湖北省武汉外国语学校高一英语上学期期末考试试题

武汉外国语学校2024-2025 学年度上学期学情调研(一)九年级英语试题第一卷(选择题共80分)第一部分听力部分第一节(共4小题，每小题1分，满分4分)听下面 4 个问题，每个问题后有三个答语，从题中所给的A、B、C 三个选项中选出最佳选项。

听完每个问题后，你都有5秒钟的时间来作答和阅读下一小题。

每个问题仅读一遍。

1. A. She likes gardening. B. She’s a nurse. C. She visited a friend.2. A. Along the river. B. After dinner. C. With my mother.3. A. It’s so cool. B. I have no brothers. C. His name is Tony.4. A. He’s fourteen. B. Matt. C. In the classroom.第二节(8 小题，每小题1分，满分8分)听下面8段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10 秒钟的时间来作答有关小题和阅读下一小题。

每段对话仅读一遍。

5. What’s the date today?A. It’s April 1st.B. It’s May 1st.C. It’s June 1st.6. What was the weather like yesterday afternoon?A. Windy.B. Rainy.C. Sunny.7. What does the girl mean?A. She was late for class yesterday.B. She didn’t see the new head teacher yesterday.C. She couldn’t find the new head teacher’s office yesterday.8. What will the woman do this afternoon?A. Give the man a lift.B. Visit the museum.C. Drop in on the man.9. What are the speakers mainly talking about?A. How to avoid bad weather.B. When to visit Mexico City.C. Whether to make a trip plan.10. What would the boy probably say next?A. What’s up?B. What a pity!C. Lucky you!11. Where does the conversation take place probably?A. B. C.12. What can we inter from the man’s words?A. The article wasn’t very good.B. The article was about an accidentC. Only part of the article was published.第三节(共13小题，每小题1分，满分13分)听下面 4 段对话或独白，每段对话或独白后几个小题，从题中所给的A、B、C 三个选项中出最佳选项。

湖北省武汉市武汉外国语学校2024届化学高一第一学期期末综合测试模拟试题考生请注意：1．答题前请将考场、试室号、座位号、考生号、姓名写在试卷密封线内，不得在试卷上作任何标记。

2．第一部分选择题每小题选出答案后，需将答案写在试卷指定的括号内，第二部分非选择题答案写在试卷题目指定的位置上。

3．考生必须保证答题卡的整洁。

考试结束后，请将本试卷和答题卡一并交回。

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、下表中,对陈述Ⅰ、Ⅱ的正确性及两者间是否具有因果关系的判断都正确的是A．A B．B C．C D．D2、将少量铁粉加入到下列溶液中，铁粉溶解，但不产生气体和沉淀的是（）A．稀H2SO4溶液B．FeCl3溶液C．CuSO4溶液D．NaCl溶液3、下列关于二氧化硅的说法错误的是( )A．二氧化硅是一种非金属氧化物B．二氧化硅不存在单个的分子C．二氧化硅中硅元素与氧元素的质量比为7∶8D．二氧化硅分子由一个硅原子和两个氧原子构成4、在下列各溶液中，所给离子一定能大量共存的是A．使酚酞试液变红的溶液：Na＋、Cl－、SO42－、Fe3＋B．小苏打溶液：K＋、SO42－、Cl－、H＋C．与铝反应产生H2的溶液中：Na＋、K＋、CO32－、Cl－D．室温下，强酸性溶液中：Na＋、Fe3＋、NO3－、SO42－5、下列关于Na、Al、Fe三种金属单质的叙述中正确的是()A．常温下，Na能被氧气氧化，Al、Fe不能B．钠能与冷水剧烈反应，Al、Fe能与沸水反应C．将三种金属单质分别投入CuSO4溶液中，都能置换出单质铜D．等质量的三种金属与足量稀硫酸反应，铝放出H2最多6、下列各项操作过程中，发生“先产生沉淀，后沉淀又溶解”现象的是①向Fe(OH)3胶体中逐滴加入过量的稀硫酸②向AlCl3溶液中通入过量的NH3③向Ba(OH)2溶液中通入过量CO2④向NaAlO2溶液中逐滴加入过量的盐酸A．①②B．①③④C．①③D．③④7、加入NaOH溶液并加热，用湿润pH试纸靠近容器口时，试纸变蓝，这是在检验A．Al3+B．HCO3－C．SO42－D．NH4+8、为检验溶液里的某种离子,进行下列实验,其中结论正确的是（）A．某溶液进行焰色反应为黄色，则该溶液中一定含有Na+，一定没有K+B．先滴加KSCN溶液无明显现象，再滴加氯水显红色，证明某溶液中含有Fe2+ C．加AgNO3溶液有白色沉淀生成，再加稀盐酸沉淀不溶解,溶液一定含Cl-D．加入稀HCl,产生使澄清石灰水变浑浊的无色气体，则溶液中一定含CO32-9、下列叙述正确的是（）A．Na2O、Na2O2组成元素相同，与CO2反应产物也相同B．将CO2通入BaCl2溶液可生成BaCO3沉淀C．将CO2通入次氯酸钙溶液可生成次氯酸D．0.12g石墨中含有6.02×1022个碳原子10、下列叙述中不正确的是( )A．氧化铝固体不溶于水，不导电，它是非电解质B．氧化铝熔点很高，是一种较好的耐火材料C．氧化铝是一种白色的固体，是冶炼铝的原料D．铝表面形成的氧化铝薄膜可防止铝被腐蚀11、在实验室中，对下列事故或药品的处理方法正确的是（）A．金属钠失火时可用水灭火B．少量的金属钠应保存在煤油中C．少量浓硫酸沾在皮肤上，立即用氢氧化钠溶液冲洗D ．有大量的氯气泄漏时，应用浸有弱碱性溶液的毛巾捂住口鼻向低处跑12、下列关于铜及其化合物的说法不正确的是（ ）A ．人类对金、银、铜、铁、铝的认识与其金属活动性顺序无关B ．将灼热的铜丝伸入盛满氯气的集气瓶中，有棕黄色的烟生成C ．蓝色硫酸铜晶体受热转化为白色硫酸铜粉末是化学变化D ．工业上可用空气、Cu 、稀硫酸来制备CuSO 413、共价键、离子键、分子间作用力都是微粒间的作用力，下列物质中只含有以上一种作用力的晶体是（ ） A ．SiO 2 B ．CCl 4 C ．S 2 D ．NaOH14、下列实验现象与对应化学方程式都正确的是（ ）A ．向Na 2SiO 3溶液中加入稀盐酸，边加边振荡，有硅酸胶体产生；Na 2SiO 3+2HCl=H 2SiO 3+2NaClB ．氢气在氯气中安静的燃烧，发出淡蓝色火焰，瓶口出现白雾；H 2+Cl 22HClC ．FeSO 4溶液中加入NaOH 溶液时，生成的白色絮状沉淀迅速变为灰绿色，最后变成红褐色；2Fe(OH)2+O 2+H 2O=2Fe(OH)3D ．钠投入水中，浮在水面熔成小球，在水面快速移动，得到的溶液显碱性；2Na+2H 2O=2NaOH+H 2↑15、国际互联网上报道：“目前世界上有近20亿人患有缺铁性贫血．”这里的铁是指A ．铁单质B ．铁元素C ．四氧化三铁D ．硫酸铁16、下列关于摩尔质量的叙述，不正确的是（ ）A ．水的摩尔质量是氢气摩尔质量的9倍B ．2mol 水的摩尔质量是1 mol 水的摩尔质量的2倍C ．磷酸的摩尔质量单位为g·mol －1时，其数值等于6.02×1023个磷酸分子的质量D ．氢气的摩尔质量单位为g·mol －1时，其数值等于氢气的相对分子质量17、将2.56gCu 和一定量的浓HNO 3反应，随着Cu 的不断减少，反应生成气体的颜色逐渐变浅，当Cu 反应完毕时，共收集到气体1.12L （标准状况），则反应中消耗HNO 3的物质的量为A ．0.05molB ．0.13molC ．1molD ．1.05mol18、下列各组离子能在酸性溶液中大量共存的是( )A ．Na +、Fe 2+、NH 4+、SO 42－B ．Na +、Ca 2+、Cl -、ClO －C ．Mg 2+、Ba 2+、CO 32－、SO 42－D ．K +、NH 4+、Cl -、HCO 3－19、下列反应的离子方程式书写正确的是A ．氯化铜溶液与铁粉反应：22Cu Fe===Fe Cu ++++B ．稀24H SO 与铁粉反应：322Fe 6H 2Fe 3H +++===+↑C ．氢氧化钡溶液与稀24H SO 反应：2244B ==a SO SO =Ba +-+↓D ．碳酸钙与盐酸反应：2322CO 2H ===H O CO -+++↑20、在下列变化中，必须加入合适的氧化剂才能实现的是A ．CuO→CuB ．SO 2→SC ．CaCO 3→CO 2D ．FeCl 2→FeCl 321、安徽省庐江县有丰富的钒矿资源——明矾，有关明矾的说法正确的是（ ）A ．明矾既可以除去水中的悬浮物，也可以杀菌消毒B ．可以通过电解明矾溶液的方法来制取金属铝C ．用酒精灯加热铝箔至熔化，铝并不滴落，说明氧化铝的熔点比铝高D ．明矾溶液与某一溶液混合有白色沉淀生成，该溶液一定是碱溶液22、下列有关说法正确的是A ．将7.8 g Na 2O 2溶于1 L 水可以配成0.2 mol·L －1的NaOH 溶液B ．定容时仰视容量瓶刻度线，所得溶液浓度偏低C ．标准状况下，22.4 L 乙醇含有的分子数目为1.0N AD ．将FeCl 3的饱和溶液滴到NaOH 溶液中可制得Fe(OH)3胶体二、非选择题(共84分)23、（14分）下图所涉及的物质均为中学化学中的常见物质，其中C 、D 均为气体单质，E 是固体单质，A 物质的焰色反应火焰为紫色，F 是黑色晶体，它们存在如图转化关系，反应中生成的水及次要产物均的已略去。

武汉外国语学校2024—2024学年度上学期期中考试高一历史试题考试时间：2014年11月20日下午4:40—6:10命题人：胡志刚满分：100分第Ⅰ卷（选择题，共60分）本卷共30小题，每小题2分，共60分。

在每小题列出的四个选项中，只有一项是符合题目要求的。

1.《荀子·儒效》记载：“(周公)兼制天下，立七十一国，姬姓独居五十三人。

”材料所述现象，对后世影响最深远的是A．稳定了西周的政治秩序B．形成了家国一体的观念C．有利于中心集权的建立D．导致了诸侯争霸的局面2.商朝灭亡后，“小邦”周还没有力气对商朝故地实行有效的限制。

为此，武王的做法是A．“封商纣子禄父殷之馀民” B．“武王征九牧之君”C．“乃褒封神农氏之后于焦” D．“于是封功臣谋士”3.《史记》记载，刘邦称帝之后以旧礼尊其父，有人劝告刘父：“今高祖虽子，人主也，太公虽父，人臣也。

奈何令人主拜人臣！如此，则威重不行。

”此后其父以尊礼待刘邦。

从文中可以看出A．宗法关系要听从君权B．刘邦违反了纲常伦理C．汉初宗法制趋于崩溃D．君臣关系等级森严4.柳宗元认为，秦末农夫起义“咎在人怨，非郡邑之制失也”；西汉七国之乱“有叛国而无叛郡”，“秦制之得亦明矣”。

下列哪种说法最符合材料原意A．郡县制与秦末农夫斗争没有关系B．七国之乱因汉初分封而爆发C．郡县制有利于中心集权D．郡县制取代分封制是历史的必定5.历史课上，探讨中国古代的管制演化，同学们征引史料，各抒己见。

