山东省菏泽一中2019-2020学年高三3月线上模拟考试试题
2020届山东省菏泽一中高三3月线上模拟考试数学试题解析
2020届山东省菏泽一中高三3月线上模拟考试数学试题一、单选题1．集合{}|12A x x =-<，1393x B x ⎧⎫=<<⎨⎬⎩⎭，则A B I 为( ) A ．()1,2 B ．()1,2-C ．()1,3D ．()1,3-答案：B计算得到{}13A x x =-<<，{}12B x x =-<<，再计算A B I 得到答案. 解：18{}13x x =-<<，{}139123x B x x x ⎧⎫=<<=-<<⎨⎬⎩⎭， 故()1,2A B =-I . 故选：B . 点评：本题考查了集合的交集运算，意在考查学生的计算能力. 2．设复数2121,()z i z x i x R =+=-∈，若12z z ⋅为实数，则x =（ ）A ．1B ．1﹣C ．1或1﹣D ．2答案：C先求得2212(1)(1)z z x x i ⋅=++-,由实数可知,其虚部为0,进而求解即可 解：解:2121,()z i z x i x R =+=-∈Q ,22212(1)(1)(1())z z i x i x x i ∴⋅=+-=++-,由12z z ⋅为实数,则210x -=,即1x =±, 故选:C 点评：本题考查已知复数的类型求参数,考查复数的乘法法则的应用3．某校高三年级的学生参加了一次数学测试，学生的成绩全部介于60分到140分之间（满分150分），为统计学生的这次考试情况，从中随机抽取100名学生的考试成绩作为样本进行统计.将这100名学生的测试成绩的统计结果按如下方式分成八组：第一组[)60,70，第二组[)70,80，第三组[)80,90，…….如图是按上述分组方法得到的频率分布直方图的一部分.则第七组的频数为（）A．8 B．10 C．12 D．16答案：A直接根据频率和为1计算得到答案.解：设第七组的频率为p，则()0.0040.0120.0160.030.020.0060.004101p+++++++⨯=，故0.008p=. 故第七组的频数为：100100.0088⨯⨯=.故选：A.点评：本题考查了频率分布直方图，意在考查学生对于频率直方图的理解和掌握.4．设函数()f x的定义域为R，满足()()2 2f x f x+=，且()()[]12,0,1ln(2),1,2x xf xx x+⎧∈⎪=⎨+∈⎪⎩则()f e=（）A．12e+B．2e C．12e-D．()ln2e+答案：B取2x e=-，代入()()2 2f x f x+=，计算得到答案.解：()()122222e ef e f e-=-=⋅=.故选：B.点评：本题考查了分段函数计算，意在考查学生的计算能力.5．在直角梯形ABCD中，4AB=，2CD=，//AB CD，AB AD⊥，E是BC的中点，则()AB AC AE ⋅+=u u u v u u u v u u u v （ ）A ．8B ．12C ．16D ．20答案：D由数量积的几何意义可得8AB AC ⋅=u u u v u u u v ，12AB AE ⋅=u u u v u u u v，又由数量积的运算律可得()AB AC AE AB AC AB AE ⋅+=⋅+⋅u u u v u u u v u u u v u u u v u u u v u u u v u u u v，代入可得结果.解：∵()AB AC AE AB AC AB AE ⋅+=⋅+⋅u u u v u u u v u u u v u u u v u u u v u u u v u u u v ，由数量积的几何意义可得：AB AC u u u v u u u v ⋅的值为AB u u u v 与AC u u u v 在AB u u u v方向投影的乘积， 又AC u u u v 在AB u u u v方向的投影为12AB =2， ∴428AB AC ⋅=⨯=u u u v u u u v ，同理4312AB AE ⋅=⨯=u u u v u u u v，∴()81220AB AC AE ⋅+=+=u u u v u u u v u u u v，故选D. 点评：本题考查了向量数量积的运算律及数量积的几何意义的应用，属于中档题.6．一个箱子中装有4个白球和3个黑球，若一次摸出2个球，则摸到的球颜色相同的概率是（ ） A ．17B ．27C ．37D ．47答案：C利用组合数计算得到基本事件总数和颜色相同的基本事件个数，由古典概型概率公式计算可得结果. 解：从箱子中一次摸出2个球共有2721C =种情况；颜色相同的共有22439C C +=种情况∴摸到的球颜色相同的概率93217p == 故选：C 点评：本题考查古典概型概率问题的求解，涉及到组合数的应用，属于基础题. 7．函数()sin()x x f x e e -=+的图象大致为（ ）A ．B ．C ．D ．答案：D判断函数为偶函数，取特殊点()00sin21f <=<，判断得到答案. 解：()00sin21f <=<，且()()f x f x -=，函数为偶函数故选：D 点评：本题考查了函数图像的判断，根据奇偶性和特殊点可以快速得到答案是解题的关键.8．设椭圆()222210x y a b a b +=>>的两焦点为12,F F ，若椭圆上存在点P ，使12120F PF ∠=o ，则椭圆的离心率e 的取值范围为( ).A ．3(0,2B ．3(0,]4C ．3[2D ．3[,1)4答案：C 解：当P 是椭圆的上下顶点时,12F PF ∠最大, 121120180,6090,F PF F PO ∴︒≤∠<︒∴︒≤∠<︒12sin 60sin sin 90,F PF ∴︒≤∠<︒113,,12c F P a F O c a ==≤<Q 则椭圆的离心率e 的取值范围为3⎡⎫⎪⎢⎪⎣⎭,故选C. 点评：本题考查了椭圆的几何意义,属于中档题目.在客观题求离心率取值范围时,往往利用图形中给出的几何关系结合圆锥曲线的定义,找出a,b,c 之间的等量关系或者不等关系,考查学生的数形结合能力,在主观题中多考查直线与圆锥曲线的位置关系,利用方程的联立和判别式解不等式求出离心率的范围.二、多选题9．空气质量指数AQI是反映空气质量状况的指数，AQI指数值越小，表明空气质量越好，其对应关系如表：AQI指数值0~50 51~100 101~150 151~200 201~300 300空气质量优良轻度污染中度污染重度污染严重污染如图是某市12月1日-20日AQI指数变化趋势：下列叙述正确的是（）A．这20天中AQI指数值的中位数略高于100B．这20天中的中度污染及以上的天数占1 4C．该市12月的前半个月的空气质量越来越好D．总体来说，该市12月上旬的空气质量比中旬的空气质量好答案：ABD根据折线图和AQI指数与空气质量对照表，结合选项，进行逐一分析即可.解：对A：将这20天的数据从小到大排序后，第10个数据略小于100，第11个数据约为120，因为中位数是这两个数据的平均数，故中位数略高于100是正确的，故A正确；对B：这20天中，AQI指数大于150的有5天，故中度污染及以上的天数占14是正确的，故B正确；对C：由折线图可知，前5天空气质量越来越好，从6日开始至15日越来越差，故C错误；对D：由折线图可知，上旬大部分AQI指数在100以下，中旬AQI指数大部分在100以上，故上旬空气质量比中旬的要好.故D 正确. 故选：ABD. 点评：本题考查统计图表的观察，属基础题；需要认真看图，并理解题意. 10．已知1a >，01c b <<<，下列不等式成立的是（ ） A ．b c a a > B ．c c a b b a+>+ C ．log log b c a a <D ．b cb ac a>++ 答案：ACD由指数函数的单调性可判断A ；由作差法和不等式的性质可判断B ；可根据换底公式，取1log log b a a b =，1log log ca a c=，运用对数函数单调性，可判断C ；运用作差法和不等式的性质，可判断D . 解：由1a >，01c b <<<，可得b c a a >，故A 正确； 由1a >，01c b <<<，c c ab b a+-+ 可得()()()0a c b cb ca bc ba b b a b b a -+--==<++ ，c c ab b a+<+ ，故B 错误； 由1a >，01c b <<<，1log log b a a b =，1log log ca a c=，则log log 0a a c b <<，则110log log a a b c<<，可得log log b c a a <，故C 正确；由1a >，01c b <<<，()()()()()0a b c b c bc ba cb cab ac a b a c a b a c a -+---==>++++++可得b cb ac a>++，故D 正确. 故选:ACD 点评：本题考查不等式基本性质和利用指数函数、对数函数单调性比较大小，属于基础题.11．已知定义域为R 的奇函数()f x ，满足()22,22322,02x f x x x x x ⎧>⎪=-⎨⎪-+<≤⎩，下列叙述正确的是（ ）A ．存在实数k ，使关于x 的方程()f x kx =有7个不相等的实数根B ．当1211x x -<<<时，恒有()()12f x f x > C ．若当(]0,x a ∈时，()f x 的最小值为1，则51,2a ⎡⎤∈⎢⎥⎣⎦D ．若关于x 的方程()32f x =和()f x m =的所有实数根之和为零，则32m =-答案：AC根据函数是奇函数，写出其解析式，画出该函数的图像，再结合选项，数形结合解决问题. 解：因为该函数是奇函数，故()f x 在R 上的解析式为：()()222,(2)2322,(20)0,122,(02)2,(2)23x x x x x f x x x x x x x ⎧<-⎪+⎪----≤<⎪⎪==⎨⎪-+<≤⎪⎪>⎪-⎩绘制该函数的图像如下所示：对A ：如图所示直线1l 与该函数有7个交点，故A 正确； 对B ：当1211x x -<<<时，函数不是减函数，故B 错误； 对C ：如图直线2:1l y =，与函数图交于()51,1,,12⎛⎫⎪⎝⎭， 故当()f x 的最小值为1时，51,?2a ⎡⎤∈⎢⎥⎣⎦，故C 正确；对D ：()32f x =时，若使得其与()f x m =的所有零点之和为0， 则32m =-，或317m =-，如图直线3l ，故D 错误. 故选：AC. 点评：本题考查由函数的奇偶性求函数解析式，以及判断方程的根的个数，以及函数零点的问题，涉及函数单调性，属综合性基础题；另，本题中的数形结合是解决此类问题的重要手段，值得总结.12．如图，矩形ABCD ，M 为BC 的中点，将ABM ∆沿直线AM 翻折成1AB M ∆，连接1B D ，N 为1B D 的中点，则在翻折过程中，下列说法中所有正确的是（ ）A ．存在某个位置，使得1CN AB ⊥; B ．翻折过程中，CN 的长是定值C ．若AB BM =，则1AM BD ⊥；D ．若1AB BM ==，当三棱锥1B AMD-的体积最大时，三棱锥1B AMD -的外接球的表面积是4π. 答案：BD对于A ，取AD 的中点为E ，连接,CE NE ，设CE MD F ⋂=.通过证明平面//CNE 平面1B AM ，得1//NF MB .假设1CN AB ⊥，得到EN CN ⊥，EN NF ⊥，这是不可能的，故A 不正确；对于B ，在CEN ∆中，由余弦定理得2NC 是定值，故NC 是定值，故B 正确；对于C ，若1AM B D ⊥，可证AM ⊥平面1ODB ，得到OD AM ⊥，此时AD MD =，由于AD MD ≠，故1AM B D ⊥不成立，故C 不正确；对于D ，只有当平面1B AM ⊥平面AMD 时，三棱锥1B AMD -的体积最大，取AD 的中点为E ，证明1EA ED EM ===，故E 就是三棱锥1B AMD -的外接球的球心，故D 正确.解：对于A ，取AD 的中点为E ，连接,CE NE ，设CE MD F ⋂=，如图1所示则1//,NE AB NE ⊄Q 平面11,B AM B A ⊂平面1B AM ，//NE ∴平面1B AM .Q 四边形AMCE 是平行四边形，//CE AM ∴，同理可证//CE 平面1B AM .又CE NE E ⋂=，且,CE NE ⊂平面CNE ，∴平面//CNE 平面1B AM .1//MB ∴平面CNE ，又1MB ⊂平面1B MD ，平面1B MD ⋂平面CNE NF =， 1//NF MB ∴.如果1CN AB ⊥，则EN CN ⊥，由于11AB MB ⊥，则EN NF ⊥， 由于三线,,NE NF NC 共面且共点，这是不可能的，故A 不正确； 对于B ，如图1，由等角定理可得1NEC MAB ∠=∠，又11,2NE AB AM EC ==， ∴在CEN ∆中，由余弦定理得：2222cos NC NE EC NE EC NEC =+-⋅⋅∠是定值，NC ∴是定值，故B 正确；对于C ，如图2所示AB BM =Q ，即11AB B M =，设O 为AM 中点，连接1B O ，则1AM B O ⊥若1AM B D ⊥，由于111B O B D B =I ，且11,B O B D ⊂平面1ODB ，AM ∴⊥平面1ODB ，OD ⊂平面1ODB ，OD AM ∴⊥，则AD MD =，由于AD MD ≠，故1AM B D ⊥不成立，故C 不正确； 对于D ，根据题意知，只有当平面1B AM ⊥平面AMD 时，三棱锥1B AMD -的体积最大，取AD 的中点为E ，O 为AM 中点， 连接1,,OE B E ME ，如图21AB BM ==Q ，1B O AM ∴⊥，Q 平面1B AM ⊥平面AMD平面1B AM ⋂平面AMD AM =，1B O ⊂平面1B AM1B O ∴⊥平面AMD ，又OE ⊂平面AMD ，1B O OE ∴⊥.又11AB B M ⊥，111AB B M ==，AM ∴=112B O AM ==，11222OE DM AM ===，11EB ∴==， 1EA ED EM ∴===.AD ∴的中点E 就是三棱锥1B AMD -的外接球的球心，球的半径为1，表面积是4π，故D 正确； 故选：BD. 点评：本题考查立体几何中的翻折问题，考查学生的空间想象能力,考查立体几何中的平行、垂直的判定定理和性质定理，考查余弦定理，属于难题.三、填空题13．函数()1xe f x x =+的图象在点()()0,0f 处的切线方程是_______________.答案：10y -=借助求导公式求出()y f x =',因为切线的斜率为()00f '=,0x =代入()1x ef x x =+求得切点,即可求出切线方程. 解：()()21xxe f x x '=+，∴()00f '=且()01f =，所以函数()1x ef x x =+的图象在()()0,0f 处的切线方程是10y -=.故答案为: 10y -=. 点评：本题考查了导数的几何意义,过曲线上一点的切线方程的求法,难度容易. 14．已知等比数列{}n a 的前n 项和为n S ，*n ∈N .若321320S S S -+=，则21a a =______. 答案：2；根据等比数列公式化简得到322a a =，3212a a a a =得到答案. 解：321320S S S -+=，故()()123121320a a a a a a ++-++=，即322a a =，32122a a a a ==. 故答案为：2. 点评：本题考查了等比数列公式，意在考查学生的计算能力.15．62x x ⎛- ⎪⎝⎭展开式的常数项为 ．（用数字作答）答案：-160 解：由6662166(2)(1)(2)(rr r r r r rr T C x C x x ---+⎛==- ⎝，令620r -=得3r =，所以62x x ⎛ ⎝展开式的常数项为33636(1)(2)160C --=-. 【考点】二项式定理.16．已知1F ，2F 分别为双曲线221927x y C -=：的左、右焦点，点A C ∈，点M 的坐标为()2,0，AM 为12F AF ∠的角平分线，则2AF =_______ 答案：6利用双曲线的方程求出双曲线的参数值；利用内角平分线定理得到两条焦半径的关系，再利用双曲线的定义得到两条焦半径的另一条关系，联立求出焦半径． 解：不妨设A 在双曲线的右支上, ∵AM 为12F AF ∠的平分线，∴11228 24AF F M AF MF ===， 又∵1226AF AF a -==，解得26AF =，故答案为6. 点评：本题考查内角平分线定理，考查双曲线的定义：解有关焦半径问题常用双曲线的定义，属于中档题.四、解答题17．ABC ∆的内角,,A B C 的对边分别为,,a b c ，已知22cos a b c B +=，c =.（1）求角C ；（2）延长线段AC 到点D ，使CD CB =，求ABD ∆周长的取值范围． 答案：(1)23π(2) (1)利用余弦定理222cos 2a c b B ac+-=化简整理再用角C 的余弦定理即可.也可以用正弦定理先边化角,再利用和差角公式求解.(2)易得ABD ∆的周长等于2a b ++再利用正弦定理将,a b 用角,A B 表示,再利用三角函数的值域方法求解即可. 解：解法一：（1）根据余弦定理得222222a c b a b c ac+-+=整理得222a b c ab +-=-，2221cos 22a b c C ab +-∴==-，()0,C π∈Q 23C π∴=（2）依题意得BCD ∆为等边三角形，所以ABD ∆的周长等于2a b ++由正弦定理2sin sin sin 2a b cA B C====， 所以2sin ,2sin a A b B ==，24sin 2sin a b A B +=+4sin 2sin()3A A π=+-)6A π=+0,3A π⎛⎫∈ ⎪⎝⎭Q ，(,)662A πππ∴+∈，1sin()(,1)62A π∴+∈，2a b \+?，所以ABD ∆的周长的取值范围是． 解法二：（1）根据正弦定理得2sin sin 2sin cos A B C B +=sin sin[()]sin()sin cos cos sin A B C B C B C B C π=-+=+=+Q ,2sin cos sin B C B ∴=-, sin 0B ≠Q ，1cos 2C ∴=-，()0,C π∈Q ， 23C π∴=（2）同解法一 点评：本题主要考查了正余弦定理求解三角形的问题,同时也考查了边角互化求解边长的取值范围问题等.属于中等题型.18．已知数列{}n a ，{}n b 满足：11a =，10b =，1443n n n b a b +=++，1434n n n a a b +=++，*n ∈N .（1）证明：数列{}n n a b +为等差数列，数列{}n n a b -为等比数列； （2）记数列{}n a 的前n 项和为n W ，求n W 及使得9n W >的n 的取值范围.答案：（1）证明见解析（2）21122nn n W ⎛⎫=+- ⎪⎝⎭；5n ≥ （1）两式相加得到()()112n n n n a b a b +++-+=，两式相减得到1112n n n n a b a b ++-=-，得到证明.（2）计算1122n n a n ⎛⎫=-+ ⎪⎝⎭，21122nn n W ⎛⎫=+- ⎪⎝⎭，解不等式得到答案.解：（1）由1434n n n a a b +=++和1443n n n b a b +=++相加得：()()11448n n n n a b a b +++=++所以()()112n n n n a b a b +++-+=，因此数列{}n n a b +是以2为公差的等差数列 由1434n n n a a b +=++和1443n n n b a b +=++相减得：()()1142n n n n a b a b ++-=-，所以1112n n n n a b a b ++-=-，1110a b -=≠，因此数列{}n n a b -是以12为公比的等比数列 （2）21n n a b n +=-，112n n n a b -⎛⎫-= ⎪⎝⎭，两式相加得：1122nn a n ⎛⎫=-+ ⎪⎝⎭所以()2111221111222212nn nn n n n W ⎡⎤⎛⎫-⎢⎥ ⎪⎝⎭+⎢⎥⎛⎫⎣⎦=-+=+- ⎪⎝⎭- 因为11022nn a n ⎛⎫=-+> ⎪⎝⎭，所以1n n W W +>又因为419916W =-<，52719232W =->， 所以使得9n W >的n 的取值范围为5n ≥. 点评：本题考查了等差数列，等比数列的证明，分组求和法，根据数列的单调性解不等式，意在考查学生对于数列公式方法的综合应用.19．如图，在三棱台ABC DEF -中，2BC EF =，G ，H 分别为AC ，BC 上的点，平面//GHF 平面ABED ，CF BC ⊥，AB BC ⊥.（1）证明：平面BCFE ⊥平面EGH ；（2）若AB CF ⊥，22AB BC CF ===，求二面角B AD C --的大小. 答案：（1）证明见解析（2）3π（1）证明GH BC ⊥，HE BC ⊥得到BC ⊥平面EGH ，得到答案.（2）分别以HG ，HB ，HE 所在的直线为x 轴，y 轴，z 轴，建立如图所示的空间直角坐标系H xyz -，计算平面ABD 的一个法向量为()0,1,1m =u r，平面ADC 的一个法向量为()1,1,0n =-r，计算夹角得到答案.解：（1）因为平面GHF ∥平面ABED ，平面BCFE ⋂平面ABED BE =， 平面BCFE ⋂平面GHF HF =，所以BE HF ∥.因为BC EF ∥，所以四边形BHFE 为平行四边形，所以BH EF =， 因为2BC EF =，所以2BC BH =，H 为BC 的中点.同理G 为AC 的中点，所以//GH AB ，因为AB BC ⊥，所以GH BC ⊥， 又HC EF ∥且HC EF =，所以四边形EFCH 是平行四边形，所以CF HE ∥， 又CF BC ⊥，所以HE BC ⊥.又HE ，GH ⊂平面EGH ，HE GH H =I ，所以BC ⊥平面EGH ， 又BC ⊂平面BCFE ，所以平面BCFE ⊥平面EGH（2）HE HB ⊥，HG HB ⊥，AB CF ⊥，CF HE ∥，//GH AB ，所以HE HG ⊥. 分别以HG ，HB ，HE 所在的直线为x 轴，y 轴，z 轴，建立如图所示的空间直角坐标系H xyz -，则()2,1,0A ，()0,1,0B ，()1,0,1D ，()0,1,0C -.设平面ABD 的一个法向量为()111,,m x y z =u r ，因为()2,0,0AB =-u u ur ，()1,1,1BD =-u u u r则1111200m AB x m BD x y z ⎧⋅=-=⎪⎨⋅=-+=⎪⎩u u u v v u u u v v ，取11y =，得()0,1,1m =u r .设平面ADC 的一个法向量为()222,,n x y z =r ，因为()1,1,1AD =--u u u r，()2,2,0AC =--u u u r则222220220n AD x y z n AC x y ⎧⋅=-=+=⎪⎨⋅=--=⎪⎩u u u v v u u u v v ，取21x =，得()1,1,0n =-r . 所以1cos ,2m n m n m n ⋅==⋅u r ru r r ur r ，则二面角B AD C --的大小为3π点评：本题考查了面面垂直，二面角，意在考查学生的空间想象能力和计算能力.20．某学校共有教职工900人,分成三个批次进行继续教育培训，在三个批次中男、女教职工人数如下表所示. 已知在全体教职工中随机抽取1名,抽到第二批次中女教职工的概率是0.16 .（1）求的值；（2）现用分层抽样的方法在全体教职工中抽取54名做培训效果的调查, 问应在第三批次中抽取教职工多少名？ （3）已知，求第三批次中女教职工比男教职工多的概率.答案：（1）144（2）12（3）49第一问中利用等概率抽样求解样本容量．可知由,解得第二问中，由于用分层抽样的方法在全体教职工中抽取54名做培训效果的调查 因此先求第三批的人数，然后按比例抽样得到第三批中抽取的人数 第三问中，结合古典概型概率公式求解得到． 解: (1)由,解得. ……………3分(2)第三批次的人数为,设应在第三批次中抽取m 名，则，解得12m =.∴应在第三批次中抽取12名. ……………6分 （3）设第三批次中女教职工比男教职工多的事件为A ，第三批次女教职工和男教职工数记为数对(,)y z ，由（2）知200,(,,96,96)y z y z N y z +=∈≥≥，则基本事件总数有：，共9个，而事件A 包含的基本事件有：(101,99),(102,98),(103,97),(104,96)共4个， ∴4()9P A =. ……………………………………12分 21．已知过抛物线的焦点，斜率为的直线交抛物线于两点，且.