(通用版)2019版高考數(shù)學(xué)二輪復(fù)習(xí) 第一部分 專題四 導(dǎo)數(shù)的綜合應(yīng)用(第一課時)“導(dǎo)數(shù)與不等式”考法面面觀講義 理(重點生含解析).doc
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專題四 導(dǎo)數(shù)的綜合應(yīng)用卷卷卷2018利用導(dǎo)數(shù)的單調(diào)性證明不等式T21(2)根據(jù)函數(shù)的極值求參數(shù)、不等式的證明T21導(dǎo)數(shù)在不等式的證明、由函數(shù)的極值點求參數(shù)T212017利用導(dǎo)數(shù)研究函數(shù)的零點問題T21(2)函數(shù)的單調(diào)性、極值、零點問題、不等式的證明T21由不等式恒成立求參數(shù)、不等式放縮T212016函數(shù)的零點、不等式的證明T21函數(shù)單調(diào)性的判斷、不等式的證明及值域問題T21函數(shù)的最值、不等式的證明T21縱向把握趨勢導(dǎo)數(shù)的綜合問題是每年的必考內(nèi)容且難度大主要涉及函數(shù)的單調(diào)性、極值、零點、不等式的證明預(yù)計2019年會考查用分類討論研究函數(shù)的單調(diào)性以及函數(shù)的零點問題導(dǎo)數(shù)的綜合問題是每年的必考內(nèi)容,涉及函數(shù)的極值、最值、單調(diào)性、零點問題及不等式的證明,且近3年均考查了不等式的證明預(yù)計2019年仍會考查不等式的證明,同時要重點關(guān)注會討論函數(shù)的單調(diào)性及零點問題導(dǎo)數(shù)的綜合問題是每年的必考內(nèi)容,涉及函數(shù)的最值、零點、不等式的恒成立及不等式的證明問題,其中不等式的證明連續(xù)3年均有考查,應(yīng)引起關(guān)注預(yù)計2019年仍會考查不等式的證明,同時考查函數(shù)的最值或零點問題橫向把握重點導(dǎo)數(shù)日益成為解決問題必不可少的工具,利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性與極值(最值)是高考的常見題型,而導(dǎo)數(shù)與函數(shù)、不等式、方程、數(shù)列等的交匯命題,是高考的熱點和難點解答題的熱點題型有:(1)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性、極值、最值;(2)利用導(dǎo)數(shù)證明不等式或探討方程根;(3)利用導(dǎo)數(shù)求解參數(shù)的范圍或值.第一課時“導(dǎo)數(shù)與不等式”考法面面觀考法一不等式的證明問題 題型策略(一)設(shè)a為實數(shù),函數(shù)f (x)ex2x2a,xR.(1)求f (x)的單調(diào)區(qū)間與極值;(2)求證:當(dāng)aln 21且x0時,exx22ax1.破題思路第(1)問求什么想什么求f (x)的單調(diào)區(qū)間與極值,想到求導(dǎo)函數(shù)f (x),然后利用不等式f (x)0及f (x)x22ax1(aln 21,x0)成立,想到證明exx22ax10成立給什么用什么通過對第(1)問的研究,求得f (x)ex2x2a的單調(diào)性與極值,仔細觀察,可發(fā)現(xiàn)(exx22ax1)ex2x2a差什么找什么需要研究函數(shù)g(x)exx22ax1的單調(diào)性或最值,利用導(dǎo)數(shù)研究即可規(guī)范解答(1)由f (x)ex2x2a(xR),知f (x)ex2.令f (x)0,得xln 2.當(dāng)xln 2時,f (x)ln 2時,f (x)0,故函數(shù)f (x)在區(qū)間(ln 2,)上單調(diào)遞增所以f (x)的單調(diào)遞減區(qū)間是(,ln 2),單調(diào)遞增區(qū)間是(ln 2,),f (x)在xln 2處取得極小值f (ln 2)eln 22ln 22a22ln 22a,無極大值(2)證明:要證當(dāng)aln 21且x0時,exx22ax1,即證當(dāng)aln 21且x0時,exx22ax10.