高考得分点基础训练 (6)
高考数学基础训练:回归分析含详解
高考数学基础训练:回归分析一、单选题1.某工厂对一批产品进行了抽样检测.右图是根据抽样检测后的产品净重(单位:克)数据绘制的频率分布直方图,其中产品净重的范围是[96,106],样本数据分组为[96,98),[98,100),[100,102),[102,104),[104,106],已知样本中产品净重小于100克的个数是36,则样本中净重大于或等于98克并且小于104克的产品的个数是.A .90B .75C .60D .452.对两个变量y 与x 进行回归分析,分别选择不同的模型,它们的相关系数r 如下,其中拟合效果最好的模型是()A .0.2B .0.8C .-0.98D .-0.73.为研究变量x ,y 的相关关系,收集得到下面五个样本点(x ,y ):x 99.51010.511y1110865若由最小二乘法求得y 关于x 的回归直线方程为 3.2y x a=-+,则据此计算残差为0的样本点是()A .(9,11)B .(10,8)C .(10.5,6)D .(11.5)4.据一组样本数据()11,x y ,()22,x y ,…,(),n n x y ,求得经验回归方程为ˆ 1.50.5yx =+,且3x =.现发现这组样本数据中有两个样本点()1.2,2.2和()4.8,7.8误差较大,去除后重新求得的经验回归直线l 的斜率为1.2,则()A .变量x 与y 具有正相关关系B .去除两个误差较大的样本点后,重新求得的回归方程仍为ˆ 1.50.5yx =+C .去除两个误差较大的样本点后,y 的估计值增加速度变快D .去除两个误差较大的样本点后,相应于样本点()2,3.75的残差为0.055.对于样本相关系数,下列说法错误的是()A .可以用来判断成对样本数据相关的正负性B .可以是正的,也可以是负的C .样本相关系数越大,成对样本数据的线性相关程度也越高D .取值范围是[]1,1-6.下列说法中正确的是A .先把高二年级的2000名学生编号:1到2000,再从编号为1到50的学生中随机抽取1名学生,其编号为m ,然后抽取编号为50,100,150,m m m +++ 的学生,这种抽样方法是分层抽样法B .线性回归直线ˆˆy bxa =+不一定过样本中心()x y C .若两个随机变量的线性相关性越强,则相关系数r 的值越接近于1D .若一组数据2,4,a ,8的平均数是5,则该组数据的方差也是57.某同学用收集到的6组数据对(),(1,2,3,4,5,6)i i x y i =制作成如图所示的散点图(点旁的数据为该点坐标),并由最小二乘法计算得到回归直线1l 的方程:µµ11y b x a =+$,相关系数为1r ,相关指数为21R :经过残差分析确定点E 为“离群点”(对应残差过大的点),把它去掉后,再用剩下的5组数据计算得到回归直线2l 的方程:µµ22y b x a =+$,相关系数为2r ,相关指数为22R .则以下结论中,正确的是()①10r >,20r >;②µ10b >,µ20b >;③µµ12b b >;④2212R R >A .①②B .①②③C .②④D .②③④8.已知变量y 关于x 的非线性经验回归方程为0.5ˆe bx y-=,其一组数据如下表所示:x 1234ye3e 4e 5e 若5x =,则预测y 的值可能为()A .152e B .112e C .7e D .5e 第II 卷(非选择题)请点击修改第II 卷的文字说明二、填空题9.高中女学生的身高预报体重的回归方程是 0.7575.5y x =-(其中x , y 的单位分别是cm ,kg ),则此方程在样本()160,46处残差的绝对值是______.10.甲、乙、丙、丁四位同学在建立变量x ,y 的回归模型时,分别选择了4种不同模型,计算可得它们的相关指数R 2分别如下表:甲乙丙丁R 20.980.780.500.85建立的回归模型拟合效果最好的同学是__________.11.在一组样本数据()11,x y ,()22,x y ,…,(),n n x y (122,,,,n n x x x ≥⋅⋅⋅不全相等)的散点图中,若所有样本点()(),1,2,3,,i i x y i n =⋅⋅⋅都在直线210x y +-=上,则这组样本数据的相关系数r 为______.12.在一组样本数据()11,x y ,()22,x y ,…,()66,x y 的散点图中,若所有样本点(),i i x y ()1,2,,6i = 都在曲线212y bx =-附近波动.经计算6112i i x ==∑,6114i i y ==∑,62123ii x==∑,则实数b 的值为________.三、解答题13.某科技公司研发了一项新产品A ,经过市场调研,对公司1月份至6月份销售量及销售单价进行统计,销售单价x (千元)和销售量y (千件)之间的一组数据如下表所示:月份i 123456销售单价i x 99.51010.5118销售量iy 111086515(1)试根据1至5月份的数据,建立y 关于x 的回归直线方程;(2)若由回归直线方程得到的估计数据与剩下的检验数据的误差不超过065.千件,则认为所得到的回归直线方程是理想的,试问(1)中所得到的回归直线方程是否理想?参考公式:回归直线方程ˆˆˆybx a =+,其中i ii 122ii 1ˆnnx y n x yb xnx==-⋅⋅=-∑∑.参考数据:5i i i 1392x y ==∑,52i i 1502.5x ==∑.14.为了巩固拓展脱贫攻坚的成果,振兴乡村经济,某知名电商平台决定为脱贫乡村的特色水果开设直播带货专场.该特色水果的热卖黄金时段为2021年7月10日至9月10日,为了解直播的效果和关注度,该电商平台统计了已直播的2021年7月10日至7月14日时段中的相关数据,这5天的第x 天到该电商平台专营店购物的人数y (单位:万人)的数据如下表:日期7月10日7月11日7月12日7月13日7月14日第x 天12345人数y (单位:万人)75849398100(1)依据表中的统计数据,请判断该电商平台的第x 天与到该电商平台专营店购物的人数y (单位:万人)是否具有较高的线性相关程度?(参考:若0.30.75r <<,则线性相关程度一般,若0.75r >,则线性相关程度较高,计算r 时精确度为0.01)(2)求购买人数y 与直播的第x 天的线性回归方程;用样本估计总体,请预测从2021年7月10日起的第38天到该专营店购物的人数(单位:万人).参考数据:521(434i iy y =-=∑,51(64i i i x x y y =--=∑65.979≈.附:相关系数()()ni i x x y y r --=∑,回归直线方程的斜率121()()()niii nii x x y y bx x ==--=-∑∑ ,截距a y bx =-$$.15.近年来,明代著名医药学家李时珍故乡黄冈市蕲春县大力发展大健康产业,蕲艾产业化种植已经成为该县脱贫攻坚的主要产业之一,已知蕲艾的株高y (单位:cm)与一定范围内的温度x (单位:℃)有关,现收集了蕲艾的13组观测数据,得到如下的散点图:现根据散点图利用y a =+或dy c x=+建立y 关于x 的回归方程,令s =1t x=得到如下数据:xyst10.15109.943.040.16113niii s ys y=-⋅∑13113iii t yt y=-⋅∑1322113ik ss=-∑1322113ii t t =-∑ 1322113ii yy =-∑13.94-2.111.670.2121.22且(i s ,i y )与(i t ,i y )(i =1,2,3,…,13)的相关系数分别为1r ,2r ,且2r =﹣0.9953.(1)用相关系数说明哪种模型建立y 与x 的回归方程更合适;(2)根据(1)的结果及表中数据,建立 y 关于x 的回归方程;(3)已知蕲艾的利润z 与x 、y 的关系为1202z y x =-,当x 为何值时,z 的预报值最大.参考数据和公式:0.21×21.22=4.4562,11.67×21.22=247.637415.7365,对于一组数据(i u ,i v )(i =1,2,3,…,n ),其回归直线方程v u αβ=+的斜率和截距的最小二乘法估计分别为 1221ni i i nii u vnu v unuβ==-⋅=-∑∑, v u αβ=-,相关系数ni i u vnu vr -⋅∑.参考答案:1.A 【解析】【详解】样本中产品净重小于100克的频率为(0.050+0.100)×2=0.3,频数为36,∴样本总数为.∵样本中净重大于或等于98克并且小于104克的产品的频率为(0.100+0.150+0.125)×2=0.75,∴样本中净重大于或等于98克并且小于104克的产品的个数为120×0.75=90.考点:频率分布直方图.2.C 【解析】【分析】由相关系数的绝对值越大,越具有强大相关性,即可求解【详解】∵相关系数的绝对值越大,越具有强大相关性,C 相关系数的绝对值最大约接近1,∴C 拟合程度越好.故选:C 3.B 【解析】【分析】先求出线性方程的样本中心点,从而可求得 3.240y x =-+,再根据残差的定义可判断.【详解】由题意可知,99.51010.511105x ++++==,111086585y ++++==所以线性方程的样本中心点为(10,8),因此有 8 3.21040aa =-⨯+⇒=,所以 3.240y x =-+,在收集的5个样本点中,(10,8)一点在 3.240y x =-+上,故计算残差为0的样本点是(10,8).故选:B 4.A 【解析】【分析】由条件可知样本中心不变,可求出新的回归直线方程,即可判断.【详解】因为重新求得的经验回归直线l 的斜率为1.2,所以变量x 与y 具有正相关关系,故A 正确;当3x =时,315055y ..=⨯+=,设去掉两个误差较大的样本点后,横坐标的平均值为x ',纵坐标的平均值为y ',则12636322n x x x x n n n ++⋅⋅⋅+--=--'==,1210510522n y y y n n n y ++⋅⋅⋅+--'==--=,因为去除两个误差较大的样本点后,重新求得回归直线l 的斜率为1.2,所以ˆ53 1.2a =⨯+,解得 1.4ˆa =,所以去除两个误差较大的样本点后的经验回归方程为ˆ 1.2 1.4yx =+,故B 错误;因为1.5 1.2>,所以去除两个误差较大的样本点后y 的估计值增加速度变慢,故C 错误;因为ˆ 1.22 1.4 3.8y=⨯+=,所以ˆ 3.75 3.80.05y y -=-=-,故D 错误.故选:A.5.C 【解析】【分析】根据相关系数的概念,依次分析各选项即可得答案.【详解】解:对于A 选项,当相关系数为正时,表明变量之间是正相关,相关系数为负数时,表明相关系数为负数,故A 选项正确;对于B ,D 选项,相关系数范围是[]1,1-,故可以为正,也可以为负,故B ,D 选项正确;对于C 选项,当相关系数为负数时,样本相关系数越大,线性相关性就越弱,故C 选项错误;故选:C6.D 【解析】A 是系统抽样,B 选项线性回归直线ˆˆy bxa =+一定过样本中心(),x y ,C 选项若两个随机变量的线性相关性越强,则相关系数r 的绝对值越接近于1,D 选项若一组数据2,4,a ,8的平均数是5,求出a ,则该组数据的方差即可求解.【详解】A 选项:先把高二年级的2000名学生编号:1到2000,再从编号为1到50的学生中随机抽取1名学生,其编号为m ,然后抽取编号为50,100,150,m m m +++ 的学生,这种抽样方法是系统抽样法,所以该选项不正确;B 选项:线性回归直线ˆˆy bxa =+一定过样本中心(),x y ,所以该选项不正确;C 选项:若两个随机变量的线性相关性越强,则相关系数r 的绝对值越接近于1,所以该选项不正确;D 选项:若一组数据2,4,a ,8的平均数是5,24854a +++=,解得6a =,则该组数据的方差是()()()()22222545658554-+-+-+-=,所以该选项正确.故选:D 【点睛】此题考查抽样方法,回归直线,相关关系的辨析,求平均数和方差,关键在于熟练掌握相关概念和公式,准确计算.7.B 【解析】【分析】根据散点图逐项进行判断即可.【详解】①:由散点图可知,,x y 之间是正相关关系,所以10r >,20r >,故①正确;②③:由散点图可知,回归直线的斜率是正数,且1l 的斜率大于2l 的斜率,所以µ10b >,µ20b >,µµ12b b >,故②③正确;④:由散点图可知,去掉“离群点”E 后,相关性更强,拟合的效果更好,所以2212R R <,故④错误;故选:B.8.C 【解析】【分析】将0.5ˆe bx y-=两边同时取对数,得ln 0.5y bx =-,设0.5z bx =-,由样本中心()x z 必在回归直线0.5z bx =-上,可求出b ,从而即可求解.【详解】解:由题意,将0.5ˆe bx y-=两边同时取对数,得ln 0.5y bx =-,设0.5z bx =-,则x1234z13451234 2.54x +++==,13453.254z +++==,由0.5z bx =-,得3.25 2.50.5b =-,解得 1.5b =,所以 1.50.5e x y -=,所以当5x =时, 1.550.57e e y ⨯-==,故选:C.9.1.5##32【解析】【分析】利用回归直线方程,求出160x =的估计值,然后求解残差的绝对值.【详解】由样本数据得到,女大学生的身高预报体重的回归方程是 0.7575.5y x =-,当160x =时, 0.7516075.544.5y =⨯-=,此方程在样本()160,46处残差的绝对值:44.546 1.5-=.故答案为:1.5.10.选甲相关指数R 2越大,表示回归模型拟合效果越好.【解析】【分析】相关指数越大,相关性越强,拟合效果越好.根据相关指数的大小即可判断.【详解】相关指数2R 越大,相关性越强,回归模型拟合效果越好,所以效果最好的是甲.【点睛】如果两个变量间的关系是相关关系,相关指数2R 越大,相关系数r 越接近1,残差平方和越接近0,都代表拟合效果越好.11.1-【解析】【分析】根据直线斜率可知两个变量负相关,结合数据点都在直线上可确定1r =-.【详解】直线210x y +-=的斜率20k =-<,∴这两个变量成负相关,0r ∴<,又所有样本点都在直线210x y +-=上,1r ∴=-.故答案为:1-.12.1723【解析】【分析】设2t x =,可得回归直线方程为12y bt =-,求出样本中心点(),t y 代入可得b 的值.【详解】令2t x =则212y bx =-即12y bt =-,6212366i i x t ===∑,61147663ii y y ====∑,因为样本中心点237,63⎛⎫ ⎪⎝⎭在回归直线12y bt =-上,所以7231362b =-,可得:1723b =,故答案为:1723.13.(1)ˆ3240y x =-+.;(2)是.【解析】【分析】(1)先由表中的数据求出,x y ,再利用已知的数据和公式求出 ,ba ,从而可求出y 关于x 的回归直线方程;(2)当8x =时,求出 y 的值,再与15比较即可得结论【详解】(1)因为()199.51010.511105x =++++=,()1111086585y =++++=,所以23925108ˆ 3.2502.5510b -⨯⨯==--⨯,得()ˆ8 3.21040a=--⨯=,于是y 关于x 的回归直线方程为 3.240ˆyx =-+;(2)当8x =时,ˆ 3.284014.4y=-⨯+=,则ˆ14.4150.60.65yy -=-=<,故可以认为所得到的回归直线方程是理想的.14.(1)具有较高的线性相关程度(2) 6.470.8y x =+,314万人【解析】【分析】(1)由已知计算相关系数r 即可.(2)由列表计算 a、b ,可得线性回归方程进一步可得解.(1)由表中数据可得3,90x y ==,所以521()10i i x x =-=∑,又55211()434,()()64i i i i i y y x x y y ==-=--=∑∑,所以()()50.970.75i i x x y y r --=>∑,所以该电商平台直播黄金时段的天数x 与购买人数y 具有较高的线性相关程度.所以可用线性回归模型拟合人数y 与天数x 之间的关系.(2)由表中数据可得()()()5152164ˆ 6.410i i i i i x x y y b x x ==--===-∑∑,则ˆˆ90 6.4370.8a y bx =-=-⨯=,所以 6.470.8y x =+,令38x =,可得 6.4387031ˆ.84y =⨯+=(万人)15.(1)用d y c x =+模型建立y 与x 的回归方程更合适;(2)10ˆ111.54y x =-;(3)当温度为20时这种草药的利润最大.【解析】【分析】(1)利用相关系数1r ,2r ,比较1||r 与2||r 的大小,得出用模型d y c x=+建立回归方程更合适;(2)根据(1)的结论求出y 关于x 的回归方程即可;(3)由题意写出利润函数ˆz ,利用基本不等式求得利润z 的最大值以及对应的x 值.【详解】(1)由题意知20.9953r =-,10.8858r =,因为121r r <<,所有用d y c x =+模型建立y 与x 的回归方程更合适.(2)因为1311322113 2.1ˆ100.2113i i i i i t y t yd tt ==-⋅-===--∑∑,ˆˆ109.94100.16111.54cy dt =-=+⨯=,所以ˆy 关于x 的回归方程为10ˆ111.54y x=-(3)由题意知11012020(111.54ˆˆ)22z y x x x =-=--20012230.8()2x x =-+2230.8202210.8≤-=,所以22.8ˆ10z≤,当且仅当20x =时等号成立,所以当温度为20时这种草药的利润最大.。
高考一轮复习基础训练《DNA是主要的遗传物质》
DNA是主要的遗传物质一、选择题1.关于染色体和DNA的关系,正确的是()A.DNA和染色体都是遗传物质B.染色体的组成成分之一是DNAC.一个染色体含有一条多核苷酸链D.DNA只存在于染色体上2.下列说法正确的是()A.一切生物的遗传物质都是DNAB.一切生物的遗传物质都是RNAC.一切生物的遗传物质是核酸D.一切生物的遗传物质是蛋白质3.噬菌体侵染细菌的实验能证明的是()A.DNA是主要的遗传物质B.蛋白质是遗传物质C.DNA是遗传物质D.DNA能产生可遗传的变异4.注射后能使小白鼠因患败血病而死亡的是()A.R型肺炎双球菌B.加热杀死后的R型肺炎双球菌C.加热杀死后的S型肺炎双球菌D.加热杀死后的S型肺炎双球菌与R型细菌混合5.T2噬菌体的结构组成为()A.蛋白质、脂类B.糖类、蛋白质C.蛋白质、DNAD.蛋白质、RNA6.用32P标记噬菌体的DNA,用35S标记细菌的蛋白质,用这种噬菌体去侵染大肠杆菌,则新生的噬菌体可含有()A.32PB.35SC.32P和35SD.二者皆无7.肺炎双球菌转化实验中,在培养有R型细菌的A、B、C、D四个试管,依次分别放人从S型活细菌中提取的DNA、DNA和DNA酶、蛋白质、多糖,经过培养,检查结果发现有R型细菌转化的是()A. B. C. D.8.噬菌体在繁殖的过程中所利用的原料是()A.自身的核苷酸和氨基酸B.自身的核苷酸和细菌的氨基酸C.细菌的核苷酸和氨基酸D.自身的氨基酸和细菌的核苷酸9.遗传物质的主要载体是()A.DNAB.染色体C.叶绿体D.线粒体10.所有病毒的遗传物质()A.都是DNAB.都是RNAC.是DNA或RNAD.是DNA和RNA11.车前草病毒是一种不含DNA的病毒,其遗传物质是()A.车前草的DNAB.车前草病毒的RNAC.车前草的RNAD.车前草病毒的蛋白质12.真核生物的遗传物质存在于()A.细胞核内B.细胞质中的细胞器内C.细胞质的基质内D.细胞核和细胞质内13.猴、噬菌体、烟草花叶病毒中参与构成核酸的碱基种类数依次是()A.4,4,5B.5,4,4C.4,5,4D.5,4,514.疯牛病病毒仅由一小团蛋白质构成,由其感染牛的脑细胞后,繁殖出的新个体仍含有与原来的一小团蛋白质,这个事实说明()A.一切生物的遗传物质是核酸B.DAN是主要的遗传物质C.RNA也可作为遗传物质D.蛋白质也可作为遗传物质15.现有两组实验数据:(1)测得豌豆中DNA有84%在染色体上,14%在叶绿体上,2%在线粒体上;(2)进一步测得豌豆染色体的组成是:DNA占36.5%,RNA 占9.6%,蛋白质占48.9%。
高考一轮复习语文基础训练专项练习题及答案
高考语文基础知识训练一1、下列词语中加点的字,读音全都正确的一项是A、星宿(某i) 馄饨(dn) 贺岁片(pin) 垂涎三尺(某in)B、粗犷(guǎng) 堤防(dī) 芝麻糊(h) 瞠目结舌(chēng)C、属意(zhǔ) 歼灭(jiān) 数来宝(hǔ) 便宜从事(pin)D、烘焙(bi) 唱和(h) 挑大梁(tiāo) 风驰电掣 (ch)1、B(A项馄饨读hn tun, C项便读bin,D项挑读tiāo)2、下列词语中,书写没有错误的一项是A、暮霭厮打脑血拴厝火积薪B、彪悍誉写蜡像馆大快朵颐C、宣泄愧怍调和油不稂不莠D、羸弱谙熟手动挡开源截流2、C(A项脑血拴应为脑血栓,B项誉写应为誊写,D项开源截流应为开源节流)3.下列各句中,标点符号使用正确的一句是A.养殖业的发展要结合市场需求,更要结合国情。
国外的养殖方式:大规模、产业化自然是好处很多,但也不能照搬。
B.各国展示的最新科技成果,包括太阳能技术、资源循环技术、固体废弃物无害化、减量化、资源化处理技术等。
C.《孟子》里有各种圣人,表现各不相同。
但有一点相同,即行一不义,杀一不辜而得天下,皆不为也(《孟子公孙丑上》)。
D.我不知道这一实验到底能不能得出最终的结论但我一定要坚持下去,直到把问题搞个水落石出,尽管我面前困难重重。
3、C(A项冒号应改为破折号,B项第一个顿号应为逗号,D项问号应为逗号)4.下列各句中,加点词语使用正确的一句是TV科教频道报道了一位老人,他经过无数次试验。
终于成功发明了具有良好保护作用的撞车防护装置。
B.人们的欣赏水平越来越高,审美趣味日趋多元,你的画作能得到大家的广泛称赞诚然很好,即便得不到人们的认可也是可以理解的。
C.早就听说贪官们多方聚敛财富,生活腐化,从中纪委公布的被查处贪官的非法财产数字来看,其贪腐之疯狂真是名不虚传。
D.许多儿童的确具有绘画天赋,虽然他们从来没有学习过素描,但笔下的人物却有鼻子有眼,形象生动逼真。
2023届高考生物一轮复习 第六单元 遗传的物质基础综合 单元检测题 基础训练A卷
2023届高考生物一轮复习第六单元遗传的物质基础综合单元检测题基础训练A卷一、单选题1.以下关于遗传变异的说法正确的是( )A.伴性遗传中亲本正交和反交的结果可能不同,不遵循孟德尔遗传规律B.一对表现正常的双亲生出白化病(常染色体隐性遗传病)孩子的概率是1/4C.遗传信息是指DNA中的碱基排列顺序D.细胞内任何一个DNA分子都可以发生基因突变,说明基因突变具有普遍性2.下列关于生物遗传物质及其传递规律探索历史的叙述,正确的是( )A.孟德尔通过豌豆的杂交实验,证明基因位于染色体上B.噬菌体侵染细菌的实验,需用含32P培养液培养噬菌体使其带上标记C.烟草花叶病毒的感染和重建实验,证明病毒的遗传物质都是RNAD.通过同位素示踪和密度梯度超速离心等技术,证明了DNA的半保留复制3.在其他条件具备的情况下,在试管中加入物质X和物质Z,可得到相应产物Y。
下列叙述正确的是( )A.若X是DNA,Y是RNA,则Z是逆转录酶B.若X是DNA,Y是mRNA,则Z是脱氧核苷酸C.若X是RNA,Y是DNA,则Z是限制性内切酶D.若X是mRNA,Y是在核糖体上合成的大分子,则Z是氨基酸4.以下与遗传物质相关的叙述,正确是( )A.豌豆细胞核的遗传物质是DNA,细胞质的遗传物质是RNAB.甲型H1N1病毒的遗传物质中含有C、H、O、N、S等元素C.T2噬菌体内,由碱基A、C、G参与组成的核苷酸共有6种D.双链DNA分子中碱基数等于磷酸基团数5.下列关于生物体内遗传信息的传递与表达的叙述,正确的是( )A.每种氨基酸都至少有两个以上的遗传密码B.遗传密码由DNA传递到RNA,再由RNA决定蛋白质C.一个DNA分子通过转录可形成许多个不同的RNA分子D.DNA聚合酶与DNA分子结合能使DNA分子的双螺旋解开6.若不考虑突变,细胞核中的遗传物质一定相同的是( )A.来自同一红眼雌果蝇卵细胞B.来自同一个玉米果穗的籽粒C.来自同一株花生上的不定芽D.来自同一株紫花豌豆的花粉7.下图为基因部分功能的示意图,下列相关叙述错误的是( )A.①过程表示基因表达中的转录过程B.②过程表示基因表达中的翻译过程C.基因指导合成的产物都是蛋白质D.基因可能发生碱基对的增添、缺失或替换8.DNA复制、转录和翻译后所形成的产物分别是( )A.RNA、RNA、蛋白质B.DNA、RNA、蛋白质C.RNA、DNA、蛋白质D.DNA、DNA、蛋白质9.下列关于遗传问题的说法正确的是( )A.孟德尔通过演绎推理证明了其假说的正确性B.玉米叶肉细胞的遗传物质中含有8种核苷酸C.一个具两对等位墓因果蝇精原细胞至少会产生两种基因型的精子D.人的胰岛素有51个肽键,指导它合成的基因至少有1534种不同的排列顺序10.下列关于“噬细体侵染细菌”实验内容的相关叙述,正确的是( )A.用含35S和32P的培养基分别培养噬菌体,可获得两种噬菌体实验材料B.用35S和32P标记的噬菌体侵染未被标记的大肠杆菌时,保温培养时间越长实验结果越可靠C.用含35S标记噬菌体侵染细菌时,若沉淀物中放射性过高,可能是搅拌不充分D.噬菌体侵染实验充分说明DNA是主要的遗传物质、蛋白质不能作为遗传物质11.下列说法正确的是( )①生物的性状是由基因控制的,性染色体上的基因都是控制性别的。
高考得分点基础训练
