山东省潍坊市第一中学2013届高三上学期12月月考测试数学文考试试题
2024届山东省潍坊市高三上学期12月普通高中学科素养能力测评英语试题
2024届山东省潍坊市高三上学期12月普通高中学科素养能力测评英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读选择Wondering what to watch this week? We have you covered. This list includes some of the best and most exciting movies you are hungering for right now.The Super Mario Bros. MovieThe most popular movie this week is the animated adventures of Super Mario. Based on the iconic video game from Nintendo, this movie features an all-star cast led by Chris Pratt and Anya Taylor-Joy. The Super Mario Bros. Movie has been a critical and commercial hit and is definitely one of the must-see movies this year.You can stream the movie on Apple TV+.InterstellarThis complex philosophical science fiction movie is one of Christopher Nolan’s best-loved movies. Telling the story of a group of astronauts searching for a new home for humanity, the film is a moving meditation on fatherhood, science, faith, and perseverance. It is regarded as one of the best science fiction films of all time.Interstellar can be streamed on Amazon Prime Video and Apple TV+ among others. Transformers: Rise of the BeastsThe seventh installment of the Transformers is a huge hit on streaming. The action-packed science fiction movie based on Hasbro toys has an impressive cast. Rise of the Beasts is set in 1994, so expect some throwback 90s references and incredible sci-fi action sequences from this thrilling Transformers movie.It can be streamed on Paramount+ and Apple TV+.Guardians of the Galaxy Vol. 3The final movie in the Guardians of the Galaxy series brings the trilogy to an emotional close with a heartfelt goodbye to the characters. This final chapter has been one of the biggest movies of the year and is directed by superhero movie master James Gunn. Guardians of the Galaxy Vol. 3 features an incredible cast that includes Chris Pratt, Zoe Saldana, Dave Bautista, etc. As an important part of Phase Five of the MCU, it is not one to be missed.The film is available to stream on Disney+.1．Which film is closely related to mankind’s fate?A．The Super Mario Bros. Movie.B．Interstellar.C．Transformers: Rise of the Beasts.D．Guardians of the Galaxy Vol. 3. 2．What do we know about Guardians of the Galaxy Vol. 3?A．It is directed by Chris Pratt.B．It is the best movie of the year.C．It strikes viewers’ heartstrings.D．It contains five chapters altogether. 3．What do the four movies have in common?A．Big names star in them.B．They are accessible online.C．They belong to movie series.D．Shooting methods are similar.When it comes to providing energy to the masses, Greg Hazle’s experience is deep. Before his retirement in 2017, his extensive corporate career spanned public service, independent energy, mining and construction materials industries. He has held roles in corporate finance and also put his training in engineering to work and helped design power projects throughout the United States and Latin America.Outside of his high-powered career, Hazle always found ways to give back. And, sometimes, others plotted those ways for him. In 2014, a board member from Boca Helping Hands (BHH) , local nonprofit, saw Hazle sing at the church they both attended. He asked around about Hazle and thought he’d be a good addition to the nonprofit’s board of trustees.“All of this was happening without me knowing,” Hazle smiled. But once Hazle learned about the nonprofit’s work, he was onboard. BHH is a community-based nonprofit in Boca Raton, Florida, which provides food, medical, financial, and job-training assistance to help people meet their basic needs and become self-sufficient.Despite his success in corporate America, Hazle always remembered his days growing up in Jamaica, when he had seen the impact of poverty, homelessness, and hunger around him. While he lived in a place that had such a reputation for wealth and self-indulgence (放纵), he was sharply aware that there were people around him — even in a place like Boca Raton — who needed help. Over the next few years, Hazle became involved in many aspects of the organization.Hazle’s preparing for his retirement at age 60 was co-occurrent with the managing director of BHH leaving around. Hazle agreed to step in and run the organization as a temporary leader with the condition that a search firm would be kept to find a new leader.