【全国百强校】四川省南充高级中学2017届高三上学期期末考试英语试题

四川省南充市高三英语上册期末试题班级：________________ 学号：________________ 姓名：______________一、听力题（每题3分）Question 1:You will hear a conversation between a student and a teacher about an upcomingproject deadline.Conversation:Teacher: Hi, John. Remember, the deadline for your history project is this Friday.Student: Oh, I thought it was next week.Teacher: No, it’s this Friday. Make sure you submit it on time.Question: What is the deadline for the student’s history project?A. This weekend.B. Next Friday.C. This Friday.D. Tomorrow.Answer: CQuestion 2:You will hear a short announcement about a book club meeting.Announcement:Attention all book lovers! Our next book club meeting will be held thisThursday evening at 7 p.m. in the school library. We’ll be discussing “The Great Gatsby” by F. Scott Fitzgerald. Don’t forget to bring your copies. Question: Where will the book club meeting take place?A. In the cafeteria.B. In the auditorium.C. In the school library.D. Online via Zoom.Answer: CQuestion 3:You will hear a conversation between two friends discussing their plans for the weekend.Conversation:Amy: What are you doing this weekend, Lily?Lily: I’m going to volunteer at the animal shelter. They always need help. Am y: That sounds great! I’m staying home to study for my English exam. Question: What is Lily planning to do this weekend?A. Study for an exam.B. Visit a museum.C. Volunteer at an animal shelter.D. Go shopping with friends.Answer: CQuestion 4:You will hear a short news report about a local event.News Report:Local residents are invited to attend the annual charity run this Sunday morning. The event starts at 8 a.m. at City Park and benefits the Children’s Hospital. Registration is free, but donations are encouraged. Question: What does the news report mainly talk about?A. A charity run for a local hospital.B. A free concert in the park.C. A job fair for volunteers.D. A sports competition for children.Answer: AQuestion 5:You will hear a student asking for advice from a tutor about improving their English listening skills.Conversation:Student: Hi, Ms. Lee. I’m really struggling with my English listening. Any tips?Tutor: Sure, try listening to English podcasts or watching English TV shows with su btitles. It’ll help you get used to the rhythm and pronunciation. Question: What advice does the tutor give to improve English listening skills?A. Reading more English books.B. Listening to English podcasts or watching TV shows with subtitles.C. Joining an English language exchange group.D. Practicing speaking with a native English speaker.Answer: B二、单选题（每题3分）1.The teacher advised us to_______reading English for half an hour every morning toimprove our language skills.A. spendB. costC. payD. takeAnswer: A2.The book,_______by the famous writer, has been a bestseller for months.A. writtenB. being writtenC. to be writtenD. having writtenAnswer: A3.Tom was so busy that he didn’t notice his friend_______into the room.A. comeB. comesC. comingD. to comeAnswer: C4.If you_______to the party last night, you would have met the famous singer.A. wentB. had goneC. goD. have goneAnswer: B5.The government is taking measures to reduce air pollution, which is_______to thehealth of the citizens.A. harmfulB. harmfullyC. benefitD. beneficialAnswer: A三、多选题（每题4分）Question 1:Which of the following statements about English learning is NOT true?A. Regular practice can significantly improve one’s language proficiency.B. Memorizing vocabulary alone is sufficient for mastering English.C. Listening to English songs and watching movies can enhance listening comprehension.D. Participating in English language clubs promotes speaking skills. Answer: BQuestion 2:The teacher emphasized the importance of_______a clear plan before starting a new project.A. makingB. makeC. to makeD. having madeAnswer: AQuestion 3:The book “Pride and Prejudice” by Jane Austen is widely considered as one of the greatest works of_______literature.A. classicalB. historicC. historicalD. historicallyAnswer: CQuestion 4:It is vital that students learn how to manage their time effectively,especially during their_______years in high school.A. seniorB. higherC. elderD. laterAnswer: AQuestion 5:The government has introduced new policies aimed at reducing pollution and promoting_______living conditions for citizens.A. betterB. goodC. bestD. wellAnswer: A四、阅读理解（每题4分）Section IV: Reading ComprehensionPassage 1: The Importance of ReadingIn today’s fast-paced world, where screens and digital devices dominate our lives, the art of reading is often overlooked. However, reading remains a vital skill that enriches our minds, broadens our horizons, and enhances our emotional intelligence.Reading exposes us to diverse perspectives, allowing us to understand and empathizewith characters from different backgrounds and experiences. It stimulates our imagination, fostering creativity and critical thinking. Furthermore, reading regularly improves vocabulary, comprehension, and writing skills, which are essential for academic success and personal development.Books, in particular, have the power to transport us to faraway lands, introduce us to fascinating historical events, and inspire us with stories of resilience and triumph. They can serve as guides in navigating life’s challenges, offering wisdom and comfort in times of need.Questions:1.What is the main idea of the passage? (4 points)•Answer: The importance of reading in today’s society.2.How does reading help broaden our horizons? (4 points)•Answer: By exposing us to diverse perspectives and characters from different backgrounds.3.What are two skills that regular reading improves? (4 points)•Answer: Vocabulary and comprehension.4.How do books inspire us? (4 points)•Answer: Books inspire us with stories of resilience, triumph, and offer guidance in navigating life’s challenges.5.What is the role of reading in enhancing emotional intelligence? (4 points)•Answer: Reading helps enhance emotional intelligence by fostering empathy and understanding of others’ experiences.Passage 2: The Benefits of Early Childhood EducationEarly childhood education, which spans from birth to age eight, lays the foundationfor lifelong learning and development. It is a crucial period when children’s brains are highly malleable and responsive to environmental stimuli.Research has shown that children who receive quality early childhood education experience numerous benefits, including improved cognitive abilities, bettersocial-emotional skills, and higher academic achievement later in life. Theseprograms often incorporate play-based learning, which fosters creativity,problem-solving, and collaboration.Moreover, early childhood education provides a safe and nurturing environment where children can explore, learn, and grow. It helps build self-confidence,resilience, and a love for learning that lasts a lifetime.Questions:1.What age range does early childhood education cover? (4 points)•Answer: From birth to age eight.2.What makes the early childhood period crucial for learning and development? (4points)•Answer: Children’s brains are highly malleable and responsive to environmental stimuli during this period.3.What type of learning is often used in early childhood education programs? (4points)•Answer: Play-based learning. two benefits of quality early childhood education. (4 points)•Answer: Improved cognitive abilities and better social-emotional skills.5.How does early childhood education contribute to building self-confidence andresilience? (4 points)•Answer: By providing a safe and nurturing environment where children can explore, learn, and grow, early childhood education helps build self-confidence andresilience.五、作文（30分）Title:“The Importance of Persistence in Achieving Goals”Instructions:Write an essay on the significance of persistence in attaining one’s goals. You shouldinclude personal experiences, if any, to support your arguments. Your essay should be no less than 150 words.Sample Essay:In the tapestry of life, the thread of persistence weaves through every successful endeavor, binding dreams to reality. The significance ofpersistence in achieving goals cannot be overstated; it is the unwavering commitment that transforms aspirations into accomplishments.As a high school student from Nanyang, Henan Province, I have come to appreciate the value of perseverance firsthand. During my preparation for the English Language Olympiad, I encountered numerous challenges—complex grammar rules, intricate vocabulary, and the pressure to excel. On several occasions, frustration threatened to dampen my enthusiasm, but it was my unyielding spirit that kept me going. I spent countless hours studying,practicing, and seeking guidance from my teachers. The journey was arduous, yet it taught me a valuable lesson: without persistence, even the most promising dreams remain unfulfilled.Moreover, history is replete with examples that underscore the importance of perseverance. Thomas Edison, after countless failures, invented the light bulb, revolutionizing the world. Mahatma Gandhi, through decades of peaceful protest, led India to independence. Their stories serve as beacons, illuminating the path to success for generations to come.In essence, persistence is the key that unlocks the door to our full potential. It fosters resilience, enabling us to bounce back from setbacks and continue striving forward. It nurtures determination, driving us to push beyond our limits and reach for the stars.In conclusion, the pursuit of goals is fraught with obstacles, but it is through persistence that we transform these challenges into stepping stones towards triumph. Let us embrace the spirit of perseverance, for it is the cornerstone upon which our dreams are built.。