（1）求抛物线的方程;（2）O 为坐标原点，C 为抛物线上一点，若 ，求的值.答案：（1）y 2＝8x .（2）λ＝0，或λ＝2.解：试题分析：第一问求抛物线的焦点弦长问题可直接利用焦半径公式，先写出直线的方程，再与抛物线的方程联立方程组，设而不求，利用根与系数关系得出，然后利用焦半径公式得出焦点弦长公式，求出弦长，第二问根据联立方程组解出的A 、B 两点坐标，和向量的坐标关系表示出点C 的坐标，由于点C 在抛物线上满足抛物线方程，求出参数值. 试题解析：(1)直线AB 的方程是y ＝2（x-），与y 2＝2p x 联立，消去y 得8x 2－10p x ＋2＝0，由根与系数的关系得x 1＋x 2＝ .由抛物线定义得|AB |＝＋p ＝9，故p=4 (2)由(1)得x 2－5x ＋4＝0，得x 1＝1，x 2＝4，从而A (1，－2)，B (4，4)．设＝(x 3，y 3)＝(1，－2)＋λ(4，4)＝(4λ＋1，4λ－2)，又y ＝8x 3，即[2(2λ－1)]2＝8(4λ＋1)，即(2λ－1)2＝4λ＋1，解得λ＝0或λ＝2. 点评：求弦长问题，一般采用设而不求联立方程组，借助根与系数关系，利用弦长公式去求；但是遇到抛物线的焦点弦长问题时，可直接利用焦半径公式，使用焦点弦长公式，求出弦长.遇到与向量有关的问题，一般采用坐标法去解决，根据联立方程组解出的A 、B 两点坐标，和向量的坐标关系表示出点C 的坐标，由于点C 在抛物线上满足抛物线方程，求出参数值.22．已知函数()sin ln(1)f x x m x =-+，且()f x 在0x =处切线垂直于y 轴． （1）求m 的值；（2）求函数()f x 在[]0,1上的最小值；（3）若2sin ln 10x x ax x e --+->恒成立，求满足条件的整数a 的最大值． （参考数据sin10.84≈，ln 20.693=） 答案：（1）1m =；（2）0；（3）2.（1）依题意，(0)0f '=，由此即可求得m 的值；（2）求导，研究函数()f x 在[0，1]上的单调性，进而得到最值； （3）先分析2a …，再证明当2a …时满足条件即可得到a 的最大值． 解：（1）因为()f x 在0x =处切线垂直于y 轴，则()00f '=因为()cos 1mf x x x '=-+，则()010f m '=-=，则1m = （2）由题意可得1()cos 1f x x x '=-+，注意到()00f '=，[]0,1x ∈则21()sin (1)f x x x ''=-++则32()cos 0(1)f x x x '''=--<+ 因此()f x ''单调递减，()010f ''=>，()11sin104f ''=-+<因此存在唯一零点()00,1x ∈使得()00f x ''=，则()f x '在()00,x 单调递增， 在()0,1x 单调递减，11(1)cos1cos 0232f π'=->-=，则()0f x '>在()0,1上恒成立 从而可得()f x 在()0,1上单调递增，则()min 00f f == （3）必要条件探路因为2sin ln 10x x ax x e --+->恒成立，令1x =，则sin1e a ≥ 因为sin1ln 232e e >>=，由于a 为整数，则2a ≤， 因此2sin 2sin ln 12ln 1x x x ax x e x x x e --+-≥--+- 下面证明2sin ()2ln 10xg x x x x e=--+->恒成立即可①当()0,1x ∈时，由（1）可知sin ln(1)x x >+，则sin 1x e x >+故22()2ln 11ln g x x x x x x x x >--++-=--，设2()ln h x x x x =--，(0,1)x ∈则2121(21)(1)()210x x x x h x x x x x--+-'=--==<，则()h x 在()0,1单调递减从而可得()()10h x h >=，由此可得()0g x >在()0,1x ∈恒成立． ②当1x >时，下面先证明一个不等式：122xe x >+，设1()22xh x e x =-- 则()2xh x e '=-，则()h x 在(),ln 2-∞-单调递减，在()ln 2,+∞单调递增因此min 1(ln 2)22ln 202h h ==-->，那么sin 12sin 2x e x >+ 由此可得2sin 212ln 12ln 2sin ()2xx x x ex x x x g x --+->--+-= 则1()222cos g x x x x '=--+，21()22sin 0g x x x''=+->因此()g x '单调递增，()(1)2cos112cos103g x g π''>=->-=， 则()g x 在()1,+∞上单调递增，因此3()(1)2sin102g x g >=-> 综上所述：a 的最大值整数值为2． 点评：本题考查利用导数研究函数的单调性，极值及最值，考查不等式的证明及先猜后证思想，考查逻辑推理能力及运算求解能力，属于难题．。
山东省菏泽第一中学2019-2020学年高三3月份质量检测地理试题(word无答案)
山东省菏泽第一中学2019-2020学年高三3月份质量检测地理试题一、单选题(★★)1 . 飞絮是杨树、柳树等植物种子成熟炸裂后飘出毛絮的自然现象，在干燥、温暖和阳光充足的天气下最易发生。
下图为北方春季一天的气温变化图。
读图完成问题。
【小题1】据图推测飞絮高发的时间段为A．2时—6时B．10时—16时C．14时—20时D．18时—22时【小题2】飞絮在上述时间段高发的主要原因是A．气温高，对流运动强B．温差大，空气不稳定C．空气湿，飞絮易扩散D．气温低，飞絮易产生(★★) 2 . 下图为某年6月琼东沿海海水表层等温线分布图（甲图）和某地理事物分布图（乙图）。
据此完成下面小题。
【小题1】甲图中琼东沿海海水表层等温线呈半环状，数值向外海递增。
形成这种分布状况，主要是因为A．大量陆地淡水的注入B．台风造成的暴雨C．大幅度的涨潮D．深层海水的上泛【小题2】根据甲图推断乙图所示地理事物最可能是A．降水量B．含沙量C．浮游生物量D．海水盐度(★★)3 . 2018年12月9日，欧洲第五大悬索桥（桥梁结构无需桥墩）——挪威哈罗格兰德大桥正式通车。
该桥由中国企业承建，绝大部分构件在中国生产，再运至挪威组装。
下图示意哈罗格兰德大桥及周边地形。
据此完成下面小题。
【小题1】对隆巴肯峡湾沿线的欧洲国际公路影响最大的自然灾害是A．冰川泥石流B．崩塌C．洪涝D．水土流失【小题2】哈罗格兰德大桥采用悬索结构主要是由于所在区域A．潮流复杂B．基岩松软C．风力较强D．水深较深(★★★★)4 . 漏刻是我国古代应用最广的计时仪器。
“漏”指漏壶，是仪器的泄水部分；“刻”，指受水壶中的刻箭，是仪器的受水部分。
漏刻计时必须经常与天文测时比对，以确定计时的起点和时间单位，白天对日晷表影位置、夜晚观测恒星位置来校正漏刻的准确度。
日晷也是我国古代长期使用的计时工具。
最常见的日晷是赤道式日晷，晷面与赤道面平行，晷针指向北极星，通过晷针日影移动来计时。
2020年高考模拟试卷山东省菏泽一中高考数学模拟试卷(3月份) 含解析
2020年高考模拟高考数学模拟试卷（3月份）一、选择题1．已知复数z1，z2在复平面内对应的点分别为（1，1），（0，1），则＝（）A．1+i B．﹣1+i C．﹣1﹣i D．1﹣i2．已知集合A＝（﹣1，3]，B＝{x|≤0}，则A∩B＝（）A．[﹣2，1）B．（﹣1，1]C．（﹣1，1）D．[﹣2，3]3．在二项式（x2﹣）5的展开式中，含x4的项的系数是（）A．﹣10B．10C．﹣5D．54．“总把新桃换旧符”（王安石）、“灯前小草写桃符”（陆游），春节是中华民族的传统节日，在宋代入们用写“桃符”的方式来祈福避祸，而现代入们通过贴“福”字、贴春联、挂灯笼等方式来表达对新年的美好祝愿，某商家在春节前开展商品促销活动，顾客凡购物金额满50元，则可以从“福”字、春联和灯笼这三类礼品中任意免费领取一件，若有4名顾客都领取一件礼品，则他们中有且仅有2人领取的礼品种类相同的概率是（）A．B．C．D．5．已知点M（2，4）在抛物线C：y2＝2px（p＞0）上，点M到抛物线C的焦点的距离是（）A．4B．3C．2D．16．在△ABC中，+＝2，+2＝0，若＝x+y，则（）A．y＝2x B．y＝﹣2x C．x＝2y D．x＝﹣2y7．已知双曲线的左、右焦点分别为F1，F2，O为坐标原点，P是双曲线在第一象限上的点，，则双曲线C的渐近线方程为（）A．B．C．y＝±x D．8．已知奇函数f（x）是R上增函数，g（x）＝xf（x）则（）A．B．C．D．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项是符合题目要求，全部选对的得5分，部分选对的得3分，有选错的0分．9．如图，正方体ABCD﹣A1B1C1D1的棱长为1，则下列四个命题正确的是（）A．直线BC与平面ABC1D1所成的角等于B．点C到面ABC1D1的距离为C．两条异面直线D1C和BC1所成的角为D．三棱柱AA1D1﹣BB1C1外接球半径为10．要得到y＝cos2x的图象C1，只要将图象C2怎样变化得到？（）A．将的图象C2沿x轴方向向左平移个单位B．的图象C2沿x轴方向向右平移个单位C．先作C2关于x轴对称图象C3，再将图象C3沿x轴方向向右平移个单位D．先作C2关于x轴对称图象C3，再将图象C3沿x轴方向向左平移个单位11．已知集合M＝{（x，y）|y＝f（x）}，若对于∀（x1，y1）∈M，∃（x2，y2）∈M，使得x1x2+y1y2＝0成立，则称集合M是“互垂点集”．给出下列四个集合：M1＝{（x，y）|y ＝x2+1}；M2＝；M3＝{（x，y）|y＝e x}；M4＝{（x，y）|y＝sin x+1}．其中是“互垂点集”集合的为（）A．M1B．M2C．M3D．M412．德国著名数学家狄利克雷（Dirichlet，1805～l859）在数学领域成就显著．19世纪，狄利克雷定义了一个“奇怪的函数”其中R为实数集，Q为有理数集．则关于函数f（x）有如下四个命题，其中真命题的是（）A．函数f（x）是偶函数B．∀x1，x2∈∁R Q，f（x1+x2）＝f（x1）+f（x2）恒成立C．任取一个不为零的有理数T，f（x+T）＝f（x）对任意的x∈R恒成立D．不存在三个点A（x1，f（x1）），B（x2，f（x2）），C（x3，f（x3）），使得△ABC 为等腰直角三角形三、填空题：本题共4小题，每小题5分，共20分．13．已知直线x﹣y+a＝0与圆O：x2+y2＝2相交于A，B两点（O为坐标原点），且△AOB 为等腰直角三角形，则实数a的值为；14．已知直线y＝x+2与曲线y＝ln（x+a）相切，则a的值为．15．l5.2019年7月，中国良渚古城遗址获准列入世界遗产名录，标志着中华五千年文明史得到国际社会认可．良渚古城遗址是人类早期城市文明的范例，实证了中华五千年文明史．考古科学家在测定遗址年龄的过程中利用了“放射性物质因衰变而减少”这一规律．已知样本中碳14的质量N随时间T（单位：年）的衰变规律满足（N0表示碳14原有的质量），则经过5730年后，碳14的质量变为原来的；经过测定，良渚古城遗址文物样本中碳14的质量是原来的至，据此推测良渚古城存在的时期距今约在5730年到年之间．（参考数据：lg2≈0.3，lg7≈0.84，lg3≈0.48）16．已知△ABC的顶点A∈平面α，点B，C在平面α异侧，且AB＝2，AC＝，若AB，AC与α所成的角分别为，则线段BC长度的取值范围为．四、解答题：本题共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．在①；②2a+c＝2b cos C；③这三个条件中任选一个，补充在下面问题中的横线上，并解答相应的问题．在△ABC中，内角A，B，C的对边分别为a，b，c，且满足，，a+c＝4，求△ABC的面积．注：如果选择多个条件分别解答，按第一个解答计分．18．设数列{a n}的前n项和为S n，已知a1＝1，S n+1﹣2S n＝1，n∈N*．（I）证明：{S n+1}为等比数列，求出{a n}的通项公式；（Ⅱ）若b n＝，求{b n}的前n项和T n，并判断是否存在正整数n使得T n•2n﹣1＝n+50成立？若存在求出所有n值；若不存在说明理由．19．《九章算术》是我国古代数学名著，它在几何学中的研究比西方早1000多年，在《九章算术》中，将底面为直角三角形，且侧棱垂直于底面的三棱柱称为堑堵（qiandu）；阳马指底面为矩形，一侧棱垂直于底面的四棱锥，鳖膈（bienao）指四个面均为直角三角形的四面体．如图在堑堵ABC﹣A1B1C1中，AB⊥AC．（I）求证：四棱锥B﹣A1ACC1为阳马；（Ⅱ）若C1C＝BC＝2，当鳖膈C1﹣ABC体积最大时，求锐二面角C﹣A1B﹣C1的余弦值．20．李克强总理在2018年政府工作报告指出，要加快建设创新型国家，把握世界新一轮科技革命和产业变革大势，深入实施创新驱动发展战略，不断增强经济创新力和竞争力．某手机生产企业积极响应政府号召，大力研发新产品，争创世界名牌．为了对研发的一批最新款手机进行合理定价，将该款手机按事先拟定的价格进行试销，得到一组销售数据（x i，y i）（i＝1，2，…，6），如表所示：单价x（千元）345678销量y（百件）7065625956t已知．（1）若变量x，y具有线性相关关系，求产品销量y（百件）关于试销单价x（千元）的线性回归方程；（2）用（1）中所求的线性回归方程得到与x i对应的产品销量的估计值．当销售数据（x i，y i）对应的残差的绝对值时，则将销售数据（x i，y i）称为一个“好数据”．现从6个销售数据中任取3个子，求“好数据”个数ξ的分布列和数学期望E （ξ）．（参考公式：线性回归方程中的估计值分别为．21．给定椭圆，称圆心在原点O，半径为的圆是椭圆C的“卫星圆”．若椭圆C的离心率，点在C上．（I）求椭圆C的方程和其“卫星圆”方程；（Ⅱ）点P是椭圆C的“卫星圆”上的一个动点，过点P作直线l1，l2，使得l1⊥l2，与椭圆C都只有一个交点，且l1，l2，分别交其“卫星圆”于点M，N，证明：弦长|MN|为定值．22．已知函数f（x）＝lnx﹣x+2sin x，f'（x）为f（x）的导函数．（Ⅰ）求证：f'（x）在（0，π）上存在唯一零点；（Ⅱ）求证：f（x）有且仅有两个不同的零点参考答案一、选择题：（共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的）1．已知复数z1，z2在复平面内对应的点分别为（1，1），（0，1），则＝（）A．1+i B．﹣1+i C．﹣1﹣i D．1﹣i【分析】由已知条件可得z1，z2，然后代入，再利用复数代数形式的乘除运算化简得答案．解：∵复数z1，z2在复平面内对应的点分别为（1，1），（0，1），∴z1＝1+i，z2＝i．∴＝．故选：D．2．已知集合A＝（﹣1，3]，B＝{x|≤0}，则A∩B＝（）A．[﹣2，1）B．（﹣1，1]C．（﹣1，1）D．[﹣2，3]【分析】求出集合A，B，由此能求出A∩B．解：∵集合A＝（﹣1，3]，B＝{x|≤0}＝{x|﹣2≤x＜1}，∴A∩B＝（﹣1，1）．故选：C．3．在二项式（x2﹣）5的展开式中，含x4的项的系数是（）A．﹣10B．10C．﹣5D．5【分析】利用二项展开式的通项公式求出第r+1项，令x的指数为4求得．解：对于，对于10﹣3r＝4，∴r＝2，则x4的项的系数是C52（﹣1）2＝10故选：B．4．“总把新桃换旧符”（王安石）、“灯前小草写桃符”（陆游），春节是中华民族的传统节日，在宋代入们用写“桃符”的方式来祈福避祸，而现代入们通过贴“福”字、贴春联、挂灯笼等方式来表达对新年的美好祝愿，某商家在春节前开展商品促销活动，顾客凡购物金额满50元，则可以从“福”字、春联和灯笼这三类礼品中任意免费领取一件，若有4名顾客都领取一件礼品，则他们中有且仅有2人领取的礼品种类相同的概率是（）A．B．C．D．【分析】有4名顾客都领取一件礼品，基本事件总数n＝34＝81，他们中有且仅有2人领取的礼品种类相同包含的基本事件个数m＝＝36，则他们中有且仅有2人领取的礼品种类相同的概率．解：从“福”字、春联和灯笼这三类礼品中任意免费领取一件，有4名顾客都领取一件礼品，基本事件总数n＝34＝81，他们中有且仅有2人领取的礼品种类相同包含的基本事件个数m＝＝36，则他们中有且仅有2人领取的礼品种类相同的概率是p＝＝．故选：B．5．已知点M（2，4）在抛物线C：y2＝2px（p＞0）上，点M到抛物线C的焦点的距离是（）A．4B．3C．2D．1【分析】由题意可知：点的坐标代入抛物线方程，求出p＝4，求得焦点F（2，0），利用抛物线的定义，即可求点M到抛物线C焦点的距离．解：由点M（2，4）在抛物线C：y2＝2px（p＞0）上，可得16＝4p，p＝4，抛物线C：y2＝8x，焦点坐标F（2，0），准线方程为x＝﹣2，点M到抛物线C的准线方程的距离为4，则点M到抛物线C焦点的距离是：4，故选：A．6．在△ABC中，+＝2，+2＝0，若＝x+y，则（）A．y＝2x B．y＝﹣2x C．x＝2y D．x＝﹣2y【分析】根据条件可判断出点D为BC的中点，并且可得出，从而可得出，并得出，进而根据向量的减法的几何意义，和向量的数乘运算即可得出，从而可得出，进而得出正确的选项．解：如图，∵，∴点D为边BC的中点，∵，∴，∴，又，∴＝＝，又，∴，∴x＝﹣2y．故选：D．7．已知双曲线的左、右焦点分别为F1，F2，O为坐标原点，P是双曲线在第一象限上的点，，则双曲线C的渐近线方程为（）A．B．C．y＝±x D．【分析】利用双曲线的定义求出m＝2a，结合向量的数量积，求出∠F1PF2，利用余弦定理求解关系式，推出渐近线方程即可．解：双曲线的左、右焦点分别为F1，F2，O为坐标原点，P是双曲线在第一象限上的点，，可得m＝2a，4a•2a cos∠F1PF2＝4a2，所以∠F1PF2＝60°，则4c2＝4a2+16a2﹣2×＝12a2，即a2+b2＝3a2，所以，所以双曲线的渐近线方程为：y＝x．故选：D．8．已知奇函数f（x）是R上增函数，g（x）＝xf（x）则（）A．B．C．D．【分析】先构造函数g（x）＝xf（x），可判断出g（x）为偶函数，x＞0时单调递增，x ＜0时，函数单调递减，距离对称轴越远，函数值越大，即可比较大小．解：由奇函数f（x）是R上增函数可得当x＞0时，f（x）＞0，又g（x）＝xf（x），则g（﹣x）＝﹣xf（﹣x）＝xf（x）＝g（x），即g（x）为偶函数，且当x＞0时单调递增，根据偶函数的对称性可知，当x＜0时，函数单调递减，距离对称轴越远，函数值越大，因为g（）＝g（log34），g（）＝g（），g（）＝g（），所以为g（）＞g（）＞g（）故选：B．二、多项选择题：本题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项是符合题目要求，全部选对的得5分，部分选对的得3分，有选错的0分．9．如图，正方体ABCD﹣A1B1C1D1的棱长为1，则下列四个命题正确的是（）A．直线BC与平面ABC1D1所成的角等于B．点C到面ABC1D1的距离为C．两条异面直线D1C和BC1所成的角为D．三棱柱AA1D1﹣BB1C1外接球半径为【分析】直接利用线面夹角的应用，异面直线的夹角的应用，三棱柱的外接球的半径的求法的应用求出结果．解：正方体ABCD﹣A1B1C1D1的棱长为1，对于选项A：直线BC与平面ABC1D1所成的角为，故选项A正确．对于选项B：点C到面ABC1D1的距离为B1C长度的一半，即h＝，故选项B正确．对于选项C：两条异面直线D1C和BC1所成的角为，故选项C错误．对于选项D：三棱柱AA1D1﹣BB1C1外接球半径r＝，故选项D正确．故选：ABD．10．要得到y＝cos2x的图象C1，只要将图象C2怎样变化得到？（）A．将的图象C2沿x轴方向向左平移个单位B．的图象C2沿x轴方向向右平移个单位C．先作C2关于x轴对称图象C3，再将图象C3沿x轴方向向右平移个单位D．先作C2关于x轴对称图象C3，再将图象C3沿x轴方向向左平移个单位【分析】直接利用三角函数关系式的恒等变换即平移变换和伸缩变换的应用求出结果．解：要得到y＝cos2x的图象C1，只要将图象C2将的图象C2沿x轴方向向左平移个单位即可，故选项A正确．或将的图象C2沿x轴方向向右平移个单位，也可得到，故选项B 正确．或先作C2关于x轴对称图象C3，再将图象C3沿x轴方向向右平移个单位，故选项C正确．故选：ABC．11．已知集合M＝{（x，y）|y＝f（x）}，若对于∀（x1，y1）∈M，∃（x2，y2）∈M，使得x1x2+y1y2＝0成立，则称集合M是“互垂点集”．给出下列四个集合：M1＝{（x，y）|y ＝x2+1}；M2＝；M3＝{（x，y）|y＝e x}；M4＝{（x，y）|y＝sin x+1}．其中是“互垂点集”集合的为（）A．M1B．M2C．M3D．M4【分析】根据题意即对于任意点P∀（x1，y1），在M中存在另一个点P′，使得．，结合函数图象进行判断．解：由题意，对于∀（x1，y1）∈M，∃（x2，y2）∈M，使得x1x2+y1y2＝0成立即对于任意点P∀（x1，y1），在M中存在另一个点P′，使得．y＝x2+1中，当P点坐标为（0，1）时，不存在对应的点P′．所以所以M1不是“互垂点集”集合，的图象中，将两坐标轴进行任意旋转，均与函数图象有交点，所以在M2中的任意点P∀（x1，y1），在M2中存在另一个点P′，使得．所以M2是“互垂点集”集合，y＝e x中，当P点坐标为（0，1）时，不存在对应的点P′．所以M3不是“互垂点集”集合，y＝sin x+1的图象中，将两坐标轴进行任意旋转，均与函数图象有交点，所以所以M4是“互垂点集”集合，故选：BD．12．德国著名数学家狄利克雷（Dirichlet，1805～l859）在数学领域成就显著．19世纪，狄利克雷定义了一个“奇怪的函数”其中R为实数集，Q为有理数集．则关于函数f（x）有如下四个命题，其中真命题的是（）A．函数f（x）是偶函数B．∀x1，x2∈∁R Q，f（x1+x2）＝f（x1）+f（x2）恒成立C．任取一个不为零的有理数T，f（x+T）＝f（x）对任意的x∈R恒成立D．不存在三个点A（x1，f（x1）），B（x2，f（x2）），C（x3，f（x3）），使得△ABC 为等腰直角三角形【分析】根据函数解析式，逐项判断即可．解：对于A，若x∈Q，则﹣x∈Q，满足f（x）＝f（﹣x）；若x∈∁R Q，则﹣x∈∁R Q，满足f（x）＝f（﹣x）；故函数f（x）为偶函数，选项A正确；对于B，取，则f（x1+x2）＝f（0）＝1，f（x1）+f （x2）＝0，1≠0，故选项B错误；对于C，若x∈Q，则x+T∈Q，满足f（x）＝f（x+T）；若x∈∁R Q，则x+T∈∁R Q，满足f （x）＝f（x+T）；故选项C正确；对于D，要为等腰直角三角形，只可能如下四种情况：①直角顶点A在y＝1上，斜边在x轴上，此时点B，点C的横坐标为无理数，则BC中点的横坐标仍然为无理数，那么点A的横坐标也为无理数，这与点A的纵坐标为1矛盾，故不成立；②直角顶点A在y＝1上，斜边不在x轴上，此时点B的横坐标为无理数，则点A的横坐标也应为无理数，这与点A的纵坐标为1矛盾，故不成立；③直角顶点A在x轴上，斜边在y＝1上，此时点B，点C的横坐标为有理数，则BC中点的横坐标仍然为有理数，那么点A的横坐标也应为有理数，这与点A的纵坐标为0矛盾，故不成立；④直角顶点A在x轴上，斜边不在y＝1上，此时点A的横坐标为无理数，则点B的横坐标也应为无理数，这与点B的纵坐标为1矛盾，故不成立．综上，不存在三个点A（x1，f（x1）），B（x2，f（x2）），C（x3，f（x3）），使得△ABC为等腰直角三角形，故选项D正确．故选：ACD．三、填空题：本题共4小题，每小题5分，共20分．13．已知直线x﹣y+a＝0与圆O：x2+y2＝2相交于A，B两点（O为坐标原点），且△AOB 为等腰直角三角形，则实数a的值为；【分析】根据题意，求出圆O的圆心以及半径，由等腰直角三角形的性质可得圆心到直线的距离d＝1，结合点到直线的距离公式，解可得a的值，即可得答案．解：根据题意，圆O：x2+y2＝2的圆心为（0，0），半径r＝，若直线x﹣y+a＝0与圆O交于A，B两点，且△AOB为等腰直角三角形，则圆心到直线的距离d＝＝1，解可得a＝；故答案为：．14．已知直线y＝x+2与曲线y＝ln（x+a）相切，则a的值为3．【分析】欲求a的大小，只须求出切线的方程即可，故先利用导数求出在切点处的导函数值，再结合导数的几何意义即可求出切线的斜率．进而求出切线方程，最后与已知的切线方程比较，从而问题解决．解：依题意得y′＝，因此曲线y＝ln（x+a）在切点处的切线的斜率等于，∴＝1，∴x＝1﹣a．此时，y＝0，即切点坐标为（1﹣a，0）相应的切线方程是y＝1×（x﹣1+a），即直线y＝x+2，∴a﹣1＝2，a＝3故答案为：3．15．