“Obviously, that’s not how it turned out,” he said.Since then, Hazle has accepted the role wholeheartedly. He feels a renewed sense of purpose in his role, which suits his personality and passions more than others he has held. “So, it just felt like a privilege that I was given this opportunity and late in my career to do what I consider to be more meaningful work than generating returns for shareholders.”4．What do we know about Hazle from the first two paragraphs?A．He is a natural musician.B．He balances work and life well.C．He sponsors churches financially.D．He is influential in many circles. 5．Why does the author talk about Hazle’s life in Jamaica?A．To recall his miserable days there.B．To highlight his current superior life.C．To justify his joining the organization.D．To introduce the motive for his success. 6．What do Hazle’s words in paragraph 5 imply?A．Take things as they are.B．In the end things will mend.C．Good things never come easy.D．Misfortune may be a blessing. 7．Which of the following best describes Hazle?A．Devoted and wise.B．Optimistic and energetic.C．Tough and generous.D．Sympathetic and responsible.In recent years, lots of American companies have gotten behind a potential climate solution called carbon capture and storage, and the government has backed it with billions of dollars in tax preferences and direct investments. The idea is to trap planet-heating carbon dioxide from the smokestacks of factories and power plants and ship it to sites via thousands of miles of new pipelines. Communities nationwide are pushing back against these pipeline construction and underground sites, arguing they don’t want the pollution running through their land.Now the U. S. Forest Service is proposing to change a rule to allow storing this carbon dioxide pollution under the country’s national forests and grasslands. “Authorizing carbon capture and storage on National Forest System (NFS) lands would support the Administration’s goal to reduce greenhouse gas emissions by 50 percent below the 2005 levels by 2030,” the proposed rule change says.Some experts, like June Sekera, a research fellow with Boston University, question the timing of the proposed rule change, given community pushback across the country topipelines planned on private land. Yet she says the Forest Service proposal to open up national parks for CO2 storage is “an end run around local towns and counties. And it’s a much simpler and way less expensive route.”In an email, Scott Owen, press officer for the Forest Service, writes that the proposed rule change would allow the Forest Service to consider proposals for carbon capture and storage projects. He writes that any proposals must still pass through a secondary screening, adding, “The Forest Service has been ‘screening’ proposals for use of NFS lands for over 20 years as a means to be increasingly consistent in our processes and also be able to reject those uses that are inconsistent with the management of the public’s land. ” He notes the Forest Service currently does not have any carbon capture project proposals under consideration. The Forest Service has opened public comments on the proposed rule change until Jan. 2, 2024.8．What does the Forest Service intend to do by changing a rule?A．Answer the appeals of communities.B．Provide legal space for carbon storage.C．Enlarge national pipeline storage capacity.D．Loosen tax burden on American companies.9．What does the underlined phrase “an end run” in paragraph 3 probably mean?A．An eventful act.B．A desperate try.C．An alternative way.D．A breathtaking race.10．What can we infer about carbon capture project from the last paragraph?A．It is still up in the air.B．It is dead in the water.C．It is widely recognized.D．It is far from satisfactory.11．What does the text mainly talk about?A．A fruitful research.B．A timely rule change.C．An authorized project.D．A controversial proposal.Nowadays, people are increasingly interacting with others in social media environments where algorithms control the flow of social information they see. People’s interactions with online algorithms may affect how they learn from others, with negative consequences including social misperceptions, conflict and the spread of misinformation.On social media platforms, algorithms are mainly designed to amplify (放大) information that sustains engagement, meaning they keep people clicking on content andcoming back to the platforms. There is evidence suggesting that a side effect of this design is that algorithms amplify information people are strongly biased (偏向的) to learn from. We call this information “PRIME”, for prestigious, in-group, moral and emotional information.In our evolutionary past, biases to learn from PRIME information were very advantageous: Learning from prestigious individuals is efficient because these people are successful and their behavior can be copied. Paying attention to people who violate moral norms is important because punishing them helps the community maintain cooperation. But what happens when PRIME information becomes amplified by algorithms and some people exploit (利用) algorithm amplification to promote themselves? Prestige becomes a poor signal of success because people can fake prestige on social media. News become filled with negative and moral information so that there is conflict rather than cooperation.The interaction of human psychology and algorithm amplification leads to disfunction because social learning supports cooperation and problem-solving, but social media algorithms are designed to increase engagement. We call it functional mismatch. One of the key outcomes of functional mismatch is that people start to form incorrect perceptions of their social world, which often occurs in the field of politics. Recent research suggests that when algorithms selectively amplify more extreme political views, people begin to think that their political in-group and out-group are more sharply divided than they really are. Such “false polarization” might be an important source of greater political conflict.So what’s next? A key question is what can be done to make algorithms facilitate accurate human social learning rather than exploit social learning biases. Some research team is working on new algorithm designs that increase engagement while also punishing PRIME information. This may maintain user activity that social media platforms seek, but also make people’s social perceptions more accurate.12．What are social media algorithms targeted at?A．Improving social environment.B．Generating PRIME information.C．Avoiding side effects of social media.D．Raising the media platform click rate. 13．Why does the author refer to “false polarization” in paragraph 4?A．To make an assumption.B．To illustrate a conclusion.C．To explain a political issue.D．To present an extreme case. 14．According to the author, algorithms will be improved so as to ________.A．boost engagement and regulate amplificationB．strengthen social learning and delete biasesC．identify biases and punish PRIME informationD．monitor media platforms and guarantee users’ privacy15．What is the best title of the text?A．PRIME information meets with misperceptionsB．Algorithms control the flow of social informationC．Social media algorithms twist human social learningD．Online algorithm designs face unexpected challengesStudies have shown that students attain more information when reading a hard-copybooks more easily, as teachers there put it. Although e-books offer various navigation tools, physical books’ ease of use is irreplaceable. Turning pages is still the best way to read.Fans of digital books may point out that e-readers have a handy “search” tool. Old-fashioned books also have a search function, in which you turn back to the opening chapter to remind yourself of the hero’s surname. They even have a “bookmark system”, which uses a device called a “bookmark”. 17 By simply turning the page, you can see new information.Electronic books offer footnote capabilities, but so do traditional books. Using a “pencil” — a favorite tool of schoolkids worldwide — you can mark important passages. This allows you to highlight significant quotes or add your own insights. The absence of a built-in pencil doesn’t mean you can’t footnote a printed book. 18 You can use a highlighter, write in the margin (页边) , or — if you’re feeling really adventurous — use a ruler to create your own “book edge” notes.19 Rarely do we get the same “new book smell” from an e-reader as we do froma freshly printed book. The concrete experience of physically turning pages and feeling the weight of a book in your hands cannot be matched by an e-reader. Actually they make excellent companions during leisure break.In conclusion, printed books have triumphed over e-books because they offer an unparalleled reading experience. The ease of use, the emotional connection, and the ability to footnote all contribute to making printed books the preferred choice for many readers. 20 A．