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首页 福建省南平市第一中学2018年第二批次自主招生实验班考试数学试题

福建省南平市第一中学2018年第二批次自主招生实验班考试数学试题.doc

福建省南平市第一中学2018年第二批次自主招生实验班考试数学试题

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2018-11-06 0人阅读 举报 0 0 暂无简介

简介:本文档为《福建省南平市第一中学2018年第二批次自主招生实验班考试数学试题doc》,可适用于综合领域

福建省南平市第一中学年第二批次自主招生实验班考试数学试题年南平一中第二批次自主招生(实验班)考试数学学科试卷考试时间:分钟满分分就读学校:姓名:考场号:座位号:一、选择题(本大题共小题每小题分共分。每小题只有一个正确答案)pr,p,pqqr实数在数轴上的位置如图化简的值为()p、q、rABCD,pr,p,p,q,pr已知为实常数则下列结论正确的是()ax,,x,A关于的方程的解是B关于的方程的解是ax,aax,axxx,C关于的方程ax,a的解是D关于的方程ax,a的解是xxx,,x,,抛物线的对称轴为直线图象如图y,axbxc(a,)abc,a,b,b,ac所示给出以下结论:a,bc,错误的结论的个数为()ABCD第题图设方程的两根为、若xx(k)xx,,x,xxxk则满足条件的整数的值有()A无数个BCD,,,,,,,,ABCDEADBE如图平行四边形中点在边上以为折痕CD,ABEAF,FDE将向上翻折点正好落在上的点若的周,FCBFC长为的周长为则的长为()第题图A(BCDb,,已知a,b都是实数且则的值为()aaba,b,,,,,,A(或B或CDRt,ABCAC,BCCCD,AB如图在中过作垂足为DAD,BC,,ABC若则的内切圆的面积为(),A(BCD,,,,,第题图,已知是正整数则当函数取得最小值时的值为()y,xxx,ABCD观察下列数的规律:则第个数是(),,,,,,?ABCD::ABCDAB,如图在四边形中B,C,BCAD,则边的长为()CD,,,A(BCD第题图二、填空题(本大题小题每小题分共分)mnmn关于的方程的解为,,m,n,mn,xxx,xx,甲、乙、丙三人在一起做“剪子、布、锤子”游戏约定每个人在每一个回合中只能随机出“剪子、布、锤子”中的一个那么在一个回合中三个人都出“锤子”的概率是ABCDAD,AB,AP矩形中,将该矩形按照如图所示位置放置在直线上然后不滑动的转动当它转动一周时()叫做一次操作则经过次这样的操作顶点A经过的路线A,A长等于,ABCAB,AC,B、CcosB,M在中若以为圆心为半径的圆经过两点则线段的长等于AM三、解答题(本大题小题共分)(本题满分分)将下列式子因式分解:()xxa,a()x,x(本题满分分)x()化简,xxx,()已知用含的式子表示axaxx(本题满分分),x,kx,,k已知函数y其中为实数,,,,xx,kx,,k,()当时在所给的网格内做出该函数图象的简图x,并利用图象求时函数的最大值k()当变化时探究函数图象与x轴的交点个数(本题满分分)ABCD如图正方形的边长为的角平分线交与点ADBDEABEBE()求的值AE()若在线段上运动如图当为何值时的值最小PBDBPEPAP第题图第题图第题备用图(本题满分分)B、CCA如图抛物线与轴交于点与轴交于两点(点在轴正半轴yxxy,axca,,ABCCBA上)为等腰直角三角形且面积为现将抛物线沿方向平移平移后的抛物线过点时EFH与轴的另一个交点为其顶点为对称轴与轴的交点为xx()求的值a、cOF、CFOFE,FCE()连接求证:PPEFHy()在轴上是否存在点当以为直径的圆交直线于点时以点为顶点QP、Q、E,EOPP的三角形与全等若存在求点的坐标若不存在请说明理由第题图第题图第题备用题年南平一中第二批次自主招生(实验班)考试数学学科答案一、选择题(每小题分共分。