高二期末检测题
浙江省杭州市2023-2024学年高二下学期数学期末检测试卷(含解析)

浙江省杭州市2023-2024学年高二下学期数学期末检测试卷考生注意:1.答题前,考生务必将自己的姓名、考生号填写在试卷和答题卡上,并将考生号条形码粘贴在答题卡上的指定位置.2.回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑.如需改动,用橡皮擦干净后,再选涂其他答案标号.回答非选择题时,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、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合,则( ){}{}31,1e M x x N x x =-<=<≤M N ⋂=A .B .C .D .{}23x x <≤{}24x x <<{}2e x x <≤{}1e x x <≤2.已知复数,则在复平面内对应的点位于( )i 31i z -=-z A .第一象限B .第二象限C .第三象限D .第四象限3.样本数据的中位数和平均数分别为( )27,30,28,34,35,35,43,40A .34,35B .34,34C .34.5,35D .34.5,344.已知直线与圆有公共点,则的可能取值为( )30kx y k --=22:1O x y +=k A .1B .C .D .131-2-5.在中,角的对边分别是,且,则ABC ,,A B C ,,a b c ()()2sin 2sin 2sin a A b c B c b C=+++( )cos A =A .B .C .D .12-1312236.已知正方体的棱长为为棱的中点,则四面体的体积为1111ABCD A B C D -2,P 1BB 1ACPD ( )A .2B C .D .837.已知,则( )4sin25α=-tan2πtan 4αα=⎛⎫+ ⎪⎝⎭A .4B .2C .D .2-4-8.已知双曲线的上焦点为,圆的圆心位于,且与的22:1C y x -=F A x C 上支交于两点,则的最小值为( ),BD BF DF+A.B CD21-二、多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求,全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知分别是定义域为的偶函数和奇函数,且,设函数()(),f x g x R ()()e xf xg x +=,则( )()()()g x G x f x =()G x A .是奇函数B .是偶函数C .在上单调递减D .在上单调递增R R 10.将函数的图象向左平移个单位长度后,所得的图象关于轴()πsin (0)3f x x ωω⎛⎫=+> ⎪⎝⎭π3y 对称,则( )A .的图象关于直线对称B .的最小值为()f x π3x =ω12C .的最小正周期可以为D .的图象关于原点对称()f x 4π52π3f x ⎛⎫- ⎪⎝⎭11.如图,有一个棱台形的容器(上底面无盖),其四条侧棱均相1111ABCD A B C D -1111D C B A 等,底面为矩形,,容器的深度为,容器壁的厚度忽略11111111m 224AB BC A B B C====1m不计,则下列说法正确的是( )A .1AA =B .该四棱台的侧面积为(2mC .若将一个半径为的球放入该容器中,则球可以接触到容器的底面0.9m D .若一只蚂蚁从点出发沿着容器外壁爬到点A 1C 三、填空题:本题共3小题,每小题5分,共15分.12.的展开式中的系数为 .(用数字作答)712x x ⎛⎫+ ⎪⎝⎭3x 13.已知椭圆的左、右焦点分别为为上一动点,则的取22224:1(0)3x y C a a a +=>12,,F F A C 12AF AF 值范围是.14.已知两个不同的正数满足,则的取值范围是.,a b 33(1)(1)a b a b ++=ab 四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.已知函数()1e 4xf x =(1)求曲线在点处的切线在轴上的截距;()y f x=()()1,1f l y (2)探究的零点个数.()f x 16.如图,在直三棱柱中,为棱上一点,111ABC A BC -12,1,AB BC AC AA M ====1CC 且.1AM BA ⊥(1)证明:平面平面;AMB ⊥1A BC (2)求二面角的大小.B AM C --17.设数列满足,且.{}n a ()122n n na n a +=+14a=(1)求的通项公式;{}n a(2)求的前项和.{}n a n n S 18.在机器学习中,精确率、召回率、卡帕系数是衡量算法性能的重要指标.科研机Q R k 构为了测试某型号扫雷机器人的检测效果,将模拟战场分为100个位点,并在部分位点部署地雷.扫雷机器人依次对每个位点进行检测,表示事件“选到的位点实际有雷”,表示事A B 件“选到的位点检测到有雷”,定义:精确率,召回率,卡帕系数()Q P A B =()R P B A =,其中.1o ee p p k p -=-()()()()()(),o e p P AB P AB p P A P B P A P B =+=+(1)若某次测试的结果如下表所示,求该扫雷机器人的精确率和召回率.Q R 实际有雷实际无雷总计检测到有雷402464检测到无雷102636总计5050100(2)对任意一次测试,证明:.()212Q R QR k Q R P AB +-=-+-(3)若,则认为机器人的检测效果良好;若,则认为检测效果一般;若0.61k <≤0.20.6k <≤,则认为检测效果差.根据卡帕系数评价(1)中机器人的检测效果.00.2k ≤≤k 19.已知抛物线的焦点为,以点为圆心作圆,该圆与轴的正、负半轴分别2:4C y x =F F x 交于点,与在第一象限的交点为.,H G C P (1)证明:直线与相切.PG C (2)若直线与的另一交点分别为,直线与直线交于点.,PH PF C ,M N MN PG T (ⅰ)证明:;4TM TN=(ⅱ)求的面积的最小值.PNT【分析】求得集合,可求{}24M x x =<<M N⋂【详解】因为,{}{}{}3124,1e M x x x x N x x =-<=<<=<≤所以.{}2e M N x x ⋂=<≤故选:C .2.B【分析】根据复数的四则运算和共轭复数的概念,以及复数的几何意义即可求解.【详解】因为,()()()()3i 1i i 342i 2i 1i 1i 1i 2z -++---====----+所以,2i z =-+故在复平面内对应的点为位于第二象限.z (2,1)-故选:B.3.D【分析】先将样本数据按从小到大进行排列,再根据样本数据的中位数、平均数概念公式进行计算即可.【详解】将样本数据按照从小到大的顺序排列可得,27,28,30,34,35,35,40,43故中位数为,343534.52+=平均数为.()12728303435354043348⨯+++++++=故选:D.4.B,求解即可.1≤【详解】由直线与圆有公共点,30kx y k --=22:1O x y +=可得圆心到直线的距离为,()0,0O 30kx y k--=1d =≤解得,所以的取值范围为.k ≤≤k ⎡⎢⎣故选:B.【分析】根据题意,利用正弦定理化简得,结合余弦定理,即可求解.222b c a bc +-=-【详解】因为,()()2sin 2sin 2sin a A b c B c b C =+++由正弦定理得,即,()()2222a b c b c b c=+++222b c a bc +-=-又由余弦定理得.2221cos 22b c a A bc +-==-故选:C.6.A【分析】设与交于点,证得平面,得到,且AC BD O AC ⊥11BDD B 113OPD V S AC =⨯中,结合,即可求解.AC =11BDD B 111111BDD B BOP B OP D P D ODD S S S S S =--- 【详解】设与交于点,在正方形中,,AC BD O ABCD AC BD ⊥又由正方体中,平面,1111ABCD A B C D -1DD ⊥ABCD 因为平面,可得,AC ⊂ABCD 1AC DD ⊥又因为且平面,所以平面,1BD DD D = 1,BD DD ⊂11BDD B AC ⊥11BDD B所以四面体的体积为,且,1ACPD 113OPD V S AC =⨯ AC =在对角面中,可得,11BDD B 111111BDD B BOP B D P OPD ODD S S S S S =-=--所以四面体的体积为.1ACPD 123V =⨯=故选:A.7.D【分析】由已知可得,利用,可求值.251tan tan 2αα+=-tan2tan 4απα⎛⎫+ ⎪⎝⎭22tan 1tan 2tan ααα=++【详解】因为,所以,2222sin cos 2tan 4sin2sin cos tan 15ααααααα===-++251tan tan 2αα+=-所以.2tan22tan 1tan tan 4ααπαα=⨯-⎛⎫+ ⎪⎝⎭221tan 2tan 2tan 41tan (1tan )1tan 2tan ααααααα-===-++++故选:D.8.B【分析】设出圆的方程与双曲线方程联立,可得,进而可得,利用两点1212,x x xx +22121x x +=间距离公式求出,并利用不等式方法求出其最小值.BF DF+【详解】由题可知.设圆,,.(F 22:()2A x a y -+=()11,B x y ()22,D x y 联立,得,则,22221()2y x x a y ⎧-=⎨-+=⎩222210x ax a -+-=212121,2a x x a x x -+==因此,故.()22212121221x x x x x x +=+-=222222121212112213y y x x x x +=+++=++=+=因为,所以,同理可得22111y x -=11BF===-.21DF =-故.)122BF DF yy +=+-又,且,故,从而22123y y +=12,1yy≥1y =≤=2y=≤=.())22121y y -≤所以)122BF DF y y +=+-2=2=2=2≥2==当时,有,,此时1a =()0,1B (D 11BF DF +=-+=所以的最小值是BF DF+故选:B.