°，∴BE=AB=1，∴AE=BE=，∴内切圆半径为，∴内切圆面积=π•（）2=；故选：A．2．（6分）若方程组的解为（a，b，c），则a+b+c=（）A．1 B．0 C．﹣1 D．2【解答】解：，②×5﹣①得：14y+3z=﹣17④，②×2﹣③得：5y+2z=﹣7⑤④×2﹣⑤×3得：13y=﹣13，解得：y=﹣1，把y=﹣1代入⑤得：z=﹣1，把y=﹣1，z=﹣1代入②得：x=2，则（a，b，c）=（2，﹣1，﹣1），则a+b+c=2﹣1﹣1=0．故选：B．3．（6分）圆O1与圆O2半径分别为4和1，圆心距为2，作圆O2的切线，被圆O1所截得的最短弦长为（）A．﹣1 B．8 C．2D．2【解答】解：∵圆O1与圆O2半径分别为4和1，圆心距为2，∴4﹣1＞2，故两圆内含，不妨设截得的弦为AB，切点为C，连接O1A，连接O1O2，O2C，∵半径确定，∴弦心距越小，则弦越长，∵AB是⊙O2的切线，∴O2C⊥AB，∴当O1、O2、C在一条线上时，弦AB最短，由题意可知OC1=2+1=3，AO1=4，在Rt△ACO1中，由勾股定理可得AC==，∴AB=2AC=2，故选：C．4．（6分）如下图，梯形ABCD中，AD∥BC，AC与BD交于O，记△AOD、△ABO、△BOC的面积分别为S1、S2、S3，则S1+S3与2S2的大小关系为（）A．无法确定B．S1+S3＜2S2C．S1+S3=2S2D．S1+S3＞2S2【解答】解：∵AD∥BC，∴△AOD∽△COB，∴=，∵△AOD与△AOB等高，∴S1：S2=AD：BC=a：b，∴S1=S2，S3=S2，∴S1+S3=（+）S2=S2，∵a≠b，∴a2+b2＞2ab，∴＞2，∴S1+S3＞2S2，故选：D．5．（6分）关于x的分式方程2k﹣4+仅有一个实数根，则实数k的取值共有（）A．1个B．2个C．3个D．4个【解答】解：方程两边都乘x（x+2）得，（2k﹣4）x（x+2）+（k+1）（x+2）=x （k﹣5），整理得，（k﹣2）x2+（2k﹣1）x+k+1=0．①当k﹣2≠0时，∵△=（2k﹣1）2﹣4（k﹣2）（k+1）=9＞0，∴一元二次方程（k﹣2）x2+（2k﹣1）x+k+1=0有两个不相等的实数根．∵关于x的分式方程2k﹣4+仅有一个实数根，而x（x+2）=0时，x=0或﹣2，∴x=0时，k+1=0，k=﹣1，此时方程﹣3x2﹣3x=0的根为x=0或﹣1，其中x=0是原方程的增根，x=﹣1是原方程的根，符合题意；x=﹣2时，4（k﹣2）﹣2（2k﹣1）+k+1=0，k=5，此时方程3x2+9x+6=0的根为x=﹣2或﹣1，其中x=﹣2是原方程的增根，x=﹣1是原方程的根，符合题意；即k=﹣1或5；②当k﹣2=0，即k=2时，方程为3x+3=0，解得x=﹣1，符合题意；即k=2．综上所述，若关于x的分式方程2k﹣4+仅有一个实数根，则实数k的取值为﹣1或5或2，共有3个．故选：C．6．（6分）两本不同的语文书、两本不同的数学书和一本英语书排放在书架上，若同类书不相邻，英语书不放在最左边，则排法的种数为（）A．32 B．36 C．40 D．44【解答】解：设从左向右位置为①，②，③，④，⑤，∵英语书不在最左边，∴最左边①有4种取法，∵同类书不相邻，∴②有3种取法，③有两种取法，④有两种取法，⑤有一种取法，共4×3×2×2×1=48，但是英语书排在第②位置时，只能是语文、英语、数学、语文、数学，或者数学、英语、语文、数学、语文，故英语书排在第②位置时只有8种情况，故种情况为48﹣8=40种，故选：C．7．（6分）若a=，则的值的整数部分为（）A．1 B．2 C．3 D．4【解答】解：∵==﹣=﹣=﹣，∴=﹣+﹣+﹣=﹣∵a=，∴==4，0＜a27＜a3=（）3=＜，∴＜1﹣a27＜1，∴1＜＜2，∴的值的整数部分为2．故选：B．8．（6分）在圆内接四边形ABCD中，∠BAD、∠ADC的角平分线交于点E，过E作直线MN平行于BC，与AB、CD交于M、N，则总有MN=（）A．BM+DN B．AM+CN C．BM+CN D．AM+DN【解答】解：如图，在NM上截取NF=ND，连结DF，AF∴∠NFD=∠NDF，∵A，B，C，D四点共圆，∴∠ADC+∠B=180°，∵MN∥BC，∴∠AMN=∠B，∴∠AMN+∠ADN=180°，∴A，D，N，M四点共圆，∴∠MND+∠MAD=180°，∵AE，DE分别平分∠BAD，∠CDA，∴∠END+2∠DFN=∠END+2∠DAE=180°，∴∠DFN=∠DAE，∴A，F，E，D四点共圆，∴∠DEN=∠DAF，∠AFM=∠ADE，∴∠MAF=180°﹣∠DAF﹣∠MND=180°﹣∠DEN﹣∠MND=∠EDN=∠ADE=∠AFM，∴MA=MF，∴MN=MF+NF=MA+ND．故选：D．9．（6分）由若干个边长为1的小正方形组成一个空间几何体（小正方形可以悬空），其三视图如图，则这样的小正方体至少应有（）A．8个B．10个C．12个D．14个【解答】解：综合三视图，我们可以得出，这个几何模型的底层至少有3个小正方体，第二层至少有3个小正方体，第三层至少有3个小正方体，则这样的小正方体至少应有3+3+3=9个，选项中10是满足条件最小的数字．故选：B．10．（6分）正方体ABCD的边长为1，点E在边AB上，BE=，BF=，动点P从E出发沿直线向F运动，每当碰到正方形的边时反弹，反弹时反射角等于入射角，而当碰到正方形顶点时沿入射路径反弹，当点P第一次返回E时，P所经过的路程为（）A．B．C．2D．【解答】解：根据已知中的点E，F的位置，可知入射角的正切值为，第一次碰撞点为F，在反射的过程中，根据入射角等于反射角及平行关系的三角形的相似可得第二次碰撞点为M，在DA上，且DM=DA，第三次碰撞点为N，在DC上，且DN=DC，第四次碰撞点为G，在CB上，且CG=BC，第五次碰撞点为H，在DA上，且AH=AD，第六次碰撞点为Z，在AB上，且AZ=AD，第七次碰撞点为I，在BC上，且BI=AD，第八次碰撞点为D，再反方向可到E，由勾股定理可以得出EF=HZ==，FM=GH=ID=，MN=NG=，ZI=，P所经过的路程为（×2+×3+×2+）×2=．故选：B．二、填空题（共8小题，每小题6分，满分48分）11．（6分）对任意实数k，直线y=kx+（2k+1）恒过一定点，该定点的坐标是（﹣2，1）．【解答】解：∵y=kx+（2k+1）∴y=k（x+2）+1，∴图象恒过一点是（﹣2，1），故答案为（﹣2，1）．12．（62，底面半径为，∠AOB=135°，经圆锥的侧面从A到B的最短距离为2．【解答】解：如右图所示，是圆锥侧面展开的一部分，∵圆锥母线长为2，底面半径为，∠AOB=135°，∴，作AD⊥SB于点D，∵SA=SB=2，∴展开的扇形所对的圆心角为，∴在Rt△SAD中，AD=SD=，∴BD=SB﹣SD=2﹣，∴AB==，故答案为：2．13．（6分）设（3x﹣2）6=a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6，那么a1+a2+a3+a4+a5+a6= 1﹣26．【解答】解：由题意可知a0=（﹣2）6，令x=1，则1=a0+a1+a2+a3+a4+a5+a6，因此a1+a2+a3+a4+a5+a6=1﹣a0=1﹣（﹣2）6=1﹣26．故答案为：1﹣26．14．（6分）如图，向正五边形ABCDE区域内均匀掷点，落在五边形FGHJK区域内的概率为．【解答】解：正五边形ABCDE，∴∠BAE=∠ABC=BCD=∠CDE∠AED=108°，AB=BC=CD=DE=AE，∴△ABC≌△ABE，∴AC=BE，同理：△ABH≌△△BCG≌△AJE，∴AH=CG=JE，∴HJ=HG，同理：FG=FK=JK=HG，∴五边形HGFKJ是正五边形，∴正五边形HGFKJ∽正五边形ACBDE，设HE=CD=a，HJ=x，由题意，△HAB∽△ABE，∴，∴x=∴落在五边形FGHJK区域内的概率为=，故答案为．15．（6分）函数y=kx﹣1与y=x2的图象交于两点（x1，y1）（x2，y2），若+=18，【解答】解：∵函数y=kx﹣1与y=x2的图象交于两点（x1，y1）（x2，y2），∴，消去y得x2﹣kx+1=0，∴x1+x2=k，x1x2=1，∴+====18，∴k（k2﹣2）﹣k=18，解答k=3．故答案为3．16．（6分）在△ABC中，∠C=90°，D、E分别是BC、CA上的点，且BD=AC，AE=CD，BE、AD相交于点P，则∠BPD=45°．【解答】解：作AF∥CD，DF∥AC，AF交DF于点F，∴四边形ACDF是平行四边形．∵∠C=90°∴四边形ACDF是矩形，∴CD=AF，AC=DF，∠EAF=∠FDB=∠AFD=90°．∵BD=AC，AE=CD∴△BDF和△AEF是等腰直角三角形，∴∠AFE=∠DFB=45°，∴∠DFE=45°，∴∠EFB=90°．∴∠EFB=∠AFD．∴△BDF∽△AEF，∵∠EFB=∠AFD，∴△ADF∽△EBF∴∠PAF=∠PEF∴∠APE=∠AFE∵∠AFE=45°∴∠APE=45°17．（6分）函数y=2+的最大值为．【解答】解：根据题意得：，解得：1≤x≤2，由柯西不等式得：y=2+≤•=×=（当且仅当2=，即x=时，取等号），故函数y=2+的最大值为．故答案为：．18．（6分）若x≥y≥z，则（2x+1）（2y+1）（2z+1）=13xyz的正整数解（x，y，z）为（45，7，1）或（19，9，1）．【解答】解：∵（2x+1），（2y+1），（2z+1）都是奇数，∴x，y，z都是奇数，∵（2x+1）（2y+1）（2z+1）=13xyz，∴（2+）（2+）（2+）=13，∵x≥y≥z，如果z≥3，那么（2+）（2+）（2+）≤（2+）2=＜13，∴z=1，∴3（2x+1）（2y+1）=13xy，化简得：xy=6（x+y）+3，则x==6+，∵39的因子有：1，3，12，39，∴y﹣6=1，3，13，39，∴y=7，9，19，45，∴x的对应只有：45，19，9，7，∵x＞y，∴正整数解（x，y，z）为：（45，7，1）或（19，9，1）．故答案为：（45，7，1）或（19，9，1）．三、解答题（共2小题，满分42分）19．（22分）正方形ABCD边长为2，与函数x=（x＞0）的图象交于E、F两点，其中E位于线段CD上，正方形ABCD可向右平移，初始位置如图所示，此时，△DEF的面积为．正方形ABCD在向右平移过程中，位于线段EF上方部分的面积记为S，设C点坐标为（t，0）（1）求k的值；（2）试写出S与t的函数关系式及自变量t的取值范围；（3）若S=2，求t的值；（4）正方形ABCD在向右平移过程中，是否存在某些位置，沿线段EF折叠，使得D点恰好落在BC边上？若存在，确定这些位置对应t的值得大致范围（误差不超过0.1）；若不存在，说明理由．=（2﹣）2=，【解答】解：（1）由题设可知S△DEF解得k=1或7（不合题意，舍去），∴k=1；（2）①如图1，当2≤t≤时，因为C点坐标为（t，0），所以E点坐标为（t，），所以DE=2﹣，而F点坐标为（，2），所以DF=t﹣，所以S=DE•DF=（2﹣）（t﹣）=t+﹣1；②如图2，当t＞时，此时OB=t﹣2，所以F点的坐标为（t﹣2，），所以AF=2﹣，所以S=•2•（DE+AF）=•2•（2﹣+2﹣）=4﹣﹣；（3）当2≤t≤时，DE和DF随t的增大而增大，S也类似，故当t=时S有最大值为＜2，所以S=2只可能发生在t＞时，令4﹣﹣=2，解得t=；（4）①如图3，当2≤t≤时，假设位置存在，由对称性知Rt△FDE∽Rt△DCD1，因为DE=D1E，则有=，其中D1C==，整理得：t（t﹣1）=4，解得t=＞，与假设矛盾，所以当2≤t≤时，不存在；②如图4，当t＞时，假设位置存在，过F作直线FG∥x轴交CD于G，由对称性可知Rt△FGE≌Rt△DCD1，DE=D1E，所以GE=D1C，而GE=﹣，整理可得t（t﹣1）（t﹣2）2=1，设y=t（t﹣1）（t﹣2）2，当t＞2时，y随t的增大而增大，取t=2.5，则y=0.9375＜1，取t=2.6，则y=1.4976＞1，利用试值法可以判断位置存在且唯一，对应的t的取值在2.5和2.6之间．20．（20分）（1）求函数y=|x﹣1|+|x﹣3|的最小值及对应自变量x的取值；（2）求函数y=|x﹣1|+|x﹣2|+|x﹣3|的最小值及对应自变量x的取值；（3）求函数y=|x﹣1|+|x﹣2|+…+|x﹣n|的最小值及对应自变量x的取值；（4）求函数y=|x﹣1|+|2x﹣1|+…+|8x﹣1|+|9x﹣1|的最小值及对应自变量x 的取值．【解答】解：（1）函数y=|x﹣1|+|x﹣3|的最小值的几何意义是数轴上x到1和3两点距离之和的最小值，∵两点之间线段最短，∴当1＜x＜3时，y min=|3﹣1|=2，（2）∵y=|x﹣1|+|x﹣2|+|x﹣3|=（|x﹣1|+|x﹣3|）+|x﹣2|，当x=2时，|x﹣2|有最小值，∴结合（1）的结论得出，当x=2时，y min=2+0=2，（3）当n为偶数时，y=|x﹣1|+|x﹣2|+…+|x﹣n|=（|x﹣1|+|x﹣n|）+（|x﹣2|+|x﹣（n﹣1）|）+…+（|x﹣|+|x﹣（+1）|），由（1）知，当＜x＜+1时，|x﹣1|+|x﹣n|有最小值n﹣1，|x﹣2|+|x﹣（n﹣1）|有最小值（n﹣1）﹣2=n﹣3，…2019年四川省成都七中自主招生考试数学试卷(含详细解析)|x﹣|+|x ﹣（+1）|有最小值1，∴当＜x＜+1时，y min=1+3+5+…+（n ﹣3）+（n﹣1）=，当n为奇数时，y=|x﹣1|+|x﹣2|+…+|x﹣n|=（|x﹣1|+|x﹣n|）+（|x﹣2|+|x ﹣（n﹣1）|）+…+（|x﹣|+|x ﹣（+1）|）+|x﹣|，由（1）知，当x=时，|x﹣1|+|x﹣n|有最小值n﹣1，|x﹣2|+|x﹣（n﹣1）|有最小值（n﹣1）﹣2=n﹣3，…|x﹣|+|x﹣（+1）|有最小值1，|x﹣|的最小值为0，∴当x=时，ymin=0+2+4+…+（n﹣3）+（n﹣1）=，（4）类似（3）的做法可知，y=|x﹣a1|+|x﹣a2|+…+|x﹣a n|，如果n为偶数时，当时，y有最小值，如果n为奇数时，当x=时，y有最小值；∵y=|x﹣1|+|2x﹣1|+…+|8x﹣1|+|9x ﹣1|=++…++|x﹣1|∴共有9+8+7+…+2+1=45项，为奇数．∴当x=时，ymin=|﹣1|+|﹣1|+…+|﹣1|+|﹣1|=第21页（共21页）。