You need to be more creative.B．You can use smart tools as you like.C．There’s still no substitute for the traditional book.D．Traditional books also excel at stirring up feelings.E．In response, one school in Australia abandoned its e-readers entirely.F．Furthermore, the ability to dog-ear pages is a bonus of physical books.G．A “progress bar” indicates what percentage of the e-book has been read.二、完形填空In 2014, Xu Yitang, a student at the National Academy of Chinese Theatre Arts, saw25．A．tough B．smooth C．delicate D．impressive 26．A．benefits B．requirements C．expenses D．responses 27．A．reminded B．informed C．warned D．convinced 28．A．shelters B．partners C．opportunities D．services 29．A．Somehow B．Nevertheless C．Therefore D．Unfortunately 30．A．extinction B．variation C．decline D．expansion 31．A．responsibility B．relief C．achievement D．belonging 32．A．consults with B．applies for C．seeks out D．acts as 33．A．find out B．note down C．show off D．speed up 34．A．revised B．contributed C．announced D．featured 35．A．assess B．follow C．relate D．invent三、语法填空阅读短文内容，在空白处填入1个适当的单词或括号内单词的正确形式。
2016届高三上学期第一次月考数学（文）试题Word版含答案
2016届高三上学期第一次月考数学（文）试题Word版含答案2016届高三上学期第一次月考数学文试卷考试时间120分钟，满分150分一、选择题：本大题共12小题，每小题5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．设集合M ＝{x |x ≥0，x ∈R }，N ＝{x |x 2<1，x ∈R }，则M ∩N 等于( ) A ．[0,1] B ．[0,1) C ．(0,1]D ．(0,1)2．已知集合A ＝{1,2}，B ＝{1，a ，b }，则“a ＝2”是“A ?B ”的( ) A ．充分不必要条件B ．必要不充分条件C ．充要条件D ．既不充分也不必要条件3．已知命题p ：所有有理数都是实数；命题q ：正数的对数都是负数，则下列命题中为真命题的是( ) A ．﹁p 或q B ．p 且q C ．﹁p 且﹁qD ．﹁p 或﹁q4．设函数f (x )＝x 2＋1，x ≤1，2x ，x >1，则f (f (3))等于( )A.15B ．3C.23D.1395．函数f (x )＝log 12(x 2－4)的单调递增区间是( )A ．(0，＋∞)B ．(－∞，0)C ．(2，＋∞)D ．(－∞，－2)6．已知函数f (x )为奇函数，且当x >0时，f (x )＝x 2＋1x ，则f (－1)等于( )A ．－2B ．0C ．1D ．27. 如果函数f (x )＝x 2－ax －3在区间(－∞，4]上单调递减，则实数a 满足的条件是( ) A ．a ≥8 B ．a ≤8 C ．a ≥4D ．a ≥－48. 函数f (x )＝a x －2＋1(a >0且a ≠1)的图像必经过点( ) A ．(0,1) B ．(1,1) C ．(2,0)D ．(2,2)9. 函数f (x )＝lg(|x |－1)的大致图像是( )10. 函数f (x )＝2x ＋3x 的零点所在的一个区间是( ) A ．(－2，－1) B ．(－1,0) C ．(0,1)D ．(1,2)11. 设f (x )＝x ln x ，若f ′(x 0)＝2，则x 0的值为( ) A ．e 2B ．eC.ln22D ．ln212. 函数f (x )的定义域是R ，f (0)＝2，对任意x ∈R ，f (x )＋f ′(x )>1，则不等式e x ·f (x )>e x ＋1的解集为( )．A ．{x |x >0}B ．{x |x <0}C ．{x |x <－1或x >1}D ．{x |x <－1或0<1}<="" p="">二、填空题：本大题共4小题，每题5分．13. 已知函数y ＝f (x )及其导函数y ＝f ′(x )的图像如图所示，则曲线y ＝f (x )在点P 处的切线方程是__________．14. 若函数f (x )＝x 2＋ax ＋b 的两个零点是－2和3，则不等式af (－2x )>0的解集是________． 15. 函数y ＝12x 2－ln x 的单调递减区间为________．16. 若方程4－x 2＝k (x －2)＋3有两个不等的实根，则k 的取值范围是________．三、解答题：解答应写出文字说明、证明过程或演算步骤17.(10分) 化简：(1)3131421413223b a b a ab b a -(a >0，b >0)；(2)(－278)23-＋(0.002)12-－10(5－2)－1＋(2－3)0.18．(12分)已知函数f (x )＝1a －1(a >0，x >0)，(1)求证(用单调性的定义证明)：f (x )在(0，＋∞)上是增函数； (2)若f (x )在[12，2]上的值域是[12，2]，求a 的值．19．（12分）已知定义在R 上的奇函数f (x )有最小正周期2，且当x ∈(0,1)时，f (x )＝2x4x ＋1.(1)求f (1)和f (－1)的值； (2)求f (x )在[－1,1]上的解析式．20．（12分）已知函数f (x )＝x 2＋2ax ＋3，x ∈[－4,6]． (1)当a ＝－2时，求f (x )的最值；(2)求实数a 的取值范围，使y ＝f (x )在区间[－4,6]上是单调函数；(3)当a ＝1时，求f (|x |)的单调区间． 21．（12分）已知函数f (x )＝x 3＋x －16. (1)求曲线y ＝f (x )在点(2，－6)处的切线的方程；(2)直线l 为曲线y ＝f (x )的切线，且经过原点，求直线l 的方程及切点坐标； 22．（12分）已知函数f (x )＝x 3－3ax －1，a ≠0. (1)求f (x )的单调区间；(2)若f (x )在x ＝－1处取得极值，直线y ＝m 与y ＝f (x )的图像有三个不同的交点，求m 的取值范围．2016届高三上学期第一次月考数学答题卡一、选择题（共12小题，每小题5分，共60分，每小题有一个正确答案）13、 14、15、 16、三、解答题17.(10分) 化简：(1)131421413223b a b a ab b a -(a >0，b >0)；(2)(－278)23-＋(0.002)12-－10(5－2)－1＋(2－3)0.18．(10分)已知函数f (x )＝1a －1x(a >0，x >0)，(1)求证(用单调性的定义证明)：f (x )在(0，＋∞)上是增函数； (2)若f (x )在[12，2]上的值域是[12，2]，求a 的值．19．（12分）已知定义在R 上的奇函数f (x )有最小正周期2，且当x ∈(0,1)时，f (x )＝2x4x ＋1.(1)求f (1)和f (－1)的值； (2)求f (x )在[－1,1]上的解析式．20．（12分）已知函数f(x)＝x3＋x－16.(1)求曲线y＝f(x)在点(2，－6)处的切线的方程；(2)直线l为曲线y＝f(x)的切线，且经过原点，求直线l的方程及切点坐标；21．（13分）已知函数f(x)＝x2＋2ax＋3，x∈[－4,6]．(1)当a＝－2时，求f(x)的最值；(2)求实数a的取值范围，使y＝f(x)在区间[－4,6]上是单调函数；(3)当a＝1时，求f(|x|)的单调区间．22．（13分）已知函数f(x)＝x3－3ax－1，a≠0.(1)求f(x)的单调区间；(2)若f(x)在x＝－1处取得极值，直线y＝m与y＝f(x)的图像有三个不同的交点，求m的取值范围．2016届高三上学期第一次月考数学文试卷参考答案1．B2.A3.D4.D5.D6.A7.A8.D9.B10.B11.B12.A13. x －y －2＝0 14. {x |－32<1}<="" p="">15. (0,1] 16. (512，34]17. 解 (1)原式＝121311113233211212633311233().a b a b abab ab a b+-++----==(2)原式＝(－278)23-＋(1500)12-－105－2＋1＝(－827)23＋50012－10(5＋2)＋1＝49＋105－105－20＋1＝－1679. 18. (1)证明设x 2>x 1>0，则x 2－x 1>0，x 1x 2>0，∵f (x 2)－f (x 1)＝(1a －1x 2)－(1a －1x 1)＝1x 1－1x 2＝x 2－x 1x 1x 2>0，∴f (x 2)>f (x 1)，∴f (x )在(0，＋∞)上是增函数． (2)解∵f (x )在[12，2]上的值域是[12，2]，又f (x )在[12，2]上单调递增，∴f (12)＝12，f (2)＝2.易得a ＝25.19. 解(1)∵f (x )是周期为2的奇函数，∴f (1)＝f (1－2)＝f (－1)＝－f (1)，∴f (1)＝0，f (－1)＝0. (2)由题意知，f (0)＝0. 当x ∈(－1,0)时，－x ∈(0,1)．