《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第7章
微积分答案
International Monetary FundMoldova and the IMF Press Release:IMF Executive Board Completes Second Review Under the Extended Credit Facility and the Extended Fund Facility Arrangements with Moldova, Approves US$79 Million Disbursement April 7, 2011Country’s Policy Intentions DocumentsE-Mail Notification Subscribe or Modify your subscription Moldova: Letter of Intent, Supplementary Memorandum of Economic and Financial Policies, and Technical Memorandum of UnderstandingMarch 24, 2011M OLDOVA:L ETTER OF I N TE N TChişinău, March 24, 2011 Mr. Dominique Strauss-KahnManaging DirectorInternational Monetary Fund700 19th Street NWWashington, DC 20431 USADear Mr. Strauss-Kahn:The economic program supported by the IMF is playing a crucial role in restoring stability and rebuilding confidence in Moldova. With growth significantly exceeding projections in 2010, GDP has broadly recovered to pre-crisis levels. Inflation is under control, and the fiscal deficit has narrowed substantially. These remarkable results were achieved notwithstanding the challenges that the economy faces: fiscal adjustment and promotion of export-led growth require profound structural reforms; rising international food and fuel prices rekindle inflation pressures; job creation lags behind and unemployment still exceeds pre-crisis levels.The program is broadly on track. All quantitative performance criteria for end-September and most indicative targets for end-December 2010 were observed. However, the difficult political environment of 2010 and unforeseen technical complications have taken their toll, and several structural benchmarks under the program were delayed. In the coming period, we will move expeditiously to implement these measures, as well as the new reforms set forth in our agreement with the IMF. The 2011 fiscal budget consistent with the program objectives will be adopted as a prior action for completion of this review. In addition, we have prepared the Annual Progress Report on the implementation of our National Development Strategy and circulated it to the IMF Executive Board for information.In consideration of our strong record of program implementation, we request the completion of the second review of the program supported by the Extended Credit Facility and the Extended Fund Facility arrangements and the associated disbursement of SDR 50 million. As the Executive Board consideration of our request falls in early April 2011, we also request waivers of applicability of the relevant end-March performance criteria. The third program review, assessing performance based on end-March 2011 performance criteria and relevant structural benchmarks, is envisaged for June 2011. Moldova remains committed to improving the well-being of the population through reforms that promote sustainable growth and reduce poverty. In the period ahead, our program will focus on maintaining the targeted pace of fiscal adjustment; reining in inflation pressures; strengthening financial stability of the banking sector; restructuring the energy sector; rolling out the long-awaitededucation and other structural reforms that would support Moldova’s reorientation toward export-led growth.We believe that the policies set forth in the attached Supplementary Memorandum of Economic and Financial Policies (SMEFP) are adequate to achieve these objectives but will take any additional measures that may become appropriate for this purpose. We will consult with the IMF on the adoption of such additional measures in advance of revisions to the policies contained in the SMEFP, in accordance with the Fund’s policies on such consultation. We will provide the Fund with the information it requests for monitoring progress during program implementation. We will also consult the Fund on our economic policies after the expiration of the arrangement, in line with Fund policies on such consultations, while we have outstanding purchases in the upper credit tranches. Sincerely yours,/s/Vladimir FilatPrime MinisterofRepublicMoldovatheGovernmentof/s/ /s/NegruţaVeaceslavValeriu LazărFinanceofDeputy Prime Minister MinisterEconomyMinisterof/s/Dorin DrăguţanuGovernorNational Bank of MoldovaAttachment: Supplementary Memorandum of Economic and Financial PoliciesUnderstandingofMemorandumTechnicalS UPPLEME N TARY M EMORA N DUM OF E CO N OMIC A N D F I N A N CIAL P OLICIESMarch 24, 20111.The present document supplements and updates the Memoranda of Economic and Financial Policies (MEFPs) signed by the authorities of the Republic of Moldova on January 14, 2010 and June 30, 2010. It accounts for recent macroeconomic developments and introduces policy adjustments, as well as additional policies necessary to achieve the objectives of the program. We remain determined to meeting our commitments made previously under the program.I. M ACROECO N OMIC D EVELOPME N TS A N D O UTLOOK2.Growth outperformed expectations in 2010, and the economic expansion is set to continue. Real GDP rebounded by 6.9 percent in 2010, more than offsetting the economic contraction of 6 percent recorded in 2009. We expect the economic growth to return to its sustainable pace of 4½-5 percent in 2011 and thereafter. Expansion of domestic demand, exports, and investment are expected to drive activity in the near term, with tailwinds from trade liberalization reforms, a more favorable external environment, and improving competitiveness.3.Barring severe external shocks, disinflation should continue in 2011-12. Despite adjustment of energy tariffs, depreciation of the leu, and higher excise rates, inflation remained under control at around 8 percent in 2010, while core inflation declined below 5 percent. Under our baseline assumptions for international food and energy prices, we expect that inflation will decline further to 7½ percent in 2011 and about 5 percent by end-2012, the medium-term target set by the NBM. However, we recognize the risk that further surges in international food and energy prices and faster than expected rebound in domestic demand can temporarily push headline inflation above the projected path.4.Strong economic recovery boosted budget revenues and helped improve the fiscal position. In 2010, revenue significantly exceeded the program projections in nominal terms, but underperformed as percent of GDP, mainly due to high contribution to growth of the largely untaxed agriculture. Expenditure targets were also comfortably met, albeit largely due to under-spending of the capital budget caused by capacity constraints. As a result, the cash budget deficit narrowed to 2½ percent of GDP in 2010, far below the program target of5.4 percent of GDP.5.After a sharp drop to single digits in 2009, the external current account deficit widened in 2010 and will remain elevated in 2011. Rising demand for consumer and investment goods has pushed the current account deficit to an estimated 12¾ percent of GDP in 2010. The same demand factors, along with higher costs of energy imports, will likely propel the deficit even higher in 2011. The elevated deficit in 2011 will be largely financed by official assistance, private capital flows, and FDI. As the economy’s borrowing space is filling up quickly, we realize that further external borrowing should proceed at a more measured pace. We expect that from 2013, thanks to our exportpromotion efforts and economic recovery in trading partners, higher exports will more than offset the rise in imports, and the current account deficit would decline towards 10 percent of GDP.6.The situation in the financial sector has improved as well, with domestic credit rebounding and nonperforming loans declining. After the decline of 2009, domestic bank credit expanded by about 13 percent in 2010, and interest rates have declined. Meanwhile, the share of nonperforming loans declined to 13.3 percent, in part reflecting write-offs. Moreover, banks maintain large liquidity and capital buffers, remaining resilient to potential risks.II. R EVISED P OLICY F RAMEWORK FOR 2011-12A. Fiscal Policy7.Building on the better-than-expected fiscal outcome in 2010, the structural fiscal adjustment will stay on course in 2011-12. Our goal is to bring down the structural fiscal deficit excluding grants—the fiscal deficit adjusted for the effects of economic cycles—from 5½ percent of GDP at end-2010 through 4½ percent of GDP in 2011 to 3½ percent of GDP by 2012. This would largely rid the budget from its dependency on exceptional foreign aid and make public finances more resilient to macroeconomic risks. In this context, we will continue to contain the unaffordable public sector wage bill and low priority current spending, while strengthening revenue through selected tax policy measures and improved tax administration. Using the created fiscal space to increase infrastructure investment and provide well-targeted social assistance to the most vulnerable will allow us to achieve our broader development goals.8.As a next step, we will adopt a 2011 budget with a deficit of 1.9 percent of GDP as a prior action. We project that the budget revenue will amount to 37¾ percent of GDP in 2011, on account of continued progress in the tax administration reform, increased excise rates on tobacco and hard liquor—in line with our EU Association agenda—and updates of selected local taxes and fees. Implementation of various structural reforms, described below, will allow us to reduce current expenditure by 1½ percent of GDP to 34½ percent of GDP. At the same time, priority social assistance spending will be safeguarded, and capital expenditure will increase to 5¼ percent of GDP. We will seek to maintain the targeted structural fiscal adjustment in case the economic outlook and budget revenue deviate from our current projections.9.With immediate fiscal pressures easing, structural reforms will help contain the large public sector wage bill while creating space for poverty reduction actions. The significant optimization efforts in the education sector (¶19) will help finance the increase of teachers’ wages planned for September 2011. During 2011, other public wage restraints will remain in place as described in Law 355, as amended in October 2009. The only exception will be made for low-income auxilliary personnel in the budget sector (with salaries below MDL 1500), whose wages will be indexed by 8.5 percent on average from July 1, 2011 to alleviate the impact of higher than expected food and fuel prices and to avoid disincentives to labor market participation. Moreover, public sectoremployment will be capped at 212,000 positions by end-2011, reflecting the effects of the education reforms, while all vacant positions in excess of that level will be eliminated in 2011.10.Greater emphasis will be placed on synchronizing fiscal consolidation efforts at the central and local levels. The local governments will be granted greater control over local tax rates and fees to allow better revenue planning. In particular, by end-March 2011, we will ensure parliamentary passage of the necessary legal amendments to remove ceilings on existing local taxes and fees. This would allow the Chişinău municipality to raise at least MDL 100 million in additional revenues to finance, among other things (discussed in ¶21), its program of granting wage supplements and heating assistance in 2011. The practice of granting these payments will be discontinued at end-2011. The Ministry of Finance will verify compliance with these commitments.11.Going forward, we will continue trimming down current spending while creating sufficient space for the large public investment needs. In 2012, we aim to reduce the budget deficit further to ¾ percent of GDP, mainly through further rationalization of current spending (1 percent of GDP), sustained by structural reforms (¶¶19-22) that will commence in 2011 and bear fruit over the medium term. Ensuring sustainability of public finances in the medium term will also require implementation of the following measures:∙To reduce spending on goods and services, we will persevere with our procurement reform, assisted by the World Bank. The reform, to be phased in during 2011, will lower the budget costs by automating the bids for delivery of goods and services in the government’scentralized procurement agency.