高考得分点基础训练(1)AFrom smartphones (智能手机) and tablet computers (平板电脑), touch screens are everywhere you look these days. As technology advances, keyboards and mice are quickly becoming a thing of the past. ___25_________ be burdened with keyboards and mice when you can have what you want with just a touch?Touch screens are electronic visual displays that allow a user to communicate directly with what is displayed on the screen, rather than ___26________ (use) a pointing device, such as a mouse. Touch screens ___27_________ (design) to respond to the touch of a finger, although an object—like a stylus (触控笔)—can also be used. The idea for the touch screen was first developed by E.A. Johnson at the Royal Radar Establishment in England. His idea was first described in a short article ___28__________ (publish) in 1965.Given the many different types of devices that use touch screens, ___29_________ is no surprise that there are several different types of touch screens. ___30__________ type of touch screen works a little differently from ___31______________ others.When touch screens first became popular, they ___32___________ only sense one point of contact at a time. Technology has advanced greatly in recent years, though. Today, many touch screen devices feature multi-touch technology. This technology allows a touch screen device to interpret quite a few points of contact at the same time.BThe Milky Way (银河) contains billions of Earth-sized planets that could support life. That’s the finding of a new study. ___33___________ makes use of data that came from NASA’s top planet-hunting telescope.___34___________ mechanical failure recently put that Kepler space telescope out of service. Kepler had played a big role in creating a census (调查统计) of planets moving around some 170,000 stars. Its data have been helping astronomers predict ___35__________ common planets are in our galaxy. The space telescope focus on hunting planets that might have conditions similar to ___36___________ on Earth.The authors of a study published in the Proceedings of the National Academy of Science now conclude ___37__________ between 14 and 30 out of every 100 stars with a mass and temperature similar to the sun may host a planet that could support life ___38__________ we know it. Such a planet would have a diameter at least as large as Earth’s, but no more than twice that big. The planet also would have to orbit in a star’s habitable (适宜居住的) zone. That’s where the surface temperature would allow any water ___39________-__ (exist) as a liquid.The new estimate of how many planets might fit these conditions comes from studying more than 42,000 stars. The scientist used those numbers to estimate the rest of the stars ___40___ the telescope could not see.▲Section B (10分)A. properlyB. achievementC. improveD. valuedE. killF. differentG. restoreH. relativelyI. motivationJ. gainedK. encouraging Psychologists take ___41___ views of how external rewards, from warm praise to old cash, affect ___42___ and creativity. Behaviorists, who study the relation between action, argue that rewards can ___43___ performance at work and school. Some other researchers who study various aspects of mental life, maintain those rewards often destroy creativity by ___44___ dependence on approval and gifts from others.The latter view has ___45___ many supporters, especially among educators. But the careful use of small monetary rewards sparks in grade- school children suggesting that ___46___ presented inducements (引诱物) indeed aid inventiveness, according to a study in the June Journal Personality and Social Psychology.“If they know they’re working for a reward and can focus on a ___47___ challenging task, they show the most creativity,”says Robert Esenberger of the University of Delaware in Newark. “But it’s easy to ___48___ creativity by giving rewards for poor performance or creating too much expectation for rewards.”A teacher who continually draws attention to rewards or who hands our high grades for ordinary ___49___ ends up with uninspired students, Esenberger holds, as an example of the latter point, he notes growing efforts at major universities to tighten grading standards and ___50___ falling grades.In earlier grades, the use of so-called token economics, in which students handle challenging problems and receive performance- based points toward valued rewards, shows promise in raising effort and creativity, the Delaware psychologist claims.AFor more than twenty years scientists have been searching for signs of life on other planets. Most of these searches have been done over the radio. The hope is that someone in outer space may be trying to get in touch with us. Scientists also have sent radio and television messages on spaceships travelling through space, on the chance that someone may be receptive to such messages. Scientists are using powerful radio telescopes to listen to signals from about 1, 000 stars, all within 100 ligh t years of the earth. In addition, they will scan the entire sky to “listen” for radio messages from more distant stars. Using a computer, they will be able to monitor more than eight channels at one time. Scientists are looking for any signal they stands out from the background noise.Of the 200 billion stars in the Milky Way galaxy (银河星系), scientists find that five percent are like our sun. Perhaps half of them have a planet like the earth. Such a planet would be a reasonable distance from the star for temperatures to be right for the evolution of life. Based on the inhabitable (that can be lived in)planets in our galaxy, most scientists agree that chances are likely that one or more of these planets support some life.However, many scientists wonder whether intelligent life exists on other planets. Some believe that twenty years of searching without any intelligible messages shows that no one is out there. They say that the evolution of intelligence comparable to ours is unlikely.Other scientists believe that our search hasn’t been long enough to rule out the possibility thatintelligent life exists in our galaxy. Although our sun family is only about five billion years old, our galaxy is about 20 billion years old. In that time, some scientists think it is likely that civilization (文明)much more advanced than ours have developed. Perhaps these civilizations send us no signals; perhaps we have not recognized the signals they have sent us. If we hope to find intelligent life, these scientists believe that we have to keep looking.( ) 66.According to the passage, how many planets in our galaxy might be inhabitable?A. 5 billion.B. 10 billion.C. 15 billion.D. 200 billion.( ) 67.The first paragraph in this passage is mainly about _________.A. how scientists are looking for signs of life on other planetsB. why scientists are looking for signs of life on other planetsC. where scientists are looking for signs of life on other planetsD. when scientists are looking for signs of life on other planets( ) 68.The underlined word “monitor”in the passage means “_________”.A. findB. followC. checkD. form( ) 69.Which of these statements is true based on the information in the passage?A. The earth is one of the oldest planets in our galaxy.B. Most scientists believe that there is intelligent life on other planets.C. Scientists are trying different ways to find signs of life on other planets.D. Scientists don’t believe that there might be life on other planets.BHomestay provides English language students with the opportunity to speak English outside the classroom and the experience of being part of a British home.What to ExpectThe host will provide accommodation and meals. Rooms will be cleaned and bedcovers changed at least once a week. You will be given the house key and the host is there to offer help and advice as well as to take an interest in your physical and mental health.Accommodation ZonesHomestays are located in London mainly in Zones 2, 3 and 4 of the transport system. Most hosts do not live in the town centre as much of central London is commercial and not residential (居住的). Zones 3 and 4 often offer larger accommodation in a less crowded area. It is very convenient to travel in London by Underground.Meal Plans Available◇Continental Breakfast◇Breakfast and Dinner◇Breakfast, Packed Lunch and DinnerIt’s important to note that few English families still provide a traditional cooked breakfast. Your accommodation includes Continental Breakfast which normally consists of fruit juice, cereal (谷物类食品), bread and tea or coffee. Cheese, fruit and cold meat are not normally part of a Continental Breakfast in England. Dinners usually consist of meat or fish with vegetables followed by dessert, fruit and coffee.FriendsIf you wish to invite a friend over to visit, you must first ask your host’s permission. You have no right to entertain friends in a family home as some families feel it is an invasion of their privacy. Self-Catering Accommodation in Private HomesAccommodation on a room—only basis includes shared kitchen and bathroom facilities and often a main living room. This kind of accommodation offers an independent lifestyle and is more suitable for the long-stay student. However, it does not provide the same family atmosphere as an ordinary homestay and may not benefit those who need to practise English at home quite as much.( ) 70.The passage is probably written for_________.A.hosts willing to receive foreign studentsB.foreigners hoping to build British cultureC.travellers planning to visit families in LondonD.English learners applying to live in English homes( ) 71.Which of the following will the host provide?A.Room cleaning. B.Medical care.C.Free transport. D.Physical training.( ) 72.What can be inferred from Paragraph 3?A.Zone 4 is more crowded than Zone 2.B.The business centre of London is in Zone 1.C.Hosts dislike travelling to the city centre.D.Accommodation in the city centre is not provided.( ) 73.According to the passage, what does Continental Breakfast include?A.Dessert and coffee.B.Fruit and vegetables.C.Bread and fruit juice.D.Cereal and cold meat.I.Translation (22分)1. 你一旦答应帮助他们,就要说话算数。
高中物理 平抛运动(附答案)高考基础训练题
平抛运动(附答案)1.在平坦的垒球运动场上,击球手挥动球棒将垒球水平击出,垒球飞行一段时间后落地.若不计空气阻力,则()A.垒球落地时瞬时速度的大小仅由初速度决定B.垒球落地时瞬时速度的方向仅由击球点离地面的高度决定C.垒球在空中运动的水平位移仅由初速度决定D.垒球在空中运动的时间仅由击球点离地面的高度决定2.质点从同一高度水平抛出,不计空气阻力,下列说法正确的是()A.质量越大,水平位移越大B.初速度越大,落地时竖直方向速度越大C.初速度越大,空中运动时间越长D.初速度越大,落地速度越大3.一个人水平抛出一小球,球离手时的初速度为v0,落地时的速度是v t,空气阻力忽略不计,下列哪个图象正确表示了速度矢量变化的过程()图4-2-194.(高考广东卷)某同学对着墙壁练习打网球,假定球在墙面上以25m/s的速度沿水平方向反弹,落地点到墙面的距离在10m至15m之间,忽略空气阻力,取g=10m/s2,球在墙面上反弹点的高度范围是()A.0.8m至1.8mB.0.8m至1.6mC.1.0m至1.6mD.1.0m至1.8m5.在高处水平抛出一物体,平抛的初速度为v0,当它的速度方向与水平方向成θ角时,物体的水平位移x与竖直位移y的关系是()A.x=y tanθB.x=2y tanθC.x=y cotθD.x=2y cotθ6.(黄冈第二次模拟)如图4-2-20所示,在一次演习中,离地H 高处的飞机以水平速度v 1发射一颗炮弹欲轰炸地面目标P ,反应灵敏的地面拦截系统同时以速度v 2竖直向上发射炮弹拦截.设拦截系统与飞机的水平距离为s ,若拦截成功,不计空气阻力,则v 1、v 2的关系应满足()A .v 1=v 2B .v 1=H s v 2C .v 1=H s v 2D .v 1=s H v 27.(江南十校模拟)如图4-2-21所示,某同学为了找出平抛运动的物体初速度之间的关系,用一个小球在O 点对准前方的一块竖直放置的挡板,O 与A 在同一高度,小球的水平初速度分别是v 1、v 2、v 3,打在挡板上的位置分别是B 、C 、D ,且AB ∶BC ∶CD =1∶3∶5,则v 1、v 2、v 3之间的正确关系是()A .v 1∶v 2∶v 3=3∶2∶1B .v 1∶v 2∶v 3=5∶3∶1C .v 1∶v 2∶v 3=6∶3∶2D .v 1∶v 2∶v 3=9∶4∶18.(温州模拟)如图4-2-22所示,从倾角为θ的斜面上的M 点水平抛出一个小球,小球的初速度为v 0,最后小球落在斜面上的N 点,则(重力加速度为g )()A .可求M 、N 之间的距离B .可求小球落到N 点时速度的大小和方向C .可求小球到达N 点时的动能D .可以断定,当小球速度方向与斜面平行时,小球与斜面间的距离最大9.如图4-2-23所示,高为h =1.25m 的平台上,覆盖一层薄冰.现有一质量为60kg 的滑雪爱好者,以一定的初速度v 向平台边缘滑去,着地时速度的方向与水平地面的夹角为45°(重力加速度g 取10m/s 2).由此可知下列各项中错误的是()A .滑雪者离开平台边缘时速度的大小是5.0m/sB .滑雪者着地点到平台边缘的水平距离是2.5mC .滑雪者在空中运动的时间为0.5sD .着地时滑雪者重力做功的瞬时功率是300W10.如图4-2-24所示,O 点离地面高度为H ,以O 点为圆心,制作一四分之一光滑圆弧轨道,小球从与O 点等高的圆弧最高点滚下后水平抛出,试求:(1)小球落地点到O 点的水平距离;(2)要使这一距离最大,R 应满足什么条件?最大距离为多少?图4-2-20图4-2-21图4-2-22图4-2-23答案:(1)2R (H -R )(2)R =H 2时,最大距离为H 11.如图4-2-25所示,从H =45m 高处水平抛出的小球,除受重力外,还受到水平风力作用,假设风力大小恒为小球重力的0.2倍,g 取10m/s 2.问:(1)有水平风力与无风时相比较,小球在空中的飞行时间是否相同?如不相同,说明理由;如果相同,求出这段时间?(2)为使小球能垂直于地面着地,水平抛出的初速度v 0为多少?图4-2-25答案:(1)相同3s (2)6m/s12.(广州、肇庆、珠海部分重点中学调研)如图4-2-26所示,在距地面高为H =45m 处,有一小球A 以初速度v 0=10m/s 水平抛出,与此同时,在A 的正下方有一物块B 也以相同的初速度v 0同方向滑出,B 与地面间的动摩擦因数为μ=0.5.A 、B 均可看作质点,空气阻力不计,重力加速度g 取10m/s 2,求:(1)A 球从抛出到落地的时间和这段时间内的水平位移;(2)A 球落地时,A 、B 之间的距离.答案:(1)3s30m(2)20m 答案:1D2D3B4A5D6D7C8ABD9D图4-2-24图4-2-26。
高考基础训练(1)
高考基础训练(1)1.下列词语中加点的字的读音完全相同的一项是()A.觊觎忌妒稽查赍恨而亡叽叽嘈嘈B.丽水狸猫黄鹂不幸罹难梨园子弟C.车棚抨击膨胀蓬门荜户令人捧腹D.蚊蚋纳罕木讷方枘圆凿千补百衲2.下列词语中,没有错别字的一组是()A.雏形首屈一指消声匿迹智者千虑必有一失B.渲泄茅塞顿开相濡以沫坐收渔人之利C.证券化险为夷固若金汤一叶障目不见泰山D.寒暄众口烁金开门揖盗山雨欲来风满楼3.依次填入下列各句横线处的词语,最恰当的一组()①愉悦的享乐、丰美的衣食、的生活,这一切若不成为为人类作贡献的条件,也不能充实人生的意义。
②柏格森《笑论》说,一切可笑都起于灵活的事物变成,生动的举止化作机械。
③贪图小利的人往往只看到自己的小圈子,打自己的小算盘,忽视了集体和国家的利益。
