湖北省武汉外国语学校高一英语上学期期末考试试题

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湖北省武汉外国语学校2024-2025 学年上学期学情调研 (一)九年级英语试题(含答案)

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含解析

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。

湖北省武汉外国语学校2024-2025学年高一上学期期中考试历史试题

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(解析版)

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含解析

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。

3．请按照题号顺序在各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试题卷上答题无效。

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

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。

湖北省武汉市武汉外国语学校2024-2025学年高三上学期10月考试数学(含答案)

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

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。

2024-2025学年湖北省武汉外国语学校高三上学期10月月考数学试题及答案

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 <<。

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武汉外国语学校2014—2015学年度上学期期末考试高一英语试题考试时间：2015年2月4日上午7:30-9:30 满分：150分第一节（共5小题；每小题1分，满分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。

每段对话仅读一遍。

1. What do we know about Tom?A. He knocks at the door before 5:30.B. He stops work at 5:30C. He will not come back today.2. When did the man probably leave the gate?A. At 12:00.B. At 12:30.C. At 1:00.3. What are the speakers talking about?A. Buying shoes.B. Their daughter.C. Their economic problems.4. What does the woman like?A. Teamwork.B. Creativity.C. Working alone.5. What will the man do?A. Keep his voice down.B. Close the door.C. Talk with the neighbors.第二节（共15小题；每小题1分，满分15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。

每段对话或独白读两遍。

听第6段材料，回答第6、7题。

6. Why didn't the man call the woman?A. He had no change.B. He wasn't in the service area.C. He was just a few minutes late.7. What will the speakers do next?A. Wait for their friends.B. Go to join their friends.C. Have the car repaired.听第7段材料，回答第8至10题。

8. What is the relationship between the speakers?A. Teacher and student.B. Boss and employee.C. Colleagues.9. Where does the man say he is coming from?A. His home.B. The hospital.C. His office.10. What does the woman think of what the man said?A. She is doubtful.B. She feels sorry for the man.C. She is delighted.听第8段材料，回答第11至13题。

11. What did the man do in the past two hours?A. He sang only one song.B. He just sat there listening.C. He sang two Elvis songs.12. What do we know about Stanley?A. He is fond of singing.B. He is a good singer.C. He is sick tonight.13. Why does the woman want to go out?A. To get some fresh air.B. To talk with the man.C. To avoid listening to Stanley's singing.听第9段材料，回答第14至16题。

14. Where does the man work?A. In a factory.B. In a hotel.C. In a shopping center.15. What's wrong with the TV?A. The picture is not clear.B. No picture comes out.C. There's no sound.16. Why is the woman surprised in the end?A. They can't adjust it.B. They sold her a used TV.C. The warranty is only two days.听第10段材料，回答第17至20题。

17. How will the weather be tonight?A. Cloudy in the west.B. The temperature will drop below zero.C. Clear both in the east and the west.18. Which word can describe the weather on Thursday?A. Fine.B. Awful.C. So-so.19. What can be said to people going to Dartmoor National Park?A. They'd better take an umbrella with them.B. They can enjoy the sunshine.C. It's too cold for them to stay out in the open.20. What will the weather be like in general?A. Dry.B. Wet.C. Foggy.第二部分：词汇知识运用（共两节，满分30分）第一节：多项选择（共10小题；每小题1分，满分10分）从A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

21. Robin Hood lived in the forest with his ____________, his “merry men”.A. representativesB. detectivesC. companionsD. landowners22. The designers of supermarkets can make customers buy one product __________ another.A. more thanB. rather thanC. other thanD. better than23. The Rare Wildlife Rescue and Br eeding Research Center in Xi’an is criticizedfor making money by __________ injured pandas.A. complainingB. displayingC. applaudingD. inspecting24. The professor has been doing some________________ research on Robin Hood, tryingto prove that he is a true figure in history.A. historicalB. historicC. apparentD. illegal25. To be _______________, the new couple chose to go on bikes instead of fancy carsto their wedding reception.A.B. sycophanticC. luxuriousD. reliableenvironmentally-friendly26. The police suggest that every family should have their old door locks changedi n t o t h e a d v a n c e d o n e s i n order to prevent ____________ from happening.A. vandalismB. burglaryC. smugglingD. arson27. Students who are late for classes often ________ into the classroom, trying to avoid being noticed.A. blowB. demandC. leadD. creep28. The __________ stole hundreds of dollars worth of jewelry from the store.A. shoplifterB. muggerC. murdererD. arsonist29. Every night School security guards ___________ will check whether the door ofeach classroom is well locked or not.A. on footB. on purposeC. on showD. on patrol30. The theft division ___________ white collar crime, property crime and anything to do withembezzlements.A. threatensB. suppliesC. investigatesD. defends第二节：完型填空（共20小题；每小题1分，满分20分）阅读下面短文，从短文后所给各题的四个选项（A、B、C和D）中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

I was shopping at a department store and had just stepped on the down escalator(电动扶梯) when I noticed a woman standing at the side. She looked about 65, and herexpression told me she was____31____. I turned towards her and asked, “Do you need help?”As the escalator continued moving I heard the woman reply ____32____, “I’m afraid.” There was a hint of doubt in her ____33____. “Want me to come ____34____ and get you?” I called to her. She____35____. But by the time I ____36____ her, the woman reconsidered, “Sorry, I don’t think I can put my feet on it.”I was suddenly____37____ that I took a more difficult task than I had expected. “I know we can do it,” I comforted her, “I can ____38____ you.” She looked down at the beast (庞然大物). Then her eyes came back to mine, looking ____39____. “This has never ____40____ to me before,” she said, as much to herse lf as to me. I felt that her sudden ____41____ had something to do with the escalator’s inhuman mechanical coldness.I took her arm. she made a little sound of ____42____ as we stepped on, but she ____43____ slightly as the escalator moved us ____44____. “I don’t understand this at all,” she said. As we near the bottom, her ____45____ was tight on my arm again, but we did fine. “I’m so ____46____…” she said. “It was nothing,” I said, “I was happy to do it.”Yes, I was very happy at that moment. I am busy ____47____ a family and working full time. I do very ____48____ to help others. For a brief moment, I had a flash of understanding into why someone would become a nurse or a social worker. When I ____49____ the woman, I felt pure, whole, and purposeful. It was a ____50____ moment than I’d had in weeks.31. A. scared B. delighted C. enjoyable D. frustrated32. A. patiently B. pleasantly C. coldly D. softly33. A. behavior B. expression C. question D. voice34. A. in B. back C. down D. out35. A. nodded B. refused C. smiled D. waved36. A. arrived B. left C. reached D. grasped37. A. disappointed B. convinced C. sure D. aware38. A. care about B. hold on to C. send for D. take control of39. A. optimistic B. proud C. doubtful D. angry40. A. attended to B. happened to C. came across D. let down41. A. fear B. decision C. attitude D. change42. A. interest B. alarm C. joy D. surprise43. A. frightened B. worried C. released D. relaxed44. A. downward B. upward C. outward D. backward45. A.B. holdC. weightD. shakepush46. A. grateful B. afraid C. brave D. embarrassed47. A. caring for B. seeking for C. reaching for D. accounting for48. A. much B. a lot C. little D. a little49. A. rescued B. contacted C. seized D. helpedC. happierD. more academic50. A. purer B. moredifficult第三部分：阅读理解（共20小题：每小题2分，满分40分）阅读下列短文，从每篇短文后所给各题的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。