∙To improve control over budget planning and execution, we have drafted a law on public finance and accountability which will introduce a rule-based fiscal framework, enhance fiscal discipline, and improve transparency. We expect the law to be passed by Parliament by end-September 2011 and used in the preparation of the 2012 budget.∙To ensure the most effective allocation of capital expenditure, we will review the list of existing and envisaged capital projects, with a view to prioritize execution on the basis oftheir viability and economic growth potential. The review will also take into account pastexecution rates and capacity for implementation.∙To ensure implementation of the recently approved tax compliance strategy, by April 30, 2011, the State Tax Service (STS) will put in place operational plans for the strategyimplementation, including audit, collection of arrears, and taxpayer service activities(structural benchmark). In addition, by September 30, 2011, we will draft and submit toParliament legislation to allow indirect assessment of individuals’ income based on theirassets and other indicators as specified in the compliance strategy. On this basis, byDecember 31, 2011, we will prepare operational plans to strengthen audit, enforcement,outreach to, and education of high-wealth individuals regarding their tax compliance.∙We will reform the outdated mechanism for sick leave benefits. By March 31, 2011, we will amend legislation to assign the financial responsibility for the first day of sick leave to theemployee and the second day to the employer, effective July 1, 2011 (structural benchmark for end-April). Further legal amendments—to accompany the passage of the 2012 budget—will increase the number of sick leave days covered by employers to 3 in 2012, 4 in 2013, and6 in 2014.∙Early retirement privileges will be gradually phased out. By March 31, 2011, we will adopt legislation that, starting July 1, 2011, would raise the statutory retirement age of civilservants, judges, and prosecutors by six months every year until it reaches the regularretirement age (structural benchmark for end-April). This legislation will also extend the requirement to pay social contributions to all persons employed in Moldova in line withbilateral treaties. Another related piece of legislation, also to be passed by March 31, 2011,will put in place a policy of increasing the years of contribution required for full pensioneligibility from 30 to 35 years (and from 20 to 25 years for military and police personnel), by6 months every year, starting July 1, 2011.∙Building on the findings and recommendations of the recent IMF TA mission, we will implement measures to rationalize the use of health care. In particular, from January 1, 2012 we will introduce a copayment of 20 lei for primary care visits for uninsured patients, tomotivate them to enroll into the health insurance system. From January 1, 2013, we willintroduce small copayments for each doctor and hospital visit (5 lei for primary care, 10 leifor specialists, and 20 lei for hospital admissions) for all other categories of patients,including those who currently receive medical services free of charge. This policy will raise revenue and deter the use of unnecessary care, thus reducing the burden on the system. Tothis end, by end-April 2011 we will prepare an action plan detailing needed legislativechanges, technical preparations, and public information campaign.B. Monetary and Exchange Rate Policies12.The N BM’s monetary policy will be focused on achieving its end-2012 inflation objective of 5 ± 1½ percent. Given the fast economic recovery, closing output gap, and inflation pressures from rising international food and energy prices, the NBM’s monetary policy stance will gradually shift from supporting the recovery to addressing inflation risks. Specifically, it should focus on anchoring expectations—thereby countering the second-round effects from surging food and energy prices—and preventing excessive credit expansion. In this context, the NBM’s recent tightening measures—the 100 basis points hike in the policy interest rate and the increase in required reserve ratio from 8 percent to 11 percent— adequately address current inflation concerns. Further tightening should be conditional on marked acceleration of credit growth or rising inflation expectations.13.At the same time, the N BM will continue to strengthen the operational and legal aspects of its monetary policy framework. Consistent with the transition to inflation targeting, theindicative target for reserve money under the program will be discontinued after March 2011. Nevertheless, the NBM will continue to monitor money growth closely as an indicator of the state of domestic demand and sharp sustained moves may warrant policy action. In parallel, the NBM will continue to further enhance its communication, research, and forecasting capacities. As regards the legal framework, by end-September 2011, the NBM will propose amendments to the central bank law to strengthen its independence in line with the international best practice and establish appropriate mechanisms of internal control over NBM’s corporate governance.14.Alongside, the N BM’s exchange rate policies will remain consistent with program objectives. Specifically, NBM interventions in the foreign exchange market will continue to aim at smoothing erratic movements, but not resist sustained depreciation pressures. Should capital inflows exceed program projections, the NBM will accelerate the pace of reserve accumulation to ensure adequate buffers against the still high external vulnerabilities.C. Financial Sector Policy15.To strengthen financial stability, we will address the quasi-fiscal liabilities stemming from recent crisis management efforts. The Government’s decision to shield from losses the depositors of Investprivatbank (IPB) that failed in 2009 was a necessary step to avoid potential panic and deposit runs. However, paying out these deposits by means of a loan from the majority state-owned Banca de Economii (BEM) to IPB—in turn, enabled by a liquidity-providing loan from the NBM—has created a burden on BEM’s balance sheet that is now inhibiting its development. To address this problem, by end-May 2011 the Government will issue to BEM a long-term bond equal to the residual face value of BEM’s loan to IPB by either purchasing this loan or—subject to agreement of BEM’s minority shareholders—recapitalizing the bank. Meanwhile, the NBM will consider a limited extension of its loan to BEM to mitigate the attendant liquidity risk, and will work with BEM and the IPB liquidator to accelerate the sale of IPB assets. The Deposit Guarantee Fund will assume the responsibility for the net cost of the payout to IPB depositors and may introduce an extraordinary deposit insurance premium to gradually reimburse the Government for the cost of the bond issued to BEM.16.To handle future risks better, we aim to put in place the remaining elements of our contingency planning framework. Recent strengthening of the bank resolution framework and the establishment of a high-level Financial Stability Committee (FSC) were followed by signing of a memorandum of understanding (MoU) between key institutions involved in responding to financial emergencies. As a next step, we aim to put in place specific contingency plans for each MoU participant by end-June 2011. These plans will establish a contingency framework based on a clear set of instruments, division of roles, responsibilities, as well as coordination channels between the involved parties.17.Looking ahead, as credit growth picks up speed, the N BM will need to strengthen its bank supervision framework by improving data collection and reducing scope for regulatoryarbitrage. To this end, the NBM, based on best international practices, will develop a new reporting system for commercial banks allowing a more detailed analysis of financial sector data. In addition, by end-September 2011, the NBM and the National Commission for Financial Markets, with assistance from the World Bank, will explore options and make proposals to consolidate all credit institutions—including banks, leasing companies, savings and credit associations, and microfinance institutions—as well as insurance companies and pension funds under a common supervisory framework. Finally, by end-September 2011, the NBM in cooperation with the World Bank will evaluate the feasibility of establishing a public credit bureau to promote information exchange and prudent lending policies by banks.18.Despite earlier delays, measures to strengthen the debt restructuring and contract enforcement frameworks are being developed and will be implemented in the coming months. The NBM has already allowed faster reclassification of restructured loans into lower-risk categories. We will now ensure by end-September 2011 parliamentary passage of the legal amendments described in the SMEFP of June 30, 2010 (¶15), to enhance the speed and predictability of collateral execution by banks and to strengthen incentives for banks to restructure nonperforming loans (structural benchmark). Furthermore, with technical assistance from the World Bank and in consultation with the IMF staff, we will seek to strengthen and simplify other aspects of the insolvency framework. Specific draft legal amendments in this area will be adopted by the Government by March 2012.D. Structural ReformsRaising Efficiency of the Public Sector19.In the coming months, we will roll out the comprehensive reform of the oversized education sector. Its main goals are to eliminate excess capacity, create a leaner and better-equipped education system with adequately trained and paid staff, and provide education that meets demands of the modern economy. The reform will seek class, school, and employment consolidation. A large part of the eventual budget savings and financial assistance from the World Bank will be used to improve school quality, secure transportation for students, and repair school bus routes. Nevertheless, the reform will save about 0.5 percent of GDP on a net permanent basis from 2013 on. Our reform strategy is based on the following elements:∙Class size optimization. By September 1, 2012, we will increase class size to 30-35 students in large schools and 25-30 students in the rest. For this purpose, we will pass legalamendments to eliminate the existing norms prescribed in the Law on Education by end-July 2011. This would reduce the number of teaching positions by 1,736, including 390 positions in 2011, and lead to estimated annual savings of about MDL 94 million.∙Optimization of the school network. Gradual consolidation of the school network through closure of schools with low enrollment and securing transportation of students to nearby“hub” schools will commence this year. Its full implementation during 2011-13 would reducethe number of teaching and non-teaching positions by 2,661 and 1,426 respectively and, when completed, will generate savings of about MDL 136 million a year. We will aim to limit the attendant transportation costs to MDL 61 million per year, and will seek grant assistance from the international financial community to defray this cost.∙Reduction of non-teaching personnel and vacant positions. As a first step, we will immediately freeze hiring of non-teaching staff and eliminate 2,400 vacant positions in thesector. Alongside, we will include in the budget law for 2011 a provision establishing wage bill ceiling for education sector, resulting in all rayons reducing personnel in educationinstitutions on average by 5 percent from their level of end 2010 (5,300 positions nationwide) before academic year 2011/12. These measures would provide savings of MDL 175 million on a full-year basis.