甲说：方镇太重，君弱臣强……惟稍夺其权，制其钱谷，收其精兵。

乙说：天下之兵，本于枢密，有发兵之权而无握兵之重。

丙说：置中书省以治内，分行省以治外，……而天下事方如指掌矣。

丁说：……青海军兴，始设军机房，领以亲王大臣。

其中涉及宋代文官体制的史料是A．甲说、丙说B．甲说、乙说C．甲说、丁说D．乙说、丙说6.元朝时右丞相铁木迭儿掌管宣政院，他的儿子也为宣政院使。

《元史·奸臣传》记载了时人的指责，称其“无功于国，尽居贵显”。

2022-2023学年湖北省武汉外国语学校高一上学期期末数学试题一、单选题1．已知{}{}2|20,Z|3<213A x x x B x x =+-==∈--<，则A B =（ ）A ．{}1B ．{}1,2C ．{}1,2-D ．{}|12x x -<<【答案】A【分析】化简集合,A B ，然后用交集运算即可得到答案【详解】因为{}{}2|202,1,A x x x =+-==-{}{}{}Z|3<213Z|1<20,1B x x x x =∈--<=∈-<=，所以{}1A B ⋂= 故选：A2．下列命题中不正确的是（ ）A ．对于任意的实数a ，二次函数2y x a =+的图象关于y 轴对称B ．存在一个无理数，它的立方是无理数C ．存在整数x 、y ，使得245x y +=D ．每个正方形都是平行四边形 【答案】C【分析】利用二次函数的对称性可判断A 选项；利用特殊值法可判断B 选项；分析可知24x y +为偶数，可判断C 选项；利用正方形与平行四边形的关系可判断D 选项.【详解】对于A 选项，对于任意的实数a ，二次函数2y x a =+图象的对称轴为y 轴，A 对；对于B 为无理数，B 对；对于C 选项，若x 、y 为整数，则2x 、4y 均为偶数，所以，24x y +也为偶数， 则245x y +=不成立，C 错；对于D 选项，每个正方形都是平行四边形，D 对. 故选：C.3．化简sin347cos148sin 77cos58+的值为（ ）A B ． C ．12D 【答案】D【分析】利用诱导公式结合两角和的正弦公式化简可得所求代数式的值.【详解】原式()()sin 27077cos 9058sin 77cos58=+++()()2sin 58cos 77cos58sin 77sin 5877sin135sin 18045sin 452=+=+==-==. 故选：D.4．已知直角三角形的面积等于250cm ，则该三角形的周长的最小值为（ ）cm . A．10+B ．20+C ．40 D ．【答案】B【分析】设两条直角边长分别为cm x、100cm x，利用勾股定理结合基本不等式可求得此三角形周长的最小值.【详解】由直角三角形的面积等于250cm 可设两条直角边长分别为cm x 、100cmx，则该直角三角形的周长为()10020cm x x +=， 当且仅当2210010000x x x x x ⎧=⎪⎪⎪=⎨⎪>⎪⎪⎩时，即当10x =时，等号成立. 故该三角形的周长的最小值为20+cm ， 故选：B5．已知函数()()()3e ,ln ,xf x xg x x xh x x x =+=+=+的零点分别为,,a b c ，则,,a b c 的大小顺序为（ ） A ．a b c >> B ．c a b >> C ．b c a >> D ．b a c >>【答案】C【分析】先判断各函数的单调性再根据零点的存在性定理求出函数零点的范围，即可得出答案. 【详解】解：因为函数3e ,ln ,,x y y x y x y x ====都是增函数，所以函数()()()3e ,ln ,xf x xg x x xh x x x =+=+=+都是增函数，又()()1110,010ef f -=-<=>，所以函数()f x 的零点在()1,0-上，即()1,0a ∈-， 因为()1110,11e e g g ⎛⎫=-+<= ⎪⎝⎭，所以函数()g x 的零点在1,1e ⎛⎫ ⎪⎝⎭上，即1,1e b ⎛⎫∈ ⎪⎝⎭，因为()00h =，所以函数()h x 的零点为0，即0c ， 所以b c a >>. 故选：C.6．在平面直角坐标系中，动点M 在单位圆上沿逆时针方向作匀速圆周运动，M 点运动的角速度为πrad/s 6，若点M的初始位置为13⎛ ⎝⎭，则经过3秒钟，动点M 所处的位置的坐标为（ ） A．133⎛⎫ ⎪ ⎪⎝⎭B．13⎛- ⎝⎭C．133⎛⎫- ⎪ ⎪⎝⎭D ．122,33【答案】C【分析】计算出运动3秒钟时动点M 转动的角，再利用诱导公式即可得解. 【详解】解：M 点运动的角速度为πrad/s 6，则经过3秒钟，转了ππ3=rad 62⨯，设点M 的初始位置坐标为()cos ,sinαα，则1cos ,sin 3αα==则经过3秒钟，动点M 所处的位置的坐标为ππcos ,sin 22αα⎛⎫⎛⎫⎛⎫++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，即()sin ,cos αα-，所以经过3秒钟，动点M 所处的位置的坐标为13⎛⎫⎪ ⎪⎝⎭.故选：C.7．已知函数()()1,04ln ,0x x f x x x x ⎧+>⎪=⎨⎪-<⎩，当1a >时，方程()()()2230f x a a f x a -++=的根的个数是（ ） A ．6 B ．5 C ．4 D ．3【答案】A【分析】解方程得()f x a =或()2f x a =，再依次解方程()f x a =，()2f x a =确定满足条件的x 的个数即可.【详解】因为()()()2230f x a a f x a -++=，所以()()()()20f x a f x a --=，所以()f x a =或()2f x a =，因为1a >，所以2a a >，当()f x a =时，若0x >，则14x a x+=，所以24410x ax -+=， 方程24410x ax -+=的判别式216160a ∆=->，方程的根为0x =>或0x =>，若0x <，则()ln x a -=，所以e a x =-，所以方程()f x a =有3个根，同理可得()2f x a =有3个根， 故方程()()()2230f x a a f x a -++=有6个根，故选：A.8．已知函数()sin 6f x x πω⎛⎫=+ ⎪⎝⎭在区间,3ππ⎛⎫ ⎪⎝⎭上单调递减，则正实数ω的取值范围是（ ）A ．302ω<≤ B ．312ω≤≤C ．413ω≤≤D ．4332ω≤≤ 【答案】C【分析】利用整体代换法求出函数()f x 的递减区间，结合集合的包含关系列出不等式组，解之即可. 【详解】由题意知，0ω>， 令322262k x k ππππωπ+≤+≤+， 解得242,Z 33k k x k ππππωωωω+≤≤+∈， 又函数()f x 在区间()3ππ,上单调递减，所以233423k k πππωωπππωω⎧+≤⎪⎪⎨⎪≤+⎪⎩，解得4612,Z 3k k k ω+≤≤+∈，当0k =时，413ω≤≤. 故选：C.二、多选题9．下列说法正确的是（ ）A ．角θ终边在第二象限或第四象限的充要条件是sin cos 0θθ⋅<B ．圆的一条弦长等于半径，则这条弦所对的圆心角等于π3C ．经过4小时，时针转了120D ．若角α和角β的终边关于y x =对称，则有π2π,Z 2k k αβ+=+∈ 【答案】ABD【分析】对于A ，利用三角函数定义结合充分条件和必要条件的定义进行判断即可；对于B ，转化求解弦所对的圆心角即可判断；对于C ，根据任意角的定义即可判断；对于D ，由角的终边得出两角的关系即可【详解】对于A ，因为角θ终边在第二象限或第四象限，此时终边上的点(),x y 的横坐标和纵坐标异号，故sin cos 0θθ⋅=<；因为sin cos 0θθ⋅<，所以sin 0cos 0θθ>⎧⎨<⎩或sin 0cos 0θθ<⎧⎨>⎩，故角θ终边上点坐标(),x y对应为：00><或00<>即00y x >⎧⎨<⎩或00y x <⎧⎨>⎩，所以角θ终边在第二象限或第四象限，综上，角θ终边在第二象限或第四象限的充要条件是sin cos 0θθ⋅<，故A 正确对于B ，圆的一条弦长等于半径，故由此弦和两条半径构成的三角形是等边三角形，所以弦所对的圆心角为π3，故B 正确；对于C ，钟表上的时针旋转一周是360︒-，其中每小时旋转3603012︒︒-=-， 所以经过4小时应旋转120︒-，故C 错误；对于D ，角α和角β的终边关于直线y x =对称，则ππ2(π)2π42k k αβ+=+=+，Z k ∈，故D 正确故选：ABD10．给出下列四个结论，其中正确的是（ ）A ．函数21log sin 2y x ⎛⎫=- ⎪⎝⎭的定义域为()π2π2π,2πZ 33k k k ⎛⎫++∈ ⎪⎝⎭ B ．函数()f x =()g x =C ．函数()2f x +的定义域为[]0,2，则函数()2f x的定义域为2,⎡⎤-⋃⎣⎦D ．函数()2f x 的最小值为2【答案】BC【分析】分别根据对数函数的性质，函数相等，抽象函数的定义域和函数的最值对四个选项逐项验证即可求解.【详解】对于A ，要使函数21log sin 2y x ⎛⎫=- ⎪⎝⎭有意义，则有1sin 02x ->，即1sin 2x >，由正弦函数的图像可知：π5π2π2π,Z 66k x k k +<<+∈，所以函数21log sin 2y x ⎛⎫=- ⎪⎝⎭的定义域为π5π(2π,2π)(Z)66k k k ++∈，故选项A 错误；对于B ，因为函数()f x [1,1]-，函数()g x =[1,1]-，定义域相同，对应法则相同，所以值域也相同，所以函数()f x =()g x =是相同的函数，故选项B 正确；对于C ，因为函数()2f x +的定义域为[]0,2，所以02x ≤≤，则224x ≤+≤，由224x ≤≤2x ≤≤或2x -≤≤()2f x 的定义域为2,⎡⎤-⋃⎣⎦，故选项C 正确；对于D ，因为函数()22f x =(2)t t ≥，则函数可化为1(2)y t t t=+≥，因为函数1y t t =+在[2,)+∞上单调递增，所以15222y ≥+=，也即函数()252f x =≥，所以函数()2f x =的最小值为52，故选项D 错误，故选：BC .11．设正数,a b 满足1a b +=，则有（ ） A ．14ab ≤B ．3314a b +≤C ．148b a b ⎛⎫⋅+≥+ ⎪⎝⎭D ．221124a b b a +≥++【答案】ACD【分析】对于A ，由基本不等式推论可判断选项；对于B ，利用分解因式结合A 分析可判断选项；对于C ，141445411a b b a a b a b a b+⎛⎫⎛⎫⋅+=⋅-+=+- ⎪ ⎪⎝⎭⎝⎭，利用基本不等式可判断选项；对于D ，()()22221223496121212b a a b b a b a b a +-+-+=+=+-++++++，利用基本不等式可判断选项. 【详解】对于A ，由基本不等式推论有()2144a b ab +≤=，当且仅当12a b ==取等号.故A 正确.对于B ，()()()23322313a b a b a b ab a b ab ab +=++-=+-=-，由A 分析可知1144ab ab ≤⇒-≥-，则331134a b ab +=-≥，当且仅当12a b ==取等号.故B 正确.对于C ，()141445454111a b b a a b a b a b a b a b +⎛⎫⎛⎫⎛⎫⋅+=⋅-+=+-=++- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭54888b a a b =++≥+=+2245a b =，即45，b a =-=-时取等号.故C 正确.对于D ，()()()()22222211122349612121212b a b a a b b a b a b a b a --+-+-+=+=+=+-++++++++ ()()()42911491126136412412a b b a b a b a ⎡⎤++⎛⎫=++++-=++-⎢⎥ ⎪++++⎝⎭⎢⎥⎣⎦1113644⎛ ≥+-= ⎝， 当且仅当()()224291a b +=+，即3255，b a ==时取等号.故D 正确. 故选：ACD12．已知函数()f x 的定义域为R ，且()1f x -为奇函数，()1f x +为偶函数，[]1,1x ∈-时，()πcos2f x x=，则下列结论正确的是（ ） A ．()f x 的周期为4B ．10132f ⎛⎫=- ⎪⎝⎭C ．()f x 在()2,4上为单调递减函数D ．方程()5log 0f x x +=有且仅有四个不同的解【答案】BCD【分析】根据题意可知函数()f x 关于()1,0-对称且关于1x =对称，结合周期函数的定义即可判断A ，根据函数的对称性结合函数的解析式即可判断B ，判断出函数在[]2,0-上的单调性，再结合函数的对称性即可判断D ，作出函数()y f x =与函数5log y x =-图象，结合图象即可判断D. 【详解】解：因为()1f x -为奇函数，所以()()11f x f x --=--，即()()2f x f x -=--， 则函数()f x 关于()1,0-对称，又()1f x +为偶函数，所以()()11f x f x -+=+， 即()()2f x f x -=+，即函数()f x 关于1x =对称， 则()()22f x f x +=--，则有()()4f x f x +=-，则()()8f x f x +=， 所以()f x 是以8为周期的周期函数，故A 错误；对于B ，104422π122cos 3333332f f f f f ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫=+=-=-=--=--=- ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭，故B 正确；对于C ，当[]1,0x ∈-时，ππ,022x ⎡⎤∈-⎢⎥⎣⎦，则函数()f x 在[]1,0-上递增，又()10f -=且函数()f x 关于()1,0-对称， 所以函数函数()f x 在[]2,0-上递增， 又因函数()f x 关于1x =对称，所以()f x 在()2,4上为单调递减函数，故C 正确； 对于D ，方程()5log 0f x x +=根的个数，即为函数()y f x =与函数5log y x =-图象交点的个数， 如图，作出两函数的图象，由图可知，两函数的图象有4个交点，即方程()5log 0f x x +=有且仅有四个不同的解，故D 正确.故选：BCD.三、填空题13．函数()()2lg 43f x x x =-+-的值域为_______________.【答案】(],0-∞【分析】求出243x x -+-的取值范围，结合对数函数的基本性质可求得函数()f x 的值域. 【详解】因为()2243211x x x -+-=--+≤，对于函数()f x ，则有20431x x <-+-≤，所以，()()(]2lg 43,0f x x x =-+-∈-∞.故答案为：(],0-∞.14．已知tan 3α=，tan 1β=，则()()cos sin αβαβ+=-____________.【答案】1-【分析】利用两角和的余弦公式、两角差的正弦公式以及弦化切可求得代数式的值. 【详解】因为tan 3α=，tan 1β=，则cos 0α≠，cos 0β≠， 所以，()()cos cos sin sin cos cos cos sin sin cos cos sin cos cos sin sin sin cos cos sin cos cos αβαβαβαβαβαβαβαβαβαβαβαβ-+-==--- 1tan tan 1311tan tan 31αβαβ--⨯===---.故答案为：1-.15．已知,0,2παβ⎛⎫∈ ⎪⎝⎭，1cos 7α=，()11cos 14αβ+=-，则sin β=___________.【分析】利用同角的三角函数的基本关系式和两角差的正弦可求sin β的值. 【详解】因为1cos 7α=，0,2πα⎛⎫∈ ⎪⎝⎭，故sin α=， 而0,2πβ⎛⎫∈ ⎪⎝⎭，故()0,αβπ+∈，而()11cos 14αβ+=-，故()sin αβ+=所以()()()sin sin sin cos cos sin βαβααβααβα=+-=+-+111714=+16．已知函数()422x xf x a a =-+-的最小值为4，则实数=a ____________．【答案】4【分析】根据指数函数的性质，结合4x 与2x 的大小，分0,01,1,1a a a a ≤<<=>四种情况讨论函数()f x 的单调性即可求解作答.【详解】当0a ≤时，函数()4223x x f x a =+⨯-在R 上单调递增，无最小值，不符合题意；当01a <<a >，有42log log log a a =>，则22444223,log ()422,log log 4223,log x x x xx x a x a f x a a x a a x a⎧--⨯+≤⎪=-+⨯-<<⎨⎪+⨯-≥⎩，显然函数()f x 在2(,log ]a -∞上单调递减，而22log log 22(log )42231a a f a a a a =--⨯+=-+<，不符合题意；当1a =时，4223,0422,()30x x x xf x x x --⨯+≤+⨯->⎧=⎨⎩，函数()f x 在(,0]-∞上单调递减，在(0,)+∞上单调递增， min ()0f x =，不符合题意；当1a >a，有422log log log a a =，则44224223,log ()422,log log 4223,log x x x xx x a x af x a a x a a x a⎧--⨯+≤⎪=-⨯+<<⎨⎪+⨯-≥⎩，函数()f x 在4(,log ]a -∞上单调递减，在2[log ,)a +∞上单调递增，当42log log a x a <<时，22)11(()x f x a -+-=，函数()f x 在42(log ,log )a a 上单调递增，则()f x 在4(log ,)a +∞上单调递增，因此44log log min 4()(log )422324a a f x f a a a ==--⨯+=-=，解得4a =，符合要求， 所以实数4a =. 故答案为：4【点睛】思路点睛：在求分段函数的最值时，应先求每一段上的最值，然后比较得最大值、最小值．四、解答题17．已知集合241|1,|212x A x B x a x a x -⎧⎫⎧⎫=≤=≤≤+⎨⎬⎨⎬-⎩⎭⎩⎭. (1)求集合RA ；(2)若A B B =，求实数a 的取值范围.【答案】(1){|1x x ≤或}3x >(2)(1,2](4,)⋃+∞【分析】（1）解分式不等式求得集合A ，进而求得R A .（2）根据B 是否为空集进行分类讨论，由此列不等式来求得a 的取值范围.【详解】（1）242431,10111x x x x x x ---≤-=≤---， 所以()()31010x x x ⎧--≤⎨-≠⎩，解得13x <≤， 所以{|13}A x x =<≤，R A ={|1x x ≤或}3x >.（2）由题意，若A B B =，则B A ⊆，①B =∅时，满足B A ⊆，此时122a a >+，解得4a >； ②B ≠∅时，12211232a a a a ⎧≤+⎪⎪>⎨⎪⎪+≤⎩，解得12a <≤；综上，a 的取值范围为(1,2](4,)a ∈⋃+∞.18．已知函数()15πcos(2)26f x x =-. (1)求函数()f x 在区间[]0,π上的单调递减区间；(2)若πcos 12α⎛⎫+= ⎪⎝⎭()f α. 【答案】(1)单调递减区间是5π11π,1212⎡⎤⎢⎥⎣⎦(2)16【分析】(1)根据余弦函数的单调区间，求出函数在整个定义域上的单调减区间，再与[]0,π取交集即可求解；(2) 令π12βα=+，则π12αβ=-，利用二倍角的余弦可得1cos 23β=-，然后将所求式子利用诱导公式化简即可求解.【详解】（1）15π15π()cos 2cos 22626f x x x ⎛⎫⎛⎫=-=- ⎪ ⎪⎝⎭⎝⎭ 令5π26t x =-，[0,]x π∈ 因为1cos 2y t =的单调递减区间是[2π,2ππ]k k +，Z k ∈， 由5π2π22ππ6k x k ≤-≤+，Z k ∈，得5π11πππ1212k x k +≤≤+，Z k ∈， 即当5π11π[π+,π]1212x k k ∈+，Z k ∈时，()f x 单调递减； 又[0,π]x ∈，0k =时[]5π11π5π11π[,]0,π,12121212⎡⎤=⎢⎥⎣⎦ 所以函数15π()cos 226f x x ⎛⎫=- ⎪⎝⎭，[0,π]x ∈的单调递减区间是5π11π[,]1212. （2）令π12βα=+，则π12αβ=-,因为πcos()12α+=，所以cos β=，则21cos 22cos 13ββ=-=-， 15π15ππ111()cos(2)cos[2()]cos(π2)cos 2262612226f ααβββ=-=--=-=-=， 19．函数()sin 2sin f x x x =+.(1)请用五点作图法画出函数()f x 在[]0,2π上的图象；（先列表，再画图）(2)设()()2m F x f x =-，[]0,2πx ∈，当0m >时，试研究函数()F x 的零点的情况.【答案】(1)答案见解析(2)答案见解析【分析】（1）将()f x 表示为分段函数的形式，然后利用列表法画出()f x 的图象.（2）由()()20m F x f x =-=转化为()y f x =与2m y =的公共点个数，对m 进行分类讨论，由此求得()F x 零点的情况.【详解】（1）3sin ,0π()sin ,π2πx x f x x x ≤≤⎧=⎨-<≤⎩， 按五个关键点列表：()sin 2sin f x x x=+ 0 3 0 1 0描点并将它们用光滑的曲线连接起来如下图所示：（2）因为()()2m F x f x =-，所以()F x 的零点个数等价于()y f x =与2m y =图象交点的个数，设2m t =，0m >，则1t >当20log 3m <<，即13t <<时，()F x 有2个零点；当2log 3m =，即3t =时，()F x 有1个零点；当2log 3m >，即3t >时，()F x 有0个零点.20．已知函数()()()2122m f x m m x m -=--∈R 为幂函数，且()f x 在()0,∞+上单调递增.(1)求m 的值，并写出()f x 的解析式； (2)令()()21g x f x x =+1,12x ⎡⎤∈-⎢⎥⎣⎦，求()g x 的值域. 【答案】(1)3m =，()2f x x =(2)11,2⎡⎤-⎢⎥⎣⎦【分析】（1）根据幂函数的定义以及单调性可得出关于实数m 的等式与不等式，求出m 的值，即可得出函数()f x 的解析式；（2）求出函数()g x 的解析式，在1,02x ⎡⎤∈-⎢⎥⎣⎦时，利用单调性求出函数()g x 的值域；当[]0,1x ∈时，换元213u x ⎡=+⎣，利用二次函数的基本性质可求得函数()g x 的值域，综合可得结果.【详解】（1）解：因为()()()2122m f x m m x m -=--∈R 为幂函数，且()f x 在()0,∞+上单调递增，则222110m m m ⎧--=⎨->⎩，解得3m =，所以，()2f x x =.（2）解：()g x x =1,12x ⎡⎤∈-⎢⎥⎣⎦.①当1,02x ⎡⎤∈-⎢⎥⎣⎦时，()g x x =-1,02⎡⎤-⎢⎥⎣⎦上单调递减， 所以()()min 01g x g ==-，()max 1122g x g ⎛⎫=-= ⎪⎝⎭，此时()11,2g x ⎡⎤∈-⎢⎥⎣⎦；②当[]0,1x ∈时，()g x x =设u =u ⎡∈⎣，可得212u x -=， ()22111111,1222y x u u u ⎡==--=--∈-⎣，此时()1,1g x ⎡∈-⎣， 综上，()g x 的值域为11,2⎡⎤-⎢⎥⎣⎦. 21．已知函数()223log 22a a f x x ax ⎛⎫=-+ ⎪⎝⎭()0,1a a >≠. (1)当2a =时，解不等式()2log 6f x <；(2)[]2,4x a a ∀∈，()1f x ≤，求实数a 的取值范围.【答案】(1){|11x x -<<或24}x << (2)2,13⎡⎫⎪⎢⎣⎭【分析】(1)根据2a =，先求出函数的定义域，在根据函数对数函数的单调性解不等式即可，最后与函数定义域取交集即可求出结果；(2)由()1f x ≤可得：223log ()log 22a a a x ax a -+≤，然后分别在01a <<和1a >两种情况下，根据对数函数的单调性进而求解.【详解】（1）当2a =时，22()log (32)f x x x =-+，要使函数有意义，则有2320x x -+>，解得：2x >或1x <，所以定义域为(,1)(2,)-∞⋃+∞.因为2()log 6f x <，即2326x x -+<，解得：14x -<<，所以不等式解集为{|11x x -<<或24}x <<.（2）由题意，[2,4]x a a ∀∈，223log ()1log 22a a a x ax a -+≤=，①当01a <<时，则有[2,4]x a a ∀∈，22322a x ax a -+≥恒成立， 设223()22a g x x ax a =-+-，对称轴为324x a a =<，()g x 在[2,4]a a 单调递增， 所以2min 3()(2)02g x g a a a ==-≥，得203a a ≤≥或，所以2[,1)3a ∈. ②当1a >时，则有[2,4]x a a ∀∈，22322a x ax a -+≤恒成立， 223()22a g x x ax a =-+-在[2,4]a a 单调递增， 所以2max 21()(4)02g x g a a a ==-≤，得2021a ≤≤，舍去. 综上，2,13a ⎡⎫∈⎪⎢⎣⎭. 22．已知函数()21ax b f x x +=+是定义域R 上的奇函数，且满足()()91210f f +=. (1)判断函数()f x 在区间()0,1上的单调性，并用定义证明；(2)已知1x ∀、()20,x ∈+∞，且12x x <，若()()12f x f x =，证明：122x x +>.【答案】(1)()f x 在()0,1上单调递增，证明见解析(2)证明见解析【分析】（1）利用奇函数的定义可求得b 的值，利用()()91210f f +=可求得a 的值，可得出函数()f x 的解析式，判断出函数()f x 在()0,1上单调递增，然后利用函数单调性的定义可证得结论成立； （2）由()()12f x f x =结合作差法可得出121=x x ，再利用基本不等式可证得结论成立.【详解】（1）解：因为函数()21ax b f x x +=+是定义域R 上的奇函数， 则()()f x f x -=-，即()2211ax b ax b x x -++=-+-+，解得0b =，则()21ax f x x =+， 又()()129122510f f a a +=+=，得1a =，所以()21x f x x =+. 函数()21x f x x =+在()0,1上单调递增，理由如下： 1x ∀、()20,1x ∈，且12x x <，即1201x x ，所以，210x x ->，1210x x -<，2110x +>，2210x +>，则()()()()()()()()()()221221211212122222221212121110111111x x x x x x x x x x f x f x x x x x x x +-+---=-==<++++++，所以()()12f x f x <，则()f x 在()0,1上单调递增. （2）证明：由题意，()()12f x f x =，则有()()()()()()21121222121011x x x x f x f x x x ---==++，因为120x x <<，所以1210x x -=，即121=x x ，所以122x x +>=，得证.。

2025届湖北省武汉外国语学校高一化学第一学期期末联考模拟试题注意事项：1． 答题前，考生先将自己的姓名、准考证号填写清楚，将条形码准确粘贴在考生信息条形码粘贴区。

2．选择题必须使用2B 铅笔填涂；非选择题必须使用0．5毫米黑色字迹的签字笔书写，字体工整、笔迹清楚。

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一、选择题（每题只有一个选项符合题意） 1、下列各组离子,在溶液中可以大量共存的是( ) A ．H +、Na +、NO 3－、OH － B ．Na +、K +、NO 3－、Cl － C ．Ca 2+、K +、AlO 2－、CO 32－D ．Ba 2+、NH 4+、HCO 3－、SO 42－2、既能通过金属单质与足量2Cl 反应得到，也能通过金属单质与酸反应得到的是()A ．2FeClB ．NaClC ．3FeClD ．2CuCl 3、下列有关Na 2CO 3和NaHCO 3的说法错误的是A ．等质量Na 2CO 3和NaHCO 3分别和足量盐酸反应，相同条件下前者生成CO 2少B ．将石灰水分别加入NaHCO 3和Na 2CO 3中，前者不生成沉淀C ．相同条件下Na 2CO 3比NaHCO 3更易溶于水D ．Na 2CO 3固体中含少量NaHCO 3，可用加热法除去 4、下列除去杂质的实验方法正确的是（ ） A ．除去CO 中少量O 2：通过灼热的Cu 网后收集气体 B ．除去K 2CO 3固体中少量NaHCO 3：置于坩埚中加热C ．除去KCl 溶液中的少量MgCl 2：加入适量NaOH 溶液，过滤D ．除去CO 2中的少量HCl ：通入饱和NaHCO 3溶液，收集气体 5、不能使干燥的有色布条褪色的是( ) A ．潮湿的氯气B ．氯水C ．次氯酸溶液D ．液氯6、除去一氧化氮中混入的少量二氧化氮，应将该混合气体通过下列试剂中的( ) A ．碳酸钠溶液B ．碳酸氢钠溶液C ．蒸馏水D ．浓硫酸7、将表面已完全钝化的铝条，插入下列溶液中，一段时间后不会有气泡冒出的是 A ．稀硫酸B ．稀盐酸C ．浓硫酸D ．氢氧化钠溶液8、通过实验得出的结论正确的是A ．某固体试样溶于水，向其中滴加NaOH 溶液，没有产生使湿润红色石蕊试纸变蓝的气体，说明原固体中无4NH +B ．某固体试样溶于稀盐酸，先滴加KSCN 溶液无明显现象，再滴加氯水后显红色，说明原固体中含有2Fe +C ．某固体试样溶于水得无色溶液，滴加少量新制氯水，再滴加少量4CCl ，振荡后静置，下层出现橙红色，说明原固体中含有Br -D ．某固体试样溶于稀盐酸，取少量溶液进行焰色反应为黄色，说明原固体为钠盐9、"NaCl+CO 2+NH 3+H 2O=NaHCO 3↓+NH 4Cl"是著名的"侯氏制碱法"的重要反应。

