微积分答案 经济数学微积分 主编张建梅 马庆华 科学出版社 广外
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.。
华理高数答案(下)
第 9 章（之 1） （总第 44 次）
教学内容:§9.1 微分方程基本概念 *1． 微分方程 2( y ) 9 y y 5xy 的阶数是
3 7
（ （D）7．
0．
解： 方程变形为
y
2 1 1 y 2 ，是一阶线性非齐次方程，其通解为 x x x
ye
2 2 1 1 x dx x dx c ( ) e dx 2 x x
1 x2
c 1 1 1 1 1 1 c ( 2 ) x 2 dx 2 c x 2 x 2 2 x x x 2 x x
y C cos 2 x 1 C sin 2 x ，实质上只有一个任意常数；
（D）中的函数确实有两个独立的任意常数，而且经验算它也确实是方程的解． *3．在曲线族 y c1e c2 e 中，求出与直线 y x 相切于坐标原点的曲线．
x x
2
解
根据题意条件可归结出条件 y(0) 0, y (0) 1，
2
解：分离变量 2 ye y dy xe 2 x dx ，两边积分就得到了通解
ey
2
1 1 1 ( xe 2 x e 2 x dx) ( xe 2 x e 2 x ) c ． 2 2 2
（3） (2 x 1)e y y 2e y 4 0 ．
ey d y dx 解： ， y 2x 1 2e 4
2
为 y y (2 x yy ) ．
2
微积分 课后习题答案精编版
习题1—1解答 1． 设y x xy y x f +=),(，求),(1),,(),1,1(),,(y x f y x xy f y x f y x f -- 解yxxy y x f +=--),(；x xy y y x f y x y x xy f x y xy y x f +=+=+=222),(1;),(;1)1,1(2． 设y x y x f ln ln ),(=，证明：),(),(),(),(),(v y f u y f v x f u x f uv xy f +++=),(),(),(),(ln ln ln ln ln ln ln ln )ln )(ln ln (ln )ln()ln(),(v y f u y f v x f u x f v y u y v x u x v u y x uv xy uv xy f +++=⋅+⋅+⋅+⋅=++=⋅=3． 求下列函数的定义域，并画出定义域的图形： （1）;11),(22-+-=y x y x f（2）;)1ln(4),(222y x y x y x f ---=（3）;1),(222222cz b y a x y x f ---=（4）.1),,(222zy x z y x z y x f ---++=解（1）}1,1),{(≥≤=y x y x D（2）{y y x y x D ,10),(22<+<=（3）⎫⎩⎨⎧++=),(22222b y a x y xD（4）{}1,0,0,0),,(222<++≥≥≥=z y x z y x z y x D4．求下列各极限： （1）22101limy x xy y x +-→→=11001=+- （2）2ln 01)1ln(ln(lim022)01=++=++→→e yx e x y y x（3）41)42()42)(42(lim 42lim000-=+++++-=+-→→→→xy xy xy xy xy xy y x y x（4）2)sin(lim )sin(lim202=⋅=→→→→x xy xy y xy y x y x5．证明下列极限不存在：（1）;lim 00yx y x y x -+→→ （2）2222200)(lim y x y x y x y x -+→→ （1）证明 如果动点),(y x P 沿x y 2=趋向)0,0( 则322lim lim0020-=-+=-+→→=→x x xx y x y x x x y x ；如果动点),(y x P 沿y x 2=趋向)0,0(，则33lim lim0020==-+→→=→y yy x y x y y x yx所以极限不存在。
微积分(二)课后题答案,复旦大学出版社_第九章
∑ ∑ 而当 b > 1时,
∞ 1 收敛，故 ∞ 1 收敛;
bn
n=1
n=1 a + bn
∑ ∑ ∑ 当
b
=
1时，
∞ n =1
1 bn
∞
= 1发散，故而由 a > 0 ,
n =1
0
<
1 a +1
<
+∞
,故
∞ n=1
a
1 + bn
也发散;
∑ 当 0 < b < 1时， lim n→∞
1 a + bn
=
∑ 而当 x = 1 时，原级数变为调 ∞ 1 ，它是发散的.
n=1 n
∑ 综上所述，当 0 < x < 1时，级数 ∞ xn 收敛.
n=1 n
（2）因为
lim Un+1 U n→∞
n
=
lim
n→∞
(n
+
1)3
⋅
⎛ ⎜ ⎝
x 2
n+1
⎞ ⎟ ⎠
n3
⋅
⎛ ⎜⎝
x 2
n
⎞ ⎟ ⎠
=
x 2
,由达朗贝尔比值判别法知，当
于是 Sn = (ln1− ln 2) + (ln 2 − ln 3) +⋯[ln n − ln(n +1)]
= ln1− ln(n +1) = − ln(n +1)
∑ 故 lim n→∞
Sn
=
−∞
,所以级数
∞ n=1
ln
n n +1
发散.
（6）∵ S2n = 0, S2n+1 = −2
马军主编第三版微积分练习册答案(第1-5单元)
《微积分》练习册参考答案练习1-1一、DDAD 无，二、1、2arcsin(1)2x k π-+；[，2、(5,2)-，3、21,0x x +≠，4、1,0,1x x x-≠；,0,1x x ≠，5、3log (1),1y x x =+>-四、20,)2(2lx x x l V <<-= 五、⎪⎪⎪⎩⎪⎪⎪⎨⎧≥≤≤--≤≤=a t a at a a t t a t t S 2,2,)(210212222， 六、（1）⎪⎩⎪⎨⎧>≤<--≤≤=1600751600100,01.0)10090100090x x x x P ，（， (2)[]⎪⎩⎪⎨⎧>≤<--≤≤=-=1600751600100,01.0)10030100030)60(x x x x x x x P x L ，（，(3)210001000==x L(元)练习1-2一、DDDBCD ，二、1、1/2，2、0；6，3、4/3，4、4，5、0，三、1、1/2，2、0，3、-1，4、1/2，5、1，6、1，7、-1/2，8、2，四、因为00lim ()lim 11;lim ()lim 11;x x x x f x x f x x --++→→→→=-=-=+=所以0lim ()lim ()x x f x f x -+→→≠，所以0lim ()x f x →不存在五、0000||||lim lim 1;lim lim 1;x x x x x x x x x x x x --++→→→→-==-==00||||lim lim ;x x x x x x -+→→≠所以0||lim x x x→不存在 七、111,0,1,lim 1x x x x e e x→+∞→+∞→∴→=当时即；0,x →当时分左右讨论，1110,,;10,,0x x x e x x e x+-→→+∞→+∞→→-∞→当时当时因此1lim xx e →不存在练习1-3一、,,,,,,,,⨯⨯⨯∨⨯⨯∨⨯二、CBCBDD 三、∞,2,0,1,532503020，四、6，41，2，-2，43，31，∞，32，0，32，31，1，五、1,1-==b a ，六、1 练习1-4 一、DCCAC 二、53，2，21-，21，21， x ,21,21-,21,32, 2,-1,3e,2-e,1-e,1-e ,3e ,e ,3-e,1练习1-5一、CCDAD 二、一；2,1,0=x ；0=b ；1，1；2十、⎪⎪⎩⎪⎪⎨⎧==<+>=1 x 1-1 x 01|| ,11||,0)(，，x x x x f ，1=x 为跳跃间断点，1-=x 不为间断点练习2-1一、,,,,,,⨯⨯⨯∨⨯⨯⨯二、B,C,B,C,D,C,C 三、1、1221y x x-'=+，2、33221522y x x -'=--，3、11ln n n y nxx x --'=+，4、22cos sin sin cos sin x x x x x xy x x--'=+，5、sin ln cos ln sin y x x x x x x'=++6、22(1ln )y x x '=-四、0lim ()lim ln(1)0,(0)0,lim ()0x x x x f x x f f x --++→→→→=+====因此f 在0x =处连续0()(0)ln(1)0(0)lim lim 100x x f x f x f x x ---→→-+-'===--，00()(0)0(0)lim lim 100x x f x f f x x +++→→-'===--，因此f 在0x =处可导练习2-2 一、1、222x y a x -'=-，2、23cos sin 222x x y '=-，3、1sin cos cos sin sin n ny n x x nx n x nx -'=- 4、csc y x '=，5、112sin cos y x x x '=-，6、1ln y x x '=，7、221y x '=+，8121x y e x-'=9、23ln33(1ln )xxy x x x '=+++，10、2sin ln(12)12y x x'=-++11、1113[]112(3)2(3)x y x x x x x '=-+-++-+，12、arctan x xy e e -'=- 二、1、122y x y y x +-'=-，2、ln 1y y x y '=-，3、1y ye y xe '=-，4、1x yx y e y e++'=- 三、1、()()()()()x x f x x f x y f e e e f e e f x '''=+，2、211(arcsin )y f xx''=-3、1()()x e x e y f e x e ex -''=++，4、22sin 2((sin )(cos ))y x f x f x '''=-四、1、50km/h 2、()()(1) 1.4/s t s t s km h ''===练习2-3一、DCDBC,DBDBC二、1、22222(1)x y x -''=+，2、1y x ''=，3、222arctan 1x y x x ''=++，4、23(64)x y e x x ''=+ 三、1、232dy bt dx at=，2、cos sin 1sin cos dy dx θθθθθθ-=-- 四、1、x y x y e y dy dx x e ++-=-，2、1()(1)(1)!