∙Increasing flexibility of labor relations in the sector. Local authorities also need support and more flexibility to be able to consolidate schools and classes. By end-July 2011, we willadopt legal amendments to the Labor Code and other enabling legislation to (i) make fixed-term (one year) contracts mandatory for teachers beyond retirement age; and (ii) allow school principals’ hiring and dismissal decisions to be based on business need and performancerather than tenure. Estimated annual savings from this measure amount to MDL 48 million. ∙Rollout of a per-student financing system. Following successful implementation of per-student financing in the pilot rayons of Cauşeni and Rişcani, the system will be expandedstarting January 1, 2012 to 9 additional rayons, as well as municipalities of Chişinău andBalţi. The system will create strong incentives to optimize schools’ financial performance. Its nationwide implementation will take place in 2013.∙Putting social protection costs in education on a means-tested basis. By end-June 2011, in consultation with the World Bank and other partners, we will conduct a thorough review ofall social expenditure in the education budget (scholarships, dormitory assistance, schoolmeals, etc.) to explore options for better targeting of such assistance to the most vulnerablegroups.In consultation with the World Bank, the Government will develop and, by end-March 2011, adopt a detailed action plan to implement this reform.20.We will reform the civil service in a way that increases efficiency without destabilizing the fiscal position. To this end, we have developed descriptions of new job functions and responsibilities for staff in central government administration along with a merit- and performance-based wage system for civil servants. Implementation of this reform will start in October 2011, and will ensure that the reform does not affect the aggregate public sector wage bill as a ratio to GDP. 21.As regards the energy sector, we will strive to achieve a stable framework for payments of current bills, pending a comprehensive sector restructuring strategy to be finalized and implemented in cooperation with the World Bank and other partners. To ensure a stablefunctioning of the sector, the Ministry of Economy, the Chişinău municipality authorities, and the key participants in the energy sector will seek to negotiate in good faith a MoU with the following key elements: (i) a monthly schedule of payments to energy suppliers that is consistent with typical collection lags in Termocom’s receivables during the heating season, (ii) full repayment of current arrears by Termocom before the following heating season; (iii) a mechanism for covering the cash gap arising from collection lags in Termocom or a bank guarantee from the Chişinău municipality backing Termocom’s adherence to the agreed payment schedule; (iv) creditors’ commitment to abstain from blocking bank accounts as long as the MoU is observed. In this context, the Chişinău municipality will budget for and pay in full its remaining debt to Termocom of MDL 64 million by end-March 2011.22.Meanwhile, we will adopt a number of legal and regulatory amendments which would help ensure cost recovery in the heating sector. By end-August 2011, we will adopt the necessary legal and/or regulatory amendments to raise the heating fee for apartments disconnected from central heating from 5 percent to 20 percent of the average heating bill. This increase is in line with regional practices and would mostly affect consumers with relatively high incomes. At the same time, the Ministry of Regional Development and Construction, the Chişinău municipality, Termocom, and the water distributor Apă Canal will seek to put an end to persistent losses caused by under-billing for hot and cold water delivery; other municipalities will seek to resolve this issue as well. And to facilitate timely collection of heating bills, by end-August 2011, we will adopt the necessary legal and/or regulatory amendments introducing a minimum payment of 40 percent of the monthly bill and setting August 1 as the deadline for settling all heating bills for the past heating season.23.With the international investment climate gradually improving, the government will accelerate the efforts to divest its noncore assets. In the first half of 2011 the government, with assistance from IFC, will put in place an advisor to review various options for private sector participation in Moldtelecom. At the same time, by mid-2011, the government will expand the list of state assets subject to privatization. This will pave the way for privatization of other large public companies. By end-September 2011, the government will approach various international financial institutions, seeking an advisor to explore options to divest Air Moldova as soon as possible. Also by end-September 2011, we shall develop a roadmap for the privatization of Banca de Economii, and, if need be, resume the engagement of the privatization advisor.Improving the Business Environment and Removing Barriers for Trade24.The wheat export ban introduced in response to dwindling grain stocks in early 2011 will be abolished as soon as possible, and we will not introduce any new barriers to trade. We plan to abolish this ban by end-April 2011, provided that domestic and regional grain shortages are alleviated. Moreover, we shall refrain from introducing any new tariff or non-tariff barriers to exports. In addition, by end-May 2011 we will conduct an assessment of the existing tariff and non-tariff barriers to trade and their consistency with Moldova’s WTO commitments with regard to market access, and will develop roadmap for their gradual elimination.。
《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第2章
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即 xn
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lim xn 0
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微积分 复旦大学出版社 曹定华主编 课后答案
微积分 复旦大学出版社 曹定华主编 课后答案
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即
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即xn 1 xn
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lim
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微分几何课后习题解答
第二章曲面论§1曲面的概念1.求正螺面={ u ,u , bv }的坐标曲线.解 u-曲线为={u ,u ,bv }=｛0,0，bv｝＋u {,,0}，为曲线的直母线；v-曲线为={,,bv }为圆柱螺线．２．证明双曲抛物面＝｛a（u+v）, b（u-v）,2uv｝的坐标曲线就是它的直母线。
证 u-曲线为={ a（u+）, b（u-）,2u}={ a, b,0}+ u{a,b,2}表示过点{ a, b,0}以{a,b,2}为方向向量的直线;v-曲线为=｛a（+v）, b（-v）,2v｝=｛a, b,0｝+v{a,-b,2}表示过点(a, b,0)以{a,-b,2}为方向向量的直线。
3．求球面=上任意点的切平面和法线方程。
解 = ，=任意点的切平面方程为即 xcoscos + ycossin + zsin - a = 0 ；法线方程为。
4．求椭圆柱面在任意点的切平面方程，并证明沿每一条直母线，此曲面只有一个切平面。
解椭圆柱面的参数方程为x = cos, y = asin, z = t , , 。
所以切平面方程为：，即x bcos + y asin － a b = 0此方程与t无关，对于的每一确定的值，确定唯一一个切平面，而的每一数值对应一条直母线，说明沿每一条直母线，此曲面只有一个切平面。
5．证明曲面的切平面和三个坐标平面所构成的四面体的体积是常数。
证，。
切平面方程为：。
与三坐标轴的交点分别为(3u,0,0),(0,3v,0),(0,0,)。
于是，四面体的体积为：是常数。
§２曲面的第一基本形式1.求双曲抛物面＝｛a（u+v）, b（u-v）,2uv｝的第一基本形式.解,∴ I = 2。
２．求正螺面={ u ,u , bv }的第一基本形式，并证明坐标曲线互相垂直。
解，，，，∴I =，∵Ｆ＝０，∴坐标曲线互相垂直。
３．在第一基本形式为I =的曲面上，求方程为u = v的曲线的弧长。
微积分第七章习题解答
第七章习题解答1.求下列函数的定义域。
()(){}1,:112222≤+--=y x y x D y x z 解：()()(){}1,4,:14ln 222222≥<+-+--=x y x y x D x y x z 解：()()()(){} ,2,1,0,122,:sin 32222±±=+≤+≤+=k k y x k y x D y x z ππ解：()()()[](){}164,:1416ln 422222222<+<---+--=y x y x D yx y x y x z 解：()(){}0,,:115><<--++=x x y x y x D yx yx z 解：()(){},0,:62>≤≤-=x x y y x D yx z 解：()()(){}222222,42,:3arcsin 7y x y x y x D y x y x z >≤+≤---=解：()()()(){}(){}94,11,1410,1,:410ln ln arcsin 82222222<+≤-≤-=>--≤---+-=y x y x y x y x y x y x D y x y x z 解：2.求下列函数的极限。
()()()()()1sin lim 1sin lim 1sin lim 10222222022220==+++++→→→→→uu u y x y x y x y x y x u x x y y 解：()()()()()001lim1lim lim lim limlim 222222222220000=+=+++=+++=++++∞→∞→∞→∞→∞→∞→∞→∞→∞→∞→∞→∞→y yx xy x y x yy x x y x y x y x y x y y y y y y x x x x x x 解：()221sin lim sin lim sin limsin lim 322220000=⋅=⋅=⋅=→→→→→→→→y u uu xy y xy xy x xy xxy y y y y u x x x 解：()022lim limlim4220222222000=⋅+=++→→→→→→yy x xy y x xy y x xy y y y x x x 解：3求下列函数的一阶偏导数。
《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第四章
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《微积分》上册部分课后习题答案
微积分上册 一元函数微积分与无穷级数第2章 极限与连续2.1 数列的极限1.对于数列n x ,若a x k →2(∞→k ),a x k →+12(∞→k ),证明:a x n → (∞→n )． 证. 0>∀ε, a x k →2 (∞→k ), Z K ∈∃∴1, 只要122K k >, 就有ε<-a x k 2; 又因a x k →+12(∞→k ), Z K ∈∃∴2, 只要12122+>+K k , 就有ε<-+a x k 12. 取{}12,2m ax 21+=K K N , 只要N n >, 就有ε<-a x n , 因此有a x n → (∞→n ). 2．若a x n n =∞→lim ，证明||||lim a x n n =∞→，并举反例说明反之不一定成立.证明： a x n n =∞→lim ，由定义有：N ∃>∀,0ε，当N n >时恒有ε<-||a x n又 ε<-≤-||||||a x a x n n对上述同样的ε和N ，当N n >时，都有ε<-||||a x n 成立 ∴ ||||lim a x n n =∞→反之,不一定成立.如取 ,2,1,)1(=-=n x nn显然 1||lim =∞→n n x ，但n n x ∞→lim 不存在.2.2 函数的极限1. 用极限定义证明:函数()x f 当0x x →时极限存在的充要条件是左、右极限各自存在且相等．证: 必要性. 若()A x f x x =→0lim , 0>∀ε, 0>∃δ, 当δ<-<00x x 时, 就有()ε<-A x f . 因而, 当δ<-<00x x 时, 有()ε<-A x f , 所以()A x f x x =+→0lim ; 同时当δ<-<x x 00时, 有()ε<-A x f , 所以()A x f x x =-→0lim .充分性. 若()A x f x x =+→0lim ,()A x f x x =-→0lim . 0>∀ε, 01>∃δ, 当100δ<-<x x 时, 就有()ε<-A x f , 也02>∃δ, 当200δ<-<x x 时, 有()ε<-A x f . 取{}21,m in δδδ=,则当δ<-<00x x 时, 就有()ε<-A x f . 所以()A x f x x =→0lim .2．写出下列极限的精确定义：（1）A x f x x =+→)(lim 0，（2）A x f x =-∞→)(lim ，（3）+∞=+→)(lim 0x f x x ，（4）-∞=+∞→)(lim x f x ，（5）A x f x =+∞→)(lim ．解：（1）设R x U f →)(:0是一个函数，如果存在一个常数R A ∈，满足关系：0,0>∃>∀δε，使得当δ<-<00x x 时，恒有ε<-|)(|A x f ，则称A 是)(x f 当+→0x x 时的极限，记作A x f x x =+→)(lim 0或 )()(0+→=x x A x f . （2）设R f D f →)(:是一函数，其中0,),,()(>>--∞⊃αααR f D .若存在常数R A ∈，满足关系：0)(,0>∈∃>∀R X ε，使得当X x -<时，恒有ε<-|)(|A x f 成立，则称A 是)(x f 当-∞→x 时的极限，记作：A x f x =-∞→)(lim 或 A x f =)()(-∞→x .（3）设R x U f →)(:0是任一函数，若0>∀M ，0>∃δ，使得当δ<-<00x x 时，恒有M x f >)(，则称当+→0x x 时)(x f 的极限为正无穷大，记作+∞=+→)(lim 0x f x x 或 +∞=)(x f )(0+→x x . （4）设R f D f →)(:是一函数，其中R f D ∈>+∞⊃ααα,0),,()(，若存在常数R A ∈，满足关系：0>∀M ，0)(>∈∃R X ，使得当X x >时，恒有M x f -<)(则称当+∞→x 时)(x f 的极限为负无穷大，记作：-∞=+∞→)(lim x f x 或 -∞=)(x f )(+∞→x .（5）设R f D f →)(:是一函数，其中R f D ∈>+∞⊃ααα,0),,()(，若存在常数R A ∈，满足关系：0,0>∃>∀X ε，使得当X x >时，恒有ε<-|)(|A x f 成立，则称A是)(x f 当+∞→x 时的极限，记作：A x f x =+∞→)(lim 或 A x f =)()(+∞→x .2.3 极限的运算法则1.求∑=∞→+⋯++Nn N n 1211lim． 解. ()()⎪⎭⎫ ⎝⎛+-=+=+=+⋯++111212211211n n n n n n n⎪⎭⎫ ⎝⎛+-=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-++⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛-=+⋯++∑=1112111312121122111N N N n Nn 21112lim 211lim1=⎪⎭⎫ ⎝⎛+-=+⋯++∴∞→=∞→∑N nN Nn N 2．求xe e xxx 1arctan11lim110-+→． 解. +∞=+→x x e 10lim , 0lim 10=-→xx e,,21arctan lim 11lim 1arctan11lim 0110110π=-+=-++++→--→→x ee x e e x xxx xxx ,21arctan lim 11lim 1arctan11lim 0110110π=-+=-+---→→→x e e x e e x x xx x x x 21arctan 11lim 110π=-+∴→x e e x xx3．设)(lim 1x f x →存在，)(lim 2)(12x f x x x f x →+=，求)(x f . 解：设 )(lim 1x f x →=A ，则A x x x f ⋅+=2)(2再求极限：A A A x x x f x x =+=⋅+=→→21)2(lim )(lim 211⇒ 1-=A∴ x x xA x x f 22)(22-=+=.4．确定a ,b ,c ，使 0)1(3)1()1(lim 2221=-+-+-+-→x x c x b x a x 成立.解：依题意,所给函数极限存在且 0)1(lim 21=-→x x∴ 0]3)1()1([lim 221=+-+-+-→x c x b x a x ⇒ 2=c∴ 上式左边=])32)(1(11[lim ))1(321(lim 21221++-+--+=-+-+-+→→x x x x b a x x x b a x x])32)(1(1)32([lim 221++---+++=→x x x x b a x同理有 0]1)32([lim 21=--++→x x b x ⇒ 21=b ∴ 163)23)(1(8)1(3lim )32)(1(1)32(21lim221221=++---=++---++-=→→x x x x x x xx a x x 故 2,21,163===c b a 为所求.2.4 极限存在准则1. 设1x =10，n n x x +=+61，( ,2,1=n )．试证数列{n x }的极限存在，并求此极限． 证: 由101=x , 4612=+=x x , 知21x x >. 假设1+>k k x x , 则有21166+++=+>+=k k k k x x x x . 由数学归纳法知, 对一切正整数n , 有1+>n n x x ,即数列{n x }单调减少. 又显然, () ,2,10=>n x n , 即{n x }有界. 故n n x ∞→lim 存在.令a x n n =∞→lim , 对n n x x +=+61两边取极限得a a +=6, 从而有062=--a a ,,3=∴a 或2-=a , 但0,0≥∴>a x n , 故3lim =∞→n n x2．证明数列 nn n x x x x ++=<<+3)1(3,3011收敛，并求其极限．证明：利用准则II ，单调有界必有极限来证明.∴301<<x ，由递推公式33312131213213)1(30111112=++<++=++=++=<x x x x x x∴ 302<<x 同理可证：30<<n x 有界又 03)3)(3(333)1(311112111112>++-=+-=-++=-x x x x x x x x x x∴ 12x x > 同理 23x x > ，… ，1->n n x x ∴数列 }{n x 单调递增，由准则II n n x ∞→lim 存在，设为A ，由递推公式有：AA A ++=3)1(3 ⇒ 3±=A （舍去负数）∴ 3lim =∞→n n x .3．设}{n x 为一单调增加的数列，若它有一个子列收敛于a ，证明a x n n =∞→lim .证明：设}{k n x 为}{n x 的一子列，则}{k n x 也为一单调增加的数列，且a x k k n n =∞→lim对于1=ε，N ∃，当N n >时有1||<-a x k n 从而||1||||||||a a a x a a x x k k k n n n +<+-≤+-=取|}|1|,|,|,max {|1a x x M N n n += ，对一切k n 都有 M x k n ≤|| 有界.由子列有界，且原数列}{n x 又为一单调增加的数列，所以，对一切n 有M x n ≤||有界，由准则II ，数列}{n x 极限存在且a x n n =∞→lim .2.5 两个重要极限1. 求]cos 1[cos lim n n n -++∞→．解： 原式 =21sin 21sin2lim nn n n n -+++-+∞→⎪⎪⎭⎫⎝⎛++=-+=-+-+-+++-=+∞→n n n n n n nn nn nn n 1110212121sin21sin2lim 2. 求)1sin(lim 2++∞→n n π．解. 原式=()()n nn n n nn n -+-=-+++∞→+∞→1sin 1lim )1sin(lim 22ππππ()()()()0111sin 1lim 222=-+⋅-+-+-=+∞→n nn n nnnn πππ3. 求x x xx )1cos 1(sinlim +∞→． 解. 原式=()[]()e t t t tttt tt xt =⎥⎦⎤⎢⎣⎡+=+=→→=22sin 2sin 10212012sin 1lim cos sin lim 令4. 设 ⎩⎨⎧+-=32)cos 1(2)(x x x x f 00≥<x x 求 20)(lim x x f x →. 解： 1lim )(lim 232020=+=++→→x x x x x f x x ，1)cos 1(2lim )(lim 2020=-=--→→x x x x f x x ∴ 1)(lim2=→xx f x .2.6 函数的连续性1. 