河南省开封市五县联考2019-2020学年高二下学期期中考试英语试题PDF高清版
2019-2020学年河南省开封市五县联考高一下学期期中考试数学试题(解析版)
【解析】利用诱导公式可得 ，再利用正弦函数的单调性即可求出 ，由 ，即可求解.
【详解】
，因为 ，所以 ，
又 ，所以 .
故答案为：
【点睛】
本题考查了三角函数的单调性比较函数值大小，需熟记三角函数的性质，属于基础题.
15．已知 ，若 ，则 ________.
【答案】
【解析】根据同角公式、二倍角的正、余弦公式计算出 和 ，再根据两角和的正弦公式可求得结果.
又 ，∴ ，
即 与 的夹角为 .
【点睛】
本题考查数量积的概念，数量积、模长、夹角的坐标运算，解题关键在于熟记数量积概念及坐标运算公式，考查计算能力，属于中档题.
20．已知函数 的图象的一部分如图所示.
（1）求函数 的解析式；
（2）当 时，求函数 的最值.
【答案】（1） ；（2）最小值 ；最大值 .
【解析】（1）由函数 的图象，求得 ， ，得到 ，再由 ，求得 ，即可得到函数 的解析式；
【详解】
因为 ，所以 ，
所以 ， ，
则
.
故答案为： .
【点睛】
本题考查了同角公式、二倍角的正、余弦公式，考查了两角和的正弦公式，属于基础题.
16．如图在平行四边形ABCD中，E，F分别为边CD，AD的中点连接AE，BF交于点G.若 ，则 ________.
【答案】
【解析】延长CD，BF交于点H，可得 ， ，从而 ，根据 即可求解.
（1）记 为 ， 的周长为 ，求 关于 的函数关系式；
（2）如何设计才能使 的周长最小？
【答案】（1） ；（2） 时， 的周长 最小.
【解析】（1）利用 ，得到 ，再利用 和 ，求得 和 ，进而得到周长的表达式；
河南省开封市五县联考2019-2020学年高二下学期期末考试生物试题(图片版)
高二生物参考答案
一、选择题（每小题1.5分，共45分）
二、非选择题
31．(1)DNA和蛋白质（1分）DNA、RNA，蛋白质（2分）氨基酸（1分）
(2)DNA、蛋白质，磷脂（2分）
(3)DNA→RNA→蛋白质（2分）
32．(1)叶绿体基质（1分）ATP和[H](二者缺一不可)（1分）C5（1分）
(2)抑制（2分）
(3)光呼吸可消耗过剩的能量(或A TP和[H])（2分）
(4)光呼吸消耗ATP，有氧呼吸产生ATP（2分）
33．（1）栖息空间和食物条件（2分）
（2）黑光灯诱捕（1分）记名计算（1分）
（3）7（1分）热带雨林中湿度较大，且植物位于草本层，接受的光照强度低
（2分，分别针对湿度和光照强度两方面各1分）
34．（1）有机物氧化分解（或有氧呼吸过程或呼吸过程）（2分）淋巴因子（2分）病毒（或COVID-19）的繁殖或对人体细胞的黏附（2分）
（2）胞吞（1分）控制合成ACE2的基因在不同细胞中的表达存在差异（2分）
（3）新型冠状病毒肺炎康复患者血浆含有抗2019-nCoV抗体，康复患者的血浆输入重型和危重型患者血液后，抗2019-nCoV抗体与2019-nCoV结合抑制病原体的增殖和对人体细胞的黏附。
（2分）
35．(1)YyRr×yyRr或YyRr×Yyrr （2分）2\5 （2分）
(2)让不耐盐的水稻植株相互受粉，选择F1中耐盐的水稻植株自交，鉴定并统计F2表现型及比例。
（3分）
若F2表现型及比例为耐盐植株∶不耐盐植株＝9∶7，说明水稻植株的耐盐性是由位于非同源染色体上的两对基因控制的。
（3分）。
河南省焦作市普通高中2024_2025学年高二英语下学期期中试题
河南省焦作市一般中学2024-2025学年高二英语下学期期中试题考生留意：1.答题前，考生务必将自己的姓名、考生号填写在试卷和答题卡上，并将考生号条形码粘贴在答题卡上的指定位置。
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第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。
每段对话后有一个小题，从题中所的A、B、C三个选项中选出最佳选项。
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每段对话仅读一遍。
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1.Where does Tom come from?A.China.B.France.C.Singapore.2.When does the man want to see Mr.Smith next Monday?AAround 9:00 a.m. B.Around 4:00 p.m. C.Around 5:00 p.m.3.What music does the woman like when she is sad?A.Pop music.B.Classical music.C.Rock music.4.What is the weather like now?A.Snowy.B.Rainy.C.Cloudy.5.What do we know about the woman?A.She is very free these days.B.She likes Yoga very much.C.She has no hobby at all.其次节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。
2019-2020学年河南省开封市五县联考高二下学期期末考试数学(理)试题(解析版)
2019-2020学年河南省开封市五县联考高二下学期期末考试数学（理）试题一、单选题1．已知集合{}2A x x =<，{}2,0,1,2B =-，则A B =（ ）A ．{}0,1B ．{}1,0,1-C ．2,0,1,2 D ．1,0,1,2【答案】A【解析】解绝对值不等式对集合A 进行化简，即可求出两个集合的交集. 【详解】解：解2x <得22x -<<，所以{}0,1A B =.故选:A. 【点睛】本题考查了集合交集的求解，属于基础题. 2．命题“2,x x R e x ∀∈>”的否定是（ ） A ．2,x x R e x ∀∈≤ B ．0200,x x R ex ∃∈>C ．0200,x x R e x ∃∈≤D ．2,x x R e x ∀∈<【答案】C【解析】根据全称命题的否定的性质进行求解即可. 【详解】命题“2,x x R e x ∀∈>”的否定是0200,x x R ex ∃∈≤.故选：C 【点睛】本题考查了全称命题的否定，属于基本题.3．下列函数中，定义域为R 且在R 单调递增的函数是（ ） A ．xy e -= B ．3y x =C ．12y x =D ．y x =【答案】B【解析】利用指数函数的基本性质可判断A 选项；利用幂函数的基本性质可判断B 、C 选项；利用绝对值函数的基本性质可判断D 选项. 【详解】对于A 选项，函数1xxy e e -⎛⎫== ⎪⎝⎭的定义域为R ，且该函数在R 上单调递减； 对于B 选项，函数3y x =的定义域为R ，且该函数在R 上单调递增； 对于C 选项，函数12y x ==[)0,+∞；对于D 选项，函数y x =的定义域为R ，且,0,0x x y x x x ≥⎧==⎨-<⎩，该函数在R 上不单调. 故选：B. 【点睛】本题考查利用函数解析式求解函数的定义域以及判断函数的单调性，属于基础题. 4．“1x <”是“ln(1)0x +<”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【答案】B【解析】根据对数不等式的性质解得()ln 10x +<，利用充分条件和必要条件的定义进行判断. 【详解】∵ln （x +1）＜0⇔0＜x +1＜1⇔﹣1＜x ＜0，∴﹣1＜x ＜0 1x ⇒<，但1x <时，不一定有﹣1＜x ＜0，如x=-3， 故“1x <”是“()ln 10x +<”的必要不充分条件， 故选B. 【点睛】本题主要考查充分条件和必要条件的应用，考查对数不等式的性质，属于基础题．5．函数()22,026lg ,0x x f x x x x ⎧-≤=⎨-+>⎩的零点的个数为（ ）A ．0B ．1C ．2D ．3【答案】C【解析】当0x ≤时，直接解方程()0f x =得x =0x >时，利用零点存在性定理判断函数零点的个数，两类情况合起来即可得选项. 【详解】当0x ≤时，直接解方程()0f x =，即220x -=，解得：x = 当0x >时，()26lg f x x x =-+为增函数，(1)40,(10)150f f =-<=>，所以()f x 在(1,10)有一零点，即()f x 在(0,)+∞有一个零点， 综上，函数()f x 有两个零点. 故选：C. 【点睛】本题考查函数的零点个数，考查零点存在性定理的应用，是基础题.6．某地区空气质量监测资料表明，一天的空气质量为优良的概率是0.8，连续两天为优良的概率是0.6，已知某天的空气质量为优良，则随后一天的空气质量为优良的概率是（ ） A ．0.45 B ．0.6C ．0.75D ．0.8【答案】C【解析】设随后一天的空气质量为优良的概率是p ，利用相互独立事件概率乘法公式能求出结果. 【详解】设随后一天的空气质量为优良的概率是p ，则0.80.6p =， 解得0.75p =， 故选：C. 【点睛】本题考查概率的求法，解题时要认真审题，注意相互独立事件概率乘法公式的合理运用，属于基础题.7．将5位同学分别保送到北京大学，上海交通大学，中山大学这3所大学就读，每所大学至少保送1人，则不同的保送方法共有（ ） A ．150种 B ．180种C ．240种D ．540种【答案】A【解析】每所大学至少保送一人，可以分类来解，当5名学生分成2，2，1时，共有12C 52C 32A 33，当5名学生分成3，1，1时，共有12C 5312C A 33，根据分类计数原理得到结果．【详解】当5名学生分成2，2，1或3，1，1两种形式，当5名学生分成2，2，1时，共有12C52C32A33=90种结果，当5名学生分成3，1，1时，共有12C5312C A33=60种结果，∴根据分类计数原理知共有90+60=150种，故选A．【点睛】求解排列、组合问题常用的解题方法：(1)元素相邻的排列问题——“捆邦法”；(2)元素相间的排列问题——“插空法”；(3)元素有顺序限制的排列问题——“除序法”；(4)带有“含”与“不含”“至多”“至少”的排列组合问题——“间接法”；（5）“在”与“不在”问题——“分类法”. 8．某地有A，B，C，D四人先后感染了传染性肺炎，其中只有A到过疫区，B确实是由A感染的.对于C难以判断是由A或是由B感染的，于是假定他是由A和B感染的概率都是12.同样也假定D由A，B和C感染的概率都是13，在这种假定下，B，C，D中都是由A感染的概率是（）A．16B．13C．12D．23【答案】A【解析】利用相互独立事件概率乘法公式能求出B，C，D中都是由A感染的概率. 【详解】∵某地有A，B，C，D四人先后感染了传染性肺炎，其中只有A到过疫区，B确实是由A感染的．对于C难以判断是由A或是由B感染的，于是假定他是由A和B感染的概率都是12，同样也假定D由A，B和C感染的概率都是13，∴在这种假定下，B，C，D中都是由A感染的概率为：1211136P=⨯⨯=，故选：A.【点睛】本题考查概率的求法，考查相互独立事件概率乘法公式等基础知识，考查运算求解能力，属于基础题.9．设3log 2a =，5log 3b =，23c =，则（ ） A ．a c b << B ．a b c <<C ．b c a <<D ．c a b <<【答案】A【解析】分别将a ,b 改写为331log 23a =，351log 33b =，再利用单调性比较即可. 【详解】 因为333112log 2log 9333ac =<==，355112log 3log 25333b c =>==， 所以a c b <<. 故选：A . 【点晴】本题考查对数式大小的比较，考查学生转化与化归的思想，是一道中档题. 10．函数()21cos 1xf x x e ⎛⎫=-⎪+⎝⎭图象的大致形状是（ ） A ． B ．C ．D ．【答案】B【解析】利用奇偶性可排除A 、C ；再由(1)f 的正负可排除D. 【详解】()21e 1cos cos 1e 1e x x xf x x x -⎛⎫=-= ⎪++⎝⎭，()1e cos()1e x x f x x ----=-=+e 1cos e 1x x x -+ ()f x =-，故()f x 为奇函数，排除选项A 、C ；又1e(1)cos101ef -=<+，排除D ，选B. 故选：B. 【点睛】本题考查根据解析式选择图象问题，在做这类题时，一般要结合函数的奇偶性、单调性、对称性以及特殊点函数值来判断，是一道基础题.11．已知命题:p 关于x 的方程210x ax ++=没有实根；命题:0q x ∀≥，20x a ->.若p ⌝和p q ∧都是假命题，则实数a 的取值范围是（ ） A ．()(),21,-∞-⋃+∞ B ．(]2,1- C ．(]1,2 D ．[)1,2【答案】D【解析】计算出当命题p 为真命题时实数a 的取值范围，以及当命题q 为真命题时实数a 的取值范围，由题意可知p 真q 假，进而可求得实数a 的取值范围.【详解】若命题p 为真命题，则240a ∆=-<，解得22a -<<； 若命题q 为真命题，0x ∀≥，20x a ->，则()min21xa <=.由于p ⌝和p q ∧都是假命题，则p 真q 假，所以221a a -<<⎧⎨≥⎩，可得12a ≤<.因此，实数a 的取值范围是[)1,2. 故选：D. 【点睛】本题考查利用复合命题、全称命题的真假求参数，考查计算能力，属于中等题.12．已知函数13,(1,0](){,()()1,1]1,(0,1]x f x g x f x mx m x x x -∈-==---+∈且在（内有且仅有两个不同的零点，则实数的取值范围是A ．91(,2](0,]42--⋃ B ．111(,2](0,]42--⋃ C ．92(,2](0,]43--⋃D ．112(,2](0,]43--⋃【答案】A 【解析】【详解】 【分析】试题分析：令，分别作出与的图像如下，由图像知是过定点的一条直线，当直线绕着定点转动时，与图像产生不同的交点.当直线在轴和直线及切线和直线之间时，与图像产生两个交点，此时或故答案选A .【考点】1.函数零点的应用；2.数形结合思想的应用.二、填空题13．设样本数据122018,,,a a a 的方差是0.01，如果有()1021,2,,2018i i b a i =-=，那么数据122018,,,b b b 的标准差为_________.【答案】1【解析】直接根据标准差的性质即可得结果. 【详解】 数据122018,,,b b b 的方差为2100.011⨯=，所以标准差为1，故答案为：1. 【点睛】本题主要考查方差、标准差，考查学生的运算能力和数学核心素养，属于基础题. 14．二项式6(2x x展开式中含2x 项的系数是________. 【答案】192-【解析】试题分析：通项为()6116322166212rrr rrr r r T C x x C x ----+⎛⎫⎛⎫=-=- ⎪⎪⎝⎭⎝⎭，所以1r =，系数为()151612192C -=-.【考点】二项式展开式. 15．设函数()()21ln 11f x x x=+-+，则使得()()12f x f x >-成立的x 的取值范围为_____________. 【答案】113xx ⎧⎫<<⎨⎬⎩⎭【解析】根据条件判断函数的奇偶性和单调性，结合函数的奇偶性和单调性的性质将不等式进行转化求解即可. 【详解】()()()()2211ln 1ln 111f x x x f x x x -=+--=+-=++，则()f x 是偶函数， 当0x ≥函数()f x 为增函数， 则()()12f x f x >-等价与()()12fx f x >-，所以12x x >-，平方得22144x x x -+>，所以23410x x -+<，所以113x <<，即不等式的解集为113x x ⎧⎫<<⎨⎬⎩⎭， 故答案为：113x x ⎧⎫<<⎨⎬⎩⎭. 