安徽省十联考(合肥市第八中学等)2022-2023学年高一上学期11月期中联考英语试题解析版

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安徽省合肥市六校联盟2023-2024学年高一上学期11月期中考试数学试题

安徽省合肥市六校联盟2023-2024学年高一上学期11月期中考试数学试题

安徽省合肥市六校联盟2023-2024学年高一上学期11月期中考试数学试题学校:___________姓名:___________班级:___________考号:___________一、单选题D .()(),41,-∞-+∞U 二、多选题9.下列各组函数中,表示同一个函数的是()A .()()2,f x x g x x ==B .()()0,1f x xg x ==C .()()()222,1x x f x g x xx-==-D .()(),01,0,1,01,0x x x x f x g x x x ⎧⎧≠≥⎪⎪==⎨⎨-<⎪⎪=⎩⎩10.下列说法正确的是()A .命题“R x ∀∈,210x +<”的否定是“R x ∃∈,使得210x +<”B .若集合{}210A x ax x =++=中只有一个元素,则14a =C .关于x 的不等式20ax bx c ++>的解集()2,3-,则不等式20cx bx a -+<的解集为11, 32⎛⎫- ⎪⎝⎭D .若函数()y f x =的定义域是[]2,3-,则函数()21y f x =-的定义域是1,22⎡⎤-⎢⎥⎣⎦11.下列命题中正确的是()A .22144x x +++的最小值为2B .函数2212x xy -⎛⎫= ⎪⎝⎭的值域为(],2-∞C .已知()f x 为定义在R 上的奇函数,且当0x >时,()22f x x x =-,则0x <时,()22f x x x=--D .若幂函数()()211m m m f x x +=+-在()0,∞+上是增函数,则1m =12.若函数()f x 同时满足:①对于定义域上的任意x ,恒有()()0f x f x +-=;②对于定义域上的任意12,x x ,当12x x ≠时,恒()()12120f x f x x x -<-,则称函数()f x 为“理想函数”,下列四个函数中能被称为“理想函数”的是()A .()f x x=-B .()3f x x=-C .()3f x x x=+D .()e ex xf x -=-三、填空题。

2023-2024学年安徽省A10联盟高一(上)期中数学试卷【答案版】

2023-2024学年安徽省A10联盟高一(上)期中数学试卷【答案版】

2023-2024学年安徽省A10联盟高一(上)期中数学试卷一、选择题(本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符含题目要求的.)1.命题“∃x∈[0,1],x3+x2>1”的否定是()A.∀x∈[0,1],x3+x2≤1B.∃x∈[0,1],x3+x2≤1C.∀x∉[0,1],x3+x2≤1D.∃x∉[0,1],x3+x2>12.已知集合M={﹣2,﹣1,0,1},N={x|x2﹣x﹣6<0,x∈N},则M∩N=()A.{1}B.{0,1}C.{﹣1,0,1}D.{﹣2,﹣1,0,1,2}3.设函数f(x)是定义在R上的偶函数,且当﹣1≤x≤0时,f(x)=x﹣1,则f(12)=()A.−12B.12C.−32D.324.二次函数f(x)=ax2+bx+c(a<0)的图象如图所示,记抛物线与x轴交点的横坐标分别为x1,x2,其中﹣2<x1<﹣1,0<x2<1,则下列说法不正确的是()A.c<2B.4a﹣2b+c<0C.4ac﹣b2<0D.a+b>05.将水注入如图所示的玻璃容器中,从空瓶到注满,单位时间内进水量相同,能正确反映该玻璃容器中水面的高度与时间关系的图象是()A.B.C .D .6.对于任意实数 x ,用[x ]表示不大于x 的最大整数,例如:[x ]=3,[2]=2,[﹣1.9]=﹣2,则“x >y ”是“[x ]>[y ]”的( ) A .充分不必要条件 B .必要不充分条件 C .充要条件D .既不充分也不必要条件7.若a >0,b >0且a +b =4,则b a+4b的最小值为( ) A .2B .83C .3D .1038.已知函数f (x )的定义域为R ,且f(x)=x 3f(1x )(x ∈(−∞,0)∪(0,+∞)),f (x )+f (y )﹣2xy =f (x +y ),则f (4)的值是( ) A .﹣20B .﹣16C .﹣12D .﹣10二、选择题(本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得2分,有选错的得0分.)9.若2∈{1,a 2﹣3a +2,a +1},则实数a 的取值可能为( ) A .0B .3C .1D .﹣110.下列命题是真命题的是( ) A .∀x ∈R ,x 2﹣4x +5>0 B .∃x ∈Q ,x 2=5C .∀x ∈R ,x +1x >2D .∃x ∈Z ,3x 2﹣2x ﹣1=011.已知幂函数f (x )的图象经过点(√2,12),则( ) A .f (x )的定义域为全体实数 B .f (x )的值域是(0,+∞)C .f (x )为偶函数D .若x 2>x 1>0,则f(x 1+x 22)<f(x 1)+f(x 2)212.已知奇函数f (x )与偶函数g (x )的定义域均为R ,在区间(a ,b )(a <b )上都是增函数,则( ) A .f (x )在区间(﹣b ,﹣a )上是增函数 B .g (x )在区间(﹣b ,﹣a )上是增函数 C .f(x)g(x)是奇函数,且在区间(a ,b )上是增函数D .f (x )+g (x )不具有奇偶性,且在区间(a ,b )上的单调性不确定 三、填空题(本题共4小题。

安徽省十联考(合肥市第八中学等)2022-2023学年高一上学期11月期中联考物理试题解析版

安徽省十联考(合肥市第八中学等)2022-2023学年高一上学期11月期中联考物理试题解析版

省十联考*合肥八中2022~2023高一上学期期中联考物理试题一、选择题(本题共10小题,共46分。

在每小题给出的四个选项中,第1~7题中只有一项符合题目要求,每小题4分,第8~10题有多项符合题目要求,全部选对的得6分,选对但不全的得3分,有选错的得0分)1.2022年8月23日,詹姆斯·韦伯太空望远镜捕捉到了令人惊叹的木星新图像,这将为科学家提供更多关于该行星内部生命的线索。

木星是太阳系八大行星中体积最大、自转最快的行星,木星的公转周期约为11.86年,自转周期约为9小时50分30秒。

下列说法正确的是()A.“9小时50分30秒”指的是时刻B.研究木星的自转时,能将木星视为质点C.研究木星的公转时,不能将木星视为质点D.比较木星和地球的速度大小时,应选择太阳为参考系2.如图所示为一静止在水平地面上的石狮子。

下列说法正确的是()A.石狮子受到地面的支持力是由于地面发生形变而产生的B.石狮子受到的重力和地面对石狮子的支持力是一对相互作用力C.石狮子受到的重力,只有受力物体,没有施力物体D.把石狮子放在地球表面上的任何地方,其所受的重力大小总是相等的3.如图甲所示,飞行背包是一种可以让使用者通过穿戴该产品后飞上天空的飞行器。