南充高中2016-2017年上学期期末考试高三英语试卷时量120分钟总分150分命题人高三英语备课组第I卷(选择题共100分)第I卷第一部分听力(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个小题；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，从每题所给的A、B、C三个选项中选出最佳选顶，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What are the speakers mainly talking about?A. A job.B. Education.C. New technology.2. What is the man doing?A. He's hiking.B. lie's running.C. He's drinking water.3. Where arc the speakers?A. Up in a tree.B. On a beach.C. On a big hill.4. How do the speakers know each other?A. From school.B. From work.C. From the restaurant5. How much will the woman pay for the new computer?A. Two hundred dollars.B. Eight hundred dollars.C. One thousand dollars.第二节听下面5段对话或独白。

每段对话或独白后有2至4个小题，.从题中所给的A、B、C三个选项中选出最佳选顶，并标在试卷的相应位置。

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

四川省南充高级中学2017届高三上学期期末考试理综合化学试题本试卷分为第I卷(选择题)和第II卷(非选择题)两部分，共38小题，共300分，共15页。

考试结束后，将本试卷和答题卡一并交回.可能用到的原子量:H-1 O-16 Al-27 S-32 Ca-407. 2016年11月14日京华时报讯：昨天，记者从环保部获悉，中国环境监测总站预测，预计14日一15日，受冷空气影响，京津冀区域污染自北向南逐步缓解，北部以良至轻度污染为主，中南部以轻至中度污染为主，局部地区可能出现短时重度污染，首要污染物为PM2.5。

下列有关描述不合理的是( )A．空气质量状况属于重度污染的地区，所有人的健康都会受到严重影响B．二氧化硫、氮氧化物、悬浮颖粒(PM)是雾霾的主要污染物，前两者是气态污染物，PM2.5才是加重雾霾天气污染的罪魁祸首C．PM2.5(可吸入颖粒物)成份很复杂，并具有较强的吸附能力，随吸入空气而进入肺部，从而影响健康D．气象环境报告中新增的PM2.5是一种新的分子【答案】D【考点定位】考查化学与生活的有关判断【名师点晴】化学与可持续发展及环境保护尽管不是教学的重点，但该内容与我们的生产、生活息息相关，因此成为历年高考的必考的热点。

试题的命题形式是常把化学知识与实际生产、生活、环境保护及科技前沿等问题结合起来，突出化学与可持续发展、环境保护的密切联系，综合考查学生分析问题、解决问题的能力，可很好的落实新课标的理念。

平时注意相关相关知识的积累和灵活应用。

8.关于三种有机物叙述错误的是(-SH 的性质类似于 - OH)A ．都能发生酯化反应B ．都能与NaOH 反应C ．甲的苯环上的一氯化物有4种D ．丙的分子式为1015C H ON ,苯环上的一氯化物有3种【答案】B考点：考查有机物结构和性质9.设阿伏加德罗常数的值为A N ，则下列说法正确的是A ．标准状况下，22.4 L 乙烯含有的共用电子对数为5A NB ．1 mol 4a N HSO 中的阳离子数为2A NC ．通常状况下，1mol NO 和0.5 mol 2O 在密闭容器中混，.生成2NO 分子数为A ND ．制取漂白粉时,标准状况下22.4 L 2Cl 参加反应，转移电子数为A N【答案】D【解析】试题分析：A 、标准状况下，22.4 L 乙烯是1mol ，含有的共用电子对数为6N A ，A 错误；B 、NaHSO 4固体含有Na +、4HSO -，1 mol NaHSO 4中的阳离子数为N A ，B 错误；C 、体系中存在2NO 2N 2O 4，生成NO 2分子数X 小于N A ，C 错误；D 、制取漂白粉时氯气与氢氧化钙生成氯化钙和次氯酸钙，氯气既是氧化剂又是还原剂，标准状况下22.4 LCl 2 参加反应，转移电子数为N A ，D 正确。

南充高中2016-2017年上学期期末考试高三理科综合试卷本试卷分为第I卷(选择题)和第II卷(非选择题)两部份，共38小题，共300分，共15页。

考试终止后，将本试卷和答题卡一并交回.可能用到的原子量:H-1 O-16 Al-27 S-32 Ca-40第I卷选择题(共21题，共126分)一、选择题:(本大题共21小题，每小题6分.在每小题给出的四个选项中，第1-17题只有一项是符合题目要求，第18-21题有多项符合题目要求。

全数选对的得6分，选对但不全的得3分，有选错的得0分。

)1.生物学是一门以实验为基础的科学，下而有关实验叙述正确的是( )A.施莱登、施旺的细胞学说指出细胞通过割裂产生新细胞B.科学家在电镜下看到细胞膜呈现出暗——亮——暗三层结构，斗胆提诞生物膜都由蛋白质——脂质——蛋白质三层结构组成C.在观看细胞内DNA和RNA散布的实验中，可利用酒精来改变细胞膜通透性，加速染色剂进入细胞D.无水乙醇在色素的提取和分离实验时起到分离色素的作用2.下列有关生命系统中方向性的描述，合理的是A.生物进化进程中，基因频率的改变是定向的B.受到刺橄时，兴奋在反射弧中的传导是双向的C.植物细胞发生质壁分离时，水分子的运动是单向的D.动物体内唾液淀粉酶的合成进程中，遗传信息的传递是双向的3.下图为基因的作用与性状的表现流程示用意，请依照图分析，不正确的选项是A.③进程中只需要mRNA，氨基酸、核糖体、酶、ATP即可完成B.某段DNA上发生了基因突变.但形成的蛋白质不必然会改变C. 囊性纤维病是基因对性状的直接操纵，使结构蛋白发生转变所致D.①进程是转录，它以DNA的一条链为模板、四种核糖核苷酸为原料合成RNA4.新华网报导，科学家发觉一种蜣螂提高了“生活品位”，不仅吃粪便，还取食包括蜈蚣在内的千足虫。

与一般蜣螂相较其部份结构也发生转变：头部较窄而长，便于进食千足虫内脏；后腿较蜷曲，便于捕猎千足虫，下列推测合理的是( )A.该蜣螂与一般蜣螂确信已经产生了生殖隔离B.与一般蜣螂相较，该蜣螂的某些基因频率发生了改变C.该蜣螂在生态系统中仅作为捕食者而存在D.该蜣螂头部和后腿结构发生转变，是人工选择的结果5.下列关于免疫调剂的叙述，错误的是A.衰老和破损细胞的清除需免疫系统参与B.自身细胞在必然条件下能够转化为抗原C.浆细胞可产生免疫活性物质而T细胞不能D.特异性免疫中发挥作用的要紧细胞是淋巴细胞6.有关植物生命活动的调剂，下列说法错误的是A.生长素的极性运输是一个耗能的进程B.农业上能够应用赤霉素打破种子的休眠C. 2, 4-D是一种具有与生长素相似的生理效应的植物激素D.植物激素调剂是基因组在一按时刻和空间上程序性表达的结果7. 2016年11月14日京华时报讯：昨天，记者从环保部得悉，中国环境监测总站预测，预计14日一15日，受冷空气阻碍，京津冀区域污染自北向南慢慢减缓，北部以良至轻度污染为主，中南部以轻至中度污染为主，局部地域可能显现短时重度污染，首要污染物为下列有关描述不合理的是( )A.空气质量状况属于重度污染的地域，所有人的健康都会受到严峻阻碍B.二氧化硫、氮氧化物、悬浮颖粒(PM)是雾霾的要紧污染物，前二者是气态污染物.PM.才是加重雾霾天气污染的罪魁罪魁C. (可吸入颖粒物)成份很复杂，并具有较强的吸附能力，随吸入空气而进入肺部，从而阻碍健康D.气象环境报告中新增的是一种新的分子8.关于三种有机物叙述错误的是(-SH的性质类似于 - OH)A.都能发生酯化反映B.都能与NaOH反映C.甲的苯环上的一氯化物有4种C H ON,苯环上的一氯化物有3种D.丙的分子式为1015N，则下列说法正确的是9.设阿伏加德罗常数的值为ANA.标准状况下， L乙烯含有的共用电子对数为5AB. 1 mol 4a N HSO 中的阳离子数为2A NC.通常状况下，1mol NO 和 mol 2O 在密闭容器中混，.生成2NO 分子数为A ND.制取漂白粉时,标准状况下 L 2Cl 参加反映，转移电子数为A N10.下列实验操作、现象与结论一致的是11.如图所示，将铁棒和石墨棒插入盛有饱和NaCl 溶液的U 型管中，下列分析正确的是A.1K 闭合，铁棒上发生的反映为222H e H +-+=↑13. 1K 闭合，石墨棒周围溶液pH 慢慢降低C. 2K 闭合，铁棒可不能被侵蚀，属于外加电流的阴极爱惜法D. 2K 闭合，电路中通过A N 个电子时，两极共产动气体12.下列说法不正确的是A.将pH=4的某酸稀释10倍，测得其pH<5 ，则该酸为弱酸B.相同pH 的3CH OONa 、 NaOH 和23Na CO 三种溶液：233()()()c NaOH c Na CO c CH COONa <<C. mo1/L 的NaHA 溶液pH=5，则溶液：2()()()c HA c H c H A -+>>D.向44NH HSO 溶液中滴加NaOH 溶液至中性，所得混合液：4()()()()c Na c NH c OH c H ++-+>>=13.在不同温度下，水济液中()c H +与()c OH -关系如图所示。

