2017年電大《高等數(shù)學(xué)基礎(chǔ)》必過考試小抄

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小抄版高等數(shù)學(xué)基礎(chǔ)歸類復(fù)習(xí)考試小抄一、單項選擇題1-1 下列各函數(shù)對中，（ C ）中的兩個函數(shù)相等．A. ， B. ，2)(xf?xg?(2)(xf?xg)(C. ， D. ，3lnln1?12?1-⒉設(shè)函數(shù) 的定義域為 ，則函數(shù) 的圖形關(guān)于（C ）對稱．)(f),(???)(fA. 坐標原點 B. 軸 C. 軸 D. xyx設(shè)函數(shù) 的定義域為 ，則函數(shù) 的圖形關(guān)于（D ）對稱．x)f?A. B. 軸 C. 軸 D. 坐標原點y?.函數(shù) 的圖形關(guān)于（ A ）對稱．2ex?(A) 坐標原點 (B) 軸 (C) 軸 (D) yxy?1-⒊下列函數(shù)中為奇函數(shù)是（ B ）．A. B. C. D. )1ln(2xy??xycos?2xa??)1ln(x?下列函數(shù)中為奇函數(shù)是（A ）．A. B. C. D. ?3xe??)1ln(?yysi?下列函數(shù)中為偶函數(shù)的是（ D ）．A B C D xysin)1(??x2xcos)1ln(2x?2-1 下列極限存計算不正確的是（ D ）．A. B. 2lim??x 0)1ln(im0???xC. D. 0sn?s?2-2 當(dāng) 時，變量（ C ）是無窮小量．A. B. C. D. xix1x1in2)l(?x當(dāng) 時，變量（ C ）是無窮小量．A B C D 0si1e?2x.當(dāng) 時，變量（D ）是無窮小量． A B C D ?xx)ln(?下列變量中，是無窮小量的為（ B ）A B C D.??1sin0x??ln10????1xe???24x??3-1 設(shè) 在點 x=1 處可導(dǎo)，則 （ D ）．)(xf ???hffh)1(2(lim0A. B. C. D. 1?)1??? )1(2f??設(shè) 在 可導(dǎo)，則 （ D ）．)(f0 xfxfh)((li00A B C D x?)2?? )(0xf?設(shè) 在 可導(dǎo)，則 （ D ）．)(f0 ??ffh)((li00A. B. C. D. )(f??)x?2? )(0f??設(shè) ，則 （ A ） A B. C. D. xfe)(?????ff1(lim0 e2e143-2. 下列等式不成立的是（D ）．A. B C. D.xxd)(cossinxd?xd?2)(lnxd?下列等式中正確的是（B ）．A. B. artn12?2)1(?C. D.xx)l( xcotta4-1 函數(shù) 的單調(diào)增加區(qū)間是（ D ）．4)(2??xfA. B. C. D. ,??,),(?),(??函數(shù) 在區(qū)間 內(nèi)滿足（A ）．5?y)6(A. 先單調(diào)下降再單調(diào)上升 B. 單調(diào)下降 C. 先單調(diào)上升再單調(diào)下降 D. 單調(diào)上升.函數(shù) 在區(qū)間（－5，5）內(nèi)滿足（ A ）62?xA 先單調(diào)下降再單調(diào)上升 B 單調(diào)下降 C 先單調(diào)上升再單調(diào)下降 D 單調(diào)上升. 函數(shù) 在區(qū)間 內(nèi)滿足（D ）．??y),2(A. 先單調(diào)下降再單調(diào)上升 B. 單調(diào)下降 C. 先單調(diào)上升再單調(diào)下降 D. 單調(diào)上升5-1 若 的一個原函數(shù)是 ，則 （D ）． A. B. C. D. )(xfx1??)(f xln21?x32.若 是 的一個原函數(shù)，則下列等式成立的是（ A ）。 )(FfA B )(aFxdxa??? )()(afbfdxba???小抄版C D)(xFf?? )()(aFbdxfba????5-2 若 ，則 （ B ）．cos???xfd)(A. B. C. D. ?inc?sc??sincx?os下列等式成立的是（D ）．A. B. )(d)(fxf??? )(xff?C. D. d?（ B ）． A. B. C. D. ?xfx)(d32 )(3xf )(32xf)(1xf1（ D ） A B C D ??fx)(2 )(2xfxfd)(1)(2fxfd)(2⒌-3 若 ，則 （ B ）．?cxF)(d??A. B. C. D. )( cx)(2cxF?)( cxF?)(1補充： , 無窮積分收斂的是 ???xefxde? d??12函數(shù) 的圖形關(guān)于 y 軸 對稱。x?10)(二、填空題⒈函數(shù) 的定義域是　?。?，+∞） 　　．)ln(392xf?函數(shù) 的定義域是 （2，3） ∪ （3，4 xy?4)ln( ]函數(shù) 的定義域是　?。ǎ?，2）xf?1若函數(shù) ，則 1 ．?????0,2)(xf ?)(f2 若函數(shù) ，在 處連續(xù)，則 　　e 　?。?????,)1)(kxf k.函數(shù) 在 處連續(xù)，則 2 ??????02sin)(xkxf ?k函數(shù) 的間斷點是　　x=0　 　?。????,sin1y函數(shù) 的間斷點是 x=3 。32?x函數(shù) 的間斷點是 x=0 ey13-⒈曲線 在 處的切線斜率是　　1/2　 　?。?(??f)2,1(曲線 在 處的切線斜率是 1/4 ．x曲線 在（0，2）處的切線斜率是 1 ．f.曲線 在 處的切線斜率是 3 ．)(3),(3-2 曲線 在 處的切線方程是　　y = 1 　?。芯€斜率是 0 xfsin?1π曲線 y = sinx 在點 (0,0)處的切線方程為 y = x 切線斜率是 1 4.函數(shù) 的單調(diào)減少區(qū)間是?。ǎ?，0 ）　　 　?。?1l(2?函數(shù) 的單調(diào)增加區(qū)間是　?。?，+∞）　 　　．exf.