编号20山西大学附中高三年级 三个二次
山西省山大附中高三语文上学期第二次月考试题(含解析)新人教版
山西大学附中2013~2014学年第二学期高三第二次考试题语文试题(考查时间:140分钟)(考查内容:与高考一致)一、论述类文本阅读(9分,每小题3分)阅读下面的文字,完成1-3题。
中国篆刻艺术是以石材为载体,以汉字为主要表现对象,并由中国古代印章镌刻技艺发展而来的中国特有的传统艺术。
古代印章是形使和授受国家机构权利、证明个人身份的凭信物,多由工匠采用金属铸造和凿制而成,距今已有三千多年的历史。
公元十三世纪之后,篆刻家开始选用石材代替金属并亲自动手刻制印章,这一技艺的改变,为篆刻艺术的发展提供了广阔的空间。
篆刻家在继承前辈优秀技艺的基础上不断创新,催生了自明清以来的皖派、浙派以至今日的流行印风等多种流派和风格,篆刻艺术界也出现了吴昌硕、齐白石以及当代的刘江、石开等诸多大师。
这些艺术家将这门古老而又弥新的传统艺术传承、发展并创造至今。
上石的书法功底、巧妙的布局能力、精湛娴熟的刀功,共同构成篆刻艺术的技艺内涵。
篆刻艺术借用契刻古代文字的方式表现中国传统文化中虚实相生、阴阳相荡的审美观念,具有抽象和神奇之美。
而今天也有用简化字入印的印章了。
篆刻艺术融多种传统文化于一身,作者要在很小的尺寸中表现中国诗、书、画的审美意境,并且真实、准确地表达其思想、情感、学养,追求布局和刀法的精微妙趣,风格各异,可谓“方寸之中表现大千世界”。
篆刻作品是人文与自然的结合,是天然石材、精制钮雕和篆刻技艺“三美结合”的产物。
篆刻艺术有着重要的文化和社会功能。
它是篆刻家品格、思想和情感的表达方式。
由于篆刻艺术集文学、美学、文字学于一身,所以它又是修身养性、提高人的综合素质、塑造人格的重要途径,同时也是人际交往、艺术及学术交流的有效方式。
随着汉文化的传播,篆刻艺术逐渐流传到日本、韩国、新加坡以及欧美等国家和地区,对促进国际文化的交流有着积极的作用。
篆刻艺术作品有着审美和收藏价值以及对人物和书画作品的印证和品鉴作用,如姓名章、收藏章等印章。
2024届山西省太原市山西大学附中高三第三次测评物理试卷含解析
2024届山西省太原市山西大学附中高三第三次测评物理试卷注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
2.答题时请按要求用笔。
3.请按照题号顺序在答题卡各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折暴、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、单项选择题:本题共6小题,每小题4分,共24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、两辆汽车a 、b 在两条平行的直道上行驶。
t=0时两车并排在同一位置,之后它们运动的v-t 图像如图所示。
下列说法正确的是( )A .汽车a 在10s 末向反方向运动B .汽车b 一直在物体a 的前面C .5s 到10s 两车的平均速度相等D .10s 末两车相距最近2、如图所示,直线a b 、和直线、c d 是处于匀强电场中的两组平行线,M 、N 、P 、Q 是它们的交点,四点处的电势分别为M N P Q ϕϕϕϕ、、、。
一质子由M 点分别运动到Q 点和P 点的过程中,电场力所做的负功相等。
下列说法正确的是( )A .直线a 位于某一等势面内,M Q ϕϕ<B .直线c 位于某一等势面内,>M P ϕϕC .若质子由M 点运动到N 点,电场力做正功D .若质子由P 点运动到Q 点,电场力做负功3、如图,吊桥AB 长L ,质量均匀分布,重G 1。
A 端由铰链支于地面,B 端由绳拉住,绳绕过小滑轮C 挂重物,重G 2。
重力作用线沿铅垂线AC ,AC=AB 。
当吊桥平衡时,吊桥与铅垂线的夹角θ为A .2arcsin 21G GB .arcsin 21G GC .2arctan 212G GD .arctan 212G G 4、下列关于温度及内能的说法中正确的是( )A .温度是分子平均动能的标志,所以两个动能不同的分子相比,动能大的温度高B .两个不同的物体,只要温度和体积相同,内能就相同C .质量和温度相同的冰和水,内能是相同的D .一定质量的某种物质,即使温度不变,内能也可能发生变化5、2019年11月23日,我国在西昌卫星发射中心用长征三号乙运载火箭,以“一箭双星”方式成功发射第五十、五十一颗北斗导航卫星。
2020届山西大学附中高三上学期第二次模块诊断 数学(文)(PDF版)
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山西大学附属中学
2019~2020 学年高三第一学期(总第二次)模块诊断 数学试题(文)参考答案
考试时间:120 分 满分:150 分 二、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.)
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山西省山西大学附中2024年高三数学第一学期期末学业水平测试模拟试题含解析
山西省山西大学附中2024年高三数学第一学期期末学业水平测试模拟试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B 铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题:本题共12小题,每小题5分,共60分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.已知函数()()1xf x k xe =-,若对任意x ∈R ,都有()1f x <成立,则实数k 的取值范围是( )A .(),1e -∞-B .()1,e -+∞C .(],0e -D .(]1,1e -2.设,a b 为非零向量,则“a b a b +=+”是“a 与b 共线”的( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件D .既不充分也不必要条件3.设()()2141A B -,,,,则以线段AB 为直径的圆的方程是( )A .22(3)2x y -+=B .22(3)8x y -+=C .22(3)2x y ++=D .22(3)8x y ++=4.等比数列{}n a 的前n 项和为n S ,若0n a >,1q >,3520a a +=,2664a a =,则5S =( ) A .48B .36C .42D .315.已知双曲线22221(0,0)x y a b a b-=>>的左焦点为F ,直线l 经过点F 且与双曲线的一条渐近线垂直,直线l 与双曲线的左支交于不同的两点A ,B ,若2AF FB =,则该双曲线的离心率为( ).A B C D 6.已知随机变量i ξ满足()()221kkk i i i P k C p p ξ-==-,1,2i =,0,1,2k =.若21211p p <<<,则( ) A .()()12E E ξξ<,()()12D D ξξ< B .()()12E E ξξ<,()()12D D ξξ> C .()()12E E ξξ>,()()12D D ξξ<D .()()12E E ξξ>,()()12D D ξξ>7.已知等差数列{}n a 的前n 项和为n S ,且282,10a a =-=,则9S =( ) A .45B .42C .25D .36以通过设计下面的实验来估计π的值:先请全校m 名同学每人随机写下一个都小于1的正实数对(),x y ;再统计两数能与1构成钝角三角形三边的数对(),x y 的个数a ;最后再根据统计数a 估计π的值,那么可以估计π的值约为( )A .4amB .2a m+ C .2a mm+ D .42a mm+ 9.设i 是虚数单位,若复数1z i =+,则22||z z z+=( )A .1i +B .1i -C .1i --D .1i -+10.函数()sin()f x x π=-223的图象为C ,以下结论中正确的是( )①图象C 关于直线512x π=对称; ②图象C 关于点(,0)3π-对称;③由y =2sin 2x 的图象向右平移3π个单位长度可以得到图象C . A .① B .①②C .②③D .①②③11.已知全集,,则( )A .B .C .D .12.某中学2019年的高考考生人数是2016年高考考生人数的1.2倍,为了更好地对比该校考生的升学情况,统计了该校2016年和2019年的高考情况,得到如图柱状图:则下列结论正确的是( ).A .与2016年相比,2019年不上线的人数有所增加B .与2016年相比,2019年一本达线人数减少C .与2016年相比,2019年二本达线人数增加了0.3倍D .2016年与2019年艺体达线人数相同二、填空题:本题共4小题,每小题5分,共20分。
山西省太原市山西大学附中2024届高三上学期12月月考(总第七次)数学试题
山西省太原市山西大学附中2024届高三上学期12月月考(总第七次)数学试题学校:___________姓名:___________班级:___________考号:___________一、单选题..C.D.年的五一劳动节是疫情后的第一个小长假,公司筹备优秀员工假期免费旅游.常见的五个旅游热门地北京、上海、广州、深圳、成都外,淄博烧烤火爆全国,山东也成为备选地之一.若每个部门从六个旅游地中选择一个旅游地,则甲、乙、丙、丁四个部门至少有三个部门所选旅游地全不相同的方法种数共有()1800B.1080C.720D.360 2020年12月17日凌晨,嫦娥五号返回器携带月球样品在内蒙古四子王旗预定区域安全着陆-嫦娥五号返回:舱之所以能达到如此高的再入精度,主要是因为它采用弹跳式返回弹道,实现了减速和再入阶段弹道调整,这与“打水漂”原理类似(如图所示A .4B .57.已知函数21()sinsin 22xf x x ωω=+-点,则ω的取值范围是A .10,8⎛⎤⎥⎝⎦B .150,,148⎛⎤⎡⋃ ⎥⎢⎝⎦⎣8.设函数()(21)xf x e x ax a =--+,其中则a 的取值范围是()A .3,12e ⎡⎫-⎪⎢⎣⎭B .33,2e 4⎡⎫-⎪⎢⎣⎭二、多选题三、填空题(1)当//AB 平面PCD 时,求PC 的长;(2)当三棱锥P COD -体积最大时,求二面角D PC -20.抽屉中装有5双规格相同的筷子,其中2双是一次性筷子,每次使用筷子时,从抽屉中随机取出1双,若取出的是一次性筷子,若取出的是非一次性筷子,则使用后经过清洗再次放入抽屉中,求:(1)在第2次取出的是非一次性筷子的条件下,第1次取出的是一次性筷子的概率;(2)取了3次后,取出的一次性筷子的个数(双)的分布列及数学期望;(3)取了(2,3,4n n =,…)次后,所有一次性筷子刚好全部取出的概率.21.已知直线l 经过椭圆()2222:10x y E a b a b +=>>的右焦点,A B ,当直线l 分别与x 轴、y 轴垂直时,线段AB 的长分别为(1)求椭圆E 的标准方程;(2)过点B 作x 轴的垂线交椭圆E 于点C (异于点,A B ADF △面积的最大值.22.已知函数()ln ,R f x ax x a =-∈.(1)讨论函数()f x 的单调性;(2)当1a =时,设()()()21x g x f x -=,求证:函数()g x 存在极大值点。
2019-2020学年山西大学附中高三(上)第二次诊断数学试卷(文科)
2019-2020学年山西大学附中高三(上)第二次诊断数学试卷(文科)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.(5分)(2019•淄博模拟)已知集合2{|20}A x Z x x =∈--…,则(z A =ð ) A .{0}B .{1}C .{0,1}D .{1-,0,1,2}2.(2019•衡水二模)复数11z i =+,2z i =,其中i 为虚数单位,则12z z 的虚部为( ) A .1-B .1C .iD .i -3.(5分)(2019•成都模拟)已知向量(3,1)a =,(3,3)b =-,则向量b 在向量a 方向上的投影为( )A .B C .1-D .14.(5分)(2019•十堰模拟)某工厂利用随机数表对生产的600个零件进行抽样测试,先将600个零件进行编号,编号分别为001,002,⋯,599,600从中抽取60个样本,如下提供随机数表的第4行到第6行:32 21 18 34 29 78 64 54 07 32 52 42 06 44 38 12 23 43 56 77 35 78 90 56 42 84 42 12 53 31 34 57 86 07 36 25 30 07 32 86 23 45 78 89 07 23 68 96 08 04 32 56 78 08 43 67 89 53 55 77 34 89 94 83 75 22 53 55 78 32 45 77 89 23 45若从表中第6行第6列开始向右依次读取3个数据,则得到的第6个样本编号( ) A .522B .324C .535D .5785.(5分)(2019•齐齐哈尔二模)函数6()22x xxf x -=+的图象大致是( )A .B .C .D .6.(5分)(2019•十堰模拟)某几何体的三视图如图所示,则该几何体的体积为( )A .116πB .73π C .136πD .83π 7.(5分)(2019•涪城区校级模拟)已知1sin()54πα-=,则3cos(2)(5πα+= )A .78-B .78 C .18D .18-8.(5分)(2019•河南模拟)下列说法正确的是( )A .设m 为实数,若方程22112x y m m+=--表示双曲线,则2m >.B .“p q ∧为真命题”是“p q ∨为真命题”的充分不必要条件C .命题“x R ∃∈,使得2230x x ++<”的否定是:“x R ∀∈,2230x x ++>”D .命题“若0x 为()y f x =的极值点,则f ’ ()0x =”的逆命题是真命题9.(5分)(2019•陕西模拟)在直三棱柱111ABC A B C -中,已知AB BC ⊥,2AB BC ==,1CC =,则异面直线1AC 与11A B 所成的角为( )A .30︒B .45︒C .60︒D .90︒10.(5分)(2015•资阳三模)已知函数()sin()(0f x A x A ωϕ=+>,0ω>,||)2πϕ<的部分图象如图所示,下列说法正确的是( )A .()f x 的图象关于直线23x π=-对称B .()f x 的图象关于点5(,0)12π-对称 C.将函数2cos 2y x x -的图象向左平移2π个单位得到函数()f x 的图象 D .若方程()f x m =在[,0]2π-上有两个不相等的实数根,则m的取值范围是(2,-11.(5分)(2019•6月份模拟)设奇函数()f x 的定义域为(,)22ππ-,且()f x 的图象是连续不间断,(,0)2x π∀∈-,有()c o s ()s i n 0f x x f x x '+<,若()2()c o s 3f m f m π<,则m 的取值范围是( ) A .(,)23ππ-B .(0,)3πC .(,)23ππ--D .(,)32ππ12.(5分)(2019•齐齐哈尔二模)已知11,10(1)(),01x f x f x x x ⎧--<<⎪+=⎨⎪<⎩…,若方程()21f x ax a -=-有唯一解,则实数a 的取值范围是( ) A .2(,)3+∞B .2[,)3+∞C .2{8}[,)3-+∞D .2{8}(,)3-+∞二、填空题(每题5分,满分20分,将答案填在答题纸上)13.(5分)(2019秋•小店区校级月考)若曲线()x x f x ae e -=+在点(0,(0))f 处的切线与直线30x y +=垂直,则a = .14.(5分)(2019•淄博模拟)已知0220x y x y -⎧⎨--⎩……,且z x y =+,则z 的最小值为 .15.(5分)(2019•齐齐哈尔二模)已知椭圆22221(0,0)x y a b a b+=>>的左焦点为F ,右顶点为A ,上顶点为B,若点F 到直线AB ,则该椭圆的离心率为 . 16.(5分)(2019•齐齐哈尔二模)在锐角ABC ∆中,角A ,B ,C 的对边分别为a ,b ,c ,若2a =,cos cos tan sin sin A C A A C +=+,则sin sin b cB C++的取值范围是 .三、解答题(解答应写出文字说明、证明过程或演算步骤.)17.(12分)(2019•淄博模拟)已知等比数列{}n a 的前n 项和为*234(),2,,4n S n N S S S ∈-成等差数列,且2341216a a a ++=.