武汉外国语学校2024—2025学年度上学期10月月考高三数学试卷命题教师： 审题教师：考试时间：2024年10月9日 考试时长：120分钟 试卷满分：150分一、单选题：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. 已知集合，，则（ ）A ．B ．C ．D ．2．复数的共轭复数是（ ）A ．B ．C ．D ．3，且，则与的夹角为（ ）A ．B ．C ．D ．4. 已知，则下列不等关系中不恒成立的是（ ）A ．B ．C ．D ．5. 将体积为1的正四面体放置于一个正方体中，则此正方体棱长的最小值为（ ）A ．3B ．C ．D ．6. 武汉外校国庆节放7天假（10月1日至10月7日），马老师、张老师、姚老师被安排到校值班，每人至少值班两天，每天安排一人值班，同一人不连续值两天班，则不同的值班方法共有（ ）种A ．114B. 120C ．126D ．1327．已知，设函数，若关于的不等式在上恒成立，则的取值范围为（ ）A ．B ．C ．D ．8. 已知函数，，函数，若为偶函数，则的值为（ ）A ．B ．C ．D ．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．下列关于概率统计的知识，其中说法正确的是（ ）A ．数据，0，2，4，5，6，8，9的第25百分位数是1B ．已知随机变量，若，，则C ．若一组样本数据（，2，…，n ）的对应样本点都在直线上，则这组样本数据的相关系数为D ．若事件M ，N 的概率满足，且，则M 与N 相互独立10. 连接抛物线上任意四点组成的四边形可能是（ ）A ．平行四边形B ．梯形C ．有三条边相等的四边形D ．有一组对角相等的四边形11. 设函数，则（ ）A ．当时，直线是曲线的切线B ．若有三个不同的零点，则C ．存在a ，b ，使得为曲线的对称轴D ．当时，在处的切线与函数的图象有且仅有两个交点 三、填空题：本题共3小题，每小题5分，共15分．12. 已知是等差数列的前n 项和，若，，则 ．13. 已知函数，写出函数的单调递减区间．14. 掷一个质地均匀的骰子，向上的点数不小于3得2分，向上的点数小于3得1分，反复掷这个骰子，（1）恰好得3分的概率为 ；（2）恰好得n 分的概率为．（用与n 有关的式子作答）{}2|230A x x x =+-≥{}|22B x x =-≤<A B = []2,1--[)1,2-[]1,1-[)1,2ii 212+-3i5-3i 5i -ib a -=c a c ⊥a b 6π3π23π56π(0,),(0,)22ππαβ∈∈()sin sin sin αβαβ+<+()sin cos cos αβαβ+<+()cos sin sin αβαβ+<+()cos cos cos αβαβ+<+33333a R ∈222,1()ln ,1x ax a x f x x a x x ⎧-+≤=⎨->⎩x ()0f x …R a[]0,1[]0,e []0,2[]1,e ()()f x f x x R =-∈，()15.5=f ()()()1g x x f x =-⋅()1+x g ()0.5-g 32.521.51-(),X B n p :()40E X =()30D X =160n =(),i i x y 1i =132y x =-+12-()()0,1P M ∈()()0,1P N ∈()()1P N M P N +=32()231f x x ax =-+0a =1y =()y f x =()f x 123,,x x x 12312x x x ⋅⋅=-x b =()y f x =02ax ≠()f x 0x x =()y f x =n S {}n a 320S =990S =6S =()()π2,0,cos 2sin ∈+=x xxx f ()x f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. （本题满分13分）已知的面积为，且满足,设和的夹角为，（1）求的取值范围；（2）求函数16.（本题满分15分）如图，已知四棱锥，，侧面为正三角形，底面是边长为4的菱形，侧面与底面所成的二面角为120°．（1）求四棱锥的体积；（2）求二面角的正弦值．17.（本题满分15分）已知函数（1）当时，求曲线在点处的切线方程；（2）若不等式恒成立，求的取值范围．18.（本题满分17分）已知椭圆的左､右焦点分别为，离心率为，且经过点A （1）求椭圆E 的方程；（2）求的角平分线所在直线的方程；（3）在椭圆E 上是否存在关于直线对称的相异两点？若存在，请找出；若不存在，说明理由．19.（本题满分17分）设使定义在区间上的函数，其导函数为.如果存在实数和函数，其中对任意的都有>0，使得，则称函数具有性质．（1）设函数，其中为实数① 求证：函数具有性质；② 讨论函数的单调性；（2）已知函数具有性质，给定，，且，若，求的取值范围．ABC ∆3360≤⋅≤AC AB AB ACθθ()2cos sin 3f πθθθθ⎛⎫=⋅+ ⎪⎝⎭ABCD P -AD PB ⊥PAD ABCD PAD ABCD ABCD P -A PB C --()2()e ln0x af x a a x-=+>a e =()y f x =()()1,1f ()2f x ≥a 2222:1(0)x y E a b a b +=>>12,F F 2352,3⎛⎫ ⎪⎝⎭21AF F ∠l l )(x f ),1(+∞)('x f a )(x h )(x h ),1(+∞∈x )(x h )1)(()('2+-=ax x x h x f )(x f )(a P )(x f 2ln (1)1b x x x +=+>+b )(x f )(b P )(x f )(x g )2(P 为正实数，设m x x x x ,),,1(,2121<+∞∈21)1(x m mx -+=α21)1(mx x m +-=β1,1>>βα12()()()()g g g x g x αβ-<-m2024-2025学年度高三10月月考数学试题参考答案一、选择题题号1234567891011答案DDBCCABDABDBCDABD二、填空题12.13. 14. （1）；（2）三、解答题15、解：（1）由题，可得，又，所以，得到或因为，所以6分（2），化简得进一步计算得，因为，故故可得13分16、解：（1）过点作垂直于平面，垂足为，连接交于，连接，则有，又，所以，因为，所以，又，所以为得中点依题侧面与底面所成的二面角为120°，即有，所以，因为侧面为正三角形，502433ππ⎛⎫⎪⎝⎭，132713425153n -⎛⎫-⋅- ⎪⎝⎭3sin 21==∆θbc S ABC θsin 6=bc 36cos 0≤=⋅≤θbc AC AB 36sin cos 60≤≤θθ33tan≥θ2πθ=()πθ,0∈,62ππθ⎡⎤∈⎢⎥⎣⎦()2cos sin 3f πθθθθ⎛⎫=⋅+ ⎪⎝⎭()21sin 24f θθθ=()1sin 223f πθθ⎛⎫=- ⎪⎝⎭,62ππθ⎡⎤∈⎢⎥⎣⎦22033ππθ⎡⎤-∈⎢⎥⎣⎦，()102f θ⎡⎤∈⎢⎥⎣⎦，P PO ABCD O BO AD E PE AD PB AD PO ⊥⊥,P PB PO =⋂POB AD 平面⊥POB PE 平面⊂PE AD ⊥PD PA =E AD PAD ABCD 32π=∠PEB 3π=∠PEO PAD所以，则，所以7分（2）如图，在平面内过点作得垂线，依题可得两两垂直，以为建立空间直角坐标系可得，，，取得中点为，则因为，所以，由（1)，，知所以，可得所成角即为二面角的平面角，求得，，则则15分17、解：（1）当时，，，所求切线方程为：，即5分（2）转化为，可得构造函数，易得在单调递增所以有，由在单调递增,故可得，即有在恒成立令，，得到，可得时，；时，，所以在时取最大值所以，得到15分323sin4=⋅=πPE 323323sin=⋅=⋅=πPE PO 38323443131=⋅⋅⋅⋅==-PO S V ABCD ABCD P ABCD O OB Ox Ox OB OP ,,Ox OB OP ,,轴轴，轴，x y z ()0,3,2A ()0,0,0P ()0,33,0B PB N ⎪⎪⎭⎫⎝⎛23,233,0N AB AP =PB AN ⊥POB AD 平面⊥AD BC //POB BC 平面⊥PB BC ⊥NA BC ,A PB C --⎪⎪⎭⎫⎝⎛-=23,23,2AN ()0,0,2=BC 72724-=-BC NA sin A PB C --=a e =1()e lnx e f x x -=+0(1)e ln 2f e =+=11()e ,(1)0x f x f x-''=-=)1(02-=-x y 2y =()2≥x f ln 2e ln ln 2a x a x +-+-≥ln 2e ln +2ln 0a x a x x x x +-+-≥+>，()e x g x x =+()g x R ()(ln 2)ln g a x g x +-≥()g x R ln 2ln a x x +-≥ln ln 2a x x ≥-+()∞+，0()2ln +-=x x x h ()011=-='xx h 1=x ()10，∈x ()0>'x h ()∞+∈，1x ()0<'x h ()x h 1=x ()ln 11a h ≥=ea ≥18、解：（1）∵椭圆E 经过点A ，∴，解得E ：；4分（2）由（1）可知，，思路一：由题意，，设角平分线上任意一点为，则得或∵斜率为正，∴的角平分线所在直线为思路二：椭圆在点A 处的切线方程为，根据椭圆的光学性质，的角平分线所在直线的斜率为，∴，的角平分线所在直线即10分（3）思路一：假设存在关于直线对称的相异两点，设，∴∴线段中点为在的角平分线上，即得∴与点A 重合，舍去，故不存在满足题设条件的相异的两点．思路二：假设存在关于直线对称的相异两点，线段中点，52,3⎛⎫⎪⎝⎭23e =222222549123a b a b c c e a ⎧⎪+=⎪⎪⎨=+⎪⎪==⎪⎩32a b c =⎧⎪=⎨⎪=⎩22195x y +=1(2,0)F -2(2,0)F 1:512100AF l x y -+=2:2AF l x =(),P x y 51210213x y x -+=-9680x y --=2390x y +-=21AF F ∠9680x y --=52,3⎛⎫⎪⎝⎭2319x y +=23k =-切21AF F ∠l 32l k =21AF F ∠34:23l y x =-9680x y --=l ()()1122,,,B x y C x y 2:3BC l y x m =-+2222195912945023x y x mx m y x m ⎧+=⎪⎪⇒-+-=⎨⎪=-+⎪⎩BC 25,39m mM ⎛⎫⎪⎝⎭21AF F ∠106803m m --=3m =52,3M ⎛⎫⎪⎝⎭l ()()1122,,,B x y C x y BC ()00,M x y由点差法，，∴，∴，与点A 重合，舍去，故不存在满足题设条件的相异的两点．17分19、解：（1）① ，∵，恒成立，∴函数具有性质；3分② 设，(i) 当即时，，，故此时在区间上递增；(ii) 当时当即时，，，故此时在区间上递增；当即时，，∴时，，，此时在上递减；时，，，此时在上递增.综上所述，当时，在上递增；当时，在上递减，在上递增.9分()()()222121()111b f x x bx x x x x +=-=-+'++1x >()()2101h x x x =>+()f x ()P b ()0f x '>()f x ()1,+∞()0f x '>()f x ()1,+∞x ⎛∈ ⎝()0f x '<()fx ⎛ ⎝()fx ∞⎫+⎪⎪⎭2b ≤()f x ()1,+∞2b >()fx ⎛ ⎝∞⎫+⎪⎪⎭2211222212122222195095195x y x x y y x y ⎧+=⎪⎪⇒+=⎨⎪+=⎪--⎩0121212120552993BC x y y x x k x x y y y -+==-=-=--+0065OM y k x ==:968052,63:5AM OM l x y M l y x --=⎧⎪⎛⎫⇒⎨⎪=⎝⎭⎪⎩()()211u x x bx x =-+>0b -≥0b ≤()0u x >0b >240b ∆=-≤02b <≤()0u x >240b ∆=->2b>1211x x ==<=>，()0u x<x ∞⎫∈+⎪⎪⎭()0u x >()0f x '<（2）由题意， ，又对任意的都有，所以对任意的都有，在上递增.10分∵，，∴①先考虑的情况即，得，此时，∴∴满足题意13分②当时，，，∴∴，∴，不满足题意，舍去16分综上所述，17分()()22()()21()1g x h x x x h x x =-+=-'()h x ()1,x ∈+∞()0h x >()1,x ∈+∞()0g x '>()g x ()1,+∞12(1)mx m x α=+-12(1)m x mx β=-+()()1212,21x x m x x αβαβ+=+-=--12x x αβ-<-()()121221m x x x x --<-01m <<1122(1)x mx m x x α<=+-<1122(1)x m x mx x β<=-+<1212()()(),()()()g x g g x g x g g x αβ<<<<12()()()()g g g x g x αβ-<-1m ≥11112(1)(1)mx m x mx m x x α--≤==++12222(1)(1)m x mx m x mx x β=--+≥=+12x x αβ≤<≤12()()()()g g x g x g αβ≤<≤12()()()()g g g x g x αβ-≥-01m <<。

2023-2024学年湖北武汉外国语学校高一上学期期末英语试题1. Have you seen my Think and Step Step UP? I ______ them since 7:30 a.m.A．have been looking for B．had been looking forC．have looked after D．had looked after2. If it had not been for the help from the Students’ Union, the WFLS 2024 New Year Party ______ so smoothly.D．went A．shouldn’t go B．would go C．wouldn’t havegone3. —Could you meet me at the airport?—I’d like to, but I’m afraid I a very important meeting when you return.A．am attending B．was attendingC．will be attending D．will have attended4. Hurry up! The concert will begin at half past eight. The performers _________ half an hour when you arrive.A． will be playing B． will have playedC． are playing D． have played5. His facial expression suggested that he ______ very impatient,which was why I made a suggestion that he _______out to take in some fresh air.A．should be/ went B．was/ wentC．were/ should go D．was/ go6. If they earlier than expected, they here now.A． had started; would be B． started; might beC． had started; would have been D． will start; might have been7. You didn’t take his advice. ________ his advice, you ________ such a mistake.A．Were you to ta ke; shouldn’t have madeB．If you had taken; would makeC．Had you taken; wouldn’t have madeD．Have you taken; won’t have made8. Later in this chapter cases will be introduced to readers __ consumer complaints have resulted in changes in the law.A．where B．whenC．who D．which9. I was born in New Orleans，Louisiana，a city ________ name will create a picture of beautiful trees and green grass in our mind.A．which B．of which C．that D．whose10. Between the two parts of the concert is an interval, ________ the audience can buy ice-cream.A．when B．whereC．that D．whichMany environmentalists and entrepreneurs are looking for ideas on how to “capture gold” ― that is, how to collect and convert plastic waste into new plastic or fuel.OK, describing plastic waste as potential “gold” may be overdoing it. But the campaigners say that publicizing the notion that plastic is worth something may help reduce the amount of waste that ends up in oceans and the bellies of sea creatures.To that end, they have set up a competition inviting members of the public to submit ideas online. Organizers will take the best ones to the Rio+20 Earth Summit in Rio de Janeiro next month, where they are planning a daylong side event called Plasticity focusing on issues related to plastic pollution.The plastic waste problem is gaining broader attention as environmentalists, scientists, manufacturers and the public become more aware of the sheer volume of the stuff that finds its way into the sea.More than 260 million metric tons of plastic are now produced per year, according to the trade association PlasticsEurope. The majority of that is not recycled. Most of it ends up in landfill, and a significant amount ends up as litter on land, in rivers and in the oceans.Technological advances have made clear that it is possible to reuse much of this plastic by turning it into fuel or new products. Yet the companies that have come up with such solutions have not achieved the economies of scale that would allow them to function profitably. Insufficient waste-collection and recycling systems in most countries also stand in the way of “trash to cash” concept, said Doug Woodring, an environmental entrepreneur in Hong Kong who is among the organizer of the Plasticity forum in Rio.Rather than breast-beating, the forum aims to highlight some of the technologies and ideas out there for collection and reuse. My personal favorite for now is a vacuum cleaner with plastic parts made from plastic waste.11. What do the campaigners like to do exactly?A．To describe plastic waste as potential “gold”.B．To invite members of the public to their forum.C．To collect ideas on how to recycle plastic waste.D．To hold a competition on how to deal with environmental pollution.12. The underlined part “trash to cash” most probably means “”.A．applying modern technology to recycling systemsB．collecting sufficient plastic waste for future useC．establishing many environmental businessesD．turning plastic waste into fuel or new products13. According to the passage, the companies that want to reuse plastic waste .A．have collected enough waste to be usedB．have no practical solutionsC．haven’t reached profitable scaleD．lack technological advancesThe tanker lay in the bay for four days, a few hundred meters from the shore. In this tideless water she lay as still and secure as if fastened to a wall. In a way, she was, for the sandy bottom held her in its grip. Twice the harbor master’s boat went out to her; the second time it brought off a number of the crew. It never occurred to the watchers on shore that the ship was in danger, she looked so calm and seaworthy. From time to time there was activity on board: when a land wind rose in the evenings, the tanker’s engines came to life. Th en the vessel shook herself and strained fiercely, but none of it did her any good. She just stayed where she was in the bay.The July sun blazed down on her flat decks. Occasionally a seaman , stripped to the waist, came out on to the deck with the movements of someone performing a complicated dance, stepping lightly, never resting on that burning metal. Once or twice he kept close to the ship’s rail, with an arm raised against the sunlight, staring at the people on the beach. Throughout the day the air rose in visible waves from the tanker’s decks. When a sea wind blew, it brought with it the heavy smell of oil. At night the ship lay in total darkness.On the fifth morning a thick bank of sea mist filled the bay. It seemed that the tanker had got away in the night and gone into harbor. But this was an illusion. Slowly, as the fog cleared a little, she came into view again but farther out. Soon two figures could be seen at work on her deck. There was the sound of hammering, of metal on metal, and then of something heavy falling on to the deck. At once the watchers on shore were half blinded by a flash of yellow light that enveloped the ship from end to end. The explosion that followed the flash was like a single crack from a giant whip. In a moment the ship, except for a dark line at water level, was lost to sight behind the flames.Two bodies were washed ashore in the bay. They were stripped to the waist, bare-footed and black with flash burns. The right arm of one body was raised to the forehead as if shielding the eyes from some bright light. The other man wore a gold chain round his neck. The tanker burned for nine days and nights.14. What prevented the tanker from sailing into harbor?A．She was waiting for a suitable tide.B．Most of her crew had gone ashore.C．She had run aground on sand.D．Her engines had broken down.15. The people who were watching from the beach _____.A．realized the trouble but could do nothing about itB．offered to help without knowing what to doC．did not know there was anything wrong with the ship D．did not want to put themselves in any danger16. Why did the seaman keep moving about?A．Because the deck was uncomfortable to stand on.B．Because that was the best way to keep his balance.C．Because he was practicing some kind of dance.D．Because he had to pretend he was working.17. How did the mist affect the situation?A．It forced the ship to move farther from the shore.B．It made the seamen’s work harder.C．It allowed the ship to move into the harbor.D．For a time it hid the ship from sight.18. The explosion occurred on the tanker when _____.A．she was unloading her oilB．the fog began to clearC．the two seamen were workingD．she was struck by lightning19. What happened to the two seamen?A．They were blown off the ship and swam ashore.B．They were killed in the explosion.C．They survived but were badly burned.D．They died shortly after reaching the beach.(1)Traveling through the country a couple of weeks ago on business, I was listening to the talk of the late UK writer Douglas Adams’ masterwork The Hitchhiker’s Guide to the Galaxy on the radio and thought- I know, I’ll pick up the next hitchhikers I see and ask them what the state of real hitching is today in Britain.(2) I drove and drove on main roads and side roads for the next few days and never saw a single one.(3) When I was in my teens and twenties, hitchhiking was a main form of long-distance transport. The kindness or curiosity of strangers took me all over Europe, North America, Asia and southern Africa. Some of the lift-givers became friends, many provided hospitality on the road.(4) Not only did you find out much more about a country than when traveling by train or plane, but there was that element of excitement about where you would finish up that night. Hitchhiking featured importantly in Western culture. It has books and songs about it. So what has happened to it?(5) A few years ago, I was asked the same question about hitching in a column of a newspaper. Hundreds of people from all over the world responded with their view on the state of hitchhiking.(6)Rural Ireland was recommended as a friendly place for hitching, as was Quebec, Canada - “if you don’t mind being criticized for not speaking French”.(7) But while hitchhiking was clearly still alive and well in some places, the general feeling was that throughout much of the west it was doomed.(8) With so much news about crime in the media, people assumed that anyone on the open road without the money for even a bus ticket must present a danger. But do we need to be so wary both to hitch and to give a lift?(9) In Poland in the 1960s, according to a Polish woman who e-mailed me, “the authorities introduced the Hitchhiker’s Booklet. The booklet contained coupons for drivers, so each ti me a driver picked somebody, he or she received a coupon. At the end of the season, drivers who had picked up the most hikers were rewarded with various prizes. Everyone was hitchhiking then.”(10) Surely this is a good idea for society. Hitchhiking would increase respect by breaking down barriers between strangers. It would help fight global warming by cutting down on fuel consumption as hitchhiker would be using existing fuels. It would also improve educational standards by delivering instant lessons in geography, history, politics and sociology.(11) A century before Douglas Adams wrote his Hitchhiker’s Guide, another adventure story writer, Robert Louis Stevenson, gave us what should be the hitchhiker’s motto: “To travel hopefully is a better thing than to arrive.” What better time than putting a holiday weekend into practice. Either put it to the test yourself, or help out someone who is trying to travel hopefully with his thumb outstretched.20. In which paragraph(s) does the writer comment on his experience of hitchhiking?A．3 and 4 B．3 C．4 and 5 D．421. According to the public’s responses to a newspaper column, ______.A．Hitchhiking is still popular in Poland.B．Hitchhiking is popular throughout the west.C．Hitchhiking is popular only in North America.D．Hitchhiking is popular in some parts of the world. 22. What is the writer’s attitude towards the practice in Poland?A．Critical. B．Stronglyfavourable. C．Somewhatfavourable.D．Unclear.23. The writer has mentioned all the following benefits of hitchhiking EXCEPT .A．Increasing one’s confidence in strangersB．Promoting mutual respect between strangersC．Enriching one’s knowledgeD．Protecting environmentOne morning a few years ago, Harvard President Neil Rudenstine overslept. For this busy man, it was a sort of alarm: after years of non-stop hard work, he might wear himself out and die an early death.Only after a week’s leave—— during which he read novels, listened to music and walked with his wife on a beach —— was Rudenstine able to return to work.In our modern life, we have lost the rhythm between action and rest. Amazingly, within this world there is a universal but silly saying: “I am so busy.”We say this to one another as if our tireless efforts were a talent by nature and an ability to successfully deal with stress. The busier we are, the more important we seem to ourselves and, we imagine, to others. To be unavailable to our friends and family, and to be unable to find time to relax—— this has become the model of a successful life.Because we do not rest, we lose our way. We miss the guide telling us where to go, the food providing is with strength, the quiet giving us wisdom.How have we allowed this to happen? I believe it is this: we have forgotten the Sabbath, the day of the week—— for followers of some religions—— for rest and praying. It is a day when we are not supposed to work, a time when we devote ourselves to enjoying and celebrating what is beautiful. It is a good time to bless our children and loved ones, give thanks, share meals, walk and sleep. It is a time for us to take a rest, to put our work aside, trusting that there are larger forces at work taking care of the world.Rest is s spiritual and biological need; however, in our strong ambition to be successful and care for our many responsibilities, we may feel terribly guilty when we take time to rest. The Sabbath gives us permission to stop work. In fact, “Remember the Sabbath” is more than simply permission to rest; it is a rule to obey and a principle to follow.24. According to Paragraph 4, a successful person is one who is believed to _______.A．be able to work without stressB．be more talented than other peopleC．be more important than anyone elseD．be busying working without time to rest25. What is the main idea of the passage?A．The Sabbath gives us permission to rest.B．We should balance work with rest.C．It is silly to say “I am busy.”D．We should be available to our family and friends.Have you ever gone to work to find that one of your co-workers is coughing and sneezing all day long? You do your best to keep a safe distance and wonder: Why did he or she come to work when they were ill? The reality for many Americans is that they do not have enough paid sick time each year to afford them the luxury of staying home because they don’t feel well.This problem doesn’t just affect the working employees who are sick, though. In an article by James Warren for Bloomberg Business Week, a second-grade school teacher, Stilli Klikizos shares about th e sick children that must stay in school all day long because their parents can’t get off work to come and get them. In the past school year, she had several children who were unable to be picked up at school who were later diagnosed with H1N1.There is a movement called the Healthy Families Act in Congress that would change this situation for many Americans. The Healthy Family Act would require employers with 15 or more employees to provide 7 paid sick days a year for their workers. These days could be used not only for days when the worker is sick, but the time can also be used when caring for others, or going to routine doctor’s appointments.Those who are against the Act argue that many businesses are struggling to make ends meet owing to recession, and point out that this is the wrong time to force employers to add an additional expense.Those who support the Healthy Families Act say that our nation can’t afford to not take these measures. When an individual goes to work sick, they are possibly infecting their co-workers, clients and customers.According to a report by Katie Couric on the CBS evening news, three fourths of low wage earners get docked when they are sick. Those individuals include daycare workers and restaurant workers, whose health can affect the health of many.26. Why do many Americans still come to work when they are sick?A．They work in high spirits.B．The cost of staying home is great.C．The cost of medical treatment is high.D．They often ignore the illness if not serious.27. In the article mentioned in this passage, James Warren intends to say ______.A．parents shouldn’t leave the sick children at schoolB．children need more thoughtful and considerate careC．adults’ not having enough paid sick time may be bad for childrenD．teachers are responsible for taking good care of children at school28. According to the Healthy Families Act, ______.A．the employees could demand their companies pay for their medical billsB．the employees can use the paid sick days to take care of their sick childrenC．the employees can use the paid sick days to take a trip so as to relax themselves.D．all the employers are required to provide 7 paid sick days a year for their workers 29. Why are some people against the Healthy Families Act?A．The nation can’t afford to do as the Act requires.B．Companies have no such duty to provide paid sick time.C．Usually one’s illness won’t infect his co-workers and customers.D．Many companies’ financial situations are not so good due to the recession.30. The term “get docked” (Para 6) probably means “______ ”.A．be fired B．get paid C．save somemoney D．lose part of wagesI dislike making school lunches.Each morning,I am in a hurry busy slicing cucumbers, washing berries,and filling water bottles,all the while feeling annoyed and even slightly angry. The lunches aren't particularly challenging to prepare.My daughters are content with the food in their lunch box.31 There's no good reason for my annoyance.And then one morning a thought suddenly came into my mind-I am so lucky.Within seconds,those four words bloomed throughout my awareness. 32 I am so lucky to live in a home with electricity,running water,and a functional refrigerator.I am so lucky to live near a grocery store with a plentiful selection of fresh food and snacks and so lucky to have enough money to afford them.I am so lucky to have two daughters who are healthy enough to eat and digest the food I send with them.I can't tell you where this sudden burst of gratitude came from,but I do know this:that small shift immediately made my morning lunch routine extremely easier.Rather than feeling impatient and annoyed,I felt calm and pleased.Rather than mentally complaining through the whole morning,I was able to appreciate my situation. 33I am so grateful for peanut butter.Thank goodness for this magical source of protein that my daughters will actually eat.And jelly,sweet,sweet,jelly.I can't forget sliced bread-oh,the magic of sliced bread!Imagine if I had to cut those slices myself each morning? 34Don't get me wrong-I still don't enjoy making lunches. 35 It gives me just enough space from my bad temper to choose a different response to whatever is going on.A video was breaking the Internet. In the video, a white truck was _______ on a highway, and people were trying to break the _______. Some viewers began to assume that it was the scene ofa(n) _______ , but the truth was quite different.The story all started on a Georgia interstate highway. While _______ a white truck, a woman called Juordin Carter noticed something wrong. The truck was going slowly on such a _______ highway, which was a dangerous situation. She looked over to the right and saw an older man driving but he bent over. Actually he had passed out, leaving his vehicle _______ .Some other drivers also noticed the man. Then they ran alongside the truck, shouting to get the driver’s _______. But it was clear that he wasn’t able to _______ , so they joined together and managed to stop the vehicle. Finding him _______, they had no choice but to break into the truck and get him out. ________ , it was easier said than done. From a stroller to a tire and even a hammer, nothing was breaking through ________ a man called Campbell took a shot at the back window. After that he kicked down the rest of the ________ , slid in and unlocked the door. After the ________ hurried to get to the scene, the driver finally received treatment and got saved.Juordin recorded the incident and shared the video online to show the ________ of the unknown drivers. It feels good to see people, especially ________, regardless of race or religion, come together to help someone else.36.A．pushed B．fixed C．stopped D．moved37.A．window B．door C．floor D．roof38.A．test B．accident C．fight D．game39.A．going after B．getting off C．focusing on D．passing by 40.A．new B．busy C．broad D．familiar 41.A．unattended B．unfounded C．unadjusted D．unsupported 42.A．permission B．attention C．rescue D．signal43.A．escape B．understand C．respond D．continue 44.A．silent B．unconscious C．confused D．uncomfortable 45.A．However B．Therefore C．Still D．Otherwise 46.A．if B．after C．as D．until47.A．crowd B．board C．glass D．tool48.A．woman B．truck C．police D．ambulance 49.A．kindness B．honesty C．ability D．intelligence 50.A．drivers B．strangers C．friends D．volunteers语法填空Newspapers, magazines, even online articles offer reading materials. More strictly speaking, reading means reading books. However, online materials are taking 51 place of books, and reading books seems to go out of fashion nowadays. 52 reading habits have changed can be felt from the amount of time young people spend 53 (sweep) through short videos on their smartphones.Some people claim short videos contain much information, and are easier to look through. But by 54 (compare) with books, short videos have fragmented (碎片化的) and disorganized contents, which could affect people’s understanding of a subject. Reading articles and short posts 55 (play) an important role in integrating knowledge and achievements. But fragmented knowledge could prevent us from acting in a 56 ( practice ) manner or thinking logically.A country’s true development is measured from such 57 (aspect): its philosophical development, its scientific development and its technological development. Cultural development, too, is important. But 58 the help of books, people cannot take the development forward. Since more and more people could 59 (bare) discover the charm of reading now, there is a need, therefore, 60 (seek) novel ways to guide them toward books.61. The explanation was simple but very unusual. A bird had _________up the snake from the ground and then dropped it on to the wires.（根据句意填空）62. This time it was the postman and he wanted me to sign for a ___________ letter! （根据句意填空）63. Very excited, the party dug a hole two feet deep. They finally found a small gold coin which was almost _________.（根据句意填空）64. After a great many loud _________, the race began. Many of the cars broke down on the course and some drivers spent more time under their cars than in them! （根据句意填空）65. Glancing at her __________, he told her that the dress was sold. The woman walked out of the shop angrily and decided to punish the assistant next day. （根据句意填空）66. The tree was planted near the church fifty years ago, but it is only in recent years that it has gained an evil __________. （根据句意填空）67. As soon as I went outside, I forgot all about Madam Bellinsky because my wife hurried towards me. “Where have you been hiding?” she asked _________.（根据句意填空）68. When the fire had at last been put out, the forest ______ ordered several tons of a special type of grass-seed which would grow quickly.69. This would solve the problem of __________, for if a train entered this tunnel, it would draw in fresh air behind it. （根据句意填空）70. Jeremy asked her why this was so and she told him that she did not like to see so many people____________ at him! （根据句意填空）请从方框中选择正确的单词，并用其正确的形式，填入到下面10个句子的空格中。