【点睛】本题主要考查不等式的求解，结合条件判断函数的奇偶性和单调性是解决本题的关键，难度中等.16．定义在R 上的偶函数()f x 满足()()2f x f x +=-，且在[]2,0-上是减函数，下面是关于()f x 的判断：（1）()0f 是函数的最大值；（2）()f x 的图像关于点()1,0P 对称；（3）()f x 在[]2,3上是减函数；（4）()f x 的图像关于直线2x =对称.其中正确的命题的序号是____________（注：把你认为正确的命题的序号都填上） 【答案】（2）（3）（4）【解析】（1）利用定义在R 上的偶函数()f x 在[]2,0-上是减函数，即可判断；（2）根据偶函数的定义和条件()()2f x f x +=-，即可判断； （3）利用函数的周期为4，()f x 在[-2，0]上是减函数，即可判断；（4）利用()()()22f x f x f x -+=--=+，可得()f x 的图象关于直线2x =对称，即可判断. 【详解】（1）∵定义在R 上的偶函数()f x 在[]2,0-上是减函数， 故()()20f f ->，()0f 不可能是函数的最大值，故错； （2）由定义在R 上的偶函数()f x 得()()f x f x -=， 又()()2f x f x +=-，故()()20f x f x ++-=，即图象关于()10，对称，故正确； （3）由于()()2f x f x +=-，则()()()42f x f x f x +=-+=， 故()f x 为周期函数，且4为它的一个周期，由在[20]-，上是减函数，可得()f x 在[2]4，上是减函数，故正确； （4）由于()()2f x f x +=-，则()()()42f x f x f x +=-+=， 又()()f x f x -=，故()()4f x f x +=-， 即图象关于直线2x =对称，故正确． 故答案为：（2）（3）（4）． 【点睛】本题主要考查了抽象函数的函数的奇偶性、周期性和对称性，考查了转化思想，属于中档题.三、解答题17．为了解某地区某种产品的年产量x （单位：吨）对价格y （单位：千元/吨）和利润z 的影响，对近五年该农产品的年产量和价格统计如下表：（1）求y 关于x 的线性回归方程ˆy bxa =+； （2）若每吨该农产品的成本为3千元，假设该农产品可全部卖出，预测当年产量为多少时，年利润z 取到最大值？（保留两位小数）参考公式：()()()1122211ˆn niii ii i nniii i x x y y x y nx ybx x xnx====---==--∑∑∑∑，ˆa y bx=-，562.7i iix y=∑.【答案】（1） 1.238.69y x =-+；（2）2.31吨.【解析】（1）计算出x 和y ，将表格中的数据代入最小二乘法公式求得b 和a 的值，由此可求得回归直线方程；（2）求得z 关于x 的函数解析式为21.23 5.69z x x =-+，利用二次函数的基本性质可求得该函数取得最大值时对应的x 值，由此可得出结论. 【详解】（1）由表格中的数据可得1234535x ++++==，7.0 6.5 5.5 3.8 2.255y ++++==，5162.7i ii x y==∑，52155i i x ==∑，所以，2152251562.7535ˆ 1.2355535i ii i i x y x ybx x==--⨯⨯===--⨯-∑∑，()5 1.2338.69a ∴=--⨯=， 因此，回归直线方程为 1.238.69y x =-+；（2）年利润()28.69 1.233 1.23 5.69z x x x x x =--=-+.当 5.692.312 1.23x =≈⨯时，z 有最大值，因此当 2.31x =吨，年利润z 最大.【点睛】本题考查利用最小二乘法求回归直线方程，同时也考查了利用回归直线方程对总体进行估计，考查计算能力，属于中等题.18．已知函数2()1f x ax bx =++(,a b 均为实数),R x ∈,(),0,()(),0.f x x F x f x x >⎧=⎨-<⎩(1)若(1)0f -=,且函数()f x 的值域为[0,)+∞,求()F x 的解析式;(2)在(1)的条件下,当[2,2]x ∈-时,()()g x f x kx =-是单调函数,求实数k 的取值范围;(3)设0mn <,0m n +>,0a >,且()f x 为偶函数,判断()()F m F n +是否大于零,并说明理由.【答案】(1)22(1),0,()(1),0.x x F x x x ⎧+>=⎨-+<⎩(2)(,2][6,)-∞-+∞(3)大于零.见解析 【解析】（1）由题得10a b -+=①，240a b -=，解方程即得解；（2）由题得222k --或222k -,解不等式得解；（3）先求出()F x 的解析式，再求出()22()()F m F n a m n +=-即得证.【详解】 解:(1)(1)0f -=,10a b ∴-+=①又函数()f x 的值域为[0,)+∞,0a ∴≠.由22424b a b y a x a a -⎛⎫=++ ⎪⎝⎭,知2404a b a -=, 即240a b -=②.解①②,得1a =,2b =.22()21(1)f x x x x ∴=++=+.22(1),0,()(1),0.x x F x x x ⎧+>∴=⎨-+<⎩(2)由(1)得22222(2)()()21(2)1124k k g x f x kx x x kx x k x x --⎛⎫=-=++-=+-+=++- ⎪⎝⎭.∵当[2,2]x ∈-时,()()g x f x kx =-是单调函数,222k -∴-或222k -,即2k -或6k , 故实数k 的取值范围为(,2][6,)-∞-+∞. (3)()()F m F n +大于零.理由如下:()f x 为偶函数,2()1f x ax =+,()221,0,()1,0.ax x F x ax x ⎧+>⎪∴=⎨-+<⎪⎩ 不妨设m n >,则0n <.由0m n +>,得0m n >->,||||m n ∴>-.又0a >, ()()()2222()()()()110F m F n f m f n an an a m n ∴+=-=+-+=->,()()F m F n ∴+大于零.【点睛】本题主要考查函数的奇偶性的应用，考查二次函数的图象和性质的应用，意在考查学生对这些知识的理解掌握水平.19．从某企业生产的某种产品中抽取500件，测量这些产品的一项质量指标值，由测量结果得如图的频率分布直方图：（1）求这500件产品质量指标值的样本平均数x 和样本方差2s （同一组数据用该区间的中点值作代表）；（2）由频率分布直方图可以认为，这种产品的质量指标值Z 服从正态分布()2,N μσ，其中μ近似为样本平均数x ，2σ近似为样本方差2s . （i ）利用该正态分布，求()175.6224.4P Z <<；（ii ）某用户从该企业购买了200件这种产品，记X 表示这200件产品中质量指标值位于区间()175.6,224.4的产品件数，利用（i ）的结果，求EX . 15012.2≈.若()2~,Z Nμσ，则()0.6827P Z μσμσ-<<+=，()220.9545P Z μσμσ-<<+=.【答案】（1）200x =，2150s =；（2）（i ）0.9545；（ii ）190.9. 【解析】（1）由平均值和方差的计算公式可计算； （2）（i ）()~200,150Z N ，根据参考值可求出； （ii ）可知()~200,0.9545X B ，即可计算EX . 【详解】（1）抽取产品的质量指标值的样本平均数x 和样本方差2s 分别为1700.021800.091900.222000.332100.242200.082300.02200x =⨯+⨯+⨯+⨯+⨯+⨯+⨯=，()()()2222222300.02200.09100.2200.33100.24200.08300.02150s =-⨯+-⨯+-⨯+⨯+⨯+⨯+⨯=；（2）（i ）由（1）知，()~200,150Z N ，从而()()175.6224.4200212.2200212.20.9545P Z P Z <<=-⨯<<+⨯=；（ii ）由（i ）知，一件产品的质量指标值位于区间()175.6,224.4的概率为0.9545， 依题意知()~200,0.9545X B ， 所以()2000.9545190.9E X =⨯=. 【点睛】本题考查由频率分布直方图估算平均值和方差，考查正态分布的概率计算，考查二项分布的均值，属于中档题.20．设函数()xxf x a ka -=-（0a >且1a ≠）是定义域为R 的奇函数.（1）若()10f >，试求不等式()()2240f x x f x ++->的解集；（2）若()312f =，且()()224x xg x a a f x -=+-，求()g x 在[)1,+∞上的最小值. 【答案】（1）()(),41,-∞-+∞；（2）2-. 【解析】由奇函数的定义求得1k =.（1）由()10f >求得1a >，判断出函数()y f x =为R 上的增函数，将所求不等式变形为()()224f x x f x +>-，可得出2340x x +->，解此不等式即可；（2）由()312f =可解得2a =，令3222x xt -=-≥，可得出()242g x t t =-+，设()242p t t t =-+，利用二次函数的基本性质求得函数()242p t t t =-+在区间3,2⎡⎫+∞⎪⎢⎣⎭上的最小值，进而可得解. 【详解】因为函数()xxf x a ka -=-是定义域为R 上的奇函数，则()()f x f x -=-，即()xx x x aka a ka ---=--，整理得()()10x x k a a --+=，由题意可知，等式()()10xxk a a--+=对任意的x ∈R 恒成立，10k ∴-=，解得1k =.（1）()10f >，210a ∴->，又0a >且1a ≠，1a ∴>，由于函数xy a =在R 上为增函数，函数x y a -=在R 上为减函数. 所以，函数()xxf x a ka -=-为R 上的增函数，由()()2240f x x f x ++->可得()()()2244f x x f x f x +>--=-，224x x x ∴+>-，即2340x x +->，解得4x <-或1x >.因此，原不等式的解集为()(),41,-∞-+∞；（2）()1312f a a =-=，整理得22320a a --=， 0a >且1a ≠，解得2a =.()()()()22222422224222xxxxxx x x g x ----∴=+--=---+，令()()221x xt x -=-≥，()242p t tt =-+.由于22x x t -=-在[)1,+∞上为增函数，所以32t ≥， ()()224222p t t t t ∴=-+=--，所以，当2t =时，()min 2p t =-，即函数()y g x =有最小值2-. 【点睛】本题考查利用函数的奇偶性求参数，利用函数的奇偶性与单调性求解函数不等式，同时也考查了指数型复合函数在区间上的最值，考查分析问题和解决问题的能力，属于中等题.21．已知直线:10L x y +-=与抛物线2yx 交于A ，B 两点.求：（1）点()1,2M -到A ，B 两点的距离之积； （2）线段AB 的长.【答案】（1）2；（2.【解析】（1）写出直线的参数方程，代入抛物线方程，可得12t t +=，122t t =-，则可得122MA MB t t ⋅=⋅=；（2）可知12AB t t =-，由t 得几何意义可得. 【详解】（1）因为直线L 过定点M ，且L 的倾斜角为34π，所以它的参数方程是 31cos 432sin 4x t y t ππ⎧=-+⎪⎪⎨⎪=+⎪⎩（t 为参数），即1222x t y t ⎧=--⎪⎪⎨⎪=+⎪⎩（t 为参数），把它代入抛物线的方程，得220t +-=，12t t +=，122t t =-，122MA MB t t ∴⋅=⋅=；（2）由参数t 的几何意义得12t t AB =-==.【点睛】本题考查直线参数方程的应用，考查参数的几何意义，属于中档题. 22．设函数()2123f x x x =-+-，x ∈R . （1）解不等式()5f x ≤； （2）若()()2g x f x m=-的定义域为R ，求实数m 的取值范围.【答案】（1）1944x x ⎧⎫-≤≤⎨⎬⎩⎭；（2）(),2-∞. 【解析】（1）根据x 的范围写出()f x ，即可判断出解集； （2）求出()f x 的最小值，令m 小于最小值即可. 【详解】（1）()344,2132,22144,2x x f x x x x ⎧-≥⎪⎪⎪=<<⎨⎪⎪-≤⎪⎩，令445x -=得94x =；令445x -=得14x =-，所以原不等式的解集是1944x x ⎧⎫-≤≤⎨⎬⎩⎭；（2）由（1）得()f x 的最小值是2，要使函数有意义，只需2m <， 即实数m 的取值范围是(),2-∞. 【点睛】本题考查含绝对值不等式的解法，属于基础题. 23．在平面直角坐标系xOy 中，曲线1C 的参数方程为22cos 2sin x y θθ=+⎧⎨=⎩（θ为参数），以原点为极点，x 轴的非负半轴为极轴，建立极坐标系，曲线2C 的极坐标方程为2224cos 4sin ραα=+. （1）求曲线1C 的极坐标方程以及曲线2C 的直角坐标方程；（2）若直线()0:L R θθρ=∈与曲线1C 、曲线2C 在第一象限交于P ，Q 两点，且2OP OQ =，点M 的坐标为()1,0，求MPQ 的面积.【答案】（1）4cos ρθ=，2214x y +=；（2）3. 【解析】（1）消去参数得到直角坐标方程，再cos x ρθ=，sin y ρθ=即可得曲线1C 的极坐标方程以及曲线2C 的直角坐标方程；（2）用极径的应用和点到直线的距离公式的应用和三角形的面积和公式的应用求出结果. 【详解】 （1）1C ：由22cos 2sin x y θθ=+⎧⎨=⎩，得()2224x y -+=即2240x y x +-=，将cos x ρθ=，sin y ρθ=代入方程得4cos ρθ=.由2224cos 4sin ραα=+得222:44C x y +=，即2214x y +=. （2）由已知得：4cos OP θ=，OQ =，并且2OP OQ =得2223sin cos sin θθθ=因为sin 0θ≠，所以21cos 3θ=，得cos θ=，sin 3θ=，从而直线的斜率k =0y -=，点()1,0MPQ ==，123MPQS ==.【点睛】本题考查的知识要点：参数方程极坐标方程和直角坐标方程之间的转换，极径的应用，点到直线的距离公式的应用，属于中档题.24．（1）已知：,a b R +∈，4a b +=，证明：111a b +≥； （2）已知,,,9a b c R a b c +∈++=，证明：1111a b c++≥，并类比上面的结论，写出推广后的一般性结论（不需证明）.【答案】（1）证明见解析；（2）证明略，推广：若212n a a a n +++=，则121111na a a ++≥. 【解析】试题分析：（1）可应用柯西不等式也可用基本不等式证明；（2）仿照（1）凑成应用柯西不等式的形式()111abc a b c ⎛⎫++++⎪⎝⎭，可得结论，由此可推广成n 个数的形式：若212n a a a n +++=，则121111na a a ++≥. 试题解析：证明：（1）根据柯西不等式：()2114a b a b ⎛⎫++≥= ⎪⎝⎭，∵4a b +=，∴111a b+≥. （2）根据柯西不等式：()21119a b c a b c ⎛⎫++++≥= ⎪⎝⎭，∵ 9a b c ++=，∴1111a b c++≥可以推广：若212n a a a n +++=，则121111na a a ++≥. 【考点】 柯西不等式．。
开封市五县联考2020_2021学年高二英语下学期期末考试试题(含答案)
河南省开封市五县联考2020-2021学年高二下学期期末考试英语试题考生注意：1.本试卷分选择题和非选择题两部分。