2019年四川省成都七中自主招生物理试卷一、单选题（本大题共11小题，共38.0分）1.以下说法中正确的是()A. 初中物理课本的宽度大约28cmB. 刚参加了中考的小明体积约55dd3C. 小明从一楼走到三楼教室克服自身重力大约做了1500J的功D. 高空中飞机飞行的速度大约25d/d2.下列关于物态变化的说法中，正确的是()A. 春天，河里冰雪消融，是升华现象B. 夏天，冰棍儿周围冒“白气”，是汽化现象C. 秋天，早晨花草上出现的小露珠是熔化现象D. 冬天，温暖的车内窗玻璃会变模糊，是因为车内水蒸气液化的缘故3.如图所示，容器中盛满水，水中放入P和Q两个小球，P球为铁球，Q球为木球，它们用细线分别系于容器的上、下底部，当容器静止时，细线均伸直处于竖直方向，现使容器以一定加速度向右匀加速运动，则此时P、Q两球相对容器()》A. 两球均向右偏移B. 两球均向左偏移C. P球向右偏移D. Q球向右偏移4.某人站在离湖岸边8m的C处，刚好能看见湖对岸的一棵树HG在水中的完整的像，如果眼距地面的高度为d.6d，湖两岸均高出湖水面lm。
湖宽50m，则该树HG的高度为()A. 10mB. 9mC. 8mD. 7m5.小球从高处下落到竖直放置的轻弹簧上(如图甲)，小球的速度v和弹簧缩短的长度△d之间的关系如图乙所示，其中A为曲线的最高点。
已知该小球重为2.2d，弹簧在受到撞击至压缩到最短的过程中始终发生弹性形变，弹簧的弹力大小与形变成正比。
下列说法正确的是()A. 从撞击轻弹簧到它被压缩至最短的过程中，小球的重力做功的功率先减小后增大B. 从撞击轻弹簧到它被压缩到最短的过程中，小球的机械能先增大后减小C. 当小球的速度为5.dd/d时，小球受到的合力为2.2dD. 从撞击轻弹簧到弹簧被压缩至最短的时候，小球受到的合力为11.22d6.在图所示的电路中，当滑动变阻器R的滑片P从B向A滑动的过程中，电压表d1、d2示数的变化量的值分别为△d1、△d2，则它们的大小相比较应该是()A. △d1<△d2B. △d1>△d2C. △d1=△d2D. 因为无具体数据，故无法比较7.如图所示，有一重力不计的方形容器，被水平力F压在竖直的墙面上处于静止状态，现缓慢地向容器内注水，直到将容器刚好盛满为止，在此过程中容器始终保持静止，则下列说法中正确的是()8.9.&A. 容器受到的摩擦力不变B. 容器受到的摩擦力逐渐增大C. 水平力F一定不变D. 水平力F必须逐渐增大10.小明在用可变焦的光学照相机(一种镜头焦距大小可根据需要发生改变的光学照相机)给小兰拍了一张半身照之后，保持相机和小兰的位置不变，又给小兰拍了一张全身照。
2019年四川省成都七中自主招生英语试卷第一部分阅读理解（共两节，满分40分）第一节（共5小题；每小题8分，满分40分）阅读下列短文，从每题所给的四个选项A．B．C和D中，选出最佳选项1. Brooke has just turned 12, with seven world records, is one of the best rock climbers in the world．Brooke comes from a climbing family．Both her parents are just past climbing champions．Her father stopped climbing some years ago, but her mother, Robyn, who won four world cup titles four years in a row, is still climbing．She runs a club for young climbers in Colorado, USA and coaches Brooke and her teammates．Sometimes it’s difficult for mother and daughter to work together so closely, but they really respect and trust each other and Brooke says her mum is a great coach．Brooke says that her mother gives her a lot of good advice and it is a very important part of her climbing life．Robyn can be strict, but she is also passionate about climbing, and she passes this passion on to her students．Robyn says Brooke has very good strong wills and is very good at making herself want to practice．This helps her when she’s facing the challenges of this difficult sport．She is also very hard﹣working．Success in rock climbing issomething you have to work at．To be a world﹣class athletes of any kind, you have to push yourself and train hard and that’s what Brooke does, every day, at the club and at home．Brooke says that climbing is always there in their lives．They even have a climbing wall in their house!But climbing isn’t only hard work．It’s fun too．Brooke loves climbing and when she’s on a high rock, she feels happy．Strangely, she says that when she looks down, she isn’t scared．All she does is to think how cool it is to be small compared to the rock．（1）It’s________for Brooke and Robyn to work together so closely．A.difficultB.uneasyC.amazingD.annoying．（2）To be a successful climber，one has to________．A.practice climbing at a very young ageB.have a great mother coach who is very strictC.face any difficult things bravely and keep on training hardD.keep pushing himself and train hard at any time and any place．（3）What’s the best title of this passage？________A.An Excellent ClimberB.A Climbing FamilyC.Climbing ChampionsD.How to Be Successful Rock Climbers．（4）Which of the sentences can the author agree with？________A.Well begun is half done．B.No pains，no gains．C.He who climbs high falls heavily．D.Doing is better than saying．2. Papa, as a son of a dirt﹣poor farmer, left school early and went to work in a factory, for education was for the rich then．So, the world became his school． With great interest, he read everything he could lay his hands on, listened to the town elders and learned about the world beyond his tiny hometown．“There’s so much to learn．”he’d say．“Though we’ re born stupid, only the stupid remain that way．”He was determined that none of his children would be denied an education．Thus, Papa insisted that we learn at least one new thing each day． Though, as children, we thought this was crazy, it would never have occurred to us to deny Papa a request． And dinner time seemed perfect for sharing what we had learned．We would talk about the news of the day; no matter how insignificant, it was never taken lightly． Papa would listen carefully and was ready with some comment, always to the point． Then came the moment﹣the time to share the day’s new learning．Papa, at the head of the table, would push back his chair and pour a glass of red wine, ready to listen．“Felice,”he’d say, “tell me what you learned today”．“I learned that the population of Nepal is ..”Silence．Papa was thinking about what was said, as if the salvation of the world would depend upon it．“The population of Nepal． Hmm．Well…”he’d say．“Get the map,and let’s see where Nepal is．”And the whole family went on a search for Nepal．This same experience was repeated until each family member had a turn． Dinner ended only after we had a clear understanding of at least half a dozen such facts．As children, we thought very little about these educational wonders． Our family, however, was growingtogether, sharing experiences and participating in one another’s education． And by looking at us, listening tous, respecting our input, affirming our value, giving us a sense of dignity, Papa was unquestionably our most influential teacher．Later during my training as a future teacher, I studied with some of the most famous educators． They were imparting what Papa had known all along﹣the value of continual learning． His technique has served me well all my life． Not a single day has been wasted, though I can never tell when knowing the population of Nepal might prove useful．（1）What do we know from the first paragraph？________A. The author’s father was born in a worker’s family．B. Those born stupid could not change their life．C. The town elders wanted to learn about the world．D. The poor could hardly afford school education．．（2）It can be learned from the passage that the author________．A. enjoyed talking about newsB. knew very well about NepalC. felt regretted about those wasted daysD. appreciated his father’s educational technique．（3）What is the greatest value of“dinner time”to the author？________A. Continual learning．