四川大学期末考试试题(闭卷
四川大学电路期末试题及答案1
+ 30°)
4 分 得：
uR 2
(t)
=
−
100 3
cos(2ω
1t
+
30°)
V
2分
uR
(t)
=
200
−
100 3
cos(2ω
1t
+
30°)
V
2分
UR =
2002
+
1 2
×
1030
2
= 201.38 V
2分
4、（15 分）
I (s)
5
+2
6
(A 卷)
学号：
5、含理想变压器电路如图 2-5 所示,欲使负载电阻 RL 获得最大功率,则 n=
PLmax 值为
。
姓名 和所获得的
2Ω 4sinωtV
1:n ••
RL = 8Ω
图 2-5
图 2-7
6、某 RC 串联电路的零输入响应 i(t) = 20 e −5000t mA ，已知 t = 0 时电容电压为 2 V，则该电路的电阻
R 为_ __，电容 C 为_ ___。
7、图 2-7 所示对称三相四线制电路中，已知星形联接负载的 R = 8Ω,ωL = 2Ω, 当相电压 uA = 2sinωtV
时，中线电流有效值等于_
___；当 uA = 50 2 sin 3ωtV 时，中线电流有效值等于_
___。
三、计算题（共 4 小题，共计 55 分）
图 2-1
2、电路如图 2-2 所示，当 R = 0 时， I =
。
图 2-2
3、图示电路 2-3 中， N 0 内不含独立电源。当U S =1V， IS =0 时，U ab =3V；当U S =0， IS =2A 时，
四川大学微积分1-2(2016)B卷
4．设空间曲面： z 1 ( x 2 y2 ) (0 z 1部分) 所围成，方向指向外侧，计算曲面积分 2
( x y)dydz ( y z)dzdx ( x z)dxdy .
5．求微分方程 y 4 y x cos x 的通解.
(1)求常数 A，以及该微分方程的通解.
(2)计算曲线积分 (0,1) 2 xydx ( Ax 2 2 y)dy 的值. (1,0)
3．设二元函数
f
(
x,
y)
xy , x2 y2
0,
( x, y) (0, 0)
.
( x, y) (0, 0)
(1)求证：二元函数 f ( x, y) 在点(0,0)处不可微.
0
2
0
确定的隐函数组，求
y(1),
z(1) .
第 1 页，共 2 页 试卷编号：
2．设空间区域是由 z x2 y2 与 z 2 x2 y2 所围成，计算三重积分
(2x y 3z)dxdydz .
3．设平面闭曲线 L： y cos x 从点 A(1,1)到 B(1,1)，计算曲线积分
四川大学期末考试试题（闭卷） （2015-2016 学年第 2 学期） B 卷
课程号：201138040 适用专业年级：
课序号： 学生人数：
课程名称：微积分（I）-2 任课教师：
成绩：
印题份数：
学号：
姓名：
考生承诺
我已认真阅读并知晓《四川大学考场规则》和《四川大学本科学生考试违纪作弊处分规定（修 订）》，郑重承诺：
2．二元函数 z
f (u, v) 具有二阶连续偏导数,
u
【期末试题】2018-2019秋线性代数(理工)
三、证明题（共 19 分）
1. （7 分）证明：向量组 ������1, ������2, ������3 线性无关的充分必要条件是向量组 ������1 + ������2, ������2 + ������3, ������3 + ������1 线性无关. 2.（6 分）设方阵 ������ 使得 ������3 = 2������, 证明 ������2 − ������ 可逆，并求 ������2 − ������ 的逆矩阵. 3.（6 分）设 ������ 阶方阵 ������ 满足 ������2 = ������. 则 ������ 是齐次线性方程组 ������������ = 0 解的充分必要条件 为：存在向量 ������ 使得 ������ = ������ − ������������.
0
相似,
则
������������ =
__________.
1 2 3 4 0 0 4 y
1 0 0
x
0
0
1
1
110 2. 若存在3维列向量不能由向量组 (0) , (������) , (2) 线性表出，则 ������ = __________.
121
1 2 4 ������1 3. 若二次型 (������1, ������2, ������3) (0 2 2) (������2) 正定，则 ������ 的取值范围为 __________.
0 0 ������ ������3 4. 设������为3阶实对称阵,������2 − ������ = 2������, ������������(������) = 0,则二次型 ������������������������ 的规范形为 __________.
四川大学期期末考试试题
四川大学期期末考试试题（2004—2005学年第2学期）课程号：课序号：课程名称：精神病学任课教师：成绩：适用专业年级：医01级五年制学生人数：印题份数：学号：姓名：注：1.试题字迹务必清晰，书写工整。
本题6 页，本页为第1页2.题间不留空，一般应题卷分开教务处试题编号：注：1.试题字迹务必清晰，书写工整。
本题 6 页，本页为第2 页2.题间不留空，一般应题卷分开教务处试题编号：注：1.试题字迹务必清晰，书写工整。
本题 6 页，本页为第3 页2.题间不留空，一般应题卷分开教务处试题编号：注：1.试题字迹务必清晰，书写工整。
本题 6 页，本页为第 4 页2.题间不留空，一般应题卷分开教务处试题编号：注：1.试题字迹务必清晰，书写工整。
本题 6 页，本页为第 5 页注：1.试题字迹务必清晰，书写工整。
本题 6 页，本页为第 6 页05年春季01级五年制精神病学考试题答案一、1.E 2.E 3.B 4.C 5.C 6.D7.E8.A9.B10.A11.A12.C13.D14.D15.B16.D17.E18.B19.C20.D二、1.1.E 1.2.E 2.1.C 2.2.B 2.3.D3.1 E 3.2.E4.1.D 4.2.E 4.3.D三、名词解释：精神疾病精神疾病指在各种生物学，心理学以及社会环境因素的影响下，大脑功能失调，导致认知，情感，意志和行为等精神活动出现不同程度障碍的疾病。
妄想：一种在病理基础上发生的歪曲的信念，毫无事实根据，病人坚信不疑。
迟发性运动障碍是抗精神病药的一种副作用。
长期大量应用抗精神病药引起的锥体外系症状。
表现为不自主的有节律的刻板式动作如咀嚼，吸吮或舞蹈样运动。
遗忘综合征其特点是识记能力障碍，时间定向能力障碍，虚构症，顺行性遗忘或逆行性遗忘。
谵妄意识清晰度降低的同时，病人产生大量的恐怖性幻觉，错觉，并产生不协调的精神运动性兴奋。
四、问答题什么是精神分裂症的阴性症状和阳性症状.精神分裂症的急性阶段，临床表现以幻觉，妄想为主，这类症状又称阳性症状，慢性阶段，临床症状主要为思维贫乏，情感淡漠，意志缺乏，孤僻内向，又称阴性症状。
四川大学期末考试试题((闭卷闭卷) (2016——2017 学年第 2 学期) A 卷
L
7．微分方程 xy′ + y = x2 满足 y(3) = 4 的特解为
．
二、解答题 (每小题 9 分，共 36 分)
1．设曲面Σ 为 z =
，求 ． ∫∫ x2 + y2 (x2 + y2 1)
(20 xy + 17 y2 )dS
Σ
2．设曲面Σ 为 z = 1 − x2 − y2 ，方向为上侧，求 ∫∫ x2dydz + y2dzdx + ． 5z3dxdy Σ 第 1 页，共 2 页
三、综合题 (每小题 9 分，共 18 分)
1．讨论函数
f
( x,
y)
=
(
2
x
2
x2 +
y 7
2
y
2
)3/
2
,
( x,
y) ≠
(0,0)在点(0,0) 处的下列性质：
0
, ( x, y) = (0,0)
（1）偏导数的存在性；（2）函数的连续性；（3）函数的可微性．
2．设
f ′( x)连续，
f (1) = 2017 ，当 x ≠ 0 时
（1） I = 2π ∫ 1t sin t5dt ； （2） I < 2π / 7 ； （3） I > π / 4 ． 0
第 2 页，共 2 页
2．在椭圆抛物面 z = x2 + y2 与平面 z = 20围成的空间区域中内置一个长方体，假设该长方
20
4
体的一个面位于z = 20上，长方体的其它面都与某个坐标平面平行，求长方体的体积的最大值．
五、证明题 (7 分)
设区域 D 为 x2 + y2 1， I = ∫∫ sin( x2 + y2 )5/2dxdy ，求证： D
药剂学__四川大学(1)--四川大学期末考试试题A(闭卷)
（2012—2013 学年第 二 学期）
课程号：50505040 课序号：0，1 课程名称：药剂学 任课教师：张志荣 等 成绩： 适用专业年级：2010 药学本科 学生人数： 印题份数： 学号： 姓名：
考试须知
四 川 大 学 学 生 参 加 由 学 校 组 织 或 由 学 校 承 办 的 各 级 各 类 考 试 ， 必 须 严 格 执 行《 四 川 大 学 考 试 工 作 管 理 办 法 》和《 四 川 大 学 考 场 规 则 》。有 考 试 违 纪 作 弊 行 为 的 ， 一 律 按 照《 四 川 大 学 学 生 考 试违纪作弊处罚条例》进行处理。
（1 分）
焦亚硫酸钠 1g
（1 分）
注射用水 加到 1000ml
制法（1 分）:
本题共 5 页，本页为第 页
4
3. 请说明处方中划线成分的作用并简述制剂制备方法。
1:100 硫酸阿托品散
处方：硫酸阿托品
1％胭脂红乳糖 0.05g
（1 分）
乳糖 99.85g
（1 分）
制法（2 分）：
4、请说明处方中划线成分的作用和制剂名称。（每空 0.5 分）
B. 液体石蜡
C.乳剂型基质
D. 水溶性基质
三、简答题（每题 3 分，共 8 题，共计 24 分） 1. 简述灭菌参数 F0 值的概念及意义？ 2. 混悬剂常用稳定剂有哪些类型？它们的主要作用是什么？ 3. 用氯化钠配制 2000ml 等渗溶液，问需要多少氯化钠?已知 1%氯化钠溶液的冰点 降低为 0.58℃。 4. 简述包合物的优点。 5. 片剂的制备方法有哪三类？简述每类方法的选用原则。
6
卡波姆 940 10g
四川大学英语2020期末考试试题
四川⼤大学“⼤大学英语（阅读与翻译）- 2” 期末考试试题（闭卷）A 卷（2019—2020学年年第 2 学期）课程号：105368020 课序号： 课程名称： 任课教师： 成绩： 适用专业年级：2019级 学生人数： 印题份数： 学号： 姓名： 第 1 ⻚页，共 11 ⻚页试卷编号：（注意：所有答案请写在答题纸上）Part I Reading Comprehension (65 points)Section A (3.5*10 = 35 points)Directions: Read the following passage and use NO MORE THAN FIVE WORDS to fill in each blank in the outline below. Remember to write down the answers on the Answer Sheet .California's Higher Education - From American Dream to Dilemma[1] For the Golden State of California, 1960 was a golden year: It was a time of rapid development, and part of this massive development was a system of public higher education.[2] California's higher education prospects of 1960 were built on a distinctive historical foundation. The idea that the state's colleges and universities could – and should – be the source of an informed, responsible citizenry and state leadership had been established by legislators and voters by World War I.[3] Robert Gordon Sproul, president of the University of California, who served from 1930 to 1958, built on this early vision. He set up six campuses statewide as part of a creation of a multi-campus system to meet California's growing demand for higher education.[4] In 1958, Clark Kerr was named president of the University of California system. With Kerr's efforts higher education became part of the California dream. In 1960 the state legislature passed the Donahoe Act, which included a 246-page report, "A Master Plan for Higher Education in California, 1960-1975." On Oct. 17, 1960, Time magazine featured Clark Kerr on its cover as the "Master Planner."[5] The California dream of higher education combined access to higher education with affordability and choice. The new California policy of no tuition was extended to all public colleges and universities statewide. Furthermore, the master plan promoted state-funded student scholarships through a state agency created in 1955, the California Student Aid Commission.[6] California came to provide a high quality of education – the best in the country. A 1966 report by the American Council on Education shows that University of California, Berkeley was the top university at the time in America for overall quality in graduate education. Between 1939 and 1968, 12 professors at UC Berkeley had received the Nobel Prize, the highest number at any university.[7] Resources were made available for the realization of the dream. As part of passing the Donahoe Act in 1960, the California state government approved US$1 billion in funding for higher education facilities. Central to its growth was an expansion of campuses. Between 1964 and 1965 the University of California built three new campuses – at SanDiego, Irvine and Santa Cruz.[8] By 1967, however, the master plan was encountering problems – it was expensive and increasingly seemed not sustainable.[9] State Senator George Deukmejian voiced Republican concerns about higher education. He argued in favor of adding a tuition charge for University of California students and endorsed Governor Ronald Reagan's new "equal education plan." The plan called for a modest tuition of $250 per year for the university and $80 per year in the state colleges.[10] The Republican reform plan included grants or loans to those who could not afford the modest tuition. It was noted that half of the enrolled students came from relatively affluent families. Only about 12 percent came from modest-income families earning $6,000 year or less.[11] Today, California's higher education system struggles with budget cuts and an uncertain future. There are concerns that the public universities are soon going to be "public no more." As education scholar Brendan Cantwell notes, even the preeminent research university, Berkeley, has been hit by budget cuts. At the same time, the state's outstanding private colleges and universities have soared in terms of academic standards, selective admissions, tuition revenues, new construction and federal research grants.[12] The master plan has struggled to keep up. It has gone through many reviews and revisions, the latest of which, in 2017, emphasized improving access and affordability. But the convergence of these trends, combined with fluctuations in the state economy and tax revenues, has turned the Californian dream of higher education into an American dilemma.(603 words)第2⻚页，共11⻚页试卷编号：（每个答案不超过5个单词！）California's Higher Education - From American Dream to Dilemma: An OutlineI.Start of the DreamA. 