高一下学期第一次月考试题
2016-2017 高一下学期期中考试生物试题考试说明：本试卷满分100分，考试时间90分钟，包括选择题和非选择题两部分，请把答案写在答题卡上面第I卷（选择题，共60分）一、单选题（本题共30道小题，每小题2分，共60分）1、下列有关孟德尔豌豆杂交实验的叙述,正确的是( )A.孟德尔研究豌豆花的构造,但无需考虑雌蕊、雄蕊的发育程度B.孟德尔在豌豆开花时进行去雄和授粉,实现亲本的杂交C.孟德尔利用了豌豆自花传粉、闭花受粉的特性D.孟德尔根据亲本中不同个体表现型来判断亲本是否纯合2、孟德尔在对一对相对性状进行研究的过程中,发现了基因的分离定律。
下列有关基因分离定律的几组比例,能说明基因分离定律实质的是( )A.的表现型比为3:1B.产生配子的比为1:1C.基因型的比为1:2:1D.测交后代的比为1:13、假设控制番茄果肉颜色的基因用D、d表示,红色和紫色为一对相对性状,且红色为显性。
杂合的红果肉番茄自交获得F1,将F1中表现型为红果肉的番茄自交得F2,下列叙述正确的是( )A.F2中无性状分离 B.F2中性状分离比为3:1C.F2红果肉个体中杂合子占2/5 D.在F2中首次出现能稳定遗传的紫果肉个体4、在“性状分离比的模拟实验”中甲、乙两个小桶中都有写有D或d的两种小球,并且各自的两种小球的数量是相等的,这分别模拟的是 ( )A.F1的基因型是Dd B.F1产生的雌雄配子数量相等C.F1产生的雌雄配子都有D和d两种,且比例D: d = 1:1D.亲本中的父本和母本各自产生D和d的配子,且比例为1:15、德尔探索遗传规律时,运用了“假说—演绎”法,该方法的基本内涵是:在观察与分析的基础上提出问题后,通过推理和想象提出解决问题的假说,根据假说进行演绎推理,再通过实验证明假说。
下列相关叙述中不正确的是 ( )A.“F2出现3:1的性状分离比不是偶然的”属于孟德尔假说的内容B.“豌豆在自然状态下一般是纯种”属于孟德尔假说的内容C.“测交实验”是对推理过程及结果进行的检验D.“体细胞中遗传因子成对存在,并且位于同源染色体上”属于假说内容6、在两对相对性状的遗传实验中,可能具有1:1:1:1比例关系的是( )①杂种自交后代的性状分离比②杂种产生配子类型的比例③杂种测交后代的表现型比例④杂种自交后代的基因型比例⑤杂种测交后代的基因型比例A.①②④B.①③⑤C.②③⑤D.②④⑤7、已知某豚鼠中毛皮粗糙对毛皮光滑为显性,黑色对白色为显性,两对性状独立遗传,用两只纯合黑色粗毛鼠与白色光毛鼠杂交得到,再相互交配得。
下列有关说法中正确的是( )A.中非亲本类型占6/16B.中能稳定遗传的个体占12/16C.能推广的新品种黑色光毛类型在中占3/16D.让中的黑色光毛类型反复自交可筛选出纯合品种8、香豌豆的花色有紫花和白花两种,显性基因C和P同时存在时开紫花。
两个纯合白花品种杂交,F1开紫花;F1自交,F2的性状分离比为紫花:白花=9:7。
下列分析不正确的是 ( )A.两个白花亲本的基因型为CCpp与ccPPB.F1测交结果紫花与白花比例为1:1C.F2紫花中纯合子的比例为1/9 D.F2中白花的基因型有5种9、果蝇灰身(B)对黑身(b)为显性,控制这一性状的基因位于常染色体上。
现将纯种灰身果蝇与黑身果蝇杂交,产生的F1 再互交产生F2,将F2中所有黑身果蝇除去,让灰身果蝇自由交配,产生F3。
问F3中灰身与黑身果蝇的比例是( )A.3:1B.5:1C.8:1D.9:110、小麦的高秆(D)对矮秆(d)为显性,有芒(B)对无芒(b)为显性。
将两种小麦杂交,后代中出现高秆有芒、高秆无芒、矮秆有芒、矮秆无芒四种表现型,且其比例为3:1:3:1,则亲本的基因型为( )A.DDBB×ddBbB.Ddbb×ddbbC.DdBb×ddBbD.DDBb×ddBB11、细胞分裂是生物体一项重要的生命活动,是生物体生长、发育、繁殖和遗传的基础。
据图分析正确的是( )A.图①表示某植株体细胞分裂,下一时期的主要特点是数目加倍B.图②表示某动物睾丸内的减数分裂,此细胞产生精子的几率是0C.图③是某高等雌性动物体内的一个细胞,一定代表的是卵细胞D.图②、③所示细胞的染色体行为分别对应于图④的、段①②③④12、性染色体XY是雄果蝇的体细胞中的一对同源染色体,细胞正常分裂的情况下,在精巢中一定含有两个Y染色体的是( )A.减数第一次分裂的初级精母细胞B.有丝分裂中期的精原细胞C.减数第一次分裂的次级精母细胞D.有丝分裂后期的精原细胞13、果蝇的红眼基因(R)对白眼基因(r)为显性,位于X染色体上;长翅基因(B)对残翅基因(b)为显性,位于常染色体上。
现有一只红眼长翅果蝇与一只白眼长翅果蝇交配,F1代的雄果蝇中约有1/8为白眼残翅。
下列叙述错误的是 ( )A.亲本雌雄果蝇的基因型依次是BbX R Xr 、BbX r YB.亲本产生的配子中含Xr的配子占1/2C.F1代产生基因型不同于双亲的几率为3/4;出现长翅雄果蝇的概率为3/16D.白眼残翅雌果蝇可能形成bbX r X r类型的次级卵母细胞14、如图是人体性染色体的模式图。
下列叙述不正确的是( )A、位于Ⅰ区基因的遗传只与男性相关B、位于Ⅱ区的基因在遗传时,后代男女性状的表现一致C、人体的初级精母细胞中只含一条X染色体D、性染色体既存在于生殖细胞中,也存在于体细胞中15、如图是一种伴性遗传病的家系图。
下列叙述错误的是( )A.该病是显性遗传病,Ⅱ-4是杂合子B.Ⅲ-7 与正常男性结婚,子女都不患病C.Ⅲ-8 与正常女性结婚,儿子都不患病D.该病在男性人群中的发病率高于女性人群16、对小白鼠性腺组织细胞进行荧光标记, 等位基因 A、a都被标记为黄色,等位基因B、b 都被标记为绿色。
若这2对基因在两对同源染色体上 , 在荧光显微镜下观察处于四分体时期的性腺组织细胞 , 荧光点出现的情况可能是()A.荧光点在1个四分体中 ,2 个黄色、2个绿色荧光点B.荧光点在1个四分体中 ,4 个黄色、4 个绿色荧光点C.荧光点在2个四分体中 , 每个四分体中2个黄色或2个绿色荧光点D.荧光点在2个四分体中 ,每个四分体中 4 个黄色或 4 个绿色荧光点17、赫尔希和蔡斯用32P标记的T噬菌体与无32P标记的大肠杆菌混合培养,一2段时间后经搅拌、离心得到了上清液和沉淀物。
下列叙述不正确的是( )A.搅拌的目的是使吸附在大肠杆菌上的噬菌体与大肠杆菌分离B.32P主要集中在沉淀物中,上清液中也能检测到少量的放射性C.如果离心前混合时间过长,会导致上清液中放射性降低D.本实验结果说明DNA在亲子代之间的传递具有连续性18、下列有关肺炎双球菌转化实验的叙述中 , 正确的说法有( )①格里菲思的肺炎双球菌转化实验说明 DNA 是主要的遗传物质②格里菲思的肺炎双球菌转化实验直接证明DNA 是遗传物质③格里菲思认为加热杀死的 S型细菌的DNA是转化因子④体内转化的实质是 S 型细菌的 DNA 可使小鼠致死A.全错B.1项C.2项D.全对19、下列关于双链DNA分子的说法,正确的是( )A.每个核糖上均连接着一个磷酸和一个碱基B.某个DNA中含有200个碱基对,则其蕴含的遗传信息种类最多有2100种C.若一条链的A:T:G:C=1:2:3:4,则另一条链相应碱基比为4:3:2:1D.DNA分子中G与C碱基对含量越高,其结构稳定性相对越强20、对如图所表示的生物学意义的描述,错误的是( )A.甲图中生物自交后产生基因型为Aadd的个体的概率为B.乙图细胞处于有丝分裂后期,该生物正常体细胞的染色体数为4条C.丙图所示家系中男性患明显多于女性患者,该病最有可能是伴X隐性遗传病D.丁图表示雄果蝇染色体组成图21、下列有关精子和卵细胞形成的说法正确的是( )A.二者形成过程中都会出现联会、四分体、同源染色体分离、非同源染色体自由组合现象B.二者形成过程中都有染色体的复制和均分,所含遗传物质均是正常体细胞的一半C.精子和卵细胞形成过程中不同的地方是精子需变形,卵细胞不需要变形,其余完全相同D.形成100个受精卵,至少需要100个精原细胞和100个卵原细胞22、减数分裂和受精作用对于维持前后代体细胞中染色体数目的恒定具有重要意义。
下列有关叙述错误的是A．减数分裂过程中细胞连续分裂两次B．减数分裂过程中染色体只复制了一次C．受精卵中的遗传物质一半来自卵细胞D．受精作用的实质是精子和卵细胞的核融合23、人类皮肤中黑色素的多少由两对独立遗传的基因(和)所控制;基因和可以使黑色素量增加,两者增加的量相等,并可以累加。
若一纯种黑人与一纯种白人婚配,肤色为中间色;若与同基因型的异性婚配,出现的基因型种类数和表现型的比例为( )A.3种,3:1B.3种,1:2:1C.9种,9:3:3:1D.9种,1:4:6:4:124、一个基因型为TtMm(这两对基因可以自由组合)的卵原细胞，在没有突变的情况下，如果它所产生的卵细胞基因组成为TM，则由该卵原细胞分裂产生的下列细胞中，基因的数目、种类表示都正确的是()A减数第一次分裂产生的极体为TTMM，减数第二次分裂产生的极体为TM B减数第一次分裂产生的极体为tm，减数第二次分裂产生的极体为tmC减数第一次分裂产生的极体为tm，减数第二次分裂产生的极体为TM或tm D减数第一次分裂产生的极体为ttmm，减数第二次分裂产生极体为TM或tm25、在家鼠中短尾()对正常尾()为显性。
一只短尾鼠与一只正常鼠交配,后代中正常尾与短尾比例相同;而短尾类型相交配,子代中有一类型死亡,能存活的短尾与正常尾之比为2 : 1,则不能存活类型的基因型可能是( )A. B. C. D.或26、减数分裂过程中染色体和DNA的数目都会发生变化（不考虑细胞质中的DNA），以下与之相关的说法中正确的是（）A.次级精母细胞中的DNA分子与正常体细胞的DNA分子数目相同B.减数第一次分裂的后期，细胞中DNA数：染色体数=1:1C.次级精母细胞中染色体的数目总是比DNA分子数目少D.减数第二次分裂后期，细胞中染色体的数目等于精细胞中染色体数27、下列关于伴性遗传方式和特点的说法中,不正确的是( )A.染色体遗传:男女发病率相当;也有明显的显隐性关系B.染色体隐性遗传:男性发病率高;若女子发病其父、子必发病C.染色体遗传:父传子、子传孙D.染色体显性遗传:女性发病率高;若男性发病其母、女必发病28、下列关于DNA分子结构的叙述中，正确的是（）①每个DNA分子中，碱基种类均有四种②在双链DNA的一条核苷酸链中A=T③相连的脱氧核糖和碱基，能够和上下相连的任一磷酸基组成脱氧核苷酸单位④双链DNA分子的两条多核苷酸链通过碱基相连接⑤脱氧核苷酸之间由磷酸核脱氧核糖相连接A．①②③B．①④⑤C．③④⑤D．①②④29、某植物的高茎(B)对矮茎(b)为显性，花粉粒长形(D)对花粉粒圆形(d)为显性，花粉粒非糯性(E)对花粉粒糯性(e)为显性，非糯性花粉遇碘液变蓝色，糯性花粉遇碘液呈棕色。
吉林省2022-2023学年高一下学期第一次月考语文试卷
高一语文试题一、现代文阅读（35分）（一）现代文阅读Ⅰ（本题共5小题，19分）阅读下面的文字，完成1--5题。
材料一：汉代许慎《说文解字》中说：“儒，柔也”。
一个“柔”字，切中要义，味道全出。
宋词专家叶嘉莹先生在一档节目中就特地提到了中国文化的“弱德之美”。
她由“儒”字的“柔”这一本义出发，加以阐释，把儒家所代表的中国文化性格多维度地彰显出来了。
“儒”字中含一个“需”字，“需”有“等待”之义。
孔子就说过：“君子藏器于身，待时而动，何不利之有？”孔子这里说的其实就是“需”的意思，它体现的显然是一种等待的姿态。
华裔英籍女作家韩素音，在描述一位华侨时说：“他是个中国人，有极好的耐心，能等待和忍耐。
”这的确典型而鲜明地体现了中国人所特有的品性。
韩素音在参观走访了中国内地之后，曾经感慨道：“我在这里重新发现了中国的弹性——它所固有的柔顺性，这使它不受外界危机的影响，同时也使它克服一次又一次动乱。
”因此，我们虽说“儒者柔也”，但并不是说柔就是软弱无力，就是废弃一切作为。
老子认为，“天下莫柔弱于水”，但是“攻坚强者莫之能胜”，这正是“天下之至柔，驰骋天下之至坚”的道理所在。
俗话说“水滴石穿”，就是“以柔克刚”的一个十分典型的例子。
其实，我们只有通过“水”的意象，才能最真切地体味到“儒”之“柔”。
柔是“水”最为突出的特性。
在中国文化中，以水喻道是有其古老传统的。
譬如，老子说“上善若水”，他还说“弱者道之用”，此所谓“弱者”指的就是水的柔弱。
他又说“水善利万物而不争”，就是说，水善于滋养万物而从不争夺，水中因此蕴含着大道理。
管子就认为：“水者何也？万物之本原也。
”如此等等，不一而足。
我们知道，水是不定形的，它被放进怎样的容器中就成为什么样子的形状，但正是因为没有一种固定不变的形状，所以才能变成一切可能的形状，这正是“道”的品格。
更为重要的是，它以隐喻和象征的方式，透露出中国文化的传统性格。
以水来比喻道的高明之处在于，它的意义是双关的：一方面确立了存在论的基本意象，让人们能够由此及彼地去领会“道”的深刻内涵；另一方面又奠定了道德论的基本取向，借助于水的“至柔”的性格来凸显道德的品性。
四川省成都市高一下学期第一次月考生物试题(解析版)
一、单选题（本大题共20小题）
1.下列最能体现孟德尔遗传规律本质的选项是（ ）
A.杂合的高茎豌豆自交，后代出现3：1的性状分离
B.纯合的黄色圆粒与绿色皱粒豌豆杂交，F1全是黄色圆粒豌豆
C.纯合的黄色圆粒与绿色皱粒豌豆杂交，F1产生四种比例相等的配子
D.纯合的黄色圆粒与绿色皱粒豌豆杂交，F2有四种表现型，比例为9：3：3：1
C.两个遗传定律发生的实质体现在F2出现了性状分离和自由组合现象
D.F2中的3∶1性状分离比依赖于雌雄配子的随机结合
【答案】D
【解析】
【分析】孟德尔采用假说演绎法得出了基因的分离定律和基因的自由组合定律。假说演绎法包括：观察实验、提出问题→做出假说→演绎推理→实验验证。
【详解】A、孟德尔研究遗传规律所用的方法是假说演绎法，A错误；
故选B。
8.孟德尔在豌豆杂交实验中，发现问题和验证假说所采用的实验方法依次是（）
A.杂交、自交和测交B.测交、自交和杂交
C.杂交、测交和自交D.测交、杂交和自交
【答案】A
【解析】
【分析】孟德尔在豌豆的杂交实验中，先通过具有一对或者两对相对性状的个体杂交产生子一代，再让子一代自交，随后提出问题，由问题提出假说，最后做测交实验，验证假说。
2.以豌豆为材料进行杂交实验。下列说法错误的是（）
A.豌豆是自花传粉且闭花授粉的二倍体植物
B.进行豌豆杂交时，母本植株需要人工去雄
C.杂合子中的等位基因均在形成配子时分离
D.非等位基因在形成配子时均能够自由组合
【答案】D
【解析】
【分析】豌豆的优点：豌豆是严格的自花闭花传粉植物，自然状态下是纯种；含有多对容易区分的相对性状。
安徽省淮北师范大学附属实验中学2022-2023学年高一下学期第一次月考数学试题(含答案解析)