B. Showing talents．C. Family get﹣together．D. Winning Papa’s approval．．（4）The author’s father can be best described as________．A. an educator expert at training future teachersB.a parent insistent on his children’s educationC.a participant willing to share his knowledgeD.a teacher strict about everything his students did．3. Some of the world’s most f amous musicians recently gathered in Paris and New Orleans to celebrate the first annual International Jazz Day． UNESCO（United Nations Educational, Scientific and Cultural Organization） recently set April 30 as a day to raise awareness of jazz music, its significance, and its potential as a unifying voice across cultures．Despite the celebrations, though, in the U．S．the jazz audience continues to shrink and grow older, and the music has failed to connect with younger generations．It’s Jason Mo ran’s job to help change ________ As the Kennedy Center’s artistic adviser for jazz, Moran hopes to widen the audience for jazz, make the music more accessible, and preserve its history and culture．“Jazz seems like it’s not really a part of the Ame rican appetite．”Moran tells National Public Radio’s reporter Neal Conan．"What I’m hoping to accomplish is that my generation and younger start to reconsider and understand that jazz is not black and white anymore．It’s actually color,and it’s actually di gital．Moran says one of the problems with jazz today is that the entertainment aspect of the music has been lost．“The music can’t be presented today the way it was in 1908 or 1958． It has to continue to move, because the way the world works is not the same, ”says Moran．Last year, Moran worked on a project that arranged Fats Waller’s music for a dance party,“just to kind of put it back in the mind that Waller is dance music as much as it is concert music, ”says Moran．“For me,it’s the re﹣contextualization．In music, where does the emotion lie? Are we, as humans, gaining any insight on how to talk about ourselves and how something as abstract as a Charlie Parker record gets us into a dialogue about our emotions and ourthoughts? Sometimes we lose sight that the music has a wider context, ”says Moran,“so I want to continue those dialogues． Those are the things I want to foster．”（1）Why did UNESCO set April 30 as International Jazz Day？________A. To protect cultural diversity．B. To recognize the value of jazz．C. To remember the birth of jazz．D. To encourage people to study music．．（2）What does the underlined word“that”in paragraph 3 refer to？________A. Jazz becoming more accessible．B. The jazz audience becoming larger．C. The production of jazz growing faster．D. Jazz being less popular with the young．．（3）What can we infer about Moran’s opinion on jazz？________A. It will disappear gradually．B. It remains black and white．C. It should keep up with the times．D.It changes every 50 years．．（4）The best title might be________．A.Celebrating the Jazz DayB.The Rise and Fall of JazzC.Exploring the Future of JazzD.The Story of a Jazz Musician．4. What makes us happy?There has long been a notion that money buys happiness．However, although“we really, really tried that for a couple of generations, it didn’t work,”said Francine Jay, author of The Joy ofLess, A Minimalist Living Guide: How toDeclutter, Organize, and Simplify Your Life．Thanks to a travel﹣inspired revelation（启发）, Jay has been happily living a simpler life for 12 years．“I always packed as lightly as possible, and found it exhilarating to get by with just a small carry﹣on bag, ”she told CNN．“I thought if it feels this great to travel lightly, how amazing would it be to live this way? I wanted to have that same feeling of freedom in my everyday life．”Jay decided to get rid of all her excess possessions and live with just the essentials．“I wanted to spend my time and energy on experiences, rather than things．”Jay is a follower of a movement called“minimalism（极简主义）”．Growing numbers of people have been attracted to this lifestyle all over the world．They share the same feeling of disappointment with modern life and a desire to live more simply．Minimalists are typically progressive and concerned about the environment, Leah Watkins, a lead researcher at Otago University in New Zealand, told Stuff magazine in March．But many simply experienced unhappiness caused by owning too many possessions．Depression with the materialism of our world isn’t new．English romantic poet William Wordsworth summed up how dispiriting this was back in I802, at the beginning of the industrial age, when he wrote, “Getting and spending, we lay waste our powers．”His preference was to go back to nature．Closer to our own times, the hippies（嬉皮士）of the 1960s also sought to“drop out”of modern life．For many minimalists, the key is to unload．Without objects, they“believe people are forced more and more into the present moment and that’s where life happens, ”wrote Stuff．But does simplicity ever feel like a sacrifice?“It’s eliminating the excess﹣unused items, unnecessary purchases﹣from your life．Well, I may have fewer possessions, but I have more space…Minimalism is making room for wh at matters most, ”said Jay．And“the real questions”, according to Duane Elgin, a US social scientist, are“what do you care about?”and“what do you value? ”He told CNN．“It’s important for people to realize minimalism isn’t simply the amount of stuf f we consume．It’s aboutour families, our work, our connection with the larger world, our spiritual dimension．It’s about how we touch the whole world．It’s a way of life．”（1）What was the author’s main purpose in writing the text？________A.To report on the trend of minimalism．B.To give tips on how to lead a happy life．C.To argue whether money buys happiness．D.To recommend one of Francine Jay’s books．．（2）What inspired Francine Jay to live a simple life？________A.A book she read．B.Her desire to keep up with modern life．C.A follower of minimalism she met on a trip．D.The pleasure she enjoyed from traveling lightly．．（3）The underlined word“eliminating”in Paragraph 9 probably means“________”A.removingB.distinguishingC.acceptingD.improving．（4）Which of the following would Duane Elgin probably agree with？________A.Minimalism is a healthy lifestyle that is in conflict with modern life．