l5.2019年7月，中国良渚古城遗址获准列入世界遗产名录，标志着中华五千年文明史得到国际社会认可．良渚古城遗址是人类早期城市文明的范例，实证了中华五千年文明史．考古科学家在测定遗址年龄的过程中利用了“放射性物质因衰变而减少”这一规律．已知样本中碳14的质量N随时间T（单位：年）的衰变规律满足（N0表示碳14原有的质量），则经过5730年后，碳14的质量变为原来的；经过测定，良渚古城遗址文物样本中碳14的质量是原来的至，据此推测良渚古城存在的时期距今约在5730年到6876年之间．（参考数据：lg2≈0.3，lg7≈0.84，lg3≈0.48）【分析】把T＝5730代入，即可求出结果，再令，两边同时取以2为底的对数，结合换底公式即可求出T的范围．解：∵，∴当T＝5730时，N＝N0，∴经过5730年后，碳14的质量变为原来的，由题意可知：，两边同时取以2为底的对数得：，∴，∴T＜6876，∴推测良渚古城存在的时期距今约在5730年到6876年之间．16．已知△ABC的顶点A∈平面α，点B，C在平面α异侧，且AB＝2，AC＝，若AB，AC与α所成的角分别为，则线段BC长度的取值范围为[]．【分析】由题意画出图形，分类求出BC的最小值与最大值即可．解：分别过B，C作底面的垂线，垂足分别为B1，C1．由已知可得，BB1＝，，AB1＝1，．如图，当AB，AC所在平面与α垂直，且B，C在底面上的射影B1，C1在A点同侧时BC长度最小，当AB，AC所在平面与α垂直，且B，C在底面上的射影B1，C1在A点两侧时BC长度最大．过C作CD⊥BB1的延长线，垂足为D，则BD＝，CD＝，则BC的最小值为，最大值为．∴线段BC长度的取值范围为[，]，故答案为：[]．四、解答题：本题共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．在①；②2a+c＝2b cos C；③这三个条件中任选一个，补充在下面问题中的横线上，并解答相应的问题．在△ABC中，内角A，B，C的对边分别为a，b，c，且满足①，，a+c＝4，求△ABC的面积．注：如果选择多个条件分别解答，按第一个解答计分．【分析】先利用正弦定理边化角，再结合两角和与差的正弦公式，求出B，再利用余弦定理求出ac，从而求出三角形的面积．解：在横线上填写“”，则由正弦定理，得．由sin A＝sin（B+C）＝sin B cos C+cos B sin C，得．由0＜C＜π，得sin C≠0．所以．又cos B≠0（若cos B＝0，则sin B＝0，sin2B+cos2B＝0这与sin2B+cos2B＝1矛盾），所以．又0＜B＜π，得．由余弦定理及，得，即12＝（a+c）2﹣ac．将a+c＝4代入，解得ac＝4．所以＝＝．18．设数列{a n}的前n项和为S n，已知a1＝1，S n+1﹣2S n＝1，n∈N*．（I）证明：{S n+1}为等比数列，求出{a n}的通项公式；（Ⅱ）若b n＝，求{b n}的前n项和T n，并判断是否存在正整数n使得T n•2n﹣1＝n+50成立？若存在求出所有n值；若不存在说明理由．【分析】（Ⅰ）将等式移项，两边加1，结合等比数列的定义和数列的递推式，可得所求；（Ⅱ）求得，再由数列的错位相减法求和，可得T n，解方程，结合函数的单调性，即可判断存在性．解：（Ⅰ）证明：∵S n+1﹣2S n＝1，∴S n+1+1＝2（S n+1）n∈N*∴{S n+1}为等比数列，∵S1+1＝2，公比为2，∴，，∴，当n≥2时，，a1＝1也满足此式，∴；（Ⅱ），，，两式相减得：，，代入，得2n﹣n﹣26＝0，令f（x）＝2x﹣x﹣26（x≥1），f'（x）＝2x ln2﹣1＞0在x∈[1，+∞）成立，∴f（x）＝2x﹣x﹣26，x∈（1，+∞）为增函数；由f（5）•f（4）＜0，所以不存在正整数n使得成立．19．《九章算术》是我国古代数学名著，它在几何学中的研究比西方早1000多年，在《九章算术》中，将底面为直角三角形，且侧棱垂直于底面的三棱柱称为堑堵（qiandu）；阳马指底面为矩形，一侧棱垂直于底面的四棱锥，鳖膈（bienao）指四个面均为直角三角形的四面体．如图在堑堵ABC﹣A1B1C1中，AB⊥AC．（I）求证：四棱锥B﹣A1ACC1为阳马；（Ⅱ）若C1C＝BC＝2，当鳖膈C1﹣ABC体积最大时，求锐二面角C﹣A1B﹣C1的余弦值．【分析】（Ⅰ）证明A1A⊥AB，结合AB⊥AC，推出AB⊥面ACC1A1，然后证明四棱锥B﹣A1ACC1为阳马．（Ⅱ）证明A1A⊥底面ABC，求出当且仅当时，取最大值，建立如图所示空间直角坐标系，求出面A1BC的一个法向量，平面A1BC1法向量，利用空间向量的数量积求解即可．【解答】（Ⅰ）证明：∵A1A⊥底面ABC，AB⊂面ABC，∴A1A⊥AB，又AB⊥AC，A1A∩AC＝A，∴AB⊥面ACC1A1，又四边形ACC1A1为矩形，∴四棱锥B﹣A1ACC1为阳马．（Ⅱ）解：∵AB⊥AC，BC＝2，∴AB2+AC2＝4，又∵A1A⊥底面ABC，∴＝，当且仅当时，取最大值，∵AB⊥AC，A1A⊥底面ABC∴以A为原点，建立如图所示空间直角坐标系，，，A1（0，0，2），，，设面A1BC的一个法向量，由得，同理得，∴二面角C﹣A1B﹣C1的余弦值为．20．李克强总理在2018年政府工作报告指出，要加快建设创新型国家，把握世界新一轮科技革命和产业变革大势，深入实施创新驱动发展战略，不断增强经济创新力和竞争力．某手机生产企业积极响应政府号召，大力研发新产品，争创世界名牌．为了对研发的一批最新款手机进行合理定价，将该款手机按事先拟定的价格进行试销，得到一组销售数据（x i，y i）（i＝1，2，…，6），如表所示：单价x（千元）345678销量y（百件）7065625956t 已知．（1）若变量x，y具有线性相关关系，求产品销量y（百件）关于试销单价x（千元）的线性回归方程；（2）用（1）中所求的线性回归方程得到与x i对应的产品销量的估计值．当销售数据（x i，y i）对应的残差的绝对值时，则将销售数据（x i，y i）称为一个“好数据”．现从6个销售数据中任取3个子，求“好数据”个数ξ的分布列和数学期望E （ξ）．（参考公式：线性回归方程中的估计值分别为．【分析】（1）由题意计算平均数和回归系数，写出线性回归方程；（2）利用所求的线性回归方程求得“好数据”的个数，知ξ的可能取值；计算对应的概率值，写出分布列，计算数学期望值．解：（1）由＝y i＝60，得×（70+65+62+59+56+t）＝60，解得t＝48，所以x i y i＝3×70+4×65+5×62+6×59+7×56+8×48＝1910，n＝6×5.5×60＝1980，＝32+42+52+62+72+82＝199，n＝6×5.52＝181.5，代入可得＝＝＝＝﹣4，＝﹣＝60﹣（﹣4）×5.5＝82，∴所求的线性回归方程为＝﹣4x+82；（2）利用（1）中所求的线性回归方程＝﹣4x+82可得，当x1＝3时，＝70；当x2＝4时，＝66；当x3＝5时，＝62；当x4＝6时，＝58；当x5＝7时，＝54；当x6＝8时，＝50；与销售数据对比可知满足|﹣y i|≤1的共有4个“好数据”：（3，70）、（4，65）、（5，62）、（6，59）；由题意知ξ的可能取值为1，2，3；计算P（ξ＝1）＝＝，P（ξ＝2）＝2•＝2×＝，P（ξ＝3）＝＝；则ξ的分布列为ξ123P数学期望为E（ξ）＝1×+2×+3×＝2．21．给定椭圆，称圆心在原点O，半径为的圆是椭圆C 的“卫星圆”．若椭圆C的离心率，点在C上．（I）求椭圆C的方程和其“卫星圆”方程；（Ⅱ）点P是椭圆C的“卫星圆”上的一个动点，过点P作直线l1，l2，使得l1⊥l2，与椭圆C都只有一个交点，且l1，l2，分别交其“卫星圆”于点M，N，证明：弦长|MN|为定值．【分析】（I）根据椭圆的离心率公式，将点代入椭圆方程，即可求得a和b的值；（Ⅱ）分类讨论，当直线的斜率存在时，设直线方程，代入椭圆方程，根据韦达定理即可求得t1•t2＝﹣1，满足条件的两直线l1，l2垂直，即可求得|MN|为定值．解：（Ⅰ）由条件可得：解得所以椭圆的方程为，…卫星圆的方程为x2+y2＝12…（II）证明：①当l1，l2中有一条无斜率时，不妨设l1无斜率，因为l1与椭圆只有一个公共点，则其方程为或，当l1方程为时，此时l1与“卫星圆”交于点和，此时经过点且与椭圆只有一个公共点的直线是y＝2或y＝﹣2，即l2为y＝2或y＝﹣2，所以l1⊥l2，所以线段MN应为“卫星圆”的直径，所以…②当l1，l2都有斜率时，设点P（x0，y0），其中，设经过点P（x0，y0）与椭圆只有一个公共点的直线为y＝t（x﹣x0）+y0，则，联立方程组，消去y，整理得，…所以…所以…所以t1•t2＝﹣1，满足条件的两直线l1，l2垂直．所以线段MN应为“卫星圆”的直径，所以综合①②知：因为l1，l2经过点P（x0，y0），又分别交其“卫星圆”于点MN，且l1，l2垂直，所以线段MN为“卫星圆”的直径，所以为定值…22．已知函数f（x）＝lnx﹣x+2sin x，f'（x）为f（x）的导函数．（Ⅰ）求证：f'（x）在（0，π）上存在唯一零点；（Ⅱ）求证：f（x）有且仅有两个不同的零点【分析】（Ⅰ）设，然后判断函数g'（x）的符号，再利用零点存在定理判断g（x）在（0，π）上是否存在唯一零点即可；（Ⅱ）分x∈（0，π），x∈[π，2π）和x∈[2π，+∞）三种情况分别考虑f（x）的零点存在情况，进一步得到结论．解：（Ⅰ）设，当x∈（0，π）时，，∴g（x）在（0，π）上单调递减．又∵，∴g（x）在上有唯一的零点．（Ⅱ）①由（Ⅰ）知，当x∈（0，α）时，f'（x）＞0，f（x）在（0，α）上单调递增；当x∈（α，π）时，f'（x）＜0，f（x）在（α，π）上单调递减；∴f（x）在（0，π）上存在唯一的极大值点，∴．∵，∴f（x）在（0，α）上恰有一个零点．∵f（π）＝lnπ﹣π＜2﹣π＜0，∴f（x）在（α，π）上也恰有一个零点；②当x∈[π，2π）时，sin x≤0，f（x）≤lnx﹣x．设h（x）＝lnx﹣x，，∴h（x）在[π，2π）上单调递减，∴h（x）≤h（π）＜0，∴当x∈[π，2π）时，f（x）≤h（x）≤h（π）＜0恒成立，∴f（x）在[π，2π）上没有零点．③当x∈[2π，+∞）时，f（x）≤lnx﹣x+2，设φ（x）＝lnx﹣x+2，，∴φ（x）在[2π，+∞）上单调递减，∴φ（x）≤φ（2π）＜0，∴当x∈[2π，+∞）时，f（x）≤φ（x）≤φ（2π）＜0恒成立，∴f（x）在[2π，+∞）上没有零点．综上，f（x）有且仅有两个零点．。
山东省菏泽市2019-2020学年高考数学三模试卷(理科)A卷
山东省菏泽市2019-2020学年高考数学三模试卷（理科）A卷姓名:________ 班级:________ 成绩:________一、选择题： (共10题；共20分)1. （2分）复数为虚数单位），则z的共轭复数是（）A . －iB . +iC . －－iD . －+i2. （2分）(2016·肇庆模拟) 已知U=R，函数y=ln（1﹣x）的定义域为M，N={x|x2﹣x＜0}，则下列结论正确的是（）A . M∩N=MB . M∪（∁UN）=UC . M∩（∁UN）=∅D . M⊆∁UN3. （2分）(2018·河北模拟) 若，则下列不等式不正确的是（）A .B .C .D .4. （2分） (2018高二上·大连期末) 若命题为真命题，则下列说法正确的是（）A . 为真命题，为真命题B . 为真命题，为假命题C . 为假命题，为真命题D . 为假命题，为假命题5. （2分）为了解某社区物业部门对本小区业主的服务情况，随机访问了100位业主，根据这100位业主对物业部门的评分情况，绘制频率分布直方图（如图所示），其中样本数据分组区间为[40，50），[50，60），[60，70），[70，80），[80，90），[90，100]．由于某种原因，有个数据出现污损，请根据图中其他数据分析，评分不小于80分的业主有（）位．A . 43B . 44C . 45D . 466. （2分）设实数x, y满足,则的取值范围是（）A .B .C .D .7. （2分） (2019高二上·扶余期中) 在空间直角坐标系中，，，，，则与平面所成角的正弦值为（）A .B .C .D .8. （2分）(2017·自贡模拟) △ABC中，∠C=90°，且CA=3，点M满足 =2 ，则• 的值为（）A . 3B . 6C . 9D . 不确定9. （2分） (2020高一上·武汉期末) 已知函数是奇函数，则的可能取值是（）A .B .C .D .10. （2分） (2020高二上·徐州期末) 已知椭圆的离心率为，过右焦点且斜率为的直线与相交于两点．若，则（）A . 1B .C .D . 2二、填空题： (共5题；共5分)11. （1分）(2017·青浦模拟) 执行如图所示的程序框图，若输入n=1的，则输出S=________．12. （1分） (2016高一下·新乡期末) 给出下列命题：①存在实数x，使sinx+cosx= ；②若α，β是第一象限角，且α＞β，则cosα＜cosβ；③函数y=sin（ x+ ）是偶函数；④函数y=sin2x的图象向左平移个单位，得到函数y=cos2x的图象．其中正确命题的序号是________（把正确命题的序号都填上）13. （1分）(2014·上海理) 若圆锥的侧面积是底面积的3倍，则其母线与底面角的大小为________（结果用反三角函数值表示）．14. （1分） (2017高二下·嘉兴期末) 已知点，圆，过点的直线l与圆交于两点，线段的中点为（不同于），若，则l的方程是________．15. （1分） (2016高二下·大丰期中) 函数y= x2﹣lnx的单调递减区间为________．三、解答题： (共6题；共55分)16. （10分） (2020高二上·吉林期末) 在中，，，已知，是方程的两个根，且．（1）求角的大小；（2）求的长．17. （15分） (2015高三上·和平期末) 如图，在三棱柱ABC﹣A1B1C1中，侧棱垂直于底面，∠BAC=90°，AB=AA1=2，AC=1，点M和N分别为A1B1和BC的中点．（1）求证：AC⊥BM；（2）求证：MN∥平面ACC1A1；（3）求二面角M﹣BN﹣A的余弦值．18. （10分）(2017·合肥模拟) 某供货商计划将某种大型节日商品分别配送到甲、乙两地销售．据以往数据统计，甲、乙两地该商品需求量的频率分布如下：甲地需求量频率分布表示：需求量456频率0.50.30.2乙地需求量频率分布表：需求量345频率0.60.30.1以两地需求量的频率估计需求量的概率（1）若此供货商计划将10件该商品全部配送至甲、乙两地，为保证两地不缺货（配送量≥需求量）的概率均大于0.7，问该商品的配送方案有哪几种？（2）已知甲、乙两地该商品的销售相互独立，该商品售出，供货商获利2万元/件；未售出的，供货商亏损1万元/件．在（1）的前提下，若仅考虑此供货商所获净利润，试确定最佳配送方案．19. （5分）(2020·山东模拟) 已知数列的前项和为，且（），数列满足，（）．（Ⅰ）求数列通项公式；（Ⅱ）记数列的前项和为，证明：．20. （5分）(2017·怀化模拟) 已知函数f（x）=x3+ax2+bx（x＞0）的图象与x轴相切于点（3，0）．（Ⅰ）求函数f（x）的解析式；（Ⅱ）若g（x）+f（x）=﹣6x2+（3c+9）x，命题p：∃x1 ，x2∈[﹣1，1]，|g（x1）﹣g（x2）|＞1为假命题，求实数c的取值范围；（Ⅲ）若h（x）+f（x）=x3﹣7x2+9x+clnx（c是与x无关的负数），判断函数h（x）有几个不同的零点，并说明理由．21. （10分） (2018高三上·静安期末) 设双曲线：，为其左右两个焦点．（1）设为坐标原点，为双曲线右支上任意一点，求的取值范围；（2）若动点与双曲线的两个焦点的距离之和为定值，且的最小值为，求动点的轨迹方程．参考答案一、选择题： (共10题；共20分)1-1、2-1、3-1、4-1、5-1、6-1、7-1、8-1、9-1、10-1、二、填空题： (共5题；共5分)11-1、12-1、13-1、14-1、15-1、三、解答题： (共6题；共55分) 16-1、16-2、17-1、17-2、17-3、18-1、18-2、19-1、20-1、21-1、21-2、。
2019-2020学年菏泽市第一中学高三英语一模试卷及答案解析
2019-2020学年菏泽市第一中学高三英语一模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AOne day when I was 12, my mother gave me an order: I was to walk to the public library, and borrow at least one book for the summer. This was one more weapon for her to defeat my strange problem inability to read.In the library,I found my way into the "Children's Room." I sat down on the floor and pulled a few books off the shelf at random. The cover of a book caught my eye. It presented a picture of a beagle. I had recently had a beagle, the first and only animal companion I ever had as a child. He was my secret sharer, but one morning, he was gone, given away to someone who had the space and the money to care for him. I never forgot my beagle. Without opening the book—Amos, the Beagle with a Plan ,1 borrowed it from the library for the summer.Under the shade of a bush, I started to read about Amos. I read very, very slowly with difficulty. Though pages were turned slowly, I got the main idea of the story about a dog who, like mine, had been separated from his family and who finally found his way back home. That dog was my dog, and I was the little boy in the book. At the end of the story, my mind continued the final scene of reunion, on and on, until my own lost dog and I were, in my mind, running together.My mother's call returned me to the real world. I suddenly realized something: I had read a book, and I had loved reading that book.I never told my mother about my “miraculous” experience that summer, but she saw a slow but remarkable improvement in my classroom performance during the next year. And years later ,she was proud that her son had read thousands of books, was awarded a PhD in literature, and authored his own books, articles, poetry and fiction. The power of the words has held.1. The author's mother told him to borrow a book in order to ________.A. let him spend a meaningful summerB. encourage him to do more walkingC. help cure him of his reading problemD. make him learn more about weapons2. The book caught the author's eye because .A. it reminded him of his own dogB. he found its title easy to understandC. it contained pretty pictures of animalsD. he liked children's stories very much3. Which one could be the best title of the passage?A. Mum's Strict Order.B. My Passion forReading.C. Reunion with My Beagle.D. The Charm of a Book.BTofight for the conservation of forest ecosystem, several ecologists including Daniel Janzen convinced Del Oro, an orange juice producer, to donate part of their forestland to a national park. In return, Del Oro was allowed to throw large amounts of waste in the form of orangepeels(皮) on a 3-hectare piece of land within the national park at no cost. Dealing with tons of leftover peels usually involved burning them or paying to have them poured into a landfill, so the proposal was very attractive.But a year later, another juice company challenged the deal in court, arguing that their competitor was "polluting a national park". They ended up winning, and the deal between Del Oro and the national park fell through.Then in 2013, while discussing possible research avenues（途径，手段）with Timothy Treuer, Daniel Janzen mentioned the orange story. Feeling interested, Treuer decided to stop by that piece of land that had been covered with fruit waste 15 years earlier. What he found shocked him."While I would walk over exposed rock and dead grass in the nearby fields, I'd have to climb through undergrowth and cut paths through walls ofvines(藤) in the orange peel site itself," said Timothy Treuer.Treuer and his team spent months picking upsamples(样品), analyzing and comparing them. They found great differences between