上海外国语大学附属浦东外国语学校2020-2021学年高二上学期第二次检测(12月)数学试题含答案

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上海外国语大学附属外国语学校2020-2021学年高二英语模拟试卷含解析

上海外国语大学附属外国语学校2020-2021学年高二英语模拟试卷含解析一、选择题1. --- Where was it _____ you interviewed the famous actor yesterday?--- It was in the hotel _____ he stayed.A. that; whereB. which; thatC. that; thatD. which; where参考答案：A2. A bank is the place _________ they lend you an umbrella in fine weather and ask for it back when it begins to rain.A. whenB. thatC. whereD. there参考答案：A3. Keep on trying! ________ practice is the groundwork for success in playing the piano.A. InstantB.Constant C. Eager D. Demanding参考答案：B略4. It is not what we do once in a while _____ shapes our lives, _____ what we do consistently.A. which; that;B. that; thatC. which; butD. that; but 参考答案：d略5. I should have bought the beautiful vase, but I ____________enough money.A. didn’t haveB. don’t haveC. haven’t hadD. hadn’t had 参考答案：a略6. The deserted island is only _______ by boat.A．suitable B．practical C．available D．accessible参考答案：D略7. _______ to sunlight for too much time will do harm to one’s skin.A. ExposedB. Having exposedC. Being exposedD. After being exposed参考答案：C8. He couldn’t________the fact that the money was found in his house.A．answer for B．leave forC．account for D．care for参考答案：C9. He found a purse on the seat and decided to return it to the owner, _______ with the address on the electricity bill.A. startedB. startingC. having startedD. to start参考答案：B【详解】考查分词做状语。