关键点睛:本题解题关键是由圆的方程与双曲线方程联立得到,再用不等式方法求22121x x +=其最小值.9.AD【分析】根据奇、偶性得到方程组求出、的解析式,从而得到的解析式,再()f x ()g x ()G x 由奇偶性的定义判断的奇偶性,利用导数判断函数的单调性.()G x 【详解】因为①,所以,()()e xf xg x +=()()e xf xg x --+-=即②,联立①②,解得,()()e xf xg x --=()()e e e e ,22x x x xf xg x --+-==所以,定义域为,又,()e e e e x x x x G x ---=+R ()()e e e e x xx xG x G x ----==-+所以是奇函数,又,()G x ()()()()()2222ee e e 40eeeexx x x xx xx G x ----+--=+'=>+所以在上单调递增,故A ,D 正确,B 、C 错误.()G x R 故选:AD10.ABD【分析】根据图象平移判断A ,根据关于直线对称可得判断B ,由周π3x =()132k k ω=+∈Z 期计算可判断C ,可先证明函数关于点对称,再由图象平移判断D.ω()f x 2π,03⎛⎫- ⎪⎝⎭【详解】对于A ,将的图象向左平移个单位长度后,关于轴对称,所以的图()f x π3y ()f x 象关于直线对称,故A 正确;π3x =对于B ,由题可知,解得,又,所以的最小()ππππ332k k ω+=+∈Z ()132k k ω=+∈Z 0ω>ω值为,故B 正确;12对于C ,若最小正周期,则,由B 项可知,不存在满足条件的,故C 错4π5T =2π52T ω==ω误;对于D ,因为,代入,得2π2ππsin 333f ω⎛⎫⎛⎫-=-+ ⎪ ⎪⎝⎭⎝⎭()132k k ω=+∈Z ,()2πsin 2π03f k ⎛⎫-=-= ⎪⎝⎭所以的图象关于点对称,将的图象向右平移个单位长度可以得到()f x 2π,03⎛⎫- ⎪⎝⎭()f x 2π3的图象,2π3f x ⎛⎫- ⎪⎝⎭则对称中心对应平移到坐标原点,故的图象关于原点对称,故D 正确.2π,03⎛⎫-⎪⎝⎭2π3f x ⎛⎫- ⎪⎝⎭故选:ABD 11.BD【分析】由勾股定理即可判断A ,由梯形的面积公式代入计算,即可判断B ,做出轴截面图形代入计算,即可判断C ,将四棱台展开,然后代入计算,即可判断D 【详解】对于A ,由题意可得,故A错误;132AA ==对于B ,梯形11ADD A =所以梯形的面积为11ADD A 242+=梯形,11ABB A=所以梯形的面积为,11ABB A 122+=故该四棱台的侧面积为,故B正确;2⨯=对于C ,若放入容器内的球可以接触到容器的底面,则当球的半径最大时,球恰好与面、面、面均相切,11ADD A 11BCC B ABCD 过三个切点的截面如图(1)所示,由题意可知棱台的截面为等腰梯形,较长的底边上的底角的正切值为,则,12212=-tan 2MPN ∠=-由于互补,故,,MPN MON ∠∠tan 2MON ∠=则,所以,从而球的半径为22tan 21tan MOPMOP ∠=-∠tanMOP ∠=,0.9=<所以将半径为的球放入该容器中不能接触到容器的底面,故C 错误;0.9cm对于D ,将平面与平面展开至同一平面,ABCD 11DCC D 如图(2),则,1AC ==将平面与平面展开至同一平面,如图(3),ABCD 11BCC B 则,145333044AC ⎛=+=< ⎝D 正确.故选:BD难点点睛:解答本题的难点在于选项D 的判断,解答时要将空间问题转化为平面问题,将几何体侧面展开,将折线长转化为线段长,即可求解.12.672【分析】利用二项式定理,求得二项展开式中的通项,把含x 的进行幂运算合并,然后令指数等于3,即可求解.【详解】因为通项为,令,得,712x x ⎛⎫+ ⎪⎝⎭77721771C (2)2C rr r r r rr T x x x ---+⎛⎫== ⎪⎝⎭72r 3-=2r =所以的系数为.3x 72272C 672-=故672.13.1,33⎡⎤⎢⎥⎣⎦【分析】先根据椭圆、、之间的关系,求出,再根据椭圆的定义,把换成a b c 12c a=1AF ,最后根据,代入即可.22a AF -[]2,AF a c a c ∈-+【详解】设椭圆的半焦距为,则,C (0)c c >12c a==,12222221AF a AF aAF AF AF -==-因为,即,[]2,AF a c a c ∈-+213,22AF a a ⎡⎤∈⎢⎥⎣⎦所以,即.2211,33a AF ⎡⎤-∈⎢⎥⎣⎦121,33AF AF ⎡⎤∈⎢⎥⎣⎦故答案为.1,33⎡⎤⎢⎥⎣⎦14.10,4⎛⎫⎪⎝⎭【分析】本题将条件式化简后结合基本不等式得出关于ab 的不等式,再构造函数并利用函数的单调性求解即可.【详解】将两边展开,33(1)(1)a b a b ++=得到,22113333a a b b a b +++=+++从而,()()221130ab a b a b ⎛⎫-+-+-= ⎪⎝⎭故,而,()130a b a b ab ⎛⎫-++-= ⎪⎝⎭a b¹故,又,130a b ab ++-=00a b >,>故,133a b ab =++>从而.321+<设函数,则,()3223g x x x=+112gg ⎛⎫<= ⎪⎝⎭观察易得在,()g x ()0,∞+12<又,所以.0,0a b >>104ab <<故答案为.10,4⎛⎫ ⎪⎝⎭关键点点睛:本题考查函数与不等式的综合,其关键是利用均值不等式构造关于ab 的不等式,再构造函数并利用函数的单调性解决问题.321+<()3223g x x x =+15.(1)12-(2)有两个零点()f x【分析】(1)求得,,利用导数的几何意()1e 4x f x '=()e 1142f ='-()e 114f =-义,求得切线方程,进而求得其在轴上的截距;y(2)得到在上递增,结合,得到,()1e 4x f x '=()0,∞+()10,104f f ⎛⎫ ⎪⎝⎭''01,14x ⎛⎫∃∈ ⎪⎝⎭使得,进而求得单调性,结合零点的存在性定理,即可求解.()00f x '=()f x【详解】(1)解析:由函数,可得,()1e 4x f x =()1e 4x f x '=()e 1142f ='-又,所以的方程为,即,()e 114f =-l ()e 1e 11424y x ⎛⎫=--+- ⎪⎝⎭e 11422y x ⎛⎫=-- ⎪⎝⎭令,可得,所以直线在轴上的截距为.0x =12y =-l y 12-(2)解:因为和上均单调递增,1e 4x y =y =()0,∞+所以在上单调递增,()1e 4x f x '=()0,∞+又因为,所以,使得,()141111e 10,1e 04442f f ⎛⎫=-=''- ⎪⎝⎭01,14x ⎛⎫∃∈ ⎪⎝⎭()00f x '=所以,当时,,在单调递减;()00,x x ∈()0f x '<()f x ()00,x 当时,,在单调递增,()0,x x ∞∈+()0f x '>()f x ()0,x ∞+又因为,()()14100111e 1e 0,110,4e 2010041044f f f ⎛⎫=->=-=- ⎪⎝⎭所以有两个零点.()f x 方法点睛:已知函数零点(方程根)的个数,求参数的取值范围问题的三种常用方法:1、直接法,直接根据题设条件构建关于参数的不等式(组),再通过解不等式(组)确定参数的取值范围2、分离参数法,先分离参数,将问题转化成求函数值域问题加以解决;3、数形结合法,先对解析式变形,在同一平面直角坐标系中作出函数的图象,然后数形结合求解.结论拓展:与和相关的常见同构模型e xln x①,构造函数或;e ln e ln e ln a a a a b b b b ≤⇔≤()lnf x x x =()e xg x x =②,构造函数或;e e ln ln e ln a a a b b a b b <⇔<()ln x f x x =()e x g x x =③,构造函数或.e ln e ln e ln a a a a b b b b ±>±⇔±>±()lnf x x x =±()e xg x x =±16.(1)证明见解析(2)4π【分析】(1)由线面垂直得到,结合勾股定理逆定理得到,证明出1AA BC ⊥BC AC ⊥平面,得到,结合题目条件证明出平面,得到面面垂直;BC⊥11AA C C AMBC ⊥AM ⊥1A BC (2)建立空间直角坐标系,设点,根据向量垂直得到方程,求出()0,0,M a ,进而求出平面的法向量,得到二面角的余弦值,得到答案.a M ⎛=⎝【详解】(1)在直三棱柱中,平面,111ABC A B C -1AA ⊥ABC ∵平面,BC ⊂ABC ∴,1AA BC ⊥∵2,1,AB BC AC ===∴,222AB AC BC =+∴,BC AC ⊥,平面,1AC AA A⋂=1,AC AA ⊂11AA C C ∴平面.BC ⊥11AA C C 平面,AM ⊂ 11AA C C ∴,AM BC ⊥,平面,11,AM A B A B BC B ⊥= 1,A B BC ⊂1A BC ∴平面.AM ⊥1A BC 又平面,AM ⊂AMB平面平面.∴AMB ⊥1A BC (2)由(1)可知两两垂直,1,,CA CB CC 如图,以点为坐标原点,所在直线分别为轴、轴、轴建立空间直角坐标C 1,,CA CB CC x y z 系,Cxyz 则.())()10,0,0,,,0,1,0C AAB设点,()0,0,M a 则.()()()1,,0,1,0,AM a BA CB AB ==-==,解得.11,30AM BA AM BA ⊥∴⋅=-+=a M ⎛=∴ ⎝设平面的法向量为,AMB (),,m x y z =则可取.0,0,m AM z m AB y ⎧⋅==⎪⎨⎪⋅=+=⎩(m = 易知为平面的一个法向量.()0,1,0n CB ==AMCcos ,m n m n m n ⋅〈〉===⋅故由图可知二面角的大小为.B AM C --4π17.(1)()12nn a n n =+⋅(2)()21224+=-+⋅-n n S n n【分析】(1)由已知可得,累乘法可求的通项公式;()122n n n a a n ++={}n a (2)由(1)可得,利用错位相减法可求的前()1212223212nn S n n =⨯⨯+⨯⨯+++⋅ {}n a 项和.n n S 【详解】(1)由题易知,且,0n a ≠()122n n n a a n ++=所以,()2341231212324251231n n n a a a a a a a a n -+⨯⨯⨯⨯⨯⨯⨯=⨯⨯⨯⨯- 所以,()()121121212n n n n n a n n a --+⋅==+⋅⨯所以也满足该式,()112,n n a n n a =+⋅所以.