卢兴江版微积分(上册)参考答案(4)

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(整理)微积分基础练习参考答案

微积分基础练习参考答案一、函数的概念和性质练习1.1 函数的定义域1、(2,3)(3,8]y D =2、 [5,1)(1,5]---3、 (1,0)(0,3]-4、(1,)+∞5、 (1,2]-6、 (1,2)练习1.2 函数的对应规则1、 A2、 D3、 34、 B 。

5、 D6、 D练习1.3 判断两函数的异同1、 C2、 B3、 A练习1.4 函数的奇偶性1、 A2、 A3、 A4、 D练习1.5 复合函数的定义和分解1、x x g f sin )]([=2、x x f g sin )]([=3、 ln ,sin 1y u u v x ===+。

4、函数由u y e =，cos u v =，1x v e =+复合而成的。

二、极限与连续练习2.1 根据基本初等函数图形求极限1、02、∞+3、∞+4、05、∞+6、∞-练习2.2 分式的极限1、∞2、13、04、-85、41练习2.3 两个重要极限1、1-e 2、 2e 3、2-e 4、e5、 16、3-e 7、e 8、 19、4110、1 11、3 12、1 练习2.4 无穷小量与无穷大量1、 A2、 B3、 D4、 A5、 D练习2.5 函数的连续性与间断点1、 (,2)(2,6)(6,)-∞--+∞2、 23、 C4、 D三、一元函数微分学练习3.1 导数的定义1、 A2、 B练习3.2 导数的几何意义1、 D2、 B3、13164y x =-+ 4、 33y x =-5、 2y x =+6、 12-，11(1)2y x -=--练习3.3 导数的四则运算法则1、12、 11+--='n n xn nx y 3、 1ln +='x y4、2ln 1x x y -='5、2sin cos cos sin x xx x x x x y -++=' 6、2121x xy -=' 7、()313+-='x y 8、B 练习3.4 复合函数求导法则1、22)1(6+='x x y 2、xy 3123--=' 3、32)1(+-='x x y4、x y 4sin 2='5、11-='x y 6、x x2cos 2sin -7、 dy=sin cos x xe dx 8、 xxe 2sin 24sin 2 9、)cos(2cos 22sin 2sin x xe xey ='10、x e x ey x x3cos 33sin 222+=' 11、322cos 3cos sin 3x x x x y +='12、xy -='121练习3.5 隐函数求导法则1、 222sin x y y x y y -'∴=- 2、522322++---='y x y x y3、(0)1y '∴=-练习3.6 对数求导法则1、(2)(ln 21)x y x x '∴=+2、)sin ln (cos sin xxx x xy x+=' 3、))12ln(sin 12cos 2()12(cos +-++='x x x x x y x4、222)65()12(6++--x x x x ＋练习3.7 高阶导数的计算1、 (0)4y ''=2、 xe x e x e y xx x +-=''2322练习3.8 求参数方程的导数1、t y tan -='2、2122-='t y四、导数的应用练习4.1 判别函数的单调性1、 C2、 (,0)-∞3、 ()+∞∞-,4、 (0,)+∞练习4.2 函数的极值和最值1、 2121=-=x x 2、有极小值41121=⎪⎭⎫ ⎝⎛y3、极大值8)1(=-y ，极小值25)3(-=y4、最大值0)3(=y ，最小值4)1(-=y5、最大值1)0(=y ，最小值4)2(-=e y练习4.3 用洛必达法则求不定型极限1、412、221－ 3、 3224、315、06、0练习4.4 经济函数的最值问题1、产量为200吨时可使平均成本达到最小，此时的总成本为1200万元。

浙江大学浙大卢兴江版微积分答案

6 定积分及其应用习题6.11. （1）e 1- （2）13 （3）122. （1）24R p （2）72（3）03. （1）1201d 1x x +ò （2）10ò （3）（i ）10d ()x a b a x +-ò 或 11d b ax b a x-ò （ii ）[]1ln ()d e a b a x x +-ò 或 1ln d e ba x xb a -ò 习题6.21. （1）112300d d x x x x >蝌（2）553233(ln )d (ln )d x x x x >蝌（3）222200sinsin d d xx x x x pp >蝌 2. （1[]222,0,1x x ?（2）提示：分析函数2()1xf x x=+在[]0,2上的最大（小）值. 3. 提示：取()()g x f x = 4. 提示：利用积分中值定理或定积分的定义证明.5. 提示：令()()F x xf x =对()F x 在10,2轾犏犏臌上用罗尔定理。

6. 提示：证明在[]0,p 内至少存在两点12,x x 使12()()0f f x x ==.习题6.31. （1）(2)sin 2x x - （2）6233e cos()x x x -（3）[][]sin ln 1sincos cos 1sinsin x x x x -+-+ （4）2221()d 2()x f t t x f x +ò（5）1()d xf t t ò2. （1）23（2）1 （3）1 （4）24p （5）13. 提示：利用夹逼定理.4. 4()sin 21f x x p =--. 5. 提示：2()y f x ⅱ= 6. 提示：利用2[()()]d 0baf x tg x x -?ò，其中t 为任意常数.7.（1）741)1)33p -++ （2）2 （3）143p - （4）26p （5）14 （6）12（7）24e --8. 提示：利用泰勒公式()()22a b a b f x f f x x 骣骣++¢琪琪=+-琪琪桫桫，x 位于x 与2a b+之间. 习题6.41. （1）15 （2）2 （3）16 （4）p （53p（6）121e骣琪-琪桫（7）24p （8）34 （9）352e 2727- （10）1ln 32- （11）3p -（12）8p（13）43p - （14）(ln 2-+ （15）()3e 15p - （16）13（提示：222101110111xx x x x x x e dx dx dx e e e ----=++++⎰⎰⎰）（17）1 （18）4π（提示：作变换2x t π=-）（19）2 （20）13（21）34p （22）当n 为偶数时：131222n n n n p ---g g L g g ；当n 为奇数时：131123n n n n ---g g L g g （23）ln 28p2. 713e-3. 提示：22()d ()d ()d a bbb a b aaf x x f x x f x x ++=+蝌?，对2()d ba b f x x +ò作变换()x a b t =+-.4. 若f 是连续偶函数，()()d xaF x f t t =ò不一定为奇函数. 例如：2311()d 13x F x x x x ==-ò5.1n （提示：对10()d x n n n t f x t t --ò作变换n nx t u -=，用洛必达法则或导数的定义.） 6. ()1cos113-（提示：用分部积分法） 7. 提示：用分部积分法 8. (0)2f =. 9.（1）2101, 1321d , 103231, 023p p p p x x p x p p p ì骣ï琪-+<-琪ï桫ïï+=-++-?íïïï+?ïïîò （2）411,01()221, 12x x x F x x x ì-+-?ï=íï-＃î10. 提示：利用()tan f x x =在0,4p 轾犏犏臌的单调性. 习题6.51.（1）2565 （2）1 （3）2p（4）163 （5）12442,633S S p p =+=- （6）92 （7）238a p （8）1ln 22 （9）1122.（1）a （2）43p3.（1）2R p （2）1ln(224+ （3）6a （4）22p 4. 1ln 32-5. 4 7. 3163a 8. （1）22x V p =，22y V p = （2）56p （3）24p （4）,33p p（5）23332325,6,7x y y a V a V a V a p p p ==== 9.2p10. 44815p11. （1）21)p （2）33211113ln 93222π⎡⎛+⎛⎫⎢ ⎪ ⎢⎥⎝⎭⎝⎭⎣⎦12. 22arcsin a a 骣+ 13. 2560g r （焦） 14. 0.5625 kg/m 2. 15. 3.675（焦） 16. 1674.667 g （焦） 17.22503h pr （焦） 18. ()343R H R H p w w +- 19. 212Mgh mgh +（焦）20.21.222k ph R k p ++ 22.()kmM a a l +，其中k 为万有引力常数 23. 22ln 12kM al a l骣琪+琪+桫，其中k 为万有引力常数习题6.61.211=-ò用矩形公式，梯形公式和抛物线公式计算(8)n = 2. 3.141592 （可利用抛物线公式计算120d 1xx +ò）3. 周长204l p q =ò，用抛物线公式计算(16)n =深其近似值为22.1035.习题6.71. （1）收敛，13 （2）发散（3）收敛，1ln 242p +（4（5 （6312p -（7）收敛，12（8）收敛，238- （9）收敛，2（10）收敛，83 （11）收敛，p （12）发散（13）收敛，79 （14）收敛，p （15）收敛，(ln 22p+（16）当1k £时发散，当1k >时，收敛于1(ln 2)1kk--2. 提示：作积分变换1xt = 3. 2a b ==- 4*.（1）收敛（2）收敛（3）发散（4）发散（5）收敛（6）收敛（7）收敛（8）发散（9）收敛（10）当1p <且1q <时收敛，其他发散. （11）收敛（12）收敛（13）当1n m >+时收敛，当1n m ?时发散（14）当12p <<时收敛，其他发散（15）当3m <时收敛，当3m ³时发散（16）当12n <<时收敛，其他发散. 5.（1）11(1)n n p +G + （2）(1)p G +6.（1）1!2m （2）12122m +⎛⎫Γ=⎪⎝⎭ （3）(1)!3m m m -g 7. （1）130（2）111,22B n 骣琪+琪桫 = 12!(21)!!n n n +⋅+。