2016-2017学年四川省南充市高级中学高三（上）期末数学试卷（理科）一、选择题：本大题共10小题，每小题5分，共50分.在每个小题给出的四个选项中，有且只有一项符合题目要求.1．cos（﹣585°）的值为（）A．B．C．D．2．如图，一个空间几何体的正视图、侧视图、俯视图均为全等的等腰直角三角形，如果直角三角形的斜边长为，那么这个几何体的体积是（）A．B．C．D．3．已知随机变量服从正态分布N（0，σ2），且P（﹣2≤ξ≤0）=0.4，则P（ξ＞2）=（）A．0.1 B．0.2 C．0.4 D．0.64．设α，β是两个不同的平面，a，b是两条不同的直线，下列四个命题中正确的命题是（）A．若a∥α，b∥α，则a∥bB．若a∥α，b∥β，a∥b，则α∥βC．若a⊥α，a⊂β，则α⊥βD．若a，b在α内的射影相互垂直，则a⊥b5．如图，该程序运行后输出的结果是（）A．6 B．8 C．10 D．126．关于实数x，y的不等式组所表示的平面区域记为M，不等式（x﹣4）2+（y﹣3）2≤1所表示的区域记为N，若在M内随机取一点，则该点取自N的概率为（）A．B．C．D．7．已知集合M={m|（m﹣11）（m﹣16）≤0，m∈N}，若（x3﹣）n（n∈M）的二项展开式中存在常数项，则n等于（）A．16 B．15 C．14 D．128．在同一平面内，下列说法：①若动点P到两个定点A，B的距离之和是定值，则点P的轨迹是椭圆；②若动点P到两个定点A，B的距离之差的绝对值是定值，则点P的轨迹是双曲线；③若动点P到定点A的距离等于P到定直线的距离，则点P的轨迹是抛物线；④若动点P到两个定点A，B的距离之比是定值，则点P的轨迹是圆．其中错误的说法个数是（）A．1 B．2 C．3 D．49．定义在R上的偶函数f（x）满足f（x）+f（x﹣1）=0，且在[﹣5，﹣4]上是增函数，A，B是锐角三角形的两个内角，则（）A．f（sinA）＞f（cosB）B．f（sinA）＜f（cosB）C．f（sinA）＞f（sinB）D．f（cosA）＞f（cosB）10．如图，已知线段PQ=，点Q在x轴正半轴，点P在边长为1的正方形OABC 第一象限内的边上运动．设∠POQ=θ，记x（θ）表示点Q的横坐标关于θ的函数，则x（θ）在（0，）上的图象可能是（）A．B．C．D．二、填空题：本大题共5小题，每小题4分，共25分.11．5人排成一列，其中甲、乙二人相邻的不同排法的种数为．（结果用数字表示）12．设函数，若，则x0的取值范围为．13．若直线l过抛物线x2=﹣8y的焦点F，且与双曲线在一、三象限的渐近线平行，则直线l截圆所得的弦长为．14．函数，数列{a n}的通项公式a n=|f（n）|，若数列从第k项起每一项随着n项数的增大而增大，则k的最小值为．15．设{a n}是集合{3p+3q+3r|0≤p＜q＜r，且p，q，r∈N*}中所有的数从小到大排列成的数列，已知a k=2511，则k=．三、解答题：本大题共5小题，共75分.解答应写出必要的文字说明或推理、验算过程.16．已知，其中A，B，C是△ABC的内角．（1）当时，求的值；（2）若，当取最大值是，求B的大小及BC边的长．17．抛掷三枚不同的具有正、反两面的金属制品A1、A2、A3，假定A1正面向上的概率为，A2正面向上的概率为，A3正面向上的概率为t（0＜t＜1），把这三枚金属制品各抛掷一次，设ξ表示正面向上的枚数．（1）求ξ的分布列及数学期望Eξ（用t表示）；（2）令a n=（2n﹣1）cos（Eξ）（n∈N+），求数列{a n}的前n项和．18．斜率为的直线l与椭圆+=1（a＞b＞0）交于不同的两点A、B．若点A、B在x轴上的射影恰好为椭圆的两个焦点．（1）求椭圆的离心率；（2）P是椭圆上的动点，若△PAB面积最大值是4，求该椭圆的方程．19．已知在三棱柱ABC﹣A1B1C1中，B1B⊥平面ABC，∠ABC=90°，B1B=AB=2BC=4，D、E分别是B1C1，A1A的中点．（1）求证：A1D∥平面B1CE；（2）设M是的中点，N在棱AB上，且BN=1，P是棱AC上的动点，直线NP与平面MNC所成角为θ，试问：θ的正弦值存在最大值吗？若存在，请求出的值；若不存在，请说明理由．20．已知函数．（1）当a=0时，求f（x）在点（1，f（1））处的切线方程；（2）是否存在实数a，当0＜x≤2时，函数f（x）图象上的点都在所表示的平面区域（含边界）？若存在，求出a的值组成的集合；否则说明理由；（3）若f（x）有两个不同的极值点m，n（m＞n），求过两点M（m，f（m）），N （n，f（n））的直线的斜率的取值范围．本题21、22、23三个选答题，每题7分，请考生任选两题作答，满分7分.如果多做，则按照所做的前两题计分.[选修4-2：矩阵与变换]21．在矩阵A的变换下，坐标平面上的点的横坐标伸长到原来的3倍，纵坐标不变．（1）求矩阵A及A﹣1；（2）求圆x2+y2=4在矩阵A﹣1的变换下得到的曲线方程．[选修4-4：坐标系与参数方程]22．在直角坐标系中，直线l的参数方程为（t为参数），在极坐标系（与直角坐标系xoy取相同的单位长度，且以原点为极点，x轴的正半轴为极轴）中，圆C的极坐标方程为ρ=4cosθ．（1）若直l线与圆C相切，求实数a的值；（2）若点M的直角坐标为（1，1），求过点M且与直线l垂直的直线m的极坐标方程．[选修4-5：不等式选讲]23．已知关于x的不等式（其中a＞0）．（1）当a=3时，求不等式的解集；（2）若不等式有解，求实数a的取值范围．2016-2017学年四川省南充市高级中学高三（上）期末数学试卷（理科）参考答案与试题解析一、选择题：本大题共10小题，每小题5分，共50分.在每个小题给出的四个选项中，有且只有一项符合题目要求.1．cos（﹣585°）的值为（）A．B．C．D．【考点】运用诱导公式化简求值．【分析】利用余弦函数为偶函数将所求式子化简，再利用诱导公式及特殊角的三角函数值化简，即可求出值．【解答】解：cos（﹣585°）=cos585°=cos=cos225°=cos=﹣cos45°=﹣故选：A2．如图，一个空间几何体的正视图、侧视图、俯视图均为全等的等腰直角三角形，如果直角三角形的斜边长为，那么这个几何体的体积是（）A．B．C．D．【考点】棱柱、棱锥、棱台的体积．【分析】由已知中的三视图可得：该几何体是一个三棱锥，代入棱锥体积公式，可得答案．【解答】解：由已知中的三视图可得：该几何体是一个三棱锥，如果直角三角形的斜边长为，则直角三角形的直角边长均为1，故几何体的体积V=×1×1×1=，故选：C3．已知随机变量服从正态分布N（0，σ2），且P（﹣2≤ξ≤0）=0.4，则P（ξ＞2）=（）A．0.1 B．0.2 C．0.4 D．0.6【考点】正态分布曲线的特点及曲线所表示的意义．【分析】本题考查正态分布曲线的性质，随机变量ξ服从正态分布N（0，σ2），由此知曲线的对称轴为y轴，|即可得出结论【解答】解：∵随机变量ξ服从正态分布N（0，σ2），且P（﹣2≤ξ≤0）=0.4，∴P（ξ＞2）=0.5﹣P（﹣2≤ξ≤0）=0.1，故选：A．