河南省开封高中2021届高一下学期第三次质量检测

合集下载

河南省开封市2021届新高考数学三月模拟试卷含解析

河南省开封市2021届新高考数学三月模拟试卷一、选择题：本题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．如图，圆O 的半径为1，A ，B 是圆上的定点，OB OA ⊥，P 是圆上的动点，点P 关于直线OB 的对称点为P '，角x 的始边为射线OA ，终边为射线OP ，将OP OP '-u u u r u u u r表示为x 的函数()f x ，则()y f x =在[]0,π上的图像大致为（）A ．B ．C ．D ．【答案】B 【解析】【分析】根据图象分析变化过程中在关键位置及部分区域，即可排除错误选项，得到函数图象，即可求解. 【详解】由题意，当0x =时，P 与A 重合，则P '与B 重合，所以||2OP OP BA '-==u u u r u u u r u u u r,故排除C,D 选项；当02x π<<时，||2sin()2cos 2OP OP P P x x π''-==-=u u u r u u u r ，由图象可知选B.故选：B 【点睛】本题主要考查三角函数的图像与性质，正确表示函数的表达式是解题的关键,属于中档题.2．如图，双曲线()2222:10,0x y C a b a b-=>>的左，右焦点分别是()()12,0,,0,F c F c -直线2bc y a =与双曲线C 的两条渐近线分别相交于,A B 两点.若12,3BF F π∠=则双曲线C 的离心率为（）A ．2B 42C 2D 23【答案】A 【解析】【分析】易得(,)22c bc B a -，过B 作x 轴的垂线，垂足为T ，在1FTB ∆中，利用1tan 3BT FT π=即可得到,,a b c 的方程. 【详解】由已知，得(,)22c bc B a -，过B 作x 轴的垂线，垂足为T ，故12cFT =，又12,3BF F π∠=所以1tan 33BT FT π==，即232bcb ac a == 所以双曲线C 的离心率21()2be a =+.故选：A. 【点睛】本题考查双曲线的离心率问题，在作双曲线离心率问题时，最关键的是找到,,a b c 的方程或不等式，本题属于容易题.3．已知函数()2sin()(0,0)3f x x A ωωπ=->>，将函数()f x 的图象向左平移3π个单位长度，得到函数()g x的图象，若函数()g x 的图象的一条对称轴是6x π=，则ω的最小值为A ．16B ．23C ．53D ．56【答案】C 【解析】【分析】【详解】将函数()f x 的图象向左平移3π个单位长度，得到函数()2sin()33g x x ωωππ=+-的图象，因为函数()g x 的图象的一条对称轴是6x π=，所以sin()1633ωωπππ+-=±，即,6332k k ωωππππ+-=+π∈Z ，所以52,3k k ω=+∈Z ，又0>ω，所以ω的最小值为53．故选C ． 4．设全集U=R ，集合()2log 41{|}A x x =-≤，()()35{|}0B x x x =-->，则()U B A =I ð（）A ．[2]5，B ．[2]3，C ．[)24，D ．[)34，【答案】D 【解析】【分析】求解不等式，得到集合A ，B ，利用交集、补集运算即得解【详解】由于2log (4)124x x -≤∴≤<故集合[)24A =， ()()350x x -->3x ∴<或5x >故集合()()35B =-∞⋃+∞，， ∴ ()[)|34U B A ⋂=，ð 故选：D 【点睛】本题考查了集合的交集和补集混合运算，考查了学生概念理解，数学运算的能力，属于中档题. 5．已知(0,)απ∈，且tan 2α=，则cos2cos αα+=（）A B ．C D 【答案】B 【解析】分析：首先利用同角三角函数关系式，结合题中所给的角的范围，求得cos α的值，之后借助于倍角公式，将待求的式子转化为关于cos α的式子，代入从而求得结果. 详解：根据题中的条件，可得α为锐角，根据tan 2α=，可求得cos α=，而22cos 2cos 2cos cos 115αααα+=+-=+-=，故选B. 点睛：该题考查的是有关同角三角函数关系式以及倍角公式的应用，在解题的过程中，需要对已知真切求余弦的方法要明确，可以应用同角三角函数关系式求解，也可以结合三角函数的定义式求解.6．根据散点图，对两个具有非线性关系的相关变量x ，y 进行回归分析，设u= lny ，v=(x-4)2，利用最小二乘法，得到线性回归方程为ˆu=-0.5v+2，则变量y 的最大值的估计值是（） A ．e B ．e 2C ．ln2D ．2ln2【答案】B 【解析】【分析】将u= lny ，v=(x-4)2代入线性回归方程ˆu=-0.5v+2，利用指数函数和二次函数的性质可得最大估计值. 【详解】解：将u= lny ，v=(x -4)2代入线性回归方程ˆu=-0.5v+2得： ()2ln 0.542y x =--+，即()20.542x y e --+=，当4x =时，()20.542x --+取到最大值2，因为xy e =在R 上单调递增，则()20.542x y e --+=取到最大值2e .故选：B. 【点睛】本题考查了非线性相关的二次拟合问题，考查复合型指数函数的最值，是基础题,.7．盒中装有形状、大小完全相同的5张“刮刮卡”，其中只有2张“刮刮卡”有奖，现甲从盒中随机取出2张，则至少有一张有奖的概率为( ) A ．12B ．35C ．710D ．45【答案】C 【解析】【分析】先计算出总的基本事件的个数，再计算出两张都没获奖的个数，根据古典概型的概率，求出两张都没有奖的概率，由对立事件的概率关系，即可求解.【详解】从5张“刮刮卡”中随机取出2张，共有2510C =种情况，2张均没有奖的情况有233C =（种），故所求概率为3711010-=. 故选:C. 【点睛】本题考查古典概型的概率、对立事件的概率关系，意在考查数学建模、数学计算能力，属于基础题. 8．如图，已知直线:l ()()10y k x k =+>与抛物线2:4C y x =相交于A ，B 两点，且A 、B 两点在抛物线准线上的投影分别是M ，N ，若2AM BN =，则k 的值是（）A ．13B ．2 C ．23D ．2【答案】C 【解析】【分析】直线()()10y k x k =+>恒过定点()10P -，，由此推导出12OB AF =，由此能求出点B 的坐标，从而能求出k 的值．【详解】设抛物线2:4C y x =的准线为:1l x =-，直线()()10y k x k =+>恒过定点()10P -，，如图过A 、B 分别作AM l ⊥于M ，BN l ⊥于N ，由2AM BN =，则2FA FB =，点B 为AP 的中点、连接OB ，则12OB AF =， ∴OB BF =，点B 的横坐标为12， ∴点B 的坐标为122B ⎛⎝，把122B ⎛ ⎝代入直线()()10y k x k =+>，解得223k =，故选：C ．【点睛】本题考查直线与圆锥曲线中参数的求法，考查抛物线的性质，是中档题，解题时要注意等价转化思想的合理运用，属于中档题.9．若平面向量,,a b c r r r，满足||2,||4,4,||3a b a b c a b ==⋅=-+=r r r r r r r ，则||c b -r r 的最大值为（）A ．523B ．523C ．2133D ．2133【答案】C 【解析】【分析】可根据题意把要求的向量重新组合成已知向量的表达，利用向量数量积的性质，化简为三角函数最值. 【详解】由题意可得：()(2)c b c a b a b -=-++-r r r r r r r，2222|2|(2)||4||444164452a b a b a b a b -=-=+⋅-⋅=+⨯-⨯=r r r rr r r r Q |2|213a b ∴-=r r2222||()[()(2)]|()(2)|c b c b c a b a b c a b a b ∴-=-=-++-=-++-r r r r r r r r r r r r r r22|||2|2|||2|cos ,2c a b a b c a b a b c a b a b =-++-+⋅-+⋅-⋅<-++>r r r r r r r r r r r r r r r35223213cos ,2c a b a b =++<-++>r r r r r55439cos ,2c a b a b =+<-++>r r r r r55439+…25543952221333(2133)+=+⨯=Q ，故选：C【点睛】本题主要考查根据已知向量的模求未知向量的模的方法技巧,把要求的向量重新组合成已知向量的表达是本题的关键点.本题属中档题.10． “角谷猜想”的内容是：对于任意一个大于1的整数n ，如果n 为偶数就除以2，如果n 是奇数，就将其乘3再加1，执行如图所示的程序框图，若输入10n =，则输出i 的（）A ．6B ．7C ．8D ．9【答案】B 【解析】【分析】模拟程序运行，观察变量值可得结论．【详解】循环前1,10i n ==，循环时：5,2n i ==，不满足条件1n =；16,3n i ==，不满足条件1n =；8,4n i ==，不满足条件1n =；4,5n i ==，不满足条件1n =；2,6n i ==，不满足条件1n =；1,7n i ==，满足条件1n =，退出循环，输出7i =．故选：B ．【点睛】本题考查程序框图，考查循环结构，解题时可模拟程序运行，观察变量值，从而得出结论． 11．设等差数列{}n a 的前n 项和为n S ，若5632a a a +=+，则7S =（） A ．28 B ．14C ．7D ．2【答案】B 【解析】【分析】根据等差数列的性质6345a a a a +=+并结合已知可求出4a ，再利用等差数列性质可得1774()772a a S a +==，即可求出结果．【详解】因为6345a a a a +=+，所以5452a a a +=+，所以42a =，所以17747()7142a a S a +===，故选：B 【点睛】本题主要考查等差数列的性质及前n 项和公式，属于基础题． 12．函数ln ||()xx x f x e=的大致图象为（） A ． B ．C ．D ．【答案】A 【解析】【分析】利用特殊点的坐标代入，排除掉C ，D ；再由1()12f -<判断A 选项正确. 【详解】1.11.1ln |1.1|( 1.1)0f e --=<，排除掉C ，D ；1211ln 122()22f e e---==122e <=Q 2e ，1()212f e ∴-=<.本题考查了由函数解析式判断函数的大致图象问题，代入特殊点，采用排除法求解是解决这类问题的一种常用方法，属于中档题.二、填空题：本题共4小题，每小题5分，共20分。