()12nn a n n =+⋅(2),①()1212223212nn S n n =⨯⨯+⨯⨯+++⋅ ,②()()2121221212n n n S n n n n +=⨯⨯++-⋅++⋅ ②-①,得.()()11212212222n n n S n n n +=+⋅-⨯⨯+⨯++⋅ 设,③1212222nn T n =⨯+⨯++⋅ 则,④()23121222122n n n T n n +=⨯+⨯++-⋅+⋅ ④-③,得,()()()1121112222222122n n n n n n T n n n ++++=⋅-+++=⋅--=-+ 所以.()()()1121122124224n n n n S n n n n n +++=+⋅--⋅-=-+⋅-18.(1);.0.625=Q 0.8R =(2)证明见解析(3)0.32【分析】(1)利用条件概率的计算公式计算即可;(2)由条件概率与互斥事件的概率公式证明即可;(3)由(2)计算出的值,判断机器人的检测效果即可.k 【详解】(1),()()()400.62564P AB Q P A B P B ====.()()()400.850P AB R P B A P A ====(2),()()()()()()1111111o e oe e P AB P AB p p p k p p P A P B P A P B ----==-=-----要证明,()212Q R QR k Q R P AB +-=-+-需证明.()()()()()()()1221P AB P AB Q R QR Q R P AB P A P B P A P B --+-=+---等式右边:()()()()()()()()||2||22||2P A B P B A P A B P B A Q R QR Q R P AB P A B P B A P AB +-+-=+-+-.()()()()()()()()()()()()()22P AB P AB P AB P AB P B P A P B P A P AB P AB P AB P B P A +-⨯⨯=+-()()()()()()()22P A P B P AB P A P B P A P B +-=+-等式左边:因为,()()()()()1P A B P AB P A P B P AB ⋃=-=+-所以()()()()()()()()()()()()()121111P AB P AB P A P B P AB P A P B P A P B P A P B P A P B --+-=⎡⎤⎡⎤------⎣⎦⎣⎦.()()()()()()()22P A P B P AB P A P B P A P B +-=+-等式左右两边相等,因此成立.()212Q R QRk Q R P AB +-=-+-(3)由(2)得,因为,0.6250.820.6250.810.320.6250.820.4k +-⨯⨯=-=+-⨯0.20.320.6<<所以(1)中机器人的检测效果一般.19.(1)证明见解析(2)(ⅰ)证明见解析;(ⅱ)163【分析】(1)根据题意,表示出直线的方程,然后与抛物线方程联立,由即可证明;PG Δ0=(2)(ⅰ)根据题意,设直线的方程为,与抛物线方程联立,即可得到点的PF 1x ty =+,N H 坐标,从而得到直线的方程,再与抛物线方程联立,即可得到点的坐标,再结合相似PH M 三角形即可证明;(ⅱ)由条件可得,再由代入计算,即可43PNTPNES S =△△12PNES EP EN = 证明.【详解】(1)由题意知,()1,0F 设,则,()2,2(0)P n n n >21PF n =+所以,所以,21GF FH n ==+()2,0G n -所以直线的斜率为,方程为.PG 1n ()21y x n n =+联立方程得,()221,4,y x n n y x ⎧=+⎪⎨⎪=⎩22440y ny n-+=因为,所以直线与相切.Δ0=PG C (2)(ⅰ)设直线的方程为,PF 1x ty =+由可得,则,又因为,所以.24,1,y x x ty ⎧=⎨=+⎩2440y ty --=4P N y y =-()2,2P n n 212,N n n ⎛⎫- ⎪⎝⎭由(1)知,点,直线的斜率为,方程为,()22,0H n +PH n -()22y n x n=---由得,由,()224,2,y x y n x n ⎧=⎪⎨=---⎪⎩224480y y n n +--=248P M y y n =--得.22444,2M n n n n ⎛⎫++-- ⎪⎝⎭作,垂足为,则,直线的方程为,NE PG ⊥E EN PM ∥EN 212y n x n n ⎛⎫=---⎪⎝⎭将直线与的方程联立,得解得.EN PG ()2212,1,y n x n n y x n n ⎧⎛⎫=--- ⎪⎪⎪⎝⎭⎨⎪=+⎪⎩11,E n n ⎛⎫-- ⎪⎝⎭所以,所以,2211441,,4,4EN n PM n n n n n ⎛⎫⎛⎫=+--=+-- ⎪ ⎪⎝⎭⎝⎭ 4PM EN =由相似三角形的性质可得.4TM TN=(ⅱ)由(ⅰ)知,所以,故,4TM TN=4TP TE=43PNT PNES S =△△因为,221111,,1,EP n n EN n n n n ⎛⎫⎛⎫=++=+-- ⎪ ⎪⎝⎭⎝⎭ 所以(当且仅当时等号成立),()323311114222PNEn S EP EN n n n +⎛⎫===+≥ ⎪⎝⎭ 1n =故,即的面积的最小值为.41633PNT PNES S =≥△△PNT 163方法点睛:利用韦达定理法解决直线与圆锥曲线相交问题的基本步骤如下:(1)设直线方程,设交点坐标为;()()1122,,,x y x y (2)联立直线与圆锥曲线的方程,得到关于(或)的一元二次方程,必要时计算;x y ∆(3)列出韦达定理;(4)将所求问题或题中的关系转化为、(或、)的形式;12x x +12x x 12y y +12y y (5)代入韦达定理求解.。
河南省郑州一中2024学年物理高二第二学期期末教学质量检测试题(含解析)

河南省郑州一中2024学年物理高二第二学期期末教学质量检测试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、单项选择题:本题共6小题,每小题4分,共24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、如图所示两半径为r的圆弧形光滑金属导轨置于沿圆弧径向的磁场中,磁场所在的平面与轨道平面垂直。
导轨间距为L,一端接有电阻R,导轨所在位置处的磁感应强度大小均为B,将一质量为m的金属导体棒PQ从图示位置(导轨的半径与竖直方向的夹角为)由静止释放,导轨及金属棒电阻均不计,下列判断正确的是A.导体棒PQ有可能回到初始位置B.导体棒PQ第一次运动到最低点时速度最大C.导体棒PQ从静止到最终达到稳定状态,电阻R上产生的焦耳热为D.导体棒PQ由静止释放到第一次运动到最低点的过程中,通过R的电荷量2、如图所示,一沙袋用无弹性轻细绳悬于O点.开始时沙袋处于静止,此后弹丸以水平速度击中沙袋后均未穿出.第一次弹丸的速度为v1,打入沙袋后二者共同摆动的最大摆角为30°.当他们第1次返回图示位置时,第2粒弹丸以水平速度v2又击中沙袋,使沙袋向右摆动且最大摆角仍为30°.若弹丸质量是沙袋质量的倍,则以下结论中正确的是( )A.v1:v2=41:42 B.v1:v2=41:83C.v2=v1D.v1:v2=42:413、用一个水平拉力F拉着一物体在水平面上绕着O点做匀速圆周运动.关于物体受到的拉力F和摩擦力f的受力示意图,下列四个图中可能正确的是()A.B.C.D.4、如图所示,两相同的木块A、B静止在水平面上,它们之间的距离为L.今有一颗子弹以较大的速度依次射穿了A、B,设子弹射穿A木块的过程中所用时为t A,子弹射穿B木块的过程中所用时间为t B.A、B停止时,它们之间的距离为s,整个过程中A、B没有相碰.子弹与木块的作用力恒定,则( )A.s=L, t A= t B B.s>L,t A<t BC.s<L,t A>t B D.s<L,t A<t B5、在平直公路上有甲、乙两辆汽车从同一位置沿着同一方向运动,它们的速度-时间图象如图所示,则( )A.甲、乙两车同时从静止开始出发B.在t=2s时乙车追上甲车C.在t=4s时乙车追上甲车D.甲、乙两车在公路上可能相遇两次6、在地球赤道上空,沿东西方向水平放置一根通以由西向东的直线电流,则此导线受到的安培力方向()A.竖直向上B.竖直向下C.由南向北D.由西向东二、多项选择题:本题共4小题,每小题5分,共20分。
福建省宁德市2023-2024学年高二下学期期末质量检测化学试题(有答案)

宁德市2023-2024学年第二学期期末高二质量检测化学试题(考试时间:75分钟 试卷总分:100分)注意:1.本学科试卷分试题卷和答题卡两部分。
试题卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)(共7页14题),全部答案必须按要求填在答题卡的相应答题栏内,否则不能得分。
2.相对原子质量:H-1 C-12 O-16 Ca-40 Ti-48第Ⅰ卷 选择题(共40分)本题包括10小题,每小题4分,共40分。
每小题只有一个选项符合题意。
1.闽东非物质文化遗产名录包含丰富的传统文化和技艺。
下列说法正确的是( ) A .福安线面制作技艺——线面主要成分是淀粉,与纤维素互为同分异构体 B .屏南红釉黄酒酿造技艺——葡萄糖在酒化酶的作用下发生水解反应生成乙醇 C .周宁硋器制作技艺——原料中的二氧化硅晶体是共价晶体D .闽东畲族服饰制作技艺——布料染色用的蓝靛(分子式为161022C H N O )是烃类物质 2.肼(24N H )是航天发射常用的高能燃料,可通过反应32422NH NaClO N H NaCl H O +++制备。