the areas covered with orange peels and those that were not. The area withorange waste had richer soil.The effect that the orange peels had on the land is probably not that surprising to people familiar withcomposting(施肥), but what is really shocking is that a judge actually thought the waste of orange "mined" a national parkand stopped it from going forward. Now that Timothy Treuer's study has received worldwide attention, this type of "ruining" is being seriously considered as a way of bringing forests back to life.4. What did Del Oro usually do with orange peels?A. Add them to fuel.B. Feed them to animals.C. Burn or bury them.D. Make them into cakes.5. What can we know about the deal between Del Oro and the national park?A. It lasted 15 years.B. It was signed by Treuer.C. It was made in about 1998.D. It was broken by Del Oro.6. What was Treuer's finding?A. Orange peels contain much fibre.B. Orange peels can make soil richer.C. Orange peels rot away in a short time.D. Orange waste ruined the national park.7. What is the author's attitude toward the judge mentioned in the last paragraph?A. Disapproving.B. Positive.C. Worried.D. Admirable.COne day when I was 5, my mother criticized me for not finishing my rice and I got angry. I wanted to play outside and not to be made to finish eating my old rice. In my angry motion to open the screen door (纱门) with my foot, I kicked back about a 12-inch part of the lower left hand corner of the new screen door. But I had no regret, for I was happy to be playing in the backyard with my toys.Today, I know if my child had done what I did, I would have criticized my child, and told him about how expensive this new screen door was, and I would have delivered a spanking (打屁股) for it. But my parents never said a word. They left the corner of the screen door pushed out, creating an opening, a crack in the defense against unwanted insects.For years, every time I saw that corner of the screen, it would remind me of my mistake from time to time. For years, I knew that everyone in my family would see that hole and remember who did it. For years, every time I saw a fly buzzing in the kitchen, I would wonder if it came in through the hole that I had created with my angry foot. I would wonder if my family members were thinking the same thing, silently blaming me every time a flying insectentered our home, making life more terrible for us all. My parents taught me a valuable lesson, one that a spanking or stern (严厉的) words perhaps could not deliver. Their silent punishment for what I had done delivered a hundred stern messages to me. Aboveall, it has helped me become a more patient person and not burst out so easily.8. When the author damaged the door, his parents _______.A. scolded him for what he had doneB. left the door unrepairedC. told him how expensive it wasD. gave him a spanking9. How did the author feel every time he saw the damaged door?A. He felt ashamed of his uncontrolled anger at that time.B. He found that his family members no longer liked him.C. He found it destroyed the happy atmosphere at his home.D. He felt he had to work hard to make up for (弥补) the damage.10. The experience may cause the author _______.A. to hide his anger away from othersB. not to go against his parents’ willC. to have a better control of himselfD. not to make mistakes in the future11. What of the following is the main idea of this passage?A. Adults should ignore their children’s bad behavior.B. Parents shouldn’t educate their children.C. What is the best way to become a more patient person?D. Silent punishment may have a better effect on educating people.DMasks that helped save lives during the Covid-19 pandemic（疫情）are proving a deadly risk for wildlife, with birds and sea creatures trapped in many facial coverings in animal habitats.Single-use masks have been found on the ground, waterways and beaches worldwide since countries required（heir use in public places to slow the pandemic's spread. Worn once, the thin protective materials can take hundreds of years to break down. "Face masks aren't going away any time soon-but when we throw them away, these items can harm the environment and the animals who share our planet," Ashley from anima! rights group PETA said.Monkeys have been found playing with used masks in the hills outsideMalaysia's capitalKuala Lumpur. And in an incident inBritain, a seagull was saved inChelmsfordafter its legs got caught in an abandoned mask for a week.However, the biggest influence is in the water. More than 1.5 billion masks made their way into the world's oceans last year, accounting for around 6200 extra tons of ocean plastic pollution, according to environmental group OceansAsia. “Masks and gloves are particularlyproblematicfor sea creatures," says George Leonard, chief scientist from NGO. "When those plastics break down in the environment, they form smaller and smaller particles （颗粒）.Those particles then enter the food chain and influence the entire ecosystem,“ he added.Campaigners have urged people to deal with masks properly after using them. OceansAsia has also called on governments to increase punishment for littering and encourage the use of washable masks.12. What bring（s）a great danger to wildlife now?A. Waste masks.B. Covid-19.C. Polluted water.D. Damaged habitats.13. What does the underlined word “problematic”in paragraph 4 mean?A. Important.B. Attractive.C. Common.D. Troubling.14. What can we infer from the text?A. Monkeys learned to wear masks from humans.B. Plastics are less harmful after becoming particles.C. Used masks have a worse effect on sea creatures.D. Waste masks arc the main ocean plastic pollution.15. How should we solve the problem from the last paragraph?A. Keep masks after they' re used.B. Call on governments to stop littering.C. Punish those who wear single-use masks.D. Put used masks in the recycling box.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年菏泽市第一中学高三英语三模试题及答案解析
2020年菏泽市第一中学高三英语三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AThank you. It’s my great honor to be given this award.You cannot imagine that I have always been a late starter. Years ago, when I was 16, I took an important exam — GCE(General Certificate of Education), which turned out to be a failure. My dad was reading my report card and saw that my position in class was 29th, but the number in class was 29. It meant that I had achieved the distinction of being bottom of my class.I wasn’t lazy, and I was really trying. You can picture how I felt. Dad put his hand on my shoulder and said, “You can only do the best you can, but whatever you decide to do, make sure you love it.” He was a really sweet guy and a great man. I knew his attempt to hide his disappointment with some of his encouraging words. I was depressed for a week, but his advice was a wake-up call.Fortunately I love working with my hands, and I was good at two things: woodwork and art, and I really loved to draw and paint. I was quite talented. Dad strongly encouraged me to go to art school, which in those days wasn’t the obvious place that a father would suggest.So I got into Hartlepool College of Art. The college was a revelation (出乎意料), the passionate teachers there, who were extremely interested in the students, not just tolerating them but actually engaging with them. It was a world apart from my schooling until then. It’s extraordinary what an enthusiastic teacher can do, drawing the student out, lighting independence, and encouraging a design of your own future, rather than waiting for something to happen. I’m honored to have become one of these passionate teachers years later.My teachers inspired me, and thanks to my dad, here I am tonight. I think I should mention all the talents I have worked with over time, and to my kids and my wife Giannina, thank you.Thank you for this great award. I shall find a very special place for it.1. How did the author feel after taking GCE?A. Happy.B. Upset.C. Tired.D. Relieved.2. What didHartlepoolCollege of Art impress the author most?A. The teachers were strict with students.B. The students set good examples for each other.C. The teachers inspired students’ passion for learning.D. The students got prepared for their lessons independently.3. The author gave this speech to ________.A. share his career choiceB. explain his teaching methodsC. describe his life experienceD. show his appreciationBMany teens may feel anxious sometimes. It’s the kind of nervousness that makes you bite your nails before a big test. We spend more time online than we should. We feel good about ourselves or bad based on how many Likes and Followers we get on social media. Young people are developing a false view of life.On the screen, we see what people want to show us. People usually only post photos where they are looking their best. They are surrounded by friends and seem that they are having a great time. No one seems sad or lonely. In short, life isfabulous. But sooner or later, our young people compare their real life to it. They find that theirs doesn’t seem as fun or exciting and grow worried that they may be missing out.No wonder teachers are reporting more anxious students. It’s reported that a lot more college students feel ―overwhelming anxiety. The percentage jumped from 50% in 2011 to 62% in 2016. Anxiety is now the most common mental-health problem in my country. It affects nearly one-third of teens and adults.Certainly, we can’t blame it on social media alone. We expect toomuch from our children and a lot of these expectations aren’t reasonable. Their schedules are packed with sports, clubs and homework. They don’t have enough free time. We want our children to succeed, and we don’t care how much it costs.As parents, we must have more balance. On one hand, we push too hard, and on the other hand, we make life too easy for children. We shouldn’t and can’t promise our children that they will always be happy. We shouldn’t try to protect them from the problems of everyday life. Let them solve the problems in person.4. What is the text mainly about?A. What causes teens’ nervousness.B. How to deal with teens’ anxiety.C. What a view of life means to people.D. How to treat social media appropriately.5. What does the underlined word “fabulous” in paragraph 2 mean?A. Wonderful.B. Encouraging.C. Anxious.D. Doubtful.6. Why does the author mention the numbers in paragraph 3?A. To draw teachers’ attention.B. To show teachers’ mental problems.C. To present the seriousness of teens’ anxiety.D. To show adults have more problems than teens.7. What should parents do to help their children out?A. Try to meet their expectations.B. Help them with their homework.C. Give them more free time to play sports.D. Allow them to solve their own problems in life.CWhen a United Kingdomsupermarket chain promised to move its farms to Net Zero by 2030, it made it clear that the effort would require working on many different fronts. From energy consumption and land - use change to methane emissions (甲烷排放), cattle farming comes with environmental challenges. So even if recent studies suggesting it's possible to cut methane emissions 80% do turn out to be accurate, there's still a very long way to go for most cattle farming to get anywhere close to truly net zero.Organic Valley, when producing milk, might be closer to that goal. It made headlines in 2019 by going 100% renewable (可再生的). Now the company is expanding on that tradition by starting a major loan initiative (贷款计划) to help its farm suppliers adopt renewable energy too.Created with Clean Energy Credit Union, the $ 1 million loan fund will deliver loans at below - marketrates. Specifically, the money will be made available to Organic Valley's 1, 700 farmer members, and can be used for a variety of projects.