上海市浦东外国语学校2020-2021学年高二上学期10月月考数学试题

上海市浦东外国语学校2020-2021学年高二上学期10月月考数学试题学校:___________姓名：___________班级：___________考号：___________一、填空题1．过点(1,3)P ，且方向向量为(1,1)d =-的直线的点方向式方程为________2．若增广矩阵23702m ⎛⎫⎪⎝⎭的线性方程组的解为21x y =⎧⎨=⎩，则实数m =________ 3．已知数列{}n a 的前n 项和为n S ，若23n S n n =+-，则n a =________4．用数学归纳法证明：21427310(31)(1)n n n n ⨯+⨯+⨯+⋅⋅⋅++=+，从n k =到1n k =+，等式左边需增加的代数式为________5．直线320x y ++=的倾斜角为__.6．设1112132D n =-，则第二行第三列元素n 的代数余子式为________7．若图中的直线1l ，2l ，3l 的斜率分别为1k ，2k ，3k ，则1k ，2k ，3k 的大小关系为________8．已知3lim 13n nn nn a a →∞-=+，则实数a 的取值范围为________9．将22111133332223a b a b a b a b a b a b +-表示成三阶行列式：________ 10．若数列{}n a 是等差数列，n S 是数列的前n 项和，则n S ，2n n S S -，32n n S S -也成等差数列.类比上述结论，若数列{}n b 是等比数列，n T 是数列的前n 项积，则对应的结论为________11．如图所示，正方形上连接等腰直角三角形，直角三角形上再连接正方形……如此无限重复下去，设正方形面积为123,,,,,n S S S S ，三角形面积为123,,,,,n T T T T .当第一个正方形的边长为2时，则这些正方形和三角形的面积的总和为______.12．取出数列{},(4)n a n ≥的任意连续四项，若其中奇数项之和，偶数项之和均为同一个常数h （如连续四项1a ，2a ，3a ，4a ，满足1324a a a a h +=+=），则称数列{},(4)n a n ≥为错位等和数列，其中常数h 是公和.若n S 表示{}n a 的前n 项和，有如下命题：（1）若一个等差数列是错位等和数列，则1n a a =；（2）若一个等比数列是错位等和数列，则2n nhS =；（3）若12a a ≠，则错位等和数列一定是最小正周期为4的周期数列；（4）在错位等和数列{}n a 中，5h =，且201320146a a +=，若n 是偶数，则104,4210,4n k n k S k n k -=-⎧=⎨=⎩；其中，真命题的序号是________二、单选题13．若{}n a 为等比数列，则“135a a a <<”是“数列{}n a 是递增数列”的（） A ．充分不必要条件 B ．必要不充分条件 C ．充要条件D ．既不充分也不必要条件14．在等式①00a ⋅=; ②00a ⋅=；③()()a b c a b c ⋅⋅=⋅⋅；④22||a a =；⑤若a b a c ⋅=⋅，则b c =；正确的个数是（）A ．0个B ．1个C ．2个D ．3个15．矩阵的一种运算a b x ax by c d y cx dy +⎛⎫⎛⎫⎛⎫⋅=⎪ ⎪ ⎪+⎝⎭⎝⎭⎝⎭，该运算的几何意义为平面上的点(,)x y 在矩阵a b c d ⎛⎫⎪⎝⎭的作用下变成点(,)ax by cx dy ++，若直线1x y +-在矩阵11a b ⎛⎫ ⎪⎝⎭的作用下变换成直线10x y --=，则+a b 的值为（） A ．1B ．2C ．-1D ．016．在平面直角坐标系中，定义11n n nn n nx x y y x y ++=-⎧⎨=+⎩（*n ∈N ）为点(,)n n n P x y 到点111(,)n n n P x y +++的变换，我们把它称为点变换，已知1(1,0)P ，222(,)P x y ，333(,)Px y ，⋅⋅⋅是经过点变换得到一组无穷点列，设112n n n n n a P P P P +++=⋅，则满足不等式122020n a a a ++⋅⋅⋅+>最小正整数n 的值为（）A ．9B ．10C ．11D ．12三、解答题17．用行列式解关于x ，y 的二元一次方程组，并对解的情况进行讨论：282(3)kx y x k y k +=⎧⎨+-=⎩18．已知直线l过点(-10y -+=的夹角为6π，求直线l 的方程 19．已知向量()3sin ,1a x =，(),1b cosx =-.（1）若a ∥b ，求tan2x 的值；（2）若()()f x a b b =+⋅，求函数()f x 的最小正周期及当0,2x π⎡⎤∈⎢⎥⎣⎦时的最大值.20．直线l 过点(2,1)P ，且分别与x ，y 轴的正半轴交于A ，B 两点，O 为原点，（1）求AOB 面积最小值时，l 的方程；（2）求||||PA PB ⋅取最小值时，l 的方程.21．如果数列{}n a 对于任意n *∈N ，都有2n n a a d +-=，其中d 为常数，则称数列{}n a 是“间等差数列”，d 为“间公差”.若数列{}n a 满足1235n n a a n ++=-，n *∈N ，()1a a a R =∈.（1）求证：数列{}n a 是“间等差数列”，并求间公差d ；（2）设n S 为数列{}n a 的前n 项和，若n S 的最小值为-153，求实数a 的取值范围；（3）类似地：非零．．数列{}n b 对于任意n *∈N ，都有2n nb q b +=，其中q 为常数，则称数列{}n b 是“间等比数列”，q 为“间公比”.已知数列{}nc 中，满足()10,c k k k Z =≠∈，11120182n n n c c -+⎛⎫=⋅ ⎪⎝⎭，N n *∈，试问数列{}n c 是否为“间等比数列”，若是，求最大的整数．．．．．k 使得对于任意N n *∈，都有1n n c c +>；若不是，说明理由.参考答案1．13 11 x y--=-【解析】【分析】先求出直线的斜率，再求出直线的点方向式方程. 【详解】因为直线的方向向量为(1,1)d=-，所以直线的斜率为11 1=--，所以直线的点方向式方程为13 11x y--=-.故答案为：13 11 x y--=-【点睛】本题主要考查直线的点方向式方程，意在考查学生对该知识的理解掌握水平. 2．2【解析】【分析】根据增广矩阵的概念直接求解.【详解】由增广矩阵为23701m⎛⎫⎪⎝⎭的线性方程组的解为21xy=⎧⎨=⎩，则0221m⨯+⨯=，得2m=.故答案为：2.【点睛】本题考查了对增广矩阵的理解与应用，属于容易题.3．1,1 2,2nn n-=⎧⎨≥⎩【解析】【分析】已知n S 与n 的关系式，利用11,1,2n n n S n a S S n -=⎧=⎨-≥⎩即可求n a 的通项公式.【详解】由已知条件，知：当1n =时，111a S ==-；当2n ≥时，2213[(1)(1)3]2n n n a S S n n n n n -=-=+---+--=；当n=1时不满足上式，∴1,12,2n n a n n -=⎧=⎨≥⎩，故答案为：1,12,2n n n -=⎧⎨≥⎩.【点睛】本题考查了根据n a 与n S 的关系求通项公式，属于基础题. 4．(1)(34)k k ++ 【解析】【分析】根据等式的结构，分别写出n k =和1n k =+时，等式的左边，对比即可求解. 【详解】当n k =时，等式的左边为：1427310(31)k k ⨯+⨯+⨯+⋅⋅⋅++，当1n k =+，等式的左边为：1427310(31)(1)(34)k k k k ⨯+⨯+⨯+⋅⋅⋅+++++，所以从n k =到1n k =+，等式左边需增加的代数式为(1)(34)k k ++. 故答案为：(1)(34)k k ++. 【点睛】本题主要考查了数学归纳法及其应用，其中根据等式的结构，分别写出n k =和1n k =+时，等式的左边是解答的关键，着重考查运算能力. 5．arctan3π-【解析】【分析】计算直线的斜率为3k =-，即tan 3θ=-，计算倾斜角得到答案. 【详解】根据题意，设直线320x y ++=的倾斜角为θ，直线320x y ++=的斜率3k =-，则有tan 3θ=-，又由0θπ≤<，则arctan3θπ=-；故答案为：arctan3π-. 【点睛】本题考查了直线的倾斜角，属于简单题. 6．1 【解析】【分析】根据行列式第m 行第n 列元素的代数余子式的定义，结合行列式运算法则求得结果. 【详解】11121320D n =-，则第二行第三列元素n 的代数余子式为()()52311123132M =-=--=，故答案为：1. 【点睛】该题考查的是有关行列式第m 行第n 列元素的代数余子式的定义，属于基础题目. 7．231k k k >> 【解析】【分析】根据题中图像，由直线斜率的性质，可直接判断出结果. 【详解】因为直线倾斜角为锐角时，斜率为正，且倾斜角越大，斜率越大；倾斜角为钝角时，斜率为负；由图像可得，10k <，230k k >>，即231k k k >>.故答案为：231k k k >>. 【点睛】本题主要考查斜率大小的判断，属于基础题. 8．()3,3- 【解析】【分析】由题意32lim lim )133()1(1n n n n n n na a a→∞→∞--==++，可知n 趋向无穷大时3()1n a +也趋向无穷大才能使极限为1，即可求a 的取值范围. 【详解】0a =时符合题意；0a ≠时，由3()1321333()1()1n nnnn n n a a a a a--==-+++， ∴32lim lim )133()1(1n n n n n n n a a a→∞→∞--==++，即31||a >. ∴(3,3)a ∈-，且0a ≠.综上可得实数a 的取值范围为()3,3-，故答案为：()3,3-. 【点睛】本题考查了数列极限，已知极限值求参数范围，属于简单题.9．112233123a b a b a b -- 【解析】【分析】根据行列式的计算方法，对角线法则即可知三阶行列式.【详解】根据行列式的对角线法则，知：11221111223333223312233a b a b a b a b a b a b a b a b a b -=+--，故答案为：112233123a b a b a b -- 【点睛】本题考查了行列式，应用对角线法则求三阶行列式形式，属于简单题. 10．n T ，2n n T T ，32nnT T 也成等比数列. 【解析】【分析】根据题中条件，类比等差数列的性质，在等比数列中研究n T ，2n n T T ，32nnT T 之间关键即可. 【详解】因为若数列{}n b 是等比数列，n T 是数列的前n 项积，则12...n n b b T b ⋅⋅=，212212212.........n nn n n n nT b b b b b b T b b b ++⋅⋅==⋅⋅⋅⋅， 3123212232122.........n nn n n n nT b b b b b b T b b b ++⋅⋅==⋅⋅⋅⋅，所以()()()()2312122232332...n n n n n n nn T b b b b b b T b b T +++=⋅⋅⋅⋅⋅⋅ ()()()()222221232...n n n n n n T b b b b T +++⎛⎫=⋅⋅⋅⋅= ⎪⎝⎭，所以n T ，2n n T T ，32nnT T 成等比数列. 故答案为：n T ，2n n T T ，32nnT T 也成等比数列. 【点睛】本题主要考查类比推理，涉及等比数列的性质，属于基础题型. 11．10 【解析】【分析】先由题意，求出1S ，1T ，得到正方形的面积123,,,,,n S S S S 构成以4为首项，以12为公比的等比数列，三角形的面积123,,,,,n T T T T 构成以1为首项，以12为公比的等比数列，根据等比数列的前n 项和公式，以及极限的运算法则，即可得出结果. 【详解】因为第一个正方形的边长为2，所以2124S ==；，其面积为：1112T ==；由题意，正方形的面积123,,,,,n S S S S 构成以4为首项，以12为公比的等比数列；所以其前n 项和为11414121181812212n n n ---⎡⎤⎛⎫⨯-⎢⎥ ⎪⎡⎤⎝⎭⎢⎥⎛⎫⎛⎫⎣⎦=-=-⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦-；三角形的面积123,,,,,n T T T T 构成以1为首项，以12为公比的等比数列；所以其前n 项和为121112121212n n --⎡⎤⎛⎫⨯-⎢⎥ ⎪⎝⎭⎢⎥⎛⎫⎣⎦=- ⎪⎝⎭-，因此这些正方形和三角形的面积的总和为:4211lim 821022n n n --→∞⎡⎤⎛⎫⎛⎫-+-=⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦. 故答案为：10. 【点睛】本题主要考查求数列的极限，熟记等比数列的求和公式，以及极限的运算法则即可，属于常考题型.12．（1）（2）（3）（4）【解析】在（1）（2）中根据等差、等比数列的性质即可知{}n a 为常数数列，即可判断正误；由43414244141n n n n n n a a a a a a ----++=+=+有4242n n a a -+=，结合已知即可判断正误；由（3）的结论及已知得126a a +=、123410a a a a +++=即可得n S ，进而可知正误. 【详解】（1）由1324a a a a h +=+=得0d =，即{}n a 为常数数列，所以1n a a =正确；（2）由1324a a a a h +=+=得1q =，所以{}n a 为常数数列，12n a h =，所以2n nhS =，正确；（3）任取四项，则4341424n n n n a a a a h ---+=+=，且4244141n n n n a a a a h --++=+=，即有4341n n a a -+=，同理4242n n a a -+=，又12a a ≠，所以错位等和数列一定是最小正周期为4的周期数列，正确；（4）由（3）及201320146a a +=，得126a a +=，又5h =，即13245a a a a +=+=，所以344a a +=，且123410a a a a +++=，而错位等和数列一定是最小正周期为4的周期数列，所以104,4210,4n k n k S k n k -=-⎧=⎨=⎩. 故答案为：（1）（2）（3）（4）【点睛】本题考查了数列新定义，综合应用了等差、等比数列的性质，以及数列的周期性，属于中档题. 13．B 【解析】【分析】根据等比数列{}n a 是递增数列，得到135a a a <<一定成立，反之不成立，结合充分条件和必要条件的判定，即可求解.若等比数列{}n a 是递增数列，可得135a a a <<一定成立；反之：例如数列{}1(1)2n n +-，此时满足135a a a <<，但数列{}n a 不是递增数列，所以“135a a a <<”是“数列{}n a 是递增数列”的必要不充分条件. 故选：B. 【点睛】本题主要考查了等比数列的单调性性，以及必要不充分条件的判定，着重考查推理与计算能力，属于基础题. 14．C 【解析】【分析】由零向量、向量数乘、点乘等概念和性质，即可判断正误，进而确定答案. 【详解】零向量与任何向量的数量积都为0，00a ⋅=错误； 0乘以任何向量都为零向量，00a ⋅=正确；向量的加减、数乘满足结合律，而向量点乘不满足结合律，()()a b c a b c ⋅⋅=⋅⋅错误；向量模的平方等于向量的平方，22||a a =正确；a b a c ⋅=⋅不一定有b c =，故错误；故选：C 【点睛】本题考核查了向量，利用向量相关概念、性质判断正误，属于基础题. 15．B 【解析】【分析】根据矩阵变换，得到x x ay y bx y=+⎧⎨=+''⎩，进而得到方程(1)(1)10b x a y -+--=，列出方程组，即可求解.设(,)P x y 是直线10x y +-=上一点，在矩阵11a b ⎛⎫⎪⎝⎭作用下变成(,)Q x y ''，则x x ay y bx y =+⎧⎨=+''⎩，又由10x y ''--=，所以()()10x ay bx y +-+-=，即(1)(1)10b x a y -+--=，所以1111b a -=⎧⎨-=⎩，解得20a b =⎧⎨=⎩，所以2a b +=.故选：B. 【点睛】本题主要考查了矩阵的运算及其应用，其中解答中熟记矩阵的运算，以及矩阵下的点的变换，求得点之间的关系式是解答的关键，着重考查推理与运算能力. 16．C 【解析】【分析】可以先求得1a （当然可求得234,,,a a a ，然后归纳出n a ，对填空、选择题这是不错的解法），然后求得22n n n a x y =+，从而可以得12n n a a +=，说明数列{}n a 是等比数列，求得通项公式n a 后求和，由2020n S >得解．【详解】由定义知1110x y =⎧⎨=⎩，2211x y =⎧⎨=⎩，330,2x y =⎧⎨=⎩，即23(1,1),(0,2)P P ．11223(0,1)(1,1)1a PP P P =⋅=⋅-=，观察可得，112,n n n n n a P P P P +++=⋅112121(,)(,)n n n n n n n n x x y y x x y y ++++++=--⋅-- 11(,)(,)n n n n y x y x ++=-⋅-2211()()n n n n n n n n n n n n y y x x y x y x x y x y ++=+=++-=+， 222222111()()2()n n n n n n n n n a x y x y x y x y +++=+=-++=+2n a =， ∴数列{}n a 是等比数列，公比为2，首项为1．∴12n na ．2112122221n n n a a a -+++=++++=-，由212020n ->，解得11n ≥．即n 的最小值为11.故答案为：C 【点睛】本题考查向量的数量积，考查等比数列的通项公式与前n 项和公式．解题关键是求出22n n n a x y =+．接着顺理成章地写出1n a +，观察两项之间的关系，问题得以解决．属于难题17．当4k ≠且1k ≠-时有唯一解6141x k k y k ⎧=⎪⎪+⎨+⎪=⎪+⎩，当4k =时，有无数组解()24x tt R y t =⎧∈⎨=-+⎩，当1k =-时，原方程组无解. 【解析】【分析】根据题意分别求行列式,,x y D D D 关于k 的表达式，再对k 的取值讨论，即可得原方程解得情况. 【详解】2(4)(1)23k D k k k ==-+-，826(4)3x D k kk ==--，8(4)(4)2y k D k k k==-+，（1）当4k ≠且1k ≠-时，0D ≠，原方程组有唯一解6141x k k y k ⎧=⎪⎪+⎨+⎪=⎪+⎩；（2）当4k =时，0x yD D D ===，原方程组有无数组解()24x tt R y t =⎧∈⎨=-+⎩；（3）当1k =-时，0D =，300x D =-≠，0y D ≠，原方程组无解. 【点睛】本题主要考查了用行列式求二元一次方程组的解和含有字母的参数的方程的解讨论等知识，属于中档题.18．10x +=或40x -+=.【解析】【分析】10y -+=的倾斜角为3π，根据题意可得直线l 的倾斜角为2π或6π，又线l 过点(-，即可求直线方程.【详解】10y -+=的倾斜角为3π，因为直线l 过点(-10y -+=的夹角为6π，所以直线l 的倾斜角为2π或6π，所以直线l 的方程为10x +=或40x +=. 【点睛】本题考查了直线的倾斜角和斜率的关系，考查了直线的一般方程，属于基础题.19．（1）（2）最小正周期为π ，最大值为32【解析】【分析】（1）由//a b 得tan 3x =-，再根据二倍角的正切公式直接求解. （2）根据平面向量的数量积以及三角函数的恒等变换，化简f （x ）即可求出T ，再根据三角函数的图象与性质，求出x ∈[0，2π]时f （x ）的最大值以及对应x 的值．【详解】解：（1）由//a b 得, cos x x =，∴tan 3x =-∴22tan tan 1tan xx x==- （2）()()23sin cos cos f x a b b x x x =+⋅=+111cos2sin 222262x x x π⎛⎫=++=++ ⎪⎝⎭∴函数()f x 的最小正周期为22T ππ== 当0,2x π⎡⎤∈⎢⎥⎣⎦ 时，72666x πππ≤+≤ ∴当262x ππ+=，即6x π=时，()max 362f x f π⎛⎫==⎪⎝⎭. 【点睛】本题考查了共线向量的坐标运算，平面向量的数量积和三角函数的恒等变换以及三角函数的图象与性质的应用问题，是综合性题目． 20．（1）240x y +-=；（2）30x y +-=. 【解析】【分析】（1）设AB 方程为1x ya b+=，点(2,1)P 代入后，应用基本不等式求出ab 的最小值等号成立条件，即得三角形OAB 面积取最小值时的直线方程．（2）设直线l 的点斜式方程，求出A ，B 两点的坐标，代入||||PA PB 的解析式，使用基本不等式，求出最小值等号成立条件即可得答案．【详解】（1）设(,0)A a 、(0B ，b )，0a >，0b >，AB 方程为1x ya b+=，点(2,1)P 代入得21212a b ab+=，8ab ∴ 当且仅当21a b=，且211a b +=，解得4a =，2b =时，等号成立，故三角形OAB 面积142S ab =，此时直线方程为：142x y+=，即240x y +-=．（2）设直线:1(2),0l y k x k -=-<，分别令0y =，0x =，得1(2A k-，0)，(0,12)-B k ．则||||4PA PB =，当且仅当21k =，即1k =±时，||||PA PB 取最小值，又0k <，1k ∴=-，这时l 的方程为30x y +-=．【点睛】本题考查直线在坐标轴上的截距的定义，直线的截距式、点斜式与一般方程的应用，以及基本不等式的应用，属于中档题．21．（1）见解析；（2）17a ≥-；（3）63. 【解析】【分析】（1）直接利用定义求出数列为间等差数列．（2）利用分类讨论思想，利用数列的前n 项和公式求出数列的和，进一步利用不等量关系求出结果．（3）利用分类讨论思想，进一步求出数列的通项公式，再利用函数的单调性求出k 的最大值．【详解】（1）若数列{a n }满足a n +a n+1＝2n ﹣35，n∈N*，则：a n+1+a n+2＝2（n+1）﹣35，两式相减得：a n+2﹣a n ＝2．故数列{a n }是“间等差数列”，公差d ＝2．（2）（i ）当n ＝2k 时，n S =（a 1+a 2）+（a 3+a 4）+…+（a n ﹣1+a n ）＝﹣33﹣29+…+（2n ﹣37）＝易知：当n ＝18时，最小值S 18＝﹣153．（ii ）当n ＝2k+1时，S n ＝a 1+（a 2+a 3）+（a 4+a 5）+…+（a n ﹣1+a n ）＝a 1+（﹣31）+（﹣29）+…+（2n ﹣37）＝，当n ＝17时最小，其最小值为S 17＝a ﹣136，要使其最小值为﹣153，则：a ﹣136≥﹣153，解得：a≥﹣17．（3）易知：c n c n+1＝2018•（）n ﹣1，则：c n+1c n+2＝2018•（）n ，两式相除得：212n n C c +=，故数列{c n }为“间等比数列”，其间等比为．，易求出数列的通项公式为：，由于c n ＞c n+1，则数列{c n }单调递减．那么，奇数项和偶数项都为单调递减，所以：k ＞0．要使数列为单调递减数列．只需c 2m ﹣1＞c 2m ＞c 2m+1，即：，k <<k 63=.【点睛】本题考查了数列的通项公式的求法及应用，数列的前n 项和的应用，函数的单调性的应用，主要考查学生的运算能力和转化能力，属于中档题．。