(1)n n nn y x ---=+，3、()2312ln ln n y x x x x -=+练习3-1一、ABBBA 二、∞，0，0，0，1，1，6，-1/2，32()ab练习3-2 一、⨯，⨯，∨，∨，⨯，⨯，⨯，二、ADBACDCC三、2,递增，（e ，+∞），（0，e ），-n-1， 1n e ---，1/2,3/2,0,0a b c d =-===，6、7略，1x e -=-，四、略，五、1、(,0),(0,2),(2,),22ln 4x -∞↑↓+∞↑=-时有极小值 2、01x x ==-时有最小值0，时有最大值e 3、(,,()-∞⋂⋃⋂+∞⋃，拐点：(0,0),()22- 4、（1）1,2x y x ==+（2）0,x y x == 5、6、8、略7、cos ,sec K x x ρ==一、21x -，12212-+x x ，32-=x y ，211x x --，c x x ++3312ln 2，二、CBDCC 三、c x x +-2325252，c x x +-arctan ，c x x ++cos sin ，c x x +-sec tan c x x +--4ln 3ln 1)43(3，c x x +--cot练习4-2一、21，21，x 2tan 21，3ln 31x-;二、DDDCC ，三、c x +--23)21(31，c a a x +ln 33，c x +3arctan 31，c e x +-1，c x+23arctan 61，c e x ++)1ln(，c x +|ln |ln ，c x +sec 练习4-3 一、CBA ，二、c x x x +--2121arcsin 21，c x ++-)1(212，c x x +-+|1)23(23|ln 312练习4-4c x x x ++-sin cos ，c n x x n n ++-++)11(ln 111，c x x x ++-)1ln(21arctan 2， c x x x +--21arccos ，c x x x +-ln ln ln ln ，c x x x x +++|)tan sec |ln tan (sec 21，c e xe e x x x x +------222，c x x e x ++)cos (sin 21练习4-4c x x +-+--|3|ln 6|2|ln 5，c x x x ++--+-|1|ln |1|ln 11， c xx x ++-|2sin 2cos |ln 2，c x x ++-)1ln(22， c x x ++-+4347)13(274)13(634， c x x +-+-22arctan 222， c e x x +-+-|1|ln ，c eex xx+-22，c x x x ++-sin 2cos 2一、1、0；2、0,2x =；3、1/2；4、5，二、DDCB三、2π，ln 22，323a ，3(1)e -，116，4，122ln 23+练习5-2一、42arctan 2-，22π-，32π+，263，121e --+，1122+，184π-，2，1，1ln 22-，二、1，π三、略练习5-3一、103，12π-，23a π，2343a π，二、1、152x V π=，863y V π=，2、24x V π=，2y V π=，三、121ln 23+，8||a ，。
华南理工大学微积分复习题参考答案
b
a
a b 时， a f (x)dx b f (x)dx ）。理解定积分的几何意义与定积分的基本性
质。掌握变上限的定积分及其导数的定理求函数的导数。掌握牛顿—莱布尼
茨公式。掌握定积分的第一、二类换元法及分部积分法。会用定积分求平面
图形的面积与旋转体的体积。会求无限区间上的广义积分。
2、 无穷级数
利用函数 1 、 ex 、 ln(1 x) 等的麦克劳林展开式将一些简单的函数展开成
1 x
x 的幂级数。
注意到无穷级数的内容不易掌握，因此复习时应有多次反复。还应注意
知识间的联系，例如常数项级数与幂级数之间，前者是后者的基础，后者是
前者的发展，两者的一些公式与方法是相通的。
3、 多元函数微积分
理解无穷级数收敛、发散以及和的概念，了解级数的基本性质（含级数
收敛的必要条件）。熟悉几何级数（即等比级数） aqn （ a 0, q 叫公比）、 n0
调和级数
n1
1
与
p
级数
n1
1 np
(
p
0)
的敛散性，掌握正项级数的比较判别法及
比值判别法。了解交错级数的莱布尼茨判别法，了解任意项级数的绝对收敛
F x, y 0 确定 y y(x) ，求 dy ；由方程 F x, y, z 0 确定 z z(x, y) ，求
dx
z , z 等等。 x y
（3）理解二元函数极值与条件极值的概念，会求二元函数的极值，了
解求条件极值的拉格朗日乘数法，会求解一些比较简单的最大值与最小值的
与条件收敛概念，以及绝对收敛与收敛的关系。
《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
f (0) 0 ,依题意知 f ( x0 ) 0 .即有 f (0) f ( x0 ) .由罗尓定理,至少存在一点 (0, x0 ) ,使
得 f ( ) 0 成立,即
a0 n n 1 a1 (n 1) n 2 … an 1 0
成立,这就说明 是方程 a0 nx n 1 a1 (n 1) x n 2 an 1 0 的一个小于 x0 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
x x x
由 e 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
x
续,在 a, b 内可导,而且 F ( a ) e f ( a ) 0, F (b) e f (b) 0, 即F ( a ) F (b) ,
(4) lim
(a x) x a x ,(a＞0)； x 0 x2
(6) lim sin x ln x ;
x 0
1 ln(1 ) x ; (7) lim x arc cot x
(9) lim(1 sin x) x ;
x 0
1
(8) lim(
x 0
ex 1 ); x ex 1
x 0
f ( x) 在 0,π 上不连续,
显 然 f ( x) 在
0, π
《微积分》课后答案(复旦大学出版社(曹定华 李建平 毛志强 著))第二章
x 0
lim f ( x) lim ex 0
x 0
所以，当 a 0 时， lim f ( x) 存在。
x 0
4. 利用极限的几何意义说明 lim sinx 不存在.
x
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 不存在。
lim xn 0
n
n.
即 xn 0
2. 证明：若 lim xn=a,则 lim ∣xn∣=|a|.考察数列 xn=(-1)n，说明上述结论反之不成立.
ne
由数列极限的定义得
lim xn k a .
t
1
xn k a
天天learn（）为您提供大学各个学科的课后答案、视频教程在线浏览及下载。
此文档由天天learn（）为您收集整理。
第二章
习题 2-1 1. 证明：若 lim xn=a,则对任何自然数 k，有 lim xn+k=a.
n n
证：由 lim xn a ，知 0 ， N1 ，当 n N1 时，有
n
xn a
n.
x 0
1 x
当 x 从小于 0 的方向无限接近于 0 时， e 的值无限接近于 0，故 lim e 0.
x 0
ww
（2）若 lim f ( x) 存在，则 lim f ( x) lim f ( x) ，
x 0 x 0
x 0
由（1）知
lim f ( x) lim ( x 2 a) lim ( x 2 a) a ，
《微积分》课后答案(复旦大学出版社(曹定华 李建平 毛志强 著))第一章
解
(1) ∵ f ( x )
1 ( x)2 1 x 2 f ( x) cos( x) cos x
∴f(x)是偶函数. (2)∵ f ( x) [( x ) ( x)]sin( x ) ( x x)( sin x) ( x x) sin x f ( x)
则
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
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:
ww
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；
w.
tt
x (l , l ) 有 F ( x) f [( x)] f ( x) f ( x) f ( x) [ f ( x) f ( x)] F ( x)
ne
1
(3)原不等式的解为 2 x 1 ,用区间表示是(-2,1).
天天learn（）为您提供大学各个学科的课后答案、视频教程在线浏览及下载。
t
(2) 原 不 等 式 可 化 为 x 1 1 或 x 1 1 , 其 解 为 x 2 或 x 0 , 用 区 间 表 示 是 (-∞,0)∪(2,+ ∞).
1 x2 ； (1) f(x)＝ cos x
(3)f(x)＝
ww
f (0) 1 02 1,
2
2
2
1 1,
1 a2 f (a) 2 a 1
复变函数与积分变(北京邮电大学)课后的习题答案
1 i 1 i 1 i 2 2 2
4、证明：当且仅当 z z 时，z 才是实数．
3
1 1 3 1 8
3 1 3
2
2
3
3
3
证明：若 z z ，设 z x iy ，
3 2 2 2 2 x x 2 y 2 2 xy 2 y x y 2x y i
③解： 2 i 3 2i 2 i 3 2i 5 13 65 ．
2 i 3 2i 2 i 3 2i 2 i 3 2i 4 7i
za 因为 L ={z: Im =0}表示通过点 a 且方 b
向与 b 同向的直线，要使得直线在 a 处与圆相切， 则 CA⊥ L ．过 C 作直线平行 L ，则有∠BCD=β， ∠ACB=90° 故 α-β=90° 所以 L 在 α 处切于圆周 T 的关于 β 的充要条件
习题二xy所以4i的一段即平面上扇形域即是以原点为焦点张口向右抛物线如图所示limlimlimlim的极限不同所以极限不存在limlimlimlim00lim00lim00lim所以fz在整个为正整数所以fz在整个xyxy时才满足cr方程
复变函数与积分变换课后答案（北京邮电大学出版社）
复变函数与积分变换 （修订版）
1 i 3 ∴ Re 1, 2
④解： ∵
3
1 i 3 Im 0． 2
2 2 2 2 π π cos isin i i 2 4 4 2 2 2
微积分(广东外语外贸大学)第六章 多元函数微积分(张建梅)
( x , y )的两个偏导数存在， 且可用下列公式计算
上一页
下一页
目 录
链式法则如图示
u
z
x
y
v
z z u z v , x u x v x
z z u z v . y u y v y
上一页 下一页 目 录
上述定理的结论可推广到中间变量多于两个的情况.