河南省开封市2021届新高考三诊数学试题含解析

河南省开封市2021届新高考三诊数学试题一、选择题：本题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1．已知函数21()(1)()2x f x ax x e a R =--∈若对区间[]01，内的任意实数123x x x 、、，都有123()()()f x f x f x +≥，则实数a 的取值范围是( ）A ．[]12， B ．[]e,4C ．[]14， D ．[)[]12,4e ⋃，【答案】C 【解析】分析：先求导,再对a 分类讨论求函数的单调区间，再画图分析转化对区间[]01，内的任意实数123x x x 、、，都有()()()123f x f x f x +≥，得到关于a 的不等式组，再解不等式组得到实数a 的取值范围. 详解：由题得()[(1)]()xxxxf x ax e x e ax xe x a e =-+-=-=-'.当a ＜1时，()0f x '<，所以函数f （x ）在[]01，单调递减，因为对区间[]01，内的任意实数123x x x 、、，都有()()()123f x f x f x +≥，所以(1)(1)(0)f f f +≥，所以111,22a a +≥ 故a≥1，与a ＜1矛盾，故a ＜1矛盾.当1≤a<e 时，函数f(x)在[0,lna]单调递增，在(lna,1]单调递减. 所以2max 1()(ln )ln ln ,2f x f a a a a a a ==-+ 因为对区间[]01，内的任意实数123x x x 、、，都有()()()123f x f x f x +≥，所以(0)(1)(ln )f f f a +≥，所以2111ln ln ,22a a a a a a +≥-+ 即211ln ln 1022a a a a a -+-≤令211()ln ln 1,(1)22g a a a a a a a e =-+-≤<，所以21()(ln 1)0,2g a a =-<'所以函数g(a)在（1，e ）上单调递减，所以max 1()(1)02g a g ==-<，所以当1≤a<e 时，满足题意.因为对区间[]01，内的任意实数123x x x 、、，都有()()()123f x f x f x +≥，所以(0)(0)(1)f f f +≥, 故1+112a ≥，所以 4.a ≤ 故 4.e a ≤≤综上所述，a ∈[]14，. 故选C.点睛：本题的难点在于“对区间[]01，内的任意实数123x x x 、、，都有()()()123f x f x f x +≥”的转化.由于是函数的问题，所以我们要联想到利用函数的性质（单调性、奇偶性、周期性、对称性、最值、极值等）来分析解答问题.本题就是把这个条件和函数的单调性和最值联系起来，完成了数学问题的等价转化，找到了问题的突破口. 2．设22(1)1z i i=+++（i 是虚数单位），则||z =（）A B ．1C ．2D 【答案】A 【解析】【分析】先利用复数代数形式的四则运算法则求出z ，即可根据复数的模计算公式求出||z ．【详解】∵22)1121(1z i i i i i=-+=+=+++，∴||z == 故选：A ．【点睛】本题主要考查复数代数形式的四则运算法则的应用，以及复数的模计算公式的应用，属于容易题．3．某中学有高中生1500人，初中生1000人为了解该校学生自主锻炼的时间，采用分层抽样的方法从高生和初中生中抽取一个容量为n 的样本.若样本中高中生恰有30人，则n 的值为（） A ．20 B ．50C ．40D ．60【答案】B 【解析】【分析】【详解】由题意，30=150015001000n⨯+，解得50n =.故选：B. 【点睛】本题考查简单随机抽样中的分层抽样，某一层样本数等于某一层的总体个数乘以抽样比，本题是一道基础题.4．定义在R 上的奇函数()f x 满足()()330f x f x --+-=，若()11f =，()22f =-，则()()()()1232020f f f f ++++=L （）A ．1-B ．0C ．1D ．2【答案】C 【解析】【分析】首先判断出()f x 是周期为6的周期函数，由此求得所求表达式的值. 【详解】由已知()f x 为奇函数，得()()f x f x -=-，而()()330f x f x --+-=，所以()()33f x f x -=+，所以()()6f x f x =+，即()f x 的周期为6.由于()11f =，()22f =-，()00f =，所以()()()()33330f f f f =-=-⇒=，()()()4222f f f =-=-=， ()()()5111f f f =-=-=-， ()()600f f ==.所以()()()()()()1234560f f f f f f +++++=，又202063364=⨯+，所以()()()()1232020f f f f ++++=L ()()()()12341f f f f +++=. 故选：C本小题主要考查函数的奇偶性和周期性，属于基础题.5．设m ，n 为直线，α、β为平面，则m α⊥的一个充分条件可以是( ) A ．αβ⊥，n αβ=I ，m n ⊥ B ．//αβ，m β⊥ C ．αβ⊥，//m β D ．n ⊂α，m n ⊥【答案】B 【解析】【分析】根据线面垂直的判断方法对选项逐一分析，由此确定正确选项. 【详解】对于A 选项，当αβ⊥，n αβ=I ，m n ⊥时，由于m 不在平面β内，故无法得出m α⊥. 对于B 选项，由于//αβ，m β⊥，所以m α⊥.故B 选项正确.对于C 选项，当αβ⊥，//m β时，m 可能含于平面α，故无法得出m α⊥. 对于D 选项，当n ⊂α，m n ⊥时，无法得出m α⊥. 综上所述，m α⊥的一个充分条件是“//αβ，m β⊥” 故选：B 【点睛】本小题主要考查线面垂直的判断，考查充分必要条件的理解，属于基础题. 6．若ABC ∆的内角A 满足2sin 23A =-，则sin cos A A -的值为（）A B ． C D ．5-3【答案】A 【解析】【分析】由2sin 22sin cos 3A A A ==-，得到1sin cos 03A A =-<，得出(,)2A ππ∈，再结合三角函数的基本关系式，即可求解. 【详解】由题意，角A 满足2sin 22sin cos 3A A A ==-，则1sin cos 03A A =-<，又由角A 是三角形的内角，所以(,)2A ππ∈，所以sin cos A A >，因为225sin cos 12sin cos 1()A A A A -=-=--=，所以sin cos 3A A -=. 故选：A. 【点睛】本题主要考查了正弦函数的性质，以及三角函数的基本关系式和正弦的倍角公式的化简、求值问题，着重考查了推理与计算能力.7．函数()sin 2sin 3f x x m x x =++在[,]63ππ上单调递减的充要条件是（）A ．3m ≤-B ．4m ≤-C．m ≤D ．4m ≤【答案】C 【解析】【分析】先求导函数，函数在[,]63ππ上单调递减则()0f x '≤恒成立，对导函数不等式换元成二次函数，结合二次函数的性质和图象，列不等式组求解可得. 【详解】依题意，2()2cos 2cos 34cos cos 1f x x m x x m x '=++=++，令cos x t =，则1[,22t ∈，故2410t mt ++≤在[1,22上恒成立；结合图象可知，1141042341042m m ⎧⨯+⨯+⎪⎪⎨⎪⨯+⨯+⎪⎩„„，解得43m m -⎧⎪⎨-⎪⎩„„故3m ≤-. 故选：C. 【点睛】本题考查求三角函数单调区间. 求三角函数单调区间的两种方法:(1)代换法：就是将比较复杂的三角函数含自变量的代数式整体当作一个角u (或t )，利用基本三角函数的单调性列不等式求解；(2)图象法：画出三角函数的正、余弦曲线，结合图象求它的单调区间. 8．设(),1,a b ∈+∞，则“a b > ”是“log 1a b <”的（） A ．充分而不必要条件 B ．必要而不充分条件【解析】【分析】根据充分条件和必要条件的定义结合对数的运算进行判断即可．【详解】∵a ，b ∈（1，+∞）， ∴a ＞b ⇒log a b ＜1， log a b ＜1⇒a ＞b ，∴a ＞b 是log a b ＜1的充分必要条件，故选C ．【点睛】本题主要考查充分条件和必要条件的判断，根据不等式的解法是解决本题的关键． 9．已知函数1()cos 22f x x x π⎛⎫=++ ⎪⎝⎭，,22x ππ⎡⎤∈-⎢⎥⎣⎦，则()f x 的极大值点为( ) A ．3π-B ．6π-C ．6π D ．3π 【答案】A 【解析】【分析】求出函数的导函数，令导数为零，根据函数单调性，求得极大值点即可. 【详解】因为()11cos 222f x x x x sinx π⎛⎫=++=- ⎪⎝⎭，故可得()12f x cosx '=-+，令()0f x '=，因为,22x ππ⎡⎤∈-⎢⎥⎣⎦，故可得3x π=-或3x π=，则()f x 在区间,23ππ⎛⎫-- ⎪⎝⎭单调递增，在,33ππ⎛⎫-⎪⎝⎭单调递减，在,32ππ⎛⎫ ⎪⎝⎭单调递增，故()f x 的极大值点为3π-.本题考查利用导数求函数的极值点，属基础题. 10．要得到函数()sin(3)3f x x π=+的导函数()f x '的图像，只需将()f x 的图像（）A ．向右平移3π个单位长度，再把各点的纵坐标伸长到原来的3倍 B ．向右平移6π个单位长度，再把各点的纵坐标缩短到原来的13倍 C ．向左平移3π个单位长度，再把各点的纵坐标缩短到原来的13倍 D ．向左平移6π个单位长度，再把各点的纵坐标伸长到原来的3倍【答案】D 【解析】【分析】先求得()'fx ，再根据三角函数图像变换的知识，选出正确选项.【详解】依题意()'553cos 33cos 33sin 33626fx x x x ππππ⎡⎤⎛⎫⎛⎫⎛⎫=+=+-=+ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦3sin 363x ππ⎡⎤⎛⎫=++ ⎪⎢⎥⎝⎭⎣⎦，所以由()sin(3)3f x x π=+向左平移6π个单位长度，再把各点的纵坐标伸长到原来的3倍得到()'f x 的图像.故选：D 【点睛】本小题主要考查复合函数导数的计算，考查诱导公式，考查三角函数图像变换，属于基础题.11．已知函数log ()a y x c =+（a ，c 是常数，其中0a >且1a ≠）的大致图象如图所示，下列关于a ，c 的表述正确的是（）A ．1a >，1c >B ．1a >，01c <<C ．01a <<，1c >D ．01a <<，01c <<【答案】D根据指数函数的图象和特征以及图象的平移可得正确的选项. 【详解】从题设中提供的图像可以看出()01,log 0,log 10a a a c c <<>+>，故得01,01c a <<<<，故选：D ．【点睛】本题考查图象的平移以及指数函数的图象和特征，本题属于基础题.12．已知1F ，2F 是椭圆与双曲线的公共焦点，P 是它们的一个公共点，且21PF PF >，椭圆的离心率为1e ，双曲线的离心率为2e ，若112PF F F =，则2133e e +的最小值为（） A．6+ B．6+C ．8D ．6【答案】C 【解析】【分析】由椭圆的定义以及双曲线的定义、离心率公式化简2133e e +，结合基本不等式即可求解.【详解】设椭圆的长半轴长为a ，双曲线的半实轴长为a '，半焦距为c ，则1ce a=，2c e a ='，设2PF m =由椭圆的定义以及双曲线的定义可得：1222m PF PF a a c +=⇒=+，2122mPF PF a a c ''-=⇒=- 则2133e e +33322633322m m c c a c c c m m c a c c c c ⎛⎫⎛⎫+- ⎪ ⎪⎝⎭⎝⎭=+=+=++'⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭68≥+= 当且仅当73a c =时，取等号.本题主要考查了椭圆的定义以及双曲线的定义、离心率公式，属于中等题. 二、填空题：本题共4小题，每小题5分，共20分。

2024届河南省开封市高三下学期第三次质量检测文综地理试题(解析版)

2024届河南省开封市高三下学期第三次质量检测文综地理试题（解析版）一、选择题：在每小题给出的四个选项中，只有一项是符合题目要求的。

净增人口数是人口自然增长数和人口迁移增长数之和。

下图为2000—2020年河南省与上海市人口变动图。

据此完成下面小题。

1. 与2000—2010年相比，2010—2020年河南省人口（）A. 人口数量下降B. 出生率提高C. 人口增幅变大D. 人口净减少2. 2010—2020年上海市人口变动幅度变化的主要原因是（）A. 住房困难B. 产业转移C. 交通拥堵D. 经济萎缩【答案】1 C 2. B【解析】【1题详解】由图可知，与2000—2010年相比，2010—2020年河南省人口净变动数明显增加，人口增幅变大，C正确；人口数量增加，A错误；无法从图中直接看出出生率是否提高，B错误；人口净增多，D错误。

故选C。

【2题详解】2010--2020年，上海市人口变动幅度减小主要原因是产业转移，劳动密集型产业向外转移，部分劳动力从上海流出，B正确；住房困难、交通拥堵可能在一定程度上减少人口迁入，但不是主要原因，A错误；上海市经济并未萎缩，D错误。

故选B。

【点睛】影响人口迁移的因素：1、自然环境因素。

如：气候、淡水资源分布、土壤、矿产、自然灾害等，这些是人类赖以生存的自然环境，在很大程度上决定了人类生活、生产的空间格局，从而决定着人类的迁移方向和规模。

2、社会经济因素。

如：经济因素、交通和通信的发展、文化教育事业的发展、婚姻和家庭等，人类为了能有更好的生活水平，满足生活态度和生活期望也会进行迁移。

3、政治因素。

如：有关人口迁移的政策、战争、政治变革等，常常引起人口迁移。

2024年2月在新加坡航展上，我国自主设计研发组装的大飞机C919实现国际航展首秀。

C919核心部件全球招标采购，零部件生产分布在全国各地，总装基地位于上海。

下图为C919飞机部分零部件厂家示意图。

据此完成下面小题。

河南省开封市普通高中2021届高三毕业班下学期第三次高考模拟考试数学(文)试题(解析版)

绝密★启用前河南省开封市普通高中2021届高三毕业班下学期第三次高考模拟考试数学(文)试题(解析版)一、选择题（共12小题,每小题5分,共60分）.1．设a,b∈R,A＝{1,a},B＝{﹣1,﹣b},若A⊆B,则a﹣b＝（）A．﹣1 B．﹣2 C．2 D．0解：a,b∈R,A＝{1,a},B＝{﹣1,﹣b},A⊆B,考试a＝﹣1,b＝﹣1,所以a﹣b＝0,故选：D．2．设复数z满足|z|＝|z﹣i|＝1,且z的实部大于虚部,则z＝（）A．B．C．D．解：设z＝a+bi,（a,b∈R）,∵复数z满足|z|＝|z﹣i|＝1,∴＝1,＝1,即x2+y2＝1,x2+y2﹣2y＝0,解得y＝,x＝±,∵z的实部大于虚部,∴x＝,∴z＝+i,故选：B．3．“方程表示双曲线”的一个必要不充分条件为（）A．m∈（﹣∞,﹣1）∪（1,+∞）B．m∈（﹣∞,﹣2）∪（1,+∞）C．m∈（﹣∞,﹣2）D．m∈（1,+∞）解：方程为双曲线时,（m+2）（m﹣1）＞0∴m∈（﹣∞,﹣2）∪（1,+∞）,∵（﹣∞,﹣2）∪（1,+∞）⊊（﹣∞,﹣1）∪（1,+∞）,“方程表示双曲线”的一个必要不充分条件为m∈（﹣∞,﹣1）∪（1,+∞）．故选：A．4．2021年开始,我省将试行“3+1+2“的普通高考新模式,即除语文、数学、外语3门必选科目外,考生再从物理、历史中选1门,从化学、生物、地理、政治中选2门作为选考科目．为了帮助学生合理选科,某中学将高一每个学生的六门科目综合成绩按比例均缩放成5分制,绘制成雷达图．甲同学的成绩雷达图如图所示,下面叙述一定不正确的是（）A．甲的物理成绩领先年级平均分最多B．甲有2个科目的成绩低于年级平均分C．甲的成绩从高到低的前3个科目依次是地理、化学、历史D．对甲而言,物理、化学、地理是比较理想的一种选科结果解：甲的成绩从高到低的前3个科目依次是地理、化学、生物（物理）,。