函數(shù) 的單調(diào)減少區(qū)間是 （－∞，－1 ） ．2?.函數(shù) 的單調(diào)增加區(qū)間是 （0，+∞） ．)(f函數(shù) 的單調(diào)減少區(qū)間是 （0，+∞） ．2xy?5-1 　　 　 　?。? . ．?deex2 ??xdsin22sitan x +C　 　?。? ??)(tan若 ，則 ?。? sin 3x　 　　?。?cf3si??)(f5-2 3 ． 0 ． ??35d)21( ????123dx0 ?exdx1ln下列積分計算正確的是（ B ）．A B C D d)(1????ex 0d)(1????xex 0d12???x小抄版0d|1???x三、計算題（一）、計算極限（1 小題，11 分）（1 ）利用極限的四則運算法則，主要是因式分解，消去零因子。（2 ）利用連續(xù)函數(shù)性質(zhì)： 有定義，則極限)(0xf )(lim00xffx??類型 1： 利用重要極限 ， ， 計算1sinl??x ksnkx?tanli1-1 求 ． 解： x5sin6lm0? 56sil5si6l00?xx1-2 求 解： 0tali3x ??x3tanlm0 31tanli10??1-3 求 解： =?i.x類型 2： 因式分解并利用重要極限 ， 化簡計算。1)(sl?ax1)sin(l??aa2-1 求 ． 解： =)1sin(lm21??xx )sin(lm21??xx2)1(.i ????2-2 解： ??21slx 21)(.)1(sinlmsin(l21 ??????xxx2-3 解： )3sin(4m??x li3si)3si(43323 ???x類型 3：因式分解并消去零因子，再計算極限3-1 解： =4586li24?x 4586li24??xx ??)1(2lixx 32li4??x3-2 31x????3335m17x?????3-3 解 4lim2x 412lim)(2li4li2 ????xxxx其他： ， 0sin21lmsin1l020 ?????xxxx 21sinlmsinl00?????xx， ?546li2x 1li2?x?54362lix 3li2?x（0807 考題）計算 ． 解： =xsin8tal0?xsin8tal0?48.sital0?x（0801 考題 . ）計算 ． 解 x2lim0 ?x2lim021li0x（0707 考題 .） = )1sin(31???? 4)3(1)sin(.l1 ?????x（二） 求函數(shù)的導(dǎo)數(shù)和微分（1 小題，11 分）（1 ）利用導(dǎo)數(shù)的四則運算法則 vu???? vu????（2 ）利用導(dǎo)數(shù)基本公式和復(fù)合函數(shù)求導(dǎo)公式x)(ln?? 1)(??aaxe ueu??.xx2cs)(otanicssi???? xexex xxx sin).(cos)( cin2ocos isiin22 ?????xxx eeeuos).(s)(sin.i 22??? xxeeeusin).(sin)(co2i22???類型 1： 加減法與乘法混合運算的求導(dǎo)，先加減求導(dǎo)，后乘法求導(dǎo)；括號求導(dǎo)最后計算 。1-1 xy)3(??解： ＝? ??322xee??????????????1322xxe????????132xe????????1-2 xylncot解：小抄版xxxxxy ????????????? ln2cs)(lnl)(cs)ln()(cot 2221-3 設(shè) ，求 ．exay?解： xexxeyxx 1scta1)(tat)(l)t( 2?????類型 2： 加減法與復(fù)合函數(shù)混合運算的求導(dǎo)，先加減求導(dǎo)，后復(fù)合求導(dǎo)2-1 ，求 解：lnsi2??y? o)(ln(si 22???????2-2 ，求 ecoxy?解： 222 csesi).(cos).(sin)(si)( xxexxx ???????2-3 ，求 ， 解：x5ln???yy 5455lnl????類型 3： 乘積與復(fù)合函數(shù)混合運算的求導(dǎo)，先乘積求導(dǎo)，后復(fù)合求導(dǎo)，求 。 解：eyxcos2? xexex sico2)(cs)( 22???其他： ，求 。x??y解： ???????? 2).(c.)(ln)o() xyx2csinl2xx?0807.設(shè) ，求 解：sie?y?2sinin cos)()( xeyx????0801.設(shè) ，求 解：2x 222)()( xxxee??????0707.設(shè) ，求 解：sin?? cos.ininsi ??0701.設(shè) ，求 解：xyecol??y xxy1.l???（三）積分計算：（2 小題，共 22 分）湊微分類型 1： ???)1(d2xx??計算 解：?xcos2 cxd????1sin)(coscos20707.計算 ． 解： ?xd1sin2 cxx??????1os)(dsin1si20701 計算 ． 解： e2 ed121xe湊微分類型 2： ???x??.計算 ． 解： xdcos cxdxx????sin2cosdcos0807.計算 ． 解：?in ?? oin2in0801.計算 解： xed cexdexex???湊微分類型 3： ， ???dln1?? )ln(1?a??計算 解：xdln cxdux|lll.計算 解： ??e12 ?????e1e1 )ln2()dl(2x5)l(1??ex5 定積分計算題，分部積分法類型 1： cxaxdaxaxdaxd aaa ???????????? 12111 )(lnlnlnln計算 解： ， e1 ?24l241)ln(ln2lxd21 exxe?? 0)l(lne1 ????e小抄版計算 解： ， ?e12dlnx2??a cxxdx?????? 1ln)1(lnl2e2)ne1e1 ??計算 解： ，dxe?1ln2a cxxdx?? 4ln2ll=de1n4)4l(ln21 ????? exxde0807 el 9421)9ln32( nd3233e1 ??? ex0707 ?ee12xllx 33類型 2 ceaxdaxa???21)(xxede21010??41042?xx???1)(1??exxxdede210102?? 