这种人他有多大的本领,我们也不能委以重任。
A.舒服呆板进而/不管 B.舒服刻板因而/尽管C.舒适刻板进而/尽管 D.舒适呆板因而/不管4.下列各句中加点的成语使用不恰当的一句()A.经营者应怀有悲天悯人的胸襟,并以正义为前提。
B.他仍然是中国最高学府的名教授,门墙桃李,此中大可物色党羽。
C.招聘又不是选美,为什么一定要女的长得人面桃花,男的身高要到达一米七五。
D.我呻吟着伸手去按那只涎皮赖脸的大脸猫闹钟。
5.下列各句中,没有语病的一句是()A.铃声响起,他急忙拿起听筒,电话那端传来一个操普通话的中年男子的声音说:“你被录取了。
”B.这位江西姑娘在51届“世界小姐”大赛中脱颖而出,获得第四名,成为本次大赛惟一进入“十佳”殊荣的亚洲选手。
C.老人在80岁的时候,还清楚地记得哥哥参加学生运动时对自己的评价:一个温情主义者。
D.“耶路撒冷”是和平之城的意思,却又是遭受劫难最多的城市,可是长期的冲突并没有使其失去迷人的魅力,从而使旅游者望而却步。
6.依次填入下面横线处的语句,与上下文衔接最恰当的一项是()“说话”艺术出现后,民间的话本如《三国志平话》等,既粗糙又不准确,;而其中的,后来在文人再创作的《三国演义》中都得到了改正与润色、丰富,从而取得了更高的艺术成就。
高考得分点基础训练 16
高考得分点基础训练(16)II.Grammar and VocabularyAMost of us can find 15 minutes or half an hour each day for some specific regular activity. For example, one famous surgeon always made ___25___ a rule to spend at least 15 minutes on general reading ___26___ he went to sleep each day. Whether he went to bed at 10 pm or 2:30 am made no difference.Nearly all “speed reading”courses have a “pacing”element—some timing device ___27___ lets the students know how many words a minutes he is reading. You can do this simply by looking at your watch every 5 or 10 minutes and ___28___ (note) down the page number you have reached.Obviously there is little point in ___29___ (increase) your reading speed if you do not understand ___30___ you are reading. If you find you have lost the thread of the story, or you can’t remember clearly the details of what was said, re-read the section or chapter.___31___ (take) four or five pages of an interesting book you happen to be reading at the time. Read them as fast as you possibly can. Do not bother about whether you understand or not. After a “lightning speed”of reading, you ___32___ (find) that your“normal”speed has increased. Most paragraphs in an article have a “topic sentence” which expresses the central ideas. The opening paragraph often suggests the general direction and content of the piece, while paragraphs that follow expand or support the first. The closing paragraph often summarizes the very essence (实质) of what has been said.25_________________26.______________ 27_______________.28.______________29.________________30._______________31._______________32_______________BBoth my parents came from towns in Mexico. I was born in Texas, and when I was four, my family moved to a housing project in East Los Angeles.Even though we struggled ___33___ (make) ends meet, my parents stressed to me and my four brothers and sisters ___34___ fortunate we were to live in a great country with limitless opportunities. They imbued (灌输) in us the concepts of family, faith and patriotism.I got my first real job when I was ten. My dad, Benjamin, ___35___ (injure) on the back working ina cardboard-box factory was retrained as a hairstylist. He rented space in a little mall and gave his shop the fancy name of Mr. Ben’s Coiffure.The owner of the shopping centre gave Dad a discount on his rent for cleaning the parking lot three nights a week, ___36___ meant getting up at 3 a.m. ___37___ (pick) up trash, Dad used a little machine that looked like a lawn mower. Mom and I emptied garbage cans and picked up litter by hand. It took two to three hours to clean the lot. I’d sleep in the car on the way home.I did this for two years, but the lessons I learned have lasted a lifetime. I acquired discipline and a strong work ethic, and learned ___38___ an early age the importance of balancing life’s competing interests—in my case, school, homework and a job. This really helped during my senior year of high school, when I worked 40 hours a week flipping burgers at a fast-food joint while ___39___ (take) a full load of percolate courses.The hard work paid off. I attended the U.S. Military Academy and went on to receive graduatedegrees in law and business from Harvard. ___40___, I joined a big Los Angeles law firm and was elected to the California state assembly. In these jobs and in everything else I’ve done, I have never forgotten those days in the parking lot. The experience taught me that there is dignity in all work and that if people are working to provide for themselves and their families that is something we should honour.33.______________ 34_______________.35.______________ 36.________________37._______________38._______________39____________ 40.__________________▲Section B (10分)A. orderB. balancedC. avoidD. amongE. bodiesF. nothingG. satisfy H. attributed I. popularity J. keeps K. followingFast food is becoming more and more popular in China, especially ___41___ children and teenagers. It seems that ___42___ is more representative of the fast pace of the modern society than the fast food. There are several reasons for its ___43___. First, it is quick and convenient. Go into a fast food restaurant, your food will be ready in a minute. And you can___44___ your hunger instantly. Precious time won’t be wasted waiting in line to ___45___ or waiting at the table for the food to arrive. Second, its popularity is also ___46___ to the clean food, excellent service and pleasant environment of the fast food restaurant.But in my opinion, fast food is not healthy enough. It doesn’t compose ___47___ diets and it is low in nutrition. Doctors suggest that we should ___48___ eating too much fast food. Though home-cooking is a bit time-wasting and ___49___ washing-up tiresomeness, it offers us delicious, healthy and nutritious meals that our ___50___ like and need. In fact, fast food is just a good choice when we are in a hurry and we should turn to it once in a while.III.Reading ComprehensionBThat cold January night, I was growing sick of my life in San Francisco. There I was, walking home at one in the morning after a tiring practice at the theatre. With opening night only a week away, I was still learning my lines. I was having trouble dealing with my part-time job at the bank and my acting at night at the same time. As I walked, I thought seriously about giving up both acting and San Francisco. City life had become too much for me.As I walked down empty streets under tall buildings, I felt very small and cold. I began running, both to keep warm and to keep away from any possible robbers. Very few people were still out except a few sad-looking homeless people under blankets.About a block from my apartment, I heard a sound behind me. I turned quickly, half expecting to see someone with a knife or a gun. The street was empty. All I saw was a shining streetlight. Still, the noise had made me nervous, so I started to run faster. Not until I reached my apartment building and unlocked the door did I realize what the noise had been. It had been my wallet falling to the sidewalk.Suddenly I wasn’t cold or tired anymore. I ran out of the door and back to where I’d heard the noise. Although I searched the sidewalk anxiously for fifteen minutes, my wallet was nowhere to be found. Just as I was about to give up the search, I heard the garbage truck pull up to the sidewalk next to me. When a voice called from the inside “Alisa Camacho?”, I thought I w as dreaming. How couldthis man know my name? The door opened, and out jumped a small red-haired man with an amused look in his eyes. “Is this what you’re looking for?”he asked, holding up a small square shape.It was nearly 3 am by the time I got into bed. I wouldn’t get much sleep that night, but I had gotten my wallet back. I also had gotten back some enjoyment of city life. I realized that the city couldn’t be a bad place as long as people were willing to help each other.( ) 70. How did the writer feel when she was walking home after work?A. Cold and sick.B. Fortunate and hopeful.C. Satisfied and cheerful.D. Disappointed and helpless.( ) 71. From the first paragraph, we learn that the writer was busy _________.A. solving her problem at the bankB. taking part in various city activitiesC. learning acting in an evening schoolD. preparing for the first night show( ) 72. On her way home the writer _________.A. lost her wallet unknowinglyB. was stopped by a garbage truck driverC. was robbed of her wallet by an armed manD. found some homeless people following her( ) 73. From the passage, we can infer that the writer would _________.A. stop working at nightB. stay on in San FranciscoC. make friends with cleanersD. give up her job at the bankCAccording to experts, the word “Hutong” originated from Mongolian language meaning “well”. In ancient times, people tended to gather and live around wells. So the original meaning of Hutong should be “a place where people gather and live”. Another explanation says that during the Yuan Dynasty, residential areas in the city were divided into many divisions. Between the smaller divisions were passageways for people to travel through. And those passageways also functioned as isolation belts against fire risks. In Mongolian language, passageways of his kind were called Hutong. But no matter what Hutong exactly means, one thing is for sure, that is, Hutong first appeared in Beijing during the Yuan Dynasty. In the early 13th century, a Mongolian tribe from the north became very stong. Led by Genghis Khan (成吉思汗), the Mongolian occupied Beijing, the capital of the Jin Dynasty. In the year 1271, Kubla Khan (忽必烈), the grandson of Genghis Khan, founded the Yuan Dynasty and set Beijing as the capital in the following year. Unfortunately, the old city was completely destroyed during the war. So they had to rebuild it. In old China, all the structures and roads were required to be symmetrical. So the city was well designed. First, they had to find a centre line, and then built a regular square city. The layout (设计) of the city was very much like a chessboard (棋盘). About 50 residential areas were constructed, with straight roads and Hutongs in between. At that time, there was a clear definition for avenue, street and Hutong. A 37-metre-wide road was called an avenue, an 18-metre-wide one was called a street, and a 9-metre-wide lane was called a Hutong. Most of today’s Hutongs were formed during the Ming and Qing Dynasty.Nobody knows exactly how many Hutongs there are in the present Beijing. But one thing is for sure, if we connected all the Hutongs together, their total length would ever be longer than the Great Wall, which is about 4,000 miles long. Or to make it clear, it could build a highway from Seattle to Boston, all across America!The main attraction of Hutong life is friendly and interpersonal communication. Children living in one courtyard play together and grow up together like one big family. So now our government is trying to preserve such Hutong area in Beijing. Without permission, nobody is allowed to tear down old houses to build high-rise apartments. We want to save it as a treasure to show later generations what Beijing used to be like.( ) 74. What is the main idea of the first paragraph?A. The origin of Hutong.B. The design of Hutong.C. The development of Hutong.D. The definition of Hutong.( ) 75. Which of the following shape is symmetrical (paragraph 1)?( ) 76. Which of the following statement is NOT true about Hutong?A. The word “Hutong” is from Mongolian language.B. Hutong originally means “a place where people gather and live.”C. Hutong first appeared in Beijing during the Yuan Dynasty.D. Hutong is about 18 meters’ wide.( ) ▲77. Why does our government try to preserve Hutong areas?A. Because children like to play Huong areas.B. Because nobody is permitted to tear down old houses.C. Because Hutong can promote the communication between neighbours.D. Because it worth a lot of money.I.Translation (22分)1. 做作业占据了每天很大部分的时间。
2023-2024学年高考数学专项复习——压轴题(附答案)