某次飞行测试中,测试员携带背包从地面起飞,其位移随时间变化关系如图乙所示,规定竖直向上为正方向。

下列说法正确的是()A.测试员在0~10s内加速上升B.测试员在20~30s内匀速下降C.在第30s末,测试员离地面最远D.测试员在0~20s内的平均速度大小为2m/s4.如图所示,飞机从静止起飞过程中,在2min内速度达180km/h;玩具车从静止启动过程中,在2s内速度达10m/s。

对于上述两个过程,下列说法正确的是()A.“180km/h”、“10m/s”均指平均速度大小B.飞机速度的变化量比玩具车速度的变化量小C.飞机速度的变化率比玩具车速度的变化率小D.玩具车在启动瞬间,速度和加速度均为零5.一名建筑工人在距地面高45m的楼层工作时,不小心将手中的砖头脱落,砖头随即做自10m/s,则砖头在落地前一秒内下落的高度为()由落体运动。

安徽省A10联盟2022-2023学年高三上学期11月阶段测试物理试题含解析

安徽省A10联盟2022-2023学年高三上学期11月阶段测试物理试题含解析

·A10联盟2023届高三上学期11月段考物理试题(答案在最后)巢湖一中合肥八中淮南二中六安一中南陵中学舒城中学太湖中学天长中学屯溪一中 宣城中学滁州中学池州一中阜阳一中灵璧中学宿城一中合肥六中太和中学合肥七中 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

满分100分,考试时间90分钟。

请在答题卡上作答。

第Ⅰ卷(选择题共40分)一、选择题:本题共10小题,每小题4分,共40分。

在每小题给出的四个选项中,第1~6题只有一项符合题目要求,第7~10题有多项符合题目要求。

全部选对的得4分,选对但不全的得2分,有选错的得0分。

1.如图所示,在直角坐标系xOy 中,某一过原点O 的直线与x 轴正方向的夹角为α。

某时刻,一个质点从坐标原点开始运动,沿x 轴正方向的分运动是初速度为零,加速度为22m /s 的匀加速直线运动,沿y 轴正方向的分运动是速度为3m/s 的匀速直线运动。

已知sin 0.6α=,则质点经过该直线时的坐标为(O 点除外)( )A.()4m,3mB.()8m,6mC.()12m,9mD.()16m,12m2.如图所示,粗细均匀的直杆倾斜固定,杆与水平面间的夹角30α=°,质量为m 的小球A 套在杆上,绕过光滑的轻质滑轮的细线一端连接在小球A 上,另一端系在杆上,滑轮下吊着质量为m 的小球B ,系在杆上的细线与杆间的夹角30θ=°,两个小球均处于静止状态,重力加速度大小为g ,则小球A 与杆间的摩擦力大小为( )A.12mg B.mgC.32mg D.2mg3.小球A 自离地面高为h 的P 点由静止释放,同时小球B 在地面竖直向上抛出,两球相遇时(到达同一高度时),小球B 向上速度大小是小球A 速度大小的2倍,重力加速度大小为g ,不计空气阻力,则小球B 抛出的初速度大小为( )D.4.如图所示,竖直细杆O 点处固定有一水平横杆,在横杆上有A 、B 两点,且OA OB =,在A 、B 两点分别用两根等长的轻质细线悬挂两个相同的小球a 和b ,将整个装置绕竖直杆匀速转动,则a 、b 两球稳定时的位置关系可能正确的是( )A. B. C. D.5.如图所示,质量为2kg 物块在光滑水平面上向右做速度为6m/s 的匀速运动,现给物块施加一个水平向左的拉力,拉力的大小F 与物块速度的大小v 成正比,即F kv =,其中3kg /s k =。

2022-2023学年安徽省合肥市六校联盟高一上学期期中联考数学试题

2022-2023学年安徽省合肥市六校联盟高一上学期期中联考数学试题

2022-2023学年安徽省合肥市六校联盟高一上学期期中联考数学试题1. 已知集合P ={x ∣y =√x +1},集合Q ={y ∣y =√x +1},则P 与Q 的关系是( )A . P =QB . P ⊆QC . Q ⊆PD . P ∩Q =∅2. 已知集合A ={x|x <0或x >2},B =N ,则集合(C R A)∩B 中元素的个数为( )A . 2B . 3C . 4D . 53. 如果a >0>b ,且a +b >0,那么以下不等式正确的个数是( )①a 2>b 2;②1a <1b ;③a 3>ab 2;④a 2b <b 3. A . 1B . 2C . 3D . 44. 若不等式x 2+kx +1<0的解集为空集,则k 的取值范围是( )A . −2≤k ≤2B . k ≤−2 ,或 k ≥2C . −2<k <2D . k <−2 ,或 k >25. 设a ∈R ,则“a >1”是“a 2>a ”的( )A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件6. 函数f(x)=x +4x+1在[−12,2]上的值域为( )A . [−3,152]B . [3,4]C . [3,152]D . [4,152]7. 已知定义在R 上的奇函数f(x)满足:当x ≥0时,f(x)=x 2+x ,若不等式f(2t)>f(m +mt 2)对任意实数t 恒成立,则实数m 的取值范围是( )A . (−∞,−1)∪(1,+∞)B . (−1,0)∪(0,1)C . (−∞,−1)D . (−1,0)8. 已知函数f(x)=x 2+ax +b(a,b ∈R)的值域为[0,+∞),若关于x 的不等式f(x)<c 的解集为(m,m +6),则实数c 的值为( )A .2B .3C .6D .99. 下列说法正确的是( )A . a >b 的一个必要不充分条件是 a −1>bB .若集合 A ={x|ax 2+x +1=0} 中只有一个元素,则 a =14C .若命题“ ∃x ∈R , x 2−ax +1≤0 ”是假命题,则实数 a 的取值范围是D .已知集合 M ={0,1} ,则满足条件 M ∪N =M 的集合 N 的个数为 4 10. 下列选项中正确的是( )A .不等式 a +b ≥2√ab 恒成立B .存在实数 a ,使得不等式 a +1a ≤2 成立 C .若 a ,b 为正实数,则 ba+ab ≥2D .若正实数 x , y 满足 x +2y =1 ,则 2x +1y ≥8 11. 下列命题中,正确的有( )A .函数 y =√x +1·√x −1 与函数 y =√x 2−1 表示同一函数B .已知函数 f(2x +1)=4x −6 ,若 f(a)=10 ,则 a =9C .若函数 f(√x −1)=x −3√x ,则 f(x)=x 2−x −2(x ≥−1)D .若函数 f(x) 的定义域为 [0,2] ,则函数 f(2x) 的定义域为 [0,4]12. 已知定义域为R 的函数f(x)在区间(4,+∞)上为减函数,且函数y =f(x +4)为偶函数,则以下错误的有( )A . f(2)>f(3)B . f(2)>f(5)C . f(3)>f(6)D . f(3)>f(5)13. 幂函数f(x)=(m 2−3m +3)x 3m−4在(0,+∞)上为减函数,则m 的值为______ ;14. 已知函数f(x)={2x ,x >0x 2+1,x ≤0,则不等式f(x)<2的解集是______.15. 计算:823−(−78)0+√(3−π)44+[(−2)6]12.__________.16. 若正实数x ,y 满足2x +y +6=xy ,则2x +y 的最小值是_______.17. 已知集合A ={x|1≤3x ≤27},B =(1,+∞).(1)求A ∪(C R B);(2)若C ={x|a −4≤x ≤a},且A ∩C =A ,求实数a 的取值范围.18. 设p :实数x 满足x 2−4ax +3a 2<0,其中a >0;q :实数x 满足2<x ≤3.若¬p 是¬q 的充分不必要条件,求正实数a 的取值范围.19. 已知函数f(x)=1−m5x +1是奇函数.(1)求实数m的值;(2)判断并证明函数f(x)在R上的单调性,并求出f(x)在区间[−2,3]上的最小值.20.已知函数f(x)=(13)ax2−4x+3.(1)若a=−1,求f(x)的单调区间(2)若f(x)有最大值3,求a的值(3)若f(x)的值域是(0,+∞),求a的值21.已知函数f(x)=ax2−|x|+2a−1(a>0).(1)请在如图所示的直角坐标系中作出a=12时f(x)的图像,并根据图像写出函数的单调区间;(2)设函数f(x)在x∈[1,2]上的最小值为g(a).①求g(a)的表达式;②若a∈[14,12],求g(a)的最大值.22. 习近平总书记一直十分重视生态环境保护,十八大以来多次对生态文明建设作出重要指示,在不同场合反复强调“绿水青山就是金山银山”,随着中国经济的快速发展,环保问题已经成为一个不容忽视的问题.某污水处理厂在国家环保部门的支持下,引进新设备,新上了一个从生活垃圾中提炼化工原料的项目.经测算,该项目月处理成本y (元)与月处理量x (吨)之间的函数关系可以近似地表示为y ={13x 3−80x 2+5040x,x ∈[120,144),12x 2−200x +80000,x ∈[144,500),且每处理一吨生活垃圾,可得到能利用的化工原料的价值为200元,若该项目不获利,政府将给予补贴.(1)当x ∈[200,300]时,判断该项目能否获利,如果获利,求出最大利润;如果不获利,则政府每月至少需要补贴多少元才能使该项目不亏损?(2)该项目每月处理量为多少吨时,才能使每吨的平均处理成本最低?。