满分150分，110分钟。
2.考生作答时，请将答案答在答题卡上。
选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；非选择题请用直径0.5毫米墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效，在试题卷、草稿纸上作答无效。
第一部分阅读理解（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和D四个选项中，选出最佳选项。
AIn 1872, the first national park was set up in the US. If s the Yellowstone National Park, which was listed as a World Heritage Site in 1978. Today the country is home to 59 national parks. Here is a list of some of the best received ones.Great Smoky Mountains National ParkThis park is a part of the Smoky Mountains. It lies at the border between the US states of North Carolina and Tennessee. It's the country's most visited national park. It was listed as a World Heritage Site in 1983. Tourists visiting the park can enjoy different activities like biking, hiking, horseback riding, fishing, etc.Grand Canyon National ParkAlmost every international tourist to the US visits this national park, making it the second most visited national park in the US. It lies in northwestern Arizona and is home to the Grand Canyon of the Colorado River. It was listed as a World Heritage Site in 1979. Covering an area of 4,926.08 square km, the park offers many activities to the visitors like driving and walking tours, hiking, biking, etc.Yosemite National ParkThis national park lies in the Sierra Nevada mountain range of Northern California. It’s known for its waterfalls, mountains, lakes, and wildlife. It was listed as a World Heritage Site in 1984. The 3,026.87-square-km-large park is home to different animals such as black bears, red foxes, etc.Cuyahoga Valley National ParkThe national park lies in northeast Ohio. Set up in 2000, it offers a lot of activities like bicycling, hiking, wildlife watching, etc. The waterfalls, caves, hills, farmlands, and more make the experience of visiting the national park a memorable experience.1. Which of the following parks is the first to become a World Heritage Site?A. Great Smoky Mountains National Park.B. Yellowstone National Park.C. Grand Canyon National park.D. Yosemite National Park.2. What do Yosemite National Park and Cuyahoga Valley National Park have in common?A. They both have waterfalls.B. They both lie between two states.C. They both allow visitors to go hunting.D. They both offer biking and hiking activities.3. What does the text mainly intend to introduce?A. Standards of becoming national parks.B. A brief history of American national parks.C. Some most popular national parks in the US.D. Benefits of visiting American national parks.BIn February 2019, my beautiful cat Xena suddenly lost the use of her back legs, and I took her straight to our vet (兽医), Chris. Xena was booked for surgery two days later. The operation was successful, and I went to see her later that day. She seemed tired, but pleased to see me and in much better shape.A few days later, I took her home, armed with medications plus instructions for her care. I had to keep her in a large cage and teach her how to walk again. And no one could tell whether Xena could walk again or not—it was largely up to her. I wasn’t expecting much, but I carried on doing as Chris had instructed.Just a week later, Xena amazed me. Not having managed to support herself on her back legs at all, she suddenly stood up all by herself. Soon she began to walk a little with her back supported. Encouraged, I booked her for some hydrotherapy(水疗), which was helpful for dogs in this situation. Most cats would panic if put in water, but my brave little cat astonished everyone, tolerating neck-deep water while walking on a machine.When I took Xena to Chris, he was delighted with her progress. Over the next few weeks, Xena moved from the cage to a room of her own, with all furniture removed so that she wouldn't run into anything or get tripped. Soon, she could manage a few steps out into our garden, though she still needed some help to walk properly without dragging her legs. Despite still being in pain, she really wanted to walk again. After three months since her surgery, she was able to run, jump and climb stairs... anything she wanted.Xena is now completely back to normal, I’d thought she might be defeated permanently by all that had happened to her. But this courageous little cat simply brushed it all off and got on with living. Many people could learn from Xena.4. From paragraph 1, it can be learned that Xena .A. was fall of energy shortly after the operationB. was given the surgery in her owner’s absenceC. got injured owi ng to her owner’s carelessnessD. received an operation on reaching the hospital5. Which word best describes the author's initial feeling about Xena’s recovery?A. Unsure.B. Inspired.C. Satisfied.D. Hopeless.6. Why did the author remove the furniture from Xena's room?A. To avoid making a mess.B. To allow her to run faster.C. To prevent her getting hurt.D. To make room for her cage.7. In which aspect does Xena set a good example to us?A. Socializing with others.B. Making proper choices.C. Being considerate to partners.D. Dealing with life challenges.CThe icy beauty of the Arctic attracts thousands of visitors every year to see its wonderful wildlife, landscape and local cultures. Visitors can take ships on a voyage along the Arctic Ocean or take flights to cities along the Arctic edge. No matter how one gets there, they should wear warm clothes and get ready to take in the attractions.“Travelling to the Arctic leaves an unforg ettable impression on the visitor. Its vast expanse and the fragility of its environment are two things that really blow away people,” said Cheryl Rosa, the director of the US Arctic Research Commission.Visitors to the Arctic Circle will have a lot of activities to choose from if they want to see all that the region has to offer. Hiking with snowshoes, dog sledding and kayaking are common activities. Visitors with sharp eyes are likely to spot polar bears. To get even closer to the animals of the sea, visitors can go polar snorkeling with seals. Of course, Arctic adventures aren't complete without viewing the wonders of the Arctic Circle’s large glaciers and icebergs as well.One of the most impressive attractions and maybe the most difficult to see is the Northern Lights, one of the seven natural wonders of the world. They are natural lights that glow a brilliant green and light up the horizon. Sometimes they appear as waves that dance across the sky in different colors.The best places to view the Northern lights are in Northern Norway, Sweden’s Abisko National Park, Iceland, America’s Alaska or Canada’s Yukon, according to space, com. Charles Deehr, an expert at the University of Alaska Fairbanks’ Geophysical Institute, recommended planning a trip between winter and spring, especially when there is a new moon.Even though the Arctic is a popular tourist destination, don't forget that people live here too. “It is recommended