dz z du z dv z dw 如 dt u dt v dt w dt
z
dz 以上公式中的导数 称为全导数. dt
上一页 下一页 目 录
u v w
t
说明: 若定理中
偏导数连续减弱为
偏导数存在, 则定理结论不一定成立. u 2v 2 2 , u v 0 2 2 例如: z f (u, v) u v
四、 内容小结 ★ 思考题 ★ 习题解答 ★ 作业
一、多元复合函数求导的链式法则
1.复合函数的中间变量均为一元函数的情形 定理6.5.1. 若函数
z f (u, v)
则复合函数
处偏导连续,
在点 t 可导, 且有链式法则
z
u
v
t
d z z d u z dv d t u d t v d t
上一页 下一页 目 录
z z u z e sin v , u x y , v x y , 例6.5.1 设 求 , . x y z z v 解: x v x
e sin v
z y
u
e cos v 1
u
z
u v
z v v y
z z u z v u v 1 2 t u t v t t t 当t 0 时， u 0 ，v 0 u du v dv , , t dt t dt
微积分 中国商业出版社经管类 课后习题答案九
微积分中国商业出版社经管类课后习题答案九微积分中国商业出版社经管类课后习题答案九[微积分](中国商业出版社经管类)课后习题答案九《微积分》(中国商业出版社经管类)课后习题答案1．确定下列微分方程的阶数；（1）dy2xy3y5；（2）(y'')25(y')4x70；dx（3）y'''2(y')22y2x5ex sinx.求解：（1）一阶（2）二阶（3）三阶2．验证下列函数是相应微分方程的解，并指出是特解还是通解.其中c,c1,c2是任意常数,1,2是常数；（1）y sin2x,y''4y0；（2）y c1cos x c2sin x,y''2y0；（3）y c1e1x c2e2x,y''(12)y'12y0，（12）；（4）y c1e x c2xe x,y''2y'2y0；（5）y ce3x,y''9y0；（6）y3e2x(2x)ex,y''3y'2y ex.解：（1）特解（2）通解（3）通解（4）通解（5）通解（6）特解3．谋以下微分方程的求解：dy1y23.(1)；（2）xy2y3xyy；2dxy(1x)22（3）xdy ydx0；（4）(x2y)dx(2x3y)dx0；2xdy2ydx(3x5y)dx(4x6y)dy0；（5）（6）求解：1.dy1y2dxxy(1x2)x24y2dx(x0).1ydy2dx2x（1x）1xdy（-）dxx1x1y11ln（1y2）lnx-ln（1x2）lnc22y（-1y2）（1x2） c.x22.dy1y2ydydxcx23121y y lny lnx cdxxy(1x2)221y2(1x2)1x2 3．x2dy y22dx0dyy21x2dx arcsiny ln(xx21)c4．(x2y)dx(2x3y)dy0p q2挑x00,y00y x则u(x,y)(x,y)(0,0)(x2y)dx(2x3y)dy123x4x y2c2213通解为:x24xy y2c22y353x5ydy5．y4x6ydx46xydydu令ux.则u xxdxdx35u4u6u2du323u x(lnx c)2du46udx22u3u1则方程可化为u xdu35u dx46u35u4u6u2du323u x(lnx c)2du46udx22u3u16.2xdy2ydx x24y2dx(x0)dyy1y()2dxx4xy u.y ux.y'u u'.xx令u u'x u1u24经计算可得14cy c2x204．求列下微分方程初值问题的直和：（1）dx4dy0,y(4)2;（2）xdx ye-xdy0,y(0)1；yxx2y2)dx xdy0(x0),y(1)02（3）(y解：（1）xdx4ydy则2y2x2cy(4)2直和1121y(x1)ex则c16直和2y2x21622212y2（2）xexdx ydy则(1x)ex cy(0)1c11直和：y2(x1)ex12ydudyy y（3）令u则u x u u2xdxdxx xu2du1dx即ln u u2lnx cx22y x y y10c0特解：ln0x25．谋以下微分方程的吉龙德或满足用户取值初始条件的直和：22ydy x1;（1）x y x1ex;（2）y'x1dx5（3）3x y'8y ex3x9;（4）y'2y xe x;（5）xy'2y sinx,y求解：（1）xdydyy y x1ex令0dxdxx1;（6）xy'y x1,y1.39则lny lnx c即y cxPR320齐次方程的吉龙德为y c x x则x c x c'x x c x x x1ex则c'xy x x x1ex2x1x1exxdx c x1x1dx c x x lnx c x22y x1中（2）y'x1522q x x1p x x15通解为：y c ec e x1dx e x1dx x12e x1dxdx2ln x12252 2ln x1e x1252e2ln x1dx c x12x12x1d x122x1c x1237（3）x2dy y2dx0y2dx x22两端分数可以得：arcsiny ln x x c(4)x2y dx2x3y dy0y12dy2y x dx3y2x3y2x令y u,y ux,dy u'x ux2u13u2存有u'x u2u13u22uu'x3u2du3u24u11dx3u2x求出u与x的方程,再将u y代入.x24xy3y2c(5)3x5y dx4x6y dy0dy3x5y y dx4x6y46x35令yx u,y ux,y'u'x udy35u u'x u dx46u35u4u6u21u'46ux6u29u316u4x求出u与x的方程,再将u y代入x y2x2y c（6）2x2y2dx•2xy3y2dy0dy2x2y2dx2xy3y2y2x22y y23x x令yx u,y ux,y'u'x u u'x u2u22u3u2求出u与x的方程,再将u y代入2x33xy23y3c6．曲线l是一条平面曲线，其上任意一点p(x,y)(x0)到坐标原点的距离恒等于曲线l在该点切线在y轴上的dT，且l经过点,0.12（1）试求曲线l的方程；（2）谋l坐落于第一象限部分的一条切线，而因切线与l以及两坐标轴所围图形的面积最轻.解：（1）x2y2y y'xx2（2）y'y令u yxdu u x u1u2dx11du dx2x uarcsiny1ln cxx1y xsin ln c x7．设l:y y(x)在点(x,y)处切线的斜率k121求解：y'y13x2y1，且曲线l过点（1，0）.试求曲线l的方x令dy2y0则lny2lnx cdx3x则y则设立吉龙德为y c x1112xy x x1edx c3x22x12x8．物体在加热的过程中温度t(t)的变化率t(t)与物体本身的温度和环境温度之差成正比，比例系数为常数k0.现在(t0)把一个温度为50度的物体放到温度始终保持恒温20度的房间内，谋此物体温度随其时间的变化规律.解：t(t)kt20k1d t20kdtt20ln t20kt ct050c ln30规律：t30e kt209．设立,1,2,,就是虚常数，且0,12，证明以下函数组是(,)上线性毫无关系：（1）e x,xe x；（2）cos x,sin x；（3）eaxcos x,eaxsin x；（4）e1x,e2x,xe2x（2）1常数线性无关xxxee xcos x ctg x常数线性无关sin x（3）eexcosβxeexsinβx ctgβx常数线性无关（4）三者线性毫无关系。
经济数学微积分1、2章 习题答案 最新版
习题一答案(A)1.1. 求下列函数的定义域：(1) 22-+=x x y ; (2) )sin(x y =;(3) 2)1lg(--=x x y ; (4) 22114xx y -+-=; (5) x xx y -++-=11lg21)1arcsin(; (6) ⎩⎨⎧><+=)0(ln )0(12x xx x y . (1)解：022≥-+x x21-≤≥x x 或∴定义域为),1[]2,(+∞--∞ .(2)解：⎩⎨⎧≥≥00)sin(x xπππ+≤≤k x k 22∴定义域为{},1,0,)12(42222=+≤≤k k x k x ππ.(3) 解：⎩⎨⎧≠->-0201x x21≠>x x 且∴定义域为),2()2,1(+∞ .(4)解： ⎩⎨⎧≠-≥-010422x x ⎩⎨⎧±≠≤≤-122x x ∴定义域为]2,1()1,1()1,2[ ---.(5) 解：⎪⎪⎩⎪⎪⎨⎧≠->-+≤-≤-01011111x xxx ⇒ ⎪⎩⎪⎨⎧≠<<-≤≤11120x x x ∴定义域为)1,0[.(6) 解：定义域为),0()0,(+∞-∞ .2已知23)(2-+=x x x f ,求)1(,1),(),1(),1(),0(+⎪⎭⎫⎝⎛--x f x f x f f f f . 解：2200)0(2-=-+=f2231)1(2=-+=f 423)1()1(2-=---=-f232)(3)()(22--=--+-=-x x x x x f231)1(2-+=xx x f 252)1(3)1()1(22++=-+++=+x x x x x f3. 已知⎩⎨⎧≥<+=1ln 113)(x x x x x f ,求)2(),1(),0(f f f .解：1103)0(=+⨯=f01ln )1(==f 2ln )2(=f4. 讨论下列函数的单调性（指出其单调增加区间和单调减少区间） (1) x x y ln +=; (2) xe y =; (3) 24x x -. 解：（1）定义域为),0(+∞，设210x x <<，0)ln (ln ln ln 1212112212>-+-=--+=-x x x x x x x x y y故在定义域内为单调增函数，单调增加区间为),0(+∞. (2) 定义域为实数R,当021<<x x 时，21x x >，021>-x x e e ，函数为减函数； 当210x x <<时，21x x <，021<-x x ee，函数为增函数.故单调减少区间为)0,(-∞,单调增加区间为),0(+∞. (3) 定义域为[]4,0，4)2(422+--=-=x x x y当20≤≤x 时，2)2(--x 为增函数，4)2(2+--x 也为增函数，当42≤≤x 时，2)2(--x 为减函数，4)2(2+--x 也为减函数.故单调增加区间为]2,0[,单调减少区间为]4,2[.5. 判别下列函数中哪些是奇函数,哪些是偶函数,哪些是非奇非偶函数. （1）2x ey -=; (2）x x y sin 2=;（3）242x x y -=; （4）2x x y -=;（5）x x y cos sin -=; （6）x xy +-=11lg; （7）)1ln(2x x y -+=; （8）x xx y cos sin +=;（9）x x xx e e e e y ---+=; （10）⎩⎨⎧≥+<-=0101x xx x y .解：（1）定义域为实数R,)()(22)(x y e e x y x x ===----，故函数为偶函数.