安徽省合肥市2022-2023学年高一上学期期中联考数学试题含解析

安徽省合肥市2022-2023学年高一上学期期中联考数学试题含解析

合肥六校联盟2022-2023学年第一学期期中联考高一年级数学试卷(答案在最后)(考试时间:120分钟满分:150分)一、选择题(本大题共8小题,每小题5分,共40分,每小题只有一个正确答案)1.已知集合{P x y ==∣,集合{Q yy ==∣,则P 与Q 的关系是()A.P Q =B.P Q ⊆C.Q P ⊆D.P Q =∅【答案】C 【解析】【分析】化简集合,P Q ,根据集合的相等、交集、子集判断即可.【详解】因为{[1,)P xy ===-+∞∣,{[0,)Q y y ===+∞∣,所以P Q =,P Q ⊆,P Q =∅ 错误,Q P ⊆正确.故选:C2.已知集合{|0A x x =<或2}x >,B =N ,则集合()A B R ð中元素的个数为()A.2B.3C.4D.5【答案】B 【解析】【分析】根据题意,结合补集、交集运算,即可求解.【详解】根据题意,可知{}R 02A x x =≤≤ð,由B =N ,得(){}R 0,1,2A B = ð,集合中有3个元素.故选:B.3.如果0a b >>,且0a b +>,那么以下不等式正确的个数是()①22a b >;②11a b<;③32a ab >;④23a b b <.A.1B.2C.3D.4【答案】C 【解析】【分析】根据不等式的性质分别进行判断即可.【详解】由0a b >>知,0b <.又0a b +>,∴0a b >->,∴()22a b >-,即22a b >.又0a >,∴32a ab >,0b < ,∴23a b b <,故①正确,③正确,④也正确,又10a >,10b<,故②错误.故选:C .4.若不等式210x kx ++<的解集为空集,则k 的取值范围是()A.22k -≤≤B.2k ≤-,或2k ≥C.2<<2k -D.2k <-,或2k >【答案】A 【解析】【分析】根据题意可得240k ∆=-≤,从而即可求出k 的取值范围.【详解】∵不等式210x kx ++<的解集为空集,∴240k ∆=-≤,∴22k -≤≤.故选:A.5.设a ∈R ,则“1a >”是“2a a >”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【答案】A 【解析】【分析】首先求解二次不等式,然后结合不等式的解集即可确定充分性和必要性是否成立即可.【详解】求解二次不等式2a a >可得:1a >或a<0,据此可知:1a >是2a a >的充分不必要条件.故选:A.【点睛】本题主要考查二次不等式的解法,充分性和必要性的判定,属于基础题.6.函数()41f x x x =++在1,22⎡⎤-⎢⎥⎣⎦上的值域为()A.153,2⎡⎤-⎢⎥⎣⎦ B.[]3,4 C.153,2⎡⎤⎢⎥⎣⎦D.154,2⎡⎤⎢⎥⎣⎦【答案】C 【解析】【分析】设1x t +=,1,32t ⎡⎤∈⎢⎥⎣⎦,则()41g t t t =+-,得到函数的单调区间,计算函数值得到值域.【详解】设1x t +=,1x t =-,1,22x ⎡⎤∈-⎢⎥⎣⎦,则1,32t ⎡⎤∈⎢⎥⎣⎦,则()41g t t t =+-,根据双勾函数性质:函数在1,22⎡⎤⎢⎥⎣⎦上单调递减,在(]2,3上单调递增,()()max 1151015max ,3max ,2232g t g g ⎧⎫⎛⎫⎧⎫===⎨⎬⎨⎬ ⎪⎝⎭⎩⎭⎩⎭,()()min 23g t g ==,故函数值域为153,2⎡⎤⎢⎥⎣⎦.故选:C.7.已知定义在R 上的奇函数()f x 满足:当0x ≥时,()2f x x x =+,若不等式()()22f t f m mt >+对任意实数t 恒成立,则实数m 的取值范围是()A.()(),11,-∞-+∞U B.()()1,00,1-U C.(),1-∞- D.()1,0-【答案】C 【解析】【分析】首先根据函数奇偶性求出函数在0x <时的解析式,即可得到()f x 在定义域上的解析式,画出函数图象,即可得到函数在定义域上单调递增,则原不等式等价于22t m mt >+对任意实数t 恒成立,对m 分两种情况讨论,当0m ≠时,则00m <⎧⎨∆<⎩,即可得到不等式组,解得即可;【详解】解:因为定义在R 上的奇函数()f x 满足:当0x ≥时,()2f x x x =+,设0x <,则0x ->,所以()()2f x f x x x =--=-+,即()22,0,0x x x f x x x x ⎧+≥=⎨-+<⎩,函数图象如下所示:由函数图象可知函数()f x 在定义域R 上单调递增,若不等式()()22f t f m mt >+对任意实数t 恒成立,即22t m mt >+对任意实数t 恒成立,即220mt t m -+<恒成立,当0m =时,20t -<不恒成立,当0m ≠时,则()22240m m <⎧⎪⎨∆=--<⎪⎩,解得1m <-;综上可得(),1m ∈-∞-故选:C8.已知函数()()2,f x x ax b a b =++∈R 的值域为[)0,∞+,若关于x 的不等式()f x c <的解集为(),6m m +,则实数c 的值为()A.2B.3C.6D.9【答案】D 【解析】【分析】利用一元二次不等式的解法求解即可.【详解】因为函数()()2,f x x ax b a b =++∈R 的值域为[)0,∞+,所以函数()()2,f x x ax b a b =++∈R 的的图象与x 轴有且仅有一个交点,所以240a b ∆=-=,即24a b =,不等式()f x c <的解集为(),6m m +,即2204a x ax c ++-<的解集为(),6m m +,所以方程2204a c x ax ++-=的两个根为12,6x m x m ==+,所以1221226(6)4x x m a a x x m m c +=+=-⎧⎪⎨=+=-⎪⎩,消去m 可得,22944a a c -=-,解得9c =.故选:D.二、多选题(本大题共4小题,每小题5分,共20.0分,在每小题给出的选项中,有多项符合题目要求,全部选对的得5分,有选错的得0分,部分选对的得3分)9.