that visitors go with tour groups that are respectful of Arctic residents(居民) and th eir culture,” said Rosa. “Too many people can disturb the small villages. Finding tour groups that work with local communities is important.”8. The underlined part “blow away” in Paragraph 2 probably means “”.A. defeatB. affectC. moveD. impress9. What can we know about the Northern Lights?A. They only produce green light at night.B. They are natural lights and hard to see.C. They seldom occur between winter and spring.D. The best time to view them is when there is a full moon.10. What can be inferred from the last paragraph?A. Visitors should choose suitable tour groups when visiting.B. Tour groups are not welcome among Arctic residents.C. The Arctic5s population has been rising in recent years.D. Tour groups should get permission from local communities.11. What is the purpose of the author in writing this passage?A. To inform visitors of local culture.B. To recommend some activities to visitors.C. To introduce a popular tourist destination.D. To share a travel experience in the Arctic.DIn the last century, a series of missions have been carried out to explore the moon—Earth’s only natural satellite. Among them, NASA’s Apollo 11 mission was groundbreaking as it succeeded in landing the first humans on the moon on July 20,1969.Several decades later, NASA announced its Artemis program. Named after the Greek goddess of the moon and twin sister of Apollo, the Artemis program will send humans to the moon by the year 2024. And this time, the moon will welcome its first female astronaut.Up until now, only 12 people, all male, have ever walked on the moon. “The last person walked on the Moon in 1972,” Bettina Inclan, NASA communications director said in a statement. “No woman has ever walked on the lunar surface.”Women, of course, have been involved in space projects and made valuable contributions. In 1963, astronaut Valentina Tereshkova from the Soviet Union became the first woman to blast off into space. However, the progress toward women's access to space flight programs has been slow. Women have been held back by various requirements and security concerns. For example, astronauts had to be test pilots with a certain amount of experience. This was a problem as the field was dominated by males at the time. Indeed, the gender bias(偏见) seemed to exist in the early space programs, The New York Times reported.In 1962, NASA wrote to a little girl who wanted to be an astronaut. The letter said, “We have no present plans to employ women on spaceflights because of the degree of scientific and flight training, and the physical characteristics, which are required.”Women have indeed made progress in this particular area, and arguably have advantages over their male colleagues. Women tend to be smaller, which means they use less oxygen and take up less space in small spacecrafts. Also, women usually have greater emotional awareness and communication skills that make them better-suited to long spaceflights.Despite challenges, women have shown that they are more than capable of joining that elite group of men who have gone to the moon. And this will truly be a remarkable moment in history.12. Which one is true about the Artemis program?A. It is named after Apollo.B. It will send 12 people to the moon by 2024.C. It succeeded in landing the first humans on the moon.D. It will send the first female astronaut to the moon.13. The letter to a little girl is mentioned to show .A. Women's space flight programs are in progress.B. Women require more flight training on spaceflights.C. the gender bias seemed to exist in the early space programs.D. female astronauts have advantages over their male colleagues.14. What’s the author5s attitude to sending female astronaut to the moon?A. ambiguous.B. hopeful.C. indifferent.D. negative.15. What’s the best title of the textA. Giant leap for women.B. NASA's Apollo 11 mission.C. The first female astronaut to the moon.D. Who is more suitable for spaceflights—male or female?第二节（共5小题，每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。
2019-2020学年开封市第一中学高三英语下学期期中考试试卷及答案解析
2019-2020学年开封市第一中学高三英语下学期期中考试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ACovid-19 has brought a great deal of trouble for all of us since March 2020. During this time, mobile phones have been the solution for the boredom and restlessness caused from staying indoors. The most downloaded apps on play store 2020 are;TikTokTikTok was the most downloaded app. With over 111.9 million downloads, TikTok has seen a huge growth in 2020, twice more than what it got in 2019. 20% of its total downloads were fromIndiaand around 9. 3% of the total downloads were in theUS.ZoomZoom was the second most installed app in the overall downloads category. With nearly 94. 6 million installs, Zoom is the most used app for online meetings and virtual classrooms. 17% of its downloads were in theUSandIndia. Offices and educational institutes were shut down and to continue working and studying from home, people relied heavily on Zoom for video conferencing and calling.WhatsAppWhatsApp ranked third in overall downloads with more than 100 million downloads. It is one of the most popular and widely used chat applications; WhatsApp also supports communication between international phone networks.FacebookIt ranked fourth in the overall downloaded list. Facebook is the world’s most popular social networking application. Facebook builds technologies that give people the power to connect with friends and family, find communities and grow businesses.1. What do we know about TikTok?A. It is an India-based app.B. It has most users inAmerica.C. It is used for growing business.D. It has doubled its download than in 2019.2. Which app is the best to turn to for online education?A. TikTok.B. Zoom.C. WhatsApp.D. Facebook.3. What function does Facebook probably serve?A. Communication.B. Training.C. Teaching.D. PaymentBMark Twain,the famous American writer,was once traveling in France.He went by trainto Dijon.He was very tired and wanted to sleep.He therefore asked the conductor to wake him up when the train came to Dijon.But first he explained he was a very heavy sleeper,“I may possibly protest(抗议)loudly when you try to wake me up,” he said to the conductor.“But don’t take any notice of what I say.Just put me off the train anyway.”Then Mark Twain went to ter,when he woke up it was night time and the train had reached Paris already.He realized at once that the conductor had forgotten to wake him up at Dijon.He was so angry that he ran to the conductor and began to shout at him.“I have never been so angry in my life,” Mark Twain said.The conductor looked at him calmly(平静地).