广东省汕头市潮南区2014-2015学年度高二第一学期期末普通高中教学质量检测文综试题(扫描版,无答案)

2024级高二第一学期阶段考试化学试卷可能用到的相对原子质量：H:1 C:12 N:14 O:16 F:19 Co:59一、单选题（本题包括16小题，每小题3分，共48分。

请将答案写在答题卷上。

） 1．当今世界面临日益加剧的能源危机，下列关于能源的描述不正确的是A ．氢气的优点是燃烧热值高，资源丰富B ．乙醇属于不行再生能源，可用作燃料C ．提高燃料利用效率是解决能源危机的方向D ．应开发太阳能、风能、生物质能等新能源 2．下列说法中正确的是A ．()()()32CaCO s =CaO s +CO g 在室温下不能自发进行，说明该反应的H 0∆>B ．-10℃的水结成冰，可用熵判据来说明反应的自发性C ．能自发进行的反应确定能快速发生D ．H 0∆<、ΔS>0的反应在低温时确定不能自发进行 3．下列各组热化学方程式中，化学反应的∆H 前者小于后者的是①C(s)+O 2(g)=CO 2(g)；∆H 1 C(s)+ 12O 2(g)=CO(g)；∆H 2 ②S(s)+O 2(g)=SO 2(g)；∆H 3 S(g)+O 2(g)=SO 2(g)；∆H 4 ③HCl(aq)+NaOH(aq)=NaCl(aq)+H 2O(l)；∆H 5 CH 3COOH(aq)+KOH(aq)=CH 3COOK(aq)+H 2O(l)；∆H 6④CaCO 3(s)=CaO(s)+CO 2(g)；∆H 7 CaO(s)+H 2O(l)=Ca(OH)2(s)；∆H 8 A ．①④B．①C．②③④D．①③ 4．下列说法正确的是A ．1 mol 盐酸与1 mol Mg(OH)2完全中和所放出的热量为57. 3kJ·mol -1B ．25℃、101 kPa 时，1 mol H 2和2 mol H 2燃烧生成液态水，求出的燃烧热相等C ．CO 是不稳定的氧化物，能接着和氧气反应生成稳定的CO 2，所以C 燃烧生成CO 确定是吸热反应D ．25℃时， 1 mol 甲烷燃烧所放出的热量为甲烷的燃烧热 5．设N A 为阿伏加德罗常数的值，下列说法正确的是A ．标准状况下，2.24LSO 3含分子数目为0.1N AB ．120gNaHSO 4固体含有H +的数目为N AC ．由于存在可逆反应2NO 2⇌N 2O 4，4.6gNO 2气体所含N 原子的数目小于0.1N AD ．标准状况下，11.2L 甲烷和乙烯混合物中含氢原子数目为2N A6．汽车尾气处理原理：2NO(g)+2CO(g)⇌N 2(g)+2CO 2(g) ΔH＜0，下列措施不利提高处理效率的是A ．提高尾气流速B ．运用更高效催化剂C ．增设尾气循环处理系统D ．运用无铅汽油，防止催化剂“中毒”7．某反应加入催化剂后，反应历程变成两个基元反应，相关能量变更如图所示(E 为正值，单位：1kJ mol -⋅)。

广东省汕头市潮阳区高中2024年高三质检（四）数学试题试卷注意事项1．考试结束后，请将本试卷和答题卡一并交回．2．答题前，请务必将自己的姓名、准考证号用0．5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置． 3．请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符．4．作答选择题，必须用2B 铅笔将答题卡上对应选项的方框涂满、涂黑；如需改动，请用橡皮擦干净后，再选涂其他答案．作答非选择题，必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答，在其他位置作答一律无效． 5．如需作图，须用2B 铅笔绘、写清楚，线条、符号等须加黑、加粗．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．已知斜率为2的直线l 过抛物线C ：22(0)y px p =>的焦点F ，且与抛物线交于A ，B 两点，若线段AB 的中点M 的纵坐标为1，则p ＝（ ） A ．1B ．2C ．2D ．42．如果实数x y 、满足条件10{1010x y y x y -+≥+≥++≤，那么2x y -的最大值为（ ）A ．2B ．1C ．2-D ．3-3．已知x ，y 满足2y x x y x a ≥⎧⎪+≤⎨⎪≥⎩，且2z x y =+的最大值是最小值的4倍，则a 的值是（ ）A ．4B ．34C ．211D ．144．执行下面的程序框图，若输出的S 的值为63，则判断框中可以填入的关于i 的判断条件是（ ）A ．5i ≤B ．6i ≤C ．7i ≤D ．8i ≤5．设等比数列{}n a 的前n 项和为n S ，则“10a <”是“20210S <”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．对某两名高三学生在连续9次数学测试中的成绩（单位：分）进行统计得到折线图，下面是关于这两位同学的数学成绩分析．①甲同学的成绩折线图具有较好的对称性，故平均成绩为130分； ②根据甲同学成绩折线图提供的数据进行统计，估计该同学平均成绩在区间内；③乙同学的数学成绩与测试次号具有比较明显的线性相关性，且为正相关； ④乙同学连续九次测验成绩每一次均有明显进步． 其中正确的个数为（ ） A ．B ．C ．D ．7．某几何体的三视图如图所示，其中正视图是边长为4的正三角形，俯视图是由边长为4的正三角形和一个半圆构成，则该几何体的体积为（ ）A ．4383π+B ．2383π+C ．343π+D ．8343π+8．已知平面向量()4,2a →=，(),3b x →=，//a b →→，则实数x 的值等于（ ） A ．6B ．1C ．32D ．32-9．如图，ABC ∆内接于圆O ，AB 是圆O 的直径，,//,,,DC BE DC BE DC CB DC CA =⊥⊥22AB EB ==，则三棱锥E ABC -体积的最大值为（ ）A ．14B ．13C ．12D ．2310．已知ABC 的内角A 、B 、C 的对边分别为a 、b 、c ，且60A =︒，3b =，AD 为BC 边上的中线，若72AD =，则ABC 的面积为（ ） A ．2534B ．1534C ．154D ．353411．历史上有不少数学家都对圆周率作过研究，第一个用科学方法寻求圆周率数值的人是阿基米德，他用圆内接和外切正多边形的周长确定圆周长的上下界，开创了圆周率计算的几何方法，而中国数学家刘徽只用圆内接正多边形就求得π的近似值，他的方法被后人称为割圆术．近代无穷乘积式、无穷连分数、无穷级数等各种π值的表达式纷纷出现，使得π值的计算精度也迅速增加．华理斯在1655年求出一个公式：π2244662133557⨯⨯⨯⨯⨯⨯=⨯⨯⨯⨯⨯⨯，根据该公式绘制出了估计圆周率π的近似值的程序框图，如下图所示，执行该程序框图，已知输出的 2.8T >，若判断框内填入的条件为?k m ≥，则正整数m 的最小值是A ．2B ．3C ．4D ．512． “8πϕ=-”是“函数()sin(3)f x x ϕ=+的图象关于直线8x π=-对称”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二、填空题：本题共4小题，每小题5分，共20分。

成田高级中学2013—2014学年度第二学期高二历史（文）质量监测（二）一、选择题 (本大题共有24小题，每小题3分，共72分。

)1．从图一到图二的变化，主要反映出的信息是：图一 图二A ．中央加强对地方的控制B ．专制主义中央集权制度达到顶峰C ．地方对中央的威胁已经消除D ．分封现象至此消除 2.汉高祖刘邦称帝后，尊旧礼，“五日一朝太公（刘邦父亲），如家人父子礼。

”后来其父以尊礼待刘邦，口中还念念有词，“帝，人主也，奈何以我乱天下法。

”这里的“天下法”是指 A ．分封制 B ．专制主义中央集权制 C ．宗法制 D ．皇位世袭制3．改革是推动社会进步和发展的重要手段，不同历史阶段的改革虽然具有不同的时代特征，但也呈现出相同之处，能够体现商鞅变法与明治维新相同之处的有 ①改革地方行政制度 ②打破原有的等级制度 ③重视发展教育 ④建立一支有战斗力的军队A ．①②③B ．②③④C ．①②④D ．①③④4.有关宋代的描述，与史实不符的是A ．地方割据问题得以解决B ．四大发明同时造福人民C ．海上丝绸之路兴盛D ．心学尚未出现5.《君主论》作者、意大利政治家马基雅维利是一位极有争议的思想家，面对道德与利益、道德与政治的冲突，他提出“为政必须会玩弄权术”的思想。

以下思想家，哪一位的思想更接近于马基雅维利的思想A.荀子B.墨子C.韩非D.董仲舒6. 形成“一粥一饭，当思来处不易；半丝半缕，恒念物力维艰。

”观念的社会背景是 A ．商品流通不畅 B ．闭关锁国政策推行 C ．小农经济盛行 D ．传统手工业产量不高7. 1975年湖北云梦发现《睡虎地秦墓竹简》，内有秦律187条，其中一条规定“有贼杀伤人冲术,偕旁人不援,百步中比野,当赀二甲”。

此条文主要体现的历史信息是 A.秦律规定见义勇为是义务 B.秦律法严苛C.秦律规定有违社会道德D.秦律鼓励百姓和睦相处 8. “儒教之所最缺点者，在专为君说法，而不为民说法。