上海市上海外国语大学附属外国语学校2020-2021学年上学期高二期末数学试卷

2020学年第一学期上外附中高二年级数学期末考试试卷一、填空题（共14小题，每题3分，满分42分） 1.计算：212 (i)n nn →∞+++=_________.2.已知正项等比数列{}n a 中，34,13231=+=a a a a ，则=+++∞→)...(lim 21n n a a a _________.3.在行列式31214053--a中，元素a 的代数余子式的值为_________.4.线性方程组的增广矩阵为⎪⎪⎭⎫⎝⎛+m m m m 124，且此方程组无解，则实数=m _________.5.若矩阵⎪⎪⎭⎫ ⎝⎛=4321A 与⎪⎪⎭⎫⎝⎛=724k B 满足BA AB =,则实数=k _________. 6.若无穷等比数列{}n a 的各项和为21，则首项1a 的取值范围是_________. 7.若向量)7,8(),0,2(--==,则b 在a 方向上的投影为_________.8.若数列{}n a 满足831-=+n n a a ，且61=a ，则数列{}n a 的通项公式为=n a _________. 9.若向量(1,2)a =与(21,3)b t t =-的夹角为锐角，则实数t 的取值范围是_________. 10.计算：=+-+)10002020lim(22n n n n _________.11.在矩形ABCD 中，2,2==BC AB ,点E 为BC 的中点，点F 在边CD 上，若2=⋅,则=⋅_________.12.若公差为d 的等差数列{}n a 的前n 项和为n S ，满足0,01312<>S S ,且283=a ,则公差d 的取值范围是_________. 13.已知ABC ∆中，3BAC π∠=，点D 为线段AB 的中点，点P 为线段CD 上一点，且满足1()3AP t AC AB t R =+∈，若ABC ∆的面积为2，则AP 的最小值为________14.已知数列{}n a 是共有k 个项的有限数列，且满足11(2,3,...,k 1),n n nna a n a +-=+=-若1224,51,0k a a a ===，则___k =二、选择题（34=12⨯分分）15.若ABC ∆所在平面内一点P 满足0PA PB PC ++=,则P 是ABC ∆的（） A.内心 B.外心 C.重心 D.垂心 16.已知ABC ∆中，点D 满足23BD DC =,若(,)AD AB AC R λμλμ=+∈，则μ的值为（）A. 23B.32C.25D.3517.设{}n a 是各项为正数的无穷数列，i A 是边长为i a 、1i a +的矩形的周长（1,2,i =⋅⋅⋅），则“数列{}n A 为等差数列”的充要条件是（） A.{}n a 是等差数列B.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅或242,,,,n a a a ⋅⋅⋅⋅⋅⋅是等差数列C.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅和242,,,,n a a a ⋅⋅⋅⋅⋅⋅都是等差数列D.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅和242,,,,n a a a ⋅⋅⋅⋅⋅⋅都是等差数列，且公差相同18.已知数列{}n a 的前n 项和为n S ，且对任意正整数n 都有13n n a S +=，则下列关于{}n a 的论断中正确的是（）A.一定是等差数列B.一定是等比数列C.可能是等差数列，但不会是等比数列D.可能是等比数列，但不会是等差数列三、解答题（6+8+10+10+12=46分） 19.（本题满分6分）用行列式解关于,x y 的方程组22(1)321x my m x y m +=⎧⎨-+=+⎩.20.（本体满分8分，第（1）小题3分，第（2）小题5分）已知两个平面向量a 与b 的夹角为3π，且1,2,a b ==记3,2m a b n ta b =-=+. （1）若m n ⊥，求实数t 的值；（2）若2t =，求m 与n 的夹角.21.（本题满分10分，第（1）小题5分，第（2）小题5分）（1）已知等差数列{}n a 满足39a =，且6732a a +=，若数列21n n a a +⎧⎫⎨⎬⎩⎭的前n 项和为n S ，求lim n n S →∞的值.（2）已知数列{}n b 的前n 项和n T 满足3+23(*)n n T b n N =∈，若t R +∈,求215lim n n nn n b t b t ++→∞++的值.22.（本大题满分10分，第（1）小题3分，第（2）题4分，第（3）小题3分）按照如下规则构造数表：第一行是：2；第二行是：21,23++；即3，5，第三行是：31,33,51,53++++即4，6，6，8；（即从第二行起将上一行的数的每一项各项加1写出，再各项加3写出）2 3 54 6 6 85 7 7 9 7 9 9 11 ……………………………………若第n 行所有的项的和为n a ．（1）求345,,a a a ；（2）试求1n a +与n a 的递推关系，并据此求出数列{}n a 的通项公式；（3）求数列{}n a 的前n 项和n S .23.（本大题满分12分，第（1）小题3分，第（2）题4分，第（3）小题5分）我们把一系列向量()1,2,,i a i n =按次序排成一列，称之为向量列，记作{}n a ，已知向量列{}n a 满足：()1,11=a ，()()11111,,2n n n n n n n a x y x y x y ----==-+()2n ≥．（1）证明：数列{}n a 是等比数列；（2）设2log n n n c a a =⋅，问数列{}n c 中是否存在最小项？若存在，求出最小项；若不存在，请说明理由；（3）设n θ表示向量1n a -与n a 间的夹角，若2n n n b θπ=，对于任意正整数n ，不等式()211log 122an a b +>-恒成立，求实数a 的范围参考答案一、填空题1.122.923.-24. 2m=- 5.3 6. 11(0,)(,1) 22 7.8- 8.1234n-⋅+ 9.1(,2)(2,)8+∞ 10.510 (8,7)-- 14.50二、选择题15.C 16.A 17.D 18.C三、解答题19 解析：226(3)(2)13mD m m m mm==-++=--+-2226(2-3)(2)213xmD m m m mm==--+=-++2224121yD mm m==+-+（1）当0D≠时，即2,3m m≠-≠，方程由唯一解，23322mxmym-⎧=⎪⎪-⎨⎪=⎪-⎩；（2）当2m=-时，0x yD D D===，方程组有无穷多组解;（3）当3m=时，0,0,0x yD D D=≠≠，方程组无解20.（1）2t=；（2）1arccos7；21.（1）335；（2）03t<<时，215limnnnnnb tb t++→∞++=3；3t=时，215limnnnnnb tb t++→∞++=24；3t >时，2215lim 5n n n n n b t t b t ++→∞+=+, 22.【解答】（1）3424,64,a a ==5160a =．（2）由题意，第n 行共有12n -项，于是有()()111123222n n n n n n n a a a a --++=+++⨯=+ 等式两边同除12n +，得11122n nn na a ++=+，即2n n a ⎧⎫⎨⎬⎩⎭为等差数列，公差为1，首项为112a = 所以2n na n =，即()*2n n a n n N =⨯∈．（3）23122232...2n n S n =⨯+⨯+⨯++⨯ （1） 23412122232...(1)22n n n S n n +=⨯+⨯+⨯++-⨯+⨯ （2）由（1）-（2）得23111222...22222n n n n n S n n +++-=++++-⨯=--⨯ 所以1(1)22n n S n +=-⨯+ 23.（1）()()22221111111122222n n n n n n n n a x y x y x y a -------=-++=+= ∴数列{}n a 是等比数列（2）12222n nn a --⎛== ⎭， 22222nn n c --∴=⋅ 假设 {}n c 中的第 n 项最小，由 12c =，20c =，210.c c ∴≤< 当3n ≥时，有0n c <，又由1n n c c +≤可得()()212222122222n nn n -+--+-⋅≤⋅，即12221n n --≥-，22112n n -⎛⎫≥ ⎪-⎝⎭. 2670n n -+≥，3n ≥或3n ≤-，5n ∴≥.…即有567c c c <<<；由1n n c c +≥，得35n ≤≤，又210c c ≤<，541c c c ∴<<<；故数列{}n c 中存在最小项，最小项是325322c -=-⋅…(3) 112cos 2n n n n na a a a θ--⋅==⋅，4n πθ∴=… 24n n b ∴=不等式化为：()32221log 121222a n n n +++>-++对任意正整数n 恒成立. 设222122n T n n n=++++. 又()122222021*******n n T T n n n n n +-=+-=->+++++，数列{}n T 单调递增 ()1min 1n T T ∴==，要使不等式恒成立，只要()311log 122a >-，120a ->， 102a ∴<<，212a a -> ，01a ∴<< 所以，使不等式对于任意正整数恒成立的a 的取值范围是()1.。