安徽省淮北师范大学附属实验中学2022-2023学年高一下学期第一次月考数学试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．有下列命题：①两个相等向量，若它们的起点相同，则终点也相同；②若||a b|=|，则a b = ；③若AB DC = ，则四边形ABCD 是平行四边形；④若m n = ，n k = ，则m k = ；⑤若//a b ，//b c，则//a c ；⑥有向线段就是向量，向量就是有向线段．其中，假命题的个数是A ．2B ．3C ．4D ．52．在三角形ABC ∆中，若点P 满足1231,3344AP AB AC AQ AB AC =+=+，则APQ ∆与ABC ∆的面积之比为（）A ．1:3B ．5:12C ．3:4D ．9:163．已知向量a ，b 满足1a = ，b = ，且a 与b的夹角为6π，则()()2a b a b +⋅-= （）A ．12B ．32-C ．12-D ．324．若向量i ，j 为互相垂直的单位向量，2a j i =- ，m b j i =+ ，且a 与b的夹角为锐角，则实数m 的取值范围是A ．1,2⎛⎫+∞ ⎪⎝⎭B ．(－∞，－2)∪12,2⎛⎫- ⎪⎝⎭C ．222,,33⎛⎫⎛⎫-⋃+∞ ⎪ ⎪⎝⎭⎝⎭D ．1,2⎛⎫-∞ ⎪⎝⎭5．设,a b均为单位向量,则“a 与b 的夹角为23π”是“||a b += 的A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件6．已知向量()1,1a = ，()1,b m = ，其中m 为实数，O 为坐标原点，当两向量夹角在0,12π⎛⎫⎪⎝⎭变动时，m 的取值范围是A ．()0,1B ．3⎛ ⎝C ．(3⎛⎫⎪ ⎪⎝⎭U D ．(A ．M ，N ，P 三点共线B ．M ，N ，Q 三点共线C ．M ，P ，Q 三点共线D ．N ，P ，Q 三点共线8．下面是如皋定慧寺观音塔的示意图，游客（视为质点）从地面D 点看楼顶点A 的仰角为30°，沿直线DB 前进51米达到E 点，此时看点C 点的仰角为45°，若23BC AC =，则该观音塔的高AB 约为（） 1.73≈）A ．8米B ．9米C ．40米D ．45米二、多选题9．下列运算正确的是（）A ．()326a a-⋅=-B ．()()223a b b a a+--=C ．()()220a b b a +-+= D ．()2362a b a b -=-10．生于瑞士的数学巨星欧拉在1765年发表的《三角形的几何学》一书中有这样一个定理：“三角形的外心、垂心和重心都在同一直线上．”这就是著名的欧拉线定理．在ABC 中，O ，H ，G 分别是外心、垂心和重心，D 为BC 边的中点，下列四个选项中正确的是（）A ．2GH OG =B ．0GA GB GC ++=C ．2AH OD=D ．ABG BCG ACGS S S == 11．下列说法正确的有（）A ．若//a b r r ，//b c，则//a cB ．若a b =，b c = ，则a c= C ．若//a b r r，则a 与b 的方向相同或相反D ．若AB 、BC共线，则A 、B 、C 三点共线12．已知ABC 是正三角形，则在下列结论中，正确的为（）A ．AB BC BC CA +=+ B ．AC CB BA BC +=+C ．AB AC CA CB +=+D ．AB BC AC CB BA CA ++=++三、双空题13．已知平面上不共线的四点O ，A ，B ，C ，若320OA OB OC -+= ，则AB =______BC ，AB BC= ______．四、填空题14．已知向量a ，b 是两个不共线的向量，且向量m a －3b 与a＋(2－m ) b 共线，则实数m 的值为___．15．如图，在四边形ABCD 中，DA DB DC ==，且DA DC DB +=，则ABC ∠=______．16．如图，在菱形ABCD 中，120ABC ∠=︒，2AB =，则BC DC += ______．五、解答题17．已知111,,()()42a ab a b a b =⋅=+⋅-= ．(1)求||b的值；(2)求向量a b - 与a b +夹角的余弦值．18．在直角梯形ABCD 中，90A ∠=︒，30B ∠=︒，AB =2BC =，点E 在线段CD上.若AE AD AB λ=+，求实数λ的取值范围.19．如图所示，A ，B ，C 为山脚两侧共线的三点，在山顶P 处测得三点的俯角分别为α，β，γ.计划沿直线AC 开通穿山隧道，请根据表格中的数据，计算隧道DE 的长度.20．已知OAB 中，点B 是点C 关于点A 的对称点，点D 是线段OB 的一个靠近B 的三等分点，设,AB a AO b ==．(1)用向量a 与b 表示向量OC ，CD；(2)若45OE OA =，求证：C ，D ，E 三点共线．21．如图，ABC 中，点D 是AC 的中点，点E 是BD 的中点，设,BA a BC c ==.（1）用a ，c 表示向量AE；（2）若点F 在AC 上，且1455BF a c =+ ，求:AF CF .22．设1e ，2e 是不共线的非零向量，且122a e e =- ，123b e e =+ ．若1243e e a ub λ-=+，求λ，u 的值．参考答案：1．C【详解】对于①，两个相等向量时，它们的起点相同，则终点也相同，①正确；对于②，若a b = ，方向不确定，则a 、b不一定相同，∴②错误；对于③，若AB DC = ，AB 、DC不一定相等，∴四边形ABCD 不一定是平行四边形，③错误；对于④，若m n = ，n k =，则m k = ，④正确；对于⑤，若//a b ，//b c，当0b = 时，//a c 不一定成立，∴⑤错误；对于⑥，有向线段不是向量，向量可以用有向线段表示，∴⑥错误；综上，假命题是②③⑤⑥，共4个，故选C.2．B【分析】由题目条件所给的向量等式，结合向量的线性运算推断P 、Q 两点所在位置，比较两个三角形的面积关系【详解】因为1233AP AB AC =+ ，所以12()()33AP AB AC AP-=-，即2BP PC = ，得点P 为线段BC 上靠近C 点的三等分点，又因为3144AQ AB AC =+ ，所以31()()44AQ AB AC AQ -=-，即3BQ QC = ，得点Q 为线段BC 上靠近B 点的四等分点，所以512PQ BC =，所以APQ ∆与ABC ∆的面积之比为512APQ ABCS PQ S BC == ，选择B 【点睛】平面向量的线性运算要注意判断向量是同起点还是收尾相连的关系再使用三角形法则和平行四边形法则进行加减运算，借助向量的数乘运算可以判断向量共线，及向量模长的关系3．A【分析】根据向量的数量积运算以及运算法则，直接计算，即可得出结果.【详解】因为1a =，b = ，且a 与b的夹角为6π，所以c 362os b b a a π=⋅=，因此()()2223122322b b a b a a b a +⋅-=+-=⋅+-= .故选：A.4．B【分析】由a 与b夹角为锐角，可得0a b ⋅ ＞且b a ，不共线，再代入向量解不等式即可得到答案．【详解】由题意可得：∵a 与b夹角为锐角，∴⋅=a b （2i j - ）()m i j ⋅+= 1-2m ＞0，且b a ，不共线∴12m ＜当a b时，可得m ＝﹣2所以实数λ的取值范围是（﹣∞，﹣2）∪（﹣2，12）．故选B ．【点睛】本题主要考查利用向量的数量积表示解决两个向量的夹角问题，当a 与b的夹角为锐角可得，0a b ⋅＞且b a ，不共线，但是学生容易忽略两个向量共线并且同向的情况.5．D【解析】按照向量的定义、充分条件和必要条件的定义，分别从充分性和必要性入手去判断即可.【详解】因为,a b 均为单位向量,且a 与b 的夹角为23π，所以||1a b +=== ,所以由“a 与b 的夹角为23π”不能推出“||a b +=若||a b +=则||a b += ==解得1cos ,2a b 〈〉= ,即a 与b 的夹角为23π,所以由“||a b += 不能推出“a 与b 的夹角为23π”.因此,“a 与b 的夹角为23π”是“||a b += 的既不充分也不必要条件.故选：D.【点睛】本题主要考查数量积的应用，考查充分条件和必要条件的应用，考查逻辑思维能力和运算能力，属于常考题.6．C【分析】设向量a 、b的起点均为O ，终点分别为A 、B ，可得出OA 与x 轴正方向的夹角为4π，设向量OB 与x 轴正方向的夹角为θ，由题意可得出63ππθ<<且4πθ≠，由tan m θ=可得出实数m 的取值范围.【详解】设向量a 、b的起点均为O ，终点分别为A 、B ，可得出OA 与x 轴正方向的夹角为4π，设向量OB 与x 轴正方向的夹角为θ，由于0,12AOB π⎛⎫∠∈ ⎪⎝⎭，则,464AOB πππθ⎛⎫⎛⎫=-∠∈ ⎪ ⎪⎝⎭⎝⎭或,443AOB πππθ⎛⎫⎛⎫=+∠∈ ⎪ ⎪⎝⎭⎝⎭.即B 在1B 与2B （不与A 重合）之间，(tan ,13m θ⎫∴=∈⎪⎪⎝⎭U ，因此，实数m 的取值范围是(3⎛⎫⎪ ⎪⎝⎭U ，故选：C.【点睛】本题考查利用向量夹角的取值范围求参数，解题时充分利用数形结合法，找到临界位置进行分析，可简化运算，考查分析问题和解决问题的能力，属于中等题.7．B【分析】利用平面向量共线定理进行判断即可．【详解】28NP a b =-+，3()PQ a b =- ，283()5NQ NP PQ a b a b a b ∴=+=-++-=+ ，5MN a b =+ ，MN NQ ∴= ，由平面向量共线定理可知，MN 与NQ为共线向量，又MN 与NQ有公共点N ，M ∴，N ，Q 三点共线，故选：B ．8．D【分析】设AC x =，根据已知条件得32BC BE x ==，52AB x =，根据ADB ∠的正切表示出BD ，再表示出DE ，由51DE =列出方程，解出x 即可得出AB 的长．【详解】解：设AC x =，根据条件可得32BC BE x ==，52AB AC BC x =+=，tan AB ADB BD ∠==，BD ∴=，3()5122DE BD BE x ∴=-=-=，18.0522x ∴=，5452AB x ∴=≈米，故选：D ．9．ABD【分析】根据向量的加减和数乘运算，即可得出结论.【详解】由题意，A 项，()326a a -⋅=- ，A 正确.B 项，()()222223a b b a a b b a a +--=+-+=，B 正确.C 项，()()22220a b b a a b b a +-+=+--=，C 错误.D 项，()2362a b a b -=- ，D 正确.故选：ABD.10．ABCD【分析】由重心的性质以及向量的加法运算法则判断选项A ；结合三角形相似及重心性质判断选项A 与C ；利用重心性质及高的比例判断选项D.【详解】在ABC 中，O ，H ，G分别是外心、垂心和重心，画出图形，如图所示．对于B 选项，根据三角形的重心性质由重心的性质可得G 为AD 的三等分点，且2GA GD =-，又D 为BC 的中点，所以2GB GC GD +=，所以20GA GB GC GD GD ++=-+= ，故选项B 正确；对于A 与C 选项，因为O 为ABC 的外心，D 为BC 的中点，所以OD BC ⊥，所以AH OD ∥，∴AHG DOG ∽，∴2GH AH AGOG OD DG===，∴2GH OG =，2AH OD =，故选项A ，C 正确；对于D ，过点G 作GE BC ⊥，垂足为E ，∴DEG DNA △∽△，则13GE DG AN DA ==，∴BGC 的面积为11112233BGC ABC S BC GE BC AN S =⨯⨯=⨯⨯⨯=△△；同理，13AGC AGB ABC S S S ==△△△，选项D 正确.故选：ABCD 11．BD【分析】取0b =可判断AC 选项的正误；利用向量相等的定义可判断B 选项的正误；利用共线向量的定义可判断D 选项的正误.【详解】对于A 选项，若0b = ，a 、c 均为非零向量，则//a b r r ，//b c成立，但//a c 不一定成立，A 错；对于B 选项，若a b =，b c = ，则a c = ，B 对；对于C 选项，若0b = ，0a ≠r r，则b 的方向任意，C 错；对于D 选项，若AB 、BC共线且AB 、BC 共点B ，则A 、B 、C 三点共线，D 对.故选：BD.12．ACD【分析】利用向量的数量积的运算律求解即可.【详解】AB BC AC += ，BC CA BA +=，而AC BA = ，故A 正确；设正三角形的边长为2a ，所以2BA BC += ，2AC CB AB a +==，所以AC CB BA BC +≠+，故B 不正确；2A B AC=+，2C A CB=+,所以AB AC CA CB+=+，故C正确；24AB BC AC AC a++==，24CB BA CA CA a++==，所以AB BC AC CB BA CA++=++，故D正确．故选:ACD.13．22【分析】先化简为()2OA OB OB OC-=-，再利用向量的减法法则化简即得解.【详解】∵320OA OB OC-+=，∴()2OA OB OB OC-=-，∴2BA CB=，∴2AB BC=，∴2ABBC=．故答案为：2，2.14．－1或3【分析】利用向量共线定理即可得出．【详解】由题意知m a－3b＝λ[a＋(2－m) b]，∴()32mmλλ=⎧⎨-=-⎩解得m＝－1或m＝3.故答案为－1或3.【点睛】本题考查了向量共线定理，属于基础题．15．120︒【分析】根据向量加法的平行四边形法则求得正确答案.【详解】因为DA DC DB+=，所以由向量的加法的几何意义可知四边形ABCD是平行四边形，又因为DA DB DC==，所以四边形ABCD是菱形，且60DAB∠=︒，所以120ABC∠=︒．故答案为：120︒16．【分析】根据向量加法运算结合菱形的性质及角度，求出模长即可【详解】如图所示，设菱形对角线交点为O ,BC DC AD AB AC +=+=．因为120ABC ∠=︒，所以60BAD ∠=︒，所以ABD △为等边三角形．又AC BD ⊥，2AB =，所以1OB =．在Rt AOB △中，AO = ，所以2BC DC AC AO +=== ．故答案为:17．(1)2；4.【分析】（1）根据11,()()2a ab a b =+⋅-= 即可求b ；（2）设向量a b + 与a b - 大角为θ,()()cos a b a b a b a b θ+⋅-=+⨯- .【详解】（1）()()2212a b a b a b +⋅-=-= ，1a = ，21||2b ∴=，b ∴= （2）22211212242a b a a b b +=+⋅+=+⨯+=,a b ∴+= 22211212142a b a a b b -=-⋅+=-⨯+= ,1a b ∴-= ，设向量a b + 与a b - 大角为θ,()()12cos a b a b a b a b θ+⋅-∴=+⨯- 18．10,2⎡⎤⎢⎥⎣⎦【解析】根据梯形的几何性质和向量的线性运算可得DE ABλ= ，可求得实数λ的取值范围.【详解】由图分析知cos30DC AB BC =-︒∵AE AD AB λ=+ ,∴AE AD AB λ-= ，即DE AB λ= ，∴DE ABλ=.又0DE ≤≤，AB =uu u r 102λ≤≤.综上，实数λ的取值范围是10,2⎡⎤⎢⎥⎣⎦.【点睛】本题考查向量的线性运算，关键在于运用梯形的几何性质得出向量间的线性关系，属于基础题.19．隧道DE 的长度为9【解析】首先利用同角三角函数的关系求出3sin 5γ=，再利用两角差的公式求出()sin 60γ︒-，在△PBC 中，利用正弦定理求出PB ，在△PAB 中，求出AB ，由DE =AB -AD -EB 即可求解.【详解】解：由4cos 5γ=，γ为锐角，可得3sin 5γ=，则()sin 60sin 60cos cos60sin γγγ︒︒︒-=-=.在△PBC 中，60BPC γ︒∠=-，PCB γ∠=，12BC =-由正弦定理可得，()3(12sin 5sin 60BC PB γγ︒-⨯==-在△PAB 中，∠PAB =45°，∠APB =75°，PB =由正弦定理可得，sin759sin452PBAB︒︒⋅==+所以DE=AB-AD-EB=9，所以隧道DE的长度为9.【点睛】本题考查了正弦定理求不可直接测量的两点间的距离，属于基础题.20．(1)OC a b=--uuu r r r，5133CD a b=+；(2)证明见解析.【分析】(1)根据向量的加法，减法，数乘运算的几何意义求解；(2)求证CE，CD共线即可.【详解】（1）因为点B是点C关于点A的对称点，所以AC AB=-，又AB a=，所以AC a=-，因为OC OA AC=+，OO A bA=-=-，所以OC a b=--uuu r r r，因为点D是线段OB的一个靠近B的三等分点，所以13BD BO=，由已知22CB AB a==，BA AB a=-=-，所以11151()2()33333 CD CB BD CB BO CB BA AO a a b a b=+=+=++=+-+=+.；（2）∵413()555CE OE OC b a b a b CD=-=-++=+=∴CE与CD平行，又∵CE与CD有公共点C，∴C，D，E三点共线.21．（1）1344AE c a=-；（2）:4:1AF CF=.【分析】（1）由于点D是AC的中点，点E是BD的中点，所以12AD AC=，1()2AE AB AD=+，而AC BC BA c a=-=-，从而可求得结果，（2）设AF ACλ=，从而可得BF BA AF BA ACλ=+=+，再用a，c表示，然后结合1455BF a c=+，可求得λ的值，从而可求得:AF CF的值【详解】（1）因为AC BC BA c a=-=-，点D是AC的中点，所以11()22AD AC c a==-，因为点E是BD的中点，所以1111113()()2222444AE AB AD AB AD a c a c a=+=+=-+-=-.（2）设AF AC λ= ，所以()(1)BF BA AF BA AC a c a a c λλλλ=+=+=+-=-+ .又1455BF a c =+ ，所以4=5λ，所以45AF AC = ，所以:4:1AF CF =.22．31u λ=⎧⎨=⎩【分析】根据向量线性运算化简已知条件，由此列方程组来求得λ，u 的值.【详解】由1243e e a ub λ-=+ ，得()()()()12121212432323e e e e u e e u e u e λλλ-=-++=++-+ ，得4233u u λλ+=⎧⎨-+=-⎩，解得31u λ=⎧⎨=⎩．。