研究函数()[]x x x g -=的连续性，并指出间断点类型. 解. n x =，Z n ∈ (整数集)为第一类 (跳跃) 间断点.2. 证明方程)0(03>=++p q px x 有且只有一个实根.证. 令()()()0,0,3>∞+<∞-++=f f q px x x f , 由零点定理, 至少存在一点ξ使得()0=ξf , 其唯一性, 易由()x f 的严格单调性可得.3．设⎪⎩⎪⎨⎧≤<-+>=-01),1ln(0 ,)(11x x x e x f x ，求)(x f 的间断点，并说明间断点的所属类型. 解. )(x f 在()()()+∞-,1,1,0,0,1内连续, ∞=-→+111lim x x e,0lim 111=-→-x x e, ()00=f , 因此,1=x 是)(x f 的第二类无穷间断点; (),lim lim 1110--→→==++e ex f x x x()()01ln lim lim 00=+=--→→x x f x x , 因此0=x 是)(x f 的第一类跳跃间断点.4．讨论nx nxn e e x x x f ++=∞→1lim )(2的连续性．解. ⎪⎩⎪⎨⎧<=>=++=∞→0,0,00,1lim)(22x x x x x e e x x x f nxnxn , 因此)(x f 在()()+∞∞-,0,0,内连续, 又()()00lim 0==→f x f x , ()x f ∴在()+∞∞-,上连续.5.设函数),()(+∞-∞在x f 内连续，且0)(lim=∞→xx f x ，证明至少存在一点ξ，使得0)(=+ξξf .证：令x x f x F +=)()(，则01]1)([lim )(lim>=+=∞→∞→x x f x x F x x ，从而0)(>xx F .由极限保号性定理可得，存在01>x 使0)(1>x F ；存在02<x 使0)(2<x F .)(x F 在],[12x x 上满足零点定理的条件，所以至少存在一点ξ使得0)(=ξF ，即0)(=+ξξf .6．讨论函数nnx x x x f 2211lim )(+-=∞→的连续性，若有间断点，判别其类型.解： ⎪⎩⎪⎨⎧-=101)(x f 1||1||1||>=<x x x ，显然 1±=x 是第一类跳跃间断点，除此之外均为连续区间.7．证明：方程)0,0(sin >>+=b a b x a x 至少有一个正根，且不超过b a +. 证明：设b x a x x f --=sin )(，考虑区间],0[b a +0)0(<-=b f ，0))sin(1()(≥+-=+b a a b a f ，当0))sin(1()(=+-=+b a a b a f 时，b a x +=是方程的根；当0))sin(1()(>+-=+b a a b a f 时，由零点定理，至少),0(b a +∈∃ξ使0)(=ξf ，即 0sin =--b a ξξ成立,故原方程至少有一个正根且不超过b a +.2.7 无穷小与无穷大、无穷小的比较1. 当0→x 时，下面等式成立吗？(1))()(32x o x o x =⋅；(2))()(2x o xx o =；(3) )()(2x o x o =． 解. （1）()()()002232→→=⋅x xx o x x o x , ()()()032→=⋅∴x x o x o x （2） ()()()0)(,00)()(2222→=∴→→=x x o x x o x x x o xxx o（3） ()2xx o不一定趋于零, )()(2x o x o =∴不一定成立(当0→x 时) 2. 当∞→x 时，若)11(12+=++x o c bx ax ，则求常数c b a ,,．解. 因为当∞→x 时，若)11(12+=++x o c bx ax , 所以01lim 111lim 22=+++=++++∞→+∞→c bx ax x x c bx ax x x , 故c b a ,,0≠任意.3．写出0→x 时，无穷小量3x x +的等价无穷小量.解： 11lim 1lim lim303630=+=+=+→→→x xx xxx x x x∴ 当0→x ，3x x +～6x第3章 导数与微分3.1 导数概念1. 设函数)(x f 在0x 处可导，求下列极限值． (1)hh x f h x f h )3()2(lim000--+→；(2)000)()(lim 0x x x xf x f x x x --→．解.（1） 原式()()()000000533)3(22)2(lim x f h x f h x f h x f h x f h '=⎥⎦⎤⎢⎣⎡⋅---+⋅-+=→（2） 原式()[]()()()()00000000)(limx f x f x x x x x x f x f x f x x x -'=----=→2．设函数R f →+∞),0(:在1=x 处可导，且),0(,+∞∈∀y x 有)()()(y xf x yf xy f += 试证：函数f 在),0(+∞内可导，且)1()()(f xx f x f '+='． 解：令1==y x ，由()()()y xf x yf xy f +=有()()121f f =得()01=f ．()+∞∈∀,0x ，()()()()()()()()()()xx f f x x f xx f x x f x x f x f x x x x xf x x f x x x f x x f x x f x f x x x x +'=+∆-⎪⎭⎫⎝⎛∆+=∆-⎪⎭⎫ ⎝⎛∆++⎪⎭⎫ ⎝⎛∆+=∆-⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛∆+=∆-∆+='→∆→∆→∆→∆111lim 11lim 1lim lim 0000 故()x f 在()+∞,0内处处可导，且()()()xx f f x f +'='1． 3.设()f x 在(,)-∞+∞内有意义，且(0)0f =，(0)1f '=， 又121221()()()()()f x x f x x f x x ϕϕ+=+，其中22()cos xx x x e ϕ-=+, 求()f x '．解： ()()()()()()()()x x f x x f x x f x x f x x f x f x x ∆-∆+∆=∆-∆+='→∆→∆ϕϕ00lim lim()()()()()()()()()001lim 0lim 00ϕϕϕϕ'+'=∆-∆+∆-∆=→∆→∆x f x f xx x f x x f x f x x ()x e x x x 22cos -+==ϕ4.设函数0)(=x x f 在处可导，且21arctan lim )(0=-→x f x e x，求)0(f '.解：由已知，必有0]1[lim )(0=-→x f x e，从而0)(lim 0=→x f x ，而0)(=x x f 在连续，故0)0(=f .于是)0(1)0()(1lim )(lim 1arctan lim200)(0f xf x f x f x e x x x x f x '=-==-=→→→. 故21)0(='f .5.设)(x f 具有二阶导数，)(,sin )()2(lim )(2x dF t xx f t x f t x F t 求⎥⎦⎤⎢⎣⎡-+=∞→.解： 令t h 1=，则)(2 sin )()2(lim)(0x f x hhxh x f h x f x F t '=⋅-+=→.从而)(2)(2)(x f x x f x F ''+'='，dx x f x x f dx x F x dF )]()([2)()(''+'='=.6．设f 是对任意实数y x ,满足方程 22)()()(xy y x y f x f x f +++= 的函数，又假设1)(lim=→xx f x ,求：（1）)0(f ；（2）)0(f '； （3）)(x f '. 解：（1）依题意 R y x ∈∀,，等式 22)()()(xy y x y f x f y x f +++=+ 成立令0==y x 有 )0(2)0(f f = ⇒ 0)0(=f（2）又 1)(lim=→x x f x ，即 )0(10)0()(lim 0f x f x f x '==--→，∴ 1)0(='f（3）xx f x x f x f x ∆-∆+='→∆)()(lim )(0x x f x x x x x f x f x ∆-∆⋅+∆⋅+∆+=→∆)()()()(lim 220 x x x x x x f x ∆∆⋅+∆⋅+∆=→∆220)()(lim ])([lim 20x x x xx f x ∆⋅++∆∆=→∆ ]1)0(22x x f +=+'=∴ 21)(x x f +='.7．设曲线)(x f y =在原点与x y sin =相切，试求极限 )2(lim 21nf nn ∞→. 解：依题意有 1)0()0(='='f y 且0)0(=f∴ 222)0()2(lim )2(lim 2121=⋅-⋅=⋅∞→∞→n nf n f n nf n n n .8．设函数)(x f 在0=x 处可导且0)0(,0)0(='≠f f ，证明1])0()1([lim =∞→nn f n f .证：n n n n f f n f f n f ])0()0()1(1[lim ])0()1([lim -+=∞→∞→.=10)0(11)0()01(lim )0()0()1(lim ===⋅-+-∞→∞→e ee f nf n f f f n f n n n .1．计算函数baxax xb ab y )()()(= （0,0>>b a ）的导数．解. a xb bx a b a x xb a b a a x b a x a b x b x b a a x x b a b a b y )(1)()()()(ln )(121⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛+⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-⎪⎭⎫⎝⎛+='-- ⎥⎦⎤⎢⎣⎡+-=x b x a a b a x x b a b b a x ln )()()( 2.引入中间变量,1)(2x x u +=计算1111ln 411arctan 21222-+++++=x x x y 的导数dx dy ．解. 引入,1)(2x x u += 得11ln 41arctan 21-++=u u u y ,于是dxdudu dy dx dy ⋅=, 又 ()()4242422111111111141121x x x u u u u du dy +-=+-=-=⎪⎭⎫ ⎝⎛--+++=,21xx dx du +=, 则()22242121121xx x x x x x dx dy ++-=+⋅⎪⎭⎫⎝⎛+-= 3.设y y x +=2，232)(x x u +=，求dudy． 解. dudxdx dy du dy ⋅= , 又()()1223,12212++=+=x x x dx du y dy dx ,得121+=y dx dy , ()x x x du dx ++=21232, 则得()()xx x y du dy +++=2121232 4．已知 2arctan )(),2323(x x f x x f y ='+-=，求=x dx dy ．解：22)23(12)2323arctan()2323()2323(+⋅+-='+-⋅+-'='x x x x x x x f y π43)23(12)2323arctan(02200=+⋅+-='=∴===x x x x x x y dxdy .1. 计算下列各函数的n 阶导数. (1) 6512-+=x x y ; (2) x e y xcos =． 解 （1）⎪⎭⎫⎝⎛+--=611171x x y ，()()()()()()⎥⎦⎤⎢⎣⎡+---=⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛+-⎪⎭⎫⎝⎛-=∴++1161117!1611171n n nn n n x x n x x y （2） ()⎪⎭⎫ ⎝⎛+=⎥⎦⎤⎢⎣⎡-=-='4cos 2sin 21cos 212sin cos πx e x x e x x e y x x x()⎪⎭⎫ ⎝⎛⋅+=⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-⎪⎭⎫ ⎝⎛+=''42cos 24sin 4cos 22πππx ex x e y xx由此推得 ()()⎪⎭⎫ ⎝⎛⋅+=4cos 2πn x eyxnn2. 设x x y 2sin 2=, 求()50y ．解 ()()()()()()()()()()"+'+=248250249150250502sin 2sin 2sin x x C x x C x x y⎪⎭⎫ ⎝⎛⋅+⋅⨯+⎪⎭⎫ ⎝⎛⋅+⋅+⎪⎭⎫ ⎝⎛⋅+=2482sin 2249502492sin 2502502sin 24950250πππx x x x xx x x x x 2sin 212252cos 2502sin 24950250⋅+⋅+-= ()[]x x x x 2cos 1002sin 212252249+-=3. 试从y dy dx '=1, 0≠'y , 其中y 三阶可导, 导出()322y y dy x d '''-=, ()()52333y y y y dy x d '''''-''= 解 y dy dx '=1 ，()()322211y y y y y dy dx y dx d dyx d '''-='⋅'-''=⋅⎪⎪⎭⎫ ⎝⎛'=∴ ()()()()()()52623333313y y y y y y y y y y y dy dx y y dx d dy x d '''''-''='⋅'''⋅'⋅''+''''-=⋅⎪⎪⎭⎫ ⎝⎛'''-=∴ 4. 设()x f 满足()()0 312≠=⎪⎭⎫⎝⎛+x xx f x f , 求()()()()x f x f x f n ,,'．解 以x 1代x ，原方程为()x x f x f 321==⎪⎭⎫ ⎝⎛，由()()⎪⎪⎩⎪⎪⎨⎧=+⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛+x x f x f x x f x f 321 312，消去⎪⎭⎫⎝⎛x f 1，求得()x x x f 12-=，且得()212xx f +='，()()()()2!111≥-=++n x n x f n n n . 5．设()arcsin f x x =，试证明()f x 满足 （1）2(1)()()0x f x xf x '''--= （2） ,1,0,0)()()12()()1()(2)1()2(2==-+--++n x f n x xf n x f x n n n（3）求()(0)n f解 （1）()211x x f -='，()()()22221112211xx xx x x x f --=-⋅--=''， ()()()012='-''-∴x f x x f x ，（2）上式两边对x 求n 阶导数得()()[]()()[]()()()()()()()()()()()()()()()[]x f n x xf x f n n x f x n x f x x f x x f x n n n n n nn⋅⋅+-⋅-⋅---+-='-''-=+++1221211021222即 ()()()()()()()()01212122=-+--++x f nx xf n x f xn n n 。
微积分曹定华版课后题答案习题详解
第二章习题2-11. 试利用本节定义5后面的注3证明：若lim n →∞x n =a ,则对任何自然数k ,有lim n →∞x n +k =a .证：由lim n n x a →∞=,知0ε∀>,1N ∃,当1n N >时,有取1N N k =-,有0ε∀>,N ∃,设n N >时此时1n k N +>有 由数列极限的定义得 lim n k x x a +→∞=.2. 试利用不等式A B A B -≤-说明：若lim n →∞x n =a ,则lim n →∞∣x n ∣=|a|.考察数列x n =-1n ,说明上述结论反之不成立.证：而 n n x a x a -≤- 于是0ε∀>,,使当时，有N n N ∃>n n x a x a ε-≤-< 即 n x a ε-<由数列极限的定义得 lim n n x a →∞=考察数列 (1)nn x =-,知lim n n x →∞不存在,而1n x =,lim 1n n x →∞=,所以前面所证结论反之不成立;3. 利用夹逼定理证明：1 lim n →∞222111(1)(2)n n n ⎛⎫+++ ⎪+⎝⎭=0； 2 lim n →∞2!n n =0. 证：1因为222222111112(1)(2)n n n n n n n n n n++≤+++≤≤=+ 而且 21lim0n n →∞=,2lim 0n n→∞=, 所以由夹逼定理,得222111lim 0(1)(2)n n n n →∞⎛⎫+++= ⎪+⎝⎭. 2因为22222240!1231n n n n n<=<-,而且4lim 0n n →∞=,所以,由夹逼定理得4. 利用单调有界数列收敛准则证明下列数列的极限存在. 1 x n =11n e +,n =1,2,…；2 x 1x n +1n =1,2,…. 证：1略;2因为12x <,不妨设2k x <,则故有对于任意正整数n ,有2n x <,即数列{}n x 有上界,又 1n n x x +-=,而0n x >,2n x <,所以 10n n x x +-> 即 1n n x x +>, 即数列是单调递增数列;综上所述,数列{}n x 是单调递增有上界的数列,故其极限存在;习题2-21※. 证明：0lim x x →fx =a 的充要条件是fx 在x 0处的左、右极限均存在且都等于a .证：先证充分性：即证若0lim ()lim ()x x x x f x f x a -+→→==,则0lim ()x x f x a →=. 由0lim ()x x f x a -→=及0lim ()x x f x a +→=知： 10,0εδ∀>∃>,当010x x δ<-<时,有()f x a ε-<,20δ∃>当020x x δ<-<时,有()f x a ε-<;取{}12min ,δδδ=,则当00x x δ<-<或00x x δ<-<时,有()f x a ε-<, 而00x x δ<-<或00x x δ<-<就是00x x δ<-<, 于是0,0εδ∀>∃>,当00x x δ<-<时,有()f x a ε-<, 所以 0lim ()x x f x a →=.再证必要性：即若0lim ()x x f x a →=,则0lim ()lim ()x x x x f x f x a -+→→==, 由0lim ()x x f x a →=知,0,0εδ∀>∃>,当00x x δ<-<时,有()f x a ε-<,由00x x δ<-<就是 00x x δ<-<或00x x δ<-<,于是0,0εδ∀>∃>,当00x x δ<-<或00x x δ<-<时,有()f x a ε-<.所以 0lim ()lim ()x x x x f x f x a -+→→== 综上所述,0lim x x →fx =a 的充要条件是fx 在x 0处的左、右极限均存在且都等于a .2. 1 利用极限的几何意义确定0lim x → x 2+a ,和0lim x -→1e x； 2 设fx = 12e ,0,,0,xx x a x ⎧⎪<⎨⎪+≥⎩,问常数a 为何值时,0lim x →fx 存在.解：1因为x 无限接近于0时,2x a +的值无限接近于a ,故2lim()x x a a →+=.当x 从小于0的方向无限接近于0时,1e x 的值无限接近于0,故10lim e 0xx -→=. 2若0lim ()x f x →存在,则00lim ()lim ()x x f x f x +-→→=, 由1知 22lim ()lim()lim()x x x f x x a x a a +--→→→=+=+=, 所以,当0a =时,0lim ()x f x →存在;3. 利用极限的几何意义说明lim x →+∞sin x 不存在.解：因为当x →+∞时,sin x 的值在-1与1之间来回振摆动,即sin x 不无限接近某一定直线y A =,亦即()y f x =不以直线y A =为渐近线,所以lim sin x x →+∞不存在;习题2-31. 举例说明：在某极限过程中,两个无穷小量之商、两个无穷大量之商、无穷小量与无穷大量之积都不一定是无穷小量,也不一定是无穷大量.解：例1：当0x →时,tan ,sin x x 都是无穷小量,但由sin cos tan xx x=当0x →时,cos 1x →不是无穷大量,也不是无穷小量;例2：当x →∞时,2x 与x 都是无穷大量,但22xx=不是无穷大量,也不是无穷小量; 例3：当0x +→时,tan x 是无穷小量,而cot x 是无穷大量,但tan cot 1x x =不是无穷大量,也不是无穷小量;2. 判断下列命题是否正确：1 无穷小量与无穷小量的商一定是无穷小量；2 有界函数与无穷小量之积为无穷小量；3 有界函数与无穷大量之积为无穷大量；4 有限个无穷小量之和为无穷小量；5 有限个无穷大量之和为无穷大量；6 y =x sin x 在-∞,+∞内无界,但lim x →∞x sin x ≠∞；7 无穷大量的倒数都是无穷小量；8 无穷小量的倒数都是无穷大量. 解：1错误,如第1题例1； 2正确,见教材§定理3；3错误,例当0x →时,cot x 为无穷大量,sin x 是有界函数,cot sin cos x x x =不是无穷大量；4正确,见教材§定理2；5错误,例如当0x →时,1x 与1x -都是无穷大量,但它们之和11()0x x+-=不是无穷大量；6正确,因为0M ∀>,∃正整数k ,使π2π+2k M >,从而ππππ(2π+)(2π+)sin(2π+)2π+2222f k k k k M ==>,即sin y x x =在(,)-∞+∞内无界,又0M ∀>,无论X 多么大,总存在正整数k ,使π>k X ,使(2π)πsin(π)0f k k k M ==<,即x →+∞时,sin x x 不无限增大,即lim sin x x x →+∞≠∞；7正确,见教材§定理5；8错误,只有非零的无穷小量的倒数才是无穷大量;零是无穷小量,但其倒数无意义; 3. 指出下列函数哪些是该极限过程中的无穷小量,哪些是该极限过程中的无穷大量. 1 fx =234x -,x →2; 2 fx =ln x ,x →0+,x →1,x →+∞； 3 fx = 1e x,x →0+,x →0-； 4 fx =2π-arctan x ,x →+∞;5 fx =1x sin x ,x →∞; 6 fx = 21xx →∞. 解：122lim(4)0x x →-=因为,即2x →时,24x -是无穷小量,所以214x -是无穷小量,因而234x -也是无穷大量; 2从()ln f x x =的图像可以看出,1lim ln ,limln 0,lim ln x x x x x x +→→+∞→=-∞==+∞,所以,当0x +→时,x →+∞时,()ln f x x =是无穷大量；当1x →时,()ln f x x =是无穷小量;3从1()e x f x =的图可以看出,110lim e ,lim e 0x xx x +-→→=+∞=, 所以,当0x +→时,1()e xf x =是无穷大量； 当0x -→时,1()e xf x =是无穷小量;4πlim(arctan)02xx→+∞-=,∴当x→+∞时,π()arctan2f x x=-是无穷小量;5当x→∞时,1x是无穷小量,sin x 是有界函数,∴1sin xx是无穷小量;6当x→∞时,21x是无穷小量,∴;习题2-41.若limx x→fx存在,limx x→gx不存在,问limx x→fx±gx,limx x→fx·gx是否存在,为什么解：若limx x→fx存在,limx x→gx不存在,则1limx x→fx±gx不存在;因为若limx x→fx±gx存在,则由()()[()()]g x f x f x g x=--或()[()()]()g x f x g x f x=+-以及极限的运算法则可得limx x→gx,与题设矛盾;2limx x→fx·gx可能存在,也可能不存在,如：()sinf x x=,1()g xx=,则limsin0xx→=,1limx x→不存在,但limx x→fx·gx=1lim sin0xxx→=存在;又如：()sinf x x=,1()cosg xx=,则π2limsin1xx→=,π21limcosx x→不存在,而0limx x→fx·gxπ2lim tanxx→=不存在;2. 若limx x→fx和limx x→gx均存在,且fx≥gx,证明limx x→fx≥limx x→gx.证：设limx x→fx=A,limx x→gx=B,则0ε∀>,分别存在1δ>,2δ>,使得当010x xδ<-<时,有()A f xε-<,当020x xδ<-<时,有()g x Bε<+令{}12min,δδδ=,则当0x xδ<-<时,有从而2A Bε<+,由ε的任意性推出A B≤即00lim()lim()x x x xf xg x→→≤.3. 利用夹逼定理证明：若a1,a2,…,a m为m个正常数,则limn →∞nma ++=A , 其中A =max{a 1,a2,…,a m }.n n n m a m A ≤++≤,即而lim n A A →∞=,1lim nn mA A →∞=,由夹逼定理得nm n a A ++=.4※. 利用单调有界数列必存在极限这一收敛准则证明：若x1＝,x 2x n +1=1,2,…,则lim n →∞x n 存在,并求该极限.证：因为12x x ==有21x x >今设1k k x x ->,则1k k x x -=>=,由数学归纳法知,对于任意正整数n有1n n x x +>,即数列{}n x 单调递增;又因为12x =<,今设2k x <,则12k x -=<=,由数学归纳法知,对于任意的正整数 n 有2n x <,即数列{}n x 有上界,由极限收敛准则知lim n n x →∞存在;设lim n n x b →∞=,对等式1n x +=两边取极限得b =即22b b =+,解得2b =,1b =-由极限的保号性,舍去,所以lim 2n n x →∞=.5. 求下列极限：1 lim n →∞33232451n n n n n +++-+；2 lim n →∞1cos n ⎡⎤⎛⎢⎥⎝⎣⎦； 3 lim n →∞4 limn →∞11(2)3(2)3n nn n ++-+-+； 5 lim n →∞1112211133n n ++++++. 解：1原式=23232433lim 11155nn n n n n→∞++=+-+；2因为lim(10n →∞=,即当n →∞时,1是无穷小量,而cos n 是有界变量,由无穷小量与有界变量的乘积是无穷小量得：lim (10n n →∞⎡⎤=⎢⎥⎣⎦；322lim(n n n→∞=而lim 0nn→∞→∞==, 2n n →∞∴==∞；41111121(1)()(2)31333limlim2(2)33(1)()13nn n n n n n n n n ++→∞→∞++-+-+==-+-+； 5111111()21111114[1()]42222lim lim lim 1111311()3[1()]3333113n n n n n n n n n ++→∞→∞→∞++-+++--===+++---.6. 求下列极限： 1 3limx →239x x --； 2 1limx →22354x x x --+； 3 lim x →∞3426423x x x ++;4 2limx π→sin cos cos 2x xx -; 5 0lim h →33()x h x h+-; 6 3lim x→7 1lim x →21n x x x n x +++--; 8 lim x →∞sin sin x x x x +-；9 lim x →+∞ 10 1lim x →313()11x x---； 11 0lim x →21(sin )x x.