(1)求数列{}n a 的通项公式;(2)若2(2)log ||n n b n a =-+,求数列1{}nb 的前n 项和n T .18.(12分)(2019•齐齐哈尔二模)如图,在直三棱柱111ABC A B C -中,E 为11A C 的中点,2AB BC ==,1C F AB ⊥,(1)求证:AB BC ⊥;(2)若1//C F 平面ABE ,且12C F =,求点A 到平面BCE 的距离.19.(12分)(2019•淄博模拟)已知椭圆2222:1(0)x y C a b a b+=>>的左,右焦点分别为1F ,2F ,离心率为12,P 是椭圆C 上的一个动点,且△12PF F(Ⅰ)求椭圆C 的方程;(Ⅱ)设斜率存在的直线2PF 与椭圆C 的另一个交点为Q ,是否存在点(0,)T t ,使得||||TP TQ =?若存在,求出t 的取值范围;若不存在,请说明理由.20.(12分)(2019•陕西模拟)某大型商场的空调在1月到5月的销售量与月份相关,得到的统计数据如表:(1)经分析发现1月到5月的销售量可用线性回归模型拟合该商场空调的月销量y (百件)与月份x 之间的相关关系.请用最小二乘法求y 关于x 的线性回归方程ˆˆˆybx a =+,并预测6月份该商场空调的销售量;(2)若该商场的营销部对空调进行新一轮促销,对7月到12月有购买空调意愿的顾客进行问卷调查.假设该地拟购买空调的消费群体十分庞大,经过营销部调研机构对其中的500名顾客进行了一个抽样调查,得到如下一份频数表:现采用分层抽样的方法从购买意愿的月份在7月与12月的这90名顾客中随机抽取6名,再从这6人中随机抽取3人进行跟踪调查,求抽出的3人中恰好有2人是购买意愿的月份是12月的概率.参考公式与数据:线性回归方程ˆˆˆybx a =+,其中1221ˆni ii nii x ynxy b xnx==-=-∑∑,5121.2i i i x y ==∑.21.(12分)(2019•6月份模拟)己知函数()()()f x x a lnx a R =-∈,它的导函数为()f x '. (1)当1a =时,求()f x '的零点;(2)若函数()f x 存在极小值点,求a 的取值范围.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.[选修4-4:坐标系与参数方程]22.(10分)(2019•河南模拟)在直角坐标系xOy 中,直线l 的参数方程为cos sin x t y t αα=⎧⎨=⎩,(t为参数),在以坐标原点为极点,x 轴正半轴为极轴的极坐标系中,曲线1:2cos C ρθ=,2:2cos()3C πρθ=-.(Ⅰ)求1C 与2C 交点的直角坐标;(Ⅱ)若直线l 与曲线1C ,2C 分别相交于异于原点的点M ,N ,求||MN 的最大值.[选修4-5:不等式选讲]23.(2019•齐齐哈尔二模)已知函数()|2||3|()f x x a x a R =+--∈. (1)若1a =-,求不等式()10f x +>的解集;(2)已知0a >,若()32f x a +>对于任意x R ∈恒成立,求a 的取值范围.2019-2020学年山西大学附中高三(上)第二次诊断数学试卷(文科)参考答案与试题解析一、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.(5分)(2019•淄博模拟)已知集合2{|20}A x Z x x =∈--…,则(z A =ð ) A .{0}B .{1}C .{0,1}D .{1-,0,1,2}【解答】解:集合2{|20}{|2A x Z x x x Z x =∈--=∈厖或1}x -…,则{0z A =ð,1}, 故选:C .2.(2019•衡水二模)复数11z i =+,2z i =,其中i 为虚数单位,则12z z 的虚部为( ) A .1-B .1C .iD .i -【解答】解:复数11z i =+,2z i =,∴1z i =-, ∴1211z i i z i-==--, 其虚部为1-. 故选:A .3.(5分)(2019•成都模拟)已知向量(3,1)a =,(3,3)b =-,则向量b 在向量a 方向上的投影为( )A .B C .1-D .1【解答】解:由投影的定义可知:向量b 在向量a 方向上的投影为:||cos ,b a b <>, 又||||cos ,a b a b a b =<>,∴3(3)3||cos ,||3a b b a b a -+<>=== 故选:A .4.(5分)(2019•十堰模拟)某工厂利用随机数表对生产的600个零件进行抽样测试,先将600个零件进行编号,编号分别为001,002,⋯,599,600从中抽取60个样本,如下提供随机数表的第4行到第6行:32 21 18 34 29 78 64 54 07 32 52 42 06 44 38 12 23 43 56 77 35 78 90 56 4284 42 12 53 31 34 57 86 07 36 25 30 07 32 86 23 45 78 89 07 23 68 96 08 0432 56 78 08 43 67 89 53 55 77 34 89 94 83 75 22 53 55 78 32 45 77 89 23 45若从表中第6行第6列开始向右依次读取3个数据,则得到的第6个样本编号() A.522B.324C.535D.578【解答】解:第6行第6列的数开始的数为808,不合适,436,789不合适,535,577,348,994不合适,837不合适,522,535,则满足条件的6个编号为436,535,577,348,522,535,则第6个编号为535,故选:C.5.(5分)(2019•齐齐哈尔二模)函数6()22x xxf x-=+的图象大致是()A.B.C.D.【解答】解:6()()22x xxf x f x---==-+,即函数()f x是奇函数,图象关于原点对称,排除,A,B,当0x>时,()0f x>,排除D,故选:C.6.(5分)(2019•十堰模拟)某几何体的三视图如图所示,则该几何体的体积为()A .116πB .73π C .136πD .83π 【解答】解:根据几何体的三视图知,该几何体是下部是圆柱、上部是半个圆锥的组合体; 画出图形如图所示;∴该几何体的体积为2211131211326V V V πππ=+=⨯⨯+⨯⨯⨯⨯=圆柱半圆锥. 故选:C .7.(5分)(2019•涪城区校级模拟)已知1sin()54πα-=,则3cos(2)(5πα+= )A .78-B .78 C .18D .18-【解答】解:1sin()54πα-=, 2233217cos(2)cos(2)cos(2)12sin ()12()555548ππππαπααα∴+=---=--=-+-=-+⨯=- 故选:A .8.(5分)(2019•河南模拟)下列说法正确的是( )A .设m 为实数,若方程22112x y m m+=--表示双曲线,则2m >.B .“p q ∧为真命题”是“p q ∨为真命题”的充分不必要条件C .命题“x R ∃∈,使得2230x x ++<”的否定是:“x R ∀∈,2230x x ++>”D .命题“若0x 为()y f x =的极值点,则f ’ ()0x =”的逆命题是真命题【解答】解:A .若方程表示双曲线,则(1)(2)0m m --<,得(1)(2)0m m -->得2m >或1m <,故A 错误,B .若p q ∧为真命题,则p ,q 同时为真命题,则p q ∨为真命题,当p 真q 假时,满足p q ∨为真命题,但p q ∧为假命题,即必要性不成立,则“p q ∧为真命题”是“p q ∨为真命题”的充分不必要条件,故B 正确,C .命题“x R ∃∈,使得2230x x ++<”的否定是:“x R ∀∈,2230x x ++…”,故C 错误,D .命题“若0x 为()y f x =的极值点,则f ’ ()0x =”的逆命题是,若()0f x '=,则0x 为()y f x =的极值点,错误,比方在3()f x x =,中,2()3f x x '=,(0)0f '=,但0x =不是极值点,故D 错误, 故选:B .9.(5分)(2019•陕西模拟)在直三棱柱111ABC A B C -中,已知AB BC ⊥,2AB BC ==,1CC =,则异面直线1AC 与11A B 所成的角为( )A .30︒B .45︒C .60︒D .90︒【解答】解:连接1AC ,1BC ,可知1BAC ∠为异面直线1AC 与11A B 所成的角.1ABC ∆为直角三角形,且1AB BC ⊥,2AB =,1BC ==∴1tan BAC ∠=160BAC ∠=︒.即异面直线1AC 与11A B 所成的角为60︒. 故选:C .10.(5分)(2015•资阳三模)已知函数()sin()(0f x A x A ωϕ=+>,0ω>,||)2πϕ<的部分图象如图所示,下列说法正确的是( )A .()f x 的图象关于直线23x π=-对称B .()f x 的图象关于点5(,0)12π-对称C .将函数2cos 2y x x -的图象向左平移2π个单位得到函数()f x 的图象D .若方程()f x m =在[,0]2π-上有两个不相等的实数根,则m 的取值范围是(2,-【解答】解:由函数的图象可得2A =,124312πππω=-,求得2ω=. 在根据五点法作图可得23πϕπ⨯+=,求得3πϕ=,∴函数()2sin(2)3f x x π=+.当23x π=-时,()0f x =,不是最值,故A 不成立. 当512x π=-时,()02f x ==-,不等于零,故B 不成立.将函数2cos22sin(2)6y x x x π=-=-的图象向左平移2π个单位得到函数5sin[2()]sin(2)266y x x πππ=+-=+的图象,故C 不成立.当[2x π∈-,0]时,22[33x ππ+∈-,]3π.2sin()sin()33ππ-=-=,sin()12π-=-,故方程()f x m =在[,0]2π-上有两个不相等的实数根时,则m 的取值范围是(2,-,故D成立; 故选:D .11.(5分)(2019•6月份模拟)设奇函数()f x 的定义域为(,)22ππ-,且()f x 的图象是连续不间断,(,0)2x π∀∈-,有()c o s ()s i n 0f x x f x x '+<,若()2()c o s 3f m f m π<,则m 的取值范围是( ) A .(,)23ππ-B .(0,)3πC .(,)23ππ--D .(,)32ππ【解答】解:奇函数()f x 的定义域为(,)22ππ-,且()f x 的图象是连续不间断,令()()cos f x g x x =,则2()cos ()sin ()cos f x x f x xg x x '+'=.因为(,0)2x π∀∈-,有()cos ()sin 0f x x f x x '=+<, 所以当(,0)2x π∈-时,()0g x '<,则()()cos f x g x x =在(,0)2π-上单调递减.又()f x 是定义域在(,)22ππ-上的奇函数,所以()()()()cos()cos f x f x g x g x x x --==-=--,则()()cos()f x g x x =也是(,)22ππ-上的奇函数并且单调递减. 又()2()cos 3f m f m π<等价于()()3cos cos 3f f m m ππ<, 即()()3g m g π<,所以3m π>,又22m ππ-<<,所以32m ππ<<.故选:D .12.(5分)(2019•齐齐哈尔二模)已知11,10(1)(),01x f x f x x x ⎧--<<⎪+=⎨⎪<⎩…,若方程()21f x ax a -=-有唯一解,则实数a 的取值范围是( ) A .2(,)3+∞B .2[,)3+∞C .2{8}[,)3-+∞D .2{8}(,)3-+∞【解答】解:令10x -<<,则011x <+<, 则(1)1f x x +=+,故11,10()1,01x f x x x x ⎧--<<⎪=+⎨⎪<⎩…,如图示:由()21ff x ax a -=-,得()(21)1f x a x =+-,函数(21)1y a x =+-恒过1(2A -,1)-,故1141312AB K +==+, 若方程()21f x ax a -=-有唯一解, 则423a >,即23a >;当12111ax a x +-=-+即图象相切时, 根据△0=,298(1)0a a a --=, 解得8a =-, 故选:D .二、填空题(每题5分,满分20分,将答案填在答题纸上)13.(5分)(2019秋•小店区校级月考)若曲线()x x f x ae e -=+在点(0,(0))f 处的切线与直线30x y +=垂直,则a = 4 .【解答】解:()x x f x ae e -=+在的导数为()x x f x ae e -'=-, 即有()f x 在0x =处的切线斜率为1k a =-, 由在0x =处的切线与直线30x y +=垂直, 即有13a -=, 解得4a =. 故答案为:4.14.(5分)(2019•淄博模拟)已知0220x y x y -⎧⎨--⎩……,且z x y =+,则z 的最小值为 4- .【解答】解:作出不等式对应的平面区域如图,由z x y =+,得y x z =-+,平移直线y x z =-+,由图象可知当直线y x z =-+,经过点(2,2)A --时,直线y x z =-+的截距最小,此时z 最小.此时z 的最小值为224z =--=-, 故答案为:4-.15.(5分)(2019•齐齐哈尔二模)已知椭圆22221(0,0)x y a b a b+=>>的左焦点为F ,右顶点为A ,上顶点为B ,若点F 到直线AB ,则该椭圆的离心率为23. 【解答】解:椭圆22221(0,0)x y a b a b +=>>的左焦点为(,0)F c -,右顶点为A ,上顶点为B ,直线AB 的方程为:1x ya b+=,即:0bx ay ab +-=点F 到直线AB ,=, 可得2222214()25255025a c a b a c +=+=-. 可得:23928360e e +-=,(0,1)e ∈,解得23e =,1813e =-(舍去), 故答案为:23. 16.(5分)(2019•齐齐哈尔二模)在锐角ABC ∆中,角A ,B ,C 的对边分别为a ,b ,c ,若2a =,cos cos tan sin sin A C A A C +=+,则sin sin b cB C++的取值范围是 ,4) .【解答】解:cos cos sin tan sin sin cos A C AA A C A+==+, 22cos cos cos sin sin sin A C A A A C ∴+=+,可得:cos2cos A B =,∴在锐角ABC ∆中,2A B =,A B C π++=,可得:3A C π+=,(0,)2C π∈,(36CA ππ-∴=∈,)3π,可得:1sin (2A ∈,2a =,∴sin sin sin b c a B C A +=∈+4).故答案为:,4). 三、解答题(解答应写出文字说明、证明过程或演算步骤.)17.(12分)(2019•淄博模拟)已知等比数列{}n a 的前n 项和为*234(),2,,4n S n N S S S ∈-成等差数列,且2341216a a a ++=. (1)求数列{}n a 的通项公式;(2)若2(2)log ||n n b n a =-+,求数列1{}nb 的前n 项和n T .【解答】解:(1)等比数列{}n a 的公比为q ,1q ≠, 前n 项和为*234(),2,,4n S n N S S S ∈-成等差数列,可得342242S S S =-,即为342111(1)(1)(1)242111a q a q a q q q q ---=----,化为2210q q --=,解得12q =-,2341216a a a ++=,即为1111111224816a a a -+-=, 解得112a =-,则1()2n n a =-,*n N ∈;(2)221(2)log ||(2)log (2)2n n nb n a n n n =-+=-+=+, 可得11111()(2)22n b n n n n ==-++, 即有前n 项和11111111(1)2324112n T n n n n =-+-+⋯+-+--++11113111(1)()22124212n n n n =+--=-+++++.18.(12分)(2019•齐齐哈尔二模)如图,在直三棱柱111ABC A B C -中,E 为11A C 的中点,2AB BC ==,1C F AB ⊥,(1)求证:AB BC ⊥;(2)若1//C F 平面ABE ,且12C F =,求点A 到平面BCE 的距离.