四川省成都市2023-2024学年高一下学期第一次月考数学试题含答案
武侯高中高2023级2023——2024下期第一次月考试题数学（答案在最后）学校：__________姓名：__________班级：__________考号：__________一、单选题1.如图，四边形ABCD 中，AB DC =，则必有（）A.AD CB= B.DO OB= C.AC DB= D.OA OC= 【答案】B 【解析】【分析】根据AB DC =，得出四边形ABCD 是平行四边形，由此判断四个选项是否正确即可．【详解】四边形ABCD 中，AB DC =，则//AB DC 且AB DC =，所以四边形ABCD 是平行四边形；则有AD CB =-，故A 错误；由四边形ABCD 是平行四边形，可知O 是DB 中点，则DO OB =，B 正确；由图可知AC DB≠，C 错误；由四边形ABCD 是平行四边形，可知O 是AC 中点，OA OC =-，D 错误．故选：B ．2.下列说法正确的是（）A.若a b ∥ ，b c ∥，则a c∥ B.两个有共同起点，且长度相等的向量，它们的终点相同C.两个单位向量的长度相等D.若两个单位向量平行，则这两个单位向量相等【答案】C 【解析】【分析】A.由0b =判断；B.由平面向量的定义判断；C.由单位向量的定义判断； D.由共线向量判断.【详解】A.当0b = 时，满足a b ∥ ，b c ∥，而,a c 不一定平行，故错误；B.两个有共同起点，且长度相等的向量，方向不一定相同，所以它们的终点不一定相同，故错误；C.由单位向量的定义知，两个单位向量的长度相等，故正确；D.若两个单位向量平行，则方向相同或相反，但大小不一定相同，则这两个单位向量不一定相等，故错误；故选：C3.若a b ，是平面内的一组基底，则下列四组向量中能作为平面向量的基底的是（）A.,a b b a --B.21,2a b a b++ C.23,64b a a b-- D.,a b a b+- 【答案】D 【解析】【分析】根据基底的知识对选项进行分析，从而确定正确答案.【详解】A 选项，()b a a b -=-- ，所以a b b a -- ，共线，不能作为基底.B 选项，1222a b a b ⎛⎫+=+ ⎪⎝⎭ ，所以12,2a b a b ++ 共线，不能作为基底.C 选项，()64223a b b a -=-- ，所以64,23a b b a --共线，不能作为基底.D 选项，易知a b a b +-，不共线，可以作为基底.故选：D4.将函数2cos 413y x π⎛⎫=-+ ⎪⎝⎭图象上各点的横坐标伸长到原来的2倍，再向左平移3π个单位，纵坐标不变，所得函数图象的一条对称轴的方程是（）A.12x π=B.6x π=-C.3x π=-D.12x π=-【答案】B 【解析】【分析】根据图像的伸缩和平移变换得到2cos(2)13y x π=++，再整体代入即可求得对称轴方程.【详解】将函数2cos 413y x π⎛⎫=-+ ⎪⎝⎭图象上各点的横坐标伸长到原来的2倍，得到2cos 213y x π⎛⎫=-+ ⎪⎝⎭，再向左平移3π个单位，得到2cos[2()]12cos(2)1333y x x πππ=+-+=++，令23x k π+=π，Z k ∈，则26k x ππ=-，Z k ∈.显然，=0k 时，对称轴方程为6x π=-，其他选项不符合.故选：B5.设a ，b 是非零向量，“a a bb =”是“a b =”的（）A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件【答案】B 【解析】【分析】根据向量相等、单位向量判断条件间的推出关系，结合充分、必要性定义即知答案.【详解】由a a b b =表示单位向量相等，则,a b 同向，但不能确定它们模是否相等，即不能推出a b =，由a b =表示,a b 同向且模相等，则a a b b = ，所以“a a bb =”是“a b =”的必要而不充分条件.故选：B6.已知向量,a b ，且2,52,72AB a b BC a b CD a b =+=-+=+，则下列一定共线的三点是（）A.,,A B CB.,,B C DC.,,A B DD.,,A C D【答案】C 【解析】【分析】利用向量的共线来证明三点共线的．【详解】2,52,72AB a b BC a b CD a b =+=-+=+，则不存在任何R λ∈，使得AB BC λ=，所以,,A B C 不共线，A 选项错误；则不存在任何R μ∈，使得BC CD μ=，所以,,B C D 不共线，B 选项错误；由向量的加法原理知242BD BC CD a b AB =+=+=．则有//BD AB ，又BD 与AB有公共点B ，所以,,A B D 三点共线，C 选项正确；44AB BC a b AC ==-++，则不存在任何R t ∈，使得AC tCD = ，所以,,A C D 不共线，D 选项错误．故选：C ．7.已知sin α＝5，且α为锐角，tan β＝－3，且β为钝角，则角α＋β的值为（）A.4π B.34π C.3π D.23π【答案】B 【解析】【分析】先求出tan α12=，再利用两角和的正切公式求出tan(α＋β)＝－1，判断出角α＋β的范围，即可求出α＋β的值.【详解】sin α，且α为锐角，则cos α5=，tan αsin 1cos 2αα==.所以tan(α＋β)＝tan tan 1tan tan αβαβ+-＝13211(3)2--⨯-＝－1.又α＋β∈3(,22ππ，故α＋β＝34π.故选：B8.筒车亦称“水转筒车”，是一种以水流作动力，取水灌田的工具，唐陈廷章《水轮赋》：“水能利物，轮乃曲成.升降满农夫之用，低徊随匠氏之程.始崩腾以电散，俄宛转以风生.虽破浪于川湄，善行无迹；既斡流于波面，终夜有声.”如图，一个半径为4m 的筒车按逆时针方向每分钟转一圈，筒车的轴心O 距离水面的高度为2m .在筒车转动的一圈内，盛水筒P 距离水面的高度不低于4m 的时间为（）A.9秒B.12秒C.15秒D.20秒【答案】D 【解析】【分析】画出示意图，结合题意和三角函数值可解出答案.【详解】假设,,A O B 所在直线垂直于水面，且4AB =米，如下示意图，由已知可得12,4OA OB OP OP ====，所以1111cos 602OB POB POB OP ∠==⇒∠=︒，处在劣弧 11PP 时高度不低于4米，转动的角速度为360660︒=︒/每秒，所以水筒P 距离水面的高度不低于4m 的时间为120206=秒，故选：D.二、多选题9.已知函数()cos f x x x =+，则下列判断正确的是（）A.()f x 的图象关于直线π6x =对称 B.()f x 的图象关于点π,06⎛⎫- ⎪⎝⎭对称C.()f x 在区间2π,03⎡⎤-⎢⎥⎣⎦上单调递增 D.当π2π,33x ⎛⎫∈-⎪⎝⎭时，()()1,1f x ∈-【答案】BC 【解析】【分析】利用辅助角公式化简函数()f x 的解析式，利用正弦型函数的对称性可判断AB 选项；利用正弦型函数的单调性可判断C 选项；利用正弦型函数的值域可判断D 选项.【详解】因为()πcos 2sin 6f x x x x ⎛⎫=+=+ ⎪⎝⎭，对于A选项，ππ2sin 63f ⎛⎫==⎪⎝⎭，故函数()f x 的图象不关于直线π6x =对称，A 错；对于B 选项，π2sin 006f ⎛⎫-== ⎪⎝⎭，故函数()f x 的图象关于点π,06⎛⎫- ⎪⎝⎭对称，B 对；对于C 选项，当2π03x -≤≤时，πππ266x -≤+≤，则函数()f x 在区间2π,03⎡⎤-⎢⎥⎣⎦上单调递增，C 对；对于D 选项，当π2π33x -<<时，ππ5π666x -<+<，则1πsin 126x ⎛⎫-<+≤ ⎪⎝⎭，所以，()(]π2sin 1,26f x x ⎛⎫=+∈- ⎪⎝⎭，D 错.故选：BC.10.下图是函数()sin()(0π)f x A x ωϕϕ=+<<的部分图像，则（）A.2πT =B.π3ϕ=C.π,06⎛⎫-⎪⎝⎭是()f x 的一个对称中心 D.()f x 的单调递增区间为5πππ,π1212k k ⎡⎤-++⎢⎥⎣⎦（Z k ∈）【答案】BCD 【解析】【分析】由图象可得πT =，由2πT ω=可求出ω，再将π12⎛⎝代入可求出ϕ可判断A ，B ；由三角函数的性质可判断C ，D .【详解】根据图像象得35ππ3ππ246124T T =-=⇒=⇒=ω，故A 错误；π12x =时，πππ22π2π1223k k ⨯+=+⇒=+ϕϕ，0πϕ<< ，π3ϕ∴=，故()π23f x x ⎛⎫=+ ⎪⎝⎭，故B 正确；因为πππ20663f ⎡⎤⎛⎫⎛⎫-=⋅-+= ⎪ ⎪⎢⎝⎭⎝⎭⎣⎦，所以π,06⎛⎫- ⎪⎝⎭是()f x 的一个对称中心，C 正确；令πππ2π22π232k x k -+≤+≤+，解得5ππππ1212k x k -+≤≤+，Z k ∈．故D 正确．故选：BCD .11.潮汐现象是地球上的海水受月球和太阳的万有引力作用而引起的周期性涨落现象.某观测站通过长时间观察，发现某港口的潮汐涨落规律为πcos 63y A x ω⎛⎫=++ ⎪⎝⎭（其中0A >，0ω>），其中y （单位：m ）为港口水深，x （单位：h ）为时间()024x ≤≤，该观测站观察到水位最高点和最低点的时间间隔最少为6h ，且中午12点的水深为8m ，为保证安全，当水深超过8m 时，应限制船只出入，则下列说法正确的是（）A.π6ω=B.最高水位为12mC.该港口从上午8点开始首次限制船只出入D.一天内限制船只出入的时长为4h 【答案】AC 【解析】【分析】根据题意可求得6π=ω，可知A 正确；由12点时的水位为8m 代入计算可得4A =，即最高水位为10m ，B 选项错误；易知ππ4cos 663y x ⎛⎫=++⎪⎝⎭，解不等式利用三角函数单调性可得从上午8点开始首次开放船只出入，一天内开放出入时长为8h ，即可判断C 正确，D 错误.【详解】对于A ，依题意π62T ω==，所以6π=ω，故A 正确；对于B ，当12x =时，ππcos 126863y A ⎛⎫=⨯++=⎪⎝⎭，解得4A =，所以最高水位为10m ，故B 错误；对于CD ，由上可知ππ4cos 663y x ⎛⎫=++⎪⎝⎭，令8y ≥，解得812x ≤≤或者2024x ≤≤，所以从上午8点开始首次开放船只出入，一天内开放出入时长为8h ，故C 正确，D 错误.故选：AC.三、填空题12.设e为单位向量，2a =r ，当,a e 的夹角为π3时，a 在e 上的投影向量为______.【答案】e【解析】【分析】利用投影向量的定义计算可得结果.【详解】根据题意可得向量a 在e 上的投影向量为22π21cos 31a e e a e e e e ee e⨯⨯⋅⋅⋅=== .故答案为：e13.已知向量a 、b 满足5a = ，4b = ，a 与b 的夹角为120，若()()2ka b a b -⊥+ ，则k =________.【答案】45##0.8【解析】【分析】运用平面向量数量积公式计算即可.【详解】因为5a = ，4b = ，a 与b的夹角为120 ，所以1cos12054102a b a b ⎛⎫⋅==⨯⨯-=- ⎪⎝⎭.因为()2ka b -⊥()a b +r r ，所以()()()()222222521610215120ka b a b kab k a b k k k -⋅+=-+-⋅=-⨯--=-=，解得45k =.故答案为：45.14.已知1tan 3x =，则1sin 2cos 2x x +=______【答案】2【解析】【分析】根据二倍角公式以及齐次式即可求解.【详解】2222222211121sin 2cos sin 2sin cos 1tan 2tan 332cos 2cos sin 1tan 113x x x x x x x x x x x ⎛⎫++⨯ ⎪+++++⎝⎭====--⎛⎫- ⎪⎝⎭.故答案为：2四、解答题15.已知1a b a == ，与b 的夹角为45︒．（1）求()a b a +⋅的值；（2）求2a b -的值【答案】（1）2（2【解析】【分析】（1）先求2,a a b ⋅ ，再根据运算法则展开计算即可；（2）先计算2b，再平方，进而开方即可.【小问1详解】因为22||1,||||cos 451122a a a b a b ==⋅=︒=⨯=所以2()112a b a a a b ++⋅=⋅=+=【小问2详解】因为22||2b b ==，所以2222|2|(2)444242a b a b a b a b -=-=+⋅=+--=所以|2|a b -=16.已知函数()222cos 1f x x x =+-.（1）求函数()f x 的最小正周期；（2）若3π,π4θ⎛⎫∈⎪⎝⎭且()85f θ=-，求cos 2θ的值.【答案】（1）π（2）410-【解析】【分析】（1）利用辅助角公式化简，求出最小正周期；（2）将θ代入可求出πsin 26θ⎛⎫+ ⎪⎝⎭，结合π26+θ的范围，求出πcos 26θ⎛⎫+ ⎪⎝⎭，因为ππ2266θθ=+-，由两角差的余弦公式求出结果.【小问1详解】()2π22cos 12cos 22sin 26f x x x x x x ⎛⎫=+-=+=+ ⎪⎝⎭，所以()f x 的最小正周期2ππ2T ==【小问2详解】()π82sin 265f θθ⎛⎫=+=- ⎪⎝⎭，所以π4sin 265θ⎛⎫+=- ⎪⎝⎭，因为3π,π4θ⎛⎫∈⎪⎝⎭，1π25π3663π,θ⎛⎫∈ ⎪⎝⎭+，所以π3cos 265θ⎛⎫+== ⎪⎝⎭，所以ππππππcos 2cos 2cos 2cos sin 2sin 666666θθθθ⎛⎫⎛⎫⎛⎫=+-=+++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭3414525210-⎛⎫=⨯+-⨯=⎪⎝⎭.17.如图，在ABC 中，6AB =，60ABC ∠=︒，D ，E 分别在边AB ，AC 上，且满足2AD DB = ，3CE EA =，F 为BC 中点.（1）若DE AB AC λμ=+，求实数λ，μ的值；（2）若8AF DE ⋅=-，求边BC 的长.【答案】（1）23λ=-，14μ=.（2）8【解析】【分析】（1）根据向量的线性运算以及平面向量的基本定理求得正确答案.（2）利用转化法化简8AF DE ⋅=-，从而求得BC 的长.【小问1详解】∵2AD DB = ，3CE EA= ，∴23AD AB = ，14AE AC = ∴1243DE AE AD AC AB =-=- ，∴23λ=-，14μ=.【小问2详解】12AF BF BA BC BA =-=- ，()1212154343412DE AC AB BC BA BA BC BA =-=-+=+ ，22115115241282412AF DE BC BA BC BA BC BC BA BA ⎛⎫⎛⎫⋅=-⋅+=-⋅- ⎪ ⎪⎝⎭⎝⎭设BC a = ，∵6AB = ，60ABC ∠=︒，221115668824212AF DE a a ⋅=-⨯⨯-⨯=- ，即2560a a --=，解得7a =-（舍）或8a =，∴BC 长为8.18.设(,)P x y 是角θ的终边上任意一点，其中0x ≠，0y ≠，并记r =cot x y θ=，sec r xθ=，csc r y θ=．（Ⅰ）求证222222sin cos tan cot sec +csc θθθθθθ+--+是一个定值，并求出这个定值；（Ⅱ）求函数()sin cos tan cot sec +csc f θθθθθθθ=++++的最小值．【答案】（Ⅰ）定值为3；（Ⅱ）min ()1f θ=-；【解析】【分析】（Ⅰ）由题可知，分别将6个三角函数分别代入，进行简单的化简，即可得到定值3；(Ⅱ)将()f x 中的未知量均用sin ,cos θθ来表示，得到1sin cos ()sin cos sin cos sin cos g θθθθθθθθθ+=+++，运用换元法设sin cos t θθ+=，化简成2()111g t t θ=-++-，再利用对勾函数的性质即可得到最值．【详解】解：（Ⅰ）222222222222222222sin cos tan cot sec +csc =y x y x r r r x y r y xθθθθθθ+--++--++2222222221113x y r y r x r x y+--⇒++=++=；（Ⅱ）由条件，1cot tan x y θθ==，1sec cos x θ=，1csc sin θθ=令()sin cos tan cot sec +csc g θθθθθθθ=++++sin cos 11sin cos +cos sin cos sin θθθθθθθθ=++++1sin cos sin cos sin cos sin cos θθθθθθθθ+=+++，令sin cos t θθ+=，则sin cos =2sin()4t πθθθ=++[2,2]∈-，1t ≠±，且21sin cos 2t θθ-=，从而2222()11t g y t t t θ==++--22(1)1t t t +=+-221111t t t t =+=-++--，令1u t =-，则21y u u =++，[21,21]u ∈---，且0u ≠，2u ≠-．