决胜3.已知函数,曲线在处的切线方程为.()2e xf x ax =-()y f x =()()1,1f 1y bx =+(1)求的值:,a b (2)求在上的最值;()f x []0,1(3)证明:当时,.0x >()e 1e ln 0x x x x +--≥4.已知函数,.()()ln 1f x x x a x =-++R a ∈(1)若,求函数的单调区间;1a =()f x (2)若关于的不等式在上恒成立,求的取值范围;x ()2f x a≤[)2,+∞a (3)若实数满足且,证明.b 21a b <-+1b >()212ln f x b <-5.椭圆的离心率是,点是椭圆上一点,过点2222:1(0)x y E a b a b +=>>22()2,1M E 的动直线与椭圆相交于两点.()0,1P l ,A B (1)求椭圆的方程;E (2)求面积的最大值;AOB (3)在平面直角坐标系中,是否存在与点不同的定点,使恒成立?存在,xOy P Q QA PAQB PB=求出点的坐标;若不存在,请说明理由.Q 6.已知函数,.()21ln 2f x a x x⎛⎫=-+ ⎪⎝⎭()()()2R g x f x ax a =-∈(1)当时,0a =(i )求曲线在点处的切线方程;()y f x =()()22f ,(ii )求的单调区间及在区间上的最值;()f x 1,e e ⎡⎤⎢⎥⎣⎦(2)若对,恒成立,求a 的取值范围.()1,x ∀∈+∞()0g x <(1)求抛物线的表达式和的值;,t k (2)如图1,连接AC ,AP ,PC ,若△APC 是以(3)如图2,若点P 在直线BC 上方的抛物线上,过点的最大值.12CQ PQ +(1)【基础训练】请分别直接写出抛物线的焦点坐标和准线l 的方程;22y x =(2)【技能训练】如图2所示,已知抛物线上一点P 到准线l 的距离为6,求点P 的坐218y x =标;(3)【能力提升】如图3所示,已知过抛物线的焦点F 的直线依次交抛物线及准()20y ax a =>线l 于点,若求a 的值;、、A B C 24BC BF AF ==,(4)【拓展升华】古希腊数学家欧多克索斯在深入研究比例理论时,提出了分线段的“中末比”问题:点C 将一条线段分为两段和,使得其中较长一段是全线段与另一AB AC CB AC AB 段的比例中项,即满足:,后人把这个数称为“黄金分割”,把CB 512AC BC AB AC -==512-点C 称为线段的黄金分割点.如图4所示,抛物线的焦点,准线l 与y 轴AB 214y x=(0,1)F 交于点,E 为线段的黄金分割点,点M 为y 轴左侧的抛物线上一点.当(0,1)H -HF 时,求出的面积值.2MH MF=HME 10.已知双曲线的一条渐近线方程的倾斜角为,焦距为4.2222:1(0,0)x y C a b a b -=>>60︒(1)求双曲线的标准方程;C (2)A 为双曲线的右顶点,为双曲线上异于点A 的两点,且.C ,M N C AM AN ⊥①证明:直线过定点;MN ②若在双曲线的同一支上,求的面积的最小值.,M N AMN(1)试用解析几何的方法证明:(2)如果将圆分别变为椭圆、双曲线或抛物线,你能得到类似的结论吗?13.对于数集(为给定的正整数),其中,如果{}121,,,,n X x x x =-2n ≥120n x x x <<<< 对任意,都存在,使得,则称X 具有性质P .,a b X ∈,c d X ∈0ac bd +=(1)若,且集合具有性质P ,求x 的值;102x <<11,,,12x ⎧⎫-⎨⎬⎩⎭(2)若X 具有性质P ,求证:;且若成立,则;1X ∈1n x >11x =(3)若X 具有性质P ,且,求数列的通项公式.2023n x =12,,,n x x x 14.已知,是的导函数,其中.()2e xf x ax =-()f x '()f x R a ∈(1)讨论函数的单调性;()f x '(2)设,与x 轴负半轴的交点为点P ,在点P()()()2e 11x g x f x x ax =+-+-()y g x =()y g x =处的切线方程为.()y h x =①求证:对于任意的实数x ,都有;()()g x h x ≥②若关于x 的方程有两个实数根,且,证明:()()0g x t t =>12,x x 12x x <.()2112e 11e t x x --≤+-15.在平面直角坐标系中,一动圆经过点且与直线相切,设该动圆圆心xOy 1,02A ⎛⎫ ⎪⎝⎭12x =-的轨迹为曲线K ,P 是曲线K 上一点.(1)求曲线K 的方程;(2)过点A 且斜率为k 的直线l 与曲线K 交于B 、C 两点,若且直线OP 与直线交//l OP 1x =于Q 点.求的值;||||AB ACOP OQ ⋅⋅(3)若点D 、E 在y 轴上,的内切圆的方程为,求面积的最小值.PDE △()2211x y -+=PDE △16.已知椭圆C :,四点中恰有三()222210x y a b a b +=>>()()1234331,1,0,1,1,,1,22P P P P ⎛⎫⎛⎫- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭点在椭圆C 上.(1)求椭圆C 的方程;(2)设直线l 不经过P 2点且与C 相交于A ,B 两点,若直线与直线的斜率的和为,2P A 2P B 1-证明:l 过定点.18.给定正整数k ,m ,其中,如果有限数列同时满足下列两个条件.则称2m k ≤≤{}n a 为数列.记数列的项数的最小值为.{}n a (,)k m -(,)k m -(,)G k m 条件①:的每一项都属于集合;{}n a {}1,2,,k 条件②:从集合中任取m 个不同的数排成一列,得到的数列都是的子列.{}1,2,,k {}n a 注:从中选取第项、第项、…、第项()形成的新数列{}n a 1i 2i 5i 125i i i <<<…称为的一个子列.325,,,i i i a a a ⋯{}n a (1)分别判断下面两个数列,是否为数列.并说明理由!(33)-,数列;1:1,2,3,1,2,3,1,2,3A 数列.2:1,2,3,2,1,3,1A (2)求的值;(),2G k (3)求证.234(,)2k k G k k +-≥答案:1.(1)极大值为,无极小值2e (2)证明见解析【分析】(1)求导,根据导函数的符号结合极值的定义即可得解;(2)构造函数,利用导数求出函数的最小值,再()21()()()2ln 12F x f x g x x x x x x =+=+->证明即可或者转换不等式为,通过构造函数可得证.()min0F x >()112ln 012x x x +->>【详解】(1)的定义域为,,()f x (0,)+∞()2(1ln )f x x '=-+当时,,当时,,10e x <<()0f x '>1e x >()0f x '<所以函数在上单调递增,在上单调递减,()f x 10,e ⎛⎫ ⎪⎝⎭1,e ⎛⎫+∞ ⎪⎝⎭故在处取得极大值,()f x 1e x =12e e f ⎛⎫= ⎪⎝⎭所以的极大值为,无极小值;()f x 2e (2)设,()21()()()2ln 12F x f x g x x x x x x =+=+->解法一:则,()2ln 1F x x x '=--令,,()()2ln 11h x x x x =-->22()1x h x x x -'=-=当时,,单调递减,当时,,单调递增,12x <<()0h x '<()h x 2x >()0h x '>()h x 又,,,(2)1ln 40h =-<(1)0h =(4)32ln 40h =->所以存在,使得,即.0(2,4)x ∈0()0h x =002ln 10x x --=当时,,即,单调递减,01x x <<()0h x <()0F x '<()F x 当时,,即,单调递增,0x x >()0h x >()0F x '>()F x 所以当时,在处取得极小值,即为最小值,1x >()F x 0x x =故,22000000(11()()12ln )222F x F x x x x x x ≥=+-=-+设,因为,2000122()p x x x =-+0(2,4)x ∈由二次函数的性质得函数在上单调递减,2000122()p x x x =-+(2,4)故,0()(4)0p x p >=所以当时,,即.1x >()0F x >()()0f x g x +>解法二:要证,即证,()0F x >()1()12ln 012p x x x x =+->>因为,所以当时,,单调递减,()124()122x p x x x x -'=-=>()1,4x ∈()0p x '<()p x 当时,,单调递增,()4,x ∞∈+()0p x '>()p x 所以,所以,即.()()4212ln 434ln 20p x p ≥=+-=->()0F x >()()0f x g x +>方法点睛:利用导数证明不等式问题,方法如下:(1)直接构造函数法:证明不等式(或)转化为证明()()f xg x >()()f xg x <(或),进而构造辅助函数;()()0f xg x ->()()0f xg x -<()()()h x f x g x =-(2)适当放缩构造法:一是根据已知条件适当放缩;二是利用常见放缩结论;(3)构造“形似”函数,稍作变形再构造,对原不等式同解变形,根据相似结构构造辅助函数.2.(1)0(2)证明详见解析(3)2a ≤【分析】(1)利用导数求得的最小值.()g x (2)根据(1)的结论得到,利用放缩法以及裂项求和法证得不等式成立.2211ln 1n n ⎛⎫+≤ ⎪⎝⎭(3)由不等式分离参数,利用构造函数法,结合导数求得的取ln (2)10xx x x a x -+--≥a a 值范围.【详解】(1)依题意,,()21ln (,0)2f x x x x t t x =-+∈>R 所以,()()()()ln 1ln 10g x f x x x x x x '==-+=-->,所以在区间上单调递减;()111x g x x x -'=-=()g x ()0,1()()0,g x g x '<在区间上单调递增,()1,+∞()()0,g x g x '>所以当时取得最小值为.1x =()g x ()11ln110g =--=(2)要证明:对任意正整数,都有,(2)n n ≥222211111111e 234n ⎛⎫⎛⎫⎛⎫⎛⎫+⋅+⋅++< ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭ 即证明,22221111ln 1111ln e234n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫+⋅+⋅++< ⎪ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦ 即证明,222111ln 1ln 1ln 1123n ⎛⎫⎛⎫⎛⎫++++++< ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 由(1)得,即()()()10f xg x g '=≥=ln 10,ln 1x x x x --≥≤-令,所以, *211,2,N x n n n =+≥∈222111ln 111n n n ⎛⎫+≤+-= ⎪⎝⎭所以222222111111ln 1ln 1ln 12323n n ⎛⎫⎛⎫⎛⎫++++++≤+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ ,()111111111122312231n n n n <+++=-+-++-⨯⨯-- 111n=-<所以对任意正整数,都有.(2)n n ≥222211111111e 234n ⎛⎫⎛⎫⎛⎫⎛⎫+⋅+⋅++< ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭ (3)若不等式恒成立,此时,ln (2)10xx x x a x -+--≥0x >则恒成立,ln 21x x x x x a x -+-≤令,()ln 21xx x x x h x x -+-=令,()()()e 10,e 10x x u x x x u x '=--≥=-≥所以在区间上单调递增,()u x[)0,∞+所以,当时等号成立,()0e 010,e 10,e 1x x u x x x ≥--=--≥≥+0x =所以,()ln e ln 21ln 1ln 212x x x x x x x x x x h x x x -+-+-+-=≥=当时等号成立,所以.ln 0,1x x x ==2a ≤利用导数求函数的最值的步骤:求导:对函数进行求导,得到它的导函数.导函数()f x ()f x '表示了原函数在不同点处的斜率或变化率.找出导数为零的点:解方程,找到使得导()0f x '=数为零的点,这些点被称为临界点,可能是函数的极值点(包括最大值和最小值),检查每个临界点以及区间的端点,并确认它们是否对应于函数的最值.3.(1),1a =e 2b =-(2);()max e 1f x =-()min 1f x =(3)证明见解析【分析】(1)利用切点和斜率列方程组,由此求得.,a b (2)利用多次求导的方法求得在区间上的单调性,由此求得在上的最值.()f x []0,1()f x []0,1(3)先证明时,,再结合(2)转化为,从0x >()()e 21f x x ≥-+()21e ln e x x x x x+--≥+而证得不等式成立.【详解】(1),()e 2x f x ax'=-∴,解得:,;()()1e 21e 1f a b f a b ⎧=-=⎪⎨=-=+'⎪⎩1a =e 2b =-(2)由(1)得:,()2e xf x x =-,令,则,()e 2x f x x '=-()e 2x h x x=-()e 2x h x '=-是增函数,令解得.()h x ()0h x '=ln 2x =∴,也即在上单调递减,()h x ()f x '()0,ln2()()0,h x h x '<在上单调递增,()ln2,+∞()()0,h x h x '>∴,∴在递增,()()ln 2ln222ln20h f ==->'()f x []0,1∴;;()()max 1e 1f x f ==-()()min 01f x f ==(3)∵,由(2)得过,()01f =()f x ()1,e 1-且在处的切线方程是,()y f x =1x =()e 21y x =-+故可猜测且时,的图象恒在切线的上方,0x >1x ≠()f x ()e 21y x =-+下面证明时,,设,,0x >()()e 21f x x ≥-+()()()e 21g x f x x =---()0x >∴,∴令,()()e 2e 2x g x x =---'()()()e 2e 2x x x g m x '--==-,()e 2x m x '=-由(2)得:在递减,在递增,()g x '()0,ln2()ln2,+∞∵,,,∴,()03e 0g '=->()10g '=0ln21<<()ln20g '<∴存在,使得,()00,1x ∈()0g x '=∴时,,时,,()()00,1,x x ∈⋃+∞()0g x '>()0,l x x ∈()0g x '<故在递增,在递减,在递增.()g x ()00,x ()0,1x ()1,+∞又,∴当且仅当时取“”,()()010g g ==()0g x ≥1x ==()()2e e 210x g x x x =----≥故,,由(2)得:,故,()e e 21x x xx+--≥0x >e 1x x ≥+()ln 1x x ≥+∴,当且仅当时取“=”,∴,1ln x x -≥1x =()e e 21ln 1x x x x x+--≥≥+即,∴,()21ln 1e e x x x x+--≥+()21e ln e x x x x x+--≥+即成立,当且仅当时“=”成立.()1ln 10e e x x x x +---≥1x =求解切线的有关的问题,关键点就是把握住切点和斜率.利用导数研究函数的单调性,如果一次求导无法求得函数的单调性时,可以考虑利用多次求导来进行求解.利用导数证明不等式恒成立,如果无法一步到位的证明,可以先证明一个中间不等式,然后再证得原不等式成立.4.(1)单调增区间为,单调减区间为;()0,1()1,+∞(2)(],2ln 2-∞(3)证明见解析【分析】(1)求导,再根据导函数的符号即可得解;(2)分离参数可得,构造函数,利用导数求出函数的最小ln 1x x a x ≤-ln (),21x xg x x x =≥-()g x 值即可得解;(3)由,得,则,要证21a b <-+21a b -<-2112()(e )e e 1a a b f x f a b ---≤=+<-+,即证,即证,构造函数()212ln f x b<-222e112ln bb b --+<-22212ln 0eb b b +-<,证明即可.()()()12ln e x h x x x x =>-()1h x <-【详解】(1)当时,,1a =()ln 1,0f x x x x x =-++>,由,得,由,得,()ln f x x '=-()0f x '>01x <<()0f x '<1x >故的单调增区间为,单调减区间为;()f x ()0,1()1,+∞(2),()ln 2,1x xf x a a x ≤∴≤- 令,ln (),21x x g x x x =≥-则,21ln ()(1)x xg x x --'=-令,则,()ln 1t x x x =-+11()1xt x x x -'=-=由,得,由,得,()0t x '>01x <<()0t x '<1x >故在递增,在递减,,()t x ()0,1()1,+∞max ()(1)0t x t ==,所以,()0t x ∴≤ln 1≤-x x 在上单调递增,,()0,()g x g x '≥∴[)2,+∞()min ()2g x g ∴=,(2)2ln 2a g ∴≤=的取值范围;a ∴(],2ln 2-∞(3),221,1b a b a <-+∴-<- 又,在上递增,11()(e )e a a f x f a --≤=+1e a y a -=+ R a ∈所以,2112()(e )e e 1a a b f x f a b ---≤=+<-+下面证明:,222e 112ln b b b --+<-即证,22212ln 0ebb b +-<令,则,21x b =>12ln 0e x x x +-<即,(2ln )e 1xx x -⋅<-令,则,()()()12ln e xh x x x x =>-()22ln 1e xh x x x x '⎛⎫=-+-⋅ ⎪⎝⎭令,则,()2()2ln 11x x x x x ϕ=-+->()()2221122()101x x x x x x ϕ---=--=<>∴函数在上单调递减,()x ϕ()1,+∞,()(1)0x ϕϕ∴<=在递减,()()0,h x h x '∴<(1,)+∞,()()1e 1h x h ∴<=-<-所以.()212ln f x b <-方法点睛:利用导数证明不等式问题,方法如下:(1)直接构造函数法:证明不等式(或)转化为证明()()f xg x >()()f xg x <(或),进而构造辅助函数;()()0f xg x ->()()0f xg x -<()()()h x f x g x =-(2)适当放缩构造法:一是根据已知条件适当放缩;二是利用常见放缩结论;(3)构造“形似”函数,稍作变形再构造,对原不等式同解变形,根据相似结构构造辅助函数.5.(1)22142x y +=(2)2(3)存在,.()0,2Q 【分析】(1)由离心率及过点列方程组求解.()2,1M,a b (2)设直线为与椭圆方程联立,将表达为的函数,由基本不l 1y kx =+1212AOB S x x =⋅- k 等式求最大值即可.(3)先讨论直线水平与竖直情况,求出,设点关于轴的对称点,证得()0,2Q B y B '三点共线得到成立.,,Q A B 'QA PAQB PB=【详解】(1)根据题意,得,解得,椭圆C 的方程为.2222222211c a a b c a b ⎧=⎪⎪⎪=+⎨⎪⎪+=⎪⎩222422a b c ⎧=⎪=⎨⎪=⎩22142x y +=(2)依题意,设,直线的斜率显然存在,()()1122,,,A x y B x y l 故设直线为,联立,消去,得,l 1y kx =+221142y kx x y =+⎧⎪⎨+=⎪⎩y ()2212420k x kx ++-=因为直线恒过椭圆内定点,故恒成立,,l ()0,1P 0∆>12122242,1212k x x x x k k +=-=-++故,()2221212221224212111214414222122AOBk S x x x x x x k k k k ⋅+⎛⎫⎛⎫=⋅=⨯-=⨯-⨯= ⎪ ⎪+⎝-+-⎝++⎭⎭- 令,所以,当且仅当,即时取得214,1t k t =+≥22222211AOB t S t t t=×=×£++1t =0k =等号,综上可知:面积的最大值为.AOB 2(3)当平行于轴时,设直线与椭圆相交于两点,如果存在点满足条件,l x ,C D Q 则有,即,所以点在轴上,可设的坐标为;||||1||||QC PC QD PD ==QC QD =Q y Q ()00,y 当垂直于轴时,设直线与椭圆相交于两点,如果存在点满足条件,l x ,M N Q 则有,即,解得或,||||||||QM PM QN PN =00221212y y --=++01y =02y =所以若存在不同于点的定点满足条件,则点的坐标为;P Q Q ()0,2当不平行于轴且不垂直于轴时,设直线方程为,l x x l 1y kx =+由(2)知,12122242,1212k x x x x k k --+==++又因为点关于轴的对称点的坐标为,B y B '()22,x y -又,,11111211QA y kx k k x x x --===-22222211QB y kx k k x x x '--===-+--.方法点睛:直线与椭圆0Ax By C ++=时,取得最大值2222220a A b B C +-=MON S 6.(1)(i );(322ln 220x y +--=(2)11,22⎡⎤-⎢⎥⎣⎦故曲线在点处的切线方程为,()y f x =()()22f ,()()32ln 222y x --+=--即;322ln 220x y +--=(ii ),,()21ln 2f x x x =-+()0,x ∈+∞,()211x f x x x x -'=-+=令,解得,令,解得,()0f x ¢>()0,1x ∈()0f x '<()1,x ∈+∞当时,,1,e e x ⎡⎤∈⎢⎥⎣⎦()()max 112f x f ==-又,,221111ln 1e 2e e 2e f ⎛⎫=-+=-- ⎪⎝⎭()2211e e ln e e 122f =-+=-+其中,()222211111e 1e 1e 20e 2e 222ef f ⎛⎫⎛⎫-=----+=--> ⎪ ⎪⎝⎭⎝⎭故,()()2min 1e e 12f x f ==-+故的单调递增区间为,单调递减区间为;()f x ()0,1()1,+∞在区间上的最大值为,最小值为;()f x 1,e e ⎡⎤⎢⎥⎣⎦12-21e 12-+(2),()21ln 22xg x a x x a ⎭-+⎛=⎪-⎫ ⎝对,恒成立,()1,x ∀∈+∞21ln 202a x x ax ⎛⎫-+-< ⎪⎝⎭变形为对恒成立,ln 122x a xa x<--⎛⎫ ⎪⎝⎭()1,x ∀∈+∞令,则,()(),1,ln x h x x x ∈=+∞()21ln xh x x -'=当时,,单调递增,()1,e x ∈()0h x '>()ln xh x x =当时,,单调递减,()e,+x ∈∞()0h x '<()ln xh x x =其中,,当时,恒成立,()10h =()ln e 1e e e h ==1x >()ln 0x h x x =>故画出的图象如下:()ln x h x x =其中恒过点122y xa a ⎛⎫ ⎪⎝=⎭--(2,1A 又,故在()210111h -'==()ln x h x x =又在上,()2,1A 1y x =-()对于2111644y x x =-+-∴点,即()0,6C -6OC =∵2114,14P m m m ⎛-+- ⎝∴点,3,64N m m ⎛⎫- ⎪⎝⎭∴,22111316624444PN m m m m m⎛⎫=-+---=-+ ⎪⎝⎭∵轴,PN x ⊥∴,//PN OC ∴,PNQ OCB ∠=∠∴,Rt Rt PQN BOC ∴,PN NQ PQ BC OC OB ==∵,8,6,10OB OC BC ===∴,34,55QN PN PQ PN==∵轴,NE y ⊥∴轴,//NE x ∴,CNE CBO ∴,5544CN EN m ==∴,2215111316922444216CQ PQ m m m m ⎛⎫+=-+=--+⎪⎝⎭当时,取得最大值.132m =12CQ PQ+16916关键点点睛:熟练的掌握三角形相似的判断及性质是解决本题的关键.8.(1)详见解析;(2)①具有性质;理由见解析;②P 1346【分析】(1)当时,先求得集合,由题中所给新定义直接判断即可;10n =A (2)当时,先求得集合, 1010n =A ①根据,任取,其中,可得,{}2021|T x x S =-∈02021t x T =-∈0x S ∈0120212020x ≤-≤利用性质的定义加以验证,即可说明集合具有性质;P T P ②设集合有个元素,由(1)可知,任给,,则与中必有个S k x S ∈12020x ≤≤x 2021x -1不超过,从而得到集合与中必有一个集合中至少存在一半元素不超过,然后利1010S T 1010用性质的定义列不等式,由此求得的最大值.P k【详解】(1)当时,,10n ={}1,2,,19,20A = 不具有性质,{}{}|910,11,12,,19,20B x A x =∈>= P 因为对任意不大于的正整数,10m 都可以找到该集合中的两个元素与,使得成立,110b =210b m =+12||b b m -=集合具有性质,{}*|31,N C x A x k k =∈=-∈P 因为可取,对于该集合中任一元素,110m =<,(),都有.112231,31c k c k =-=-*12,N k k ∈121231c c k k -=-≠(2)当时,集合,1010n ={}()*1,2,3,,2019,2020,1010N A m m =≤∈ ①若集合具有性质,那么集合一定具有性质.S P {}2021|T x x S =-∈P 首先因为,任取,其中.{}2021|T x x S =-∈02021t x T =-∈0x S ∈因为,所以.S A ⊆{}01,2,3,,2020x ∈ 从而,即,所以.0120212020x ≤-≤t A ∈T A ⊆由具有性质,可知存在不大于的正整数,S P 1010m 使得对中的任意一对元素,都有.s 12,s s 12s s m -≠对于上述正整数,从集合中任取一对元素,m {}2021|T x x S =-∈112021t x -=,其中,则有.222021t x =-12,x x S ∈1212t t s s m --≠=所以,集合具有性质P ;{}2021|T x x S =-∈②设集合有个元素,由(1)可知,若集合具有性质,S k S P 那么集合一定具有性质.