【解答】(1)证明:1CC ⊥平面ABC ,AB ⊂平面ABC , 1CC AB ∴⊥,又1AB C F ⊥,111CC C F C =,AB ∴⊥平面11BCC B ,又BC ⊂平面11BCC B ,AB BC ∴⊥.(2)过F 做//FM AC 交AB 于M ,连接EM , 1//EC AC ,1//FM EC ∴,1//C F 平面ABE ,1C F ⊂平面1EMFC ,平面1EMFC ⋂平面ABE EM =, 1//C F EM ∴,∴四边形1EMFC 是平行四边形,112FM EC AC ∴==, FM ∴是ABC ∆的中位线.112CF BC ∴==,1CC2EB EC BC ∴===,22EBC S ∆∴设A 到平面EBC 的距离为d ,则13A BEC V d -==又112232A BEC E ABC V V --==⨯⨯⨯2d ∴=,即A 到平面EBC 的距离为2.19.(12分)(2019•淄博模拟)已知椭圆2222:1(0)x y C a b a b+=>>的左,右焦点分别为1F ,2F ,离心率为12,P 是椭圆C 上的一个动点,且△12PF F(Ⅰ)求椭圆C 的方程;(Ⅱ)设斜率存在的直线2PF 与椭圆C 的另一个交点为Q ,是否存在点(0,)T t ,使得||||TP TQ =?若存在,求出t 的取值范围;若不存在,请说明理由.【解答】解:(Ⅰ)椭圆离心率为12,当P 为C 的上顶点时,△12PF F∴22212122c a c b a b c ⎧=⎪⎪⎪⨯⨯=⎨⎪=+⎪⎪⎩2a ∴=,b =,1c =. 故椭圆C 的方程为:22143x y +=.(Ⅱ)设直线PQ 的方程为(1)y k x =-,当0k ≠时,(1)y k x =-代入22143x y +=,得:2222(34)84120k x k x k +-+-=;设1(P x ,1)y ,2(Q x ,2)y ,线段PQ 的中点为0(N x ,0)y ,212024234x x k x k +==+,120023(1)234y y ky k x k +-==-=+, 即22243(,)3434k kN k k -++,||||TP TQ =,∴直线TN 为线段PQ 的垂直平分线;TN PQ ∴⊥,则1TN PQ k k =-.所以2223431443kt k k k k --+=-+,213434k t k k k ⇒==++,当0k >时,因为34k k +…∴t ∈.当0k <时,因为34k k+-…∴[t ∈. 当0k =时,0t =符合题意. 综上,t的取值范围为[. 20.(12分)(2019•陕西模拟)某大型商场的空调在1月到5月的销售量与月份相关,得到的统计数据如表:(1)经分析发现1月到5月的销售量可用线性回归模型拟合该商场空调的月销量y (百件)与月份x 之间的相关关系.请用最小二乘法求y 关于x 的线性回归方程ˆˆˆybx a =+,并预测6月份该商场空调的销售量;(2)若该商场的营销部对空调进行新一轮促销,对7月到12月有购买空调意愿的顾客进行问卷调查.假设该地拟购买空调的消费群体十分庞大,经过营销部调研机构对其中的500名顾客进行了一个抽样调查,得到如下一份频数表:现采用分层抽样的方法从购买意愿的月份在7月与12月的这90名顾客中随机抽取6名,再从这6人中随机抽取3人进行跟踪调查,求抽出的3人中恰好有2人是购买意愿的月份是12月的概率.参考公式与数据:线性回归方程ˆˆˆybx a =+,其中1221ˆni ii nii x ynxy b xnx ==-=-∑∑,5121.2i i i x y ==∑.【解答】解:(1)1(12345)35x =++++=,1(0.60.8 1.2 1.6 1.8) 1.25y =++++=,∴221.253 1.2ˆ0.325553b-⨯⨯==-⨯, 则ˆ 1.20.3230.24a=-⨯=, 于是y 关于x 的回归直线方程为ˆ0.320.24yx =+. 当6x =时,ˆ0.3260.24 2.16y=⨯+=(百台); (2)现采用分层抽样的方法从购买意愿的月份在7月与12月的这90名顾客中随机抽取6名,则购买意愿为7月份的抽4人记为a ,b ,c ,d ,购买意愿为12月份的抽2人记为A ,B .从这6人中随机抽取3人的所有情况为(a ,b ,)c 、(a ,b ,)d 、(a ,b ,)A 、(a ,b ,)B 、(a ,c ,)d 、(a ,c ,)A 、(a ,c ,)B 、(a ,d ,)A 、(a ,d ,)B 、(a ,A ,)B 、(b ,c ,)d 、(b ,c ,)A 、(b ,c ,)B 、(b ,d ,)A 、(b ,d ,)B 、(b ,A ,)B 、(c ,d ,)A 、(c ,d ,)B 、(c ,A ,)B 、(d ,A ,)B ,共20种,恰好有2人是购买意愿的月份是12月的有(a ,A ,)B 、(b ,A ,)B 、(c ,A ,)B 、(d ,A ,)B ,共4种,故所求概率为41205P ==. 21.(12分)(2019•6月份模拟)己知函数()()()f x x a lnx a R =-∈,它的导函数为()f x '. (1)当1a =时,求()f x '的零点;(2)若函数()f x 存在极小值点,求a 的取值范围. 【解答】解:(1)()f x 的定义域为(0,)+∞, 当1a =时,()(1)f x x lnx =-,1()1f x lnx x'=+-. 易知1()1f x lnx x'=+-为(0,)+∞上的增函数, 又f '(1)1110ln =+-=,所以1x =是()f x '的零点. (2)()1x a af x lnx lnx x x-'=+=-+, 令()1a g x lnx x =-+,则221()a x a g x x x x+'=+=. ①当0a =时,()1f x lnx '=+,令()0f x '>,得1x e >;令()0f x '<,得10x e<<, 所以()f x 在1(0,)e上单调递减,在1(,)e +∞上单调递增,符合题意.②当0a >时,()0g x '>,所以()g x 在(0,)+∞上单调递增. 又1()0g ae e =-<,1()11(1)0a a a a g e a a e e=-+=+->,所以()g x 在(0,)+∞上恰有一个零点0x ,且当0(0,)x x ∈时,()()0f x g x '=<;当0(x x ∈,)+∞时,()()0f x g x '=>,所以0x 是()f x 的极小值点,符合题意. ③当0a <时,令()0g x '=,得x a =-.当(0,))x a ∈-时,()0g x '<;当(,)x a ∈-+∞时,()0g x '>,所以()()2()min g x g a ln a =-=+-.若()2()0g a ln a -=+-…,即当2a e --…时,()()()0f x g x g a '=-厖恒成立, 即()f x 在(0,)+∞上单调递增,无极值点,不符合题意. 若()2()0g a ln a -=+-<,即当20e a --<<时,(1)1(1)01ag a ln a a-=-+->-, 所以()(1)0g a g a --<,即()g x 在(,)a -+∞上恰有一个零点1x ,且当1(,)x a x ∈-时,()()0f x g x '=<;当1()x x ∈+∞时,()()0f x g x '=>, 所以1x 是()f x 的极小值点,符合题意.综上,可知2a e ->-,即a 的取值范围为2(e --,)+∞.请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.[选修4-4:坐标系与参数方程]22.(10分)(2019•河南模拟)在直角坐标系xOy 中,直线l 的参数方程为cos sin x t y t αα=⎧⎨=⎩,(t为参数),在以坐标原点为极点,x 轴正半轴为极轴的极坐标系中,曲线1:2cos C ρθ=,2:2cos()3C πρθ=-.(Ⅰ)求1C 与2C 交点的直角坐标;(Ⅱ)若直线l 与曲线1C ,2C 分别相交于异于原点的点M ,N ,求||MN 的最大值. 【解答】解:(Ⅰ)由2cos ρθ=,得22cos ρρθ=, 则曲线1C 的直角坐标方程为222x y x +=, 由2cos()3πρθ=-,得2cos sin ρρθθ=+,则曲线2C的直角坐标方程为220x y x +--=.由222220x y x x y x ⎧+=⎪⎨+-=⎪⎩,解得00x y =⎧⎨=⎩或32x y ⎧=⎪⎪⎨⎪=⎪⎩, 故1C 与2C 交点的直角坐标为(0,0),3(2;- 21 - (Ⅱ)不妨设0απ<…,点M ,N 的极坐标分别为1(ρ,)α,2(ρ,)α. ∴12|||||2cos 2cos()|3MN πρραα=-=--|2cos (cos )||cos |2|cos()|3παααααα=-==+. ∴当23πα=时,||MN 取得最大值2. [选修4-5:不等式选讲]23.(2019•齐齐哈尔二模)已知函数()|2||3|()f x x a x a R =+--∈.(1)若1a =-,求不等式()10f x +>的解集;(2)已知0a >,若()32f x a +>对于任意x R ∈恒成立,求a 的取值范围.【解答】解:(1)当1a =-吋,函数()|21||3|f x x x =---, 当12x …时,()12(3)2f x x x x =-+-=--, 不等式()10f x +>化为210x --+>,解得1x <-; 当132x <<时,()21(3)34f x x x x =-+-=-, 不等式()10f x +>化为3410x -+>,解得1x >,取13x <<; 当3x …时,()21(3)2f x x x x =---=+,不等式()10f x +>化为210x ++>,解得3x >-,取3x …; 综上所述,不等式()10f x +>的解集为{|1x x <-或1}x >;(2)当0a >吋,若2a x -…,则()2(3)3f x x a x x a =--+-=---, 此时()()322min a a f x f =-=--,则5()3322f x a a +->…,解得1a >; 若32a x -<<,则()2(3)33f x x a x x a =++-=+-, 此时1()()322a f x f a >-=--,则5()3322f x a a +>->,解得1a >; 若3x …,则()2(3)3f x x a x x a =+--=++, 此时()min f x f =(3)6a =+,则()3462f x a a ++>…恒成立;综上所述,不等式()32f x a +>对任意x R ∈恒成立时,a 的取值范围是1a >.。
山西大学附中2021-2021学年高三第一学期(总第二次)语文试题及参考答案
山西大学附中2021-2021学年高三第一学期(总第二次)语文试题及参考答案山西高校附中2021-2021学年高三第一学期(总其次次)模块诊断语文试题一、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1~3题。
戉:早期王权的象征王者,一国之主。
其形三横一竖,为何这样的构形?汉字早期在表达这个王字时,以器具"戉"的象形来指代"王",表明白"戉"即"王"的含义。
不过这个原初的字义,好像很快就被忘却了。
两周到西汉时期对"王"字的解释,臆断成分很重。
孔子曰:一贯三为王。
"董仲舒曰:"三画而连其中谓之王。
"孔子和董仲舒都没有将这个字解释精确。
甲骨文的发觉,为推定"王"字的本初意义供应了证据。
文史学家吴其昌说,戊、戉、戍、成、咸诸字皆由石斧的外形演化而来,其锋刃左右旁向者衍为上述各字,其锋刃向下时则衍为工、士、壬、王诸字。
这个斧头的外形,竟然造就了如此多的字形,斧头的方向判定了字的意义,让我们见识了古人造字的意趣。
考古学家林沄有专文《说王》,论"王"字本像无柄且刃缘向下的斧钺之形,本表示军事统率权,后来这军事统率权的象征演化为王的权杖。
"戉"的象形,是王字定形的基础,这已经成为古文字学家的共识。
历史学家徐中舒也说戉的写法,"像刃部下向之斧形,以主刑杀之斧钺象征王者之权威"。
甲骨文存在的商代,钺已经是青铜质,而戉字的消失却是更早时代的事,良渚文化陶器上的刻画就是证据。
良渚文化玉戉的背后,或许已经有了王权的定义吧。
戉这个字,可以给相关的字形字义更多的提示。
有讨论者论"辛",说最早的甲骨文"辛",是一把执行最严峻刑法的割人肉用的锐利小刀,三面有刃,字的下尖或左偏或右偏,表示刀锋歪斜,类似雕刻刀。
2021届山西大学附属中学高三英语第二次联考试题及参考答案
2021届山西大学附属中学高三英语第二次联考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AWashington D.C. SightseeingWith the information below, you’re not missing anything in D.C.! Click Here to find the perfect hotel for your stay as well.The Old Town Trolley TourIt offers something for the whole family. Not only will it give them something fun to do, but it will give them a history lesson. This tour will last about three hours and it’s proper for people of all ages.African American History TourBe sure to take this tour because African Americans have had an important role in the making of our country. Take this historical four-hour tour, where you will visit some important sites including Museum of African American History and Culture.Comedy Walks Washington D.C.This is a great experience allowing you to enjoy the capital in a new way. The walking tour lasts for about one hour and thirty minutes, which takes place in less than a mile journey from the starting place.D.C. Twilight TourCheck out the D.C. Twilight Tour for a unique view of some of the most famous sites! What makes this two-hour guided tour truly unique is that you can view many wonderful sites at night time!1.Which tour is recommended to a tourist who is fond of hiking?A.The Old Town Trolley TourB.African American History Touredy Walks WashingtonD.C. D.D.C. Twilight Tour2.Which tour lasts longest?A.The Old Town Trolley TourB.African American History Touredy Walks WashingtonD.C. D.D.C. Twilight Tour3.Where will you read this text most likely?A.In a guidebook.B.In a magazine.C.In a newspaper.D.On the Internet.BLight pollution is a significant but overlooked driver of the rapid decline of insect populations, according to the most comprehensive review of the scientific evidence to date.Artificial light at night can affect every aspect of insects' lives, the researchers