4．设α，β是两个不同的平面，a，b是两条不同的直线，下列四个命题中正确的命题是（）A．若a∥α，b∥α，则a∥bB．若a∥α，b∥β，a∥b，则α∥βC．若a⊥α，a⊂β，则α⊥βD．若a，b在α内的射影相互垂直，则a⊥b【考点】空间中直线与平面之间的位置关系．【分析】在A中，a与b相交、平行或异面；在B中，α与β相交或平行；在C中，由面面垂直的判定定理得α⊥β；在D中，a与b相交、平行或异面．【解答】解：由α、β、γ是三个不同的平面，a、b是两条不同的直线，知：在A中，若a∥α，b∥α，则a与b相交、平行或异面，故A错误；在B中若a∥α，b∥β，a∥b，则α与β相交或平行，故B错误；在C中，若a⊥α，a⊂β，则根据平面与平面垂直的判定定理，可得α⊥β，故C正确；在D中，若a，b在平面α内的射影互相垂直，则a与b相交、平行或异面，故D 错误．故选：C．5．如图，该程序运行后输出的结果是（）A．6 B．8 C．10 D．12【考点】程序框图．【分析】经过观察为当型循环结构，按照循环结构进行执行，当满足执行条件时跳出循环，输出结果即可．【解答】解：模拟程序的运行，可得A=12，s=0不满足条件A≤3，执行循环体，S=2，A=10不满足条件A≤3，执行循环体，S=4，A=8不满足条件A≤3，执行循环体，S=6，A=6不满足条件A≤3，执行循环体，S=8，A=4不满足条件A≤3，执行循环体，S=10，A=2满足条件A≤3，退出循环，输出S的值为10．故选：C．6．关于实数x，y的不等式组所表示的平面区域记为M，不等式（x﹣4）2+（y﹣3）2≤1所表示的区域记为N，若在M内随机取一点，则该点取自N的概率为（）A．B．C．D．【考点】几何概型．【分析】由题意知本题是一个几何概型，分别求出对应的面积，即可得到结果．【解答】解：关于实数x，y的不等式组所表示的平面区域记为M，面积为=8，不等式（x﹣4）2+（y﹣3）2≤1所表示的区域记为N，且满足不等式组，面积为，∴在M内随机取一点，则该点取自N的概率为=，故选A．7．已知集合M={m|（m﹣11）（m﹣16）≤0，m∈N}，若（x3﹣）n（n∈M）的二项展开式中存在常数项，则n等于（）A．16 B．15 C．14 D．12【考点】二项式定理的应用．【分析】化简集合M，求出二项式的通项公式，化简整理后，令x的指数为0，对照M中的元素，即可得到答案．【解答】解：集合M={m|（m﹣11）（m﹣16）≤0，m∈N}={11，12，13，14，15，16}，（x3﹣）n（n∈M）的二项展开式的通项公式为Tr+1==，令3n﹣5r=0，则n=，由于n∈M，则n=15．故选B．8．在同一平面内，下列说法：①若动点P到两个定点A，B的距离之和是定值，则点P的轨迹是椭圆；②若动点P到两个定点A，B的距离之差的绝对值是定值，则点P的轨迹是双曲线；③若动点P到定点A的距离等于P到定直线的距离，则点P的轨迹是抛物线；④若动点P到两个定点A，B的距离之比是定值，则点P的轨迹是圆．其中错误的说法个数是（）A．1 B．2 C．3 D．4【考点】轨迹方程．【分析】利用椭圆，双曲线、抛物线的定义，即可得出结论．【解答】解：①平面内与两定点距离之和为常数的点的轨迹是椭圆，如果距离之和等于两点间的距离，轨迹表示的是线段，不表示椭圆，所以①不正确；②平面内与两定点距离之差绝对值为常数的点的轨迹是双曲线，这个常数必须小于两定点的距离，此时是双曲线，否则不正确，所以②不正确；③当定点位于定直线时，此时的点到轨迹为垂直于直线且以定点为垂足的直线，只有当定点不在直线时，轨迹才是抛物线，所以③错误；④若动点P到两个定点A，B的距离之比是定值，则点P的轨迹是圆，也可以是直线，故不正确．故选D．9．定义在R上的偶函数f（x）满足f（x）+f（x﹣1）=0，且在[﹣5，﹣4]上是增函数，A，B是锐角三角形的两个内角，则（）A．f（sinA）＞f（cosB）B．f（sinA）＜f（cosB）C．f（sinA）＞f（sinB）D．f（cosA）＞f（cosB）【考点】抽象函数及其应用．【分析】首先根据A、B是锐角三角形的两个内角，结合y=cosx在区间（0，）上是减函数，证出sinA＞cosB．然后根据偶函数f（x）满足f（x+1）=﹣f（x），可得函数f（x）是周期为2的函数，且f（x）在[0，1]上是减函数．最后根据f（x）在[0，1]上是减函数，结合锐角三角形中sinA＞cosB，得到f（sinA）＜f（cosB）．【解答】解：∵A、B是锐角三角形的两个内角，∴A+B＞，可得A＞﹣B，∵y=cosx在区间（0，）上是减函数，＞A＞﹣B＞0，∴sinA＞sin（﹣B）=cosB，即锐角三角形的两个内角A、B是满足sinA＞cosB，∵函数f（x）满足f（x+1）=﹣f（x），∴f（x+2）=﹣f（x+1）=﹣[﹣f（x）]=f（x），可得函数f（x）是周期为2的函数．∵f（x）在[﹣5，﹣4]上是增函数，∴f（x）在[﹣1，0]上也是增函数，再结合函数f（x）是定义在R上的偶函数，可得f（x）在[0，1]上是减函数．∵锐角三角形的两个内角A、B是满足sinA＞cosB，且sinB、cosA∈[0，1]∴f（sinA）＜f（cosB）．故选：B10．如图，已知线段PQ=，点Q在x轴正半轴，点P在边长为1的正方形OABC 第一象限内的边上运动．设∠POQ=θ，记x（θ）表示点Q的横坐标关于θ的函数，则x（θ）在（0，）上的图象可能是（）A．B．C．D．【考点】函数的图象．【分析】当θ∈（0，）时，求得x（θ）=1+，图象是上凸的．当θ∈[，）时，求得x（θ）=cotθ+1，图象是下凹的．结合所给的选项，可得结论．【解答】解：当θ∈（0，）时，PA=tanθ，AQ==，x（θ）=1+，它的图象是上凸的．当θ∈[，）时，PA=1，OA=cotθ，AQ===1，x（θ）=cotθ+1，它的图象是下凹的．结合所给的选项，故选：A．二、填空题：本大题共5小题，每小题4分，共25分.11．5人排成一列，其中甲、乙二人相邻的不同排法的种数为48．（结果用数字表示）【考点】排列、组合的实际应用．【分析】根据题意，分2步用捆绑法进行分析：①、将甲乙二人看成一个元素，考虑其顺序，②、二人排好后，与剩余三人全排列，分别用排列、组合数公式计算每一步的情况数目，由分步计数原理计算可得答案．【解答】解：根据题意，分2步进行分析：①、将甲乙二人看成一个元素，考虑其顺序，有A22=2种排法；②、二人排好后，与剩余三人全排列，有A44=24种情况，则一共有2×24=48种不同排法；故答案为：48．12．设函数，若，则x0的取值范围为x0＞．【考点】其他不等式的解法．【分析】x＞，f（x）=lnx|=1，利用，可得x0的取值范围．【解答】解：x＞，f（x）=lnx|=1，∵，，∴x0＞，故答案为x0＞．13．若直线l过抛物线x2=﹣8y的焦点F，且与双曲线在一、三象限的渐近线平行，则直线l截圆所得的弦长为2．【考点】抛物线的简单性质．