1 _______________________ formed by World War IB.Robert Gordon Sproul’s 2_______________________II.The Dream YearsA.Clark Kerr’s 3_______________________B. 4 _______________________ of higher educationC.California’s 5_______________________ of educationD.Funding and 6_______________________III.7_______________________ EncounteredA.George Deukmejian’s argument for a 8 _______________________B.9 _______________________ to poor studentsIV.Realities Today: the DilemmaA.Struggles with 10 _______________________B.Struggles with the master planSection B (2*15 = 30 points)Directions:There are 6 passages in this section. Each passage is followed by some questions or unfinished statements. For each of them there are four choices marked A), B), C) and D). You should decide on the best choice and write it down on the Answer Sheet.Passage OneQuestions 11 to 15 are based on the following passage.[1]What do Justin Bieber, Taylor Swift, Katy Perry, Usher and Lady Gaga have in common apart from being highly popular, and rich, music singers? The answer is they all owe some of their success to people behind the scenes from a relatively small country in northern Europe – Sweden. Ever since ABBA exploded onto the world stage in 1974 by winning the European song contest with their easy-to-sing-along hit “Waterloo”, Swedish groups, and more importantly the people who make the groups what they are, have had a disproportionate influence on modern music. To understand why this should be so it is important to see how Sweden nurtures its musical talent.第3 ⻚页，共 11 ⻚页试卷编号：[2] All Swedish children are encouraged and expected to learn how to appreciate music and play an instrument from kindergarten onwards. After-school music lessons are inexpensive and instruments may be rented. Music education is supported by the government as may be seen by the recent establishment of the Academy of Music and Business (AMB) in southern Sweden by two music producers and teachers, which receives funds for each of the students enrolled in the school. The school aims to develop both creativity and commercial success in young musicians by preparing them for the enormous changes the music industry is facing. Note the second half of the school’s name is “Business.” This is a clear recognition of the fact that in Sweden music making is about making money. In 2011, Sweden’s foreign music sales earned the country US $135 million, which placed it third in the world after the USA and the UK and the highest in the world on a per capita basis.[3] So who are the people responsible for making money out of music and helping the big name stars achieve such popularity? The man credited with starting Sweden’s creative push into modern music is Dag V olle aka Denniz Pop. It is claimed that he discovered the Swedish group Ace of Base in the early 1990s when the group’s demo tape got jammed in his car stereo and he had to listen to it over and over again. He loved the song so much that he decided to release it to the public. Their new title, “All That She Wants” became a smash hit. He subsequently opened a recording studio, Cheiron Studios, and it was there that he produced the debut album for one of the biggest boy bands of the 1990s, Backstreet Boys. V olle worked collaboratively with, and mentored, a number of young songwriters until his death in 1998. Perhaps the best known of these today is Max Martin.[4]Max Martin, whose real name is Martin Karl Sandberg, has written a string of hits for some of the biggest recording artists in the world today. These include Kelly Clarkson’s “Since U Been Gone,” Usher’s “DJ Got Us Fallin’ in Love,” and Britney Spears’ “Baby One More Time.” Other prominent Swedish songwriters and producers include Andreas Carlsson, who co-wrote “Bye Bye Bye” for NSYNC, Shellback, Bloodshy and RedOne, who co-wrote and produced two of Lady Gaga’s biggest hits “Poker Face” and “Bad Romance.” If you have not heard of any of these people maybe it is because the Swedes famously do not like standing out from the crowd, but prefer to work their magic behind the scenes.第4⻚页，共 11⻚页试卷编号：[5] Aside from writing and producing hits, Swedes are also activein other areas of the music business: as sound and video engineers and technicians, the people who make the music sound the way it does; choreographers and directors who put together the all-important videos that accompany songs and often do so much to shape the image of the artist, especially when seen by millions on YouTube; Sweden also has a great number of music teachers and voice coaches, managers, marketers and promoters. The fans who buy the star’s music may be unaware of these teams of people working in collaboration in the background, but these professionals are all crucial for the ultimate success of the artist’s work.[6] One reason why Sweden is so active in the music field may come down to its adaptability. Ola Hakansson, a singer whose career spans fifty years, says, “We’re a small country, so we can’t set any trends, but we’re really good at picking up on other people’s, and we’re really good at mimicking.” A case in point is ABBA, whose records mix jazz, blues, and keyboard style with local folk music, and have sold over 380 million copies.[7] Another reason is that nearly all Swedes are fluent in the music world’s lingua franca – English. In order to be globally successful songs need to be written and performed in English. Swedish groups are well placed to capitalize on their linguistic prowess. Since ABBA, other Swedish artists making waves in the music world include Roxette, The Cardigans, Avicii and Robyn.[8] So the next time you buy a CD or download a song by a band or singer that you like, why not check out who wrote it, who produced it, who directed the dance moves on the video, and who manages the artist. You may not now be so surprised to find that someone from Sweden is behind the effort to create the perfect pop song!(875 words)11. What is the exact meaning of "disproportionate" in Para. 1 ?A)Big country, small influence.B)Small country, big influence.C)Less population, more money.D)More population, less money.第5⻚页，共 11 ⻚页试卷编号：12. Which country ranked FIRST in foreign music sales business in 2011?A)Sweden.B)Switzerland.C)UK.D)USA.13. Who is the best-known Swedish songwriter according to the passage?A)Dag V olle aka Denniz Pop.B)Martin Karl Sandberg.C)Andereas Carlsson.D)Shellback.14. Which of the following is one of the reasons for Sweden’s being active in the musicfield?A)Swedes like standing out from the crowd.B)Swedes do all business related to music.C)Swedes are good at imitating others.D)Swedes have talent in music.15. Which of the following is true according to the passage?A)Swedish groups always adapt to audience’s tastes.B)ABBA is good at picking up on blues.C)The lingua franca in Sweden is music.D)Swedish people’s language skills give them an edge.Passage TwoQuestions 16 to 19 are based on the following passage.[1] We’ll sit tables apart at near-empty restaurants, wear medical masks in public, keep a six-foot distance from others while shopping, and submit to routine temperature checks at work. These scenes may look more like an episode of Black Mirror, but the world’s new reality amid the coronavirus pandemic (COVID-19) is forcing the global society to confront issues beyond the scope of science fiction to revive communities slowly, safely and sustainably.[2] New York may have the Empire State Building, lit up in festive hues to mark holidays big and small, but in Washington’s Petworth neighborhood, residents look to the front lawn of a blue rowhouse to celebrate the passage of time. Pirate ships and undead swashbucklers for Halloween. Cartoonish turkeys for Thanksgiving. Pink and red hearts for Valentine’s Day. Red, white and blue everything for the Fourth of July. Bright rainbows and unicorns to honor Pride.第6⻚页，共 11 ⻚页试卷编号：[3] Hardly a holiday goes uncelebrated. Even the small ones.[4] So when the District announced its stay-at-home order, meant to slow the spread of the novel coronavirus, it was no surprise to neighbors that homeowners Curtis Gilbert and Chris Rowland transformed their Easter Bunny display into a flashy pandemic warning sign.[5] The couple measured a six-foot arrow and propped it up between two rabbits, ensuring they were socially distanced as they held speckled Easter eggs. They wanted the crowds on the green grass of Sherman Circle across the street to see: Even the Easter Bunny was keeping his social distance. “We wanted it to be a friendly reminder that it is a pandemic,” Gilbert said. “But with all the horrible news, we were hoping it would give people a chance to giggle, too.”[6] Even the mayor took notice. On Easter Sunday, D.C. Mayor Muriel E. Bowser tweeted a photo of the bunnies. “Happy Easter!” she wrote, after changing her Twitter username to include the hashtag #StayHomeDC.[7] As the pandemic death toll has continued to climb — and states struggle with how best to keep their residents safe — Gilbert and Rowland said they didn’t feel it was right to jump into their usual rotation of holiday trimmings. Instead they continued on the pandemic theme, hoping it would catch the eye of someone who might need a reminder to wear a face mask or keep their distance from others.