微分几何课后答案

r r r r r r r r r 量，且 r (t ) · n = 0 。两次求微商得 r ' · n = 0 ， r ' ' · n = 0 ，即向量 r ， r ' ， r ' ' r r r r 垂直于同一非零向量 n ，因而共面，即（ r r ' r ' ' ）=0 。 r r r r r r r r r r r r 反之, 若（ r r ' r ' ' ）=0，则有 r × r ' = 0 或 r × r ' ≠ 0 。若 r × r ' = 0 ，由上题
}
1．求圆柱螺线 x =a cos t ， y =a sin t ，
解 r ' ={
r
-a sin t ,a cos t ,b}, r ' ' ={-a cos t ,- a sin t ,0
y − a sin t a cos t − a sin t
r
所以曲线在任意点的密切平面的方程为
x − a cos t − a sin t − a cos t
r r r r | r '×r ' ' | 2a 2 cosh t 1 = r '×r ' ' = a{− sinh t , cosh t ,−1} ，所以 k = r 3 = 3 | r '| 2a cosh 2 t ( 2a cosh t )
29
微分几何主要习题解答
τ=
r r r (r ' , r ' ' , r ' ' ' ) a2 1 = = r r 2 4 2 (r '×r ' ' ) 2a cosh t 2a cosh 2 t

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.。

经管类微积分(上)参考答案

经管类《微积分》（上）习题参考答案第一章函数习题一一、1．否； 2．是； 3．是； 4.否.二、1．)[()5,33,2⋃； 2．()πππ+k k 2,2； 3． 2,24>-<<-x x 或；4．[]a a -1,； 5.[]2,0； 6.222+-x x . 三、1．奇函数；2．奇函数． 3.（略）四、1(略)；2．212+x ； 3．11-+x x ．五、1．x v v u u y sin ,,ln 2===；2．x x u e y u ln ,==；3．1525++⋅x x ．六、50500,,)50(8.050)(>≤<⎩⎨⎧-+=x x x a a ax x R ．第二章极限与连续习题一一、 1．0,1,1,0； 2．e e e e ,,,231- 二、1．1； 2．0； 3．21； 4．4.三、1. (略)； 2.证明（略），极限为2 四、()1lim 0=+→x f x ，()1lim 0-=-→x f x ，()x f x 0lim →不存在. 五、都不存在. 六、15832.5,32.4,221.3,1.2,0.1 1.8,3.7,.6e .七、2,1==b a 八、2.4,32.3,21.2,2.1-习题二一、()().1,1.4,,22,1.3,2.2,.1+∞⋃第一类二、1．为可去间断点1=x ，为第二类间断点2=x ； 2.为跳跃间断点1=x . 三、2ln ,2==b a ．四、0,0,10,00,1)(=⎪⎩⎪⎨⎧>=<-=x x x x x f 为()x f 的跳跃间断点。