武汉外国语学校2024—2025学年度上学期10月月考高三数学试卷考试时间：2024年10月9日 考试时长：120分钟 试卷满分：150分一、单选题：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. 已知集合{}2|230A x x x =+-≥，{}|22B x x =-≤<，则A B = （ ）A. []2,1--B. [)1,2- C. []1,1- D. [)1,22. 复数2i12i-+的共轭复数是（ ）A. 3i 5- B. 3i 5 C. i- D. i3. 若2b a = ，=- c a b ，且c a ⊥，则a 与b 的夹角为（ ）A.π6B.π3C.2π3D.5π64. 已知π(0,)2αβ∈∈，则下列不等关系中不恒成立的是（ ）A. ()sin sin sin αβαβ+<+ B. ()sin cos cos αβαβ+<+C ()cos sin sin αβαβ+<+ D. ()cos cos cos αβαβ+<+5. 将体积为1的正四面体放置于一个正方体中，则此正方体棱长的最小值为（ ）A. 3B.C.D.6. 武汉外校国庆节放7天假（10月1日至10月7日），马老师、张老师、姚老师被安排到校值班，每人至少值班两天，每天安排一人值班，同一人不连续值两天班，则不同的值班方法共有（ ）种A. 114B. 120C. 126D. 1327. 已知a R ∈，设函数222,1,()ln ,1,x ax a x f x x a x x ⎧-+=⎨->⎩…若关于x 的不等式()0f x …在R 上恒成立，则a 的取值范围为A. []0,1 B. []0,2 C. []0,e D. []1,e 8. 已知函数()(),R f x f x x =-∈，()5.51f =，函数()()()1g x x f x =-⋅，若()1g x +为偶函数，则()0.5g -的值为（ ）.A. 3B. 2.5C. 2D. 1.5二、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. 下列关于概率统计的知识，其中说法正确的是（ ）A. 数据1-，0，2，4，5，6，8，9的第25百分位数是1B. 已知随机变量(),X B n p ，若()40E X =，()30D X =，则160n =C. 若一组样本数据(),i i x y （1i =，2，…，n ）的对应样本点都在直线132y x =-+上，则这组样本数据的相关系数为12-D. 若事件M ，N 的概率满足()()0,1P M ∈，()()0,1P N ∈且()()1P N M P N +=，则M 与N 相互独立10. 连接抛物线上任意四点组成的四边形可能是（ ）A. 平行四边形B. 梯形C. 有三条边相等的四边形D. 有一组对角相等的四边形11. 设函数32()231f x x ax =-+，则（ ）A. 当0a =时，直线1y =是曲线()y f x =的切线B. 若()f x 有三个不同的零点123,,x x x ，则12312x x x ⋅=-⋅C. 存在,a b ，使得x b =为曲线()y f x =的对称轴D. 当02ax ≠时，()f x 在0x x =处的切线与函数()y f x =的图象有且仅有两个交点三、填空题：本题共3小题，每小题5分，共15分．12. 已知n S 是等差数列{}n a 的前n 项和，若320S =，990S =，则6S =____________．13. 已知函数()()sin ,0,2π2cos xf x x x=∈+，写出函数()f x 的单调递减区间____________．14. 掷一个质地均匀的骰子，向上的点数不小于3得2分，向上的点数小于3得1分，反复掷这个骰子，（1）恰好得3分的概率为____________；（2）恰好得n 分的概率为____________．（用与n 有关的式子作答）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. 已知ABC ∆的面积为3，且满足0AB AC ≤⋅≤ 设AB 和AC的夹角为θ，（1）求θ的取值范围；（2）求函数()2πcos sin 3fθθθθ⎛⎫=⋅+- ⎪⎝⎭值域．16. 如图，已知四棱锥P ABCD -，PB AD ⊥，侧面PAD 为正三角形，底面ABCD 是边长为4菱形，侧面PAD 与底面ABCD 所成的二面角为120︒．（1）求四棱锥P ABCD -的体积；（2）求二面角A PB C --的正弦值．17. 已知函数f(x)=a e x−2+ln ax (a >0)（1）当e a =时，求曲线y =f (x )在点(1,f (1))处切线方程；（2）若不等式()2f x ≥恒成立，求a 的取值范围．18. 已知椭圆2222:1(0)x y E a b a b+=>>的左､右焦点分别为12,F F ，离心率为23，且经过点52,3A ⎛⎫ ⎪⎝⎭（1）求椭圆E 的方程；（2）求12F AF ∠的角平分线所在直线l 的方程；（3）在椭圆E 上是否存在关于直线l 对称的相异两点？若存在，请找出；若不存在，说明理由．19. 设()f x 使定义在区间(1,)+∞上的函数，其导函数为()f x '.如果存在实数a 和函数()h x ，其中()h x 对任意的(1,)x ∈+∞都有()h x >0，使得()()()21f x h x x ax '=-+，则称函数()f x 具有性质()P a ．（1）设函数()f x 2ln (1)1b x x x +=+>+，其中b 为实数① 求证：函数()f x 具有性质()P b ；② 讨论函数()f x 单调性；（2）已知函数()g x 具有性质(2)P ，给定1212,(1,),,x x x x ∈+∞<设m 为正实数，12(1)mx m x α=+-，12(1)m x mx β=-+，且1,1αβ>>，若12()()()()g g g x g x αβ-<-，求m 的取值范围．的的的的武汉外国语学校2024—2025学年度上学期10月月考高三数学试卷考试时间：2024年10月9日 考试时长：120分钟 试卷满分：150分一、单选题：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. 已知集合{}2|230A x x x =+-≥，{}|22B x x =-≤<，则A B = （ ）A. []2,1--B. [)1,2- C. []1,1- D. [)1,2【答案】D 【解析】【分析】根据一元二次不等式求集合A ，即可得交集.【详解】由题意可得：{}(][)2|230,31,A x x x =+-≥=-∞-+∞U ，且{}|22B x x =-≤<，所以A B = [)1,2.故选：D.2. 复数2i12i-+的共轭复数是（ ）A. 3i 5- B. 3i5C. i -D. i【答案】D 【解析】【分析】先根据复数的除法求解，再根据共轭复数的概念求解.【详解】因为()()()()2i 12i 2i5i i 12i 12i 12i 5----===-++-，所以其共轭复数是i .故选：D.3. 若2b a = ，=- c a b ，且c a ⊥，则a 与b 的夹角为（ ）A.π6B.π3C.2π3D.5π6【答案】B 【解析】【分析】根据向量垂直列方程，结合向量数量积的运算以及向量夹角的知识求得正确答案.【详解】因为c a ⊥，所以()22cos ,0a c a a b a a b a a b a b ⋅=⋅-=-⋅=-⋅⋅= ，由于2b a = ，所以212cos ,0,cos ,2a a a a b a b -⋅⋅== ，由于0,πa b ≤≤ ，所以π,3a b = .故选：B4. 已知ππ(0,),(0,)22αβ∈∈，则下列不等关系中不恒成立的是（ ）A. ()sin sin sin αβαβ+<+ B. ()sin cos cos αβαβ+<+C. ()cos sin sin αβαβ+<+ D. ()cos cos cos αβαβ+<+【答案】C 【解析】【分析】由两角和的正弦、余弦公式展开后结合不等式的性质可判断ABD ，举反例判断C ．【详解】,αβ都是锐角，则sin (0,1),cos (0,1),sin (0,1),cos (0,1)ααββ∈∈∈∈，sin()sin cos cos sin sin sin αβαβαβαβ+=+<+，A 正确；sin()sin cos cos sin cos cos αβαβαβαβ+=+<+，B 正确；15αβ==︒时，cos()cos30αβ+=︒=，sin15︒====，sin sin sin15sin15αβ+=︒+︒=>C 错误；()cos cos cos sin sin cos cos cos cos cos αβαβαβαβααβ+=-<<<+，D 正确．故选：C ．5. 将体积为1的正四面体放置于一个正方体中，则此正方体棱长的最小值为（ ）A. 3B.C.D.【答案】C 【解析】【分析】反向思考，求出边长为a 的正方体的最大内接正四面体的体积，结合条件，即可求解.【详解】反向思考，边长为a 的正方体，其最大内接正四面体的体积为33311141323a a a -⨯⨯⨯==，得到33a =，解得a =故选：C.6. 武汉外校国庆节放7天假（10月1日至10月7日），马老师、张老师、姚老师被安排到校值班，每人至少值班两天，每天安排一人值班，同一人不连续值两天班，则不同的值班方法共有（ ）种A. 114 B. 120 C. 126 D. 132【答案】A 【解析】【分析】依据值班3天的为分类标准，逐类解决即可.【详解】因为有三位老师值班7天，且每人至少值班两天，每天安排一人值班，同一人不连续值两天班，所以必有一人值班3天，另两人各值班2天.第一类：值班3天在(1,3,5)、(1,3,6)、(1,4,6)、(2,4,7)、(2,5,7)、(3,5,7)时，共有1113226C C C 72⨯=种不同的值班方法；第二类：值班3天在(1,3,7)、(1,5,7)时，共有11322C C 12⨯=种不同的值班方法；第三类：值班3天在(1,4,7)时，共有111322C C C 12=种不同的值班方法；第四类：值班3天在(2,4,6)时，共有1234C C 18=种不同的值班方法；综上可知三位老师在国庆节7天假期共有72121218114+++=种不同的值班方法.故选：A7. 已知a R ∈，设函数222,1,()ln ,1,x ax a x f x x a x x ⎧-+=⎨->⎩…若关于x 的不等式()0f x …在R 上恒成立，则a 的取值范围为A. []0,1 B. []0,2 C. []0,e D. []1,e 【答案】C 【解析】【分析】先判断0a ≥时，2220x ax a -+≥在(,1]-∞上恒成立；若ln 0x a x -≥在(1,)+∞上恒成立，转化为ln xa x≤在(1,)+∞上恒成立．【详解】∵(0)0f ≥，即0a ≥，（1）当01a ≤≤时，2222()22()22(2)0f x x ax a x a a a a a a a =-+=-+-≥-=->，当1a >时，(1)10f =>，故当0a ≥时，2220x ax a -+≥在(,1]-∞上恒成立；若ln 0x a x -≥在(1,)+∞上恒成立，即ln xa x≤在(1,)+∞上恒成立，令()ln xg x x=，则2ln 1'()(ln )x g x x -=，当,x e >函数单增，当0,x e <<函数单减，故()()min g x g e e ==，所以a e ≤．当0a ≥时，2220x ax a -+≥在(,1]-∞上恒成立；综上可知，a 的取值范围是[0,]e ，故选C ．【点睛】本题考查分段函数的最值问题，关键利用求导的方法研究函数的单调性，进行综合分析．8. 已知函数()(),R f x f x x =-∈，()5.51f =，函数()()()1g x x f x =-⋅，若()1g x +为偶函数，则()0.5g -的值为（ ）A. 3B. 2.5C. 2D. 1.5【答案】D 【解析】【分析】由()1g x +为偶函数，推得()()2g x g x =-，再由()()()1g x x f x =-⋅，求得()f x 关于(1,0)对称，结合()()f x f x =-，推得(4)()f x f x -=，得到()f x 是周期为4的周期函数，根据(5.5)1f =，得到(2.5)1f =，进而求得(0.5)g -的值，得到答案.【详解】因为函数()1g x +为偶函数，可()g x 的图象关于1x =对称，所以()()2g x g x =-，由()()()1g x x f x =-⋅，可得()()()()112x f x x f x -⋅=-⋅-，即()()20f x f x +-=，所以函数()f x 关于(1,0)对称，又因为()()f x f x =-，所以()f x 是定义在R 上的偶函数，所以()()2(2)f x f x f x =--=--，所以()4[(2)2](2)[()]()f x f x f x f x f x -=--=--=-=，即(4)()f x f x -=，所以函数()f x 是周期为4的周期函数，所以(5.5)(1.54)(1.5)( 2.5)(2.5)1f f f f f =+==-==，则(0.5)(2.5)(2.51)(2.5) 1.5g g f -==-=.故选：D.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. 下列关于概率统计知识，其中说法正确的是（ ）A. 数据1-，0，2，4，5，6，8，9的第25百分位数是1B. 已知随机变量(),X B n p ，若()40E X =，()30D X =，则160n =C. 若一组样本数据(),i i x y （1i =，2，…，n ）的对应样本点都在直线132y x =-+上，则这组样本数据的相关系数为12-D. 若事件M ，N 的概率满足()()0,1P M ∈，()()0,1P N ∈且()()1P N M P N +=，则M 与N 相互独立【答案】ABD 【解析】【分析】根据百分位数的定义计算判断A ，由二项分布的数学期望与方差公式计算可判断B ，根据相关系数的定义可判断C, 根据相互独立事件及条件概率的概率公式计算可判断D.【详解】对于选项A ，8个数据从小到大排列，由于825%2⨯=，所以第25百分位数应该是第二个与第三个的平均数0+2=12，故A 正确；对于选项B ，因为(),X B n p ，()40E X =，()30D X =，所以40(1)30np np p =⎧⎨-=⎩，解得1,1604p n ==，故B 正确；对于选项C ，因为样本点都在直线132y x =-+上，说明是负相关且线性相关性很强，所以相关系数为1-，故C 错误.的对于选项D ，由()()1P N M P N +=，可得()()1P N M P N =-，即()()()N P NM P P M =，即()()()N P NM P P M =，所以M 与N 相互独立，故D 正确；故选：ABD.10. 连接抛物线上任意四点组成的四边形可能是（ ）A. 平行四边形B. 梯形C. 有三条边相等的四边形D. 有一组对角相等的四边形【答案】BCD 【解析】【分析】根据题意作出相应的图形，结合抛物线的性质逐项分析判断.【详解】对于选项A ：作两条平行线与抛物线均相交，根据抛物线的性质可知：截得的弦长一定不相等，所以所得的四边形不可能为平行四边形，故A 错误；对于选项C ：任作一条直线垂直与抛物线的对称轴，交抛物线与,A B 两点，则OA OB =，再以A 圆心，OA 为半径作圆，该圆以抛物线必有一个异于坐标原点的交点C ，此时可得OA OB OC ==，符合题意，故C 正确；对于选项B ：任作两条直线垂直与抛物线的对称轴，分别与交抛物线交于,A B 和,C D ，此时AB CD ≠，即ABCD 为梯形，故C 正确；对于选项D ：如图，以AC 为直径作圆，与抛物线交于,,,A B C D ，此时90ABC ADC ∠=∠=︒，符合题意，故D 正确；故选：BCD.11 设函数32()231f x x ax =-+，则（ ）A. 当0a =时，直线1y =是曲线()y f x =的切线B. 若()f x 有三个不同的零点123,,x x x ，则12312x x x ⋅=-⋅C. 存在,a b ，使得x b =为曲线()y f x =的对称轴D. 当02ax ≠时，()f x 在0x x =处的切线与函数()y f x =的图象有且仅有两个交点【答案】ABD 【解析】【分析】根据曲线的切线、函数的零点、曲线的对称轴，直线和曲线的交点个数等知识对选项进行分析，从而确定正确答案.【详解】A 选项，当0a =时，()321f x x =+，令()260f x x ='=解得0x =，且()01f =，此时()f x 在0x =处的切线方程为10y -=，即1y =，正确.B 选项，()()322()231,666f x x ax f x x ax x x a '=-+=-=-，.要使()f x 有三个零点，则0a ≠，若32()231f x x ax =-+有三个不同的零点123,,x x x ，则()()()()1232f x x x x x x x =---()()32123122313123222x x x x x x x x x x x x x x x =-+++++-，通过对比系数可得1231231212x x x x x x -=⇒=-，正确.C 选项，若存在,a b ，使得x b =为曲线()y f x =的对称轴，则()()2f x f b x =-，即()()323223122321x ax b x a b x -+=---+，即3232232223162412212123x ax b b x bx x ab ab ax -=-+--+-，即()3222364330x bx b x b ab a b -+--+=，此方程不恒为零，所以不存在符合题意的,a b ，使得x b =为曲线()y f x =的对称轴，错误.D 选项，当02a x ≠时，()322()231,66f x x ax f x x ax =-+=-'，则()322000000()231,66f x x ax f x x ax =-+=-'，所以()f x 在0x x =处的切线方程为()()()3220000023166y x ax x ax x x --+=--，()()()2320000066231y x ax x x x ax =--+-+，由()()()232000003266231231y x ax x x x ax y x ax ⎧=--+-+⎪⎨=-+⎪⎩，消去y 得()()323220000023123166x ax x ax x ax x x -+=-++--①，由于()()()333322000002222x x x x x x x xx x -=-=-++，()()()222200003333ax ax a x x a x x x x -+=--=--+，所以①可化为()()()()()()2220000000023660x x x xx x a x x x x x ax x x -++--+---=，提公因式0x x -得()()()()22200000023660x x x xx x a x x x ax ⎡⎤-++-+--=⎣⎦，化简得()()()220000223430x x x x a x x ax ⎡⎤-+---=⎣⎦，进一步因式分解得()()2002430x x x x a -+-=，解得010234,2a x x x x -==，由于02a x ≠，所以020x a -¹，所以()0001203234630222x a a x x a x x x ----=-==≠，所以12x x ≠，所以当02a x ≠时，()f x 在0x x =处的切线与函数y =f (x )的图象有且仅有两个交点，正确.故选：ABD 【点睛】关键点点睛：D 选项的解答涉及到切线与曲线交点的个数，利用联立方程组和因式分解的方法，最终得出交点个数的结论，过程完整而严谨．三、填空题：本题共3小题，每小题5分，共15分．12. 已知n S 是等差数列{}n a 的前n 项和，若320S =，990S =，则6S =____________．【答案】50【解析】【分析】设{}n a 首项为1a ，公差为d ，后由等差数列求和公式可得答案.【详解】设{}n a 首项为1a ，公差为d ，由题，则111503320993690109a a d a d d ⎧=⎪+=⎧⎪⇒⎨⎨+=⎩⎪=⎪⎩.则6161550S a d =+=.故答案为：5013. 已知函数()()sin ,0,2π2cos x f x x x =∈+，写出函数()f x 的单调递减区间____________．【答案】2π4π33⎛⎫⎪⎝⎭，【解析】【分析】利用导数判断函数的单调性即可.【详解】()()()()222cos 2cos sin 2cos 12cos 2cos x x xx f x x x +++'==++，()0,2πx ∈，令()()22cos 102cos x f x x +'==+，即2cos 10x +=，解得2π3x =或4π3x =.当2π0,3x ⎛⎫∈ ⎪⎝⎭时，()0f x '>，则()f x 在2π0,3⎛⎫ ⎪⎝⎭上单调递增；当2π4π,33x ⎛⎫∈ ⎪⎝⎭时，()0f x '<，则()f x 在2π4π,33⎛⎫ ⎪⎝⎭上单调递减；当4π,2π3x ⎛⎫∈ ⎪⎝⎭时，()0f x '>，则()f x 在4π,2π3⎛⎫ ⎪⎝⎭上单调递增.综上可知，函数()f x 的单调递减区间为2π4π,33⎛⎫⎪⎝⎭.故答案为：2π4π,33⎛⎫ ⎪⎝⎭.14. 掷一个质地均匀的骰子，向上的点数不小于3得2分，向上的点数小于3得1分，反复掷这个骰子，（1）恰好得3分的概率为____________；（2）恰好得n 分的概率为____________．（用与n 有关的式子作答）【答案】 ①. 1327 ②. 13425153n -⎛⎫-⨯- ⎪⎝⎭【解析】【分析】因为一次得2分，另一次得1分或三次的1分时恰好得3分，进而利用独立重复试验的概率可求（1）；令n P 表示“恰好得n 分”的概率，不出现n 分的唯一情况是得到1n -分以后再掷出一次不小于3的情况，则有1213n n P P --=，进而利用构造等比数列可求（2）.【详解】（1）掷一个质地均匀的骰子，向上的点数不小于3的概率4263=,掷一个质地均匀的骰子，向上的点数小于3的概率2163=.因为一次得2分，另一次得1分或三次得1分时恰好得3分，所以恰好得3分的概率等于21023********C +C ==3332727+⎛⎫⋅⨯⋅ ⎪⎝⎭.（2）令n P 表示“恰好得n 分”的概率，不出现n 分的唯一情况是得到1n -分以后再掷出一次不小于3的情况，因为“不出现n 分”的概率是1n P -，所以“恰好得到1n -分”的概率是1n P -.因为“掷一次得2分”的概率是23，所以有1213n n P P --=，即1213n n P P -=-+，则构造等比数列{}n P λ+，设()123n n P P λλ-=-++，即13532n n P P λ--=-，则513λ-=，35λ=-，所以1323535n n P P -⎛⎫-=-- ⎪⎝⎭，又113P =，1313453515P -=-=-，所以35n P ⎧⎫-⎨⎬⎩⎭是首项为415-，公比为23-的等比数列，即13425153n n P -⎛⎫-=-⨯- ⎪⎝⎭，13425153n n P -⎛⎫=-⨯- ⎪⎝⎭.故恰好得n 分的概率为13425153n -⎛⎫-⨯- ⎪⎝⎭.故答案为：（1）1327；（2）13425153n -⎛⎫-⨯- ⎪⎝⎭.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. 已知ABC ∆的面积为3，且满足0AB AC ≤⋅≤ 设AB 和AC 的夹角为θ，（1）求θ的取值范围；（2）求函数()2πcos sin 3f θθθθ⎛⎫=⋅+- ⎪⎝⎭的值域．【答案】（1）ππ,62⎡⎤⎢⎥⎣⎦ （2）10,2⎡⎤⎢⎥⎣⎦【解析】【分析】（1）根据题意由三角形面积公式可得6cos 0sin θθ≤≤，继而可得tan θ≥或π2θ=，结合θ的范围即可求解；（2）利用和差公式、降幂公式、倍角公式及辅助角公式化简可得1π()sin 223f θθ⎛⎫=- ⎪⎝⎭，由（1）所求的θ的范围可得π23θ-的范围，继而即可求得值域.小问1详解】由题1sin 32ABC S bc θ∆==，【可得6sin bc θ=，又0cos AB AC bc θ≤⋅=≤ ，所以6cos 0sin θθ≤≤得到tan θ≥或π2θ=，因为()0,πθ∈，所以ππ,62θ⎡⎤∈⎢⎥⎣⎦.【小问2详解】()2πcos sin 3f θθθθ⎛⎫=⋅++ ⎪⎝⎭21cos (sin cos 2θθθθ=⋅+21sin 24θθ=+11cos 2sin 242θθ+=-1πsin 223θ⎛⎫=- ⎪⎝⎭，因为ππ,62θ⎡⎤∈⎢⎥⎣⎦，故π2π20,33θ⎡⎤-∈⎢⎥⎣⎦，故可得()10,2f θ⎡⎤∈⎢⎥⎣⎦.16. 如图，已知四棱锥P ABCD -，PB AD ⊥，侧面PAD 为正三角形，底面ABCD 是边长为4的菱形，侧面PAD 与底面ABCD 所成的二面角为120︒．（1）求四棱锥P ABCD -的体积；（2）求二面角A PB C --的正弦值．【答案】（1）（2【解析】【分析】（1）作出四棱锥P ABCD -的高，并计算出高的长度，进而计算出四棱锥P ABCD -的体积.（2）建立空间直角坐标系，利用向量法来求得二面角A PB C --的余弦值，进而计算出正弦值.【小问1详解】过点P 作PO 垂直于平面ABCD ，垂足O ，连接BO 交AD 于E ，连接PE ，因为AD ⊂平面ABCD ，PO AD ⊥，又PB AD ⊥，又,,PO PB P PO PB =⊂ 平面POB ，所以AD ⊥平面POB ，因为,PE BE ⊂平面POB ，所以AD PE ⊥，AD BE ⊥，又PA PD =，所以E 为AD 得中点，所以4BD BA ==，因为侧面PAD 与底面ABCD 所成的二面角为120︒，即有120PEB ∠=︒，所以60PEO ∠=︒，因为侧面PAD 为正三角形，所以4sin 60PE =⋅︒=sin 603PO PE =⋅︒==，所以1144333P ABCD ABCD V S PO -=⋅⋅=⋅⋅=.【小问2详解】在平面ABCD 内过点O 作OB 的垂线Ox ，依题可得,,OP OB Ox两两垂直，为以,,OP OB Ox 为z 轴，y 轴，x 轴建立空间直角坐标系，可得()A ，()0,0,3P，()B，()C -，取PB 得中点为N，则32N ⎛⎫ ⎪ ⎪⎝⎭，因为AP AB =，所以AN PB ⊥，由（1）AD ⊥平面POB ，//BC AD ，知⊥BC 平面POB ，PB ⊂平面POB ，所以BC PB ⊥，可得,BC NA 所成角即为二面角A PB C --的平面角，记为θ，求得32,2NA ⎛⎫=- ⎪ ⎪⎝⎭，()4,0,0BC =-，则cos ,NA BC NA BC NA BC ⋅===⋅则sin θ==17. 已知函数()()2e ln 0x a f x a a x-=+>（1）当e a =时，求曲线y =f (x )在点(1,f (1))处的切线方程；（2）若不等式()2f x ≥恒成立，求a 的取值范围．【答案】（1）2y =（2）ea ≥【解析】【分析】（1）根据导数的几何意义，根据导数求切线的斜率，再代入点斜式方程，即可求解；（2）首先根据指对公式，变形不等式为e ln a +x−2+ln a +x−2≥ln x +e ln x ，x >0，再构造函数()e x g x x =+，结合函数的单调性，转化为不等式ln 2ln a x x +-≥恒成立，再利用参变分离，转化为函数最值问题，即可求解.【小问1详解】当e a =时，()1e e ln x f x x -=+，()01e ln e 2f =+=，()()11e ,10x f x f x-=-'='，所求切线方程为：20(1)y x -=-，即2y =；【小问2详解】()2f x ≥转化为ln 2e ln ln 2a x a x +-+-≥，可得e ln a +x−2+ln a +x−2≥ln x +e ln x ，x >0，构造函数()e x g x x =+，易得()g x 在R 单调递增，所以有()(ln 2)ln g a x g x +-≥，由()g x 在R 单调递增，故可得ln 2ln a x x +-≥，即有ln ln 2a x x ≥-+在()0,∞+恒成立，令()ln 2h x x x =-+，()110h x x-'==，得到1x =，可得()0,1x ∈时，ℎ′(x )>0；()1,x ∞∈+时，()0h x '<，所以ℎ(x )在1x =时取最大值，所以()ln 11a h ≥=，得到e a ≥.18. 已知椭圆2222:1(0)x y E a b a b+=>>的左､右焦点分别为12,F F ，离心率为23，且经过点52,3A ⎛⎫ ⎪⎝⎭（1）求椭圆E 的方程；（2）求12F AF ∠的角平分线所在直线l 的方程；（3）在椭圆E 上是否存在关于直线l 对称的相异两点？若存在，请找出；若不存在，说明理由．【答案】（1）22195x y += （2）9680x y --=（3）不存在，理由见解析【解析】【分析】（1）根据椭圆经过的点的坐标以及离心率解方程组可求得椭圆E 的方程；（2）思路一：利用角平分线上的点的性质，由点到直线距离公式整理可得结论；思路二：求得椭圆在点A 处的切线方程，再由椭圆的光学性质可得平分线所在直线方程；（3）思路一：假设存在关于直线l 对称的相异的两点，联立直线与椭圆方程可得线段BC 中点52,3M ⎛⎫ ⎪⎝⎭与点A 重合，假设不成立；思路二：利用点差法求出65OM k =，联立直线方程可得点52,3M ⎛⎫ ⎪⎝⎭与点A 重合，不合题意，可得结论.【小问1详解】椭圆E 经过点52,3A ⎛⎫ ⎪⎝⎭，23e =可得222222549123a b a b c c e a ⎧⎪+=⎪⎪⎪=+⎨⎪⎪==⎪⎪⎩，解得32a b c =⎧⎪=⎨⎪=⎩因此可得椭圆E 的方程为22195x y +=；【小问2详解】由（1）可知，1(2,0)F -，2(2,0)F 思路一：由题意可知1:512100AF l x y -+=，2:2AF l x =，如下图所示：设角平分线上任意一点为P (x,y )，则51210213x y x -+=-得9680x y --=或2390x y +-=又易知其斜率为正，∴12F AF ∠的角平分线所在直线为9680x y --=思路二：椭圆在点52,3A ⎛⎫ ⎪⎝⎭处的切线方程为2319x y +=，23k =-切根据椭圆的光学性质，12F AF ∠的角平分线所在直线l 的斜率为32l k =，所以12F AF ∠的角平分线所在直线34:23l y x =-，即9680x y --=【小问3详解】思路一：假设存在关于直线l 对称的相异两点B (x 1,y 1),C (x 2,y 2)，设2:3BC l y x m =-+，联立2219523x y y x m ⎧+=⎪⎪⎨⎪=-+⎪⎩可得229129450x mx m -+-=，∴线段BC 中点为25,39m m M ⎛⎫⎪⎝⎭在12F AF ∠的角平分线上，即106803m m --=，解得3m =；因此52,3M ⎛⎫ ⎪⎝⎭与点A 重合，舍去，故不存在满足题设条件的相异的两点．思路二：假设存在关于直线l 对称的相异两点B (x 1,y 1),C (x 2,y 2)，线段BC 中点()00,M x y ，由点差法可得22112222195195x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩，即22221212095x x y y --+=；∴0121212120552993BC x y y x x k x x y y y -+==-=-=--+，因此0065OM y k x ==，联立:96806:5AM OM l x y l y x --=⎧⎪⎨=⎪⎩可得52,3M ⎛⎫ ⎪⎝⎭与点A 重合，舍去，故不存在满足题设条件相异的两点．19. 设()f x 使定义在区间(1,)+∞上的函数，其导函数为()f x '.如果存在实数a 和函数()h x ，其中()h x 对任意的(1,)x ∈+∞都有()h x >0，使得()()()21f x h x x ax '=-+，则称函数()f x 具有性质()P a ．的（1）设函数()f x 2ln (1)1b x x x +=+>+，其中b 为实数① 求证：函数()f x 具有性质()P b ；② 讨论函数()f x 的单调性；（2）已知函数()g x 具有性质(2)P ，给定1212,(1,),,x x x x ∈+∞<设m 为正实数，12(1)mx m x α=+-，12(1)m x mx β=-+，且1,1αβ>>，若12()()()()g g g x g x αβ-<-，求m 的取值范围．【答案】（1）①证明见解析；②答案见解析（2）01m <<【解析】【分析】(1)①对()f x 求导，可得ℎ(x)=1x (x +1)2>0恒成立，即可函数()f x 具有性质()P b ；②设u (x )=x 2−bx +1(x >1)，f ′(x )与()u x 符号相等，对b 讨论，可知f ′(x )符号，即可得出函数()f x 的单调区间；(2)对()g x 求导，()()()()()22211g x h x x x h x x ='=-+-，分析可知()g x '其在(1,)+∞恒成立，对m 讨论，再根据αβ，与12,x x 大关系进行讨论，验证是否满足条件，可求解m 的取值范围.【小问1详解】① ()()()()222121111b f x x bx x x x x +=-=-+'++，所以1x >，ℎ(x )=1x (x +1)2>0恒成立，则函数()f x 具有性质()P b ；② 设u (x )=x 2−bx +1(x >1)，(i) 当0b -≥即0b ≤时，()0u x >，()'0f x >，故此时()f x 在区间(1,)+∞上递增；(ii) 当0b >时当240b ∆=-≤即02b <≤时，()0u x >，()'0f x >，故此时()f x 在区间(1,)+∞上递增；当240b ∆=->即2b >时，1211x x ==<=>，，所以x ⎛∈ ⎝时，()0u x <，()0f x '<，此时()f x 在⎛ ⎝上递减；x ∞⎫∈+⎪⎪⎭时，()0u x >，()0f x '<，此时()f x 在∞⎫+⎪⎪⎭上递增.综上所述，当2b ≤时，()f x 在(1,)+∞上递增；当2b >时，()f x 在⎛ ⎝上递减，在∞⎫+⎪⎪⎭上递增.【小问2详解】由题意，()()()()()22211g x h x x x h x x ='=-+-，又()h x 对任意的,(1)x ∈+∞都有()0h x >，所以对任意的,(1)x ∈+∞都有()0g x '>，()g x 在(1,)+∞上递增. 所以12(1)mx m x α=+-，12(1)m x mx β=-+，因为()()1212,21x x m x x αβαβ+=+-=--先考虑12x x αβ-<-的情况即()()121221m x x x x --<-，得01m <<，此时1122(1)x mx m x x α<=+-<，1122(1)x m x mx x β<=-+<所以1212()()(),()()()g x g g x g x g g x αβ<<<<所以12()()()()g g g x g x αβ-<-满足题意当1m ≥时，11112(1)(1)mx m x mx m x x α--≤==++，12222(1)(1)m x mx m x mx x β=--+≥=+，所以12x x αβ≤<≤所以12()()()()g g x g x g αβ≤<≤，则12()()()()g g g x g x αβ-≥-，不满足题意，舍去综上所述，01m <<。