[8] Gilbert and Rowland — who have been together for 25 years — were never much for holiday displays before 2013, when they moved into the house. But after the move, the couple realized they would likely get a stream of Halloween trick-or-treaters. They decided to embrace it, putting up a small display of skeletons to signal to neighborhood kids they were open for business.[9] The next year, the skeletons got a pirate ship, assembled with leftover wood from a backyard project. The house quickly became a destination for kids and their families. Although it’s not clear when they will be able to host a similar gathering without fears of clustered crowds and spreading disease, the couple said they hope the whimsical pandemic displays send a similar message to their community: we’re all in this together, even if we have to stand six feet apart.(521 words)16. Why does the author mention different festive celebrations in Para. 2?A)To present how much people in New York love celebrating holidays.B)To show that New York can continue to be a metropolitan city.C)To explain holiday decoration services people should have held.D)To suggest that there could be no festive hues to mark holidays over the months.第7⻚页，共 11 ⻚页试卷编号：17. What did the couple do for their community during the pandemic according to thepassage?A)They texted messages to their community that we were all together for thepandemic.B)They warned clustered crowds against the pandemic by asking them to stay home.C)They decorated their house to signal to neighborhood kids they loved the kids verymuch.D)They reminded people of social distancing by connecting two Easter bunnies withthe warning sign of six-foot length.18. Which of the following statements is not TRUE?A)D.C mayor acknowledged the couple’s behavior by tweeting the bunnies picture.B)The couple have been proud of themselves on delivering holiday cheers year byyear since they were together.C)The couple considered it’s not right to jump into the usual holiday displays in thepandemic situation.D)The neighborhood kids and their families used to gather in the couple’s house inholiday seasons.19. The word “whimsical” in Para. 9 is closest in meaning to_____.A)Fanciful DisplaysB)Weird displaysC)Impractical displaysD)Capricious displaysPassage ThreeQuestions 20 is based on the following passage.Campus organizer: Before last year’s campus cleanup, only 69 of the students signed up to participate, but then well over 100 actually participated. This year, 75 students have signed up to participate. Since our campus cleanup will be a success if we have at least 100 participants, we can be confident that this year’s cleanup will be a success.20. The reasoning in the campus organizer’s argument is most vulnerable to criticism on thegrounds that the argument ________.A)confuses a condition that is required for an outcome with one that is sufficient for thatoutcomeB) generalizes about the outcome of an event based on a single observation of a similarsituationC) overlooks the possibility that the cleanup will attract participants who are not studentsat campusD) defines a term in such a way as to ensure that whatever the outcome, it will beconsidered a positive outcome第8⻚页，共 11 ⻚页试卷编号：Passage FourQuestions 21 is based on the following passage.Psychologist: In our study, participants who were offered the opportunity to purchase a Starbucks mug were not willing to pay more than $10. If, however, they were given a very similar mug and asked immediately afterwards how much they would be willing to sell it for, most of them held out for more than $10.21. Which one of the following, if true, most helps to resolve the apparent discrepancydescribed above?A）A person’s assessment of the value of an object depends on his or her evaluation of the inherent properties of the object.B）The amount a person is willing to spend on an object is determined by the amount that object sold for in the past.C）People tend to value an object that they do not own less than they value a very similar object that they already own.D）People are more likely to undervalue objects they have been given than objects they have purchased.Passage FiveQuestions 22 is based on the following passage.Global use of the Internet has led to an increase in certain crimes such as information theft and to new crimes like hacking. This seems due, at least in part, to the impersonal nature of the Internet. People seem to feel more free to harm others through the Internet than in person, suggesting that people feel less morally constrained when they use the Internet. For this reason, steps should be taken to educate people about the ethical use of the Internet.22. Which one of the following principles, if valid, most helps to justify the reasoning inthe argument?A) When new technologies emerge, society needs to formulate new ethical guidelines tocover the use of those technologies.B) The more educated that people are about the ethical usage of the Internet, the greaterthe amount of harm that they can do to others.C) People feel morally constrained from doing an action only if that action causes harmto others.D) Education about the ethical use of a tool increases one’s sense of moral responsibilityregarding its use.第9 ⻚页，共 11 ⻚页试卷编号：Passage SixQuestions 23 to 25 are based on the following passage.A Professor of life science will assign each of five students—Li Hua, Han Meimei, Li Ming, Xiao Long, and Xiao Fan—to exactly one of two research teams, the Jiang’an team and the Huaxi team. One team will have two members, and the other will have three members. One member of each team will be designated as facilitator. The assignment must satisfy the following conditions:•Li Hua is assigned to a different team than Xiao Fan is.•Li Ming is assigned to Jiang’an team.•Han Meimei is not a facilitator.•Xiao Fan is a facilitator.23. Which one of the following could be an accurate listing of the members and facilitatorsof the two research teams?A) Jiang’an team: Han Meimei, Li Ming(facilitator), Xiao Fan;Huaxi team: Li Hua, Xiao Long(facilitator)B) Jiang’an team: Han Meimei, Xiao Long, Xiao Fan(facilitator)Huaxi team: Li Hua(facilitator), Li MingC) Jiang’an team: Li Ming, Xiao Fan( facilitator)Huaxi team: Li Hua, Han Meimei(facilitator), Xiao LongD) Jiang’an team: Han Meimei, Li Ming, Xiao Fan(facilitator);Huaxi team: Li Hua(facilitator), Xiao Long24. Which one of the following must be true?A) Li Hua is assigned to Huaxi team.B) Xiao Fan is assigned to Jiang’an team.C) Li Hua and Xiao Long are not both facilitators.D) Neither Li Hua nor Han Meimei is a facilitator.25. If Xiao Long is assigned to Jiang’an team, then which one of the following must betrue?A) Han Meimei is assigned to Huaxi team.B) Li Hua is assigned to Jiang’an team.C) Xiao Fan is assigned to Jiang’an team.D) Xiao Long is a facilitator.第10 ⻚页，共 11 ⻚页试卷编号：Part II Writing (35 points)Directions:In this part, you are allowed about 30 minutes for a Xu Writing （续写） of no fewer than 120 words based on the following passage taken from the textbook. (Attention: part of the original article has been adapted in the following passage for the convenience of all students to write about.)Scoring Rubric（评分参考）:•Coherence and cohesion with the given passage （续写内容与前⽂连贯度）（40%）;•Verbal dexterity （语⾔表达，包括语⾔⾃然流畅度和词汇丰富性等）（60%）.When I was growing up I was told, as many students are, to do what I am best at. But I didn’t know what that was. Most people think that when you are good at something, it comes easily to you. But this is what I discovered: just because a subject is difficult to learn, it does not mean you are not good at it. You just have to grit your teeth and work harder to get good at it. Once you do, there’s a strong chance you will enjoy it more than anything else.In eighth grade I took algebra. On one test I got only 36 percent of the answers correct. I failed the next one, too. I started to think, maybe I’m just no good at this. I was lucky enough to have a teacher who didn’t take my bad grades as a judgement of my abilities, but simply as an indication that I should study more. He pulled me aside and told me he knew I could do better. He let me retake the tests, and I pulled my grade up to an A.I studied a lot in college, too. I had moments of panic while sitting underneath the buzzing fluorescent lights in the engineering library on Saturday afternoons, when I worried that I was not able to fully understand thermodynamics. But the guys in my classes had to work just as hard, and I knew that I couldn’t afford to lose confidence in myself. The ones that say that math comes more naturally to men or language would be easier to women are based on a faulty premise: that you can judge a person’s abilities separate from the cultural cues that they have received since they are infants. No man is an island. No woman, is either.Why are we so quick to limit ourselves? It may be true that some people favor the right side of their brain, and others the left. Instead of translating our differences into hard and fast conclusions about the human brain, why can’t we focus instead on how incredibly flexible we are?(357 words)第11 ⻚页，共 11 ⻚页试卷编号：。
大学计算机基础 四川大学期末考试题.