广东省部分学校2024-2025学年高二上学期第一次联考数学试卷学校:___________姓名：___________班级：___________考号：___________一、单选题1．已知()()2,1,3,1,1,1a b =-=- ，若()a a b λ⊥-，则实数λ的值为（）A ．2-B ．143-C ．73D ．22．P 是被长为1的正方体1111ABCD A B C D -的底面1111D C B A 上一点，则1PA PC ⋅的取值范围是（）A ．11,4⎡⎤--⎢⎥⎣⎦B ．1,02⎡⎤-⎢⎥⎣⎦C ．1,04⎡⎤-⎢⎥⎣⎦D ．11,42⎡⎤--⎢⎥⎣⎦3．已知向量()4,3,2a =- ，()2,1,1b = ，则a 在向量b上的投影向量为（）A ．333,,22⎛⎫ ⎪⎝⎭B ．333,,244⎛⎫ ⎪⎝⎭C ．333,,422⎛⎫ ⎪⎝⎭D ．()4,2,24．在棱长为2的正方体1111ABCD A B C D -中，E ，F 分别为棱1AA ，1BB 的中点，G 为棱11A B 上的一点，且()102A G λλ=<<，则点G 到平面1D EF 的距离为（）AB C ．3D 5．已知四棱锥P ABCD -，底面ABCD 为平行四边形，,M N 分别为棱,BC PD 上的点，13CM CB =，PN ND =，设AB a =，AD b =，AP c = ，则向量MN 用{},,a b c 为基底表示为（）A ．1132a b c++B ．1162a b c-++C ．1132a b c -+D ．1162a b c--+ 6．在四面体OABC 中，空间的一点M 满足1146OM OA OC λ=++ ．若,,MA MB MC共面，则λ=（）A ．12B ．13C ．512D ．7127．已知向量()()1,21,0,2,,a t t b t t =--=，则b a - 的最小值为（）AB C D8．“长太息掩涕兮，哀民生之多艰”，端阳初夏，粽叶飘香，端午是一大中华传统节日.小玮同学在当天包了一个具有艺术感的肉粽作纪念，将粽子整体视为一个三棱锥，肉馅可近似看作它的内切球（与其四个面均相切的球，图中作为球O ）.如图：已知粽子三棱锥P ABC -中，PA PB AB AC BC ====，H 、I 、J 分别为所在棱中点，D 、E 分别为所在棱靠近P 端的三等分点，小玮同学切开后发现，沿平面CDE 或平面HIJ 切开后，截面中均恰好看不见肉馅.则肉馅与整个粽子体积的比为（）.A ．π9B ．π18C ．π27D ．π54二、多选题9．如图，在棱长为2的正方体1111ABCD A B C D -中，E 为1BB 的中点，F 为11A D 的中点，如图所示建立空间直角坐标系，则下列说法正确的是（）A ．13DB =B ．向量AE 与1AC uuu r 所成角的余弦值为5C ．平面AEF 的一个法向量是()4,1,2-D ．点D 到平面AEF 10．在正三棱柱111ABC A B C -中，1AB AA =，点P 满足][1([0,1,0,])1BP BC BB λμλμ=+∈∈，则下列说法正确的是（）A ．当1λ=时，点P 在棱1BB 上B ．当1μ=时，点P 到平面ABC 的距离为定值C ．当12λ=时，点P 在以11,BC B C 的中点为端点的线段上D ．当11,2λμ==时，1A B ⊥平面1AB P 11．布达佩斯的伊帕姆维泽蒂博物馆收藏的达・芬奇方砖在正六边形上画了具有视觉效果的正方体图案，如图1，把三片这样的达・芬奇方砖拼成图2的组合，这个组合再转换成图3所示的几何体.若图3中每个正方体的棱长为1，则（）A ．122CG AB AA =+ B ．直线CQ 与平面1111D C B A 所成角的正弦值为23C ．点1C 到直线CQ 的距离是3D ．异面直线CQ 与BD 三、填空题12．正三棱柱111ABC A B C -的侧棱长为2，底面边长为1，M 是BC 的中点．在直线1CC 上求一点N ，当CN 的长为时，使1⊥MN AB ．13．四棱锥P ABCD -中，PD ⊥底面ABCD ，底面ABCD 是正方形，且1PD =，3AB =，G 是ABC V 的重心，则PG 与平面PAD 所成角θ的正弦值为.14．坡屋顶是我国传统建筑造型之一，蕴含着丰富的数学元素.安装灯带可以勾勒出建筑轮那，展现造型之美.如图，某坡屋顶可视为一个五面体，其中两个面是全等的等腰梯形，两个面是全等的等腰三角形.若25m AB =，10m BC =，且等腰梯形所在平面、等腰三角形所在平面与平面ABCD 的夹角的正切值均为5，则该五面体的所有棱长之和为.四、解答题15．如图，在长方体1111ABCD A B C D -中，11,2AD AA AB ===，点E 在棱AB 上移动.(1)当点E 在棱AB 的中点时，求平面1D EC 与平面1DCD 所成的夹角的余弦值；(2)当AE 为何值时，直线1A D 与平面1D EC 所成角的正弦值最小，并求出最小值.16．如图所示，直三棱柱11ABC A B C -中，11,92,0,,CA CB BCA AA M N ︒==∠==分别是111,A B A A 的中点．(1)求BN 的长；(2)求11cos ,BA CB的值．(3)求证：BN ⊥平面1C MN ．17．如图，在四棱维P ABCD -中，平面PAD ⊥平面ABCD ，PA PD ⊥，PA PD =，AB AD ⊥，1AB =，2AD =，AC CD ==(1)求直线PB 与平面PCD 所成角的正切值；(2)在PA 上是否存在点M ，使得//BM 平面PCD ？若存在，求AMAP的值；若不存在，说明理由．18．如图1，在边长为4的菱形ABCD 中，60DAB ∠=︒，点M ，N 分别是边BC ，CD 的中点，1AC BD O ⋂=，AC MN G ⋂=．沿MN 将CMN 翻折到PMN 的位置，连接PA ，PB ，PD ，得到如图2所示的五棱锥P ABMND -．(1)在翻折过程中是否总有平面PBD ⊥平面PAG ？证明你的结论；(2)若平面PMN ⊥平面MNDB ，线段PA 上是否存在一点Q ，使得平面QDN 与平面PMN 所成Q 的位置；若不存在，请说明理由．19．如图，四棱锥P ABCD -中，四边形ABCD 是菱形，PA ⊥平面,60ABCD ABC ∠= ，11,,2PA AB E F ==分别是线段BD 和PC 上的动点，且()01BE PFBD PC λλ==<≤．(1)求证：//EF 平面PAB ；(2)求直线DF 与平面PBC 所成角的正弦值的最大值；(3)若直线AE与线段BC交于M点，AH PM于点H，求线段CH长的最小值．参考答案：题号12345678910答案C BADDDCBBCDBCD题号11答案BC1．C【分析】利用两个向量垂直的性质，数量积公式即求得λ的值．【详解】 向量()()2,1,3,1,1,1a b =-=-若()a a b λ⊥-，则2()(419)(213)0a a b a a b λλλ⋅-=-⋅=++-++=，73λ∴=．故选：C ．2．B【分析】建立空间直角坐标系，写出各点坐标，同时设点P 的坐标为(),,x y z ，用坐标运算计算出1PA PC ⋅，配方后可得其最大值和最小值，即得其取值范围．【详解】如图，以点D 为坐标原点，1,,DA DC DD 所在直线分别为,,x y z 轴，建立空间直角坐标系，则1,0,0，()10,1,1C ，设(),,P x y z ，01x ≤≤，01y ≤≤，1z =，()1,,1PA x y ∴=--- ，()1,1,0PC x y =--，()()2222111111222PA PC x x y y x x y y x y ⎛⎫⎛⎫∴⋅=----=-+-=-+--⎪ ⎪⎝⎭⎝⎭，当12x y ==时，1PA PC ⋅ 取得最小值12-，当0x =或1，0y =或1时，1PA PC ⋅取得最大值0，所以1PA PC ⋅ 的取值范围是1,02⎡⎤-⎢⎥⎣⎦.故选：B.3．A【分析】根据投影向量公式计算可得答案.【详解】向量a 在向量b上的投影向量为()()()2242312333cos ,2,1,12,1,13,,222b a b a a b b b b ⋅⨯+⨯-⎛⎫⋅⋅=⋅=⋅== ⎪⎝⎭r r rr r r r r r .故选：A.4．D【分析】建立空间直角坐标系，由点到平面的距离公式计算即可．【详解】以D 为坐标原点，DA 所在直线为x 轴，DC 所在直线为y 轴，1DD 所在直线为z 轴，建立如图所示的空间直角坐标系，则()2,,2G λ，()10,0,2D ，()2,0,1E ，()2,2,1F ，所以()12,0,1ED =- ，()0,2,0= EF ，()0,,1EG λ=．设平面1D EF 的法向量为(),,n x y z = ，则12020n ED x z n EF y ⎧⋅=-+=⎪⎨⋅==⎪⎩，取1x =，得()1,0,2n =r，所以点G 到平面1D EF的距离为EG n d n ⋅== ，故选：D ．5．D【分析】利用空间向量的线性运算结合图形计算即可.【详解】由条件易知()11113232MN MC CD DN BC BA DP AD BA AP AD =++=++=++-()11113262b ac b a b c =-+-=--+.故选：D 6．D【分析】根据给定条件，利用空间向量的共面向量定理的推论列式计算即得.【详解】在四面体OABC 中，,,OA OB OC不共面，而1146OM OA OB OC λ=++ ，则由,,MA MB MC ，得11146λ++=，所以712λ=.故选：D 7．C【分析】计算出b a -=≥ .【详解】因为()()1,21,0,2,,a t t b t t =--=，所以b a -=当0t =时，等号成立，故ba -.故选：C.8．B【分析】设1PFCF ==，易知PA PB AB AC BC =====，且23FG =，设肉馅球半径为r ，CG x =，根据中点可知P 到CF 的距离4d r =，sin 4dPFC r PF∠==，根据三角形面积公式及内切圆半径公式可得1x =，结合余弦定理可得1cos 3PFC ∠=，进而可得3PC =，sin 3PFC ∠=，可得内切球半径且可知三棱锥为正三棱锥，再根据球的体积公式及三棱锥公式分别求体积及比值.【详解】如图所示，取AB 中点为F ，PF DE G ⋂=，为方便计算，不妨设1PF CF ==，由PA PB AB AC BC ====，可知3PA PB AB AC BC =====，又D 、E 分别为所在棱靠近P 端的三等分点，则2233FG PF ==，且AB PF ⊥，AB CF ⊥、PF CF F = ，PF ，CF ⊂平面PCF ，即AB ⊥平面PCF ，又AB ⊂平面ABC ，则平面PCF ⊥平面ABC ，设肉馅球半径为r ，CG x =，由于H 、I 、J 分别为所在棱中点，且沿平面HIJ 切开后，截面中均恰好看不见肉馅，则P 到CF 的距离4d r =，sin 4d PFC r PF∠==，12414233GFC r S r =⋅⋅⋅=△，又2132GFC rS x ⎛⎫=++⋅ ⎪⎝⎭ ，解得：1x =，故22241119cos 223213CF FG CG PFC CF FG +-+-∠===⋅⋅⋅⋅，又2222111cos 21132P PF CF PC PC F F C P F C +-+⋅-∠=⋅=⋅⋅，解得PC =，sin 3PFC ∠=，所以：4sin 31rPFC ∠==，解得6r =，343V r =π=球，由以上计算可知：P ABC -为正三棱锥，故111sin 4332ABC V S d AB AC BAC r =⋅⋅=⋅⋅⋅∠⋅粽11432332627=⋅⋅⋅⋅⋅⋅=，=.故选：B.9．BCD【分析】先写出需要的点的坐标，然后利用空间向量分别计算每个选项即可.【详解】由题可知，2,0,0,()0,0,0D,()2,2,1E,()1,0,2F,()12,2,2B,()10,2,2C，所以1DB==A错误；()0,2,1AE=,()12,2,2AC=-，所以111·cos,AE ACAE ACAE AC=B正确；()0,2,1AE=,()1,0,2AF=-，记()4,1,2n=-，则0,0AE AFn n==，故,AE AFn n⊥⊥，因为AE AF A⋂=,,AE AF⊂平面AEF，所以()4,1,2n=-垂直于平面AEF，故选项C正确；B =2,0,0，所以点D到平面AEF的距离·21DA ndn===，故选项D正确；故选：BCD10．BCD【分析】对于A，由1CP BP BC BBμ==-即可判断；对于B，由[]11,0,1B P BP BB BCλλ=-=∈和11//B C平面ABC即可判断；对于C，分别取BC和11B C的中点D和E，由BP BD=+1BBμ即1DP BBμ=即可判断；对于D，先求证1A E⊥平面11BB C C，接着即可求证1B P⊥平面1A EB，进而即可求证1A B⊥平面1AB P.【详解】对于A，当1λ=时，[]1,0,1CP BP BC BBμμ=-=∈，又11CC BB=，所以1CP CCμ=即1//CP CC，又1CP CC C=，所以1C C P、、三点共线，故点P在1CC上，故A错误；对于B ，当1μ=时，[]11,0,1B P BP BB BC λλ=-=∈，又11B C BC =，所以111B P B C λ= 即111//B P B C ，又1111B B C P B = ，所以11B C P 、、三点共线，故点P 在棱11B C 上，由三棱柱性质可得11//B C 平面ABC ，所以点P 到平面ABC 的距离为定值，故B 正确；对于C ，当12λ=时，取BC 的中点11,D B C 的中点E ，所以1//DE BB 且1DE BB =，BP BD =+[]1,0,1BB μμ∈ ，即1DP BB μ= ，所以DP E D μ= 即//DP DE，又DP DE D ⋂=，所以D E P 、、三点共线，故P 在线段DE 上，故C 正确；对于D ，当11,2λμ==时，点P 为1CC 的中点，连接1,A E BE ，由题111A B C △为正三角形，所以111A E B C ⊥，又由正三棱柱性质可知11A E BB ⊥，因为1111BB B C B = ，111BB B C ⊂、平面11BB C C ，所以1A E ⊥平面11BB C C ，又1B P ⊂平面11BB C C ，所以11A E B P ⊥，因为1111B C BB CC ==，所以11B E C P =，又111π2BB E B C P ∠=∠=，所以111BB E B C P ≌，所以111B EB C PB ∠=∠，所以1111111π2PB C B EB PB C C PB ∠+∠=∠+∠=，设BE 与1B P 相交于点O ，则1π2B OE ∠=，即1BE B P ⊥，又1A E BE E = ，1A E BE ⊂、平面1A EB ，所以1B P ⊥平面1A EB ，因为1A B ⊂平面1A EB ，所以11B P A B ⊥，由正方形性质可知11A B AB ⊥，又111AB B P B = ，11B P AB ⊂、平面1AB P ，所以1A B ⊥平面1AB P ，故D 正确.故选：BCD.【点睛】思路点睛：对于求证1A B ⊥平面1AB P ，可先由111A E B C ⊥和11A E BB ⊥得1A E ⊥平面11BB C C ，从而得11A E B P ⊥，接着求证1BE B P ⊥得1B P ⊥平面1A EB ，进而11B P A B ⊥，再结合11A B AB ⊥即可得证1A B ⊥平面1AB P .11．BC【分析】A 选项，建立空间直角坐标系，写出点的坐标，得到122AB AA CG +≠ ；B 选项，求出平面的法向量，利用线面角的夹角公式求出答案；C 选项，利用空间向量点到直线距离公式进行求解；D 选项，利用异面直线夹角公式进行求解.【详解】A 选项，以A 为坐标原点，1,,DA AB AA所在直线分别为,,x y z 轴，建立空间直角坐标系，则()()()()()()10,0,0,0,1,0,0,0,1,1,1,2,0,1,2,1,1,0A B A G Q C ----，()()()110,1,1,1,1,1,1,0,0B C D --，()()()10,2,2,0,1,0,0,0,1CG AB AA =-==，则()()()1220,2,00,0,20,2,2AB AA CG +=+=≠，A 错误；B 选项，平面1111D C B A 的法向量为()0,0,1m =，()()()0,1,21,1,01,2,2CQ =---=-，设直线CQ 与平面1111D C B A 所成角的大小为θ，则2sin cos ,3CQ m CQ m CQ m θ⋅===⋅，B 正确；C 选项，()10,0,1CC =，点1C 到直线CQ 的距离为3d ==，C 正确；D 选项，()()()1,0,00,1,01,1,0BD =--=--，设异面直线CQ 与BD 所成角大小为α，则cos cos ,6CQ BD CQ BD CQ BDα⋅=====⋅，D 错误.故选：BC 12．18/0.125【分析】根据正三柱性质建立空间直角坐标系，利用向量垂直的坐标表示可得结果.【详解】取11B C 的中点为1M ，连接1,MM AM ，由正三棱柱性质可得11,,AM MM BM MM AM BM ⊥⊥⊥，因此以M 为坐标原点，以1,,AMBM MM 所在直线分别为x 轴，y 轴，z 轴建立空间直角坐标系，如下图所示：易知()11,0,0,0,,2,0,0,022A B M ⎛⎫⎛⎫ ⎪ ⎪ ⎪⎝⎭⎝⎭，设CN 的长为a ，且0a >，可得10,,2N a ⎛⎫- ⎪⎝⎭；易知1110,,,,,2222MN a AB ⎛⎫⎛⎫=-=- ⎪ ⎪ ⎪⎝⎭⎝⎭若1⊥MN AB ,则1112022MN AB a ⋅=-⨯+= ,解得18a =，所以当CN 的长为18时，使1⊥MN AB .故答案为：1813．23【分析】建立空间直角坐标系，求出平面PAD 的一个法向量m 及PG，由PG 与平面PAD 所成角θ，根据sin cos ,m PG m PG m PGθ⋅==⋅即可求解.【详解】因为PD ⊥底面ABCD ，底面ABCD 是正方形，所以,,DA DC DP 两两垂直，以D 为坐标原点，,,DA DC DP的方向分别为,,x y z 轴的正方向，建立如图所示空间直角坐标系，则()0,0,0D ，()0,0,1P ，()3,0,0A ，()3,3,0B ，()0,3,0C ，则重心()2,2,0G ，因而()2,2,1PG =- ，()3,0,0DA = ，()0,0,1DP =，设平面PAD 的一个法向量为(),,m x y z =，则300m DA x m DP z ⎧⋅==⎪⎨⋅==⎪⎩ ，令1y =则()0,1,0m = ，则22sin cos ,133m PG m PG m PG θ⋅====⨯⋅，故答案为：23.14．117m【分析】先根据线面角的定义求得5tan tan EMO EGO ∠=∠，从而依次求EO ，EG ，EB ，EF ，再把所有棱长相加即可得解．【详解】如图，过E 做EO ⊥平面ABCD ，垂足为O ，过E 分别做EG BC ⊥，EM AB ⊥，垂足分别为G ，M ，连接OG ，OM ，由题意得等腰梯形所在的面、等腰三角形所在的面与底面夹角分别为EMO ∠和EGO ∠，所以5tan tan EMO EGO ∠=∠．因为EO ⊥平面ABCD ，⊂BC 平面ABCD ，所以EO BC ⊥，因为EG BC ⊥，EO ，EG ⊂平面EOG ，EO EG E = ，所以⊥BC 平面EOG ，因为OG ⊂平面EOG ，所以BC OG ⊥，同理，OM BM ⊥，又BM BG ⊥，故四边形OMBG 是矩形，所以由10BC =得5OM =，所以EO 5OG =，所以在直角三角形EOG 中，EG =在直角三角形EBG 中，5BG OM ==，8EB ==，又因为55255515EF AB =--=--=，所有棱长之和为2252101548117⨯+⨯++⨯=．故答案为：117m15．(2)当2AE =时，直线1A D 与平面1D EC 【分析】（1）以D 为坐标原点，1,,DA DC DD 所在直线为坐标轴建立空间直角坐标系，求得平面1D EC 的一个法向量，平面1DCD 的一个法向量，利用向量法可求平面1D EC 与平面1DCD 所成的夹角的余弦值；（2）设AE m =，可求得平面1D EC 的一个法向量，直线的方向向量1DA，利用向量法可得sin θ=.【详解】（1）以D 为坐标原点，1,,DA DC DD 所在直线为坐标轴建立如图所示的空间直角坐标系，当点E 在棱AB 的中点时，则1(0,0,1),(1,1,0),(0,2,0),(0,0,0),(1,0,0)E C D A D ，则1(1,1,1),(1,1,0),(1,0,0)ED EC DA =--=-=，设平面1D EC 的一个法向量为(,,)n x y z =，则1·0·0n ED x y z n EC x y ⎧=--+=⎪⎨=-+=⎪⎩ ，令1x =，则1,2y z ==，所以平面1D EC 的一个法向量为(1,1,2)n =，又平面1DCD 的一个法向量为(1,0,0)DA =，所以·cos ,·DA n DA n DA n=== 所以平面1D EC 与平面1DCD（2）设AE m =，则11(0,0,1),(1,,0),(0,2,0),(0,0,0),(1,0,1)E m C D A D ，则11(1,,1),(1,2,0),(02),(1,0,1)ED m EC m m DA =--=--≤≤=，设平面1D EC 的一个法向量为(,,)n x y z =，则1·0·(2)0n ED x my z n EC x m y ⎧=--+=⎪⎨=-+-=⎪⎩ ，令1y =，则2,2x m z =-=，所以平面1D EC 的一个法向量为(2,1,2)n m =-，设直线1A D 与平面1D EC 所成的角为θ，则11||sin ||||n DA n DA θ===令4[2,4]m t -=∈，则sin θ=当2t =时，sin θ取得最小值，最小值为5.16．(2)10(3)证明见解析【分析】（1）建立空间直角坐标系，求出相关点坐标，根据空间两点间距离公式，即得答案；（2）根据空间向量的夹角公式，即可求得答案；（3）求出1C M ，1C N，BN 的坐标，根据空间位置关系的向量证明方法，结合线面垂直的判定定理，即可证明结论.【详解】（1）如图，建立以点O 为坐标原点，CA 、CB 、1CC 所在直线分别为x 轴、y 轴、z轴的空间直角坐标系．依题意得(0,1,0),(1,0,1)B N ，∴BN == （2）依题意得，()()()()111,0,2,0,1,0,0,0,0,0,1,2A B C B ，∴1(1,1,2)BA =- ，1(0,1,2)CB =，113BA CB =⋅，1BA1CB所以11111cos ,BA CB BA CB BA CB ⋅=⋅（3）证明：()()()10,0,2,0,1,0,1,0,1C B N ，11,,222M ⎛⎫⎪⎝⎭．∴111,,022C M ⎛⎫= ⎪⎝⎭ ，()11,0,1C N =- ，()1,1,1BN =-，∴1111(1)10022C M BN ⋅=⨯+⨯-+⨯= ，1110(1)(1)10C N BN ⋅=⨯+⨯-+-⨯=，∴1C M BN ⊥ ，1C N BN ⊥，即11,C M BN C N BN ⊥⊥，又1C M ⊂平面1C MN ，1C N ⊂平面1C MN ，111= C M C N C ，∴BN ⊥平面1C MN ．17．(2)存在点M ，使得//BM 平面PCD ，14AM AP =.【分析】（1）取AD 的中点为O ，连接,PO CO ，由面面垂直的性质定理证明⊥PO 平面ABCD ，建立空间直角坐标系求解直线PB 与平面PCD 所成角的正切值即可；（2）假设在PA 上存在点M ，使得()01PM PA λλ=≤≤，由线面平行，转化为平面的法向量与直线的方向向量垂直，求解参数即可.【详解】（1）取AD 的中点为O ，连接,PO CO ，因为PA PD =，所以PO AD ⊥，又平面PAD ⊥平面ABCD ，平面PAD ⋂平面ABCD AD =，PO ⊂平面PAD ，所以⊥PO 平面ABCD ，又AC CD =，所以CO AD ⊥，PA PD ⊥，2AD =，所以1PO =，AC CD ==2CO =，所以以O 为坐标原点，分别以,,OC OA OP 所在的直线为,,x y z 轴建立空间直角坐标系，0,0,1，()2,0,0C ，()0,1,0A ，()1,1,0B ，()0,1,0D -，所以()2,0,1PC =- ，()0,1,1PD =--，()1,1,1PB =- ，设平面PCD 的一个法向量为 =s s ，则00PC m PD m ⎧⋅=⎪⎨⋅=⎪⎩，200x z y z -=⎧⎨--=⎩，令1,x =则2,2z y ==-，所以()1,2,2m =-，设直线PB 与平面PCD 所成角为θ，sin cos ,m PB m PB m PB θ⋅====，所以cos 3θ==，所以tan θ所以直线PB 与平面PCD所成角的正切值2.（2）在PA 上存在点M ，使得()01PM PA λλ=≤≤，所以()0,1,1PA =- ，所以()0,,PM PA λλλ==-，所以()0,,1M λλ-，所以()1,1,1BM λλ=---，因为//BM 平面PCD ，所以BM m ⊥ ，即()()121210λλ---+-=，解得34λ=，所以存在点M ，使得//BM 平面PCD ，此时14AM AP =.18．(1)总有平面PBD ⊥平面PAG ，证明详见解析(2)存在，Q 是PA 的靠近P 的三等分点，理由见解析.【分析】（1）通过证明BD ⊥平面PAG 来证得平面PBD ⊥平面PAG .（2）建立空间直角坐标系，利用平面QDN 与平面PMN 所成角的余弦值来列方程，从而求得Q 点的位置.【详解】（1）折叠前，因为四边形ABCD 是菱形，所以AC BD ⊥，由于,M N 分别是边BC ，CD 的中点，所以//MN BD ，所以MN AC ⊥，折叠过程中，,,,,MN GP MN GA GP GA G GP GA ⊥⊥⋂=⊂平面PAG ，所以MN ⊥平面PAG ，所以BD ⊥平面PAG ，由于BD ⊂平面PBD ，所以平面PBD ⊥平面PAG .（2）存在，理由如下：当平面PMN ⊥平面MNDB 时，由于平面PMN 平面MNDB MN =，GP ⊂平面PMN ，GP MN ⊥，所以GP ⊥平面MNDB ，由于AG ⊂平面MNDB ，所以GP AG ⊥，由此以G 为空间坐标原点建立如图所示空间直角坐标系，依题意可知())(),2,0,,0,1,0,P D B N PB --=- ()A，(PA = ，设()01PQ PA λλ=≤≤ ，则(()(),0,3,0,GQ GP PQ GP PA λ=+=+=+-= ，平面PMN 的法向量为()11,0,0n = ，()(),DQ DN ==，设平面QDN 的法向量为()2222,,n x y z = ，则()2222222200n DQ x y z n DN y ⎧⋅=-++=⎪⎨⎪⋅=+=⎩ ，故可设()21n λλ=--+ ，设平面QDN 与平面PMN 所成角为θ，由于平面QDN 与平面PMN所成角的余弦值为13，所以1212cos n n n n θ⋅==⋅解得13λ=,所以当Q 是PA 的靠近P 的三等分点时，平面QDN 与平面PMN 所成角的余弦值为13.19．(1)证明见解析(2)8(3)5【分析】（1）根据条件建立合适的空间直角坐标系，利用空间向量证明线面关系即可；（2）利用空间向量研究线面夹角，结合二次函数的性质计算最大值即可；（3）设BM tBC = ，利用空间向量基本定理及三点共线的充要条件得出AH ，利用向量模长公式及导数研究函数的单调性计算最值即可.【详解】（1）由于四边形ABCD 是菱形，且60ABC ∠= ，取CD 中点G ，则AG CD ⊥，又PA ⊥平面ABCD ，可以A 为中心建立如图所示的空间直角坐标系，则()()()()()2,0,0,,,0,0,1,B C D P G -，所以()()()1,,2,0,1PC BD BP =-=-=- ，由()01BE PF BD PCλλ==<≤，可知,,BE BD PF PC EF EB BP PF BD BP PC λλλλ==∴=++=-++ ()42,0,1λλ=--，易知()AG = 是平面PAB 的一个法向量，显然0EF AG ⋅= ，且EF ⊄平面PAB ，即//EF 平面PAB；（2）由上可知()()()1,,DP PF DF λλλλ+==+-=+- ，设平面PBC 的一个法向量为(),,n x y z =r，则200n BP x z n PC x z ⎧⋅=-+=⎪⎨⋅=+-=⎪⎩，令1x =，则2,3z y ==，2n ⎛⎫= ⎪ ⎪⎝⎭，设直线DF 与平面PBC 所成角为α，则sin cos ,n DF n DF n DF α⋅==⋅ ，易知35λ=时，()2min 165655λλ-+=，即此时sin α取得最大值8；（3）设()(](),0,0,12,0BM t BC t t AM AB BM t ==-∈⇒=+=- ，由于,,H M P 共线，不妨设()1AH xAM x AP =+- ，易知AM AP ⊥，则有()()22010AH PM AH AM AP x AM x AP ⋅=⋅-=⇒--= ，所以22114451x t t AM ==-++ ，则()()2CH CA AH t x x =+=--- ，即()()2222454454655445t CH t t x t x t t --=-+-++=+-+ 记()(]()2450,1445t f t t t t --=∈-+，则()()()2228255445t t f t t t --+'=-+，易知22550t t -+>恒成立，所以()0f t '<，即()f t 单调递减，所以()()min 9155f t f CH ≥=-⇒==.。