2022-2023学年湖北省武汉外国语学校高一（下）期末数学试卷一、单选择：本大题共8个小题，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．复数z 满足z （1+i ）＝2i （i 为虚数单位），则z 的虚部为（ ） A ．1B ．﹣1C ．﹣iD ．i2．关于用统计方法获取数据，分析数据，下列结论错误的是（ ） A ．调查一批炮弹的杀伤半径，合理的调查方式为抽样调查B ．抽签法适用于总体中个体数较少，样本量也较小的抽样，随机数法适用于总体中个体数较多，但样本量较小的抽样C ．若数据x 1，x 2，x 3，…，x n 的平均数为x ，则数据y i ＝ax i ﹣b （i ＝1，2，3，…，n ）的平均数为ax −bD ．若甲、乙两组数据的标准差满足s 甲＜s 乙，则可以估计乙比甲更稳定 3．若tan α＝2，则sin2α2+cos 2α的值为（ ）A ．47B ．411C ．211D ．274．已知向量a →=(1，1)，b →=(1，−1)，则“(a →+λb →)⊥(a →+μb →)”是“λμ＝﹣1”的（ ）条件． A ．充要条件 B ．充分不必要 C ．必要不充分D ．既不充分也不必要5．一个正四面体的棱长为2，则它的外接球与内切球体积之比为（ ） A ．3：1B ．√3：1C ．9：1D ．27：16．如图，△A 'O 'B '是水平放置的△AOB 的直观图，但部分图象被墨汁覆盖，已知O '为坐标原点，顶点A '、B '均在坐标轴上，且△AOB 的面积为9，则O 'B '的长度为（ ）A ．34B ．3√22C ．32D ．3√247．已知I 为△ABC 的内心，且满足4IA →+3IB →+3IC →=0→，若△ABC 内切圆半径为2，则其外接圆半径的大小为（ ） A ．92B ．3C ．94D ．48．已知一个长方体的封闭盒子，从同一顶点出发的三条棱长分别为3，4，5，盒内有一个半径为1的小球，若将盒子随意翻动，则小球达不到的空间的体积是（ ） A ．36−203πB ．32−223πC ．60﹣12πD ．60−403π二、多选择：本大题共4个小题，每小题5分，共20分.多选或不选得0分，漏选得2分. 9．已知A ，B 是相互独立事件，且P(A)=12，P(B)=12，下列说法正确的是（ ） A ．A ，B 可能是互斥事件B ．P(AB)=14C ．P （A ∪B ）=34D ．由于P （A ）+P （B ）＝1，则A ，B 可能是相互对立事件10．△ABC 中，角A ，B ，C 所对的边为a ，b ，c ，下列叙述正确的是（ ） A ．若a cos B ＝b cos A ，则△ABC 是等腰三角形 B ．若a cosA=b cosB=c cosC，则△ABC 一定是等边三角形C ．若A ＞B ，则cos A ＜cos BD ．若2b ≥a +c ，则B ∈(0，π3]11．若正四棱柱ABCD ﹣A 1B 1C 1D 1的底面棱长为4，侧棱长为3，且M 为棱AA 1的靠近点A 的三等分点，点P 在正方形ABCD 的边界及其内部运动，且满足MP 与底面ABCD 的所成角θ≥π4，则下列结论正确的是（ ）A ．点P 所在区域面积为π4B ．有且仅有一个点P 使得MP ⊥PC 1 C ．四面体P ﹣A 1CD 1的体积取值范围为[6，8] D ．线段|PC 1|长度最小值为√1712．已知a ＞0，b ＞0，下列命题中正确的是（ ）A ．若ab ﹣a ﹣2b ＝0，则a +2b ≥8B ．若a +b ＝2，则ba +4b≥5C ．若a +b ＝1，则√2a +4+√b +1≤2√3D ．若1a+1+1b+2=13，则ab +a +b ≥14+6√6三、填空题（本题4小题，每题5分，共20分）13．已知A （2，3），B （4，﹣3），点P 在线段AB 的延长线上，且|AP →|=43|PB →|，则点P 的坐标为 ． 14．函数f(x)=log 2(2sinx −√3)的单调增区间是 ．15．在直角△ABC 中，AB ＝AC ＝2，AC 上有一动点P （异于B ，C ），将△ABP 沿AP 折起，使得三棱锥A ﹣PBC 的顶点A 在底面PBC 上的投影M 恰好落在线段BC 上，则BM 长度的范围 ． 16．已知锐角△ABC 中，a 2−b 2c 2=2cosC ，则a2bcos 2B的取值范围 ．四、解答题（本题共6题，总分70分）17．（10分）如图，棱长为2的正方体ABCD ﹣A 1B 1C 1D 1中，E ，F 分别为棱AB ，BC 的中点． （1）求作过D 1，E ，F 三点的截面（写出作图过程）； （2）求截面图形的面积．18．（12分）某校为了解学生每日行走的步数，在全校3000名学生中随机抽取200名，给他们配发了计步手环，统计他们的日行步数，按步数分组，得到频率分布直方图如图所示．（1）求a 的值，并求出这200名学生日行步数的样本众数、中位数、平均数；（2）学校为了鼓励学生加强运动，决定对步数大于或等于13000步的学生加1分，计入期末三好学生评选的体育考核分，估计全校每天获得加分的人数．19．（12分）设△ABC 的内角A 、B 、C 的对边长分别为a 、b 、c ，cos （A ﹣C ）+cos B ＝1，b 2＝ac ． （1）求B ；（2）若b ＝2，求△ABC 的周长．20．（12分）甲、乙两名射击运动员进行射击比赛，每轮比赛甲、乙各射击一次，已知甲中靶的概率为34，乙中靶的概率为45，每轮比赛中甲、乙两人射击的结果互不影响，求下列事件的概率：（1）第一轮射击中恰好有一人中靶； （2）经过两轮射击，两人共中靶3次．21．（12分）如图，四棱锥S ﹣ABCD 的底面是正方形，SD ⊥平面ABCD ，SD ＝2a ，AD =√2a ，点E 是SD 上的点，且DE ＝λa （0＜λ≤2），（1）若λ＝1，求AC 和BE 所成角的余弦值；（2）设二面角C ﹣AE ﹣D 的大小为θ，直线BE 与平面SCD 所成的角为φ，求出tanφtanθ的最大值，并指出此时的λ取值22．（12分）已知函数f(x)=log 2(4x +1)+ax 是偶函数． （1）求实数a 的值； （2）若函数g （x ）＝22x +2﹣2x+m •2f（x ）的最小值为﹣3，求实数m 的值；（3）当k 为何值时，讨论关于x 的方程[f （x ）﹣1+k ][f （x ）﹣1﹣4k ]+2k 2+k ＝0的根的个数．（请写出详细解答过程）2022-2023学年湖北省武汉外国语学校高一（下）期末数学试卷参考答案与试题解析一、单选择：本大题共8个小题，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．复数z 满足z （1+i ）＝2i （i 为虚数单位），则z 的虚部为（ ） A ．1B ．﹣1C ．﹣iD ．i解：由z （1+i ）＝2i ，得z =2i 1+i =2i(1−i)(1+i)(1−i)=2+2i2=1+i ，∴z 的虚部为1． 故选：A ．2．关于用统计方法获取数据，分析数据，下列结论错误的是（ ） A ．调查一批炮弹的杀伤半径，合理的调查方式为抽样调查B ．抽签法适用于总体中个体数较少，样本量也较小的抽样，随机数法适用于总体中个体数较多，但样本量较小的抽样C ．若数据x 1，x 2，x 3，…，x n 的平均数为x ，则数据y i ＝ax i ﹣b （i ＝1，2，3，…，n ）的平均数为ax −bD ．若甲、乙两组数据的标准差满足s 甲＜s 乙，则可以估计乙比甲更稳定解：对于A ，因为调查一批炮弹的杀伤半径，具有破坏性，所以合理的调查方式为抽样调查，故A 正确； 对于B ，抽签法适用于总体中个体数较少，样本量也较小的抽样，随机数法适用于总体中个体数较多，但样本量较小的抽样，故B 正确；对于C ，根据平均数的性质，可知数据y i ＝ax i ﹣b （i ＝1，2，3，…，n ）的平均数为a x −b ，故C 正确； 对于D ，若甲、乙两组数据的标准差满足s 甲＜s 乙，则可以估计甲比乙更稳定，故D 错误． 故选：D ． 3．若tan α＝2，则sin2α2+cos 2α的值为（ ）A ．47B ．411C ．211D ．27解：因为tan α＝2， 所以sin2α2+cos 2α=2sinαcosα2sin 2α+3cos 2α=2tanα2tan 2α+3=2×22×22+3=411．故选：B ．4．已知向量a →=(1，1)，b →=(1，−1)，则“(a →+λb →)⊥(a →+μb →)”是“λμ＝﹣1”的（ ）条件． A ．充要条件 B ．充分不必要 C ．必要不充分D ．既不充分也不必要解：由a →=(1，1)，b →=(1，−1)，可得a →+λb →=（1+λ，1﹣λ），a →+μb →=（1+μ，1﹣μ）， 若(a →+λb →)⊥(a →+μb →)，则有（a →+λb →）•（a →+μb →）＝0， 即（1+λ）（1+μ）+（1﹣λ）（1﹣μ）＝0，整理得λμ＝﹣1，反之，当λμ＝﹣1时，得a →+λb →=（1−1μ，1+1μ），a →+μb →=（1+μ，1﹣μ），此时有（a →+λb →）•（a →+μb →）＝（1−1μ）（1+μ）+（1+1μ）（1﹣μ）＝0， 故“(a →+λb →)⊥(a →+μb →)”是“λμ＝﹣1”的充要条件． 故选：A ．5．一个正四面体的棱长为2，则它的外接球与内切球体积之比为（ ） A ．3：1B ．√3：1C ．9：1D ．27：1解：正四面体P ﹣ABC 中，取BC 中点D ，连接AD ，则AD ⊥BC ， 过点P 作PE ⊥AD 于点E ，则PE ⊥平面ABC ，外接球球心O 在PE 上，连接OA ，则OA ＝OP ＝R ， 因为正四面体的棱长为2，所以BD =CD =1，AD =√AB 2−BD 2=√3，则AE =23AD =2√33，PE =√PA 2−AE 2=√4−43=2√63．OE =PE −PO =2√63−R ， 由勾股定理得OE 2+AE 2＝AO 2，即(2√63−R)2+(2√33)2=R 2，解得R =√62； 设内切球球心为O 1，则O 1在PE 上，过点O 1作O 1H ⊥PD 于点H ，则O 1E ＝O 1H ＝r ， 故PO 1=2√63−r ，PD =√3，DE =13AD =√33， 因为△PO 1H ∽△PDE ， 所以PO 1PD=O 1H ED，即2√63−r √3=√33，解得r =√66，故它的外接球与内切球半径之比为R ：r =√62：√66=3：1， 体积之比为27：1． 故选：D ．6．如图，△A 'O 'B '是水平放置的△AOB 的直观图，但部分图象被墨汁覆盖，已知O '为坐标原点，顶点A '、B '均在坐标轴上，且△AOB 的面积为9，则O 'B '的长度为（ ）A ．34B ．3√22C ．32D ．3√24解：因为△AOB 的面积为9， 所以它的直观图△A 'O 'B '的面积为9×12√2=92√2，即12×6×O ′B ′×sinπ4=2√2，解得O ′B ′=32．故选：C ．7．已知I 为△ABC 的内心，且满足4IA →+3IB →+3IC →=0→，若△ABC 内切圆半径为2，则其外接圆半径的大小为（ ） A ．92B ．3C ．94D ．4解：设BC 的中点为D ，由4IA →+3IB →+3IC →=0→，知4IA →=−3（IB →+IC →）＝﹣6ID →， 所以A ，I ，D 三点共线，又AI 为∠BAC 的平分线，所以△ABC 为等腰三角形，且AB ＝AC ， 所以ID 为内切圆的半径，即ID ＝2，所以IA ＝3，AD ＝5，设内切圆与AB 相交于点E ，则IE ＝ID ＝2，且IE ⊥AB ， 所以sin ∠BAD =IE IA =23，cos ∠BAD =√53，tan ∠BAD =5，所以sin ∠BAC ＝sin2∠BAD ＝2sin ∠BAD cos ∠BAD =4√59， 在Rt △ABD 中，BD ＝AD tan ∠BAD ＝2√5， 所以BC ＝2BD ＝4√5，在△ABC 中，由正弦定理知，2R =BC sin∠BAC =4√5459=9，所以△ABC 外接圆半径R =92． 故选：A ．8．已知一个长方体的封闭盒子，从同一顶点出发的三条棱长分别为3，4，5，盒内有一个半径为1的小球，若将盒子随意翻动，则小球达不到的空间的体积是（ ） A ．36−203πB ．32−223πC ．60﹣12πD ．60−403π解：在正方体的 8 个顶点处的立方体空间内，小球不能到达的空间为：8[13−18（43π×13]＝8−43π，除此之外，以正方体的12条棱边长为1的12个正四棱柱空间内小球不能到达的空间共 （1−π4）×（4×1+4×2+4×3）＝24﹣6π， 其它空间小球均能到达，故小球不能到达的空间体积为：8−43π+24﹣6π＝32−22π3． 故选：B ．二、多选择：本大题共4个小题，每小题5分，共20分.多选或不选得0分，漏选得2分. 9．已知A ，B 是相互独立事件，且P(A)=12，P(B)=12，下列说法正确的是（ ）A．A，B可能是互斥事件B．P(AB)=14C．P（A∪B）=34D．由于P（A）+P（B）＝1，则A，B可能是相互对立事件解：∵A，B是相互独立事件，且P(A)=12，P(B)=12，故P（AB）＝P（A）P（B）=14，故A错误；P（A B）＝P（A）•（1﹣P（B））=12×12=14，故B正确；P（A∪B）＝P（A）+P（B）﹣P（AB）=12+12−14=34，故C正确，D错误．故选：BC．10．△ABC中，角A，B，C所对的边为a，b，c，下列叙述正确的是（）A．若a cos B＝b cos A，则△ABC是等腰三角形B．若acosA =bcosB=ccosC，则△ABC一定是等边三角形C．若A＞B，则cos A＜cos BD．若2b≥a+c，则B∈(0，π3]解：对于A：因为a cos B＝b cos A，由正弦定理可得sin A cos B﹣sin B cos A＝0，即sin（A﹣B）＝0，在三角形中，可得A﹣B＝0，即A＝B，这时不知道C是否和A，B相等，所以△ABC不一定是等腰三角形，所以A不正确；B中，acosA =bcosB=ccosC，由正弦定理可得sinAcosA=sinBcosB=sinCcosC，整理可得sin A cos B﹣cos A sin B＝0，即sin （A﹣B）＝0，可得A＝B，同理可得A＝C，所以A＝B＝C，即△ABC为等边三角形，所以B正确；C中，π＞A＞B，所以y＝cos x为单调递减，所以cos A＜cos B，所以C正确；D中，因为2b≥a+c，两边平方可得4b2≥a2+c2+2ac，由余弦定理可得4b2＝4（a2+c2﹣2ac cos B），所以4（a2+c2﹣2ac cos B）≥a2+c2+2ac恒成立，整理可得cos B≤3(a2+c2)−2ac8ac恒成立，而3(a2+c2)−2ac8ac≥6ac−2ac8ac=12，所以cos B≤12，即B∈[π3，π），所以D不正确．故选：BC．11．若正四棱柱ABCD﹣A1B1C1D1的底面棱长为4，侧棱长为3，且M为棱AA1的靠近点A的三等分点，点P 在正方形ABCD 的边界及其内部运动，且满足MP 与底面ABCD 的所成角θ≥π4，则下列结论正确的是（ ）A ．点P 所在区域面积为π4B ．有且仅有一个点P 使得MP ⊥PC 1 C ．四面体P ﹣A 1CD 1的体积取值范围为[6，8] D ．线段|PC 1|长度最小值为√17解：A ．由线面角的定义可知，∠MP A ＝θ＝45°，即MA ＝AP ＝1，故点P 所在区域为以A 为圆心，1为半径的圆在正方形ABCD 内部部分（包含边界弧长）， 即圆的14，面积为14π×12=14π，A 正确；如图，设点P 的轨迹与AD ，AB 交于点E ，F ，B ．不妨点P 与点F 重合，此时PC 1=√FB 2+BC 2+C 1C 2=√34，由余弦定理得：cos ∠MFC 1=MF 2+C 1F 2−C 1M 22MF⋅C 1F =2+34−362×√2×√34=0，则∠MFC 1=π2，同理可得：∠MEC 1=π2，故不止一个点P 使得MP ⊥PC 1，B 错误； C ．如图，AA 1⊥平面ABCD ，BC ⊂平面ABCD ，所以AA 1⊥BC ，且AB ⊥BC ，AA 1∩AB ＝A ，AA 1，AB ⊂平面ABB 1A 1，所以BC ⊥平面ABB 1A 1， BC ⊂平面A 1CD 1，所以平面A 1CD 1⊥平面ABB 1A 1， 且平面A 1CD 1∩平面ABB 1A 1＝A 1B ，因为AE ∥A 1D 1，AE ⊄平面A 1CD 1，A 1D 1⊂平面A 1CD 1， 所以AE ∥平面A 1CD 1，所以点A ，E 到平面A 1CD 1的距离相等，如图，当点P 在点E 处时，此时点P 到平面A 1CD 1的距离最大，最大距离为AH =3×45=125， 此时四面体P ﹣A 1CD 1的体积为13S △A 1CD 1⋅AH =13×12×4×5×125=8，当P 与点F 重合时，此时点P 到平面A 1CD 1的距离最小，最小距离为FK ，因为△BFK ∽△BAH ，所以FK =34AH ，所以最小体积为34×8=6，故四面体P ﹣A 1CD 1的体积取值范围为[6，8]，C 正确；D ．