四川大学期末考试试题(闭卷一、选择题(每题分,共4040分每题11分,共1.微型计算机中,运算器、控制器和内存储器的总称是【】。
(A主机(BMPU(CCPU(DALU2.微型计算机,ROM是【】。
(A顺序存储器(B只读存储器(C随机存储器(D高速缓冲存储器3.下列设备中,只能作输出设备的是【】。
(A磁盘存储器(B键盘(C鼠标器(D打印机4.微型计算机中存储数据的最小单位是【】。
(A字节(B字(C位(DKB5.微型计算机中,I/O设备的含义是【】。
(A输入设备(B输出设备(C输入输出设备(D控制设备6.数字字符“1”的ASCII码的十进制表示为49,那么数字字符“8”的ASCII码的十进制表示为【】。
(A56(B58(C60(D547.键盘当中用来输入上档字符的是【】。
ENTER (A回车键回车键ENTERBackSpace 退格键BackSpace(B退格键(C换档键Shift 换档键ShiftTab 制表键Tab(D制表键8.左手的中指应放在键盘上【】位置。
(AS(BF(CJ(DD9.512个汉字占用【】个字节。
(A512B(B512KB(C1024KB(D1KB10.MIPS来衡量的计算机性能指标是【】(A处理能力(B运算速度(C存储容量(D可靠性11.世界上第一台电子数字计算机取名为【】(AUNIVAC(BEDSAC(CENIAC(DEDVAC12.计算机的发展阶段通常是按计算机所采用的什么来划分的?【】(A内存容量(B电子器件(C程序设计语言(D操作系统13.现代计算机之所以能自动地连续进行数据处理,主要是因为【】(A采用了开关电路(B采用了半导体器件(C具有存储程序的功能(D采用了二进制14.在软件方面,第一代计算机主要使用【】(A机器语言(B高级程序设计语言(C数据库管理系统(DBASIC和FORTRAN15.个人计算机简称PC机。
这种计算机属于【】(A微型计算机(B小型计算机(C超级计算机(D巨型计算机16.微IBMPC/XT采用的微处理器是由【】公司生产的(AIBM(BInter(CHP(DDigital17.人们常说486微机、586微机,其中的数字指的是【】(A硬盘的型号(B软盘的型号(C显示器的型号(D微处理器的型号18.巨型计算机指的是【】(A重量大(B体积大(C功能强(D耗电量大19.二进制数01100100转换成十六进制数是【】(A64(B63(C100(D14420.下面不属于网络拓扑结构的是【】(A环形结构(B总线结构(C层次结构(D网状结构21.操作系统是一种【】。
四川大学期末考试试题(闭卷(闭卷)
四川大学 期末 考试试题考试试题（（闭卷闭卷））（2016——2017学年第 1 学期学期））课程号:204127030 课序号: 1-4 课程名称:微生物学 任课教师: 孙群、徐恒、张杰、赵建 成绩: 适用专业年级:2015生命学院各专业 学生人数: 137 印题份数: 148 学号: 姓名: 考 生 承 诺我已认真阅读并知晓《四川大学考场规则》和《四川大学本科学生考试违纪作弊处分规定（修订）》，郑重承诺：1、已按要求将考试禁止携带的文具用品或与考试有关的物品放置在指定地点；2、不带手机进入考场；3、考试期间遵守以上两项规定，若有违规行为，同意按照有关条款接受处理。
考生签名考生签名：：一、填空（25分，每空1分）1. 上世纪五十年代我国学者汤飞凡教授分离的 研究成果是一项具有国际领先水平的开创性成果。
2. 在鉴别EMB 培养基上，在反射光下大肠杆菌菌落呈现的颜色是 。
3. 病毒的核酸类型是极其多样的，总的来说，动物病毒以 和 居多，植物病毒以 居多，而噬菌体以 居多。
4. 要获得金黄色葡萄球菌原生质体，可用 酶去掉细胞壁，而要获得酵母菌的原生质体，则要用 酶去掉细胞壁。
5. 列举一种益生菌，中文名： 英文名： 。
6. 常见的食品保存方法包括 、 、 。
7. 葡萄糖效应是指在培养基中葡萄糖没有被利用完之前，大肠杆菌 操纵子一直被阻遏。
8. 利用细菌营养缺陷型的回复突变来检测环境或食品中化学物质致突变性的简便有效方法是 。
9. 抗原一般应同时具备两个特性：免疫原性和免疫反应性。
缺乏 而有 的物质，称为半抗原或不完全抗原。
10. 大多数对蛋白质合成有抑制作用的抗生素利用了原核与真核细胞在 上差异11. Woese 等将生物分为 、 和 三域。
12. 常用的微生物菌种保藏方法包括 、 、 、 。
第 1 页，共 3 页试卷编号：二、下列拉丁文译成中文下列拉丁文译成中文：：（5分，每题1分） 1. Streptococcus pneumoniae2. Salmonella typhimurium3. Bacillus thuringiensis4. Saccharomyces cerevisiae5. Cyanobacteria三、名词解释名词解释（（30分，每题5分）1. 亚病毒2. 灭菌3. 抗原决定簇4. 光复活作用5. 干扰素6. 最低抑制浓度（MIC ）四、问答题问答题（（40分，四选三四选三，，第一题必选第一题必选））1. 木质纤维素尤其是木质素难以被降解，如何理解木生真菌对木质纤维素的降解和利用。
川大期末考试题库及答案
川大期末考试题库及答案一、单项选择题（每题2分，共20分）1. 四川大学的校训是（）。
A. 海纳百川，有容乃大B. 厚德载物，自强不息C. 明德新民，止于至善D. 博学笃志，切问近思答案：A2. 四川大学的前身之一是（）。
A. 华西医科大学B. 成都科技大学C. 四川联合大学D. 成都大学答案：A3. 四川大学的校庆日是每年的（）。
A. 5月4日B. 9月29日C. 10月1日D. 11月11日答案：B4. 四川大学的校歌是（）。
A. 《四川大学校歌》B. 《华西医科大学校歌》C. 《成都科技大学校歌》D. 《四川联合大学校歌》答案：A5. 四川大学的校徽中的主要元素是（）。
A. 书卷和凤凰B. 书卷和龙C. 书卷和熊猫D. 书卷和莲花答案：A6. 四川大学的校花是（）。
A. 梅花B. 荷花C. 桂花D. 牡丹答案：C7. 四川大学的校树是（）。
A. 松树C. 银杏树D. 榕树答案：C8. 四川大学的校史馆位于（）。
A. 望江校区B. 华西校区C. 江安校区D. 龙泉校区答案：A9. 四川大学的图书馆藏书量超过（）万册。
A. 300B. 400C. 500答案：C10. 四川大学的现任校长是（）。
A. 李言荣B. 谢和平C. 李向群D. 张林答案：A二、多项选择题（每题3分，共15分）11. 四川大学的前身包括（）。
A. 四川中西学堂B. 华西协和大学C. 成都工学院D. 成都科技大学答案：ABD12. 四川大学的国家级重点学科包括（）。
A. 数学B. 材料科学与工程C. 化学工程与技术D. 生物学答案：ABCD13. 四川大学的国家级重点实验室包括（）。
A. 国家生物医学材料工程技术研究中心B. 国家高分子材料工程研究中心C. 国家水污染控制工程技术研究中心D. 国家能源新材料工程技术研究中心答案：ABC14. 四川大学的国家级教学团队包括（）。
A. 数学教学团队B. 材料科学与工程教学团队C. 化学工程与技术教学团队D. 生物学教学团队答案：ABCD15. 四川大学的国家级精品课程包括（）。
四川大学期末考试试题(闭卷
四川大学期末考试试题（闭卷、开卷、半开卷）（2007-2008学年第1学期）课程号：30403030 课程名称：计算机图形学（A卷）任课教师：陈蓉，代术成适用专业年级：计算机科学技术学号：姓名：一、单项选择题（本大题共10小题，每小题3分，共30分）提示：在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。
错选、多选或未选均无分1）灰度等级为256级，分辨率为1024*1024的显示模式，至少需要的帧缓存容量为___B____bit。
A、7MB、8MC、10MD、16M2) ___C___是在高于显示分辨率的较高分辨率下用点取样方法计算，然后对几个像素的属性进行平均得到较低分辨率下的像素属性。
实际上是把显示器看成是比实际更细的网格来增加取样率。
A、提高显示分辨率B、图像分割C、过取样（supersampling）D、区域取样（areasampling）3）用一个n位的整数表示一个位串，用它控制线型时，可以n个像素为周期进行重复显示。
若Patten=11100101，而i表示画线程序中的第i个像素，则画线程序中的SETPIXEL（X，Y，COLOR）可改写为___C__A、if(pattern[i%4])setixel(x,y,color);B、if(pattern[i%6])setixel(x,y,color);C、if(pattern[i%8])setixel(x,y,color);D、if(pattern[i%12])setixel(x,y,color);4、点P 的齐次坐标为(8,6,2)，其对应的空间坐标为__D ____。
A 、（8，6，2） B 、（8，6） C 、（4，3，1） D 、（4，3）5)在多边形的逐边裁剪法中,对于某条多边形的边(方向为从端点S 到端点P)与某条裁剪线(窗口的某一边)的比较结果共有以下四种情况,分别需输出一些顶点.请问哪种情况下输出的顶点是错误的____A ____。
四川大学阶段期中期末考试试题(开卷闭卷)
int main()
{
int i, x;
for(i=1; i<=50; i++)
{
x=i;
if(++x%2==0)
if(x%3==0)
if(x%7==0)
cout<<i<<"";
}
cout<<endl;
return0;
}
A.27B.28C.41D.42
18.以下叙述中错误的是（）。
}
A. ABB. ABLMNPC. LBD. LBLMNP
28.下列程序的输出结果是（）。
#include<iostream>
usingnamespacestd;
int main()
{
char str[ ]="quert?", *p=str;
while(putchar(*p++)!='?');
return0;
#include <string.h>
using namespace std;
int main( )
{
char a[80]="AB",b[80]= "LMNP";
int i=0;
strcat(a, b);
while(a[i++] != '\0')
b[i]=a[i];
puts(b);
return 0;
{
int red;
int green;
int blue;
};
C.