said. "We strongly believe artificial light at night — in combination with habitat loss, chemical pollution.invasive (入侵的) species, and climate change — is driving insect declines, " the scientists concluded after assessing more than 150 studies.Insect population collapses have been reported around the world, and the first global scientific review published in February,said widespread declines threatened to cause a "catastrophic collapse of nature's ecosystems".There are thought to be millions of insect species, most still unknown to science, and about half are active at night. Those active in the day may also be disturbed by light at night when they are at rest.The most familiar impact of light pollution is moths (飞蛾) flapping around a bulb, mistaking it for the moon. Some insects use the polarisation of light to find the water they need to breed, as light waves line up after reflecting from a smooth surface. But artificial light can scupper (使泡汤) this. Insects areimportant prey (猎物) for many species, but light pollution can tip the balance in favour of the predator if it traps insects around lights. Such increases in predation risk were likely to cause the rapid extinction of affected species, the researchers said.The researchers said most human-caused threats to insects have analogues in nature, such as climate change and invasive species. But light pollution is particularly hard for insects to deal with.However, unlike other drivers of decline, light pollution is ly easy to prevent. Simply turning off lights that are not needed is the most obvious action, he said, while making lights motion-activated also cuts light pollution. Shading lights so only the area needed is lit up is important. It is the same with avoiding blue-white lights, which interfere with daily rhythms. LED lights also offer hope as they can be easily tuned to avoid harmful colours and flicker rates.4. What is discussed in the passage?A. Causes of declining insect populations.B. Consequences of insect population collapses.C. Light pollution: the key bringer of insect declines.D. Insect declines: the driver of the collapsed ecosystem.5. What is the 5th paragraph mainly about?A. How light travels in space.B. How light helps insects find food.C. How the food chain is interrelated.D. How light pollution affects insects.6. What does the underlined word"analogues"in Paragraph 6probably mean?A. Selective things.B. Similar things.C. Variations.D. Limitations.7. What is the purpose of the last paragraph?A. To offer solutions.B. To give examples.C. To make comparisons.D. To present arguments.CBill Gates on how to fight future pandemicsWHEN HISTORIANS write the book on the covid-19 pandemic, what we've lived through so far will probably take up only the first third or so.The bulk of the story will be what happens next.I believe that humanity will beat this pandemic, but only when most of the population is vaccinated(接种疫苗).Until then, life will not return to normal.As the pandemic slows in developed nations,itwill accelerate in developing ones.Their experience,however,will be worse.In poorer countries,where fewer jobs can be done remotely,distancing measures won't work as well.The virus will spread quickly,and health systems won't be able to care for the infected.Wealthy nations can help.But people in rich and poor places alike will be safe only once we have an effective medical solution for this virus,which means a vaccine.My hope is that,by the second half of 2021,facilities around the world will be manufacturing a vaccine.If that's the case,it will be a history-making achievement: the fastest humankind has ever gone from recognizing a new disease to immunizing(免疫)against it.Apart from this progress in vaccines,two other big medical breakthroughs will emerge from the pandemic.One will be in the field of diagnostics.The next time a novel virus crops up,people will probably be able to test for it at home.Researchers could have such a test ready within a few months of identifying a new disease.The third breakthrough will be in antiviral drugs.We haven't been as effective at developing drugs to fight viruses as we have those to fight bacteria.But that will Researchers will develop large diverse libraries ofantivirals,which they'll be able to scan trough and quickly find effective treatments for novel viruses.All three technologies will prepare us for the next pandemic by allowing us to intervene(干预)early when the number of cases is still very low.Our progress won't be in science alone.It will also be in our ability to make sure everyone benefits from that science.In the years after 2021,I think we'll learn from the years after 1945. With the end of the Second World War, leaders built international institutions like the UN to prevent more conflicts.After covid-19, leaders will prepare institutions to prevent the next pandemic.These will be a mix of national,regional and global organizations.I expect they will participate in regular"germ games”in the same way as armed forces take part in War games.These will keep us ready for the next time a novel virus jumps from bats or birds to humans.I hope wealthy nations include poorer ones in these preparations,especially by devoting more foreign aid to building up their primary health-care systems.This pandemic has shown us that viruses don't obey border laws and that we are all connected biologically by a network of microscopic germs,whether we like it or not.The best analogy(类比)for today might be November 10th 1942.Britainhad just won its first land victory of the war,and Winston Churchill declared in a speech: “This is not the end.It is not even the beginning of the end.But it is,perhaps,the end of the beginning.”8. What are the three technologies that will prepare us for the next pandemic?①manufacturing a vaccine fast②diagnosing a virus at home③developing antiviral drugs④allowing us to intervene earlyA. ①②③B. ①②④C. ①③④D. ②③④9. As far as poorer countries areconcerned,which of the following is TRUE according to this passage?A. pandemic disease is more likely to begin in poorer countriesB. Working from home can work well in poorer countries.C. Health systems are sufficient to care for the infected in poorer countries.D. Virus will cross borders if poorer countries fail to contain it.10. Why is the Second World War mentioned in Para.8?A. The fight against the COVID-19 is similar to the Second World War.B. People are suffering just as they were in the Second World War.C. We should cooperate globally just as we did after the Second World War.D. Countries are fighting each other like in the Second World War.11. What is the tone of this passage?A. pessimisticB. optimisticC. neutralD. indifferentDIn the natural habitat, a binge-watcher is a strange sub-species of modern human beings. They are alone and are often found lying on their bed or sofa, still as a rock, looking searchingly into their laptop or at the TV. They rarely get up, only taking occasional breaks for those urgent calls of nature. Unlike so many others of their species, they don’t sleep at the end of every day. They stay up late and are often found to have red eyes.This, of course, is a little bit exaggerated, but for many of us, binge-watching a