2020-2021学年上海外国语大学附属外国语学校高三英语第二次联考试卷及答案解析

2020-2021学年上海外国语大学附属外国语学校高三英语第二次联考试卷及答案解析第一部分阅读（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ASheffieldLincoln College of EnglishClasses for foreign students at all levels.3 months, 6 months, 9 months and one year course.Open all year.Small class (at most 12 students).Library, language laboratory and listening center.Accommodation (住宿)with selected families.25 minutes from London.Course fees for English for one year are￡1,380 with reduction for shorter periods of study.1.This passage is probably taken from _______.A.an advertisementB.a noticeC.a posterD.a piece of news2.Who will be accepted by this college?A.Both foreign and native students.B.Only foreign beginners and the advanced.C.Foreign students from beginners to the advanced.D.Only foreign students advanced.3.While you stay there, who will take care of you?A.Your parents.B.Your classmates.C.The school where you study.D.The family you have chosen.BImaginary friends in childhood refer to the invisible beings that a child gives a personality to and plays with for over three months.Crabbycrab（蟹）appeared on a holiday in Norway by running out of my four-year-old son Fisher's ear after a night of tears from an earache. Like other childhood imaginary friends, Crabby should be a sign thatFisher's mind is growing and developing positively. Indeed, research shows that imaginary friends can help develop children's social skills.Research has shown that the positive effects of having imaginary friends as a child continue into adulthood. Adolescents who remember their imaginary playmates have been found to use more activecoping（应对）styles, such as seeking advice from loved ones rather than bottle things up inside. Even adolescents with behavioral problems who had imaginary friends as children have been found to have better coping skills through the teenage years.Scientists thinkthis could be because these teens have been able to adjust themselves to the social world with imagination rather than choose to be involved in relationships with more difficult classmates. It could also be because the imaginary friends help to reduce these adolescents,loneliness.These teens are also more likely to seek out social connections -they tend to turn to others for advice. Current research by Tori Watson is taking this evidence and looking at how adolescents who have imaginary friends as children deal withbullying（欺凌）at school. It is found that teens who remember their imaginary friends are better at dealing with bullying.While we know a lot about childhood imaginary friends such as Crabby Crab and the positive effects they can have, there is still a lot to learn about imaginary friends.4. What is Crabby crab?A. It is a crab Fisher caught inNorway.B. It is Fisher's imaginary friend.C. It is a toy Fisher like much.D. It is a cause of earache.5. Why do children with imaginary friends have better coping skills?A. Imaginary friends help improve their adjustment.B. Having imaginary friends makes them smarter.C. They have rich imagination.D. They are no longer alone.6. What will a child with imaginary friends probably do if he is bullied?A. Escape from the bully.B. Fight with the bully bravely.C. Keep silent about being bullied.D. Ask a parent or a teacher for help.7. What is the author's attitude towards the effect of imaginary friends?A. Concerned.B. Doubtful.C. Optimistic.D. Indifferent.CSomeday soon an emoji (表情符号)might really save lives.Hiroyuki Komatsu is a Google engineer who suggested adding a series of new emojis to the standard emoji library. It could help those with food allergies (过敏)understand what they are eating anywhere inthe world. Emojis should cover characters representing major food causing allergies. They make people understand what are used in foods even in foreign countries and safely select meals.Emojis are universal because they are chosen and developed by the Unicode Consortium, a non-profit company that oversees, develops and maintains how text is represented. This is in regards to all software products and standards. It's thanks to the Unicode Standard that when you text a friend six pizza emojis, they’ll see those six pizza pieces on their phone. This is true regardless of whether they use an iPhone or an Android.Because emojis are everywhere and visual(视觉的),they could be helpful for restaurants and food packaging designers. They can communicate whether a product is made with common causing-allergy food. But as Komatsu’s advice argues, many of the most common causing-allergy foods are missing or poorly represented by the present emoji library. For example, there is an emoji for octopus, but nothing for squid. There is a loaf of bread that could symbolize grain, but a picture of wheat could be clearer. The emojis can be more direct when symbolizing foods.It’s not uncommon for the Unicode Consortium to add new emojis to the library: several food-related emojis were put into use last June, including some long-waited food emojis. Apple included support for multiracial emojis in a recent iOS update. An artist even recreated Moby-Dick in emoji characters. Some might be sorry for the continuing death of the written word if Komatsu’s suggestion is accepted, but look on the bright side: if you ever see that happy poop on a box, you’ll know to stay away.8. How will emojis save lives according to the text?A. By showing what the food contains visually.B. By telling the safest places in the world.C. By teaching people how to treat allergies.D. By adding standard emojis about safety.9. What does paragraph 3 mainly tell us?A. Emojis have the same meanings around the world.B. The Unicode Consortium is a non-profit company.C. What emojis represent is different in different places.D. Different mobile operating systems have different emojis.10. What can be the reason for Komatsu’s advice?A. Emojis are easy to mix up.B. Present emojis are not enough.C. Emojis can't interest most users.D. Emojis can't represent foods directly.11. What is the author’s attitude to Komatsu's suggestion?A. Doubtful.B. Worried.C. Supportive.D. Uninterested.DSam, I say to myself as I start across the bridge, you must stop these thoughts and start thinking about what to do now that you have lost your falcon, Frightful.Life, my friend Ban do once said, is meeting problems and solving them whether you are an amoeba or a space traveller. I have a problem. I have to provide my younger sister Alice and myself with meat. Fish, nuts, and vegetables are good and necessary, but they don't provide enough fuel for the hard physical work we do. Although we have venison now, I can't always count on getting it. So far this year, our venison has been only road kill from in front of Mrs Strawberry's farm.I decide to take the longest way home, down the flood plain of the West Branch of Delaware to Spillkill, my own name for a fast stream that cascades down the south face of the mountain range I'm on. I need time to think. Perhaps Alice and I should be like the early Eskimos. We should walk, camp and hunt, and when the seasons change, walk on to new food sources. But I love my tree and my mountaintop.Another solution would be to become farmers, like the people of the Iroquois Confederacy who once lived here. They settled in villages and planted corm and squash, bush beans and berries. We already grow groundnuts in the damp soil and squash in the poor land. But the Iroquois also hunted game. I can't do that anymore.I'm back where I started from.Slowly I climb the Spillkill. As I hop from rock to rock beneath shady basswoods and hemlocks, I hear the cry of the red-tailed hawk who nests on the mountain crest. I am reminded of Frightful and my heart aches. I canalmost hear her call my name, Cree, Cree, Cree, Car-ree.Maybe I can get her back if I beg the man who is in charge of the peregrines at the university. “But it's the law,” he would say. I could write to the president of the United States and ask him to make an exception of Alice and me. That won't work. The president swore to uphold the Constitution and laws of the United States when he took office.I climb on. I must stop thinking about the impossible and solve the problem of what to do now. I must find a new way to provide for us. Frightful is going to be in good hands at the university, and she will have young.I smile at the thought of little Frightfuls and lift my reluctant feet.When I am far above the river, I take off my clothes and moccasins and bathe in a deep, clear pool until I am refreshed and thinking more clearly. Climbing up the bank, I dress and sit down. I breathe deeply of the mountain air and try to solve my problem more realistically.12. What does this excerpt main describe?A. Delicate mental activities.B. Unique story environment.C. Everchanging story events.D. Complicated character relationship.13. What is Sam's first worry?A. How to get back quicklyB. How to get enough venison.C. How to ensure the safety of Frightful.D. How to provide meat for Alice and himself.14. What do we know about Frightful?A. He left Sam and Alice due to lack of food.B. He helped Sam hunt before being taken away.C. He is living with the red-tailed hawk happily.D. He has given birth to babies in the university.15. Which of the following can best describe Sam?A. Humorous.B. Aggressive.C. Responsible.D. Unrealistic.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。