四川大学电路考试期末试题及答案3
四川大学期末考试试题(闭B卷) （2007 ——2008 学年第二学期）课程号：30332250 课序号：0 -3 课程名称：电路原理任课教师：成绩：适用专业年级：07电类学生人数：印题份数：学号：姓名：2 题间不留空，一般应题卷分开教务处试题编号：3务必用A4纸打印本题4 页，本页为第2 页教务处试题编号：本题4 页，本页为第3 页教务处试题编号：本题4 页，本页为第4 页教务处试题编号：B 卷答案一、选择题(每个选择题3分,共计15分)1、B ；2、B ；3、D ；4、C ；5、B 二、填空题(每空3分,共计30分)1、40V ；2、1；3、32812U U −=，−+=−25912U U ；2，-14、电阻；5Ω；5、4：－2：1 ； 0.75：1：－16、C.D.1°u sC ()0−7、(e e e )()105102−−−−−tttt t ε三、分析计算（本题共4小题，共计55分）1、（10分）简化电路12IΩ5分I R R x x =+−=1222122分欲使121272×=R x则 R x =2Ω3分2、（15分）设R 左边的戴维南等效网络为电压源oc U 串内阻o R方程 =×+=×+2020)20(5.221010(2ooc 2o oc R U R U6分得V 30oc =U ，Ω=10o R4分得到Ω=30R 时，W 875.16=P5分3、（15分）()() −=−−=−+2j 2j 12j 11331I I I I &&&&10.分()A j221−=I & 5分4、（15分）L 2、C 对三次谐波谐振 3分 得 ()u t t 1123=sin ω V 3分 表V 1读数为2V 1分 ()u t t 21123=+()sin ω V 3分 表V 2读数为3V 1分 ()i t t =21sin3ω A 3分 表A 读数为2A 1分。
大学计算机基础 四川大学期末考试题
（C）数据共享性、数据独立性和冗余度小 （D）数据共享性和数据独立性 40. 关于信息和数据，下面正确的说法是【 】 （A）信息与数据，只有区别没有联系 （B）信息是数据的载体 （C）同一信息用同一数据表示形式 （D）数据处理本质上就是信息处理
信子网组成。 10. 数据库管理系统不属于系统软件。各位老师、同学共同切磋或发送邮件至 chenjiehua@。第14页共32页
四、简答题(每题 5 分，共 20 分)
1. 计算机是由哪五大部件组成的？请简要说明这些部件的具体功能。 2. 计算机系统是由哪两大系统组成的？请简要说明这两大系统的关系。 3. 什么是操作系统？简述操作系统一般应该具有哪五大管理功能？ 4. 请简要说明数据庫系统组成的四个方面。
6. 数字字符“1”的 ASCII 码的十进制表示为 49，那么数字字符“8”的 ASCII 码的十进 制表示为【 】。 （A）56 （B）58 （C）60 （D）54
7. 键盘当中用来输入上档字符的是【 】。 （A）回车键 ENTER
本文档素材取自于各种资料，不是陈杰华原创，欢迎各位老师、同学共同切磋或发送邮件至 chenjiehua@。第2页共32页
二、填空题(每题 1 分，共 30 分)
1. 将各种数据转换成为计算机能处理的形式并输送到计算机中去的设备统称为【 】。 2. 一个非零的无符号二进制整数，若在其右边末尾加上两个“0”形成一个新的无符号二
进制整数，则新的数是原来数的【 】倍。 3. 以国标码为基础的汉字机内码是两字节编码，每个字节的最高位为【 】。 4. 磁盘驱动器属于【 】。 5. 总线分为地址总线、数据总线和【 】总线。 6. 为区分内存中不同存储单元可为每个存储单元分配一个编号，称为内存【 】。 7. 计算机硬件系统由【 】、存储器、输入设备和输出设备五大部件组成。 8. 微机可以配置不同的显示系统，在 CGA、EGA 和 VGA 标准中，显示性能最好的一种是
四川大学2015-2016控制工程基础期末考试题2015A及答案解析
四川大学期末考试试题（闭卷）（2015——2016学年第1学期）A卷课程号：302099030课序号：01/02课程名称：控制工程基础任课教师：傅波、陈昆成绩：适用专业年级：机制13学生人数：280印题份数：学号：姓名：考生承诺我已认真阅读并知晓《四川大学考场规则》和《四川大学本科学生考试违纪作弊处分规定（修订）》，郑重承诺：1、已按要求将考试禁止携带的文具用品或与考试有关的物品放置在指定地点；2、不带手机进入考场；3、考试期间遵守以上两项规定，若有违规行为，同意按照有关条款接受处理。
考生签名：一、选择题（36分）1.机械工程控制论实质上是_________。
A.研究系统输入和输出两者之间的动态关系B.研究机械工程领域中广义系统的动力学问题C.只研究机床在切削加工中的动力学问题D.重点研究制造企业的生产管理问题2.一个闭环的控制系统主要由控制部分和_____组成。
A.放大运算环节B.测量环节C.被控对象D.执行环节3.对控制系统的基本要求是____。
A.系统的稳定性B.响应的快速性C.响应的准确性D.A、B、C都对4.当输入与输出均已知，求出系统的结构和参数，即建立系统的数学模型，此即_____。
A.最优设计问题B.最优控制问题C.系统辨识问题D.系统分析问题5.系统的时间响应按振动来源可分为________。
A.零输入响应与零状态响应B.自由响应与强迫响应C..瞬态响应与稳态响应D.脉冲响应与阶跃响应6.下列哪种措施对提高系统的稳态精度没有效果_______。
A.增加积分环节B.提高系统的开环增益C.增加延时环节D.引入PID校正器7.对于开环最小相位系统，系统闭环稳定的充要条件是：开环频率特性不包围____这一点，幅相频率特性越接近这一点，系统稳定程度_______。
A.(+1,j0)、越差B.(-1,j0)、越差C.(+1,j0)、越好D.(-1,j0)、越好8.若最小相位系统的幅值裕度K g>1，则下列说法正确的是______。
信息安全数学基础期末试卷
B) (a, b, c)
C) (a, c )(b, c) D) (a, b)(b, c)
5. 2017 年 1 月 9 日是星期一（为第 0 天），之后的第 220170109 天是（ ）。
A) 星期二
B) 星期三
C) 星期四
D) 星期五
6. 11 的原根有（ ）个。
A) 1 B) 2
C) 3 D) 4
7. a，b 为互素的整数，则存在唯一的整数 s，t，使得 sa tb 1 。（ ）
8. 设 m 是正整数，如果 ad { bd（mod m），则 a { b（mod m）。（ ）
9. 设 a 是整数，若 x 遍历正整数模 m 的完全剩余系，则 ax 也遍历模 m 的完全剩余系。（ ）
Hale Waihona Puke 10. 设 p，q 是不同的素数，则M( pq) ( p 1)(q 1) 。（
）
11. n 是整数，如果 a2 { b2 (mod n) ，而 n 不整除 (a b) 和 (a b) ，则 n 为合数。（ ）
§ a2 · ¨ ¸1 12. 设 (a, p) 1，则 © p ¹ 。（ ）
13. 设整数 m>1， (a, m) 1，则整数 d 满足 ad { 1 (mod m) 的必要条件是 ordm(a） d 。（ ）
5. 求解 x2≡ 2 (mod 17)。
四、 证明题（25 分，5 分/小题，需写出证明过程）
1. 证明：形如 4k+1 和 4k-1 的素数都有无限个。
2. 证明：对素数 P， P 都是无理数。 3. 证明：设 P1 d P2 d P3 是素数，n 是正整数，若 P1P2P3|n ，则 P1 d n1/3 ， P2 d（n/2）1/2 。
四川大学期末考试试题（闭卷A）
四川大学期末考试试题（闭卷A）（2014——2015学年第一学期）课程号：102105020课序号：01, 02课程名称：英语经贸文写作任课教师：谢关平成绩：适用专业年级：2012级国贸学生人数：160 印题份数：170 学号：姓名：注：1试题字迹务必清晰，书写工整。
本题8 页，本页为第 1 页2 题间不留空，一般应题卷分开教务处试题编号：3务必用A4纸打印本题8 页，本页为第 2 页教务处试题编号：本题8 页，本页为第 3 页教务处试题编号：本题8 页，本页为第 4 页教务处试题编号：V. Fill in the contract form in English with the particulars given in the following letters and/ortelexes(20%):（1）Incoming LetterVancouver，July 25，2014LI DU TEXTILE IMP.EXP CORP.，Beijing，ChinaRe：COTTON BATH TOWELSDear Sirs，A customer of our firm，who is one of the biggest importers in Canada，is in the market for the subject commodity. We would therefore ask you to make us an offer as soon as possible.When offering，kindly quote your lowest priced on the basis of CFR Vancouver including our commission of 3% with indication of colors，assortments，method of packing and any additional information you consider necessary. We shall appreciate it if you arrange for shipment to be made as early as possible by a direct steamer to Vancouver.We are looking forward to receiving your offer.Yours faithfully，VANCOUVER TRADINC CO.