{}2021|T x x S =-∈P 任给,,则与中必有一个不超过.x S ∈12020x ≤≤x 2021x -1010所以集合与中必有一个集合中至少存在一半元素不超过.S T 1010不妨设中有个元素不超过.S 2k t t ⎛⎫≥ ⎪⎝⎭12,,,t b b b 1010由集合具有性质,可知存在正整数.S P 1010m ≤使得对中任意两个元素,都有.S 12,s s 12s s m -≠所以一定有.12,,,t b m b m b m S +++∉ 又,故.100010002000i b m +≤+=121,,,b m b m b m A +++∈ 即集合中至少有个元素不在子集中,A t S 因此,所以,得.20202k k k t +≤+≤20202k k +≤1346k ≤当时,取,{}1,2,,672,673,,1347,,2019,2020S = 673m =则易知对集合中的任意两个元素,都有,即集合具有性质.S 12,y y 12673y y -≠S P 而此时集合S 中有个元素,因此,集合元素个数的最大值为.1346S 1346解新定义题型的步骤:(1)理解“新定义”——明确“新定义”的条件、原理、方法、步骤和结论.(2)重视“举例”,利用“举例”检验是否理解和正确运用“新定义”;归纳“举例”提供的解题方法.归纳“举例”提供的分类情况.(3)类比新定义中的概念、原理、方法,解决题中需要解决的问题.9.(1),10,8⎛⎫ ⎪⎝⎭18y =-(2)或()42,4()42,4-(3)14a =(4)或51-35-【分析】(1)根据焦点和准线方程的定义求解即可;(2)先求出点P 的纵坐标为4,然后代入到抛物线解析式中求解即可;(3)如图所示,过点B 作轴于D ,过点A 作轴于E ,证明,推BD y ⊥AE y ⊥FDB FHC ∽出,则,点B 的纵坐标为,从而求出,证明16FD a =112OD OF DF a =-=112a 36BD a =,即可求出点A 的坐标为,再把点A 的坐标代入抛物线解析式AEF BDF ∽123,24a ⎛⎫ ⎪⎝+⎭-中求解即可;(4)如图,当E 为靠近点F 的黄金分割点的时候,过点M 作于N ,则,MN l ⊥MN MF=先证明是等腰直角三角形,得到,设点M 的坐标为,则MNH △NH MN=21,4m m ⎛⎫⎪⎝⎭过点B 作轴于D ,过点BD y ⊥由题意得点F 的坐标为F ⎛ ⎝1FH =当E 为靠近点F 的黄金分割点的时候,过点∵在中,Rt MNH △sin MHN ∠∴,∴是等腰直角三角形,45MHN ︒=MNH △双曲线方程联立,利用韦达定理及题目条件可得,后由题意可得AM AN ⋅= ()()222131t t m -+=-所过定点坐标;②结合①及图形可得都在左支上,则可得,后由图象可得,M N 213m <,后通过令,结合单调性229113m S m +=-223113m λλ⎛⎫+=≤< ⎪⎝⎭()423313f x x x x ⎛⎫=-≤< ⎪⎝⎭可得答案.【详解】(1)设双曲线的焦距为,C 2c 由题意有解得.2223,24,,ba c c ab ⎧=⎪⎪=⎨⎪=+⎪⎩1,3,2a b c ===故双曲线的标准方程为;C 2213y x -=(2)①证明:设直线的方程为,点的坐标分别为,MN my x t =+,M N ()()1122,,,x y x y 由(1)可知点A 的坐标为,()1,0联立方程消去后整理为,2213y x my x t ⎧-=⎪⎨⎪=+⎩x ()222316330m y mty t --+-=可得,2121222633,3131mt t y y y y m m -+==--,()212122262223131m t tx x m y y t t m m +=+-=-=--,()()()()222222222121212122223363313131m t m t m t x x my t my t m y y mt y y t t m m m -+=--=-++=-+=----由,()()11111,,1,AM x y AN x y =-=-有()()()1212121212111AM AN x x y y x x x x y y ⋅=--+=-+++,()()()()22222222222222222132331313131313131t t t t t t m t t t m m m m m m -----++-=--++===------由,可得,有或,AM AN ⊥0AM AN ⋅=1t =-2t =当时,直线的方程为,过点,不合题意,舍去;1t =-MN 1my x =-()1,0当时,直线的方程为,过点,符合题意,2t =MN 2my x =+()2,0-②由①,设所过定点为121224,31x x x x m +==-若在双曲线的同一支上,可知,M N 有12240,31x x x m +=<-关键点睛:求直线所过定点常采取先猜后证或类似于本题处理方式,设出直线方程,通过题一方面:由以上分析可知,设椭圆方程为一方面:同理设双曲线方程为()22221y m x a b +-=,()2222221b x a k x m a b -+=化简并整理得()(2222222112ba k x a mk x a m ---+一方面:同理设抛物线方程为(22x p y =,()212x p k x n =+化简并整理得,由韦达定理可得12220pk x x pn --=2,2x x pk x x pn +=⋅=-(2)构造,故转化为等价于“对任()()()()()13131931x x xx f x k k g x f x +--==+++()()()123g x g x g x +>意,,恒成立”,换元后得到(),分,和1x 2x 3R x ∈()()11k g x q t t -==+3t ≥1k >1k =三种情况,求出实数k 的取值范围.1k <【详解】(1)由条件①知,当时,有,即在R 上单调递增.12x x <()()12f x f x <()f x 再结合条件②,可知存在唯一的,使得,从而有.0R x ∈()013f x =()093x x f x x --=又上式对成立,所以,R x ∀∈()00093x x f x x --=所以,即.0001393x x x --=0009313x x x ++=设,因为,所以单调递增.()93x x x xϕ=++()9ln 93ln 310x x x ϕ'=++>()x ϕ又,所以.()113ϕ=01x =所以;()931x x f x =++(2)构造函数,()()()()()13131931x x xx f x k k g x f x +--==+++由题意“对任意的,,,1x 2x 3R x ∈均存在以,,为三边长的三角形”()()()11113x f x k f x +-()()()22213x f x k f x +-()()()33313x f x k f x +-等价于“对任意,,恒成立”.()()()123g x g x g x +>1x 2x 3R x ∈又,令,()111313x x k g x -=+++1131231333x x x x t ⋅=++≥+=当且仅当时,即时取等号,91x=0x =则(),()()11k g x q t t -==+3t ≥当时,,因为且,1k >()21,3k g x +⎛⎤∈ ⎥⎝⎦()()122423k g x g x +<+≤()3213k g x +<≤所以,解得,223k +≤4k ≤即;14k <≤当时,,满足条件;1k =()()()1231g x g x g x ===当时,,因为且,1k <()2,13k g x +⎡⎫∈⎪⎢⎣⎭()()122423k g x g x ++<≤()3213k g x +<≤所以,即.2413k +≤112k -≤<综上,实数k 的取值范围是.1,42⎡⎤-⎢⎥⎣⎦复合函数零点个数问题处理思路:①利用换元思想,设出内层函数;②分别作出内层函数与外层函数的图象,分别探讨内外函数的零点个数或范围;③内外层函数相结合确定函数交点个数,即可得到复合函数在不同范围下的零点个数.13.(1)14x =(2)证明过程见解析(3),()112023k n k x --=1k n≤≤【分析】(1)由题意转化为对于,都存在,使得,其中(),m a b =(),n c d =0m n ⋅= ,选取,,通过分析求出;,,,a b c d X ∈()1,,2m a b x ⎛⎫== ⎪⎝⎭ ()(),1,n c d d ==- 14x =(2)取,,推理出中有1个为,则另一个为1,即,()()11,,m a b x x == (),n c d =,c d 1-1X ∈再假设,其中,则,推导出矛盾,得到;1k x =1k n <<101n x x <<<11x =(3)由(2)可得,设,,则有,记11x =()11,m s t =()22,n s t =1212s t t s =-,问题转化为X 具有性质P ,当且仅当集合关于原点对称,得到,,s B s X t X s t t ⎧⎫=∈∈>⎨⎬⎩⎭B ,共个数,由对称性可知也有个数,(){}234,0,,,,n B x x x x -∞=---- ()1n -()0,B +∞ ()1n -结合三角形数阵得到,得到数列为首项为1的等比123212321n n n n n n x x x x x x x x x x -----===== 12,,,n x x x 数列,设出公比为,结合求出公比,求出通项公式.q 2023n x =【详解】(1)对任意,都存在,使得,,a b X ∈,c d X ∈0ac bd +=即对于,都存在,使得,其中,(),m a b =(),n c d =0m n ⋅= ,,,a b c d X ∈因为集合具有性质P ,11,,,12x ⎧⎫-⎨⎬⎩⎭选取,,()1,,2m a b x ⎛⎫== ⎪⎝⎭ ()(),1,n c d d ==-则有,12x d -+=假设,则有,解得,这与矛盾,d x =102x x -+=0x =102x <<假设,则有,解得,这与矛盾,1d =-12x --=12x =-102x <<假设,则有,解得,这与矛盾,1d =12x -+=12x =102x <<假设,则有,解得,满足,12d =14x -+=14x =102x <<故;14x =(2)取,,()()11,,m a b x x == (),n c d =则,()10c d x +=因为,所以,即异号,120n x x x <<<< 0c d +=,c d 显然中有1个为,则另一个为1,即,,c d 1-1X ∈假设,其中,则,1k x =1k n <<101n x x <<<选取,,则有,()()1,,n m a b x x ==(),n s t =10n sx tx +=则异号,从而之中恰有一个为,,s t ,s t 1-若,则,矛盾,1s =-11n x tx t x =>≥若,则,矛盾,1t =-1n n x sx s x =<≤故假设不成立,所以;11x =(3)若X 具有性质P ,且,20231n x =>由(2)可得,11x =设,,则有,()11,m s t =()22,n s t =1212s t t s =-记,则X 具有性质P ,当且仅当集合关于原点对称,,,s B s X t X s t t ⎧⎫=∈∈>⎨⎬⎩⎭B 注意到是集合中唯一的负数,1-X 故,共个数,(){}234,0,,,,n B x x x x -∞=---- ()1n -由对称性可知也有个数,()0,B +∞ ()1n -由于,已经有个数,123421n n n n n nn n n n x x x x x x x x x x x x ----<<<<<< ()1n -对于以下三角形数阵:123421n n n n n n n n n n x x x x x xx x x x x x ----<<<<<< 1111123421n n n n n n n n x x x x xx x x x x --------<<<<< ……3321x x x x <21x x 注意到,123211111n n n x x x x x x x x x x -->>>>> 所以有,123212321n n n n n n x x x x x x x x x x -----===== 从而数列为首项为1的等比数列,设公比为,12,,,n x x x q 由于,故,解得,2023n x =112023n nx q x -==()112023n q -=故数列的通项公式为,.12,,,n x x x ()112023k n k x --=1k n ≤≤集合新定义问题,命题新颖,且存在知识点交叉,常常会和函数或数列相结合,很好的考虑了知识迁移,综合运用能力,对于此类问题,一定要解读出题干中的信息,正确理解问题的本质,转化为熟悉的问题来进行解决,要将“新”性质有机地应用到“旧”性质上,创造性的解决问题.14.(1)答案见解析(2)①证明见解析;②证明见解析【分析】(1)求出的导数,结合解不等式可得答案;()e 2x f x ax'=-(2)①,利用导数的几何意义求得的表达式,由此构造函数,()y h x =()()()F x g x h x =-利用导数判断其单调性,求其最小值即可证明结论;②设的根为,求得其表达式,()h x t=1x '并利用函数单调性推出,设曲线在点处的切线方程为,设11x x '≤()y g x =()0,0()y t x =的根为,推出,从而,即可证明结论.()t x t=2x '22x x '≥2121x x x x ''-≤-【详解】(1)由题意得,令,则,()e 2x f x ax'=-()e 2x g x ax=-()e 2x g x a'=-当时,,函数在上单调递增;0a ≤()0g x '>()f x 'R 当时,,得,,得,0a >()0g x '>ln 2x a >()0g x '<ln 2x a <所以函数在上单调递减,在上单调递增.()f x '(),ln 2a -∞()ln 2,a +∞(2)①证明:由(1)可知,令,有或,()()()1e 1x g x x =+-()0g x ==1x -0x =故曲线与x 轴负半轴的唯一交点P 为.()y g x =()1,0-曲线在点处的切线方程为,()1,0P -()y h x =则,令,则,()()()11h x g x '=-+()()()F x g x h x =-()()()()11F x g x g x '=--+所以,.()()()()11e 2e x F x g x g x '''=-=+-()10F '-=当时,若,,1x <-(],2x ∈-∞-()0F x '<若,令,则,()2,1x --()1()e 2e x m x x =+-()()e 30xm x x '=+>故在时单调递增,.()F x '()2,1x ∈--()()10F x F ''<-=故,在上单调递减,()0F x '<()F x (),1-∞-当时,由知在时单调递增,1x >-()()e 30x m x x '=+>()F x '()1,x ∈-+∞,在上单调递增,()()10F x F ''>-=()F x ()1,-+∞设曲线在点处的切线方程为()y g x =()0,0令()()()()(1e x T x g x t x x =-=+当时,2x ≤-()()2e x T x x =+-'()()2e xn x x =+-设,∴()()1122,,,B x y C x y 1x 又1211,22AB x AC x =+=+依题意,即,则,0bc <02x >()()220220004482x y c x x b =+---因为,所以,2002y x =0022x b c x -=-所以,()()00000242248122424S b c x x x x x -⋅=-++≥-⋅+=-=-当且仅当,即时上式取等号,00422x x -=-04x =所以面积的最小值为8.PDE △方法点睛:圆锥曲线中最值或范围问题的常见解法:(1)几何法,若题目的条件和结论能明显体现几何特征和意义,则考虑利用几何法来解决;(2)代数法,若题目的条件和结论能体现某种明确的函数关系,则可首先建立目标函数,再求这个函数的最值或范围.16.(1)2214x y +=(2)证明见解析(3)存在,7,,777⎛⎫⎛⎫-∞- ⎪ ⎪ ⎪ ⎪⎝+∞⎝⎭⎭ 【分析】(1)根据椭圆的对称性,得到三点在椭圆C 上.把的坐标代入椭圆234,,P P P 23,P P C ,求出,即可求出椭圆C 的方程;22,a b (2)当斜率不存在时,不满足;当斜率存在时,设,与椭圆方程联立,利():1l y kx t t =+≠用判别式、根与系数的关系,结合已知条件得到,能证明直线l 过定点;21t k =--()2,1-(3)利用点差法求出直线PQ 的斜率,从而可得直线PQ 的方程,与抛物线方程联14PQ k t =立,由,及点G 在椭圆内部,可求得的取值范围,设直线TD 的方程为,0∆>2t 1x my =+与抛物线方程联立,由根与系数的关系及,可求得m 的取值范围,进而可求得直线11DA TB k k =的斜率k 的取值范围.2l【详解】(1)根据椭圆的对称性,两点必在椭圆C 上,34331,,1,22P P ⎛⎫⎛⎫- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭又的横坐标为1,4P ∴椭圆必不过,()11,1P ∴三点在椭圆C 上.()234330,1,1,,1,22P P P ⎛⎫⎛⎫- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭把代入椭圆C ,()3231,20,1,P P ⎛⎫- ⎪ ⎪⎝⎭得,解得,222111314b a b ⎧=⎪⎪⎨⎪+=⎪⎩2241a b ⎧=⎨=⎩∴椭圆C 的方程为.2214x y +=(2)证明:①当斜率不存在时,设,,:l x m =()(),,,A A A m y B m y -∵直线与直线的斜率的和为,2P A 2P B 1-∴,221121A A P A P B y y k k m m m ----+=+==-解得m =2,此时l 过椭圆右顶点,不存在两个交点,故不满足.②当斜率存在时,设,,,:l y kx t =+1t ≠()()1122,,,A x y B x y 联立,消去y 整理得,22440y kx tx y =+⎧⎨+-=⎩()222148440k x ktx t +++-=则,,122814kt x x k -+=+21224414t x x k -=+则()()()()222112************111111P A P B x y x y x kx t x kx t y y k k x x x x x x -+-+-++---+=+==,()()()()()()12121222222448218114141144411142t k k kx x t tk t k t k k t t x t x x x +-+=--⋅+-⋅-++===--+-+又,∴,此时,1t ≠21t k =--()()222222644144464161664k t k t k t k ∆=-+-=-+=-故存在k ,使得成立,0∆>∴直线l 的方程为,即21y kx k =--()12y k x +=-∴l 过定点.()2,1-(3)∵点P ,Q 在椭圆上,所以,,2214P P x y +=2214Q Q x y +=两式相减可得,()()()()04PQ P Q P Q P Q y xy x x x y y +-++-=又是线段PQ 的中点,()1,G t -∴,2,2P Q P Q x x x x t+=-=∴直线PQ 的斜率,()144P Q P QP Q P QPQ x x k ty y x y y x +==-=--+∴直线PQ 的方程为,与抛物线方程联立消去x 可得,()114y x t t =++()22164410y ty t -++=由题可知,∴,()2161210t ∆=->2112t >又G 在椭圆内部,可知,∴,故,2114t +<234t <213124t <<设,,由图可知,,221212,,,44y y A y B y ⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭223434,,,44y y T y D y ⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭2134,y y y y >>∴,()2121216,441y y t y y t +==+当直线TD 的斜率为0时,此时直线TD 与抛物线只有1个交点,不合要求,舍去,设直线TD 的方程为,与抛物线方程联立,消去x 可得,()10x my m =+≠2440y my --=∴,34344,4y y m y y +==-由,可知,即,11//ATB D 11DA TB k k =3142222234214444y y y y y y y y --=--∴,即,1342y y y y +=+1243y y y y -=-∴,()()221212343444y y y y y y y y +-=+-∵,()()()()()222212124161641161210,128y y y y t t t +-=-+=-∈∴,解得,即,()()223434416160,128y y y y m +-=+∈27m <()7,7m ∈-∴直线TD 即的斜率.2l 771,77,k m ⎛⎫⎛⎫=∈-∞- ⎪ ⎪ ⎪ ⎪⎝+∞⎝⎭⎭ 思路点睛:处理定点问题的思路:(1)确定题目中的核心变量(此处设为),k (2)利用条件找到与过定点的曲线的联系,得到有关与的等式,k (),0F x y =k ,x y (3)所谓定点,是指存在一个特殊的点,使得无论的值如何变化,等式恒成立,()00,x y k 此时要将关于与的等式进行变形,直至找到,k ,x y ()00,x y ①若等式的形式为整式,则考虑将含的式子归为一组,变形为“”的形式,让括号中式k ()k ⋅子等于0,求出定点;②若等式的形式是分式,一方面可考虑让分子等于0,一方面考虑分子和分母为倍数关系,可消去变为常数.k 17.(1)1y =-(2)2ln23-+【分析】(1)由题意,将代入函数的解析式中,对函数进行求导,得到1m =()f x ()f x 和,代入切线方程中即可求解;()1f '()1f (2)得到函数的解析式,对进行求导,利用根的判别式以及韦达定理对()g x ()g x 进行化简,利用换元法,令,,可得,12122()()y x x b x x =--+12x t x =01t <<2(1)ln 1t y t t -=-+根据,求出的范围,构造函数,对进行求导,利用导数得到322m ≥t 2(1)()ln 1t h t tt -=-+()h t 的单调性和最值,进而即可求解.()h t 【详解】(1)已知(为常数),函数定义域为,()ln f x x mx =-m (0,)+∞当时,函数,1m =()ln f x x x =-可得,此时,又,11()1x f x x x -'=-=()=01f '()11=f -所以曲线在点处的切线方程为,即.()y f x =()()1,1f (1)0(1)y x --=⨯-1y =-(2)因为,函数定义域为,22()2()2ln 2g x f x x x mx x =+=-+(0,)+∞可得,222(1)()22x mx g x m x x x -+=-+='此时的两根,即为方程的两根,()0g x '=1x 2x 210x mx -+=因为,所以,由韦达定理得,,322m ≥240m ∆=->12x x m +=121=x x 又,所以1212lnx x b x x =-121212121212ln 22()()()()xx y x x b x x x x x x x x =--=--++-,11211211222212()ln 2ln 1x x x x x x x x x x x x --=-=⨯-++令,,所以,12x t x =01t <<2(1)ln 1t y t t -=-+因为,整理得,2212()x x m +=22212122x x x x m ++=因为,则,121=x x 2221212122x x x x m x x ++=等式两边同时除以,得,12x x 212212=x x m x x ++可得,因为,212t m t ++=322m ≥所以,,152t t +≥()()2252=2210t t x x -+--≥解得 或,则,12t ≤2t ≥102t <≤不妨设,函数定义域为,2(1)()ln 1t h t t t -=-+10,2⎛⎤⎥⎝⎦可得,22(1)()0(1)t h t t t -'=-<+所以函数在定义域上单调递减,()h t 此时,min 12()()ln223h t h ==-+故的最小值为.12122()()y x x b x x =--+2ln23-+利用导数求解在曲线上某点处的切线方程,关键点有两点,第一是切线的斜率,第二是切点。
【每日一练】经典高考数学基础训练(6)(含参考答案)
【每日一练】经典高考数学基础训练(6)(含参考答案)一、选择题:1.已知全集U ={1,2,3,4,5},集合M ={1,2,3},N ={3,4,5},则M ∩(ðU N )=( )A.{1,2}B.{4,5}C.{3}D.{1,2,3,4,5}2. 复数z=i 2(1+i)的虚部为( ) A.1 B. i C. -1D. - i3.如图是一个几何体的三视图,则该几何体的体积为( ) A.π3B.π37C.π320D.π4.在等比数列}{n a 中,32-=a ,64-=a ,则8a 的值为( ) A .–24B .24C .±24D .–125.在四边形ABCD 中,“DC AB 2=”是“四边形ABCD 是梯形”的( )A .充要条件B.充分不必要条件C.必要不充分条件D.既不充分也不必要条件6. 方程062=-+x e x的解一定位于区间( ) A .(1,2)B .(2,3)C .(3,4)D .(5,6)7.如图所示,墙上挂有一边长为a 的正方形木板,它的四个角的空白部分都是以正方形的顶点为圆心,半径为2a的圆弧,某人向此板投镖,假设每次都能击中木板,且击中木板上每个点的可能性都一样,则他击中阴影部分的概率是( ) A .41π-B .4π C .81π-D .与a 的取值有关8. 在三角形ABC 中,CBBC AB A sin sin ,7,5,120则===的值为( )A .58 B .85 C .35D .539.设⎩⎨⎧<+-≥--=0,620,12)(2x x x x x x f ,若2)(>t f ,则实数t 的取值范围是( )A .),4(1,(+∞⋃--∞) B.),3(2,(+∞⋃-∞)C .),1(4,(+∞⋃--∞) D.),3(0,(+∞⋃-∞)10.设α表示平面,b a ,表示直线,给定下列四个命题: ①αα⊥⇒⊥b b a a ,// ②αα⊥⇒⊥b a b a ,// ③αα//,b b a a ⇒⊥⊥ ④b a b a //,⇒⊥⊥αα其中正确命题的个数有( )A.1个B.2个C.3个D.4个二、填空题:11.已知,x y 满足约束条件50,0,3,x y x y x -+≥⎧⎪+≥⎨⎪≤⎩则y x z +=2的最小值为 .12. 右面是一个算法的程序框图,当输入的值x 为20时,则其输出的结果是 .13.若一个圆的圆心在抛物线24x y -=的焦点处,且此圆与直线0143=-+y x 相切,则圆的方程是 .14. 对任意实数x 、y ,定义运算x *y =ax +by +c xy ,其中a 、b 、c 为常实数,等号右边的运算是通常意义的加、乘运算.现已知2*1=3,2*3=4,且有一个非零实数m ,使得对任意实数x ,都有x *m =2x ,则m = . 三、解答题已知(sin ,cos )a x x =,)cos ,(cos x x b =,f (x )=b a ∙⑴ 求f (x )的最小正周期和单调增区间; ⑵ 如果三角形ABC 中,满足f (A )=12,求角A 的值.答案一、选择题题号 1 2 3 4 5 6 7 8 9 10 答案ACBABAADDB二、填空题 13. 25-14. 0 15. 161)161(22=++y x 16. 3 三、解答题:本题考查向量、二倍角和辅助角公式、三角函数性质和三角形的有关性质,要求学生能运用所写的知识解决实际问题.满分12分解:⑴f (x )= sin x cos x +x 2cos ………1分 =21x 2sin +x 2cos 2121+………2分 =22sin(2x+4π)+21………3分 最小正周期为π,…………………4分单调增区间[k π-83π,k π+8π](k ∈Z )……………………6分 ⑵由21)(=A f 得sin(2A+4π)=0, …………7分4π<2A+4π<49π,……………9分 ∴2A+4π=π或2π∴A =83π或87π…………………… 12分。