五、()()+∞⋃∞-,00,. 六、左不连续；右连续. 七、,.4,.3,.2,2ln .1623e e e - 八、九、十 (略).第二章测验题一、B A C A D .5,.4,.3,.2,.1.二、21.4,2.3,2.2,2.1-e .三、.31.4,3.3,1.2,61.1.四、x x x x p ++=232)(.五、为第二类间断点为可去间断点处连续21,1,2,,1===-=x x x x ．六、.3,21==b a 七、（略）. 八、a .第三章导数与微分习题一一、),0(.2),(,)(2,)(.1000f x f x f x f '''')(),(1.3000000x x x y y x x x y y --=--=- 二、00,,2)(<>⎩⎨⎧='x x x e x f x 三、)0(2)(g a f ='. 四、处连续且可导0=x ．五、()的有理数；互质与且)2(,201n m mna a ≠> ()互质）的有理数与且n m mna a 2(,1212-≠>. 习题二一、,ln 1.3,1.2,622ln 2.123x xx x x -++- )2(42,)2(42.422ππππππ-=---=-x y x y . )(4)(2.5222x f x x f ''+'二、2)1()sin 3(cos sin cos 2.1x x e x x e x x +-+-；x x x x x x x x cos sin ln cos 2sin .2+-+； 211arcsin 2.3xx -⋅； 21)ln (ln .4x x n x n --；a a x x x ax a a a 21211sec ln .5+⋅+-；6.x x exx 1tan 1sec 221sec 22⋅⋅⋅-； )(87略-.三、1．()x f x f '⋅)(2； 2．)()(222x x x x x e f e e e f xe '+.四、00,,11)12()(222=≠⎪⎪⎩⎪⎪⎨⎧+-='x x x e x x f x . 五、(略) 习题三一、()dx x x x 1ln .1+； ()dx e e f x x '.2；x e x e x x x ln ln ,arctan ),13sin(31,61,2.36+；4. ppQ -+2;252. 二、1．)sin ln (cos sin xxx x x x +⋅； 2．⎥⎦⎤⎢⎣⎡-----+-+------)5(51)4(54)3(53)2(5211)5()4()3()2()1(5432x x x x x x x x x x 三、1．()184-==p dpdQ，54.04-≈=P EP ED经济意义：当价格从4上升%1时，需求量从59下降%54.0；()246.04≈=P EP ER，价格从4上涨%1时总收益将从263增加%46.0.四、1．dx x x x x ⎥⎦⎤⎢⎣⎡--+-2222211cot )1(2)11ln(sin ．五、212x +. 第三章测验题一、,1.3,1.2,)1(21.1arctan =⋅+--y dx e x x x π21)1()1(2.4xx f x f '-, 2ln 21.5-.二、..3,.2,.1C D D 三、1．yyxey e +-2； 2．0； 3．[]()0,,02121cos )(sin )()(),0(2=≠⎪⎪⎩⎪⎪⎨⎧''++-+'=''=x x g x xx g x x g x x f g a第四章中值定理与导数的应用习题一一、1．不满足，没有； 2．1； 3．满足，914； 4．4,1--.；5.不存在二、三、四、五(略)六、1.6,ln .5,21.4,21.3,0.2,21.1a -. 七、连续. 八、1.习题二一、1．单减，凹的； 2．)4,1(；3．0,0==x y ；4．29,23-；5. ac b 32≤.6.e p 1=二、单增区间为[]2,0；单减区间为]()[∞+⋃∞-,20,. 三、拐点为()7,1-；凹区间为)[∞+,1；凸区间为[]1,0．四、0,3,3,1==-==d c b a .五(略)六、为极大值3)3(,2==πf a .七、20000=Q ，最大利润()34000020000=L 元. 八、5.9元，购进140件时，最大利润490元. 九、十（略）.第四章测验题一、..3;.2;.1A B B 二、()0.4;2,1.3;3.2;1.1=x三、.1.2;61.1-四、.1;0;3==-=c b a 五、获利最大时的销售量()t x -=425，当2=t 政府税收总额最大，其税收总额为10万元.六、()1证明略； ()254.06≈=P EP ER，经济意义：当价格从6上涨%1时，总收益从156增加%54.0.第五章不定积分习题一一、1．dx x f )(，C x f +)(，)(x f ，C x f +)(； 2．C ； 3．C x +2； 4．32x. 二、1．C x x +-arctan ； 2．C x e x +-2；3．C x x +-sec tan ； 4．C x +tan 21．三、1ln +=x y .四、12)(2+-=x x x G .习题二一、1．C e x x ++-tan tan ； 2．C x f +--)1(212； 3．C x F ++)12(； 4．C x f +--）2cos 3(31．二、1．C x +|ln ln |ln ； 2．C x ++-|1cos |ln 2； 3．C e x +arctan ；4．C x +--21)32(312； 5．C x x x +---------999897)1(991)1(491)1(971；6．C e xx ++1； 7．C x x +-32)cos (sin 23； 8．C e x x ++-)1ln(； 9．C x x ++-)9ln(292122； 10．C x +)arctan(sin 212； 11．C x+-arcsin 1；12．C x x ++-+ln 12)ln 1(3223； 13．()()()C x x x +++++-+11ln 313123313132；14．C e x+-1arctan 2； 15．C xx ++61611ln； 16．C x x x +-+22211arccos 21. 习题三一、1．C x e x ++-)1(；2．C x xf +)(； 3．C x f x f x +'-'')()(； 4．C e xe x x +-2．二、1．C x x x x +++-)1ln(6161arctan 31223； 2．C e xe x x +------11；3．C x x x x x ++-2ln 2ln 2； 4．C x x x x++++-)6ln 6ln 3(ln 123；5．C x x e x ++-)22(33323； 6．()()[]C x x x++ln sin ln cos 2；7．C x x x x x +--+2arcsin 12)(arcsin 22； 8．C x x x x ++-sin 4cos )24(； 9．C x x x +-+arctan )1(； 10．C x x x x x +++-+221ln 1ln ．三、C x x x +-++21)arcsin 1(．四、C x x x x ++-+arctan 22)1ln(2．五、)1(21x x +．习题四1．C x x x x x x +--+-+++|1|ln 3|1|ln 4||ln 82131232．C x x x x +-+-+-arctan 21)1ln(41|1|ln 21||ln 2第六章定积分及其应用习题一一、a b a b -+-）（3331二、1．≥， 2．≥ 三、（提示：用定积分性质6证）四、1．412x x +； 2．81221213x x x x +-+； 3．3； 4．21； 5．28-x ； 6．]41,0(； 7．yx e y 2cos 22．五、)(x f 在0=x 处有极小值0)0(=f ．六、1．6π； 2．4； 3．38．七、1．1； 2．2八、4π．九、)1ln(e +十（略）．习题二一、1．)(sin x f ； 2．)0(arctan )1(arctan f f -； 3．)]()([2122a F b F -； 4．3243π；5．0； 6．)()(a x f b x f +-+； 7．8； 8．0二、1．34-π； 2．32ln 22+； 3．a )13(-； 4．34； 5．22； 6．214-π； 7．)11(2e -； 8．)2(51-πe ．三、四（略）五、（提示：令x t -=2π）； 4π.六、()1,11=-=-a e x f x . 七、x x sin cos -. 八、x 2ln 21.习题三一、1．332； 2．2ln 23-； 3．67； 4．49．二、62221,21-=⎪⎭⎫ ⎝⎛=S a ．三、2ln 214+-x ．四、1．π145； 2．24π； 3．ππ564,727．五、10/100Q Qe -．六、31666．七、1．2； 2．2ln 21．。

微积分各章习题及详细答案(供参考)

微积分各章习题及详细答案(供参考)第一章函数极限与连续一、填空题1、已知 f (sin x) 1cos x ，则 f (cos x)。

2(4 3x)22、 lim2)。

xx(1 x3、 x 0 时， tan x sin x 是 x 的阶无量小。

4、 lim xksin10 建立的 k 为。

xx5、 lim e x arctan xx6、 f ( x)ex1, xb,7、 limln( 3x1)x 06x。

x 0在 x 0处连续，则 b 。

x 0。

8、设 f (x) 的定义域是 [ 0,1] ，则 f (ln x) 的定义域是 __________ 。

9、函数 y 1 ln( x 2) 的反函数为 _________。

10、设 a 是非零常数，则 lim (xa) x ________ 。

xx a111、已知当 x 0时， (1 ax 2 ) 3 1与 cosx 1 是等价无量小，则常数 a ________。

12、函数 f ( x)arcsin3x的定义域是 __________ 。

1 x13、 lim ( x 22x 2 2)____________ 。

x14、设 lim (x2a ) x 8 ，则 a________。

xx a15、 lim ( n n 1)( n 2n) =____________ 。

n二、选择题1、设 f ( x), g(x) 是 [ l , l ] 上的偶函数， h( x) 是 [ l , l ] 上的奇函数，则中所给的函数必为奇函数。

（Ａ） f ( x) g( x) ；（Ｂ） f ( x) h( x) ；（ C ） f (x)[ g(x) h( x)] ；（ D ） f ( x) g( x) h(x) 。

2、1 x3x( x)，( x)1x ，则当时有。

1 x1（Ａ）是比高阶的无量小；（Ｂ）是比低阶的无量小；（ C ）与是同阶无量小；（ D ）~。

3、函数 f (x)1 x 1 ,x 0( x1) 在 x0处连续，则 k3 1 x 1 。

卢兴江版微积分(上册)参考答案(4)

4 微分中值定理及导数应用习题4.11.（1）4373，()f 为最小值。

（2）,()2f 为最大值。

（3）1，()f 为最大值。

2.（1）(1)1f ，(2)4f ，3(2)(1)()()3221f f f f ；（2）(0)30,(0)0,()0363ff ff ff；（3）()14f ，()14f，()()444()(arccos)2()44f f f f ；（4）(1)(1)(1)1,(1)1,()(0)01(1)f f f f f f .3.2. 4. 提示：利用Lagrange 定理. 5. 提示：用反证法.6. 提示：利用Rolle 定理.7. 提示：对()()1f x F x x在0,1上用罗尔定理 8. 提示：利用Lagrange 定理. 9. 提示：f 在,a b 上有界. 10. 提示：证明()0f x .11.（1）不能，理由见（2）; （2）112，233，323. 12. 4.13. （1）提示：利用“()0f x 则()f x C （常数）”的结论。

（2）提示：令22()1tan sec f x x x ，证明()0f x .14（1）提示：和差化积或直接用拉格朗日定理; （2）提示：利用Lagrange 定理.习题4.21. 提示：利用函数单调性定义和拉格朗日定理。

2.（1）单调减少. （2）单调增加. （3）单调增加. （4）单调增加.3.（1）在1(,)2内单调增加，在1(,)2内单调减少；（2）在,1或1,内单调减少，在1,1内单调增加；（3）当0时，f 单调减少；当0α>时，f 在(0,)单调增加，在(,)单调减少；（4）在,1或0,1内单调减少，在1,0或1,内单调增加.4. 提示：设()()F x xf x ，证明F 在12(,)x x 内必取到F 在12,x x 上的最小值或者最大值.5.（3）提示：令()n f x x ，在,b a 上用拉格朗日定理。