所以，(,122][322,)y ∈-∞-⋃++∞．从而()221f y θ=≥-，即min ()221f θ=-．19.已知函数()2000ππ2sin sin 2sin 266f x x x x C ωωω⎛⎫⎛⎫=+++-+ ⎪ ⎪⎝⎭⎝⎭（R C ∈）有最大值为2，且相邻的两条对称轴的距离为π2（1）求函数()f x 的解析式，并求其对称轴方程；（2）将()f t 向右平移π6个单位，再将横坐标伸长为原来的24π倍，再将纵坐标扩大为原来的25倍，再将其向上平移60个单位，得到()g t ，则可以用函数()sin()H g t A t B ωϕ==++模型来模拟某摩天轮的座舱距离地面高度H 随时间t （单位：分钟）变化的情况．已知该摩天轮有24个座舱，游客在座舱转到离地面最近的位置进仓，若甲、乙已经坐在a ，b 两个座舱里，且a ，b 中间隔了3个座舱，如图所示，在运行一周的过程中，求两人距离地面高度差h 关于时间t 的函数解析式，并求最大值．【答案】（1）()π2sin 26f x x ⎛⎫=- ⎪⎝⎭，ππ32k x =+，Z k ∈（2）ππ()50sin 126f x t ⎛⎫=-⎪⎝⎭，50【解析】【分析】（1）由二倍角公式与两角和与差的正弦公式化简得()0π2sin 216f x x C ω⎛⎫=-++ ⎪⎝⎭，再结合最值及周期即可得解析式；（2）由正弦型函数的平移变换与伸缩变换得变换后的解析式为ππ50sin 60122y t ⎛⎫=-+ ⎪⎝⎭，则ππ50sin 126h H H ⎛⎫=-==- ⎪⎝⎭甲乙，再求最值即可.【小问1详解】()00001cos 2π22sin 2cos 2cos 2126x f x x C x x C ωωωω-=⨯++=-++0π2sin 216x C ω⎛⎫=-++ ⎪⎝⎭，所以2121C C ++=⇒=-，因为相邻两条对称轴的距离为π2，所以半周期为ππ22T T =⇒=，故002ππ12=⇒=ωω，()π2sin 26f x x ⎛⎫=- ⎪⎝⎭令ππππ2π6232k x k x -=+⇒=+，Z k ∈【小问2详解】()f t 向右平移π6得到π2sin 22y t ⎛⎫=- ⎪⎝⎭，将横坐标伸长为原来的24π倍，得到ππ2sin 122y t ⎛⎫=- ⎪⎝⎭，将纵坐标扩大为原来的25倍，得到ππ50sin 122y t ⎛⎫=- ⎪⎝⎭，再将其向上平移60个单位，得到ππ50sin 60122y t ⎛⎫=-+ ⎪⎝⎭游客甲与游客乙中间隔了3个座舱，则相隔了2ππ4243⨯=，令ππ50sin 60122H t ⎛⎫=-+ ⎪⎝⎭甲，则π5π50sin 60126H t ⎛⎫=-+ ⎪⎝⎭乙，则πππ5π50sin sin 122126h H H t t ⎛⎫⎛⎫=-=--- ⎪ ⎪⎝⎭⎝⎭甲乙π1πcos 12212t t =-ππ50sin 126t ⎛⎫=- ⎪⎝⎭，π12ω=，24T =，024t ≤≤，故πππ11π61266t -≤-≤，当πππ1262t -=或3π82t ⇒=或20时，max 50h =。
广东省汕头市潮阳实验学校2023-2024学年高一下学期第一次月考英语试题
汕头市潮阳实验学校2023-2024第二学期第一次月考试题高一英语命题人：赖泽锋审题人：刘兰芬李丽芳本试卷8页，满分120分。
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第一部分阅读（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。
AFestivals are a great way to experience a destination in a different way. Whether you are looking for a unique cultural difference or the experience of sheer joy, there is something here for everyone in this article.St. Patrick's DayDublin, Ireland & New York, the USA, March 17thIt has been a cultural and religious celebration held in memory of the death date of Saint Patrick, the foremost patron saint of Ireland, since 1762. On that day the whole city turns green and many Irish make traditional bread. It is also one of the most fun days of the year, when the whole city turns into a big green party.South by Southwest（SXSW）Austin, the USA, MarchFounded in 1987, SXSW has been praised by music fans and the media worldwide as one of the top 10 music festivals in the world. It is not only a music feast but also dedicated to the integration of technology and films. You can also enjoy free food, drinks and music. Sounds good?King's DayAmsterdam, Netherlands, April 27dhKing's Day may well be the best party in Holland and Amsterdam turns a very bright color of orange on April 27th. People celebrate King Willem-Alexander's birthday with music, street parties, flea markets, and fun fairs. The king himself travels through the country with his family.Just for LaughsMontreal, Canada, July 14th—30thFor comedy lovers there is no other festival in the world better than Montreal's Just for Laughs. The festival concentrates most of its shows in the Latin Quarter. During the day street performers delight the crowds and at night the city comes alive with comedy all over the city.1. Which of the four festivals mainly involves music and films?A. St. Patrick's Day.B. South by Southwest.C. King's Day.D. Just for Laughs.2. What do St. Patrick's Day and King's Day have in common?A. They are both religious festivals.B. They are celebrated in the same month.C. They are in honour of the birthday of a great person.D. They both feature a color.3. What can we learn about Just for Laughs?A. It lasts for a monthB. It provides free food for all.C. It's the best comedy feast.D. It is concerned about the royal family.BFor the past 13 years, Martin Burrows has been working as a long-distance truck driver. Spending up to five nights a week on the road can be a lonely business, leaving him with plenty of time to notice his surroundings. “I kept seeing more rubbish everywhere and it was getting on my nerves. I decided I had to do something about it,” he says. One day, he stopped his vehicle, took out a trash bag and started picking up the garbage. The satisfaction after clearing a small area was remarkable.Before his time on the road, Burrows spent over two decades in the military as a vehicle driver. His service saw him stationed throughout Europe and also on tours in Afghanistan. After returning to civilian life, he was diagnosed with PTSD（创伤后应激障碍）and had a mental health crisis in 2017. His involvement in fundraising for Help for Heroes led him to meet a man who used model-building as a distraction from PTSD. Burrows realized that his act of roadside cleanup had a similar calming effect on his mental well-being.By 2019, Burrows had begun using his free time on the road to regularly clean up garbage. A passerby encouraged him to set up a Facebook group, which he called Truckers Cleaning Up Britain. “I was worried I'd be the laughing stock of my town for putting videos and photos up of me cleaning but people started to join,” he says. “I was amazed. The local council stepped in and gave me litter-picking supplies and we're up to almost 3,000 members now.”Since truckers are so often on the move, the Facebook page acts as a means of raising awareness rather than a platform for organizing cleanups. Burrows expressed his intention to continue the cleanup efforts as long as his physical condition allowed, as he still found joy in the process.4. What initially caused Burrows to pick up roadside garbage?A. He wanted to kill time by picking up litter.B. He aimed to raise fund for soldiers with PTSD.C. He felt annoyed to see the increasing rubbish.D. He received the assignment from his employer.5. How did collecting roadside garbage affect Burrows' PTSD?A. It resulted in his embarrassment.B. It increased his sense of isolation.C. It worsened his stress and anxiety.D. It brought him comfort and relief.6. What concerned Burrows when he started Truckers Cleaning Up Britain?A. He feared being teased for his action.B. He was lacking in advanced cleanup tools.C. He was unsure about the group's development.D. He worried about the local council's disapproval.7. What can be a suitable title for the text?A. A Joyful V olunteer ExperienceB. A Trucker's Cleanup InitiativeC. A Fighting Hero against PTSDD. A Platform for EnvironmentalistsCWhen most people think of drones（无人机）, they think of technology and fun. Safe to say, few people would think about farming. However, a group of students from York College of Pennsylvania have been building a drone that will not only help local farmers but the environment, as well.Samantha Gotwalt and Blayde Reich, two senior Mechanical Engineering majors at York College in the group, both found the work to be quite fascinating. According to Samantha, the idea came from a York College professor, who has worked with drones, and wanted to get students involved with a project beneficial to the community. “We really want to help farming and agriculture. I's super-important to America and our economy,” Blayde says. “We want to help the smaller farmers, and one of the perks is not having to spend their money on fertilizer and pesticides（杀虫剂）.”The idea is to design and build a drone that will take video imagery of the fields to determine what is needed to produce the best crop, while saving money and sparing the environment by reducing pollutants in the water runoff . Ideally, that data gained will help the farmers better determine what chemicals they need and what they don't.However, finding the right equipment for the project was a challenge, starting with what drone the team would design for this particular usage. Samantha says she researched durability and control of drones to help make the proper determination.