(2）定义域为实数R,)(sin )sin()()(22x y x x x x x y -=-=--=-，故为奇函数.（3）定义域为实数R,)(2)(2)()(2424x y x x x x x y =-=---=-，故函数为偶函数.（4）定义域为实数R,函数2x x y -=为非奇非偶函数. （5）非奇非偶函数 （6）定义域为011>+-xx，0)1)(1(>+-x x ，即11<<-x ， 0111lg 11lg )()(==+-+-+=+-lg xxx x x y x y ，即)()(x y x -=-y ，故函数为奇函数. （7）定义域为实数R,01ln )1ln()1ln()()(22==-+++=+-x x x x x y x y ，)()(x y x -=-y ，故函数为奇函数.（8）定义域为),0()0,(+∞-∞ ，)(cos sin )cos()sin()(x y x xx x x x x y =+=-+--=-，故函数为偶函数. （9）定义域为),0()0,(+∞-∞ ，)()(x y ee e e e e e e x y xx xx x x x x -=-+-=-+=-----，故函数为奇函数. （10）)(01010101)(x y x xx x x x x x x y =⎩⎨⎧>+≤-=⎩⎨⎧≥--<-+=-，故函数为偶函数.6. 设)(x f 在),(+∞-∞内有定义,证明：)()(x f x f -+为偶函数,而)()(x f x f --为奇函数.证明：令)()()(x f x f x g -+=，)()()(x f x f x h --=，)()()()(x g x f x f x g =+-=-，)(x g 为偶函数， )()()()(x h x f x f x h -=--=-，)(x h 为奇函数.7. 判断下列函数是否为周期函数,如果是周期函数,求其周期： （1）x x y cos sin +=; （2）x x y cos =; （3）)32sin(+=x y ; （4）x y 2sin =; （5）x y 2sin 1+=; （6）xy 1cos =. 解：（1）)4sin(2)cos 22sin 22(2π+=+=x x x y故函数周期为π2.（2）无周期 （3）周期为ππ==22T（4）22cos 1sin 2xx y -==，周期为ππ==22T（5）设)22sin(1)(2sin 12sin 1T x T x x y ++=++=+= , 解得π=T 2 ，2/π=T .（6）无周期8. 讨论下列函数是否有界：（1）221xx y +=; （2）2x e y -=; （3）x y 1sin=; （4）x y -=11; （5）xx y 1cos =.解：（1）1122≤+=xx y ，故函数有界.（2）02≥x ,02≤-x ,102≤<-x e ,故函数有界.（3）11sin≤x，函数有界. （4）xy -=11无界. （5）xx y 1cos =无界.9. 设21)(x x x f -=,求)(cos x f .解：x x x x x f cos sin cos 1cos )(cos 2=-=10. 已知⎩⎨⎧>-≤+=0102)(2x x x x x f ,求)1(+x f 及)()(x f x f -+.解：⎩⎨⎧->-≤++=⎩⎨⎧>+-+≤+++=+1132011)1(012)1()1(22x xx x x x x x x x f⎩⎨⎧<--≥+=-0102)(2x x x x x f ⎩⎨⎧>-≤+=0102)(2x x x x x f⎪⎩⎪⎨⎧>++=<+-=-+01041)()(22x x x x x x x x f x f 11. 已知x x x f -=3)(,x x 2sin )(=ϕ,求)]([x f ϕ,)]([x f ϕ. 解：x x x f 2sin )2(sin )]([3-=ϕ，)(2sin )]([3x x x f -=ϕ 12. (1) 已知 2211xx x x f +=⎪⎭⎫ ⎝⎛+,求)(x f .(2)已知2ln )1(222-=-x x x f ,且x x f ln )]([=ϕ,求)(x ϕ.解：(1) 2)1(12-+=⎪⎭⎫ ⎝⎛+xx x x f ,2)(2-=∴x x f (2)令12-=x t ，11ln)(-+=t t t f ，xx x x f ln 1)(1)(ln ))((=-+=ϕϕϕ，x x x x =-+=-+1)(211)(1)(ϕϕϕ11112)(-+=+-=x x x x ϕ13. 在下列各题中,求由给定函数复合而成的复合函数,并确定定义域： （1）21,x u u y +==； （2）2,ln ,4xv v u u y ===； （3）x v v u u y 21,sin ,3+===;（4）222,tan ,arctan x a v v u u y +===. 解：（1）21x y +=，),(+∞-∞∈x （2）2ln4x y =,由02>x，),0(+∞∈x（3）)21(sin 3x y +=，),(+∞-∞∈x（4）)](arctan[tan 222x a y +=，由2/)(22ππ+≠+k x a ，有⎭⎬⎫⎩⎨⎧∈-+≠∈Z R k a k x x x ,2,22ππ14. 指出下列各函数是由哪些简单函数复合而成的？ （1）x y alog =; （2）x e y -=arctan ;（3）x y 2sin ln =; （4）⎪⎭⎫⎝⎛-=2212arcsin x xy .解： （1）x y alog =，x u = （2）u y arctan =，v e u =，x v -=（3）u y ln =，2v u =，x v sin = （4）2u y =，v u arcsin =，212x x v -= 15. 求下列反函数及反函数的定义域：（1）)31ln(x y -=,)0,(-∞=f D ； （2）29x y -=,]3,0[=f D ;（3）22-+=x x y ,),2()2,(+∞-∞= f D ; （4）2xx e e y --=,),(+∞-∞=f D ;（5）⎩⎨⎧≤<--≤<-=21)2(210122x x x x y . 解：（1）由)31ln(x y -=解得3/)1(ye x -=,故)1(31x e y -=,),0(1+∞=-f D （2）由29x y -=解得29y x -=，故29x y -=,]3,0[1=-f D（3）由22-+=x x y 解得1)1(2-+=y y x ,故1)1(2-+=x x y ,),1()1,(1+∞-∞=- f D （4）由2x x e e y --=同乘解得x e 解得12++=y y e x ,故)1ln(2++=x x y ,),(1+∞-∞=-f D（5）可解得⎩⎨⎧≤<--≤<-+=2122112/)1(y yy y x故⎪⎩⎪⎨⎧≤<--≤<-+=212211)1(21x x x x y ,]2,1(1-=-f D16. 某玩具厂每天生产60个玩具的成本为300元,每天生产80个玩具的成本为340元,求其线性成本函数,并求每天的固定成本和生产一个玩具的可变成本.解：设玩具的线性成本函数为bx a x C +=)(，则有⎩⎨⎧+=+=b a b a 8034060300 解得⎩⎨⎧==2180b a ，所以x x C 2180)(+= 故固定成本为180(元/每天),可变成本为2(元/每个).17. 某公司全年需购某商品2000台,每台购进价为5000元,分若干批进货.每批进货台数相同,一批商品售完后马上进下一批.每进货一次需消耗费用1000元,商品均匀投放市场（即平均年库存量为批量的一半）,该商品每年每台库存费为进货价格的%4.试将公司全年在该商品上的投资总额表示为批量的函数.解：设批量为x ，投资总额为y ，则x xy 1001021067+⨯+= 18. 某饲料厂日产量最多为m 吨,已知固定成本为a 元,每多生产1吨饲料,成本增加k 元.若每吨化肥的售价为p 元,试写出利润与产量x 的函数关系式.解：设利润为)(x L ，则a x k p x L --=)()( (元) ,],0[m x ∈19. 生产某种产品,固定成本为3万元,每多生产1百台,成本增加1万元,已知需求函数为p Q 210-=（其中p 表示产品的价格,Q 表示需求量）,假设产销平衡,试写出：（1）成本函数；（2）收入函数；（3）利润函数.解：(1) 3)(+=Q Q C (万元)(2) 2215)10(21)(Q Q Q Q P Q Q R -=⋅--=⋅= (万元) (3) 3421)()()(2-+--=Q Q Q C Q R Q L (万元) 20. 某酒店现有高级客房60套,目前租金每天每套200元则基本客满,若提高租金,预计每套租金每提高10元均有一套房间会空出来,试问租金定为多少时,酒店房租收入最大？收入多少元？这时酒店将空出多少套高级客房？解：设每套资金为x 元，酒店房租总收入为y 元，则有16000)400(101)1020060(2+--=--=x x x y ，故400=x 元/套,收入最大，为16000元, 这时酒店将空出20套高级客房.（B ）1. 设x x f x x f =-⎪⎭⎫⎝⎛-+)(212212,求)(x f . 解：令2212-+=x x t ，得2212-+=t t x ，有2212221221)(-+=⎪⎭⎫ ⎝⎛-+-t t t t f t f ,即2212221221)(-+=⎪⎭⎫ ⎝⎛-+-x x x x f x f 又()x x f x x f =--+21)2212(，可解得()11322-++=x x x x f 2. 设下面所考虑的函数都是定义在区间),(l l -上的,证明：（1）两个偶函数的和是偶函数,两个奇函数的和是奇函数；（2）两个偶函数的乘积是偶函数,两个奇函数的乘积是偶函数,偶函数与奇函数的乘积是奇函数.证明：设)(1x f 和)(2x f 为偶函数，)(1x g 和)(2x g 为奇函数， （1）设)()()(21x f x f x f +=)()()()()()(2121x f x f x f x f x f x f =+=-+-=-故)(x f 为偶函数，得证. 设)()()(21x g x g x g +=)()()()()()(2121x g x g x g x g x g x g -=--=-+-=-故)(x g 为奇函数，得证.（2）设)()()(21x f x f x h ⋅=)()()()()()(2121x h x f x f x f x f x h =⋅=-⋅-=-故)(x h 为偶函数，得证. 设)()()(21x g x g x I ⋅=[][])()()()()()(2121x I x g x g x g x g x I =-⋅-=-⋅-=-故)(x I 为偶函数，得证. 设)()()(11x g x f x J ⋅=[])()()()()()(1111x J x g x f x g x f x J -=-⋅=-⋅-=-故)(x J 为奇函数，得证.3. 设函数)(x f 和)(x g 在D 上单调增加,试证函数)()(x g x f +也在D 上单调增加.