設(shè)g(x)exx22ax1(x0)則g(x)ex2x2a,由(1)知g(x)ming(ln 2)22ln 22a.又aln 21,則g(x)min0.于是對xR,都有g(shù)(x)0,所以g(x)在R上單調(diào)遞增于是對x0,都有g(shù)(x)g(0)0.即exx22ax10,故exx22ax1.題后悟通思路受阻分析本題屬于導(dǎo)數(shù)綜合應(yīng)用中較容易的問題,解決本題第(2)問時,易忽視與第(1)問的聯(lián)系,導(dǎo)函數(shù)g(x)ex2x2a的單調(diào)性已證,可直接用,若意識不到這一點,再判斷g(x)的單調(diào)性,則造成解題過程繁瑣,進而造成思維受阻或解題失誤技法關(guān)鍵點撥利用單調(diào)性證明單變量不等式的方法一般地,要證f (x)g(x)在區(qū)間(a,b)上成立,需構(gòu)造輔助函數(shù)F(x)f (x)g(x),通過分析F(x)在端點處的函數(shù)值來證明不等式若F(a)0,只需證明F(x)在(a,b)上單調(diào)遞增即可;若F(b)0,只需證明F(x)在(a,b)上單調(diào)遞減即可 對點訓(xùn)練1已知函數(shù)f (x)xln x,g(x)(x21)(為常數(shù))(1)若曲線yf (x)與曲線yg(x)在x1處有相同的切線,求實數(shù)的值;(2)若,且x1,證明:f (x)g(x)解:(1)f (x)ln x1,g(x)2x,則f (1)1,從而g(1)21,即.(2)證明:設(shè)函數(shù)h(x)xln x(x21),則h(x)ln x1x.設(shè)p(x)ln x1x,從而p(x)10對任意x1,)恒成立,所以當(dāng)x1,)時,p(x)ln x1xp(1)0,即h(x)0,因此函數(shù)h(x)xln x(x21)在1,)上單調(diào)遞減,即h(x)h(1)0,所以當(dāng),且x1時,f (x)g(x)成立題型策略(二)已知函數(shù)f (x)aexbln x,曲線yf (x)在點(1,f (1)處的切線方程為yx1.(1)求a,b;(2)證明:f (x)0.破題思路第(1)問求什么想什么求a,b的值,想到建立關(guān)于a,b的方程組給什么用什么題目條件中給出函數(shù)f (x)在點(1,f (1)處的切線方程,可據(jù)此建立關(guān)于a,b的方程組第(2)問求什么想什么要證f (x)0,想到f (x)的最小值大于0差什么找什么需求f (x)的最小值,因此只要利用導(dǎo)數(shù)研究函數(shù)f (x)的單調(diào)性即可規(guī)范解答(1)函數(shù)f (x)的定義域為(0,)f (x)aex,由題意得f (1),f (1)1,所以解得(2)證明:由(1)知f (x)exln x(x0)因為f (x)ex2在(0,)上單調(diào)遞增,又f (1)0,所以f (x)0在(0,)上有唯一實根x0,且x0(1,2)當(dāng)x(0,x0)時,f (x)0,從而當(dāng)xx0時,f (x)取極小值,也是最小值由f (x0)0,得ex02,則x02ln x0.故f (x)f (x0)e x02ln x0x022 20,所以f (x)0.題后悟通思路受阻分析本題屬于隱零點問題解決第(2)問時,常因以下兩個原因造成思維受阻,無法正常解題(1)f (x)0在(0,)上有解,但無法解出;(2)設(shè)出f (x)0的零點x0,即f (x)的最小值為f (x0),但是不能將函數(shù)f (x0)轉(zhuǎn)化成可求最值的式子,從而無法將問題解決當(dāng)遇到既含有指數(shù)式,又含有對數(shù)式的代數(shù)式需判斷其符號時,常需應(yīng)用這種技巧,把含有指數(shù)式與對數(shù)式的代數(shù)式轉(zhuǎn)化為不含有指數(shù)式與對數(shù)式的代數(shù)式,從而可輕松判斷其符號技法關(guān)鍵點撥利用最值證明單變量不等式的技巧利用最值證明單變量的不等式的常見形式是f (x)g(x)證明技巧:先將不等式f (x)g(x)移項,即構(gòu)造函數(shù)h(x)f (x)g(x),轉(zhuǎn)化為證不等式h(x)0,再次轉(zhuǎn)化為證明h(x)min0,因此,只需在所給的區(qū)間內(nèi),判斷h(x)的符號,從而判斷其單調(diào)性,并求出函數(shù)h(x)的最小值,即可得證對點訓(xùn)練2已知函數(shù)f (x).