B.Minimalism limits people’s freedom to enjoy their lives to the fullest．C.Minimalism enables people to reflect on what truly counts in their lives．D.Minimalism means people have to sacrifice some pleasure to live simply．5. Did you know that people who live in different parts of China have different habits and preferences?For example, people from southern China prefer to eat vegetables, while people from northern China like to cat meat．But what causes these differences?According to a new study published in the journal Cell in October, gene variations（基因变异）might be responsible for these differences, Xinhua reported．In the study, researchers from Chinese genome﹣sequencing（基因组测序）firm BGI collected genetic informationfrom 141.431Chinese women．The women came from 31 provinces and comprised 36 ethnic minority groups． The researchers found that there are six gene frequencies that are different among people from both northern and southern China．They found that natural selection has played an important role in the ways that people living in different regions of China have evolved, affecting their food preferences, immunities（免疫性）to illness and physical traits（特点）, The New York Times reported．The researchers reported that a variation of the gene FADS2 is more commonly found in northern people than it is in southern people．It helps people metabolize（新陈代谢）fatty acids（脂肪酸）, which suggests a diet that is rich in meat．According to Xinhua, this is due to climate differences．Northern China is at a higher latitude（纬度）, which means it’s cold and dry throughout the whole year．This weather is difficult to grow vegetablesin．Therefore, northerners tend to eat more meat．The study also found differences in the immune systems of both groups．Most people in southern China carry thegene CR1, which protects against malaria（疟疾）．This is because malaria was once quite common in southern China．Inorder to survive, the genes of people in the south evolved to fight against this disease．However, people in the south are also more ________ to certain blood﹣borne illnesses, as they lack the genes to stop them．Genes can also cause physical differences between northerners and southerners．Most northerners have theABCCl1gene, which causes dry earwax（耳屎）, less body odor and fewer sweat secretions（分泌）, The New York Times reported．These physical differences are also more beneficial to living in cold environments．Southerners are less likelyto have this gene, as it did not evolve in their population．（1）What’s the new study mainly about？________A.The genes of Chinese minority groups．B.Habits of people from different regions of China．C.Physical differences between northerners and southerners．D.Differences in the genes of people from different parts of China．．（2）A variation of the gene FADS2 can help our body________．A.store fatB.digest meatC.fight diseaseD.control body temperature．（3）The underlined word“vulnerable”is closest in meaning to________．A.unprotectedB.uninterestedC.generousD.responsible．（4）Compared with people in southern China，most northerners________．a．are immune to malariab．catch blood﹣borne illnesses easilyc．have less body odord．sweat less frequentlyA.abB.beC.bdD.cd．第二节（每小题10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，选项中有两项为多余选项．1. Public Speaking TrainingGet a coach（1）_______，so get help．Since there are about a billion companies out there all ready to offer you public speaking training and courses，here are some things to look for when deciding the training that’s right for you．Focus on positivesAny training you do to become more effective at public speaking should always focus on the positive aspects of what you already do well．Nothing can hurt confidence more than being told that you aren’t doing well．（2）_______．so good public speaking training should develop those instead of telling you what you shouldn’t do．（3）_______If you find a public speaking course that looks as though it’s going to give you lots of dos and don’ts，walk away！Your brain is so full of what you’re going to be talking about．（4）_______．As far as we’re concerned，there are basically no hard and fast rules about public speaking．Your audience can be your friends．You are a special person not a cloneMost importantly，good public speaking training should treat you as a special one，with your own personal habits．（5）选项（A．B．C和D），选出可以填入空白处的最佳选项。
一.选择题：本大题共12个小题，每小题3分，共36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.D.6Z B = 40° ，Z ACD = :120° ，则Z A 等于（） A. 60°B. 70°C. 80°D.90° 下列计算中， 正确的是（ )A. 20B . / 3、2(a )a 6 C.9B. 9C. 12D. 155.把不等式2x< 4的解集表示在数轴上，正确的是（成都七中实验学校自主招生考试试题数学试题注意事项： 1. 本试题分第I 卷和第U 卷两部分.第I 卷为选择题36分；第U 卷为非选择题114 分；全卷共150分.考试时间为120分钟.2. 本试卷的选择题答案用2B 铅笔涂在机读卡上，非选择题在卷丄上作答.3. 考生务必将自己的姓名及考号写在密封线以内指定位置 —4. 非选择题必须在指定的区域内作答，不能超出指定区域或在非指定区域作答，否则 答案无效.卷I （选择题，共36分）-2 0A右 --- *--2 0C33.4.如图 2，在口 ABC 中，AC 平分Z DAB AB = 3, 则口 ABC 的周长为（）A. 6D. a aa图26.如图3，在5 X5正方形网格中，一条圆弧经过A B, C三点，那么这条A.点PB.点M C .点R D.点Q… 2 ------------------7.若x 2x寸y 3 0，则xy的值为（）A. 6 或0B. 6 或0C. 5 或0D. 8 或图39•如图4,已知边长为1的正方形ABC ,E 为CD 边的中点，动点P 在正方形ABC 边上沿A B C E 运动，设点P 经过的路程 为x ，△ APE 的面积为y ，则y 关于x 的函数的图象大致为（A. 7B. 8C. 9D. 1011.如图6,已知二次函数y ax 2bx c 的图像如图所示，则下列6个代数式ab, ac, a b c, a b c,2a b,2a b 中其值为正的式子个数为（)A. 1个 B . 2个 C . 3个D.4个&已知0 a b,x , a b , b, y b , b a,则x, y 的大小关系是()A. x y B . x = y C . x y D .与a 、b 的取值有关10.如图5,两个正六边形的边长均为1,其中一个正六边形 一边恰在另一个正六边形的对角线上，则这个图形（阴影部 分）外轮廓线的周长是（） 12•将正方体骰子（相对面上的点数分别为1和6、2和5、3和4）放置于水平桌面上，如图 7-1 .在图7-2中，将骰子向右翻滚90°,然后在桌面上按逆时针方向旋转90°,则完成一次变换.若 骰子的初始位置为图7-1所示的状态，那么按上述规则连续完成10次变换后，骰子朝上一面的点 数是（）图7-1图7-2卷□（非选择题,共114分）二.填空题：本大题共6个小题，每小题4分，共24分.将答案直接填写在题中横线 上.13.5的相反数是 _______ .14. 如图8,矩形ABC 的顶点A, B 在数轴上，CD = 6, 点A 对应的数为1，则点B 所对应的数为 _______ .E CO 2.5 x (A ) O2.5 x(B )(C )O2.5 x(D )）16.已知x = 1是一元二次方程x 2mx n 0的一个根，则22a 0.15. _____________________________________________ 如图9,有五张点数分别为2, 3, 7, 8, 9的扑克牌， 从中任意抽取两张，则其点数之积是偶数的概率为 m 2 2mn n 2的值为 __________17. 把三张大小相同的正方形卡片A, B, C 叠放在一 个底面为正方形的盒底上，底面未被卡片覆盖的部分用 阴影表示.若按图10-1摆放时，阴影部分的面积为S ; 若按图10-2摆放时，阴影部分的面积为S 2,则S ____ S 2（填 “〉” “V” 或“=”）.18. 南山中学高一年级举办数学竞赛，A B 、C D E 五位同学得了 前五名，发奖前，老师让他们猜一猜各人的名次排列情况• A 说：B 第三名，C 第五名； B 说：E 第四名，D 第五名； C 说：A 第一名，E 第四名； D 说：C 第一名，B 第二名； E 说：A 第三名，D 第四名.