解：23333311(1)limlim lim 9(3)(3)36x x x x x x x x x →→→--===--++2211lim(54)0,lim(23)1x x x x x →→-+=-=-3344226464lim lim 03232x x x x x x x x→∞→∞++==++； 4π2ππsincos sin cos 22lim1cos 2cos πx x xx →--==-； 5[]223300()()()()lim limh h x h x x h x h x x x h x h h→→⎡⎤+-+++++-⎣⎦= 222lim ()()3h x h x h x x x →⎡⎤=++++=⎣⎦；633(23)92)x x x →→+-=343x x →→===；72211(1)(1)(1)limlim 11n n x x x x x n x x x x x →→+++--+-++-=--1123(1)2n n n =++++=+； 8sin lim0x x x →∞=无穷小量1x与有界函数sin x之积为无穷小量sin 1sin lim lim 1sin sin 1xx x x x x xx x x→∞→∞++∴==--； 922limlimx x→+∞=limlim1x x ===；101lim x →313()11x x---231(1)3lim 1x x x x →++-=- 11当0x →时,2x 是无穷小量,1sinx是有界函数,∴它们之积21sinx x 是无穷小量,即201lim sin 0x x x →⎛⎫= ⎪⎝⎭;习题2-5求下列极限其中a ＞0,a ≠1为常数： 1. 0limx →sin 53x x; 2. 0lim x →tan 2sin 5xx ; 3. 0lim x →x cot x ;4. 0lim x→; 5. 0lim x →2cos5cos 2x x x -; 6. lim x →∞1xx x ⎛⎫⎪+⎝⎭; 7. 0lim x →()cot 13sin xx +; 8. 0lim x →1x a x-; 9. 0lim x →x x a a x --;10. lim x →+∞ln(1)ln x x x +-; 11. lim x →∞3222xx x -⎛⎫⎪-⎝⎭; 12.lim x →∞211xx ⎛⎫+ ⎪⎝⎭; 13. 0limx →arcsin x x ; 14. 0lim x →arctan xx; .解：1. 000sin 55sin 55sin 55lim lim lim 335353x x x x x x x x x →→→===；2. 000tan 2sin 221sin 25lim lim lim sin 5cos 2sin 55cos 22sin 5x x x x x x x x x x x x x→→→== 0205021sin 252lim lim lim 5cos 22sin 55x x x x x x x x →→→==； 3. 0000lim cotlim cos lim limcos 1cos01sin sin x x xx x xx x x x x x →→→→=⋅==⨯=；4. 0000sin22limlim22x x x x x x x→→→→=== 0sin2221222xx →===； 5. 2200073732sin sin sin sin cos5cos 2732222lim lim lim (2)732222x x x x x x x x x x x x x →→→⎡⎤-⎢⎥-==-⋅⋅⋅⋅⎢⎥⎢⎥⎣⎦0073sin sin 212122limlim 732222x x x x x x →→=-⋅=-；6. 111lim lim lim 111e (1)xxx x x x x x x x x →∞→∞→∞⎛⎫ ⎪⎛⎫=== ⎪ ⎪++⎝⎭ ⎪+⎝⎭； 7. 3cos cos 1cot sin 3sin 0lim(13sin )lim(13sin )lim (13sin )xx xxx x x x x x x →→→⎡⎤+=+=+⎢⎥⎣⎦8.令1xu a =-,则log (1)a x u =+,当0x →时,0u →,111ln log elimlog (1)a ua u a u →===+. 9. 000(1)(1)11lim lim lim x x x x x x x x x a a a a a a x x xx ---→→→⎛⎫------==+ ⎪-⎝⎭ 利用了第8题结论01limln x x a a x→-=； 10. ln(1)ln 11limlim lnx x x x xx x x→+∞→+∞+-+=⋅ 1111lim ln(1)lim lim ln(1)0x x x x x x x→+∞→+∞→+∞=+=+=； 11. 22223211lim lim 1lim 1222222x xxxxx x x x x x x --→∞→∞→∞⎡⎤-⎛⎫⎛⎫⎛⎫=+=+⎢⎥ ⎪ ⎪ ⎪---⎝⎭⎝⎭⎝⎭⎢⎥⎣⎦1232lim e 22xx x x -→∞-⎛⎫∴= ⎪-⎝⎭； 12. 1221222111ln (1)lim ln(1)2211lim(1)lim (1)lim eex x xxx xx x x xx x x x x →∞⎡⎤++⎢⎥⎣⎦→∞→∞→∞⎡⎤+=+==⎢⎥⎣⎦2121lim lim ln(1)0lne 0e e e 1xx x x x→∞→∞+⋅====；13.令arcsin x u =,则sin x u =,当0x →,0u →,000arcsin 1limlim 1sin sin limx u u x u u x u u→→→===；14.令arctan x u =,则tan x u =,当0x →,0u →,00000arctan 1lim lim lim cos lim limcos 1sin tan sin x u u u u x u u u u u xu u u→→→→→====. 习题2-61. 证明： 若当x →x 0时,αx →0,βx →0,且αx ≠0,则当x →x ０时,αx ～βx 的充要条件是0lim x x →()()()x x x αβα-＝０. 证：先证充分性.若0lim x x →()()()x x x αβα-＝０,则0lim x x →()(1)()x x βα-＝0, 即0()1lim 0()x x x x βα→-=,即0()lim 1()x x x x βα→=. 也即0()lim 1()x x x x αβ→=,所以当0x x →时,()()x x αβ. 再证必要性：若当0x x →时,()()x x αβ,则0()lim 1()x x x x αβ→=, 所以0lim x x →()()()x x x αβα-＝0lim x x →()(1)()x x βα-＝0()1lim ()x x x x βα→-=011110()lim ()x x x x αβ→-=-=. 综上所述,当x →x 0时,αx ～βx 的充要条件是0lim x x →()()()x x x αβα-＝０. 2. 若βx ≠0,0lim x x →βx =0且0lim x x →()()x x αβ存在,证明0lim x x →αx =0. 证：0000()()lim ()lim ()lim lim ()()()x x x x x x x x x x x x x x x αααββββ→→→→==0()lim 00()x x x x αβ→== 即 0lim ()0x x x α→=. 3. 证明： 若当x →0时,fx =ox a ,gx =ox b ,则fx ·gx ＝o a b x+,其中a ,b 都大于0,并由此判断当x →0时,tan x －sin x 是x 的几阶无穷小量.证: ∵当x →0时, fx =ox a ,gx =ox b ∴00()()lim(0),lim (0)a bx x f x g x A A B B x x →→=≠=≠ 于是: 0000()()()()()()lim lim lim lim 0a b a b a b x x x x f x g x f x g x f x g x AB x x x x x +→→→→⋅=⋅=⋅=≠ ∴当x →0时, ()()()a b f x g x O x +⋅=,∵tan sin tan (1cos )x x x x -=-而当x →0时, 2tan (),1cos ()x O x x O x =-=,由前面所证的结论知, 3tan (1cos )()x x O x -=,所以,当x →0时,tan sin x x -是x 的3阶无穷小量.4. 利用等价无穷小量求下列极限：1 0lim x →sin tan ax bx b ≠0；2 0lim x →21cos kx x-； 3 0lim x→； 4 0lim x→5 0lim x →arctan arcsin x x ；6 0lim x →sin sin e e ax bx ax bx-- a ≠b ； 7 0limx →ln cos 2ln cos3x x ； 8 设0lim x →2()3f x x-＝100,求0lim x →fx . 解 00sin (1)lim lim .tan x x ax ax a bx bx b→→== 8由20()3lim 100x f x x →-=,及20lim 0x x →=知必有0lim[()3]0x f x →-=, 即 00lim[()3]lim ()30x x f x f x →→-=-=, 所以 0lim ()3x f x →=. 习题2-7１.研究下列函数的连续性,并画出函数的图形：1 fx = 31,01,3,12;x x x x ⎧+≤<⎨-≤≤⎩ 2 fx ＝,111,1 1.x x x x -≤<⎧⎨<-≥⎩，或 解: 1300lim ()lim(1)1(0)x x f x x f ++→→=+== ∴ fx 在x =0处右连续,又11lim ()lim(3)2x x f x x ++→→=-= ∴ fx 在x =1处连续.又 22lim ()lim(3)1(2)x x f x x f --→→=-== ∴ fx 在x =2处连续.又fx 在0,1,1,2显然连续,综上所述, fx 在0,2上连续.图形如下:图2-12 11lim ()lim 1x x f x x --→→==∴ fx 在x =1处连续.又11lim ()lim 11x x f x -+→-→-== 故11lim ()lim ()x x f x f x -+→-→-≠ ∴ fx 在x =-1处间断, x =-1是跳跃间断点.又fx 在(,1),(1,1),(1,)-∞--+∞显然连续.综上所述函数fx 在x =-1处间断,在(,1),(1,)-∞--+∞上连续.图形如下:图2-22. 说明函数fx 在点x 0处有定义、有极限、连续这三个概念有什么不同又有什么联系 略.3.函数在其第二类间断点处的左、右极限是否一定均不存在试举例说明.解:函数在其第二类间断点处的左、右极限不一定均不存在. 例如0(),010x x f x x x x ≤⎧⎪==⎨>⎪⎩是其的一个第二类间断点,但00lim ()lim 0x x f x x --→→==即在0x =处左极限存在,而001lim ()lim x x f x x++→→==+∞,即在0x =处右极限不存在. 4.求下列函数的间断点,并说明间断点的类型：1 fx = 22132x x x -++；2 fx ＝sin sin x x x+； 3 fx = ()11x x+； 4 fx = 224x x +-； 5 fx = 1sinx x . 解: 1由2320x x ++=得x =-1, x =-2∴ x =-1是可去间断点,x =-2是无穷间断点.2由sin x =0得πx k =,k 为整数.∴ x =0是跳跃间断点.4由x 2-4=0得x =2,x =-2.∴ x =2是无穷间断点,x =-2是可去间断点. 5 001lim ()lim sin 0,()x x f x x f x x→→==在x =0无定义 故x =0是fx 的可去间断点.5.适当选择a 值,使函数fx = ,0,,0x e x a x x ⎧<⎨+≥⎩在点x =0处连续.解: ∵f 0=a ,要fx 在x =0处连续,必须00lim ()lim ()(0)x x f x f x f +-→→==. 即a =1.6※.设fx = lim x →+∞x xx x a a a a ---+,讨论fx 的连续性. 解: 22101()lim lim sgn()10100x x xx x x a a x a aa f x x x a a a x --→+∞→+∞-<⎧--⎪====>⎨++⎪=⎩ 所以, fx 在(,0)(0,)-∞+∞上连续,x =0为跳跃间断点. 7. 求下列极限：1 2lim x →222x x x +-； 2 0lim x→； 3 2lim x →ln x -1; 4 12lim x →5 lim x e→ln x x . 解: 222222(1)lim 1;2222x x x x →⨯==+-+- 习题2-81. 证明方程x 5-x 4-x 2-3x =1至少有一个介于1和2之间的根.证: 令542()31f x x x x x =----,则()f x 在1,2上连续,且 (1)50f =-<, (2)50f =>由零点存在定理知至少存在一点0(1,2),x ∈使得0()0f x =.即 542000031x x x x ---=, 即方程54231x x x x ---=至少有一个介于1和2之间的根.2. 证明方程ln １＋e x -2x =0至少有一个小于1的正根.证: 令()ln(1)2e x f x x =+-,则()f x 在(,)-∞+∞上连续,因而在0,1上连续, 且 0(0)ln(1)20ln 20e f =+-⨯=>由零点存在定理知至少存在一点0(0,1)x ∈使得0()0f x =.即方程ln(1)20e xx +-=至少有一个小于1的正根.3※. 设fx ∈C -∞,+∞,且lim x →-∞fx =A , lim x →+∞fx =B , A ·B ＜0,试由极限及零点存在定理的几何意义说明至少存在一点x 0∈－∞,＋∞,使得fx 0＝０.证: 由A ·B <0知A 与B 异号,不防设A >0,B <0由lim ()0,lim ()0x x f x A f x B →-∞→+∞=>=<,及函数极限的保号性知,10X ∃>,使当1x X <-,有()0,f x >20X ∃<,使当2x X >时,有()0f x <.现取1x a X =<-,则()0f a >,2x b X =>,则()0f b <,且a b <,由题设知()f x 在[,]a b 上连续,由零点存在定理,至少存在一点0(,)x a b ∈使0()0f x =, 即至少存在一点0(,)x ∈-∞+∞使0()0f x =.4.设多项式P n x ＝x n +a 11n x-+…+a n .,利用第3题证明： 当n 为奇数时,方程P n x =0至少有一实根.证: 122()1n n n n a a a P x x x x x ⎛⎫=++++ ⎪⎝⎭()lim 10n nx P x x →∞∴=>,由极限的保号性知. 0X ∃>,使当X x >时有()0nn P x x>,此时()n P x 与n x 同号,因为n 为奇数,所以2X n 与-2X n 异号,于是(2)n P X -与(2)n P X 异号,以()n P x 在[2,2]X X -上连续,由零点存在定理,至少存在一点0(2,2)X X X ∈-,使0()0n P x =,即()0n P x =至少有一实根.。
微积分曹定华版课后题答案习题详解
第9章习题9-11． 判定下列级数的收敛性：(1) 115n n a ∞=⋅∑（a ＞0）； (2)∑∞=-+1)1(n n n ;(3) ∑∞=+131n n ; (4)∑∞=-+12)1(2n nn; (5) ∑∞=+11ln n n n ; (6)∑∞=-12)1(n n;(7) ∑∞=+11n n n ; (8)(1)21n n nn ∞=-⋅+∑． 解：（1）该级数为等比级数，公比为1a ,且0a >,故当1||1a <,即1a >时，级数收敛，当1||1a≥即01a <≤时，级数发散.（2）Q n S =+++L∴1n ∞=∑发散.（3）113n n ∞=+∑是调和级数11n n ∞=∑去掉前3项得到的级数，而调和级数11n n∞=∑发散，故原级数113n n ∞=+∑发散. （4）Q 1112(1)1(1)222n n nn n n n ∞∞-==⎛⎫+--=+ ⎪⎝⎭∑∑ 而1112n n ∞-=∑,1(1)2m nn ∞=-∑是公比分别为12的收敛的等比级数，所以由数项级数的基本性质知111(1)22n n n n ∞-=⎛⎫-+ ⎪⎝⎭∑收敛，即原级数收敛.（5）Q lnln ln(1)1nn n n =-++ 于是(ln1ln 2)(ln 2ln 3)[ln ln(1)]n S n n =-+-+-+L故lim n n S →∞=-∞,所以级数1ln1n nn ∞=+∑发散. （6）Q 2210,2n n S S +==-∴lim n n S →∞不存在，从而级数1(1)2n n ∞=-∑发散.（7）Q 1lim lim10n n n n U n→∞→∞+==≠∴ 级数11n n n ∞=+∑发散. （8）Q (1)(1)1, lim 21212n n n n n n U n n →∞--==++∴ lim 0n x U →∞≠，故级数1(1)21n n nn ∞=-+∑发散.2． 判别下列级数的收敛性，若收敛则求其和：(1) ∑∞=⎪⎭⎫ ⎝⎛+13121n n n ； (2) ※∑∞=++1)2)(1(1n n n n ; (3) ∑∞=⋅12sin n n n π; (4)πcos2n n ∞=∑． 解：Q （1）1111, 23n n n n ∞∞==∑∑都收敛，且其和分别为1和12，则11123n n n ∞=⎛⎫+ ⎪⎝⎭∑收敛，且其和为1+12=32.（2）Q11121(1)(2)212n n n n n n ⎛⎫=-+ ⎪++++⎝⎭1lim 4n n S →∞=故级数收敛，且其和为14. （3）πsin 2n U n n =,而πsinππ2lim lim 0π222n n n U n→∞→∞=⋅=≠，故级数1πsin2n n n ∞=⋅∑发散. （4）πcos 2n n U =,而4lim limcos2π1k k k U k →∞→∞==,42lim limcos(21)π1k k k U k +→∞→∞=+=-故lim n n U →∞不存在，所以级数πcos2n n ∞=∑发散.3※． 设1nn U∞=∑ (U n ＞0)加括号后收敛，证明1nn U∞=∑亦收敛．证：设1(0)nn n UU ∞=>∑加括号后级数1n n A ∞=∑收敛，其和为S .考虑原级数1n n U ∞=∑的部分和1n k k S U ∞==∑，并注意到0(1,2,)k U k >=L ，故存在0n ,使又显然1n n S S +<对一切n 成立，于是，{}n S 是单调递增且有上界的数列，因此，极限lim n n S →∞存在，即原级数1nn U∞=∑亦收敛.习题9-21． 判定下列正项级数的收敛性：(1) ∑∞=++1n n n )2)(1(1; (2)∑∞=+1n n n1; (3) ∑∞=++1n n n n )2(2; (4)∑∞=+1n n n )5(12；(5) 111nn a ∞=+∑ (a ＞0); (6) ∑∞=+1n nb a 1(a , b ＞0); (7)()∑∞=--+1n a n a n22(a ＞0); (8)∑∞=-+1n n n 1214； (9) ∑∞=⋅1n nn n 23； (10) ※∑∞=1n nn n !; (11) ∑∞=+⋅⋅⋅⋅+⋅⋅⋅⋅1n n n )13(1074)12(753ΛΛ; (12)∑∞=1n nn3；(13) ※∑∞=1n n n 22)!(2； (14)∑∞=⎪⎭⎫ ⎝⎛+1n nn n 12; (15)∑∞=1πn nn3sin2; (16) ∑∞=1πn n n n 2cos 32．解：（1）因为211(1)(2)n n n <++而211n n ∞=∑收敛，由比较判别法知级数11(1)(2)n n n ∞=++∑收敛.（2）因为lim 10n n n U →∞==≠，故原级数发散.（3）因为21(1)(1)1n n n n n n n +>=+++，而111n n ∞=+∑发散，由比较判别法知，级数12(1)n n n n ∞=++∑发散.（4321n<=,而1n ∞=是收敛的p -级数3(1)2p =>,由比较判别法知,级数n ∞=.（5）因为111lim lim lim(1)111n n n n n n n na a a a a →∞→∞→∞+==-++ 而当1a >时，11n n a ∞=∑收敛，故111nn a ∞=+∑收敛; 当1a =时，11n n a∞=∑=11n ∞=∑发散，故111nn a∞=+∑发散;当01a <<时1lim101n n a →∞=≠+，故1lim1nn a →∞+发散; 综上所述，当01a <≤时，级数1lim 1n n a →∞+发散，当1a >时，1lim 1nn a →∞+收敛. （6）因为1lim lim lim(1)1n n n n n n n n b aa b a b a bb→∞→∞→∞+==-++ 而当1b >时, 11n n b ∞=∑收敛，故11nn a b ∞=+∑收敛; 当1b =时，1111n n n b ∞∞===∑∑发散，故而由0a >, 101a <<+∞+,故11nn a b ∞=+∑也发散; 当01b <<时，11lim 0n n a b a →∞=≠+故11n n a b ∞=+∑发散; 综上所述知，当01b <≤时，级数11n n a b ∞=+∑发散;当b >1时，级数11nn a b∞=+∑收敛. （7）因为lim lim 1n n n→∞→∞=而11n n ∞=∑发散，故级数10)n a ∞=>∑发散. （8）因为434431121lim lim 1212n n n n n n n n→∞→∞++-==-而311n n ∞=∑收敛，故级数21121n n n ∞=+-∑收敛.（9）因为1113233lim lim lim 1(1)232(1)2n n n n n n n n nU n n U n n +++→∞→∞→∞⋅⋅==>+⋅+由达朗贝尔比值判别法知，级数132nnn n ∞=⋅∑发散. （10）因为11(1)!1lim lim lim(1)1(1)!n n n n n n n nU n n e U n n n ++→∞→∞→∞+=⋅=+=>+，由达朗贝尔比值判别法知，级数1!nn n n ∞=∑发散. （11）因为1357(21)(23)4710(31)limlim 4710(31)(34)357(21)n n n nU n n n U n n n +→∞→∞⋅⋅⋅⋅+⋅+⋅⋅⋅⋅+=⋅⋅⋅⋅⋅+⋅+⋅⋅⋅⋅+L L L L232lim1343n n n →∞+==<+,由达朗贝尔比值判别法知原级数收敛.（12）因为111311lim lim lim 1333n n n n n n nU n n U n n ++→∞→∞→∞++=⋅==<，由达朗贝尔比值判别法知，级数13n n n ∞=∑收敛.（13）因为22221221(1)[(1)!]2(1)lim lim lim (!)22n n n n n n n nU n n U n +++→∞→∞→∞++=⋅= 由2212121(1)2(1)1lim lim lim 222ln 22ln 2x x x x x x x x x +++→∞→+∞→+∞+++==⋅⋅2121lim 022(ln 2)x x +→+∞==⋅知2121(1)lim lim 012n n n n n U n U ++→∞→∞+==< 由达朗贝尔比值判别法知，级数221(!)2n n n ∞=∑收敛.（14）因为1lim 1212n n n n →∞==<+，由柯西根值判别法知级数121nn n n ∞=⎛⎫ ⎪+⎝⎭∑收敛. （15）因为ππ2sinsin 33lim lim 1π2π33n n nn n n n n→∞→∞==⋅ 而112233nn n n n ∞∞==⎛⎫= ⎪⎝⎭∑∑是收敛的等比级数，它的每项乘以常数π后新得级数12π3n n n ∞=⋅∑仍收敛，由比较判别法的极限形式知，级数1π2sin3n n n ∞=∑收敛. （16）因为2πcos 322n nn n n ≤而与（12）题类似地可证级数12n n n ∞=∑收敛，由比较判别法知级数1πcos 32nn n n ∞=∑收敛.2． 试在(0，+∞)内讨论x 在什么区间取值时，下列级数收敛：(1) ∑∞=1n nn x ; (2)nn x n ∑∞=⎪⎭⎫⎝⎛123． 解：（1）因为11lim lim lim 11n n n n n n nU x n nxx U n x n ++→∞→∞→∞=⋅==++由达朗贝尔比值判别法知，当1x >时，原级数发散;当01x <<时，原级数收敛; 而当1x =时，原级数变为调11n n∞=∑，它是发散的. 综上所述，当01x <<时，级数1nn x n ∞=∑收敛.（2）因为1313(1)2limlim 22n n n n n nx n U xU x n ++→∞→∞⎛⎫+⋅ ⎪⎝⎭==⎛⎫⋅ ⎪⎝⎭,由达朗贝尔比值判别法知，当12x >即2x >时，原级数发散;当012x<<即02x <<时，原级收敛.而当12x =即 2x =时，原级数变为31n n ∞=∑，而由3lim n n →∞=+∞知31n n ∞=∑发散，综上所述，当02x <<时，级数31()2n n xn ∞=∑收敛.习题9-31． 判定下列级数是否收敛，如果是收敛级数，指出其是绝对收敛还是条件收敛：(1) ∑∞=--1121)1(n nn ; (2)11(1)2(1)2n n nn ∞-=-+-⋅∑; (3) ∑∞=12sin n n nx; (４) 111π(1)sin πn n n n∞+=-∑; (5) ∑∞=-⎪⎭⎫ ⎝⎛-11210121n n n ； (6)∑∞=+-1)1(n n x n ; (7) ∑∞=⋅1!)2sin(n n n x ．解：（1）这是一个交错级数121n U n =-, 1lim lim 021n n n U n →∞→∞==-, 1112121n n U U n n +=>=-+ 由莱布尼茨判别法知11(1)21nn n ∞=--∑. 又1111(1)2121n n n n n ∞∞==-=--∑∑，由1121lim 12n n n→∞-=,及11n n ∞=∑发散，知级数1121n n ∞=-∑发散，所以级数11(1)21nn n ∞=--∑条件收敛. （2）因为2111(1)211(1)22(1)2n n n n n ----+-=+-⋅-⋅,故 而112n n ∞=∑收敛，故132n n ∞=∑亦收敛，由比较判别法知11(1)2(1)2n n nn ∞-=-+-⋅∑收敛，所以级数11(1)2(1)2n n n n ∞-=-+-⋅∑绝对收敛.（3）因为22sin 1,nx n n ≤而级数211n n ∞=∑收敛，由比较判别法知21sin n nxn ∞=∑收敛，因此，级数21sin n nxn ∞=∑绝对收敛.（4）因为121ππ|(1)sin |sin πlimlim 11πn n n n n n n n+→∞→∞-==而211n n∞=∑收敛，由比较判别法的极限形式知，级数111π|(1)sin |πn n n n∞+=-∑收敛，从而级数11π(1)sin πn n n+-绝对收敛. （5）因为212121111111210210210n n n n n n ----≤+=+,而级数112nn ∞=∑收敛的等比级数1()2q =;由比值判别法，易知级数211110n n ∞-=∑收敛，因而21111210n n n ∞-=⎛⎫+ ⎪⎝⎭∑收敛，由比较判别法知级数21111210n n n ∞-=-∑收敛，所以原级数21111210n n n ∞-=-∑绝对收敛. （6）当x 为负整数时，级数显然无意义；当x 不为负整数时，此交错级数满足莱布尼茨判别法的条件，故它是收敛的，但因11n x n ∞=+∑发散，故原级数当x 不为负整数时仅为条件收敛. （7）因为sin(2)1!!n x n n ⋅≤由比值判别法知11!n n ∞=∑收敛(Q 1(1)!lim 01!n n n →∞+=)，从而由比较判别法知1sin(2)!n n x n ∞=⋅∑收敛，所以级数1sin(2)!n n x n ∞=⋅∑，绝对收敛.2． 讨论级数∑∞=--111)1(n pn n 的收敛性(p ＞0)． 解：当1p >时，由于11111(1)n p p n n n n ∞∞-==-=∑∑收敛，故级数111(1)n p n n ∞-=-∑绝对收敛. 