“You are not half so angry as the American whom I put off the train atDijon,” he said.4. Mark Twain knew that he was a heavy sleeper,so ________.A. he protested loudly to the conductorB. he did not sleep before he arrived inDijonC. he told the conductor to wake him up no matter how loudly he might protestD. he slept lightly that time5. The conductor didn’t wake up Mark Twain atDijonbecause ________.A. he didn’t take Mark Twain’s words seriouslyB. he forgot Mark Twain’s words when the train came toDijonC. he did not want to bear his protestD. he mistook another American traveler for Mark Twain6. The American whom the conductor put off the train ________.A. did not want to get off atDijonB. wanted to get off atParisC. wanted to get off atDijonD. did not want to get off atParis7. Which of the following is TRUE?A. The conductor didn’t take Mark Twain’s words seriously.B. The conductor did take Mark Twain’s words seriously.C. The conductor was a heavy sleeper.D. Mark Twain must get off atParis.CWhen the COVID-19 hit and supermarket shelves were empty, Chris Hall and Stefanny Lowey decided they no longer wanted to rely on others for food. The couple, who live on Pender Island in BritishColumbia, Canada, decided to start a year-long challenge where they wouldn't buy a single thing to eat. Instead they would grow, raise or catch everything—right down to sugar, salt and flour. Now, five months in, they say the challenge has changed their lives.Chris, 38, said, “It has always been something that we have wanted to do. We have had a garden and grown vegetables for a long time already. When the COVID-19 hit, it gave us that extra push that we needed to do it. We were both out of work when we started, and with the reality check of grocery stores running out of items, it gave us even more motivation to see if we could look after ourselves.”The pair spent the months before building a house for chickens, ducks and turkey as well as studying as much as possible to figure out where they would get all the things they needed. Chris adds, “We had to learn so many new things like how to grow mushrooms, process our Stevia plants, and harvest salt from the ocean. We spent a lot of time reading and studying online to figure out all the things we were going to need to do.”Now after five months, they both feel its been going well but Chris admits the first few weeks were difficult. “The first three weeks were very challenging as our bodies adjusted to cutting out coffee, wine and sugar all on the same day,” he says. “After three weeks our energy levels balanced out and our wishes reduced and now we feel great.” Now February has ended. As they come through winter, they feel positive about continuing with this way of living, with their challenge officially ending in August.8. Why did the pair decide to produce foods on their own?A. They were isolated by Pender Island.B. They couldn't afford to buy them because they were out of work.C. They believed it's good for their health.D. They could hardly buy them in shops.9. Which words can be used to describe the couple?A. Rich and generous.B. Helpful and positive.C. Optimistic and self-dependent.D. Motivated and brave.10. What can we learn from the last paragraph?A. Their challenge may last about eleven months in total.B. They were discouraged by the difficulty at first.C. They had difficulty because they wanted more.D. They couldn't adjust their bodies to the hard work after three weeks.11. In which column may you read such a passage?A. Sports.B. Agriculture.C. Lifestyle.D. Business.DAbout 12 years ago, Sandy Cambron noticed her mother, Pearl Walker, had become quiet after she moved into a nursing home for patients of Alzheimer's disease inKentucky.“We tried everything — photos, old stories — but nothing worked,” she said. “It was really hard for everyone to see how she had changed.” Then one day whileSandywas in a toy store, she had an idea: Why not givePearla baby doll so she could feel as if she were caring for something again? And why not give one to all the other care center seniors?As soon asSandygavePearlthe doll, her mother's face lit up. “She started talking again and she never went anywhere without that baby,”Sandysaid. “She took 'baby' to the dining room with her and slept with her in her arms every night. When she passed away a year later, we even buried her with that well-loved baby doll.”In the following 10 years, Sandy and her husband, Wayne Cambron, continued to buy dolls and hand them out to the elderly of care centers near their home every New Year. Now Pearl's Memory Babies is anonprofit (非营利) organization that has contributed more than 300 dolls to old people with Alzheimer's disease at nursing homes since February 2018.Last year,Sandyposted New Year’s photos on Facebook, all of which are about seniors reacting to dolls thatshe and Wayne sent to a local nursing home. The post was shared more than 210,000 times overnight. People gave almost $15,000 online. That helped the group buy many dolls.“The dolls offer treatment and comfort,” said Elise Hinchman, who works at a care center inKentucky, "Someseniors cry when they get a doll. And they always rock and talk to their dolls. People with Alzheimer might lose their memories, but they don't lose their ability to love.”12. How did Pearl change after she moved to a nursing home?A. She hardly talked.B. She lost hope in life.C. She felt bored.D. She forgot everything.13. How didPearlreact after receiving the doll?A. She began to cry.B. She was very delighted.C. She played it with a baby.D. She was unconcerned about it.14. What's paragraph 5 mainly about?A. What Sandy did for her mother.B. People's reaction toSandy's post.C. The rising needs for baby dolls.D. HowSandybecame a celebrity.15. What can be a suitable title for the text?A. Good News for Alzheimer's PatientsB. How to Remove Alzheimer's DiseaseC. Baby Dolls Cheer up Alzheimer's PatientsD. Daughter Helps Mother Recover Memory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。
开封市五县联考2019-2020学年高二下学期期中考试数学(文)试题含解析
故பைடு நூலகம்:D
【点睛】本题考查合情推理的应用，属于基础题。
10. 已知表中数据y与x有较好的线性关系，通过计算得到y关于x的线性回归方程为 ，则相应于下列各点的残差中绝对值最小的是( ）
x
2
4
6
8
10
y
4
6
9
10
12。5
A。 （2，4） B。 （4，6） C. (8，10) D. （10,12。5）
4。 以下成语的语境为合情推理的是( )
A。 坐井观天 B. 管中窥豹 C. 开门见山 D。 一叶障目
【答案】B
【解析】
【分析】
由成语的意思结合合情推理的定义判定即可。
【详解】A为眼光狭小,看到的有限；
C意为说话写文章直截了当;
D意为被局部或暂时的现象所迷惑，不认清事物的全貌或问题的本质；
所以A，C，D都没有推理过程；
A。 设 为双曲线的左焦点， 为双曲线的右顶点， 为双曲线虚轴上的一个顶点，当 时，该双曲线的离心率为2
B。 设 为双曲线的左焦点, 为双曲线的右顶点， 为双曲线虚轴上的一个顶点,当 时，该双曲线的离心率为4
C. 设 为双曲线的左焦点， 为双曲线的右顶点， 为双曲线虚轴上的一个顶点，当 时,该双曲线的离心率为2
其次,直线不在平面内又包括“相交”和“平行”两种,
故选：C．
【点睛】本题主要考查直线与平面的位置关系,还考查了理解辨析的能力，属于基础题。
3。 已知两个变量 ， 线性相关，且根据观测到的数据 计算样本平均数得 , ,则根据这组观测数据算得的线性回归方程不可能是（ ）．
A. B.
C。 D。
【答案】D
开封市五县联考2019-2020学年高二下学期期中考试数学(理)试题含解析
【点睛】本题考查合情推理的判定，属于基础题.
5.已知曲线 过点 ，则该曲线在 处的切线的方程是（ )
A。 B。 C. D。
【答案】D
【解析】
【分析】
由曲线上的一点坐标求得参数m，在有导数的几何意义求得切线的斜率,最后由直线的点斜式方程表示直线方程即可.
【详解】因为曲线 过点 ,
所以 ，解得 ，即 ，
7.用0，1,2,3，4，5可以组成无重复数字且能被2整除的三位数的个数是（ ）
A。 50 B. 52 C. 54 D. 56
【答案】B
【解析】
【分析】
特殊元素优先考虑,即优先考虑个位数是0的情况,再考虑不是0的情况，最后将所有结果加起来即可。
【详解】能被2整除的三位数是偶数,
当个位数是0时，有 种情形；
令 ，得 ，
所以展开式中 的系数为 .