(1)若f (x)在區(qū)間(,2上為單調(diào)遞增函數(shù),求實數(shù)a的取值范圍;(2)若a0,x01,設(shè)直線yg(x)為函數(shù)f(x)的圖象在xx0處的切線,求證:f (x)g(x)解:(1)易得f (x),由題意知f (x)0對x(,2恒成立,故x1a對x(,2恒成立,1a2,a1.故實數(shù)a的取值范圍為(,1(2)證明:若a0,則f (x).函數(shù)f (x)的圖象在xx0處的切線方程為yg(x)f (x0)(xx0)f (x0)令h(x)f (x)g(x)f (x)f (x0)(xx0)f (x0),xR,則h(x)f (x)f (x0).設(shè)(x)(1x)ex0(1x0)ex,xR,則(x)ex0(1x0)ex.x01,(x)0,(x)在R上單調(diào)遞減,而(x0)0,當(dāng)xx0時,(x)0,當(dāng)xx0時,(x)0,當(dāng)xx0時,h(x)0,當(dāng)xx0時,h(x)0,h(x)在區(qū)間(,x0)上為增函數(shù),在區(qū)間(x0,)上為減函數(shù),xR時,h(x)h(x0)0,f (x)g(x)構(gòu)造函數(shù)證明雙變量函數(shù)不等式若ba0,求證:ln bln a.破題思路證明:ln bln a,想到如下思路:(1)構(gòu)造以a為主元的函數(shù),利用導(dǎo)數(shù)求解(2)考慮到ln bln aln ,設(shè)t,化為只有一個因變量t的函數(shù)求解(3)原不等式右邊可分開寫,觀察此式兩邊,發(fā)現(xiàn)其與f (x)ln x有關(guān),故先研究f (x)的單調(diào)性,從而得解規(guī)范解答法一:主元法(學(xué)生用書不提供解題過程)構(gòu)造函數(shù)f (x)ln bln x,其中0xb,則f (x).0xb,f (x)a0,故f (a)f (b)0,即ln bln a.法二:整體換元法(學(xué)生用書不提供解題過程)令t(t1),構(gòu)造函數(shù)f (t)ln t,則f (t).t1,t210,t22t112210,則f (t)0,f (t)在(1,)上單調(diào)遞增,故f (t)f (1)0,即ln 0,從而有l(wèi)n bln a.法三:函數(shù)不等式的對稱性(學(xué)生用書提供解題過程)原不等式可化為ln bln a,則構(gòu)造函數(shù)f (x)ln x(bxa0),則f (x)0,f (x)ln x在(a,b)上單調(diào)遞增,即f (b)f (a),則ln bln a,故ln bln a.題后悟通思路受阻分析由于題目條件少,不能正確分析要證不等式的特點,并構(gòu)造相應(yīng)的函數(shù)將問題轉(zhuǎn)化,從而導(dǎo)致無從下手解決問題技法關(guān)鍵點撥證明雙變量函數(shù)不等式的常見思路(1)將雙變量中的一個看作變量,另一個看作常數(shù),構(gòu)造一個含參數(shù)的輔助函數(shù)證明不等式(2)整體換元對于齊次式往往可將雙變量整體換元,化為一元不等式(3)若雙變量的函數(shù)不等式具有對稱性,并且可以將兩個變量分離開,分離之后的函數(shù)結(jié)構(gòu)具有相似性,從而構(gòu)造函數(shù)利用單調(diào)性證明對點訓(xùn)練3(2019屆高三黃岡模擬)已知函數(shù)f (x)ln xex(R)(1)若函數(shù)f (x)是單調(diào)函數(shù),求的取值范圍;(2)求證:當(dāng)0x11.