“We are focused on a whole systems approach to renewable energy, and I'm excited to launch this energy loan fund. From the farm to the shelf, I see renewable energy playing a bigger role in organic food,” said Bob Kirchoff, Organic Valley CEO.“Organic Valley is already helping to protect the environment through organic farming practices, and now they re going one step further by supporting the introduction of renewable energy projects for their farmer members,” said Blake Jones, volunteer board chair of Clean Energy Credit Union. “In addition to the environmental benefits, we re eager to help family farmers throughout the world to lower their energy costs.”The world is not short of examples of farmers innovating in the field of renewable energy. What's encouraging about Organic Valley's announcement is the idea of a national brand putting its marketing and financing weight behind such efforts and, hopefully, creating consumer demand that pushes the rest of the industry in this direction too.8. What does paragraph 1 indicate about going net zero for cattle farming?A It is not easy to achieve B. It is common in the UKC. It is an impractical goalD. It meets no challenges9. What is Organic Valley's tradition according to the text?A. Helping farm suppliersB. Using renewable energyC. Having a loan initiativeD. Making headlines annually10. What is Organic Valley's initiative mainly intended to do?A. Reform organic farming practicesB. Make use of environmental benefitsC. Help farmers decrease energy costsD. Shrink the group of farmer members11. What is the author's attitude towards Organic Valley's initiative?A. WorriedB. DoubtfulC. AmbiguousD. PositiveDThe regular world presented to us by our five senses — you could call it reality 1.0 — is not always the most user-friendly. We get lost in unfamiliar cities; we meet people whose language we don’t understand. So why not try the improved version: augmented reality（AR）or reality 2.0 ? AR technology adds computer-produced images on the real world via a mobile phone camera or special video glasses.Early forms of AR are already here — smart phones can deliver information about nearby ATMs and restaurants and other points of interest. But that’s just the beginning. A few years from now the quantity of information available will have increased hugely. You will not only see that there’s a Chinese restaurant on the next block, but you will be able to see the menu and read reviews of it.This is where the next revolution in computing will take place: in the interface（界面）between the real world and the information brought to us via the Internet. Imagine bubbles floating before your eyes, filled with cool information about anything and everything that you see in front of you.Let’s jump ahead to ten years from now. A person trying to fix a car won’t be reading a book with pictures; he will be wearing a device that projects animated 3D computer graphics onto the equipment under repair, labelling parts and giving step-by-step guidance.The window onto the AR world can be a smart phone or special video glasses. But in ten years’ time these will have been replaced by contact lenses(隐形眼镜) with tiny LEDs, which present something at a readable distance in front of eyes. So a deaf person wearing these lenses will be able to see what people are saying.The question is, while we are all absorbed in our new augmented reality world, how willwe be communicating with each other?12. What is the text mainly about?A. The relationship between reality 1.0 and reality 2.0.B. Different forms of the AR technology.C. The next information technology revolution.D. The popularity of the AR technology.13. Which of the following will AR technology support according to the text?A. To pay for things online conveniently.B. To play online games merrily.C. To offer information efficiently.D. To communicate with others socially.14. What are Contact lenses with tiny LEDs used for?A. Show texts and images.B. Protect people’s eyes.C. Help deaf people communicate.D. Replace smart phone.15. What’s the author’s attitude towards the AR technology?A. Indifferent.B. Critical.C. Concerned.D. Favourable.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。
2019-2020学年菏泽市第一中学高三生物模拟试卷及参考答案
2019-2020学年菏泽市第一中学高三生物模拟试卷及参考答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 如图是雄性哺乳动物体内处于分裂某时期的一个细胞的染色体示意图。
相关叙述不正确的是（）A. 该个体的基因型为AaBbDdB. 该细胞正在进行减数分裂C. 该细胞分裂完成后只产生2种基因型的精子D.A、a和D、d基因的遗传遵循自由组合定律2. 下列关于生物体生命活动调节的叙述，不正确的是（）A.神经系统对躯体运动和内脏活动都存在分级调节B.神经系统的某些结构也能分泌激素，参与体液调节C.如果母鸡出现了在清晨打鸣的现象，与性激素的调控有关D.神经递质和抗体，都可以作为信息分子调节生物体的生命活动3. 除草剂敏感型的小麦经辐射获得了抗性突变体,敏感和抗性是一对相对性状。
关于突变体的叙述,正确的是A.若为基因突变所致,则再经诱变不可能恢复为敏感型B.若为一对碱基缺失所致,则该抗性基因一定不能编码肽链C.若为染色体片段缺失所致,则该抗性基因一定是隐性基因D.若为染色体易位所致,则四分体内一定发生过非姐妹染色单体片段交换4. 下列有关人类对遗传物质探索历程的叙述，错误的是（）A.格里菲思的肺炎双球菌体内转化实验证明了S型细菌中存在一种转化因子B.艾弗里及其同事的肺炎双球菌体外转化实验证明DNA是遗传物质C.赫尔希和蔡斯的实验证明了DNA是遗传物质，蛋白质不是遗传物质D.烟草花叶病毒侵染烟草的实验证明RNA是遗传物质5. 基因组稳定性是维持一切生命活动的基础，然而，多种外源和内源因素作用下产生的广泛DNA损伤和复制压力，构成了基因组不稳定的主要来源。
人体中有一种名为ATR激酶的蛋白质负责启动细胞对基因组不稳定的响应和修复，一旦感应到DNA损伤和复制压力会迅速活化，全局性地调控基因组的稳定。
下列叙述错误的是（）A.A TR激酶的合成场所是核糖体，合成时需要A TP提供能量B.在DNA复制过程中若A TR激酶丢失会增加分裂间期的基因组的不稳定性C.DNA复制时由于基因突变导致的DNA损伤，不一定引起生物性状的改变D.某些环境因素使基因中磷酸和核糖之间的化学键断裂会活化A TR激酶6. 图是根据生物的相似或不同点进行分类的, 下列选项中不是此图的分类依据的是A.有无叶绿体B.有无染色体C.有无细胞壁D.有无色素7. 下列有关显微镜操作的叙述，正确的有（）A.标本染色较深，观察时可选用凹面反光镜和调大光圈B.显微镜的放大倍数是指物像的面积或体积的放大倍数C.若要换用高倍物镜观察，需先升镜筒，以免镜头破坏玻片标本D.显微镜下所成的像是倒立放大的虚像，若在视野中看到细胞质顺时针流动，则实际上细胞质是逆时针流动8. 下丘脑在人体生命活动过程中有重要的调节作用，下列分析错误的是()A.由a造成，引起下丘脑分泌①促甲状腺激素释放激素，最后使甲状腺激素分泌增多B.由b造成，引起下丘脑释放①抗利尿激素，最后使细胞外液渗透压降低C.由c造成，下丘脑通过神经的作用，可以促使两种激素①肾上腺素、胰高血糖素分泌增多，最后使血糖升高D.由d引起，通过下丘脑的体温调节作用，可以引起①甲状腺激素增多，导致体内产热量增加9. 以下二倍体生物的细胞中含有两个染色体组的是( )①有丝分裂中期细胞①有丝分裂后期细胞①减数第一次分裂中期细胞①减数第二次分裂中期细胞①减数第一次分裂后期细胞①减数第二次分裂后期细胞A. ①①①B.①①①C.①①①①D.①①①①10. 水分子中氢原子以共用电子对与氧原子结合。
2019-2020学年菏泽市曹县一中高三语文三模试题及答案
2019-2020学年菏泽市曹县一中高三语文三模试题及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字,完成下面小题材料一当今是数字化的信息时代,阅读载体趋于多元化发展,网络小说电子条志、电子书等数字化阅读方式层出不穷,成为了和传统阅读并驾齐驱的阅读方式,如今几乎大部分经典畅销书籍都能在网络上找到其电子版本,甚至是有声图书,虽然目前纸质书籍并没有受到重创,仍然有相当部分读者愿意购买纸质图书,正如英国情报学家K麦克格雷所指出的:没有任何一种媒介可以完全取代另外一种媒介,总的情形是相互补充并逐步统一起来以解决一个特定的交流问题。
但毫无疑问,新的阅读方式已经分流了传统阔读的部分受众。
(摘编自《当今国民阔读习惯与趋势探计》)材料二阅读年龄与阅读类别图中国人习惯的阅读方式图材料三当今社会,人们习惯于畅游互联网,有时间读书似乎成了一种奢求。
虽然精神食粮不再乏,各类书籍品种丰富,应有尽有,遗憾的是,如今我们却很难再见到人人都爱读书的场景。
读书的习惯都去哪儿了,引人深思“全民阅读”今年已是第五次写入政府工作报告,倡导全民阅读,建设学习型社会”成为其中的重要内容。
据不完全统计,日前全国已有400多个城市建立了区域的阅读节、阅读月,开展了传统文化讲学、经典诵读、亲子阅读等主题阅读活动;由多家出版社出版的50多种图书入选中国出版政府奖、中华优秀出版物奖和国家有关部门推荐的中国好书。
全民阅读的倡导,能够让我们的老百性通过多读书,读好书,去获得更多的幸福感和得感,从而增强文化自信,进而达到全民悦读。
“少年强、青年强则中国强。
”金民阅读更应该从娃蛙抓起,让孩子们从小把古代经典嵌在脑子里,薪火相传,与时俱进,推陈出新。
(摘编自新华网)材料四陶明在《五柳先生传》中曾写到:“好读书,不求甚解;每有会意,便欣然忘食。
“陶渊明读书时注意抓住重点、去繁就简和独立考。
实际上,他追求的是读书会意,着重领会书中深含的旨意,而不死抠个别字句。
2019-2020学年菏泽市第一中学高三英语三模试题及答案解析
2019-2020学年菏泽市第一中学高三英语三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AExciting Adventure Options to Choose From!BIRD WALK(Any time of year)-Join us for a private bird walk through our sanctuary(保护区）。
The Bent's grasslands, trees and woods provide great habitat(栖息地）for birds moving from one place to another, such as Warblers, Vireos, Indigo Buntings, Thrushes, Orioles, and more. This walk will be made to the members in your party.Suitable for ages 5 and upProgram Fee:$150NATURE HIKE(Any time of year)-Take a private hike with Bent of the River! Your personal guide will show you notable habitats and wildlife around the center trails. Nature is exciting and always changing, so you never know what we will find along the way! This program is ideal for people who want to enjoy beautiful scenery while hiking.Suitable for ages 8 and upProgram Fee:$150POMPERAUG RIVER EXPLORATION(June and July only)-Many fascinating creatures live in and around the Pomperaug River! During this recreational(休闲的)program, an Audubon naturalist will share the human and natural history of the river and teach you how to catch fish and animals. The Bent will supply you with necessary tools, such as nets, containers, and field guides. Once the animals are caught, we will observe and identify(确定身份)them and learn how they can help show the health of the river before we put them back to the wild.Suitable for ages 8 and upProgram Fee:$150OWL PROWL(January and February only)-Enjoy a special guided adventure in search of one of the most beloved groups of birds-owls(猫头鹰）！