当PC 取最小值时，线段|PC 1|长度最小，由三角形两边之和大于第三边知：当A ，P ，C 三点共线时，PC 取得最小值，即|PC|min =4√2−1， 则|PC 1|min =√(4√2−1)2+32=√42−8√2，D 错误． 故选：AC ．12．已知a ＞0，b ＞0，下列命题中正确的是（ ） A ．若ab ﹣a ﹣2b ＝0，则a +2b ≥8B ．若a +b ＝2，则ba +4b≥5C ．若a +b ＝1，则√2a +4+√b +1≤2√3D ．若1a+1+1b+2=13，则ab +a +b ≥14+6√6解：对于A ，因为a ＞0，b ＞0，ab ﹣a ﹣2b ＝0，所以ab ＝a +2b ，所以1=1b +2a， 所以a +2b ＝（a +2b ）（1b +2a ）＝4+ab +4ba ≥4+2√ab ⋅4ba =4+4＝8，当且仅当ab =4b a，即b ＝2，a ＝4时取“＝”，故A 正确；对于B ，若a +b ＝2，ba+4b=b a+2a+2b b=b a+2ab+2≥2√b a ⋅2ab +2＝2√2+2，当且仅当b a=2a b，即a ＝2√2−2，b ＝4﹣2√2时取“＝”，故B 错误；对于C ，由a +b ＝1，a ＞0，b ＞0，由柯西不等式得：（√2a +4+√b +1）2﹣（√a +2•√2+√b +1•1）2≤（a +2+b +1）（2+1）＝12， 所以√2a +4+√b +1≤2√3，当且仅当√a+2√2=√b+11，即a =23，b =13时取等号，故C 正确；对于D ，由1a+1+1b+2=13，得3（b +2）+3（a +1）＝（a +1）（b +2），化简得ab ＝a +2b +7，所以a =2b+7b−1， 因为a ＞0，b ＞0，所以b ＞1，所以ab +a +b =2a +3b +7=4b+14b−1+3b +7=3(b −1)+18b−1+14≥2√3(b −1)⋅18b−1+14=6√6+14， 当且仅当3(b −1)=18b−1，即b =√6+1时取等号， 所以ab +a +b ≥14+6√6，故D 正确． 故选：ACD ．三、填空题（本题4小题，每题5分，共20分）13．已知A （2，3），B （4，﹣3），点P 在线段AB 的延长线上，且|AP →|=43|PB →|，则点P 的坐标为 （10，﹣21） ．解：∵点P 在线段AB 的延长线上，且|AP →|=43|PB →|，∴AP →=43BP →，设P （x ，y ），且A （2，3），B （4，﹣3）， ∴（x ﹣2，y ﹣3）=43（x ﹣4，y +3）， ∴{x −2=43(x −4)y −3=43(y +3)，解得{x =10y =−21， ∴P （10，﹣21）． 故答案为：（10，﹣21）．14．函数f(x)=log 2(2sinx −√3)的单调增区间是 （π3+2kπ，π2+2kπ]，k ∈Z ．解：令t ＝2sin x −√3＞0，得sin x ＞√32，则π3+2kπ＜x ＜2π3+2kπ，k ∈Z ．函数t ＝2sin x −√3在（π3+2kπ，π2+2kπ]，k ∈Z 上为增函数，又f （x ）＝log 2t 是定义域内的增函数，∴函数f(x)=log 2(2sinx −√3)的单调增区间是（π3+2kπ，π2+2kπ]，k ∈Z ．故答案为：（π3+2kπ，π2+2kπ]，k ∈Z ．15．在直角△ABC 中，AB ＝AC ＝2，AC 上有一动点P （异于B ，C ），将△ABP 沿AP 折起，使得三棱锥A ﹣PBC 的顶点A 在底面PBC 上的投影M 恰好落在线段BC 上，则BM 长度的范围 （√2，2） ． 解：在Rt △ABC 中，AB ＝AC ＝2，BC =√22+22=2√2， 折叠后的图形如图所示，∵三棱锥的顶点A 在底面BCD 的射影M 在线段BC 上，∴AM ⊥BC ， ∴在Rt △AMB 中，BM ＜AB ＝2，在△ABC 折叠前，如图，作AM 1⊥BC 于点M 1，在折叠过程中，随着点D 不断靠近点C ，点M 也不断靠近点M 1， ∴BM ＞BM 1＝AB cos45°＝2×√22=√2， ∴BM ∈（√2，2），∴BM 长度的范围为（√2，2）． 故答案为：（√2，2）． 16．已知锐角△ABC 中，a 2−b 2c 2=2cosC ，则a2bcos 2B的取值范围 （1，43） ．解：锐角△ABC 中，a 2−b 2c 2=2cos C ，由正弦定理得sin 2A ﹣sin 2B ＝2sin C 2cos C ＝sin2C sin C ， 又因为sin 2A ﹣sin 2B ＝sin 2A ﹣sin 2A sin 2B ﹣sin 2B +sin 2A sin 2B ＝sin 2A （1﹣sin 2B ）﹣sin 2B （1﹣sin 2A ） ＝sin 2A cos 2B ﹣sin 2B cos 2A＝（sin A cos B +cos A sin B ）（sin A cos B ﹣cos A sin B ） ＝sin （A +B ）sin （A ﹣B ） ＝sin C sin （A ﹣B ）， 所以sin （A ﹣B ）＝sin2C ，当A ﹣B ＝2C 时，因为A +B +C ＝π，所以2A ＝π+C ，因为△ABC 是锐角三角形，所以A ＜π2，2A ＜π，所以A ﹣B ＝2C 不成立； 当2C +A ﹣B ＝π时，因为A +B +C ＝π，所以C ＝2B ， 因为△ABC 是锐角三角形，所以C ＝2B ＜π2，B +C ＝3B ＞π2， 所以π6＜B ＜π4，所以C ＝2B ∈（π3，π2），所以A ∈（π4，π2），由正弦定理得，a2bcos 2B=12•sinAsinBcos 2B=12•sin3B sinBcos 2B=12•sin(B+2B)sinBcos 2B=12•sinBcos2B+cosBsin2BsinBcos 2B=12•sinB(cos 2B−sin 2B)+2cos 2BsinB sinBcos 2B =12•3cos 2B−sin 2B cos 2B=12（3﹣tan 2B ），又因为tan B ∈（√33，1），所以tan 2B ∈（13，1），所以3﹣tan 2B ∈（2，83），所以a2bcos 2B的取值范围是（1，43）．故答案为：（1，43）．四、解答题（本题共6题，总分70分）17．（10分）如图，棱长为2的正方体ABCD ﹣A 1B 1C 1D 1中，E ，F 分别为棱AB ，BC 的中点． （1）求作过D 1，E ，F 三点的截面（写出作图过程）； （2）求截面图形的面积．解：（1）在正方体ABCD ﹣A 1B 1C 1D 1中，画直线EF 与DA ，DC 的延长线分别交于点M ，N ，连接D1M，D1N，分别与棱AA1，CC1交于点G，H，连接EG，FH，如图1，抹去MA，ME，MG和NC，NF，NH得过D1，E，F三点的正方体ABCD﹣A1B1C1D1的截面五边形D1GEFH，如图2，（2）在正方体ABCD﹣A1B1C1D1中，AB＝2，E，F分别为棱AB，BC的中点，由（1）及图1知，ME=NF=EF=√2，即MN=3√2，AM=AE=1，则DM＝3，D1N=D1M=√DM2+DD12=√13，等腰△D1MN底边MN上的高ℎ=√D1M2−(12MN)2=√13−(322)2=√342，△D1MN的面积SΔD1MN =12MN⋅ℎ=12×3√2×√342=3√172，由AG∥DD1，得MGMD1=MAMD=13，即有MGMD1=13=MEMN，因此△MEG∽△MND1，于是S△MEG=19S△D1MN，同理S△NFH=19S△D1MN，所以截面五边形D1GEFH的面积S=S△D1MN −2S△NFH=79S△D1MN=7√176．18．（12分）某校为了解学生每日行走的步数，在全校3000名学生中随机抽取200名，给他们配发了计步手环，统计他们的日行步数，按步数分组，得到频率分布直方图如图所示．（1）求a 的值，并求出这200名学生日行步数的样本众数、中位数、平均数；（2）学校为了鼓励学生加强运动，决定对步数大于或等于13000步的学生加1分，计入期末三好学生评选的体育考核分，估计全校每天获得加分的人数．解：（1）2×（0.005+0.055+0.09+0.15+a +0.08+0.015+0.005）＝1， 解得a ＝0.1，这300名员工日行步数x 的样本众数为8+102=9，因为2×（0.005+0.055+0.09）＝0.3＜0.5，2×（0.005+0.055+0.09+0.15）＝0.6＞0.5， 所以中位数落在区间[8，10），设中位数为m ， 则0.3+（m ﹣8）×0.15＝0.5，解得m =283， 所以300名员工日行步数x 的样本中位数为283这300名员工日行步数x 的样本平均数为x =2×0.005×3+2×0.055×5+2×0.09×7+2×0.15×9+2×0.1×11+2×0.08×13+2×0.015×15+2×0.005×17＝9.44；（2）步数大于或等于13000步的频率为2×（0.015+0.005）+（14﹣13）×0.08＝0.12， 估计全校获得加分的人数为3000×0.12＝360人．19．（12分）设△ABC 的内角A 、B 、C 的对边长分别为a 、b 、c ，cos （A ﹣C ）+cos B ＝1，b 2＝ac ． （1）求B ；（2）若b ＝2，求△ABC 的周长．解：（1）由于cos （A ﹣C ）+cos B ＝1，整理得cos （A ﹣C ）﹣cos （A +C ）＝1，化简得：cos A cos C +sin A sin C ﹣cos A cos C +sin A sin C ＝1， 所以2sin A sin C ＝1，由于b 2＝ac ，利用正弦定理sin 2B ＝sin A sin C ，所以sin 2B =12，故sin B =√22； 故B =π4或3π4．（2）由（1）得：B =π4或3π4．①当B =π4时，据余弦定理b 2＝a 2+c 2﹣2ac cos B ，所以4=a 2+c 2−2ac ⋅√22， 整理得a 2+c 2−√2ac =4， 由于ac ＝b 2＝4，故(a +c)2=a 2+c 2+2ac =4+4√2+8=12+4√2， 所以a +c =√12+4√2．故a +b +c ＝2+√12+4√2．②当B =3π4时，据余弦定理b 2＝a 2+c 2﹣2ac cos B ，所以4=a 2+c 2+2ac ⋅√22， 整理得a 2+c 2+√2ac =4，所以a 2+c 2=4−4√2＜0（出现矛盾） 所以a +c =√12+4√2．故a +b +c ＝2+√12+4√2．20．（12分）甲、乙两名射击运动员进行射击比赛，每轮比赛甲、乙各射击一次，已知甲中靶的概率为34，乙中靶的概率为45，每轮比赛中甲、乙两人射击的结果互不影响，求下列事件的概率：（1）第一轮射击中恰好有一人中靶； （2）经过两轮射击，两人共中靶3次．解：（1）设“甲中靶”为事件A ，“乙中靶”为事件B ，所以第一轮射击中恰好有一人中靶的概率P 1＝P （A B ）+P （A B ）=34×15+14×45=720． （2）经过两轮射击，两人共中靶3次分两种情况：甲中靶2次，乙中靶1次；甲中靶1次，乙中靶2次，其中甲中靶2次，乙中靶1次的概率为34×34×45×15×2=72400，甲中靶1次，乙中靶2次的概率为34×14×2×45×45=96400，所以P 1=72400+96400=168400=2150． 21．（12分）如图，四棱锥S ﹣ABCD 的底面是正方形，SD ⊥平面ABCD ，SD ＝2a ，AD =√2a ，点E 是SD 上的点，且DE ＝λa （0＜λ≤2），（1）若λ＝1，求AC 和BE 所成角的余弦值；（2）设二面角C ﹣AE ﹣D 的大小为θ，直线BE 与平面SCD 所成的角为φ，求出tanφtanθ的最大值，并指出此时的λ取值解：（1）连接BD，因为SD⊥面ABCD，AC⊂面ABCD，所以AC⊥SD，因为底面ABCD是正方形，所以AC⊥BD，又SD∩BD＝D，所以AC⊥面SBD，又BE⊂面SBD，所以AC⊥BE，所以直线AC和直线BE所成的角为直角，其余弦值为0．（2）如图，连接BE，CE，AE，因为SD⊥面ABCD，BC⊂面ABCD，所以SD⊥BC，因为底面ABCD为正方形，所以BC⊥CD，又SD∩CD＝D，所以BC⊥面SCD，所以φ＝∠BEC，tanφ=BC CE，过点D作DF⊥AE交AE于F，连接AF，因为SD⊥面ABCD，CD⊂面ABCD，所以SD⊥CD，因为底面ABCD为正方形，所以CD⊥AD，又SD∩AD＝D，所以CD⊥面SAD，又AE ⊂面SAD ， 所以CD ⊥AE ，因为DF ⊥AE ，CD ∩DF ＝D ， 所以AE ⊥面CDF ， 又CF ⊂面CDF ， 所以AE ⊥CF ， 所以θ＝∠CFD ，tan θ=CD DF， 因为AD =√2a ，DE ＝λa ， 所以tanφtanθ=BC CE •DFCD =DFCE=AD⋅DE AEAE=AD⋅DE AE 2=AD⋅DEAD 2+DE 2=√2a⋅λa (√2a)2+(λa)2=√2λ2+λ2=√22λ+λ≤√222=12，当且仅当2λ=λ，即λ=√2时，取等号，所以tanφtanθ的最大值为12．22．（12分）已知函数f(x)=log 2(4x +1)+ax 是偶函数． （1）求实数a 的值； （2）若函数g （x ）＝22x +2﹣2x+m •2f（x ）的最小值为﹣3，求实数m 的值；（3）当k 为何值时，讨论关于x 的方程[f （x ）﹣1+k ][f （x ）﹣1﹣4k ]+2k 2+k ＝0的根的个数．（请写出详细解答过程）解：（1）f(−x)=log 2(4−x +1)−ax =f(x)=log 2(4x +1)+ax ⇒2ax +log 2(4x +1)−log 2(4−x +1)=0，而log 2(4x+1)−log 2(4−x+1)=log 24x +14−x +1=log 2(4x +1)⋅4x (4−x +1)⋅4x =log 2(4x +1)⋅4x4x+1=log 24x =2x ， ∴2ax +2x ＝0⇒a ＝﹣1； （2）f(x)=log 2(4x +1)−x ， ∴2f(x)=2log 2(4x +1)−x=4x+12x =2x +2−x ，故函数g （x ）＝22x +2﹣2x+m （2x +2﹣x ）的最小值为﹣3，令2x +2﹣x ＝t （t ≥2），故h （t ）＝t 2+mt ﹣2（t ≥2）的最小值为﹣3，等价于{−m2≤2ℎ(2)=2m +2=−3，或{−m2＞2ℎ(−m 2)=−m 24−2=−3， 解得m =−52；（3）由f(x)=log 2(4x+1)−x =log 24x+12x =log 2(2x +12x )，令φ(x)=2x +12x (x ≥0)，x 2＞x 1≥0， 有φ(x 2)−φ(x 1)=(2x 2+12x 2)−(2x 1+12x 1)=2x 2−2x 1+12x 2−12x 1=2x 2−2x 1+2x 1−2x22x 1+x 2=(2x 2−2x 1)(2x 1+x2−1)2x 1+x2． 由x 2＞x 1≥0，有2x 2−2x 1＞0，2x 1+x 2＞20=1，可得φ（x 2）＞φ（x 1），可知函数φ（x ）为增函数，故当x ≥0时，函数f （x ）单调递增， 由函数f （x ）为偶函数，可知函数f （x ）的增区间为[0，+∞），减区间为（﹣∞，0）， 令n ＝f （x ）﹣1，有n ≥f （0）﹣1＝log 22﹣1＝0，方程[f （x ）﹣1+k ][f （x ）﹣1﹣4k ]+2k 2+k ＝0（记为方程①）可化为（n +k ）（n ﹣4k ）+2k 2+k ＝0， 整理为n 2﹣3kn ﹣2k 2+k ＝0（记为方程②），Δ＝9k 2﹣4（﹣2k 2+k ）＝17k 2﹣4k ， ①当Δ＜0时，有0＜k ＜417，此时方程②无解，可得方程①无解； ②当Δ＝0时，k ＝0时，方程②的解为n ＝0，可得方程①仅有一个解为x ＝0； k =417时，方程②的解为n =617，可得方程①有两个解； ③当Δ＞0时，可得k ＞417或k ＜0， 1°当方程②有零根时，k =12，此时方程2还有一根为n =32，可得此时方程①有三个解；2°当方程②有两负根时，{x 1+x 2=3k ＜0x 1x 2=k −2k 2＞0，可得{k ＜00＜k ＜12，不可能；3°当方程②有两正根时，可得0＜k ＜12，又由Δ＞0，可得417＜k ＜12，此时方程①有1个根；4°当方程2有一正根一负根时，x 1x 2=k −2k 2＜0，可得k ＞12或k ＜0， 又由Δ＞0，可得k ＞12或k ＜0，此时方程①有两个根， 由上知：当k ＝0时，方程①有一个根；当0＜k＜417时，方程①没有根；当k=417或k＜0或k＞12时，方程①有两个根；当k=12时，方程①有三个根；当417＜k＜12时，方程①有四个根．第21页（共21页）。