由f (x )是奇函数，∴f (x )＝－f (－x )＝－2－x4－x ＋1＝－2x4x ＋1，综上，在[－1, 1]上，f (x )＝2x4x ＋1，x ∈(0，1)，－2x 4x＋1，x ∈(－1，0)，0，x ∈{－1，0，1}.20．解 (1)当a ＝－2时，f (x )＝x 2－4x ＋3＝(x －2)2－1，∵x ∈[－4,6]，∴f (x )在[－4,2]上单调递减，在[2,6]上单调递增，∴f (x )的最小值是f (2)＝－1，又f (－4)＝35，f (6)＝15，故f (x )的最大值是35. (2)∵函数f (x )的图像开口向上，对称轴是x ＝－a ，∴要使f (x )在[－4,6]上是单调函数，应有－a ≤－4或－a ≥6，即a ≤－6或a ≥4. (3)当a ＝1时，f (x )＝x 2＋2x ＋3，∴f (|x |)＝x 2＋2|x |＋3，此时定义域为x ∈[－6,6]，且f (x )＝?x 2＋2x ＋3，x ∈(0，6]，x 2－2x ＋3，x ∈[－6，0]，∴f (|x |)的单调递增区间是(0, 6]，单调递减区间是[－6,0]．21.解 (1)可判定点(2，－6)在曲线y ＝f (x )上．∵f ′(x )＝(x 3＋x －16)′＝3x 2＋1.∴f ′(x )在点(2，－6)处的切线的斜率为k ＝f ′(2)＝13. ∴切线的方程为y ＝13(x －2)＋(－6)，即y ＝13x －32.(2)法一设切点为(x 0，y 0)，则直线l 的斜率为f ′(x 0)＝3x 20＋1，∴直线l 的方程为y ＝(3x 20＋1)(x －x 0)＋x 30＋x 0－16，又∵直线l 过点(0,0)，∴0＝(3x 20＋1)(－x 0)＋x 30＋x 0－16，整理得，x 30＝－8，∴x 0＝－2，∴y 0＝(－2)3＋(－2)－16＝－26，k ＝3×(－2)2＋1＝13. ∴直线l 的方程为y ＝13x ，切点坐标为(－2，－26.) 法二设直线l 的方程为y ＝kx ，切点为(x 0，y 0)，则k＝y0－0x0－0＝x30＋x0－16x0又∵k＝f′(x0)＝3x20＋1，∴x30＋x0－16x0＝3x2＋1，解之得x0＝－2，∴y0＝(－2) 3＋(－2)－16＝－26，k＝3×(－2)2＋1＝13.∴直线l的方程为y＝13x，切点坐标为(－2，－26)．22.解(1)f′(x)＝3x2－3a＝3(x2－a)，当a<0时，对x∈R，有f′(x)>0，∴当a<0时，f(x)的单调增区间为(－∞，＋∞)．当a>0时，由f′(x)>0，解得x<－a或x>a.由f′(x)<0，解得－a<x<a，< p="">∴当a>0时，f(x)的单调增区间为(－∞，－a)，(a，＋∞)，单调减区间为(－a，a)．(2)∵f(x)在x＝－1处取得极值，∴f′(－1)＝3×(－1)2－3a＝0，∴a＝1.∴f(x)＝x3－3x－1，f′(x)＝3x2－3，由f′(x)＝0，解得x1＝－1，x2＝1.由(1)中f(x)的单调性可知，f(x)在x＝－1处取得极大值f(－1)＝1，在x＝1处取得极小值f(1)＝－3.∵直线y＝m与函数y＝f(x)的图像有三个不同的交点，结合如图所示f(x)的图像可知：实数m的取值范围是(－3,1)．</x<a，<>。
山东省邹城一中2013届高三12月份月考试题 数学文
邹城一中高三数学(文史类)月度质量检测试题2012.12本试卷分第I 卷和第II 卷两部分，共4页。
满分150分。
考试时间120分钟。
注意事项：1.答题前，考生务必用直径0.5毫米黑色墨水签字笔将自己的姓名、准考证号和科类填写在答题卡上和试卷规定的位置上。
2.第I 卷每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号，答案不能答在试卷上。
3.第II 卷必须用0.5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域内相应的位置，不能写在试卷上；如需改动，先划掉原来的答案，然后再写上新的答案；不能使用涂改液、胶带纸、修正带。
不按以上要求作答的答案无效。
4.填空题请直接填写答案，解答题应写出文字说明、证明过程或演算步骤。
第I 卷（共60分）一、选择题：本大题共12个小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的。
1.设集合{}{}{}()B C A B A U U 则,2,1,2,2,1,2,1,0,1,2--==--=等于（ ）A.{}1B.{}1,2C.{}2D.{}0,1,22. 已知函数⎪⎩⎪⎨⎧<+=>=)0(1)0()0(0)(2x x x x f ππ，则)))1(((-f f f 的值等于（ ）A.12-πB.12+πC.πD.0 3.命题“,x x R e x ∃∈<”的否定是（ ）A.,x x R e x ∃∈>B.,x x R e x ∀∈≥C.,x x R e x ∃∈≥D.,x x R e x ∀∈>4．在各项均为正数的等比数列{}n a 中，,12,1253+=-=a a 则2326372a a a a a ++=（ ）A ．4B ．6C ．8 D．8-5.已知向量),(0,1),(2,a b c k a b c k ===+=若与垂直则（ ） A ．3 B ．4 C ．-3 D ．-4 6．一个空间几何体的正视图、侧视图都是面积为32，且一个内角为60°的菱形，俯视图为正方形，那么这个几何体的表面积为( )A ．2 3B ．4 3C ．4D ．87．△ABC 中，内角A 、B 、C 的对边分别为a 、b 、c ，且222222c a b ab =++，则△ABC 是( )A ．钝角三角形B ．直角三角形C ．锐角三角形D ．等边三角形8．设x 、y 满足24,1,22,x y x y x y +≥⎧⎪-≥-⎨⎪-≤⎩则z x y =+（ ）A ．有最小值2，最大值3B ．有最小值2，无最大值C ．有最大值3，无最大值D ．既无最小值，也无最大值9．将函数sin y x =的图象向左平移)20(πϕϕ≤≤个单位后，得到函数sin()6y x π=-的图象，则ϕ等于（ ）A ．6π B ．56π C ．76π D ．116π10. 函数1lg|1|y x =+的大致图象为（ ）11．已知双曲线22221(0,0)x y a b a b-=>>的两条渐近线均与22:650C x y x +-+=相切，则该双曲线离心率等于（ ）A．5B．2C ．32 D．512. 已知定义在R 上的函数)(x f y =满足下列三个条件：①对任意的R x ∈都有);()2(x f x f -=+②对于任意的2021≤<≤x x ，都有),()(21x f x f <③)2(+=x f y 的图象关于y 轴对称.则下列结论中，正确的是（ ）A ．)7()5.6()5.4(f f f <<B ． )5.6()7()5.4(f f f <<C ．)5.6()5.4()7(f f f <<D ． )5.4()5.6()7(f f f <<第II 卷（非选择题 共90分）二、填空题：本大题共4个小题，每小题4分，共16分.将答案填在题横线上.13.已知过抛物线y 2＝4x 焦点F 的直线交该抛物线于A 、B 两点，|AF |＝2，则|BF |＝______.14.如图，长方体ABCD －A1B 1C 1D 1中，AA 1＝AB ＝2，AD ＝1， 点E 、F 、G 分别是DD 1、AB 、CC 1的中点． 直线A 1E 与GF 所成角等于__________．15.设直线ax －y ＋3＝0与圆(x －1)2＋(y －2)2＝4相交于A 、B 两点，且弦AB 的长为23，则a ＝________. 16.下列命题：（1）若函数）a x x x f ++=2lg()(为奇函数，则1=a ； （2）函数x x f sin )(=的周期π=T ； （3）方程x x sin lg =有且只有三个实数根； （4）对于函数x x f =)(，若2)()()2(0212121x x f x x f x x +<+<<，则. 其中真命题的序号是__________（写出所有真命题的编号）三、解答题：本大题共6个小题.共74分.解答应写出文字说明，证明过程或演算步骤.17.（本小题满分12分）已知集合22{|280},{|(23)(3)0,}A x x x B x x m x m m m R =--≤=--+-≤∈ （1）若[2,4],A B = 求实数m 的值；（2）设集合为R ，若R A C B ⊆，求实数m 的取值范围。
文科高考数学立体几何大题求各类体积方法
A BCD PA B CDP文科高考数学立体几何大题求各类体积方法【三年真题重温】1.【2011⋅新课标全国理，18】如图，四棱锥P ABCD -中，底面ABCD 为平行四边形，∠DAB =60，2AB AD =，PD ⊥底面ABCD ． (Ⅰ) 证明：PA ⊥BD ；(Ⅱ) 若PD AD =，求二面角A PB C --的余弦值． 2.【2011 新课标全国文，18】如图，四棱锥P ABCD -中，底面ABCD 为平行四边形．60,2,DAB AB AD PD ∠==⊥底面ABCD ．(Ⅰ) 证明：PA BD ⊥；(Ⅱ) 设1PD AD ==，求棱锥D PBC -的高．根据DE PB PD BD ⋅=⋅，得32DE =．即棱锥D PBC -的高为32．3.【2010 新课标全国理，18】如图，已知四棱锥P-ABCD 的底面为等腰梯形，AB CD,AC ⊥BD ，垂足为H ，PH 是四棱锥的高 ，E 为AD 中点.（1） 证明：PE ⊥BC（2） 若∠APB=∠ADB=60°，求直线PA 与平面PEH 所成角的正弦值【解析】命题意图：本题主要考查空间几何体中的位置关系、线面所成的角等知识，考查空间想象能力以及利用向量法研究空间的位置关系以及线面角问题的能力.4.【2010 新课标全国文，18】如图，已知四棱锥P ABCD -的底面为等腰梯形，AB ∥CD ,AC BD ⊥,垂足为H ，PH 是四棱锥的高。
（Ⅰ）证明：平面PAC ⊥ 平面PBD ; （Ⅱ）若6AB =,APB ADB ∠=∠=60°,求四棱锥P ABCD -的体积。
5.【2012 新课标全国理】（本小题满分12分）如图，直三棱柱111ABC A B C -中，112AC BC AA ==， D 是棱1AA 的中点，BD DC ⊥1（1）证明：BC DC ⊥1（2）求二面角11C BD A --的大小。