下列说法正确的是()A.a b >的一个必要不充分条件是1a b->B .若集合2{|10}A x ax x =++=中只有一个元素,则14a =C.若命题“x ∃∈R ,210x ax -+≤”是假命题,则实数a 的取值范围是{|22}a a -<<D.已知集合{}0,1M =,则满足条件M N M ⋃=的集合N 的个数为4【答案】CD 【解析】【分析】对于A :直接利用不等式的性质,结合充分条件和必要条件分析判断;对于B :分0a =和0a ≠两种情况,结合二次方程分析判断;对于C :根据命题真假的判定,结合恒成立问题分析求解;对于D :由M N M ⋃=可得N M ⊆,结合包含关系分析求解.【详解】对于选项A :当1a b ->,即1a b >+时,则a b >,即充分性成立;当a b >时,例如1,0a b ==,则10a b -==,即必要性不成立;所以a b >的充分不必要条件为1a b ->,故A 错误;对于选项B :若集合2{|10}A x ax x =++=中只有一个元素,当0a =时,集合{}{}|101+===-A x x ,只有一个元素,符合题意;当0a ≠时,由140a ∆=-=,解得14a =;综上所述:0a =或14a =,故B 错误;对于选项C :命题“x ∃∈R ,210x ax -+≤”是假命题,则命题“x ∀∈R ,210x ax -+>”是真命题,所以240a ∆=-<,解得22a -<<,故C 正确;对于选项D :因为M N M ⋃=,则N M ⊆,且集合{}0,1M =,则满足条件的集合N 为:{}0或{}1或{}0,1或∅,故集合N 的个数为4,故D 正确.故选:CD .10.下列选项中正确的是()A.不等式a b +≥恒成立B.存在实数a ,使得不等式12a a+≤成立C.若a ,b 为正实数,则2b a a b+≥D.若正实数x ,y 满足21x y +=,则218x y+≥【答案】BCD 【解析】【分析】根据基本不等式的条件与“1”的用法等依次讨论各选项即可得答案.【详解】解:对于A 选项,当0,0a b <<时不成立,故错误;对于B 选项,当a<0时,()112a a a a ⎡⎤⎛⎫+=--+-≤ ⎪⎢⎥⎝⎭⎣⎦,当且仅当1a =-等号成立,故正确;对于C 选项,若a ,b 为正实数,则0,0b a a b >>,所以2b a a b +≥,当且仅当a b =时等号成立,故正确;对于D 选项,由基本不等式“1”的用法得()212142448y x x y x y x y x y ⎛⎫+=++=++≥+= ⎪⎝⎭,当且仅当2x y =时等号成立,故正确.故选:BCD11.下列命题中,正确的有()A.函数y =与函数y =B.已知函数(21)46f x x +=-,若()10f a =,则9a =C.若函数1)-=-f x ,则()()221f x x x x =--- D.若函数()f x 的定义域为[]0,2,则函数()2f x 的定义域为[]0,4【答案】BC 【解析】【分析】A.两函数的定义域不同,故不是同一函数,所以A 错误;解方程组21446109x a x x a +==⎧⎧⇒⎨⎨-==⎩⎩,故B 正确;求出()()221f x x x x =--- ,故C 正确;函数()2f x 的定义域为[]0,1,故D 错误.【详解】解:()f x =的定义域是10{|}{|1}10x x x x x +⎧=⎨-⎩,()g x =的定义域是2{|10}x x - {|1x x = 或1}x - ,两函数的定义域不同,故不是同一函数,所以A 错误;函数(21)46f x x +=-,若()10f a =,则214610x a x +=⎧⎨-=⎩,所以49x a =⎧⎨=⎩,故B 正确;若函数)211)1)2fx -=-=--,则()()221f x x x x =--- ,故C 正确;若函数()f x 的定义域为[]0,2,则函数()2f x 中,022x ≤≤,所以01x ≤≤,即函数()2f x 的定义域为[]0,1,故D 错误.故选:BC12.已知定义域为R 的函数()f x 在区间()4,+∞上为减函数,且函数()4y f x =+为偶函数,则以下错误的有()A.()()23f f >B.()()25f f >C.()()36f f >D.()()35f f >【答案】ABD 【解析】【分析】根据函数的单调性和奇偶性求解.【详解】因为函数()4y f x =+为偶函数,所以()4(4)f x f x +=-+,所以函数()f x 关于直线4x =对称,又因为函数()f x 在区间()4,+∞上为减函数,所以函数()f x 在区间(),4-∞上为增函数,因为234<<,所以()()23f f <,A 错误;因为()5(3)f f =,()()23f f <,所以()()25f f <,B 错误;因为()6(2)f f =,()()23f f <,所以()()36f f >,C 正确;()()35f f =,D 错误;故选:ABD.三、填空题(本大题共4小题,每小题5分,共20分)13.幂函数()234()33m f x m m x -=-+在()0+∞,上为减函数,则m 的值为______;【答案】1【解析】【分析】由题意可得m 2﹣3m +3=1,求得m 值,再满足3m ﹣4<0即可.【详解】∵函数f (x )=(m 2﹣3m +3)x 3m ﹣4是幂函数,∴m 2﹣3m +3=1,即m 2﹣3m +2=0,解得m =1或m =2.又幂函数f (x )=(m 2﹣3m +3)x 3m ﹣4在(0,+∞)上为减函数,∴3m ﹣4<0,即m 43<,故m =1.故答案为:1.【点睛】本题考查幂函数的性质,明确m 2﹣3m +3=1是关键,是基础题.14.已知函数22,0()1,0x x f x x x ⎧>=⎨+≤⎩,则不等式()2f x <的解集是______.【答案】()1,1-【解析】【分析】根据分段函数的解析式,分0x >,0x ≤求出对应的解集,再求并集即可得到不等式的解集.