上海市上海外国语大学附属上外高中2020-2021学年高二上学期期中数学试题

上海市上海外国语大学附属上外高中2020-2021学年高二上学期期中数学试题学校:___________姓名：___________班级：___________考号：___________一、填空题1．等比数列{}n a 中：473,2a a π=则29sin()a a =__________ 2．等差数列{}n a 中：1d =，若13599...99a a a a ++++=，则100S =__________ 3．已知等差数列{}n a 中，110a =，当且仅当5n =时，前n 项和n S 取得最大值，则公差d 的取值范围时________4．“x =是“,,a x b ”成等比数列的_______条件；5．数列{}n a 中：372,1,a a ==且数列11n a ⎧⎫⎨⎬+⎩⎭是等差数列，则数列{}n a 的通项公式为______6．已知数列{}n a 满足317n a n =-，则数列{}n a 的前n 项和n T =__________7．数列{}n a 中：*1111,30(,2),n n n n a a a a a n N n --=+-=∈≥则10a =__________ 8．记实数等比数列{}n a 的前 n 项和为n S 若1010S =，3070S =，则40S =_______.9．已知数列{}n a 是等比数列，其前n 项和为1*3()n n S k n N -=+∈，则常数k =__________10．已知{}n a 是首项为a ，公差为1的等差数列，1n n n a b a +=，若对任意的*n N ∈，都有5n b b ≤成立，则实数a 的取值范围是________11．已知数列{}n a 满足()*1115,2n n a a a n n N+=-=∈，则n a n 的最小值为_______ 12．已知1()1f x x=+，各项均为正数{}n a 满足11a =，2()n n a f a +=，若20182020a a =，则2011a a +的值为__________二、单选题13．如果数列{}n a 是等比数列，那么下列数列中不一定是等比数列的是（）A ．1n aB ．C ．{}1n n a a +⋅D ．{}1n n a a ++14．等比数列{}n a 的前n 项积为n T ，并且满足1100910101,10,a a a >->10091010101a a -<-，现给出下列结论：①20221;a >②1009101110a a -<；③1010T 是n T 中的最大值；④使1n T >成立的最大自然数n 是2019，期中正确的结论个数是（）A ．1B ．2C ．3D ．415．已知数列{}n a 是等比数列，其公比为q ，则“1q >”是“数列{}n a 为单调递增数列”的（）A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件16()1n n N *<+∈，某同学用数学归纳法证明的过程如下：（1）当1n=11<+，不等式成立.（2）假设当()n k k N *=∈1k <+成立，当1n k =+时，()11k =<==++.∴当1n k =+时，不等式成立，则上述证法（）A ．过程全部正确B ．1n =验得不正确C ．归纳假设不正确D ．从n k =到1n k =+的推理不正确三、解答题17．等差数列{}n a 中，设其前n 项和为n S ，若153,,62k a a ==-，且前k 项和5k S =- （1）求此数列的公差d（2）当n 为何值时，n S 取得最大值18．已知等差数列{}n a 的公差0d ≠，{}n a 中的部分项组成的数列1k a 、2k a 、、n k a 、恰好为等比数列，其中11k =，25k =，517k =，求数列{}n k 的通项公式.19．某林场现有木材存量为a ，每年以25%的增长率逐年递增，但每年年底要砍伐的木材量为b ，经过n 年后林场木材存有量为y（1）求()y f n =的解析式（2）为保护生态环境，防止水土流失，该地区每年的森林木材存量不应少于79a ，如果1972b a =，那么该地区会发生水土流失吗？若会，要经过几年？（取lg 20.30=） 20．已知数列{}n a 的前n 项和n S 满足条件()231n n S a =-,其中n ∈N *.(1)证明:数列{}n a 为等比数列;(2)设数列{}n b 满足3log n n b a =,若n n n c a b =,求数列{}n c 的前n 项和n T .21．已知数列{}{},n n a b 满足：1121,1,41n n n n n b a a b b a +=+==- （1）求：1234,,,b b b b ，（2）猜想数列{}n b 的通项公式，并用数学归纳法证明；（3）若1223341...,n n n S a a a a a a a a +=++++且4n n aS b <对于*n N ∈恒成立，求实数a 的取值范围参考答案1．-1【解析】【分析】根据等比数列下标和性质，结合正弦函数求值，即可求得.【详解】因为数列{}n a 是等比数列，故294732a a a a π==故29sin()a a =312sinπ=-. 故答案为：-1.【点睛】本题考查等比数列下标和性质，以及特殊角的正弦值.2．248【分析】利用数列前n 项和的性质，求出偶函数项的和，再求解出100S .【详解】因为13599...99a a a a ++++=根据等差数列的性质：241009950149a a a d +++=+=故10099149248S =+=.故答案为：248.【点睛】本题考查等差数列前n 项和的性质，属基础题. 3．5,22⎛⎫-- ⎪⎝⎭【分析】根据等差数列的前n 项和的公式，讨论其单调性，结合题意，即可求得范围.【详解】因为数列中110a =，故其前n 项和21022n d d S n n ⎛⎫=+- ⎪⎝⎭是关于n 的二次函数，且n N +∈. 因为当且仅当5n =时，前n 项和n S 取得最大值故只需该二次函数的对称轴范围在911,22⎛⎫ ⎪⎝⎭，即110911,222d ⎛⎫-∈ ⎪⎝⎭，解得5,22d ⎛⎫∈-- ⎪⎝⎭故答案为：5,22⎛⎫-- ⎪⎝⎭. 【点睛】本题考查等差数列前n 项和的函数属性，属基础题.4．既不充分也不必要【分析】通过反例，结合等比中项的定义，进行推理和说明.【详解】当0a x ==时，由x =,,a x b ”成等比数列，故不充分；当“,,a x b ”成等比数列，且0,0,0a b c <<<时，得不到x =. 故答案为：既不充分也不必要.【点睛】本题考查充要条件的判定，本质是关于等比数列的考查，属基础题.5．195n n -+ 【分析】根据条件，求出等差数列11n a ⎧⎫⎨⎬+⎩⎭的通项公式，再推导出n a 即可. 【详解】因为数列11n a ⎧⎫⎨⎬+⎩⎭是等差数列，记11n n b a =+，故n b 为等差数列，由题可知：37371111,1312b b a a ====++ 故可得数列n b 的公差为：()7311424b b -=，首项114b = 故51241n n n b a +==+ 故可得：2419155n n a n n -=-=++ 故答案为：195n n -+. 【点睛】本题考查等差数列基本量的求解，属基础题.6．22313,52233180,622n n n n T n n n ⎧-≤⎪⎪=⎨⎪-+≥⎪⎩【分析】找出数列n a 与0的大小关系的临界值，根据n 的取值范围，进行分类讨论即可.【详解】因为317n a n =-，故1314n a n +=-，则13n n a a +-=，故该数列是一个首项为-14，公差为3的等差数列，设数列{}n a 的前n 项和为2331,22n n S S n n =- 令0n a <，解得5n ≤，故该数列的前5项都是负数，从第6项开始都为正数.若6n ≥，则()123456n n T a a a a a a a =-++++++ 55n S S S =-+-52n S S =-()()()3114101462n n n -=⨯-+--+ 23318022n n =-+若5n ≤，则233122n n n T S n =-=-+ 故答案为：22313,52233180,622n n n n T n n n ⎧-≤⎪⎪=⎨⎪-+≥⎪⎩. 【点睛】本题考查绝对值数列的前n 项和，其难点在于分类讨论.7．128【分析】根据递推公式，构造等差数列，进而求解通项公式.【详解】因为1111,30n n n n a a a a a --=+-= 故1113n n a a --=，故1n a ⎧⎫⎨⎬⎩⎭是首项为1，公差为3的等差数列，则132n n a =-，解得132n a n =-，即可得10128a =. 故答案为：128. 【点睛】本题考查构造数列求解通项公式，属经典题型.8．150【解析】【详解】记110b S =，22010b S S =-，33020b S S =-，44030b S S =-设 q 为{}n a 的公比，则 b 1、b 2、b 3、b 4构成以10r q =为公比的等比数列．于是，()()22301231701101.S b b b b r rr r ==++=++=++ 从而，260r r +-=. 解得2r =或3r =-（舍去）.故()2340101150.S r r r =+++=9．1-3【分析】根据等比数列前n 项和的性质，即可求得参数.【详解】因为数列{}n a 是等比数列，故()1111111n n n a q a a S q q q q -==-+--- 故13n n S k -=+=133n k ⨯+，故可得13k =-. 故答案为：13-.【点睛】本题考查等比数列前n 项和的性质，属数列基础题. 10．()-4-3，【分析】根据1n n na b a +=的通项公式，讨论其单调性，从而找到参数的取值范围. 【详解】由题可知：()111n b n a =+-- 当(),1n a ∈-∞-和()1,a -+∞，该数列单调递减.若10a -≤，则该数列在n N +∈时，单调递减，不可能满足题意；若10a ->，要满足题意，只需：()14,5a -∈解得：()4,3a ∈--.故答案为：()-4-3，. 【点睛】本题考查数列的单调性，要用函数的角度去看待数列的单调性，同时也要注意其与函数不同的点.11．274【分析】利用累加法求得通项公式，再求解n a n ，讨论其最小值. 【详解】因为12n n a a n +-=，故可得：211221n n n n a a a a a a n n ----+-++-=- 解得：215n a n n =-+，则n a n =151n n +- 结合对勾函数的单调性，可知：当n =4时，取得最小值，最小值为：274. 故答案为：274. 【点睛】本题考查累加法求通项公式，涉及数列的最小值，属综合基础题. 12．326【分析】根据递推公式，可得奇数项的值；再根据20182020a a =即可求得偶数项的值.【详解】2()n n a f a +=，故可得211n n a a +=+ 因为11a =，故可得35791235,,,2358a a a a ==== 11813a =，又因为20182020201811a a a ==+ 化简得22018201810a a +-=2018a =或2018a =（舍），又因为2018201611a a =+，解得2016a =，同理201820162014201220a a a a a ====⋯== 则2011a a +=81132+=326故答案为：326. 【点睛】本题考查由递推公式求数列中的某几项，属基础题；关键的解题步骤是，要发现数列的规律. 13．D【分析】根据等比数列的性质，进行判断即可.【详解】取等比数列()1nn a =-，则数列10n n a a ++=不是等比数列，故D 错误；对其它选项，均满足等比数列通项公式的性质.故选：D.【点睛】本题考查等比数列的通项公式，属基础题.14．A【分析】由1100910101,10,a a a >->可得公比的范围，结合数列的单调性，对四个选项进行逐一判断即可.【详解】因为1009101010a a ->，故可得2201711a q >，因为11a >，故可得0q >，因为10091010101a a -<-，若1q >，则1009a 和1010a 均大于1，与已知矛盾，故()0,1q ∈，因此10091a >，10101a <，数列{}n a 是个递减数列.对①，因为数列是个减数列，且10101a <，故20221a <，故①错误；对②，2100910111010a a a =，因为10101a <，故100910111a a <，故②正确；对③，101010091010T T a =⨯，因为11a >，数列是减数列，且10091a >，10101a <故n T 中最大值为1009T ，故③错误；对④，()10092018100910101T a a =>，()2019201910101T a =<，故1n T >成立的最大自然数n 是2018，故④错误.综上所述，只有1个正确.故选：A .【点睛】本题综合考查等比数列的下标和性质，以及数列的单调性，属中档题.15．D【分析】由题意结合等比数列的性质、增减性可得“1q >”与“数列{}n a 为单调递增数列”之间的推出关系，由充分条件和必要条件的概念即可得解.【详解】在等比数列{}n a 中，当10a <，1q >时，数列{}n a 为单调递减数列，则充分性不成立；当10a <，01q <<时，满足数列{}n a 为单调递增数列，此时1q >不成立，则必要性不成立；综上所述，“1q >”是“数列{}n a 为单调递增数列”的既不充分也不必要条件. 故选：D ．【点睛】本题考查了等比数列的性质和充分条件、必要条件的判定，解决此类问题要从充分性和必要性两方面进行判定，属于基础题.16．D【分析】根据数学归纳法证明的基本过程可得出结论.【详解】在1n k =+时，没有应用n k =时的假设，即从n k =到1n k =+的推理不正确.故选：D.【点睛】本题考查数学归纳法，考查对数学归纳法证明过程的理解，属于基础题.17．（1）16-（2）5n =或6 【分析】（1）根据数列的基本量，列方程进行求解即可；（2）由（1）可用公式求得n S ，根据二次函数的单调性，即可求得.【详解】（1）设该数列的公差为d ，根据题意： ()53162k d +-=-，()15562k k d k -+=- 解得115,6k d ==-. 故该数列的公差为16-. （2）由（1）可知，21111212n S n n =-+ 该二次函数对称轴为：5.5，又n Z ∈故当5n =或者6n =时，n S 取得最大值.【点睛】本题考查用基本量求解等差数列的通项公式，以及求等差数列前n 项和的最值.18．1231n n k -=⋅-，n *∈N【分析】先由数列{}n k a 前三项成等比数列，得出1a 与d 的等量关系，并求出数列{}n k a 的公比，利用等差数列和等比数列的通项公式得出关于n k 的等式，解出即可.【详解】由题意可知，1a 、5a 、17a 成等比数列，则25117a a a =，即()()2111416a d a a d +=+，化简得21181616a d d a d +=，0d ≠，12a d ∴=，()()()11211n a a n d d n d n d ∴=+-=+-=+，56a d ∴=，等比数列{}n k a 的公比为51632a d q a d===，111132323n n n n k k a a d d ---∴=⋅=⋅=⋅，由等差数列的通项公式得()1n k n a k d =+，1123n n k -∴+=⋅，解得1231n n k -=⋅-.因此，数列{}n k 的通项公式为1231n n k -=⋅-，k *∈N .【点睛】本题考查等差数列的通项公式，同时也考查了等比数列的定义，在解题时要根据条件确定出等差数列首项和公差之间的等量关系，并利用等差数列和等比数列的通项公式建立等式求解，考查计算能力，属于中等题.19．（1）()554144n n f n a b ⎡⎤⎛⎫⎛⎫=--⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦（2）会；8年后【分析】（1）根据前三年木材存量，归纳出解析式，再用数学归纳法进行证明即可；（2）根据（1）中所求函数关系式，结合参考数据，解不等式即可.【详解】（1）1年后，木材存量()514f a b =-， 2年后，木材存量()25555214444f a b b a b ⎛⎫⎛⎫⎛⎫=--=-+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭3年后，木材存量()3255531444f a b ⎡⎤⎛⎫⎛⎫=-++⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦根据以上数据归纳推理得：()15555514144444n n n n f n a b a b -⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=-+++=--⎢⎥⎢⎥ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦用数学归纳法证明如下：①当1n =时，()514f a b =-，显然成立； ②假设当(),2,n k k k N =≥∈时，()554144k k f k a b ⎡⎤⎛⎫⎛⎫=--⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦成立，则当1n k =+时，()()55551414444k k f k f k b a b b ⎧⎫⎡⎤⎪⎪⎛⎫⎛⎫+=-=---⎢⎥⎨⎬ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎪⎪⎣⎦⎩⎭115554444k k a b b ++⎡⎤⎛⎫⎛⎫=---⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦ 11554144k k a b ++⎡⎤⎛⎫⎛⎫=--⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦ 即证，当n N +∈时，()554144n n f n a b ⎡⎤⎛⎫⎛⎫=--⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦（2）当1972b a =时，若该地区今后发生水土流失，则木材存量必须小于79a 则551974144729n n a a a ⎡⎤⎛⎫⎛⎫--⨯<⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦，解得554n ⎛⎫> ⎪⎝⎭ 两边取对数得554nlg lg > 即()1257522132lg lg n lg lg lg ->=≈-- 故：经过8年后，该地区就会发生水土流失.【点睛】本题考查归纳推理，以及用数学归纳法证明，涉及不等式的求解，属函数应用综合题. 20．（1）见解析；（2）()121334n nn T +-+=. 【解析】试题分析：(1)利用()12n n n a S S n -=-≥化简,易得结论;(2)结合(1)可得3n n c n =⋅,利用错位相减法,结合等比数列的前n 项和公式求解即可.试题解析：(1)证明:由题意得()()11322n n n n n a S S a a n --=-=-≥, ∴13n n a a -=,∴()132n n a n a -=≥, 又()111312S a a =-=,解得13a =,∴数列{}n a 是首项为3,公比为3的等比数列.(2)由(1)得3n n a =,则3n n a =,∴3n n n n c a b n ==⋅,设()1231132333133n n n T n n -=⋅+⋅+⋅++-⋅+⋅,()23413132333133n n n T n n +=⋅+⋅+⋅++-⋅+⋅.∴()12311313233333313nn n n n T n n ++--=++++-⋅=-⋅-.∴()121334n n n T +-+=. 点睛：用错位相减法求和应注意的问题(1)要善于识别题目类型，特别是等比数列公比为负数的情形；(2)在写出“S n ”与“qS n ”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“S n －qS n ”的表达式；(3)在应用错位相减法求和时，若等比数列的公比为参数，应分公比等于1和不等于1两种情况求解.21．（1）12343456,,,4567b b b b ====；（2）23n n b n +=+，证明见详解；（3）1a < 【分析】（1）通过赋值，结合已知条件，即可求得；（2）根据数列的规律，进行归纳总结，再遵循数学归纳法的证明过程即可证明；（3）先求n S ，将问题转换为恒成立问题，再求最值即可.【详解】（1）()()()121112n n n n n n n n n b b b b a a a b b +===--+-12n b =- 因为1113,44a b ==，故234456,,567b b b === （2）由（1）猜想23n n b n +=+ ①当1n =时，134b =，显然成立 ②假设当()1,n k k k N +=≥∈时成立，即23k k b k +=+则当1n k =+时，1113(1)2224(1)323k k k k b k b k k k ++++====+-+++-+即证当1n k =+时候，猜想成立；综上所述：23n n b n +=+对任意正整数都成立. （3）因为113n n a b n =-=+，故： 12231111114556(3)(4)444(4)n n n n S a a a a a a n n n n +=++⋯+=+++=-=⨯⨯++++ 22(1)(36)8443(3)(4)n n an n a n a n aS b n n n n +-+---=-=++++ 若4n n aS b <对于*n N ∈恒成立，则只需满足2(1)(36)80a n a n -+--<恒成立即可当1a =时，380n --<恒成立满足题意；当1a >时，显然不可能成立；当1a <时，对称轴3231102121a a a -⎛⎫-⋅=--< ⎪--⎝⎭故()f n 在[)1,+∞单调递减，故()()1max f n f ==2(1)(36)8(1)(36)8a n a n a a -+--=-+-- 4150a =-<解得154a <，又1a <，故当1a <时，满足题意.综上所述，(],1a ∈-∞时，4n n aS b <对于*n N ∈恒成立.【点睛】本题考查数学归纳法证明通项公式，涉及裂项求和，恒成立问题的转化，属综合题.。