，LTD.（2）Outgoing LetterBeijing，August 2，2014VANCOUVER TRADING CO.，LTD.Vancouver，CanadaDear Sirs，Thank you for your letter of July 25 inquiring for COTTON BATH TOWELS. We are glad to send you under separate cover our quotation sheet No. AC-8115 for your reference. Sample cuttings of each design have been dispatched today by air parcel.本题8 页，本页为第 5 页教务处试题编号：学号：姓名We expect our quotation will reach you in due time and assure you of our best and prompt attention to your requirements at any time.Your early reply will be highly appreciated.Yours faithfully，LI DU TEXTILE IMP/EXP CORP.（3）INCOMING TelexAugust 10，2014RE: COTTON BATH TOWELSTKS FR UR LTR AUG 2 AS WELL AS SMPL CUTTINGS OF THE SUBJECT GDS. ON EXAMG UR SMLPLS，OUR CSTMARS HV FAN INTRST IN ART NO.G303 N WISH TO PLACE TLR ORDR FR 1，000 DZ IN WHITE N YELLOW COLOR EQUALLY ASSORTED FR SHPMT OCT/NOV. AS USUAL，OUR SIGHT IRRE - LC WL B OPND IN UR FAVOR 30 DAYS BFR SHPMT TIME. RGDS.（4）Outgoing letterBeijing，August 11，2014VANCOUVER TRADING CO.，LTD.Vancouver，CanadaDear Sirs，Thank you for your order of August 10 for 1，000 dozen COTTON BATH TOWELS. We are now making you the following offer，subject to your confirmation reaching us not later than August 20：1，000 dozen of Art. No. G3030 COTTON BATH TOWELS （details as per your telex of August 10）at CAN $ 24 per dozen CFRC2 Vancouver for shipment from any Chinese mainland port in October. Please note that，there is no direct steamer available for Vancouver in October，we find it only possible to ship the goods with transshipment at Hong Kong. The goods are to be packed in cartons each containing 5/10 dozen at buyers’option. We look forward to your early acceptance.本题8 页，本页为第 6 页教务处试题编号：学号：姓名Yours faithfully，LI DU TEXTILE IMP/EXP CORP.（5）Incoming TelexAugust 15，2014TK U FR UR LTR AUG 11. UR OFR 1，000 DZ COTTON BATH TOWELS ACEPTD. V R NOW ARRANGING WITH OUR BK FR RELEVANT LC. AS TO PACKG，V PREFER CTNS CONTAINING 5 DZ. PLS SEND CONTRACT ASAP. RGDS.SALES CONTRACTCONTRACT NO.98 – 110SELLERS：BUYERS：This Contract is made by and between the Buyers and the Sellers，whereby the Buyers agree to buy and the Sellers agree to sell the under - mentioned commodity according to the terms and conditions stipulated below：Commodity：Specifications：Quantity：Unit Price：Packing：Shipping Mark：Insurance：Time of Shipment：Port of Shipment：本题8 页，本页为第7 页教务处试题编号：学号：姓名本题8 页，本页为第8 页教务处试题编号：。
四川大学阶段期中期末考试试题(开卷闭卷)
2、不带手机进入考场；
3、考试期间遵守以上两项规定，若有违规行为，同意按照有关条款接受处理。
考生签名：
1.某炼油厂根据计划每季度需供应合同单位汽油15万 (吨）、煤油12万 、重油12万 。该厂从A,B两处运回原油提炼，已知两处原油成分如表所示。又如从A处采购原油每 价格（包括运费、下同）为200元,B处原油每 为310元。试求：(a)选择该炼油厂采购原油的最优决策；(b)如A处价格不变,B处降为290元/ ，则最优决策有何改变？
A/%
B/%
含汽油
含煤油
含重油
其他
15
20
50
15
50
30
15
5
2.为确保飞行的安全，飞机上的发动机每半年必须强迫更换进行大修。某维修厂估计某种型号战斗机从下一个半年算起的今后三年内每半年发动机的更换需要量分别为：100,70,80,120,150,140.更换发动机时可以换上新的，也可以用经过大修的旧的发动机。已知每台新发动机的购置费为10万元，而旧发动机的维修有两种方式：快修，每台2万元，半年交货（即本期拆下来送修的下批即可用上）；慢修每台1万元，但需一年交货（即本期拆下来送修的需下下批才能用上）。设该厂新接受该项发动机更换维修任务，又知这种型号战斗机三年后将退役，退役后这种发动机将报废。问在今后三年的每半年内，该厂为满足维修需要各新购，送去快修和慢修的发动机数各多少，使总的维修费用为最省？（将此问题归结为运输问题，只列出产销平衡表与单位运价表，不求数值解）。
四川大学期末考试试题（开卷）
（2016——2017学年第1学期）B卷
课程号：303146020课序号：01课程名称：最优化方法任课教师：曾晓东成绩：
四川大学期末考试试题(闭卷)2018-2019春微积分
四川大学期末考试试题（闭卷）（2018——2019 学年第 2 学期） A 卷课程号：201138040 课序号：01~50 课程名称：微积分（I）-2 任课教师：成绩：第 1 页，共 2 页第 2 页，共 2 页⎨2018-2019 春微积分（I ）-2 A 卷参考答案一．填空题(每小题 4 分，共 5 分)1．曲面 z = 2x 3 - y 2 在点(1,1,1)处的切平面方程是 ( 6x - 2 y - z- 3 = 0 ).解: dz = 6x 2dx - 2 ydy , 6x - 2 y - z - 3 = 02. 设= lnarctan y, 则 x=( 3 ).x = 2, y =1 5解 :∂z = 1 2x - 1 - y =x + y , ∂z =3∂x 2 x 2+y2y 2x21 +x2x 2 + y 2∂x x = 2, y =1 53 设 f ( x , y ) =+ ⎰⎰ x 2+ y 2≤1f (x , y )dxdy ,则⎰⎰x 2+ y2≤1f ( x ,y )dxdy =2π ().3(π-1)解:f ( x , y )dxdy = k = k )dxdy = 2π k π ⎰⎰⎰⎰ 3 +x 2 + y 2 ≤1x 2 + y 2 ≤14. f (x ) 是周期为2π的周期函数， f (x ) = ⎧cos x - 2, -π≤ x < 0, ⎩sin x + 2, 0 < x < π设 s (x ) 是其傅里叶级数的和函数, 则 s (π) = ( -1).2解: s (π) = 1 [ lim f ( x ) + lim f ( x )] = - 12 x →-π+x →π-2 5. 二阶微分方程 y y ''+ y '2= 1 的通解是(y 2 = x 2 + c x + c ).∂z ∂xy 2 = 2 ( x + c )dx = x 2 + ⎰ yy '=x + c ,12二．计算题(每小题 7 分，共 28 分)6.设e z + x - 2xy + z - 1 = 0 确定的函数 z = z ( x , y ) ，∂z ∂ 2z求(1) ∂y, (2)∂y ∂x. ( 0,0) 解：(1)方程两边同时对 y 求偏导：(1) (e z+ 1)∂z- 2x = 0,∂y…………………………………………..(2 分)∂z=2x, .............................. (1 分)∂y e z + 1(2) 方程 两边再同时对 x 求偏导：∂z ∂z ∂ 2z (2) e z⋅ + (e z + 1) - 2 = 0 …………………………..