语文基础训练练习参考答案
语文基础训练练习参考答案一参考答案1C 2A 3C 4A 5C6(1)早春;欣喜。
残春;惆怅(或惋惜)(2)所写都非眼前实景,运用了衬托(或虚实结合,或想像联想)的手法。
杨诗用想像中的盛春时的喧闹反衬早春的清新可喜。
王诗想像春色在邻家,表现希望春天不曾远去(或希望把春留住)的惜春心情。
7、科技和人文不可偏废8、草正面:坚韧不拔的性格使人在各种环境下都能顽强生存反面:随风摇摆,永远缺乏自己的主张。
二参考答案1.D 2.A 3.B 4.C 5.B(A“加强”与“紧迫感与主动性”搭配不当,将“加强”改为“增强”。
C多重否定不当,应将“谁能说……呢?”改为“未必没有一定的好处”。
D句式杂糅,“是因为……”与“……所造成的”不能混用)6.初夏天气的总体特点是清和(或清新、和暖)(2分)这首诗“散”与“不散”是指在写景上是散的,有雨景,有晴景,有远景,有近景,有清风,有阳光;但作者对晴朗风物的喜悦之情是贯穿始终的。
(4分)7.①爆竹(1分)②本诗拟人手法,抓住爆竹的特点,托物言志,表达了作者雄心壮志。
(答题要点:拟人、托物言志、雄心壮志)8 我的老师,您是我心灵的灯光,我的梦想是长大后我就成了你,像你一样托起明天的太阳。
三参考答案1. C2. A3. B4. A (A.差可告慰:大体上还可告慰。
“差”作“稍微”、“尚”解。
B.叹为观止:赞扬事物尽善尽美,好到无以复加的地步。
这里言“交通混乱”,褒词贬用了。
C. 曾几何时:表示时间过去没多久,必须有一个相对参照。
句中将“没多久”丢掉了,作了相反的理解,用于三年之前的过去了。
D. 密云不雨:满天浓云而不下雨,比喻事情正在酝酿,尚未发作。
)5. C ( A. 缺少与“为”相呼应的介词,在“最有效”前加“以”。
B. 语序不当,据上下文意,应为“大学毕业生的父母及其家庭情况对其毕业选择会有一定影响”。
D. 搭配不当,应为“心脏病的发病率急剧上升”;)6①善用衬托(或映衬、陪衬):因有梅花的陪衬,窗前月才?有一番韵味,不仅是嗅觉,视觉上也使人大觉不相同。
2020届高考英语3500词汇读练测基础训练06(含解析)(最新整理)
基础训练06一、选词填空(从晨读材料中选用恰当的词并用其正确的形式填空)1.When he __________,he found himself lying in the hospital and being looked after by an old woman。
【答案】awoke【句意】当他醒来时,他发现自己躺在医院里,由一个老太太的护理着。
2.He came to my class every week, but his __________ suggested that he was not really interested.【答案】attitude【句意】他每周都来上我的课,但他的态度表明他并不是真的感兴趣。
3。
It never occurred to me that I had to __________ an important meeting in Shanghai the next day。
【答案】attend【句意】我突然想起明天我不得不去上海参加一个重要的会议.4。
—Why have you moved here?—I only __________ to enjoy the farming life here。
【答案】attempt【句意】“你为什么搬到了这里?"“我只是尝试享受这里的田园生活。
”5。
As a great scientist,little joy can equal that of being __________ the Nobel Prize。
【答案】awarded【句意】作为一个伟大的科学家,任何喜悦都赶不上获得诺贝尔奖。
6。
There was an __________ moment when she didn’t k now whether to shake his hand or kiss his cheek.【答案】awkward【句意】她不知道该和他握手还是吻他的面颊时很尴尬。
高考物理一轮复习 必拿分基础训练_
取夺市安慰阳光实验学校【人教通用版】高考物理一轮复习必拿分基础训练(15)一、选择题A.拉力的水平分力等于桌子所受的合力B.拉力的竖直分力小于桌子所受重力的大小C.拉力与摩擦力的合力大小等于重力大小D.拉力与重力的合力方向一定沿水平方向解析:本题考查共点力平衡.由于桌子做匀速直线运动,故受力平衡,所受合力为零,拉力在水平方向的分量不为零,选项A错误;由于拉力在水平方向的分量与摩擦力平衡,所以地面对桌子的支持力不为零,选项B正确;拉力与摩擦力的合力应等于重力与支持力的合力大小,选项C错误;由于拉力大小和方向均未明确,拉力与重力的合力方向不确定,选项D错误.答案:B2.如右图所示,一运送救灾物资的直升飞机沿水平方向匀速飞行.已知物资的总质量为m,吊运物资的悬索与竖直方向成θ角.设物资所受的空气阻力为F阻,悬索对物资的拉力为F,重力加速度为g,则( )A.F阻=mgsin θB.F阻=mgtan θC.F=mgcos θ D.F =mg tan θ解析:救灾物资匀速飞行,受力平衡,它受到向下的重力mg,向右的阻力F 阻和沿细绳斜向上的拉力,可得F阻=mgtan θ,B项正确.答案:B3.如右图所示,斜面小车M静止在光滑水平面上,一边紧贴墙壁.若再在斜面上加一物体m,且M 、m 相对静止,小车后来受力个数为( )A.3 B.4C.5 D.6解析:对M和m整体,它们必受到重力和地面支持力.对小车因小车静止,由平衡条件知墙面对小车必无作用力,以小车为研究对象.如右图所示,它受四个力;重力Mg,地面的支持力FN1,m对它的压力FN2和静摩擦力Ff,由于m静止,可知Ff和FN2的合力必竖直向下,故B项正确.答案:B4.如右图所示,在倾角为53°的斜面上,用沿斜面向上5 N的力拉着重4 N的木块向上做匀速运动,则斜面对木块的总作用力的方向是( )A.垂直斜面向上B.水平向左C.沿斜面向下D.竖直向上解析:木块受到重力、支持力、拉力和摩擦力而处于平衡状态,因沿斜面向上的拉力在竖直方向的分力为Fsin 53°=5×0.8 N=4 N=mg,沿斜面向上的拉力在水平向右方向的分力为Fcos 53°=5×0.6 N=3 N,由平衡条件知斜面对木块的总作用力的方向是水平向左,故选项B正确.答案:B5.密绕在轴上的一卷地膜用轻绳一端拴在轴上,另一端悬挂在墙壁上A点,如右图所示,当逆时针缓慢向下用力F抽出地膜时,整卷地膜受的各个力要发生变化,不计地膜离开整卷时对地膜卷的粘扯拉力和地膜卷绕轴转动时的摩擦力,但在D点地膜与墙壁间有摩擦力,随着地膜的不断抽出,下述分析正确的是( )A.悬挂地膜的轻绳上的拉力在增大B.地膜对墙的压力在增大C.拉动地膜的力在减小D.地膜卷受墙的支持力与轻绳的拉力的合力不变解析:当地膜不断被抽出过程中,OD逐渐减小,∠OAD逐渐减小,由于地膜质量不断减小,由共点力平衡可知轻绳拉力逐渐减小,选项A、B均错误;因地膜卷对墙壁的压力逐渐减小,由Ff=μFN可知摩擦力逐渐减小,选项C正确;由于支持力逐渐减小,∠OAD逐渐减小,根据三角形定则可知地膜卷受墙的支持力与轻绳的拉力的合力减小,选项D错误.答案:C6.如图甲所示,在圆柱体上放一物块P ,圆柱体绕水平轴O缓慢转动,从A转至A′的过程,物块与圆柱体保持相对静止,则图乙反映的是该过程中( ) A.重力随时间变化的规律 B.支持力随时间变化的规律C.摩擦力随时间变化的规律 D.合外力随时间变化的规律解析:在圆柱体缓慢转动的过程中,物块P的重力是定值,不随时间发生变化,A错;物块P受三个力的作用,竖直向下的重力G,沿半径指向外的支持力FN,沿切线方向的静摩擦力Ff,因圆柱体缓慢移动,所以物块P在任意位置所受合力为零,D错;对三力正交分解,设重力G与支持力FN方向所夹锐角为θ,则FN=mgcos θ,Ff=mgsin θ,从A转至A′的过程中,θ先减小后增大,所以FN先增大后减小,B对;而Ff先减小后增大,C错.答案:B7.如右图所示,物体A静止在倾角为30°的斜面上,现将斜面倾角由30°增大到37°,物体仍保持静止,则下列说法中正确的是( )A.A对斜面的压力不变B.A对斜面的压力增大C.A受到的摩擦力不变D.A受到的摩擦力增大解析:物体A受力分析如图所示,将重力沿平行于斜面方向和垂直于斜面方向分解.则静摩擦力Ff=mgsin θ,Ff随θ的增大而增大;斜面对物体A的支持力FN=mgcos θ,由牛顿第三定律,A对斜面的压力F′N=FN=mgcos θ,随θ的增大而减小.答案:D8.如右图所示,质量分别为mA和mB的物体A、B用细绳连接后跨过滑轮,A静止在倾角为45°的斜面上.已知mA=2mB,不计滑轮摩擦,现将斜面倾角由45°增大到50°,系统保持静止.下列说法正确的是( )A.细绳对A的拉力将增大B.A对斜面的压力将减小C.A受到的静摩擦力不变D.A受到的合力将增大解析:对A受力分析如图所示,由物体的平衡条件得:FN-Gcos θ=0,Gsin θ-Ff-F =0,F =G2若θ从45°增大为50°,则有FN 减小,Ff 增大.物体A 受到的合力仍为0.答案: B9.如右图所示,一个木块放在固定的粗糙斜面上,今对木块施一个既与斜面底边平行又与斜面平行的推力F ,木块处于静止状态,如将力F 撤消,则木块( )A .仍保持静止B .将沿斜面下滑C .受到的摩擦力大小不变D .受到的摩擦力方向不变解析: 有力F 作用时,木块在斜面内的受力如右图,且Ff =F2+mgsin θ2,当撤去力F 后,木块只受mgsin θ和F′f,且F′f<Ff 故仍静止,摩擦力的方向变为沿斜面向上.答案应为A.答案: A二、非选择题10.如右图所示,在倾角为37°的固定斜面上静置一个质量为5kg 的物体,物体与斜面间的动摩擦因数为0.8.求:(1)物体所受的摩擦力;(sin 37°=0.6,cos 37°=0.8)(2)若用原长为10 cm ,劲度系数为3.1×103 N/m 的弹簧沿斜面向上拉物体,使之向上匀速运动,则弹簧的最终长度是多少?(取g =10 m/s2) 解析:(1)物体静止在斜面上受力分析如右图所示,则物体受到的静摩擦力Ff =mgsin 37°代入数据得Ff =5×10×sin 37°N=30 N ,摩擦力方向为沿斜面向上.(2)当物体沿斜面向上被匀速拉动时,如下图所示,弹簧拉力设为F ,伸长量为x ,则F =kxF =mgsin 37°+F 滑F 滑=μmgcos 37°弹簧最终长度l =l0+x ,由以上方程解得l =12 cm.答案: (1)30 N 方向沿斜面向上 (2)12 cm11.如下图所示,A 、B 两物体叠放在水平地面上,已知A 、B 的质量分别为mA=10 kg ,mB =20 kg ,A 、B 之间,B 与地面之间的动摩擦因数均为μ=0.5.一轻绳一端系住物体A ,另一端系于墙上,绳与竖直方向的夹角为37°,今欲用外力将物体B 匀速向右拉出,求所加水平力F 的大小,并画出A 、B 的受力分析图.(取g =10 m/s2,sin 37°=0.6,cos 37°=0.8)解析: A 、B 的受力分析如下图所示.对A 应用平衡条件FTsin 37°=Ff1=μFN1①FTcos 37°+FN1=mAg ②联立①、②两式可得:FN1=3mAg4μ+3=60 NFf1=μFN1=30 N对B用平衡条件F=F′f1+Ff2=F′f1+μFN2=Ff1+μ(FN1+mBg)=2Ff1+μmBg=160 N.答案:160 N滚动训练(一)(本栏目内容,在学生用书中以活页形式分册装订!)一、选择题1.一个物体从静止开始做匀加速直线运动,以T为时间间隔,在第三个T内的位移是3 m,第三个T终了时的瞬时速度是3 m/s,则( )A.物体的加速度是3 m/s2B.物体在第一个T终了时的瞬时速度为1 m/sC.时间间隔为1 sD.物体在第一个T时间内的位移为1 m答案:B2.(2013·浙江嘉兴基础测试)如右图所示,倾角为30°,重为80 N的斜面体静止在水平面上.一根弹性轻杆一端垂直固定在斜面体上,杆的另一端固定一个重为2 N的小球,小球处于静止状态时,下列说法正确的是( )A.斜面有向左运动的趋势B.地面对斜面的支持力为80 NC.球对弹性轻杆的作用力为2 N,方向竖直向下D.弹性轻杆对小球的作用力为2 N,方向垂直斜面向上解析:把小球、杆和斜面作为整体受力分析可知,仅受重力和地面的支持力,且二力平衡,故A、B错;对小球受力分析知,只受竖直向下的重力和杆给的竖直向上的弹力(杆对小球的力不一定沿杆),故C对,D错.答案:C 3.某人欲估算飞机着陆时的速度,他假设飞机停止运动前在平直跑道上做匀减速运动,飞机在跑道上滑行的距离为x,从着陆到停下来所用的时间为t,则飞机着陆时的速度为( )A.xtB.2xtC.x2tD.xt到2xt之间的某个值解析:根据匀变速直线运动位移公式x=v t、平均速度公式v=v0+v2,得v0=2xt.答案:B4.如右图所示,轻绳上端固定在天花板上的O点,下端悬挂一个重为10 N的物体A,B是固定的表面光滑的圆柱体.当A静止时,轻绳与天花板的夹角为30°,B受到绳的压力是( )A.5 N B.10 NC.5 3 N D.10 3 N解析:对物体A,根据平衡条件可知,绳子拉力FT=10 N,B受到两个方向夹角为120°的等大的两个拉力FT的作用,合力沿着角平分线,故合力的大小应等于FT=10 N,B对.答案:B5.一辆轿车和一辆卡车在同一公路上均由静止开始同时做匀加速直线运动,加速度大小分别为3 m/s2和7 m/s2,两车能达到的最大速度均为30 m/s,刚开始运动时两车距离为20 m,轿车车身全长5 m,卡车车身全长20 m ,则两车的错车时间为( ) A .1.1 s B .1.0 s C .1.2 s D .1.7 s答案: B6.物体A 的质量为1 kg ,置于水平地面上,物体与地面的动摩擦因数为μ=0.2.从t =0开始物体以一定初速度v0向右滑行的同时,受到一个水平向左的恒力F =1 N 的作用,则能反映物体受到的摩擦力Ff 随时间变化的图象是(取向右为正方向,g =10 m/s2)( )解析: 物体开始运动,受到滑动摩擦力为2 N ,方向向左,后来Ff <μmg 物体静止,受到静摩擦力为1 N ,方向向右,故A 正确;本题要对运动过程清楚,分析受力从而确定过程,过程不清楚容易误选.答案: A7.一滑块以某一速度从斜面底端滑到顶端时,其速度恰好减为零.若设斜面全长L ,滑块通过最初34L 所需时间为t ,则滑块从斜面底端到顶端所用的时间为( )A.43tB.53tC.32t D .2t 解析: 假设存在逆过程,即为初速度是零的匀加速直线运动,将全过程分为位移均为L/4的四个阶段,根据匀变速直线运动规律,其时间之比为1∶(2-1)∶(3-2)∶(2-3),根据题意可列方程:2-1+3-2+2-31+2-1+3-2+2-3=tt′,t′=2t. 答案: D8.如右图所示,A 、B 两物体通过轻细绳跨过定滑轮相连接,已知物体A 的质量大于物体B 的质量,开始它们处于静止状态.在水平拉力F 的作用下,使物体A 向右做变速运动,同时物体B 匀速上升.设水平地面对物体A 的支持力为FN ,对A 的摩擦力为Ff ,绳子对A 的拉力为FT ,那么在物体B 匀速上升的过程中,FN 、Ff 、FT 的大小变化情况是( )A .FN 、Ff 、FT 都增大B .FN 、Ff 增大,FT 不变C .FN 、Ff 、FT 都减小D .FN 、FT 减小,Ff 不变解析: 该题考查连接体和动态平衡,物体B 匀速上升过程,说明物体B 受力平衡,绳子的拉力FT 与物体B 的重力大小相等,保持不变,隔离A 分析,A 受重力、地面的支持力FN 、绳子的大小不变、方向变化的拉力FT ,滑动摩擦力Ff 和外力F ,在竖直方向上,合力不变,拉力FT 的竖直分力逐渐减小,则物体A 与地面之间的弹力FN 变大,滑动摩擦力Ff 变大,故选B. 答案: B9.如右图所示,一质量为M 、倾角为θ的斜面体在水平地面上,质量为m 的小木块(可视为质点)放在斜面上,现用一平行斜面、大小恒定的拉力F 作用于小木块,拉力在斜面所在的平面内绕小木块旋转一周的过程中,斜面体和木块始终保持静止状态.下列说法中不正确的是( )A .小木块受到斜面的最大摩擦力为F +mgsin θB .小木块受到斜面的最小摩擦力可能为零C .斜面体受到地面的最大摩擦力为FD .斜面体受到地面的最小摩擦力为零 答案: D10.如下图所示,物体m 通过定滑轮牵引另一水平面上的物体沿斜面匀速下滑,此过程中斜面仍静止,斜面质量为M ,则水平地面对斜面体( ) A .无摩擦力 B .有水平向左的摩擦力 C .支持力为(M +m)g D .支持力小于(M +m)g解析: 设斜面夹角为θ,对M 、m 整体分析受力可得平衡方程:FTcos θ=Fμ静,FTsin θ+FN =(M +m)g ,故D 正确. 答案: D 二、非选择题11.(2013·合肥模拟)测速仪安装有超声波发射和接收装置,如右图所示,B 为测速仪,A 为汽车,两者相距335 m ,某时刻B 发出超声波,同时A 由静止开始做匀加速直线运动.当B 接收到反射回来的超声波信号时,A 、B 相距355 m ,已知声速为340 m/s ,求汽车的加速度大小.解析: 设超声波往返的时间为2t ,根据题意汽车在2t 时间内位移为12a(2t)2=20 m ,①所以超声波追上A 车时,A 车前进的位移为12at2=5 m ,②所以超声波在2t 内的路程为2×(335+5) m ,由声速340 m/s 可得t =1 s ,代入①式得,a =10 m/s2.答案: 10 m/s212.(2013·湖北八校二次联考)某公共汽车的运行非常规则,先由静止开始匀加速启动,当速度达到v1=10 m/s 时再做匀速运动,进站前开始匀减速制动,在到达车站时刚好停住.公共汽车在每个车站停车时间均为Δt=25 s ,然后以同样的方式运行至下一站.已知公共汽车在加速启动和减速制动时加速度大小都为a =1 m/s2,而所有相邻车站间的行程都为x =600 m .有一次当公共汽车刚刚抵达一个车站时,一辆电动车刚经过该车站一段时间t0=60 s ,已知该电动车速度大小恒定为v2=6 m/s ,而且行进路线、方向与公共汽车完全相同,不考虑其他交通状况的影响,试求:(1)公共汽车从其中一车站出发至到达下一站所需的时间t ;(2)若从下一站开始计数,公共汽车在刚到达第n 站时,电动车也恰好同时到达此车站,n 为多少?解析: (1)设公共汽车启动时加速所用的时间为t1 t1=v1/a得t1=10 s 设加速启动时行驶的路程为x1 x1=12at21得x1=50 m上面所求时间和路程同时也是减速制动所用的时间和路程,设汽车每次匀速行驶所经过的路程为x2x2=x -2x1得x2=500 m设匀速行驶所花时间为t2t2=x2/v1得t2=50 s所以公共汽车在每两站之间运动所经历的时间为t=2t1+t2=70 s.(2)设电动车到达第n站所用的总时间为TT=n(t+Δt)+t0所以有v2T=nx代入数据可求得n=12.答案:(1)70 s (2)12。
【精品含答案】高考一轮复习5.3线段的定比分点与平移基础训练题(理科)
2009届高考一轮复习5.3线段的定比分点与平移基础训练题(理科)注意:本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
满分100分,考试时间45分钟。
第Ⅰ卷(选择题部分 共36分)一、选择题(本大题共6小题,每小题6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) 1.(2007·辽宁高考)若函数)x (f y =的图象按向量a 平移后,得到函数2)1x (f y -+=的图象,则向量=a ( )(A ))2,1(-- (B ))2,1(-(C ))2,1(- (D ))2,1(2. 设点P 在有向线段AB 的延长线上,且|BP |4|AP |=,则点A 分BP 所成的比为( )(A )45- (B )32-(C )43- (D )34- 3. 将函数x 1y =的图象按向量a 平移后,得到1x 12y ++=的图象,则( ) (A ))2,1(a = (B ))2,1(a -= (C ))2,1(a -= (D ))2,1(a --= 4.(易错警示题)已知)1,2(P 1-,)5,0(P 2且点P 在21P P 延长线上,使P 2P 21=,则点P 的坐标是( ) (A ))11,2(- (B ))3,34( (C ))3,32( (D ))7,2(- 5.(2008·长春模拟)若把一个函数的图象按)2,3(a -π-= 平移后,得到函数x cos y =的图象,则原图象的函数解析式是( )(A )2)3x cos(y -π+= (B )2)3x cos(y -π-= (C )2)3x cos(y +π+= (D )2)3x cos(y +π-= 6. 已知点)2,6(M 1和)7,1(M 2,直线7mx y -=与线段21M M 的交点M 分有向线段21M M 的比为2:3,则m 的值为( )(A )23- (B )32- (C )41 (D )4第Ⅱ卷(非选择题部分 共64分)二、填空题(本大题共3小题,每小题6分,共18分。
高中物理 量子论初步 原子核(附答案)高考基础训练题
量子论初步原子核(附答案)1.人类对光的本性认识经历了曲折的过程.下列关于光的本性的陈述正确的是()A.牛顿的“微粒说”与爱因斯坦的“光子说”本质上都是一样的B.任何一个运动着的物体,都具有波动性C.麦克斯韦预言了光是一种电磁波D.光波是概率波2.在演示光电效应的实验中,原来不带电的一块锌板与灵敏验电器相连,用弧光灯照射锌板时,验电器的指针就张开一个角度,如图15-1-5所示,这时()A.锌板带正电,指针带负电B.锌板带正电,指针带正电C.锌板带负电,指针带正电图15-1-5 D.锌板带负电,指针带负电3.某单色光照射某金属时不能产生光电效应,则下述措施中可能使金属产生光电效应的是()A.延长光照时间B.增大光的强度C.换用波长较短的光照射D.换用频率较低的光照射4.(创新应用题)物理学家做了一个有趣的实验:在双缝干涉实验中,在光屏处放上照相底片,若减弱光流的强度,使光子只能一个一个地通过狭缝,实验结果表明,如果曝光时间不太长,底片上只出现一些不规则的点子;如果曝光时间足够长,底片上就会出现规则的干涉条纹,对这个实验结果有下列认识,正确的是()A.曝光时间不长时,出现不规则的点子,表现出光的波动性B.单个光子通过双缝后的落点无法预测C.干涉条纹中明亮的部分是光子到达机会较多的地方D.只有大量光子的行为才能表现出光的粒子性5.(2009年高考上海卷)光电效应的实验结论是:对于某种金属()A.无论光强多强,只要光的频率小于极限频率就不能产生光电效应B.无论光的频率多低,只要光照时间足够长就能产生光电效应C.超过极限频率的入射光强度越弱,所产生的光电子的最大初动能就越小D .超过极限频率的入射光频率越高,所产生的光电子的最大初动能就越大6.A 和B 两种单色光均垂直照射到同一条直光纤的端面上,A 光穿过光纤的时间比B 光穿过的时间长,现用A 和B 两种光照射同种金属,都能发生光电效应,则下列说法正确的是()A .光纤对B 光的折射率大B .A 光打出的光电子的最大初动能一定比B 光的大C .A 光在单位时间内打出的电子数一定比B 光的多D .B 光的波动性一定比A 光显著7.已知一束可见光a 是由m 、n 、p 三种单色光组成的.检测发现三种单色光中,n 、p 两种色光的频率都大于m 色光;n 色光能使某金属发生光电效应,而p 色光不能使该金属发生光电效应.那么,光束a 通过三棱镜的情况是下图中的()图15-1-68.如图15-1-7所示是某金属在光的照射下产生的光电子的最大初动能E k 与入射光频率ν的关系图象.由图象可知()A .该金属的逸出功等于EB .该金属的逸出功等于hν0C .入射光的频率为2ν0时,产生的光电子的最大初动能为ED .入射光的频率为ν0/2时,产生的光电子的最大初动能为E /29.分别用波长为λ和34λ的单色光照射同一金属板,发出光电子的最大初动能之比为1∶2,以h 表示普朗克常量,c 表示真空中的光速,则此金属板的逸出功为()A.hc 2λB.hc 3λC.34hcλD.hλ5c10.一种X 射线,每个光子具有4×104eV 的能量,此X 射线的波长是多少?一个电子具有多少电子伏特能量时,其德布罗意波长与上述X 射线的波长相等?图15-1-7(电子的质量m=9.1×10-31kg)11.波长为λ=0.17μm的紫外线照射至金属筒上能使其发射光电子,光电子在磁感应强度为B的匀强磁场中,做最大半径为r的匀速圆周运动,已知r·B=5.6×10-6T·m,光电子质量m=9.1×10-31kg,电荷量e=1.6×10-19C,求:(1)光电子的最大动能;(2)金属筒的逸出功.12.(西安模拟)如图15-1-8所示,一伦琴射线管,K为阴极可产生电子,阴极K与对阴极A外加电压U AK=30kV.设电子离开K极时速度为零,通过电压加速后而以极大的速度撞到对阴极A上而产生X射线,假定电子的全部动能转为X射线的能量.求:(1)电子到达A极时的速度是多大?(2)从A极发出的X射线的最短波长是多少?(3)若电路中的毫安表的示数为10mA,则每秒从A极最多图15-1-8能辐射出多少个X光子?(已知电子的质量m e=9.1×10-31kg,电子的电荷量e=1.6×10-19C,普朗克常量h=6.6×10-34J·s)答案1BCD2B3C4BC5AD6BD7A8ABC9B10答案:3.1×10-11m 1.6×103eV11答案:(1)4.41×10-19J(2)7.3×10-19J12答案:(1)1.0×108m/s(2)4.1×10-11m(3)6.25×1016个。
高考英语阅读理解基础训练(6)
定额市鞍钢阳光实验学校四川省雅安市2014高考英语阅读理解金榜题名基础训练(6)及答案阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。