河南省开封市2021届新高考数学第三次调研试卷含解析

河南省开封市2021届新高考数学第三次调研试卷一、选择题：本题共12小题，每小题5分，共60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．若θ是第二象限角且sinθ =1213，则tan()4πθ+= A ．177-B ．717- C ．177D ．717【答案】B 【解析】由θ是第二象限角且sinθ =1213知：25cos 1sin 13θθ=--=-，5t n 1a 2θ-=．所以tan tan 457tan()41tan tan 4517πθθθ+︒+==--︒．2．已知抛物线24y x =的焦点为F ，P 为抛物线上一点，(1,1)A ，当PAF ∆周长最小时，PF 所在直线的斜率为（） A ．43-B ．34-C ．34D ．43【答案】A 【解析】【分析】本道题绘图发现三角形周长最小时A,P 位于同一水平线上，计算点P 的坐标，计算斜率，即可．【详解】结合题意，绘制图像要计算三角形PAF 周长最小值，即计算PA+PF 最小值，结合抛物线性质可知，PF=PN ，所以PF PA PA PN AN AG +=+≥≥，故当点P 运动到M 点处，三角形周长最小，故此时M 的坐标为1,14⎛⎫ ⎪⎝⎭，所以斜率为1041314k -==--，故选A ．【点睛】本道题考查了抛物线的基本性质，难度中等．3．若,x y 满足320020x y x y x y --≤⎧⎪-≥⎨⎪+≥⎩，且目标函数2(0,0)z ax by a b =+>>的最大值为2，则416a b +的最小值为( ) A ．8 B ．4C ．22D ．6【答案】A 【解析】【分析】作出可行域，由2(0,0)z ax by a b =+>>，可得22a z y x b b =-+.当直线22a z y x b b=-+过可行域内的点()1,1B 时，z 最大，可得22a b +=.再由基本不等式可求416a b +的最小值. 【详解】作出可行域，如图所示由2(0,0)z ax by a b =+>>，可得22a zy x b b=-+. 平移直线22a z y x b b =-+，当直线过可行域内的点B 时，2zb最大，即z 最大，最大值为2. 解方程组3200x y x y --=⎧⎨-=⎩，得()1,1,11x B y =⎧∴⎨=⎩. 22(0,0)a b a b ∴+=>>.22224164424424248a b a b a b a b +∴+=+≥⨯===，当且仅当244a b =，即12,1222a a b a b b =⎧=⎧⎪⎨⎨+==⎩⎪⎩时，等号成立.416a b ∴+的最小值为8.故选：A . 【点睛】本题考查简单的线性规划，考查基本不等式，属于中档题. 4．已知函数()()614,7,7x a x x f x ax -⎧-+≤=⎨>⎩是R 上的减函数，当a 最小时，若函数()4y f x kx =--恰有两个零点，则实数k 的取值范围是（） A ．1(,0)2-B ．1(2,)2- C ．(1,1)- D ．1(,1)2【答案】A 【解析】【分析】首先根据()f x 为R 上的减函数，列出不等式组，求得112a ≤<，所以当a 最小时，12a =，之后将函数零点个数转化为函数图象与直线交点的个数问题，画出图形，数形结合得到结果. 【详解】由于()f x 为R 上的减函数，则有()1001714a a a a ⎧-<⎪<<⎨⎪≤-+⎩，可得112a ≤<，所以当a 最小时，12a =，函数()4y f x kx =--恰有两个零点等价于方程()4f x kx =+有两个实根，等价于函数()y f x =与4y kx =+的图像有两个交点．画出函数()f x 的简图如下，而函数4y kx =+恒过定点()0,4，数形结合可得k 的取值范围为102k -<<．故选：A. 【点睛】该题考查的是有关函数的问题，涉及到的知识点有分段函数在定义域上单调减求参数的取值范围，根据函数零点个数求参数的取值范围，数形结合思想的应用，属于中档题目.5．甲在微信群中发了一个6元“拼手气”红包,被乙､丙､丁三人抢完,若三人均领到整数元,且每人至少领到1元,则乙获得“最佳手气”(即乙领到的钱数多于其他任何人)的概率是（） A ．13B ．310C ．25D ．34【答案】B 【解析】【分析】将所有可能的情况全部枚举出来,再根据古典概型的方法求解即可. 【详解】设乙,丙,丁分别领到x 元,y 元,z 元,记为(,,)x y z ,则基本事件有(1,1,4),(1,4,1) ,(4,1,1),(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1),(2,2,2),共10个,其中符合乙获得“最佳手气”的有3个,故所求概率为310, 故选：B. 【点睛】本题主要考查了枚举法求古典概型的方法,属于基础题型.6．已知函数()()()2ln 14f x ax x ax =-+-，若0x >时，()0f x ≥恒成立，则实数a 的值为（）A ．2eB ．4eC ．2e - D ．4e- 【答案】D 【解析】【分析】通过分析函数()ln 10y ax x =->与()240y x ax x =+->的图象，得到两函数必须有相同的零点t ，解方程组2ln 1040at a at -=⎧⎨+-=⎩即得解.【详解】如图所示，函数()ln 10y ax x =->与()240y x ax x =+->的图象，因为0x >时，()0f x ≥恒成立，于是两函数必须有相同的零点t ，所以2ln 1040at a at -=⎧⎨+-=⎩24at t e =-=，解得4a e- 故选：D 【点睛】本题主要考查函数的图象的综合应用和函数的零点问题，考查不等式的恒成立问题，意在考查学生对这些知识的理解掌握水平.7．将函数()2sin(3)(0)f x x ϕϕπ=+<<图象向右平移8π个单位长度后，得到函数的图象关于直线3x π=对称，则函数()f x 在,88ππ⎡⎤-⎢⎥⎣⎦上的值域是（）A ．[1,2]-B ．[2]C ．2⎡⎤-⎢⎥⎣⎦D ．[2]【答案】D 【解析】【分析】由题意利用函数sin()y A x ωϕ=+的图象变换规律，三角函数的图象的对称性，余弦函数的值域，求得结果. 【详解】解：把函数()2sin(3)(0)f x x ϕϕπ=+<<图象向右平移8π个单位长度后，可得32sin 38y x πϕ⎛⎫=-+⎪⎝⎭的图象；再根据得到函数的图象关于直线3x π=对称，33382k πππϕπ∴⨯-+=+，k Z ∈， 78πϕ∴=，函数7()2sin 38f x x π⎛⎫=+⎪⎝⎭.在,88ππ⎡⎤-⎢⎥⎣⎦上，753,824x πππ⎡⎤+∈⎢⎥⎣⎦，sin 382x π⎡⎤⎛⎫∴-∈-⎢⎥ ⎪⎝⎭⎣⎦，故()2sin 3[8f x x π⎛⎫=-∈ ⎪⎝⎭，即()f x 的值域是[2]，故选：D. 【点睛】本题主要考查函数sin()y A x ωϕ=+的图象变换规律，三角函数的图象的对称性，余弦函数的值域，属于中档题．8．曲线(2)x y ax e =+在点(0,2)处的切线方程为2y x b =-+，则ab =（） A ．4- B ．8-C ．4D ．8【答案】B 【解析】【分析】求函数导数，利用切线斜率求出a ，根据切线过点(0,2)求出b 即可. 【详解】因为(2)x y ax e =+，所以(2)xy e ax a '=++，故0|22x k y a ='==+=-，解得4a =-，又切线过点(0,2)，所以220b =-⨯+，解得2b =，所以8ab =-，故选：B 【点睛】本题主要考查了导数的几何意义，切线方程，属于中档题.9．已知集合A ＝{y|y =}，B ＝{x|y ＝lg （x ﹣2x 2）}，则∁R （A∩B ）＝（）A ．[0，12） B ．（﹣∞，0）∪[12，+∞） C ．（0，12）D ．（﹣∞，0]∪[12，+∞）【答案】D 【解析】【分析】求函数的值域得集合A ，求定义域得集合B ，根据交集和补集的定义写出运算结果. 【详解】集合A ＝{y|y =}＝{y|y≥0}＝[0，+∞）； B ＝{x|y ＝lg （x ﹣2x 2）}＝{x|x ﹣2x 2＞0}＝{x|0＜x 12＜}＝（0，12）， ∴A∩B ＝（0，12）， ∴∁R （A∩B ）＝（﹣∞，0]∪[12，+∞）. 故选：D. 【点睛】该题考查的是有关集合的问题，涉及到的知识点有函数的定义域，函数的值域，集合的运算，属于基础题目.10．中，如果，则的形状是（） A ．等边三角形 B ．直角三角形C ．等腰三角形D ．等腰直角三角形【答案】B 【解析】【分析】化简得lgcosA ＝lg＝﹣lg2，即，结合，可求，得代入sinC＝sinB ，从而可求C ，B ，进而可判断.【详解】由，可得lgcosA ＝＝﹣lg2，∴，∵，∴，，∴sinC ＝sinB ＝＝，∴tanC ＝，C＝，B ＝.故选：B 【点睛】本题主要考查了对数的运算性质的应用，两角差的正弦公式的应用，解题的关键是灵活利用基本公式，属于基础题．11．已知抛物线24y x =的焦点为F ，准线与x 轴的交点为K ，点P 为抛物线上任意一点KPF ∠的平分线与x 轴交于(,0)m ，则m 的最大值为( ) A ．322- B ．33C ．23D ．22-【答案】A 【解析】【分析】211(1)4mmx x-=+++，求出等式左边式子的范围，将等式右边代入，从而求解．【详解】解：由题意可得，焦点F （1，0），准线方程为x ＝−1，过点P 作PM 垂直于准线，M 为垂足，由抛物线的定义可得|PF|＝|PM|＝x＋1，记∠KPF的平分线与x轴交于(m,0),(1m1)H-<<根据角平分线定理可得||||||=||||||PF PM FHPK PK KH=，211(1)4mmx x-∴=+++，当0x=时，0m=，当0x≠时，212,142(1)4112x xxx⎡⎫=∈⎪⎢⎪++⎣⎭+++，211032221mmm-∴≤<⇒<≤-+，综上：0322m≤≤-．故选：A．【点睛】本题主要考查抛物线的定义、性质的简单应用，直线的斜率公式、利用数形结合进行转化是解决本题的关键．考查学生的计算能力，属于中档题．12．某市气象部门根据2018年各月的每天最高气温平均数据,绘制如下折线图,那么,下列叙述错误的是( )A．各月最高气温平均值与最低气温平均值总体呈正相关B ．全年中,2月份的最高气温平均值与最低气温平均值的差值最大C ．全年中各月最低气温平均值不高于10°C 的月份有5个D ．从2018年7月至12月该市每天最高气温平均值与最低气温平均值呈下降趋势【答案】D 【解析】【分析】根据折线图依次判断每个选项得到答案. 【详解】由绘制出的折线图知：在A 中，各月最高气温平均值与最低气温平均值为正相关，故A 正确；在B 中，全年中，2月的最高气温平均值与最低气温平均值的差值最大，故B 正确；在C 中，全年中各月最低气温平均值不高于10℃的月份有1月，2月，3月，11月，12月，共5个，故C 正确；在D 中，从2018年7月至12月该市每天最高气温平均值与最低气温平均值，先上升后下降，故D 错误. 故选：D. 【点睛】本题考查了折线图，意在考查学生的理解能力. 二、填空题：本题共4小题，每小题5分，共20分。

2021届河南省开封市高考物理三模试卷附答案详解

2021届河南省开封市高考物理三模试卷一、单选题（本大题共5小题，共30.0分）1.在电磁波谱中，频率不同的电磁波具有不同的特征．真空中，红外线、紫外线和X射线的波长由长到短排列顺序正确的是()A. X射线、紫外线、红外线B. X射线、红外线、紫外线C. 红外线、紫外线、X射线D. 紫外线、红外线、X射线2.一辆小汽车以12m/s的速度匀速行驶，当车头距路口停车线18m，时，绿灯还有3s时间将熄灭，司机因经验不足而刹车．在这3s内，汽车运动的v−t图象如图所示，下列说法正确的是()A. 小车在这3s内的平均速度为6m/sB. 小车刹车时的加速度大小为4m/s2C. 在绿灯熄灭时，该车的车头己越过停车线6mD. 至少应提前0.4s刹车才能在绿灯熄灭前不超停车线3.物体静止于光滑水平面上，力F作用于物体上的O点，现要使合力沿着OO′方向，如图所示，则必须同时再加一个力F′，如F和F′均在同一水平面上，则这个力的最小值为()A. FcosθB. FsinθC. FtanθD. Fcotθ4.如图所示，质量为m的竖直光滑圆环A的半径为R，固定在质量为m的木板B上，木板B的左右两侧各有一竖直挡板固定在地面上，B不能左右运动，在环的最低点静止放有一质量为m的小球C，现给小球一水平向右的瞬时速度v，使小球C在圆环内侧做圆周运动，某时刻木板离开水平面。

重力加速度为g，则小球的初速度v应满足()A. v≥√6gRB. v≥√7gRC. v≥√5gRD. v≥√3gR5.在如图所示的U−I图像中，直线a为某电源的路端电压与电流的关系，直线b为某电阻元件的电压与电流的关系。

现用该电源直接与元件连接成闭合电路。

则此时A. 电源将其它能转化为电能的功率为18WB. 该元件的电阻为2ΩC. 该元件发热功率为6WD. 电源外电路与内电路消耗功率之比为2:3二、多选题（本大题共5小题，共27.0分）6.图中虚线是某电场的一组等势面．两个带电粒子从P点沿等势面的切线方向射入电场，粒子仅受电场力作用，运动轨迹如实线所示，a、b是实线与虚线的交点．下列说法正确的是()A. 两粒子的电性相反B. a点的场强小于b点的场强C. a点的电势高于b点的电势D. 与P点相比两个粒子的电势能均减小7.四个质量相同的物体a、b、c、d，a做匀速直线运动、b做匀速圆周运动，c做匀减速直线运动，d做平抛运动，其中()A. 速度保持不变的仅是物体aB. 加速度保持不变的仅是物体bC. 动能保持不变的仅是物体a、bD. 动量保持不变的仅是物体b、c、d8.如图所示，两块相对的平行金属板M、N与电池相连接，N板接地，在距两板等远的一点P固定一个带正电的点电荷，如果将Ｍ板向上平移一小段距离，则A. 点电荷受到的电场力减小B. 点电荷受到的电场力增大C. 点电荷的电势能减小D. 点电荷的电势能保持不变9.下列说法正确的是()A. 液体表面层分子间距离大于液体内部分子间距离，故液体表面存在张力B. 悬浮在液体中的固体小颗粒会不停的做无规则的运动，这种运动是分子热运动C. 把很多小的单晶体放在一起，就变成了非晶体D. 恒温时压缩气体，气体分子与单位面积容器壁的碰撞次数变多E. 在绝热条件下压缩气体，气体的内能一定增加10.一列简谐横波沿x轴正方向传播，t=0时刻的波形如图所示。