2020-2021年上海市上外附中高二上期末

2020学年第一学期上外附中高二年级数学期末考试试卷一、填空题（共14小题，每题3分，满分42分）1.计算：212 (i)n nn→∞+++=_________. 【答案】122.已知正项等比数列{}n a 中，34,13231=+=a a a a ，则=+++∞→)...(lim 21n n a a a _________.【答案】923.在行列式31214053--a 中，元素a 的代数余子式的值为_________.【答案】2-4.线性方程组的增广矩阵为⎪⎪⎭⎫⎝⎛+m m m m 124，且此方程组无解，则实数=m _________.【答案】2m =- 5.若矩阵⎪⎪⎭⎫ ⎝⎛=4321A 与⎪⎪⎭⎫ ⎝⎛=724k B 满足BA AB =,则实数=k _________. 【答案】36.若无穷等比数列{}n a 的各项和为21，则首项1a 的取值范围是_________. 【答案】11(0,)(,1)227.若向量)7,8(),0,2(--==b a ,则在方向上的投影为_________. 【答案】8-8.若数列{}n a 满足831-=+n n a a ，且61=a ，则数列{}n a 的通项公式为=n a _________. 【答案】1234n -⋅+9.若向量(1,2)a =与(21,3)b t t =-的夹角为锐角，则实数t 的取值范围是_________. 【答案】1(,2)(2,)8+∞10.计算：=+-+)10002020lim(22n n n n _________. 【答案】51011.在矩形ABCD 中，2,2==BC AB ,点E 为BC 的中点，点F 在边CD 上，若2=⋅AF AB ,则=⋅BF AE _________.12.若公差为d 的等差数列{}n a 的前n 项和为n S ，满足0,01312<>S S ,且283=a ,则公差d 的取值范围是_________. 【答案】(8,7)--13.已知ABC ∆中，3BAC π∠=，点D 为线段AB 的中点，点P 为线段CD 上一点，且满足1()3AP t AC AB t R =+∈，若ABC ∆的面积为2，则AP 的最小值为________【解析】本题为上海市2020届浦东新区高三下二模11题.因为P 为CD 上一点，设DP λDC =，则()(1)AP AD DP AD λDC AD λAC AD λAC λAD =+=+=+-=+-，又1233AP t AC AB t AC AD =+=+，由平面向量分解定理可得213λtλ=⎧⎪⎨-=⎪⎩，从而求得13t =；由1||sin 23ABC πS AB AC ∆==‖||||6AB AC =， 2222111||||||2||||cos 3393πAP AC AB AC AB AC AB ⎡⎤=+=++⎢⎥⎣⎦，211||[2||||||||]||||293AP AC AB AC AB AC AB ≥+==，当且仅当||||AC AB =时取等号，所以||AP14.已知数列{}n a 是共有k 个项的有限数列，且满足11(2,3,...,1),n n nna a n k a +-=-=-若1224,51,0k a a a ===，则___k =【答案】因为11(2,,1)n n nna a n k a +-=-=⋯-，所以11,(2,,1)n n n n a a a a n n k +--=-=⋯-，所以32212a a a a -=-，43323a a a a -=-，54434a a a a -=-，，112(1)k k k k a a a a k ----=--，将以上不等式相加，并代入1224,51,0k a a a ===得224500k k --=，解得50k =或49k =-（舍去），所以50k =. 二、选择题（34=12⨯分分）15.若ABC ∆所在平面内一点P 满足0PA PB PC ++=,则P 是ABC ∆的（ C ） A.内心 B.外心 C.重心 D.垂心16.已知ABC ∆中，点D 满足23BD DC =,若(,)AD AB AC R λμλμ=+∈，则μ的值为（ A ）A.23 B.32 C.25 D.3517.设{}n a 是各项为正数的无穷数列，i A 是边长为i a 、1i a +的矩形的周长（1,2,i =⋅⋅⋅），则“数列{}n A 为等差数列”的充要条件是（ D ） A.{}n a 是等差数列B.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅或242,,,,n a a a ⋅⋅⋅⋅⋅⋅是等差数列C.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅和242,,,,n a a a ⋅⋅⋅⋅⋅⋅都是等差数列D.1321,,,,n a a a -⋅⋅⋅⋅⋅⋅和242,,,,n a a a ⋅⋅⋅⋅⋅⋅都是等差数列，且公差相同18.已知数列{}n a 的前n 项和为n S ，且对任意正整数n 都有13n n a S +=，则下列关于{}n a 的论断中正确的是（ C ）A.一定是等差数列B.一定是等比数列C.可能是等差数列，但不会是等比数列D.可能是等比数列，但不会是等差数列三、解答题（6+8+10+10+12=46分） 19.（本题满分6分）用行列式解关于,x y 的方程组22(1)321x my m x y m +=⎧⎨-+=+⎩.【解析】226(3)(2)13mD m m m m m ==-++=--+- 2226(2-3)(2)213x mD m m m m m ==--+=-++2224121y D m m m ==+-+（1）当0D ≠时，即2,3m m ≠-≠，方程由唯一解，23322m x m y m -⎧=⎪⎪-⎨⎪=⎪-⎩；（2）当2m =-时，0x y D D D ===，方程组有无穷多组解; （3）当3m =时，0,0,0x y D D D =≠≠，方程组无解20.（本题满分8分，第（1）小题3分，第（2）小题5分）已知两个平面向量a 与b 的夹角为3π，且1,2,a b ==记3,2m a b n ta b =-=+. （1）若m n ⊥，求实数t 的值；（2）若2t =，求m 与n 的夹角. 【答案】（1）2t =；（2）1arccos 7.21.（本题满分10分，第（1）小题5分，第（2）小题5分）（1）已知等差数列{}n a 满足39a =，且6732a a +=，若数列21n n a a +⎧⎫⎨⎬⎩⎭的前n 项和为n S ，求lim n n S →∞的值.（2）已知数列{}n b 的前n 项和n T 满足3+23(*)n n T b n N =∈，若t R +∈,求215limn n nn n b t b t ++→∞++的值. 【答案】（1）335；（2）03t <<时，215lim n n n n n b t b t++→∞++=3； 3t =时，215lim n n n n n b t b t ++→∞++=24； 3t >时，2215lim 5n n n n n b t t b t++→∞+=+.22.（本大题满分10分，第（1）小题3分，第（2）题4分，第（3）小题3分）按照如下规则构造数表：第一行是：2；第二行是：21,23++；即3，5，第三行是：31,33,51,53++++即4，6，6，8；（即从第二行起将上一行的数的每一项各项加1写出，再各项加3写出）23 54 6 6 85 7 7 9 7 9 9 11 ……………………………………若第n 行所有的项的和为n a ．（1）求345,,a a a ；（2）试求1n a +与n a 的递推关系，并据此求出数列{}n a 的通项公式；（3）求数列{}n a 的前n 项和n S .【解析】（1）3424,64,a a ==5160a =．（2）由题意，第n 行共有12n -项，于是有()()111123222n n n n n n n a a a a --++=+++⨯=+ 等式两边同除12n +，得11122n nn na a ++=+，即2n n a ⎧⎫⎨⎬⎩⎭为等差数列，公差为1，首项为112a = 所以2n n a n =，即()*2n n a n n N =⨯∈．（3）23122232...2nn S n =⨯+⨯+⨯++⨯ （1）23412122232...(1)22n n n S n n +=⨯+⨯+⨯++-⨯+⨯ （2）由（1）-（2）得23111222...22222nn n n n S n n +++-=++++-⨯=--⨯所以1(1)22n n S n +=-⨯+.23.（本大题满分12分，第（1）小题3分，第（2）题4分，第（3）小题5分）我们把一系列向量()1,2,,i a i n =按次序排成一列，称之为向量列，记作{}n a ，已知向量列{}n a 满足：()1,11=a ，()()11111,,2n n n n n n n a x y x y x y ----==-+()2n ≥．（1）证明：数列{}n a 是等比数列；（2）设2log n n n c a a =⋅，问数列{}n c 中是否存在最小项？若存在，求出最小项；若不存在，请说明理由；（3）设n θ表示向量1n a -与n a 间的夹角，若2n n n b θπ=，对于任意正整数n ，不等式()211log 122a n ab +>-恒成立，求实数a 的范围【解析】（1）()()22221111111122222n n n n n n n n a x y x y x y a -------=-++=+= 所以数列{}n a 是等比数列；（2）因为122222n nn a --⎛⎫== ⎪ ⎪⎭，所以22222nn n c --=⋅，假设{}n c 中的第n项最小，由12c =，20c =，210.c c ≤< 当3n ≥时，有0n c <，又由1n n c c +≤可得()()212222122222n nn n -+--+-⋅≤⋅，即12221n n --≥-，22112n n -⎛⎫≥ ⎪-⎝⎭. 2670n n -+≥，3n ≥+3n ≤（舍），所以5n ≥. 即有567c c c <<<；由1n n c c +≥，得35n ≤≤，又210c c ≤<，所以541c c c <<<；故数列{}n c 中存在最小项，最小项是325322c -=-⋅（3）因为112cos 2n n n n na a a a --⋅==⋅θ，所以4n =πθ，所以24n nb =，不等式化为：()32221log 121222a n n n +++>-++对任意正整数n 恒成立.设222122n T n n n=++++. 又因为()122222021*******n n T T n n n n n +-=+-=->+++++，数列{}n T 单调递增所以()1min 1n T T ==，要使不等式恒成立，只要()311log 122a >-，因为120a ->，所以102a <<，212a a -> ，所以01a <<，所以，使不等式对于任意正整数恒成立的a 的取值范围是()1.。