下列说法正确的是( )A .3NH 的电子式:B .24N H 的结构式:C .2H O 的VSEPR 模型:D .用电子式表示NaCl 的形成过程:3.化学推动着社会的进步和人类的发展,下列有关说法错误的是( ) A .液晶是液体和晶体的混合物,可用作电子表、电视的显示屏B .α−氰基丙烯酸正丁酯对人体几乎无毒性,能与人体组织强烈结合,常用作医用胶C .利用红外光谱实验可确定青蒿素分子中含有的部分基团D .利用冠醚不同大小的空腔适配不同大小的碱金属离子,可进行分离、传递特定离子 4.设A N 为阿伏加德罗常数的值,下列说法正确的是( ) A .264.6gC H O 中含有C H −键的数目一定为A 0.6NB .1mol 乙二醇(22HOCH CH OH )中含3sp 杂化的原子数为A 4NC .标准状况下,311.2LCHCl 含有的分子数为A 0.5ND .1mol 甲酸和足量乙醇充分反应后,生成甲酸乙酯的分子数为A N 5.反应()23222CO 2NH H O CO NH ++高压△可应用于工业合成尿素。
河南省部分重点高中2023-2024学年高二上学期期末质量检测语文试题(含答案)

2023~2024学年度高二1月质量检测语文全卷满分150分,考试时间150分钟。
注意事项:1.答题前,先将自己的姓名、准考证号填写在试卷和答题卡上,并将条形码粘贴在答题卡上的指定位置。
2.请按题号顺序在答题卡上各题目的答题区域内作答,写在试卷、草稿纸和答题卡上的非答题区域均无效。
3.选择题用2B铅笔在答题卡上把所选答案的标号涂黑;非选择题用黑色签字笔在答题卡上作答;字体工整,笔迹清楚。
4.考试结束后,请将试卷和答题卡一并上交。
一、现代文阅读(35分)(一)现代文阅读I(本题共5小题,19分)阅读下面的文字,完成1~5题。
①赵树理和汪曾祺的作品都有地地道道的中国味,赵树理的“山药蛋味”和汪曾祺的“高邮味”都有丰富的内涵和深厚的文化底蕴。
考察赵、汪作品两种中国味的丰富内涵,特别是发掘它们的文化底蕴,就显得十分有必要。
②赵树理由于从小受民间故事、鼓词等中国俗文化的熏陶,形成了以“俗”为主的大众化的审美心理结构和审美趣味,而且在创作理念上又承诺毕生做“地摊文学家”为农民写作,从而使作品的艺术结构形式具有一个突出特点:故事化。
而汪曾祺则与之恰好相反,他的小说的结构形式是散文化,他指出:“我不善于讲故事,也不喜欢太像小说的小说,即故事性很强的小说。
故事性太强了,我觉得就不太真实。
”这样的理论主张和创作实践,使他的小说更接近于散文,或者说是一种介乎散文与小说之间的新品种。
汪曾祺在小说创作上的这种现实主义的美学主张,与他受到了中国雅文化中庄子散文的影响,接受了鲁迅、契诃夫等大师的美学主张直接关联。
③赵小说的故事化,主要表现在小说故事性强,甚至故事套故事,有悬念,一个故事套一个故事地写下去,脉络格外分明。
为此,常用保留关节的方法,按下一个故事暂不表另起一个头绪,使用“扣子”制造悬念,吸引读者非一口气卒读不可。
而汪的小说结构与赵的相反,简直与散文区别不大,完全是散文化的结构。
他很重视小说的气氛,并认为只要写出了气氛可以不讲故事,没有情节,甚至不必直接写人物的性格、心理活动。
2024学年河南安阳市林虑中学高二化学第二学期期末达标检测试题(含解析)

2024学年河南安阳市林虑中学高二化学第二学期期末达标检测试题考生请注意:1.答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内,不得在试卷上作任何标记。
2.第一部分选择题每小题选出答案后,需将答案写在试卷指定的括号内,第二部分非选择题答案写在试卷题目指定的位置上。
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考试结束后,请将本试卷和答题卡一并交回。
一、选择题(每题只有一个选项符合题意)1、下列物质属于有机物的是()A.Na2CO3B.CaC2C.CO(NH2)2D.CO2、有机物H是一种广谱高效食品防腐剂,如图所示是H分子的球棍模型,下列有关说法正确的是A.有机物H的分子式为C9H11O3B.有机物H的官能团有羟基、羰基和醚键C.有机物H能发生加成反应、取代反应和氧化反应D.1mol有机物H最多可与1molNaOH反应3、用VSEPR 模型预测下列分子或离子的立体结构,其中正确的是()A.H2O 与BeCl2为V 形B.CS2与SO2为直线形C.SO3与CO32-为平面三角形D.BF3与PCl3为三角锥形4、下列关于元素性质的有关叙述中不正确...的是A.C、N、O、F的原子半径依次减小B.KOH、NaOH、LiOH的碱性逐渐减弱C.P、S、Cl、Ar的最高正价依次升高D.Na、Mg、Al、Si的最外层电子数依次增加5、下列有关铁及其化合物的说法中正确的是()A.铁锈的主要成分是Fe3O4B.铁与水蒸气在高温下的反应产物为Fe2O3和H2C.除去FeCl2溶液中的FeCl3杂质可以向溶液中加过量铁粉,然后过滤D.因为Fe3+具有氧化性,所以可用来做净水剂6、某同学配制一定物质的量浓度的NaOH溶液时,造成所配制溶液浓度偏高的原因是()A.转移时容量瓶未干燥B.定容时俯视刻度线C.没有洗涤烧杯和玻璃棒D.用带游码的托盘天平称2.4 g NaOH时误用了“左码右物”方法7、下表中根据实验操作和现象所得出的结论正确的是()选项实验操作实验现象结论A 蘸有浓氨水的玻璃棒靠近浓X溶液有白烟产生X一定是HClB 将Cl2持续通入淀粉-KI溶液中溶液先变蓝色,后褪色氯气先表现氧化性,后表现漂白性C向强酸性溶液Y中加入Ba(NO3)2溶液,静置后再加入铁氰化钾溶液只有白色沉淀产生Y中一定含有SO42-,一定不含Fe2+D将无色气体X通入品红溶液中,然后加热开始时品红溶液褪色,加热后又恢复红色X中一定含有SO2A.A B.B C.C D.D8、下列化合物中的所有碳原子可以在同一平面上的是()A.B.C.CH2=CHCH=CHCH=CH2D.(CH3)2CH-C≡C-CH=CHCH39、化学使人们的生活越来越美好,下列过程没有涉及化学反应的是A.用四氯化碳萃取碘水中的碘B.用氯化铁溶液刻制印刷电路C.用漂白粉漂白纸张D.用油脂和氢氧化钠溶液制取肥皂10、将一块铝箔用砂纸打磨表面后,放置一段时间,在酒精灯上加热至熔化,下列说法正确的是( )A.熔化的是铝B.熔化的是Al2O3C.熔化物滴落D.熔化物迅速燃烧11、习主席在十九大报告中指出:“绿水青山就是金山银山。
辽宁省盘锦市高级中学2025届高二化学第二学期期末检测试题含解析

辽宁省盘锦市高级中学2025届高二化学第二学期期末检测试题请考生注意:1.请用2B铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。
写在试题卷、草稿纸上均无效。
2.答题前,认真阅读答题纸上的《注意事项》,按规定答题。
一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项)1、下列各组离子能大量共存,当加入相应试剂后会发生化学变化,且发生反应的离子方程式正确的是选项离子组加入试剂加入试剂后发生的离子方程式A Fe2+、NO3-、NH4+NaHSO4溶液3Fe2++NO3-+4H+═3Fe3++NO↑+2H2OB Ca2+、HCO3-、Cl-少量NaOH溶液Ca2++2HCO3-+2OH-═2H2O+CaCO3↓+CO32-C AlO2-、HCO3-通入少量CO22AlO2-+3H2O+CO2═2Al(OH)3↓+CO32-D NH4+、Al3+、SO42-少量Ba(OH)2溶液2NH4++SO42-+Ba2++2OH═BaSO4↓+2NH3•H2OA.A B.B C.C D.D2、下列各组物质中,均为强电解质的是A.NaOH、KNO3、BaSO4B.NH4Cl、CH3COOH、K2SC.NH3·H2O、KOH 、CaCO3D.HClO、NaCl、KOH3、化学与生活、生产密切相关。
下列说法错误的是A.MgO的熔点很高,可作优良的耐火材料,工业上也用其电解冶炼镁B.明矾水解形成的胶体能吸附水中悬浮物,可用于水的净化C.燃放的焰火是某些金属元素焰色反应所呈现出来的色彩D.铝比铁活泼,但铝制品比铁制品在空气中耐腐蚀4、下列离子方程式书写正确的是A.H2S 溶液显酸性:H2S+2H2O2H3O++S2-B.将少量二氧化硫通入次氯酸钠溶液中:SO2+H2O+3ClO-=SO42-+Cl-+2HC1OC.向Ca(HCO3)2溶液中加入澄清石灰水:Ca2++2HCO3-+2OH-=CaCO3↓+CO32-+2H2OD.向FeI2溶液中通入少量Cl2:2Fe2++Cl2=2Fe3++2Cl-5、某气体的摩尔质量为M g·mol-1,N A表示阿伏加德罗常数,在一定的温度和压强下,体积为V L的该气体所含有的分子数为x 。
郑州第一中学2024学年高二化学第一学期期末教学质量检测试题含解析

郑州第一中学2024学年高二化学第一学期期末教学质量检测试题注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
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一、选择题(每题只有一个选项符合题意)1、已知CuS、PbS、HgS 的溶度积分别为l.3xl0﹣36、9.0x10﹣29、6.4x10﹣23.下列推断不正确的是A.向含Pb2+、Cu2+、Hg2+的浓度均为0.010 mol•L﹣1的溶液中通人硫化氢气体,产生沉淀的顺序依次为PbS、CuS、HgSB.在硫化铅悬浊液中滴几滴硝酸铜溶液,会生成硫化铜C.在含Hg2+、Cu2+、Pb2+的溶液中滴加硫化钠溶液,当c(S2﹣)=0.001 mol•L﹣1时三种金属离子都完全沉淀D.硫化钠是处理废水中含上述金属离子的沉淀剂2、25 ℃时,将1.0 L w mol·L-1的CH3COOH溶液与0.1 mol的NaOH固体混合,充分反应后向混合液中通(加)入HCl 气体或NaOH固体。
溶液pH随通(加)入HCl或NaOH的物质的量的变化如下图。
下列叙述正确的是()A.a、b、c对应的混合液中,水的电离程度由大到小的顺序是a>b>cB.若忽略体积变化,则25 ℃时,CH3COOH的电离平衡常数K=×10-7 mol·L-1C.加入NaOH过程中,逐渐减小D.加入NaOH过程中,c(Na+)·c(CH3COO-)逐渐减小3、一种植物生长调节剂的分子结构如图所示。