广东省汕头市潮南区2024届高一化学第一学期期末监测模拟试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。

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一、选择题(共包括22个小题。

每小题均只有一个符合题意的选项）1、氢氧化铁胶体稳定存在的主要原因是A．胶粒直径小于1 nm B．胶粒不停地作无规则运动C．胶粒带正电荷D．胶粒不能通过半透膜2、在无色透明溶液中，下列各组离子能大量共存的是A．Cu2+、K＋、Cl－、NO3－B．Ag＋、Na＋、NO3－、Cl－C．Mg 2＋、Na＋、SO42－、Cl－D．Ba2＋、NH4＋、Cl－、CO32－3、关于酸、碱、盐的下列说法中错误的是()A．酸在水溶液中电离产生的阳离子全部是氢离子B．碱在水溶液中电离产生的阴离子全部是氢氧根离子C．盐在水溶液中电离一定有金属阳离子产生D．酸和盐在水溶液中电离都可能有含氧酸根离子产生4、以下是几种常用基本物理量的名称与符号的对应关系，其中不正确的是( )A．物质的量mol B．质量m C．时间t D．电流强度I5、下列实验操作正确的是（）用铂丝分别蘸取溶液，在酒精灯外焰上灼D 检验Na2CO3与K2CO3溶液烧，直接观察火焰的颜色A．A B．B C．C D．D6、某元素的原子结构示意图为，下列关于该元素的说法中，错误的是A．它的阳离子带3个单位正电荷B．它是一种金属元素C．它的阳离子有10个质子D．其原子核外有13个电子7、下列实验装置不能．．达到实验目的的是A．图1：制备并观察氢氧化亚铁B．图2：证明过氧化钠与水反应放热C．图3：验证NaHCO3和Na2CO3的热稳定性D．图4：验证酸性强弱H2SO4>H2CO3>H2SiO38、CN－为拟卤素离子，(CN)2性质和卤素类似。

出题人：李秀妮汕头市潮南区职业技术教育中心 审题人：张晓娜汕头市潮南区职业技术学校 2014—2015学年度第一学期期末考试1416、1417班《卫生保健》试题班级：______________ 姓名：____________ 座号：______ 得分：_________（考试时间90分钟）5．影响幼儿心理健康的因素有_______因素、_______因素、_______因素。

6．在日常生活中要帮助幼儿学习社会交往技能，对幼儿进行_______教育，引导幼儿多设身处地地为别人着想。

7．幼儿午睡前，教师的准备工作要做到“三要”：一要提醒幼儿_______；二要要求幼儿不做剧烈活动；三要要求幼儿安静地上床。

8．晨检的重点内容包括：一_______、二_______、三_______、四_______。

12．幼儿体温超过（ ）就属于高热。

A ．37.5℃以上B ．38℃以上C ．38.5℃以上D ．39℃以上 13．幼儿烫伤以后，急救方法是（ ）。

A ．用冷水冲淋或浸泡 B ．涂碘酒 C ．涂酱油 D ．涂清凉油 14．幼儿的手被割伤，处理方法是（ ）。

A ．直接消毒伤口 B ．直接包扎 C ．止血即可 D ．先止血，后消毒，再包扎 15．幼儿脚扭伤后的处理方法是（ ）。

A ．冷敷患处B ． 热敷患处C ．先冷敷，一天后再改用热敷D ．先热敷，一天后再改用冷敷 16．判断幼儿心理活动是否正常，必须（ ）。

A ．看幼儿的某一行为表现B ．看幼儿偶尔的行为表现C ．多方面地观察、调查和分析比较D ．轻易下结论 17．幼儿出现打人、骂人、咬人、踢人等行为属于（ ）。

A ．攻击性行为B ．多动症C ．习惯性口腔动作D ．行为障碍18．幼儿在没有言语器官损伤或智力发展障碍情况下表现出沉默不语，问之不答的现象为（ ）。

A ．口吃 B ．语言发育迟缓 C ．选择性缄默 D ．习惯性口腔动作 19．对幼儿进行性教育时应注意（ ）。

汕头金山中学2014——2015年度第一学期月考高二物理出卷者：宋青一．单选题（3分×6=18分）1．水平放置的一根直导线通有水平向右的电流，在其正上方的小磁针N极将（）A．向右平移B．向左平移C．垂直纸面向外转D．垂直纸面向里转2．如图所示的电路中，合开关S，将滑动变阻器的滑片P向右移，下列说法中正确的是（）A．电流表A的示数变小B．电压表V l的示数不变C．电压表V2的示数变小D．电压表V l与电压表V2的示数之和保持不变3．平行板电容器充电后与电源断开，当两极板距离增大时，则（）A．电容器的电容不变B．电容器极板的电量不变C．电容器极板间的电压不变D．电容器两极板间电场强度变大4．如图所示，当导线棒MN在外力作用下沿导轨向右运动时，流过R的电流方向是（）A．由A→B B．由B→A C．无感应电流D．无法确定5．图中的D为置于电磁铁两极间的一段通电直导线，电流方向垂直于纸面向里．在电键S接通后，导线D所受磁场力的方向是（）A．向上B．向下C．向左D．向右6．如图所示，在一个小的圆形区域O内有一垂直于纸面向内的匀强磁场，当磁场的磁感应强度B增加时，那么它在该区域的右侧P点感应出的电场强度的方向是（）A．在纸面内向上B．在纸面内向下C．垂直纸面向里D．垂直纸面向外二．双选题（4分×6=24分）7．下列说法正确的是（）A．安培发现了电流的磁效应B．奥斯特发现了电荷之间的相互作用规律C．法拉第发现了电磁感应定律D．楞次找到了判断感应电流方向的方法8．如图所示，图中两组曲线中实线代表电场线（方向未画出）、虚线代表等势线，点划线是一个带电粒子仅在电场力作用下从A到B的运动轨迹，下列说法正确的是（）A．粒子一定带正电B．粒子的动能一定是越来越大C．粒子的电势能一定是越来越小D．A点的电势一定高于B点的电势9．关于回旋加速器中电场和磁场作用的叙述，正确的是（）A．电场和磁场都对带电粒子起加速作用B．只有电场对带电粒子做功的C．磁场只对带电粒子起偏转作用D．带电粒子在磁场中的运动周期会随运动半径的增大而增大10．如图是质谱仪工作原理示意图．带电粒子被加速电场加速后，进入速度选择器．速度选择器内相互正交的匀强磁场和匀强电场的强度分别为B和E．平板S上有可让粒子通过的狭缝P和记录粒子位置的胶片A1A2．S下方有磁感应强度为B0的匀强磁场．下列表述正确的是（）A．质谱仪是分析同位素的重要工具B．速度选择器中的磁场方向垂直纸面向外C．速度选择器只能一种电性，且速度等于的粒子D．带电量相同的粒子打在胶片上的位置越靠近狭缝P，粒子的质量越大11．磁流体发电是一项新兴技术．如图所示，平行金属板之间有一个很强的磁场，将一束含有大量正、负带电粒子的等离子体，沿图中所示方向喷入磁场．图中虚线框部分相当于发电机．把两个极板与用电器相连，则（）A．用电器中的电流方向从A到BB．用电器中的电流方向从B到AC．若只增强磁场，发电机的电动势增大D．若只增大喷入粒子的速度，发电机的电动势增小12. 如图所示，a、b、c为三个完全相同的灯泡，L为自感线圈（自感系数较大，电阻不计），E为电源，S为开关．闭合开关S，电路稳定后，三个灯泡均能发光．则（）A．断开开关瞬间，c熄灭，稍后a、b同时熄灭B．断开开关瞬间，流过b的电流方向改变C．闭合开关，a、b、c同时亮D．闭合开关，a、b同时先亮，c后亮三．实验题（每个空2分，共18分）13．某同学使用多用电表粗略测量一定值电阻的阻值：（1）表内电池的极与红表笔相连；（2）先把选择开关旋到“×1k”挡位，测量时指针偏转如图所示。

2023-2024学年广东省汕头市潮阳实验学校高二上学期期中考试英语试题The Junction CampgroundThe Junction Campground is a great spot for river-based camping along one of Australia’s best-known rivers, near Grafton. Paddling the beautiful Nymboida River is a great way to explore the rainforest-lined riverbanks and rocky cliffs of Nymboi-Binderay National Park.Craigmhor Mountain RetreatSet in 1, 000 acres of wilderness, Craigmhor is a perfect base for bush walking, mountain biking, fishing, and bird watching. Within easy driving distance of Sydney, Craigmhor is the ideal location for overseas visitors or urban people to experience the Australian bush.The Stone CottageSaid to be Albury’s oldest house, the Stone Cotta ge warmly welcomes visitors with its exposed stone walls and an open fireplace. Set well back from the street in a spacious enclosed garden (ideal for pets), the two-bedroom self-contained house is perfect for small families and couples.Located behind the main home, the Stone Cottage’s Kitchen Cottage turns on the charm as much as the larger house. Private from the main house, the one-bedroom self-contained cottage has a veranda (游廊) which looks out over an established herb garden, ideal for couples.Instead of asking guests to wake at a certain time for a prepared breakfast, guests of both cottages receive a selection of breakfast cuisines as they like, allowing sleep-ins and lazy mornings.Pets are welcome.Country Barn RetreatSituated in a peaceful location and perfect for short stays, Elaine and John offer this amnazing self-contained two-bedroom Country Barn Retreat, overlooking rural views and surounded by lovely gardens.Sleeping up to four adults in this comfortable cottage where you can relax by the cosy wood fire in winter.Fully equipped kitchen. Lounge and dining area with air-conditioning and fans to keep it cool in summer. Pet friendly, however, you would need to make arrangements with Elaine or John first. 1. Where will you go if you’re a bird lover?A．The Stone Cottage.B．Craigmhor Mountain Retreat.C．The Junction Campground.D．Country Barn Retreat.2. What can be learned about the Stone Cottage?A．It lacks an open fireplace.B．It’s building a herb garden.C．It has three bedrooms in all.D．It provides a well-prepared breakfast.3. What’s special about the last two destinations?A．Pets are allowed.B．Fires are forbidden.C．Both have a veranda.D．Both are near rivers.I used to live in Southern Africa and recently went back to Botswana to visit the Central Kalahari Game Reserve with my friend Oscar． Why? Because there's nothing quite like the vast African wilderness...Our jeep moved slowly along the too-hot-to-touch sand． With no air conditioner and an engine which might break down at any time， it was tough work． The dry heat was fierce and the only break came from using our precious water for brief bucket showers．With no supplies available within the park—it was a sort of no man's land which was inhabited by the occasional cow — it is necessary to bring all your fuel， food and water with you．We camped in a small place in the center of the woods． At the height of the dry season， leaves in the trees had all fallen down， shade was in scant supply， so we placed ourselves near a leaf less tree． As the sun set， we drove towards the watering hole a few miles away， and were reminded just why we'd chosen to spend weeks suffering this exercise on ourselves． A lone lion paused his lapping to acknowledge our arrival． He yawned， and then carried on drinking．That night， sleeping on the roof of the jeep， we felt his shouting as much as heard it ． Those vast African skies offered the only indication that the outside world hadn't come to a state of quietness， as bright satellite whipped across the blackness．In the morning we saw that the lion had walked in the road overnight， his paw prints pressed into yesterday's tyre tracks．We set off for another morning inching through the sand; another day of changing landscapes and raw， wild beauty; another night lit by shooting stars．4. What can we learn about the author?A．He slept in a hotel at night．B．He is now living in Botswana．C．He enjoys the wildness of nature．D．He didn't take enough food and water．5. What did they think of their journey?A．Amazing and hard．B．Terrible and annoying．C．Comfortable and safe．D．Pleasant and easy．6. What does the underlined word “scant” in paragraph 4 mean?A．Inadequate．B．unbending．C．Abundant．D．Ready．7. What can be a suitable title for the text?A．Shining Stars and A Lone Lion B．Travelling in the Southern AfricaC．My Great Escape into the Wild D．Camping in Too-Hot-to-Touch Sand Jessica Damiano moved into a new home in the spring of 2005. At the time, she was delighted to see a flowering plant growing in the garden.She did not know the name of the beautiful plant. But that did not matter to her. She loved it!Two years later,Damiano graduated from Cornell University’s master gardener program. She worked as a gardening writer for a local newspaper. That is when she found out the truth. She discovered that her favorite plant was called purple loose strife and that it was considered “invasive (入侵的) ” in her home state of New York. However, she told herself that the plant was not spreading on the property and that it was not out of control.Then she learned more. Some plants known to be invasive are what Damiano calls “wolves in sheep’s clothing”. This means they may seem harmless and well-contained in the garden but become harmful in other places.Birds eat the seeds of invasive plants and spread them to other places. Those seeds grow into new plants that outgrow native plants. This is because invasive plants often are not eaten by local wildlife, which would otherwise keep them under control. If not controlled, invasive plants grow larger and push out native plants that provide food and shelter for birds,insects,and small animals. This harms the local environment.Many state environmental agencies ban the sale and use of plants found to be harmful to human or ecological health. But some invasive plants are not officially considered invasive. Others may be listed as invasive in one area but not another. And some invasive plants continue to be sold in stores. So, what is a gardener to do? Damiano says to avoid any plants sold as “vigorous”, “ fast-spreading”, “quick-climbing” or a “rapid self-sower”. She warns that these are all sellers’ code words (暗号) for invasive plants. Next, learn about your local area: Which plants are invasive? And which plants are native? Ask government agencies, universities and colleges, and environmental groups. You can also do your own research at libraries or online.In the end, Damiano replaced that plant with the native and equally beautiful blazing star plant. This plant has been growing happily without problems in her garden for the past 15 years.8. How did Damiano react when she first learned purple loose strife was“invasive”?A．She felt regretful for loving it. B．She reported it to the government.C．She didn’t take the f act seriously. D．She got rid of the plant immediately.9. Why can invasive plants grow better than native plants?A．They need less water. B．They have fewer natural enemies.C．They adapt to the environment quicker. D．They have longer roots and bigger leaves.10. What does the author mainly intend to express in paragraph 5?A．Policies about invasive plants vary in different areas.B．Some invasive plants are not harmful.C．Invasive plants are rarely seen in stores.D．The government’s attitude to invasive plants is improper.11. Which is a method the gardener can use to avoid planting invasive plants?A．Give up any plants that sell well in the local area.B．Choose plants that can provide food for wild animals.C．Refuse to say the code words of sellers for invasive plants.D．Refer to books for information on which plants are invasive.A laser-powered robotic climber has won $900,000 in a competition designed to stimulate technology for a future elevator to space.Building a space elevator would require fixing a cable on the ground near Earth’s equator and deploying (部署) the other end thousands of kilometers into space. The centrifugal (离心的) force due to Earth’s spin would keep the cable tight so that a robot could climb it and release payloads into orbit.Building a space elevator would make for cheaper trips into space than is possible using rockets. To achieve that end, NASA offered $2 million in prize money in a competition called the Power Beaming Challenge, in which robotic climbers, powered wirelessly from the ground, attempt to climb up a cable as fast as possible.Now, a robotic climber has made a prize-winning ascent worth $900,000, making it the first to win money in the competition, which has occurred annually since 2005. The winning climber was built by a team called Laser Motive, based in Seattle, Washington. Like the other two vehicles in the competition, it used solar cells to absorb energy from a ground-based infrared (红外的) laser.On Wednesday, Laser Motive fired up its laser, powering the climber to climb 900 meters up a cable. The climber reached the top in just over 4 minutes, for an average speed of 3.7meters per second. The team’s climber repeated the achievement at a slightly higher speed of 3.9 meters per second on Thursday. On Friday, two other teams failed in their final attempted climbs. That means Laser Motive will receive the entire $900,000 NASA set aside for climbers that could make the climbfaster than 2 meters per second. The remaining $1.1million in NASA prize money was reserved for climbs faster than 5 meters per second, which none of the competitors was able to achieve.Though a space elevator remains a distant prospect, NASA is thrilled about its technological advances. It is now interested in wireless power transmission for other applications, like sending power to lunar rovers travelling in shadowed craters (阴影环形山), where solar energy is unavailable.12. What do we know about the Power Beaming Challenge?A．It offered prize money to support NASA.B．It was powered wirelessly from the ground.C．It was a competition first launched in 2005.D．It was the first to win money in the competition.13. What is the best achievement that Laser Motive has got in the competition?A．3.7 meters per second. B．3.9 meters per second.C．2 meters per second. D．5 meters per second.14. Why is NASA thrilled about the technological advances of space elevator?A．Because it may stimulate unexpected applications.B．Because it demonstrates the possibility of space traveling.C．Because it can send lunar rovers to travel in shadowed craters.D．Because it can generate power when solar energy is unavailable.15. What is the best title for the text?A．“Space Elevator” Wins $900,000 NASA PrizeB．Laser Motive Unveils Its Secret TechnologyC．A Robotic Climber Rides an Elevator to SpaceD．Space Elevator Stimulates the Advance of TechnologyAll apps collect data as you use them. Following these four steps will help ensure you’re not oversharing. For starters, download apps only from the App Store, not from random websites. 16 Before you accept an app’s terms and conditions, look at what information it’s planning to collect and think twice if it’s asking for too many permissions. 17 Also, most apps don’t need to know your location, but for those that do, you can choose to enable location services only when using the particular app, another smart privacy safeguard.Sign up using an e-mail address you’ve set up just for things like app permissions and e-mail newsletters. 18 This way, if there is a security breach (缺口), your exposure is contained to things connected to that address.19 This means a long one (at least ten characters) with a mixture of letters, numbers, and special characters — and a unique password for each app. According to security experts, a good trick is to create a memorable “passphrase” by creating a series of random words, and then substitute numbers or special characters (i. e. ,@ for at) for some of the letters. 20 Go ahead and write them down —just store your cheat sheet in a secure location (not your wallet or phone case!).Throughout history,female scientists,engineers and mathematicians have changed the world. But while their_______have been massive, their names and their stories have_______been publicized. Physicist Jess Wade wanted to_______the stories of great scientific pioneers who may be overlooked and she_______a unique way to do that: writing Wikipedia pages.Wade told CBS News that while the_______of women in science is small in comparison to that of men, she has always had_______in the field. When she was a graduate student,Wade_______ an inspiring woman in science,Kim Cobb,who is a climate scientist at Brown University."So when I met her I thought she is a(n)_______person and I need to learn more about her. When I did a little search, I couldn't_______any information about her. What I________was a Wikipedia page but it wasn't there."That's how Wade got the________—to start writing Wikipedia pages for diverse people in science who don't yet have them.In her free time, Wade searches the Internet to________information,and then she gets to work writing Wikipedia pages. She's written more than 1,700 so far.Wade is now making a________for herself in science,and she knows she wouldn't be here without those who came before her.“We all have a role to________in making science a more diverseand________place," she said.21.A．ambitions B．potentials C．achievements D．demands22.A．naturally B．permanently C．rarely D．regularly 23.A．share B．choose C．read D．appreciate 24.A．thought little of B．came up with C．passed down D．left out25.A．goal B．progress C．responsibility D．number26.A．concepts B．role models C．safety rules D．followers 27.A．referred to B．turned down C．ran across D．took in28.A．awesome B．ordinary C．serious D．patient29.A．remember B．identify C．offer D．find30.A．desired B．accepted C．recommended D．acknowledged 31.A．help B．order C．belief D．idea32.A．collect B．post C．manage D．correct33.A．suggestion B．will C．name D．request34.A．make B．play C．select D．decide35.A．fairer B．safer C．quieter D．wider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。