show is how we consume our entertainment. With streaming services bringing seemingly endless content to the tips of our finger, creatorstailortheir shows to our needs and tastes, while their marketing team sells it to us as the next most bingeable show. I can’t help but wonder if this way of consuming television does us any good.We’re advised to get 7 to 8 hours of sleep per day, but staying up all night to finish shows like ‘Breaking Bad’ and ‘The Fall’, which Netflix says are binged the fastest, won’t result in 8 hours of uninterrupted sleep. Netflix’s CEO Richard Hastings told analysts at a conference, “Think about it… when you watch a show from Netflix and you get addicted to it, you stay up late at night. We’re competing with sleep.” And it isn’t only the amount of sleep we get; the quality matters too! A study published in the Journal for Clinical Sleep Medicine reported that those who binged television more often were found to have poorer sleep quality. The mental arousal we get from watching TV doesn’t lend itself to peaceful sleep.Besides, binging TV can cause weight gain. For every extra hour of TV watched, there was a 2% increase in the prevalence of obesity, according to a study conducted by Harvard that linked TV watching to obesity in children and adolescents.While it is true that there’s growing evidence that binging isn’t good for us, the results aren’t all hopeless. Binging, occasionally, might have some benefits. For many people, binging is a good way to socialize. It gives people something to talk about at parties and with their friends. Fans of popular shows often hold viewing parties where people can binge-watch shows together and then discuss what they just experienced! This interaction can create a sense of community for many.For others, binging might just be relaxing. After a long day, a few hours of Friends—still one of the most popular shows on any streaming platform—can ease the stress of a long day. A study followed 240 people through their binging and recorded their stress hormones. They noted for some people that their stress hormone levels decreased during their binging experience.With binging, there is not a “good” or “bad” answer. Like many things, the key is moderation. Watching TV can be relaxing, but only if it doesn’t stop you from exercising, taking care of your health, and fulfilling your social responsibilities.12. What does the underlined word “tailor” in Paragraph 2 probably mean?A. Study.B. Sell.C. Adjust.D. Promote.13. What can we learn from the passage?A. The amount of sleep matters more than the quality.B. There’s no real harm in binge-watching a TV series.C. Binging may help keep stress hormone levels stable.D. Binge-watching may help increase social connections.14. What is the author’s attitude towards binge-watching?A. Supportive.B. Unconcerned.C. ObjectiveD. Doubtful.15. Which of the following would be the best title for the passage?A. Who Is to Blame for Binge-Watching?B. Is Binge-Watching Good or Bad for You?C. Is Binge-Watching Getting out of Control?D. HowDoes Binge-Watching Affect Your 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。
2020-2021学年山西大学附属中学高三英语第三次联考试题及答案
2020-2021学年山西大学附属中学高三英语第三次联考试题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AWhen it comes to the greatest inventions of the world, China’s name is sure to be mentioned several times. There are hundreds of things which were invented by the Chinese.AlcoholShocked? I was because when they said alcohol I thought about either the US or the UK. China had never crossed my mind. In China, alcohol was made by two legendary persons named Yi Di and Du Kang who belonged to the Xia Dynasty. This period was about 2000 BC - 1600 BC. Research says that in ancient China, beer with 4% alcoholic content was widely consumed by people.TeaChina is the proud inventor of tea which was first drunk by Shen Nong, a Chinese emperor around 2737 BC. Tea production was rapidly developed, making tea a popular drink during the Tang and Song Dynasties.The Mechanical ClockHave you ever wondered what on earth we would be doing without any idea of time? A clock really is an invention without which things were incomplete. The credit of making the first mechanical clock goes to ancient China. The first mechanical clock was invented by Yi Xing in the Tang Dynasty. This was during 618 and 907.Silk FabricSilk, the favorite fabric of many girls out there, is also a Chinese invention. Although we all know that silk is made by silkworms, it was Chinese people who first invented a way to harvest the silk and then use it to make clothes. The oldest silk which has been found so far is in Henan Province and dates back to 3630 BC.1.We can learn from Paragraph 2 that ________.A.people in the US like alcoholB.beer was popular in the Xia DynastyC.Yi Di and Du Kang invented alcohol by accidentD.the author didn’t know alcohol is a Chinese invention2.The author asks the question in Paragraph 4 to show ________.A.ancient Chinese inventors were wiserB.many things in our lives are incompleteC.ancient Chinese people never wasted timeD.the invention of the mechanical clock is important3.Which of the following invention has a longer history?A.Tea.B.Alcohol.C.Silk fabric.D.The mechanical clock.BHoneybees can’t swim, and when their wings are wet, they can’t fly, either. But Chris Roh and other researchers at the California Institute of Technology found that when bees drop into bodies of water, they can use their wings toproduce little waves and slide toward land-like surfers who create and then ride their own waves.As with many scientific advances-IsaacNewton’s apple or Benjamin Franklin’s lightning bolt-Dr. Roh’s experiment began with a walk. Passing Caltech’s Millikan Pond in 2016, he observed a bee on the water’s surface producing waves. He wondered how an insect known for flight could push itself through water.Dr. Roh and his co-worker, Morteza Gharib, used butterfly nets to collect localPasadenahoneybees and observed their surf-like movements. The researchers used a wire to restrict each bee’s bodily movement, allowing close examination of their wings. They found that the bee bends its wings at a 30-degree angle, pulling up water and producing a forward force. Bees get trapped on the surface because water is roughly three times heavier than air. But that weight helps to push the bee forward when its wings move quickly up and down. It’s a tough exercise for the bees, which the researchers guess could handle about 10 minutes of the activity.The researchers said the surf-like movement hasn’t been documented in other insects and most semiaquatic insects use their legs forpropulsion, which is known as water-walking. It may have evolved in bees, they-predicted.Dr. Roh and Dr. Gharib have imagined many practical applications for bees’ surfing. One plan is to use their observations to design robots able to travel across sky and sea. “This could be useful for search and rescues, or for getting samples of the surface of the ocean, if you can’t send a boat or helicopter,” Dr. Gharib said.4. What does the author intend to show by mentioningNewtonandFranklin?A. Roh’s admiration for them.B. Roh’s chance discovery about bees.C. Their outstanding talent for science.D. Their similar achievements in discovery.5. What plays the most vital role in a bee’s moving forward on water?A. The air weight.B. Its leg extension.C. The water movement.D. Its continuous wingbeat.6. What does the underlined word “propulsion” in Paragraph 4 mean?A. Fast flightB. Driving force.C. Pulling speed.D. Explosive power.7. What does the text mainly tell us?A. Honeybees can surf to safety.B. Bees help scientists make inventions.C. Insects can adapt to the environment.D. Nature is a helpful guide for discovery.CTyphoons can be deadly — in 2013, Typhoon Haiyan, the strongest ever recorded, was responsible for 6,340 deaths—and cost billions in damages. Current forecast modelscan only predict these storms 10 days in advance, at most, and they cannot precisely predict how intense the storms will become.An international team of researchers has developed a model that analyzes nearly a quarter of Earth’s surface and atmosphere in order to better predict the conditions that birth typoons.