2020届上海外国语大学附属浦东外国语学校高三英语二模试题及答案解析

2020届上海外国语大学附属浦东外国语学校高三英语二模试题及答案解析第一部分阅读（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AA medical capsule robot is a small,often pill-sized device that can do planned movement inside the body after being swallowed or surgically inserted. Most models use wireless electronics or magnets or a combination of the two to control the movement of the capsule. Such devices have been equipped with cameras to allow observation and diagnosis, with sensors that “feel,” and even with mechanical needles that administer drugs.But in practice, Biomechatronics engineer Pietro Valdastri has found that developing capsule models from scratch (从头开始) is costly, time-consuming and requires advanced skills. “The problem was we had to do them from scratch every time,” said Valdastri in an interview. “And other research groups were redeveloping those same modules from scratch, which didn’t make sense.”Since most of the capsules have the same parts of components: a microprocessor, communication submodules, an energy source, sensors, and actuators (致动器), Valdastri and his team made the modular platform in which the pieceswork in concertand can be interchanged with ease. They also developed a flexible board on which the component parts are snapped in like Legos. The board can be folded to fit the body of the capsule, down to about 14 mm. Additionally, they compiled (编译) a library of components that designers could choose from, enabling hundreds of different combinations. They arranged it all in a free online system. Designers can take the available designs or adapt them to their specific needs.“Instead of redeveloping all the modules from scratch, people with limited technological experience can use our modules to build their own capsule robots in clinical use and focus on their innovation,” Valdastri said.Now, the team has designed a capsule equipped with a surgical clip to stop internal bleeding. Researchers at Scotland’s Royal Infirmary of Edinburg have also expressed interest in using the system to make a crawling capsule that takes images of the colon(结肠). One research group, led by professors at the Institute of Digestive Disease of the Chinese University of HongKong, is making a swimming capsule equipped with a camera that pushes itself through the stomach.One limitation of Valdastri’s system is that it’s only for designing models. Researchers can confirm their hypotheses (假设) and do first design using the platform, but will need to move to a custom approach to developtheir capsules further and make them practical for clinical use.1. According to the passage, Valdastri and his team created the platform to ________.A. adopt the latest technologiesB. make their robots dream come trueC. help build specialized capsule robotsD. do preciser observation and diagnosis2. What does the underlined phrase “work in concert” mean in Para.3?A. Perform live.B. Run independently.C. Act in a cooperative way.D. Carry on step by step.3. What can be learnt from the passage?A. Valdastri’s system can’t provide a complete capsule creation.B. The modular platform is more useful than a custom approach.C. The capsules can move in human’s body automatically.D. It costs more to module the capsules on the board.BDid you know that horses talk? Well, they do, and you can lean to understand “horse talk” if you pay close attention to the horses you see.When horses live in the wild, other animals try to eat them, so a lot of horse talk is about staying alive. Even now, when most horses live on farms, they watch for danger. For this reason, never walk behind a horse. If you surprise it, the horse might mistake you for a mountain lion or wolf and give a dangerous kick.By watching the ears of a horse, you can get clues to what it's hearing. A horse can tum each ear in a different direction. For a wild horse, this trick is important for survival. The horse can hear something sneaking up behind it while also checking out a threatening noise in front. When a horse lets its ears down, it's feeling safe and relaxed. If horses becomeisolated, they neigh, or “whinny,” calling for company. They're saying, “Where are you? I'm over here!” If a horse snorts(哼) while holding its head high and staring at something, it's saying, “That looks dangerous. Get ready to run!” When two horses meet, they put their noses together and smell each other's breath. It's their way of asking, “Are you a friend?” Horses nicker,too. Nickering is a quiet sort of sound. This friendly noise means they're feeling secure and saying, “Clad to see you.”In the wild, horses live in herds, with all members watching for danger. In a herd, only one horse is the leader, the “boss hoss”. The “boss hoss” is usually an older female. She watches for threats and teaches younger horseshow to behave. However, others may want her job. When that happens, she pins her ears back against her head and may even bite or kick to get challengers to back off. She's using body language to say, “Hey, I'm in charge here!” All horses know that the one who makes others move is the leader. Horses relate to people that way, too.Horses have a language of their own. Now you know a bit of what they might be saying.4. What is the general idea of this passage?A. Horses can talk with their owners.B. You can know a bit of horses' language.C. Horses can “talk” in their own way.D. Other animals can also learn language.5. What does the underlined word “isolated” in Paragraph 3 mean?A. Lonely.B. Glad.C. Sad.D. Frightened.6. What will a horse do when he wants to know whether another horse is a friend?A. He will let his ears down.B. He and another horse will put their noses together and smell each other's breath.C They both will hold their heads high and stare at something.D. He will give another horse a dangerous kick.7. How will the “boss hoss” deal with her challengers?A. By warning or fighting.B. By watching for threats.C. By teaching younger horses how to behave.D. By relating to people.CMove over, helicopter parents. “Snowplow (扫雪机) parents” are the newest reflection of an intensive (强化的) parenting style that can include parents booking their adult children haircuts, texting their college kids to wake them up so they don’t sleep through a test, and even calling their kids’ employers.Helicopter parenting the practice of wandering anxiously near one’s children, monitoring their every activity, is so 20th century. Some rich mothers and fathers now are more like snowplows: machines moving ahead, clearing any difficulties in their children’s path to success, so they don’t have to suffer failure, frustration (挫折) or lose opportunities.It starts early, when parents get on wait lists for excellent preschools before their babies are born and try to make sure their kids never do anything that may frustrate them. It gets more intense when school starts: running forgotten homework to school or calling a coach to request that their children make the team.Rich parents may have more time and money to devote to making sure their children don’t ever meet with failure, but it’s not only rich parents practicing snowplow parenting. This intensive parenting has become the most welcome way to raise children, regardless of income, education, or race.Yes, it’s a parent’s job to support the children, and to use their adult wisdom to prepare for the future when their children aren’t mature enough to do so. That’s why parents hide certain toys from babies to avoid getting angry or take away a teenager’s car keys until he finishes his college applications.But snowplow parents can take it too far, some experts say. If children have never faced a difficulty, what happens when they get into the real world?“Solving problems, taking risks and overcoming frustration are key life skills,” many child development experts say, “and if parents don’t let their children experience failure, the children don’t acquire them.”8. What do we know about snowplow parenting?A. It appeared before helicopter parenting.B. It costs parents less than helicopter parenting.C. It was a typical phenomenon of the 20th century.D. It provides more than enough services for children.9. What is mainly discussed about snowplow parenting in Paragraph 4?A. Its cost.B. Its benefits.C. Its popularity.D. Its ending.10. Why does the author mention parents’ taking away car keys?A. To show teenagers are no better than babies.B. To advise teenagers not to treat their cars as toys.C. To advise parents not to buy cars for their teenagers.D. To show it’s appropriate to help children when necessary.11. What’s the possible result of snowplow parenting according to the experts?A. Children lacking problem-solving ability in reality.B. Children mastering more key life skills than parents.C. Children gaining great success in every aspect of life.D. Children meeting no problems or frustration after growing up.DHave you ever done something that was really dangerous just because you thought it was safe？Maybe you did a dangerous trick on your bicycle or skateboard because you were wearing a helmet and thought you couldn’t get hurt. The psychology(心理) of this sort of behavior is called the Peltzman Effect, named after Sam Peltzman, professor of economics at the University of Chicago. Peltzman believes that those moments when people think they are the safest are the times when they act most dangerously.Peltzman said that people drove more dangerously when they wore seat belts(安全带) . Driving a large four-wheel drive vehicle has a similar effect on drivers’ behavior. Because drivers of large vehicles sit up higher and can see better, they feel they can make better judgments when they drive. They are better protected in accidents,so they act more dangerously. This makes driving morehazardousto other drivers.The Peltzman Effect isn’t just limited to driving. In 1972, the American Food and Drug Administration (FDA) passed a law requiring child safety caps on most medicine bottles. The safety caps were designed to prevent children from accidentally taking the medicine, especially painkillers such as aspirin. Requiring safety caps sounded like a great idea, but there was an unexpected side effect. Because the safety caps are so hardto take off, some people leave them off altogether．Worse, some parents leave the bottles where kids can reach them because they feel that it is safe because of the cap. A study on the Peltzman Effect showed that more than 3,500 children have been harmedby aspirin because of the safety caps.The Peltzman Effect describes how we’re likely to take more risks and act more dangerously when we feel safest. What’s more, the effects of these behaviors can be quite different from what we expect.12. What is the Peltzman Effect？A. People behave less safely when they feel safe．B. People feel safest when they are under protection．C. Something that seems dangerous turns out to be safe．D. People who act dangerously are likely to be together．13. What does the underlined word“hazardous”in Paragraph 2 mean？A. Interesting．B. Expensive．C. Dangerous．D. Important．14. Medicine bottles with safety caps ________．A. are required throughout the worldB. meet the demands of the Peltzman EffectC. sell well in the worldD. are not completely safe15. What would be the best title for the text？A. Unsafe Safety MeasuresB. Types of Decision MakingC. People’s Fear of Taking RisksD. Different Behaviors of People in Danger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。

2020-2021学年上海师范大学附属外国语中学高三英语二模试卷及参考答案

2020-2021学年上海师范大学附属外国语中学高三英语二模试卷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AAre you looking for some new and exciting places to take your kids to? Try some of these places.★Visit art museums.They offer a variety of activities to excite your kids' interest. Many offer workshops for making hand-made pieces, traveling exhibits, book signings by children's favorite writer, and even musical performances and other arts.★Head to a natural history museum.This is where kids can discover the past from dinosaur(恐龙) models to rock collections and pictures of stars in the sky. Also, ask what kind of workshops and educational programs are prepared for kids and any special events that are coming up.★Go to a Youtheater.Look for one in your area offering plays for child and family visitors.Pre-show play shops are conducted by area artists and educators where kids can discover the secret about performing arts Puppet(木偶)making and stage make-up are just a couple of the special offerings you might find.★Tryhands-on science.Visit one of the many hands-on science museums aroundthe country.These science play-lands are great fun for kids and grown-ups alike. They'll keep your child mentally and physically active the whole day through while pushing buttons, experimenting, and building. When everyone is tired, enjoy a fun family science show, commonly found in these museums.1.If a child is interested in the universe, he probably will visit _____.A.a YoutheaterB.an art museumC.a hands-on science museumD.a natural history museum2.What does "hands-on science" mean in the last paragraph?A.Science games designed by kids.B.Learning science by doing things.C.A show of kids' science work.D.Reading science books.3.Where does this text probably come from?A.A museum guide.B.A tourist map.C.A science textbook.D.A news report.BAt the age of 50, Nina Schoen expects to have a long lifeahead of her, but has thought a lot about death—and why people are so reluctant to talk about it: “It’s going to happen to all of us,” she says, “but it should be a more positive experience than the fear we put into it.”When she first heard about a new end-of-life process that turns the body into compost (堆肥), “I was really moved by the idea,” says Schoen, who became one of the first to reserve a spot with a Seattle-based company called Recompose, the county’s first funeral home to offer human composting.Last year Recompose began transforming bodies to soil, more formally known as natural organic reduction. Before that, end-of-life options in the U.S. were limited to burial or cremation (火化), both of which come with environmental costs—U.S. cremations alone dump 1.7 billion pounds of carbon dioxide into theatmosphere every year.Katrina Spade pioneering the composting movement has spent a decade developing the process in hopes of offering people a greener option for death care. “I wondered, ‘What if we had a choice that helps the planet rather than harms it?’” Spade tells PEOPLE. “To know that the last gesture you’ll make will be gentle and beneficial and it just feels like the right thing to do.”After she had her own two sons, she began to wonder what she might do with her body after death. A friend who knew her interest in the topic reminded her that farmers sometimes compost the body of cows, and thatsparkedan idea for her theory: “If you can compost a cow, you can probably compost a human,” she thought, and she set about designing a facility to do just that.“This is about giving people another choice,” Spade says. “At first, people react with shock—‘You really can do that?’ But so many people today are looking at their impact on the Earth. This is a popular thing because when you die, you can give back to the planet.”4. How do people react when it comes to death according to paragraph 1?A. They are unwilling to comment.B. They can face it without fear.C. They feel it a positive experience.D. They would like to compost their bodies.5. What can we know about the company Recompose?A. Its CEO is Katrina Spade.B. It is located in Seattle.C. It was founded to resist cremation.D. It has spent 10 years composting bodies.6. What does the underlined word “sparked” probably mean?A. Changed.B. Compromised.C. Quitted.D. Inspired.7. What message does the author seem to convey in the text?A. A little things in our life can bring in big outcomes.B. We human beings should do all we can to help the earth.C. Composting is so popular that we should reserve a spot soon.D. We should reject burial because of its harm to environment.CWhat do you think of 80s pop music? Do the names George Michael, Madonna and Michael Jackson sound familiar? Well, these are just some of the names that were well-known in the music scene of the 80s and early 90s. The 80s pop musicscene was an important step to the popularity (普及) of present-day music. A new wave in the music scene was introduced, which made such music styles as punk rock, rap music and the MTV popular. Although it was an end to the old 60s and 70s styles, it was also the beginning of something big. The popularity of music videos meant that artists now replaced their guitar-based music with visual displays. A new wave of artists came on the scene and the entire industry developed quickly.The most famous 80s pop music video is Michael Jackson’s Thriller. Introduced in 1982, few people can forget the video not only because of its never-be-foreseen images, but also because of the popularity it received. Think of how 80s pop music changed the lives of people who grew up in the 80s. Ask a young man today to tell you the names of the “New Kids on the Block” and he will start talking about the neighbor kids who just moved in. These are not the answers you might have heard in the 80s. Though today’s young men do not recognize how cool 80s pop music was, most people will always remember it for what it was and these are happy memories they will always love.Some of the 80s pop music legends (传奇人物) include Madonna, U2, AeroSmith and of course the King of Pop Michael Jackson. Let’s not forget Prince, Tina Turner, Phil Collins and Motown’s Lionel Ritchie. Some of these musicians played music that has stood the test of time. Undoubtedly, the 80s pop music scene will live on for many more years to come.8. What is the text mainly about?A. The characters of 80s pop music.B. What made 80s pop music popular.C. 80s pop music’s steps to popularity.D. The effects of 80s pop music.9. 80s pop music mainly includes the following styles EXCEPT ________.A. guitar-based musicB. the MTVC. rap musicD. punk rock10. Michael Jackson’s Thriller impressed people so deeply mainly because ________.A. it changed the lives of peopleB. he sang it in a special styleC. it was made into a music videoD. it left people with happy memories11. The purpose of the last paragraph is to tell readers that ________.A. 80s pop music is and will remain popularB. 80s pop music has many faultsC. 80s pop music is now out of dateD. we shouldn’t forget the great musicians of the 80sDPhotographer Rebecca Douglas has always been fascinated by the night sky. Her love for stars has taken theU.K.resident on “star walking” trips toIcelandand into theArctic, where she steps out onto darkened trails to capture twinkling stars and glowing planets in her images.Hiking at night isn’t uncommon. Plenty of people hike after dark to get to campsites or watch the sunrise from a mountaintop. Star walking goes a step further by blending hiking with stargazing. Rather than heading to an observatory or setting up a telescope in your backyard,star walking takes you on a brief journey to look at the stars from different viewpoints.Whether you’re in the mountainside or by the lake with stars reflecting on the water, star walking is often much more dynamic than traditional stargazing.What’s more, star walking is good for you. There are plenty of studies that show the health benefits of being in nature. Spending at least two hours a week outdoors, particularly while engaging in what involves “effortless attention”, can decrease blood pressure, heart rate, and stress levels.So how does an aspiring star walker get started? It doesn’t take much more than a sturdy pair of boots.While telescopes and binoculars obviously have their uses, people are encouraged to start with naked-eye stargazing. Using only the eyes allows one to get lost in the infinite expanse of space and lets the mind go.It is advised that one read up on the night sky before heading out. Free mobile apps, such as Star Walk 2, can help identify celestial bodies（天体）and are easy to use—simply point your phone at the sky to reveal a map. Websites like Sky & Telescope and NASA’s Space Place cover the basics, have in-depth explanations on stars, and offer advice on equipment. Space Place posts monthly skywatching updates, so you can plan outings around events such as meteor showers.In the United States, national parks are great options for inexperienced night hikers. Many offer guided outings that explain the importance of protecting night skies. Those with a good number of walks under their belt may want to try GlacierNational Park in Montana.If national parks and other dark-sky designated areas are out of range, check local astronomy clubs and observatories for guided sessions. Sites like the International Astronomical Union are useful for finding resources by area.At the end of the day, the best advice is to take it slow and enjoy the journey.“With all of the chaos（混乱）that’s happening around us, the one constant has been the night sky,” explains Douglas, who hasbeen exploring popular places nearby, long after the crowds have gone home for the day. “Walking is quite a mindful way of looking up and being reminded that, although everything feels so different, some things are still the same.”12. According to the passage, star walking refers to ________.A. going to an observatoryB. looking at stars in mountainsC. combining hiking and stargazingD. setting up a telescope in the backyard13. One of the reasons why people go on a star walking is that _______.A. it’s easier to identify celestial bodiesB. it is good for physical and mental healthC. they can enjoy the journey without crowdsD. they can raise awareness of protecting night skies14. According to the passage, a star walking beginner is advised to _______.A. prepare a pair of strong bootsB. start by observing with telescopesC. join an astronomy club or an observatoryD. find guided outings with the help of mobile apps15. The main purpose of the passage is to ________.A. excite people’s interest in star walkingB. recommend some places for star walkingC. explain the health benefits of star walkingD. introduce the preparations for star walking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。