下列说法不正确的是( )A.该物质含有3种官能团B.该物质属于有机物C.该物质属于多官能团化合物 D.该物质属于脂环烃4、下列装置中,能构成原电池的是()A.B.C.D.5、下列事实不能用勒夏特列原理解释的是A.红棕色的NO2加压后颜色先变深后变浅B.温度过高对合成氨不利C.钢铁在潮湿的空气中容易生锈D.常温下,将1mLpH=3的醋酸溶液加水稀释至l00mL,测得其pH<56、能够快速、微量、精确的测定出有机物的相对分子质量的物理方法是( )A.质谱法B.红外光谱法C.紫外光谱法D.核磁共振氢谱法7、下列说法或表示方法不正确的是()A.盖斯定律实质上是能量守恒定律的体现B.在稀溶液中:H+(aq)+OH-(aq)=H2O(l);△H = –57.3kJ/mol,若将含0.5mol H2SO4的浓硫酸与含1molNaOH的溶液混合,放出的热量大于57.3kJC.由C(石墨)→C(金刚石);△H = +73 kJ/mol,可知石墨比金刚石稳定D.在101kPa时,1molH2完全燃烧生成液态水,放出285.8kJ热量,氢气燃烧的热化学方程式表示为:2H2(g)+O2(g)=2H2O(l);△H = –285.8kJ/mol8、在含有浓度均为0.01mol•L-1的Cl-、Br-、I-离子的溶液中,缓慢加入AgNO3稀溶液,析出三种沉淀的先后顺序是()A.AgCl、AgBr、AgI B.AgI、AgBr、AgClC.AgBr、AgCl、AgI D.三种沉淀同时析出9、某溶液中存在的离子有Na+、H+、S2-、HS-、OH-,则下列有关该溶液的叙述错误的是A.离子浓度可能是c(Na+) >c(S2-)>c(OH-) >c(HS-) >c(H+)B.组成该溶液的溶质只能是Na2S 或NaHS 或二者的混合物C.离子物质的量的关系是2n(S2-) +n(HS-)+n(OH-)=n(Na+)+n(H+)D.该溶液可能呈中性、碱性或酸性10、在由水电离的c(H+)=10-14mol·L-1的溶液中,一定能大量共存的离子组是A.K+、Fe2+、S2-、MnO4-B.Al3+、Cl-、NH4+、SO42-C.Na+、NO3-、Cl-、SO42-D.Na+、AlO2-、Br-、Cl-11、暖宝宝贴有散热均衡、发热持久、直接撕开外包装便可发热等优点,它能保持51~60℃约12个小时,它采用铁的“氧化放热”原理,利用微孔透氧的技术,采用完全隔绝空气的方式,使其发生原电池反应,铁粉在原电池中( )原材料名:铁粉、水、活性炭、蛭石、食盐A.作正极B.得到电子 C.被还原D.作负极12、将金属钠放在燃烧匙中加热使其燃烧,不可能出现的现象是( )A.金属钠熔化为小球B.火焰呈黄色C.燃烧后得到白色固体D.燃烧后得到淡黄色固体13、下列说法中,不正确的是A.用惰性电极电解饱和食盐水或熔融氯化钠时,阳极的电极反应式均为2Cl--2e-===Cl2↑B.酸性介质或碱性介质的氢氧燃料电池的正极反应式均为O2+2H2O+4e-===4OH-C.精炼铜和电镀铜时,与电源负极相连的电极反应式均为Cu2++2e-===CuD.钢铁发生吸氧腐蚀和析氢腐蚀的负极反应式均为Fe-2e-===Fe2+14、关于下列两种物质的说法,正确的是A.核磁共振氢谱都有3个吸收峰B.都不能发生消去反应C.都能与Na反应生成H2D.都能在Cu作催化剂时发生氧化反应15、下列关于苯乙烷()的叙述,正确的是①能使酸性高锰酸钾溶液褪色②可以发生加聚反应③可溶于水④可溶于苯⑤能与浓硝酸发生取代反应⑥所有原子可能共面A.①④⑤B.①②⑤⑥C.①②④⑤⑥D.①②③④⑤⑥16、原子序数在数值上等于A.该原子的质量数B.原子核内中子数C.原子核内的质子数D.该元素的相对原子质量二、非选择题(本题包括5小题)17、有机物A可用作有机合成、食品、化妆品、医药的杀菌防腐剂。
云南省昆明市2023-2024学年高二下学期期末质量检测数学试题

云南省昆明市2023-2024学年高二下学期期末质量检测数学试题一、单选题1.已知复数z 满足i 1i z =-,则z =( )AB C .2D 2.已知向量()()1,2,1,3a b λλ=+=-r r,若a b r r ∥,则λ=( ) A .6-B .5-C .4-D .3-3.已知命题:,20x p x ∃∈<R ,命题()2:0,,ln 0q x x ∞∀∈+>,则( )A .p 和q 都是真命题B .p ⌝和q 都是真命题C .p 和q ⌝都是真命题D .p ⌝和q ⌝都是真命题 4.已知函数()y f x =,x ∈R 且()03f =,()()()0.520.51f n f n =-,*n ∈N ,则()f x 的一个解析式为( )A .()32xf x =⋅B .()132x f x -=⋅ C .()34xf x =⋅ D .()134x f x -=⋅5.某人连续投一枚骰子4次,记录向上的点数得到一组样本数据,若该组样本数据的平均数为2,则( )A .极差可能为5B .中位数可能为3C .方差可能为1D .众数可能为46.已知F 为抛物线()2:20C y px p =>的焦点,过C 上一点P 作圆()2222x y r -+=的两条切线,切点分别为,F A ,若PF PA ⊥,则p =( ) A .12B .23C .1D .437.已知正四棱台的体积为143则该球的表面积为( ) A .20π B .25π C .36π D .50π8.函数()1cos22f x x =+,2π,3x t t ⎛⎫∈+ ⎪⎝⎭,则下列说法错误的是( )A .R t ∃∈,使得()f x 为偶函数B .R t ∃∈,使得曲线()y f x =为中心对称图形C .R t ∀∈,()f x 存在极值D .R t ∀∈,()f x 存在两个零点二、多选题9.等比数列{}n a 的公比为q ,前n 项和为n S ,{}n S 为等差数列,则( ) A .1q = B .n n S na =C .{}n n a S +为等差数列D .{}n n a S 为等比数列10.已知函数()()()1f x x x x a =--,R a ∈,则下列说法正确的是( )A .若()36f =,则()11f '=-B .若()20f >,则2a <C .若()f x 在10,3⎛⎫⎪⎝⎭上单调递增,则a 的范围为0a >D .函数()f x 有两个极值点11.已知双曲线22:18y E x -=的左、右焦点分别为12,F F ,点P 在E 的右支上,则F 列说法正确的是( )A .若12PF F △的周长为24,则12PF F △的面积为48B .221212PF PF -≥ C.120tan PF F ∠≤<D .若12F PF ∠为锐角,则点P 的纵坐标范围是()(),88,∞∞--⋃+三、填空题12.已知函数()()13,02,0x x f x f x x ⎧⎪≤=⎨->⎪⎩,则()3f =.13.已知1sin cos ,tan 3tan 3αβαβ==,则()sin αβ+=.14.甲、乙两人先后在装有m 颗黑球的1号盒子与装有n 颗白球的2号盒子(*,m n m <∈N ,*n ∈N )轮流取球,规定每次取球可以从某一盒子中取出任意多颗(至少取1颗),或者在两个盒子中取出相同颗数的球(至少各取1颗),最后不能按规则取的人输.例如:当1,2m n ==时,甲先手不论如何取球,乙后手取球均有必定获胜的策略.若8m n +=,且后手取球者有必定获胜的策略,则满足条件的一组数组(),m n 可以为.四、解答题15. ABC V 的内角A ,B ,C 的对边分别为a ,b ,c ,已知222sin sin sin sin A B C A B +=.(1)求C ;(2)若1c =, ABC V ,求 ABC V 的周长. 16.如图,在四棱锥P ABCD -中,11,,,33AB AD CO AD AO AD PM PD ⊥⊥==u u u u u u r u u u r u u u ru r .(1)证明://CM 平面PAB ;(2)若直线PO ⊥平面,1,2,ABCD OA AB OC OD OP =====,求平面PBC 与平面PCD 的夹角的大小.17.已知椭圆()2222:10x y E a b a b +=>>的短轴长为2.(1)求E 的方程;(2)过点(作直线l 与椭圆E 相交于,A B 两点,若AB l 的方程.18.如图,甲、乙、丙、丁四名同学分别站在一个正方形的四个顶点进行传球训练,每次由一人随机将球传给另外三人中的一人,任意一人持球时,传给位于相邻顶点同学的概率为p ,传给位于对角线顶点同学的概率为q ,传球3次为一轮.(1)已知第一次由随机一名同学将球传出,若p q =,设事件A 为“一轮中每人各持一次球”. (i )求p 及事件A 的概率;(ii )设三轮传球中,事件A 发生的次数为X ,求X 的分布列与数学期望;(2)已知第一次由甲将球传出,在一轮传球中,乙、丙两人,谁两次持球的可能性更大? 19.已知函数()y f x =的定义域为I ,设0x I ∈,曲线()y f x =在点()()00,x f x 处的切线交x 轴于点()1,0x ,当1n ≥时,设曲线在点()(),n n x f x 处的切线交x 轴于点()1,0n x +,依次类推,称得到的数列{}n x 为函数()y f x =关于'0x 的“N 数列”,已知()()2ln 1f x x x =-+.(1)若{}n x 是函数()y f x =关于01x =的“N 数列”,求1x 的值;(2)若()(){},n g x f x a '=是函数()y g x =关于034a =-的“N 数列”,记2log 21n nb a =+.(i )证明:{}n b 是等比数列;(ii )证明:()()12121sin ln log ,2,n n i b n n i ++=<-≥∈∑N .。