“We are flying over the field and we want to have enough efficiency and go relatively slow enough that our pictures turn out well and fly low enough that it is not using up all of its power,” she says. “The fields are a couple hundred acres（英亩）, so you need your drone to be able to fly the length of that field.”Blayde says the team continues to learn a great deal of information that will help the farmers and the environment.8. What does the underlined word “perks” in paragraph 2 refer to?A. challenges.B. features.C. benefits.D. solutions.9. What particular usage is the drone designed for?A. Spreading proper quantities of pesticides.B. Helping to determine the chemicals needed.C. Assisting to monitor the state of crops.D. Measuring the areas of the fields.10. What technical issue of the drone shall be tackled?A. Its camera capacity.B. Its data collection ability.C. Its durability and control.D. Its material and efficiency.11. What can best describe the students?A. Disciplined and realistic.B. Experienced and reliable.C. Humble and reserved.D. Responsible and creative.DAi-Da sits behind a desk, a paintbrush in her hand. She looks up at the person posing for her, and then back down as she applies another drop of paint onto the canvas（画布）. A lifelike portrait is taking shape. If you didn't know a robot produced it, this portrait could pass as the work of a human artist.Ai-Da produces portraits of sitting subjects using a robotic hand attached to her lifelike figure. She's also able to talk, giving detailed answers to questions about her artistic process and attitudes towards technology. She even gave a TEDx talk titled “The Intersection of Art and AI” in Oxford several years ago. Ai-Da's creators have also been experimenting with having her write and perform her own poetry.But how are we to interpret Ai-Da's output? Should we consider her paintings and poetry original and creative? Are these works actually art?What discussions about AI and creativity often overlook is the fact that creativity isn't an absolute quality that can be defined, measured and reproduced objectively. When we describe an object—for instance, a child's drawing—as being creative, we project our own assumptions about culture onto it. Indeed, art never exists in isolation. It always needs someone to give it “art” status. And the criteria for whether you think something is art are formed by both your expectations and broader cultural conceptions.If we extend this line of thinking to AI, it follows that no Al application or robot can objectively be “creative”. It is always we—humans—that decide whether works created by AI are art.Some may see robot-produced paintings as something coming from creative computers, while others may be skeptical, given the fact that robots act on clear human instructions. In any case, attribution（归属）of creativity never depends on technical arrangement alone—no computer is objectively creative. Rather, the attribution of computational creativity is largely inspired by contexts of reception. Through particular social information, some people are inspired to think of AI output as art, systems as artists, and computers as creators. Therefore, as with any piece of art, your appreciation of AI output ultimately depends on your own interpretation.12. What can we learn about Ai-Da?A. She has a complex many-sided personality.B. She beat others in the debate on art and ALC. She is capable of drawing high-quality portraits.D. She can write poems without being programmed.13. What fact do discussions about AI and creativity often ignore?A. That art is content-based.B. That art can take many forms.C. That creativity is closely related to cultures.D. That creativity is often measured subjectively.14. What idea does the author want to convey in the last paragraph?A. Every coin has two sides.B. Great minds think alike.C. Four eyes see more than two.D. Beauty is in the eye of the beholder.15. What would be the best title for the text?A. Is AI-created Art Really Art?B. Will People Accept AI Artists?C. Can We Use Al to Create Portraits?D. Do We Need to Improve AI's Creativity?第二节（共5小题；每小题2.5分，满分12.6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。
广东省汕头市潮阳启声学校2023-2024学年高一下学期第一次月考英语试题
广东省汕头市潮阳启声学校2023-2024学年高一下学期第一次月考英语试题一、阅读理解Four books that will inspire you to travel the worldThere's truly nothing like travel when it comes to gaining perspectives and exposing yourself to other cultures. To get you in the adventuring mood, we asked Amazon Senior Editor Chris Schlep to help you come up with a list of books that transport readers to another time and place. Below, see his list of four books that will inspire you to travel the world.ITAL Y: Beautiful Ruins by Jess WalterThis book by the popular author Jess Walter is a love story that begins on the Italian Coast in the early 60s and eventually concludes in contemporary Hollywood's screen. As the settings shift from Italy to Edinburgh and Los Angeles, you will find yourself longing to go as well. Buy it on Amazon. Price: $28.90SEATTLE: Where'd You Go, Bernadette by Maria SempleMaria Semple's first novel is not exactly a love story in Seattle, but if you read it, you just might want to come here to see if people are really as selfinvolved as the characters in her book. What really shines through is the strange storytelling and the laughs. Buy it on Amazon. Price: $26.60ENGLAND: Wolf Hall by Hilary MantelYou can't travel to Thomas Cromwell's England without a time machine, but reading Mantel's prizewinning novel is the next best thing. It will make you long to see the ancient buildings and green grass of the English countryside, most of which are still there. Buy it on Amazon. Price: $25.10NANTUCKET: Here's to Us by Elin HilderbrandElin Hilderbrand has built a writing career out of writing about her hometown island of Nantucket. Her latest book is Here's to Us, which, perhaps not surprisingly, is a great beach book. Buy it on Amazon. Price: $30.801．Which book has been produced into a film according to the text?A．Here's to Us.B．Wolf Hall.C．Beautiful Ruins.D．Where'd You Go, Bernadette.2．What is the feature of the Where'd You Go, Bernadette?A．Its low price.B．Its characters.C．Its content about love.D．Its storytelling and laughs.3．Why is Here's to Us suitable for reading on the beach?A．