4130)4222????exx（0801 考題） ?e(10?x類型 3： caxadaxaxd ?????? sin1cocoscssin 2xx?? sin1in1ico20sin?xd 02)sico(s20 ??x??20cos?xd 120)cosin(si20 ???? ?? xx??? cxdsin421coin??20s?xd 402)i1cos(s20 ?? ??????xx2 200 011coin|in|4d? ????四、應(yīng)用題（1 題，16 分）類型 1： 圓柱體上底的中心到下底的邊沿的距離為 l，問當(dāng)?shù)装霃脚c高分別為多少時，圓柱體的體積最大？解：如圖所示，圓柱體高 與底半徑 滿足 hr22rh?圓柱體的體積公式為 lV)(π2???求導(dǎo)并令 0)3(π2???l得 ，并由此解出 ．lr6即當(dāng)?shù)装霃?，高 時，圓柱體的體積最大．lr36h類型 2：已知體積或容積，求表面積最小時的尺寸。2-1（0801 考題） 某制罐廠要生產(chǎn)一種體積為 V 的有蓋圓柱形容器，問容器的底半徑與高各為多少時用料最??？解：設(shè)容器的底半徑為 ，高為 ，則其容積rh22.,.rhr??表面積為 rSπ2π????， 由 得 ，此時 。24rV?? 0?S32V3π4Vrh由實際問題可知，當(dāng)?shù)装霃?與高 時可使用料最省。3π?r一體積為 V 的圓柱體，問底半徑與高各為多少時表面積最?。? 解： 本題的解法和結(jié)果與 2-1 完全相同。生產(chǎn)一種體積為 V 的無蓋圓柱形容器，問容器的底半徑與高各為多少時用料最??？解：設(shè)容器的底半徑為 ，高為 ，則無蓋圓柱形容器表面積為 rh l小抄版，令 ， 得 rVrhS2π2π??02π???rVS，Vr?,3由實際問題可知，當(dāng)?shù)装霃?與高 時可使用料最省。3πr?rh2-2 欲做一個底為正方形，容積為 32 立方米的長方體開口容器，怎樣做法用料最省？（0707 考題）解： 設(shè)底邊的邊長為 ，高為 ，用材料為 ，由已知 ， ，xhy322?Vhx2xh表面積 ，xVy422??令 ，得 ， 此時 =2042???xV63 ,4x2由實際問題可知， 是函數(shù)的極小值點，所以當(dāng) ， 時用料最省。?h欲做一個底為正方形，容積為 62.5 立方米的長方體開口容器，怎樣做法用料最?。拷猓?本題的解法與 2-2 同，只需把 V=62.5 代入即可。類型 3 求 求曲線 上的點，使其到點 的距離最短．kxy?2 )0,(aA曲線 上的點到點 的距離平方為,aaxL?????2)()(， 0? kx??23-1 在拋物線 上求一點，使其與 軸上的點 的距離最短． xy42 )0,3(解：設(shè)所求點 P（x，y），則滿足 ，點 P 到點 A 的距離之平方為y4)()3(22令 ，解得 是唯一駐點，易知 是函數(shù)的極小值點，0????L1x1?x當(dāng) 時， 或 ，所以滿足條件的有兩個點（1，2）和（1，－2）1?3-2 求曲線 上的點，使其到點 的距離最短．xy2 ),(A解：曲線 上的點到點 A（2，0） 的距離之平方為x)()(?????令 ，得 ， 由此 ， ??L122?xy?y即曲線 上的點（1， ）和（1， ）到點 A（2，0）的距離最短。xy2?08074 求曲線 上的點，使其到點 A（0，2）的距離最短。解： 曲線 上的點到點 A（0，2）的距離公式為 2xy?222 )()(??????yyxd與 在同一點取到最大值，為計算方便求 的最大值點，2d3)(1???y令 得 ，并由此解出 ，0)(2?23y6??x即曲線 上的點（ ）和點（ ）到點 A（0，2）的距離最短x?,6,2?ag an employment tribunal clai Emloyment tribunals sort out disagreements between employers and employees. You may need to make a claim to an employment tribunal if: you don't agree with the disciplinary action your employer has taken against you your employer dismisses you and you think that you have been dismissed unfairly. For more informu, take advice from one of the organisations listed under Fur ther help. Employment tribunals are less formal than some other courts, but it is still a legal process and you will need to give evidence under an oath or affirmation. Most people find making a claim to an employment tribunal challenging. If you are thinking about making a claim to an employment tribunal, you should get help straight away from one of the organisations listed under Further help. ation about dismissal and unfair dismissal, see Dismissal. You can make a claim to an employment tribunal, even if you haven't appealed against the disciplinary action your employer has taken against you. However, if you win your case, the