广东省汕头市潮南区2024学年物理高二下期末达标检测试题注意事项1．考试结束后，请将本试卷和答题卡一并交回．2．答题前，请务必将自己的姓名、准考证号用0．5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置．3．请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符．4．作答选择题，必须用2B铅笔将答题卡上对应选项的方框涂满、涂黑；如需改动，请用橡皮擦干净后，再选涂其他答案．作答非选择题，必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答，在其他位置作答一律无效．5．如需作图，须用2B铅笔绘、写清楚，线条、符号等须加黑、加粗．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、下列叙述中符合史实的是A．玻尔理论很好地解释了氢原子的光谱B．汤姆孙发现电子，表明原子具有核式结构C．卢瑟福根据α粒子散射实验的现象，提出了原子的能级假设D．贝克勒尔发现了天然放射现象，并提出了原子的核式结构He)，经过同一电压的加速电场后，它们的速度大小之比为：（）2、原来都是静止的质子和α粒子(氦核42A．2:2B．1:2C．2:1D．1:13、如图所示，A、B两物体质量分别为m A、m B，且m A＞m B，置于光滑水平面上，相距较远．将两个大小均为F的力，同时分别作用在A、B上经过相同距离后，撤去两个力，两物体发生碰撞并粘在一起后将（）A．停止运动B．向左运动C．向右运动D．运动方向不能确定4、关于光电效应，下列说法正确的是A．极限频率越大的金属材料逸出功越大B．只要光照射的时间足够长，任何金属都能产生光电效应C．入射光的频率越高，单位时间内逸出的光电子数就越多D．入射光的光强越大，逸出的光电子的最大初动能越大5、杂技表演过程中，演员将小球以初速度v0=10m/s竖直向上抛出，忽略空气阻力，重力加速度g=10m/s2，下列说法正确的是A．抛出后经过1s，小球落回出发点B ．抛出后经过1s ，小球的加速度大小为零C ．抛出后经过2s ，小球的位移为零D ．抛出后经过2s ，小球的位移最大6、2018年6月5日，是第47个世界环境日，“绿水青山就是金山银山”，中国向世界发出“绿色治理”的铿锵之音，中国承诺到2020年碳排放量下降40%～45%；为了实现负责任大国的承诺,我国将新建核电站项目．目前关于核电站获取核能的基本核反应方程可能是( )A ．238234492902U Th He →+ B ．23519013619203854010U n Sr Xe n +→++ C ．2424011121Na Mg e -→+ D ．1441717281N He O H +→+二、多项选择题：本题共4小题，每小题5分，共20分。

潮南区2022-2023学年度第二学期期初高三摸底考试物理试卷本试卷共6页，16小题，满分100分。

考试用时75分钟。

注意事项：1．答题前，考生务必用黑色字迹的钢笔或签字笔将自己姓名、考号、座位号等相关信息填写在答题卡指定区域内。

2．选择题每小题选出答案后，把答案填涂在答题卡指定区域内，不能答在试卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液．不按以上要求作答的答案无效。

4．考生必须保持答题卡的整洁。

第Ⅰ卷（选择题）一、单项选择题：本题共7小题，每小题4分，共28分。

在每小题给出的四个选项中，只有一项符合题目要求。

1．新冠疫情下，无数医务工作者化身“大白”挺身而出，如图是一位“大白”手里拿着核酸采样管静止在空中，则（）A．“大白”对采样管的压力大小等于采样管的重力大小B．“大白”对采样管的摩擦力大小等于采样管的重力大小C．“大白”对采样管拿得越紧，采样管受到的摩擦力会越大D．“大白”对采样管的作用力大小等于采样管的重力大小2．如图所示，半球形陶罐和陶罐内的物块（视为质点）绕竖直转轴OO 从静止开始缓慢加速转动，物块相对陶罐始终保持静止，下列说法正确的是（）A．物块的线速度可能先增大后减小B．某时刻物块可能只受两个力C．物块一定有上滑的趋势D．物块一定有下滑的趋势3．在学习机械波相关知识后，两名同学分别乘坐静止在湖面的甲、乙两艘小船，两船水平距离20m。

某时刻，一列水波从甲船向乙船传播，每艘船在1min时间内上下浮动30次，已知甲船在波峰时，乙船在波谷，两船间恰好还有2个波峰，以下说法正确的是（）A．水波的周期为1s B．水波的波长为8mC．水波的波速为8m/s D．水波经过一段时间，甲乙两船将靠近4．我国首颗量子科学实验卫星“墨子”已于酒泉卫星发射中心成功发射。