【分析】求出抛物线的焦点和双曲线的渐近线方程，求得直线l的方程，求出圆心到直线的距离，运用弦长公式即可得到弦长．【解答】解：抛物线x2=﹣8y的焦点F为（0，﹣2），双曲线双曲线在一三象限的渐近线为y=x，则直线l的方程为：y=x﹣2，圆（x﹣4）2+y2=4的圆心为（4，0），半径为2，则圆心到直线的距离d==，则弦长为2=2，故答案为：2．14．函数，数列{a n}的通项公式a n=|f（n）|，若数列从第k项起每一项随着n项数的增大而增大，则k的最小值为3．【考点】数列的函数特性．【分析】x≥4时，利用导数研究函数的单调性即可得出．【解答】解：f（1）=，f（2）=，f（3）=﹣，x≥4时，f（x）＞0，f（4）=，x≥4时，f′（x）=++＞0，因此函数f（x）单调递增，f（x）≥f（4）＞0．a4＞a3，因此a n单调递增．∴数列从第3项起每一项随着n项数的增大而增大，则k的最小值为3．故答案为：3．15．设{a n}是集合{3p+3q+3r|0≤p＜q＜r，且p，q，r∈N*}中所有的数从小到大排列成的数列，已知a k=2511，则k=50．【考点】计数原理的应用．【分析】a k=2511，可得p=4，q﹣p=1，r﹣p=3，从而q=5，r=7，用列举法求解即可．【解答】解：0≤p＜q＜r，且p，q，r∈Na n=3p+3q+3r=3p（1+3q﹣p+3r﹣p），a k=2511，∴p=4，q﹣p=1，r﹣p=3，∴q=5，r=7，∴（p，q，r）=（4，5，7）（4，5，7）（3，5，7）（3，4，7）（2，5，7）（2，4，7）（2，3，7）（1，5，7）（1，4，7）（1，3，7）（1，2，7）（0，5，7）（0，4，7）（0，3，7）（0，2，7）（0，1，7）（4，5，6）（3，5，6）（3，4，6）（2，5，6）（2，4，6）（2，3，6）（1，5，6）（1，4，6）（1，3，6）（1，2，6）（0，5，6）（0，4，6）（0，3，6）（0，2，6）（0，1，6）（3，4，5）（2，4，5）（2，3，5）（1，4，5）（1，3，5）（1，2，5）（0，4，5）（0，3，5）（0，2，5）（0，1，5）（2，3，4）（1，3，4）（1，2，4）（0，3，4）（0，2，4）（0，1，4）（1，2，3）（0，2，3）（0，1，3）（0，1，2）∴（5+4+3+2+1）×2+（4+3+2+1）+（3+2+1）+（2+1）+1=50，故答案为：50三、解答题：本大题共5小题，共75分.解答应写出必要的文字说明或推理、验算过程.16．已知，其中A，B，C是△ABC的内角．（1）当时，求的值；（2）若，当取最大值是，求B的大小及BC边的长．【考点】平面向量数量积的运算；正弦定理．【分析】（1）由角A可得的坐标，代入向量模的公式计算的值；（2）由数量积的坐标运算得到，利用辅助角公式化积，可得当A=时，取得最大值，求出对应的B值，再由正弦定理求得BC边的长．【解答】解：（1）当时，，∴；（2）=．∴当A=时，取得最大值，此时B=，根据正弦定理：，得．17．抛掷三枚不同的具有正、反两面的金属制品A1、A2、A3，假定A1正面向上的概率为，A2正面向上的概率为，A3正面向上的概率为t（0＜t＜1），把这三枚金属制品各抛掷一次，设ξ表示正面向上的枚数．（1）求ξ的分布列及数学期望Eξ（用t表示）；（2）令a n=（2n﹣1）cos（Eξ）（n∈N+），求数列{a n}的前n项和．【考点】离散型随机变量的期望与方差；数列的求和；离散型随机变量及其分布列．【分析】（1）通过求出ξ=0、1、2、3时相应的概率，进而求出ξ的分布列及数学期望Eξ；（2）通过（1）、化简可知a n=（﹣1）n（2n﹣1），进而分n为奇数、偶数两种情况讨论即可求出S n．【解答】解：（1）依题意，ξ的可能取值为0、1、2、3，P（ξ=0）=••（1﹣t）=，P（ξ=1）=••（1﹣t）+••（1﹣t）+••t=，P（ξ=2）=••（1﹣t）+••t+••t=，P（ξ=3）=••t=，∴ξ的分布列为：数学期望Eξ=0•+1•+2•+3•=；（2）由（1）可知a n=（2n﹣1）cos（•）=（2n﹣1）cos（nπ）=（﹣1）n（2n﹣1），当n为偶数时，S n=[（﹣1）+3]+[（﹣5）+7]+…+[﹣（2n﹣3）+（2n﹣1）]=2•=n；当n为奇数时，S n=[（﹣1）+3]+[（﹣5）+7]+…+[﹣（2n﹣5）+（2n﹣3）]+[﹣（2n﹣1）]=2•﹣（2n﹣1）=n﹣1﹣2n+1=﹣n；综上所述，S n=（﹣1）n•n．18．斜率为的直线l与椭圆+=1（a＞b＞0）交于不同的两点A、B．若点A、B在x轴上的射影恰好为椭圆的两个焦点．（1）求椭圆的离心率；（2）P是椭圆上的动点，若△PAB面积最大值是4，求该椭圆的方程．【考点】椭圆的简单性质；椭圆的标准方程．【分析】（1）画出图形，结合图形，得出直线与椭圆两交点坐标，根据两点间的斜率公式，求出离心率e；（2）由（1）知，设出椭圆的标准方程+=1，求出|AB|的值，利用三角形的面积求出高h；再求点P到直线的最大距离d，由此求出c即可．【解答】解：（1）由题意知：直线与椭圆两交点的横坐标为﹣c，c，纵坐标分别为﹣，，∴由=转化为：2b2=2（a2﹣c2）=ac即2e2+e﹣2=0，解得e=，e=﹣（负根舍去），∴椭圆的离心率为e=；（2）∵P是椭圆上的动点，当△PAB的面积最大值是4时，有|AB|h=4，∵e=，∴b=c，∴a=c；∴设椭圆的方程为+=1，则|AB|=c，∴三角形PAB的高为h=；又直线为y=x，即x﹣2y=0；则点P（ccosθ，csinθ）到直线的距离表示为d==≤，令=，解得c=2，∴椭圆的方程为+=1．19．已知在三棱柱ABC﹣A1B1C1中，B1B⊥平面ABC，∠ABC=90°，B1B=AB=2BC=4，D、E分别是B1C1，A1A的中点．（1）求证：A1D∥平面B1CE；（2）设M是的中点，N在棱AB上，且BN=1，P是棱AC上的动点，直线NP与平面MNC所成角为θ，试问：θ的正弦值存在最大值吗？若存在，请求出的值；若不存在，请说明理由．【考点】直线与平面平行的判定；直线与平面所成的角．【分析】（1）法一（几何法）：连结BC1，与B1C交于点O，连结EO，DO，推导出四边形A1EOD是平行四边形，从而A1D∥EO，由此能证明A1D∥平面B1CE．法二（向量法）：建立空间直角坐标系B﹣xyz，利用向量法能证明A1D∥平面B1CE．（2）建立空间直角坐标系B﹣xyz，利用向量法求出存在符合题意的点P，且=．【解答】证明：（1）证法一（几何法）：连结BC1，与B1C交于点O，连结EO，DO，在△B1BC1中，DO B1B，在四边形B1BA1A中，A1E B1B，∴A1E DO，∴四边形A1EOD是平行四边形，∴A1D∥EO∵A1D⊄平面B1CE，EO⊂平面B1CE，∴A1D∥平面B1CE．