【详解】由题意,()22,01,0x x f x x x ⎧>=⎨+≤⎩,不等式()2f x <,当0x >时,由22x <,解得1x <,所以01x <<;当0x ≤时,由212x +<,解得11x -<<,所以10-<≤x ;综上,不等式()2f x <的解集为()1,1-.故答案为:()1,1-.【点睛】本题考查一元二次不等式的解法,分段函数,以及指数函数的性质的应用,考查分类讨论思想.15.计算:02163278(2)8⎛⎫⎡⎤--+-= ⎪⎣⎦⎝⎭__________.【答案】π8+##8π+【解析】【分析】由分数指数幂和根式互化、幂的乘方计算即可求解.【详解】由题意()()122633233782213π28⎛⎫⎡⎤---=-+-+ ⎪⎣⎦⎝⎭()221π38π8=-+-+=+.故答案为:π8+.16.若正实数x ,y 满足26x y xy ++=,则2x y +的最小值是_______.【答案】12【解析】【分析】首先右边边是xy 的形式,左边是2x y +和常数的和的形式,考虑把左边也转化成似xy 的形式,使形式统一.可以猜想到应用基本不等式a b +≥转化后变成关于xy 的方程,可把xy 看成整体换元后求最小值.【详解】解析:解法一:0x >,0y >,()21122222x y xy x y +⎛⎫∴=⋅⋅≤⋅ ⎪⎝⎭,()()21262628x y x y xy x y ∴++=++=≤+,()()2282480x y x y ∴+-+-≥,令2x y t +=,则0t >,且28480t t --≥,()()1240t t ∴-+≥,12t ∴≥,即212x y +≥,当且仅当2x y =,即3x =,6y =时取等号.2x y ∴+的最小值是12.解法二:由0x >,0y >,26x y xy ++=得6xy ≥,(当且仅当2x y =时取等号),即260-≥,0∴≥,又0>,≥即18xy ≥(当且仅当3x =,6y =时取等号),∴xy 的最小值是18,26x y xy +=-,∴2x y +的最小值是12.故答案为12.【点睛】本题主要考查了用基本不等式a b +≥解决最值问题的能力,以及换元思想和简单一元二次不等式的解法,属基础题.四、解答题(本大题共6小题,共70.0分,第17题10分,其余5题分别12分,解答应写出文字说明,证明过程或演算步骤)17.已知集合{}1327xA x =≤≤,()1,B =+∞.(1)求()R A B ⋃ð;(2)若{}4C x a x a =-≤≤,且A C A = ,求实数a 的取值范围.【答案】(1)()(],3R A B ⋃=-∞ð;(2)34a ≤≤.【解析】【分析】(1)首先解指数不等式得到集合A ,再根据补集、并集的定义计算可得;(2)由A C A = ,得A C ⊆,则403a a -⎧⎨⎩ ,由此能求出实数a 的取值范围.【详解】解:(1)由1327x ≤≤,得03333x ≤≤,所以03x ≤≤,所以[]0,3A =,由()1,B =+∞,得(],1R B =-∞ð,所以()(],3R A B ⋃=-∞ð(2)由A C A = ,得A C ⊆,所以403a a -≤⎧⎨≥⎩,解得43a a ≤⎧⎨≥⎩,所以34a ≤≤.18.设p :实数x 满足22430x ax a -+<,其中0a >;q :实数x 满足23x <≤.若p ⌝是q ⌝的充分不必要条件,求正实数a 的取值范围.【答案】(]1,2【解析】【分析】利用一元二次不等式的解法和充分必要条件的概念求解.【详解】由22430x ax a -+<,可得()(3)0x a x a --<,因为0a >,所以3a a >,所以不等式的解为3a x a <<,即p :3a x a <<,则p ⌝:x a ≤或3x a ≥,q ⌝:2x ≤或3x >,因为p ⌝是q ⌝的充分不必要条件,所以233a a ≤⎧⎨>⎩,解得12a <≤,经检验,2a =时,满足题意,所以正实数a 的取值范围是(]1,2.19.已知函数()151x m f x =-+是奇函数.(1)求实数m 的值;(2)判断并证明函数()f x 在R 上的单调性,并求出()f x 在区间[]2,3-上的最小值.【答案】(1)2m =(2)1213-【解析】【分析】(1)利用奇函数的定义求解;(2)利用函数单调性的定义证明,并根据单调性与最值的关系求最小值.【小问1详解】因为函数()151x m f x =-+的定义域为R ,且为奇函数,所以()01011m f =-=+,解得2m =,所以()2151x f x =-+,经检验,2m =时,()225115151xx x f x -⋅-=-=-++,所以()2522(51)()1120515151x x x x x f x f x ⋅+-+=-+-=-=+++,即()()f x f x -=-,满足函数()f x 为奇函数,所以2m =.【小问2详解】判断:函数()f x 在R 上单调递增,证明如下:任意1212,,x x x x ∈<R ,()()()()()121212*********(51)2(51)2(55)()11515151515151x x x x x x x x x x f x f x +-+--=--+==++++++,因为1212,,x x x x ∈<R ,所以1212510,510,55x x x x +>+><,所以()()12122(55)05151x x x x -<++,即()12()f x f x <,所以函数()f x 在R 上单调递增,所以min 2212()(2)15113f x f -=-=-=-+.20.已知函数()24313ax x f x -+⎛⎫⎪⎝⎭=.(1)若1a =-,求()f x 的单调区间(2)若()f x 有最大值3,求a 的值(3)若()f x 的值域是()0,∞+,求a 的值【答案】(1)函数()f x 的单调递增区间是()2,-+∞,单调递减区间是(),2-∞-;(2)1;(3)0.【解析】【分析】(1)根据复合函数单调性判断,结合指数函数、二次函数性质判断()f x 单调区间;(2)由(1)及题设知02(1a g a>⎧⎪⎨=-⎪⎩,即可求参数值;(3)根据复合函数的值域,结合指数函数、二次函数性质确定参数值即可.