6.【2012 新课标全国文】（本小题满分12分）如图，三棱柱ABC －A 1B 1C 1中，侧棱垂直底面，∠ACB=90°，AC=BC=12AA 1，D 是棱AA 1的中点(Ｉ)证明：平面BDC 1⊥平面BDC（Ⅱ）平面BDC 1分此棱柱为两部分，求这两部分体积的比。
高三数学：专题11 圆锥曲线 文(教师版)
山东省2013届高三数学各地市最新模拟理数试题精品分类汇编专题11 圆锥曲线文（教师版）一、选择题：1．（山东省济南市2013年1月高三上学期期末文12）已知椭圆方程22143x y+=,双曲线的焦点是椭圆的顶点,,则双曲线的离心率C. 2D. 32．(山东省德州市2013届高三上学期期末校际联考文7)过点P（0，2）的双曲线C的一个焦点与抛物线216x y=-的焦点相同，则双曲线C的标准方程是（）A．221124x y-=B．221204x y-=C．221412y x-=D．221420y x-=3. (山东省济宁市2013届高三1月份期末测试文5)已知圆22670x y x+--=与抛物线()220y px p=>的准线相切，则p的值为A.1B.2C.12D.44. (山东省济宁市2013届高三1月份期末测试文9)已知双曲线的方程为()222210,2x y a b ab-=>>，双曲线的一个焦点到一条渐近线的距离为3c （其中c 为双曲线的半焦距长），则该双曲线的离心率为A.3222D.525. (山东省烟台市2013届高三上学期期末文7)已知点P 是抛物线x 2=4y 上的动点，点P 在直线y+1=0上的射影是点M ，点A 的坐标(4,2)，则P A P M +的最小值是A.C.3D.2【答案】A【解析】抛物线的焦点坐标(1,0)F ,准线方程为1y =-。
根据抛物线的定义可知P M P F =,所以P A P M P A P F AF +=+≥，即当A,P,F 三点共线时，所以最小值为=选A.6. (山东省烟台市2013届高三上学期期末文8)已知与向量v=(1,0)平行的直线l 与双曲线2214xy -=相交于A 、B 两点，则A B 的最小值为A.2 C.4 D.7.(山东省潍坊市2013年1月高三上学期期末考试A 卷文9)已知双曲线()0,012222>>=-b a by ax 的一条渐近线的斜率为2，且右焦点与抛物线x y 342=的焦点重合，则该双曲线的离心率等于（A ）2（B ）3（C ）2（D ）238. （山东省泰安市2013届高三上学期期末文11）以双曲线22163xy-=的右焦点为圆心且与双曲线的线相切的圆的方程是A.(22x y -+=B.(223x y -+=C.()223x y -+=D.()2233x y -+=9. （山东省青岛即墨市2013届高三上学期期末考试文12）抛物线)0(42>=p px y 与双曲线)0,0(12222>>=-b a by ax 有相同的焦点F ，点A 是两曲线的交点，且AF ⊥x 轴，则双曲线的离心率为 A.215+ B.12+ C.13+ D.2122+10.（山东省潍坊一中2013届高三12月月考测试文）设12,F F 分别是椭圆22221x y ab+=()0a b >>的左、右焦点，与直线y b =相切的2F 交椭圆于点E ，E 恰好是直线EF 1与2F 的切点，则椭圆的离心率为A.2B.3C.3D.411.(山东省实验中学2013届高三第三次诊断性测试文)椭圆191622=+yx的焦距为A.10B.5C.7D.72【答案】D【解析】由题意知2216,9a b ==，所以2227c a b =-=，所以c =，即焦距为2c =，选D.12.(山东省兖州市2013届高三9月入学诊断检测文)若m 是2和8的等比中项，则圆锥曲线221yx m+=的离心率是 （ ）A 2B ．C 2D 2213.(山东省实验中学2013届高三第一次诊断性测试文)已知双曲线22221(0,0)x y a b ab-=>>的两条渐近线均与22:650C x y x +-+=相切，则该双曲线离心率等于A ．5B ．2C ．32D ．514.(山东省聊城市东阿一中2013届高三上学期期初考试)过椭圆22221x y ab+=(0a b >>)的左焦点1F 作x 轴的垂线交椭圆于点P ，2F 为右焦点，若1260F P F ∠= ，则椭圆的离心率为 （ ）A ．2B 3C ．12D ．13二、填空题：15. （山东省泰安市2013届高三上学期期末文13）若双曲线221yx m-=的一个焦点与抛物线28y x =的焦点重合，则m 的值为__________.16.(山东省青岛一中2013届高三1月调研考试文)过抛物线2x =2py(p>0)的焦点F 作倾斜角030的直线，与抛物线交于A 、B 两点（点A 在y 轴左侧），则A FB F的值是___________．17.(山东省青岛一中2013届高三1月调研考试文)如图所示, C是半圆弧x2+y2=1(y≥0)上一点, 连接AC并延长至D, 使|CD|=|CB|, 则当C点在半圆弧上从B点移动至A点时,D 点的轨迹是_______的一部分，D点所经过的路程为.18.(山东省实验中学2013届高三第一次诊断性测试文)已知点P 是抛物线24y x =上的动点，点P 在y 轴上的射影是M ，点A 的坐标是（4，a ），则当||4a >时，||||PA PM +的最小值是 。
2024届山东省潍坊市高三上学期12月普通高中学科素养能力测评英语试题(3)
2024届山东省潍坊市高三上学期12月普通高中学科素养能力测评英语试题(3)一、听力选择题1. What are the speakers talking about?A．School kids.B．A flower shop.C．A special day.2. What are the speakers talking about?A．Gifts for Jason.B．A baseball game.C．The woman's retirement.3. When does the bookshop close probably?A．At 5:15.B．At 5:30.C．At 6:00.4. Where are the speakers probably?A．In a hotel.B．In a restaurant.C．At home.5. What are the speakers mainly talking about?A．A shopping list.B．Preparations for a picnic.C．Children’s clothes.二、听力选择题6. 听下面一段较长对话，回答以下小题。
1. Where are the speakers probably?A．In a cafe.B．In a park.C．In a bookstore.2. What mainly makes the man interested in the book?A．Its themes.B．Its characters.C．Its author.7. 听下面一段较长对话，回答以下小题。
1. How many guests will be invited?A．50.B．80.C．100.2. Which of the following will cost the woman another £90?A．The carpet.B．The chairs.C．The lighting.3. When will the tent be set up?A．On June 5th.B．On June 6th.C．On June 7th.8. 听下面一段较长对话，回答以下小题。
2023年山东省潍坊市高密一中、三中、四中高二(上)月考数学试卷(12月份)+答案解析(附后)
2021-2022学年山东省潍坊市高密一中、三中、四中高二（上）月考数学试卷（12月份）一、单选题（本大题共8小题，共40分。
在每小题列出的选项中，选出符合题目的一项）1.若直线经过两点，且倾斜角为，则m的值为( )A. 2B.C. 1D.2.抛物线的焦点坐标是( )A. B. C. D.3.如图，空间四边形OABC中，，，，点M在线段OC上，且，点N为AB中点，则( )A.B.C.D.4.若圆A：与圆B：相切，则的值为( )A. 3B. 9C. 3或7D. 9或495.为了落实“精准扶贫”工作，县政府分派5名干部到3个贫困村开展工作，每个贫困村至少安排一名干部，则分配方案的种数有( )A. 540B. 240C. 150D. 1206.点到双曲线C：渐近线的距离为1，则双曲线的离心率为( )A. 2B.C.D. 47.在《九章算术》中，将四个面都是直角三角形的四面体称为鳖臑，在鳖臑中，平面BCD，，且，M为AD的中点，则异面直线BM与CD夹角的余弦值为( )A. B. C. D.8.设P是椭圆上一点，M，N分别是圆：和：上的点，则的最大值为( )A. 13B. 10C. 8D. 7二、多选题（本大题共4小题，共20分。
在每小题有多项符合题目要求）9.已知方程，则( )A. 时，方程表示椭圆B. 时，所表示的曲线离心率为C. 时，方程表示焦点在y轴上的双曲线D. 时，所表示曲线的渐近线方程为10.已知m，n是两条不同的直线，，是两个不同的平面，给出下列命题中正确的是( )A. 若，，则B. 若，，则C. 若，，，则D. 若，，则11.如图所示，一个底面半径为4的圆柱被与其底面所成的角的平面所截，截面是一个椭圆，则下列正确的是( )A. 椭圆的长轴长为8B. 椭圆的离心率为C. 椭圆的离心率为D. 椭圆的一个方程可能为12.平面内到定点的距离比到直线l：的距离大1的动点的轨迹为曲线C，则( )A. 曲线C的方程为B. 点P是该曲线上的动点，其在x轴上的射影为点Q，点A的坐标为，则的最小值为5C. 过点F的直线交曲线C于A，B两点，若，则D. 点M为直线上的动点，过M作曲线C的两条切线，切点分别为，，则三、填空题（本大题共4小题，共20分）13.已知圆：与圆：相交，则它们交点所在的直线方程为__________ .14.已知双曲线：，与共渐近线的双曲线过，则的方程是__________.15.中国古代桥梁的建筑艺术，有不少是世界桥梁史上的创举，充分显示了中国劳动人民的非凡智慧.如图，一个抛物线型拱桥，当水面离拱顶2m时，水面宽若水面下降1m，则水面宽度为__________.16.设，为椭圆的两个焦点，点P在C上，e为C的离心率．若是等腰直角三角形，则__________；若是等腰钝角三角形，则e的取值范围是__________.四、解答题（本大题共6小题，共70分。