“The target problem of this study is how to foretell the birth of typhoons,” said paper author Mingkui Li, associate professor in the Key Laboratory of Physical Oceanography in the Ocean University of China and the Pilot National Laboratory for Marine Science and Technology (QNLM). “We specifically address three aspects: the beginning time, inner pressure and maximum wind speed.”The researchers also accounted for the influence of one variable(变量) on another, such as wind speed on sea surface temperature. This influence is well understood and accounted for in climate predictions and in weather forecasts, butithas not been fully applied in understanding how long-term climate affects day-to-day weather, according to Li. “We aimed to provide insights on the time scale that can be used to forecast typhoons in advance.”From their study, the researchers determined that a model with the ability to better understand the relationship between warm sea surface temperatures and weak wind movement— conditions that favor typhoon formation—could improve typhoon predictability.“Our goal is to develop a 10 to 30-day prediction system that will lead to seamless(无缝的) weather-climate predictions.” Shaoqing Zhang, paper author and professor in the Key Laboratory of Physical Oceanography, said.8. What is the problem with the present forecast system?A. It cannot foretell storms in advance.B. It is ineffective in accuracy and timeliness.C. It costs too much and causes great damages.D. It can hardly predict the intensity of typhoons.9. What does the underlined word “it” in paragraph 4 refer to?A. The variable.B. The climateC. The temperature.D. The influence.10. What is the purpose of the study?A. To advance the prediction system.B. To figure out the three main aspects.C. To know how climate affects daily weather.D. To understand the influence of the variables.11. Where is this text most likely from?A. Awork diary.B. A travel guidebook.C. A science magazine.D. A fantasy fiction.DDragon boating is a team sport that has its root in ancient China. The boats are decorated with a dragon head and tail. In recent years cancer survivor groups have got involved in the sport to help make friends and help rebuild their lives.On a recent Saturday morning, a group of 20 women were on a boat in the Anacostia River in Washington DC. They moved their paddles(船桨)in rhythm to the call of a coach. The women belong to the dragon boat team GoPink! DC, which trains weekly. It also races against other breast cancer survivor teams in dragon boat festivals. As a result, GoPink! DC won medals in this Washington dragon boat festival.Lydia Collins joined five years ago after finding out she had breast cancer. “I was diagnosed with breast cancer.I was demoralized because of my illness - I lost all interest in life and wouldn't even get out of bed to eat. But now I love the team spirit. I just love everything about it. It is like a floating support group on the water.”The paddles are breast cancer survivors and their supporters. Annette Rothemel helped establish(建立)the group in 2006. She is a researcher with the National Institutes of Health as well as a breast cancer survivor. “It is sort of an easy entry sport because on the same boat people at different levels can be doing the same sport.” But Ms Rothemel saysdragon boating can be physically demanding, especially for someone who is sick and getting treatment for cancer.“It’s hard but I think you have to challenge yourself in life. This is something I look forward to. I get to be out here with my sisters and supporters that understand what I’m going through and help motivate me. So it makes me stronger and it makes me feel better,” another cancer survivor Rhonda Hartzel said.Annette Rothemel says the cancer survivors feel a sense of sisterhood and share good times when they paddle together. She says both feelings are treasured by the team.12. What do the underline wordsdemoralizedIn para.3 probably mean?A. depressedB. anxiousC. astonishedD. awkward13. What can we know about Lydia Collims from the text?A. she helps establish Go Pink !DCB. she tries to find a cure for the cancerC. she benefits from the dragon boat raceD. she gives up hope because of her illness14. How can the dragon boat race help the cancer survivorsA. forget their tough experiencesB. recover physically and mentallyC. get rid of the pains of their cancerD. enjoy their rest life without sufferings15. What does the text tell us about Annette Rothemel?A. she is an expert in studying the cause of the cancerB. she helps the cancer survivors in financial difficultiesC. she believes there is a healthful result from the dragon boat raceD. she thinks it unwise for the patient to join in the dragon boat race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学年山西大学附中高三(上)第二次诊断数学试卷(文科)
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山西省山西大学附中2020届高三数学上学期第二次模块诊断试题理【含答案】
考试时间:120 分 满分:150 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,
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山西大学附中2020届高三上学期第二次模块诊断 历史 Word版含答案
山西大学附中2019-2020学年高三第一学期(总第二次)模块诊断历史试题考试时间:80分钟考查范围:古代史一.选择题(共计42小题,每题1.5分,共计63分)1.《礼记》载:“殷人尊神,率民以事神,先鬼而后礼。
”《尚书·召诰》曰:“肆(尽,极)惟王其疾(急速)敬德!王其德之用,祈天永命。
”这一变化反映出A. 王权失去神权色彩B. 天命观被彻底否定C. 敬德决定国家兴衰D. 治国渐趋理性务实2.西周初年各地发生叛乱,周公东征,平定了殷地的“三监之乱”,征服了殷商在东方的残余势力后,大行分封齐、鲁、燕、卫等国。
这说明西周A.分封制是周公执政时期的制度创新 B.分封制是当时巩固统治的重要手段C.实现了对东方广大地区的垂直管理 D.政治中心实现由西部向东部的转移3.《左传·昭公七年》载:“天有十日,人有十等。
下所以事上,上所以共神也。
故王臣公,公臣大夫,大夫臣士,士臣皂。
”材料反映我国古代早期政治制度的基本特点是A.神权与王权相结合B. 实行严格的等级制度C.实行嫡长子继承制D. “国”与“家”相结合4.据史料记载,秦代的官吏须经过封建国家考核后正式委任才能任职,若未任命而敢先行行使职权,或相“听以遣之”,即受私人派遣,都要依法治罪。
这表明秦代A.中央官僚体系发生变革 B.选官的方法和标准较严苛C.任命官吏遵循一定程序 D.任命和考核官吏密切结合5.东晋高僧慧远曾说:“道法之与名教,如来之与尧、孔,发致虽殊,潜相影响,出处诚异,终期则同。
”东晋道士葛洪著《抱朴子》有内、外两篇,“其内篇言神仙方药,鬼怪变化。
养生延年,禳邪却祸之事,属道家;其外篇言人间得失,世事臧否,属儒家。
”这说明A.儒、道、佛出现合流的趋势 B.佛、道挑战了儒学正统地位C.儒、道、佛的终极目标一致 D.东晋最终实现了三教合归儒6.读表1,由表中所给的数据可知中国历代粮食亩产量变化一览表单位:市斤/市亩表1(资料余也非《中闰历代粮食平均亩产量考略》)A.秦汉时期开始形成“南稻北粟”格局 B.魏晋时期农业重心出现南移趋势C.唐宋时期国家分裂导致农业经济衰退 D.明清时期北方农业税赋高于南方7.上表为不同史籍关于东晋义熙六年同一事件的历史叙述。
山西省太原市山西大学附属中学高三第二次联考新高考语文试卷及答案解析
山西省太原市山西大学附属中学高三第二次联考新高考语文试卷请考生注意:1.请用2B铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。
写在试题卷、草稿纸上均无效。
2.答题前,认真阅读答题纸上的《注意事项》,按规定答题。
1、阅读下面的文字,完成下列各题。