Because it's about the author’s hometown island.B．Because it needs a time machine.C．Because it's about ancient buildings.D．Because it exposes yourself to other cultures.As I walked along the Edgware Road, I felt as though the world was closing in on me. All the sounds I take for granted, had gone. I had entered a world of silence. This unsettling experience occurred a few weeks ago when I agreed to go deaf for a day to support the work of the charity Hearing Dogs for Deaf People, for which I am an ambassador.When I managed to take a cab to the office of my manager, Gavin, I couldn’t hear what the taxi driver was saying to me. Conversation was impossible. Then, when I reached the office, I had to ring the intercom (对讲机) five times as I couldn’t hear a response.Everybody said I was shouting at them ---I simply wasn’t aware of how loudly I was speaking as I couldn’t hear my own voice. Gavin kept telling me my phone was ringing, but I didn’t realize. I was too busy trying to concentrate on reading his lips. And when he tried to tell me a code to put into my phone, I had to keep asking him to repeat it, more slowly. Eventually he lost his patience and snapped at me: “Just give me the phone!” I was shocked.People couldn’t be bothered to repeat themselves, so they kept trying to do things for me that I was perfectly capable of doing myself. I felt I’d lost control.Being deaf for a day was extraordinarily tiring. I had to work so hard to “listen” with my eyes, get people’s attention and use my other senses to make up for my lack of hearing. It was a huge, exhausting effort.Until that experience, I didn’t realize how much I took my own hearing for granted, or the sorts of emotions and experiences deaf people go through. If a deaf person asks you to repeatsomething, never think: “It doesn’t matter.” It does matter.4．Why did the author focus on reading Gavin’s lips?A．He didn’t want to bother Gavin to repeat what he was saying.B．He wanted to be aware of what the code was.C．He attempted to get the code into the phone by himself.D．By doing this he could understand what Gavin was saying.5．What advice does the author give in the passage?A．Repeat things as slowly as possible for the deaf.B．Speak at the top of your voice if you can’t hear others speaking.C．Take your own hearing for granted.D．Do as many things as possible for the deaf.6．What can be inferred from the passage?A．It’s boring to live in a world of silence.B．Many ordinary people just take hearing for granted until they lose it.C．There are many other ways to help the deaf understand others.D．The author has to use gestures to communicate with his friends.7．What can be the best title of the passage?A．Listening with Eyes B．Helping the DeafC．Being Deaf for a Day D．The Importance of Reading LipsYou’ve heard an old Chinese saying before: Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime. You may even be nodding your head in agreement right now. However, we can have a different understanding about it.When a person is starving, that’s not the time to fill their head with knowledge but to first give the person a fish-eradicating their hunger-and only then teach them to fish. Far too often, people ignore this common sense first step. They see someone who is struggling, and they rush to offer wisdom. “Let me tell you what I’d do in your position,” a well-meaning individual might offer.But few of us understand the anxiety, confusion and uncertainty that come with overwhelming need. People meeting with personal disasters don’t have the ability to think straight.Their nerves may be shot. Their confidence may be non-existent. Can you imagine what it would be like to be in that person’s situation?Rushing to offer a struggling person long-term advice is a waste of time. Instead, it makes far more sense to help them regain their emotional balance. Once their ears, heart and mind open, you’ll have an opportunity to teach a new skill.But how do we know whether a person needs a fish before a fishing lesson? Two things: One is the ability to pay attention. We need to know whether the other person is open and receptive, or looking at the world through narrowed eyes? You can’t just take their words but have to look at how the person acts and what they don’t say. The other is empathy (共情).The more successful you are, the harder it is to imagine what it must be like to be the opposite. Try to create a safe environment for emotional acceptance before the fishing lesson.8．What is the function of the first paragraph?A．To illustrate a concept.B．To bring in a new viewpoint.C．To introduce the background.D．To put up an example.9．What does the underlined word “eradicating” mean in paragraph 2?A．Getting rid of.B．Paying attention to.C．Making up for.D．Putting up with.10．What should we do first for those in disasters?A．Get them to think straight.B．Enhance their confidence.C．Satisfy their primary needs.D．Teach them a new skill.11．Which does the author probably agree with according to the last paragraph?A．Live and learn.B．Never teach a fish to swim.C．Put yourself in others’ shoes.D．Don’t teach old dogs new tricks.Paper is one of our oldest, simplest and most important inventions. But it also presents a danger to the world in two significant ways. First, the making of paper requires the loss of millions of trees each year. Between 2001 and 2019, the world lost 386 million hectares of forest. Of the trees that were cut down, 42% went to paper production. And worldwide use of paper is expected to double in the next 40 years. Clearly, the planet cannot sustain such a high rate of forest loss.The second great problem with paper is what happens once it is no longer useful. A huge quantity of wastepaper ends up in dumps and landfills (垃圾填埋场), where it can produce harmful gases. Paper in landfills leads to the release (释放) of methane, a gas that is a significant contributor in global warming.One simple solution can greatly reduce both of these problems: paper recycling. Paper is mainly made from cellulose (纤维素), which makes up the cell walls of trees and many other plants. Because of its structure, cellulose can be used repeatedly in papermaking. So far, trees are the only source of cellulose that can fill the massive demand for paper products. Therefore, recycling paper is simply one of the best ways to save trees.Thanks to advances in processing, recycled paper isn’t the dull-colored stuff many of us are familiar with any more. It now can offer the same print performance as non-recycled paper.Effective recycling requires a consistent effort. The way to begin is with education and understanding. Once enough people realize the need for recycling, more effective recycling systems can be carried out. The massive loss of trees affects everybody on earth. Everyone should do their part to recycle paper and encourage government and industries to do the same. 12．What does the author want to express in paragraph 1?A．Consequences resulting from forest loss.B．The significance of paper in daily life.C．The disadvantages of current paper production.D．The severe situation caused by papermaking.13．Why is paper in landfills harmful?A．