We will be prowling for owls on awalk through the grassland and forests in hopes of seeing one of the three owl species known to live in Connecticut: the Great-horned Owl, Barred Owl, or Eastern Screech-Owl.Evening eventSuitable for ages 10 and upProgram Fee:$2251.Which of the programs is suitable for the Browns with a girl of five years old?A.BIRD WALK.B.NATURE HIKE.C.POMPERAUG RIVER EXPLORATION.D.OWL PROWL.2.What will you do with the fish you catch in POMPERAUG RIVER EXPLORATION?A.Find out their health.B.Do a scientific research.C.Cook them as food on the table.D.Set them free back to the river.3.Whom is this text written for?A.Students.B.Teachers.C.Scientists.D.Adventurers.BWhy isn’t science better? Look at career incentive(激励).There are oftensubstantial gaps between the idealized and actual versions of those people whose work involves providing a social good. Government officials are supposed to work for their constituents. Journalists are supposed to provide unbiased reporting and penetrating analysis. And scientists are supposed to relentlessly probe the fabric of reality with the most rigorous and skeptical of methods.All too often, however, what should be just isn’t so. In a number of scientific fields, published findings turn out not toreplicate(复制), or to have smaller effects than, what was initially claimed. Plenty of science does replicate — meaning the experiments turn out the same way when you repeat them -but the amount that doesn’t is too much for comfort.But there are also waysin which scientists increase their chances of getting it wrong. Running studies with small samples, mining data for correlations and forming hypotheses to fit an experiment’s results after the fact are just some of the ways to increase the number of false discoveries.It’s not like we don't know how to do better. Scientists who study scientific methods have known about feasible remedies for decades. Unfortunately, their advice often falls ondeaf ears.Why? Why aren't scientific methods better than they are? In a word: incentives. But perhaps not in the way you think.In the 1970s, psychologists and economists began to point out the danger in relying on quantitative measures for social decision-making. For example, when public schools are evaluated by students’ performance on standardized tests, teachers respond by teaching “to the test”. In turn, the test serves largely as of how well the school can prepare students for the test.We can see this principle—often summarized as “when a measure becomes a target, it ceases to be a good measure”—playing out in the realm of research. Science is a competitive enterprise. There are far more credentialed (授以证书的) scholars and researchers than there are university professorships or comparably prestigious research positions. Once someone acquires a research position, there is additional competition for tenure grant funding, and support and placement for graduate students. Due to this competition for resources, scientists must be evaluated and compared. How do you tell if someone is a good scientist?An oft-used metric is the number of publications one has in peer-reviewed journals, as well as the status of those journals. Metrics like these make it straightforward to compare researchers whose work may otherwise be quite different. Unfortunately, this also makes these numbers susceptible to exploitation.If scientists are motivated to publish often and in high-impact journals, we might expect them to actively try to game the system. And certainly, some do—as seen in recent high-profile cases of scientific fraud(欺诈). If malicious fraud is the prime concern, then perhaps the solution is simply heightened alertness.However, most scientists are, I believe, genuinely interested in learning about the world, and honest. The problem with incentives is that they can shape cultural norms without any intention on the part of individuals.4. Which of the following is TRUE about the general trend in scientific field?A. Scientists are persistently devoted to exploration of reality.B. The research findings fail to achieve the expected effect.C. Hypotheses are modified to highlight the experiments' results.D. The amount of science that does replicate is comforting.5. What doesdeaf earsin the fourth paragraph probably refer to?A. The public.B. The incentive initiators.C. The peer researchers.D. The high-impact journal editors.6. Which of the following does the author probably agree with?A. Good scientists excel in seeking resources and securing research positions.B. Competition for resources inspires researchers to work in a more skeptical way.C. All the credentialed scholars and researchers will not take up university professorships.D. The number of publication reveals how scientists are bitterly exploited.7. According to the author, what might be a remedy for the fundamental problem in scientific research?A. High-impact journals are encouraged to reform the incentives for publication.B. The peer-review process is supposed to scale up inspection of scientific fraud.C. Researchers are motivated to get actively involved in gaming the current system.D. Career incentives for scientists are expected to consider their personal intention.CMark 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.8. 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 time9. 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 Twain10. 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 atParis11. 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.DOne-year-old Tallulah turned purple and stopped moving after the sweet became stuck in her throat. Her mum Leigh-Anne said the drama began during a visit to her grandma’s house when her grandparents gave her older kids some sweets.“Then at about 4:45 pm, Tallulah started to choke—we all went into a panic.”“It seemed like it went on for ages. Not one of us knew what to do.”“I rang an ambulance while my grandma and granddad tried to get the sweet to come up.”“Tallulah was panicking at first but then she started to go purple—she almost had no oxygen left in her.”With her daughter limp (无力的) and time running out, Leigh—Anne knew she couldn’t afford to wait for the ambulance to arrive.“The only thing I could think was to go out into the street.” She said.“I rushed out and screamed for someone to help while my grandma rushed out crying with Tallulah.”At exactly the moment, Caitlin, who is studying public services atRedcarCollege, was passing byQueen Street. She said, “I was waiting to go to work when I heard someone screaming for help, so I ran straight over.”The 17-year-old girl added, “Something just clicked and I went into auto mode. The little girl was completely limp, so I checked her airways and tilted (使倾斜) her over and started hitting her back. I turned her round and tapped on her chest, then after what felt like forever she coughed up the sweet and spat it out.As soon as she started crying I felt a huge relief. I was just so pleased I was able to help.”Caitlin was taught her lifesaving skills when she joined the Army Cadets four years ago.12. When did Tallulah get choked?A. While eating sweets.B. While enjoying a drama.C. While having a meal.D. While taking some medicine.13. Why did the family go out into the street?A. To buy some needed tools.B. To search for timely help.C. To get a breath of fresh air.D. To wait for the ambulance to arrive.14. Which of the following can best describe Caitlin?A. Brave and selfless.B. Kind and energetic.C. Determined and generous.D. Quick-thinking and helpful.15. What may be the best title for the text?A. First aid skill sounds important.B. Screaming for help makes sense.C. Eating sweets endangers baby girl.D. Heroic teenager saves baby girl’s life.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。
2019-2020学年菏泽市单县一中高三英语三模试题及参考答案
2019-2020学年菏泽市单县一中高三英语三模试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ACome and enjoy Vivaldi's TheFour Seasonsperformed by live musicians!Tickets△Zone A Sating (Excellent Visibility, $75)△Zone B Seating (Great Visibility, $60)△Zone C Seating (Good Visibility, $45)△Zone D Seating (Restricted Visibility, 30)Zone A and Zone B audiences will get the chance to take pictures with the performers on the stage after the show.Highlights* A beautiful venue bathed in candlelight.*Classical music performance by the Angel Strings quartet*A safe and socially-distanced event, ensuring you are comfortable and at ease.General Info*Dates and times: Various dates, at 6:30 pm and 8:30 pm (select during purchase).*How long: 65 minutes. Doors open 45 minutes before the start time. We recommend you arrive at least 30 minutes before the start of the event, as late entry is not permitted.*Where: Events on Oxlade*Age requirement: Must be 8 years old or older to attend. Anyone under the age of 16 must be accompanied by an adult.*Please note: The 6:30 pm seating will take place during daylight hours outdoors, and the space will not be that dark. In the case of rain, the event will be moved to the indoor area of the venue.DescriptionWhether you're looking for a beautifully unique classical music performance or a romantic candlelit experience, this performance is for you. You don't need to know all things about Vivaldi to enjoy the evening; simply sit back and admire the wonderful atmosphere and the pieces you'll hear.Join our musicians for an evening under the stars, and prepare to be taken into the clouds with Vivaldi' s most treasured masterpieces!1.What can someone with a $45 ticket do?A.Perform on the stage.B.Enjoy good visibility.C.Select a seat in Zone B.D.Take photos with the musicians.2.What should potential audiences keep in mind?A.Arrive at the venue on time.B.Learn about Vivaldi in advance.C.The performance lasts 45 minutes.D.The event will be canceled if it rains.3.What do we know about the 8:30 p.m. performance?A.It welcomes children under the age of 8.B.Its performers differ on different dates.C.Its stage will be decorated with candles.D.It will be shown in the indoor area of the venue.BBob, a Burroughs junior high school football player, always had his mom cheering him on. He didn’t play exciting positions. He played as a linebacker(中后卫球员). Sadly, he often found himself at the bottom of the piles, where everyone would jump onto each other at the end of every play. Bob's mom realized it was hard for her son to hear her cheering. She hadto find a solution, but couldn't find one.Then one day the coach from the school team asked him if he wanted to join the team. Bob wasecstatic, because he was only a ninth grader. His mom was also excited, since she loved football and especially loved watching her son play. She kept considering a way for him to hear her. A cowbell! That was it. Now from the bottom of the piles Bob would hear his mom shaking her cowbell crazily, knowing she was there for him.Bob's team finally made it to the state championship game. What exciting time to play at Busch Stadium under the lights! This experience made Bob appreciate all the years that his mom had sacrificed everything to get him to practice every day, to wash his uniforms, and to never miss a game. He had to do something.On the night of the state championship game, the loudspeakers introduced Bob, and as he walked onto the field his mom shook the cowbell, hard. However, it didn’t sound right. She looked its inside, and found a note saying,”Thank you, Mom.” Bob had left her a note expressing his appreciation for always being with him, filling her heart with warmth.Finally Burroughs claimed the title of State Champion. While others were cheering and admiring the statechampionship cup, Bob' s mom clutched(紧握) her cowbell happily.Years later, Bob’s mom died. While digging through her belongings, he found the cowbell with the note. Bob took it to his mom's funeral and rang it, whispering, "Thank you, Mom.”4. What does the underlined word "ecstatic” mean in Paragraph 2?A. CuriousB. DisappointedC. AnxiousD. Delighted5. Why did Bob's mom want to get a cowbell?A. To amuse her cow.B. To teach her cowC. To attract his attention.D. To make him hear her6. What can be learned from Paragraph 3?A. Bob's mom devoted much time to himB. Bob was the best player in his teamC. Bob owed his success to his coachD. Bob's mom was a football player at college7. Why didn’t the cowbell sound as usual that night?A. The mother was very weak.B. Bob had put a note inside the bell.C. The weather became terrible suddenlyD. The bell had been broken deliberates.CA new study has discovered that meditation (冥想) and oxygen sport together reduce depression. The Rutgers University study found that this mind and body combination, done twice a week for only two months,reduced the symptoms for a group of students by 40 percent.