2021届开封高级中学高三英语三模试题及参考答案

2021届开封高级中学高三英语三模试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AAmid the coronavirus outbreak, the U. S. Department of Homeland Security recommends having at least a two - week supply of water and food.PotatoesShelf life：2 to 5 weeks if stored in a cool, dry, dark placeYukon Gold, red, and fingerling potatoeswill last from two to three weeks. Larger white potatoes can last for three to five weeks. Sweet potatoes have about the same shelf life. Don't store them next to onions, however. The two might go together well in cooking, but raw, each gives off gases and moisture that might cause the other to spoil faster.Tea※Shelf life：6 to 12 months past "sell - by" dateDried tea leaves, whether loose (in a sealed container) or in teabags (in an unopened box) can easily last a year or more if they' re not subjected to damp or humidity. However, the tea does tend to lose flavor over time.Peanuts● Shelf life：1 to 2 monthsPeanuts in their shell, especially when kept cool and dry, are perfectly happy in the cupboard for as long as two months.Canned fruits and vegetables● Shelf life：1 to 2 years past "sell - by” dateCanning is an extremely efficient means of preserving food. Generally speaking, if canned foods aren't subjected to extreme heat, their contents should stay good for two years or more. Be aware, however, of dented cans or those with swollen tops, which may indicate the presence of bacteria inside.1. Which can go bad faster if stored with onions?A. Potatoes.B. Tea.C. Peanuts.D. Canned fruits and vegetables.2. What is special about tea?A. The flavor of tea can always remain the same.B. Tea leaves are better to be preserved in an open jar.C. Tea leaves should be kept away from the state of being wet.D. The maximum length of time that tea can be stored is 6 months.3. What may shorten the "sell - by” date of canned foods ?A. Shapes of cans.B. Categories of foods.C. Decline of the temperature.D. Exposure to high temperature.BDogs are often referred to as “man's best friend”. But MacKenzie, a four-pound Chihuahua (奇瓦瓦狗), who was named winner of the 2020 American Hero Dog competition, is making the world a better place for humans and animals alike. Often called the “Oscars for dogs”, the award recognizes dogs who make great contributions to society.This year's competition attracted over 400 competitors from across the country. While all were impressive, it was tiny MacKenzie who won the judges' hearts. Born at a rescue shelter in Hilton, New York, in 2013, she had a cleft palate (腭裂) that required her to be tube fed for the first year of her life. A life-saving operation, performed in 2014, gave her the ability to eat and drink independently, enabling the tiny dog to focus on doing what she loved most: taking care of others.The seven-year-old Chihuahua is now gainfully “employed” by the Mid Foundation, a Rochester, New York-based non-profit organization that shelters and cares for animals born with disabilities. MacKenzie's official job is “to provide love and care for baby rescue animals born with birth defects”. The Chihuahua is good at her job and hasnurturedmany different species-from puppies to kittens to turkeys, squirrels, birds and even a goat. She acts as their mother and teaches them how to socialize, play, and have good manners.In addition to her role as an animal caretaker, MacKenzie also has the important job of greeting the foundation's volunteers and friends. The incredible dog, who has lost her ability to bark, also visits area schools to help children understand physical disabilities in both animals and people. Her heart-warming and inspiring story makes MacKenzie worthy of America's top dog honor!4. What made MacKenzie American Hero Dog?A. Being man's best friend.B. Her struggle with disabilities.C. Rescuing animals with disabilities.D. Her contributions to a better world.5. What can we infer about MacKenzie from Paragraph 2?A. Her growth path was not easy.B. She was deserted by her owner.C. She was operated on at two years old.D. She still needs taking care of by others.6. What does the underlined word “nurtured” in Paragraph 3 mean?A. Trained.B. Comforted.C. Tended.D. Abused.7. Which can be a suitable title for the text?A. MacKenzie—The Most Hard-working DogB. MacKenzie—America's “Most Heroic Dog”C. Chihuahua—Inspiration of Positive EnergyD. Chihuahua—Appeal for Animals' ProtectionCParents and math teachers regularly asked by their school-aged charges whether math matters inreal life now have an answer.In a study published by the National Bureau of Economic Research this week, Harvard Kennedy School Policy Professor Joshua Goodman took a look at what happened to students whose high schools were required in the 1980s to increase the minimum level of coursework required to graduate. What he found is that the students were more likely to increase the number of math courses they took as a result of the change in standards and that translated into higher earnings down the line.Put simply: About 15 years after they graduated, the high school graduates who went to school when these changes took effect saw their average earnings increase about 10% for every extra year of math coursework. The findings may add fuel to the steady drum of education experts, policy makers and others calling for an increased focus on science and math education.The increase in required math courses didn’t necessarily produce rocket scientists, Goodman notes, because the extra coursework wasn’t at a particularly high level. But becoming familiar with and practicing ly basic math skills allowed high school graduates to pursue and excel at jobs that required some level of computational knowledge, he said.Goodman acknowledged that the earnings boost for the students connected with the up tick （上升）in math education may be dependent on the state of the economy. When Goodman checked in the late 1990s and early 2000s on the earnings of the students who graduated in the late 1985，he found that their earnings increased significantly if they took more math.8. What does Joshua Goodman’s study find?A. High school graduates earn more.B. High school students prefer math.C. Math matters a lot in one’s income.D. Math is a compulsory course in school.9. Why is increase in required math courses useful?A. It produced rocket scientists.B. It makes high school students smarter.C. It meets the requirements of the government.D. It enables high school graduates to get better jobs.10. What might be the author’s purpose in writing the text?A. To give a warning to policy makers.B. To seek help for high school students’ problems.C. To show the great importance of learning math.D. To tell a struggling history of high school students.11. What could be the best title for the text?A. Efforts matter mostB. Math makes moneyC. Earnings of different graduatesD. Changes of high school coursesDLas Vegascity inNevadais built in a desert. The city may be known to the outside world for its partying. But officials have found that there are 21 square kilometers of useless grass. The grass is never laid on, played on or even stepped on. The grass is only there to look nice.Now, the city is asking theNevadastate legislature to ban useless grass. It is trying to become the first place inAmericato ban that kind of grass often seen between streets, in housing developments and in office parks.It is estimated that useless grass makes up 40% of all the grass inLas Vegasand it needs a lot of water to survive. Grass needs four times more water than dry climate plants like cactus. By tearing out the grass, the city could reduce yearly water usage by 15%.In 2003, the Southern Nevada Water Authority banned developers from planting grass in front of new homes. It also offered homeowners $ 30 for each square meter of grass they tear out. But fewer people are now using the program. Water usage has increased in southernNevadaby 9% since 2019. And last year,Las Vegaswent a record 240 days without major rainfall. The Colorado River provides much ofNevada's drinking water. The river could losemore water as climate change affects it.Water officials in other dry cities said water usage needs to be reduced. But they fear the reaction to reforms like the ones inLas Vegasif their communities do not accept them. Cynthia Campbell is the water resources adviser for the city ofPhoenixinArizona. “There might come a point when city restrictions get too severe for some residents. They'll say that is the point of no return for them,”Campbellsaid. “For some people, it's a pool. For some people, it's grass.”12. Why doesLas Vegascity try to ban useless grass?A. To protect the local people.B. To beautify the city.C. To reduce water usage.D. To reduce waste.13. What program was carried out inLas Vegasin 2003?A. Allowing planting grass before new houses.B. Encouraging the residents to tear out grass.C. Praising those who signed on the program.D. Awarding those who reduced water usage.14. What is implied inCampbell's words in the last paragraph?A. Many residents won't follow the ban.B. Reaction to the reform will vary personally.C. Other measures should be taken to protect wetter.D. Water officials should take many factors into account.15. What is the best title for the text?s VegasPlans to Ban Useless GrassB. A Method Is Adopted to SaveLas VegasC. Choices between Beauty and PracticeD. Grass Is Important but Useless inLas Vegas第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2021年开封高中高三英语三模试卷及答案

2021年开封高中高三英语三模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AMy Biggest ChallengeAs a writer on an adventure sports magazine, I’dalways fought shy of doing the adventurous stuff myself, preferring instead to observe the experts from a safe distance and relay their experiences to readers in the form of written language. Thus, when I was challenged to take part in a mountain climb in aid of raising money for charity-and to write about it afterwards-I was unwilling, to say the least.I was lucky enough to have a brilliant climbing coach called Keith, who put me through my paces after my daily work. He gave me knowledge about everything from the importance of building muscle groups to how to avoid tiredness through nutrition. It quickly became apparent that the mechanics of climbing were more complex than I had imagined. There was the equipment and techniques I’d never even heard of, all of which would come in handy on the snow-capped peak I’d be climbing.Aware of the challenge, Keith made a detailed action plan and I forced myself to stick to it, doing a daily workout at the gym and going on hikes with a heavy pack. I perfected my technique on the climbing wall and even went to climb the mountains to get vital experience. My self-belief increased alongside my muscle power and I became confident about finishing the climb.All too soon I was on a plane to my destination. On that day, when I looked up at the mountain, I thought of abandoning it. But then I remembered all the hard work I’d done and how disappointed Keith would be if I gave up at the last minute-not to mention letting down the charity and the sense of failure I’d experience myself. With a deep breath I gathered my equipment and headed out into the sunshine to meet the rest of the group.And as I sit here now, tapping away on my laptop, I’m amazed at the details in which I can recall every second of the climb: the burning muscles, the tiredness, the minor problems along the way. Could I have been better prepared? Possibly. Would I be back for another go? Thankfully not. The feeling of being excited when I stood on top of the world is a never-to-be-repeated experience but one I will enjoy forever nevertheless.1. At the beginning of the activity, the author revealed his ______.A. disappointment in the coming adventure.B. expectation of writing about his experience.C. lack of enthusiasm for the challenge he’d been offered.D. curiosity about taking part in the mountain climb for charity.2. What did the author realise during his climb training?A. The knowledge about climbing was really confusing.B. The equipment was the key factor to reach the peak.C. Climbing was much more complicated than expected.D. Hard training was far more important than making plans.3. How did the author feel after he succeeded in climbing the mountain?A. He was relieved that he wouldn’t have to do it again.B. He was well satisfied that he had done his best for it.C. He was surprised that he had managed to complete it.D.He was regretful that it wasn’t as smooth as imagined.BBrian Hamilton's life changed in a prison when he went there with his friend, Reverend Robert J. Harris, who often went to local prisons to do ministry work. During the visit,Hamiltonstarted talking to one of the prisoners and asked what he was going to do when he got out. “He said he was going to get a job,”Hamiltonrecalls. “I thought to myself, wow, that’s going to be difficult with a criminal background.”The conversation madeHamiltonconsider how to help those who came out from prison. Finally in 2008, 16 years after that initial conversation,Hamiltoncreated Inmates to Entrepreneurs, a nonprofit organization that helps people with criminal backgrounds start their own small businesses.At the time,Hamiltonwas building his own company, a software technology company for the banking industry. As his company grew, so didHamilton’s time devoted to giving lessons to prisoners. He averaged three to four courses a month at prisons throughoutNorth Carolina.Eventually,Hamiltondecided to shift his focus to his true passion. In May 2019, he sold his company and focused on helping those who were imprisoned. His online courses will be set next year. “By March 1, 2022, anyone will be able to access the courses, either to become a certificated instructor or to access it for themselves as a prisoner or part of the general population,”Hamiltonexplained. In addition, he visits middle schools and presents the course to at-risk students as a preventative measure against crime.The free course is funded by the recently established Brian Hamilton Foundation, which offers assistance to military members as they return to civilian life and provides loans o small businesses. “Starting up a business isn'tfor everyone, but if we make opportunities available, and let people know that other people care about them, it makes a difference.”Hamiltonsaid.4. Why did Brian Hamilton went to a prison?A. He accompanied his friend.B. He took lessons in the prison.C. He wanted to get a job in the prison.D. He had a friend who was in prison.5. What can be inferred about Inmates to Entrepreneurs?A. It often assists military members.B It provides loans to small businesses.C. Its course has been largely broadened.D. It is an organization intended for business men.6. According to the author, which of the following best describesHamilton?A. He is a man who always changes his mind.B. He has a sense of social responsibility.C. He is good at running a big company.D. He makes money by giving lessons.7. What is the main idea of the text?A. A man made a fruitless visit to the prison.B. A man sold his business to teach prisoners.C. A man realized his dream of being a teacher.D. A man successfully created two organizations.CLike all cultural institutions, galleries and art fairs are adapting to a new reality.Art Basel Hong Kong, Asia's biggest contemporary-art fair, was cancelled because of covid-19, but anyone he who had planned to visit last week could enjoy an experimental alternative：the viewing room. At the click of a keyboard) you could enter an overall view but private visual salon, without having to brave the airless Hong Kong Convention and Exhibition Centre.There, on one webpage, was Jeff Koons riffing(翻唱)on Botticelli's "Primavera" in a tribute to the history of painting at David Zwirner Gallery. Ota Fine Arts offered one collector the chance to acquire an "infinity(无限)room“, one of the most Instagrammed artworks of recent years- the creation of the fantasies,nonagenarian (九十多岁) Japanese artist, Yayoi Kusama. White Cube presented a large collection of international works by Andreas Gursky (German), Theaster Gates (American) and Beatriz Milhazes (Brazilian). But not every artist, gallery and form showed to equal advantage in this alternative fair. Not surprisingly, simple two-dimensional works in brightcolours came across best No sculpture or conceptual art was includedBesides depth and texture, there are aspects of gallery hopping that a website is unlikely to copy. One isserendipity—the sense of wandering between artworks and encountering the unexpected. Another is sociability. Art is a communion between artist and viewer, but galleries and fairs are also places to swap opinions and share enthusiasms.Thereare ways to compensate (or these inevitable(不可避免的)disadvantages. As they shut their physical doors, some of the world's nest galleries and museums are offering advanced interactive visits, 360 degree videos and walk around tours of their collections, allwithout queues and high ticket prices. One of the best is laid on by the Rijksmuseum in Amsterdam；its tour allows visitors to view its Vermeers and Rembrandts, including the magnicent "Night Watch”, far more closely than would normally he possible. Another standout offering is from the Museu de Arte de SPaulo, which has an even broader collection. On its virtual platform, its pain tings, spanning 700 years, appear to be hanging in an open-plan space, seemingly hung on glass panels, or "crystal easels” as the. museum calls them, ideal for close-up inspection,8. From the paragraph 3, we may knowA. "Primavera" was originally painted by JeKoonsB.“innity room” was the most popular artworks of recent years on social mediaC. White Cube displayed a great many works by artists from different nationsD. Various paintings and statues showed on the webpage are well-received9. What does the underlined word "serendipity" in Paragraph 4 probably mean?A. surpriseB. treasureC. expectationD. compromise10. What is the purpose of writing the last paragraph?A.To introduce some popular art show in the world.B. To show that there are some alternative ways to promote the display during covid-19 epidemic.C. To prove that paintings hung on glass panels can be idealfor close-up inspection,D. To release some information about some extraordinary art shows.11. What is the best title?A. Art beyond the internet.B. Fascinating art.C. Art under covid-19.D. gallery hoppingDA crew（全体成员）of six teenage girls completed a nine-day sailing trip in the US recently, after havingseasickness and strong winds.For the past three years, the Sea Cadet teenagers whoset sail were all male. Roger Noakes, who captained（担任队长）the boat, said this was the first time he’d taken out an all-female crew.The girls asked for an all-girls trip in August this year. The crew set sail along with three adults, Noakes and two Sea Cadet representatives. The original plan was for the girls to sail 24 hours a day in rotating shifts（轮流换班）along the coast and then return. Things turned out differently, however. “The first night was difficult because the wind was really hard. The waves were going up and down,” said Abby Fairchild,16. “Everybody got seasick.” Noakes gave the girls the choice of just sailing in the bay and not going into open water. “But they decided they were going.”The teenagers then sailed a long way overnight and slept in shifts. “We’ve learned everything from controlling the boat to putting up the sails while we have rough seas,” said 15-year-old Olivia Wilcox.The teenagers stopped on land in Massachusetts. They didn’t make it to their original destination（目的地）in Maine, where they were supposed to have a celebratory dinner, due to the weather and winds. They said they weren’t disappointed, however, as they’d learned a lot. “They learned about boating, and above all, they built confidence and character,” said Noakes.12. What was special about the Sea Cadet trip this year?A. It was the longest sailing trip ever.B. It was the first all-female-crew sailing trip.C. It was the most dangerous sailing trip ever.D. It was the first sailing trip for teenagers.13. What happened on the crew’s first day of the trip?A. They all felt sick on the boat.B. Some of them were hurt.C. Their boat was out of control.D. They went into open water by mistake.14. Which of the following best describes these young sailors?A. Strong-minded and having a strong sense of teamwork.B. Hard-working and having great leadership skills.C. Understanding and creative.D. Adventurous and skillful.15. According to Noakes, what was the sailors’ greatest benefit from the trip?A. They knew the sea better.B. They made many friends.C. They got excellent sailing skills.D. They developed good personalities.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