老师说:每个名次都有人猜对，试判断获得第一至第五名的依次为 ____ •三、解答题（本大题共7个小题，共90分•解答应写出文字说明、证明过程或演算步骤）19. （1）（本小题满分8分）解方程:（2）（本小题满分8分）先化简再求值:宀)「，其中2a 2 4a 3a 2 4a 4 a 220. （本小题满分12分）甲、乙两校参加区教育局举办的学生英语口语竞赛，两校参赛人数相等•比赛 结束后，发现学生成绩分别为7分、8分、9分、10分（满分为10分）.依据统计数据绘制了如下尚 不完整的统计图表.分数 7分 8分9分 10分 人数118甲校成绩统计表 乙校成绩扇形统计图图 11-1(1) 在图11-1中，“7分”所在扇形的圆心角 (2) 请你将图11-2的统计图补充完整.(3) 经计算，乙校的平均分是8.3分，中位数是8分， 请写出甲校的平均分、中位数；并从平均分和中位数的角度分析哪 个学校成绩较好.(4) 如果该教育局要组织8人的代表队参加市级团体 赛，为便管理，决定从这两所学校中的一所挑选参赛选手， 请你分析，应选哪所学校？ 21. (本小题满分12分)如图12,在直角坐标系中，矩形OAB 的顶点0与 坐标原点重合，顶点A C 分别在坐标轴上，顶点B 的坐标为(4, 2).过点D (0, 3)和E (6, 0)的直 线分别与AB BC 交于点M N(1) 求直线DE 的解析式和点M 的坐标； (2) 若反比例函数y m (x >0)的图象经过点Mx求该反比例函数的解析式，并通过计算判断点N 是否在该函数的图象上;(3) 若反比例函数y m (x >0)的图象与△MN 有公共点，请直接写出m 的取值范围.x22. (本小题满分12 分)某仪器厂计划制造A B 两种型号的仪器共80套,该公司所筹资金不少于209(万元，但不超过2096万元,且所筹资金全部用于制造仪器，两种型号的制造成本和售价如下表：AB 成本(万兀/套) 25 28 售价(万兀/套)3034(1) 该厂对这两种型号仪器有哪几种制造方案？ (2) 该厂应该选用哪种方案制造可获得利润最大？(3) 根据市场调查，每套B 型仪器的售价不会改变，每套A 型仪器的售价将会提高a 万元(a >0),且 所制造的两种仪器可全部售出，问该厂又将如何制造才能获得最大利润？乙校成绩条形统计图图 11-2y0 C x图1290 , AD = 6, BC = 8, AB 3 3，点 M 是23. (本小题满分12分)在图13-1至图15-3中，直线MN 与线段AB 相交 于点 O /1= Z 2 = 45°.(1) 如图13-1，若AO = 0B 请写出AO 与 BD 的数量关系和位置关系；(2) 将图13-1中的MN 绕点0顺时针旋转得到 图 13-2，其中 AO = 0B 求证：AC = BD AC 丄 BD(3) 将图13-2中的0B 拉长为A0的 k 倍得到 图13-3，求BD 的值.AC24. (本小题满分12分)如图14,在直角梯形ABC 中, AD/BCB BC 的中点•点P 从点M 出发沿MB 以每秒1个单位长的速度向点B 匀速运动，到达点B 后立刻以原速度沿BM 返回；点Q 从点M 出发以每秒1个单位长的速度在射线MC 上匀速运动.在点P, Q 的运动过程中，以PC 为边作等边三角形EPQ 使它与梯形ABC 在射线BC 的同侧•点P, Q 同时出发，当点P 返回到点M 时停止运动，点Q 也随之停止. 设点P, Q 运动的时间是t 秒(t >0). (1) 设PQ 的长为y ,在点P 从点M 向点B 运动的过程中， 写出y 与t 之间的函数关系式(不必写t 的取值范围).(2) 当BP = 1时，求△EPQf 梯形ABC 重叠部分的面积. (3) 随着时间t 的变化，线段AD 会有一部分被△ EP®盖，被覆盖线段的长度在某个时刻会达到最大值，请回答：该最 大值能否持续一个时段？若能，直接写出t 的取值范围； 若不能，请说明理由. • •(备用图)图1425. (本小题满分14分)如图15,抛物线y ax2 bx c(a 0)经过x轴上的两点A^Q)、B(x2,0)和y轴上的点C(0, 3)，e P的圆心P在y轴上，且经过B、C两点，若b ,3a，AB 2-、3 .求：(1)抛物线的解析式；(2)D在抛物线上，且C、D两点关于抛物线的对称轴对称，问直线BD是否经过圆心P ?并说明理由；(3)设直线BD交e P于另一点E，求经过点E和e P的切线的解析式.图15由已知得a22a3，代入上式的原式22320.解：(1) 144; ................... 3分(2)如图1; ........... 6分2011 年数学参考答案、选择题题号123456789101112答案「D C B C A D B C A B B C二、填空题13.、5 14.5 15. 上16.1 17. = 18. C 、B A E、D.10三、解答题19. (1)解：X 1 2(x 1) , X 3 •经检验知，x 3是原方程的解. ............. 8分(2)解：原式[a 2乞^]电'a(a 2) (a 2)2 a 4(a 2)(a 2) a(a 1) a 22a(a 2) a 4a24 a2a a 2a(a 2)2 a 4a 4 a 2a(a 2)2a 41a(a 2)1a22a................. 6分(3) ...................................................................................... 甲校的平均分为8.3分，中位数为7分；.................................................... 8分由于两校平均分相等，乙校成绩的中位数大于甲校的中位数，所以从平均分和中位数角度上判断，乙校的成绩较好. ............ 9分乙校成绩条形统计图12分又•••点N 在BC 边上，B (4, 2), •••点N 的横坐标为4. (4) 因为选8名学生参加市级口语团体赛，甲校得10分的有8人，而乙校得10分的只有5人，所以应选甲校.21•解：(1)设直线DE 的解析式为y kx b , •/点 D , E 的坐标为(0, 3)、(6, 0),3 b, 0 6k b.解得kb 3.y2X 3 .点M 在 ABi 上，B(4, 2),而四边形OAB 是矩形, •••点M 的纵坐标为2.1又•••点M 在直线y3上,21 •2 =2x 3 .• • x = 2.A M(2, 2). ...................... ........... 4分(2)V ym (x >0) 经过点M(2, 2),x• m 4.4 y x.................. 5分•••点N 的横坐标为4. 1•••点N 在直线y —x 3上，2 • y 1 .• N(4, 1). ........... 8 分4•••当 x 4 时，y == 1, x4•••点N 在函数y 的图象上. ............... 9分x(3) 4< m W 8. ......................... 12分 22.解：(1)设A 种型号的仪器造x 套,则B 种型号的仪器造(80-x)套， 由题意得：2090 25x28 80 x 2096解之得：48 x 50 ...................... 2分 所以x=48、49、50三种方案：即：A 型48套, B 型32套；A 型49套，B 型31 套；A 型50套, B 型30套。
2019年四川省成都七中自主招生物理试卷一、单项选择题：本题共6小题，每小题3分，共18分．每小题给出的四个选项中，只有一项符合题目要求．1．（3分）以下说法中正确的是（）A．初中物理课本的宽度大约28cmB．刚参加了中考的小明体积约55dm3C．小明从一楼走到三楼教室克服自身重力大约做了1500J的功D．高空中飞机飞行的速度大约25m/s2．（3分）下列关于物态变化的说法中，正确的是（）A．春天，河里冰雪消融，是升华现象B．夏天，冰棍儿周围冒“白气”，是汽化现象C．秋天，早晨花草上出现的小露珠是熔化现象D．冬天，温暖的车内窗玻璃会变模糊，是因为车内水蒸气液化的缘故3．（3分）如图所示，容器中盛满水，水中放入P和Q两个小球，P球为铁球，Q球为木球，它们用细线分别系于容器的上、下底部，当容器静止时，细线均伸直处于竖直方向，现使容器以一定加速度向右匀加速运动，则此时P、Q两球相对容器（）A．两球均向右偏移B．两球均向左偏移C．P球向右偏移D．Q球向右偏移4．（3分）某人站在离湖岸边8m的C处，刚好能看见湖对岸的一棵树HG在水中的完整的像，如果眼距地面的高度为l.6m，湖两岸均高出湖水面lm。
湖宽50m，则该树HG的高度为（）A．10m B．9m C．8m D．7m5．（3分）小球从高处下落到竖直放置的轻弹簧上（如图甲），小球的速度v和弹簧缩短的长度△x之间的关系如图乙所示，其中A为曲线的最高点。
已知该小球重为2.2N，弹簧在受到撞击至压缩到最短的过程中始终发生弹性形变，弹簧的弹力大小与形变成正比。
下列说法正确的是（）A．从撞击轻弹簧到它被压缩至最短的过程中，小球的重力做功的功率先减小后增大B．从撞击轻弹簧到它被压缩到最短的过程中，小球的机械能先增大后减小C．当小球的速度为5．lm/s时，小球受到的合力为2.2ND．从撞击轻弹簧到弹簧被压缩至最短的时候，小球受到的合力为11.22N6．（3分）在图所示的电路中，当滑动变阻器R的滑片P从B向A滑动的过程中，电压表V1、V2示数的变化量的值分别为△U1、△U2，则它们的大小相比较应该是（）A．△U1＜△U2B．△U1＞△U2C．△U1＝△U2D．因为无具体数据，故无法比较二、不定项选择题：本题共12小题，每小题4分，共48分．每小题给出的四个选项中，有一项或多项符合题目要求．全部选对的得4分，选对但不全的得2分，有选错的得0分．7．（4分）如图所示，是一种漏电保护装置的设计图，图中a为双线并绕的线圈与铁芯，b为衔铁与触片组成开关，C为带锁止钩的金属簧片（锁止钩的作用是：当开关b断开后，需人工复位才能闭合），当线圈中火线与零线中的电流不相等时a具有磁性，吸引衔铁B，断开火线。
2019年四川省成都七中自主招生英语试卷第一部分阅读理解（共两节，满分40分）第一节（共5小题；每小题8分，满分40分）阅读下列短文，从每题所给的四个选项A．B．C和D中，选出最佳选项1．（8分）Brooke has just turned 12，with seven world records，is one of the best rock climbers in the world．Brooke comes from a climbing family．Both her parents are just past climbing champions．Her father stopped climbing some years ago，but her mother，Robyn，who won four world cup titles four years in a row，is still climbing．She runs a club for young climbers in Colorado，USA and coaches Brooke and her teammates．Sometimes it's difficult for mother and daughter to work together so closely，but they really respect and trust each other and Brooke says her mum is a great coach．Brooke says that her mother gives her a lot of good advice and it is a very important part of her climbing life．Robyn can be strict，but she is also passionate about climbing，and she passes this passion on to her students．Robyn says Brooke has very good strong wills and is very good at making herself want to practice．This helps her when she's facing the challenges of this difficult sport．She is also very hard﹣working．Success in rock climbing issomething you have to work at．To be a world﹣class athletes of any kind，you have to push yourself and train hard and that's what Brooke does，every day，at the club and at home．Brooke says that climbing is always there in their lives．They even have a climbing wall in their house！But climbing isn't only hard work．It's fun too．Brooke loves climbing and when she's on a high rock，she feels happy．Strangely，she says that when she looks down，she isn't scared．All she does is to think how cool it is to be small compared to the rock．（1）It's for Brooke and Robyn to work together so closely．A．difficultB．uneasyC．amazingD．annoying（2）To be a successful climber，one has to．A．practice climbing at a very young ageB．