证明：设D x x ∈<21,[][][][]0)()()()()()()()(12121122>+-+=+-+x g x g x f x f x g x f x g x f∴函数)()(x g x f +也在D 上单调增加.4. 设函数)(x f 在区间],[b a 和],[c b 上单调增加,试证)(x f 在区间],[c a 上仍单调增加.证明: 设[]c a x x ,21∈<,若c x x ≤<21，由题意有)()(12x f x f >， 若21x x b <≤，由题意有)()(12x f x f >， 若21x b x <≤，则)()()(12x f b f x f ≥>，若21x b x ≤<，则)()()(12x f b f x f >≥， 综上，)(x f 在区间],[c a 上仍单调增加.5. 设函数)(x f 和)(x g 在D 上有界,试证函数)()(x g x f ±和)()(x g x f ⋅在D 上也有界.证明：由题)(x f 和)(x g 在D 上有界，即对D x ∈∀，0,021>>∃M M ，有1)(M x f ≤,2)(M x g ≤,则21)()(M M x g x f +≤+,21)()(M M x g x f ⋅≤⋅ 即函数)()(x g x f ±和)()(x g x f ⋅在D 上有界. 6. 证明函数x x y sin =在),0(+∞上无界.证明：对任意0>M ，都存在02[,]x M M π∈+使得1sin 0=x ，则M x x x >=000sin ，即函数x x y sin =在),0(+∞上无界.7. 设)(x f 为定义在),(l l -的奇函数,若)(x f 在),0(l 内单调增加,证明)(x f 在)0,(l -内也单调增加.证明：设)0,(21l x x -∈<，则),0(12l x x ∈-<-，)()()()()()(211212x f x f x f x f x f x f ---=-+--=-)(x f 在),0(l 内单调增加,∴0)()(12>-x f x f ，∴)(x f 在)0,(l -内也单调增加.8. 已知函数)(x f 满足如下方程：0,)1()(≠=+x xcx bf x af其中c b a ,,为常数,且b a ≠,求)(x f ,并讨论)(x f 的奇偶性.解：由已知，xc x bf x af =+)1()(， 令xt 1=，则有ct t bf t af =+)()1(，即cx x bf x af =+)()1(可解得)()(22xabx a b c x f --= ， 而)()(x f x f -=-,故)(x f 是奇函数.习题二答案(A)1. 观察判别下列数列的敛散性；若收敛,求其极限值：(1) nn u 31=; (2) 11ln +=n u n; (3) 212nu n +=; (4) 11+-=n n u n ;(5) nn u n πsin 1=; (6) n u n n )1(-=;(7) nn u )1(3-=; (8) πn nu n cos 1=. 解：(1) 收敛于0; (2) 发散; (3) 收敛于2; (4) 收敛于1; (5) 收敛于0; (6) 收敛于0; (7) 发散; (8) 收敛于0.2. 利用数列极限的分析定义证明下列极限： (1) 011lim=++∞→n n ; (2) 1311lim =⎪⎭⎫ ⎝⎛-+∞→n n ;(3) 532513lim =+++∞→n n n ; (4) 071lim =⎪⎭⎫⎝⎛-+∞→nn .（1）证明：0>∀ε，不妨设1<ε，要使ε<+=-110n u n 成立，只需112->εn 成立，因此取⎥⎦⎤⎢⎣⎡=21εN ，则当N n >时，有ε=<+=-N n u n 1110，所以011lim=++∞→n n .（2）证明：0>∀ε，要使ε<=-n u n 311成立，只需ε31>n 成立，因此取131+⎥⎦⎤⎢⎣⎡=εN ，则当N n >时，有ε<<=-N n u n 31311，即1)311(lim =-+∞→nn . （3）证明：0>∀ε，不妨设101<ε，取152251+⎥⎦⎤⎢⎣⎡-=εN ，则当N n >时，有ε<+<+=-)25(51)25(5153N n u n ，所以532513lim =+++∞→n n n .（4）证明：0>∀ε，不妨设1<ε，取11log 7+⎥⎦⎤⎢⎣⎡=εN ，则当N n >时，有ε<<=-N n n u 71710，所以071lim =⎪⎭⎫⎝⎛-+∞→nn .3. 求下列数列的极限：(1) 98124lim 22++-+∞→n n n n ; (2) 529lim 2+++∞→n n n n ; (3) nn n n n -+-++∞→32lim; (4) )5(lim 2n n n n -++∞→;(5) )11()311)(211(lim 222nn ---+∞→ ; (6) nnn 5151131311lim+++++++∞→ ; (7) )1sin (sin lim --+∞→n n n ; (8) nnn n n 1)4321(lim ++++∞→;(9) ⎪⎪⎭⎫ ⎝⎛+++⋅+⋅+∞→)1(1321211lim n n n ; (10) 11)1(6)1(6lim +++∞→-+-+n n nn n . （1）＝98124lim22++-+∞→n n n n 21/98/1/24lim 222=++-+∞→n n n n n （2）＝529lim2+++∞→n n n n 235219lim =+++∞→nn n （3）nn n n n -+-++∞→32lim32)2(3)3(2lim)2)(3)(3()3)(2)(2(lim =++++=++++-+++++-+=+∞→+∞→n n n n n n n n n n n n n n n n n n（4） )5(lim 2n n n n -++∞→2555lim 5)5)(5(lim2222=++=++++-++∞→+∞→n n n n n n n n n n n n n n n ＝ （5）因为 n n n n n n n 11)11)(11(112+⋅-=+-=-， 所以)11()311)(211(lim 222nn ---+∞→2121lim 11454334322321lim=+=+-⨯⨯⨯⨯⨯+∞→+∞→n n nn n n n n ＝（6）＝nnn 5151131311lim +++++++∞→ 565/1113/111lim =-÷-+∞→n（7）)1sin (sin lim --+∞→n n n21cos )1(21sin 2lim 21cos21sin2lim =-+-+=-+--+∞→+∞→n n n n n n n n n n ＝ （8）＝nnn n n 1)4321(lim ++++∞→4)43()42()41(1lim 41=⎥⎦⎤⎢⎣⎡++++∞→nn n n n （9） ⎪⎪⎭⎫⎝⎛+++⋅+⋅+∞→)1(1321211lim n n n 1)111(lim )111()3121()211(lim 4=+-=⎥⎦⎤⎢⎣⎡+-+-+-=+∞→+∞→n n n n n（10）=-+-++++∞→11)1(6)1(6lim n n n n n 61)6)1(1()6)1(61(lim 111=-+-+++++∞→n n n nn4. 判断下列结论是否正确,为什么？(1) 设数列}{n u ,当n 越来越大时,A u n -越来越小,则A u n n =+∞→lim ;(2) 设数列}{n u ,当n 越来越大时,A u n -越来越接近于零,则A u n n =+∞→lim ;(3) 设数列}{n u ,若对+∈∃>∀Z N ,0ε,当N n >时,有无穷多个n u 满足ε<-A u n ,则A u n n =+∞→lim ;(4) 设数列}{n u ,若对0>∀ε,}{n u 中仅有有限个n u 不满足ε<-A u n ,则A u n n =+∞→lim ;(5) 若}{n u 收敛,则k n n n n u u ++∞→+∞→=lim lim (k 为正整数);(6) 有界数列}{n u 必收敛； (7) 无界数列}{n u 必发散； (8) 发散数列}{n u 必无界.解： (1) 错; (2) 错; (3) 错; (4) 正确; (5)正确; (6) 错; (7) 正确; (8) 错.5. 利用函数极限的分析定义证明下列极限：(1) 539lim22=--→x x x ; (2) 0)21(lim =+∞→x x ; (3) 1)32(lim 2=-→x x ; (4) 02lim 2=-+→x x .证明：（1）0>∀ε，取εδ=，当δ<-<20x 时，有εδ=<-=--2392x x x ，故 539lim 22=--→x x x .（2）0>∀ε，不妨设1<ε，取ε1log 2=M ，则当M x >时，有ε=<M x )21()21(，故0)21(lim =+∞→x x .（3）0>∀ε，取2/εδ=，当δ<-<20x 时，有εδ=<-=--222132x x ，故 1)32(lim 2=-→x x .（4）0>∀ε，取2εδ=，当δ<-<20x 时，有εδ=<-=-22x x ，故 02lim 2=-+→x x .6. 下列函数什么过程中是无穷小量,什么过程中是无穷大量？(1) 21xy =; (2) )2ln()1(+-=x x y ; (3) xe y -=; (4) 2tan x y =;(5) xy -=112; (6) 12322-+-=x x x y . 解：(1) ∞→x 无穷小量,0→x 无穷大量;(2) 1→x 无穷小量,1-→x 无穷小量,+-→2x 无穷大量,+∞→x 无穷大量;(3) +∞→x 无穷小量 ,-∞→x 无穷大量;(4) πk x 2→(k 为整数)无穷小量 ,ππ+→k x 2(k 为整数)无穷大; (5) +→1x 无穷小量,-→1x 无穷大量; (6) 2→x 无穷小量,1-→x 无穷大量. 7. 求下列函数的极限：(1) 852)3)(sin 6(lim 32+--+∞→x x x x x x ; (2) 732523lim 42+--+∞→x x x x x ; (3) 12102)12()31(lim +-∞→x x x x ; (4) )2(lim 22++-∞→x x x x ; (5) 125lim 3++∞→x x x ; (6) 2)2sin(lim --∞→x x x ;(7) )1(lim 33x x x -+∞→; (8) xx x 1lim2++∞→; (9) xx x 1lim2+-∞→;(10) )49(lim +-++∞→x x x a a (0>a 且1≠a )解：（1）0852)3)(sin 6(lim 32＝+--+∞→x x x x x x （2）=+--+∞→732523lim 42x x x x x 23732523lim 432=+--+∞→xx x x x （3）=+-∞→12102)12()31(limx x x x 1210121012121210223)21()31(lim )12()31(lim =+-=+-∞→∞→x x x x x x x x x（4）)2(lim 22++-∞→x x x x122lim2)2)(2(lim 222222222-=+--=+-+-++=-∞→-∞→x x x xx x x x x x x x x x x（5）∞=++∞→125lim3x x x （6）02)2sin(lim=--∞→x x x（7）)1(lim 33x x x -+∞→)1(11lim)1(1))1(1)(1(lim32333232333232333233=-+-⋅-=-+-⋅--+-⋅--+=∞→∞→x x x x x x x x x x x x x x x x（8）11lim2=++∞→xx x （9）11lim2-=+-∞→xx x （10）当10<<a 时，=+-++∞→)49(lim x x x a a 149=-当1>a 时，)49(lim +-++∞→x x x a a049)49)(49(lim=+++++++-+=+∞→xxx x x x x a a a a a a8. 