河南省开封市2021届新高考第三次质量检测物理试题含解析

河南省开封市2021届新高考第三次质量检测物理试题一、单项选择题：本题共6小题，每小题5分，共30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．下列说法正确的是（）A ．某金属能发生光电效应，当入射光的颜色不变而增大光照强度时，逸出的光电子的最大初动能也增大B ．若利用黄光和蓝光分别在同一装置研究光电效应，用蓝光照射时的遏止电压大于用黄光照射时的遏止电压C ．换用频率小的光照射，但入射光的强度不变，则逸出的光电子的最大初动能不变D ．换用频率小的光照射，但入射光的强度不变，则从光照射到金属表面上到发射出电子的时间明显减少【答案】B【解析】【详解】A ．金属发生光电效应时逸出的光电子最大初动能满足0k E h W ν=-，由此可知光电子的最大初动能与入射光的频率和金属的逸出功有关，与入射光的强度无关，故A 错误；B ．根据光电效应方程0k E h W ν=-和能量守恒k c E eU =，联立解得0c eU h W ν=-由于黄光的频率小于蓝光，故黄光的遏止电压小于蓝光的遏止电压，故B 正确；C ．减小入射光的频率，逸出的光电子的最大初动能将减小，故C 错误；D ．光电效应具有瞬时性，所以无论入射光的频率和强度如何变化，从光照射到金属表面上到发射出电子的时间都是瞬时的，故D 错误。

故选B 。

2．下列说法正确的是（）A ．爱因斯坦在1900年首次把能量子的概念引入物理学B ．单色光照射金属表面发生光电效应时，入射光越强，单位时间内发射的光电子数越多C ．一个氢原子从量子数n=3的激发态跃迁到基态时最多可辐射3种不同频率的光子D ．玻尔的原子理论能够解释氦原子的光谱【答案】B【解析】【分析】【详解】A ．普朗克在1900年把能量子引入物理学，破除了“能量连续变化”的传统观念，故A 错误；B ．单色光照射金属表面发生光电效应时，入射光越强，则入射光子的数目越多，所以单位时间内发射的光电子数越多，故B 正确；C ．根据玻尔理论，一个氢原子从量子数n=3的激发态跃迁到基态时最多可辐射2 种不同频率的光子，故C 错误；D ．玻尔第一次将量子观念引入原子领域，提出了定态和跃迁的概念，很好地解释了氢光谱，但不能够解释氦原子的光谱，故D 错误。

河南省开封市2021届新高考三诊物理试题含解析

河南省开封市2021届新高考三诊物理试题一、单项选择题：本题共6小题，每小题5分，共30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．在平直公路上有甲、乙两辆汽车从同一位置沿着同一方向运动，它们的速度－时间图象如图所示，则()A．甲、乙两车同时从静止开始出发B．在t＝2s时乙车追上甲车C．在t＝4s时乙车追上甲车D．甲、乙两车在公路上可能相遇两次【答案】C【解析】由图像可知，乙车比甲车迟出发1s，故A错误．根据速度时间图线与时间轴围成的面积表示位移，知t=2s 时，甲车的位移比乙的位移大，则知该时刻乙车还没有追上甲车，故B错误．在0-4s内，甲车的位移x甲=12×8×4m=16m，乙车的位移x乙=12×（1+3）×8m=16m，所以x甲=x乙，两者又是从同一位置沿着同一方向运动的，则在t=4s时乙车追上甲车，故C正确．在t=4s时乙车追上甲车，由于t=4s时刻以后，甲车的比乙车的速度大，两车不可能再相遇，所以两车只相遇一次，故D错误．故选C．点睛：解决本题的关键是要理解速度时间图线表示的物理意义，知道图线与时间轴围成的面积表示位移，相遇时两车的位移相等．2．用两个相同且不计重力的细绳，悬挂同一广告招牌，如图所示的四种挂法中，细绳受力最小的是（）A．B．C．D．【答案】B【解析】【分析】【详解】由题意可知，两绳子的拉力的合力与广告招牌的重力大小相等，方向相反，是相等的，根据力的平行四边形定则，可知当两绳子的夹角越大时，其拉力也越大。

因此对比四个图可知，B图的拉力方向相同，拉力的大小最小，故B正确，ACD均错误。

故选B。

3．关于近代物理学，下列说法正确的是A ．光电效应现象揭示了光具有波动性B ．—群氢原子从n ＝4的激发态跃迁时，最多能辐射6种不同频率的光子C ．卢瑟福通过α粒子散射实验证实原子核由质子和中子组成D ．氡的半衰期为3.8天，若取4个氡原子核，经过7.6天后一定剩下1个氡原子核【答案】B【解析】光电效应现象揭示了光具有粒子性，故A 错误；—群氢原子从n ＝4的激发态跃迁时，最多能辐射246C =，即6种不同频率的光子，故B 正确；卢瑟福通过α粒子散射实验证实原子的核式结构模型，故C 错误；半衰期只适用大量原子核，对极个别原子核没有不适用，故D 错误．所以B 正确，ACD 错误．4．如图所示，一理想变压器的原副线圈匝数分别为n 1和n 2，原线圈输入电压保持不变，副线圈输出端通过开关S 接电阻R 1和滑动变阻器R ，电流表为理想电表，下列说法正确的是（）A ．若S 断开，则副线圈输出端电压为零B ．若S 闭合，滑动片P 向上滑动，则两电流表示数都变小C ．若S 闭合，滑动片P 向上滑动，则电流表A 1示数变大，A 2示数变小D ．若S 闭合，滑动片P 向上滑动，则电流表A 1示数不变，A 2示数变小【答案】B【解析】【分析】【详解】A ．理想变压器的变压关系与副线圈是否是通路或断路无关，根据理想变压器的电压规律变形2211n U U n = 匝数比恒定，输入电压不变时，输出电压就保持不变，A 错误；BCD ．S 闭合，滑片P 上滑时，电阻值变大，则电流表A 2示数变小，又因为流过定值电阻R 1的电流不变，所以电流表A 1示数也变小，B 正确，CD 错误。

2021届开封高级中学高三英语三模试卷及答案

2021届开封高级中学高三英语三模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ASongkran, ThailandThis festival marks the Thai New Year, which is celebrated in the second week of April. This is the hottest time of the year in Thailand, so it makes sense that the biggest Songkran tradition is to throw water on people. Thais of all ages join in water fights, using buckets, hoses (软管)and water guns to celebrate the event.Boryeong Mud Festival, South KoreaDon't expect to come home clean after experiencing the Boryeong Mud Festival in South Korea. Famous for its mineral-rich mudflats, visitors rush to Boryeong in July to make themselves in mud, swim in grey pools and enjoy the party. There are even mudslides and a mud skiing competition for those who are seeking the extreme mud adventure!La Tomatina, SpainThe festival dates back to a parade in which some naughty teenagers knocked one performer off his stilts (高跷)and caused a fight of throwing vegetables accidentally. It was once banned until 1957 when the locals held a protest with a funeral. They carried a coffin containing a huge tomato as bands played a funeral march. In the following decades, La Tomatina has become a popular event. If you join the event, be aware that you squash (压扁) the tomato before throwing it. Have a great time but avoid causing any injury.Dia de los Muertos, MexicoBeginning at the midnight of October 31 and lasting through November 2,it is a festival when families gather together to remember those loved ones who have died, aiming to help them on their spiritual journey. On these days, Mexican families prepare special tables in their homes. On top of them they'll put photos of the dead and their favorite food. They also visit the graves of their beloved ones to show their respect to the dead.1.If you want to experience a special new year in a country, you can choose to visit ________.A.ThailandB.South KoreaC.SpainD.Mexico2.What may happen to you as a visitor in La Tomatina?A.You can perform stilts.B.Your can play funeral music in bands.C.You may be covered with mud.D.You may be attacked with tomatoes.3.Which festival is similar to the Tomb Sweeping Festival in China?A.Songkran.B.Boryeong Mud Festival. Tomatina.D.Dia de los Muertos.BHave you ever noticed that some people can eat what they want and stay in shape, while others carefully watch what they eat and still put on weight? How annoying! However, scientists are beginning to believe that some people get the ability to stay thin from their parents.Scientists say that when some people eat, their bodies will naturally store the excess energy from food as fat. These people gain weight if they eat more than they need each day. Meanwhile, other people eat more than they need but their bodies are able to “burn off” the extra food without making fat, so they rarely have weight problems.To show that this is true, scientists have experimented on laboratory mice. They gave the mice a special diet with a lot of fat. Some mice gained weight while the other mice stayed thin, even though both groups of mice ate the same amount of food and got the same amount of exercise. Scientists concluded that weight gain seemed to be influenced by genetic (基因的) factors. They also believe they have now identified the genes that may cause this tendency. These genes, according to the scientists, get passed down the generations. Armed with this knowledge, they believe it may be possible to develop medicines thatcan target the genes and stop people from gaining weight.The study of weight gain is becoming more important as the numbers of people who are overweight continues to grow. In theUnited States, over seventy percent of the population have weight problems. This means they are at risk of illness such as high blood pressure and heart disease.Current medicines for weight control do not work very well because they can have serious side effects, such as heart problems. However, it takes a long time to develop and thoroughly test new drugs. Therefore, despite this breakthrough in the study of weight-loss drugs, scientists and health professionals all agree that currently the best way to control weight gain is to have a healthy, balanced diet and an active lifestyle with daily exercise. This will not only help people avoid becoming overweight, but also help them stay healthy and energetic.4. According to the scientists, some people ________.A. are able to stay thin because of genetic factorsB. can eat what they want and stay healthy foreverC. usually keep doing exercise to lose their weightD. don’t eat anything every day but still put on weight5. Some people don’t need to worry about weight problems because ________.A. they eat many vegetablesB. they eat less than othersC. they don’t store any energy from food as fatD. they can burn off the extra food without making fat6. What can we learn from the passage?A. Fewer people will have weight problems in the future.B. People with weight problems may suffer from heart trouble.C. We now have good medicine for people with weight problems.D. The result of the experiments on mice and humans are different.7. What does the last paragraph imply?A. Current medicine solve weight problems well.B. We should exercise regularly and eat properly.C. Taking exercise every day may cause heart problems.D. Testing new drugs is hard because of the side effects.CMy school appeared on the news last week because we had made an important change in our local area. Our class had planted a large garden in what was once only a vacant lot. It was a lot of work but it was all worth it. I got blisters(水泡) from digging, and we all got insect bites, too.I learned a lot about gardening and collaboration(合作), and then I learned about the media. Our teacher telephoned the TV station and informed them of what we had accomplished. She spoke with the producer. The producer checked with the directors, but they said there were plenty of stories similar to ours. They wanted to know what was special about our particular garden, since many schools plant them.The teacher explained that, after going on the Internet to learn about the prairie(大草原), we had made a prairie garden. We had gone to a prairie and gotten seeds from the plants, and then we planted them. We did not water the garden, but we did weed it. We decided to let nature water it with rain, since that was how prairies grew in the past. We sent a picture of the garden to the news station. In the picture, the grass was so high that it stood taller than the fourth grade students.As a result, the producer sent a reporter to our school. He interviewed the headmaster and asked him manyquestions about the garden. After that, they interviewed us, and we explained to them what we had learned through this project.That night, we watched the news, and there we were. The news reporter told our story. It was only two minutes long, but it was us. We were famous. All that work, all those blisters, it was worth it. We knewthatwhen we saw the garden every day, but now we knew that the whole city thought so, too.8. What seemed to be the TV directors’ initial reaction to the garden?A. They were excited.B. They were surprised.C. They were worried.D. They were uninterested.9. What is special about the garden?A. Weeds were allowed to spread naturally.B. The grass grew faster than common grass.C. The seeds came from the plants of a prairie.D. Underground water was used for the plants.10. What does the underlined word “that” refer to in the last paragraph?A. We got blisters on our hands.B. Our hard work was worthwhile.C. The garden would be famous.D. The project would be finished.11. How did the author feel about the project?A. Annoyed.B. Curious.C. Proud.D. Regretful.DPigeons inLondonhave a bad reputation. Some people call them flying rats. And many blame them for causing pollution with their droppings. But now the birds are being used to fight another kind of pollution in this city of 8.5 million.“The problem for air pollution is that it’s been largely ignored as an issue for a long time,” says Andrea Lee, who works for the London-based environmental organization Client Earth. “People don’t realize how bad it is, and how it actually affects their health.”London’s poor air quality is linked to nearly 10,000 early deaths a year. Lee says, citing(引用)a report released by the city manager last year. If people were better informed about the pollution they’ re breathing, she says, they could pressure the government to do something about it.Nearby, on a windy hill inLondon’s Regent’s Park, an experiment is underway that could help—the first week of flights by the Pigeon Air Patrol. It all began when Pierre Duquesnoy, the director for DigitasLBi, a marketing firm,won a London Design Festival contest last year to show how a world problem could be solved using Twitter. Duquesnoy, fromFrance, chose the problem of air pollution.“Basically, I realized how important the problem was,” he says. “But also I realized that most of the people around me didn’t know anything about it.” Duquesnoy says he wants to better measure pollution, while at the same time making the results accessible to the public through Twitter.“So”, he wondered, “how could we go across the city quickly collecting as much data as possible?” Drones were his first thought. But it’s illegal to fly them overLondon. “But pigeons can fly aboveLondon, right?” he says. “They live—actually, they are Londoners as well. So, yeah, I thought about using pigeons equipped with mobile apps. And we can use not just street pigeons, but racing pigeons, because they fly pretty quickly and pretty low.”So it might be time for Londoners to have more respect for their pigeons. The birds may just be helping to improve the quality of the city’s air.12. What can we infer aboutLondon’s air quality from Paragraph 2?A. Londoners are very satisfied with it.B. The government is trying to improve it.C Londoners should pay more attention to it.D. The government has done a lot to improve it.13. Duquesnoy attended the London Design Festival to _________.A. entertain Londoners.B. solve a world problem.C. design a product for sale.D. protect animals like pigeons.14. Why did Duquesnoy give up using drones to fly acrossLondon?A. Because they are too expensive.B. Because they fly too quickly.C. Because they are forbidden.D. Because they fly too high.15. Which can be the best title for the text?A. Clean air inLondon.B. London’s dirty secret.C. London’s new pollution fighter.D. Causes of air pollution inLondon.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2021届开封高中高三语文第三次联考试题及答案