微积分习题解答

教材习题参考答案习题一答案(A)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) xxx y -++-=11lg21)1arcsin(; (6) ⎩⎨⎧><+=)0(ln )0(12x xx x y . 【解】(1) 022≥-+x x21-≤≥x x 或∴定义域为),1[]2,(+∞--∞ .(2) ⎩⎨⎧≥≥00)sin(x xπππ+≤≤k x k 22∴定义域为{},1,0,)12(42222=+≤≤k k x k x ππ.(3) ⎩⎨⎧≠->-0201x x21≠>x x 且∴定义域为),2()2,1(+∞ .(4) ⎪⎩⎪⎨⎧≠-≥-010422x x⎩⎨⎧±≠≤≤-122x x ∴定义域为]2,1()1,1()1,2[ ---.(5) ⎪⎪⎩⎪⎪⎨⎧≠->-+≤-≤-01011111x x x x ⇒ ⎪⎩⎪⎨⎧≠<<-≤≤11120x x x ∴定义域为)1,0[. (6) 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(+∞，任取),0(,21+∞∈x x ，不妨设210x x <<，0)ln (ln ln ln 1212112212>-+-=--+=-x x x x x x x x y y故函数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, 任取R x x ∈21,，当021<<x x 时，21x x >，021>-x x ee ，函数为单调减函数；当210x x <<时，21x x <，021<-x x ee ，函数为单调增函数.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 e y -=; (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 xxy cos sin +=;（9）xx xx e e e e y ---+=; （10）⎩⎨⎧≥+<-=0101x x x 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 ， 01lg 11lg 11lg )()(==+-+-+=+-xxx x x y x y ，即)()(y x y x -=-，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 ，)()(y x y x -=-，故函数为奇函数.（8）定义域为),0()0,(+∞-∞ ，)(cos sin )cos()sin()(x y x xxx 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 xx 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）)cos 22sin 22(2x x y +=)4sin(2π+=x 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）周期为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≤+=x x 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⎩⎨⎧>-≤+=012)(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 x y . 【解】（1）x y a log =，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 xv -=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 -=,故反函数为)1(31x e y -=,),0(1+∞=-f D （2）由29x y -=解得29y x -=，故反函数为29x y -=,]3,0[1=-f D（3）由22-+=x x y 解得1)1(2-+=y y x ,故反函数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 ,故反函数为)1ln(2++=x x y ,),(1+∞-∞=-f D（5）可解得⎩⎨⎧≤<--≤<-+=2122112/)1(y yy y x故反函数为⎪⎩⎪⎨⎧≤<--≤<-+=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())1(3)1(22-++=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x t 1=，则有ct t bf t af =+)()1(，即cx x bf xaf =+)()1( 可解得)()(22xabx a b c x f --= ，而)()(x f x f -=-,故)(x f 是奇函数.三教材习题选解或提示1.观察判别下列数列的敛散性；若收敛，求其极限值：（5）1sinn u nnπ=解：1231111sin ,sin ,222336sin 0,u u u πππ======，数列{}nu 收敛，1sin 0lim lim n n n u n n π→∞→∞== （7）(1)3nn u -=解：123411,3,,3,33u u u u ====，数列{}nu 发散．（8）1cos n u n nπ=解：12341111,,,,234u u u u =-==-=，数列{}nu 收敛，1cos 0lim lim n n n u n n π→∞→∞== 2.利用数列极限的分析定义证明下列极限：（4）17lim n n→+∞⎛⎫- ⎪⎝⎭证明：0ε∀>，不妨设1ε<，取71log 1N ε⎡⎤+⎢⎥⎣⎦=，则当n N >时，有11077n N n u ε-=<<，所以107lim n n→+∞⎛⎫-= ⎪⎝⎭3.求下列数列极限：（5）22211111123lim n n →+∞⎛⎫⎛⎫⎛⎫--- ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭解：222132411111111223323lim lim n n n n n nn →+∞→+∞⎛⎫⎛⎫⎛⎫-+---=⋅⋅⋅⋅⋅⋅ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭12lim n n n →+∞+= 12= （7）(sinsin limn →+∞-解：(cos 22sinsin 2sinlimlimn n →+∞→+∞-=1cos22sinlimn →+∞=0=(8) ()121234limn n n n →+∞+++解：()112123123441444limlim n n n n nnn n n→+∞→+∞⎡⎤⎛⎫⎛⎫⎛⎫⎢⎥+++=+++ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦1231144412344412341444lim n n n n n n n n n n n →+∞⎡⎤⎛⎫⎛⎫⎛⎫⎢⎥++ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦⎛⎫⎛⎫⎛⎫++⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎧⎫⎡⎤⎪⎪⎛⎫⎛⎫⎛⎫⎨⎬=+++⎢⎥ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎪⎪⎢⎥⎣⎦⎩⎭44e ==（9）()11112231lim n n n →+∞⎛⎫ ⎪+++ ⎪⋅⋅+⎝⎭ 解：()1111111141122312231lim lim n n n n n n →+∞→+∞⎛⎫⎡⎤⎛⎫⎛⎫⎛⎫ ⎪⎢⎥+++=-+-++- ⎪ ⎪⎪ ⎪⋅⋅++⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦⎝⎭111lim n n →+∞⎛⎫=- ⎪+⎝⎭1=7.求下列函数的极限：（7）(lim x x →∞解：((2limlim x x xx x x →∞→∞⎛- =1limx →∞=0=（10）lim x →+∞解：当01a <<时，1lim x →+∞=-=当1a >时，limlimxx→+∞→+∞==8.求下列函数的极限：（2）11lim1--→nmx xx解：()()()()1111111111lim limmmn nx xx x xxx x x x--→→-+++-=--+++mn=（3）11lim31--→xxx解：))111x x→→=23=（10）211limnxx x x nx→+++--解：211limnxx x x nx→+++--()()()()()21 111111111lim limnn x xx x xx x xx-→→-++++-⎡⎤==+++++++⎣⎦-()1122n nn+=+++=9.求下列各题中的常数a 和b ：（2）2151lim x x bx ax →++=-解：()()221111lim lim x x x bx ax bx a x x →→++++=--()211101lim lim x x x bx ax x →→++=-=-10a b ∴++=，即 1b a =--又 ()2211111lim lim x x x a x a x bx ax x →→-++++=--()()()2111111lim lim x x x x a x bx ax x x →→--++=-=--15a =-=故 6,7a b == 10.