求下列函数的极限：(1) )153(lim 22--→x x x ; (2) 11lim 1--→n m x x x (n m ,为正整数);(3) 11lim31--→x x x ; (4) ⎪⎭⎫⎝⎛---→121lim 21x x xx ;(5) 22lim 2-→x xx ; (6) 3152lim 23--+→x x x x ;(7) 2211limx x x +-→; (8) ⎪⎭⎫⎝⎛+-++--→x x x x x x 212112lim ;(9) x x xx -----→111lim 1; (10) 1lim 21--+++→x nx x x n x .解：（1）1)153(lim 22＝--→x x x（2）＝11lim 1--→nm x x x nm x x x x x x n m x =+++-+++---→)1)(1()1)(1(lim 111（3）=--→11lim31x x x 32)1)(1)(1()1)(1)(1(lim33233231=+++-+++-→x x x x x x x x x （4）=⎪⎭⎫ ⎝⎛---→121lim 21x x xx 2312lim 12)1(lim 22121=--+=--+→→x x x x x x x x （5）由022lim2=-→xx x ，有∞=-→22lim 2x xx（6）=--+→3152lim 23x x x x 83)3)(5(lim 3=--+→x x x x（7）=+-→2211limx x x 2)1(1)11(lim 2220-=+-++→x x x x （8）=⎪⎭⎫⎝⎛+-++--→x x x x x x 212112lim 4)1(2lim 22lim 1221=-=+-+--→-→x x x x x x x x x （9）=-----→xx x x 111lim 11111lim 1-=---→x x（10）1lim 21--+++→x nx x x n x2)1(21)1()1(1[lim 1)1()1()1(lim 1121+=+++=+++++++=--++-+-=-→→n n n x x x x x x x n x n x9. 求下列各题中的常数a 和b :(1) 1112lim 23=⎪⎪⎭⎫ ⎝⎛++-+∞→x x b ax x ; (2) 51lim 21=-++→x abx x x ;(3) k b ax x x x =--+++∞→)1(lim 2(k 为已知常数).解：(1)因为11)2(11222323+-+++-=++-+x b ax bx x a x x b ax 若1)112(lim 23=++-+→∝x x b ax x ，则02=-a ，1=b ，即2=a ，1=b . （2）因为01lim )1(lim 1)1(lim )(lim 2112121=-++-=-++⋅-=++→→→→xa bx x x x a bx x x a bx x x x x x所以01=++b a ，即a b --=1511))(1(lim 1)1(lim 1lim 12121=-=---=-++-=-++→→→a xa x x x a x a x x a bx x x x x 故6=a ，7-=b . （3）因为kbax x x b x ab x a b ax x x x x =++++-+-+-=--++∝+→∝+→11)21()1(lim)1(lim 22222因此012=-a ，0>a ，k a ab =+-121，求得1=a ，k b -=21.10. 求下列函数极限： (1) x x x 3arcsin 4arctan lim0→; (2) xxx 3sin 2tan lim 0→;(3) x x x 1sin lim ∞→; (4) 2)4sin(lim 22--→x x x ;(5) 2220)cos 1(tan lim x x x x -→; (6) )1cos 1(lim 2xx x -∞→;(7) 30sin 1tan 1limxx x x +-+→; (8) x x xx x sin 3sin 2lim 0+-→;(9) x x x x 2sin 5tan lim0-→; (10) hxh x h sin )sin(lim 0-+→.解：（1）3434lim 3arcsin 4arctan lim 00＝＝x x x x x x →→（2）3232lim 3sin 2tan lim 00==→→x x x x x x（3）=∞→x x x 1sin lim 1/1)/1sin(lim =∞→xx x（4）=--→2)4sin(lim22x x x 424lim 22=--→x x x （5）=-→2220)cos 1(tan limx xx x 4)2/(lim 2240=→x x x （6）=-∞→)1cos1(lim 2x x x 2121lim 22=⋅∞→xx x（7）=+-+→30sin 1tan 1limx xx x 4121lim 212)cos 1(tan lim 32030=⋅=-→→xxx xx x x x （8）=+-→x x xx x sin 3sin 2lim041/)(sin 3/)(sin 2lim0=+-→x x x x x （9）=-→x xx x 2sin 5tan lim03252sin lim 5tan lim 00=-=-→→xx x x x x（10）=-+→h x h x h sin )sin(lim 0x hh x h h cos ]2/)2cos[()2/sin(2lim 0=+→11. 求下列函数极限： (1) xx x)11(lim -∞→; (2) x x x 2cot 20)(sec lim →;(3) 121011lim +→⎪⎭⎫⎝⎛+xx x ; (4) xx x x ⎪⎭⎫⎝⎛+-∞→22lim ;(5) 311lim +∞→⎪⎭⎫⎝⎛-+x x x x ; (6) xx x-→111lim .解：（1）=-∞→x x x )11(lim 1)1()11(lim ---∞→=-e xx x（2）=→xx x 2cot20)(sec lim e x xx =+→2tan120)tan 1(lim（3）121011lim +→⎪⎭⎫⎝⎛+xx x21)1(21100210)11(lim 11lim )11(lim -+⋅+→→→=+-=+⋅+=exxx x x x x x x x x（4）xx x x ⎪⎭⎫⎝⎛+-∞→22lim42)4(422)4(42)241(lim )241(lim )241(lim --∞→-⋅+-∞→--⋅+-∞→=+-⋅+-=+-=e x x x x x x x x （5）=⎪⎭⎫⎝⎛-++∞→311lim x x x x 212213)121(lim )11(lim )11(lim e x x x x x x x x x x =-+=-+⋅-++⋅-∞→∞→∞→（6）=-→xx x111lim 1)1(111)11(lim --⋅-→=-+e x x x12. 求下列函数极限： (1) )6sin(sin 21lim6ππ--→x x x ; (2) xxx 251ln lim0+→;(3) )21ln()31ln(lim x x x ++-∞→; (4) 1arcsin lim 20--→x x e xx ;(5) )1ln(121lim2x x x x ---→; (6) x e x x 21lim3sin 0-→;(7) xx x 1)tan 21(lim ++→; (8) xxx e x 10)(lim +→;(9) x x x x 3)421ln(lim 20+-→; (10) )4tan()2tan(lim 4x x x -⋅→ππ;(11) xx x 1)sin 1(lim -→; (12) x x x 2cot 10)(cos lim +→.解：（1）令6π-=x t ，6π→x 时0→t ，原式化为)6sin(sin 21lim6ππ--→x x x3)sin 3(lim cos 1lim ]2/)(cos 2/)(sin 3[21limsin )6/sin(21lim0000-=-+-=+-=+-→→→→tt t t tt t tt t t t t π＝（2）＝x x x 251ln lim0+→45252122/)51ln(lim 0=⋅=+→x x x（3）＝)21ln()31ln(limx x x ++-∞→0)23(lim 23lim ==-∞→-∞→xx x x x （4）＝1arcsin lim2--→xx e x x 1lim220-=-→x x x （5）=---→)1ln(121lim 20x x x x 12/)2(lim 220=--→xx x（6）＝xe x x 21lim3sin 0-→6123/)(sin lim 0-=-→x x x（7）xx x 1)tan 21(lim ++→ 2tan 2limtan 210tan 22tan 2100)tan 21(lim )tan 21(lim e x x xxx x x x x x x =+=+=+→++⋅→⋅→（8）=+→xxx e x 1)(lim 2111)11(lim e e x x e x e x x x x x =-++-+⋅-+→（9）=+-→x x x x 3)421ln(lim2032324lim 20-=-→x x x x （10）令x t -=4π，4π→x 时0→t ，)4tan()2tan(lim 4x x x -⋅→ππ21tan 2tan 1lim 2cot lim )22/tan(lim 2000=-⋅==⋅-=→→→t t t t t tt t t t π（11）=-→xx x 10)sin 1(lim 1sin sin 10)sin 1(lim --⋅-→=-e x xx x x（12）xx x 2cot 10)(cos lim +→21tan 1cos 1cos 10cot 022)1cos 1(lim cos lim cos lim --⋅-→→→=-+=⋅=ex xx xx x x xx x13. 证明：0)2(1)1(11lim 222=⎥⎦⎤⎢⎣⎡+++++∞→n n n n . 