have a great mother coach who is very strictC．face any difficult things bravely and keep on training hardD．keep pushing himself and train hard at any time and any place（3）What's the best title of this passage？A．An Excellent ClimberB．A Climbing FamilyC．Climbing ChampionsD．How to Be Successful Rock Climbers（4）Which of the sentences can the author agree with？A．Well begun is half done．B．No pains，no gains．C．He who climbs high falls heavily．D．Doing is better than saying．2．（8分）Papa，as a son of a dirt﹣poor farmer，left school early and went to work in a factory，for education was for the rich then．So，the world became his school．With great interest，he read everything he could lay his hands on，listened to the town elders and learned about the world beyond his tiny hometown．"There's so much to learn．"he'd say．"Though we' re born stupid，only the stupid remain that way．"He was determined that none of his children would be denied an education．Thus，Papa insisted that we learn at least one new thing each day．Though，as children，we thought this was crazy，it would never have occurred to us to deny Papa a request．And dinner time seemed perfect for sharing what we had learned．We would talk about the news of the day；no matter how insignificant，it was never taken lightly．Papa would listen carefully and was ready with some comment，always to the point．Then came the moment﹣the time to share the day's new learning．Papa，at the head of the table，would push back his chair and pour a glass of red wine，ready to listen．"Felice，"he'd say，"tell me what you learned today"．"I learned that the population of Nepal is .."Silence．Papa was thinking about what was said，as if the salvation of the world would depend upon it．"The population of Nepal．Hmm．Well…"he'd say．"Get the map，and let's see where Nepal is．"And the whole family went on a search for Nepal．This same experience was repeated until each family member had a turn．Dinner ended only after we hada clear understanding of at least half a dozen such facts．As children，we thought very little about these educational wonders．Our family，however，was growing together，sharing experiences and participating in one another's education．And by looking at us，listening to us，respecting our input，affirming our value，giving us a sense of dignity，Papa was unquestionably our most influential teacher．Later during my training as a future teacher，I studied with some of the most famous educators．They were imparting what Papa had known all along﹣the value of continual learning．His technique has served me well all my life．Not a single day has been wasted，though I can never tell when knowing the population of Nepal might prove useful．（1）What do we know from the first paragraph？A．The author's father was born in a worker's family．B．Those born stupid could not change their life．C．The town elders wanted to learn about the world．D．The poor could hardly afford school education．（2）It can be learned from the passage that the author．A．enjoyed talking about newsB．knew very well about NepalC．felt regretted about those wasted daysD．appreciated his father's educational technique（3）What is the greatest value of"dinner time"to the author？A．Continual learning．B．Showing talents．C．Family get﹣together．D．Winning Papa's approval．（4）The author's father can be best described as．A．an educator expert at training future teachersB．a parent insistent on his children's educationC．a participant willing to share his knowledgeD．a teacher strict about everything his students did3．（8分）Some of the world's most famous musicians recently gathered in Paris and New Orleans to celebrate the first annual International Jazz Day．UNESCO（United Nations Educational，Scientific and Cultural Organization）recently set April 30 as a day to raise awareness of jazz music，its significance，and itspotential as a unifying voice across cultures．Despite the celebrations，though，in the U．S．the jazz audience continues to shrink and grow older，and the music has failed to connect with younger generations．It's Jason Moran's job to help change that．As the Kennedy Center's artistic adviser for jazz，Moran hopes to widen the audience for jazz，make the music more accessible，and preserve its history and culture．"Jazz seems like it's not really a part of the American appetite．"Moran tells National Public Radio's reporter Neal Conan．"What I'm hoping to accomplish is that my generation and younger start to reconsider and understand that jazz is not black and white anymore．It's actually color，and it's actually digital．Moran says one of the problems with jazz today is that the entertainment aspect of the music has been lost．"The music can't be presented today the way it was in 1908 or 1958．It has to continue to move，because the way the world works is not the same，"says Moran．Last year，Moran worked on a project that arranged Fats Waller's music for a dance party，"just to kind of put it back in the mind that Waller is dance music as much as it is concert music，"says Moran．"For me，it's the re﹣contextualization．In music，where does the emotion lie？Are we，as humans，gaining any insight on how to talk about ourselves and how something as abstract as a Charlie Parker record gets us into a dialogue about our emotions and our thoughts？Sometimes we lose sight that the music has a wider context，"says Moran，"so I want to continue those dialogues．Those are the things I want to foster．"（1）Why did UNESCO set April 30 as International Jazz Day？A．To protect cultural diversity．B．To recognize the value of jazz．C．To remember the birth of jazz．D．To encourage people to study music．（2）What does the underlined word"that"in paragraph 3 refer to？A．Jazz becoming more accessible．B．The jazz audience becoming larger．C．The production of jazz growing faster．D．Jazz being less popular with the young．（3）What can we infer about Moran's opinion on jazz？A．It will disappear gradually．B．It remains black and white．C．It should keep up with the times．D．It changes every 50 years．（4）The best title might be．A．Celebrating the Jazz DayB．The Rise and Fall of JazzC．Exploring the Future of JazzD．The Story of a Jazz Musician4．（8分）What makes us happy？There has long been a notion that money buys happiness．