求下列函数极限：（7）x →解：30limx x →()3tan 1cos 2lim x x x x →-= 2301122lim x x x x →⋅= 14= 12.求下列函数极限：（1）612sin sin 6limx xx ππ→-⎛⎫- ⎪⎝⎭ 解：令 ,,066t x x t ππ=-→→原式=012sin 6sin limt t t π→⎛⎫-+ ⎪⎝⎭000cos 1221cos lim lim lim t t t t t t t t →→→⎤-+⎥-⎛⎣⎦==+ ⎝= （B ） 3.设数列{}n u1,2,,n u -+证明：lim n nu →+∞存在，并求此极限值．证明：首先证{}n u 单调增加．21u u =>=；设1n n u u ->，则1n n u u +=>=，由数学归纳法可知，{}nu 单调增加．其次，证{}n u 有界．由 n u =<可得，2n u <，1n ≥．即{}n u 有上界．因此，由极限存在性定理可知，lim n n u →+∞存在，设lim n n u θ→+∞=，由n u =，两边求极限，得2a θθ=+，故 2θ=或者1θ=-．由于0θ>，所以2lim n n u θ→+∞==．4.讨论函数()21lim nx nx n x x e f x e→+∞+=+的连续性，若有间断点，判别其类型．解：0x >时，()22211lim lim nn x x n n xx n n x x e x e x f x x ee --→+∞→+∞++===++ 0x <时，()21lim nxn x n x x e f x x e→+∞+==+ 综上：()2,,00x x f x x x ⎧⎪=⎨⎪⎩<>5.设()()1ln lim t t t f x e x t →+∞=+ ()0x > ，讨论()f x 的连续性解：当0x e <<时，()11ln 1ln 1lim lim limt t t t tt t t x x e t t t e x t tt⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦→+∞→+∞→+∞⎡⎤⎛⎫⎛⎫+⎢⎥⎪ ⎪⎝⎭⎢⎥⎝⎭⎣⎦==+++=当x e >时，()11ln ln 1ln ln lim lim limt t t t t t t t e e x t x x x e x x t tt⎡⎤⎢⎥⎢⎥⎢⎥⎣⎦→+∞→+∞→+∞⎡⎤⎛⎫⎛⎫+⎢⎥⎪ ⎪⎝⎭⎢⎥⎝⎭⎣⎦==+++=当x e =时，()11ln 2ln lim lim t t tt t e e x t t →+∞→+∞=+= 综上：()1,0ln ,x ef x x x e ⎧=⎨⎩<≤> ，()f x 在(),-∞+∞上连续．9.设()f x 在上连续，12a x x b <<<，且1k 与2k 是任意正常数，证明：在(),a b 内至少存在一点ξ，使得()()()112212k f x k f x f k k ξ+=+证明：令()()112212()()k f x k f x F x f x k k +=-+()F x 在(),a b 上连续，()()211212()k F x f x f x k k ⎡⎤-⎣⎦=+ ()()122112()k F x f x f x k k ⎡⎤-⎣⎦=+ 若()()12f x f x = ，则12()()F x F x =0=，即1x 和2x 为所求．若()12()f x f x ≠ ，则12()()0F x F x ⋅<，由零值定理可知，至少存在一点12,()(,)x x a b ξ∈⊂，使得()F ξ0=，即()()112212()k f x k f x f k k ξ+=+．(三)教材习题选解或提示（A ）2.设函数()x f 在0x 处可导，求下列极限：()4 ()()x x x f x x f x ∆∆+-∆+→∆0005lim.解：()()xx x f x x f x ∆∆+-∆+→∆0005lim()()()()x x x f x f x f x x f x ∆∆+-+-∆+=→∆000005lim()()x x f x x f x ∆-∆+=→∆0005lim+()()x x x f x f x ∆∆+-→∆000lim()()x x f x x f x ∆-∆+=→∆55lim5000()()xx f x x f x ∆-∆+-→∆000lim ()04x f '=.3. 求曲线3x y =在点()1,1出的切线方程和法线方程.解：函数3x y =的导数为23x y ='，所以()31='y ，则切线方程为 ()131-=-x y法线方程为311--=-x y . 6. 若函数()x f y =是可导函数，试证：()1 ()x f y =为奇函数，则()x f '为偶函数；()2 ()x f y =为偶函数，则()x f '为奇函数，且()00='f .证：（1）()x f y =为奇函数，则()()x f x f -=-，将此式两边同时对x 求导，得()()x f x f '-=-'-，即 ()()x f x f '=-' 则()x f '为偶函数.（2）()x f y =为偶函数，则()()x f x f =-，()00=f ，将此式两边同时对x 求导，得()()x f x f '=-'-，即 ()()x f x f '-=-' 则()x f '为奇函数.函数()x f y =是可导函数，则()0f '存在，且()='-0f ()0+'f =()0f '()='-0f ()()xf x f x 0lim 0--→=()x x f x -→0lim()='+0f ()()xf x f x 0lim 0-+→=()x x f x +→0lim =()x x f x ---→0lim =()x x f x -→-0lim =()0-'-f ，于是可得 ()00='f .9. 利用对数求导法求下列函数的导数：()1 xx x y ⎪⎭⎫⎝⎛+=1解：先在两边取对数，得()[]x x x y +-=1ln ln ln等式两边分别对x 求导得()⎪⎭⎫ ⎝⎛+-++-='x x x x x y y 1111ln ln 1 从而得()y x x x x x y ⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-++-='1111ln ln()⎥⎦⎤⎢⎣⎡⎪⎭⎫ ⎝⎛+-++-='x x x x x y 1111ln ln xx x ⎪⎭⎫ ⎝⎛+1.11. 求由下列方程确定的隐函数()x y y =的导数：()4y xe y -=1解：方程两边对x 求导，得y xe e y y y '--='，从中可求得yyxee y -='1. 13. 求下列函数的n 阶导数：()3x y 2sin =解：x x x y 2sin cos sin 2=='⎪⎭⎫ ⎝⎛+=''22sin 2πx y⎪⎭⎫ ⎝⎛⨯+='''222sin 4πx y()⎪⎭⎫ ⎝⎛+=-22sin 21πn x y n n15. 求由下列方程确定的隐函数()x y y =的微分：()1 xy y =sin解：把隐函数xy y =sin 两边同时取微分，得ydx xdy ydy +=cos ，可解得 dx xy ydy -=cos .（B ）1. 设函数()()().x a x x f φ-=如果()x φ在a x =处连续，那么a x =处是否可导？如果()x φ在a x =处有定义，但不连续，又有怎样的结果？解：()()()x a f x a f a f x ∆-∆+='→∆0lim()xx a x x ∆∆+∆=→∆φ0lim ()x a x ∆+=→∆φ0lim 所以当a x =点为()x φ的连续点或者可去间断点时，()x f 可导，否则不可导.2. 讨论函数()⎪⎩⎪⎨⎧=≠=001sinx x xx x f α在0=x 点处的连续性和可导性.解：当0≥α时，xx x 1sinlim 0α→0=，所以函数连续，当0<α时，xx x 1sin lim 0α→不存在，所以函数不连续，讨论()0f 'x x x x 1sinlim0α→∆=x x x 1sin lim 10-→=α，因此当1≥α时，xx x 1sin lim 10-→α0=，所以函数在0=x 处可导，当1<α时，xx x 1sin lim 0α→不存在，所以函数在0=x 处不可导.3．求曲线e xy e y=+在点()1,0处的切线方程.解：把方程e xy e y=+的两端同时对x 求导，得0='++'y x y e y y于是得yex y y +-='，得()e y 10-=' 切线方程为x e y 11-=-4．证明曲线1=xy 上任意一点的切线与两个坐标轴围成的三角形面积恒等于2.证：任取曲线上一点⎪⎪⎭⎫ ⎝⎛001,x x ，该点切线斜率为()()2001x x y -=' 该点切线方程为()()020011x x x x y --=-切线与x 轴交点坐标为()0,20x ，与y 轴交点坐标为⎪⎪⎭⎫ ⎝⎛02,0x，所以曲线1=xy 上任意一点的切线与两个坐标轴围成的三角形面积恒等于2.5．讨论()⎪⎩⎪⎨⎧=≠+=00011x x e x x f x 的()0'-f 及()0'+f 及()0f '的存在性.解：()xexf xx 101lim 0+='-→-1=()xexf xx 101lim 0+='+→+0=所以()0f '不存在.6．火车以h km 100的速度向东行驶，汽车以h km 80的速度向北行驶，初始汽车在火车正北km 50，求一小时后两车的相离速度. 