Believe it or not, optical illusion(视错觉) can cut highway crashes.Japan is a case in point. It has reduced car crashes on some roads by nearly 75 percent using a simple optical illusion. But stripes, called chevrons(人字形), painted on the roads make drivers think that they are driving faster than they really are, and thus drivers slow down.Now the American Automobile Association Foundation for Traffic Safety in Washington D.C. is planning to repeat Japan’s success. Starting next year, the foundation will paint chevrons and other patterns of stripes on selected roads around the country to test how well the patterns reduce highway crashes.Too great speed plays a major role in as much as one fifth of all serious traffic accidents, according to the foundation. To help reduce those accidents, the foundation will conduct its tests in areas where speed-related risks are the greatest curves, exit slopes, traffic circles, and bridges. Some studies suggest that straight, horizontal bars(水平杆)painted across roads can initially cut the average speed of drivers in half. However, traffic often returns to full speed within months as drivers become used to seeing the painted bars.Chevrons, scientists say, not only give drivers the impression that they are driving faster than they really are but also make a lane appear to be narrower. The result is a longer lasting reduction in highway speed and the number of traffic accidents.36. The passage mainly discusses ________.A. a new type of optical illusionB. a new pattern for painting highwaysC. a new way of training driversD. a new way of highway speed control【答案】D【解析】主旨大意题。
江苏高考英语3500词汇读练测精品系列(解析版) (6)
2020届江苏高考英语3500词汇读练测精品系列(基础训练)19一、选词填空(从晨读材料中选用恰当的词并用其正确的形式填空)1.Such problems as high blood pressure and heart trouble are __________ of old age.【答案】features【句意】比如高血压和心脏病这样的问题是老年人的普遍特点。
2.Madagascar is the most __________ place I have ever been to【答案】fascinating【句意】马达加斯加是我去过的最迷人的地方。
3.Passengers __________ their seat belts immediately after taking their seats.【答案】fasten【句意】所有的乘客在坐下来后都要系好安全带。
4.Fine violins by skilled old Italians can __________ hundreds of thousands of dollars, and prices still keep rising. 【答案】fetch【句意】由熟练的老意大利人制作的上好的小提琴可以卖得上千美元,并且价格还在持续上涨。
5.It’s high time that we __________ against spreading __________ information on the Internet.【答案】fought, false【句意】是我们与互联网上传播虚假信息作斗争的时候了。
6.With the application of 5G technology come profound changes in almost all __________ throughout the world. 【答案】fields【句意】随着5G技术的应用,世界几乎所有领域都发生了深刻的变化。
新高考古代诗歌鉴赏基础训练:选择题(典题示例+强化训练+答案解析)
新高考古代诗歌鉴赏基础训练:选择题(典题示例+强化训练+答案解析)【典题示例】1.选出对李白《蜀道难》有关内容的赏析不正确的一项()A.开篇极言蜀道之难,以感情强烈的咏叹点出主题,为全诗奠定奔放雄奇的感情基调。
B.诗人以夸张的笔墨描绘了自古以来秦蜀之间被高山峻岭阻挡的情形。
由秦入蜀,必经太白峰,自古只有高飞的鸟儿能从其低缺处飞过。
C.千里翱翔的黄鹤不能飞渡,轻疾敏捷的猕猴也不能攀援,不言而喻,人的行走就更难了。
诗人的夸张、衬托之笔极写蜀道的高不可攀。
D.青泥岭为唐人人蜀要道,诗人以写实的手法着重写其山路的萦回和山势的峻危来表现人行其上的情状和畏惧心理,捕捉了手扪星辰、呼吸紧张、抚胸长叹等细节加以摹写,困危之状如在目前。
【答案】D【解析】【详解】本题考查学生赏析文本内容的能力。
D.“诗人以写实的手法着重写其山路的萦回和山势的峻危”错误,攀援青泥岭的细节并非写实而是想象。
故选D。
2.选出对白居易《琵琶行》一诗有关诗句赏析不正确的一项()A.“浔阳江头夜送客,枫叶荻花秋瑟瑟”,第一句包含了时间、地点、人物、事件,第二句通过环境描写有力地烘托了秋夜送客的萧瑟落寞之感。
B.“醉不成欢惨将别,别时茫茫江浸月”,第一句点明了送别的基调:“惨”。
第二句用环境描写来烘托人物的心境,可谓情景交融:与友分别心绪难平,心情和眼前月色一样,一片茫然。
C.“东船西舫悄无言,唯见江心秋月白”,情景交融。
与开头“夜”“船”“月”相呼应,烘托了琵琶女高超的演奏技巧及演奏的悲凉气氛,同时也给读者留下了回味的广阔空间。
D.“座中泣下谁最多?江州司马青衫湿”,该句运用了反问和借代修辞手法将诗人的感情推向了高潮。
“江州司马青衫湿”一句实为画龙点睛之笔,同时也是全诗的主题即上段“同是天涯沦落人……”的形象化说明。
【答案】D【解析】本题主要考查赏析诗歌的能力。
D.“座中泣下谁最多?江州司马青衫湿”,要问在座之中谁流的眼泪最多?我江州司马泪水湿透青衫衣襟。
高考得分点基础训练 13
高考得分点基础训练(13)II.Grammar and VocabularyAThe flu season doesn’t officially start until next month, and continues until March. Yet signs touting flu shots are all over. It’s too early ___25___________ (roll) up the sleeve and get one, isn’t it? Not at all, says Dr. Derek Van Amerongen, medical director of Chicago-based Humana Vitality. Actually, the earlier the better. Now is the perfect time, because then you ward off the flu season, according to Van Amerongen. “Everyone should take the flu seriously and get the vaccine,” says Van Amerongen. “The shots are a safe, inexpensive way to prevent ___26___ (get) sick.”And ___27______________ (forget) those misconceptions that the shot causes the flu. “The shot contains only dead cultures of the virus,”V an Amerongen says. “It is a matter of coincidence ___28_________ someone gets the flu after getting the shot.”Even if a person had the shot and later came down with the flu, there is an upside. “The flu shot is not perfect,”said Van Amerongen, “but it is the only ___29___________ (prove) protection against the virus and if someone should get the flu after having the shot it will be much milder.”No new strains of the flu have surfaced this season; however, the ones that exist Dr. Van Amerongen says, “are dangerous enough.”Last year 40,000 people died from the virus. The young and elderly are ___30________ most susceptible to get the virus and typically have the worse eases. He recommends everyone age 6 months and older ___31_____ have the vaccine. 25_________________26.______________ 27_______________.28.______________29.________________30._______________31._______________BIs social media making us smarter? You bet it is, argues a new study.Academic newspaper Globe and Mail claims that social media—especially Twitter—is making us hyper literate. But how could our feed, ___32_________ probably contains a mixture of fun tweets, hard news, and satirical parody accounts, actually be making us smarter? Think about it: Twitter gives us access to massive amounts of information, right at our fingertips. ___33___ also introduces us to new words, ideas and opinions. From 1986 to 2006, the average length of a freshman’s essay ___34__________ (increase) from 422 words to 1038 plus the essays were far more complex, and students used better evidence to support their point. ___35_______ (scroll) your Twitter feed in your free time may seem like totally no big deal, ___36________ maybe you should now refer to it as studying—absorbing all the information improves your writing and critical thinking skills! Plus, each time you rush to think of a witty reply ___37___ a tweet or add your commentary to an event you’re live-tweeting (like the VMAs or Grammys), it actually helps you become a better thinker, faster thinker, claims Globe and Mail. “There’s powerful evidence ___38__________ digital tools are helping young people write and think far ___39__________ (good) than in the past,”the article states. The Globe and Mail claim that, thanks ___40__________ Twitter, people aren’t just consuming media they’re “talking back to it.”32_________________33.______________ 34_______________.35.______________36.________________37._______________38._______________39____________40.________▲Section B (10分)A. expressedB. specializedC. averageD. speciallyE. provedF. vacantG. worked H. history I. hunters J. suggestion K. upThere is a new type of small advertisement becoming increasingly common in newspaper classified columns. It is sometimes placed among “situations ___41___”, although it does not offer anyone a job, and sometimes it appears among “situations wanted”, although it is not placed by someone looking for a job either. What it does is to offer help in applying for a job. “Contact us before writing your application”, or “Make use of our long experience in preparing your job history”, is how it is usually ___42___. The growth and success of such a(n) ___43___ service is, of course, a reflection on the current high levels of unemployment. It is also an indication of the growing importance of job history, with the ___44___ that it may now qualify as an art form itself.There was a time when those job ___45___ simply wrote letters of application. “Just put down your name, address, age and whether you have passed any exams”, was about the ___46___ level of advice offered to young people applying for their first jobs when I left school. The letter was really just for a start, it was explained, everything else could and should be saved for the interview. And in those days of full employment, the technique ___47___. The letter proved that you could write and were available for work. Your eager face and intelligent replies did the rest.Later, as you moved ___48___ the ladder, something slightly more sophisticated (老练的) was called for. The advice then was to put something in the letter which would make you different from the rest. “Your search is over. I am the person you are looking for,”was a widely used trick that occasionally succeeded. Or it might be some special feature ___49___ designed for the job in view.There is no doubt, however, that it is the increasing number of applicants with university education that has led to the greater importance of the job ___50___.III.Reading ComprehensionAAndrew Carnegie, once the world’s richest person, was born in 1835 to a weaver’s family in Scotland. As a child, he was expected to follow his father’s profession. But the industrial revolution destroyed the weavers’craft, and the family had to leave for new possibilities in America.In 1848 the Carnegies arrived in Pittsburgh, then the iron-manufacturing centre of the country. Young Carnegie took odd jobs at a cotton factory and later worked as a messenger boy in the telegraph office. He was often asked to deliver messages to the city theatre, where he would stay to watch plays by great playwrights. He also spent most of his leisure hours in a small library that a local benefactor (捐助者) made available to working boys.After the Civil War, Carnegie saw great potential in the iron industry. He devoted himself to the replacement of wooden bridges with stronger iron ones and earned a fortune. He further introduced a new steel refining process to convert iron into steel. By 1900, Carnegie Steel produced more of the metal than all of Great Britain.However, Carnegie often expressed his uneasiness with the businessman’s life. Wishing to spend more time receiving instruction and reading systematically, he once wrote, “If I were to continue much longer in the state of being overwhelmed by business cares and with most of my thoughts wholly upon the way to make more money in the shortest time, it would degrade me beyond hope of permanent recovery.”The strong desire for intellectual (智力的) pursuit led him to sell his company and retire at 64.Fond of saying that “the man who dies rich dies disgraced,” Carnegie then turned his attention t o giving away his fortune. He abhorred charity; instead, he used his money to help others help themselves. He established over 2,500 public libraries, and sponsored numerous cultural, educational and scientific institutions. By the time he died in 1919, he had given away 350 million dollars.( ) 66. Andrew Carnegie moved to the United States because _________.A. his father was offered a good job in PittsburghB. he did not want to follow his father’s professionC. there were serious political problems in ScotlandD. his family could not make a good living in their hometown( ) 67. When did Carnegie begin to show his interest in artistic and intellectual pursuit?A. After he retired from his business.B. When he was a young boy back in Scotland.C. After he earned his fortune from his iron business.D. When he worked as a messenger boy in Pittsburgh.( ) 68. Which of the following best characterizes how Carnegie managed his business?A. He was willing to make new changes.B. He set out to beat all the other competitors.C. He was happy to make more money in the shortest time.D. He did not hesitate in making investments in his hometown.