故答案为：
【点睛】本题考查了根据通项公式求项的系数，属于基础题。
16。设函数 ，观察 ， ， , ,…，根据以上事实,由归纳推理可得:当 且 时， ______。
【答案】
【解析】
【分析】
对四个分母中 的系数和常数进行归纳，找出规律可得答案。
【详解】观察知：四个等式等号右边的分母为 , ， ， ，
4.本卷命题范围：人教版选修2—2，选修2-3。
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。
1。已知复数 ,则 （ ）
A.2B.4C.6D。8
【答案】A
【解析】
【分析】
由复数的除法运算整理已知复数,再由共轭复数概念表示，最后由复数的几何意义求复数的模长即可。
【详解】因为 ,则 ,
【解析】
【分析】
【解析】河南省开封市五县联考2019-2020学年高二上学期期中考试数学(理)试题
开封五县联考高二期中考试数学（理科）考生注意：1.本试卷分第I 卷（选择题）和第Ⅱ卷（非选择题）两部分.满分l50分，考试时间120分钟.2.考生作答时，请将答案答在答题卡上.第I 卷每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑；第Ⅱ卷请用直径0.5毫来黑色墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效，在试题卷、草稿纸上作答无效.3.本卷命题范围：必修5第二、三章，21第二章双曲线结束.第I 卷（选择题共60分）一、选择题：本大题共12个小题，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.下列命题是真命题的是（ ） A. x ∀∈R ，20x > B. 0x ∃∈R ，020x < C. 0x ∃∈R ，200x ≥ D. x ∀∈R ，21x ≥【答案】C 【分析】根据基本初等函数的值域来对各选项中的特称或全称命题的真假进行判断. 【详解】对于选项A ，x ∀∈R ，20x ≥，A 选项错误；对于B 选项，x ∀∈R ，20x >，所以，不存在0x ∈R ，使得020x <，B 选项错误； 对于C 选项，x ∀∈R ，20x ≥，所以，0x ∃∈R ，200x ≥，C 选项正确； 对于D 选项，x ∀∈R ，20x >，D 选项错误. 故选：C.【点睛】本题考查全称命题和特称命题真假的判断，常用逻辑推证法或特例法来进行判断，考查推理能力，属于基础题.2.双曲线22:184x y C -=的焦点坐标为（ ）A. ()2,0±B. ()0,2±C. ()± D. (0,±【答案】C 【分析】判断双曲线C 的焦点位置，计算出双曲线的半焦距，即可得出双曲线C 的焦点坐标.【详解】由题意知，双曲线C 的焦点在x =因此，双曲线C 的焦点坐标为()±. 故选：C.【点睛】本题考查双曲线焦点坐标的求解，要判断出双曲线焦点的位置，考查计算能力，属于基础题.3.在等比数列{}n a 中，3512a a =，则4a =（ ）A. B. ± C. D. ±【答案】D 【分析】利用等比中项的性质可求出4a 的值.【详解】由等比中项的性质得243512a a a ==，因此，4a =±.故选：D.【点睛】本题考查等比中项的计算，在解题时还要注意所求项的符号，考查计算能力，属于基础题.4.“04x <<”是“2log 1x <”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【答案】B 【分析】解不等式2log 1x <，利用集合的包含关系可对两条件之间的关系进行判断.【详解】由2log 1x <得02x <<，故“04x <<”是“2log 1x <”的必要不充分条件. 故选：B.【点睛】本题考查必要不充分条件的判断，一般转化为两集合的包含关系，同时也可以逻辑关系来进行判断，考查推理能力，属于基础题.5.已知椭圆()2222:10x y C a b a b+=>>的上顶点为A ，左、右两焦点分别为1F 、2F ，若12AF F ∆为等边三角形，则椭圆C 的离心率为（ ）A.12B.2C.13D.3【答案】A 【分析】由题意得出1212AF AF F F ==，可得出2a c =，从而可求出椭圆C 的离心率. 【详解】设椭圆C 的焦距为2c ，由于12AF F ∆为等边三角形，则12122AF AF AF AF a ⎧=⎪⎨+=⎪⎩，12AF AF a ∴==，由题意可得2a c =，因此，椭圆C 的离心率为12c a =. 故选：A.【点睛】本题考查椭圆离心率的计算，同时也考查了椭圆定义的应用，考查计算能力，属于基础题.6.若双曲线()222210,0x y a b a b-=>>的实轴长、虚轴长、焦距成等差数列，则双曲线的渐近线方程是（ ）A. 34y x =?B. 54y x =±C. 45y x =±D. 43y x =±【答案】D 【分析】设双曲线的焦距为()20c c >，由题意得出关于a 、b 、c 的关系式，求出a 、b 的等量关系，即可求出双曲线的渐近线方程.【详解】设双曲线的焦距为()20c c >，根据实轴长、虚轴长、焦距成等差数列，得2222b a c c a b =+⎧⎨=+⎩， 则()224b c a =+，即()()2224c ac a -=+，即()4c a c a -=+，35c a ∴=，则53c a =，43b a ∴===. 因此，双曲线的渐近线方程为43y x =±. 故选：D.【点睛】本题考查双曲线渐近线方程的求解，解题的关键就是根据题中条件得出a 、b 的等量关系，考查运算求解能力，属于中等题. 7.已知0a >，0b >，则()641a b a b ⎛⎫++ ⎪⎝⎭的最小值为（ ） A. 32 B. 36C. 39D. 45【答案】B 【分析】 将代数式()641a b a b ⎛⎫++ ⎪⎝⎭展开，然后利用基本不等式可求出该代数式的最小值. 【详解】a >Q ，b >，由基本不等式得()41641620b aa b a b a b ⎛⎫++=++ ⎪⎝⎭2036≥=，当且仅当2b a =时，等号成立.因此，()641a b a b ⎛⎫++ ⎪⎝⎭的最小值为36. 故选：B.【点睛】本题考查利用基本不等式求最值，在利用基本不等式时要注意“一正、二定、三相等”条件的成立，考查计算能力，属于中等题.8.已知椭圆()2222:10x y C a b a b+=>>两焦点间的距离为A，则椭圆C 的标准方程为（ ）A. 22142x y +=B. 22164x y +=C. 22186x y +=D. 22153x y +=【答案】B 【分析】利用椭圆的定义求出a ，再结合半焦距求出b 的值，从而可得出椭圆C 的标准方程. 【详解】由题意知，椭圆C的焦点坐标为()， 由椭圆的定义得2a ==))11=+=，a ∴=2b ==.因此，椭圆C 的标准方程为22164x y +=.故选：B.【点睛】本题考查椭圆标准方程的求解，在涉及椭圆的焦点时，可以充分利用椭圆的定义来求解，考查计算能力，属于中等题. 9.已知等差数列{}n a 的前n 项和为n S ，若1513S S =，则36aa =（ ） A.112B.113C.335D.225【答案】C 【分析】利用等差数列的前n 项和公式以及等差中项的性质可得出36a a 的值.【详解】由等差数列的前n 项和公式得()()1553111116552311112a a S a a a S a +===+，解得36335a a =. 故选：C.【点睛】本题考查利用等差数列前n 项和公式以及等差中项性质的应用，考查计算能力，属于中等题.10.设命题:p 若函数()()32xf x a =--是减函数，则1a <，命题:q 若函数()224g x x ax =++在[)2,+∞上是单调递增，则2a <-.那么下列命题为真命题的是（ ）A. p q ∧B. p ⌝C.()p q ⌝∨D.()p q ⌝∧【答案】D 【分析】先判断出命题p 、q 的真假，然后利用复合命题的真假判断出各选项中命题的真假.【详解】若函数()()32xf x a =--是减函数，则321a ->，解得1a <，命题p 为真命题； 若函数()224g x x ax =++在[)2,+∞上是单调递增，其对称轴为直线x a =-，则2a -≤，解得2a ≥-，命题q 为假命题. 因此，p q ∧为假，p ⌝为假，()p q ⌝∨为假，()p q ⌝∧为真.故选：D.【点睛】本题考查复合命题真假的判断，同时也考查了指数函数与二次函数的单调性，解题的关键就是判断出各简单命题的真假，考查推理能力，属于中等题.11.设A 、B 分别为双曲线()2222:10,0x y C a b a b-=>>的左、右顶点，P 、Q 是双曲线C 上关于x 轴对称的不同两点，设直线AP 、BQ 的斜率分别为m 、n ，若1mn =-，则双曲线C 的离心率e 是（ ）C. 2【答案】A 【分析】设点()00,P x y ，则点()00,Q x y -，由点P 在双曲线C 上得出()2220220a ay b x =-，然后利用斜率公式得出221b a=，由此可计算出双曲线C 的离心率.【详解】设点()00,P x y .则()00,Q x y -，00AP y m k x a ∴==+，00BQ y n k x a==--， 则2220y m n a x ⋅=-，又2200221x y a b -=，即()2220220a a y b x =-，()2222022220b x a b a mn a x a-∴==--， 由1mn =-有a b =，c e a ===C.故选：A.【点睛】本题考查双曲线离心率的计算，同时也考查双曲线方程的应用，解题的关键在将点横坐标与纵坐标通过点的坐标满足双曲线方程建立等式求解，考查运算求解能力，属于中等题.12.已知椭圆2222915:x y C a a+=，点P 为椭圆C 上位于第一象限一点，O 为坐标原点，过椭圆左顶点A 作直线//l OP ，交椭圆于另一点B ，若12AB OP =，则直线l 的斜率为（ ）【答案】A 【分析】设点11(,)B x y ，()22,P x y ，由题意得出12AB OP =u u u r u u u r ，可得出12121212x x a y y ⎧=-⎪⎪⎨⎪=⎪⎩，然后将点B 、P 的坐标代入椭圆方程，得出2x 、2y ，即可求出直线l 的斜率.【详解】由题知(),0A a -，设11(,)B x y ，()22,P x y .则12AB OP =u u u r u u u r ，可得()112211,,22x a y x y ⎛⎫+= ⎪⎝⎭，1212x x a ∴=-，1212y y =，Q 点P 、B 都在椭圆C 上，2222222222595159522x y a y x a a ⎧+=⎪∴⎨⎛⎫⎛⎫-+=⎪ ⎪ ⎪⎝⎭⎝⎭⎩，解得24a x =，243y =，因此，直线l 的斜率为22453343y x a =⋅=. 故选：A.【点睛】本题考查直线斜率的求解，同时也考查了直线与椭圆的综合问题，在涉及平行线截椭圆所得弦长的比例关系时，可转化为共线向量比的问题求解，考查运算求解能力，属于中等题.第Ⅱ卷（非选择题共90分）二、填空题：本大题共4小题，每小题5分，共20分.13.命题“0,x R ∃∈0sin 1x >”的否定为 . 【答案】【详解】因为特称命题的否定是全称命题，先改变量词，再否定结论， 所以命题“0,x R ∃∈0sin 1x >”的否定为，故答案为.14.已知实数x 、y 满足约束条件102101x y x y x -+≥⎧⎪++≥⎨⎪≤⎩，则3z x y =-的最小值为__________．【答案】3- 【分析】作出不等式组所表示的可行域，平移直线3z x y =-，观察该直线在x 轴上的截距最小时对应的最优解，代入目标函数即可得出结果.【详解】作出不等式组102101x y x y x -+≥⎧⎪++≥⎨⎪≤⎩所表示的可行域如下图所示：联立21010x y x y ++=⎧⎨-+=⎩，解得1x y =-⎧⎨=⎩，得点()1,0A -，平移直线3z x y =-，当直线3z x y =-经过可行域的顶点()1,0A -时，该直线在x 轴上的截距取最小值，此时，目标函数3z x y =-取得最小值()min 3103z =⨯--=-. 故答案为：3-.【点睛】本题考查简单的线性规划问题，考查线性目标函数的最值问题，一般要作出可行域，利用数形结合思想来求解，考查数形结合思想的应用，属于中等题.15.在数列{}n a 中，14a =，132n n a a +=-，若对于任意的*n ∈N ，()125n k a n -≥-恒成立，则实数k 的最小值为__________． 【答案】127【分析】利用待定系数法得出13nn a -=，然后由参变量分离法得出253nn k -≥，并利用单调性的定义分析数列253nn -⎧⎫⎨⎬⎩⎭，求出该数列的最大项的值，从而可得出实数k 的取值范围. 【详解】由132n n a a +=-有()1131n n a a +-=-，故数列{}1n a -为首项为3，公比为3的等比数列，可得13nn a -=.不等式()125n k a n -≥-可化为253nn k -≥， 令()()*253nn f n n -=∈N ，当12n ≤≤时()0f n <；当3n ≥时，()0f n >. 故当3n ≥时，()()()1143232510333n n n n n n f n f n ++---+-=-=≤，故()()1327f n f ≤=， 127k ∴≤，因此，实数k 的最小值是127. 故答案为：127.【点睛】本题考查利用待定系数法求数列通项，同时也考查了数列不等式的解法，在解题时一般利用参变量分离法转化为数列的最值来求解，考查分析问题和解决问题的能力，属于中等题.16.已知点A 、B 为椭圆22:14x C y +=的左、右顶点，点M 为x 轴上一点，过M 作x 轴的垂线交椭圆C 于P 、Q 两点，过M 作AP 的垂线交BQ 于点N ，则BMNBMQS S ∆∆=_______． 【答案】45【分析】 设点(),Pm n ，则(),0M m ，(),Q m n -，写出直线MN 和BQ的方程，联立这两条直线的方程，求出点N 的坐标，即可得出BMNBMQS S ∆∆的值. 【详解】如下图所示，设(),Pm n ，则(),0M m ，(),Q m n -，由题设知2m ≠±且0n ≠，直线AP 的斜率2AP n k m =+，直线MN 斜率2MN m k n+=-.∴直线MN 的方程为()2m y x m n +=--，直线BQ 的方程为()22ny x m=--. 联立()()222m y x m n n y x m +⎧=--⎪⎪⎨⎪=-⎪-⎩，解得()22244N n m y m n -=--+.又点P 在椭圆C 上，得2244m n -=，45N y n ∴=-. 又1225BMN N S BM y BM n ∆=⋅=⋅，12BMQ S BM n ∆=⋅，45BMN BMQ S S ∆∆∴=. 故答案为：45. 【点睛】本题考查椭圆中三角形的面积比的计算，解题的关键就是要求出点的坐标，同时也要注意点的坐标满足椭圆方程，结合等式进行计算，考查运算求解能力，属于中等题. 三、解答题：本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤. 17.