It releases smelly gases.B．It results in global climate change.C．It pollutes the nearby land.D．It may lead to fire accidents.14．What can we know about paper-recycling?A．It produces cellulose to make more paper.B．The structure of cellulose makes it possible.C．The color of recycled paper is different.D．It produces cellulose without using trees. 15．What’s the purpose of this text?A．To introduce paper recycling technology.B．To stress the threat of global warming.C．To appeal to people to recycle paper.D．To describe the considerable need for paper.You’ll make new friends in each stage of your life. Some of them will come and go, while others will last for the rest of your life. 16There are many benefits of having strong friendships. According to experts, many people regret not keeping friendships going and end up living a life with no close friends or even enduring mental and physical sufferings. 17 People who have healthy friendships tend to enjoy life more and may even live longer.18 There are a lot of ways to make new friends that can stay with you for the rest of your life. Look for people who share things with you. If you have kids, join a mom’s group, or sign your child up for classes where you’ll naturally meet other parents. 19 You share your career and will have a lot to talk about while also having built-in opportunities to spend time with each other.Sometimes it’s a good idea to let a friendship go, even if you’ve been friends for a long time. If one party isn’t making an effort to keep things going, it can lead to feelings of hurt and betrayal, and it might be time to let things cool off and pursue other friendships. 20 Besides, it can encourage you to focus on the partnerships that are healthy and where both of you are committed to keeping it going for your entire life.A．Lifelong friendships are what most people desire.B．You don’t have to be mean or harsh to get this done.C．If you are lacking in good friendships, it’s not too late.D．You can also make friends with people you work with.E．So making time for your friendships is vital to a healthy life.F．Sign up for a dancing class to meet those with the same passion.G．While this can be sad, it can actually improve your quality of life.二、完形填空The train had been long delayed. Running out of 21 , Andy put down his book and looked out. He found the 22 at once: It was raining hard.He lay down and fell asleep but was soon woken up by a woman. She handed him his bagand 23 that it had slipped to the floor. He gratefully took it back and opened it, 24 to see his mother’s scarf and some sandwiches inside.Andy's thoughts drifted (飘) to when he was 25 . His mother had insisted on putting her scarf in. “If it rains, it may get cold.” He remembered feeling 26 and had taken it out. But it was still here.27 , Andy realized he was burning with fever. Feeling helpless, he called his mom. “Take a 28 . I have put in medicine, just 29 ,” she suggested. Touched by his mother’s 30 , he took the medicine and soon fell deep asleep in the 31 of the scarf.Andy woke up later feeling much better. Then he noticed the woman, who’d 32 him pick up his bag earlier, 33 holding a baby in her arms, both shaking. Their clothes did little against the cold wind.Without thinking twice, Andy wrapped his mother’s scarf around the baby. To his 34 , the child soon fell asleep in the love of not one, but two 35 .21．A．luck B．patience C．time D．energy 22．A．train B．truth C．cause D．notice 23．A．insisted B．explained C．apologized D．admitted 24．A．surprised B．ready C．thankful D．frightened 25．A．checking B．leaving C．planning D．packing 26．A．hurt B．annoyed C．ashamed D．puzzled 27．A．Lately B．Finally C．Suddenly D．Instantly 28．A．look B．pill C．rest D．sandwich 29．A．for safety B．on purpose C．in case D．by accident 30．A．calmness B．confidence C．comfort D．concern 31．A．warmth B．memory C．smell D．touch 32．A．helped B．pleased C．disturbed D．greeted 33．A．carefully B．casually C．gently D．tightly 34．A．relief B．amazement C．mind D．advantage 35．A．arms B．scarfs C．passengers D．mothers三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
重庆市巫溪县尖山中学校2022-2023学年高一下学期第一次月考化学试题
尖山中学高2025届高一下化学月考试卷（一）姓名：班级：得分：考试范围：必修一、必修二第五章；考试时间：75分钟；命题人：向月审题人：向东可能用到的相对原子质量：C—12O—16Ca—40Cu—64I—127注意事项：1．答题前填写好自己的姓名、班级、考号等信息2．请将答案正确填写在答题卡上第I 卷（选择题）一、单选题(每小题3分，共42分)1．2022年11月29日，神舟十五号载人飞船成功发射，我国6名航天员首次实现太空会师。
下列说法错误的是A ．镁铝合金可用作飞船零部件材料B ．SiO 2可用作航天器的太阳能电池板C ．活性炭可用于吸附航天舱中异味D ．N 2H 4可用作飞船推进器的燃料2．下列除杂的操作方法，正确的是A ．NH 3中混有少量水蒸气：通过装有浓硫酸的洗气瓶B ．SiO 2中混有少量大理石：滴加盐酸充分反应后过滤C ．食盐中混有少量NH 4Cl ：加过量氢氧化钾溶液后加热蒸干D ．硝酸中混有少量硫酸：加适量BaCl 2后再过滤3．下列说法错误的是A ．SiO 2是不溶于水的酸性氧化物B ．水晶属于硅酸盐材料C ．制普通玻璃的原料主要成分是纯碱、石灰石和石英D ．陶瓷、水泥、玻璃属于传统无机非金属材料4．实验室中，下列试剂的保存方法有误．．的是A ．AgNO 3固体保存在棕色试剂瓶内B ．Na 保存在石蜡中C ．HF 溶液保存在塑料瓶中D ．浓硝酸保存在棕色广口瓶内5．下列各组中的物质在指定的分散系中能大量共存的是A ．空气：HCl 、CO 2、NO 、NH 3B ．水玻璃：K +、H +、NO 3-、Cl -C ．pH=10的溶液：SO 23-、Na +、S 2-、Ca 2+D ．氨水：Na +、NH 4+、NO 3-、Al(OH)36．硫和氮及其化合物对人类生存和社会发展意义重大，但硫氧化物和氮氧化物造成的环境问题也日益受到关注，下列说法错误的是A ．工业废气中的2SO 可采用生石灰法进行脱除B ．汽车尾气中的主要大气污染物为NO 、2SO 和 2.5PM C ．豆科植物直接吸收空气中的2N 作为肥料，实现氮的固定D ．NO 、2NO 和2SO 均为有刺激性气味的有毒气体7．化学是研究物质及其变化的科学。
2022-2023学年上海市新川中学高一年级下册学期第一次月考数学试题【含答案】
2022-2023学年上海市新川中学高一下学期第一次月考数学试题一、填空题1．的终边经过点,则的正切值为________.α()5,12-α【答案】125-【分析】直接根据正切函数的广义定义带入即可算出.【详解】.1212tan 55y x α-===-故答案为: .125-2．已知是第二象限角，，则________．α1sin 3α=πsin 2α⎛⎫+=⎪⎝⎭【答案】【分析】根据诱导公式，结合同角的三角函数关系式进行求解即可.【详解】因为是第二象限角，，α1sin 3α=所以πsin cos 2αα⎛⎫+==== ⎪⎝⎭故答案为：3．已知角终边上一点，则________．α()2,3P -()()πcos sin π23πcos πcot 2αααα⎛⎫+- ⎪⎝⎭=⎛⎫++ ⎪⎝⎭【答案】【分析】根据三角函数定义及诱导公式化简即可得解.【详解】由诱导公式知，，()()πcos sin πsin sin 2sin 3πcos (tan )cos πcot 2ααααααααα⎛⎫+- ⎪-⋅⎝⎭===--⋅-⎛⎫++ ⎪⎝⎭因为角终边上一点，α()2,3P -所以sin α所以原式sin α=-=故答案为：4化成的形式___________.cos x x -sin()(0,02)A x A ϕϕπ+>≤<【答案】112sin 6x π⎛⎫+ ⎪⎝⎭，再由诱导公式将其转化为cos 2sin(6x x x π-=-的形式即可.sin()(0,02)A x A ϕϕπ+>≤<，1cos cos )2(sin cos cos sin 2sin()2666x x x x x x x πππ-=-=-=-.112sin()2sin[2(2sin()666x x x ππππ-=+-=+故答案为：.112sin()6x π+5．化简________．()()()()sin 70cos 10cos 70sin 170αααα︒+︒+-︒+︒-=【分析】根据诱导公式以及两角和的正弦公式进行化简，即可求得答案.【详解】由题意可得()()()()sin 70cos 10cos 70sin 170αααα︒+︒+-︒+︒-()()()()sin 70cos 10cos 70sin 10αααα=︒+︒+-︒+︒+()()7010]sin 6sin[0αα︒+-︒+=︒==6．若，则_______________．1cos()3αβ-=22(sin sin )(cos cos )αβαβ+++=【答案】83【解析】原式展开，利用、两角差的余弦公式，化简整理，即可得答案.22sin cos 1αα+=【详解】222222(sin sin )(cos cos )sin +sin 2sin sin cos cos 2cos cos αβαβαβαβαβαβ+++=++++=.22sin sin 2cos 282cos()2323cos αβαβαβ++=+-=+=故答案为：83【点睛】本题考查同角三角函数的关系，两角差的余弦公式，考查计算化简的能力，属基础题.7．已知，,则________.2tan()5αβ+=1tan()44πβ-=tan()4πα+=【答案】322【分析】由，再结合两角差的正切公式求解即可.()()44ππααββ+=+--【详解】解：因为，,2tan()5αβ+=1tan()44πβ-=又，()()44ππααββ+=+--所以=，tan()tan()4tan()tan[()()]441tan()tan()4παββππααββπαββ+--+=+--=++-213542122154-=+⨯故答案为.322【点睛】本题考查了两角差的正切公式及考查了角的拼凑，重点考查了观()()44ππααββ+=+--察能力及运算能力，属中档题.8．已知则________．1sin cos 3αα+=2πcos 4α⎛⎫-=⎪⎝⎭【答案】118【分析】由两角差余弦公式可得，结合条件可求.πππcos cos cos sin sin444ααα⎛⎫-=+ ⎪⎝⎭2πcos 4α⎛⎫- ⎪⎝⎭【详解】因为πππcos cos cos sin sin444ααα⎛⎫-=+ ⎪⎝⎭所以，)πcos cos sin 4ααα⎛⎫-+ ⎪⎝⎭又，1sin cos 3αα+=所以，2π111cos 42918α⎛⎫-=⨯=⎪⎝⎭故答案为：.1189．中，，，________．ABC 60A ∠=︒75C ∠=︒a =ABC S = 【分析】根据正弦定理可求得c ，再求出B ，根据三角形面积公式即可求得答案.【详解】因为sin 75sin(4530)sin 45cos30cos 45sin 30︒=︒+=︒+︒在中，由正弦定理可得，ABC sin ,sin sin sin a c a C c A C A =∴===因为，，故，60A ∠=︒75C ∠=︒45B ∠=︒所以，11sin 22ABC S ac B ===10．