证明：222221)2(1)1(11)2(1n n n n n n n +≤++++≤+ 又由01lim )2(1lim22=+=++∞→+∞→n n n n n n所以0])2(1)1(11[lim 222=+++++∞→n n n n 14．求下列函数的间断点,并判断类型.)1( 1212)(11+-=xxx f ; (2) ()x x x x f 21)1ln()(2--=;(3) ⎪⎩⎪⎨⎧-=-≠+-=10111)(2x x xx x f ; +++++++++++++++++++++++ (4) ⎪⎩⎪⎨⎧≥+<≤+<=23212416)(2x x x x x x f 解：（1）11212lim/1/10-=+--→x x x 12/112/11lim 1212lim /1/10/1/10=+-=+-++→→xxx x x x即0=x 为跳跃间断点.（2）0)21()1ln(lim20=--→x x x x ，即0=x 为可去间断点. 0)21()1ln(lim 20=--→x x x x ，即21=x 为无穷间断点. （3）211lim 21=+--→xx x ，即1-=x 为可去间断点.（4）10)(lim 2=-→x f x ，7)(lim 2=+→x f x ，即2=x 为跳跃间断点. 15. 讨论下列函数的连续性:(1) ⎪⎩⎪⎨⎧≥<=)0(0)0(1sin)(2x x xx x f ;(2) ⎪⎩⎪⎨⎧=≠=)0(1)0(sin )(x x xx x f ;(3) ⎪⎩⎪⎨⎧=≠=003)(1x x x f x. 解：(1) 0<x 时，xx x f 1sin)(2=连续， 0>x 时，0)(=x f 连续，0=x 时，)(lim 0)(lim 0_x f x f x x +→→==连续， 所以)(x f 在),(+∞-∞连续.（2） 0<x 时，x xx f sin )(-=连续， 0>x 时，xxx f sin )(=连续，0=x 时，1)sin (lim )(lim 00-=-=--→→xxx f x x ，1sin lim )(lim 00==++→→xxx f x x ， 所以)(x f 在0=x 处不连续.（3）0≠x 时，xx f 13)(=连续，03lim )(lim 10==--→→x x x x f ，∞==++→→xx x x f 103lim )(lim ， 所以)(x f 在0=x 处不连续. 16. 确定常数b a ,使下列函数连续：(1) ⎪⎩⎪⎨⎧+<<--=其他53541)(2bxa x x x f ; (2) ⎪⎪⎩⎪⎪⎨⎧>+=-<=01sin 010sin 1)(x b x x x a x xx x f .解：(1)若)(x f 在54-=x ，53=x 处连续，则有2)54()54(1lim)(lim x bx a x x -=++--→-→，)(lim 1lim)53(2)53(bx a x x x +=-+-→→，即⎪⎪⎩⎪⎪⎨⎧=+=-54535354b a b a ，解得⎪⎪⎩⎪⎪⎨⎧==7175b a （2）)(x f 在0=x 处连续，1sin 1lim 0-=-→a x x x ，1)sin 1(lim 0-=++→a b x xx ， 有11=-a ，1-=a b ，解得2=a ，1=b . 17. 试证方程0133=--x x 在区间)2,1(内至少有一个实根. 证明：令13)(3--=x x x f ，)(x f 在)2,1(连续，03)1(<-=f ，01)2(>=f ，由零点定理知，至少存在一点)2,1(0∈x ，使得0)(0=x f 成立，即 方程0133=--x x 在区间)2,1(内至少有一个实根. 18. 试证方程2-=xe x 在区间)2,0(内至少有一个实根. 证明：令2)(+-=xe x xf ，)(x f 在)2,0(连续，01)0(>=f ，04)2(2<-=e f ，由零点定理知，至少存在一点)2,0(0∈x ，使得0)(0=x f 成立，即 方程2-=xe x 在区间)2,0(内至少有一个实根.(B)1. 求极限)31ln()21ln(lim x x x +++∞→.解：)31ln()21ln(lim x x x +++∞→3ln 2ln )3/11ln(3ln )2/11ln(2ln lim)3/11ln(3ln )2/11ln(2ln lim=++++=+++++∞→+∞→x xx x x x x x x x ＝2. 设nn n n n u n ++++++=2222211 ,求n n u +∞→lim . 解：因为1212122++++≤≤++++n n u n n n n ， 212/)1(lim 21lim 22=++=++++∝+→∝+→nn n n n n n n n ， 2112/)1(lim 121lim 22=++=++++∝+→∝+→n n n n n n n ， 由极限存在定理可知，21lim =∝+→n n u . 3. 设数列}{n u ：,2,,222,22,21-++++n u ,证明：n n u +∞→lim 存在,并求此极限值.证明：首先证}{n u 单调增加。
《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
微积分中国商业出版社_课后习题答案详解二
解：1） lim (x + h)3 − h3 = lim x3 + 3x2h + 3xh2 + h3 − x3 = lim (3x2 + 3xh + h2) = 3x2
h→0
h
h→0
h
h→0
2) lim xn −1 = n
x→1 x −1
3）
⎛ xl→im+∞⎜⎜⎜ arctan
x
+
1
2x
⎞ ⎟ ⎟⎟
（7） lim
2x +1 − 3 ;
x→4 x − 2 − 2
（2） lim xn −1 ;
x→1 x −1
（4）
⎛ xli→m1⎜⎜⎝
x
x −1
−
1 x2 −
x
⎞ ⎟⎟ ; ⎠
（6） lim
1− x −3 ;
x→∞ 2 + 3 x
（8） lim ( x2 + x + 1 − x2 − x − 3) .
n→∞ 5 + 9( 3)n
=
1 5
5
9．下列数列{xn} ，当 n → ∞ 时是否是无穷小量？
（1）
xn
=
1050 3n
;
[ ] （2） xn = 1+ (−1)n 1 ; n
（3） xn = n n .
解：1)是无穷小量 因为 lim xn = 0
n→∞
2)是,因为 lim xn = 0 ( n 为奇数或者偶数)
x → −1+
（4） lim 10x = 0 .
x → −∞
解：1)对于任意给定的 ξ ,要使 δψξ 成立,只要使 4 n > 1 即
数学分析梅加强。答案
数学分析梅加强.答案[篇一：南京大学基础数学考研参考书目]思想政治理论②201 英语一③627 数学分析④801 高等代数复试科目:2305 通信与信息系统专业综合参考书目：《数学分析》梅加强著,高等教育.《高等代数》丘维声编,科学.复试参考书目:《实变函数与泛函分析概要〔第一册〕》〔第二版〕郑维行、王声望编,高等教育.《常微分方程教程》丁同仁、李承治编,高等教育.《代数学引论》聂灵沼、丁石孙编著,高等教育.《概率论基础》李贤平著,高等教育.《数值计算方法〔上、下册〕》林成森编著,科学.参考资料：《南京大学801高等代数考研专业课复习全书》聚英南大〔含真题与答案解析〕《2017南京大学801高等代数考研专业课历年真题与答案解析》[篇二：国内常见数学教材评价.doc]orich,数学分析〔两卷〕作者是s.p.novikov的学生,写本书的时候还很年轻.研究也作的很好,20##国际数学家大会上几何组作过45分钟报告.说句实话,要是把这两卷学下来〔包括习题〕,可能许多博导也做不到.如果作为教材去学,确实不容易,清华数学专业就用的这个,听说第二卷也比较困难.但用来自学还是很好的张筑生数学分析新讲<共三册>这个张老师是十年动乱后的较早期的北大博士之一, 20##2月因病去世,基础绝对过硬,还写过《微分动力系统》与《微分拓扑新讲》两本书,做过几年imo的领队或教练第一册的最后介绍万有引力的证明,其实这个内容也应该教授给工科学生.和国内大多教材差不多,可惜没有习题.邹应数学分析作者是武汉大学的,书学的法国.可惜我没见过他,当我知道他的时候,已经去了.可以用来参考,当然包括习题.我知道它曾经是武大中法班的教材,我的许多老师应该就是受的它的教育常庚哲,史济怀数学分析教程〔两册〕第一作者曾经是imo的领队或教练,中国科技大学的.内容选材和处理都很好,被称为经典.习题也不错,稍微有点难.l.loomis,s.sternberg advanced calculus这两位都是美国数学学派的人,当然其祖上也来自德国.作为研究生的教材,其实适合所有方向的学生.它本来就是mit的研究生教材齐民友重温微积分齐老师是绝对的院士水平,多本名著的译作,近来很关心本科教学.作为为高年级的参考书是很适合的,读过后会很有收益的.尤其是会学到许多新的知识s.m.nikolski 数学分析教程〔两卷〕很长寿的老一辈数学家,已经105岁了 .研究领域是逼近论v.i.smirnov 高等数学教程〔五卷〕圣彼得堡学派的传人.这两部俄罗斯教材的特点是比较全面,一般不易做为教材,但做为参考是很不错的m.fitzpatrick 高等微积分作者倒是没多大名气,但这部马里兰大学的微积分教材很值的借鉴.推荐理由当然也有个人因素,因为我对马大很熟悉k.kodaira 微积分入门陶哲轩实分析小平是日本的数学之神,相信大家对他很熟悉；小陶被誉为世界最聪明的数学家,是奥数培养起来的,想想国内的奥数教育,虽然也有些年头了,但没见什么成效.这两部教材有点象,很注重数学基础,但小陶的书缺少多元积分这部分很有用的内容,可能更适合准备多年学数学的学生梅加强数学分析讲义richard courant 微积分和数学分析引论梅老师是科大少年班的优秀学生,现在南京大学,这是一部很好的数学分析教材,不过有机会得问问梅老师,为什么没有正式出版,看看那么多烂教材都出来骗人,觉得有点遗憾.courant是世界级的应用数学大师,hilbert的得意门生,自己也有许多得意门生.强人易惹人,richard与商人和官方有密切的关系,因此招惹了不少人.他把自己的女儿<二婚>嫁给了moser,侵占了《数学是什么》另一作者robinson<著名的女数理逻辑学家,因其姐姐reid是hilbert的传记作者和richard熟识>的,后因robinson多方努力才使其名字见于书中.这两本书里对许多问题的处理很有特色,还有些有趣而且有用的例子和习题.我自己在教学中就吸纳了不少他们的处理办法和例子陈天权数学分析讲义3卷陈老师据说是当年北大的大才子,毕业后去了##大学,我上大学的那年他已经去了清华,没有听过他的课.被大家称为国内唯一可与v.a.zorich,数学分析比肩的分析教材高等代数--线性代数-空间解析几何-近世代数-数论postnikov 解析几何学与线性代数<第一学期>postnikov 解析几何学与线性代数<第二学期>作者水平应该很高,反正他的学生s.p.novikov是很有名气,他也研究拓扑.书写的绝对好.这套书还有一些分册,但只能找到俄语版.解析几何可以说很重要,但学起来又觉的没什么内容.学会第一本应该就可以了.