2021届开封高中高三语文第三次联考试题及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

材料一：在经历了一周的地月转移、近月制动、环月飞行后，嫦娥五号探测器于12月1日23时11分降落在月面预选着陆区。

嫦娥五号的落月过程经历了主动减速、快速调整、接近、悬停避障、缓速下降和自由下落段，看似轻松的平稳降落背后，蕴藏着众多科研人员的智慧和积淀。

嫦娥五号落月的过程也是为后续上升器月面起飞选择“发射场”的过程。

相较于嫦娥三号、四号，嫦娥五号对着陆点的位置精度和平整度的要求是空前的，需要着陆区域内既没有太高的凸起，也没有太深的凹坑，坡度要符合任务要求。

可以说，落月的过程就是边飞行边找寻落点。

为了实现“选址正确，落得准确”，嫦娥五号采用了中国航天科技集团五院502所已在嫦娥三号和四号上应用的“粗精接力避障”的方式，即在制导导航与控制系统的指挥下，着陆器和上升器组合体先大推力反向制动快速减速，然后快速调整姿态并对预定落区地形进行拍照识别，避开大障碍，实现“粗避障”。

之后，组合体在飞到距离月面100米时悬停，并再次对选定区域精确拍照，实现“精避障”，再斜向下飘向选定的着陆点，在移动到着陆点正上方后开始竖直下降，到距离月面较近时关闭发动机，利用着陆腿的缓冲实现软着陆。

在嫦娥五号的落月过程中，还有两个精妙的设计。

一是虽然表面上看是着陆器“背着”上升器软着陆到了月面上，但实际上，整个落月过程中，着陆器都借助了“外脑”和“外眼”分别是上升器月面起飞时要用的“最强大脑”中央控制计算机和通过“看星星”确定自己姿态的星敏感器，这是设计人员根据“上升器全程陪同着陆器”的实际想出的妙招，既节约成本又减轻重量，二是由于距离月面较近时主发动机激起的月尘会污染星敏感器，影响上升器月面起飞，科研人员专门设计了盖子，在距离月面一定的高度时把星敏感器的镜头盖起来，待落月之后月尘散去再通知星敏感器把盖子打开。