“We are excited by the findings because we saw such a meaningful improvement in both clinically depressed and non-depressed students,” said lead author Dr. Brandon Alderman. “It is the first time that both of these two behavioral ways have been looked at together for dealing with depression.”Researchers believe the two activities have an interactive effect on combatingdepression. Alderman and Dr. Tracey Shors discovered that a combination of mental and physical training (MAP) enabled students with major depressive disorder not to let problems or negative thoughts defeat them.Rutgers researchers say those who participated in the study began with 30 minutes of focused attentionmeditation followed by 30 minutes of oxygen sport. They were told that if their thoughts drifted to the past or the future they should refocus on their breathing, enabling those with depression to accept moment-to-moment changes in attention.Shors, who studies the productionof new brain cells in the hippocampus—part of the brain involved in memory and learning—says scientists have shown in animal models that oxygen sport exercise keeps a large number of certain cells alive.The idea for the human intervention (干预) came fromher laboratory studies, she says, with the main goal of helping individuals acquire new skills so that they can learn to recover from stressful life events.By learning to focus their attention and exercise, people who are fighting depression can acquire new learning skills that can help them process information and reduce the overwhelming recollection of memories from the past, Shors says.“We know these treatments can be practiced over a lifetime and that they will be effective in improving mental health.” said Alderman. “The good news is that this intervention can be practiced by anyone at any time and at no cost.”8. What made the research so different?A. Adopting a way of meaningful talk.B. Combining the two behavioral ways to treat depression.C. Treating depression with special medicine.D. Comparing the depressed with the non-depressed.9. The underlined word “combating” in Paragraph 3 can be replaced by ______.A. fightingB. identifyingC. distinguishingD. examining10. What did the participants do in the research?A. They did oxygen sport half an hour before thinking.B. They thought quietly and then took exercise.C. They took exercise longer than they thought.D. They took exercise while thinking quietly.11. What is Shors’ main purpose of her studies?A. To find out certain brain cells of humans.B. To study the production of new brain cells.C. To offer people a new method to treat stress.D. To decide the links between stress and exercise.DIf you've ever had a dog, you know just howdeep a connection you can develop with “man's best friend”. But a dog's life is much shorter than humans, about 12 to 15 years long, which means every dog owner has to go through the heartbreaking moment when their loving pet passes away.Why not make a clone of that dog then? This is the solution offered by a South Korean company, Sooam Biotech Research Foundation. The company has already successfully cloned at least 400 dogs, mostly for US customers, ever since it pioneered the technique in 2005. Now, Sooam Biotech has introduced its business toUKdog owners as well, offering them dogs that look just like their lost ones.To clone a dog, researchers first need to take a skin cell from a living dog or one that has just died. Meanwhile,another dog is selected to supply an egg. Researchers then replace the DNA in the egg with that from the skin cell and implant the egg into the womb (子宫) of a female dog. The egg grows into a puppy over the following two months. The whole process takes less than a day, but it comes at a shockingly high price — around ￡63,000.But if you can't afford it now, you can also save the cell in a laboratory andaccess it at a later date.However, magical as cloning might sound, there is no guarantee that the cloned dog will be a perfect copy of the original one. Just like identical twins of humans, they share the exactly same DNA but there will still be small differences between them. “The spots on a Dalmatian (斑点狗) clone will be different, for example” Insung Hwang, head of Sooam Biotech, told The Guardian.Dog owners will also have to accept the fact that personality is not “cloneable”. Apart from genes, personality is also determined by upbringing and environment, which are both random elements that cloning technologies simply cannot overcome, Professor Tom Kirkwood atNewcastle University,UK, told The Telegraph.Perhaps bringing our dogs back by cloning is not the best way to remember them after all.Kirkwood, a dog owner himself, pointed out, “An important aspect of our relationship with them is coming to terms with the pain of letting go.”12. What service does Sooam Biotech Research Foundation offer?A. Making copies of pet dogs.B. Giving pet dogs identical twinsC. Helping dogs give birth to more puppies.D.Helping dog owners love their dogs more.13. Which order is correct in the dog cloning process?a. An egg is taken from another dog.b. A skin cell is taken from the pet dog.c. The egg grows into a puppy in two months.d. The egg is placed in the womb of a female dog.e. The DNA in the egg is replaced by the DNA from the skin cell.A.a→d→b→e→c.B. a→e→b→d→cC. b→a→d→e→c.D. b→a→e→d→c.14. What can we learn about dog cloning from the passage?A. It has not been put into practice until recently.B. It is very popular among US andUKpet owners.C. It might not give the owners an exactlysame dog.D. It is very expensive and usually takes half a year to complete.15. What doesKirkwoodthink of dog cloning?A. He disagrees with it.B. He supports it.C. He is curious about it.D. He thinks it unbelievable.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。
2019-2020学年菏泽市第一中学高三生物第三次联考试卷及答案解析
2019-2020学年菏泽市第一中学高三生物第三次联考试卷及答案解析一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. PM2．5是指大气中直径≤2．5μm的颗粒物，富含大量有毒、有害物质，易通过肺部进入血液。
目前PM2．5已成为空气污染指数的重要指标。
下列有关PM2．5的推测正确的是（）A.PM2．5进入人体肺泡中即进入了人体的内环境B.颗粒物中的一些酸性物质进入人体血液将导致血浆最终呈酸性C.PM2．5可能成为过敏原，其诱发的过敏反应属于免疫缺陷症D.颗粒物进入呼吸道引起咳嗽属于非条件反射2. 灵菌红素具有抗肿瘤和抑制T细胞介导的免疫反应的功能。
科研人员在橡胶树内生菌ITBBB5-1中发现一个新细胞器——马阔囊胞，该囊胞位于杆状细菌末端。
电镜观察发现，灵菌红素位于马阔囊胞和其分泌的胞外小囊泡中。
下列说法错误的是（）A.橡胶树内生菌ITBBB5-1的核膜包含不连续的四层磷脂分子B.新型细胞器马阔囊胞可能具有储存和分泌灵菌红素的功能C.马阔囊胞分泌灵菌红素过程体现了生物膜的结构特点D.灵菌红素在癌症治疗和器官移植中具有潜在的应用价值3. 赫尔希和蔡斯的T2噬菌体侵染大肠杆菌实验证实了DNA是遗传物质，下列关于该实验的叙述正确的是（）A. 实验需分别用含32P和35S的培养基培养噬菌体B. 搅拌目的是使大肠杆菌破裂，释放出子代噬菌体C.35S标记噬菌体的组别，搅拌不充分可致沉淀物的放射性增强D.32P标记噬菌体的组别，放射性同位素主要分布在上清液中4. 下列关于植物激素的叙述，正确的是（）A.植物体内不同的腺体能够分泌不同的激素B.可向未完成受粉的水稻喷洒适宜浓度的生长素溶液以减少损失C.缺乏氧气会影响植物体内生长素的极性运输D.植物横放状态下茎的弯曲生长体现了生长素的两重性5. 肉瘤病毒为逆转录病毒，它携带有逆转录酶和整合酶（化学本质为蛋白质），能使RNA（甲）逆转录成双链DNA（乙）。
山东省菏泽市2019届高三3月一模考试英语试题含答案
第Ⅰ卷第一部分听力（共两节，满分30分）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，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
例：How much is the shirt?A.￡19.15.B.￡9.18.C.￡9.15.答案是C。
1.Where did this conversation most probably take place?A.In a restaurant.B.In a shop.C.In a vegetable market.2.What did the man do last night?A.He went to visit a friend.B.He went to say goodbye to his friend at the airport.C.He went to another city with his friend.3.What can we learn about the man?A.He enjoys using e-mails.B.He often receives letters from friends.C.He never writes letters to his friends.4.What time does the next plane to London leave?A.At 10:00.B.At 11:00.C.At 12:00.5.What is the man’s pr oblem?A.He isn’t feeling wel l.B.He is caught in bad weather.C.He feels very cold.第二节（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学年菏泽市第一中学高三生物第三次联考试题及参考答案
2019-2020学年菏泽市第一中学高三生物第三次联考试题及参考答案一、选择题：本题共15小题，每小题2分，共30分。
每小题只有一个选项符合题目要求。
1. 水稻细胞和人的口腔上皮细胞中共有的细胞器是（）A.叶绿体、线粒体和中心体B.叶绿体、线粒体和高尔基体C.线粒体、内质网和高尔基体D.线粒体、内质网和中心体2. 电子显微镜下观察水稻根细胞，可观察到的结构有（）A.①①①①①B.①①①C.①①①①D.①①①①3. 关于人体内环境稳态的叙述，错误的是( )①血浆渗透压与蛋白质含量有关，与无机盐离子含量无关①人吃酸性食品会导致体内的pH降低①每个人的体温在一天中是保持不变的①内环境稳态的维持需要多种器官、系统的协调作用，并且人体维持内环境稳态的调节能力有限①严重腹泻、呕吐，只需要补充足够的水，不用补充Na＋A.①①①①B.①①C.①①①①D.①①①①4. 一对同源染色体上的DNA分子之间一般不同的是A.相邻核苷酸之间的连接方式B.碱基种类C. (A＋G)/(T＋C)的比值D.碱基序列5. 下图表示人体内的细胞与外界环境进行物质交换的过程，下列叙述正确的是（）A.从外界环境摄入的O2进入肝脏细胞的途径为外界环境→呼吸系统→A→①→①→肝脏细胞B.内环境是细胞代谢的主要场所C.毛细淋巴管壁细胞生活的环境是①①D.肾炎导致①渗透压增加，①减少6. 蛋白质是生命活动的主要体现者，下列叙述错误的是（）A.蛋白质的特定功能都与其特定的结构有关B.唾液淀粉酶进入胃液后将不再发挥催化作用C.细胞膜上的某些蛋白质起着物质运输的作用D.蛋白质的结构一旦改变将失去生物学活性7. 在生命系统的各个层次中，能完整地表现出各项生命活动的最小的层次是（）A. 个体B. 细胞C. 种群和群落D. 各种化合物8. 交感神经和副交感神经是神经系统的重要组成部分，下列有关它们的叙述正确的是（）A.它们包括传入神经和传出神经B.它们都属于中枢神经系统中的自主神经系统C.它们通常共同调节同一器官，且作用通常相反D.交感神经使胃肠蠕动加强，副交感神经使胃肠蠕动减弱9. 在如图所示的玻璃容器中，注入一定浓度的NaHCO3溶液并投入少量的新鲜绿叶碎片，密闭后，设法减小液面上方的气体压强，会看到叶片沉入水中。
2019-2020学年菏泽市鄄城一中高三英语三模试卷及答案