【小问1详解】当1a =-时,()24313x x f x --+⎛⎫= ⎪⎝⎭,令()243g x x x =--+,由()g x 在(),2-∞-上单调递增,在()2,-+∞上单调递减,而13xy ⎛⎫= ⎪⎝⎭在R 上单调递减,所以()f x 在(),2-∞-上单调递减,在()2,-+∞上单调递增,即()f x 的单调递增区间是()2,-+∞,单调递减区间是(),2-∞-.【小问2详解】令()243g x ax x =-+,()()13g x f x ⎛⎫= ⎪⎝⎭,由于()f x 有最大值3,所以()g x 应有最小值1-,因此必有0234()1a a g aa >⎧⎪-⎨==-⎪⎩.解得1a =,即()f x 有最大值3时,a 为1.【小问3详解】由指数函数的性质知,要使()13g x y ⎛⎫= ⎪⎝⎭的值域为()0,∞+,应使()243g x ax x =-+的值域为R ,因此只能0a =(因为若0a ≠,则()g x 为二次函数,其值域不可能为R ),故a 的值为0.21.已知函数()()2||210f x ax x a a =-+->.(1)请在如图所示的直角坐标系中作出12a =时()f x 的图像,并根据图像写出函数的单调区间;(2)设函数()f x 在[]1,2x ∈上的最小值为()g a .①求()g a 的表达式;②若11,42a ⎡⎤∈⎢⎥⎣⎦,求()g a 的最大值.【答案】(1)图象见解析,增区间()()1,0,1,-+∞,减区间()(),1,0,1-∞-;(2)①()132,211121,442163,04a a g a a a a a a ⎧->⎪⎪⎪=--≤≤⎨⎪⎪-<<⎪⎩;②12-.【解析】【分析】(1)12a =时,()21||2f x x x =-,画出函数图象,根据图象即可得出单调区间;(2)①[]1,2x ∈时,()()2210f x ax x a a =-+->,讨论对称轴的范围,根据二次函数的单调性求解;②11,42a ⎡⎤∈⎢⎥⎣⎦时,()1214g a a a=--,根据单调性即可求出.【详解】(1)12a =时,()21||2f x x x =-,函数图象如图:增区间()()1,0,1,-+∞;减区间()(),1,0,1-∞-.(2)①因为[]1,2x ∈,所以()()2210f x ax x a a =-+->.若112a <,即12a >时,()f x 在[]1,2上单调递增,所以()()min 132f x f a ==-;若1122a≤≤,即1142a ≤≤时,()f x 在11,2a ⎡⎤⎢⎥⎣⎦上递减,在1,22a ⎡⎤⎢⎥⎣⎦上递增,所以()min 112124f x f a a a ⎛⎫=-⎪⎝⎭=-;若122a>,即10a 4<<时,()f x 在[]1,2上单调递减,所以()()min 263f f x a ==-,综上()132,211121,442163,04a a g a a a a a a ⎧->⎪⎪⎪=--≤≤⎨⎪⎪-<<⎪⎩;②11,42a ⎡⎤∈⎢⎥⎣⎦时,()1214g a a a=--,因为12,4y a y a ==-在11,42a ⎡⎤∈⎢⎥⎣⎦单调递增,所以()1214g a a a =--在11,42a ⎡⎤∈⎢⎥⎣⎦单调递增,所以()g a 的最大值为1122g ⎛⎫=-⎪⎝⎭.【点睛】关键点睛:本题考查含参二次函数最值的求解以及函数最值问题,解题的关键是讨论二次函数对称轴的位置,再结合二次函数的单调性求解,对于函数最值问题,解题的关键是求出函数的单调性,利用单调性求出最值.22.习近平总书记一直十分重视生态环境保护,十八大以来多次对生态文明建设作出重要指示,在不同场合反复强调“绿水青山就是金山银山”,随着中国经济的快速发展,环保问题已经成为一个不容忽视的问题.某污水处理厂在国家环保部门的支持下,引进新设备,新上了一个从生活垃圾中提炼化工原料的项目.经测算,该项目月处理成本y (元)与月处理量x (吨)之间的函数关系可以近似地表示为[)[)3221805040,120,144,3120080000,144,500,2x x x x y x x x ⎧-+∈⎪⎪=⎨⎪-+∈⎪⎩且每处理一吨生活垃圾,可得到能利用的化工原料的价值为200元,若该项目不获利,政府将给予补贴.(1)当[200,300]x ∈时,判断该项目能否获利,如果获利,求出最大利润;如果不获利,则政府每月至少需要补贴多少元才能使该项目不亏损?(2)该项目每月处理量为多少吨时,才能使每吨的平均处理成本最低?【答案】(1)不会,政府每月至少需要补贴5000元才能使该项目不亏损(2)400吨【解析】【分析】(1)当[200,300]x ∈时,由项目获利为221120020080000(400)22S x x x x ⎛⎫=--+=-- ⎪⎝⎭求解;(2)由生活垃圾每吨的平均处理成本[)[)21805040120,1443180000200144,5002x x x y x x x x ⎧-+∈⎪⎪=⎨⎪-+∈⎪⎩求解.【小问1详解】解:当[200,300]x ∈时,该项目获利为S ,则221120020080000(400)22S x x x x ⎛⎫=--+=-- ⎪⎝⎭,∴当[200,300]x ∈时,0S <,因此,该项目不会获利,当300x =时,S 取得最大值5000-,所以政府每月至少需要补贴5000元才能使该项目不亏损;【小问2详解】由题意可知,生活垃圾每吨的平均处理成本为:[)[)21805040120,1443180000200144,5002x x x y x x x x ⎧-+∈⎪⎪=⎨⎪-+∈⎪⎩,当[120,144)x ∈时,2211805040(120)24033y x x x x =-+=-+,所以当120x =时,y x 取得最小值240;当[144,500)x ∈时,1800002002y x x x =-+200400200200≥-=-=,当且仅当800002x x =,即400x =时,y x 取得最小值200,因为240200>,所以当每月处理量为400吨时,才能使每吨的平均处理成本最低.。