山东省潍坊市2024-2025学年高三上学期开学调研监测考试数学试题
高三开学调研监测考试数学试题2024.9本试卷共4页.满分150分.考试时间120分钟.注意事项：1.答题前，考生务必在试题卷､答题卡规定的地方填写自己的准考证号､姓名.2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其它答案标号､回答非选择题时，将答案写在答题卡上､写在本试卷上无效3.考试结束，考生必须将试题卷和答题卡一并交回.一､单项选择题：本大题共8个小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.复数12i12i -+的虚部是（）A.45 B.45-C.35D.35-【答案】B 【解析】【分析】根据复数的除法运算和虚部的概念即可得到答案.【详解】()()()212i 12i 144i 34i 12i 12i 12i 555----===--++-，则其虚部为45-.故选：B.2.设集合{}{}21,2,4,50A B xx x m ==-+=∣，若{}2A B = ，则B =（）A.{}2,3- B.{}2,6- C.{}2,3 D.{}2,6【答案】C 【解析】【分析】由交集可得6m =，再解方程可得集合B ；【详解】因为{}2A B = ，所以2B ∈，代入250x x m -+=，可得6m =，所以方程变为2560x x -+=，可解得2x =或3，所以{}2,3B =，故选：C.3.已知向量,,a b c在正方形网格中的位置如图所示，若网格纸上小正方形的边长为2，则()a b c a b +⋅+⋅=（）A.0B.3C.6D.12【答案】D 【解析】【分析】建立合适的直角坐标系，写出相关向量计算向量数量积即可.【详解】以两向量公共点为坐标原点建立如图所示直角坐标系，则()4,2a = ，()4,2b =- ，()0,2c =，则()()()()()8,00,24,24,212a b c a b +⋅+⋅=⋅+⋅-=.故选：D.4.坡屋顶是我国传统建筑造型之一，蕴含着丰富的数学元素.如图，某坡屋顶可视为一个五面体，其中两个面是全等的等腰梯形，还有两个面是全等的等腰三角形，若25m,10m AB BC ==，且等腰梯形所在平面､等腰三角形所在平面与平面ABCD 的夹角均为45 ，则该五面体的体积为（）A.3375mB.31625m 3C.3545m D.3625m 【答案】B 【解析】【分析】作出图形，结合二面角的定义分别求出15m 2PO FO BC ===，最后利用五面体的体积为2倍的四棱锥ADHG F -的体积加上三棱柱FGH EMN -的体积求出结果即可；【详解】如图，作FG AB ⊥于G ，//HG DA ，连接FH ；同理作EM AB ⊥于M ，//BC MN ，连接EN ，取AD 中点P ，连接OP ，再作FO GH ⊥于O ，因为等腰梯形所在平面､等腰三角形所在平面与平面ABCD 的夹角均为45 ，因为FG AB ⊥，房顶的底面为矩形，//HG DA ，所以HG AB ⊥，又AD 中点P ，PF AD ⊥，且FG FH =，所以FO GH ⊥，所以OP AD ⊥，FP AD ⊥，所以由二面角的定义可得45FGH FPO ∠=∠=︒，因为10m BC =，所以15m 2PO FO BC ===，因为FO GH ⊥，FO PO ⊥，PO GH O ⋂=，且,PO GH ⊂底面ABCD ，所以FO ⊥底面ABCD ，所以该五面体的体积为2倍的四棱锥ADHG F -的体积加上三棱柱FGH EMN -的体积，即112150016252510510515375323233AG GH FO GH FO GM ⋅⨯⨯+⋅⨯=⨯⨯⨯+⨯⨯⨯=+=，故选：B .5.已知圆22:20C x y x +-=，则过点()3,0P 的圆C 的切线方程是（）A.()132y x =±- B.()23y x =±-C.()333y x =±-D.)3y x =-【答案】C 【解析】【分析】首先说明点在圆外，再设点斜式方程，利用圆心到直线距离等于半径得到方程，解出即可.【详解】将()3,0P代入圆方程得22302330+-⨯=>，则该点在圆外，22:20C x y x +-=，即()22:11C x y -+=，则其圆心为()1,0，半径为1，当切线斜率不存在时，此时直线方程为3x =，显然不合题意，故舍去，则设切线方程为：()3y k x =-，即30kx y k --=，1=，解得33k =±，此时切线方程为()333y x =±-.故选：C.6.数列中，112,2n n a a a +==+，若19270k k k a a a +++++= ，则k =（）A.7B.8C.9D.10【答案】C 【解析】【分析】先求等差数列求出通项，再求和得出参数.【详解】因为112,2n n a a a +=-=，所以()2122n a n n =+-⨯=,()()()91910102218291027022k k k k k a a k k a a a k +++++++++===+⨯= ,所以9k =.故选：C.7.设412341010x x x x ≤<<<≤，随机变量1ξ取值1234,,,x x x x 的概率均为14，随机变量2ξ取值123234341412,,,3333x x x x x x x x x x x x ++++++++的概率也均为14，若记()1D ξ，()2D ξ分别是12,ξξ的方差，则（）A.()()12D D ξξ>B.()()12D D ξξ=C.()()12D D ξξ< D.()1D ξ与()2D ξ的大小不确定【答案】A【解析】【分析】先由期望和方差公式表示出()1D ξ，()2D ξ，再比较公式中不同部分的大小，然后再由基本不等式比较即可；【详解】()123414x x x x E ξ+++=，()()()()()()()()()()()()()2222111213141222221234123411141244D xE x E x E x E x x x x x x x x E E ξξξξξξξ⎡⎤=-+-+-+-⎢⎥⎣⎦⎡⎤=+++-++++⎣⎦，()()1232343411234412211433334x x x x x x x x x x x x x x x x E E ξξ+++++++++++⎛⎫=+++==⎪⎝⎭，()()()()()222212323434141221111143333x x x x x x x x x x x x D E E E E ξξξξξ⎡⎤++++++++⎛⎫⎛⎫⎛⎫⎛⎫=-+-++⎢⎥⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦()()()2222212323434141212341112443333x x x x x x x x x x x x x x x x E E ξξ⎡⎤⎛⎫++++++++⎛⎫⎛⎫⎛⎫⎛⎫=+++-++++⎢⎥ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎝⎭⎣⎦，所以只需比较22222341x x x x +++与22221232343414123333x x x x x x x x x x x x ++++++++⎛⎫⎛⎫⎛⎫⎛⎫+++ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭的大小，因为()2222234123121323122239x x x x x x x x x x x x ++⎛⎫=+++++ ⎪⎝⎭，所以22221232343414123333x x x x x x x x x x x x ++++++++⎛⎫⎛⎫⎛⎫⎛⎫+++ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭()()222212341213232434141322222229x x x x x x x x x x x x x x x x ⎡⎤=+++++++++⎣⎦，①因为412341010x x x x ≤<<<≤，所以2212122x x x x <+，2213132x x x x <+，2223232x x x x <+，2224242x x x x <+，2234432x x x x <+，2214142x x x x <+，所以①()()()2222222222221234123412341369x x x x x x x x x x x x ⎡⎤<+++++++=+++⎣⎦，所以()()12D D ξξ>，故选：A.。
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山东省潍坊一中2013届高三上学期12月月考测试文试题
第I 卷（选择题，共60分）
一、选择题：本大题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的. 1.已知集合{}(){}1,2,3,4,5,,,,A B x y x A y A x y A ==挝-?，则B 中所含元素的个
数为 A.3
B.6
C.8
D.10
2.已知两非零向量,,a b 则“a b a b ?”是“a 与b 共线”的
A.充分不必要条件
B.必要不充分条件
C.充要条件
D.既不充分也不必要条件
3.已知直线()()1:3410l k x k y -+-+=与()2:23230l k x y --+=平行，则k 的值是 A.1或3
B.1或5
C.3或5
D.1或2
4.平面直角坐标系中，已知两点()()3,1,1,3A B -，若点C 满足12OC OA OB l l =+（O 为原点），其中12,R l l Î，且121l l +=，则点C 的轨迹是 A.直线
B.椭圆
C.圆
D.双曲线
5.已知函数()()2
4,f x x y g x =-=是定义在R 上的奇函数，当0x >时，()2log g x x =，则函数()()f x g x ×的大致图象为
6.各项为正数的等比数列{}n a 中，2311
,
,2a a a 成等差数列，则4534a a a a ++的值为
A.