河南省开封市2021届高三三模文科数学答案

河南省开封市2021届高三三模文科数学试题参考答案1．D 【思路点拨】根据集合的包含关系，结合集合的性质求参数a 、b ，即可求-a b .【解析】由A B ⊆知：A B =，即11a b =-⎧⎨-=⎩，得11a b =-⎧⎨=-⎩，∴0a b -=. 故选：D.2．B 【思路点拨】利用复数的差的模的几何意义，求出复数对应点所在象限，再排除不合题意的选项.【解析】设(,)z x yi x y R =+∈，复数z 对应复平面内点(,)P x y ， z 的实部大于虚部，即x y >， ∴排除选项C 、D.1z =且1z i -=，则P 在以原点为圆心的单位圆上运动，且P 在以(0,1)为圆心的单位圆上运动. 如图.法一：点P 在两圆交点A 或B 处，即第一或第二象限，排除选项A. 法二：当点P 在A 处时，0,0x y <>，不合题意，即点P 在第一象限，故选：B.【名师指导】两个复数差的模的几何意义是：两个复数在复平面上对应的点的距离. 3．A 【思路点拨】由题设命题可得(1)(2)0m m -+>，即可求m 的范围，由推出关系知：所求得m 的范围包含于必要不充分条件的m 范围.【解析】由方程22112x y m m -=-+表示双曲线，知：(1)(2)0m m -+>，∴()(),21,m ∈-∞-+∞，故它的一个必要不充分条件为()(),11,m ∈-∞-+∞.故选：A.4．C 【思路点拨】根据雷达图，判断甲各科成绩与年级平均分的高低，以及各科成绩的高低，进而可确定理想的选科组合，即可判断各选项的正误.【解析】A ：由图知：甲的物理成绩领先年级平均分1.5分左右，比化学、地理要高，正确； B ：其中有政治、历史比年级平均分低，正确；C ：甲的成绩从高到低的前3个科目依次是地理、化学、物理或生物，错误；D ：由C 知：物理、化学、地理对于甲是比较理想的一种选科结果，正确；故选：C.5．B 【思路点拨】根据sin 2sin 2αα=，利用二倍角的正弦公式求得cos 4α=，再利用二倍角的余弦公式求解.【解析】因为sin 2sin 2αα=，所以sin 22sin cos sin sin ααααα==解得cos 4α=，所以223cos 22cos 1214αα=-=⨯-=-⎝⎭，故选：B6．B 【思路点拨】由函数的导数()0f x '=求极值点，将极值点代入()f x 可得方程，进而求得a 值.【解析】由已知得：2(1)()()x e x a f x x a +-'=+()x a ≠-，令()0f x '=，有1x a =-，且1x a <-上递减，1x a >-上递增，∴()f x 的极小值为1(1)a f a e --==，即112a -=，得12a =. 故选：B.7．C 【思路点拨】由已知递推关系得122n n a -=，根据123n a a a a =有(1)1542n n -=，即可求n 值.【解析】由题意知：122n n a -=，∴(1)41232n n n a a a a -==∴(1)1542n n -=且*n N ∈，可得6n =. 故选：C.8．C 【思路点拨】由函数零点代入解析式待定系数ϕ、ω.【解析】由图象可知，由(0)0f =得cos 0ϕ=，又0ϕπ<<，解得2ϕπ=. 则()4cos 4sin 2x xx x ee f x πωω⎛⎫+ ⎪⎝⎭==-，法一：由(1)0f =得sin 0ω=，解得()k k Z ωπ=∈，又当(0,1)x ∈，(0,)x ωω∈时，恒有()0f x <，即sin 0x ω>恒成立，故0ωπ<≤，1k ∴=，即ωπ=，则2ωϕ=. 法二：由sin 0x ω=，解得()k x k Z πω=∈，故两相邻零点的距离为πω，由图象可知1πω=，则ωπ=，则2ωϕ=. 故选：C.【名师指导】已知函数图象待定解析式，一是从函数的特征点入手，代入点的坐标从而待定系数，如函数的零点、极值点、与纵轴的交点、已知横纵坐标的点等等；二是从函数的特征量入手，找到等量(不等量)关系待定系数(范围)，如函数的周期、对称轴、切线斜率、图象上两点间的距离、相关直线所成角等等.9．D 【思路点拨】根据三视图还原直观图后可求解.【解析】解：对①：如图，则31151-111=326V =⨯⨯⨯⨯，故①正确；对②：如图，则3112=1-2111=323V ⨯⨯⨯⨯⨯，故②正确；对③：如图AB 和CD 在直观图中所对应的棱分别为EF 和GF ，由EFG 为正三角形，所以AB 和CD 在直观图中所对应的棱所成的角为3π，故③正确；对④：如图，平面//ABCD 平面111B C D ，平面1//ADD 平面11BCC B ，平面1//ABB 平面11DCC D ，平面11//AB D 平面1BC D ，故④正确. 故选：D.【名师指导】方法点睛：由三视图画出直观图的步骤和思考方法 1、首先看俯视图，根据俯视图画出几何体地面的直观图； 2、观察正视图和侧视图找到几何体前、后、左、右的高度； 3、画出整体，然后再根据三视图进行调整.10．A 【思路点拨】由题设分析知第二次和第三次的球必在乙、丙两人，即传球顺序为甲→乙→丙→甲或者甲→丙→乙→甲，并求出所有可能传球可能情况，根据古典概型求概率即可.【解析】由题设，知：要使第四次由甲传球，则第三次球不在甲，又由甲开始传球，所以第二次球也不在甲，∴第二次和第三次的球必在乙、丙两人，故传球顺序：甲→乙→丙→甲或者甲→丙→乙→甲；而所有可能的传球顺序有32种，∴第四次仍由甲传球的概率为32124=. 故选：A.11．D 【思路点拨】设20a m =>，把指数式改为对数式，利用对数的运算求解．【解析】设25a b c z m ===，则0m >且1m ≠，25log ,log ,log z a m b m c m ===，2510111111log 2log 5log 10log log log log m m m z a b m m m m+=+=+===， 10log log z m m =，所以10m =．故选：D ．12．C 【思路点拨】在12PF F △中，由正弦定理可得211221sin sin PF PF PF F PF F =，结合已知条件得到12a PF c PF =，设点00(,)P x y ，得到1020,PF a ex PF a ex =+=-，整理得到0()(1)()(1)a c a a e x e c a e e --==+-，根据椭圆的几何性质可得0x a >-，化简得到2210e e +->，即可求解.【解析】在12PF F △中，由正弦定理可得211221sin sin PF PF PF F PF F =,又由2112sin sin PF F c PF F a∠=∠，即1221sin sin a cPF F PF F =∠∠，即12a PF c PF =, 设点00(,)P x y ，可得1020,PF a ex PF a ex =+=-，则00()()a a ex c a ex +=-，解得0()(1)()(1)a c a a e x e c a e e --==+-，由椭圆的几何性质可得0x a >-，即(1)(1)a e a e e ->--，整理得2210e e +->，解得1e <或1e >，又由(0,1)e ∈，所以椭圆的离心率的取值范围是1,1).故选：C.【名师指导】方法点拨：在12PF F △中，由正弦定理和结合已知条件得到12a PF c PF =，设点00(,)P x y ，结合椭圆的焦半径公式，得到0()(1)()(1)a c a a e x e c a e e --==+-，根据椭圆的几何性质可得0x a >-，列出关于离心率e 的不等式是解答的关键. 13．0【思路点拨】直接根据等差数列的通项公式即可得结果. 【解析】设等差数列{}n a 的公差为d ，因为5732a a =，所以()()113426a d a d +=+，解得10a=，故答案为：0.14．2【思路点拨】由向量数量积的几何意义有||cos ,3a a b <>=可得//a b ，根据向量平行有a b λ=即可求t 值.【解析】由题意知：||cos ,3a a b <>=，而()1,2a =，∴cos ,1a b <>=，即a,b 0<>=，故//a b ，∴a b λ=，易得1t λ=⎧⎪⎨=⎪⎩，即122t λ⎧=⎪⎨⎪=⎩. 故答案为：2.15由正弦定理求a ，由余弦定理求bc ，再应用三角形面积公式求ABC 的面积.【解析】由题意，结合正弦定理知：2sin aA=，即a = 又由余弦定理知：22222()21cos 222b c a b c a bc A bc bc +-+--===，而b c +=，∴综上，有：3322bc =，即1bc =，∴1sin 2ABC S bc A ==△.故答案为：3. 16．92； 86π【思路点拨】画出几何体的图形，取BC 的中点D ，连结SD ，AD ，作SO ⊥平面ABC ，垂足O 在AD 上，然后求解该六面体的体积．当该六面体内有一球，且该球的体积取最大值时，球心为O ，且该球与SD 相切，求出球的半径即可求解该球体积的最大值．【解析】解：该六面体是由两个全等的正四面体组合而成，正四面体的棱长为3，如图，在棱长为3的正四面体S ABC -中，取BC 的中点D ，连结SD ，AD ，作SO ⊥平面ABC ，垂足O 在AD 上，则22331363AD SD OD AD SO SD OD ====-= 所以该六面体的体积为113392223632S ABC V V -==⨯⨯⨯．当该六面体内有一球，且该球的体积取最大值时，球心为O ，且该球与SD 相切，过球心O 作OE SD ⊥，则OE 就是球的半径，因为SO OD SD OE ⨯=⨯，所以球的半径366332SO ODOE SD⨯=，该球体积的最大值为346863V π=⨯=⎝⎭．．【名师指导】关键点点睛：（1）解题关键是求出正四面体的高SO ；（2）问解题的关键是利用截面将空间问题平面化，从而找到并求出球的半径OE . 17．【思路点拨】（1）在△ABD 中，由余弦定理2222cos AD AB BD AB BD B =+-⋅⋅，即可求AD .（2）在ABC 中，由正弦定理sin sin AB ACC B=，即可求sin C .【解析】（1）在△ABD 中，AB =4B π=，3BD =，由余弦定理得：2222co 25s 9AD AB BD AB BD B =+-⋅=⋅+-=，∴AD =（2）在ABC 中，AB =，AC =4B π=，由正弦定理得：sin sin AB AC C B=，即sin sin 4C π=，∴sin 4C =． 18．【思路点拨】（1）由已知有AB AP ⊥，CD PD ⊥，根据线面垂直的判定可得CD ⊥平面PAD ，再由面面垂直的判定可证平面PAD ⊥平面ABCD ．（2）设E 为AD 中点，连接PE ，EB ，由已知有PE ⊥平面ABCD ，由（1）易证ABCD 是矩形，由已知线段的长度，结合勾股定理求相关线段长并确定P BCD V -、PBDS ，根据等体积法有C PBD P BCD V V --=，即可求C 到平面BDP 的距离．【解析】（1）证明：由90BAP CDP ∠=∠=︒，得：AB AP ⊥，CD PD ⊥，由//AB CD ，即CD AP ⊥，又APPD P =，∴CD ⊥平面PAD ，又CD ⊂平面ABCD ， ∴平面PAD ⊥平面ABCD ．（2）设E 为AD 中点，连接PE ，由PA PD =， ∴PE AD ⊥，由（1）知：平面PAD ⊥平面ABCD ，面PAD 面ABCD AD =，PE ⊂面PAD ，∴PE ⊥平面ABCD ， ∵//AB CD ，AB CD =， ∴ABCD 是平行四边形，由（1）知：CD ⊥平面PAD ，AD ⊂平面PAD ， ∴CD AD ⊥，即ABCD 是矩形，由90APD ∠=︒，22AD BC ==，2PE =，∴由上知：11114222232323P BCD V CB CD PE -⎛⎫⎛⎫=⨯⨯⨯⨯=⨯⨯⨯⨯= ⎪ ⎪⎝⎭⎝⎭，连接EB ，在△PEB 中，6EB =，2PE =，可得22PB =，在△PBD 中，2PD =，23DB =，22PB =，所以90BPD ∠=︒， ∴112222222PBDSPB PD =⨯⨯=⨯⨯=，设点C 到平面PBD 的距离为h ，由C PBD P BCD V V --=，有142233h ⨯=． ∴2h =，即点C 到平面PBD 的距离为2．【名师指导】关键点点睛：（1）应用线面、面面垂直的判定证明面面垂直；（2）应用等体积法求点面距.19．【思路点拨】（1）由频率直方图得到(]0,10内的频率，由频率即为对应区间的概率即可求区间(]0,10内的概率；（2）由（1），结合已知可得样本中听力为优秀的学生人数，由样本中各组人数的比例关系即可估计总体中听力为优秀的学生人数.（3）由题设，列出所有2Y ≤情况下1a ，2a ，3a ，4a 的组合数量，并写出所有情况的组合数量，应用古典概型求概率即可.【解析】（1）根据频率分布直方图知，样本中测试值在区间(]0,10内的频率为()10.060.080.02510.80.2-++⨯=-=，以频率为概率，从总体的500名学生中随机抽取1人，估计其测试值在区间(]0,10内的概率为0.2．（2）由（1）知：样本中听力为优秀的学生人数为0.25046⨯-=， ∴估计总体中听力为优秀的学生人数为65006050⨯=．（3）当11a =时，序号1a ，2a ，3a ，4a 的情况为6种：分别记为()1,2,3,4，()1,2,4,3，()1,3,2,4，()1,3,4,2，()1,4,2,3，()1,4,3,2，同理，当12,3,4a =时，序号1a ，2a ，3a ，4a 的情况也分别为6种， ∴序号1a ，2a ，3a ，4a 所有的情况总数为24种．当Y 0=时，11a =，22a =，33a =，44a =，当123412342Y a a a a =-+-+-+-=时，1a ，2a ，3a ，4a 的取值为11a =，22a =，34a =，43a =，或11a =，23a =，32a =，44a =，或12a =，21a =，33a =，44a =， ∴2Y ≤时，序号1a ，2a ，3a ，4a 对应的情况为4种，即()412246P Y ≤==．【名师指导】关键点点睛：（1）应用频率确定指定样本区间中的人员被抽到的概率.（2）根据样本中指定区间人数的所占比例，估计总体中对应区间的人数. （3）应用列举法求古典概型的概率.20．【思路点拨】（1）由向量的坐标表示，列方程求抛物线参数p ，写出抛物线方程.（2）设直线:2l y x m =+，()11,A x y ，()22,B x y ，联立抛物线方程，应用韦达定理求12x x +，12x x ，根据等差中项的性质，结合抛物线的定义求参数m ，进而由21||FB FA x x -=-=.【解析】（1）由题设知：,02p F ⎛⎫⎪⎝⎭，设点()00,P x y ，由()0,2FP =-，即()00,0,22p x y ⎛⎫-=- ⎪⎝⎭， ∴02p x =，02y =-，代入22y px =，得24p =，又0p >， ∴2p =，则抛物线C 的方程为24y x =．（2）设直线:2l y x m =+，则224y x m y x=+⎧⎨=⎩，消去y 得：()224440x m x m +-+=，满足()22441632160m m m ∆=--=-+>，即12m <，设点()11,A x y ，()22,B x y ，则121x x m +=-，2124m x x =，若FA ，FP ，FB 成等差数列，则2FA FB FP +=，即1224x x ++=，即34m -=，即1m =-．∴此时直线l 与抛物线C 联立方程为24810x x -+=，即122x x +=，1214x x =，又∵公差d 满足212d FB FA x x =-=-，而21x x -==，∴2d =2d =±．【名师指导】关键点点睛：（1）由向量的坐标表示求抛物线参数，写出抛物线方程.（2）联立直线与抛物线方程，应用韦达定理、等差中项的性质及抛物线的定义求数列公差即可.21．【思路点拨】（1）由题设得()()10mxf x x x-'=>，讨论0m ≤、0m >时()'f x 的符号，进而判断()f x 的单调性，即可确定满足()f x 有两个零点的情况，结合零点存在性定理求m 的范围.（2）由题设若120x x >>易得1212ln ln x x m x x -=-，由()12121212ln ln 1x x f x x x x x x -'+=-+-应用分析法知：要证结论只需证11222ln101x x x x +->+，令121x t x =>，设()2ln 11t t t ϕ=+-+则应用导数证明()0t >φ即可.【解析】（1）已知函数()ln f x x mx =-有两个零点，()()110mxf x m x x x-'=-=>， ①当0m ≤时，0fx，则()f x 在0,上单调递增，至多有一个零点；②当0m >时，10x m<<时，0f x，则()f x 在10,m ⎛⎫⎪⎝⎭上单调递增；1x m>时，0f x，则()f x 在1,m ⎛⎫+∞ ⎪⎝⎭上单调递减，∴()f x 在1x m =处取得最大值，即有1ln 10f m m ⎛⎫=--> ⎪⎝⎭得：10m e <<，此时，有2111e m m>>>，而()10f m =-<，2112ln 0f m m m ⎛⎫=--< ⎪⎝⎭，∴由零点存在性定理可知，()f x 在11,m ⎛⎫ ⎪⎝⎭和211,m m ⎛⎫⎪⎝⎭上各有一个零点．综上所述，m 的取值范围是10,e ⎛⎫ ⎪⎝⎭．（2）∵1x ，2x 是()f x 的两个零点，不妨设120x x >>， ∴11ln 0x mx -=①，22ln 0x mx -=②， ①-②得：1212ln ln x x mx mx -=-，即有1212ln ln x x m x x -=-，由()1f x m x'=-，有()1212121212ln ln 11x x f x x m x x x x x x -'+=-=-++-， ∴要证()120f x x '+<，即证121212ln ln 1x x x x x x ->-+，即证121212ln ln x x x x x x -->+，即证1121221ln 01-->+x x x x x x ，即证11222ln 101x x x x +->+，令121x t x =>，设()2ln 11t t t ϕ=+-+，()()22101t t t t ϕ+'=>+， ∴()t ϕ在1t >时单调递增，则()()10t ϕϕ>=，即()120f x x '+<得证．【名师指导】关键点点睛：（1）分类讨论思想，利用导数研究函数单调性，结合零点存在性定理及已知零点的个数求参数范围；（2）利用分析法，将所要证的不等式转化为证明函数不等式恒成立即可. 22．【思路点拨】（1）利用极坐标与直角坐标互化公式即可求解；（2）将直线l 的参数方程与曲线C 的直角坐标方程联立，利用一元二次方程的根与系数的关系以及直线参数方程中t 的几何意义即可求解.【解析】解：（1）曲线C 的极坐标方程可化为2222sin 3ρρθ+=，即2233x y +=，化简得曲线C 的直角坐标方程为2213x y +=.（2）联立直线l 的参数方程与曲线C 的直角坐标方程得：()()22cos 32sin 3t t αα++=，化简得()2212sin 12sin 90t t αα+++=，则12212sin 12sin t t αα+=-+，1229012sin t t α=>+，且()22144sin 3612sin 0αα∆=-+>，22sin 10α->，则有sin 2α⎛⎤∈ ⎥ ⎝⎦，则12212sin 12sin PA PB t t αα+=+=+，令sin 2m α⎛⎤=∈ ⎥ ⎝⎦，有1212PA PB m m⎡+=∈⎣+，所以PA PB +的取值范围为⎡⎣.【名师指导】关键点点睛：由直线l 的参数方程与P 点坐标知，PA PB +可利用t 的几何意义求解.23．【思路点拨】（1）根据绝对值三角不等式进行求解即可；（2）根据余弦函数的值域性质，结合辅助角公式、正弦型函数的性质进行求解即可. 【解析】解：（1）由已知可得()()()11111222y f x g x x x x x ⎛⎫=+=-+----= ⎪⎝⎭≥，当且仅当()1102x x ⎛⎫--- ⎪⎝⎭≤即112x ≤≤时等号成立，所以函数()()y f x g x =+的最小值为12. （2）因为cos 1θ≤，所以由已知15sin cos 122θθ-+->，原不等式可化为13sin cos 22θθ-->，当1sin 2θ≥时，即5,66ππθ⎡⎤∈⎢⎥⎣⎦时，原不等式化为sin cos 2)24πθθθ->⇒->，)4πθ-≤，所以此时无解，当1sin 2θ<时，50,,266ππθπ⎡⎫⎛⎫∈⎪⎪⎢⎣⎭⎝⎭，原不等式化为sin cos 1θθ+<-，即sin 42πθ⎛⎫+< ⎪⎝⎭，所以57444πππθ<+<，32ππθ<<，综上所述，3,2πθπ⎛⎫∈ ⎪⎝⎭为原不等式的解集.。

2021届河南省开封市高三下学期第三次模拟考试理综物理试题

2021届河南省开封市高三下学期第三次模拟考试理综物理试题一、单选题 (共6题)第(1)题如图所示为某示波管内的聚焦电场，实线和虚线分别表示电场线和等势线，两电子分别从a、b两点运动到c点，设电场力对两电子做的功分别为和，a、b点的电场强度大小分别为和，则（）A．，B．，C．，D．，第(2)题电场线的形状可以用实验来模拟，把头发屑悬浮在蓖麻油里，加上电场，头发屑按照电场的方向排列起来，如图所示，关于此实验，下列说法正确的是（）A．a图是模拟两等量同种电荷的电场线B．b图一定是模拟两等量正电荷的电场线C．a图中A、B应接高压起电装置的两极D．b图中A、B应接高压起电装置的两极第(3)题运动员以图示的姿势静止于水平地面上，则运动员（）A．一定受到摩擦力B．对地面的压力就是重力C．受到的支持力和重力是一对平衡力D．受到的支持力是由于脚掌形变产生的第(4)题某正弦式交变电流随时间变化的图像如图所示。

下列说法正确的是（）A．此交变电流的有效值为B．时，线圈位于中性面C．时，穿过线圈的磁通量最大D．时，穿过线圈的磁通量变化率最大第(5)题“嫦娥三号”着陆器和月球车首次使用了同位素核电池，该电池将放射性同位素衰变时释放的能量通过温差热电转换器转化为电能，在恶劣的月球环境中支持月球车低速移动及与地球间不间断通讯。