当01p <≤时，由于111,(1)n n p pu u n n +=>=+ lim 0n n u →∞=,由莱布尼茨判别法知交错级数111(1)n p n n ∞-=-∑收敛，然而，当01p <≤时，11111(1)n p p n n n n ∞∞-==-=∑∑发散，故此时，级数111(1)n p n n ∞-=-∑条件收敛.综上所述，当01p <≤时，原级数条件收敛;当p >1时，原级数绝对收敛.3※． 设级数∑∞=12n na及∑∞=12n nb都收敛，证明级数∑∞=1n n n b a 及()∑∞=+12n n n b a 也都收敛．证：因为2222||||110||222n n n n n n a b a b a b +≤≤=+ 而由已知1nn a ∞=∑及21n n b ∞=∑都收敛，故221111,22n n n n a b ∞∞==∑∑收敛，从而2211122n n n a b ∞=⎛⎫+ ⎪⎝⎭∑收敛，由正项级数的比较判别法知1n nn a b∞=∑也收敛，从而级数1n nn a b∞=∑绝对收敛.又由222()2,n n n n n n a b a a b b +=++及2211,n n n n a b ∞∞==∑∑,以及1n n n a b ∞=∑收敛，利用数项级数的基本性质知，221(2)nn n n n aa b b ∞=++∑收剑，亦即21()n n n a b ∞=+∑收敛.习题9-41． 指出下列幂级数的收敛区间：(1) ∑∞=0!n nn x (0!=1)； (2)∑∞=0!n nn x n n ； (3) ∑∞=⋅022n n nnx ； (4)∑∞=++-01212)1(n n nn x ． (5) ∑∞=⋅+02)2(n n nn x ； (6)∑∞=-0)1(2n n nx n． 解：（1）因为111(1)!limlim lim 011!n n n n na n p a n n +→∞→∞→∞+====+，所以收敛半径r =+∞，幂级数1!n n x n ∞=∑的收敛区间为(,)-∞+∞.（2）因为-111lim lim lim 1e 11n nn n n n na n p a n n +→∞→∞→∞⎛⎫===-= ⎪++⎝⎭，所以收敛半径1e r p ==. 当x =e 时，级数01!!e n n n n n n n n x n n ∞∞===∑∑，此时11(1)n n n u e u n+=+，因为1(1)nn +是单调递增数列，且1(1)n n+<e 所以1n n u u +>1,从而lim 0n n u →∞≠，于是级数当x =e 时，原级数发散.类似地，可证当x =-e 时，原级数也发散(可证lim ||0n n u →∞≠),综上所述，级数!nnn n x n∞=∑的收敛区间为(-e,e).（3）因为2111limlim ()212n n n n a n p a n +→∞→∞===+,所以收敛半径为r =2. 当2x =时，级数221012n n n n x n n∞∞===⋅∑∑是收敛的p 一级数（p =2>1）;当x =-2时，级数22011(1)2n nn n n x n n ∞∞===-⋅⋅∑∑是交错级数，它满足莱布尼茨判别法的条件，故它收敛.综上所述，级数202nn n x n∞=⋅∑的收敛区间为[-2,2].（4）此级数缺少偶次幂的项，不能直接运用定理2求收敛半径，改用达朗贝尔比值判别法求收敛区间.令21(1)21n nn x u n +=-+，则22121lim lim 23n n n nu n x x u n +→∞→∞+=⋅=+.当21x <时，即||1x <时，原级数绝对收敛.当21x >时，即||1x >时，级数0||n n u ∞=∑发散，从而210(1)21n nn x n +∞=-+∑发散，当1x =时，级数变为01(1)21nn n ∞=-+∑;当1x =-时，级数变为11(1)21n n n ∞+=-+∑;它们都是交错级数，且满足莱布尼茨判别法的条件，故它们都收敛.综上所述，级数210(1)21n nn x n +∞=-+∑的收敛区间为[-1,1].（5）此级数为（x +2）的幂级数. 因为11limlim 2(1)2n n n n a n p a n +→∞→∞===+. 所以收敛半径12r p==，即|2|2x +<时，也即40x -<<时级数绝对收敛.当|2|2x +>即4x <-或0x >时，原级数发散.当4x =-时，级数变为01(1)nn n ∞=-∑是收敛的交错级数， 当x =0时，级数变为调和级数11n n ∞=∑,它是发散的.综上所述，原级数的收敛区间为[-4,0).（6）此级数（x -1）的幂级数 故收敛半径12r =. 于是当1|1|2x -<即1322x <<时，原级数绝对收敛. 当1|1|2x ->即12x <或32x >时，原级数发散. 当32x =时，原级数变为01n n∞=∑是调和级数，发散. 当12x =时，原级数变为11(1)n n n ∞=-∑，是收敛的交错级数. 综上所述，原级数的收敛区间为13,22⎡⎫⎪⎢⎣⎭. 2． 求下列幂级数的和函数： (1) ∑∞=-1)1(n nn n x ； (2) ∑∞=-1122n n nx ； (3) n n x n n ∑∞=+1)1(1； (4) ∑∞=+0)12(n n xn ．解：（1）可求得所给幂级数的收敛半径r =1. 设1()(1)nnn x S x n ∞==-∑,则 1111()(1)(1)1n n n n n n x S x x n x ∞∞-=='⎡⎤'=-=-=-⎢⎥+⎣⎦∑∑ 又当x =1时，原级数收敛，且()S x 在x =1处连续.（2）所给级数的收敛半经r =1，设211()2n n S x nx∞-==∑,当||1x <时，有 于是22222()1(1)x x s x x x '⎛⎫== ⎪--⎝⎭又当1x =±时，原级数发散.故 2122122 (||1)(1)n n x nx x x ∞-==<-∑ （3）可求所给级数的收敛半径为1.令1111()(0)(1)(1)n n n n x x s x x n n x n n +∞∞====≠++∑∑ 令11()(1)n n x g x n n +∞==+∑,则111()1n n g x x x ∞-=''==-∑ 所以0()ln(1)d ln(1)ln(1)xg x x x x x x x =--=+---⎰; 所以1()11ln(1),||1,S x x x x ⎛⎫=+--< ⎪⎝⎭且0x ≠. 当1x ±时，级数为11(1)n n n ∞=+∑和11(1)(1)n n n n ∞=-+∑,它们都收敛.且显然有(0)0S =. 故111ln(1)(1,0)(0,1)()00,1x x S x x x x ⎧⎛⎫+--∈-⋃⎪ ⎪=⎝⎭⎨⎪=±⎩. （4）可求得所给级数的收敛半径为r =1且1x ±时，级数发散，设10()n n S x nx ∞-==∑,则001()d .1xn n s x x x x∞===-∑⎰ 于是211()()1(1)S x x x '==--,即1211(1)n n nx x ∞-==-∑. 所以1101(21)2n n n n n n n xx nx x ∞∞∞-===+=+∑∑∑ 3． 求下列级数的和： (1) ∑∞=125n n n ； (2)∑∞=-12)12(1n n n ； (3) ∑∞=--112212n n n ； (4) 1(1)2n n n n ∞=+∑． 解：（1）考察幂级数21n n n x ∞=∑，可求得其收敛半径1r = ，且当1x ±时，级数的通项2n n u n x =，2lim ||lim n n n u n →∞→∞==+∞，因而lim 0n n u →∞≠，故当1x ±时，级数21n n n x ∞=∑发散，故幂级数21n n n x ∞=∑的收敛区间为（-1，1）.设21() (||1)n n S x n x x ∞==<∑，则211()n n S x x n x ∞-==∑ 令2111()n n S x n x∞-==∑,则11011()d x n n n n S x x nx x nx ∞∞-====∑∑⎰. 再令121()n n S x nx∞-==∑,则201()d 1x n n x S x x x x∞===-∑⎰. 故221()(||1)1(1)x S x x x x '⎛⎫==< ⎪--⎝⎭，从而有120()d (1)x x S x x x =-⎰. 于是 213()() (||1)(1)x x S x xS x x x +==<- 取15x =，则223111()11555()5532115n n n S ∞=+===⎛⎫- ⎪⎝⎭∑. （2）考察幂级数21121n n x n ∞=-∑，可求得收敛半径r =1，设 令21111()21n n S x x n ∞-==-∑,则221211()1n n S x x x ∞-='==-∑. 即 1111()(0)ln (,(0)0)21x S x S s x+-==-. 于是 111()ln ,(||<1)21x S x x x +=-，从而取x =则11(21)21n n S n ∞===--∑（3）考察幂级数211(21)n n n x∞-=-∑，可求得其级数半经为r =1,因为 令2111()2n n S x nx∞-==∑，则221201()d 1x n n x S x x xx ∞===-∑⎰.所以212222() (||1)1(1)x x S x x x x '⎛⎫==< ⎪--⎝⎭,于是 取12x =，得 3212111()121102212291()2n n n S ∞-=+-⎛⎫=== ⎪⎛⎫⎝⎭- ⎪⎝⎭∑. （4）考察幂级数1(1)n n n n x∞=+∑，可求得其收敛半径r =1. 设1()(1) (||1)n n S x n n xx ∞==+<∑ 则121011()d xn n n n S x x nx x nx ∞∞+-====∑∑⎰. 又设111()n n S x nx∞-==∑则101()d 1x n n x S x x x x∞===-∑⎰. 从而121()1(1)x S x x x '⎛⎫== ⎪--⎝⎭, 取12x =，则 习题9-51． 将下列函数展开成x 的幂级数：(1) 2cos 2x ; (2) 2sin x ； (3) 2x x -e ; (4) 211x -； (5)πcos()4x -． 解：（1）2201cos 11cos (1)2222(2)!n n n x x x n ∞=+==+-∑ （2）2101sin (1) ()2(21)!2n n n x x x n +∞=⎛⎫=--∞<<+∞ ⎪+⎝⎭∑ （3）22210011e ()(1) ()!!x n n n n n x x x x x n n ∞∞-+===-=--∞<+∞∑∑ （4）211111211x x x ⎡⎤=+⎢⎥--+⎣⎦（5）πππcos cos cos sin sin 444x x x ⎛⎫-=+ ⎪⎝⎭ 2． 将下列函数在指定点处展开成幂级数，并求其收敛区间： (1)x -31,在x 0＝１； (2) cos x,在x 0=3π; (3) 3412++x x ,在x 0=1; (4) 21x, 在x 0＝３． 解：（1）因为11113212x x =⋅---,而 0111 (||112212nn x x x ∞=--⎛⎫=< ⎪-⎝⎭-∑即13x -<<). 所以100111(1) (13)3222nnn n n x x x x ∞∞+==--⎛⎫=⋅=-<< ⎪-⎝⎭∑∑. 收敛区间为：（-1，3）.（2）πππ2π2cos cos ()cos cos()sin sin()333333x x x x ⎡⎤=+-=---⎢⎥⎣⎦ 收敛区间为(,)-∞+∞.（3）211111111()1143213481124x x x x x x =-=⋅-⋅--++++++ 由112x -<且114x -<得13x -<<，故收敛区间为（-1，3） （4）因为011113(1)()333313n n n x x x ∞=-=⋅=-⋅-+∑ 而21011(3)(1)3n n n n x x x ∞+=''⎡⎤-⎛⎫=-=-- ⎪⎢⎥⎝⎭⎣⎦∑ 由313x -<得06x <<. 故收敛区间为（0，6）.。
《微积分》课后答案第7章(复旦大学版)解析
第七章
习题 7-1 1. 略. 2． 求点(a,b,c)关于(1) 各坐标面；(2) 各坐标轴；(3) 坐标原点的对称点的坐标． 解：（1）点（a,b,c）关于 xoy 面的对称点是(a,b,-c); 关于 xoz 面的对称点是（a,-b,c）; 关于 yoz 面的对称点是(-a,b,c); （2）点（a,b,c）关于 x 轴的对称点是（a,-b,-c）; 关于 y 轴的对称点是（-a,b,-c）; 关于 z 轴的对称点是（-a,-b,c）; (3)点（a,b,c）关于原点的对称点是（-a,-b,-c）; 3． 自点 P0(x0, y0, z0)分别作各坐标面和坐标轴的垂线，写出各垂足的坐标．
2．试用向量证明：如果平面上一个四边形的对角线互相平分，则该四边形是平行 四边形． 证：（如上题图），依题意有 AM MC, DM MB. 于是 AB AM MB MC DM DC. 故 ABCD 是平行四边形. 3．已知向量 a=i-2j+3k 的始点为(1，3，-2)，求向量 a 的终点坐标． 解：设 a 的终点坐标为( x, y, z ),则
0 ( x0 , y0 , z0 ) 作 xoy 面的垂线，垂足坐标是 ( x0 , y0 , 0) ; 解：自点 P
作 xoz 面的垂线，垂足是 ( x0 , 0, z0 ) ; 作 yoz 面的垂线，垂足是 (0, y0 , z0 ); 自点 P 0 ( x0 , y0 , z0 ) 作 x 轴的垂线，垂线是 ( x0 , 0, 0);
解得 b , c
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38 5 38 ,故所求点的坐标为 0, , . 3 3 3
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《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第三章
第三章习题3-11．设s =12gt ２，求2d d t s t =．解：22221214()(2)2lim lim 22t t t g g ds s t s dt t t t →→=-⨯-==--21lim (2)22t g t g →=+=2．设f (x )=1x，求f '(x 0)(x 0≠0)．解：1211()()()f x x x x--'''===00201()(0)f x x x '=-≠3．试求过点（3，8）且与曲线2y x =相切的直线方程。
解：设切点为00(,)x y ，则切线的斜率为002x x y x ='=，切线方程为0002()y y x x x -=-。
由已知直线过点(3,8),得00082(3)y x x -=-(1)又点00(,)x y 在曲线2y x =上，故200y x =(2)由(1),(2)式可解得002,4x y ==或004,16x y ==，故所求直线方程为44(2)y x -=-或168(4)y x -=-。
也即440x y --=或8160x y --=。
4．下列各题中均假定f ′(x 0)存在，按照导数定义观察下列极限，指出A 表示什么：(1)0limx ∆→00()()f x x f x x-∆-∆=A ;(2)f (x 0)=0,0limx x →0()f x x x-＝A ；(3)0limh →00()()f x h f x h h+--=A ．解：(1)0000000()()[()]()limlim ()x x f x x f x f x x f x f x xx →-→--+--'=-=-- 0()A f x '∴=-(2)000000()()()limlim ()x x x x f x f x f x f x x x x x →→-'=-=--- 0()A f x '∴=-(3)000()()limh f x h f x h h→+-- 00000[()()][()()]lim h f x h f x f x h f x h→+----=000000()()[()]()lim lim h h f x h f x f x h f x h h →-→+-+--=+-000()()2()f x f x f x '''=+=02()A f x '∴=5．求下列函数的导数：(1)y；(2)y；(3)y2．解：(1)12y x==11221()2y x x -''∴===(2)23y x-=225133322()33y x x x ----''∴==-=-=(3)2152362y x x xx-==15661()6y x x -''∴===6．讨论函数y在x =0点处的连续性和可导性．解：00(0)x f →==000()(0)0lim lim 0x x x f x f x x →→→--===∞-∴函数y =在0x =点处连续但不可导。
《微积分》课后答案(复旦大学出版社(曹定华 李建平 毛志强 著))第六章 定积分
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4. 估计下列各积分值的范围：
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《微积分》课后答案(复旦大学出版社(曹定华 李建平 毛志强 著))第四章
习题 4-11.验证函数f (x )=lnsin x 在[π5π,66]上满足罗尔定理的条件，并求出相应的ξ，使f ′(ξ)=0. 解: 显然()ln sin f x x =在5π,66x ⎡⎤⎢⎥⎣⎦上连续,在π5π,66⎛⎫⎪⎝⎭内可导,且π5π()()ln 266f f ==-,满足罗尓定理的条件. 令cos ()cot 0sin x f x x x '===,则π2x = 即存在ππ5π(,)66ξα=∈,使()0f ξ'=成立.2. 下列函数在指定区间上是否满足罗尔定理的三个条件?有没有满足定理结论中的ξ ? [][][]2(1)()1,;(2)(),;1,10,21sin ,0π(3)()0,π1,0e x f x f x x x x f x x =-=--<≤⎧=⎨=⎩解: (1) 2()1e x f x =-在[]1,1-上连续,在()1,1-内可导,且(1)1,(1)1,e e f f -=-=-即 (1)(1)f f -=() f x ∴在[]1,1-上满足罗尓定理的三个条件. 令 2()20ex f x x '==得 0x =,即存在0(1,1)ξ=∈-,使()0f ξ'=.(2) 101()1112x x f x x x x -≤<⎧==-⎨-≤≤⎩显然()f x 在(0,1),(1,2)内连续,又1111(10)lim ()lim(1)0,(10)lim ()lim(1)0,(10)(10)(1)0,即x x x x f f x x f f x x f f f --++→→→→-==-=+==-=-=+==所以()f x 在1x =处连续,而且22(00)lim ()lim (1)1(0),(20)lim ()lim (1)1(2),x x x x f f x x f f f x x f ++--→→→→+==-==-==-==即()f x 在0x =处右连续,在2x =处左连续,所以()f x 在[]0,2 上连续.又w ww .t tl ea rn .n et1111()(1)1(1)lim lim 1,11()(1)1(1)lim lim 111x x x x f x f xf x x f x f xf x x --++-→→+→→--'===-----'===--(1)(1)() f f f x -+''∴≠∴在1x =处不可导,从而()f x 在(0,2)内不可导. 又 (0)(2)1f f == 又由 101()112x f x x -<<⎧'=⎨<<⎩知 ()0f x '≠ 综上所述,函数()f x 满足罗尓定理的条件(1),(3)不满足条件(2),没有满足定理结论的ξ. (3) 由0(00)lim sin 0(0)1x f x f +→+==≠=知()f x 在0x =不右连续, () f x ∴在[]0,π上不连续, 显然()f x 在()0,π上可导,又(0)1,(π)0f f ==,即(0)(π)f f ≠,且()cos (0,π) f x x x '=∈,取π(0,π)2ξ=∈,有π()cos cos 02f ξξ'===. 综上所述,函数()f x 满足罗尓定理的条件(2),不满足条件(1),(3),有满足定理结论的ξ,ξ=π2.3. 不用求出函数()(1)(2)(3)f x x x x =---的导数，说明方程()0f x '=有几个实根，并指出它们所在的区间.解: 显然()f x 在[]1,2上连续,在()1,2内可导,且(1)(2)0f f ==,由罗尓定理知,在()1,2内至少存在一点1ξ,使1()0f ξ'=,即()0f x '=在()1,2内至少有一个实根.同理 ()0f x '=在()2,3内也至少有一个实根2ξ.又()0f x '=是二次方程,最多有两个实根,故()0f x '=有两个实根,分别在区间()1,2和()2,3内.4. 验证拉格朗日中值定理对函数3()2f x x x =+在区间［0，1］上的正确性.解: 显然3()2f x x x =+在［0，1］上连续,在()0,1内可导,满足拉格朗日中值定理的条件.若令2(1)(0)()32310f f f x x -'=+==- 则3x =±,取3ξ=,即存在w ww .t tl ea rn .n et(0,1)3ξ=∈,使得(1)(0)()10f f f ξ-=-成立. 从而拉格朗日中值定理对函数3()2f x x x =+在［0，1］上成立.5. 已知函数f (x )在［a ,b ］上连续，在(a ,b )内可导，且f (a )=f (b )=0,试证：在(a ,b )内至少存在一点ξ，使得 f (ξ)+f ′(ξ) = 0，ξ(∈a ,b ). 证: 令()()e xF x f x =,则()()()e e xxF x f x f x ''=+由e x在(),-∞+∞上连续,可导,()f x 在[],a b 上连续,在(),a b 内可导,知()F x 在[],a b 上连续,在(),a b 内可导,而且()()0,()()0,()()e e 即abF a f a F b f b F a F b =====,由罗尓定理至少存在一点(,)a b ξ∈使()0F ξ'=.即 ()()0e e f f ξξξξ'+= 而0e ξ≠ 故 ()()0f f ξξ'+=即在(),a b 内至少存在一点ξ,使得()()0f f ξξ'+=.6.若方程10110n n n a x a x a x --+++= 有一个正根x 0,证明方程12011(1)0n n n a nx a n x a ---+-++=必有一个小于0x 的正根.证: 令1011()…nn n f x a x a xa x --=+++,显然()f x 在[]00,x 连续,在()00,x 内可导,且(0)0f =,依题意知0()0f x =.即有0(0)()f f x =.由罗尓定理,至少存在一点0(0,)x ξ∈,使得()0f ξ'=成立,即12011(1)0…n n n a n a n a ξξ---+-++=成立,这就说明ξ是方程12011(1)0n n n a nx a n x a ---+-++= 的一个小于0x 的正根. 7. 设f (a ) = f (c ) = f (b ),且a ＜c ＜b , f ″(x )在［a ,b ］上存在，证明在(a ,b )内至少存在一点ξ，使f ″(ξ) = 0.证: 显然()f x 分别在[],a c 和[],c b 上满足罗尓定理的条件,从而至少存在1(,)a c ξ∈,2(,)c b ξ∈,使得12()()0f f ξξ''==.w ww .t tl ea rn .n et又由题意知()f x '在[]12,ξξ上满足罗尓定理的条件,从而至少存在一点12(,)(,)a b ξξξ∈⊂,使得()0f ξ''=.即在(,)a b 内至少存在一点ξ,使()0f ξ''=. 习题4-2 1.利用洛必达法则求下列极限： (1) sin 3limtan 5x x x π→; (2) 0e 1lim (e 1)x x x x x →---; (3)lim m m n n x a x a x a →--； (4) 20()lim x xx a x a x →+-,(a ＞0)； (5) 0ln lim cot x xx +→； (6) 0lim sin ln x x x +→;(7) 1ln(1)lim arccot x x x →+∞+; (8) 0e 1lim(e 1x x x x →--;(9) 10lim(1sin )xx x →+; (10) 2lim (arctan )πx x x →+∞(11) csc 03e lim(2x x x x →-+ ; (12) 2120lim e x x x →;(13) lim )x x →+∞-; (14) 1101lim (1)e xxx x →⎡⎤+⎢⎥⎣⎦.解:222000011sin 33cos33(1)limlim lim cos3cos 5tan 55sec 5533(1)(1)5511(2)limlim lim (1)111lim 22(3)lim lim limπππe e e e e e e e e x x x x x xx x x x x xx x x x m m m n n n x a x a x a x x x x x x x x x x x x a mx x a nx →→→→→→→--→→→==⋅=⋅-⋅-=----==--+++==+-==-.m n m n m m x a n n --= 2002220()ln ln()()(4)lim lim 21()()()ln ln()()lim2x xxxx x x x x x x a x a a a x a x a a x x xa x a x a x a a a x a x a x a x →→→⎡⎤+-++⎢⎥+-+⎣⎦=⎡⎤++++-++⎢⎥+++⎣⎦=w ww .t tl ea rn .n et[]2000221()ln ln 012 aa a a aa a a a ++-⋅+==2200000000001ln sin 2sin cos (5)lim lim lim lim cot csc 12sin 0cos 001ln sin (6)lim sin ln lim lim lim tan csc csc cot sin lim lim tan 100 x x x x x x x x x x x x x x x x x x x x x x x xx x x xxx x ++++++++++→→→→→→→→→→==-=--=-⋅====-⋅-=-⋅=-⨯=222221111ln(1111(7)limlim lim lim 111cot 11arc x x x x xx x x x x x x x x →+∞→+∞→+∞→+∞-++++====+-++ 20002200001(1)(8)lim()lim lim 1(1)21443limlim 12022e e e e e e e e e e e e e e e e e e e x x x x x x x x x x x x xxxx x x x x x x x x x x xx x →→→→→-----==-------====+-++00022cos 11ln(1sin )cos 1sin ln(1sin )lim limlim 11sin 12112ln(arctan )arctan 1limlim 112ln(arctan )(9)lim(1sin )lim 2(10)lim (arctan )lim πππee =eee ee eeπx x x x x xx xx x xxxxx x x x x x x x xxx x x x →→→→+∞→+∞++++→→⋅⋅+-→+∞→+∞+========2221lim12lim(1)arctan (1)arctan πeeex x x x x xx→+∞→+∞--+-+===020033lnln322csc ln lim csc 2sin sin 0002(2)(3)33(2)limlim 1(3)(2)cos cos 3(11)lim()lim lim 21e e e e e e e e eee ee exxxx x x x x x x x x e e ex x x x xxxxx x x x x x x x xxx →→→---+++→→→+-+--⋅----+--+-===+====2222111122000221()(12)lim lim lim lim 11()e e ee x xx x x x x x x x x x→→→→'⋅====∞'w ww .t tl ea rn .n et202211ln(1)1ln(1)1limlim lim 0(13)lim )lim1111lim31(14)lim (1) ee ee x x x x x x x x xx xxx x x x x →→→+∞→+∞+-+-→=++===⎡⎤===+⎢⎥⎣⎦00111211lim2(1)2eex x xx →→-+--+==2.