第二本是线性代数和部分初等微分几何,内容讲的很清晰.a.i.kostrikin,代数学引论<共三卷>这三卷都值得一读,尤其是第二,三卷,作者毕竟是前苏联通讯院士.他是纯粹的俄罗斯学派的传人,其祖上是俄本土数学大家chebevshev.s的学生,这个沙老师作为苏联人,居然有点反对十月革命,结果被学校停了职,也不知道解体后的情况怎么样,水平是很高.克老师这么优秀的人物,可惜没有培养学生.书很好,但学起来不容易,有些抽象,其实这已经是作者的简化版了.m．artin 代数s．lang 线性代数导论很害羞的法国人,不过这个色狼很能写书,把他写的书都学会了,也成了大半个全能数学家了.把这两位放一块是因为他们有关系,色狼是m．artin 的父亲e．artin 的学生,m是以严厉著称的代数几何学家扎老师的学生,据说在扎老师那学习很难毕业,不过他的学生可真是争气.e．artin 是哥学派的,据说他的文章不多,才50多篇,但每篇都是精品.第一本是非常优秀的本科教材,美国几个名校都用.作者是地道的代数几何学家,但教材里看不出作者的倾向,是所有教师的榜样,就是要敢于讲授自己不从事的领域的内容.s．lang是出色的数学家,优秀的教师,它的这本书曾经很畅销n.jacobson lectures on abstract algebra〔三卷〕是个犹太人,代数方面的权威,但被pontryagin贬的一塌糊涂,本来是国际数学联盟主席的候选,但被庞瞎子抵制下去了. 上面俄罗斯人写了三卷,美国人也写这么多,可见代数的重要.作为教科书其实不太适合,有点太代数了.但参考是可以的rotman 高等近世代数作者写过好几本代数方面的著作,要追究其师源,居然是物理学大家maxwell,当然他也是伟大的数学家.rotman所有的书都有个最大的优点,就是介绍名词的来历.学了它应该会对数学有更深刻的认识.hardy g.h., wright e.m. an introduction to the theory of numbers〔中文〕 hardy的大名在数学界应该很响,看起来挺帅的一个英国人,但他老自己觉的自己丑.wrigh是他的学生.很优秀的数论教材华罗庚数论导引华老师是个天才,包括学识和领导才能.这本书的选材不错,比较适合作为教材serre j.-p a course in arithmetic布学派第二代的领袖,研究范围很广,荣誉得了一大堆.有名的数学大奖他都拿了.作为法国数学的代表人物,和阿老师有争论.最大的特点就是薄,内容还不少,难怪他老获奖.atiyah m., macdonald i.g. introduction to mutative algebra20世纪后半叶英国数学的代表,不但自己出色,其学生也很优秀.第二作者这个麦当劳就是他的学生,也在国际数学家大会上做过报告大久保进群论引论这个作者不熟悉,但日本人写的东西还是不错的,很干脆,解释的不错.看来一定是自己学的时候用了不少功夫klingenberg 线性代数与几何老柏林学派的传人,现在很难见到德国的教科书了,或许是因为德国的数学稍有没落吧.我们国内也有许多工科开设的线性代数课程也叫这个名字,但我们的所谓几何太不象几何了.强烈希望所有学数学的学生学点真的几何a.j.khinchin 连分数莫斯科学派第二代中的优秀数学家写了许多部优秀的著作,这一本更以其精致透彻而受到大家的青睐[篇三：南京大学运筹学与控制论考研考试科目]思想政治理论②201 英语一③627 数学分析④801 高等代数2106 近世代数；2110 概率论；2111 计算方法参考书目：《数学分析》梅加强著,高等教育.《高等代数》丘维声编,科学.参考资料：《南京大学801高等代数考研专业课复习全书》聚英南大〔含真题与答案解析〕《2017南京大学801高等代数考研专业课历年真题与答案解析》《2017南京大学627数学分析考研专业课历年真题与答案解析》。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
上一页
下一页
目 录
7. (2) 计算积分 I
y x
1 2 1 4
dy 1 e dx 1 dy
2 2
y
y x
1
y y
e dx .
y x
解 e d x 不能用初等函数表示
先改变积分次序.
原式
y x
y x2
I
1
1 2
1
1 2
dx
x x
2
e dy
x
y x
dxdy ,
解 由对称性，可只考虑第一象限部分 ，
D 4 D1
D1
D
注意：被积函数也要有对称性. 2 2 4 sin( x y )
x2 y2
2
dxdy
D1
sin(
2
x2 y2 ) x y
2
dxdy
4 d
0
2
1
sin r rd r 4. r
上一页 下一页 目 录
下一页
目 录
6.
(1)
上一页
下一页
目 录
6(2)
上一页
下一页
目 录
6(3) 改变积分
1 0
dx
2 x x2 0
f ( x , y )d y d x
1
2
2 x 0
f ( x , y )d y 的次序.
解 积分区域如图
y 2 x
y 2x x2
原式
1 0
dy
xy
t 解: 两个隐函数方程两边对 x 求导, 得
u
x y z
解得
因此
e x ( x z) z 1 sin( x z ) y du e x ( x z) f1 f 2 1 f3 x dx sin( x z )
上一页 下一页 目 录
x x
§6.6 部分习题答案
上一页 下一页 目 录
5.设某工厂生产甲产品数量 S(吨)与所用两种原料 A、 的数量 x,y(吨)间的关系式 S ( x , y ) 0 .005 x y ， B
2
现准备向银行贷款 150 万元购原料，已知 A,B 原料 每吨单价分别为 1 万元和 2 万元，问怎样购进两种 原料，才能使生产的数量最多? 解 按题意，即求函数 S ( x , y ) , y )dxdy
b b
[
a
b
b a
f ( x ) f ( y )dy ]dx
[ f ( y )dy ] f ( x )dx
a a
f ( y )dy
a
b
b a
[ f ( x )dx ]2 f ( x )dx
a
b
上一页
下一页
目 录
上一页
上一页
下一页
目 录
§6.3 部分习题答案 (3).
上一页
下一页
目 录
2(1)
上一页
下一页
目 录
3.
上一页
下一页
目 录
4.
上一页
下一页
目 录
5(1)
上一页
下一页
目 录
§6.4 部分习题答案
2.
(4).
上一页
下一页
目 录
4.
上一页
下一页
目 录
5.
上一页
下一页
目 录
§6.5 部分习题答案 1.
目 录
x3 y 6. 证明 lim 6 2不存在． x0 x y y 0
证
y kx 3 , 取
x3 y x 3 kx 3 k lim 6 , 2 lim 2 6 2 6 x0 x y x0 x k x 1 k 3 y 0
y kx
其值随k的不同而变化， 故极限不存在．
2L x1 x 2
8, C
2L x 2
2
20
B 2 AC 64 80 16 0, 且 A 4 0,
故点 ( 0.75 ,1.25 )为极大值点， 由问题的实际意义可知 ：它为最大点
即此时最优广告策略是用0.75万元作电台广告， 用1.25万元作报纸广告.
6 x 2 y 2 2 y r 2 sin
( x y )d x d y
2 2 D
x 3y 0 1
3
6
d
4 sin 2 sin
r rd r 15 ( 3 ). 2
2
上一页
下一页
目 录
LOGO
广东外语外贸大学
上一页
下一页
目 录
4. 某公司可通过电台及报纸两种方式做销售某商品 的广告.根据统计资料，销售收入 R(万元)与电台 广告费用 x 1 (万元)及报纸广告费用 x 2 (万元)之间 的关系有如下的经验公式：
R 15 14 x1 32 x 2 8 x1 x 2 2 x1 10 x 2
上一页
下一页
目 录
4.
上一页
下一页
目 录
6.
上一页
下一页
目 录
8.
已知 ln
dy y . x y a r c t a n，求 x dx
2 2
y 解 令 F ( x , y ) ln x y arctan , x
2 2
x y y x , Fy ( x, y) 2 , 则 Fx ( x , y ) 2 2 2 x y x y
解得 x 100 , y 25
因 仅 有 一 个 驻 点 ， 且 最大 值 一 定 存 在 ， 故 驻 点(100, 25 ) 为 最 大 值 ， 最 大 值S (100, 25 ) 0 .005 100 2 25 125吨 ， 即 购 进 A 原 料100吨 ， B 原 料 25吨 ， 可 使 生 产 量 达 到 最大 值 1250吨 .
( 2 )做拉格朗日函数 F ( x 1 , x 2 , ) L( x 1 , x 2 ) ( x 1 x 2 1 .5 )
上一页 下一页 目 录
15 13 x 1 31 x 2 8 x 1 x 2 2 x 1 10 x 2
2
2
( x 1 x 2 1 .5 )
0 y R2 x2 ( x, y ) D : 0 xR
R
y
则所求体积为
x2 z 2 R2 x
R2 x2 d x
R2 x2 0
8 8
R 0
R
0
dy
16 3 (R x ) d x R 3
2 2
上一页 下一页 目 录
11.
上一页
下一页
3 1 x ( e e )d x e e. 8 2
上一页 下一页 目 录
8(1)
上一页
下一页
目 录
9(1)
上一页
下一页
目 录
10. 解: 设两个直圆柱方程为
x2 y2 R2 , x2 z 2 R2
z
R
O
利用对称性, 考虑第一卦限部分,
其曲顶柱体的顶为 z R 2 x 2
F 13 8 x 2 4 x 1 0 x 1 F 由 31 8 x 1 20 x 2 0 x 2 x 1 x 2 1 .5 解得 x1 0, x 2 1.5
即广告费1.5万元全部用于报纸广告，可使利润最大.
目 录
12.
上一页
下一页
目 录
§6.9 部分习题答案
（2）
上一页
下一页
目 录
（1）
上一页
下一页
目 录
（2）
上一页
下一页
目 录
3.(1)
x y D {( x , y ) | 1 x 2 y 2 4 } . 其中积分区域为
2 2 D
计算二重积分
sin(
x y )
2 2
2 y
2
1 1 y
f ( x , y )d x .
上一页
下一页
目 录
6. (4)
上一页
下一页
目 录
7.(1) 计算
1 0
dx e dy.
y2 x
1
解 e dy 不能用初等函数计算
y2
只能用 Y - 型.
I
1 0
1 0
dy e dx
y2 0
y
1 ye dy ( e 1) 2
5.
计算 ( x y )d x d y ，其中 D 为由圆
2 2 D
x 2 y 2 2 y ， x 2 y 2 4 y 及直线 x 3 y 0 ， y 3 x 0 所围成的平面闭区域.
解
y 3x 0 2
3
x 2 y 2 4 y r 4 sin
在条件 x 2 y 150下的最大值
作拉格朗日函数
F ( x , y , ) 0 .005 x 2 y ( x 2 y 150 )
上一页 下一页 目 录
F x 0.01 xy 0 F 由 0.005 x 2 2 0 y x 2 y 150 0
Fx dy x y . y x dx Fy
上一页