已知半衰期为88年，衰变方程为，、、X的结合能分别为、、，一次衰变释放能量为△E ，下列说法正确的是（）A．衰变发出的射线是高速氦核流，能穿透几毫米厚的铝板B．一次衰变释放的能量C ．经过88年，同位素核电池内的剩余25%D．若一静止核衰变释放的能量全部转化为和X 的动能，则动能为第(6)题如图所示，有一束平行于等边三棱镜截面ABC 的复色光从空气射向AB 边的中点D ，经三棱镜折射后分为a 、b 两束单色光，单色光a 偏折到BC 边的中点E ，单色光b 偏折到F 点，则下列说法正确的是（）A ．a 光的折射率大于b 光的折射率B ．在棱镜中a 光的传播速度大于b 光的传播速度C ．分别通过同一双缝干涉装置，a 光的相邻亮条纹间距大D ．若a 光恰好能使某种金属发生光电效应，则b 光也能使该金属发生光电效应二、多选题 (共4题)第(1)题B 超成像的基本原理是探头向人体发射一组超声波，遇到人体组织会产生不同程度的反射。

2021届河南省开封市高三下学期第三次模拟考试理综物理试题

2021届河南省开封市高三下学期第三次模拟考试理综物理试题一、单选题 (共6题)第(1)题若有一颗“宜居”行星，其质量为地球的p倍，半径为地球的q倍，则该行星卫星的环绕速度是地球卫星环绕速度的（）A ．倍B．倍C．倍D．倍第(2)题如图所示，实线表示某静电场的电场线，虚线表示该电场的等势面．下列判断正确的是（）A．、两点的场强相等B．、两点的场强相等C．、两点的电势相等D．、两点的电势相等第(3)题如图甲所示是我国 2023 年4月 12 日取得新突破的全超导托卡马克核聚变实验装置（EAST），该装置成功实现稳态高约束模式等离子体运行403 s，装置原理图如图乙所示。

该装置可使用和为核燃料进行热核聚变反应，核反应方程分别为：①；②。

则（）A．X为B．强磁场通过洛伦兹力约束运动的YC．核反应①中质量亏损较大D．核反应②中的比结合能较更大第(4)题我国比亚迪集团采用减小厚度、增加长度的结构创新方案推出了“刀片电池”，可以在同样的空间内装入更多电池。

如图所示，一款车型装配了120块“刀片电池”。

每块“刀片电池”的容量是200 A·h，平均工作电压是3.2 V。

该车型采用充电电压为600 V的快充充电桩时，充电效率为80%，充满电需要的时间为1 h。

该车型每行驶100km平均能耗是13 kW·h。

则该车型的汽车充电桩（）A．充满电后续航里程是640 kmB．充满电后储存的电能是640 W·hC．单块电池充满电后储存的电荷量是200CD．快充充电桩的平均充电电流是 160 A第(5)题如图（a），在均匀介质中有和四点，其中三点位于同一直线上，垂直．时，位于处的三个完全相同的横波波源同时开始振动，振动图像均如图（b）所示，振动方向与平面垂直，已知波长为．下列说法正确的是（）A．这三列波的波速均为B．时，处的质点开始振动C．时，处的质点向轴负方向运动D．时，处的质点与平衡位置的距离是第(6)题把线圈、电容器、电源和单刀双掷开关按照图示连成电路。

2021届河南省开封市高三下学期第三次模拟考试理综物理试题

2021届河南省开封市高三下学期第三次模拟考试理综物理试题一、单选题 (共6题)第(1)题北京正负电子对撞机，是我国的一座重大科学装置。

正负电子对撞机是一种高能物理实验设备，它的作用是通过将正电子和负电子加速到接近光速，然后让它们碰撞，以探索宇宙的基本结构和性质。

动能大小均为的一对正负电子相向运动发生对撞并发生湮灭，产生一对光子。

已知正、负电子的静止质量均为m，光在真空中的传播速度为c，则所产生的一个光子的动量大小为（）A．B．C．D．第(2)题如图所示，水平板上有质量m=1.0kg的物块，受到随时间t变化的水平拉力F作用，用力传感器测出相应时刻物块所受摩擦力F f的大小．取重力加速度g=10m/s2．下列判断正确的是（）A．5s内拉力对物块做功为零B．4s末物块所受合力大小为4.0NC．物块与木板之间的动摩擦因数为0.4D．6s～9s内物块的加速度的大小为2.0m/s2第(3)题2023年亚运会在杭州举行。

有关运动项目的描述，下列说法正确的是（）A．甲图中跳水运动员在空中运动到最高点时加速度为零B．乙图中研究羽毛球运动员击球动作时，运动员可视为质点C．丙图中撑杆跳运动员在撑杆起跳上升过程中，运动员始终处于超重状态D．丁图中跨栏运动员在100米跨栏比赛中获得第一名，该运动员全程平均速度最大第(4)题如图（a）所示，太阳系外行星M、N均绕恒星Q做同向匀速圆周运动。

由于N的遮挡，行星M被Q照亮的亮度随时间做如图（b）所示的周期性变化，其中为N绕Q运动的公转周期。

则两行星M、N运动过程中相距最近时的距离与相距最远时的距离之比为（）A．B．C．D．第(5)题2020 年 9 月，中国发布“双碳战略”，计划到 2030 年实现碳达峰、2060 年实现碳中和。

碳排放问题的治本之策是转变能源发展方式，加快推进清洁替代和电能替代，一座小型水电站向山下村镇供电的示意图如图所示，升压变压器T1 与降压变压器T2 都= 250 V，两个变压器的匝数比，，输电线电阻R = 20W，输是理想变压器。

2021届河南省开封市高三下学期第三次模拟考试理综物理试题

2021届河南省开封市高三下学期第三次模拟考试理综物理试题一、单选题 (共6题)第(1)题明代方以智在《物理小识》中记载：“凡宝石面凸则光成一条，有数棱则必有一面五色”，这描述的是光的色散现象。

如图，半圆是一宝石的横截面，是其直径，是圆弧上的一点。

在横截面所在的平面，一束光自点射入宝石，折射为、两束单色光。

下列说法正确的是（）A．宝石对光的折射率比对光的折射率大B．在宝石中光的传播速度比光的传播速度小C．若仅增大光在P点的入射角，光可能在上发生全反射D．用同一双缝干涉装置做实验，光的干涉条纹间距比光的干涉条纹间距大第(2)题一个均匀带正电的绝缘球体，单位体积内电荷量为ρ，球心为O，半径为r，内部有一个球形空腔，腔内的电场为匀强电场，空腔球心为P，半径为0.5r，且OP处于同一水平面上，OP连线延长线上有一点A，，静电力常量为k。

已知：均匀带电球壳在其内部产生的场强为0。

则下列说法正确的是（）A．空腔内匀强电场方向可能垂直OP向上B．该球体是一个等势体C．A处场强大小为D．空腔内场强大小为第(3)题2020年12月我国科学家在量子计算领域取得了重大成果，构建了一台76个光子100个模式的量子计算机“九章”，它处理“高斯玻色取样”的速度比目前最快的超级计算机“富岳”快一百万亿倍。

关于量子，下列说法正确的是（）A．是计算机运算的一种程序B．表示运算速度的一个单位C．表示微观世界的不连续性观念D．类似于质子、中子的微观粒子第(4)题石墨烯是目前世界上已知的强度最高的材料，它的发现使“太空电梯”缆线的制造成为可能，人类将有望通过“太空电梯”进入太空。

已知悬梯沿地球半径方向延伸到太空，现假设有一“太空电梯”的轿厢悬在赤道上空某处，相对悬梯静止，且做匀速圆周运动，如图所示，那么关于“太空电梯”，下列说法正确的是（）A．电梯轿厢悬停在同步卫星轨道时处于完全失重状态B．电梯轿厢在悬梯不同位置悬停，运动周期随高度增大而增大C．电梯轿厢在悬梯不同位置悬停，加速度与轿厢离地球球心距离的二次方成反比D．任意相等时间内轿厢所受合力冲量不为零且大小相等第(5)题如图所示，抛物线a和直线b分别是在平直公路上行驶的汽车a和b的位置—时间图像，时刻对应抛物线的顶点．下列说法正确的是（）A．在时刻，两车速率相等B．在时间内，b车做匀变速直线运动C．在时间内，时刻两车相距最远D．在时间内，a与b车的平均速度相等第(6)题下列说法中正确的是（）A．用三棱镜观察太阳光谱是利用光的干涉现象B．肥皂泡在阳光下出现彩色条纹，这是光的衍射现象C．用光导纤维传送图像信息，这其中应用到了光的全反射现象D．通过两支夹紧的笔杆间缝隙看发白光的灯丝能观察到彩色条纹，这是光的偏振现象二、多选题 (共4题)第(1)题如图所示，沿x轴正方向传播的一列简谐横波在某时刻的波形图为正弦曲线，其波速为200m/s，下列说法中正确的是（）A．此简谐波可能为次声波B．图示时刻质点b的加速度正在增大C．若此波遇到另一波并发生稳定干涉现象，则该波所遇到的波的频率为50HzD．若发生明显衍射现象，则该波所遇到的障碍物或孔的尺寸一定大于4mE．从图示时刻开始，经过0.01s，质点a通过的路程为0.4m第(2)题如图所示，匝数为100的矩形导线圈处于磁感应强度大小为的水平匀强磁场中，线圈的面积为、电阻为。

2021届开封高中高三语文第三次联考试卷及答案解析

2021届开封高中高三语文第三次联考试卷及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

拿去主义“拿来主义”是鲁迅的一篇杂文的题目，现在已成为引进、吸收外来文化的一个专用名词，同样重要的是“拿去主义”，虽然未经鲁迅述及，却也值得一提。

在“文化热”中，加强对外文化交流的呼声越来越高。

可是在有些同志的心目中，所谓“交流”，就是指“拿”，至于“拿去”，对不起，意识不强，也许是没有考虑。

这实际上是单向流程，而不是双向流程。

“来而不住，非礼也。

”中国人是礼仪之邦，总不能光拿人家的啊！有些同志认为我国的民族文化落后、陈旧，对是否有输出价值和竞争能力表示怀疑。

其实，这些同志未熟谙西方的心理和行情。

须知东西方文化各有所长，各有所短，互为补充，不可代替，从而构成世界文化整体。

中国文化，诸如长城、秦始皇兵马俑、汉代墓葬中的金缕玉衣、唐诗、宋词、《红楼梦》等等闻名世界，这是众所周知的。

即使是未被我们重视的所谓“糟粕”，西方人的评价也常常出乎意外。

例如老庄哲学和《易经》，我们认为是完全过了时的东西，而目下在西方却大为走运。

许多科学家对目前科学技术的巨大进步无法进行概括，他们找来找去，终于在老庄的“道”和《易经》的爻象中找到了合适的语言，并给予很高的评价。

这里说明一个问题，具有五千年历史的中国传统文化，毕竟是一座内容丰富的宝库，我们大可不必妄自菲薄，说得一无是处。

但是，历史在发展，时代在前进。

如果我们拿出去的永远是一些陈年百代的古董、祖传的遗物，那也不能使我们今天的炎黄子孙脸上增光。

对西方读者来说，他们既想了解中国的昨天，当然也想了解中国的今天和明天。

所以拿出去新的文化成果，才是今天中国人的光荣职责。

我们非常高兴地看到，近几年来，西方学者对中国当代文化艺术有了较多的关注，就以文学方面来说，王蒙、冯骥才、谌容、张贤亮、王安忆等作家的作品已陆续有了各种译本，被介绍到西方，有的还有专著论述，成为新一代的汉学家的研究课题。

2021届开封高级中学高三生物第三次联考试题及答案

2021届开封高级中学高三生物第三次联考试题及答案一、选择题：本题共15小题，每小题2分，共30分。

每小题只有一个选项符合题目要求。

1. 下列关于孟德尔的遗传定律及其研究过程的分析，错误的是A.提出问题是建立在豌豆纯合亲本杂交和F1自交实验的基础上B.为了验证作出的假设是否正确，孟德尔设计并完成了测交实验C.孟德尔所作假设的核心内容是“性状是由位于染色体上的基因控制的”D.孟德尔遗传定律只适用于进行有性生殖的真核生物的核遗传2. 下面是关于脂质的叙述，正确的是（）A. 脂肪由C、H、O三种元素组成，是构成膜的主要成分B. 性激素的化学本质是蛋白质，对维持生物体的生殖过程起着重要的调节作用C. 脂肪只存在于动物的脂肪细胞中，而其他部位和植物细胞中没有D. 企鹅体内的脂肪有减少热量散失、维持体温恒定的作用3. 在利用鸡血进行“DNA的粗提取与鉴定”的实验中，下列叙述正确的是A. 用蒸馏水将NaCl溶液浓度调至0.14 mol/L，滤去析出物B. 调节NaCl溶液浓度或加入木瓜蛋白酶，都可以去除部分杂质C. 将丝状物溶解在2 mol/LNaCl溶液中，加入二苯胺试剂即呈蓝色D. 由于DNA对高温耐受性较差，故需向DNA滤液中加入冷酒精4. 下列关于人体内环境及其稳态的叙述，正确的是（）A.人体的内环境即指体液，血细胞生活的内环境是血液B.血红蛋白、神经递质、生长激素、葡萄糖、尿素和O2都是内环境的成分C.H2CO3/HCO3-对血浆pH相对稳定有重要作用D.稳态的维持只需要消化系统、呼吸系统和循环系统三个系统的参与5. 如图取新鲜的神经—肌肉标本(生理盐水湿润)，a、d分别是放置在传出神经和骨骼肌上的电极，可刺激神经和骨骼肌；b是放置在传出神经上的电流计，用于记录神经兴奋电位；c为神经与肌细胞形成的突触。

正常时，用a刺激神经会引起骨骼肌收缩，下列判断分析正确的是（）A.如果用d刺激神经，在b处不能记录到电位变化，表明传出神经受损B.如果用a刺激神经，在b处记录到电位变化，骨骼肌不收缩，表明部位c受损C.如果用d刺激骨骼肌，骨骼肌不收缩，表明骨骼肌受损D.若本结构完好，用a或d刺激，在c的任何一点都可检测到电位变化6. 将生长状态一致的某种植物幼苗分成甲、乙两组，分别移入适宜的营养液中在光下培养，并给甲组的营养液适时通入空气，乙组不进行通气处理。