tribunal may reduce any compensation awarded to you as a result of your failure to appeal. Remember that in most cases you must make an application to an employment tribunal within three months of the date when the event you are complaining about happened. If your application is received after this time limit, the tribunal will not usually accept i. If you are worried about how the time limits apply to you If you are being represented by a solicitor at the tribunal, they may ask you to sign an agreement where you pay their fee out of your compensation if you win the case. This is known as a damages-based agreement. In England and Wales, your solicitor can't charge you more than 35% of your compensation if you win the case. You are clear about the terms of the agreement. It might be best to get advice from an experienced adviser, for example, at a Citizens Advice Bureau. To find your nearest CAB, including those that give advice by e-mail, click on nearest CAB. For more information about making a claim to an employment tribunal, see Employment tribunals. The (lack of) air up there Watch m Cay man Islands-based Webb, the head of Fifa's anti-racism taskforce, is in London for the Football Association's 150th anniversary celebrations and will attend City's Premier League match at Chelsea on Sunday. "I am going to be at the match tomorrow and I have asked to meet Ya ya Toure," he told BBC Sport. "For me it's about how he felt and I would like to speak to him first to find out what his experience was." Uefa has opened disciplinary proceedings against CSKA for the "racist behaviour of their fans" during City's 2-1 win. Michel Platini, president of European football's governing body, has also ordered an immediate investigation into the referee's actions. CSKA said they were "surprised and disappointed" by Toure's complaint. In a statement the Russian side added: "We found no racist insults from fans of CSKA." Age has reached the end of the beginning of a word. May be guilty in his seems to passing a lot of different life became the appearance of the same day; May be back in the past, to oneself the paranoid weird belief disillusionment, these days, my mind has been very messy, in my mind constantly. Always feel oneself should go to do something, or write something. Twenty years of life trajectory deeply shallow, suddenly feel something, do it.一字開頭的年齡已經(jīng)到了尾聲。或許是愧疚于自己似乎把轉(zhuǎn)瞬即逝的很多個不同 的日子過成了同一天的樣子；或許是追溯過去，對自己那些近乎偏執(zhí)的怪異信念的醒 悟，這些天以來，思緒一直很凌亂，在腦海中不斷糾纏?？傆X得自己似乎應(yīng)該去做點 什么，或者寫點什么。二十年的人生軌跡深深淺淺，突然就感覺 到有些事情，非做不可了。The end of our life, and can meet many things really do?　 　 而窮盡我們的一生，又能遇到多少事情是真正地非做不可？ During my childhood, think lucky money and new clothes are necessary for New Year, but as the advance of the age, will be more and more found that those things are optional; Junior high school, thought to have a crush on just means that the real growth, but over the past three years later, his writing of alumni in peace, suddenly found that isn't really grow up, it seems is not so important; Then in high school, think don't want to give vent to out your inner voice can be in the high school children of the feelings in a period, but was eventually infarction when graduation party in the throat, later again stood on the pitch he has sweat profusely, looked at his thrown a basketball hoops, suddenly found himself has already can't remember his appearance. Baumgartner the disappointing news: Mission aborted. r plays an important role in this mission. Starting at the ground, conditions have to be very calm -- winds less than 2 mph, with no precipitation or humidity and limited cloud cover. The balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. It will climb higher than the tip of Mount Everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. As he crosses the boundary layer (called the tropopause),e can expect a 小抄版lot of turbulence. We often close ourselves off when traumatic events happen in our lives; instead of letting the world soften us, we let it drive us deeper into ourselves. We try to deflect the hurt and pain by pretending it doesn’t exist, but although we can try this all we want, in the end, we can’t hide from ourselves. We need to learn to open our hearts to the potentials of life and let the world soften us.生活發(fā)生不幸時，我們常常會關(guān)上心門；世界不僅沒能慰藉我們，反倒使我們更加消沉。我們假裝一切仿佛都不曾發(fā)生，以此試圖忘卻傷痛，可就算隱藏得再好，最終也還是騙不了自己。既然如此，何不嘗試打開心門，擁抱生活中的各種可能，讓世界感化我們呢？ Whenever we start to let our fears and seriousness get the best of us, we should take a step back and re-evaluate our behavior. The items listed below are six ways you can open your heart more fully and completely.當(dāng)恐懼與焦慮來襲時，我們應(yīng)該退后一步，重新反思自己的言行。下面六個方法有助于你更完滿透徹地敞開心扉。Whenever a painful situation arises in your life, try to embrace it instead of running away or trying to mask the hurt. When the sadness strikes, take a deep breath and lean into it. When we run away from sadness that’s unfolding in our lives, it gets stronger and more real. We take an emotion that’s fleeting and make it a solid event, instead of something that passes through us.當(dāng)生活中出現(xiàn)痛苦的事情時，別再逃跑或隱藏痛苦，試著擁抱它吧；當(dāng)悲傷來襲時，試著深呼吸，然后直面它。如果我們一味逃避生活中的悲傷，悲傷只會變得更強烈更真實——悲傷原本只是稍縱即逝的情緒，我們卻固執(zhí)地耿耿于懷 By utilizing our breath we soften our experiences. If we dam them up, our lives will stagnate, but when we keep them flowing, we allow more newness and greater experiences to blossom.深呼吸能減緩我們的感受。屏住呼吸，生活停滯；呼出呼吸，更多新奇與經(jīng)歷又將拉開序幕。