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2009学年第一学期高二英语(牛津版)期末测试卷I. Grammar and vocabulary1. Mike was very tired_____ he played basketball all afternoon.A. if B as soon as C because D. before2. ______ he has limited knowledge, the old worker has a lot of experience.A Unless B. Since C. As D. Although3. ________ I know, the computer can never take the place of the human brain.A. As forB. As long asC. So far forD. As far as4. He was walking along the sands ______ he saw a big foot-print in the sand.A. while B when C. as D after5. It won‟t be long______ you regret what you‟ve done.A. afterB. sinceC. when D before6. He always gets up early_______ he can catch the early bus.A. becauseB. so thatC. in caseD. if only7. Child _____ he is, he can speak four different languages.A. asB. althoughC. if D before8. She still failed in the final exam_______ hard she worked.A. whatever B wherever C. however D. whenever9. Great Changes ________in Shanghai since 2000.A. have been taken placeB. took placeC. had taken placeD. have taken place10. ---Oh, it‟s you! I ______ you.---- I‟ve just had my hair cut.A. don‟t recognizeB. didn‟t recognizeC. haven‟t recognizedD. hadn‟t recognized11. The broken bike needs______.A. to repairB. repairedC. being repairedD. repairing12. He is tired because he _____ the living room all day.A. paintedB. had paintedC. has been paintingD. has painted13. Mr. Li has a good _______ of spoken French.A. comment B command C care D course14. They offer me some _______materials.A. supplementaryB. supposedC. suspendedD. supply15. Jack and his famil y are on _____, they‟ve gone to Beijing for Olympics.A vocationB visionC vacation D. volleyball16. Both “can” and “could” may _______ a future possibilityA. look forward toB. in addition toC. add toD. refer toII. VocabularyA. gentleB. slopesC. compete inD. dyingE. a variety ofF. skiingG. definitely H combination I resort J desiresIf you are ___17___ to look for a winter adventure and want to try some of the finest and mostchallenging____18___ in the world, then you should try a visit to Kitzbuhel. Located in southern Austria, Kitzbuhel is a paradise(天堂) for skiers, Even though the altitude of the city and its surrounding Alps is not enough to guarantee snow, the good weather and breathtaking scenery make Kitzbuhel a world-class ski __19___.The world‟s best and fastest skiers gather here once a year to ___20____ the downhill races that everyone___21___ to win. But there is no need to worry if you have never skied before. There are plenty of ___22__ slopes and Kitzbuhel is well known for its many ski schools.Skiing is not the only thing you can do in Kitzbuhel. The town offers many shops, cafes and ____23____ entertainment. A walk around Kitzbuhel is a feast for the eyes---- a lively ____24__of old village culture and the excitement of international tourist area. Should you have enough energy left after a day on the ____25__, you can take a dip in the pool, work out at the gym or go dancing in one of the city‟s many hotels an d clubs.III.ClozeThe most unforgettable thing, of course is my trip to England last year. This was my first time to go abroad, so it really made me __26____.When I walked around Oxford, I saw quite a lot of ancient buildings on both sides of the streets. One of the mostfamous tourist ____27__ in Oxford is Christ(基督)Church. There is a huge hall in Christ Church, which was used as the dining hall at Hogwarts in the Harry Potter___28__. I felt really excited when I heard this.Another place that impressed me ___29__during my trip was a place called Stratford-upon-A von. Maybe you haven‟t ___30___it, but I‟m sure you have heard about William Shakespeare—the great man‟s ___31___. In Stratford-upon-A von, I could see Shakespeare‟s Birthplace, Shakespeare Coun tryside Museum, The Royal Shakespeare Theatre, and a lot of places.I went to Shakespeare‟s Birthplace. It was the house ___32__ Shakespeare was born and spent his early years. Its sitting room, hallway (走廊), bedrooms, and kitchen were furnished (布置)in 16th-century ___33__.I really enjoyed this nice trip to England, __34___I learnt a lot. I‟m very happy that I could visit such an interesting __35___. I want to go there again in the future.26. A. excited B. appointed C. exciting D. appointing27. A. views B. scenes C. sites D. interests28. A. novels B. records C. films D. songs29. A. deep B. deeply C. high D. highly30. A. thought about B. heard about C. listened to D. considered as31. A. hometown B. office C. play D. theatre32. A. who B. that C. when D. where33. A. style B. site C. shape D. shade34. A. for which B. which C. from which D. in which35. A. countryside B. town C. country D. summerIV.Reading.A.The most frightening words in the English language are, “Our computer is down.” Y ou hear it more and more when you are on business. The other day I was at the airport waiting for a ticket to Washington and the girl in the ticket office said, “I‟m sorry, I can‟t sell you a ticket. Our computer is down.”