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2021年1月上海外国语大学附属浦东外国语学校高二第一学期第二次检测数学试卷总分：150分，时间：120分钟一、填空题（本题共12小题，前6题每题4分，后6题每题5分，共计54分）1．抛物线24y x =的准线方程为___________． 2．经过点1235A B （，）,（，）的直线的点方向式方程为___________．3．圆2214x y -+=（）关于直线10x y -+=对称的圆的方程为___________． 4．过点12A -（，）且与原点距离最大的直线的倾斜角为___________． 5．已知椭圆2212516x y +=的左右焦点为12,F F ，直线l 经过点1F 且与椭圆交于A ，B 两点，则2ABF 的周长为___________．6．已知双曲线2222100x y a b a b-=>>（，）的一条渐近线方程是2y x =，它的一个焦点与抛物线220y x =的焦点相同，则此双曲线的方程是___________．7．过点(11P ，）作一条直线与椭圆22142x y +=交于M ，N 两点，且P 恰为线段MN 的中点，则该直线的方程为___________．8．设定点32A （，），点P 是抛物线24y x =上任意一点，当PA PF +的值最小时，点P 的坐标为___________． 9．椭圆221259x y +=与双曲线221115x y -=的焦点相同，请将此命题推广到一般情形，使已知命题成为推广后命题的一个特例___________．10．已知直线/与直线43160x y -+=垂直，且它被圆2224200x y x y +-+-=所截得的线段长为8，则直线l 的方程为___________．11．若关于x 的方程x m +=m 的取值范围为___________． 12．记椭圆221441x ny n +=+围成的区域（含边界）为123n n Ω=⋯（，，），当点,x y （）分别在12,,ΩΩ⋯上时，x y +的最大值分别是12,M M ⋯，则lim n n M →+∞=___________．二、选择题（本大题共4小题，每小题5分，共计20分）13．若m R ∈则方程221x y m +=所表示的曲线一定不是（）A ．直线B ．圆C ．抛物线D ．双曲线 14．已知P 为双曲线2214x y -=上一点，12,F F 为两焦点，若1260F PF ∠=︒，则12F PF 的面积为（） A． BC．3 D．315．设点M ，N 均在双曲线22:143x y C -=上运动，12,F F 是双曲线C 的左、右焦点，则122MF MF MN +-的最小值为（）A． B ．4 C． D ．以上都不对16．已知111,P a b （）与222,P a b （）是直线1y kx =+（k 为常数）上两个不同的点，则关于x 和y 的方程组112211a x b x a x b y +=⎧⎨+=⎩的解的情况是（） A ．无论12,,k P P 如何，总是无解 B ．无论12,,k P P 如何，总有唯一解C ．存在12,,k P P ，使之恰有两解D ．存在12,,k P P ，使之有无穷多解三、解答题（本大题共5小题，共计14+14+14+16+18=76分）17．已知两条直线1120l t x y t -+-=:（）和2:40l x ty t ++-=，当t 为何值时，1l 与2l （1）平行？（2）重合？（3）垂直？18．为了进一步开发上海市旅游项目，大治河上的旧桥梁全邮改造成新的桥梁以便旅游船只可以顺利通行．某一新建圆形拱桥的跨度为40米，拱顶离水面距高10米现有一艘平顶型旅游船只，船宽为8米，船在水面以上的高度为9米，请问该船只能否安全从桥下通过．19．已知直线:2l y kx =-与抛物线2:20C x py p =->（）交于A ，B 两点，O 为坐标原点， 412OA OB +=--（，）．（1）求直线/和抛物线C 的方程；（2）抛物线上一动点P 从A 到B 运动时，求ABP 面积的最大值．20．在平面直角坐标系xOy 中，已知向量1,1m x y n x y =+=-（）,（,），且而4m n +=，动点,A x y （）的轨迹为C ．（1）求曲线C 的标准方程；（2）若M ，N 是曲线C 上关于x 轴对称的任意两点，设40P -（，），连接PM 交曲线C 于另一点E ，求证：直线NE 过定点B ，并求出点B 的坐标；（3）在（2）的条件下，过点B 的直线交曲线C 于S ，T 两点，求OS OT ⋅的取值范围．21、如图，已知双曲线221:12x C y -=，曲线2:1C y x =+．P 是平面内一点，若存在过点P 的直线与12C C 、都有公共点，则称P 为“12C C -型点”（1）在正确证明1C 的左焦点是“12C C -型点”时，要使用一条过该焦点的直线，试写出一条这样的直线的方程（不要求验证）；（2）设直线y kx =与2C 有公共点，求证1k >，进而证明原点不是“12C C -型点”；（3）求证：圆2212x y +=内的点都不是“12C C -型点”．上海外国语大学附属浦东外国语学校高二第一学期第二次检测数学试卷总分：150分，时间：120分钟一、填空题（本题共12小题，前6题每题4分，后6题每题5分，共计54分）1．【答案】1x =- 2．【答案】1223x y --= 3．【答案】22124x y ++-=（）（） 4．【答案】1arctan 25．【答案】20 6．【答案】221520x y -=7．【答案】230x y +-= 【提示】点差法 8．【答案】()1,29．【答案】椭图2222116x y a a +=-与双曲线2222116x y k k -=-的焦点相同 10．【答案】34200x y ++=或34100x y +-=11．【答案】[)1,1-⋃ 12．【答案】()max x y +==lim n n n M →∞==二、选择题（本大题共4小题，每小题5分，共计20分）13．C 14．B 15．B 16．B三、解答题（本大题共5小题，共计14+14+14+16+18=76分）17．【答案】（1）1t =- （2）2t = （3）13t = 18．【答案】能【提示】建系19．（1）由222y kx x py =-⎧⎨=-⎩得2240x pkx p +-=，设1122,,A x y B x y （）,（），则21212122424x x pk y y k x x pk +=-+=+-=--，（），因为21212224412OA OB x x y y pk pk +=++=---=--（，）（，）（，），所以2242412pk pk -=-⎧⎨--=-⎩，解得12p k =⎧⎨=⎩，所以直线l 的方程为22y x =-，抛物线C 的方程为22x y =-；（2）设2,2m P m ⎛⎫- ⎪⎝⎭，则点P 到直线AB 的距离d =，1122ABP S AB d =⋅=))222244828m m m m=+-=++-=+-易得2m⎡⎤∈-⎣⎦，故当2m=-时，ABPS取得最大值20．（1）设1210,10F F-（，）（，），则|124m n AF AF+=+=，故点,A x y（）的轨迹为以121010F F-（，）,（，）为焦点的椭圆，且24,1a c==，所以2a b===，所以曲线C的标准方程为22143x y+=；（2）由题意得直线PM的斜率存在，设直线PM的方程为4y k x=+（），由224143y k xx y=+⎧⎪⎨+=⎪⎩（），得2222433264120k x k x k+++-=（），设1222,M x y E x y（，）,（），则11N x y-（），且22111222326412,4343k kx y x xk k-+=-=++．根据对称性，直线NE过的定点B在x轴上，直线NE的方程为()212221y yy y x xx x+-=--，令0y=，得()()()()()2212112211222121214444y x x x k x x k xx y x yx xy y y y k x k x-⋅++⋅++=-==+++++，即()()()2212122212128241282418323224k kx x x xxx x k k--++===-++-++，故直线NE过定点1,0B-（）；（3）当直线ST的斜率不存在时，其方程为1x=-，此时此时3351,,1,,224S T OS OT⎛⎫⎛⎫---⋅=-⎪ ⎪⎝⎭⎝⎭,4当直线ST 的斜率存在时，设直线ST 的方程为1y m x =+（），由()221143y m x x y ⎧=+⎪⎨+=⎪⎩得22224384120m x m x m +++-=（），设3344,,,S x y T x y （）（），则223434228412,4343m m x x x x m m -+=-=++．所以2343434341)1)OS OT x x y y x x m x x ⋅=+=+++（(()()()2222343422512533514,4344443m m x x m x x m m m +⎡⎫=++++=-=--∈--⎪⎢++⎣⎭，综上所述，OS OT ⋅的取值范围是54,4⎡⎤--⎢⎥⎣⎦． 21．（1）由于1C的左焦点为F （），过F的直线x =1C交于2⎛± ⎝⎭，与2C交于()1，故1C ，的左焦点是“12C C -型点”；（2）由于直线y kx =与2C 有公共点，则1y kx y x =⎧⎪⎨=+⎪⎩，即(11k x -=），若方程组有解，必须1k >，若直线y kx =与1C 有公共点，则2222y kxx y =⎧⎨-=⎩，即22122k x -=（）．若方程组有解，必须21<2k ，故直线y kx =至多与曲线1C 和2C 中的一条有交点，故原点不是“12C C -型点”；（3）显然过圆2212x y +=内一点的直线l 若与曲线1C 有公共点，则斜率必存在，根据对称性，不妨设直线l 的斜率存在且与曲线2C 交于点,1)0t t t +≥（（），则直线:1l y t k x t -+=-（）（），即(1)0kx y t kt -++-=，若直线l 与固2212x y +=2，化简得()()2211<12t tk k +-+，若直线l 与曲线1C 有交点，则22112y kx kt t x y =-++⎧⎪⎨-=⎪⎩ 得()22212(1)1102k x k t kt x t kt -++-++-+=（），若212k ≠，则()222214(1)41102k t kt k t kt ⎛⎫⎡⎤=+-+-+-+≥ ⎪⎣⎦⎝⎭．即22121t kt k +-≥-（），所以()2221211<12k t kt k -≤+-+（）．解得21k <，但此时()()2210,111,1<12t t k k ≥+-≥+⎡⎤⎣⎦与()()2211<12t tk k +-+矛盾，故无解，若212k =，则221112t kt k +-<+（）（）也矛盾，无解，综上，若直线l 与圆2212x y +=内部有公共点，则此直线不可能同时与曲线1C 和2C 有公共点，即圆2212x y +=内的点都不是“12C C -型点”．。