材料一2016年7月,国家版权局、国家互联网信息办公室、工业和信息化部、公安部在京联合召开“剑网2016”专项行动新闻通气会,下发《关于开展打击网络侵权盗版“剑网2016”专项行动的通知》,启动“剑网2016”专项行动。
国家版权局版权管理司司长于慈珂通报了“剑网2016”专项行动的工作部署。
专项行动从2016年7月开始,利用5个月的时间,突出整治未经授权非法传播网络文学、新闻、影视等作品的侵权盗版行为,保障有关权利人的合法权益;重点查处通过智能移动终端第三方应用程序、电子商务平台、网络广告联盟、私人影院等平台的侵权盗版行为,维护网络版权正常秩序;进一步規范网络音乐、网络云存儲空间、网络转载新闻作品的版权秩序。
(选自《中国知识产权》)材料二2017年8月﹣11月,青海省版权局、互联网信息办公室、公安厅、通信管理局联合组织开展打击网络侵权盗版”剑网2017”专项行动。
此次专项行动聚焦新闻出版影视行业的网络版权保护及电子商务平台和移动互联网应用程序领成的版权整治,以规范影视和新闻作品网络版权秩序为重点,以查办大案要为抓手,严厉打击各类网站、移动客户端、“自媒体”传播侵权盗版作品行为,集中整治电子商务平台、移动互联网应用程序商店版权秩序,巩固网络文学、网络音乐、网络云存儲空间、网络广告联盟版权治理成果,强化互联网企业的主体责任,维护良好的网络版权秩序。
(选自《中国防伪报道》)材料三:昨天,国家版权局、国家互联网信息办公室、工业和信息化部、公安部联合启动“剑网2018”专项行动,旨在打击网络侵权盗版。
2020届山西大学附中高三上学期第二次模块诊断 数学(理)(PDF版)
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2021届山西大学附属中学高三英语第二次联考试卷及答案
2021届山西大学附属中学高三英语第二次联考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AI truly thought thatI might die that day. Had I not seen three bears and a few wolves over the last couple of days near the road, I might have just lain down and called it quits. What in the world was I thinking, dragging my bike up to Yellowstone and thinking I could ride from West Yellowstone to Old Faithful in the snow? No less!I had planned this trip for more than two months, and I wasn't about to give up so easily. The first few miles were beautiful. Ten miles in, I started sucking wind. Fifteen miles and my legs felt like they were made of lead. By twenty miles, my lungs were burning and felt like there was nothing left in the tank. That was when I turned around and saw my husband and three children cheering me on in the van behind me. I knew I couldn't quit because I tell my children all the time that just because something is hard doesn't mean that you stop doing it. I had to live what I'd been preaching(说教). That thought got me up that mountain and to the end of the ride.The importance of that ride was apparent after only a week. My eight-year-old daughter Emalee wanted to ride in a twelve-mile charity ride. That day was cold as well. She was the youngest rider. About four miles into the ride, she started feeling cold. The chill(寒冷)was making her muscles cramp a little and she began to struggle. By six miles, she had tears running down her face. It broke my heart to see her suffering like that.I told herthat she didn't have to finish. She said that she wouldn't quit. I told her how I had wanted to quit the week before, but perseverance had gotten me to the end and I knew she could do it, too.The look on her face as she pulled into the finish was priceless. She threw her arms around me and said, “You didn't quit, Mommy, so neither did I.”1. What motivated the author to finish her ride?A. Her family's encouragement.B. Her own instruction for children.C. The beautiful scenery and weather.D. The careful and thorough preparation.2. By describing the conditions Emalee met, the author wanted to________.A. show her guilt for her daughterB. prove that riding is a tough taskC. persuade her daughter to give upD. indicate she was proud of Emalee3. What does the author want to tell us according to the story?A. Never say die.B. Do nothing by halves.C. Children are what the mothers are.D. Every mother's child is handsome.BYour best friend that follows you around when the sun comes out - your shadow - doesn’t serve an important function like your heart or brain, but what if you could use shadows to create electricity? When using solar panels (电池板) that are powered by light, shadows can be boring because it means electricity can’t be created. However, researchers from the National University of Singapore have engineered a way to create power from the shadows present everywhere.A team of the university created a machine that can collect energy from shadows. It is created by placing a thin coating of gold onto silicon (硅). Like in a normal solar panel, when put in light, the silicon electrons (电子) become energized and the energized electrons then jump from the silicon to the gold. The voltage (电压) of the part of the machine that is placed in the light increases to the dark part and the electrons in the machine flow from high to low voltage. They are sent through an external circuit (外电路) creating a current that can be used to power another machine. The greater the contrast between light and dark, more energy is provided by the machine.The team isworking on improving the performance of the machine, borrowing approaches from solar panels to gather light. Increasing the amount of light the machines can receive allows them to better make use of shadows, as well as developing shadow energy collecting panels that can successfully gather from indoor lighting. The team is also researching the use of other materials other than gold to drop the price of the machine, meaning they would be more cost effective and easier to apply in society.Shadows are present everywhere and perhaps one day in the future we will be able to collect energy from them by placing the shadow-effect energy machine around the world in places that have been considered unfit for solar panels to work, or indoors. “A lot of people think that shadows are useless,” Tan says, but “anything can be useful, even shadows.”4. What is Paragraph 1 mainly about?A. Your best friend always stays with you after the sunrise.B. The shadow has the same function as the heart and brain.C. Shadows can stop solar panels from creating electricity.D. Researchers have found a way to create power from shadows.5. What is the key working principle of the machine mentioned in the text?A. The silicon produces electricity when it is in the light.B. The gold produces power with the help of the silicon.C. The energized electrons flow from high to low voltage.D. An external circuit creates current using another machine.6. How does the team improve the performance of the machine?A. Using solar panels in the machine.B. Increasing the amount of light received.C. Developing light energy collecting panels.D. Bringing down the price of gold.7. Which of the following is the best place to apply the machine?A. A gym.B. A park.C. A farm.D. A playground.CHi, Momis a hit inChina. Since its release in early February the movie has earned more than 5 billion yuan at the box office. It is currently the second-highest-grossing film ever released in the country, and may yet overtakeWolf Worrior2, an action film from 2017.Written and directed by Jia Ling, the film is adapted from her comedy routine from 2016 which explored her relationship with her mother, who died in an accident when Ms. Jia was 19. At a time when many people have been separated from their families, the themes of grief and filial piety (孝道) inHi, Mombring agreement.The film’s impressive box office income is also a reminder of the health of the industry inChinacompared withHollywood. In the West cinemas remain closed due to lockdown or operate at limited capacity; inChina, where many restrictions have been lifted, between 50% and 75% of seats are available to book. During the Spring Festival, Chinese cinemas made an estimated $1.5 billion in ticket sales – 71% of takings at the American box office in all of 2020.With cinemas shut, film-sales agent stopped bringingHollywoodmovies to market. That has given locally made films