However，although"we really，really tried that fora couple of generations，it didn't work，"said Francine Jay，author of The Joy ofLess，A Minimalist Living Guide：How to Declutter，Organize，and Simplify Your Life．Thanks to a travel﹣inspired revelation（启发），Jay has been happily living a simpler life for 12 years．"I always packed as lightly as possible，and found it exhilarating to get by with just a small carry﹣on bag，"she told CNN．"I thought if it feels this great to travel lightly，how amazing would it be to live this way？I wanted to have that same feeling of freedom in my everyday life．"Jay decided to get rid of all her excess possessions and live with just the essentials．"I wanted to spend my time and energy on experiences，rather than things．"Jay is a follower of a movement called"minimalism（极简主义）"．Growing numbers of people have been attracted to this lifestyle all over the world．They share the same feeling of disappointment with modern life anda desire to live more simply．Minimalists are typically progressive and concerned about the environment，LeahWatkins，a lead researcher at Otago University in New Zealand，told Stuff magazine in March．But many simply experienced unhappiness caused by owning too many possessions．Depression with the materialism of our world isn't new．English romantic poet William Wordsworth summed up how dispiriting this was back in I802，at the beginning of the industrial age，when he wrote，"Getting and spending，we lay waste our powers．"His preference was to go back to nature．Closer to our own times，the hippies（嬉皮士）of the 1960s also sought to"drop out"of modern life．For many minimalists，the key is to unload．Without objects，they"believe people are forced more and more into the present moment and that's where life happens，"wrote Stuff．But does simplicity ever feel like a sacrifice？"It's eliminating the excess﹣unused items，unnecessary purchases﹣from your life．Well，I may have fewer possessions，but I have more space…Minimalism is making room for what matters most，"said Jay．And"the real questions"，according to Duane Elgin，a US social scientist，are"what do you care about？"and"what do you value？"He told CNN．"It's important for people to realize minimalism isn't simply the amount of stuff we consume．It's about our families，our work，our connection with the larger world，our spiritual dimension．It's about how we touch the whole world．It's a way of life．"（1）What was the author's main purpose in writing the text？A．To report on the trend of minimalism．B．To give tips on how to lead a happy life．C．To argue whether money buys happiness．D．To recommend one of Francine Jay's books．（2）What inspired Francine Jay to live a simple life？A．A book she read．B．Her desire to keep up with modern life．C．A follower of minimalism she met on a trip．D．The pleasure she enjoyed from traveling lightly．（3）The underlined word"eliminating"in Paragraph 9 probably means""A．removingB．distinguishingC．acceptingD．improving（4）Which of the following would Duane Elgin probably agree with？A．Minimalism is a healthy lifestyle that is in conflict with modern life．B．Minimalism limits people's freedom to enjoy their lives to the fullest．C．Minimalism enables people to reflect on what truly counts in their lives．D．Minimalism means people have to sacrifice some pleasure to live simply．5．（8分）Did you know that people who live in different parts of China have different habits and preferences？For example，people from southern China prefer to eat vegetables，while people from northern China liketo cat meat．But what causes these differences？According to a new study published in the journal Cell in October，gene variations（基因变异）might be responsible for these differences，Xinhua reported．In the study，researchers from Chinese genome﹣sequencing（基因组测序）firm BGI collected genetic informationfrom 141.431Chinese women．The women came from 31 provinces and comprised 36 ethnic minority groups．The researchers found that there are six gene frequencies that are different among people from both northern and southern China．They found that natural selection has played an important role in the ways that people living in different regions of China have evolved，affecting their food preferences，immunities（免疫性）to illness and physical traits（特点），The New York Times reported．The researchers reported that a variation of the gene FADS2 is more commonly found in northern people than it is in southern people．It helps people metabolize（新陈代谢）fatty acids（脂肪酸），which suggests a diet that is rich in meat．According to Xinhua，this is due to climate differences．Northern China is at a higher latitude（纬度），which means it's cold and dry throughout the whole year．This weather is difficult to grow vegetables in．Therefore，northerners tend to eat more meat．The study also found differences in the immune systems of both groups．Most people in southern China carry thegene CR1，which protects against malaria（疟疾）．This is because malaria was once quite common in southern China．In order to survive，the genes of people in the south evolved to fight against this disease．However，people in the south are also more vulnerable to certain blood﹣borne illnesses，as they lack the genes to stop them．Genes can also cause physical differences between northerners and southerners．Most northerners have the ABCCl1gene，which causes dry earwax（耳屎），less body odor and fewer sweat secretions（分泌），The New York Times reported．These physical differences are also more beneficial to living in cold environments．Southerners are less likely to have this gene，as it did not evolve in their population．（1）What's the new study mainly about？A．The genes of Chinese minority groups．B．Habits of people from different regions of China．C．Physical differences between northerners and southerners．D．Differences in the genes of people from different parts of China．（2）A variation of the gene FADS2 can help our body．A．store fatB．digest meatC．fight diseaseD．control body temperature（3）The underlined word"vulnerable"is closest in meaning to．A．unprotectedB．uninterestedC．generousD．responsible（4）Compared with people in southern China，most northerners．a．are immune to malariab．catch blood﹣borne illnesses easilyc．have less body odord．sweat less frequentlyA．abB．beC．bdD．