“墨子”由火箭发射至高度为500km的预定圆形轨道。

试卷类型：A汕头市2023-2024学年高二下学期期末普通高中教学质量监测数学注意事项：1.答题前，先将自己的姓名､准考证号填写在试题卷和答题卡指定位置.2.选择题的作答：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，写在试题卷､草稿纸和答题卡上的非答题区域均无效.3.非选择题的作答：用黑色签字笔直接答在答题卡上对应的答题区域内.4.考试结束后，请将本试题卷和答题卡一并上交.第Ⅰ卷选择题一､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，只有一项是符合题目要求的.1.已知32i -+是关于x 的实系数方程220x px q ++=的一个根，则q 的值为（ ）A.26B.-26C.13D.-132.若空间中四条不同的直线1l ，2l ，3l ，4l 满足12l l ⊥，23l l ⊥，34l l ⊥，则下面结论正确的是（ ）A.14l l ⊥B.14l l ∥C.1l ，4l 既不垂直也不平行 D.1l ，4l 的位置关系不确定3.已知1tan 3α=-，则sin 2α=（ ）A.35 B.35- C.35± D.45±4.已知{}n a 为等差数列，135105a a a ++=，24699a a a ++=，则20a =（ ）A.1B.33C.65D.-15.对于变量Y 和变量x 的成对样本观测数据，用一元线性回归模型()()20,Y bx a eE e D e σ=++⎧⎨==⎩得到经验回归模型ˆˆˆy bx a =+，对应的残差如图所示，则模型误差（）A.满足一元线性回归模型的所有假设B.不满足一元线性回归模型的()0E e =的假设C.不满足一元线性回归模型的()2D e σ=的假设D.不满足一元线性回归模型的()0E e =和()2D e σ=的假设6.通过随机询问某中学110名学生是否爱好跳绳，得到如下22⨯列联表.已知()()()()()22n ad bc a b c d a c b d χ-=++++，()210.8280.001Pχ≥=，根据小概率值0.001α=的独立性检验，以下结论正确的是（ ）性别跳绳男女合计爱好402060不爱好203050合计6050110A.爱好跳绳与性别有关B.爱好跳绳与性别有关，这个结论犯错误的概率不超过0.001C.爱好跳绳与性别无关D.爱好跳绳与性别无关，这个结论犯错误的概率不超过0.0017.在ABC 中，π4B =，BC 边上的高等于13BC ，则cos A =（ ）C. D.8.某海湾拥有世界上最大的海潮，其高低水位之差可达到15m .在该海湾某一固定点，大海水深d （单位：m ）与午夜24：00后的时间t （单位：h ）的关系由函数()104cos d t t =+表示，则上午9：00潮水的涨落速度为（精确到0.01m /h ，参考数据：33sin 30.140.0027≈≈）（ ）A.3.00B.-1.64C.1.12D.-2.15二､多选题：本题共3小题，每小题6分，共18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6分，部分选对的得部分分，有选错的得0分.9.已知点O ､N ､P 在ABC 所在平面内，则（）A.若OA OB OC ==，则点O 是ABC 的外心B.若0NA NB NC ++=，则点N 是ABC 的重心C.若PA PB PB PC PC PA ⋅=⋅=⋅，则点P 是ABC 的内心D.若0AB AC BC AB AC⎛⎫⎪+⋅= ⎪⎝⎭，则ABC 是等腰三角形10.已知函数()ππsin sin cos 66f x x x x a ⎛⎫⎛⎫=++-++ ⎪ ⎪⎝⎭⎝⎭的最大值为1，则（ ）A.1a =-B.()f x 的最小正周期为2πC.()f x 在π,π4⎛⎫⎪⎝⎭上单调递减D.()f x 的图象按向量π,16a ⎛⎫=-⎪⎝⎭平移，所得图象过原点11.已知点()2,3P --和以点Q 为圆心的圆()()22129x y -+-=，以PQ 为直径，点Q '为圆心的圆与圆Q 相交于A ､B 两点，则（ ）A.圆Q '的方程为()()()()12230x x y y -++-+=B.PA 与PB 两条直线中，有一条直线的斜率不存在C.直线AB 的方程为3560x y +-=D.线段AB第II 卷非选择题三､填空题：本题共3小题，每小题5分，共15分.12.写出()81x +的展开式中系数最大的项：__________.13.已知一正四面体状木块V ABC -的棱长为3，点P 为侧面VAC 的重心，过点P 将木块锯开，使截面平行于直线VB 和AC ，则截面周长为__________.14.设椭圆()222210x y a b a b +=>>的离心率为e ，双曲线22221x y a b -=e 的取值范围是__________.四､解答题：本题共5小题，共77分.解答应写出文字说明､证明过程或演算步骤.15.（本小题满分13分）已知等比数列{}n a 的前n 项和为n S ，且132n n a S +=+，*n ∈N .（1）求数列{}n a 的通项公式；（2）在n a 与1n a +之间插入n 个数，使这2n +个数组成一个公差为n d 的等差数列，在数列{}n d 中是否存在3项m d ､k d ､p d （其中m ､k ､p 成等差数列）成等比数列？若存在，求出这样的3项；若不存在，请说明理由.16.（本小题满分15分）在长方体1111ABCD A B C D -中，点E ､F 分别在棱1BB ､1DD 上，且1AE A B ⊥，1AF A D ⊥.（1）求证：1AC ⊥平面AEF ；（2）当3AD =，4AB =，15AA =时，求平面AEF 与平面11D B BD 的夹角的余弦值.17.（本小题满分15分）已知函数()()e 211x x f x x -=-.（1）作出()y f x =的大致图象，并说明理由；（2）讨论函数()12e 1x a g x x =---的零点个数.18.（本小题满分17分）甲公司现有资金200万元，考虑一项投资计划，假定影响投资收益的唯一因素是投资期间的经济形势：若投资期间经济形势好，投资有25%的收益率；若投资期间经济形势不好，投资有10%的损益率.如果不执行该投资计划，损失为1万元.现有如下两个方案，方案一执行投资计划；方案二聘请投资咨询公司乙分析投资期间的经济形势，聘请费用为5000元，若投资咨询公司乙预测投资期间经济形势好，则执行投资计划；若投资咨询公司乙预测投资期间经济形势不好，则不执行该计划.以往的资料表明，投资咨询公司乙预测不一定正确.投资期间经济形势好，咨询公司乙预测经济形势好的概率是0.8；投资期间经济形势不好，咨询公司乙预测经济形势不好的概率是0.7.假设根据权威资料可以确定，投资期间经济形势好的概率是0.4，经济形势不好的概率是0.6.（1）求投资咨询公司乙预测投资期间经济形势好的概率；（2）根据获得利润的数学期望的大小，甲公司应该执行哪个方案？说明理由.19.（本小题满分17分）抛物线具有光学性质：由其焦点F 发出的光线经抛物线上的点M （不同于抛物线的顶点）反射后，反射光线平行于抛物线的对称轴.由光路可逆知，反之也成立.（1）已知平行于x 轴的光线l 从点()(),20P m m >发出，经抛物线22y x =上的点A 反射后，再经该抛物线上另一点B ，最后沿BQ 方向射出，若射线BP 平分ABQ ∠，求实数m 的值；（2）光线被抛物线上某点反射，其实是被抛物线在该点处的切线反射.对于一般的抛物线()220y px p =>，请证明上述抛物线的光学性质.汕头市2023-2024学年高二下学期期末普通高中教学质量监测数学科参考答案与评分标准第I 卷题号1234567891011答案ADBACCDBABDABABD1.【解析】实系数一元二次方程的两根互为共轭复数，由韦达定理得2|32i |132q=-+=；2.【解析】利用长方体易得；3.【解析】2222sin cos 2tan 3sin2sin cos tan 15ααααααα===-++；4.【解析】1353353a a a a ++==，同理433a =，故公差2d =-，所以204161a a d =+=；5.【解析】由残差图的点没有均匀分布在水平带状区域内可知：不满足()2e D σ=的假设；6.【解析】计算得20.0017.810.828χα≈<=，说明没有充分证据作此推断；7.【解析】作AD BC ⊥于D ，设BC a =，则2,,33a a AD BD CD AB AC =====，故由余弦定理可求得Cos A ；8.【解析】由导数的意义知，上午9:00潮水的涨落速度为()()()()()2294sin94sin 634sin6Cos3Cos6sin342sin31sin 312sin 3sin3d ⎡⎤=-=-+=-+=--+-⎣⎦'()344sin 33sin3=-()440.002730.14 1.64;=⨯⨯-⨯≈-9.【解析】由外心定义，A 正确；设D 是AB 中点，由0NA NB NC ++= 得2NC ND =-，B 正确；由PA PB PB PC ⋅=⋅ 得()0PB PC PA PB AC ⋅-=⋅=，即PB AC ⊥，同理，PC AB ⊥，故点P 是ABC 的垂心，C 错误；设AB ACAF AB AC=+，则AF 为BAC ∠的平分线，又AF BC ⊥，故D 正确；10.【解析】化简得()π2sin 6f x x a ⎛⎫=++ ⎪⎝⎭，故21a +=，A 正确；显然，B 正确；π6u x =+在π,π4⎛⎫ ⎪⎝⎭上递增，且5π7π,126u ⎛⎫∈⎪⎝⎭，而sin u 在5π7π,126⎛⎫⎪⎝⎭上没有单调性，故C 错误；设()f x 的图象按向量π,16a ⎛⎫=-⎪⎝⎭平移，得到函数()g x 的图象，则()π2sin 3g x x ⎛⎫=+ ⎪⎝⎭，D 错误；11.【解析】设点(),M x y 为圆Q '上任一点，由0MP MQ ⋅=知，A 正确；显然，PA 与PB 为圆Q 的切线，若有一条的斜率不存在，则其方程必为2x =-，它到圆心Q 的距离为3，与圆Q 半径相等，符合题意，故B 正确；圆Q 与圆Q '的方程相减得直线AB 的方程为3540x y +-=，故C 错误；圆心Q 到直线AB，所以AB ==，故D 正确；第II 卷12.【解析】8(1)x +的展开式中系数最大的项也即是二项式系数最大的项，即4458T C x =；13.【解析】由线面平行的性质定理知，截面的两组对边分别与AC 和VB 平行，与AC 平行的边长为2，与VB 平行的边长为1，故周长为6；14.【解析】依题意，0b a <<，故e ⎫=⎪⎪⎭；15.【答案】（1）设等比数列{}n a 的公比为q ，则当1n =时：1132a q a =+，①当2n =时：()211132a q a a q =++，②由①②解得：12,4a q ==，所以数列{}n a 的通项公式121242n n n a --=⨯=；（2）设数列{}n d 中存在3项m k p d d d ､､成等比数列，则2k m p d d d =⋅，因为2113211n n n n a a d n n -+-⨯==++，所以2212121323232111k m p k m p ---⎛⎫⨯⨯⨯=⋅ ⎪+++⎝⎭，即()()()22242223232(1)11m p k k m p +--⨯⨯=+++；又因为m k p ､､成等差数列，所以2k m p =+，所以()()2(1)11k m p +=++，化简得22k k mp m p +=++，所以2k mp =，又m k p ､､各不相等，所以222()4m p k mp k +=<=，矛盾.从而假设不成立，故在数列{}n d 中不存在3项,,m k p d d d 成等比数列.16.【答案】（1）证明：因为()()110AC AE A B BC AE BC AE BC AB BE ⋅=+⋅=⋅=⋅+=，所以1AC AE ⊥，因为()()110AC AF A D DC AF DC AF DC AD DF ⋅=+⋅=⋅=⋅+= ，所以1AC AF ⊥，又AE AF A ⋂=，故1AC ⊥平面AEF ；（2）以点D 为原点，分别以直线1DA DC DD ､､为x y z ､､轴，建立空间直角坐标系，则()()13,4,0,0,0,5DB DD ==设平面11DBB D 的法向量为(),,n x y z =，则150340n DD z n BD x y ⎧⋅==⎪⎨⋅=+=⎪⎩，取()4,3,0n =- ，由（1）知：()13,4,5A C =--是平面AEF 的一个法向量所以，111cos ,n A C n A C n A C⋅==⋅，设平面AEF 和平面11D B BD 的夹角为θ，则1cos cos ,n A C θ==.17.【答案】（1）()f x 的定义域为{}1xx ≠∣，且()()2e 23(1)x x x f x x -=-'，由()0f x '=得：0x =或32x =，列表得：x(),0∞-0()0,131,2⎛⎫ ⎪⎝⎭323,2∞⎛⎫+ ⎪⎝⎭()f x '+--+()f x极大值极小值所以，()f x 的递增区间为(),0∞-与3,2∞⎛⎫+⎪⎝⎭，递减区间为()0,1与31,2⎛⎫⎪⎝⎭，()f x 的极大值为()01f =，极小值为3234e 2f ⎛⎫= ⎪⎝⎭，当x ∞→-时，()0f x →，且0x <时，()0f x >，当x 从1的左侧无限趋近1时，()f x ∞→-，当x 从1的右侧无限趋近1时，()f x ∞→+又10,2f ⎛⎫=⎪⎝⎭所以函数()y f x =的大致图象如图所示：（2）令()120e 1x a g x x =--=-得：()()e 211x x a f x x -==-，由（1）知，当()32,01,4e a ∞⎧⎫∈-⋃⎨⎬⎩⎭时，()y g x =恰有1个零点；当()320,14e ,a ∞⎛⎫∈⋃+ ⎪⎝⎭时，()y g x =恰有2个零点；当321,4e a ⎛⎫∈ ⎪⎝⎭时，()y g x =没有零点.18.【答案】（1）记B =“投资期间经济形势好”，A =“投资咨询公司预测投资期间经济形势好”，则()()0.4,0.6P B P B ==，()0.8P A B =∣，()()110.70.3,P A B P A B =-=-=∣∣由全概率公式得：()()()()()P A P B P A B P B P A B =+∣∣0.40.80.60.30.5;=⨯+⨯=（2）设采取方案一获得利润X 万元，则X 的分布列是X50-20P 0.40.6设采取方案二获得利润Y 万元，则Y 的所有可能取值为20.5, 1.5,49.5--，(20.5)()((0.18P Y P BA P B P A B =-===∣，( 1.5)(1()10.50.5P Y P A P A =-==-=-=，()()()()49.50.32P Y P BA P B P A B ====∣，Y ∴的分布列为：Y -20.5-1.549.5P0.180.50.32()()500.4200.68,20.50.18 1.50.549.50.3211.4E X E Y ∴=⨯-⨯==-⨯-⨯+⨯=，()(),E X E Y <∴ 甲公司应该选择方案二.19.【答案】（1）依题意可知，直线l 的方程为2y =，由222y y x =⎧⎨=⎩得：()2,2A ，又1,02F ⎛⎫ ⎪⎝⎭，所以43AB k =，故直线AB 的方程为4132y x ⎛⎫=- ⎪⎝⎭，由()2413222y x y x x ⎧⎛⎫=-⎪ ⎪⎝⎭⎨⎪=≠⎩得：11,82B ⎛⎫- ⎪⎝⎭，则2081BP k m =-，设直线BP 的倾斜角为θ，由2222tan 4tan21tan 13BP AB BP k k k θθθ====--得12BP k =或-2（舍去）所以201812m =-，故418m =；（2）设直线()0y kx b k =+≠与拋物线22(0)y px p =>相切于点M ，由22y kx b y px=+⎧⎨=⎩得：()222220k x kb p x b +-+=，故222Δ(22)40kb p k b =--=，整理得2kb p =，从而(),2,,0b M b F kb k ⎛⎫ ⎪⎝⎭，进而()21,2b MF k b k ⎛⎫=-- ⎪⎝⎭，取直线MF 的一个方向向量()211,2n k k =-- ，直线()0y kx b k =+≠的一个方向向量为()1,m k =，焦点F 发出的光线经点M 反射，设反射光线斜率为k '，取其一个方向向量为()21,n k '= ，故12cos ,cos ,0m n m n += ，即：=整理得：()2120k k k k ⎡⎤-+⎣'=⎦'，因为1n 与2n 不共线，所以()2120k k k '-+≠，从而0k '=，所以由抛物线焦点F 发出的光线经拋物线反射后，反射光线平行于抛物线的对称轴.。

2024~2025学年度第一学期九年级期中考试数学试卷（S ）说明：1、本卷满分120分；2、考试时间120分钟；3、答案请写在答题卷上．一、选择题（每小题3分，共30分）1.关于的一元二次方程（为实数）根的情况是（ ）A.有两个相等的实数根B.有两个不相等的实数根C.没有实数根D.不能确定2.已知二次函数，当时，随增大而增大，则实数的取值范围是（ ）A. B. C. D.3.下列四幅图案是四所大学校徽的主体标识，其中是中心对称图形的是（ ）A. B.C. D.4.二次函数图象的顶点所在的象限是（ ）A.第一象限B.第二象限C.第三象限D.第四象限5.是一元二次方程的一个根，则代数式的值是（ ）A. B.2017 C. D.20256.某商品原价200元，连续两次降价后售价为148元，下列所列方程正确的是（ ）A. B.C. D.7.如图，是一个中心对称图形，为对称中心，若，，，则的长为（ ）B.D.48.若直角三角形的两边长分别是方程的两根，则该直角三角形的面积是（ ）A.6B.12C.12D.6x 220x kx --=k 2(1)y a x =-0x >y x a 0a >1a >1a ≠1a <2(1)2y x =-++m 220x x ++=2222021m m +-2017-2025-%a 2200(1%)148a +=()22001%148a -=200(12%)148a -=2200(1%)148a -=A 90C ∠=︒60BAC ∠=︒1BC =CC '27120x x -+=9.已知抛物线，则当时，函数的最大值为（ ）A. B. C.0 D.210.如图，抛物线经过正方形的三个顶点，，，点在轴上，则的值为（ ）A. B. C. D.二、填空题（每小题3分，共15分）11.已知关于的方程有一个根1，那么__________.12.若二次函数的图象与轴有且只有一个交点，则的值为________.13.如图，在正方形中，，E 为的中点，连接，将绕点按逆时针方向旋转得到，连接，则的长为_________.14.在平面直角坐标系中，将抛物线先绕原点旋转，再向下平移5个单位，所得到的抛物线的顶点坐标是_________.15.观察下列图形规律：当_________时，图形“”的个数是“”的个数的2倍.三、解答题（一）（每小题7分，共21分）16.用配方法解一元二次方程：17.如图，在中，，点、点分别为、的中点，连结，将绕点旋转得到.试判断四边形的形状，并说明理由.221y xx =--03x ≤≤2-1-2y axc =+OABC A B C B y a c 1-2-3-4-x 20ax bx c ++=a b c ++=2(1)42y a x x a =--+x a ABCD 4AB =AB DE DAE △D 90︒DCF △EF EF 221y xx =+-180︒n =∆∙2213x x+=ABC △2AB BC =D E AB AC DE ADE △E 180︒CFE ∆BCFD18.已知开口向上的抛物线经过点.（1）确定此拋物线的解析式；（2）当取何值时，有最小值，并求出这个最小值.四、解答题（二）（每小题9分，共27分）19.如图，在边长均为1个单位长度的小正方形组成的网格中，点，点，点均为格点（每个小正方形的顶点叫做格点）.【实践与操作】（1）作点关于点的对称点；（2）连接，将线段绕点顺时针旋转得点对应点，画出旋转后的线段；【应用与计算】（3）连接，求出四边形的面积.20.如图，二次函数（为常数）的图象的对称轴为直线.（1）求的值.（2）向下平移该二次函数的图象，使其经过原点，求平移后图象所对应的二次函数的表达式。

广东省汕头潮南区2005-2006学年度第二学期期终质检高二语文试卷本试卷分两部分，共8页，满分为150分，考试用时150分钟。

第一部分阅读鉴赏（75分）一、诗文阅读鉴赏（共40分）（一）古诗文默写（8分）1．补写出下列古诗词中空缺的部分。

（任选4小题，8分）（1）三顾频繁天下计，两朝开济老臣心。

，。

（杜甫《蜀相》）（2）______________，______________。

情人怨遥夜，竟夕起相思。

（张九龄《望月怀远》）（3），。

江东子弟多才俊，卷土重来未可知。

（杜牧《题乌江亭》）（4）拟把疏狂图一醉，对酒当歌，强乐还无味。

，。

(柳永《蝶恋花》)（5），。

竹杖芒鞋轻胜马，谁怕? （苏轼《定风波》）（6）众里寻他千百度，，。

（辛弃疾《青玉案》）（二）古诗阅读（12分）阅读描写庐山瀑布的两首诗和苏轼评价它们的一首诗，完成2～4题。

(一)日照香炉生紫烟，遥看瀑布挂前川。

飞流直下三千尺，疑是银河落九天。

——李白(二)虚云落泉千仞直，雷奔入江不暂息。

今古长如白练飞，一条界破青山色。

——徐凝(三)帝遣银河一律垂，古来惟有谪仙词。

飞流溅沫知多少，不与徐凝洗恶诗。

——苏轼2．根据诗歌意象填空。

（3分）（1）李白诗中的“紫烟”，即徐凝诗中的“”。

（2）徐凝诗中的“落泉”，即李白诗中的“”。

（3）李白诗中的“银河”，即徐凝诗中的“”。

3．以下诗句运用了什么修辞手法，有何表达效果 (4分)（1）飞流直下三千尺：（2）今古长如白练飞：4．苏轼诗对李徐二诗作了评价，试指出苏轼对二诗的看法，并说说你是否同意他的看法，为什么？（5分）（三）文言文阅读（20分）阅读下面文言文，完成5～7题。

项王乃复引兵而东，至东城，乃有二十八骑。

汉骑追者数千人。

项王自度不得脱。

谓其骑曰：“吾起兵至今八岁矣，身七十余战，所当者破,所击者服,未尝败北,遂霸有天下；然今卒困于此，此天之亡我，非战之罪也。

今日固决死，愿为诸君快战，必三胜之，为诸君溃围、斩将、刈旗，令诸君知天亡我，非战之罪也。

东莞市2014—2015学年度第一学期教学质量自查九年级历史试卷一、选择题：每题只有一个正确答案，每题3分，25题，共75分。

1．文艺复兴的先驱但丁抨击教会的罪恶与贪婪，赞美了人的价值和理性的伟大，支持这一观点的有力证据出自A．《神曲》中的诗句B．《蒙娜丽莎》的微笑C．米开朗琪罗的雕塑 D．《哈姆雷特》的独白2．“他是画家、雕塑家、数学家，还是一名哲学家……（他）以新奇的目光重新审视一切的勇气和热情，才能使人类精神从经院的暗室，冲向色彩缤纷的新天地。

”他的画作中体现的时代精神是A．人文主义B．社会主义C．理性主义D．专制主义3．亚当·斯密说：“美洲的发现，经由好望角前往东印度群岛的航道的发现，是人类历史上所记载的最伟大的、最重要的事件。

”这一“事件”是指A．独立战争B．新航路开辟C．西进运动D．郑和下西洋4．同一段历史，不同的角度看待就会有不同的表达和诠释。

对15-17世纪欧洲的海外探险活动，下列选项明显缺乏历史依据的表述是A．全球化时代B．大航海时代C．大发现时代D．大工厂时代5．1689年，威廉三世在伦敦威斯敏寺里，除双手接受议会代表奉送的王冠，还接受了一部世界历史上著名的宪法性文件。

这一文件是A．《大宪章》B．《权利法案》C．《商法典》D．《人权宣言》6．《华盛顿传》一书中不仅提到了华盛顿独立战争时期的英勇睿智，还描述了华盛顿在美国建立后A．领导莱克星顿起义B．攻占巴士底狱C．担任两届美国总统D．实施极权统治7．“两百年以后的今天，美国人越来越深切地认识到，没有杰斐逊的思考，也同样不会有美国的存在。

”杰斐逊的思考集中体现在A．《大宪章》B．《权利法案》C．《民法典》D．《独立宣言》8．“（农奴制改革）后还不到一年，贵族就上报了1000多起骚乱。

”农民不满改革的主要理由是A．农民不可以选择职业B．农民须高价赎买土地的使用权C．农民不可以拥有财产D．地主可以任意交换或买卖农民9．“林肯获胜后几个星期之内，联邦的解体已经开始，这一过程迅速导致两派美国人之间的一场漫长而血腥的战争”，这一战争是A．独立战争B．鸦片战争C．南北战争D．美西战争10．“人类自己创造的这场前所未有的大灾难的缘起，包括欧洲主要国家经济上的竞争、殖民地的争夺、相互冲突的联盟体系和势不两立的民族主义愿望”。