湖北省武汉外国语学校高一期末历史试卷一、单选题 (45分)1.下面是世界古代某一帝国形势图。

下列相关叙述，与这一形势图相符的是( )A. 建立起了世界上第一个地跨亚非欧三大洲的帝国B. 糅合各种文化，引领“希腊化世界”的文化走向C. 从共和国到帝国，成为一个囊括地中海的大帝国D. 首次实现了两河流域的统一，把势力伸展到地中海东岸2.12~13世纪，西欧庄园主的奢侈品消费日益增加，为此他们不仅用货币地租逐渐取代劳役地租，甚至部分庄园主将庄园农产品，如粮食、羊毛等大量投入市场销售，以此来获取更多的货币。

这些现象的出现 ( )A. 导致了资本主义萌芽的产生B.表明庄园经济已走向崩溃C. 提高了封建主社会经济地位D.有助于中世纪城市的发展3. 印度西海岸城市果阿为深水良港，也是印度西海岸的贸易中心和北上麦加朝圣的出发点。

后来罗马教廷在果阿成立主教区，其辖区从好望角一直延伸到日本的广大地区。

耶稣会也在果阿建立了传教中心。

当时的果阿被称“东方的罗马”。

果阿的变迁折射出 ( )A. 印度人口结构的重大改变B.大帆船贸易航线的开辟C. 天主教在印度的全面流行D.全球贸易网的逐步形成4. “楚弓楚得”的典故出自《说苑·至公》。

欧洲学者在17世纪用拉丁文给《论语》作注时，认为中国人心胸宽广、有君子之德，援引了这一典故，但将其中的“弓”换成了西方骑士惯用的“盾”。

这说明 ( )A.文献转译曲解原著价值立场B.文明交融推动世界文化繁荣C.文教推广带有个体主观经验D.文化借鉴基于特定历史语境5. 1750~1770年, 英国出现“收费公路热”, 当时收费公路总里程近24140公里。

到18世纪晚期，伦敦、英格兰中西部和北部的大城市及其周边道路都已成为收费公路。

这一现象出现的主要原因是A.社会经济的变革B. 城市治理的改善C.垄断资本的扩张D. 工业革命的完成6.圣西门提倡创立牛顿宗教以取代基督教神学体系，他把“新基督教”作为他所畅想的“实业制度”赖以生存的基础，强调用“基督教道德的神圣性”去支撑未来社会的精神世界，使所有人都“为改进最穷苦阶级的精神和物质生活而工作”。

武汉外国语学校2021-2022学年度下学期期末考试高二英语试题考试时间：2022年6月24日命题人：高二英语组审题人：高二英语组满分：150分第一部分：听力(共两节，满分20分)第一节(共5小题；每小题1分，满分5分)1.What does the woman mean?A.She enjoyed the movie.B.She didn’t watch the movie.C.She didn’t like the movie.2.When does the man finish work on Monday?A.At5:00p.m.B.At6:00p.m.C.At4:00p.m.3.What can we learn from the woman?A.The staff don’t like weekends.B.The staff are all very busy.C.They don’t serve steak.4.What does the man ask the woman to do?A.Watch his bag.B.Call his friend.C.Go to the washroom.5.What does the woman think of the apartments in New York City?A.They are expensive.B.They are valuable.C.They are too small.第二节(共15小题；每小题1分，满分15分)听第6段材料，回答第6、7题。

6.What prevented the woman’s car from starting?A.The faulty engine.B.The dead battery.C.The lack of petrol.7.What will the woman do?A.Return her battery.B.Buy a new car.C.Get some petrol.听第7段材料，回答第8、9题。

湖北省武汉市外国语学校2024-2025学年八年级上学期10月月考数学试题一、单选题1．如图，已知A D ∠=∠，12∠=∠，那么要得到ABC DEF ≌△△，还应给出的条件是（ ）A ．EB ∠=∠ B ．ED BC = C ．AB EF =D ．AF CD = 2．小明不慎将一块三角形的玻璃碎成如图所示的四块（图中所标1、2、3、4），你认为将其中的哪一块带去，就能配一块与原来大小一样的三角形玻璃？应该带去（ ）A ．第1块B ．第2块C ．第3块D ．第4块 3．已知ABC DEF ≌△△，6cm BC EF ==，ABC V 的面积为18平方厘米，则EF 边上的高是（ ）A ．3cmB ．6cmC ．8cmD ．9cm4．如图，在四边形ABCD 中，CB CD =，90ABC ADC ∠=∠=︒，35BAC ∠=︒，则B C D ∠的度数为（ ）A ．145°B ．130°C ．110°D ．70°5．尺规作图中蕴含着丰富的数学知识和思想方法．如图，为了得到MBN PAQ ∠=∠，在用直尺和圆规作图的过程中，得到ACD BEF ≌△△的依据是（ ）．A ．SASB ．SSSC ．ASAD ．AAS6．如图为6个边长相等的正方形组成的图形，则∠1+∠2+∠3的大小是（ ）A ．90°B ．120°C ．135°D ．150°7．如图，已知线段AB ＝20米，MA ⊥AB 于点A ，MA ＝6米，射线BD ⊥AB 于B ，P 点从B 点向A 运动，每秒走1米，Q 点从B 点向D 运动，每秒走3米，P 、Q 同时从B 出发，则出发x 秒后，在线段MA 上有一点C ，使△CAP 与△PBQ 全等，则x 的值为（ ）A ．5B ．5或10C ．10D ．6或108．如图，在Rt ABC △中，90BAC ∠=︒，ABC ∠的角平分线交AC 于点D ，DE BC ⊥于点E ，若ABC V 与CDE V 的周长分别为13和3，则AB 的长为（ ）A ．10B ．16C ．8D ．59．如图，ABC V 中，AD 是角平分线，BE 是ABD △的中线，若ABE V 的面积是2.553AB AC ==，，，则ABC V 的面积是（ ）A ．5B ．6.8C ．7.5D ．810．如图，在ABC V 中，90ACB ∠=︒，AC BC =，AD 平分BAC ∠，CE AD ⊥交AB 于E ，点G 是AD 上的一点，且45ACG ∠=︒，连BG 交CE 于P ，连DP ，下列结论：①AC AE =，②CD BE =，③2BG DP AD +=，④PG PE =，其中正确的有（ ）A ．①②③B ．①②④C ．①③④D ．①②③④二、填空题11．一个三角形的三条边长分别为6，7，x ，另一个三角形的三条边长分别为y ，6，4，若这两个三角形全等，则x y +=．12．在ABC V 中，86AB AC ==，，则BC 边上的中线AD 的取值范围是．13．如图，在ABC V 中，AB AC =，BF CD =，BD CE =，65FDE ∠=︒，则A ∠的度数是．14．如图，直线 1l ，2l ，3l 分别过正方形ABCD 的三个顶点A ，D ，C ，且相互平行，若 1l ，2l 的距离为 1，2l ，3l 的距离为2， 则正方形的边长为．15．如图，B 、C 、E 三点在同一条直线上，CD 平分ACE ∠，DB DA =，DM BE ⊥于M ，若2AC =，32BC =，则CM 的长为．16．如图：在△ABC 中，∠ACB ＝90°，点D 在边AB 上，AD ＝AC ，点E 在BC 边上，CE ＝BD ，过点E 作EF ⊥CD 交AB 于点F ，若AF ＝2，BC ＝8，则DF 的长为三、解答题17．如图，已知12AB AC AD AE =∠=∠=，，．求证：BAD CAE V V ≌．18．如图，D 、C 、F 、B 四点在一条直线上，AB DE =，AC BD ⊥，EF BD ⊥，垂足分别为点C 、点F ，CD BF =．求证：AB DE ∥．19．已知，如图AB AE =，B E ∠=∠，BC ED =，AF 平分BAE ∠，求证：AF CD ⊥．20．如图，在Rt ABC △中，90ABC ∠=︒，在Rt DBE V 中，90DBE ∠=︒，AB DB =，BAC BDE ∠=∠．连接CD ，连接AE 交BD 于F ，点F 恰好是AE 的中点，求证：2CD BF =．21．如图是由小正方形组成的66⨯网格，每个小正方形的顶点叫做格点，点A 、B 、C 、D 都是格点，点P 是线段AB 上一点．仅用无刻度的直尺在给定网格中完成画图，画图过程用虚线表示．(1)在图1中，画出ABC V 的中线AM 和高线BN ；(2)在图2中，在边AC 上取一点E ，使得=45ABE ∠︒；(3)在图3中，在线段AD 上取一点Q ，使得AQ AP =．22．在ABC V 中，AE 、BF 是角平分线，交于O 点．(1)如图1，AD 是高，50BAC ∠=︒，70C ∠=︒，直接写出DAC ∠和BOA ∠的度数．(2)如图2，若OE OF =，AC BC ≠，求C ∠的度数．(3)如图3，若90C ∠=︒，8BC =，6AC =，10AB =，直接写出AOB S V ．23．如图，已知AC BC =，点D 是BC 上一点，ADE C ∠=∠．图1 图2(1)如图1，若90C ∠=︒，135DBE ∠=︒，求证：①EDB A ∠=∠②DA DE =(2)如图2，请直接写出DBE ∠与C ∠之间满足什么数量关系时，总有DA DE =成立． 24．ABE V 和ACF △始终有公共角A ∠，连接BC ，EF ，BE ，CF 相交于点O ．(1)如图1，若ABE ACF ∠=∠，BE CF =，求证：ABE ACF V V ≌．(2)如图2，若ABE ACF α=∠=∠，且CE CF =，求CBE ∠的度数（用含α的式子表示）(3)如图3，若BE CF =，过点C 作CD AB ∥且CD AB =，连接DO 并延长交AC 于点G ，过点G 作GH CF ⊥于点H ，请直接写出OGH ∠与COE ∠的关系为：_____________．。

2025届湖北省武汉市武汉外国语学校高三英语第一学期期末综合测试试题注意事项:1．答题前，考生先将自己的姓名、准考证号码填写清楚，将条形码准确粘贴在条形码区域内。

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4．作图可先使用铅笔画出，确定后必须用黑色字迹的签字笔描黑。