(2 分)∂x ∂y ∂x ∂y把 x =0,y =0 代入原方程可得 z = 0 ，( 0,0) = 2xe z + 1 = 0 , …(1 分)∂ 2 z=2=再代入方程(2), 得 ∂x ∂y e z +11 ................ (1 分)7. 空间闭曲面∑ 由 x 2 + y 2 = 1, z = 0和z = 4 + y 围成, 求它的表面积.解: S = π+⎰x 2 + y 2 =1(4 + y )ds +⎰⎰x 2 + y 2 ≤11dxdy…………………………..(4 分)= π+ 8π+2π=（9 + 2）π ……………………………..…………..(3 分)8. 计算曲线积分 I = ⎰L(e x sin y - 2 y )dx + (e x cos y - 3)dy ,其中 L 为由点 A:(2,0)到点 B;(1,2)再到原点 O (0,0)的折线段.解： ∂P =∂y ∂Q = ∂x ∂ (e x sin y - 2 y ) = e x cos y - 2 ,∂y∂(e x cos y - 3) = e x cos y ,∂x ∴ ∂P ≠ ∂y∂Q ∂x , 曲线积分与路径有关 .................... (1 分)∂z ∂y ( 0,D⎰⎰ =⎰⎰ ( ∂Q - ∂P )dxdy = 2⎰⎰ dxdy = 4,………………………… (3 分)ABOA ∂x ∂y D⎰OA= 20 ⋅ dx + (e x- 3) ⋅ 0 ⋅ dx = 0 , ............................ (1 分) 0因此∴ I =⎰AMOA- ⎰OA= 4- 0 = 4. ………………………… (1 分)⎰⎨⎩xy9. 设可导函数φ(x ) 满足φ(x ) c os x + 2 xφ(t )sin t d t = x +1, 求φ(x ) .解：设 y =φ(x ) , 两边对 x 求导,得y 'cos x + ysin x = 1其通解为 y = c cos x +sin x .………………………… (2 分)………………………… (4 分)因为 x = 0时, y = 1, 得c = 1. 所以 y = cos x + sin x .…………(1 分)三. 解答题(每小题 9 分，共 27 分)⎧ x k y10. 设二元函数 f ( x , y ) = ⎪ 4 x 2 + y 2, ( x , y ) ≠ (0, 0) . ⎪ 0,( x , y ) = (0, 0)(1) 当k 为何值时 f ( x , y ) 在点(0,0)处连续;(2) 当k 为何值时 f ( x , y ) 在点(0,0)处可微. 解：(1)令 x = ρcos θ, y = ρsin θ,lim f ( x , y ) = lim ρk -1 cos θsin θx →0 y →0ρ→0 1+ 3cos 2 θ 因为∀θ, cos θsin θ 有界,1+ 3cos 2 θ所以当k > 1时 f ( x , y ) 在点(0,0)处连续 ............. (3 分)(2)根据偏导数定义f (0, 0) = lim h →0f (0, 0) = lim k →0f (h , 0) - f (0, 0) = 0 hf (0, k ) - f (0,0)= 0 k…………………………… (2 分)讨论极限ρ→0 ρ= lim ρk -2 cos θsinθρ→0 1+3cos2 θ所以当k > 2 时f ( x, y) 在点(0,0)处可微............ (2 分)1+1+∞n =0⎩ 1 21 211. 设 f (x )=1x1+x 2∞+ arctan x , (1)把 f (x ) 展开成 x 的幂级数; (2)求 f (2019) (0) .n n解:= ∑(-1)x ,n =0x ∈(-1,1)…………………… (1 分)∴ x = 1+ x 2x ∑ n =0 (-1)nx 2n = ∑ n =0 (-1)nx 2n+1, x ∈(-1,1)…………………… (2 分)1 1 1 ∞ n 2narctan x = ⎰0 2 dx = ⎰0 ( ∑(-1) xn =0)dx= ∑(-1)nx 2n +1 , x ∈[-1,1]…………………… (3 分)n =02n +1f (x ) = x+ arctan x = ∞2n + 2 (-1)n x 2n +11+ x 2f (2019) (0) = 2019!a2019∑ 2n+1= -2019! 2020= -2020 ⋅ 2018!2019…………………… (1 分)…………………… (2 分)⎧ y ' - 3 y ' + 2 y = - xe x12 求初值问题⎨y (0) =y '(0) = 1的特解.解：对应特征方程为r 2 - 3r + 2 = 0 ，其特征根为r = 1, r = 212所以齐次方程的通解为 y = C e x+ C e 2 x ................................. (3 分)因为λ= 1 是单根, 设非齐次方程特解为 y * = x (ax +b )e x……… (2 分)代入原方程，化简得a =1, b = 1 . 所以 y * = ( 122x 2 + x )e x ， .... (2 分)从而原方程的通解为 y = C e x+ C e 2 x+ ( 1 x 2 + x )e x…………………… (1 分)∞ ∞由y(0) =y '(0) = 1 , 得c1=-1, c2= 2所以初值问题的特解为y = ( 12x 2 +x + 2)e x -e 2 x ........................... (1 分)四. 证明题(7 分)13.证明级数∑n=1(-1)n-11条件收敛于ln 2 .n∞n n ⎰ 1 2∞ n -1 1 ∞ 1证明: ∑(-1)n =1 =∑ 是调和级数, 发散; n =1 …………………… (1 分)又由莱布尼茨判别法,交错级数∑(-1) n =1 n -1 1 收敛, n…………………… (1 分)则∑(-1)n =1 n -1 1 条件收敛. n …………………… (1 分)考虑幂级数 s (x ) = ∑(-1)n =1n -1 1x n n ,收敛半径为 1, ....................................... (1 分)在绝对收敛区间为(-1,1)内,s '(x ) = ∑(-1)n =1 n -11 (x nn ) ' = ∑(-1) n =1 n -1x n -1 = 1 , 1+ x …………………… (3 分) s (x ) =x 1 dx =ln(1 + x ) 0 1+ x因为 x=1 时, 级数收敛,则 s (1) = lim ln(1+ x ) = ln 2. x →1-…………………… (1 分)五．应用题(每小题 9 分，共 18 分)14. 求曲面 x 2 + y 2 - 4x - 4 y + 7 = 0 和平面 x + z = 8 交线上的点到 y 轴的最长距离和最短距离.解：点( x , y , z ) 到 y 轴的距离为. …………………… (2 分)令F = x 2 + z 2 + λ( x 2 + y 2 - 4x - 4 y - 7) + λ( x + z - 8) , (2 分) ⎧F = 2x + λ(2x - 4) + λ = 0⎪ x 1 2 ⎪F y = λ1 (2 y - 4) = 0 F = 2z + λ = 0 …………………… (2 分)⎨ z 2 ⎪F = x 2 + y 2 - 4x - 4 y - z - 7 = 0 ∞ ∞ ∞ ∞ ∞⎪ λ1 ⎪F = x + z - 8 = 0⎩ λ2解得(x , y , z ) = (1, 2, 7)或(3, 2, 5) ....................................................................(1分) 将这两个点分别代入目标函数，可得最大值5 和最小值 .……… (2 分) 34 2∑121 2 2 15. 设空间曲面∑： z 2 = x 2 + y 2 (1 ≤ z ≤ 2部分) ，方向指向外侧，计算曲面积分 I = ⎰⎰ cos( y + z )dydz + y 2dzdx + ( x + z 2 )dxdy .解：作辅助面∑1： z = 1, ( x , y ) ∈ D : x 2 + y 2 ≤ 1 ,方向指向下侧;辅助面∑2： z = 2, ( x , y ) ∈ D : x 2 + y 2 ≤ 4 ,方向指向上侧 (2 分) ∑+∑1+∑2 围成空间闭区域Ω ,方向指向外侧, 根据高斯公式, 有⎰⎰∑+∑ +∑ = ⎰⎰⎰Ω (2 y + 2z )dxdydz = 2⎰⎰⎰Ωzdxdydz = ⎰ ⎰⎰ ⎰2 2 15 2 zdz dxdy = 2π 1 D z 1 z ⋅ z dz = π, 2…………………… (4 分) ⎰⎰∑⎰⎰∑ = -⎰⎰x 2 + y 2 ≤ 1 ( x + 1)dxdy = -π,= ⎰⎰x 2 + y 2 ≤ 4 ( x + 4)dxdy =16π,…………………… (1 分) …………………… (1 分) ∴ I = ⎰⎰1 1 2∑+∑1 +∑2-⎰⎰∑-⎰⎰∑2=15π+π- 16π=-15π.2 2.……… (1 分)。
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四川大学期末考试试题（闭卷、开卷、半开卷）（2007-2008学年第1学期）课程号：30403030 课程名称：计算机图形学（A卷）任课教师：陈蓉，代术成适用专业年级：计算机科学技术学号：姓名：一、单项选择题（本大题共10小题，每小题3分，共30分）提示：在每小题列出的四个备选项中只有一个是符合题目要求的，请将其代码填写在题后的括号内。