设命题()()2:2110p x a x a a --+-<，命题:0211q x ≤-≤，若p 是q 的必要不充分条件，求实数a 的取值范围. 【答案】31,2⎛⎫⎪⎝⎭【分析】解出命题p 、q 中的不等式，由题意得出()1,11,2a a ⎡⎤-⎢⎥⎣⎦Ü，由此可得出关于实数a 的不等式组，解出即可得出实数a 的取值范围.【详解】解不等式()()22110x a x a a --+-<，即()()10x a x a ⎡⎤---<⎣⎦，解得1a x a -<<.解不等式0211x ≤-≤，解得112x ≤≤. 所以，:1p a x a -<<，1:12q x ≤≤， 由于p 是q 的必要不充分条件，则()1,11,2a a ⎡⎤-⎢⎥⎣⎦Ü，1121a a ⎧-<⎪∴⎨⎪>⎩，解得312a <<.因此，实数a 的取值范围是31,2⎛⎫⎪⎝⎭.【点睛】本题考查利用必要不充分条件关系求参数的取值范围，一般转化为集合的包含关系，考查化归与转化思想的应用，属于中等题.18.在平面直角坐标xOy 中，()13,0F -，()23,0F ，点P 是平面上一点，使12PF F ∆的周长为16.（1）求点P 的轨迹方程； （2）求12PF PF ⋅的最大值.【答案】（1）()22102516x y y +=≠；（2）25. 【分析】（1）由题意得出121210PF PF F F +=>，由椭圆的定义可知点P 的轨迹是以点1F 、2F 为焦点的椭圆（去掉左右端点），设点P 的轨迹方程为()222210,0x y a b y a b+=>>≠，求出a 、b 的值，可得出点P 的轨迹方程；（2）利用1210PF PF +=，并利用基本不等式可求出12PF PF ⋅的最大值.【详解】（1）由题知121216PF PF F F ++=，126F F =，121210PF PF F F ∴+=>， 由椭圆的定义可知，动点P 的轨迹是以点1F 、2F 为焦点的椭圆（去掉左右端点），设动点P 的轨迹方程为()222210,0x y a b y a b+=>>≠，则210a =，5a =，3=，得4b =，因此，动点P 的轨迹方程为()22102516x yy +=≠； （2）由（1）可知1210PF PF +=，由基本不等式得21212252PF PF PF PF ⎛+⎫⋅≤= ⎪⎝⎭， 当且仅当125PF PF ==时等号成立，因此，12PF PF ⋅的最大值为25.【点睛】本题考查利用椭圆的定义求轨迹方程，同时也考查了利用椭圆的定义和基本不等式求最值，考查运算求解能力，属于中等题.19.已知等差数列{}n a 的前n 项和为n S ，有1055S =，410S =. （1）求数列{}n a 的通项公式； （2）令11n n n b a a +=，记数列{}n b 的前n 项和为n T ，证明：112n T ≤<. 【答案】（1）n a n =；（2）证明见解+析. 【分析】（1）设等差数列{}n a 的公差为d ，根据题意列出关于1a 和d 的方程组，解出这两个量，再利用等差数列的通项公式可求出数列{}n a 的通项公式； （2）由（1）得知()11111n b n n n n ==-++，然后利用裂项法求出数列{}n b 的前n 项和n T ，即可证明出112n T ≤<. 【详解】（1）设数列{}n a 的公差为d ，有111045554610a d a d +=⎧⎨+=⎩，解得111a d =⎧⎨=⎩，有()11n a n n =+-=，因此，数列{}n a 的通项公式为n a n =；（2）由（1）知()11111n b n n n n ==-++，有1111111122311n T n n n ⎛⎫⎛⎫⎛⎫=-+-++-=- ⎪ ⎪ ⎪++⎝⎭⎝⎭⎝⎭L ， 由*n ∈N ，有11012n <≤+，故有111121n ≤-<+，由上知112n T ≤<.【点睛】本题考查等差数列通项公式的求解，同时也考查了裂项求和法的应用，在求解等差数列的通项公式时，一般要建立首项和公差的方程组，利用方程思想求解，考查计算能力，属于中等题.20.双曲线()2222:10,0x y C a b a b-=>>的左、右焦点分别为1F 、2F ，点12⎫⎪⎭，()1-在双曲线C 上.（1）求双曲线C 的标准方程；（2）直线l 过点2F 且与双曲线C 交于A 、B 两点，且AB的中点的横坐标为-l 的方程.【答案】（1）2214x y -=；（2）y x =.【分析】 （1）将点12⎫⎪⎭、()1-的坐标代入双曲线C 的方程，求出a 、b 的值，即可得出双曲线C 的标准方程；（2）由题意知，直线l 的斜率存在，设点()11,A x y 、()22,B x y ，并设直线l的方程为(y k x =，将直线l 的方程与双曲线的方程联立，结合韦达定理求出k 的值，即可得出直线l 的方程.【详解】（1）由题意有22225114811a ba b ⎧-=⎪⎪⎨⎪-=⎪⎩，解得21a b =⎧⎨=⎩，因此，双曲线C 的标准方程为2214x y -=；（2）由题意知，直线l 的斜率存在，设点()11,A x y 、()22,B x y .并设直线l的方程为(y k x =-，联立方程(2214x y y k x ⎧-=⎪⎨⎪=⎩，消去y 整理得()()2222144200k x x k -++=-，12x x +==-得k =l的方程为y x =. 【点睛】本题考查双曲线方程的求解，同时也考查了中点弦问题，可以利用点差法，也可以将直线方程与双曲线方程联立，结合韦达定理求解，考查运算求解能力，属于中等题.21.已知数列{}n a 的前n 项和为n S ，16a =，1133n n n a a ++=+.（1）证明：数列3n n a ⎧⎫⎨⎬⎩⎭为等差数列；（2）求n S ；（3）对任意*m ∈N 将数列3n n a ⎧⎫⎨⎬⎩⎭中落入区间()23,3m m 内的项的个数记为m b ，求数列{}m b 的前m 项和m T .【答案】（1）证明见解+析；（2）1213344n n ++⨯-；（3）22134338m m m ++-⨯+-. 【分析】（1）利用等差数列的定义结合递推公式1133n n n a a ++=+，可证明出数列3n n a ⎧⎫⎨⎬⎩⎭是等差数列； （2）求出数列{}n a 的通项公式为()13nn a n =+⨯，然后利用错位相减法求出数列{}n a 的前n项和n S ；（3）令2313m m n <+<，得出23131m m n -<<-，可得出2331m mm b =--，然后利用分组求和法求出数列{}m b 的前m 项和m T .【详解】（1）由1111333333n n n n n n n n n a a a a +++++-=-1133nn n na a ⎛⎫=+-= ⎪⎝⎭，且123a =， 故数列3n n a ⎧⎫⎨⎬⎩⎭是首项为2，公差为1的等差数列； （2）由（1）知()11133n n a a n n =+-=+，有()13n n a n =+⨯， 由()2233313nn S n =⨯+⨯+++⨯L ， 有()23132333313nn n S n n +=⨯+⨯++⨯++⨯L ，作差有()23122333313nn n S n +-=⨯++++-+⨯L ，得()()23123333313nn n S n +-=+++++-+⨯L有()()()1121332313231313n n n n S n n ++--=+-+⨯⋅--++=， 因此，1213344n n n S ++=⨯-； （3）对任意*m ∈N ，若2313m m n <+<，得23131m m n -<<-，得2332m m n ≤≤-， 故()223231331mm m m m b =--+=--，有()()()()24229193133333331913m mm m m T m m--=+++-+++-=----L L ()()93913182m m m =----22134338m m m ++-⨯+=-. 【点睛】本题考查利用定义证明等差数列，同时与考查了错位相减法与分组求和法，要熟悉这些数列求和法对数列通项的要求，考查计算能力，属于中等题.22.已知椭圆(222:12x y C a a +=>的右焦点为F ，P 是椭圆C 上一点，PF x ⊥轴，2PF =. （1）求椭圆C 的标准方程；（2）若直线l 与椭圆C 交于A 、B 两点，线段AB 的中点为M ，O 为坐标原点，且OM =求AOB ∆面积的最大值.【答案】（1）22182x y +=；（2）2. 【分析】（1）设椭圆C 的焦距为()20c c >，可得出点,2c ⎛ ⎝⎭在椭圆C 上，将这个点的坐标代入椭圆C 的方程可得出2234c a =，结合222a c =+可求出a 的值，从而可得出椭圆C 的标准方程；（2）分直线AB 的斜率不存在与存在两种情况讨论，在AB x ⊥轴时，可得出AB而求出AOB ∆的面积；在直线AB 斜率存在时，设直线AB 的方程为y kx t =+，设点()11,A x y 、()22,B x y ，将直线AB的方程与椭圆方程联立，利用韦达定理结合OM =得出()2222214116k t k+=+，计算出AB 与AOB ∆的高，可得出AOB ∆面积的表达式，然后可利用二次函数的基本性质求出AOB ∆面积的最大值.【详解】（1）设椭圆C 的焦距为()20c c >，由题知，点,P c ⎛ ⎝⎭，b =则有222212c a ⎛ ⎝⎭+=，2234c a ∴=，又22222a b c c =+=+，28a ∴=，26c =， 因此，椭圆C 的标准方程为22182x y +=；（2）当AB x ⊥轴时，M 位于x 轴上，且OM AB ⊥，由OM =AB =12AOB S OM AB ∆=⋅= 当AB 不垂直x 轴时，设直线AB 的方程为y kx t =+，与椭圆交于()11,A x y ，()22,B x y ，由22182x y y kx t ⎧+=⎪⎨⎪=+⎩，得()222148480k x ktx t +++-=. 122814kt x x k -∴+=+，21224814t x x k -=+，从而224,1414kt t M k k -⎛⎫ ⎪++⎝⎭已知OM =()2222214116k t k+=+.()()()22222212122284814141414kt t AB kx x x x k k k ⎡⎤--⎛⎫⎡⎤=++-=+-⨯⎢⎥ ⎪⎣⎦++⎝⎭⎢⎥⎣⎦Q ()()()222221682114k t k k -+=++.设O 到直线AB 的距离为d ，则2221t d k =+，()()()222222221682114114AOBk t t S k k k ∆-+=+⋅++. 将()2222214116k t k+=+代入化简得()()2222219241116AOBk k Sk ∆+=+.令2116k p +=，则()()()22222211211192414116AOBp p k k S p k ∆-⎛⎫-+ ⎪+⎝⎭==+211433433p ⎡⎤⎛⎫=--+≤⎢⎥ ⎪⎢⎥⎝⎭⎣⎦.当且仅当3p =时取等号，此时AOB ∆的面积最大，最大值为2. 综上：AOB ∆的面积最大，最大值为2.【点睛】本题考查椭圆标准方程的求解，同时也考查了直线与椭圆中三角形面积最值的计算，一般将直线方程与椭圆方程联立，利用韦达定理设而不求法来求解，同时在计算最值时，常用函数的基本性质以及基本不等式进行求解，考查运算求解能力，属于难题.。
河南省周口市中英文学校2019-2020学年高二下学期期中考试(6月)英语试题+Word版含答案
周口中英文学校2019-2020学年下期高二期中考试英语试题本试卷分选择题和非选择题两部分。
满分120，考试用时120分钟第I卷第二部分阅读理解（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，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
AA woman was sitting by herself in a movie theatre.The movie she was about to see was a musical version(版本）of a very successful book that had sold a million copies that year.As the woman was waiting for the movie to begin，she noticed that the theatre was very crowded but the two seats next to her were empty.Suddenly a large man carrying a big fur coat walked down the aisle(通道）and sat down,placing the fur coat on the seat next to her.When the lights went out,the fur coat began to move，and the woman realized it was not a coat but a large furry dog.He was sitting up in the seat,watching the movie screen with great interest.As soon as the movie started，the dog began to nod his head and beat his paws(爪）in perfect time to the music.When the movie was over，the woman turned to the man and said,“Excuse me，sir,but I’ve never seen such a well-behaved dog in a movie theatre before.Does he often go to the movies?”“Oh，yes,”replied the man.“And he seemed to enjoy everything so much,”she said“It was just amazing!”