解：两车距离函数为()()228050100t t S ++=()()2280501006400400010000t t t t dt dS ++++=，于是可知26920401==t dtdS.7．计算以方程yxe y -=1确定的隐函数()x y y =的二阶导数.解：把方程两边对x 求导，得y xe e y yy'--='，yy xee y +-='1，把y xe e y yy '--='两边对x 求导得 ()y xe y xe y e y e y y y y y ''-'-'-'-=''2，整理可得()()3322112y yy yxe xe xe e y +-+=''.8．求函数211x y -=的n 阶导数.解：⎪⎭⎫⎝⎛++-=x x y 111121()()⎥⎦⎤⎢⎣⎡+--='22111121x x y()()⎥⎦⎤⎢⎣⎡++-=''33121221x x y ()()()()⎥⎦⎤⎢⎣⎡+-+-=++111!11!21n nn n x n x n y . 9.设曲线()nx x f =在点()1,1处的切线与x 轴的交点为()0,n x ，计算()n n x f ∞→lim .解：曲线在在点()1,1处的切线斜率为()n f ='1，因此，切线方程为()11-=-x n y ，曲线与x 轴的交点为⎪⎭⎫⎝⎛-0,11n ，因此()n n x f ∞→lim =e n nn 111lim =⎪⎭⎫⎝⎛-∞→. 10.已知()2arctan ,2323x x f x x f y ='⎪⎭⎫⎝⎛+-=，计算=x dx dy .解：=dx dy ()2223122323arctan +⎪⎭⎫⎝⎛+-x x x 所以0=x dxdy π43=.(三)教材习题选解或提示（A ）2.不用求出函数()()()()321---=x x x x x f 的导数，说明()x f '有几个根及所在区间.解：()()()()321---=x x x x x f 的导数为三次多项式，则()0='x f 最多有三个解，因为()()()()3210f f f f ===，根据罗尔定理，可知存在()1,01∈ξ使得()01='ξf ；存在()2,12∈ξ使得()02='ξf ；存在()3,23∈ξ使得()03='ξf .3. 证明方程0535=+++x x x 有且仅有一个实根. 证：设函数()535+++=x x x x f ，则()x f 在R 上连续.由于()372-=-f ，()50=f ，所以存在一点1x ()0,2-∈，使得()01=x f .假设0535=+++x x x 除1x 外还有一根2x 0≠.不妨假设21x x <，则()()21x f x f =.()x f 在闭区间[]21,x x 上连续，在开区间()21,x x 内可导.因此，有()()21,,0x x f ∈='ξξ而()113524≥++='x x x f ，矛盾,得证.4. 设1,0>>>n b a ，证明：()()b a na b a b a nbn n n n -<-<---11.证：设函数()nx x f =，在区间[]b a ,上应用拉格朗日定理，得1-=--n nn n ab a b ξ ()b a ,∈ξ因为()b a ,∈ξ，所以111---<<n n n nb n naξ，所以11--<--<n n n n nb ab a b na，得()()b a na b a b a nb n n n n -<-<---11.6.设函数()x f 在[]a ,0上连续，在()a ,0内可导，且()0=a f ，证明：至少存在一点()a ,0∈ξ，使得()()0='+ξξξf f .证：设函数()()x xf x F =，因为()()00==a F F ，可知()x F 在区间[]a ,0满足罗尔定理，则有()0='ξF ()a ，0∈ξ，即()()0='+ξξξf f()a ，0∈ξ.7.若方程01110=+++--x a x a x a n n n 有一个正根0x x =，证明：方程()0112110=++-+---n n n a x n a nxa 必有一个小于0x 的正根.证：设函数()x a x a x a x F n n n 1110--+++= ，()00=F ，则可知()x F 在区间[]0,0x 满足罗尔定理，可知()x F 在区间[]0,0x 满足罗尔定理，则有()0='ξF ()00x ，∈ξ，即()0112110=++-+---n n n a n a n a ξξ，()00x ，∈ξ，方程()0112110=++-+---n n n a x n a nx a 必有一个小于0x 的正根.8.设函数()x f 在[]b a ,上连续，在()b a ,内可导，并且有()()b f a f =0=.试证：至少存在一点()b a ,∈ξ，使得()()0=-'ξξf f . 证：设函数()()xex f x F -=， ()()0==b F a F ，可知()x F 在区间[]b a ,满足罗尔定理，则有()0='ξF ()b a ,∈ξ，即()()[]0=-'-ξξξe f f ，可得，至少存在一点()b a ,∈ξ，使得()()0=-'ξξf f . 9.求下列极限：()1 ()x x x +→1ln lim 0； ()2 x e e x x x sin lim 0-→-；()3 301cos lim x x x x +-→； ()4 x b a xx x -→0lim ()0,>b a ；()5 x arc x x cot 11ln lim ⎪⎭⎫⎝⎛++∞→； ()6 2120lim x x e x →；()7 ⎪⎭⎫ ⎝⎛--→x x x x ln 11lim 1； ()8 ()xx x sin 0tan lim +→；()9 xx xx x sin sin lim +-∞→； ()10 x x x x x e e e e --+∞→+-lim ；()11 x x x a ⎪⎭⎫ ⎝⎛+∞→1lim ； ()12 xx x tan 01lim ⎪⎭⎫⎝⎛+→.解：()1 ()x x x +→1ln lim 0=1111lim 0=+→x x ；()2 x e e x x x sin lim 0-→-= 2cos lim0=+-→xe e xx x ； ()3 301cos lim xx x x +-→=23121sin lim xxx x +--→∞=；()4 x b a x x x -→0lim =ba bb aa x x x ln 1ln ln lim0=-→； ()5 x arc x x cot 11ln lim ⎪⎭⎫ ⎝⎛++∞→=1111111lim22=+-⎪⎭⎫⎝⎛-++∞→x x x x ； ()6 2120limx x e x → ==→2101lim 2x e xx ∞=-⎪⎭⎫ ⎝⎛-→33101212lim 2x x e x x ；()7 ⎪⎭⎫ ⎝⎛--→x x x x ln 11lim 1=()x x x x x x ln 11ln lim1-+-→=xx x x 1-1ln ln lim 1+→=∞=-→211x 11limxx x ；()8()xx x sin 0tan lim +→=()xx x esin tan ln 0lim +→=xx x etan ln sin lim 0+→=x x x esin 1tan ln lim0+→=x x x e1tan ln lim0+→=2201sec tan 1limxx x x e -+→=10=e ；()9 x x x x x sin sin lim +-∞→=1sin 1sin 1lim=+-∞→xx x xx ； ()10 x x x x x e e e e --+∞→+-lim = 111lim22=+---+∞→xxx e e ； ()11 xx x a ⎪⎭⎫⎝⎛+∞→1lim =xx a x e ⎪⎭⎫⎝⎛+∞→1ln lim =⎪⎭⎫⎝⎛+∞→x a x x e1ln lim =xx a x e 11ln lim ⎪⎭⎫⎝⎛+∞→=22111limxx a x ax e-⎪⎭⎫⎝⎛-+∞→=ae ；()12 xx x tan 01lim ⎪⎭⎫⎝⎛+→= xx x etan 1ln 0lim ⎪⎭⎫⎝⎛→+=xx x etan 11lnlim 0+→=xx x e11lnlim0+→=101lim220==--+→e ex x x x .10．确定下列函数单调区间：()1 29323+--=x x x y ； ()4 x e x y -=.解：()1 29323+--=x x x y ，令09632=--='x x y ，得3,121=-=x x ，列表讨论](1,-∞-和[)+∞,3为函数()x f 的单调增加区间，[]3,1-为函数()x f 的单调减少区间；()4 x e x y -=，令01=-='xe y ，得0=x ，当0<x 时，0>'y ；当0>x 时，0<'y ，因此(]0,∞-为单调增加区间，[)+∞,0单调减少区间.11.证明下列不等式:()1 当0>x 时，x x +>+121解：设函数()=x f x x+-+121，()x x f +-='12121，当0>x 时，函数单调增加，有()()00=>f x f ，即x x+>+121. 13.求下列函数的最值:()1 []4,1,3223-∈-=x x x y解：令x x y 662-='=0，得1,021==x x ，()()()()804,11,00,51=-==-=-f f f f ，函数的最大值为()804=f ，函数最小值为()51-=-f .18.设某厂生产某种产品x 个单位时，其销售收入()x x R 3=，成本函数为()1412+=x x C .