a chance to excel. Since 2013, domestic films have taken around 60% of total box office earnings. In2021, that number may get closer to 100%.8. What can be inferred from the first paragraph aboutHi, Mom?A. It will be a success.B.Wolf Worrior2 will be released after it.C. It is better than Wolf Worrior 2.D. It may be the first-highest-grossing film inChina.9. Which statement is NOT true according to the text?A. The film industry inChinais developing better than that inHollywood.B. The film is based on a real story.C. You can’t go to cinema because there are many restrictions.D. Many people can’t see their families often nowadays.10. What is the author’s attitude towards local films?A. NegativeB. PositiveC. IndifferentD. Skeptical11. What can be a suitable title for the text?A. Local Movies Arise.B. China Film Industry.C. Comedy Time.D.Hi, Mom-China’s Latest Hit.DThe far side of the moonis a strange and wild region, quite different from the familiar and mostly smooth face we see nightly from our planet. Soon this rough space will have even stranger features: it will be crowded with radio telescopes.Astronomers are planning to make the moon's distant side our newest and best window on the cosmic(宇宙的) dark ages, a mysterious era hiding early marks of stars and galaxies. Our universe was not always filled with stars. About 380,000 years after the big bang, the universe cooled, and the first atoms of hydrogen formed. Gigantic hydrogen clouds soon filled the universe. But for a few hundred million years, everything remained dark, without stars. Then came the cosmic dawn: the first stars flickered, galaxies came into existence and slowly the universe's largescale structure took shape.The seeds of this structure must have been present in the darkage hydrogen clouds, but the era has been impossible toprobeusing optical(光学的) telescopes—there was no light. And although this hydrogen producedlongwavelength(or lowfrequency) radio emissions,radio telescopes on Earth have found it nearly impossible to detect them. Our atmosphere either blocks or disturbs these faint signals; those that get through are drowned out by humanity's radio noise.Scientists have dreamed for decades of studying the cosmic dark ages from the moon's far side. Now multiple space agencies plan lunar missions carrying radiowavedetecting instruments—some within the next three years—and astronomers' dreams are set to become reality.“If I were to design an ideal place to do lowfrequency radio astronomy, I would have to build the moon,”says astrophysicist Jack Burns of the University of Colorado Boulder. “We are just now finally getting to the place where we're actually going to be putting these telescopes down on the moon in the next few years.”12. What's the purpose of building radio telescopes on the moon?A To research the big bang. B. To discover unknown stars.C. To study the cosmic dark ages.D. To observe the far side of the moon.13. What does the underlined word “probe” in Paragraph 3 possibly mean?A. Explore.B. Evaluate.C. Produce.D. Predict.14. Hydrogen radio emissions can't be detected on Earth because ________.A. there was no light in the dark agesB. they cannot possibly get through our atmosphereC. gigantic hydrogen clouds no longer fill the universeD. radio signals on Earth cause too much interference15. What can we infer from theunderlined sentence in the last paragraph?A. Scientists have to rebuild the moon.B. We will finally get to the moon's distant side.C. The moon is a perfect place to set up radio telescopes.D. A favorable research environment will be found on the moon.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。
2023年山西大学附中中考数学二模试卷及答案解析
2023年山西大学附中中考数学二模试卷一、选择题(每题3分,本大题共10小题,共30分。
在每小题列出的选项中,选出符合题目的一项)1.(3分)如图,矩形ABCD中,AB=3,BC=8,点P为矩形内一动点,且满足∠PBC=∠PCD,则线段PD的最小值为()A.5B.1C.2D.32.(3分)如图,抛物线y=ax2+bx+c(a≠0)交x轴于点A,B,交y轴于点C.若点A坐标为(﹣4,0),对称轴为直线x=﹣1,则下列结论错误的是()A.二次函数的最大值为a﹣b+c B.a+b+c>0C.b2﹣4ac>0D.2a+b=03.(3分)已知二次函数y=﹣x2+x+c(c<0),当自变量为x1时,其函数值y1大于零;当自变量为x1﹣1与x1+1时,其函数值分别为y2,y3,则()A.y2>0,y3>0B.y2>0,y3<0C.y2<0,y3<0D.y2<0,y3>0 4.(3分)如图,在⊙O中,AB是直径,CD⊥AB,∠ACD=60°,OD=2,那么DC的长等于()A.B.C.2D.45.(3分)二次函数y=ax2+bx+c(a,b,c是常数,a≠0)的自变量x与函数值y的部分对应值如下表:x…﹣2﹣1012…y=ax2+bx+c…t m﹣2﹣2n…且当x=﹣时,与其对应的函数值y>0.有下列结论:①abc>0;②﹣2和3是关于x的方程ax2+bx+c=t的两个根;③0<m+n<.其中,正确结论的个数是()A.0B.1C.2D.36.(3分)如图,将一个小球从斜坡的点O处抛出,小球的抛出路线可以用二次函数y=4x﹣x2刻画,斜坡可以用一次函数y=x刻画,下列结论错误的是()A.当小球抛出高度达到7.5m时,小球距O点水平距离为3mB.小球距O点水平距离超过4米呈下降趋势C.小球落地点距O点水平距离为7米D.斜坡的坡度为1:27.(3分)关于x的一元二次方程有一个根是﹣1,若二次函数的图象的顶点在第一象限,设t=2a+b,则t的取值范围是()A.B.C.D.8.(3分)如图,AB是⊙O的直径,点C和点D是⊙O上位于直径AB两侧的点,连接AC,AD,BD,CD,若⊙O的半径是13,BD=24,则sin∠ACD的值是()A.B.C.D.9.(3分)如图,在⊙O中,OA=2,∠C=45°,则图中阴影部分的面积为()A.﹣B.π﹣C.﹣2D.π﹣210.(3分)如图,在⊙O中,AB切⊙O于点A,连接OB交⊙O于点C,过点A作AD∥OB交⊙O于点D,连接CD.若∠B=40°,则∠OCD为()A.15°B.20°C.25°D.30°二、填空题(每题3分,本大题共5小题,共15分)11.(3分)喜迎二十大,“龙舟故里”赛龙舟.丹丹在汨罗江国际龙舟竞渡中心广场点P处观看200米直道竞速赛.如图所示,赛道AB为东西方向,赛道起点A位于点P的北偏西30°方向上,终点B位于点P的北偏东60°方向上,AB=200米,则点P到赛道AB 的距离约为米(结果保留整数,参考数据:≈1.732).12.(3分)在数学实践活动课上,某兴趣小组测量操场上篮球筐距地面的高度如图所示,已知篮球筐的直径AB约为0.45m,某同学站在C处,先仰望篮球筐直径的一端A处,测得仰角为42°,再调整视线,测得篮球筐直径的另一端B处的仰角为35°.若该同学的目高OC为1.7m,则篮球筐距地面的高度AD大约是m.(结果精确到1m).(参考数据:tan42°≈0.9,tan35°=0.7,tan48°≈1.1,tan55°≈1.4)13.(3分)下列关于抛物线y=x2﹣2mx+m2+m+1(m为常数)的结论:①抛物线的对称轴为直线x=m;②抛物线的顶点在直线y=x+1上;③抛物线与y轴的交点在原点的上方;④抛物线上有两点A(x1,y1),B(x2,y2),若x1<x2,x1+x2>2m,则y1<y2.其中正确结论的序号是.14.(3分)如图,在正方形ABCD中,BC=4,以B为圆心,BC长度为半径画弧AC,在AB上有一点E,将BC沿CE折叠后,点B恰好落在弧AC上的点F处,弧CF的长度.15.(3分)如图,正方形ABCD的边长为2,点E是BC边上一点,以AB为直径在正方形内作半圆O,将△DCE沿DE翻折,点C刚好落在半圆O的点F处,则CF的长为.三、解答题(本大题共8小题,共55分。
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山西大学附中高中数学(高三)导学设计 编号20
二次函数、二次不等式、二次方程
1.若关于x 的不等式0)(322>++-a x a a x 的解集为}|{2
a x a x x ><或,则实数a 的取值范围是
A .0≤a
B . 1≥a
C .10≤≤a
D .10<<a
2.a b +<<10,若关于x 的不等式2()x b ->2()ax 的解集中的整数恰有3个,则
A .01<<-a
B .10<<a
C .31<<a
D .63<<a
3.若函数()2(2)(21)f x m x mx m =-+++的两个零点分别在区间(1,0)-和区间()1,2
内,则实数m 的取值范围是 A.11(,)24- B.11(,)42- C.11(,)42 D.11
[,]42
4.已知函数2()22(4)1f x mx m x =--+,()g x mx =,
若对于任一实数x ,()f x 与()g x 至少有一个为正数,则实数m 的取值范围是
A .(0,2)
B .(0,8)
C .(2,8)
D .(,0)-∞
5.函数2()(0)f x ax bx c a =++≠的图象关于直线2b x a
=-对称.据此可推测,对任意的非零实数p n m c b a ,,,,,,关于x 的方程[]2()()0m f x nf x p ++=的解集都不可能是 A. {}1,2 B {}1,4 C {}1,2,3,4 D {}
1,4,16,64
6.已知函数()()()()222222,228.f x x a x a g x x a x a =-++=-+--+设()()(){}()()(){}{}()12max ,,min ,,max ,H x f x g x H x f x g x p q ==表示,p q 中的较大值,{}min ,p q 表示,p q 中的较小值,记()1H x 得最小值为,A ()2H x 得最大值为B ,则A B -=
A .2216a a --
B .2216a a +-
C .16-
D .16
7.若不等式y y a x x 2222--≥++对一切实数y x ,都成立,则实数a 的取值范围是____
8.已知))()((1)(b a b x a x x f <---=,并且)(,n m n m <是方程0)(=x f 的两根,则
实数n m b a ,,,的大小关系是____________.
9.若方程2210ax x ++=至少有一个负数根,则实数a 的取值范围_______________.
10.已知关于x 的方程016)82(22=-+--m x m x 的两实根21,x x 满足2123x x <<
,则实数m 的取值范围__________.
11.若方程0422=+-ax x 在区间(]2,1上有且仅有一个实根,则实数a 的取值范围是__
12.已知关于x 的方程02cos sin 2
=-+a x a x 有实数解,则实数a 的取值范围为___
13.若函数]1,0[12)(2在区间a ax x x f -++-=上的最大值为2,则实数a 的值是______. 14.已知关于x 的方程03)3()13)(1(3
112=⋅----+++x x x m m 有两个不同的实数根,则
实数m 的取值范围是______.
15.如果二次函数1)3()(2
+-+=x m mx x f 的图象与x 轴的交点至少有一个在原点的右侧,试求实数m 的取值范围.
16.已知a 是实数,函数()a x ax x f --+=3222,如果函数()x f y =在区间[]1,1-上有零点,求a 的取值范围.
17.已知二次函数)0()(2
≠+=a bx ax x f 满足条件:)3()1(x f x f -=-且方程
x x f 2)(=有等根.
(1)求)(x f 的解析式;(2)是否存在实数n m ,)(n m <,使)(x f 定义域和值域分别为],[n m 和]4,4[n m ,如果存在,求出n m ,的值;如果不存在,说明理由.。