5．保持卡面清洁，不要折暴、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。

第一部分（共20小题，每小题1.5分，满分30分）1．Not until the bus disappeared in the distance ________ her school bag was gone. A．Tina had found B．had Tina foundC．did Tina find D．Tina found2．—The food here is nice enough.—My friend ________me a right place.A．introduces B．introducedC．had introduced D．was introducing3．---Where is my Chinese book? I remember I put it here yesterday.---You _________ it in the wrong place.A．must put B．should have putC．might have put D．might put4．Despite the fact that his scores were good, they were hardly as excellent as a student with his intelligence .A．achieved B．had achievedC．would achieve D．should have achieved5．---My son is addicted to computer games. He is hopeless，isn't he？---Yes，_____________he is determined to give up and start all over.A．if B．unlessC．though D．so6．—Amazingly, I’ve managed to start my own shop online!—________ I told you it was easy.A．There you are! B．Believe it or not.C．How come? D．Y ou got me there!7．Not until the end of the performance ________ the chance to take photos with the respectable actor.A．the audience got B．the audience had gotC．did the audience get D．had the audience got8．—What happened to the young trees we planted last week?—The trees ________ well, but I didn’t water them.A．might grow B．needn’t have grownC．would have grown D．would grow9．______ flag-raising ceremony was held at the Golden Bauhinia Square on July 1 to celebrate ______ 17th anniversary of Hong Kong’s return to China.A．A; / B．A; theC．The; the D．/; the10．The shocking news made me realize ______ terrible problems we would face.A. that B．how C．what D．why11．A new ________ bus service to Tianjin Airport started to operate two months ago. A．common B．usualC．regular D．ordinary12．— Who recommended Nancy for the post?— It was James ______ admiration for her was obvious.A．who B．that C．whose D．whom13．It was only after a family related conversation ______________ I found out she was actually my distant cousin.A．when B．thatC．which D．who14．--- Hello, Tom. This is Mary speaking.--- What a coincidence! I_________ about you.A．just thought B．was just thinkingC．have just thought D．would just think15．It is not only blind men who make such stupid mistakes. People who can see sometimes act__________.A．just foolishly B．less foolishly C．as foolishly D．so foolishly 16．Among the crises that face humans ________ the lack of natural resources.A．is B．are C．is there D．are there 17．Whether to favor urban development or the preservation of historical sites is especially controversial in China, where there exists rich history, diversified tradition and cultural ________.A．surplus B．deposits C．accounts D．receipts18．It’s second time in five days that he has asked me for higherpay.A．不填；a B．a;the C．the；a D．the；the19．In that remote area, the trees _____ by the volunteers are growing well. A．planted B．planting C．being planted D．to plant20．You will have to stay at home all day ______ you finish all your homework.A．if B．unless C．whether D．because第二部分阅读理解（满分40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。

湖北省武汉市武汉外国语学校2024-2025学年八年级上学期期中英语试题一、单项选择1．—Frank, do remember to take the report to the teacher’s office tomorrow morning.—Oh, I have almost forgotten, but ________.A．it’s my pleasure B．no problem C．no need D．never mind 2．—Jason, let’s take a break and have a cup of tea in the tearoom.—________? We have been working for hours.A．Pardon B．What for C．Why not D．How come 3．—Excuse me, may I use the Wi-Fi here?—________. Here is the key to it. You can stay here as long as you can.A．Never mind B．I’m afraid not C．That’s a deal D．Be my guest 4．—A boy with glasses at the railway station gave J.K Rowling the ________ for Harry Potter.—Oh, that’s interesting.A．tradition B．information C．inspiration D．condition5．—It seems that you appreciate Peter very much.—He used his great sense of humor to ________ my spirits.A．open B．lift C．increase D．rise6．—Many children and women were killed in the air attack and people had to leave their homes in the middle east.—What sad news! People there must have a ________ of a world with love and peace.A．perspective B．message C．vision D．tale7．—It’s said that the latest Yueju Opera, New Dragon Gate Inn, has been a real hit.—That’s true. It interests young people to have a ________ look at the traditional Chinese art form.A．careful B．fresh C．friendly D．quick8．—Do you know where the last two paintings went?—A billionaire bought ________ of them. He thought they were worth.A．either B．each C．none D．both9．— You’ve got it all wrong. We need to ________ a time to have a talk now!— What about tomorrow? I am too busy today.A．put up B．fix up C．use up D．look up10．The poem “Grown from the same trees, why boil us so hot?” tells us that ________.A．we can’t pollute by burning trees B．we need more trees to prevent the sun from boiling usC．as brothers, we shouldn’t hurt each other D．people like to hurt their neighbours when they grow up二、完形填空阅读下面短文，从短文后各题所给的A、B、C、D 四个选项中，选出可以填入空白处的最佳选项。

武汉外国语学校2025届高三10月月考英语试题第一部分听力 (共两节，满分30分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节 (共5小题: 每小题1.5分, 满分7.5分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man advise the woman to do?A. Buy a new refrigerator.B. Clean the refrigerator.C. Have the refrigerator fixed.2. What does the man plan to do tomorrow?A. Have a rest.B. Attend a meeting.C. Watch a match.3. How does the woman sound?A. Angry.B. Disappointed.C. Excited.4. What does the woman mean?A. She isn't feeling well.B. She likes gymnastics a lot.C. She is unable to join the team.5. When does the conversation probably take place?A. In the morning.B. At noon.C. In the afternoon.第二节 (共15小题: 每小题1.5分, 满分 22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。

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

湖北省武汉外国语学校2022-2023学年高一下学期期末考试英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读理解At the start of the 20th century, an American engineer named John Elfreth Watkins made predictions about life today. His predictions about slowing population growth, mobile phones and increasing height were close to the mark. But he was wrong in one prediction: that everybody would walk 10 miles a day.Today, in Australia, most children on average fall 2, 000 steps short of the physical activity they need to avoid being overweight. In the early 1970s, 40 per cent of children walked to school, while in 2010, it was as low as 15 percent.The decline is not because we have all become lazy. Families are pressed for time, many with both parents working to pay for their house, often working hours not of their choosing, living in car-dependent neighborhoods with limited public transport.The other side of the coin is equally a deprivation: for health and well-being, as well as lost opportunities (机会) for children to get to know their local surroundings. And for parents there are lost opportunities to walk and talk with their young scholar about their day.Most parents will have eagerly asked their child about their day, only to meet with a “good”, quickly followed by “I’m hungry”. This is also my experience as a mother. But somewhere over the daily walk more about my son’s day comes out. I hear him making sense of friendship and its limits. This is the unexpected and rare parental opportunity to hear more.Many primary schools support walking school-bus routes (路线), with days of regular,parent-accompanied walks. Doing just one of these a few times a week is better than nothing. It can be tough to begin and takes a little planning-running shoes by the front door, lunches made the night before, umbrellas on rainy days and hats on hot ones-but it's certainly worth trying.1．Why does the author mention Watkins' predictions in the first paragraph?A．To make comparisons.B．To introduce the topic.C．To support her argument.D．To provide examples.2．What has caused the decrease in Australian children’s physical activity?A．Plain laziness.B．Health problems.C．Lack of time.D．Security concerns.3．Why does the author find walking with her son worthwhile?A．She can get relaxed after work.B．She can keep physically fit.C．She can help with her son's study.D．She can know her son better.There is an old Chinese proverb that states “One generation plants the trees; another gets the shade,” and this is how it should be with mothers and daughters. The relationship between a mother and a daughter is sometimes confusing. The relationship can be similar to friendship. However, the mother and daughter relationship has unique characteristics that distinguish it from a friendship. These characteristics include a hierarchy (等级) of responsibilities and unconditional love, which preclude mothers and daughters from being best friends.Marina, 27 years old, said, “I love spending time with my mom, but I wouldn’t consider her my best friend. Best friends don’t pay for your wedding. Best friends don’t remind you how they carried you in their body and gave you life! Best friend: don’t tell you how wise they are because they have been alive at least 20 years longer than you.” This doesn’t mean that the mother and daughter relationship can’t be very close and satisfying.While some adult relationships are still troubled, many find them to be extremely rewarding. This generation of mothers and adult daughters has a lot in common, which increases the likelihood of shared companionship. Mothers and daughters have always shared the common experience of being homemakers, responsible for maintaining and passing on family values and traditions. Today contemporary mothers and daughters also share the experience of the workforce and technology, which may bring them even closer together.Best friends may or may not continue to be best friends, but for better or worse, the mother and daughter relationship is permanent, even if for some unfortunate reason they aren’t speaking. The mother and child relationship is closer than any other. There is not an equal relationship. Daughters should not feel responsible for their mother’s emotional well-being. It isn’t that they don’t care deeply about their mothers. It’s just that they shouldn’t be burdened with their mother’s well-being.The mother and daughter relationship is a relationship that is not replaceable by any other. Mothers never stop being mothers, which includes frequently wanting to protect their daughters and often feeling responsible for their happiness. Mothers always “trump (胜过)” friends.4．What does the underlined word “preclude” in paragraph 1 probably mean?A．differ B．prevent C．benefit D．change 5．What do we know from the text?A．The mother and daughter relationship can be replaced by a best friend.B．A mother’s love brings her and her daughter a close friendship.C．The mother and daughter relationship goes beyond best friends’ friendship.D．Marina has a troubled relationship with her mother.6．How does the author mainly prove his statements?A．By listing data.B．By giving explanations.C．By quoting sayings.D．By giving examples.7．What is the best title for the text?A．How to Be a Good Mother and Daughter?B．Who Is a Mother’s Best Friend?C．Mothers or Friends?D．Can a Mother Be a Daughter’s Best Friend?Skin: The Body’s CanvasIf you could take off your skin and lay it flat, it would cover an area of about 1.9 square meters. Skin is, by far, the body’s largest organ. Covering almost the entire body, skin protects us from a variety of external forces. For example, it protects us from extremes of temperature, damaging sunlight, harmful chemicals, and dangerous infections. Skin is also packed with nerves, which keeps the brain in touch with the outside world. The health of our skin and its ability to perform its protective functions are crucial to our well-being. However, the appearance of our skin is equally - if not more - important to many people on this planet.Take skin color, for example. Your genes determine your skin’s color, but for centuries, humans have tried to lighten or darken their skin in an attempt to be more attractive. In the 1800s, white skin was desirable for many Europeans. Skin this color meant that its owner was a member of the upper class and did not have to work in the sun. Among darker-skinned people in some parts of the world, products used to lighten skin are still popular today. During the 20th century, attitudes toward light skin shifted in the opposite direction in other cultures, as cities grew and work moved indoors. Tanned skin began to indicate leisure time and health. In many places today, tanning on the beach or in a salon remains popular, even though people are more aware of the dangers of UV rays.Identity and StatusJust as people have altered their skin’s color to denote wealth and beauty, so too have cultures around the globe marked their skin to indicate cultural identity or community status. Tattooing, for example, has been carried out for thousands of years. Leaders in places including ancient Egypt, Britain, and Peru wore tattoos to mark their status, or their bravery. Today, among the Maori people of New Zealand as well as in cultures in Samoa, Tahiti, and Borneo, full-facial tattoos are still used to identify the wearer as a member of a certain family. These tattoos can also symbolize the person’s achievements in life.In Japan, tattooing has been practiced for thousands of years, but was outlawed in the 19th century. Although there are no laws against it today, tattoos are still strongly associated with criminals -particularly the yakuza, or the Japanese mafia, who are known for their full-body tattoos. The complex design of a yakuza member’s tattoo usually includes symbols of character traits that the wearer wants to have. The process of getting a full-body tattoo is both slow and painful and can take two years or more to complete.In some cultures, scarring -a marking caused by cutting or burning the skin -is practiced, usually among people who have darker skin on which a tattoo would be difficult to see. For many men in West Africa, for instance, scarring is a rite of passage -an act that symbolizes that a male has matured from a child into an adult. In Australia, among some native peoples, cuts are made on the skin of both men and women when they reach age 16 or 17. Without these, they were traditionally not permitted to trade, sing ceremonial songs, or participate in other activities.Not all skin markings are permanent, though. In countries such as Morocco and India, women decorate their skin with colorful henna designs for celebrations such as weddings and important religious holidays. The henna coloring, which comes from a plant, fades and disappears over time.In recent years in many industrialized nations, tattooing, henna(散沫花染料) body art, and, to a lesser degree, scarring have been gaining in popularity. What makes these practices appealing to those living in modern cities? According to photographer Chris Rainier, whose book Ancient Marks examines body markings around the globe, people are looking for a connection with the traditional world. “There is a whole sector of modern society - people in search of identity, people in search of meaning,” says Rainier. “Hence, there has been a huge explosion of tattooing and body marking.” Rainier reasons that it’s “mankind wanting identity,wanting a sense of place and a sense of culture within their community.”8．What is the main idea of paragraph1 ?A．Skin covers about 20 square feet.B．Skin can be damaged by sunlight’s ultraviolet rays.C．Skin is a very important part of the body.D．Skin contains many nerve cells.9．What is the purpose of paragraph 2?A．To tell why skin color was socially important in the 1800sB．To explain changing attitudes towards skin colorC．To make the connection between dark skin and working outdoorsD．To explain why indoor tanning salons are popular10．In paragraph 5, what is a rite of passage?A．a ceremony when children get tattoosB．a time when women sing ceremonial songsC．a special holiday on the full moonD．an event marking an important transition in life11．Which statement would Chris Rainier mostly likely agree with?A．People who get tattoos are seeking identity and tradition.B．Employers are not supportive of workers with body art.C．People looking for identity should wear ethnic clothing.D．In industrialized nations, fewer people have body markingsThe World’s Favorite DrugIt’s 1:45 a.m., and 21-year-old Thomas Murphy is burning the midnight oil, studying for an important engineering exam he has at 2:00 in the afternoon later today. To stay awake and alert, he’s had two cups of coffee in the last three hours and is now downing a popular energy drink —one that has two to three times the amount of caffeine as a similar sized can of soda. Many students like Murphy, as well as marathon runners, airline pilots, and long-distance travelers, owe their energy to one of humankind’s oldest stimulants(刺激物): caffeine.The power to counter physical fatigue and increase alertness is part of the reason caffeine ranks as the world’s most popular mood-altering drug. It is found not only in sodas, energy drinks, coffee, and tea, but in diet pills, pain relievers (like aspirin), and chocolate bars.Many societies around the world have also created entire rituals around the use of caffeine. For example, there’s the cafe culture of France, the tea ceremony in Japan, and the morning cup of coffee or tea that marks the start of the day in many cultures.Caffeine is present in many of the foods and drinks we consume, but is it good for us? Charles Czeisler, a scientist and sleep expert at Harvard Medical School, believes that caffeine causes us to lose sleep, which he says is unhealthy. “Without adequate sleep—the typical eight hours—the human body will not function at its best, physically, mentally, or emotionally.” Too often, Czeisler says, we consume caffeine to stay awake, which later makes it impossible for us to get the rest we need.Health risks have also been tied to caffeine consumption. Over the years, studies have attributed higher rates of certain types of cancer and bone disease to caffeine consumption. To date, however, there is no proof that caffeine actually causes these diseases.A number of scientists, including Roland Griffiths—a professor at the Johns Hopkins School of Medicine in the United States—believe that regular caffeine use causes physical dependence. Heavy caffeine users, Griffiths says, exhibit similar behaviors. For example, their moods fluctuate from high to low, they get mild to severe headaches, or they feel tired or sad when they can’t have a caffeinated drink. To minimize or stop these feelings, users must consume caffeine - a behavior Griffiths says is characteristic of drug addiction.Despite these concerns, the general opinion in the scientific community is that caffeine is not dangerous when consumed in moderation. This means having one or two small cups of coffee (about 300 milligrams of caffeine) per day, for example. Furthermore, a lot of current research contradicts long-held negative beliefs about caffeine, and suggests that it may, in fact, have health benefits. For instance, studies have shown that caffeine can help ease muscle pain. Because it is a stimulant, caffeine can also help improve one’s mood. Research has also shown that some caffeinated drinks—specifically certain teas—have disease-fighting chemicals that can help the body fight a number of illnesses, including certain types of cancer.In addition, as a type of mental stimulant, caffeine increases alertness, memory, and reaction speed. Because it fights fatigue, it facilitates performance on tasks like driving, flying, and solving simple math problems. And while it is true that caffeine can increase blood pressure, the effect is usually temporary and therefore not likely to cause heart trouble. This is especially true if caffeine is consumed in moderation. Moreover, despite its nearlyuniversal use, caffeine has rarely been abused. “With caffeine, overuse tends to stop itself,” says Jack Bergman, a specialist at Harvard Medical School. If you consume too much, “you get ... uncomfortable, and you don’t want to continue.”Caffeine’s behavioral effects are real, but most often mild. Getting that burst of energy, of course, is why many of the world’s most popular drinks contain caffeine. Whether it’s a student drinking coffee before class or a businessperson enjoying tea with lunch, humankind’s favorite stimulant is at work everyday all over the world.12．What is the main idea of paragraph 1?A．Engineering students must study all night before their exams.B．Caffeine is a stimulant that helps people be alert and energetic.C．Energy drinks give marathon runners an advantage in races.D．Airline pilots drink a lot of coffee to stay awake on long flights.13．What is the purpose of paragraph 5?A．to describe mood fluctuationsB．to explain people’s feelings when they can’t have caffeineC．to present results of recent medical researchD．to claim that caffeine is addictive14．In the third sentence of paragraph 6, what does it refer to?A．muscle painB．stimulantsC．caffeineD．general opinion15．According to the passage, which is the best summary of the general medical opinion about caffeine and health?A．Certain teas may protect a person from cancer.B．If consumed in moderation, caffeine is not dangerous.C．Caffeine in pain pills can relieve sore muscles.D．People who drink a lot of coffee tend to have mood swings.二、七选五Nothing stays the same for long: things and people change.I grew up on a small farm, where a flock of sheep wandered around the surrounding mountains. My father was not highly educated, but he was smart. He was a man made of leather and chewing tobacco who rarely tried to talk with my brother or me. He was quiet and distant, I might say. 16One day I came home and his car was already there. 17 In fact, when he came home, he went to the barn (谷仓) to labour even more. I still remember hiding around the corner and stealing a look at my father lying bitterly on the bed that day. Multiple myeloma, I learned, is a type of blood cancer. 18 For the last year of my father’s life, his entire day consisted of rising from his bed and walking to his chair to sit and think alone.He was predictably in that chair on his own when I came in. 19 He told me about his life, his heartbreaks and his loves. It was as if a pipe had burst, his inner self rushing out to me in a great flood. He had been speaking for maybe an hour or more when I realized that he was doing more than telling. He was asking to be understood in a way that he had never done before.20 I realize, though, that if he hadn’t, I might never have come to know him and love him.A．My father never missed work.B．I did not like him very much.C．He was skillful at any farm work.D．He became better after some special treatment.E．I’m certainly not glad that my father got sick.F．As the disease develops, the person who has it shrinks.G．What followed still moves me these decades later三、完形填空schools.When he was called up for 25 , “my first thought was, what about my kids?” says Kohut. “The last thing these students need is a 26 in their teaching.”When on duty, Kohut’s days began in the morning with teaching his elementary class remotely, 27 the lesson mere minutes before his Guard shift started at 10 am. Later in the day, during his break, he would go 28 to teach his middle school students from the back of a Humvee.Music has always been a driving force in Kohut’s life. He was a(n) 29 saxophone player throughout high school, studied music in college, and ultimately earned his doctorate in music composition.“What I really wanted was to 30 .” Says Kohut. “My mom, a single mother, was a music teacher. She was such a good role model.”Kohut’s 31 duty has caught the attention of parents at Canterbury Woods Elementary.“I just wanted to share how 32 I am with Dr. Kohut this week,” Susi Brittain wrote in an e-mail to Leipzig. “what he did seems so 33 the expectations of a teacher in these circumstances.”Kohut insists he is not doing anything extraordinary. But during the long and sometimes stressful hours of standing guard, Kohut said his teaching time offers 34 .“It brings about unity and 35 ,” he says. “And that’s what the world needs right now.”21．A．napping B．talking C．eating D．cooking 22．A．front B．seat C．back D．end 23．A．what B．how C．when D．where 24．A．taught B．learned C．played D．exercised 25．A．rights B．obligation C．army D．duty 26．A．disruption B．burden C．disturbance D．turbulence 27．A．interrupting B．finishing C．starting D．inserting 28．A．inside B．outside C．online D．offline 29．A．enthusiastic B．basic C．standard D．usual 30．A．study B．teach C．show D．perform 31．A．daily B．common C．double D．single32．A．interested B．connected C．concerned D．impressed 33．A．beyond B．within C．against D．below 34．A．gratitude B．money C．comfort D．ease 35．A．appreciation B．peace C．music D．teamwork四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

湖北省武汉市外国语学校2013-2014学年高一下学期期末考试英语试题第一部分：听力（共两节，满分30分）第一节（共5小题；每题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.Where will the woman go first?A.To the beach.B.To the bank.C.To the bathroom.2.What does the woman mean?A.The man forgot to do his hair.B.The man forgot to put on a tie.C.The man is wearing clothes that don‟t match.3.How does the woman probably feel?A.Annoyed and unconcerned.B.Hungry and impatient.C.Surprised and excited.4.Why was the woman worried?A.The man lost his phone.B.The man would be back very late.C.The man didn‟t answer the phone.5.Who did the woman want to call?A.JamesB.DrakeC.Daniel.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

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

每段对话或独白读两遍。