广东省潮南区2024-2025学年高三上学期摸底考试英语试题一、阅读理解The Most Beautiful Train Stations in the WorldTokyo Station Marunouchi BuildingThe wide-open entrance square on the Marunouchi side of Tokyo’s main train station would look more at home in Amsterdam or Paris. Kingo Tatsuno’s classical European design took over six years to complete, finally opening in 1914. The statin has received several facelifts since the Second World War, most recently in 2012. It’s one of the biggest and busiest train stations in the world.Beijing West Railway StationThe idea of building this station in the capital of China was put forward as early as 1959 but didn’t come to reality until 1996. It was the largest station in Asia at the time, although Shanghai’s Hongqiao Station has since surpassed it. The station itself has a unique architectural (建筑的) style, with the main body of the building having quite a functional Russian look, but the rooftop is decorated with three amazing Chinese pagodas.Duoliang Station, TaiwanThis is the smallest station on this lit by a distance. In fact we’re not really choosing Duoliang for architectural reasons. This station is all about location, location, location. Surrounded by hi top greenery on one side and amazing Pacific Ocean scenery on the other, it’s one of the most scenic spots in all of Taiwan. This isn’t even a working station anymore, but the platforms have been adapted into viewing spots, extremely popular with local train spotters.Stazione Milano Centrale in MilanMilan’s central train station looks pretty much exactly as you’d imagine it. There are a mix of styles at play here, from Art Nouveau to Art Deco, but the entire building is filled with a classic sense of Roman monumentality. The outer look is guarded by two statues, while its insides are equally impressive, as grand stonewalls curve (弯曲) into the station’s glass ceilings.1．What do Tokyo Station and Beijing West Railway Station have in common?A．Both have European features.B．Both were built in the early 1990s.C．Both have been rebuilt several times.D．Both are the largest domestic stations.2．What do train spotters favor most about Duoliang Station?A．Its distance.B．Its small size.C．Its surrounding scenery.D．Its architectural style.3．What makes Stazione Milano Centrale impressive?A．Roman buildings B．Its various styles.C．Statues on the walls D．The curving ceilings.It all started with a simple question: “Can I paint your portrait?”One day in the summer of 2015, Peterson was relaxing in his living room, reading the book Love Does, about the power of love in action, when his quiet was disturbed by a homeless man outside his apartment. Inspired by the book’s compassionate message, Peterson made a decision; he was going to go outside and introduce himself.In that first conversation, Peterson learned that the man’s name was Matt Faris. He’d moved to Southern California from Kentucky to pursue a career in music, but he soon fell on hard times and ended up living on the street for more than a decade. “I saw beauty on the face of a man who hadn’t shaved in probably a year, because his story, the life inside of him, inspired me,” Peterson recalled. Therefore, Peterson asked if he could paint Faris’ portrait. Faris agreed.Peterson’s connection with Faris led him to form Faces of Santa Ana, a nonprofit organization focused on befriending and painting portraits of members of the community who are unhoused. Peterson sells the paintings for money, splitting the proceeds and putting half into a “love account” for his model. He then helps people use the money to get back on their feet.Many of Peterson’s new friends use the donations to secure immediate necessities — medical care, hotel rooms, food. Faris used the funds from his portrait to record an album, fulfilling his musical dreams. Another subject, Kimberly Sondoval, had never been able to financially support her daughter. She asked, “Can I use the money to pay my daughter’s rent?” When the check was delivered, “They both wept in my arms,” Peterson recalls.Peterson has painted 41 of these portraits himself. But there’s more to the finished productsthan the money they bring to someone who’s down and out. He’s discovered that the buyers tend to connect to the story of the person in the painting, finding similarities and often friendship with someone they might have otherwise overlooked or stereotyped.“People often tell me, ‘I was the one that would cross the street, but I see homeless people differently now,” Peterson says. “I didn’t know that would happen.”4．What made Peterson start a conversation?A．The curiosity about strangers.B．The touching story of Matt Faris.C．The disturbance by a homeless man.D．The sympathetic message in Love Does. 5．What do we know about Faces of Santa Ana?A．It pays the homeless salaries.B．It is an official nonprofit organization.C．It tries to restore the lives of Peterson’s models.D．It spends all the money on helping the unhoused.6．What does the underlined phrase “get back on their feet” in Paragraph 4 mean?A．regain their financial stability and independence B．physically recover from an injury C．become more self-confident D．return to their original home7．How might a buyer view the homeless after buying a portrait?A．lazy and poor B．odd but inspiringC．disturbing and untidy D．pitiful but respectableIs forgiveness against our human nature? To answer our question, we need to ask a further question: What is the essence of our humanity? For the sake of simplicity, people consider two distinctly different views of humanity. The first view involves dominance and power. In an early paper on the psychology of forgiveness, Droll (1984) made the interesting claim that humans’ essential nature is more aggressive than forgiving allows. Those who forgive are against their basic nature, much to their harm. In his opinion, forgivers are compromising their well-being as they offer mercy to others, who might then take advantage of them.The second view involves the theme of cooperation, mutual respect, and even love as the basis of who we are as humans. Researchers find that to fully grow as human beings, we need both to receive love from and offer love to others. Without love, our connections with a wide range ofindividuals in our lives can fall apart. Even common sense strongly suggests that the will to power over others does not make for harmonious interactions. For example, how well has slavery worked as a mode of social harmony?From this second viewpoint of who we are as humans, forgiveness plays a key role in the biological and psychological integrity of both individuals and communities because one of the outcomes of forgiveness, shown through scientific studies, is the decreasing of hatred and the restoration of harmony. Forgiveness can break the cycle of anger. At least to the extent the people from whom you are estranged accept your love and forgiveness and are prepared to make the required adjustments. Forgiveness can heal relationships and reconnect people.As an important note, when we take a Classical philosophical perspective, that of Aristotle, we see the distinction between potentiality and actuality. We are not necessarily born with the capacity to forgive, but instead with the potential to learn about it and to grow in our ability to forgive. The actuality of forgiving, its actual appropriation in conflict situations, develops with practice.8．What is Droll’s idea about forgiveness?A．People should offer mercy to others.B．People who forgive can have their own welfare affected.C．Forgiveness depends on the nature of humanity.D．Aggressive people should learn to forgive.9．What does the example in Paragraph 2 illustrate?A．To forgive is to love.B．To fight is to grow.C．To dominate is to harm.D．To give is to receive.10．What is the writer’s attitude toward forgiveness?A．Objective.B．Reserved.C．Favorable.D．Skeptical. 11．What is message of the last paragraph?A．Forgiveness is in our nature.B．Forgiveness grows with time.C．Actuality is based on potentiality.D．It takes practice to forgive.Do you have a brain for math? New research indicates that levels of two key neurotransmitters (脑神经传递素) — glutamate (谷氨酸) and gamma-aminobutyric acid(GABA)can predict mathematical abilities, suggesting brain chemistry may be playing a role in those who find math easy.The new study, published in the journal PLOS Biology, recruited 255 subjects extending a range of six-year olds in primary school to university students. The research focused on glutamate and GABA, known to play a role in brain plasticity (可塑性) and learning. Based on prior research, the focus was on two brain regions linked with mathematical abilities — the left intraparietal sulcus (IPS 顶叶内沟) and the left middle frontal gyrus (MFG 脑额中回).The results were interestingly different. In the youngest subjects high GABA levels and low glutamate levels in the left IPS were consistently associated with high math skills. But in the older university group the exact opposite was seen: low GABA and high glutamate were linked with strong mathematical abilities. Levels of both neurotransmitters in the MFG did not associate with math skills.The group was tested twice over 18 months, allowing the researchers to see if these neurotransmitter levels could predict mathematical ability into the future. And it worked, with neurotransmitter levels effectively predicting one’s success on math tests completed a year and a half later.Another recent study from the same research team looked specifically at GABA levels in MFG of 14 to 18 year olds. That research indicated MFG GABA levels could effectively predict whether a student was still studying maths or had ceased that subject years prior.Cohen Kadosh, one of the researchers working on the study, says this may indicate math education can help stimulate the development of key brain regions. Further research will work on whether certain learning interventions can help those children less interested in math so these brain regions still get the developmental workout they need.“Not every adolescent enjoys maths so we need to investigate possible alternatives, such as training in logic and reasoning that engage the same brain area as maths,” says Cohen Kadosh. 12．What is the new study aimed at?A．Exploring mental development of the subjects.B．Finding the tie between brain chemistry and math.C．Testing the link between brain regions.D．Revealing the structure of brain.13．What can be learned from the findings?A．The levels of GABA decide one’s math skills.B．Low MFG glutamate means poor math ability.C．Neurotransmitters in the MFG affect math skills.D．Math education may help with brain development.14．What can better math education according to Cohen Kadosh?A．Studying more possible options.B．Tracing slow learners’ early learning.C．Training math learners respectively.D．Developing key relevant brain areas.15．What can be the best title of the text?A．Factors Affecting Math SkillsB．Ways to Promote Math EducationC．Brain Activities Involved in Math StudyD．Math Ability Predicted by NeurotransmittersEveryone knows that early to bed and early to rise is good for your health. Having maintained adequate sleep, people can keep themselves refreshed and dynamic. 16 Here are some steps to take to become a morning person on a freezing day.Seek out as much natural light as possible. 17 When you wake up, go outside for a relaxed walk around the block, or sit out back while you have a cup of coffee. To avoid being in a sleep state all along, you’d better rethink the black-out curtains you might use to make a very dark sleeping environment.18 If you fly straight into getting up at your desired time every day, you’ll feel tired during the transition within a few weeks. For some people, those first few days of exhaustion from switching to a new schedule aren’t safe and hard to bear. In that case, gradually ease into the early-bird life. They are supposed to shift half an hour, wait a few days, shift another half an hour, wait a few days, and then shift another half an hour. 19De-stress and relax in the evenings. A consistent bedtime isn’t as crucial as sticking to the same wake up time every day, but it’s still important to make sure you’re getting enough sleep.20 So you’ll probably need to inch your bedtime forward as you transition to a new schedule. Start at least an hour before you hit the bed and cut back on stuff related to daily life.With all these tips, you’re not dreading it if you think of getting up early on a chilly winter. A．Cultivate the habit gradually.B．That’s a little easier for people to tolerate.C．The sun is the driver of our internal clock.D．Get rid of the previous sleep patterns instantly.E．Therefore, developing a healthy sleeping habit is of great significance.F．Without sufficient hours of sleep, the mind and body can’t properly recharge. G．However, it usually takes a lot of willpower to get out of bed early on a winter’s day.二、完形填空I know next to nothing about baseball. When in the field during P. E. as a boy, I 21 the ball wouldn’t come to me. Catching it seemed a(n) 22 task for me. Yet I later became the father of two boys, Will and Tim (8 and 5), who were both 23 in baseball. When playing with Will, thanks to my weak arm, the ball often 24 before he could catch it. After many 25 one day, he said, “How am I ever going to 26 the major leagues?” “You got the 27 dad, kid,” I thought. Baseball isn’t my cup of tea.Still, I wanted to give my kids 28 on the field. The official Little League in my area was super-competitive and primarily 29 older boys. What if we had something more 30 , welcoming boys and girls, and younger ones?The idea echoed in my mind for days, forcing me to pluck up courage to call the department 31 , inquiring about starting a baseball league. What if they said no? I got butterflies in my stomach. But to my relief, they responded 32 . Our league was 33 . And I became the most unlikely baseball commissioner (专员) ever.Many years have passed. Now Will and Tim are 34 themselves, and I can’t wait to see what they’ll have to do. I know better than anyone: Parenthood calls you to do the most 35 things.21．A．suggested B．prayed C．begged D．demanded22．A．impossible B．effortless C．easy D．straightforward 23．A．weak B．skillful C．disappointed D．interested 24．A．dropped B．rose C．bounced D．disappeared 25．A．catches B．misses C．tests D．dangers 26．A．deserve B．admit C．make D．reach 27．A．wrong B．strong C．right D．poor 28．A．advice B．happiness C．confidence D．goals 29．A．refused B．admired C．required D．targeted 30．A．challenging B．low- key C．reliable D．appealing 31．A．in charge B．off duty C．under control D．at hand 32．A．eagerly B．doubtfully C．proudly D．positively 33．A．rejected B．held C．stopped D．founded 34．A．coaches B．fathers C．players D．officials 35．A．interesting B．rewarding C．unlikely D．meaningful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。

汕头市2023-2024学年度普通高中毕业班期末调研测试英语注意事项：1.答题前，考生务必将自己的姓名、考生号、考场号和座位号填写在答题卡上。

2.作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑：如需要改动，用橡皮擦干净后，再选涂其它答案。

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试题序号从“21”开始。

3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

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考试结束后，将试卷和答题卡一并交回。

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 Hotels in Edinburgh Close to the Must-See SightsEdinburgh is home to some of the best hotels in Scotland.Here are some selected hotels,ranging from iconicfive-star institutions to new apartment hotels.The Balmoral HotelFor luxury accommodation,the Balmoral hotel offers five-star service and fantastic views of Edinburgh Castle.It is in the city center and features the iconic clock tower.The Balmoral is probably the most famous hotel in the city and is famous for its hospitality.It is also popular with locals for afternoon tea.Wilde Apart HotelWilde Apart hotel is a modern,stylish hotel.The hotel features well-appointed（设备齐全的）serviced apartments that have fully equipped small kitchens,free WiFi,and car parking nearby.The apartments are ideal for those who are staying in the city for more than a night or two and would like to have a small kitchen to enjoy breakfast or lunch.The Radisson Collection HotelThe Radisson Collection Hotel overlooks the iconic Royal Mile.It’s just a short walk from attractions such as Edinburgh Castle and Holy rood Park,and it’s within easy reach of local restaurants,shops,bars,and galleries. Everything you will want to see and do in Edinburgh is close to hand,and it’s five minutes from Waverly Train Station and airport links.The hotel’s design is modern.Scotsman HotelOriginally built as The Scotsman newspaper offices in1905,Scotsman Hotel has been carefully restored into nine floors of rooms and suites with commanding views over the city.Located at the top end of North Bridge,the hotel occupies the prime real estate spot(黄金地段)in Edinburgh.Easy access to Waverley Train Station and airport shuttles,the hotel is probably the most well-appointed accommodation in terms of transport links.We hope you love the Edinburgh hotels we recommend!21.Which hotel suits customers who love cooking by themselves?A.Scotsman Hotel.B.Wilde Apart Hotel.C.The Balmoral Hotel.D.The Radisson Collection Hotel.22.What makes the Scotsman Hotel different from the others?A.Its special origin.B.Its central situation.C.Its commanding views.D.Its convenient transportation.23.Where is this text probably taken from?A.A travel journal.B.A tour report.C.A tourist website.D.A hotel research.BI went to Yuanhe Middle School in Xihaigu,Ningxia Hui Autonomous Region,a place that used to be one of the poorest regions in China,to participate in an English class with village children.Most kids in the school come from poor village homes and are“left behind”children一kids whose parents work in the cities to make a better living.After class,we chatted a lot,and they wondered about everything,using“outside world”as a term to describe the wider globe for many times.Taking care of rural children is essential.The school offers boarding services and free breakfast and lunch to the students,who can also video chat with their parents daily.To better care for the village children,the school provides outdoor hiking,technology,art,and foreign language program s to help them understand the outside world.After all, high-quality education to disadvantaged village children is the key to poverty-alleviation（扶贫）efforts.During my trip in Xihaigu,I also visited Ningxia Normal University,where over70%of its students major in education.In2013,local and central governments launched public-funded programs to recruit students interested in becoming teachers in Ningxia’s rural villages and districts.By2023,3,500students were enrolled in the program, many of whom came from disadvantaged village households in Xihaigu.I had the opportunity to speak with Yang,a public-funded student from the once-poor Xihaigu region.After finishing school,he planned to stay in the Xihaigu area as a teacher to teach the kids knowledge.Yang told me that knowledge is the key to leading a better life as well as understanding the world.A lack of knowledge is one of the main factors of being poor.Money could help the poor for a while but not once for all.He hopes that by sharing knowledge with the kids in his hometown,he can help them broaden their horizons and develop an objective perspective on the world.He envisions a future where these children can use their knowledge to make a better life,stop people’stereotypes(固有印象)about China and share Xihaigu’s stories with the rest of the world.24.Why did the kids repeatedly mention“outside world”?A.Out of curiosity about the unknown world.B.With the hope of changing the world outside.C.To show their experiences about the wider globe.D.For lack of confidence towards their inner world.25.What is paragraph3mainly about?A.The key to poverty-alleviation efforts.B.The aid to the kids on life and education.C.The teaching conditions of the school.D.The high-quality education of the school.26.What can we infer from the author’s trip in Ningxia Normal University?A.Students in this university can be free of charge.B.The authorities support Ningxia’s rural areas by stressing education.C.The public-funded programs target at the students from poor regions.D.Money is less important than knowledge for the people in Xihaigu’s villages.27.Which of the following may agree with Yang’s opinion?A.Learn and live.B.Sharp tools make good work.C.Hang on to your dreams.D.Poverty-alleviation starts with education.CAs archaeologists（考古学家）examined ancient tombs in Turfan in western China,they discovered some surprisingly well-preserved and familiar relics.Though hardened from over1,000years,there sat little dumplings.Exactly who invented dumplings remains a mystery.But some scholars suspect they were first spread by nomadic（游牧的）Turki c peoples living in western China and Central Asia.This is thought to be the case because “manti,”meaning“dumpling”or“steamed bun”in many Turkic languages,appears to be the root word for dumpling in several other languages.Ancient Turkic people probably filled their dumplings with meat.But it’s unclear when this practice began,or whether they learned the art of dumpling-making from others.However this happened,dumplings certainly gathered steam in ancient China.Dumplings continued to take off and diversify in China over the next thousand years.Instead of the traditional meat filling,some communities chose vegetarian（素食）dumplings.People developed new cooking methods.The relationship between Chinese dumplings and those in other areas is tricky to trace,but food historians have made their best guesses based on available clues.Besides Turkic tribes,some scholars believe that the Mongol Empire also contributed to the spread of dumplings, perhaps introducing them to parts of Eastern Europe.These dumplings could have come by way of China or directly from some of the Turkic peoples the Mongols hired to run their empire.One theory is that this gave rise to dumplings like pelmeni in Russia,pierogi in Poland and vareniki in Ukraine.The Mongol Empire also controlled Korea and might have likewise introduced dumplings ter,after Chinese dumpling varieties were introduced to more countries,English speakers began calling them dumplings,which means“little lumps”.During the Second World War, Chinese“jiaozi”were brought to Japan.So what about the Italian dumpling-like pasta?Some historians think it might be brought by Arab conquerors.It’s unlikely that all dumpling dishes came from the same root tradition.However,we can appreciate the mysterious historical web that made dumplings so various.28.What made Turkic peoples suspected to first spread dumplings?A.The languages they used.B.Their eating habits.C.Their dumpling-making skills.D.The newly found tombs.29.What does the phrase“gathered steam”in paragraph2mean?A.Originated.B.Stabilized.C.Got well-cooked.D.Became popular.30.What does paragraph4focus on?A.The spreading process of dumplings.B.The possible origins of dumplings.C.Differences between various dumplings.D.Reasons for the popularity of dumplings.31.Which of the following best describe dumplings according to the passage?A.Delicious.B.Diverse.C.Unusual.D.Regional.DSonia Kleindorfer was a new director of the Konrad Lorenz Research Center in Austria.At her institute,Konrad Lorenz was a famous Austrian zoologist who could correctly name each kind of goose,which made Kleindorfer feel a certain amount of pressure.“I can do five,but when the next five come,I start to have a mental meltdown,”she says. So she contacted a more technically-minded colleague and asked him:Could he write a program to distinguish these faces?He said,yes,but he’d need a database of geese photos to work with.Kleindorfer got her team out there,snapping picture s of the geese from every angle.After building the database,they wrote a piece of facial recognition Al that could ID a goose,by looking at specific feature s of its beak（喙）.After a couple of years,the team reports that their goose recognition software is now about97%accurate.“Geese have such drama—there are archrivals（劲敌）,and jealousy and retribution（报答）,”Kleindorfer says. To find out how faces figured into this drama,she presented the geese with full-sized pictures of themselves,their partners,or another member of the flock.She showed evidence that geese seemed to recognize photos of their partners and friends,but not themselves.For further study,Sonia Kleindorfer hopes birdwatchers will someday be able to snap a picture of a goose,ID it,and share its location with scientists.But she adds,just remember,her new research suggests that bird watching goes both ways:Geese can remember faces too.“If you are ever not kind to a goose,”she warns,“that goose may find you again.”Kleindorfer thinks that facial recognition is going to play a really important role in conservation and ecology.“We need more computer scientists trained in behavioral ecology and we need more conservation scientists trained in computer science,”she says.“But working together,I think we can do this.”32.What was the problem of Sonia Kleindorfer at work?A.She was always stressed.B.She was not able to count the geese.C.She suffered a mental illness.D.She couldn’t recognize all the geese.33.How does the program distinguish the geese?A.By snapping pictures of the geese.B.By presenting the full-sized pictures.C.By identifying the beaks in the photos.D.By building the database of the geese.34.What is the finding of the new research?A.Birdwatching is a dangerous activity.B.Geese can locate and find human beings.C.Geese have the ability of facial recognition.D.Birdwatchers can snap a picture of a goose.35.What does Kleindorfer stress in the last paragraph?A.Science training.puter science.C.Photo-taking skills.D.Cross-subject study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。