安徽省十联考(合肥市第八中学等)2022-2023学年高一上学期11月期中联考地理试题解析版

安徽省十联考(合肥市第八中学等)2022-2023学年高一上学期11月期中联考地理试题解析版

省十联考*合肥八中2022~2023高一上学期期中联考地理试题一、选择题:本大题共15小题,每小题3分,共45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。

英仙座流星雨是北半球三大流星雨之一,每年固定时间稳定出现,是最活跃、最常被观测到的流星雨。

英仙座流星雨这一天象与彗星有关,当彗星接近太阳时,表面物质升华形成流星体,当地球的运行轨道与彗星轨道相交时,这些流星体受地球引力的影响闯入地球大气层并且燃烧,形成流星雨。

下图为太阳系结构示意图。

据此完成下面小题。

1.形成英仙座流星雨的流星体归属的最低级别的天体系统是()A.河外星系B.太阳系C.银河系D.可观测宇宙2.与彗星相比,地球()A.宇宙环境较安全B.体积较大C.主要由气体构成D.距太阳较近在日全食时,太阳的周围镶着一个红色的环圈,上面跳动着鲜红的火舌,这种火舌状物体叫做日珥,日珥是主要的太阳活动现象之一。

图为日珥景观图。

据此完成下面小题。

3.日珥通常出现在()A.日冕层B.光球层C.色球层D.太阳内部4.日珥爆发时()A.太阳活动明显减弱B.每次都能被肉眼观测到C.会破坏太阳风流动D.会喷射大量带电粒子琥珀是松柏科、南洋杉科等植物的树脂化石。

2018年7月19日,来自中国、加拿大、美国和澳大利亚的科学家团队宣布,他们首次在琥珀中发现蛇类标本,一条0.99亿年前的小蛇,定名缅甸晓蛇。

下图为缅甸晓蛇琥珀景观图。

据此完成下面小题。

5.缅甸晓蛇生存时期()A.是重要的成矿时期B.蕨类植物高度繁盛C.爬行动物恐龙繁盛D.哺乳动物快速发展6.推测缅甸晓蛇琥珀形成时的地理环境最可能是()A.炎热干燥的陆地B.寒冷的高山湖泊C.广阔的热带浅海D.温湿茂密的森林下图为大气的垂直分层示意图。

据此完成下面小题。

7.大气垂直分层中容易发生云、雨、雾、雪等天气现象的是()A.①B.②C.③D.④8.某航天器穿越②层时,可能遇到的现象是()A.夜晚极光灿烂美丽B.多云、雷雨等天气频发C.大气对流运动旺盛D.22~27km形成臭氧层9.某航天器从②层顶部降落到地面,经历的气温变化是()A.逐渐降低B.先降低,再升高C.逐渐升高D.先升高,再降低读某地区海平面等压线分布示意图(单位:百帕),完成下面小题。

安徽省合肥市八校联考2022-2023学年高一上学期集中练习英语试卷(含答案)

安徽省合肥市八校联考2022-2023学年高一上学期集中练习英语试卷(含答案)

2022—2023学年度第一学期集中练习2高一英语试卷(试卷满分150分,考试时间120分钟)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。

1. What is the woman doing?A. Asking the way.B. Telling the way.C. Getting in the way.2. What does the man show great concern for?A. His studies.B. World affairs.C. His health.3. What does the man suggest the woman do?A. Have independent thought.B. Show respect for her teacher.C. Reach an agreement with her teacher.4. What does the woman want to do?A. Hire more staff.B. Interview the guard’s family.C. Know more about the guard.5. What does the man mean?A. He agrees with the woman.B. The woman should be polite.C. The woman’s concern is useless.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。

每段对话或独白后有几个小题,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。

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省十联考*合肥八中2022~2023高一上学期期中联考英语试题考生注意:1.本试卷分选择题和非选择题两部分。

满分150分,考试时间120分钟。

2.答题前,考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目写清楚。

3.考生作答时,请将答案答在答题卡上。

选择题每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题卡区域书写的答案无效,在试题卷、草稿纸上作答无效。

.............................第一部分听力(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。

听完每段对话后,你都有10秒钟的诗句来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Where does the conversation probably take place?A. In a shop.B. At home.C. In a bank.2. What will the woman do after arriving home?A. Draw a map of Italy.B. Make pizza with the man.C. Drive the man to school.3. Where the man go first after the class?A. His home.B. The woman’s house.C. The library.4. What is Tony now?A. A teacher.B. A reporter.C. A writer.5. When will the flight leave?A. At 4:30 pm.B. At 7:00 pm.C. At 8:30 pm.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。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6、7题。

6. Why does the man hold a party for Sheila?A. For her birthday.B. For her graduation.C. For her marriage.7. Where does Sheila live now?A. In Houston.B. In New York.C. In Dallas.听第7段材料,回答第8、9题。

8. When did Mrs. Philips’ pains start?A. On Tuesday.B. On Wednesday.C. On Thursday.9. What does the man suggest Mrs. Philips do next Wednesday?A. Stay at home.B. Buy the medicine.C. Visit him again.听第8段材料,回答第10至12题。

10. How does the man feel at first?A. Impatient.B. Confident.C. Unsure.11. What topic does the woman advise the man to talk about?A. How to choose materials.B. How to draw pictures.C. How to take photos.12. What is the probable relationship between the speakers?A. Friends.B. Teacher and student.C. Shop assistant and customer.听第9段材料,回答第13至16题。

13. Why does Paul refuse to go to Russia with his parents?A. The weather there is cold.B. He has much work to do.C. He needs to meet the woman.14. How many people will be there at the woman’s house on Thursday?A. Four.B. Five.C. Eight.15. Who has already volunteered for the dessert?A. The woman’s sister.B. The woman’s mother.C. The woman’s brother.16. What will Paul bring to the woman’s house?A. A bottle of wine.B. Some cakes.C. Some pies.17. Who is the speaker talking to?A. Tourists.B. Students.C. Parents.18. What will the students and teachers do on Thursday morning?A. Visit a library.B. Attend a concert.C. Hold a chess game.19. What is a must for the activity on Thursday morning?A. Chess.B. A book.C. A signed note.20. What does the speaker mainly talk about?A. Plans for a school trip.B. A series for a school trip.C. Schedules before winter vacation.第二部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分,满分37.5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。