“As a matter of fact,it surprised me,too，”said the maa“He hated the book!”21.What kind of movie was the woman about to see?A.A movie about a clever dog.B.A movie about two film-goers.C.A musical movie based on a book.D.A movie telling of a popular book.22.What can we conclude from Para4?A.The man enjoyed the movie very muchB.The dog made noise and disturbed othersC The woman enjoyed the movie very much.D.The dog was bright enough to enjoy the movie.23.What did the woman do when the movie was over?A.She communicated with the man happily.B.She praised the man for bringing his dog.C.She punished the dog for making noise.D.She complained about the dog angrily.BThere was once a jar(罐子)of fresh,clean water.Every drop of water in the jar felt immensely proud of being so clear and pure.Day after day they would congratulate each other on how clean and beautiful they were.That was,until one day when one of the drops got bored with his ultra-clean existence.He wanted to try what it was like being a dirty drop.The other drops tried to talk him out of it,but he stuck to his guns.Hardly realizing, when the drop came back all dirty,he turned all the other drops in the jar into dirty drops,too.They tried to get clean again,but couldn't.They tried everything to shake off the dirtiness.Finally,much later,someone dipped the jar in a fountain,and only when a lot of clean water entered the jar,did the drops regain their old transparency and purity. Now they all know that if they all want to be nice clean drops,then each and every one of them has to stay clean,even if they find it difficult.That drop that succeeded in being dirty has realized that correcting the mistake of one single drop needs a lot of work for everyone else.The same happens with us and our friends.If we want to live in a jar of clean water, each one of us will have to be a clean drop.None of us should try being the dirty drop who spoils everything.How about you?What are you?A clean drop?24.One of the drops made a change because he_______.A.got tired of his state of existing.B.was less beautiful than other drops.C.wanted to dirty othersD.was too proud a drop25.We can infer from Para.2that the drop was________.A.confidentB.braveC.dangerousD.stubborn26.Realizing the trouble he caused,the drop that succeeded in being dirty must have felt very________.A.pleasedB.importantC.sorryD.proud27.What’s the writer’s purpose of writing this article?A.To blame the drop for his foolish mistake.B.To advise readers to be“a clean drop”.C.To tell readers how to work together effectively.D.To tell readers the way to be“a clean drop”.CWinslow Homer was the second of the three sons of Henrietta Benson Homer and Charles Savage Homer.He was born in Boston,Massachusetts in1836and grew up in Cambridge,Massachusetts.His father was an importer of tools and other goods.His mother was a painter.Winslow got his interest in drawing and painting form his mother. But his father also supportedhis son’s interest.Once,on a business trip to London,Charles Winslow bought a set of drawing facsimiles(摹本)for his son to copy.Young Winslow used these to develop his early skill.Winslow’s older brother Charles went to Harvard University in Cambridge, Massachusetts.The family expected Winslow would go,too.But,at the time,Harvard didn’t teach art.So Winslow’s father found him a job as an assistant in the trade of making and preparing pictures for print media.At19,Winslow learned the process of lithography(平版印刷术),which was the only formal training that Winslow ever received in art.In1859,Winslow Homer moved to New York City to work for Harper’s Weekly. Homer also started to paint seriously.He hoped to go to Europe to study painting.But, something cropped up that would change the direction of Winslow Homer’s artistic work.Harper’s Magazine would send him to draw pictures of the biggest event in American history since independence.It was the American Civil War between the Union and the rebel(反叛的)southern states.Winslow Homer went to Washington,D.C.,in1861.He drew pictures of the campaign of Union Army Major General George B.McClellan the next year.His pictures of the war showed many ways that conflicts affect people.28.Who had the same interest as Winslow according to the text?A.His father.B.His mother.C.His teacher.D.His brother29.Why didn’t Winslow go to Harvard University?A.He didn’t want to go there.B.He wasn’t admitted to Harvard University.C.He couldn’t learn art in Harvard UniversityD.His family had no money to send him there.30.The underlined phrase“cropped up”in the third paragraph is closest in meaning to “____”.A.appearedB.remainedC.followedD.interrupted31.How does the text develop?A.By comparison.B.In order of time.C.By giving examples.D.By questioning the points.DOne ambitious high school student knew exactly how to show his family that he got accepted into his dream college—by surprising them on Christmas Day.Barrington Lincoln,class president at Lutheran High School North in Ferguson,got accepted into Morehouse College in Atlanta back on Dec.15.But for months,he knew he wanted to make the news known in a big way to his mother and aunt,Lincoln told ABC News.“I thought of the surprise in October,”he said,adding that once he got accepted he would purchase two school T-shirts from Morehouse’s online store On Christmas Day,in a now video on Twitter,Lincoln,17,gifted his mom Lisa McDonald and his aunt Shirley Gray the T-shirts with the school’s name on it.When the two sisters opened the gifts,they had no idea what they meant.In the video,Lincoln nudges(用肘轻推)his family,“You know what that means right?I got in!”While McDonald falls heavily on the couch in delight,Gray asks,“You’ve been holding out on us?”“I didn’t expect anything like that,”McDonald told ABC News.“It’s so satisfying to see him get the return on his studying.”“He always wanted to be the first in line and help everybody stay in line,”his mother added.McDonald said it’s especially sweet since she had to work an extra job to afford his private school after his father,a former Marine,passed away in2015.Lincoln was only 15.“All kids need to have quality education,”the mother added.“I am putting an investment(投资)in his future.”32.What did Lincoln choose two T-shirts for his mother and aunt mainly for?A.To show he loves them forever.B.To give them aChristmas gift.C.To tell them his academic success.D.To show he had grown upalready.33.How did Lincoln’s mother and aunt feel when they saw his presents?A.Puzzled.B.Satisfied.C.Disappointed.D.Embarrassed.34.Which of the following can replace the underlined sentence“You’ve beenholding out on us”?A.You've been telling a lie to the two of us.B.You've been longing to tell us the truth.C.You've been playing a joke with two of us.D.You've been keeping it a secret from us.35.What can we infer from the text?A.Lincoln's parents could hardly afford his education.B.Lincoln's mother felt her efforts paid off at last.C.Lincoln bought the T-shirts in the local supermarket.D.The video had been popular before December15.第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
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