51
2
- B.
51
2
+ C.
15
2-
D.
5151
22
-+或 7.在ABC D 中，角A,B,C 所对的边分别为,,,a b c S 表示ABC D 的面积，若
()2
221cos cos sin ,4
a B
b A
c C S b c a +==
+-，则B ?
A.30°
B.45°
C.60°
D.90°
8.若圆2
2
:2430C x y x y ++-+=关于直线260ax by ++=对称，则由点(),a b 向圆所作的切线长的最小值是
A.2
B.3
C.4
D.6
9.已知直线a 和平面,,,,l a a a b a b a b ?怂，
且a 在,a b 内的射影分别为直线b 和c ，则b 和c 的位置关系是 A.相交或平行 B.相交或异面
C.平行或异面
D.相交、平行或异面
10.设12,F F 分别是椭圆22
221x y a b
+=()0a b >>的左、右焦点，与直线y b =相切的2F 交
椭圆于点E ，E 恰好是直线EF 1与
2F 的切点，则椭圆的离心率为
A.
3
2
B.
33
C.
53 D.
54
11.四棱锥P ABCD -的三视图如右图所示，四棱锥P ABCD -的五个顶点都在一个球面上，E 、F 分别是棱AB 、CD 的中点，直线EF 被球面所截得的线段长为22，则该球表面积为 A.12p B.24p
C.36p
D.48p
12.定义在()(),,f M m n p =，其中M 是ABC D 内一点，m 、n 、p 分别是MBC D 、
MCA
D 、
MAB
D 的面积，已知
ABC
D 中，
()
123,
30,,,2A B
A
C B A C f N x y
骣÷ç??=÷ç÷ç桫，则14x y
+的最小值是 A.8 B.9
C.16
D.18
（非选择题 共90分）
题号 二 三
总分 17 18 19 20 21 22 得分
二、填空题：本大题共4小题，每小题4分，共16分。
请将答案填写在试题的横线上。
13.设O 为坐标原点，()1,2A ，若点(),B x y 满足22221012,
1 2.
x y x y x y ìï+--+?ïï
ï＃íïï＃ïïî则OA OB ×取得最小值时，点B 的坐标是________.
14.若函数()()
()23log 4a f x ax -=+在[]1,1-上是单调增函数，则实数a 的取值范围是____.
得分 评卷人
15已知正四棱柱ABCD-A 1B 1C 1D 1的高为22，外接球的体积是323
p
，则A 、B 两点的球面距离为____________. 16.给出以下五个命题： ①命题“2
,10x R x
x "?+>”的否定是：“2
,10x R x x $?+<”.
②已知函数()cos f x k x =?的图象经过点,13
P p
骣÷ç÷ç÷ç桫，则函数图象上过点P 的切线斜率等于
3-.
③1a =是直线1y ax =+和直线()21y a x =--垂直的充要条件.
④函数()1
312
x
f x x 骣÷ç=-÷ç÷ç桫在区间()0,1上存在零点. ⑤已知向量()1,2a =-与向量()1,b m =的夹角为锐角，那么实数m 的取值范围是
1,2骣÷ç-?÷ç÷
ç桫. 其中正确命题的序号是________.
三、解答题：本题共6小题，共74分.解答应写出文字说明、证明过程或演算步骤.
17.（本题12分） 已知向量()cos sin ,sin a x x x w w w =-，()
cos sin ,23cos b x x x w w w =--，设函数
()()f x a b x R l =??的图象关于直线x p =对称，其中,w l 为常数，且1,1.2
w 骣÷çÎ÷ç÷ç桫 （I ）求函数()f x 的最小正周期；
（II ）若()y f x =的图象经过点,04
p 骣÷ç÷ç÷ç桫，求函数()f x 在区间30,5
p
轾犏
犏臌上的取值范围.
得分 评卷人
18.（本题12分） 各项均为正数的数列{}n a 中，前n 项和2
12
n n a S 骣+÷
ç=÷ç÷ç桫. （1）求数列{}n a 的通项公式； （2）若12231
111n n k a a a a a a +++鬃?<恒成立，求k 的取值范围；
19.（本题12分） 设函数()()()101x x f x a k a a -=-->?且是定义域为R 的奇函数. （1）求k 值；
（2）若()10f <，试判断函数单调性并求使不等式()
()240f x tx f x ++-<恒成立的取值范围..
20.（本题12分） 如图，四棱锥P ABCD -的底面是边长为1的正方形，侧棱PA ^底面ABCD.且2PA =，E 是侧棱PA 上的动点。
（1）求三棱锥C-PBD 的体积；
（2）如果E 是PA 的中点，求证PC//平面BDE ；
（3）是否不论点E 在侧棱PA 的任何位置，都有BD CE ^？证明你的结论..
21.（本题12分）
已知函数()3
2x f x a
=图象上斜率为3的两条切线间的距离为()210,5f x 的导数为()f x ¢，函数()()()
3bf x g x f x x
¢=-+。
（1）若函数()1g x x =在有极值，求()g x 的解析式；
（2）若函数()[]1,1g x -在是增函数，且()24b mb g x -+?在[]1,1-上都成立，求实数m 的取值范围.
22.（本题14分）
已知椭圆22
22:1x y E a b +=()0a b >>的左顶点为A ，左、右焦点分别为F 1、F 2，且圆C ：
223360x y x y ++
--=过A ，F 2两点.
（1）求椭圆E 的方程；
（2）直线BC 过坐标原点，与椭圆E 相交于B,C,点Q 为椭圆E 上的一点，若直线QB,QC 的斜率,QB QC k k 存在且不为0，求证：QB QC k k ×为定值；
（3）设直线PF 2的倾斜角为a ，直线1PF 的倾斜角为b ，当23
p
b a -=时，证明：点P 在一定圆上.。