AThe Best Four Soccer Clubs in TorontoToronto is home to a number of soccer clubs. If you are interested in soccer, let’s take a look at four best soccer clubs in Toronto.Cherry Beach Soccer ClubCherry Beach Soccer Club, which first opened its door in 2009, now has 56 house leagues, 10 competitive teams and six adult teams. They provide 30,000 children and young people in the downtown area with high-quality and low-cost soccer lessons. It offers a recreational house league, development, and competitive soccer for those aged five to thirty.Phone: (416)367-4359North Soccer ClubNorth Soccer Club, which was established in 1980, is a welcoming environment for everyone who enjoys soccer or wants to try something new. You don’t need to be concerned about your qualifications or experience. However, to take part in practice, you must first complete a health check. It offers a Special Olympics program for players with intellectual disabilities. The fee range is from $230 to $270.Phone:(416)924-9911Power Soccer ClubPower Soccer Club, founded in 1996 by a former Galway United player, has grown up to be one of Canada’s most famous soccer schools. You will for sure progress in your skill development with a active and engaging training sessions. The fee can range from $210 to $315.North Y ork Hearts Soccer ClubNorth York Hearts Soccer Club was founded in 1990. This club exposes everyone who loves sports to the history of soccer with the help of a group of skilled sports coaches and professors. The fee range is from $210 to $320.Phone: (416)650-574321. Which club was founded in the 21st century?A. North Soccer Club.B. Cherry Beach Soccer Club.C. Power Soccer Club.D. North York Hearts Soccer Club.22. What is required for people to join North Soccer Club?A. Rich experience.B. A wise mind.C. A health check.D. A curious heart.23. To learn the history of soccer, you should call ________ to sign up.A. (416)367-4359B. (416)650-5743C. (416)425-6062D. (416)924-9911B“Twinkle, twinkle, little star, how I wonder what you are …”The sweet sound of the well-known childhood song came out of the village office in Gaozhuang in Zhongwei, Nningxia Hui autonomous region. Inside, a small group of children were learning to play the guitar with Song Hao, deputy curator(副馆长) of Zhongwei Museum.Song is 48 years and was born into an artistic family. He received strict musical training as a child and was quite good at playing the guitar and composing music. He had a wish to do what he could with his own music to make the world a brighter place for the children. In June last year, he found an opportunity to turn that wish into reality.He was chosen to take part in rural vitalization(振兴) work in Gaozhuang for two years, and in his spare time, he started teaching village children to play the guitar.“There’s something really magical about music,” he said, remembering that when he played for the children, many had twinkles in their eyes.Song asked them if they wanted to learn to play, and they all said “Yes”, so he started a free class to teach them on weekends and during the summer and winter holidays. He paid for 20 guitars and music textbooks from his own pocket and began to teach them the basics, like how to pluck(弹拨) the strings and read music.The children loved the classes so much that they would arrive at the village office early to wait for Song. Over the past six months, a number of children have learned to play entire pieces, and Song said that he could feel the happiness and confidence that learning to play music had brought to them.Shen Mingyi, Shen Huihong’s father, spoke highly of Son g and his guitar class, saying that his daughter had become more extroverted as a result.24. What did Song Hao do during his spare time in Gaozhuang?A. He held concerts in the village.B. He went sightseeing with village school.C. He helped to build the village school.D. He taught village children to play the guitar.25. What made the children go to the guitar classes early?A. The school rules.B. The teacher’s plan.C. Their love for the classes.D. Their parents’ requirements.26. What does the underlined word “extroverted” mean in the last paragraph?A. Polite.B. Outgoing.C. Anxious.D. Confused.27. Which of the following can best describe Song Hao?A. Selfless.B. Humorous.C. Honest.D. Brave.CThe history of Azerbaijani cinema dates back to the end of the 19th century, during which, many excellent movies have been filmed, which are still watched and admired by everyone.These movies were filmed in many beautiful places: from the comfortable and winding streets of the Icharishahar(Old City)—UNESCO World Heritage Site in Baku—to the beautiful fruit gardens and forests of beautiful areas of Azerbaijan.Here, on these streets, one of the most famous Azerbaijani films, Arshin Mal Alan, directed by Rza Tahmasib, was filmed.Filmed in 1945 and characterized by its amazing plot, beautiful music, great singing and acting, Arshin Mal Alan quickly became a box office sensation(轰动).It remains an important cultural touchstone across Eurasia and has been translated into 86 languages and shown 136 countries.The film is about pure love, women’s sights, and a young people’s dreams for a modern way of life.The plot tells us about a rich young man named Asker. One day, he decides to get married. However, according to the Eastern rule, a bride(新娘)needs to cover her face before the wedding. But soon he finds a solution to this problem. On the advice of his friend Suleyman, Asker dresses up a street seller of cloth. By selling goods, these sellers can enter houses where the women and girls choosing their cloth don’t cover their faces. Now Asker can go into any yard and choose a bride.The film has a lot of entering moments, songs and humour. The black-and white version Amphibian Man and The Diamond Arm, you can see the streets well known to every Azerbaijani.28. What can we know about Arshin Mal Alan?A. It was filmed before 1945.B. It can be viewed in black and white only.C. It is an action film with amazing plot.D. It is popular around the world.29. What is the Eastern rule about marriage in Arshin Mal Alan?A. A man must be rich in order to marry his bride.B. A man must dress up as a seller at the wedding.C. A man must buy a house for his bride before the wedding.D. A man mustn’t see his bride face before the wedding.30. Why does Asker dress up as a seller of cloth?A. To choose a bride.B. To make a film.C. To work in the street.D. To earn more money.31. What is the text mainly about?A. The street culture in Azerbaijan.B. The history of black-and-white TV.C. A famous movie filmed in Old City-Icharishabar.D. A love story between Asker and his bride.DEveryone has “down days”. Maybe it’s because of the bad weather, or the disappointing grades on a difficult test, and some days teenagers just act uninterested in life or school. But these symptoms(症状) often pass quickly, as teens move on to new school subjects, or meet with friends to prevent themselves from thinking what troubles them at the moment. But if a teenager displays symptoms of sadness for more than two lasting weeks, it might point to something serious.As teenagers develop, they push new boundaries(边界), complain about rules and look for more free rights from their parents. According to the online Health Guide on Adolescent Development, parents must be lasting figures in their teenagers’ life, providing safe boundaries for teens to grow, even if the teenagers act like these boundaries are unwanted.Parents need to provide rules, while also remaining flexible(灵活的) and respectful of the growing tee ns’ need for freedom. For example, teenagers will often feel frustrated, embarrassed, and even angry that thought they want freedom, they still need to ask their parents for an agreement to go to a friend’s house, or need their mothers to take them to scho ol.The US Department of Education says that parents should respect and support their teen’s choices as long as those choices won’t have long-term harmful effects. For example, even if a parent doesn’t enjoy the music his or her teen listens to, it’s unlik ely that the choice of music will prevent that teen from entering a good college, or lead to health problems. However, if that teen is drinking alcohol and driving, parents must get through strict punishments to teach that there are bad results for poor choices that come with increased freedom.32. Why do teen’s feelings of bad days usually disappear quickly?A. Their teachers help them.B. They take some medicine.C. Their parents talk with them.D. They change their attention.33. What does the example in paragraph 3 show?A. Freedom must be given at anytime.B. Teens are mad at being controlled.C. Teens need both freedom and proper rules.D. Rules must be absolutely strict for teens.34. What should parents do about their teens’ choices?A. Support their helpful hobbies.B. Tell them which college to attend.C. Cancel their after-school activities.D. Get them away from singing pop songs.35. What is the best title for the text?A. How to Be With Growing TeensB. Causes of Teens’ SadnessC. Teens’ Wo rries About Strict RulesD. The Importance of Making Friends With Teens第二节(共5小题;每小题2.5分,满分12.5分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

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