证法二（向量法）：如图，建立空间直角坐标系B﹣xyz，由已知得A（4，0，0），C（0，2，0），B1（0，0，4），C1（0，2，4），D（0，1，4），E（4，0，2），则=（﹣4，1，0），=（0，2，﹣4），=（4，0，﹣2），设平面B1CE的一个法向量=（x，y，z），则，取x=1，得=（1，4，2），∵=﹣4+4=0，且A1D⊄平面B1CE，∴A1D∥平面B1CE．解：（2）设存在符合题意的点P．如图，建立空间直角坐标系B﹣xyz，由已知得A（4，0，0），C（0，2，0），M（2，0，3），N（1，0，0），则=（﹣1，0，﹣3），=（﹣1，2，0），=（﹣4，2，0），设平面MNC的一个法向量=（x，y，z），则，取x=6，得=（6，3，﹣2），设=，（0≤λ≤1），则==（3﹣4λ，2λ，0），由题设得sinθ=|cos＜＞|===，设t=1﹣λ（0≤λ≤1），则λ=1﹣t，且0≤t≤1，∴sinθ=，当t=0时，sinθ=0，当0＜t≤1时，sinθ==≤=．∴当且仅当，即t=时，sinθ取得最大值，此时λ=．∴存在符合题意的点P，且=．20．已知函数．（1）当a=0时，求f（x）在点（1，f（1））处的切线方程；（2）是否存在实数a，当0＜x≤2时，函数f（x）图象上的点都在所表示的平面区域（含边界）？若存在，求出a的值组成的集合；否则说明理由；（3）若f（x）有两个不同的极值点m，n（m＞n），求过两点M（m，f（m）），N （n，f（n））的直线的斜率的取值范围．【考点】利用导数求闭区间上函数的最值；利用导数研究曲线上某点切线方程．【分析】（1）求出函数的导数，计算f（1），f′（1）的值，求出切线方程即可；（2）法一：根据﹣2lnx≤0，设φ（x）=﹣2lnx，则问题等价于x∈（0，2]时，φ（x）max≤0，通过讨论a的范围，求出函数的最大值，从而求出a的范围即可；法二：由﹣2lnx≤0得，a≤2xlnx，令φ（x）=2xlnx，（0＜x≤2），则a≤[φ（x）]min，根据函数的单调性求出函数的最小值，从而求出a的范围即可；（3）求出函数f（x）的导数，求出a的范围，表示出直线MN的斜率，结合换元思想以及函数的单调性求出斜率k的范围即可．【解答】解：（1）a=0时，f（x）=x﹣2lnx，f′（x）=1﹣，∴f（1）=1，f′（1）=﹣1，∴求出直线方程是y﹣1=﹣（x﹣1），即y=﹣x+2；（2）由题意得：0＜x≤2时，f（x）≤x，即﹣2lnx≤0，设φ（x）=﹣2lnx，则问题等价于x∈（0，2]时，φ（x）max≤0，φ′（x）=﹣，（i）当a≥0时，φ′（x）＜0，不合题意，（ii）当a＜0时，①﹣∈（0，2）时，φ（x）在（0，﹣）上递增，在（﹣，2）上递减，φ（x）max=φ（﹣）=﹣2﹣2ln（﹣）≤0，此时，a∈（﹣4，﹣]；②﹣∈[2，+∞）时，φ（x）在（0，2]递增，φ（2）=﹣2ln2≤0，此时，a∈（﹣∞，﹣4]；综上，存在实数a组成的集合{a|a≤﹣}；方法二：由题意f（x）≤x，对x∈（0，2]恒成立，即﹣2lnx≤0对x∈（0，2]恒成立，由﹣2lnx≤0得，a≤2xlnx，令φ（x）=2xlnx，（0＜x≤2），则a≤[φ（x）]min，φ′（x）=2（lnx+x•）=2（lnx+1），当0＜x＜时，φ′（x）＜0，当＜x＜2时，φ′（x）＞0，∴φ（x）在（0，2]上的最小值是φ（）=﹣，故a≤﹣为所求；（3）由f′（x）==0（x＞0），得x2﹣2x﹣a=0，（x＞0），由题意得：，解得：﹣1＜a＜0，k MN===2﹣，设t=，（m＞n），则k MN=2﹣（t＞1），设g（t）=lnt，（t＞1），则g′（t）=，设h（t）=t﹣﹣2lnt（t＞1），则h′（t）=1+﹣=＞0，∴h（t）在（1，+∞）递增，∴h（t）＞h（1）=0即g（t）＞0，∴g（t）在（1，+∞）递增，t→+∞时，g（t）→+∞，设Q（t）=lnt﹣（1﹣），（t＞1），则Q′（t）=＞0，∴Q（t）在（1，+∞）递增，∴Q（t）＞Q（1）=0，即lnt＞1﹣，同理可证t﹣1＞lnt，∴t+1＞＞，当t→1时，t+1→2，→2，∴t→1时，g（t）→2，∴直线MN的斜率的取值范围是（﹣∞，0）．本题21、22、23三个选答题，每题7分，请考生任选两题作答，满分7分.如果多做，则按照所做的前两题计分.[选修4-2：矩阵与变换]21．在矩阵A的变换下，坐标平面上的点的横坐标伸长到原来的3倍，纵坐标不变．（1）求矩阵A及A﹣1；（2）求圆x2+y2=4在矩阵A﹣1的变换下得到的曲线方程．【考点】矩阵与向量乘法的意义；逆矩阵的意义．【分析】（1）由题意求出A=，再求出△=|A|=3，由此能求出A﹣1．（2）由=，得，由此能求出圆x2+y2=4在矩阵A﹣1的变换下得到的曲线方程．【解答】解：（1）∵在矩阵A的变换下，坐标平面上的点的横坐标伸长到原来的3倍，纵坐标不变，∴A=，∵△=|A|=3，∴A﹣1=．（2）由=，得，∴，代入x2+y2=4，得9x'2+y'2=4，∴圆x2+y2=4在矩阵A﹣1的变换下得到的曲线方程为9x2+y2=4．[选修4-4：坐标系与参数方程]22．在直角坐标系中，直线l的参数方程为（t为参数），在极坐标系（与直角坐标系xoy取相同的单位长度，且以原点为极点，x轴的正半轴为极轴）中，圆C的极坐标方程为ρ=4cosθ．（1）若直l线与圆C相切，求实数a的值；（2）若点M的直角坐标为（1，1），求过点M且与直线l垂直的直线m的极坐标方程．【考点】参数方程化成普通方程；简单曲线的极坐标方程．【分析】（1）直线l的参数方程为（t为参数），消去参数t化为普通方程．圆C的极坐标方程为ρ=4cosθ，即ρ2=4ρcosθ，利用互化公式可得直角坐标方程．利用点到直线的距离公式，根据直l线与圆C相切的性质即可得出a．（2）由直线l的方程为：3x﹣4y﹣a=0，利用相互垂直的直线斜率之间的关系可得：直线m的斜率为﹣．再利用点斜式可得直线m的方程，把代入可得极坐标方程．【解答】解：（1）直线l的参数方程为（t为参数），消去参数化为普通方程：3x﹣4y﹣a=0．圆C的极坐标方程为ρ=4cosθ，即ρ2=4ρcosθ，化为：x2+y2﹣4x=0，即（x﹣2）2+y2=4，可得圆心C（2，0），半径r=2．∵直l线与圆C相切，∴=2，化为：|a﹣6|=10，解得a=16或﹣4．（2）∵直线l的方程为：3x﹣4y﹣a=0，∴斜率为，∴直线m的斜率为﹣．∴直线m的点斜式为：y﹣1=﹣（x﹣1），化为4x+3y﹣7=0，把代入可得极坐标方程：4ρcosθ+3ρsinθ﹣7=0．[选修4-5：不等式选讲]23．已知关于x的不等式（其中a＞0）．（1）当a=3时，求不等式的解集；（2）若不等式有解，求实数a的取值范围．【考点】绝对值不等式的解法．【分析】（1）通过讨论x的范围得到关于x的不等式组，解出即可；（2）求出f（x）的最大值，得到关于a的不等式组，解出即可．【解答】解：（1）a=3时，|x﹣1|﹣|2x﹣1|＞﹣1，∴或或，解得：﹣1＜x＜1，故不等式的解集是（﹣1，1）；（2）f（x）=，∴f（x）∈（﹣∞，]，∴f（x）的最大值是，∵不等式有解，∴＞a，解得：a＞．2017年3月9日。