( ) 69. The underlined word “abhorred”is closest in meaning to “_________”.A. hatedB. enjoyedC. practicedD. encouragedBOne of India’s top engineering schools has restricted Internet access in its boarding houses, saying addiction to surfing, gaming and blogging was affecting students’ performance, making them lonely and even suicidal.Authorities at the best Indian Institute of Technology (IIT) in Mumbai said students had stopped socializing and many were late for morning classes or slept through them. “Now, a student doesn’t even know who lives two doors away from him because he is so busy on the Internet,” said Prakash Gopalan, dean of Student Affairs. “The old dormitory culture of companionship and socializing among students is gone. This is not healthy in our opinion.”IIT-Mumbai, with about 5,000 students, is one of seven IITs across India which is considered to be among the finest engineering schools in the world. They are also a talent pool for global technology giants. But their hard courses, tough competition and lonely campus lifestyle have taken an effect on students. Depressive and dysfunctional (不正常的) lifestyles are known to becommon among IIT students, and at least nine have committed suicide in the past five years. Students have unlimited free Internet access in their boarding houses to help them in their studies, but many also use it to surf, chat, download movies and music, blog and for gaming.“Starting Monday, Internet access will be banned between 11 pm and 12:30 pm at IIT-Mumbai’s 13 boarding buildings to encourage students to sleep early and to try and force them out of their shells,”Gopalan said. But the move has not gone down well with students who say they hate their lives being regulated. “Now they will say we need to listen to a lullaby (摇篮曲) to go to sleep,”said Rajiv, an electronics student.( ) 70. Which of the following is NOT the possible effect of free Internet access on students?A. There has been a decline in students’ lessons.B. Participation in social activities has gone down.C. Some students feel lonely and even suicidal.D. Students don’t even know his classmates.( ) 71. What measures have been taken in IIT-Mumbai?A. Students have unlimited free Internet access in their dorms.B. Students are forbidden to surf the Internet.C. Internet access is unavailable in deep night.D. Students must go to bed before 11 pm.( ) 72. In the passage the students are told to _________.A. sleep through their lessonsB. break away from the InternetC. go to sleep along with musicD. stop surfing and make friends( ) 73. We may infer from the last paragraph that _________.A. the banning order causes some complaints among the studentsB. all electronics students hate the banning orderC. more students prefer listening to music to surfing the InternetD. there is no Internet access on the IIT campus even sinceI.Translation (22分)1. 一些家长担忧,虚拟现实对年轻人的影响害处大于好处。
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高考得分点基础训练(6)II.Grammar and VocabularyAWhen I succeeded in ___25________ (become) a part-time employee of Nokia China last summer, many friends asked me how I survived the interview.I once asked myself the same questions. Many of my peers also ___26_______ (apply) for the job, including some very competitive and intelligent students ___27________ (study) in famous universities.But why did the interviewer pick me instead of them?Finally, curiosity pushed me ___28_______ (ask) the interviewers soon ___29________ we became colleagues. The answer was that I appeared confident but humble, responsible and communicative. They evaluated people not just on their academic certificates, ___30________ on the base of their qualities and abilities. I happened to be the right person.To be frank, I once felt ashamed of being a student from ___31________ unknown college, and I think this may apply to some of you. I thought that my future was ruined. It was only at the time of my successful interview ___32_______-__ I finally understood the famous saying—“Y ou decide where you go.”25_________________26.______________ 27_______________.28.______________29.________________30._______________31._______________ 32______________BSports are more than games. The ___33________ (defeat) can be the winner next time if some improvement ___34_________ (make). We can learn from sports on how to deal with our daily affairs ___35_________ (well).Sports ___36_________ bring us a lot of fun and enjoyment. Meanwhile, sports are a profitable industry. However, sports can mean more. In___37________ to appreciation of excellent athletes’ performances, people are also attracted by sportsmanship, including fairness, fair play and respect for rules.Sports are a stimulation of real life. Sport had ___38_________ origin in the daily activities such as hunting, planning and even war. The competitions and conflicts are simplified by a series of sports rules.Now sport is a part of our lives. ___39__________ an exciting game brings us is not only a pleasant period but also the influences ___40________________ the heroes have on us in the sports field. We can learn better devotion to our own arena, the real life. That’s the essence of sports.33.______________34_______________.35.______________ 36.________________37._______________38._______________39____________ 40.__________________▲Section B (10分).A. extendB. phenomenonC. noticesD. probablyE. criticalF. moodG. tiedH. forgivingI. increasesJ. realize K. heavilyWhen our hair is healthy, shiny and in place just the way we like it, all is right with the world. But when it’s behaving more like a two-year-old in the grocery store just before nap time, it’s another story.Aside from moments of total frustration with the bathroom mirror, the effects of a bad hair day can ___41___ well past morning dressing time. In fact, research links bad hair days with entire bad ___42___ days.Are we really so vain that our hair can affect our whole day?Yes, says Dr. Marianne LaFrance, Professor of Psychology and of Women’s and Gender Studies at Yale University. According to LaFrance’s study, ___43___ in self-doubt, social insecurities and self-critical thoughts emerge when people feel their hair is awkward. “Individuals ___44___ their capabilities to be significantly lower than others when experiencing bad hair,”LaFrance stated in the research analysis.And while both women and men are affected by the ___45___ of bad hair days, LaFrance concluded that women tend to feel more disgraced, embarrassed, or self-conscious, while men feel more nervous, unconfident and unsociable.Psychotherapist (精神治疗医师) Heather Turgeon agrees, “Our culture ___46___ emphasizes appearance. It affects each person differently, and how we look to the outside world is ___47___ to our mood and self image.”Turgeon says being in a low-confidence-hair-day mood is hard, because you feel like everyone ___48___, and you’re sure that’s the first thing they see.She also says that when we’re feeling low we’re more likely to think we don’t look good, even if our hair looks exactly the same. “It’s___49___ worse in your head than it is in other people’s eyes,”says Turgeon. “We think people notice our small flaws, but generally they are more ___50___ than we are to ourselves.”III.Reading ComprehensionPassage AHave you ever seen any students whose trousers hang so low you can see their underwear? What do you think of that? Fashionable? Some of today’s teenagers are big fans of such a look. But recently this trend has been at the centre of an argument in Italian middle schools. The headmaster of a school in central Italy has asked students to stop wearing low-rise jeans that expose underwear and parts of the body. His request came after a class trip, when he saw one boy’s baggy trousers slide to his feet. He pointed out that this way of dressing is not suitable for school. But in Italy, a nation that takes fashion very seriously, the suggestion caused a debate among parents, teachers and students. The issue is whether the headmaster’s request will limitstudents’ freedom or whether dress in Italian schools is too casual. A parents’ group praised the move in favour of good taste, while others advised schools to stop worrying about fashion and fix up old school buildings. “We do not want to kick fashion out,”the headmaster explained, “but extremes (极端) of fashion like this are not right in school.”Many other schools have now requested that their students also stop wearing such trousers. Most students have simply ignored the request. Ludovica Gaudio, 14, wore extremely low trousers exposing orange underwear in class. It was cold, so she wore a matching orange scarf. Another 14-year-old said she would probably respect the request, simply for practical reasons. “I don’t really feel comfortable in those sort of jeans,”said Sarah Lattanzi, “in winter, when dressed like that, it’s quite cold and I am af raid my stomach will ache.”Choose the best answer.( ) 1. What led to the argument in Italian middle schools?A. Students’ craze for fashions.B. Students’ exposing Clothes.C. Students’ ignoring dress codes (着装规则)D. Students’ underwear.( ) 2. Which of the following supports the headmaster’s request?A. Fashion should be taken seriously.B. Fashion should not be followed in school.C. Students should have their freedom in choosing what they wear.D. Students should be encouraged to have good taste in clothes.( ) 3. The opinion against the headmaster is that _________.A. schools should ignore the students’ exposing clothesB. fashion should not be followed in schoolC. the way studnets dress in Italian schools is too casualD. low-rise jeans can do harm to youngster’s health( ) 4. The purpose of this story is to show that _________.A. wearing fashionablly in school is under attack in other countries, tooB. dress code is necessary even in a country like ItalyC. a debate over the way students dress in Italian middle schoolsD. Italian students react differently to schools’ requestsBSocial customs and ways of behaving change. Things which were considered impolite many years ago are now acceptable. Just a few years ago,it was considered impolite behaviour for a man to smoke on the street.No man who thought of himself as being a gentleman would make a fool of himself by smoking when a lady was in the room.Customs also differ from country to country.Does a man walk on the left or the right of a woman in your country? Or doesn’t it matter? What about table manners? Should you use both hands while you are eating? Should you leave one in your lap (膝部), or on the table?The Americans and British not only speak the same language but also share a large number ofsocial customs. For example, in both America and England people shake hands when they meet each other for the first time. Also,most Englishmen will open a door for a woman or offer their seat to a woman, and so will most Americans. Promptness (准时) is important both in England and America—That is if a dinner invitation is for seven o’clock, the dinner guest either arrives close to that time or calls up to explain his delay.The important thing to remember about social customs is not to do anything that might make other people feel uncomfortable—especially if they are your guests. There is an old story about a man who gave a very formal dinner party. When the food was served, one of the guests started to eat his peas with a knife. The other guests were amused or shocked, but the host calmly picked up his knife and began eating the same way. It would have been bad manners to make his guest feel foolish or uncomfortable.( ) 70. Can social customs and behaviour change?A. Yes, they can.B. No, they can’t.C. Surely!they can’t.D. Sorry!I don’t know.( ) 71. It was considered impolite _________ a few years ago.A. to eat with both handsB. to leave a hand when eatingC. to smoke in the roomD. to smoke on the street( ) 72. It is impolite _________ in both America and England.A. for people to shake hands when they meet each other for the first timeB. for a man to open a door for a womanC. for a person to be late for a friend’s engagementD. for a man to offer his seat to a woman( ) 73. If your guest has bad table manners at your dinner party, you should _________.A. tell him not to do soB. do as he didC. not do the same thingD. be amused or shockedI.Translation (22分)1.假如你能尽快回信,我将不胜感激。