“If your computer is down, just write me out a ticket.”“I can‟t write you out a ticket. The computer is the only one allowed to do so.”I looked down on the computer and every passenger was just standing there drinking coffee and staring at the black screen. Then I asked her, “What do all you people do?”“We give the computer the information about your trip, and then it tells us whether you can fly with us or not.”“So when it goes down, you go down with it.”“That‟s good, sir.”“How long will the computer be down?” I wanted to know.“I have no idea. Sometimes it‟s down for 10 minutes, sometimes for two hours. There‟s no way we can find out without asking the computer, and since it‟s down it won‟t answer us.”After the girl told me they had no backup(备用) computer, I said. “Let‟s forget the computer. What about your planes? They‟re still flying, aren‟t they?”“I couldn‟t tell without asking the computer.”“Maybe I could just go to the gate and ask the pilot if he‟s flying to Washington, ” I suggested.“I wouldn‟t know what gate to send you to. Even if the pilot was going to Washington, he couldn‟t take you if you didn‟t have a ticket.”“Is there any other airline flying to Washington within the next few hours?”“I wouldn‟t know, ” she said, pointing at the dark screen. “Only …IT‟ knows. …It‟ can‟t tell me.”By this time there were quite a few people standing in lines. The word soon spread to other travelers that the computer was down. Some people went white, some people started to cry and still others kicked their luggage.36. The best title for the article is _______.A. When the Computer Is DownB. The Most Frightening WordsC. The Computer of the AirportD. Asking the Computer37. What could the girl in the ticket office do for the passengers without asking the computer?A. She could sell a ticket.B. She could write out a ticket.C. She could answer the passengers‟ questions.D. She could do noth ing.38. Why do you think they had not a backup computer?A. Because it was easy down.B. Because it was very expensive.C. Because it was not advanced enough.D. Because it was not as big as the main computer.39. The last paragraph suggests that _______.A. a modern computer won‟t be down.B. computers can take the place of humansC. sometimes a computer may bring suffering to peopleD. there will be great changes in computersB.At the moment, it may be difficult to imagine, but many people believe that, by the year 2100, we will live on the planet Mars. Our own planet, Earth, is becoming more and more crowed and polluted. Luckily, we can start again and build a better world on Mars. Here is what life could be like.First of all, transport should be much better. At present, our spaceships are too slow to carry large numbers of people to Mars—it takes months. However, by 2100, spaceship can travel at half the speed of light. It might take us two or three days to get to Mars!Secondly, humans need food, water and air to live. Scientists should be able to develop plants that can be grown on Mars. These plants will produce the food and air that we need. However, can these plants produce water for us? There is no answer now.There is a problem for us to live on Mars. Mars pulls us much less than the Earth does. This will be dangerous because we could easily jump too high and fly slowly away into space. To prevent this, humans on Mars have to wear special shoes to make themselves heavier.Life on Mars will be better than that on the Earth in many ways, people will have more space. Living in a large building with only 10 bedrooms is highly possible. Many people believe that robot will do most of our work, so we have more time for our hobbies.There will probably be no school on Mars. Every student will have a computer at home which is connected to the Internet. They can study, do their homework and take exams in online schools. Each student will also have their own online teacher called “e-teacher”.However, in some ways, life on Mars may not be better than that on the Earth today. Food will not be the same—meals will probably be in the form of pills and will not be as delic ious as they are today. Also, space travel will make many people feel ill. The spaceship will travel fast but the journey to Mars will probably be very uncomfortable.40. According to the passage, by the year 2100, some of us will possibly live ______.A. on MarsB. on the moonC. under the seaD. in the sun41. So far, how long will it take us from the Earth to Mars by spaceship?A. Two or three days.B. A few months.C. A few days.D. A