设 21lim 1x x mx nx →++-=5,求常数m ,n 的值.解: 1lim(1)0, x x →-= 而21lim 51x x mx nx →++=-21lim()0 x x mx n →∴++= 且21()lim5(1)x x mx n x →'++='- 即 10m n ++= 且 1lim(2)5x x m →+=即 1m n +=- 且 25m +=于是得 3,4m n ==-. 3.验证极限sin lim x x xx→∞+存在，但不能由洛必达法则得出. 解: sin 1limlim(1sin )1x x x x x x x→∞→∞+=+=,极限存在,但若用洛必达法则,有sin lim lim(1cos )x x x xx x→∞→∞+=+因lim cos x x →∞不存在,所以不能用洛必达法则得出.4.设f (x )二阶可导，求2()2()()limh f x h f x f x h h →+-+-.解: 这是型未定式,利用洛必达法则有 [][]200000()2()()()()limlim2()()()()1lim 21()()1()()11lim lim ()()2222().h h h h h f x h f x f x h f x h f x h h hf x h f x f x h f x hf x h f x f x h f x f x f x h h f x →→→→→''+-+-+--=''''-+---=''''+---''''=+=+-''= w ww .t tl ea rn .n et5.设f (x )具有二阶连续导数，且f (0) = 0,试证g (x ) = (),0'(0),0f x x x f x ⎧≠⎪⎨⎪=⎩可导，且导函数连续. 证: 当0x ≠时,2()()()()()f x xf x f x g x x x '-''==当0x =时,由200000()(0)()(0)()(0)lim lim lim 00()(0)1()(0)1lim lim (0)2202x x x x x f x f g x g f x xf x x x x f x f f x f f x x →→→→→'-'--==--''''--''===- 即 1(0)(0)2g f '''=所以 2()(),0()1(0),02xf x f x x xg x f x '-⎧≠⎪⎪'=⎨⎪''=⎪⎩由(),()f x f x '的连续性知()g x '在0x ≠处连续,又20000()()()()()lim ()limlim 211lim ()(0)(0)22x x x x xf x f x f x xf x f x g x x xf x fg →→→→'''''-+-'=='''''=== 故()g x '在0x =处连续,所以()g x '在(),-∞+∞内处处连续.综上所述,(),0()(0),0f x xg x x f x ⎧≠⎪=⎨⎪'=⎩可导,且导函数连续. 习题4-31.求函数f (x ) =e x x 的n 阶马克劳林公式.解:()()(1),()(1)(2),()()ٛ… x x x x x x k x f x e xe e x f x e x e e x f x e k x '=+=+''=++=+=+w ww .t tl ea rn .n et()()(0)1(0),(1,2,3,)!!(1)!k k f k fk k k k k ∴====- 又 (0)0f =321(1)()(01)2!(1)!(1)!n x n x x e n x f x x x x n n θθθ+++∴=+++++<<-+ 2.当01x =-时，求函数f (x ) = 1x的n 阶泰勒公式. 解:()()[]23()2341()1()112212!3!!()(1),()(1),()(1),,()(1)!(1)(1)!(1)(1)!1,(0,1,2,)!!(1)()(1)1(1)111(1)ٛ … n n n n n n n n n n n n f x f x f x f x x x x x n f n f n n n n x f x x x x x θ-++++''''''=-=-=-=-∴-=-⋅=----==-=+∴=-+-⎡⎤+++++++⎣⎦-++(01)θ<<3.按(4)x -的乘幂展开多项式432()53 4.f x x x x x =-+-+ 解: 函数432()534f x x x x x =-+-+,根据泰勒公式按(4)x -的幂的展开式是2(4)34(4)()(4)(4)(4)(4)2!(4)(4)(4)(4)3!4! f f x f f x x f f x x '''=+-+-'''+-+- 而[][][]432324244(4)(4)454434456,(4)21,41523(4)137,123022!2(4)111,24303!3!(4)12414!4!x x x f f x x x f x x f x f ====-⨯+-⨯+=-'==-+-''==-+'''==-=⨯=所以,234()5621(4)37(4)11((4)(4)f x x x x x =-+-+-+-+-.4.利用泰勒公式求下列极限: w ww .t tl ea rn .n et(1) 30sin limx x x x →-; (2) 21lim ln(1)x x x x →+∞⎡⎤-+⎢⎥⎣⎦. 解: (1) 利用泰勒公式,有34sin ()3!x x x o x =-+所以 343300430()sin 3!lim lim 1()1lim()66x x x x o x x x x x o x x→→→--==-= (2) 利用泰勒公式,有221111ln(1(2o x x x x+=-+,所以222222221111lim lim ln(1(())21()1111lim lim .(1222x x x x x x x x o x x x x o x x o x x →+∞→+∞→+∞→+∞⎡⎤⎡⎤=-+--+⎢⎥⎢⎥⎣⎦⎣⎦⎡⎤⎢⎥⎡⎤==-=-⎢⎥⎢⎥⎣⎦⎢⎥⎣⎦习题4-4 1. 求下面函数的单调区间与极值： (1)32()26187f x x x x =---； (2)()ln f x x x =-； (3)23()1(2)f x x =--； (4)()(4)f x x x =-. 解: (1) 2()612186(1)(3),f x x x x x '=--=+-令()0f x '=得驻点121,3,x x =-=-在()(),,13,-∞-+∞上,()0f x '>,在()1,3-上()0f x '< ∴ ()f x 在(,1],[3,)-∞-+∞上单调增加,在[]1,3-上单调减少.当 1x =-时, ()f x 有极大值,极大值为(1)3f -=, 当 3x =时, ()f x 有极小值,极小值为(3)61f =-.w ww .t tl ea rn .n et(2) 11()1x f x x x-'=-=,令()0f x '=得驻点1x = 在()0,1上,()0f x '<;在()1,+∞上,()0f x '> ∴ ()f x 在(0,1]上单调递减;在[1,)+∞上单调递增. 当1x =时,()f x 有极小值,极小值为(1)1f =.(3) ()()0f x f x ''=≠ 但当2x =时,()f x '不存在, 在(,2)-∞上,()0f x '>;在(2,)+∞上,()0f x '<,∴ ()f x 在(,2]-∞上单调递增;在[2,)+∞上单调递减.当2x =时, ()f x 有极大值,极大值为(2)1f =.(4) 2240()40x xx f x x xx ⎧-≥=⎨-+<⎩ ,则 240()240x x f x x x ->⎧'=⎨-+<⎩且当 0x =时,()f x '不存在,又令()0f x '=得2x =在(,0),(2,)-∞+∞上,()0f x '>,在(0,2)上()0f x '<∴ ()f x 在(,0],[2,)-∞+∞上单调递增;在[0,2]上单调递减;当0x =时,()f x 有极大值,极大值为(0)0f =; 当2x =时, ()f x 有极小值,极小值为(2)4f =-. 2. 试证方程sin x = x 只有一个根.证: 显然0x =是方程sin x x =得一个根(亦可将()sin f x x x =-运用零点定理).令()sin f x x x =-,则()cos 10f x x '=-≤,而()0f x '=的点不是单调区间的分界点,故()f x 在(,)-∞+∞内单调下降,所以()f x 在(,)-∞+∞内只有一个零点,即方程sin x x =只有0x =一个根. 3. 已知()([0,))f x C ∈+∞，若f (0) = 0, f ′(x )在[0,)+∞内存在且单调增加，证明()f x x在［0,＋∞)内也单调增加. 解: 0 x ∀>,由题意知()f x 在[]0,x 上满足拉格朗日中值定理的条件,利用拉格朗日中值定理得,(0,) x ξ∃∈,使w ww .t tl ea rn .n et()(0)()f x f xf ξ'-=, 因 ()f x '在[0,)+∞单调增加,且(0)0f =,所以()()()f x xf xf x ξ''=≤ 即 ()()0xf x f x '-≥令 ()()(0) f x F x x x=>,则 2()()()0xf x f x F x x '-'=≥ 所以()F x 单调递增,即 ()f x x在(0,)+∞内单调增加.4. 证明下列不等式： (1) 1+12xx ＞0； (2)2ln(1)(0)2 x x x x x -<+<>.证: (1) 令1()12f x x =+,则1()(12f x '=, 当 0x >时1,()0f x '<>即()f x 单调递增,从而 ()(0)0f x f >=,故112x +>. (2) 令 2()ln(1)2x f x x x =+-+,则 21()111x f x x x x '=-+=++ 当 0x >时,有()0f x '>,即()f x 单调递增,从而()(0)0f x f >= ,即2ln(1)2x x x +>-又令 ()ln(1)g x x x =-+,则1()111xg x x x'=-=++ 当 0x >时,()0g x '>,即 ()g x 单调递增,从而()(0)0g x g >=,即ln(1)x x >+.综上所述,当0x >时有2ln(1)2x x x x -<+<. 5. 试问a 为何值时，f (x ) = a sin x +13sin ３x 在x =3π处取得极值?是极大值还是极小值?并求出此极值. 解: ()cos cos3f x a x x '=+w ww .t tl rn et若3πx =为极值点,则cos cos 03ππa +=,所以2a =.又()2sin 3sin 3,()03πf x x x f ''''=--=<故函数在3πx =处取得极大值,极大值为(3πf =.习题4 - 51. 某个体户以每条10元的价格购进一批牛仔裤，设此批牛仔裤的需求函数为402Q P =-,问该个体户应将销售价定为多少时，才能获得最大利润? 解: 利润2()10260400L P PQ Q P P =-=-+-,()460L P P '=-+,令 ()0L P '=得 P =15所以应将销售价定为每条15元,才能获得最大利润.2.设 f (x ) = cx α (c ＞0,0＜α＜1)为一生产函数，其中c 为效率因子，x 为投入量，产品的价格P 与原料价格Q 均为常量，问：投入量为多少时可使利润最大? 解: 依题意,总利润()()()L x Pf x Q x P cx Qx α=-=⋅-则 1()L x Pc xQ αα-'=-令 ()0L x '=得 11Q x Pc αα-⎛⎫=⎪⎝⎭所以,投入量为11Q Pc αα-⎛⎫⎪⎝⎭时利润最大.3. 某产品的成本函数为23()156C Q Q Q Q =-+，(1) 生产数量为多少时，可使平均成本最小?(2) 求出边际成本，并验证边际成本等于平均成本时平均成本最小. 解: (1) 2()()156C Q C Q Q Q Q==-+ 令 260()Q C Q '=-=⎡⎤⎣⎦得Q =3 故 生产数量3Q =时,可使平均成本最小. (2) 2()15123MC C Q Q Q '==-+当 3Q =时,15123396MC =-⨯+⨯= 2()156336C Q =-⨯+=w ww .t tl ea rn .n et即边际成本等于平均成本时平均成本最小. 4. 已知某厂生产Q 件产品的成本为C ＝25000＋2000Q ＋1402Q （元). 问：(1) 要使平均成本最小，应生产多少件产品?(2) 若产品以每件5000元售出，要使利润最大，应生产多少件产品? 解: (1) 平均成本 250001()200040C Q Q Q =++ 边际成本1()200020C Q Q '=+. 当()()C Q C Q '=时,平均成本最小,由()()C Q C Q '=即2500011200020004020Q Q Q ++=+ 得1000Q =(负值不合题意已舍去).所以要使平均成本最小,应生产1000件产品. (2)221()5000()500025000200040130002500040L Q Q C Q Q Q Q Q Q =-=---=-+-令 1()3000020L Q Q '=-+=, 得60000Q =(件) 所以应生产60000件产品.5. 某厂全年消耗(需求)某种钢材5170吨，每次订购费用为5700元，每吨钢材单价为2400元，每吨钢材一年的库存维护费用为钢材单价的13.2%，求： (1) 最优订购批量； (2) 最优批次； (3) 最优进货周期； (4) 最小总费用.解: 由题意 215170,5700,1,240013.2%316.8 R C T C ====⨯= 则(1)最优订购批量*431.325q === (2)最优批次 5170*12*431.325R n q ==≈(次) w ww .t tl ea rn .n et(3)最优进货周期 36530.452*12T t n ===(天) (4)最小总费用*136643.9E ==≈(元)6. 用一块半径为R 的圆形铁皮，剪去一圆心角为α的扇形后，做成一个漏斗形容器，问α为何值时，容器的容积最大?解: 设漏斗的底面半径为r ,高为h ,为了计算方便令2ϕπα=-,则2,,2ππR r R r h ϕϕ==== 漏斗的容积2322123(83)πππV hr V ϕϕ==<<'=-令 0V '=得10ϕ=(舍之),2ϕ=,34222237),40,9πππV V ϕϕϕ''=-+-⎫''=-<⎪⎭故当ϕ=时漏斗得容积最大.由2πϕα=-得2π2ππ3α=-=-, 所以,当2ππ3α=-时,容积最大. 7. 工厂生产出的酒可即刻卖出，售价为k ；也可窖藏一个时期后再以较高的价格卖出.设售价V 为时间t 的函数V = k ，(k ＞0)为常数.若贮存成本为零，年利率为r ，则应何时将酒售出方获得最大利润(按连续复利计算). 解: ()e rtrtA t k k -=⋅=令()0rt r A t k ⎫'-==⎪⎭得214t r = 所以,应窖藏214r 时以后售出可获得最大利润. 8. 若火车每小时所耗燃料费用与火车速度的三次方成正比，已知速度为20km/h ，每小时的w w.t tl ea rn .n et燃料费用40元，其他费用每小时200元，求最经济的行驶速度. 解: 设火车每小时所耗燃料费为Q ,则 3Q kv = (k 为比例常数) 依题意得 34020k =⋅, 解得 1200k =, 又设火车行驶()km s 后,所耗费用为, 32200(200)()s E kv kv s v v=+⋅=+ 令 2200()0100v E s v'=-=, 得27.14v =≈ (km/h), 所以,最经济得行驶速度为27.14 km/h.习题 4-61. 讨论下列函数的凸性，并求曲线的拐点：(1) y =2x －3x ； (2) y = ln(1+2x )；(3) y = x e x； (4) y = 4(1)x +＋e x；(5) y =2(3)x x +； (6) y=arctan e x. 解: (1)223,126,0.3令 得 y x x y x y x '=-''''=-==当13x <时,0y ''>; 当13x >时,0y ''<,且12()327f = 所以,曲线23y x x =-在1(,3-∞内是下凸的,在1(,)3+∞内是上凸的,点12(,)327是曲线的拐点.(2) 222222222(1)222(1),1(1)(1)x x x x x y y x x x +-⋅--'''===+++, 令0y ''=得,121,1x x =-=,这两点将定义域(,)-∞+∞分成三个部分区间,列表考察各部分区间上二阶导数得符号.x (,1)-∞-1-(1,1)-1(1,)+∞y' -0 + 0 -y上凸ln 2 下凸ln 2上凸所以,曲线2ln(1)y x =+在(,1)-∞-及(1,)+∞内是上凸的,在(1,1)-内是下凸的,点w ww .t tl ea rn .n et(1,ln 2)±是曲线的拐点.(3) 324(1),12(1)0x xy x e y x e '''=++=++> 所以,曲线在定义域(,)-∞+∞内处处下凸,没有拐点.(4) 343212,(3)(3)x x y y x x --'''==++,令 0y ''=得6x = 当 6x <时,0y ''<,当6x >时,0y ''>;又2(6)27f =,函数的定义域为(,3)(3,)-∞--+∞ ;所以曲线在(,3),(3,6)-∞--内上凸,在(6,)+∞内下凸,点2(6,)27是拐点. (6)arctan 2arctan arctan arctan 2222221112(12)(1)(1)(1)x x x x y e x x x e y e e x x x '=⋅+-''=⋅-⋅=+++令 0y ''= 得 12x =当 12x <时,0y ''>,当12x >时,0y ''<,且 1arctan 21(2e f =,所以曲线在1(,)2-∞内向下凸,在1(,)2+∞内向上凸,点1arctan 21(,)2e 是拐点.2. 利用函数的凸性证明下列不等式：(1)e e 2x y+＞2e x y+， x ≠y ; (2) x ln x +y ln y ＞(x +y )ln2x y+,x ＞0,y ＞0,x ≠y .证: (1) 令()e x f x =,则()e x f x '=,()0e xf x ''=>,所以函数()f x 的曲线在定义域(,)-∞+∞内是严格下凸的,由曲线下凸的定义有: ()(),()()22x y f x f y x y f x y ++∀≠<≠ 即 22e e ex y x y ++< 即 2()2e e e x yx yx y ++>≠.w ww .t tl ea rn .n et(2) 令()ln f x x x =,则1()1ln ,()f x x f x x'''=+=当 0x >时,恒有()0f x >,所以()f x 的曲线在(0,)+∞内是严格下凸的,由曲线下凸的定义有, 0,0,,x y x y ∀>>≠有()()()22f x f y x y f ++>即ln ln ()ln222x x y x y x y+++> 即 ln ln ()ln 2x yx x y y x y ++>+.3. 当a ，b 为何值时，点(1，3)为曲线y =a 3x +b 2x 的拐点. 解: 因为32y ax bx =+是二阶可导的,所以在拐点处0y ''=,而 232,62y ax bx y ax b '''=+=+ 所以 620a b +=又拐点(1,3)应是曲线上的点,所以3a b +=解方程6203a b a b +=⎧⎨+=⎩ 得 39,22a b =-=所以当39,22a b =-=时,点(1,3)为曲线32y ax bx =+的拐点. 4. 求下列曲线的渐近线：(1) y = ln x ; (2)y =22x -; (3) y = 23x x -； (4) y = 221x x -.解: (1) 0lim lim ln x x y x ++→→==-∞,所以ln y x =有垂直渐近线 0x =. 又 lim x y →+∞=+∞,但1ln lim limlim 01x x x y xx y x x→+∞→+∞→+∞====,lim (0)x y x →+∞-⋅=∞, 所以不存在水平或斜渐近线.(2) 220x x -=,所以有水平渐近线0y =,又22lim 0x x x y x -→∞→∞== ,所以没有斜渐近线,w ww .t tl ea rn .n et又函数22x y -=没有间断点,因而也没有垂直渐近线.(3) 221limlim 0331x x xxx x →∞→∞==--,所以有水平渐近线0y =,又函数23x y x ==-有两个间断点x x ==,且22,,33x x x x x x=∞=∞--所以有两条垂直渐近线x =x =又 21lim lim 3x x y x x →∞→∞==∞-,所以没有斜渐近线.(4) 2lim lim 21x x x y x →∞→∞==∞- ,所以没有水平渐近线,又 函数221x y x =-有间断点12x =,且212lim 21x x x →=∞-,所以有垂直渐近线12x =.又 1limlim 212x x y x x x →∞→∞==- 2111lim()lim()lim 22122(21)4x x x x x y x x x x →∞→∞→∞-=-==--所以有斜渐近线1124y x =+. 5.作出下列函数的图形：(1) f (x ) =21xx+； (2) ()2arctan f x x x =- (3) ()2,(0,)e xf x x x -=∈+∞. 解: (1) (i) 定义域为(,)-∞+∞.()()f x f x -=- ,故曲线关于原点对称.(ii) 21lim limlim012x x x x y x x→∞→∞→∞===+ ,故曲线有渐近线0y =. (iii) 222222121,(1)(1)x x x x y x x +-⋅-'==++22223322423232(1)(1)2(1)222442(3)(1)(1)(1)x x x x x x x x x x x y x x x -+--⋅+⋅---+-''===+++, w ww .t tl ea rn .n et令0y '=即210x -=得驻点1x =±,又使0y ''=的点为0,x =.作图如下:图4-1(2) (i) 定义域为(,)-∞+∞.又 ()arctan y x x x y -=-+=-,故为奇函数.(ii) 2arctan lim ,limlim (1)1,x x x y xy x x→±∞→±∞→±∞=∞=-=πlim ()lim (2arctan )(2)()π2x x y x x →±∞→±∞-=-=-±= 所以有渐近线πy x = .(iii) 222211,11x y x x -'=-=++2222222(1)(1)24,(1)(1)x x x x xy x x +--⋅''==++令 0y '=得驻点1x =±,又使0y ''=的点为0x =. 列表如下:w ww .t tl e a rn .e图4-2(3) (i) 定义域为(,)-∞+∞,且()((,))f x C ∈-∞+∞. (ii) ()2(1),()2(2),e e xxf x x f x x --'''=-=-由()0f x '=得1x =,由()0f x ''=得2x =,把定义域分为三个区间 (,1),(1,2),(2,);-∞+∞(iv) lim ()0x f x →+∞=,故曲线()y f x =有渐近线0y =,lim ()x f x →+∞=-∞.(v) 补充点(0,0)并连点绘图,如图所示:图4-3w wl ea rn .n et。
《微积分》各章习题及详细答案
第一章 函数极限与连续之勘阻及广创作一、填空题1、已知x x f cos 1)2(sin +=，则=)(cos x f 。
2、=-+→∞)1()34(lim 22x x x x 。
3、0→x 时，x x sin tan -是x 的阶无穷小。
4、01sin lim 0=→xx k x 成立的k 为。
5、=-∞→x e x x arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续，则=b 。
7、=+→xx x 6)13ln(lim0。
8、设)(x f 的定义域是]1,0[，则)(ln x f 的定义域是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 是非零常数，则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时，1)1(312-+ax 与1cos -x 是等价无穷小，则常数________=a 。
12、函数xx x f +=13arcsin)(的定义域是__________。
13、lim ____________x →+∞=。
14、设8)2(lim =-+∞→xx ax a x ，则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 是],[l l -上的偶函数，)(x h 是],[l l -上的奇函数，则中所给的函数必为奇函数。
（Ａ）)()(x g x f +；（Ｂ）)()(x h x f +；（C ）)]()()[(x h x g x f +；（D ）)()()(x h x g x f 。
2、xxx +-=11)(α，31)(x x -=β，则当1→x 时有。
（Ａ）α是比β高阶的无穷小； （Ｂ）α是比β低阶的无穷小； （C ）α与β是同阶无穷小； （D ）βα~。
实用文档之《微积分》各章习题及详细答案
实用文档之"第一章 函数极限与连续"一、填空题1、已知x x f cos 1)2(sin +=，则=)(cos x f 。
2、=-+→∞)1()34(lim22x x x x 。
3、0→x 时，x x sin tan -是x 的 阶无穷小。
4、01sin lim 0=→xx kx 成立的k 为 。
5、=-∞→x e xx arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续，则=b 。
7、=+→xx x 6)13ln(lim 0 。
8、设)(x f 的定义域是]1,0[，则)(ln x f 的定义域是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 是非零常数，则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时，1)1(312-+ax 与1cos -x 是等价无穷小，则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域是__________。
13、lim ____________x →+∞=。
14、设8)2(lim =-+∞→xx ax a x ，则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题 1、设)(),(x g x f 是],[l l -上的偶函数，)(x h 是],[l l -上的奇函数，则 中所给的函数必为奇函数。
（Ａ）)()(x g x f +；（Ｂ）)()(x h x f +；（C ）)]()()[(x h x g x f +；（D ）)()()(x h x g x f 。
2、xxx +-=11)(α，31)(x x -=β，则当1→x 时有 。
（Ａ）α是比β高阶的无穷小； （Ｂ）α是比β低阶的无穷小； （C ）α与β是同阶无穷小； （D ）βα~。
《微积分》各章习题及详细答案
第一章 函数极限与连续一、填空题1、已知x x f cos 1)2(sin +=,则=)(cos x f 。