每小题只有一个正确答案)序号答案BDADCABCDA二、填空题(每小题分共分)mn,或x,m,n三、解答题(本题满分分)解:()=„„„„„„„„„„„„„„„„„分xxa,axax,a()解法一:„„„„„„分x,x,x,x,x,x,x,x,„分,xx,,(x,),(x,),,xx,,(x,)xx,„„„„„„„„„„„„„„„„„„„„„„分,x,x„„„„„„„分解法二:x,xx,x,xx,x,x,„„„„„„„„„„„„„„„„„„„„分,x,xx,„„„„„„„„„„„„„„„„„„„„„„分,x,x((本题满分分)xx,,,解:()„„„„„„„„„„„„„„„„„分xx,x,,xxxx,,x,a()由得且axxxx,a„„„„„„„„„„„„„„„„„„„„„„„„„„分?x,,?x,,x,,x,,a„„„„„„„„„分,,xxx,,xa,,,„„„„„„„„„„分?axxx(结论没有有理化扣分)(本题满分分)()能够正确通过描点画出函数图象给分x,当时函数的最大值为„„„„„„„„„„„分()结合图象可知:k,k,或时函数图象与轴有个交点„„„分xk,k,或时函数图象与轴有个交点„„„分x,k,时函数图象与轴有个交点„„„„„分x(本题满分分)CB解:()延长与交于点„„„„„„„„„„„„„„„„„„„„分DEFADFCADE,EFB„„„„„„„„„„„„„„„„„„分??又„„„„„„„„„分ADE,BDEBDE,EFBBD,BF??„„„„„„„„„„„„„„„„„„„„„„„„„分,ADE,BFE?BEBFBD,,,„„„„„„„„„„„„„„„„„„„„„„分?AEADADACACBD()连接,则与互相垂直平分ECAP,PCBDP连接交于,则EPAP,EPPC,ECEPAP,此时的值最小„„分?BPBEBP,x,DCP?,?,BEP设,DPDCBEBE,由()知„„„„„„„分,?AEDCxx,,解得„„„„„„„分,?,xEPAPBP,,当时的值最小。„„„„„分?OA,cAy)解:(抛物线与轴交于点A,c,c,则y,axca,??,ABCOA,OB,OC,c为等腰直角三角形??c,,c,c,解得(舍负根)„„„„„„分?C,y,ax?a,a,,C,将代入中得得y,axc,a,,综上:„„„„„„„„„„„„„分()设直线的解析式为ABy,kxbb,k,,,将、代入得解得A(,)B(,,),,,kb,b,,,则直线的解析式为„„„„„„„„„„„„„„„„„„„„„分ABy,xC设抛物线沿方向平移平移后的抛物线过点时顶点为BAFF(t,t)?可设平移后抛物线的解析式为y,,x,tt?将代入得解得(舍去)„„„„„分,,tt,C,t,t,平移后抛物线的解析式为y,,x,„„„„„„„„„分F(,)??„„„„„„„„„„„„„„„„„„„„„„„„分OF,,?CH,HEH对称轴与轴的交点为,x??OE,从而,„„„„„„„„„„„„„„„„„„„分OH,,OC,?OE,OFFEO,OFE,„„„„„„„„„„„„„„„„„„„分??FC,FEFCE,FEO又,??OFE,FCE„„„„„„„„„„„„„„„„„„„„„„„„„分?P()假设存在点满足题意PEFH以为直径的圆交直线于点„„„„„„„„„„„分QPQ,QE??:,EOPPOE,若以点为顶点的三角形与全等P、Q、E?则有或者„„„„„„„„„„分EQ,EO,PQ,EO,设分两种情况:P,mQ,nHE,,,n,若EQ,EO,?n,n,,解得此时„„„„分?Q,,PPK,FHK过作直线垂足为则PQ,n,mmn,m,,,由OP,PQ则有化简得?m,n,mn,,,,P,,„„„„„„„„„„„„„„„„„„„„„„„„„分?,,,,若,则,PQ,EO,QE,OP,n,m,()m,,()代入()得„„„„„„分n,?,m,n,m(),m,,代入()得解得„„„„„„„„„„„„分m,P,,?,,,,综上存在满足条件的点坐标为或„„„„„„分P,,,,,,,,

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福建省南平市第一中学2018年第二批次自主招生实验班考试数学试题

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