few years.42. In the future, the students on Mars______.A. will study mainly through the InternetB. will study mainly in a classroomC. will meet each other face to face every dayD. needn‟t study at allC.I fell in love with England because it was quaint (古色古香的)—all those little houses, looking terribly old-fashioned but nice, like doll‟s house. I loved the countryside and the pubs, and I loved London. I‟ve slightly changed my mind after seventeen years because I think it‟s ugly now.Things have changed. For everybody, England meant gentleman, fair play, and good manners. The fair play is going, unfortunately, and so are the gentlemanly attitudes and good manners—people shut doors heavily and politeness is disappearing.I regret that there are so few comfortable meeting places. Y ou are forced to live indoors. In Paris I go out much more, to restaurants and nightclubs. To meet friends here usually has to be in a pub, and it can be difficult to go there alone as a woman. The cafés are not terribly nice.As a woman, I feel unsafe here. I spend a bomb on taxis because I will not take public transport after 10 p.m. I used to use it, but now I‟m afraid.The idea of family seems to be more or less non-existent in England. My family is well united and that‟s typically French. In Middleses I had a neighbour who is 82 now. His family only lived two miles away, but I took him to France for Christmas once because he was always alone.43. The writer doesn‟t like London because she ______.A. is not used to the life there nowB. has lived there for seventeen yearsC. prefers to live in an old-fashioned houseD. has to be polite to everyone she meets there44. Where do people usually meet their friends in England?A. In a café.B. In a restaurant.C. In a nightclub.D. In a pub.45. The underlined word “it” in Paragraph 4 refers to “______”.A. a taxiB. the moneyC. a bombD. public transportDLiu Xiang was the first Asian hero to win the men‟s 110m hurdles at the Olympic Games.When he crossed the finishing line far ahead of the others in a world record—equaling time of 12.91 seconds, many fans burst into tears. It was Liu‟s first Olympics and he won the gold medal. In 2006, he set the new record of 12.88.When he won gold medals, he said in tears, “I never thought I could run so fast, and I never thought I would break the Oly mpic record, I am very, very excited. This is a wonder. I‟m proud not just for myself and for Chinese but for Asia. My race went wonderfully.” Liu added, “It is an amazing experience being the Olympic champion. I want to thank many people. We Chinese have shown the world we can run as fast as anybody else.”When Liu was a Grade Four student, he was chosen by the Junior Sports School to practise jumping. But when a bone test showed that he would not become a tall man, Liu was advised to give up practising after one year although he had won the national championship for his age group. Liu‟s parents also wanted him to give it up and study more at school, but Liu decided to continue.In 1998 when Liu Xiang, aged 15, met the famous coach Sun Haiping, his career c hanged greatly. Sun believed Liu was a rising star and he visited Liu‟s parents several times, finally they were persuaded to let Liu practise the 110m hurdles.But Liu thinks 2004 is just the beginning, and he expects to be at his peak in Beijing in 2008. Liu Xiang said, “For some players, it‟s just a job. For me, it‟s what I love.”46. At what age did Liu Xiang win the Olympic championship?A. 12.B. 15.C. 21.D. 22.47. What does the underlined word “persuaded” mean?A. supportedB. made sb. decide to do somethingC. educatedD. prevented48. Liu Xiang didn‟t give up sports because ______.A. a bone test showed that he wouldn‟t become tall enough to be a high jumperB. his parents stopped him from jumpingC. he was satisfied after winning the national jumping champion for his age groupD. the famous coach Sun Haiping changed his and his parents‟ ideas49 What does Liu Xiang think of being a sports player?A. It is just a job for him.B. It can bring him a lot of money.C. It is what he loves.D. It can bring him a lot of honour (荣誉).50. From the passage, we can know that ______.A. Liu Xiang was not the first Chinese to win the sprint gold medal in AthensB. if Liu Xiang reached the finishing line in 13 seconds, he would break the world recordC. the Chinese can run as fast as the people in the other countriesD. Liu Xiang didn‟t want to take part in the Olympic Games in 2008V. T ranslation1.毫无疑问,那个年轻人将接管公司。