四川省南充高级中学2017-2018学年高二上学期期中考试英语试题第一部分听力（共两节，20小题，每小题1.5分，满分30分）第一节听下面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罝。

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鹿。

每段对话仅读一遍。

1. What is the woman going to do tomorrow?A. Go fishing.B. Go swimming.C. Go to the hospital.2. How does the man find his job?A. It is interesting and fun.B. Its pay is good.C. It is always tiling.3. What are the speakers talking about?A. A story.B. A book.C. A movie.4. Where does the conversation probably take place?A. At the railway station.B. At the bus stop.C. At the airport.5. Where is the man going?A. The classroom.B. The cinema.C. The library.第二节听下面5段对话或独白。

选出最佳选项。

请听第6段材料，回答第6至7题。

6. Why is the man going to the United States?A. To travel.B. To visit his friends.C. To attend a conference.7. When will the man probably leave for the United States?A. In mid-February.B. In mid-January.C. In early January.请听第7段材料，回答第8至9题。

2019-2020学年四川省南充高级中学高三英语上学期期末试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AThe history ofpop art begins with abstract art. Pop art emerged from the foundation of abstract art in the 1950s, first gaining recognition in Great Britain, then establishing itself in the United States in the 1960s.In the 1930s and 1940s, abstract art was greatly popular, but people began to hate this art form. Most abstract art produced in this era could be found in art galleries or the homes of the elite(名流), not in the homes of everyday people. Pop art sprung onto the scene as the people's art.Some art critics say pop art is a rebellion against abstract art; others say it is an extension of abstract art. You can see elements of abstract art in many pop art prints, especially those that consist of a collage（拼贴画）of images. However, some pop art pieces have nothing to do with abstract art, looking more like a photo of a popular consumer item, which impressed people deeply.Richard Hamilton and Eduardo Paolozzi were two of the first contemporary pop art pioneers recognised in Great Britain. They were a part of the Independent Group, an organised group of British artists who wanted to challenge ruling modernist approaches to culture. They recognised the value of modern advertising and comic book images. They used these popular images in art as a social commentary, often building irony and humour into their artworks, thus creating a new form of art. These pieces were mass-produced and sold at affordable prices.When pop art took off in the United States, it expanded to include a celebration of kitsch and the common images found in movies and television. Andy Wharhol is perhaps the most famous American pop art artist, known for his prints of the actress Marilyn Monroe, the singer Elvis Presley and of Campbell's soup cans. Andy Wharhol was inspired by images from advertisements and common consumer items. He also made a series of paintings which showed images of car accidents.1. Why did people dislike abstract art?A. People found it hard to understand.B. Ordinary people couldn't afford it.C. People disliked enjoying it in galleries.D. It wasn't based on everyday life.2. What impression do pop art works leave on people?A. They are a rebellion against abstract art.B. They have many elements of abstract art.C. They are created so real.D. They are products of advertisers.3. The Independent Group was set up to ________.A. introduce a new art form to the worldB. develop modern advertising furtherC. create new comic book imagesD. make British culture better knownBLearning to say “yes, and”When I first heard about the improvisation (即兴交流) class, I was hesitating. As a quiet and shy girl, I feared improvising in front of strangers. However,I knew I wanted to work as a science communicator after finishing my Ph.D., so it seemed like a perfect opportunity to learn how to speak and communicate with others effectively. I signed up, knowing the experience would give me help.During our first class, we learned an important concept of improvisation: “yes, and.” It means that, as improvisers, we’d better accept what fellow performers say. If someone says that rhinos (犀牛) are librarians, for example, then rhinos are librarians. We do not question the logic; we say “yes” and then continue with the scene as if nothing is wrong.The first few scenes were hard, but as weeks turned into months, I became more comfortable andeven started to enjoy our classes. I became better at listening, relating to my conversation partners, and communicating clearly in the moment. Once when I was giving a presentation about my science, an audience member surprised me with a question that didn’t grow out of the information I’d presented. Instead of getting confused and nervous, I took the “yes, and” approach—accepting the question and letting my mind focus on why it was asked. That helped me find an appropriate answer. I got pretty excited about it.The benefits of improvisation go beyond communication. Before attending the class, I would get stuck when my experiments produced unexpected data, thinking that I had made a mistake. But now, instead of getting discouraged, I will stay open to the possibility that the results are real, keep exploring the data and end up identifying a new type of cell—one that isn’t behaving as expected.I think all scientists can benefit from this lesson. If the data say rhinos are librarians, then it’s worth findingout whether rhinos are, in fact, librarians. As scientists, our job isn’t to challenge data that support a preconceived (先入为主的) story, but to say “yes, and.”4. Why did the author attend the improvisation class?A. To get a different experience.B. To finish her Ph.D. at university.C. To give up her job as a science communicator.D. To improve her speaking and communicating ability.5. What was the author’s change after attending the improvisation class?A. She formed her own idea quickly.B. She came up with lots of creative responses.C. She paid more attention to the logic of answers.D. She became a good listener before giving an opinion.6. The author mentions applying the “yes, and” approach to her scientific experiments to ______.A. explain the process of using the methodB. prove the benefits of the improvisation classC. share her own research experiences with readersD. attract fellow scientists to attend the improvisation class7. What can be inferred about scientists from the last paragraph?A. They should attend the improvisation class.B. They should question all preconceived ideas.C. They should carry on research by admitting earlier data.D. They should try to improve their professional knowledge.CParents and math teachers regularly asked by their school-aged charges whether math matters inreal life now have an answer.In a study published by the National Bureau of Economic Research this week, Harvard Kennedy School Policy Professor Joshua Goodman took a look at what happened to students whose high schools were required in the 1980s to increase the minimum level of coursework required to graduate. What he found is that the students were more likely to increase the number of math courses they took as a result of the change in standards and that translated into higher earnings down the line.Put simply: About 15 years after they graduated, the high school graduates who went to school when these changes took effect saw their average earnings increase about 10% for every extra year of math coursework. The findings may add fuel to the steady drum of education experts, policy makers and others calling for an increased focus on science and math education.The increase in required math courses didn’t necessarily produce rocket scientists, Goodman notes, becausethe extra coursework wasn’t at a particularly high level. But becoming familiar with and practicing ly basic math skills allowed high school graduates to pursue and excel at jobs that required some level of computational knowledge, he said.Goodman acknowledged that the earnings boost for the students connected with the up tick （上升）in math education may be dependent on the state of the economy. When Goodman checked in the late 1990s and early 2000s on the earnings of the students who graduated in the late 1985，he found that their earnings increased significantly if they took more math.8. What does Joshua Goodman’s study find?A. High school graduates earn more.B. High school students prefer math.C. Math matters a lot in one’s income.D. Math is a compulsory course in school.9. Why is increase in required math courses useful?A. It produced rocket scientists.B. It makes high school students smarter.C. It meets the requirements of the government.D. It enables high school graduates to get better jobs.10. What might be the author’s purpose in writing the text?A. To give a warning to policy makers.B. To seek help for high school students’ problems.C. To show the great importance of learning math.D. To tell a struggling history of high school students.11. What could be the best title for the text?A. Efforts matter mostB. Math makes moneyC. Earnings of different graduatesD. Changes of high school coursesDEveryone can be angry. But if you take the time to actually examine your anger instead of just “feeling” angry, you’ll have a better understanding of yourself. Knowing why you feel so angry can provide you with somesurprising answers. These answers can enable you to suddenly grow spiritually and mentally.I can give you a personal example. I went to a meeting once and I was verbally attacked over an application I supported at my workplace. Various people went on and on about how terrible this system was and that it never worked. That didn’t bother me that much. I was used to that but one of the comments that was said was, “Your job is pointless.” This really upset meand at the time, I was ly furious(发怒地) with that comment.I was so angry and upset that they would treat me that way. Once I took the time to think about what was really making me so mad. I learned a lot. I realized that the comment was more truethan I wanted to admit. In the big scheme of things, my job was pointless. It wasn’t what I really wanted to be doing with my life and this was what frustrated me the most. Here I was pouring part of my heart and soul into a job I didn’t even really want to do. I was using it as a crutch(拐杖) because I didn’t have the confidence in myself to take the scary road towards what I really wanted to do. As soon as I realized that, a lot of my anger just melted away. I also realized that I needed to start focusing on what I really wanted to do.I now consider this incident as a great gift It got me back on track to moving in the direction I wanted to go with my life. I probably wouldn’t be where I am today if I hadn’t taken the time to figure out why I was really so angry.12. What made the author angry at the meeting?A. The system of the company was terrible.B. Someone said his job is insignificant.C. The policy of the company didn’t work.D. The application he supported was of no effect.13. What brought down the anger of the author?A. The apology someone made at the meeting.B. The crutch he used to take the scary road.C. His awareness of the fact that the comment was true.D. The courage he had to overcome the challenges.14. Why did the author consider the incident as a gift?A. It brought him back on track to the goal.B. It helped him get promoted to a higher position.C. It helped him change his character since then.D. It provided him with confidence in his career.15. What can be the best title of the passage?A. How to Cope with Verbal Attack in a CompanyB. Avoid Being Pointless at WorkC. Anger Is Harmful to HealthD. Understanding Yourself Better Through Anger第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