2023-2024学年广东省汕头市高一上册期末数学模拟试题一、单选题1．集合{}13A x x =<<，集合{4B x x =或2}x <，则集合()R A B = ð（）A ．RB ．[2,3)C ．(1,4]D ．∅【正确答案】C【分析】先求得{|24}R B x x =≤≤ð，结合集合并集的运算，即可求解.【详解】由题意，集合{4B x x =或2}x <，可得{|24}R B x x =≤≤ð，又由{|13}A x x =<<，所以(){|14}(1,4]R A B x x =<≤= ð.故选：C.2．设0.30.77,0.3,ln 0.3a b c ===，则,,a b c 的大小关系为（）A ．c<a<b B ．c b a<<C ．a b c<<D ．a c b<<【正确答案】B【分析】由指数函数和对数函数的单调性比较大小.【详解】因为0.30771>=，所以1a >；因为0.7000.30.31<<=，所以01b <<；因为ln 0.3ln10<=，所以0c <，所以c b a <<.故选：B.3．在下列区间中，方程20x x +=的解所在的区间是（）A ．(2,1)--B ．(1,0)-C ．(0,1)D ．(1,2)【正确答案】B【分析】根据函数零点存在定理求解.【详解】设()2x f x x =+，且10(1)210,(0)200f f --=-<=+>，且()f x 为增函数，根据函数零点存在定理知，方程20x x +=在区间(1,0)-内有唯一的解.故选：B.4．函数cos y x x =-的部分图像是A ．B ．C ．D ．【正确答案】D【分析】根据函数cos y x x =-的奇偶性和函数值在某个区间上的符号，对选项进行排除，由此得出正确选项.【详解】∵cos y x x =-是奇函数，其图像关于原点对称，∴排除A,C 项；当0,2x π⎛⎫∈ ⎪⎝⎭时，cos 0y x x =-<，∴排除B 项.故选D.本小题主要考查函数图像的识别，考查函数的单调性，属于基础题.5．下列结论中正确的是（）A ．当02x <≤时，1x x-无最大值B ．当3x ≥时，11x x +-的最小值为3C ．当0x >且1x ≠时，1lg 2lg x x+≥D ．当0x <时，1x x+1≤-【正确答案】D【分析】利用1y x x=-在(0,2]单调递增，可判断A ；利用均值不等式可判断B ，D ；取0.1x =可判断C【详解】选项A ，由1,y x y x==-都在(0,2]单调递增，故1y x x =-在(0,2]单调递增，因此1y x x =-在(0,2]上当2x =时取得最大值32，选项A 错误；选项B ，当1x >时，10x ->，故11111311x x x x +=-++≥+=--，当且仅当111x x -=-，即2x =时等号成立，由于3x ≥，故最小值3取不到，选项B 错误；选项C ，令0.1,lg 1x x ==-，此时1lg 0lg x x+<，不成立，故C 错误；选项D ，当0x <时，0x ->，故11[()(2x x x x+=--+-≤-，当且仅当1x x =，即=1x -时，等号成立，故1x x+1≤-成立，选项D 正确故选：D6．将函数πcos 23y x ⎛⎫=- ⎪⎝⎭向左平移ϕ个单位后所得图象关于原点对称，则ϕ的值可能为（）A ．π6-B ．π6C ．π3D ．5π12【正确答案】D【分析】根据图象平移结论求出平移后函数解析式，根据奇函数的性质可求出ϕ的值，检验可得结果．【详解】平移后得到函数解析式为()()3ππcos 2cos 223g x x x ϕϕ⎡⎤⎛⎫=+-=+- ⎪⎢⎥⎣⎦⎝⎭，因为()g x 图象关于原点对称，即()g x 是奇函数，所以()00g =，故πcos 203ϕ⎛⎫-= ⎪⎝⎭所以ππ2πZ 3,2k k ϕ-=+∈，所以()π5πZ 122k k ϕ=+∈，当0k ＝时，125πϕ=，当125πϕ=时，()5ππcos 2sin 263g x x x ⎛⎫=+-=- ⎪⎝⎭，()()()sin 2sin 2x x g x g x =--=--=，所以()g x 为奇函数，满足要求，故选：D ．7．函数sin()y A x ωϕ=+的部分图象如图所示，则A ．2sin(2)6y x π=-B ．2sin(2)3y x π=-C ．2sin(+)6y x π=D ．2sin(+)3y x π=【正确答案】A【详解】试题分析：由题图知，2A =，最小正周期2[(36T πππ=--=，所以22πωπ==，所以2sin(2)y x ϕ=+.因为图象过点(,2)3π，所以22sin(2)3πϕ=⨯+，所以2sin()13πϕ+=，所以22()32k k Z ππϕπ+=+∈，令0k =，得6πϕ=-，所以2sin(2)6y x π=-，故选A.三角函数的图象与性质【名师点睛】根据图象求解析式问题的一般方法是：先根据函数=sin()y A x h ωϕ++图象的最高点、最低点确定A ，h 的值，由函数的周期确定ω的值，再根据函数图象上的一个特殊点确定φ值．8．若2233x y x y ---<-，则（）A ．ln(1)0y x -+>B ．ln(1)0y x -+<C ．ln ||0x y ->D ．ln ||0x y -<【正确答案】A【分析】将不等式变为2323x x y y ---<-，根据()23t tf t -=-的单调性知x y <，以此去判断各个选项中真数与1的大小关系，进而得到结果.【详解】由2233x y x y ---<-得：2323x x y y ---<-，令()23t tf t -=-，2x y = 为R 上的增函数，3x y -=为R 上的减函数，()f t ∴为R 上的增函数，x y ∴<，0y x ->Q ，11y x ∴-+>，()ln 10y x ∴-+>，则A 正确，B 错误；x y -Q 与1的大小不确定，故CD 无法确定.故选：A.本题考查对数式的大小的判断问题，解题关键是能够通过构造函数的方式，利用函数的单调性得到,x y 的大小关系，考查了转化与化归的数学思想.二、多选题9．下列几种说法中，正确的是（）A ．“x y >”是“22x y >”的充分不必要条件B ．命题“Z x ∀∈，20x >”的否定是“0Z x ∃∈，200x ≤”C ．若不等式20x ax b +-<的解集是(2,3)-，则20ax x b -+>的解集是(3,2)-D ．“,0()3k ∈-”是“不等式23208kx kx +-<对一切x 都成立”的充要条件【正确答案】BC【分析】利用充分必要条件的定义可判断A ；由命题的否定可判断B ；由不等式的解法可判断C ；由不等式恒成立求出k 的取值范围，再由充分必要条件的定义可判断D ．【详解】对于A ，x ＞y 不能推出x 2＞y 2，例如x ＝﹣1，y ＝﹣2，x 2＞y 2也不能推出x ＞y ，例如x ＝﹣2，y ＝﹣1，故“x ＞y ”是“x 2＞y 2”的既不充分也不必要，故A 错误；对于B ，命题“Z x ∀∈，20x >”的否定是“0Z x ∃∈，200x ≤”，故B 正确；对于C ，若不等式x 2+ax ﹣b ＜0的解集是（﹣2，3），则﹣2，3是方程x 2+ax ﹣b ＝0的两个根，由根与系数的关系可得﹣a ＝﹣2+3，﹣b ＝﹣6，可得a ＝﹣1，b ＝6，所以ax 2﹣x +b ＞0即为﹣x 2﹣x +6＞0，即x 2+x ﹣6＜0，解得﹣3＜x ＜2，可得不等式ax 2﹣x +b ＞0的解集为（﹣3，2），故C 正确；对于D ，不等式23208kx kx +-＜对一切x 都成立，当k ＝0时，不等式38-＜0恒成立，当k ≠0时，0,Δk <＝k 2﹣4×2k ×（38-）＜0，解得﹣3＜k ＜0，综上，k ∈（﹣3，0]，所以“k ∈（﹣3，0）”是“不等式23208kx kx +-＜对一切x 都成立”的充分不必要条件，故D 错误．故选：BC ．10．下列结论正确的是（）A ．76π-是第三象限角B ．若圆心角为3π的扇形的弧长为π，则该扇形的面积为32πC ．若角α的终边上有一点()3,4P -，则3cos 5α=-D ．若角α为锐角，则角2α为钝角【正确答案】BC【分析】A 中，由象限角的定义即可判断；B 中，由弧长公式先求出半径，再由扇形面积公式即可；C 中，根据三角函数的定义即可判断；D 中，取30α=︒即可判断.【详解】选项A 中，75266πππ-=-+，是第二象限角，故A 错误；选项B 中，设该扇形的半径为r ，则3r ππ⋅=，∴3r =，∴2133232S ππ=⨯⨯=扇形，故B 正确；选项C 中，5r ==，cos 53x r α==-，故C 正确；选项D 中，取30α=︒，则α是锐角，但260α=︒不是钝角，故D 错误．故选：BC ．11．已知函数()πsin 26f x x ⎛⎫=- ⎪⎝⎭，则下列说法正确的是（）A ．直线4π3x =是函数()f x 图象的一条对称轴B ．函数()f x 在区间π7π,412⎡⎤⎢⎥⎣⎦上单调递减C ．将函数()f x 图像上的所有点向左平移π6个单位长度，得到函数πsin 26y x ⎛⎫=+ ⎪⎝⎭D ．若()π6f x a f ⎛⎫-> ⎪⎝⎭对任意的π0,2x ⎡⎤∈⎢⎥⎣⎦恒成立，则10a <-.【正确答案】AC【分析】利用三角函数对称轴的性质即可验证选项A ，利用函数的单调性即可验证选项B ，利用图像平移的特性验证选项C ，将问题转化为求最值即可得D 选项.【详解】函数()πsin 26f x x ⎛⎫=- ⎪⎝⎭，对于A ：4π8ππsin 1336f ⎛⎫⎛⎫=-= ⎪ ⎪⎝⎭⎝⎭，故A 正确；对于B ：由于π7π,412x ⎡⎤∈⎢⎥⎣⎦，所以ππ2,π63x ⎡⎤-∈⎢⎥⎣⎦，故函数在该区间上有增有减，故B 错误；对于C ：将函数π()sin(2)6f x x =-的图像上的所有点向左平移π6个单位，得到函数sin 2sin(2)666y x x ⎡ππ⎤π⎛⎫=+-=+ ⎪⎢⎥⎝⎭⎣⎦的图像，故C 正确；对于D ：函数()π6f x a f ⎛⎫-> ⎪⎝⎭，整理得π1sin(2)62a x <--，即求出函数()π1sin(262g x x =--的最小值即可，由于π0,2x ⎡⎤∈⎢⎥⎣⎦，所以ππ5π2666x ⎡⎤-∈-⎢⎣⎦，故当0x =时取得最小值1-，故1a <-，故D 不正确．故选：AC ．三、单选题12．已知函数()23,02ln ,0x x x f x x x ⎧+-≤=⎨-+>⎩，令()()h x f x k =-，则下列说法正确的（）A ．函数()f x 的单调递增区间为()0,∞+B ．当()4,3k ∈--时，()h x 有3个零点C ．当2k =-时，()h x 的所有零点之和为1-D ．当(),4k ∈-∞-时，()h x 有1个零点【正确答案】D【分析】画出()f x 的图像，然后逐一判断即可.【详解】()f x 的图像如下：由图像可知，()f x 的增区间为()1,0,0,2⎛⎫∞ ⎪⎝⎭-+，故A 错误当()4,3k ∈--时，如图当1334k -<<-时，()y f x =与y k =有3个交点，当134k =-时，()y f x =与y k =有2个交点，当1344k -<<-时，()y f x =与y k =有1个交点，所以当()4,3k ∈--时()y f x =与y k =有3个交点或2个交点或1个交点，即()h x 有3个零点或2个零点或1个零点，故B 不正确；当2k =-时，由2232x x +-=-可得12x =-由2ln 2x -+=-可得1x =所以()h x 的所有零点之和为1212-=-故C 错误；当(),4k ∈-∞-时，由B 选项可知：()y f x =与y k =有1个交点，即()h x 有1个零点，故D 正确；故选：D 四、填空题13．函数()()311log 4,23,2x x x f x x -⎧+-<=⎨≥⎩，则()1f f =⎡⎤⎣⎦______.【正确答案】3首先求出()311log 32f =+=，再将2代入对应的解析式即可求解.【详解】由()()311log 4,23,2x x x f x x -⎧+-<=⎨≥⎩，所以()311log 32f =+=，所以()()211233f f f -===⎡⎤⎣⎦，故3本题考查了求分段函数的函数值，属于基础题.14．已知函数()223f x x ax =-+在区间[]28,是单调递增函数，则实数a 的取值范围是______．【正确答案】2a ≤【分析】求出二次函数的对称轴，即可得()f x 的单增区间，即可求解.【详解】函数()223f x x ax =-+的对称轴是x a =，开口向上，若函数()223f x x ax =-+在区间[]28,是单调递增函数，则2a ≤，故2a ≤．15．已知函数()f x 是定义在R 上的周期为2的奇函数．当01x <<时，()2f x x=，则()()944f f -+=______．【正确答案】8-【分析】根据()()2(),()(),00f x f x f x f x f +==-=-解决即可.【详解】由题知，函数()f x 是定义在R 上的周期为2的奇函数，所以()()2(),()(),00f x f x f x f x f +==-=-因为当01x <<时，()2f x x=，所以()()()9118444f f f -=-=-=-，()()()4200f f f ===，所以()()9484f f -+=-.故8-16．已知π3sin()35x -=，则πcos()6x +=______．【正确答案】35##0.6【分析】已知角与所求角之间存在和为π2的情况，诱导公式求解即可.【详解】根据题意可得ππ2ππ3cos()cos ()sin()6335x x x ⎡⎤+=--=-=⎢⎥⎣⎦.故答案为.35五、解答题17．（1）已知π02α<<，sin α，求tan α的值；（2）若tan 4α=，求()()()πsin π2cos 2sin cos παααα⎛⎫+-+ ⎪⎝⎭--++的值.【正确答案】（1）1tan 2α=；（2）43【分析】（1）根据同角三角函数的商关系，代入计算，即可得到结果；（2）根据题意，先由诱导公式化简，然后将式子化为齐次式，即可得到结果.【详解】（1）∵π02α<<，sin α∴cos α===∴sin 1tan cos 2ααα==（2）∵tan 4α=，∴()()()πsin π2cos sin 2sin tan 42sin cos πsin cos tan 13αααααααααα⎛⎫+-+ ⎪-+⎝⎭===--++--18．已知集合{}3A x a x a =≤≤+，集合105x B xx +⎧⎫=>⎨⎬-⎩⎭，集合{}26720C x x x =-+<.(1)若A B B ⋃=，求实数a 的取值范围；(2)命题:p x A ∈，命题:q x C ∈，若p 是q 的必要不充分条件，求实数a 的取值范围.【正确答案】(1)4a <-或5a >(2)7132a -≤≤【分析】（1）根据分式不等式的解法求出集合B ，利用集合间的基本关系即可求得a 的取值范围；（2）根据必要不充分条件的定义可得C A ，由一元二次不等式的解法求出集合C ，利用集合间的基本关系即可求出a 的取值范围.【详解】（1）解：解不等式105x x +>-得1x <-或5x >，所以{1B x x =<-或}5x >，因为A B B ⋃=，所以A B ⊆所以31a +<-或5a >，解得4a <-或5a >，所以实数a 的取值范围为4a <-或5a >.（2）解：p 是q 的必要不充分条件，所以CA ，解不等式()()267221320x x x x -+=--<，得1223x <<，所以1223C x x ⎧⎫=<<⎨⎬⎩⎭，所以12a ≤且233a +≥，解得7132a -≤≤，所以实数a 的取值范围7132a -≤≤.19．已知函数()π2sin 1(0)3f x x ωω⎛⎫=++> ⎪⎝⎭的最小正周期为π.(1)求π6f ⎛⎫ ⎪⎝⎭的值；(2)求函数()f x 的单调递减区间：(3)若π0,2x ⎡⎤∈⎢⎥⎣⎦，求()f x 的最值.【正确答案】(1)π16f ⎛⎫= ⎪⎝⎭(2)π7ππ,π,Z 1212k k k ⎡⎤++∈⎢⎥⎣⎦(3)最大值为3，最小值为1【分析】（1）由最小正周期，求得ω，得到()f x ，再求6f π⎛⎫ ⎪⎝⎭；（2）整体代入法求函数的单调递减区间；（3）由x 的取值范围，得到π23x +的取值范围，可确定最值点，算出最值.【详解】（1）由最小正周期公式得：2ππω=，故2ω=，所以()π2sin 213f x x ⎛⎫=++ ⎪⎝⎭，所以πππ2sin 211663f ⎛⎫⎛⎫=⨯++= ⎪ ⎪⎝⎭⎝⎭.（2）令ππ3π2π22π,Z 232k x k k +≤+≤+∈，解得π7πππ,Z 1212k x k k +≤≤+∈，故函数()f x 的单调递减区间是π7ππ,π,Z 1212k k k ⎡⎤++∈⎢⎥⎣⎦.（3）因为π0,2x ⎡⎤∈⎢⎣⎦，所以ππ4π2,333x ⎡⎤+∈⎢⎣⎦，当ππ232x +=，即π12x =时，()f x 的最大值为3，当π4π233x +=，即π2x =时，()f x的最小值为1.20．已知幂函数()y f x =的图像经过点()4,16M .(1)求()f x 的解析式：(2)设()()1f x g x x+=，利用定义证明函数()g x 在区间[)1,+∞上单调递增.【正确答案】(1)()2f x x=(2)证明见解析【分析】（1）待定系数法求解即可，（2）利用定义证明即可.【详解】（1）设()a f x x =，则416a =，得2a =，所以()2f x x =.（2）由（1）得()211x g x x x x+==+设[)12,1,x x ∞∈+，且12x x <，∴()()12121211g x g x x x x x ⎛⎫-=+-+ ⎪⎝⎭121211x x x x =-+-()211212x x x x x x -=-+()121211x x x x ⎛⎫=-- ⎪⎝⎭()1212121x x x x x x -=-∵[)12,1,x x ∞∈+，且12x x <，∴12120,1x x x x -<>∴()()120g x g x -<即()()12g x g x <∴函数()g x 在区间[)1,+∞上单调递增.21．某医学研究所研发一种药物，据监测，如果成人在2h 内按规定的剂量注射该药，在注射期间，血液中的药物含量呈线性增加；停止注射后，血液中的药物含量呈指数衰减，每毫升血液中的药物含量()y g μ与服药后的时间t （h ）之间近似满足如图所示的曲线，其中OA 是线段，曲线段AB 是函数()2,0,,t y ka t a k a =≥>是常数的图象，且()()2,8,4,2A B.(1)写出注射该药后每毫升血液中药物含量y 关于时间t 的函数关系式；(2)据测定：每毫升血液中药物含量不少于1g μ时治疗有效，如果某人第一次注射药物为早上8点，为保持疗效，第二次注射药物最迟是当天几点钟？(3)若按（2）中的最迟时间注射第二次药物，则第二次注射后再过1.5h ，该人每毫升血液中药物含量为多少g μ1.4≈）？【正确答案】(1)4,02132,22t t t y t ≤≤⎧⎪=⎨⎛⎫⨯> ⎪⎪⎝⎭⎩(2)13点(3)()6.35g μ【分析】(1)根据函数图象分段求解函数解析式即可；(2)根据题意列出不等式，求解出答案即可；(3)分别求解出第二次注射后每毫升血液中含第一次和第二次服药后的剩余量，相加即为结果.【详解】（1）当02t ≤≤时，4y t =，当2t ≥时，把()()2,8,4,2A B 代入2y ka =（2,0,,t a k a ≥>是常数）得：2482ka ka ⎧=⎨=⎩，解得：1232a k ⎧=⎪⎨⎪=⎩，∴4,02132,22t t t y t ≤≤⎧⎪=⎨⎛⎫⨯> ⎪⎪⎝⎭⎩.（2）设第一次注射药物后最迟过t 小时注射第二次药物，其中2t >.则13212t⎛⎫⨯≥ ⎪⎝⎭，解得：5t ≤，∴第一次注射药物5h 后开始第二次注射药物，即最迟13点注射药物.（3）第二次注射药物1.5h 后每毫升血液中第一次注射药物的含量： 6.5113224y ⎛⎫=⨯= ⎪⎝⎭每毫升血液中第二次注射药物的含量：241.56y g μ=⨯=，所以此时两次注射药物后的药物含量为：()6 6.354g μ+≈22．已知函数()41log 2x a x f x +=（0a >且1a ≠）.(1)试判断函数()f x 的奇偶性；(2)当2a =时，求函数()f x 的值域；(3)已知()g x x =-[]14,4x ∀∈-，[]20,4x ∃∈，使得()()122f x g x ->，求实数a 的取值范围.【正确答案】(1)()f x 是偶函数(2)[)1,+∞(3)()1,2【分析】（1）根据偶函数的定义可判断出结果；（2）根据基本不等式以及对数函数的单调性可求出结果；（3）将[][]124,4,0,4x x ∀∈-∃∈，使得12()()2f x g x ->，转化为min [()]f x min [()2]g x >+，利用换元法求出min [()2]g x +，分类讨论a ，利用函数()f x 的单调性求出()f x 的最小值，代入可求出结果.【详解】（1）因为41()log 02x a xf x a +=>且1)a ≠，所以其定义域为R ，又4114()log log ()22x xa a x x f x f x --++-===，所以函数()f x 是偶函数；（2）当2a =时，241()log 2x x f x +=，因为20x >,4112222x x x x =+≥+,当且仅当21x =，即0x =时取等，所以241()log 2x x f x +=2log 21≥=,所以函数()f x 的值域为[1,)+∞.（3）1[4,4]x ∀∈-，2[0,4]x ∃∈，使得12()()2f x g x -≥，等价于min [()]f x min [()2]g x >+，令t =，[0,4]x ∈，[0,2]t ∈，令2()22h t t t =-+，则()2g x +在[0,4]上的最小值等于()h t 在[0,2]上的最小值，()h t 在[0,1]上单调递减，在[1,2]上单调递增，所以()h t 在[0,2]上的最小值为(1)1h =，所以min [()]1f x ≥.因为()f x 为偶函数，所以()f x 在[4,4]-上的最小值等于()f x 在[0,4]上的最小值，设41()2x x v x +=，则()log ()a f x v x =，任取1204x x ≤<≤，1212124141()()22x x x x v x v x ++-=-12121(22)(1)2x x x x +=--，因为1204x x ≤<≤，所以1222x x <，12220x x -<，120x x +>，1221x x +>，121102x x +->，所以12121(22)(102x x x x +--<，12()()v x v x <，所以41()2x x v x +=在[0,4]上为单调递增函数，当01a <<时，函数()log ()a f x v x =在[0,4]上为单调递减函数，所以4min 441()(4)log 2a f x f +==257log 16a =，所以257log 116a ≥，得25716a ≥（舍）；当1a >，函数()log ()a f x v x =在[0,4]上为单调递增函数，所以m in ()f x (0)f =log 2a =，所以log 21>a ，12a <<.综上得：实数a 的取值范围为()1,2.。