边长为10，14，16的三角形中最大角与最小角的和为________．【答案】##2π3120【分析】利用余弦定理求得最大角与最小角的和的补角即可.【详解】解：设边长为10，14，16分别对应边a ，b ，c ，由余弦定理得：,2222221016141cos 2210162a c b B ac +-+-===⨯⨯因为，()0,B π∈所以，则，3B π=23A C π+=故三角形中最大角与最小角的和为，2π3故答案为：2π311．在中，边，，则角的取值范围是________________.ABC ∆2BC =AB C 【答案】0,3π⎛⎤ ⎝⎦【分析】利用余弦定理构建方程，利用判别式可得不等式，从而可求角的取值范围．C 【详解】由题意，设，由余弦定理得，AC b =2222cos AB AC BC AC BC C =+-⋅⋅即，即，，2344cos b b C =+-24cos 10b b C -+=216cos 40C ∴∆=-≥或，1cos 2C ∴≥1cos 2C ≤-，不可能为钝角，则，AB BC < C ∴1cos 2C ≥又，.0C >03C π∴<≤因此，角的取值范围是.C 0,3π⎛⎤ ⎥⎝⎦故答案为：.0,3π⎛⎤ ⎥⎝⎦【点睛】本题考查余弦定理的运用，考查解不等式，解题的关键是利用余弦定理构建方程，利用判别式得不等式，属于中等题.12．已知，存在实数，使得对任意，总成立，则的最小值是0θ>ϕn N *∈()cos cos8n πθϕ+<θ______．【答案】27π【分析】作出单位圆，根据终边位置可得；结合，即可求得最n θϕ+4πθ>2N πθ*∈()2k N k πθ*=∈小值.【详解】作出单位圆如图所示，由题意知：的终边需落在图中阴影部分区域，n θϕ+，即，()()188n n ππθϕθϕθ⎛⎫∴++-+=>--⎡⎤ ⎪⎣⎦⎝⎭4πθ>对任意，总成立，，即，n N *∈()cos cos 8n πθϕ+<2N πθ*∴∈()2k N k πθ*=∈又，，.4πθ>1,2,3,4,5,6,7k ∴=min 27πθ∴=故答案为：.27π【点睛】关键点点睛：本题考查三角函数中的恒成立问题的求解，解题关键是能够根据三角函数定义，结合单位圆，确定角的终边的位置，进而利用位置关系构造不等式求得所求变量所满足的范围.二、单选题13．下列命题中，正确的是（ ）A ．第二象限角大于第一象限角；B ．若是角终边上一点，则()(),20P a a a ≠αsin α=C．若，则、的终边相同；sin sin αβ=αβD ．．tan x =ππ,Z 3x x k k ⎧⎫=-∈⎨⎬⎩⎭【答案】D【分析】取特例可判断AC ，根据三角函数的定义判断B ，利用周期解出三角方程的解集判断D.【详解】因为象限角不能比较大小，如是第二象限角，是第一象限角，故A 错误；100α=︒400β=︒因为是角终边上一点，所以，()(),20P a a a ≠α|r a==所以B 错误；sin α==当时，满足，但、的终边不相同，故C 错误；π2π,33αβ==sin sin αβ=αβ当上的解为，故在定义域上的解为，tan x =ππ(,)22-π3-ππ,Z 3x x k k ⎧⎫=-∈⎨⎬⎩⎭故D 正确.故选：D14．化简 ）A ．B ．C ．D ．2sin 22sin 2-2sin 24cos 2-2sin 24cos2-+【答案】C【分析】根据正弦、余弦的二倍角公式即可求解．【详解】又2sin 2cos 22cos 2==-+因为，所以，即原式22ππ<<sin 20,cos 20><2sin 24cos 2=- 故选C【点睛】本题考查正弦、余弦的二倍角公式，属于基础题．15．中，设，则的形状为（ ）ABC 21cos cos cos 2CA B -=ABC A ．直角三角形B ．锐角三角形C ．等腰三角形D ．钝角三角形【答案】C 【分析】先将降幂扩角,再将利用诱导公式换成,再利用和角公式展开即可2cos 2Ccos C ()cos A B -+得出结论.【详解】由得21cos cos cos 2C A B -=1cos 1cos cos 2CA B +-=整理得,因为,12cos cos cos A B C -=πA B C ++=所以()()cos cos πcos cos cos sin sin C A B A B A B A B=-+=-+=-+⎡⎤⎣⎦所以12cos cos cos cos sin sin A B A B A B -=-+所以()1cos cos sin sin cos A B A B A B =+=-又因为,所以,即.(),0,πA B ∈0A B -=A B =所以为等腰三角形.ABC 故选:C.16．设a ，，，若对任意实数x 都有，则满足条件的有R b ∈[)0,2πc ∈()π2sin 3sin 3x a bx c ⎛⎫-=+ ⎪⎝⎭序实数组的组数为（ ）()a b c ，，A ．1组；B ．2组；C ．4组；D ．无数组．【答案】C【分析】由题意得出，，然后对、的取值进行分类讨论，结合题中等式求出的值，3b =2=a a b c 即可得出正确选项.【详解】由题意知，函数与函数的最大值相等，最小值也相等，2sin 3π3y x ⎛⎫=- ⎪⎝⎭()sin y a bx c =+则，2=a 函数与函数的最小正周期相等，则，2sin 3π3y x ⎛⎫=- ⎪⎝⎭()sin y a bx c =+3b =当，时，由于，则，2a =3b =()2sin 32sin 33πx x c ⎛⎫-=+ ⎪⎝⎭()π2πZ 3c k k =-+∈由于，此时，；02πc ≤<5π3c =当，时，，2a =3b =-()()2sin 32sin 32sin 33πx x c x c π⎛⎫-=-+=-+ ⎪⎝⎭则，得，，此时，；()ππ2πZ 3c k k -=-∈()4π2πZ 3c k k =-∈02πc ≤< 4π3c =当，时，，2a =-3b =()()2sin 32sin 32sin 33πx x c x c π⎛⎫-=-+=++ ⎪⎝⎭则，得，，则；()ππ2πZ 3c k k +=-∈()()213c k k Z ππ=--∈02c π≤< 23c π=当，时，，2a =-3b =-()()π2sin 32sin 32sin 33x x c x c ⎛⎫-=--+=- ⎪⎝⎭则，得，，则.()π2πZ 3c k k -=-∈()π2πZ 3c k k =-∈02πc ≤< π3c =因此，满足条件的有序实数组的组数为组.()a b c ，，4故选：C ．三、解答题17．已知，，都是锐角，求的值．cos αsin βαβαβ+【答案】π4αβ+=【分析】利用同角三角函数的基本关系求得，的值，再利用两角和的余弦公式求出sin αcos β的值，可得的值.()cos αβ+αβ+【详解】因为，cos α=sin β=αβ所以sin α==cos β==所以()cos cos cos sin sin αβαβαβ+=-==因，为都是锐角，所以，．所以，αβπ02α<<π02β<<0παβ<+<所以．π4αβ+=18．证明：()sin 211tan 1sin 2cos 212θθθθ+=+++【答案】证明见解析【分析】根据二倍角公式以及同角三角函数之间的基本关系即可得出证明.【详解】证明：由二倍角公式以及可得，22sin 22sin cos cos 2cos sin θθθθθθ==-，22sin cos 1θθ+=222sin 212sin cos sin cos sin 2cos 212sin cos 2cos θθθθθθθθθθ+++=+++()()2sin cos sin cos 2cos sin cos 2cos θθθθθθθθ++==+1sin cos 2cos cos θθθθ⎛⎫=+ ⎪⎝⎭()1tan 12θ=+得证．19．设点P 是以原点为圆心的单位圆上的一个动点，它从初始位置出发，沿单位圆按顺时()01,0P 针方向转动角后到达点，然后继续沿着单位圆按顺时针方向转动角到达点，若π02αα⎛⎫<< ⎪⎝⎭1P π32P点的纵坐标为，求点的坐标．2P 35-1P【答案】【分析】由三角函数的定义可得，利用两角差的正弦、余弦公式可求得、π3sin 35α⎛⎫--=-⎪⎝⎭sin α的值，即可得出点的坐标.cos α1P 【详解】由三角函数的定义可知，点的纵坐标为，即，2P π3sin 35α⎛⎫--=-⎪⎝⎭π3sin 35α⎛⎫-+=- ⎪⎝⎭故．因为，则，π3sin 35α⎛⎫+= ⎪⎝⎭π02α<<ππ5π336α<+<若，不符合题意；πππ332α<+<πsin 13α⎛⎫<+< ⎪⎝⎭若，则，符合题意．ππ5π236α≤+<1πsin 123α⎛⎫<+≤⎪⎝⎭故．所以．ππ5π236α≤+<π4cos 35α⎛⎫+==-⎪⎝⎭所以ππ1ππcos cos cos 33233αααα⎡⎤⎛⎫⎛⎫⎛⎫=+-=+++=⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦．ππ1ππsin sin sin 33233αααα⎡⎤⎛⎫⎛⎫⎛⎫=+-=+-+=⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦而()cos cos αα-==()sinsin αα-=-=所以点的坐标为．1P 20．在中，角A ，B ，C 对应边为a ，b ，c ，其中．ABC 2b =(1)若，且，求边长c ；120A C +=︒2a c =(2)若，求的面积．15,sin A C a A =︒-=ABC ABC S 【答案】(2)3【分析】（1）利用正弦定理以及三角恒等变换的知识求得.c （2）利用正弦定理、两角和的正弦公式以及三角形的面积公式求得正确答案.【详解】（1）依题意，，2a c =由正弦定理得，即，sin 2sin A C =()sin 1202sin C C︒-=，1sin 2sin ,tan 2C C C C +==由于，所以，则，0120C ︒<<︒30C =︒90,60A B =︒=︒由正弦定理得.sin ,sin sin sin c b b Cc C B B====（2）依题意，，sin a A =由正弦定理得，sin sin A C A =由于，，所以，15180A ︒<<︒sin 0A>sin C =由于，所以为锐角，所以，150A C -=︒>C 45C =︒则，60,75A B =︒=︒()sin 75sin 4530sin 45cos30cos 45sin 30︒=︒+︒=︒︒+︒︒=由正弦定理得，sin ,sin sin sin c b b Cc C B B====)21==所以.)11sin 221322ABC S bc A ==⨯⨯=△21．在某海滨城市附近海面有一台风，据监测，当前台风中心位于城市（如图）的东偏南O 方向300千米的海面处，并以20千米/时的速度向西偏北45°方向移动，台风侵袭(cos θθ=P的范围为圆形区域，当前半径为60千米，并以10千米/时的速度不断增大，问几个小时后该城市开始受到台风的侵袭？受到台风的侵袭的时间有多少小时？【答案】12小时后该城市开始受到台风侵袭，受到台风的侵袭的时间有12小时.【分析】设经过小时台风中心移动到点时，台风边沿恰好在城，由题意得，t Q O，在中，300,20,r()6010OP PQ t OQ t t ====+cos 45a θθ==-︒4sin 5a θ==POQ ∆由余弦定理得：.2222cos OQ OP PQ OP PQ a =+-⋅【详解】解：设经过小时台风中心移动到点时，台风边沿恰好在城，t Q O 由题意得，300,20,r()6010OP PQ t OQ t t====+cos 45a θθ==-︒4sin 5a θ∴==由余弦定理得：2222cos OQ OP PQ OP PQ a=+-⋅即2224(6010)300(20)230020t 5t t +=+-⨯⨯⨯即2362880t t -+=解得，1212,24t t ==2112t t -=答：12小时后该城市开始受到台风侵袭，受到台风的侵袭的时间有12小时.【点睛】本题主要考查了余弦定理在实际生活中的应用，需熟记定理内容，属于基础题.。
2023-2024学年福建省莆田第十五中学高一下学期第一次月考英语试题
2023-2024学年福建省莆田第十五中学高一下学期第一次月考英语试题1. What did Karen do last night?A．She stayed at home. B．She went to a party. C．She saw a movie.2. What is the probable relationship between Mary and the woman?A．Strangers. B．Former schoolmates. C．Employer and employee.3. When will the woman probably go to Chicago?A．Today. B．Tomorrow. C．The day after tomorrow.4. Where are the speakers?A．At a store. B．At the airport. C．At the post office.5. What are the speakers talking about?A．A kind of food. B．A history lesson. C．An ancient dynasty.听下面一段较长对话，回答以下小题。
6. What does the woman want Peter to do?A．Drive her to the airport. B．Help her with the bag. C．Call a taxi for her.7. At what time will t he woman’s flight take off?A．2:00 pm. B．3:00 pm. C．4:00 pm.听下面一段较长对话，回答以下小题。
8. How did the speakers feel about visiting the park?A．Tired. B．Happy. C．Disappointed.9. What does Amy plan to do tomorrow?A．Study for an exam. B．Attend a party. C．Go to the movies.10. What kind of movie will the speakers watch?A．A romantic movie. B．A comedy. C．A horror movie.听下面一段较长对话，回答以下小题。
高一下学期第一次月考(物理)试题含答案
高一下学期第一次月考（物理）（考试总分：100 分）一、单选题（本题共计9小题，总分37分）1.（4分）列车在通过桥梁、隧道的时候，要提前减速。
假设列车的减速过程可看作匀减速直线运动，下列与其运动相关的物理量(位移x、加速度a、速度v、动能E k)随时间t变化的图像，能正确反映其规律的是( )2.（4分）2010年诺贝尔物理学奖授予英国曼彻斯特大学科学家安德烈·海姆和康斯坦丁·诺沃肖洛夫，以表彰他们在石墨烯材料方面的卓越研究。
石墨烯是目前世界上已知的强度最高的材料，它的发现使“太空电梯”缆线的制造成为可能，人类将有望通过“太空电梯”进入太空。
现假设有一“太空电梯”悬在赤道上空某处，相对地球静止，如图所示，那么关于“太空电梯”，下列说法正确的是( )A．“太空电梯”各点均处于完全失重状态B．“太空电梯”各点运行周期随高度增大而增大C．“太空电梯”上各点线速度与该点离地球球心距离的开方成反比D．“太空电梯”上各点线速度与该点离地球球心距离成正比3.（4分）如图所示，一铁球用细线悬挂于天花板上，静止垂在桌子的边缘，细线穿过一光盘的中间孔，手推光盘在桌面上平移，光盘带动细线紧贴着桌子的边缘以水平速度v匀速运动，当光盘由A位置运动到图中虚线所示的B位置时，细线与竖直方向的夹角为θ，此时铁球( )A．竖直方向速度大小为v cosθB．竖直方向速度大小为v sinθC．竖直方向速度大小为v tanθD．相对于地面速度大小为v4.（4分）如图所示，小球从静止开始沿光滑曲面轨道AB滑下，从B端水平飞出，撞击到一个与地面呈θ＝37°的斜面上，撞击点为C。
已知斜面上端与曲面末端B相连。
若AB的高度差为h，BC间的高度差为H，则h与H的比值等于(不计空气阻力，sin37°＝0.6，cos37°＝0.8)( )A．34B．43C．49D．945.（4分）浙江省诸暨陈蔡镇是我省有名的板栗产地。