2019-2020学年菏泽市鄄城一中高三英语三模试卷及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ATop Four Places to Visit in and aroundBaliUbudUbud is pretty away from the beaches. It is considered the cultural center of Bali, where you will discover the art and spirit ofBaliand learn about Balinese religion through paintings, dance, and other art forms. You could drink a beer while watching a local band perform, but most people come here for the cultural appreciation.Nusa LembonganThis tiny island off the coast of Bali is where many people miss their journey plans, but it is a must-see, It is the perfect place to stay for a couple of nights, The roads are not good, but the island itself is so beautiful and quiet.SanurSanur is an especially relaxing beach town. It's close to the airport and Kuta. It's also the starting place to explore Nusa Lembongan.KutaThebeachofKutais great for surfing. It is crowded and you won't get far without someone trying to get you to rent a surfboard or buy something. The main town is a mixture of market stands and shopping malls containing the biggest brands, from Zara to Ralph Lauren. The familiarity of the western style shops and restaurants won't give anyone too much of a culture shock.1. Which place are you probably advised to experience Balinese culture?A. Nusa Lembongan.B. Sanur.C. Ubud.D. Kuta.2. What is special about Kuta?A. It is a beach town near the airport.B. It has a convenient transportation means on it.C. It is a shopping center and famous for surfing.D. It is a place where you can learn Balinese history.3. Where is the text probably taken from?A. An official document.B. A travel magazine.C. A science report.D. A history book.BNarasimha Das is on his way to feed 169,379 hungry children. Das is in charge of a kitchen in Vrindaban. The town is about a three-hour drive fromIndia’s capital,New Delhi. Das gets to work at 3:00 a.m. Thirty workers are already working to make tens of thousands of rounds of bread. It will be brought to 1,516 schools in and around Vrindaban.A Growing ProblemGoing to school is difficult for more than 13 million children inIndia. They must go to work instead, or go hungry. That’s whyIndiabegan the Mid-Day Meal Scheme, the largest school-lunch program in the world. A free lunch encourages children to come to school and gives them the energy they need for learning. The program began in the 1960s.The kitchen in Vrindaban is run by the Akshaya Patra Foundation. It is one of the lunch program’s biggest partners. “Just $11.50 can feed one child for an entire year,” said Madhu Sridhar, president of the Akshaya Patra Foundation.Lunch Is Served!The Akshaya Patra food truck arrives atGopalgarhPrimary School. Since the program started, the number of underweight children has gone down. The children get foods they need — as long as they finish what’s on their plates.4. What does Narasimha Das do？A. A waiter.B. A salesman.C. A cook.D. A shopkeeper.5. The kitchen in Vrindaban supplies food to ________.A. the poorB. school childrenC. college studentsD. the old6. Why is it difficult for children to go to school inIndia?A. Because they have to work to make money.B. Because there are not enough schools.C. Because there are not enough teachers.D. Because their parents refuse to send them to school.7. Which of the following about the Mid-Day Meal Scheme is NOT true?A. It is to encourage children to go to school.B. It has been carried out for about 50 years.C. It is run by Narasimha Das.D. It is the largest school-lunch program in the world.CImaginary 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.8. 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.9. 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.10. 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.11. What is the author's attitude towards the effect of imaginary friends?A. Concerned.B. Doubtful.C. Optimistic.D. Indifferent.DMark Bertram lost the tips of two fingers at work in 2018 when his hand became trapped in a fan belt. “It’s life-changing but it’s not life-ending,”he says.After two surgeries and occupational therapy, Bertram decided to ask Eric Catalano, a tattoo artist, to create fingernail tattoos. The idea made everyone in the studio laugh—until they saw the final result. “The mood changed,” Catalano recalls from his Eternal Ink Tattoo Studio in Hecker, Illinois. “Everything turned from funny to wow.”Catalano posted a photo of the tattoos, and it eventually was viewed by millions of people around the world. The viral photo pushed Catalano, 40, further into the world of paramedical tattooing. Now people who want to cover their life-altering scars come from as far away as Ireland to visit his shop.Leslie Pollan, a dog breeder, was bitten on the face by a puppy. She underwent countless surgeries but those gave her no hope. She ultimately traveled six hours for a session with Catalano. HecamouflagedPollan’s lip scar, giving her back confidence.Though he is now known for his talent with intricate fingernail, Catalano uses the techniques he picked up years ago while helping breast cancer survivors. Those tattoos are among the most common paramedical requests. His grandmother had breast cancer, and her battle with the disease is one reason Catalano is so dedicated to helping those with the diagnosis.Catalano performs up to eight reconstructive tattoos each “Wellness Wednesday”. While he charges $100 per regular tattoo, he doesn’t charge for paramedical tattoos: A GoFundMe page established last year brought in more than $16,000, allowing Catalano to donate his work.“Financially, it doesn’t make sense,” Catalano says. “But every time I see emotions from my customers, I am 100 percent sure this is something that I can’t stop doing.”12. How did people in the studio react to Bertram’s idea at first?A. They took it lightly.B. They found it creative.C. They were confused.D. They were impressed.13. What does the underlined word “camouflaged” in Paragraph 4 probably mean?A. Exposed.B. Hid.C. Ignored.D. Removed.14. What does Catalano say about his work with paramedical tattoos?A. It is flexible.B. It is demanding.C. It is profitable.D. It is rewarding.15. Which of the following can best describe Catalano?A. Humorous and experienced.B. Devoted and generous.C. Cooperative and grateful.D. Professional and tolerant.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。
山东省菏泽市第一中学2019届高三模拟语文试卷 含解析
2019年山东省菏泽市第一中学高三模拟语文试题2019/6/2一、现代文阅读（36分）（一）论述类文本阅读（本题共3小题，9分）阅读下面的文字，完成各题我国合同法颁布至今不过20年，但合同制度在我国则是历史悠久。
在中华传统法律文化中，合同一般被称为契约。
如今存世的传统契约文书，是中国先民们留下的文化速产，传统契约中展现的契约观念，至今仍直接或间接地影响着人们对合同的看法。
史料表明，我国传统契约实践有2000年以上的历史。
目前能够解读出来的最早传世契约资料铭刻于青铜礼器上，记录了西周贵族之间就土地、奴隶等财产进行交换的行为。
《周礼》还记录了先秦时期使用竹木制作傅别、书契和质剂三种契约券书的方法，这些古老的契约应用于当时买卖、借贷等交易行为。
汉晋时期，人们依然以竹木制作交易券书，内容简单直接，满足了当时人们订立合同的需要。
东晋以后，纸张开始应用于契约书写。
吐鲁番出土和敦煌发现的纸质契约跨越了我国北朝、唐和五代时期，记录了近600年间买卖、借贷、租佃等丰富的契约行为。
历史还记载，北宋时为了减少契约纠纷，曾出现过由官方审定并印制的榜样契约。
从徽州等地发现的数以万计的传统契约来看，南宋以来契约已经广泛应用于人们生活的方方面面。
尽管我国传统社会中没有现代意义上的合同法，但契约制度并不缺乏，仅从文献记载看，传统契约制度包括国家法律和社会习愤两个方面。
法律方面，唐代及其后各代法典对不同类别契约所需要满足的交易条件都有明确规定。
社会习惯方面，历代官箴、村规民约以及家法族规对传统契约制度也有不同程度的记载。
此外，传统契约制度还直接体现在流传下来的契约文本中。
契约往往由职业或半职业的代书人书写。
代书人为了方便，会根据当时的制度规定，结合缔约习惯制定各种契约契式，并装订成册。
随着印刷技术的发展，明清时的士人开始收集整理契式文本，印成书并广为传播。
在缔约时，交易者会要求一并移交上手老契，并将其与新订契约、纳税凭证等相关文书粘附在一起。
2019-2020学年菏泽市郓城第一中学高三语文三模试卷及答案解析
2019-2020学年菏泽市郓城第一中学高三语文三模试卷及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
我们时代的塑胶跑道迟子建哈尔滨对于我来说，是一座埋藏着父辈眼泪的城，在后辈的写作者眼里，它可以是一只血脚印，也可以是一颗露珠。
我十七岁前的行迹，在连绵的大兴安岭山脉。
山脉像长长的线，日月之光是闪亮的针，把我结结实实缝在它的怀抱中。
初春的风认识我，夏日的溪流认得我，秋天时的薄冰认得我，冬天生产队的牛马认得我，我是大自然围场里的一个小小生物。
我对哈尔滨最早的认知，是从父亲的回忆中来的。
父亲童年不幸，我奶奶去世早，爷爷便把父亲从帽儿山，送到哈尔滨的四弟家，而他四弟是在兆麟公园看门的，多子多女，生活拮据。
父亲读中学时寄宿，他常在酒醉时讲他去食堂买饭，不止一次遭遇因没有续上伙食费而被停伙的情景。
贫穷和饥饿的滋味，被父亲过早地尝到了。
父亲功课不错，小提琴拉得也好，但家里没钱供他继续求学，中学毕业后，他没跟任何人商量，独自报名来参加大兴安岭的开发建设。
父亲这一去，直到1986年因病辞世，近三十年没回过哈尔滨。
而他留给我的哈尔滨故事，多半浸透着眼泪。
1990年，我从大兴安岭师范学校调转到哈尔滨工作。
每次去兆麟公园，我都会忧伤满怀，想着这曾是父亲留下足迹的地方啊。
初来哈尔滨，我的写作与这座城市少有关联，虽是它的居民，但更像个过客。
直到上世纪末我打造《伪满洲国》，哈尔滨作为历史主场景，我无法回避，所以开始读城史，在作品中尝试建构它。
但它始终没有以强悍的主体风貌在我作品中独立呈现过。
二十年过去了，我在哈尔滨生活日久，了解愈深，很想对它进行一次酣畅淋漓的文学表达。
2019年，我开始了《烟火漫卷》的写作。
写完第二章，我随作协代表团访欧，虽然旅途中没有续写，但笔下的人物和故事，一路跟着我漂洋过海，始终在脑海沉浮升腾。
首站去的是曾到访过的挪威的卑尔根，令我吃惊的是，这座城市少有变化，几乎每个标志性建筑物，还是我记忆中的模样。
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（2）现用分层抽样的方法在全体教职工中抽取54名做培训效果的调查,问应在第三批次中抽取教职工多少名？
（3）已知 ，求第三批次中女教职工比男教职工多的概率.
21．已知过抛物线 的焦点，斜率为 的直线交抛物线于 两点，且 .
（1）求抛物线的方程;
（2）O为坐标原点，C为抛物线上一点，若 ，求 的值.
故B正确；
对C：由折线图可知，前5天空气质量越来越好，从6日开始至15日越来越差，
故C错误；
对D：由折线图可知，上旬大部分AQI指数在100以下，中旬AQI指数大部分在100以上，
故上旬空气质量比中旬的要好.故D正确.
故选：ABD.
【点睛】
本题考查统计图表的观察，属基础题；需要认真看图，并理解题意.
，故 ，即 ， .
故答案为： .
【点睛】
本题考查了等比数列公式，意在考查学生的计算能力.
15．-160
【解析】
【分析】
【详解】
由 ，令 得 ，所以 展开式的常数项为 .
考点：二项式定理.
16．6
【解析】
【分析】
利用双曲线的方程求出双曲线的参数值；利用内角平分线定理得到两条焦半径的关系，再利用双曲线的定义得到两条焦半径的另一条关系，联立求出焦半径．
平面 平面 ， 平面
平面 ，又 平面 ， .
又 ， ， ， ，
， ，
.
的中点 就是三棱锥 的外接球的球心，球的半径为 ，
表面积是 ，故D正确；
故选：BD.
【点睛】
本题考查立体几何中的翻折问题，考查学生的空间想象能力,考查立体几何中的平行、垂直的判定定理和性质定理，考查余弦定理，属于难题.
13．
【解析】
由数量积的几何意义可得： 的值为 与 在 方向投影的乘积，
又 在 方向的投影为 =2，
∴ ，同理 ，
∴ ，
故选D.
【点睛】
本题考查了向量数量积的运算律及数量积的几何意义的应用，属于中档题.
6．C
【解析】
【分析】
利用组合数计算得到基本事件总数和颜色相同的基本事件个数，由古典概型概率公式计算可得结果.
【详解】
由 和 相减得： ，
所以 ， ，因此数列 是以 为公比的等比数列
（2） ， ，两式相加得：
所以
因为 ，所以
又因为 ， ，
所以使得 的n的取值范围为 .
【点睛】
本题考查了等差数列，等比数列的证明，分组求和法，根据数列的单调性解不等式，意在考查学生对于数列公式方法的综合应用.
19．（1）证明见解析（2）
空气质量
优
良
轻度污染
中度污染
重度污染
严重污染
如图是某市12月1日-20日AQI指数变化趋势：
下列叙述正确的是（）
A．这20天中AQI指数值的中位数略高于100
B．这20天中的中度污染及以上的天数占
C．该市12月的前半个月的空气质量越来越好
D．总体来说，该市12月上旬的空气质量比中旬的空气质量好
10．已知 ， ，下列不等式成立的是（）
对B：当 时，函数不是减函数，故B错误；
对C：如图直线 ，与函数图交于 ，
故当 的最小值为1时， ，故C正确；
对D： 时，若使得其与 的所有零点之和为0，
则 ，或 ，如图直线 ，故D错误.
故选：AC.
【点睛】
本题考查由函数的奇偶性求函数解析式，以及判断方程的根的个数，以及函数零点的问题，涉及函数单调性，属综合性基础题；另，本题中的数形结合是解决此类问题的重要手段，值得总结.
本题主要考查了正余弦定理求解三角形的问题,同时也考查了边角互化求解边长的取值范围问题等.属于中等题型.
18．（1）证明见解析（2） ；
【解析】
【分析】
（1）两式相加得到 ，两式相减得到 ，得到证明.
（2）计算 ， ，解不等式得到答案.
【详解】
（1）由 和 相加得：
所以 ，因此数列 是以2为公差的等差数列
9．ABD
【解析】
【分析】
根据折线图和AQI指数与空气质量对照表，结合选项，进行逐一分析即可.
【详解】
对A：将这20天的数据从小到大排序后，第10个数据略小于100，第11个数据约为120，
因为中位数是这两个数据的平均数，故中位数略高于100是正确的，故A正确；
对B：这20天中，AQI指数大于150的有5天，故中度污染及以上的天数占 是正确的，
22．已知函数 ，且 在 处切线垂直于 轴．
（1）求 的值；
（2）求函数 在 上的最小值；
（3）若 恒成立，求满足条件的整数 的最大值．
（参考数据 ， ）
参考答案
1．B
【解析】
【分析】
计算得到 ， ，再计算 得到答案.
【详解】
， ，
故 .
故选： .
【点睛】
本题考查了集合的交集运算，意在考查学生的计算能力.
A． B． C． D．
11．已知定义域为R的奇函数 ，满足 ，下列叙述正确的是（）
A．存在实数k，使关于x的方程 有7个不相等的实数根
B．当 时，恒有
C．若当 时， 的最小值为1，则
D．若关于 的方程 和 的所有实数根之和为零，则
12．如图，矩形 ， 为 的中点，将 沿直线 翻折成 ，连接 ， 为 的中点，则在翻折过程中，下列说法中所有正确的是（）
10．ACD
【解析】
【分析】
由指数函数的单调性可判断 ；由作差法和不等式的性质可判断 ；可根据换底公式，取 ， ，运用对数函数单调性，可判断 ；运用作差法和不等式的性质，可判断 .
【详解】
由 ， ，可得 ，故 正确；
由 ， ， 可得 ， ，故 错误；
由 ， ， ， ，则 ，则 ，可得 ，故 正确；
由 ， ， 可得 ，故 正确.
【详解】
不妨设A在双曲线的右支上,
∵ 为 的平分线，
∴ ，
又∵ ，解得 ，故答案为6.
【点睛】
本题考查内角平分线定理，考查双曲线的定义：解有关焦半径问题常用双曲线的定义，属于中档题.
17．(1) (2)
【解析】
【分析】
(1)利用余弦定理 化简整理再用角 的余弦定理即可.也可以用正弦定理先边化角,再利用和差角公式求解.
(2)易得 的周长等于 ,再利用正弦定理将 用角 表示,再利用三角函数的值域方法求解即可.
【详解】
解法一：（1）根据余弦定理得
整理得 ，
，
（2）依题意得 为等边三角形，所以 的周长等于
由正弦定理 ，
所以 ，
， ，
，
，
所以 的周长的取值范围是 ．
解法二：（1）根据正弦定理得
,
,
，
，
，
（2）同解法一
【点睛】
16．已知 ， 分别为双曲线 的左、右焦点，点 ，点 的坐标为 ， 为 的角平分线，则 _______
17． 的内角 的对边分别为 ，已知 ， .
（1）求角C；
（2）延长线段 到点D，使 ，求 周长的取值范围．
18．已知数列 ， 满足： ， ， ， ， .
（1）证明：数列 为等差数列，数列 为等比数列；
A．8B．10C．12D．16
4．设函数 的定义域为R，满足 ，且 则 （）
A． B． C． D．
5．在直角梯形 中， ， ， ， ， 是 的中点，则 （）
A． B． C． D．
6．一个箱子中装有4个白球和3个黑球，若一次摸出2个球，则摸到的球颜色相同的概率是（）
A． B． C． D．
7．函数 的图象大致为（）
A．存在某个位置，使得 B．翻折过程中， 的长是定值
C．若 ，则 ；D．若 ，当三棱锥 的体积最大时，三棱锥 的外接球的表面积是 .
13．函数 的图象在点 处的切线方程是_______________.
14．已知等比数列 的前n项和为 ， .若 ，则 ______.
15． 展开式的常数项为． （用数字作答）
故选:
【点睛】
本题考查不等式基本性质和利用指数函数、对数函数单调性比较大小，属于基础题.
11．AC
【解析】
【分析】
根据函数是奇函数，写出其解析式，画出该函数的图像，再结合选项，数形结合解决问题.
【详解】
因为该函数是奇函数，故 在R上的解析式为：
绘制该函数的图像如下所示：
对A：如图所示直线 与该函数有7个交点，故A正确；
【详解】
对于A，取 的中点为 ，连接 ，设 ，如图 所示
则 平面 平面 ， 平面 .
四边形 是平行四边形， ，同理可证 平面 .
又 ，且 平面 ， 平面 平面 .
平面 ，又 平面 ，平面 平面 ，
.
如果 ，则 ，由于 ，则 ，
由于三线 共面且共点，这是不可能的，故 不正确；
对于B，如图 ，由等角定理可得 ，又 ，
【分析】
借助求导公式求出 ,因为切线的斜率为 , 代入 求得切点,即可求出切线方程.
【详解】
，∴ 且 ，所以函数 的图象在 处的切线方程是 .
故答案为: .
【点睛】
本题考查了导数的几何意义,过曲线上一点的切线方程的求法,难度容易.
14．2；
【解析】
【分析】
根据等比数列公式化简
【解析】
【分析】
对于A，取 的中点为 ，连接 ，设 .通过证明平面 平面 ，得 .假设 ，得到 ， ，这是不可能的，故 不正确；对于B，在 中，由余弦定理得 是定值，故 是定值，故 正确；对于C，若 ，可证 平面 ，得到 ，此时 ，由于 ，故 不成立，故 不正确；对于D，只有当平面 平面 时，三棱锥 的体积最大，取 的中点为 ，证明 ，故 就是三棱锥 的外接球的球心，故D正确.