潮南区2020-2021学年度第一学期期末普通高中教学质量监测高二英语试题本试卷共三部分，共8页，满分135分（120分*1.125）。

考试时间120分钟。

注意事项：1. 答题前，考生在答题卡上务必用直径0.5毫米黑色墨水签字笔将自己的姓名、准考证号填写清楚。

2. 单项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，如需改动，用橡皮擦干净后，再选涂其它答案。

3. 非选择题必须用黑色字迹钢笔或签字笔作答，答题前必须先填好答题纸密封线内各项内容。

答案必须写在答题纸上各题目制定区域内相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

第一部分阅读理解（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，从每题所给的四个选项（A,B,C和D）中，选出最佳选项，并在答题卡上将该项涂黑。

AThe Lake District Attractions GuideDalemain Mansion & Historic GardensHistory, Culture & Landscape（景观）. Discover and enjoy 4 centuries of history, 5 acres of celebrated and award-winning gardens with parkland walk. Owned by the Hasell family since 1679, home to the International Marmalade Festival. Gifts and antiques, plant sales, museums & Mediaeval Hall Tearoom.Open：29 Mar-29 Oct，Sun to Thurs.Tearoom，Gardens & Gift Shop：10.30-17.00（16.00 in Oct）.House：11.15-16.00（15.00 in Oct）Town: Pooley Bridge & PenrithAbbot Hall art Gallery & MuseumThose viewing the quality of Abbot Hall’s temporary exhibitions may be forgiven for thinking they are in a city gallery. The impressive permanent collection includes Turners and Romneys and the temporary exhibition programme has Canaletto and the artists from St Ives.Open: Mon to Sat and Summer Sundays. 10.30 -17.00 Sumner.10.30 -16.00 Winter.Town：KendalTullie House Museum & Art GalleryDiscover, explore and enjoy award-winning Tullie House, where historic collections, contemporary art and family fun are brought together in one impressive museum and art gallery. There are four fantastic galleries to visit from fine art to interactive fun, so there’s something for everyone!Open: High Season 1 Apr – 31 Oct: Mon to Sat 10.00 – 17.00, Sun 11.00 – 17.00.Low Season 1 Nov – 31 Mar: Mon to Sat 10.00 – 16.30, Sun 12.00 – 16.30.Town: CarlisleDove Cottage & The Wordsworth MuseumDiscover William Wordsworth’s inspirational home. Take a tour of his Lakeland cottage, walk through his hillside garden and explore the riches of the collection in the Museum. Visit the shop and relax in the café. Exhibitions, events and family activities throughout the year.Open: Daily, 09.30 – 17.30 (last admission 17.00).Town: Grasmere1. When is the House at Dalemain Mansion & Historic Gardens open on Sundays in July?A. 09.30—17.30.B. 10.30—16.00.C. 11.15—16.00.D. 12.00—16.302. What can visitors do at Abbot Hall Art Gallery & Museum?A. Enjoy Romney’s works.B. Have some interactive fun.C. Attend a famous festival.D. Learn the history of a family3. Where should visitors go if they want to explore Wordsworth’s life?A. Penrith.B. Kendal.C. Carlisle.D. Grasmere.BResearchers in Hong Kong showed that when the patients at nursing home with pain enjoyed jokes, funny books and videos, and singing and dancing on a weekly basis, their awareness of pain and loneliness decreased significantly. They also felt happier and more satisfied with life.Others have confirmed that laughter can be associated with increased blood flow, improved immune response, lower blood sugar levels, and better sleep. You don't have to write a joke to obtain the benefits. Merely experiencing humor will do the trick.But there might be an even stronger reason that a sense of humor is born into the human body. Not only does humor make us smarter and healthier, but it may also make us more attractive to others. Because creating and appreciating jokes both require us to make connections between many separate pieces of information, having a sense of humor shows that we are knowledgeable and that we know how to think about it in new ways.It is amazing that intelligence is related to humor. A University of New Mexico study of 400 college students found that those who scored highest on intelligence tests also scored high on humor ability.Humor has several unique powers. It forces people to consider different perspectives. It brings people together. If they are laughing together at something, they must agree with each other on some level. It reduces the pain associated with life's difficulties. It exercises your brain. And it makes you happy. If a healthy sense of humor can make you smarter and happier, then one thing is clear: Finding time in your day for a good joke or two is no laughing matter. To start laughing, check the funniest jokes of all time.4.What does the underlined phrase "do the trick" in paragraph 2 mean?A. Form a joke.B. Make no difference.C. Achieve a good result.D. Leave a deep impression.5.What can we learn about people with a sense of humor?A. They are born healthy.B. They have wide knowledge.C. They tend to reject new thingsD. They like to collect information.6.Which of the following agrees with the author's opinion?A. Humor helps get rid of bad reputation.B. College students usually have humor ability.C. A sense of humor is only developed after birth.D. Intelligent people have a good sense of humor.7.What does the author advise us to do in the last paragraph?A. Enjoy good jokes.B. Treat laughter seriously.C. Develop unique powers.D. Be brave to overcome all difficulties.CIf you are fond of learning languages, you must start learning Chinese. English is the most widely spoken language in the world and it is a more powerful language in all fields. But it is expected that Chinese, which is one of the six official languages of the United Nations (UN), will be the most important language in the coming years. What are reasons behind the rapid spread of this language and why should you learn it?My desire to learn languages was the reason why I studied many Latin languages and in the process, I deepened my knowledge of the languages and literature. However, I did not feel self-sufficient (自给自足的) from this knowledge. I gained a lot of information about the Western world. But my thinking was always about Asian civilizations. I always felt I needed to learn Chinese to be a global citizen because “without learning Chinese, we see with one eye”.My contact with many international organizations and government institutions (机构) made me believe that Chinese is one of the most important languages of our time. After visiting the world’s most celebrated capitals such as Paris and London, I discovered that Chinese language can be seen everywhere in these places. In the subway in Paris, you will hear instructions in French, English and Chinese. In London, for example, the Chinese language was introduced for instruction in schools.China is an important political (政治的) and economic country because it is making great economic progress that has never been seen before. The world is watching China with great surprise, and this peaceful Chinese rise makes us decide to focus on learning the Chinese language and knowing more about Chinese culture.To be a global citizen these days, I would advise you to learn Chinese. It will add more beauty to your life and allow you to better understand Chinese civilization.8. What can we learn about languages in the future according to the first paragraph?A. Some of them will disappear.B. More official languages will be added to the UN.C. English will still have an advantage over others.D. Chinese language will probably be second to none.9. What do the examples in Paragraph 3 show?A. Paris is a wonderful capital city worth a visit.B. Chinese learning is very popular in schools in London.C. People in some western countries like speaking Chinese.D. Chinese language is playing an important role in the world.10. What mainly causes Westerners to start learning Chinese according to the text?A. The rise of Chinese economy.B. Their love for language learning.C. Their curiosity about Asian culture.D. The long history of Chinese civilization.11. What could be the best title for the text?A. How to be a global citizenB. Chinese — the language you must learnC. What you should know about Chinese cultureD. Chinese economy — the most powerful engineDWith the young unable to afford to leave home and the old at risk of isolation(孤独), more families are choosing to live together.The doorway to peace and quiet, for Nick Bright at least, leads straight to his mother-in-law, she lives on the ground floor, while he lives upstairs with his wife and their two daughters.Four years ago they all moved into a three-storey Victorian house in Bristol - one of a growing number of multigenerational families in the UK living together under the same roof. They share a front door and a washing machine, but Kathryn Whitehead has her own kitchen, bathroom, bedroom and living room on the ground floor.“We floated the idea to my mum of sharing at a house,” says Rita Whitehead. Kathryn cuts in: “We spoke more with Nick because I think it’s a big thing for Nick to live with his mother-in-law.”And what does Nick think? “From my standpoint, it all seems to work very well. Would I recommend it? Yes, I think I would.”It’s hard to tell exactly how many people agree with him, but research indicates that the numbers have been rising for some time. Official reports suggest that the number of households with three generations living together had risen from 325,000 in 2002 to 419,000 in 2013.Other varieties of multigenerational family are more common. Some people live with their elderly parents; many more adult children are returning to the family home, if they ever left. It is said that about 20% of 25-34-year-olds live with their parents, compared with 16% in 1991.The total number of all multigenerational households in Britain is thought to be about 1.8 million.Stories like that are more common in parts of the world where multigenerational living is more firmly rooted. In India, particularly outside cities, young women are expected to move in with their husband’s family when they get married.12. Who mainly uses the ground floor in the Victorian house in Bristol?A. Nick.B. Rita.C. KathrynD. The daughters.13. What is Nick’s attitude towards sharing the house with his mother-in-law?A. Positive.B. Carefree.C. TolerantD. Unwilling.14. What is the author’s statement about multigenerational family based on?A. Family traditions.B. Financial reports.C. Published statistics.D. Public opinions.15. What is the text mainly about?A. Lifestyles in different countries.B. Conflicts between generations.C. A housing problem in Britain.D. A rising trend of living in the UK.第二节（共5小题；每小题2.5分，满分12.5分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。