求使总利润达到最大的产量x . 解：总利润为()14132--=x x x L ，()223x x x L -='，得驻点39=x ，当39=x 时，总利润最大.20.当a 、b 为何值时，点()3,1为曲线23bx ax y +=的拐点？解：()31=f ，即3=+b a ，()0261=+=''b a f ，得29,23=-=b a .（B ）2.已知函数()x f 在[]10，上连续，在()10，内可导，且()()11,00==f f ，()x f 是x 的非线性函数.试证：在()10，内至少存在一点ξ，使得()1>'ξf .证：()x f 是x 的非线性函数，则至少有一点()1,00∈x ，使得()00x x f ≠，不妨设()00x x f >，则在()0,0x 满足拉格朗日中值定理，即()()()ξf x f x f '=--00001>，其中()0,0x ∈ξ()1,0⊂.5.设函数()x f 在闭区间[]A ,0上连续，且()00=f .如果()x f '存在且为增函数()()A x ,0∈.试证：函数()()x f xx F 1=也是增函数. 证：()()()x f xx f x x F 211-'='，当0>x ， ()x f 在区间()x ,0满足拉格朗日中值定理，则有()()()x f xx f ,0,∈'=ξξ, ()()()011>'-'='ξf x x f x x F ,函数()()x f xx F 1=是增函数.9.设()x f 在0=x 处二阶可导，且二阶导数连续，已知()31201lim e x x f x x xx =⎪⎭⎫ ⎝⎛+++→，求()()()0,0,0f f f '''及()xx x x f 101lim ⎪⎭⎫ ⎝⎛+→. 解：()()⎪⎭⎫⎝⎛++→→=⎪⎭⎫ ⎝⎛+++201lim 1201lim x x f x xx x e x x f x x 3e =, 则()2lim 20=→x x f x ，()22lim 0='→x x f x ，()22lim 0=''→x f x ，则()()()10,00,00=''='=f f f ，()xx x x f 101lim ⎪⎭⎫ ⎝⎛+→=()2lim 20e e x x f x =⎪⎭⎫⎝⎛→.三、不定积分习题解答（A ）7. 用换元法求下列不定积分：(3)3221x x x++⎰d 解32222222222(1)22()111121111ln(1)2arctan 22d d d d d d x x x x x x x x xx x x x x x x x x x xx x x C ++-+==-+++++=-+++=-+++⎰⎰⎰⎰⎰⎰(8)2d x解2123333315)(5)(5)3d x x d x x C C-=--=-+=⎰（(9)d x 解1d d x x =+⎰（）32213C x =++（）(11)x d解d d d x x x ==⎰2C ==+(16)341d x x x x ++⎰解334244444111114211111d d d d d x x x x x x x x x x x x x x +=+=++++++⎰⎰⎰⎰⎰（）+4211ln(1)arctan 42x x C =+++ (21)sin cos 1sin 2d x xx x -+⎰解222sin cos sin cos sin cos 1sin 2sin cos 2sin cos (cos sin )d d d x x x x x xx x xx x x x x x x ---==++++⎰⎰⎰211(cos sin )cos sin (cos sin )d x x C x x x x =-+=+++⎰(22)d x x 解22112d d d d()x x x x x x =-=--⎰⎰（ 332211133=()x x C --+ (26)sin cos 25d x x x ⎰解422sin cos sin cos sin (1sin )2522d dsinx dsinx x x x x x x x ==-⎰⎰⎰46357121(sin 2sin sin )sin sin sin sin 3572d x x x x x x x C=-+=-++⎰ (27)4sin d xx⎰解22222344sin cos 1(csc cot csc )cot cot 3sin sin d d d x x x x x x x x x x C x x +==+=--+⎰⎰⎰(28)tan 3d x x ⎰解 221tan (sec 1)tan tan ln cos 23d d x x x x x x x C =-=++⎰⎰(29)11cos d x x +⎰解2211sec tan 1cos 2222cos 2d d d x x xx x C x x ===++⎰⎰⎰ (30)3sin()sin(3)44x x x ππ++⎰d 解31sin()sin(3)cos(2)cos(4)4422d d x x x x x x ππππ⎡⎤++=--+⎢⎥⎣⎦⎰⎰ 111sin(2)sin(4)2224x x C ππ⎡⎤=--++⎢⎥⎣⎦(31)24arctan 1d x xx x -+⎰解2224arctan 114arctan arctan (1)211d d d x x x x x x x x -=-+++⎰⎰⎰ 2212(arctan )ln(1)2x x C =-++(32)221sin 2cos d x x x +⎰解22211)2sin 2cos 12cot d x x x C x x x==-+++⎰ (33)2tan(3)cos (3)d x x x ++⎰ 解22tan(3)1tan(3)tan(3)tan (3)2cos (3)d d x x x x x C x +=++=+++⎰⎰ (34)d t t te e -+⎰解 22111d d d()=arctan()t t ttt t t t e t e e C e e e e -==++++⎰⎰⎰（）(35)1d x x e -⎰解 11111d d d x xx x x x x e e e x x e e e ⎛⎫-+==- ⎪-⎝⎭⎰⎰⎰-- 1(1)ln 11d x x xe x e x C e =--=--+⎰-(36)x 解n(d dl x x =+322n()3l x C ⎡⎤=++⎣⎦8. 用换元法求下列不定积分：(2)xx ⎰解设 431,(3)22t x t dx t dt ==-=则，438413)2(3)2d d d x x t t t t t t t =-⋅⋅=-⎰⎰⎰（9513139595t t C C =-+=+(4)d x解设 2,22t x t dx tdt ==+=则，2223121(2)2d dt 2dt t x t t t t +⎛⎫=⋅=+⎪++⎝⎭⎰⎰2()2(arctan22t C C =++=+6)221(1)d x x +⎰解设 2tan ,sec x t dx t dt ==则()22224111sec cos 1cos 22(1)sec d d d d x t t t t t t x t =⋅==++⎰⎰⎰⎰2111sin 2arctan 2221x t t C x C x ⎛⎫⎛⎫=++=++ ⎪ ⎪+⎝⎭⎝⎭(8)2x d解设 sin ,cos x t dx tdt ==则()222sin 1cos sin 1cos 2cos 2d dt d d t x t t t t t t =⋅==-⎰⎰⎰111sin 2arcsin 222t t C x C ⎛⎫=-+=-+ ⎪⎝⎭（(9)x解设222,ln(1),1tt x t dx dt t ==-=-则 222122111d d d t x t t t t ⎛⎫==+ ⎪--⎝⎭⎰⎰112ln 2ln(1)21t t C x C t ⎛-⎫=++=-++ ⎪+⎝⎭9. 用分部积分法求下列不定积分： (2)2ln(3)d x x +⎰解222223ln(3)ln(3)ln(3)2133d d d x x x x x x x x x x x x ⎛⎫+=+-⋅=+-- ⎪++⎝⎭⎰⎰⎰2ln(3)2x x x C =+-++(4)2sin d x x x ⎰解 2222sin cos cos cos d d d x x x x x x x x x =-=-+⎰⎰⎰22cos 2cos cos 2sin x x x x dx x x x d x=-+=-+⎰⎰()2cos 2sin sin x x x x x dx =-+-⎰2cos 2sin 2cos x x x x x C =-+++(7)d x ⎰解设2,,2t x t dx tdt ===则()2222()d d d d t t t t t t x e t t te t te e t te e C =⋅==-=-+⎰⎰⎰⎰2()C =-+(9)32(ln )d x x x ⎰解32244241111(ln )ln )ln 2ln 444d d(d x x x x x x x x x x x==-⋅⋅⎰⎰⎰ 42342411111ln ln ln ln ()42424d d x x x x x x x x x =-=-⎰⎰ 44244211111ln ln (8ln 4ln 1)424432x x x x x x dx x x C x ⎛⎫=--⋅=-++ ⎪⎝⎭⎰ (10)2sin d x x x ⎰解 22111sin (1cos 2)cos 2242d d d x x x x x x x x x x =⋅-=-⎰⎰⎰ 221111sin 2sin 2sin 24444d x x x x x x x dx =-=--⎰⎰（）。