四川南充2017年上学期9月检测考试（满分 150 分，时间 100 分钟）第一卷（共 75 分）一.单项填空：（每小题 1 分，共 15 分）1．The students are hard-working and ready to help others should be praised.A. whoB. whomC. whichD. whose 2．—What’s the noise outside?—Some children hide-and-seek in the yard.A. were playingB. have playedC. are playingD. played3．We are tired. Let’s stop .A. have a restB. to have a restC. having a restD. had a rest4．I have just bought five stamps. One is German stamp, are American stamps.A.otherB. othersC.the otherD..the others 5．Harry Potter is a film we all like it.A. so good, thatB. such, thatC. so good ,a ndD. such, and6．My house when you visited Beijing last time, so I didn ’t invite you to myhome.A. was being decoratedB. is being decoratedC. has been decoratedD. was decorated7．She had to drop out of school her family couldn’t aff ord the cost of hereducation.A. althoughB. unlessC. becauseD. so8．The plane hasn’t arrived yet. Could you tell me ?.A.that the plane arrivesB. how is the plane lateC. when will the plane arriveD. why the plane is late 9．It will take a lot of time and to finish the task on time.A. powerB. pressureC. struggleD. energy 10．--- Do you know Jack well? ---Certainly, we friends since ten years ago..A.wereB. have madeC.have beenD. have become11．--- “Have you seen the film?”He asked me .A. had I seen the filmB.have I seen the filmC.if I had seen the filmD. whether I have seen the film四川南充2017年上学期9月检测考试（满分 150 分，时间 100 分钟）第一卷（共 75 分）一.单项填空：（每小题 1 分，共 15 分）1．The students are hard-working and ready to help others should be praised.A. whoB. whomC. whichD. whose2．—What’s the noise outside?—Some children hide-and-seek in the yard.A. were playingB. have playedC. are playingD. played3．We are tired. Let’s stop .A. have a restB. to have a restC. having a restD. had a rest4．I have just bought five stamps. One is German stamp, are American stamps.A.otherB. othersC.the otherD..the others5．Harry Potter is a film we all like it.A. so good, thatB. such, thatC. so good ,a ndD. such, and6．My house when you visited Beijing last time, so I didn’t invite you to my home.A. was being decoratedB. is being decoratedC. has been decoratedD. was decorated7．She had to drop out of school her family couldn’t afford the cost of her education.A. althoughB. unlessC. becauseD. so8．The plane hasn’t arrived yet. Could you tell me ?.A.that the plane arrivesB. how is the plane lateC. when will the plane arriveD. why the plane is late9．It will take a lot of time and to finish the task on time.A. powerB. pressureC. struggleD. energy10．--- Do you know Jack well? ---Certainly, we friends since ten years ago..A.wereB. have madeC.have beenD. have become11．--- “Have you seen the film?”He asked me .A. had I seen the filmB.have I seen the filmC.if I had seen the filmD. whether I have seen the film。

四川省南充高级中学2017届高三语文上学期期末考试试题不分版本南充高中2016-2017学年上学期期末考试高三语文试卷(时间:120分钟，总分值:150分)本试卷分第I卷(阅读题)和第II卷〔表达题)两局部，其中第I卷第三.、四题为选考题，其它题为必考题。

考生作答时，将答案答在答题卡上，在本试卷上答题无效。

考试结束后，将本试卷和答题卡一并交回。

考前须知：1.答题前，考生务必先将自己的姓名、准考证号填写在答题卡上，认真核对条形码上的姓名、准考证号，并将条形码粘贴在答题卡的指定位置上。

2.答题时使用0.5毫米的黑色中性(签字)笔或碳素笔书写，字体工整，笔迹清楚。

3.请按照题号在各题的答题区域(黑色线框)内作答，超出答题区域书写的答案无效4.保持卡面清活，不折叠.不破损。

5.做选考题时，考生按照题目要求作答，并用2B铅笔在答题卡上把所选题目对应的题号涂黑。

第I卷阅读题甲必考题一、现代文阅读(9分，每题3分)阅读下面的文字，完成1一3题。

中国人的传统家庭苏叔阳个人、家庭(扩大至家族)、社会的关系是任何社会都要重视的根本问题。

从历史的实践来看，西方似乎是个——社会，缺乏“家庭〞的一环，而中国那么是家庭(家族)——社会，缺乏至少淡化了“个人〞一环。

人类社会的进展似乎在提醒我们，这三者少了哪一环都不行。

理想的社会生活方式应当是：个人——家庭——社会都和谐、有序。

中国的传统生活方式，在理论层面上也是主张个人——家庭——社会和谐有序的，到达这理想的良策是“修身齐家治国平天下〞，而在实践的层面上却有许多导病在中国传统文化中，家庭是社会的根本单元。

整个社会被看作是家庭的延伸和扩大，而个人只是家庭的一分子，常常不被看作是“独立的单元〞。

在中国传统家庭中，父子的血像关系又是核心局部，所以，中国的家庭实际上是一个以父子为中心的血缘关系的小群体。

旧式大家庭有“四世同堂〞或“五世同堂〞，二世同堂者已经有新式家庭的味道，不过，其奋斗目标依旧是“四世同堂〞。

南充市高三期末统考（含听力）英语试题I. 听力(共 30 分)1.【此处可播放相关音频，请去附件查看】What will the man do first?A. Have dinner with the woman.B. Have an evening class.C. Have a snack.2.【此处可播放相关音频，请去附件查看】Why does the man look terrible?A. He got the flu.B. He studied late last night.C. He had difficulty in sleeping.3.【此处可播放相关音频，请去附件查看】What do we know about the sisters?A. They are different in character.B. They have a lot in common.C. Their voices are beautiful.4.【此处可播放相关音频，请去附件查看】What does the woman suggest the man do?A. Have a talk with his boss.B. Stick to what he did.C. Give up the job.5.【此处可播放相关音频，请去附件查看】What does the woman mean?A. The comer is a better place for the plants.B. The man should water the plants less.C. The plants may need more light.听下面一段较长对话，回答以下小题。

【此处可播放相关音频，请去附件查看】6. What does the man do?A. A pilot.B. A scientist.C. An office clerk.7. What are the speakers talking about?A. Their jobs.B. Their dreams.C. Their hobbies.听下面一段较长对话，回答以下小题。