cd第二节（每小题10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，选项中有两项为多余选项．6．（10分）Public Speaking TrainingGet a coach（1），so get help．Since there are about a billion companies out there all ready to offer you public speaking training and courses，here are some things to look for when deciding the training that's right for you．Focus on positivesAny training you do to become more effective at public speaking should always focus on the positive aspects of what you already do well．Nothing can hurt confidence more than being told that you aren't doing well．（2）．so good public speaking training should develop those instead of telling you what youshouldn't do．（3）If you find a public speaking course that looks as though it's going to give you lots of dos and don'ts，walk away！Your brain is so full of what you're going to be talking about．（4）．As far as we're concerned，there are basically no hard and fast rules about public speaking．Your audience can be your friends．You are a special person not a cloneMost importantly，good public speaking training should treat you as a special one，with your own personal habits．（5）．Your training course should help you bring out your personality，not try to turn you into someone you're not．A．You aren't like anybody elseB．You already do lots of things wellC．Turn your back on too many rulesD．Check the rules about dos and don'tsE．Whatever the presentation，public speaking is toughF．The one thing you don't want is for them to fall asleepG．So trying to force a whole set of rules into it will just make things worse第二部分英语知识运用完形填空（共1题；每小题30分，满分30分）阅读下面短文，从短文后各题所给的四个选项（A．B．C和D），选出可以填入空白处的最佳选项。
成都七中2019年外地生自主招生考试数 学(时间 120 分钟，满分 150 分)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;2.回答选择题时，选出每小题答案后，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 皮擦擦干净后，再选涂其它答案标号．答非选择题时，将答案写在答题卡上，写在本试卷上无效．3.考试结束后，将本试卷和答题卡一并交回.一、选择题(每小题只有一个正确答案，每小题 5 分，共 60 分)1．若M ＝5x 2－12xy ＋10y 2－6x －4y ＋13(x ，y 为实数)，则M 的值一定为（ A ）A ．非负数B ．负数C ．正数D ．零分析：配方:M ＝(2x －3y )2＋(x －3)2＋(y －2)2≥0，当x ＝3，y ＝2取等号. 注意：此类题目要注意几个非负数是否能同时取到等号！比如：M ＝2x 2－4xy ＋5y 2－6x －4y ＋13＝(x －2y )2＋(x －3)2＋(y －2)2中，三个非负数不能同时取等，因此采用这种配方因式不能确定其最小值！正确的配方形式应该是：M ＝2(x －2y ＋32)2＋3(y －53)2＋16≥16，即y ＝53，x ＝196时，M 取得最小值16.2．将一个棱长为 m ( m ＞2且m 为正整数)的正方体木块的表面染上红色，然后切成m 3个棱长为1的小正方体，发现只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的 12 倍，则m 等于（ C ）A ．16B ．18C ．26D ．32分析：只有一个表面染有红色的小正方体的数量为6(m －2)2，恰有两个表面染有红色的小正方体的数量12(m －2)，根据只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的12倍，即可得到m 的值．解：将一个棱长为m (m ＞2且m 为正整数)的正方体木块的表面染上红色，然后切成m 3个棱长为1的小正方体，则只有一个表面染有红色的小正方体的数量为6(m －2)2， 恰有两个表面染有红色的小正方体的数量12(m －2)，∵只有一个表面染有红色的小正方体的数量是恰有两个表面染有红色的小正方体的数量的12倍， ∴6(m －2)2＝12×12(m －2)， 解得m 1＝26，m 2＝2(舍去)．3．已知6a 2－100a ＋7＝0,7b 2－100b ＋6＝0，且ab ≠1，则ab的值为（ D ）A ．503B ．67C ．1007D ．76分析：显然由方程7b 2－100b ＋6＝0,可得6(1b )2－100(1b )＋7＝0，又ab ≠1，∴a ，1b 是方程6x 2－100x ＋7＝0的两个不相等的实数根，∴a b ＝76.注意：此类题目一定要注意是否有类似于“ab ≠1”这样的限制条件！若无，则必须分一元二次方程“有两个相等的实数根”和“有两个不相等的实数根”两种情况讨论！ 4．若a ＝32＋3＋5，b ＝2＋6－10，则ab 的值为（ B ）A ．12B ．14C ．12＋3D ．16＋10分析：∵b ＝2＋6－10＝2(2＋3－5)，∴a b ＝32＋3＋5·12(2＋3－5)＝14．注意：此类分子或分母中含有多个a 的代数和的题目，一般都要习惯性的思考能否利用因式分解的方法进行化简！5．满足｜ab ｜＋｜a －b ｜－1＝0的整数对(a ，b )共有（ C ）A ．4个B ．5个C ．6个D ．7个分析：由｜ab ｜＋｜a －b ｜－1＝0，得｜ab ｜＋｜a －b ｜＝1，∵a ，b 都是整数，∴⎩⎨⎧｜ab ｜＝1，｜a －b ｜＝0．或⎩⎨⎧｜ab ｜＝0，｜a －b ｜＝1．故共有6组解．详解：∵|ab |＋|a －b |＝1， ∴0≤|ab |≤1，0≤|a －b |≤1， ∵a ，b 是整数，∴|ab |＝0，|a －b |＝1或|a －b |＝0，|ab |＝1． ①当|ab |＝0，|a －b |＝1时， Ⅰ，当a ＝0时，b ＝±1，∴整数对(a ，b )为(0，1)或(0，－1)， Ⅱ，当b ＝0时，a ＝±1，∴整数对(a ，b )为(1，0)或(－1，0)， ②当|a －b |＝0，|ab |＝1时， ∴a ＝b ，∴a 2＝b 2＝1，∴a ＝1，b ＝1或a ＝－1，b ＝－1， ∴整数对(a ，b )为(1，1)或(－1，－1)，即：满足|ab |＋|a －b |＝1的所有整数对(a ，b )为(0，1)或(0，－1)或(1，0)或(－1，0)或(1，1)或(－1，－1)． ∴满足|ab |＋|a －b |－1＝0的整数对(a ，b )共有6个．6．在凸四边形ABCD 中，E 为BC 边的中点，BD 与AE 相交于点O ，且BO ＝DO ，AO ＝2EO ，则S △ACD ︰S △ABD 的值为（ D ）A ．25B ．13C ．23D ．12方法一：过点B 作BF ∥AD 交AE 延长线于F ，连接OC ，先证明△FOB ≌△AOD ，再证明△BEF ≌△CEO ，可得AD ∥OC ，可得S △ACD ＝S △AOD ，由S △ABD ＝2S △AOD ，可得S △ACD ︰S △ABD ＝1︰2． 解：如图，过点B 作BF ∥AD 交AE 延长线于F ，连接OC ， ∴∠F ＝∠DAO ．∵BO ＝DO ，∠BOF ＝∠DOA ， ∴△FOB ≌△AOD (AAS )，∴FO ＝AO ． ∵AO ＝2EO ，∴FO ＝2EO ，∴EO ＝EF ． ∵E 为BC 边的中点，∴BE ＝CE ．∵∠BEF ＝∠CEO ，∴△BEF ≌△CEO (SAS )， ∴∠BFE ＝∠COE ，∴BF ∥OC ， ∴AD ∥OC ，∴S △ACD ＝S △AOD ． ∵BD ＝2OD ，∴S △ABD ＝2S △AOD ， ∴S △ABD ＝2S △ACD ， ∴S △ACD ︰S △ABD ＝1︰2． 方法二：连接OC ，∵E 为BC 边的中点，BO ＝DO ， ∴OE 是△BCD 的中位线， ∴OE ∥CD ，且OE ＝12CD ，∴S △ACD ＝S △DOC ＝12S △BDC ＝12×4S △BOE ＝2S △BOE ，∵AO ＝2EO ，∴S △BOE ＝12S △ABO ＝12×12S △ABD ＝14S △ABD ，∴S △ACD ＝2×14S △ABD ＝12S △ABD ，∴S △ACD ︰S △ABD ＝1︰2．7．从1到2019连续自然数的平方和，即12＋22＋32＋…＋20192的个位数字是（ A ）A ．0B ．1C ．5D ．9分析：由公式12＋22＋32＋…＋n 2＝n (n ＋1)(2n ＋1)6，当n ＝2019时，显然尾数为0．另解：由题中可以看出，个位的数字是以10为周期变化的，用2019÷10，计算一下看看有多少个周期即可． 解：以2为指数的幂的末位数字是1，4，9，6，5，6，9，4，1，0依次循环的， ∵2019÷10＝201…9，(1＋4＋9＋6＋5＋6＋9＋4＋1＋0)×201＋(1＋4＋9＋6＋5＋6＋9＋4＋1) ＝45×201＋45 ＝9045＋45 ＝9090，∴12＋22＋32＋42＋…＋20192的个位数字是0．8．已知x ＋y ＋z ＝0，且1x ＋1＋1y ＋2＋1z ＋3＝0，则代数式(x ＋1)2＋(y ＋2)2＋(z ＋3)2的值为（ D ）A ．3B ．4C ．16D ．36分析：设x ＋1＝a ，y ＋2＝b ，z ＋3＝c ，则已知条件可转化为： a ＋b ＋c ＝x ＋y ＋z ＋6＝6，1a ＋1b ＋1c ＝0，由1a ＋1b ＋1c＝0，得ab ＋bc ＋ca ＝0， ∴(x ＋1)2＋(y ＋2)2＋(z ＋3)2＝a 2＋b 2＋c 2＝(a ＋b ＋c )2－2(ab ＋bc ＋ca )＝62－2×0＝36．9．将一枚六个面编号分别为1，2，3，4，5，6的质地均匀的正方体骰子先后投掷两次，记第一次掷出的点数为a ，第二次掷出的点数为b ，则使关于x ，y 的方程组⎩⎨⎧ax ＋by ＝2，2x ＋y ＝3，只有正数解的概率为（ B ）A ．112B ．16C ．518D ．1336分析：首先分两种情况：①当a －2b ＝0时，方程组无解；②当a －2b ≠0时，方程组的解为由a ，b 的实际意义为1，2，3，4，5，6可得．把方程组两式联合求解可得x ＝3b －22b －a ，y ＝4－3a 2b －a ，再由x ，y 都大于0可得x ＝3b －22b －a ＞0，y ＝4－3a2b －a ＞0，求出a ，b 的范围，列举出a ，b 所有的可能结果，然后求出有正数解时，所有的可能，进而求出概率． 解：①当a －2b ＝0时，方程组无解；②当a －2b ≠0时，方程组的解为由a ，b 的实际意义为1，2，3，4，5，6可得． 易知a ，b 都为大于0的整数，则两式联合求解可得x ＝3b －22b －a ，y ＝4－3a2b －a， ∵使x ，y 都大于0则有x ＝3b －22b －a ＞0，y ＝4－3a2b －a＞0， ∴解得a ＜43，b ＞23，或者a ＞43，b ＜23，∵a ，b 都为1到6的整数，∴可知当a 为1时b 只能是1，2，3，4，5，6；或者a 为2，3，4，5，6时b 无解， 这两种情况的总出现可能有6种：(1，1)(1，2)(1，3)(1，4)(1，5)(1，6)， 又掷两次骰子出现的基本事件共6×6＝36种情况，故所求概率为＝636＝16．10．方程3a 2－8a －3b －1＝0，当a 取遍0到5的所有实数值时，则满足方程的整数b 的个数是（ C ）A ．12个B ．13个C ．14个D ．15个方法一：首先将已知条件变形成用含a 的代数式表示b ，然后把含a 的代数式配方，再根据a 的取值求出b 的取值范围，由于是求b 的整数的个数，所以再找b 的取值范围内的整数解即可． 解：∵3b ＝3a 2－8a －1∴b ＝a 2－8a 3－13＝(a －43)2－199，∵0≤a ≤5， ∴－43≤a －43≤113，∴0≤(a －43)2≤1219，∴－199≤(a －43)2－199≤1029，即－199≤b ≤343，∴－219≤b ≤1113，∴整数b ＝－2，－1，0，1，…，11，共14个．方法二：由3b ＝3a 2－8a －1，得b ＝a 2－8a 3－13＝(a －43)2－199，因此b 是关于a的二次函数，其图象是一条抛物线，当a 取遍0到5的所有实数值时，求整数b 的个数就是求b 的最大值与最小值之间的整数的个数．解：作出b ＝a 2－8a 3－13＝(a －43)2－199的图象（草图即可））（注意0≤a ≤5）。