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4 微分中值定理及导数应用习题4.11.（1）x ，()f x 为最小值。

（2）,()2f p x x =为最大值。

（3）1x =，()f x 为最大值。

2.（1）(1)1f =，(2)4f =，3(2)(1)()()3221f f f f x -ⅱ===-；（2）(0)30,(0)0,()03603f f f f f f pp px p 骣-琪琪骣骣桫ⅱ琪琪=====琪琪桫桫-；（3）()14f p =，()14f p -=-，()()444()(arccos )2()44f f f f p p x p p p --ⅱ===--；（4）(1)(1)(1)1,(1)1,()(0)01(1)f f f f f f x --ⅱ=--=-===--.3. 2x =.4. 提示：利用Lagrange 定理.5. 提示：用反证法.6. 提示：利用Rolle 定理.7. 提示：对()()1f x F x x=+在[]0,1上用罗尔定理 8. 提示：利用Lagrange 定理. 9. 提示：f 在[],a b 上有界.10. 提示：证明()0f x ¢=. 11.（1）不能，理由见（2）; （2）112x =，2x =，323x =. 12. 4p x =.13. （1）提示：利用“()0f x ¢=则()f x C º（常数）”的结论。

（2）提示：令22()1tan sec f x x x =+-，证明()0f x º.14（1）提示：和差化积或直接用拉格朗日定理; （2）提示：利用Lagrange 定理.习题4.21. 提示：利用函数单调性定义和拉格朗日定理。

2.（1）单调减少. （2）单调增加. （3）单调增加. （4）单调增加.3.（1）在1(,)2- 内单调增加，在1(,)2+ 内单调减少；（2）在(),1-?或()1,+内单调减少，在()1,1-内单调增加；（3）当0a £时，f 单调减少；当0α>时，f 在(0,)a 单调增加，在(,)a + 单调减少；（4）在(),1-?或()0,1内单调减少，在()1,0-或()1,+内单调增加.4. 提示：设()()F x x f x h =-，证明F 在12(,)x x 内必取到F 在[]12,x x 上的最小值或者最大值.5.（3）提示：令()n f x x =，在[],b a 上用拉格朗日定理。

6.（2）提示：32(tan )tan 113x x x x 骣¢琪=+>++琪桫（3）更强的结果为：357911131512173840241tan 31531528351559251228559535x x x x x x x x x >+++++++习题4.31.（1）－1 （2）+（3）32 （4）23（5）3 （6）2e （7）3 （8）e （9）－2 （10）2 （11）0 （12）0 （13）0 （14）12（15）1 （16）16e- （17）e （18）1 （19）e 2- （20）12- 2.（1）a （2）2e - （3）13e习题4.41.（1）57223422arcsin 3(1)(1)15(1)3!4!x x x x ξξξξ--⎡⎤=+++-+-⎣⎦，,0 1.x ξθθ=<< （2）21(1)(1)(1)()1(1)2!!(1)!n n x x e x f x e x n n ξ+⎡⎤---=+-++++⎢⎥+⎣⎦ ，其中x 位于1和x 之间. （3）24611sin sin 1()()()2!24!26!2x x x x p p x p =-++-+-+，其中x 位于x 与2p-之间.（4）246()()cos ()1(1)2!4!6!x x f x x ππξ--=-+---，其中x 位于p 与x 之间. （5）234()(1)3(1)7(1)3(1)f x x x x x =--+-+-+-.（6）623457()1(1)x f x x x x x x x =++++++-，其中x 位于0与x 之间.（7）23222331111311351()(4)(4)(4)22422!423!4f x x x x =--+--- 92444135741(4)24!44x x -骣-琪++-琪桫 , 其中x 在x 与4之间.（8）35()3!5!2x x e e f x x ξξ-+=++⋅，其中x 位于0与x 之间. （9）2111()1(1)(1)(1)(1)(1)n n n n f x x x x x x++=--+-+--++-+ , 其中x 在x 与-1之间.2.（1）2 （2）124 （3）112- （4）12 （5）163. 3211()1(1),0128f x x x q q -=+-+<<，误差3211188E x q -=+<.4. 99!97!-5. 364()tan (sec sec ),013x f x x x x x q q q ==+-<<. 6. （1）342()2!3!(1)!nn x x x P x x x n =+++++- . （2）4622462221,2,0,1,2,2!3!!()1,21,1,2,2!3!(1)!k n k x x x x n k k k P x x x x x n k k k -⎧-+-++==⎪⎪=⎨⎪-+-++=-=⎪-⎩7.提示：利用()f x 在x a =点的n 阶泰勒公式。

8.提示：利用21n -阶的带拉格朗日余项的泰勒公式。

9.提示：利用()f x 在x b =点的1n -阶泰勒公式，然后将x a =代入。

10.提示：利用f 在x 点处带拉格朗日余项的一阶泰勒公式。

11.提示：(,)d a b $ ，使()0f d =，再在[],c d 上对f 用拉格朗日公式。

12.（1）1.6484375，4.5×10-4。

习题4.51（1）极大值00x y == ，极小值11x y ==-；（2）极大值12x y =-=-；极小值12x y==；（3）无极值；（4）极小值00x y==；（5）极大值3243242k x k y pp p p +=+=，极小值7247242k x k y pp p p +=+=- ，k Î . （6）极小值2e 2ex y-==-. 2. （1）max 12y =, min 13y =-; （2）max min 2,12y y p p ==-; （3）max 54y =, min 5y =- （4）max min 1,0e y y == ; （5）max2y p =, min 2y p=- ; （6）max min 0,8y y ==-; （7）max 35y =, min 1y =- ; （8）max min 22y y ==-; （9）max e y =, min 1ey =-. 3（3）提示：令sin ()xf x x=，利用f 的单调性。

4. 当n 为偶数时，无极值；当n 为奇数时，有极大值01x y ==.5.6. 底半径∶高 = 1∶27.2V R 8.水厂应建在甲城与乙城到岸的垂足之间，离甲城5013骣琪-琪桫公里处。

9.矩形在第一象限的顶点坐标为骣，其他顶点坐标由对称性可得，此时矩形面积最大，最大值为2ab .10. 11ni i x x n ==å.习题4.61（1）在(,1)-?上向下凹，在(1,)-+ 上向上凹，拐点为(1,1)-;（2）在,33骣琪琪-?? 琪琪桫桫向上凹；在33骣琪-琪桫向下凹，拐点为55,3939骣琪琪---琪琪桫桫;（3）在(,1)- 上向下凹，在(1,)+ 上向上凹，拐点为(1,0); （4）在(),1- 向下凹；在 (1,)+ 向上凹，无拐点.2. 提示：先证当()0f x ⅱ>时，有1212()()22x x f x f x f 骣++琪<琪桫 3.（1）0y =，1x =-，2x = （2）1,1x y x =-=- （3）1e x =-，1ey x =+ （4）0y = （5）0x =，3y x =+ 4. 略。

5. 三个拐点同位于直线430x y -+=上。

6. 曲率为2. 7.曲率为1.8. 在(,0),(0,)a b 处的曲率分别为22,a bb a. 9. 所求抛物线为2211228y x x p p =-++-，曲率圆方程为2212x y p 骣琪-+=琪桫.10. 提示：(0)(0)0f f ¢==.。