2、=-+→∞)1()34(lim22x x x x 。
3、0→x 时,x x sin tan -就是x 的 阶无穷小。
4、01sin lim 0=→xx kx 成立的k 为 。
5、=-∞→x e xx arctan lim 。
6、⎩⎨⎧≤+>+=0,0,1)(x b x x e x f x 在0=x 处连续,则=b 。
7、=+→xx x 6)13ln(lim 0 。
8、设)(x f 的定义域就是]1,0[,则)(ln x f 的定义域就是__________。
9、函数)2ln(1++=x y 的反函数为_________。
10、设a 就是非零常数,则________)(lim =-+∞→xx ax a x 。
11、已知当0→x 时,1)1(312-+ax 与1cos -x 就是等价无穷小,则常数________=a 。
12、函数xxx f +=13arcsin )(的定义域就是__________。
13、lim ____________x →+∞=。
14、设8)2(lim =-+∞→xx ax a x ,则=a ________。
15、)2)(1(lim n n n n n -++++∞→=____________。
二、选择题1、设)(),(x g x f 就是],[l l -上的偶函数,)(x h 就是],[l l -上的奇函数,则 中所给的函数必为奇函数。
(Ａ))()(x g x f +;(Ｂ))()(x h x f +;(C))]()()[(x h x g x f +;(D))()()(x h x g x f 。
2、xxx +-=11)(α,31)(x x -=β,则当1→x 时有 。
(Ａ)α就是比β高阶的无穷小; (Ｂ)α就是比β低阶的无穷小; (C)α与β就是同阶无穷小; (D)βα~。
《微积分》课后答案（复旦大学出版社（曹定华李建平毛志强著））第1章
第一章习题1-11.用区间表示下列不等式的解2(1)9;(2)1;1(3)(1)(2)0;(4)00.011 x x x x x ≤>--+<<<+解(1)原不等式可化为(3)(3)0x x -+≤,其解为33x -≤≤,用区间表示是[-3,3].(2)原不等式可化为11x ->或11x -<-,其解为2x >或0x <,用区间表示是(-∞,0)∪(2,+∞).(3)原不等式的解为21x -<<,用区间表示是(-2,1).(4)原不等式可化为0.0110.0110x x -<+<⎧⎨+≠⎩即 1.010.991x x -<<-⎧⎨≠⎩用区间表示是(-1.01,-1)∪(-1,-0.99).2.用区间表示下列函数的定义域:1(1)(2)arcsin(1)lg(lg );1(3).ln(2) y y x x xy x ==-+=-解(1)要使函数有意义,必须2010x x ≠⎧⎨-≥⎩即011x x ≠⎧⎨-≤≤⎩所以函数的定义域为[-1,0)∪(0,1].(2)要使函数有意义,必须111lg 0x x x -≤-≤⎧⎪>⎨⎪>⎩即0210x x x ≤≤⎧⎪>⎨⎪>⎩所以函数的定义域是12x <≤,用区间表示就是(1,2].(3)要使函数有意义,必须2650ln(2)020x x x x ⎧--≥⎪-≠⎨⎪->⎩即6112x x x -≤≤⎧⎪≠⎨⎪<⎩所以函数的定义域是-6≤x <1,用区间表示就是[-6,1).3.确定下列函数的定义域及求函数值f (0),f),f (a )(a 为实数),并作出图形(1)1,0,2,011,12x x y x x x ⎧<⎪⎪=⎨≤<⎪⎪<≤⎩；(2)y＝211,12x x x ⎧≤⎪⎨-<<⎪⎩解(1)函数的定义域(){|0}{|01}{|12}{|112}(,1)(1,2]或D f x x x x x x x x x =<≤<<≤=<<≤=-∞10(0)200,1,()201112a af ff a a a a ⎧<⎪⎪=⨯===⎨≤<⎪⎪<≤⎩,图1-1图1-2(2)函数的定义域(){|1}{|12}{|2}(2,2)D f x x x x x x =≤<<=<=-221(0)1,11,()112a f ff a a a ≤===-==-<<⎪⎩4.设1,1()1,1x f x x ⎧≤⎪=⎨->⎪⎩,求f (f (x )).解当|x |≤1时,f (x )=1,f (f (x ))=f (1)=1;当|x |>1时,f (x )=-1,f (f (x ))=f (-1)=1,综上所述f (f (x ))=1(x ∈R ).5.判定下列函数的奇偶性：(1)f (x )＝21cos x x-；(2)f (x )＝(x 2＋x )sin x ；(3)f (x )＝1e ,0e 1,0x xx x -⎧-≤⎨->⎩解(1)∵221()1()()cos()cos x x f x f x x x----===-∴f (x )是偶函数.(2)∵222()[()()]sin()()(sin )()sin ()f x x x x x x x x x x f x -=-+--=--=--≠且()()f x f x -≠-,∴f (x )是非奇非偶函数.(3)当x <0时,-x >0,()1(1)()ee xx f x f x ---=-=--=-;当x ≥0时,-x ≤0,()()11(1)()ee e x x xf x f x ---=-=-=--=-,综上所述,x ∀∈R ,有f (-x )=-f (x ),所以f (x )是奇函数.6.设f (x )在区间(-l ,l )内有定义,试证明：(1)f (-x )+f (x )为偶函数；(2)f (-x )-f (x )为奇函数.证(1)令()()()F x f x f x =-+(,)x l l ∀∈-有()[()]()()()()F x f x f x f x f x F x -=--+-=+-=所以()()()F x f x f x =-+是偶函数;(2)令()()()F x f x f x =--,(,)x l l ∀∈-有()[()]()()()[()()]()F x f x f x f x f x f x f x F x -=----=--=---=-所以()()()F x f x f x =--是奇函数.7.试证：(1)两个偶函数的代数和仍为偶函数；(2)奇函数与偶函数的积是奇函数.证(1)设f (x ),g (x )均为偶函数,令()()()F x f x g x =±则()()()()()()F x f x g x f x g x F x -=-±-=±=,所以()()f x g x ±是偶函数,即两个偶函数的代数和仍为偶函数.(2)设f (x )为奇函数,g (x )为偶函数,令()()()F x f x g x =⋅,则()()()()()()F x f x g x f x g x F x -=-⋅-=-=-,所以()()f x g x ⋅是奇函数,即奇函数与偶函数之积是奇函数.8.求下列函数的反函数:22(1)2sin 3;(2);212101,(3)()2(2)1 2.xx y x y x x f x x x ==+-≤≤⎧=⎨--<≤⎩解(1)由2sin 3y x =得1arcsin 32yx =所以函数2sin 3y x =的反函数为1arcsin (22)32xy x =-≤≤.(2)由221x x y =+得21xy y =-,即2log 1y x y =-.所以函数221xx y =+的反函数为2log (01)1x y x x=<<-.(3)当01x ≤≤时,由21y x =-得1,112yx y +=-≤≤;当12x <≤时,由22(2)y x =--得22x y =<≤;于是有1112212y y x y +⎧-≤≤⎪=⎨⎪<≤⎩,所以函数22101()2(2)12x x f x x x -≤≤⎧=⎨--<≤⎩的反函数是1112()212xx f x x +⎧-≤≤⎪=⎨⎪<≤⎩.9.将y 表示成x 的函数,并求定义域:222(1)10,1;(2)ln ,2,sin ;(3)arctan ,().为实数u v y u x y u u v x y u u v a x a ==+======+解(1)211010ux y +==,定义域为(-∞,+∞);(2)sin ln ln 2ln 2sin ln 2vxy u x ====⋅定义域为(-∞,+∞);(3)arctan y u ===(a 为实数),定义域为(-∞,+∞).习题1-21.下列初等函数是由哪些基本初等函数复合而成的?(1)y=；(2)y =sin 3ln x ;(3)y =tan 2x a；(4)y =ln ［ln 2(ln 3x )］.解(1)令arcsin x u a =,则y =再令x v a =,则arcsin u v =,因此y =是由基本初等函数arcsin ,xy u v v a ===复合而成的.(2)令sin ln u x =,则3y u =,再令ln v x =,则sin u v =.因此3sin ln y x =是由基本初等函数3,sin ,ln y u u v v x ===复合而成.(3)令2tan u x =,则uy a =,再令2v x =,则tan u v =,因此2tan x y a=是由基本初等函数2,tan ,u y a u v v x ===复合而成.(4)令23ln (ln )u x =,则ln y u =,再令3ln(ln )v x =则2u v =,再令3ln w x =,则ln v w =,再令ln t x =,则3w t =,因此23ln[ln (ln )]y x =是由基本初等函数2ln ,,ln ,y u u v v w ===3,ln w t t x ==复合而成.2.设f (x )的定义域为［0,1］,分别求下列函数的定义域：(1)f (x 2)；(2)f (sin x );(3)f (x +a ),(a ＞0)；(4)f (e x +1).解(1)由f (x )的定义域为[0,1]得0≤x 2≤1,于是-1≤x ≤1,所以f (x 2)的定义域为[-1,1].(2)由f (x )的定义域为[0,1]得0≤sin x ≤1,于是2k π≤x ≤(2k +1)π,k ∈z ,所以f (sin x )的定义域为[2k π,(2k +1)π],k ∈Z .(3)由f (x )的定义域为[0,1]得0≤x+a ≤1即-a ≤x ≤1-a 所以f (x+a )的定义域为[-a ,1-a ].(4)由f (x )的定义域为[0,1]得0≤e x +1≤1,解此不等式得x ≤-1,所以f (e x +1)的定义域为(-∞,-1].3.求下列函数的表达式：(1)设ϕ(sin x )=cos 2x +sin x +5,求ϕ(x );(2)设g (x -1)=x 2+x +1,求g (x );(3)设1()f x x +=x 2+21x,求f (x ).解(1)法一:令sin t x =,则222cos 1sin 1x x t =-=-,代入函数式,得:22()156t t t t t ϕ=-++=+-,即2()6x x x ϕ=++.法二:将函数的表达式变形得:22(sin )(1sin )sin 56sin sin x x x x xϕ=-++=+-令sin t x =,得2()6t t t ϕ=+-,即2()6x x x ϕ=+-.(2)法一:令1t x =-,则1x t =+,将其代入函数式,得22()(1)(1)133g t t t t t =++++=++即2()33g x x x =++.法二:将函数表达式变形,得22(1)(21)(33)3(1)3(1)3g x x x x x x -=-++-+=-+-+令1x t -=,得2()33g t t t =++,即2()33g x x x =++.(3)法一:令1x t x +=,两边平方得22212x t x++=即22212x t x+=-,将其代入函数式,得2()2f t t =-,即2()2f x x =-.法二:将函数表达式变形,得222111222f x x x x x x ⎛⎫⎛⎫⎛⎫=-=-++++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭令1x t x+=,得2()2f t t =-,即2()2f x x =-.4.设f (x )为奇函数,证明：若f (x )在x =0有定义,则f (0)=0.证∵f (x )为奇函数,且f (x )在x =0处有定义,∴(0)(0)f f -=-又(0)(0)f f -=于是(0)(0)f f =-即2(0)0,(0)0f f =∴=.5.证明：狄利克雷函数是周期函数,任何一个正有理数均是它的周期,但无最小正周期.证狄利克雷函数1,,()0,当为有理数时当为无理数时.x D x x ⎧=⎨⎩设T 是任一正有理数,x ∀∈R ,当x 为有理数时,x+T 为有理数,于是()1D x T +=,又()1D x =,所以()()D x T D x +=;当x 为无理数时,x+T 为无理数,于是()0D x T +=,又()0D x =,所以()()D x T D x +=.综上所述,x ∀∈R 有()()D x T D x +=,所以()D x 是周期函数,任何一个正有理数均是它的周期,又设P 是任一无理数,x P ∃=-∈R ,使()(0)1D x P P +==,而()0D x =,故()()D x P D x +≠,即无理数不是()D x 的周期;因为不存在最小的正有理数,所以()D x 无最小正周期.习题1-31.设销售商品的总收入是销售量x 的二次函数,已知x =0,2,4时,总收入分别是0,6,8,试确定总收入函数TR(x ).解设2()TR x ax bx c =++,由已知(0)0,(2)6,(4)8TR TR TR ===即04261648c a b c a b c =⎧⎪++=⎨⎪++=⎩解得1240a b c ⎧=-⎪⎪⎨=⎪⎪=⎩所以总收入函数21()42TR x x x =-+.2.设某厂生产某种产品1000吨,定价为130元/吨,当一次售出700吨以内时,按原价出售；若一次成交超过700吨时,超过700吨的部分按原价的9折出售,试将总收入表示成销售量的函数.解设销售量为x ,实际每吨售价为P 元,由题设可得P 与x 间函数关系为1307001177001000x P x ≤⎧=⎨<≤⎩,总收入130700()130700(700)1177001000TR x x x x x ≤⎧=⎨⨯+-⨯<≤⎩,即130700()91001177001000TR xx x x x ≤⎧=⎨+<≤⎩.3.已知需求函数为105QP =-,成本函数为C =50+2Q ,P 、Q 分别表示价格和销售量.写出利润L 与销售量Q 的关系,并求平均利润.解由题设知总收入2()105Q R Q PQ Q ==-,则总利润()221()()()8505021055Q L Q R Q C Q Q Q Q Q ⎛⎫=-=-=--+- ⎪⎝⎭,平均利润()150()85L Q AL Q Q Q Q==--.4.已知需求函数Q d 和供给函数Q s ,分别为Q d =100233P -,Q s =-20+10P ,求相应的市场均衡价格.解当d s Q Q =时供需平衡,由d s Q Q =得1002201033P P -=-+,解得5P =所以市场均衡价格5P =.。
微积分第七章习题答案
第七章习题答案习题7.11. A 在第一卦限 ，B 在第五卦限，C 在第三卦限，D 在ｙ轴负向，E 在xoy 面上2. 证明：2227,7,AB AC BC AB AC BC ===+=是直角三解形。
习题7.21. (2,0,0)A =2. 直线3.（1） 222222134y 134x y z x z y ++=++=绕x 轴 绕轴 （2）222222()1y 1x y z x y z -+=+-=绕x 轴 绕轴4.椭球面5. 222216y 216y z x z -=+=平行于x 轴 3绕轴 3习题7.31.（1）()2222,1x y x y a b ⎧⎫+≤⎨⎬⎩⎭(2 ) (){},,1x y x y x y >-≠且2. （1）(2,3)13f -= (2 ) 3212124(,)f x y x xy y =-+ 习题7.4（1）否(2 ) 否2.（1）0 （2）1 （3）2 （4）03. 不连续习题7.5 1.12222222212(1)(1);ln(1)(1)sec sec 1(2);tan tan 1(3);1(4)ln ;;ln y yz z z x x x z z y x x x x yy y z y z x x y y x x y xx xz y z x x x y y x y x u y u z u y y y y y x x y x z x --∂∂=+=++∂∂∂∂=-=∂∂∂-∂==∂+∂+∂∂∂=-==∂∂∂2. (0,1)1,(0,1)2x y f f ''=-=3. 1128111;1arcsin 2x x y y z z x y ====∂∂==+∂∂4.略5. (1,1,1)df dx dy =-6.（1） (1,1,1)du dx dz =-（2）3132222222222()[()()]dz xy x y dx x y y x y dy ---=-+++-+习题7.6 1. 22cos 33123222(1)(62sin )23(2)sec (3)(34)2t t e t t t t t t t +--++-+2. 22222222arctan 2222arctan (1)(sin 2sin )cos (cos sin 2)sin (sin 2sin )(sin )(cos sin 2)cos (2)arctan arctan y y x x y z uv v u v v u v uv v v uz uv v u v u v u v uv v u v v z y x y e x x x y x y z y e y x ∂=-+-∂∂=--+-∂∂=-∂++∂=∂2222y x x y x x y x y +++3. 1122221112323311(1);();()(2)2;2(3);;u u x u y f f f f x y y y z z zz z f x f y x yu u u f f y f yz f x f xz f xy x y z ∂∂∂''''==-+=-∂∂∂∂∂''==∂∂∂∂∂''''''=++=+=∂∂∂ 4. 2;z z x y x x y∂∂=+=∂∂代入，可证。
《微积分》课后答案(复旦大学出版社(曹定华_李建平_毛志强_著))第7章
M 3 (0, 0,5) ，则点 M (4, 3,5) 到 x 轴，y 轴，z 轴的距离分别为：
d x | MM 1 | (4 4) 2 (3 0) 2 (5 0) 2 34. d y | MM 2 | (4 0) 2 (3 3) 2 (5 0) 2 41. d z | MM 3 | (4 0) 2 (3 0) 2 (5 5) 2 5.
5． 在 yＯz 面上，求与三个已知点 A(3，1，2)，B(4，-2，2)和 C(0，5，1)等距离的点． 解：设所求点 P (0, b, c) ,则 | PA || PB || PC | 即 9 (b 1) (c 2) 16 (b 2) (c 2)
2 2 2 2
| AB || AC | ,且 | AB |2 | AC |2 | BC |2
所以 ABC 是等腰直角三角形. 习题 7-2 1．在平行四边形 ABCD 内，设 AB a , AD b ，M 为对角线的交点，试用向量 a 和 b 表示向量 MA, MB, MC 和 MD ． 解： （如图） DC AB a, BC AD b,
2．试用向量证明：如果平面上一个四边形的对角线互相平分，则该四边形是平行 四边形． 证： （如上题图） ，依题意有 AM MC , DM MB. 于是 AB AM MB MC DM DC. 故 ABCD 是平行四边形. 3．已知向量 a=i-2j+3k 的始点为(1，3，-2)，求向量 a 的终点坐标． 解：设 a 的终点坐标为( x, y, z ),则 源自
a ( x 1)i ( y 3) j ( z 2)k ,
而 a i 2 j 3k , 从而有
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3 . 2
4． 在 xＯy 坐标面上求向量 a，使其垂直于向量 b=4i-3j+5k,且｜a｜=2｜b｜． 解：设向量 a ( x, y, 0) ,由 a b 得 a b 0 即 4x 3y 0 , 由 | a | 2 | b | 得 解方程组
(6,10, 2) (6, 6, 6) (16, 4, 12) (16, 0, 20)
5．已知两点 M1(0，1，2)和 M2(1，-1，0)，求向量 M 1M 2 ，并求 M 1M 2 及与 M 1M 2 平 行的单位向量． 解： M 1M 2 (1 0)i (1 1) j (0 2)k i 2 j 2k (1, 2, 2)
2．试用向量证明：如果平面上一个四边形的对角线互相平分，则该四边形是平行 四边形． 证： （如上题图） ，依题意有 AM MC , DM MB. 于是 AB AM MB MC DM DC. 故 ABCD 是平行四边形. 3．已知向量 a=i-2j+3k 的始点为(1，3，-2)，求向量 a 的终点坐标． 解：设 a 的终点坐标为( x, y, z ),则
即与 M 1M 2 平行的单位向量为 ,
1 3
2 2 1 2 2 , 或 , , . 3 3 3 3 3
习题 7-3
) 1． 已知 a =2, b =1, (a,b
解： （1） a a | a | 4
2
，求(1) a·a,(2) a·b,(3) (2a+3b)·(3a-b)． 3 ) 2 1 cos π 1 （2） a a | a | | b | cos(a,b 3
AC AB BC a b BD BA AD b a
于是： MA AM 图 7-1
微积分 复旦大学出版社 曹定华主编 课后答案
x 1 1 y 3 2 z 2 3
解得
x 2 y 1 z 1
所以，向量 a 的终点坐标为（2，1，1）. 4．已知向量 a=(3,5,-1),b=(2,2,2),c=(4,-1,-3)，求 2a-3b+4c． 解： 2a 3b 4c 2(3,5, 1) 3(2, 2, 2) 4(4, 1, 3)
b c 1 2 1 (6) (4) (1) 0
从而 c a 且 c b ，又显然 a 与 b 不平行，于是 c 垂直于 a 与 b 确定的平面，而已知平面 平行于向量 a 与 b,边就平行于 a 与 b 确定的平面，所以 c 也垂直于这个已知平面. 7． 求垂直于向量 a=3i+6j+8k 和 x 轴的单位向量． 解：设所求向量为 xi yj zk ,依题意有：
| AB || AC | ,且 | AB |2 | AC |2 | BC |2
所以 ABC 是等腰直角三角形. 习题 7-2 1．在平行四边形 ABCD 内，设 AB a , AD b ，M 为对角线的交点，试用向量 a 和 b 表示向量 MA, MB, MC 和 MD ． 解： （如图） DC AB a, BC AD b,
x 2 y 2 16 9 25 10 2 ,即 x 2 y 2 200 .
得
4 x 3 y 0 x y 200
2 2
x 6 2 y 8 2
或
x 6 2 y 8 2
所以所求向量 a 为 (6 2,8 2, 0) 或 ( 6 2, 8 2, 0) . 5． 求以 A(1，2，3)，B(3，4，5)，C(2，4，7)为顶点的△ABC 的面积 S． 解： AB (2, 2, 2), AC (1, 2, 4)
微积分 复旦大学出版社 曹定华主编 课后答案
第七章
习题 7-1 1. 略. 2． 求点(a,b,c)关于(1) 各坐标面；(2) 各坐标轴；(3) 坐标原点的对称点的坐标． 解： （1）点（a,b,c）关于 xoy 面的对称点是(a,b,-c); 关于 xoz 面的对称点是（a,-b,c）; 关于 yoz 面的对称点是(-a,b,c); （2）点（a,b,c）关于 x 轴的对称点是（a,-b,-c）; 关于 y 轴的对称点是（-a,b,-c）; 关于 z 轴的对称点是（-a,-b,c）; (3)点（a,b,c）关于原点的对称点是（-a,-b,-c）; 3． 自点 P0(x0, y0, z0)分别作各坐标面和坐标轴的垂线，写出各垂足的坐标． 解：自点 P0 ( x0 , y0 , z0 ) 作 xoy 面的垂线，垂足坐标是 ( x0 , y0 , 0) ; 作 xoz 面的垂线，垂足是 ( x0 , 0, z0 ) ; 作 yoz 面的垂线，垂足是 (0, y0 , z0 ); 自点 P0 ( x0 , y0 , z0 ) 作 x 轴的垂线，垂线是 ( x0 , 0, 0); 作 y 轴的垂线，垂足是 (0, y0 , 0); 作 z 轴的垂线，垂足是 (0, 0, z0 ). 4． 求点 M(4，-3，5)到各坐标轴间的距离． 解：自点 M 分别作 x 轴， y 轴， z 轴的垂线，垂足分别为 M 1 (4, 0, 0) ， M 2 (0, 3, 0) ，
| M 1M 2 | 12 (2) 2 (2) 2 3
MM 2 1 1 2 2 与 M 1M 2 平行的单位向量 e (1, 2, 2) , , 3 | MM 2 | 3 3 3
6． 试证明以三点 A(4，1，9)，B(10，-1，6)，C(2，4，3)为顶点的三角形是等腰直角三 角形． 证： | AB |
(10 4) 2 (1 1) 2 (6 9) 2 36 4 9 7
| AC | (2 4) 2 (4 1) 2 (3 9) 2 4 9 36 7 | BC | (2 10) 2 (4 1) 2 (3 6) 2 64 25 9 7 2
3
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微积分 复旦大学出版社 曹定华主编 课后答案
(3)(a b) (a b) (a b) a (a b) (b) a a b a a b b b 2(a b) 2(5i j 7k ) 10i 2 j 14k
5． 在 yＯz 面上，求与三个已知点 A(3，1，2)，B(4，-2，2)和 C(0，5，1)等距离的点． 解：设所求点 P (0, b, c) ,则 | PA || PB || PC | 即 9 (b 1) (c 2) 16 (b 2) (c 2)
2 2 2 2
4
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3x 6 y 8 z 0 1 x 0 y 0 z 0 2 2 2 x y z 1
M 3 (0, 0,5) ，则点 M (4, 3,5) 到 x 轴，y 轴，z 轴的距离分别为：
d x | MM 1 | (4 4) 2 (3 0) 2 (5 0) 2 34. d y | MM 2 | (4 0) 2 (3 3) 2 (5 0) 2 41. d z | MM 3 | (4 0) 2 (3 0) 2 (5 5) 2 5.
解：(1) a b (3, 1, 2) (1, 2, 1) 3 1 (1) 2 ( 2) ( 1) 3
i
j
k
a b 3 1 2 5i j 7 k 1 2 1
（2） (2a ) 3b 6a b 18
a 2b 2(a b) 2(5i j 7k ) 10i 2 j 14k
1 1 AC (a b) 2 2 1 1 1 MB BM BD (b a) (a b) 2 2 2 1 1 MC AC (a b) 2 2 1 1 MD BD (b a ) 2 2
(3)(2a 3b) (3a b) (2a 3b) 3a (2a 3b) b 6a a 9b a 2a b 3b b 6a a 7 a b 3b b 6 4 7 1 3 1 28
2． 设 a=3i-j-2k,b=i+2j-k,求 (1) a·b 及 a×b； (2) (-2a)·3b 及 a×2b； (3) (a+b)×(a-b)．
i j k AB AC 2 2 2 4i 6 j 2k 1 2 4
S
1 2 1 4 (6) 2 22 14 . AB AC 2 2
6． 设一平面平行于向量 3i+j 和向量 i+j-4k，证明向量 2i-6j-k 垂直于这个平面． 证:令 a 3i j , b i j 4k , c 2i 6 j k , 则 a c 3 2 1 ( 6) 0 (1) 0
3． 已知 a，b，c 为单位向量，且满足 a+b+c=0,计算 a·b+b·c+c·a． 解： 0 ( a b c) (a b c )
a a a b a c b a b b b c c a c b c c | a |2 | b |2 | c |2 2(a b b c c a) 3 2(a b b c c a )