错选、多选或未选均无分1）灰度等级为256级，分辨率为1024*1024的显示模式，至少需要的帧缓存容量为___B____bit。
A、7MB、8MC、10MD、16M2) ___C___是在高于显示分辨率的较高分辨率下用点取样方法计算，然后对几个像素的属性进行平均得到较低分辨率下的像素属性。
实际上是把显示器看成是比实际更细的网格来增加取样率。
A、提高显示分辨率B、图像分割C、过取样（supersampling）D、区域取样（areasampling）3）用一个n位的整数表示一个位串，用它控制线型时，可以n个像素为周期进行重复显示。
若Patten=11100101，而i表示画线程序中的第i个像素，则画线程序中的SETPIXEL（X，Y，COLOR）可改写为___C__A、if(pattern[i%4])setixel(x,y,color);B、if(pattern[i%6])setixel(x,y,color);C、if(pattern[i%8])setixel(x,y,color);D、if(pattern[i%12])setixel(x,y,color);4、点P 的齐次坐标为(8,6,2)，其对应的空间坐标为__D ____。
A 、（8，6，2） B 、（8，6） C 、（4，3，1） D 、（4，3）5)在多边形的逐边裁剪法中,对于某条多边形的边(方向为从端点S 到端点P)与某条裁剪线(窗口的某一边)的比较结果共有以下四种情况,分别需输出一些顶点.请问哪种情况下输出的顶点是错误的____A ____。
A ：S 和P 均在可见的一侧,则输出S 和P .B ：S 和P 均在不可见的一侧,则不输出顶点.C ：S 在可见一侧,P 在不可见一侧,则输出线段SP 与裁剪线的交点.D ：S 在不可见的一侧,P 在可见的一侧,则输出线段SP 与裁剪线的交点和P .6)扫描线多边形填充算法中，对于扫描线同各边的交点的处理具有特殊性。
穿过某两条边的共享顶点的扫描线与这两条边的交点数只能计为___B __交点：A 、0 个B 、1个C 、2个D 、3个7、如果观察方向（视线方向）为Z 轴负向，观察向量可设为V=(0,0,-1)，则对场景中的图形表平面可判定其可见性。
令某平面的法向量为N=(A,B,C)。
当___A __时，该平面可判定为后向面（Back-Face ）即是观察时不可见的面。
Ａ、C<=0 Ｂ、C>=0 C 、A>=0 D 、B<=08、多边形面的平面方程为：Ax+By+Cz+D=0。
投影后，若扫描线上起始点的深度值为，则该面的扫描线上所有后继点的深度值计算公式为__B __ A ）z(x+1,y)=z(x,y)+A/C B ）z(x+1,y)=z(x,y)-A/C C ）z(x+1,y)=z(x,y)+C/A D ）z(x+1,y)=z(x,y)-C/A9）当观察光照下的光滑物体表面时，在某个方向上看到高光或强光，这个现象称为__B __ Ａ、漫反射 Ｂ、镜面反射 C 、环境光 D 、折射CDBy Ax y x z ---=),(10）、绘制样条曲线时，如果控制点中的任一个发生了变动，则整条曲线都将受到影响的是_A __曲线： Ａ、自然三次样条 Ｂ、Ｈermite 插值样条 Ｃ、Cardinal 样条 Ｄ、Ｋorchanek-Bartels 样条二、判断题（本大题共5小题，每小题2分，共10分）提示：正确打 ，错误打 ，并分别简述理由。
１、显示处理器的主要任务是将应用程序给出的图形定义数字化为一组像素强度值，并存放在帧缓存中，这个数字化过程称为扫描转换。
对２、绕多边形的边界，计算相邻边界向量的叉乘可识别出该多边形是凸还是凹多边形。
如果叉乘结果全部为正则为凹多边形；若有正有负，则为凸多边形。
错（凸，凹）3、使用查色表可以提供合理的能够同时显示的颜色数，而无须大容量的帧缓冲器。
这时，帧缓冲器中存放的是真正的颜色编码。
错（颜色编码改为索引（或地址））4、某种颜色，在GRB 颜色模型下坐标值（1,0.7,0.8），在CMY 颜色模型下也是（1,0.7,0.8） 错（0,0.3,0.2）5、透视投影变换后，图形中的不平行于观察平面的各组平行线的延长线，能够汇聚成最多3个灭点。
错（可以是无数个灭点）三．计算推导题目（本大题共2小题，每小题10分，共20分）。
1．给定四点P1(0,0,0),P2(1,1,1),P3(2,-1,-1),P4(3,0,0)。
用其作为特征多边形来构造一条三次贝塞尔曲线段，请写出该曲线的参数化表达式，并计算参数为1、2/3时曲线上点的值。
答: 三次贝塞尔曲线的公式为： 4分当t=1时，根据端点性质，它就是P(1)=P1=(1,1,1) 3分 当t=2/3时，x=(1/27)*0+3*(2/3)*(1/9)*1+3*(4/9)*(1/3)*2+(8/27)*3=233,323,223,133,0)()1(3)()1(3)()1()(tt BEN t t t BEN t t t BEN t t BEN =-=-=-=0,,11,000, )0()0()0( )0()0(P BEN P BEN P BEN P BEN P p n n n n n nk n k k =+++==∑= 33,323,213,103,03322120333,)()()()( [0,1]t )1(3)1(3)1( )()(P t BEN P t BEN P t BEN P t BEN P t P t t P t t P t t BENP t p k k k +++=∈+-+-+-==∑=Y=(1/27)*0+3*(2/3)*(1/9)*1+3*(4/9)*(1/3)*(-1)+(8/27)*0= —2/9Z=(8/27)*0+3*(2/3)*(1/9)*1+3*(4/9)*(1/3)*(-1)+(8/27)*0= —2/9 3分P(2/3)=(2,-2/9,-2/9)评分标准：按步骤给分。
若写出计算的表达式后计算结果错误只扣1分。
2．用Liang-Barsky线段裁剪方法，使用窗口（0，0）（2，2）裁剪以下线段，要求写出计算步骤和裁剪结果。
a)线段A（1,-2）B(1，2)解:x1=1, x2=1, y1= -2, y2=2,xmin=0,xmax=2, ymin=0, ymax=2dx=X2-x1=0, dy=y2-y1= 4P1= -dx= 0 q1=x1-xmin= 1P2=dx=0 q2=xmax-x1=1P3= -dy= -4 q3=y1-ymin= -2 t3=1/2P4=dy=4 q4=ymax-y1=4 t4=1 4分P1,p4<0, tmin=MAX(0,t3)=MAX(0,1/2)=1/2 2分P3,P4>0, tmax=MIN(1,t4)=MIN(1,1)=1 2分Tmin<tmax,有裁剪结果，将参数带入x=x1+dx*t,y=y1+dy*t裁剪结果是线段：（1，0）（1，2）2分四．变换题（本大题共3小题，每小题10分，共30分）。
提示：用列向量表示，注意矩阵乘的顺序。
用齐次坐标表示变换矩阵。
不要求计算出最后结果，但是每个矩阵要表示出来。
1．二维空间中，图形绕点(-1，-2)，顺时针旋转50度的变换矩阵。
2．在XOY二维平面坐标系中有点P(4，1)和点O’(3，4)。
现以O’P作为Y’轴正向建立新坐标系X’O’Y’（都是右手坐标系），请写出图形由XOY到X’O’Y’的坐标变换矩阵。
3．设投影参考点为（0，0，d）,投影面为xoy平面，请推导投影变换矩阵。
答案：1．二维空间中，图形绕点(-1，-2)，顺时针旋转50度的变换矩阵。
1 0 -1 cos(-50) –sin(-50) 0 1 0 10 1 -2 sin(-50) cos(-50) 0 0 1 20 0 1 0 0 1 0 0 1评分标准：每一个矩阵写正确得3分，顺序正确1分。
2、在XOY二维平面坐标系中有点P(6，1)和点O’(3，5)。
现以O’P作为X’轴正向来建立新坐标系X’O’Y’（都是右手坐标系），请写出图形由XOY到X’O’Y’的坐标变换矩阵。
计算向量o’p为（3，-4），则单位向量为u=（ 3/5，-4/5），计算得到u=(-4/5,-3/5)3/5 -4/5 0 1 0 -3-4/5 -3/5 0 0 1 -50 0 1 0 0 1评分标准：单位向量计算正确得4分，每一个矩阵写正确得2分，顺序正确2分。
3．设投影参考点为Pprp（0，0，d）,投影面为xoy平面，请推导投影变换矩阵1）当投影中心点是（0，0，d）时，根据两点(Pprp和P)可求得PP’射线的参数方程为X’= (x-0) u+0Y’=(y-0) u+0Z’= (z-d) u + d投影面为xoy平面，则Z’=0，带入可得到u=d/(d-z)X’= x *d/(d-z)Y’= y*d /(d-z)Z’= 0令h=1-z/d，可以得到齐次坐标表示的投影变换矩阵为1，0，0，00，1，0，00，0，0，00，0，-1/d 1评分标准：写出直线参数方程得5分，推导步骤完整，矩阵写正确得5分。
五、编程及分析题（本大题共1小题，每题10分，共10分），1．请根据中点圆生成算法思想，对圆x2+y2=R2，推出第一象限中从y=0到y=x这段弧的生成算法。
要求推导出主要的计算公式，并写出算法（描述性算法）。
提示这一段上，y的变换率比x大。
解答：构造函数F(X,Y)=Y2+X2-R2。
对于圆上的点，F(X,Y)=0；对于圆外的点，F(X,Y)>0；而对于圆内的点，F(X,Y)<0。
1）当前点为（Xi, Yi），下一个代定点为（X i-1,Y i+1），以Y每次增加1，Y i+1= Yi +1，X i-1需要判定。
中点M的坐标为：M(,Xi-0.5, Yi +1)当F(XM, YM)<=0时，取Pu(Xi，Yi +1,)当F(XM, YM)>0时，取Pd(Xi-1，Yi +1,)2）决策参数：d=F(XM,YM)=F(Xi-0.5,Yi+1)=(xi-0.5)2+(yi+1)2-R2当d≤0时，下一点取Pu(Xi，Yi +1)；当d>0时，下一点取Pd(Xi-1，Yi +1)。