解:(1)函數(shù)f (x)的定義域為(0,),f (x)ln xex,f (x)ex,函數(shù)f (x)是單調(diào)函數(shù),f (x)0或f (x)0在(0,)上恒成立,當(dāng)函數(shù)f (x)是單調(diào)遞減函數(shù)時,f (x)0,0,即xex0,xex,令(x),則(x),當(dāng)0x1時,(x)1時,(x)0,則(x)在(0,1)上單調(diào)遞減,在(1,)上單調(diào)遞增,當(dāng)x0時,(x)min(1),.當(dāng)函數(shù)f (x)是單調(diào)遞增函數(shù)時,f (x)0,0,即xex0,xex,由得(x)在(0,1)上單調(diào)遞減,在(1,)上單調(diào)遞增,又(0)0,x時,(x)0,0.綜上,的取值范圍是0,)(2)證明:由(1)可知,當(dāng)時,f (x)ln xex在(0,)上單調(diào)遞減,0x1f (x2),即ln x1ex1ln x2ex2,e1x2e1x1ln x1ln x2.要證e1x2e1x11,只需證ln x1ln x21,即證ln1,令t,t(0,1),則只需證ln t1,令h(t)ln t1,則h(t),當(dāng)0t1時,h(t)0,即ln t1,故原不等式得證考法二恒成立與能成立問題 題型策略(一)已知函數(shù)f (x)xln x,若對于所有x1都有f (x)ax1,求實數(shù)a的取值范圍破題思路求什么想什么求實數(shù)a的取值范圍,想到建立關(guān)于實數(shù)a的不等式給什么用什么題目條件中,已知f (x)ax1,即xln xax1,想到將不等式轉(zhuǎn)化為xln xax10或aln x差什么找什么缺少xln xax1的最小值或ln x的最小值,利用導(dǎo)數(shù)求解即可規(guī)范解答法一:分離參數(shù)法(學(xué)生用書不提供解題過程)依題意,得f (x)ax1在1,)上恒成立,即不等式aln x在x1,)恒成立,亦即amin,x1,)設(shè)g(x)ln x(x1),則g(x).令g(x)0,得x1.當(dāng)x1時,因為g(x)0,故g(x)在1,)上是增函數(shù)所以g(x)在1,)上的最小值是g(1)1.故a的取值范圍是(,1法二:構(gòu)造函數(shù)法(學(xué)生用書提供解題過程)當(dāng)x1時,有f (1)a1,即a10,得a1.構(gòu)造F(x)f (x)(ax1)xln xax1,原命題等價于F(x)0在x1上恒成立F(x)min0,x1,)由于F(x)ln x1a0在x1,)上恒成立,因此,函數(shù)F(x)在1,)上單調(diào)遞增,所以F(x)minF(1)1a0,得a1.故a的取值范圍是(,1題后悟通(一)思路受阻分析求解本題時,直接作差構(gòu)造函數(shù)或分離參數(shù)后構(gòu)造函數(shù)求a的取值范圍,其關(guān)鍵是正確求解所構(gòu)造函數(shù)的最值,這也是大多數(shù)同學(xué)不會求解或不能正確求解最值而導(dǎo)致無法繼續(xù)解題或解題失誤的地方(二)技法關(guān)鍵點撥分離參數(shù)法解含參不等式恒成立問題的思路與關(guān)鍵(1)分離參數(shù)法解含參不等式恒成立問題的思路用分離參數(shù)法解含參不等式恒成立問題是指在能夠判斷出參數(shù)的系數(shù)的正負的情況下,可以根據(jù)不等式的性質(zhì)將參數(shù)分離出來,得到一個一端是參數(shù),另一端是變量表達式的不等式,只要研究變量表達式的最值就可以解決問題(2)求解含參不等式恒成立問題的關(guān)鍵是過好“雙關(guān)”轉(zhuǎn)化關(guān)通過分離參數(shù)法,先轉(zhuǎn)化為f (a)g(x)(或f (a)g(x)對xD恒成立,再轉(zhuǎn)化為f (a)g(x)max(或f (a)g(x)min)求最值關(guān)求函數(shù)g(x)在區(qū)間D上的最大值(或最小值)問題(三)解題細節(jié)提醒有些含參不等式恒成立問題,在分離參數(shù)時會遇到討論的麻煩,或者即使分離出參數(shù),但參數(shù)的最值卻難以求出,這時常利用導(dǎo)數(shù)法,借助導(dǎo)數(shù),分析函數(shù)的單調(diào)性,通過對函數(shù)單調(diào)性的分析確定函數(shù)值的變化情況,找到參數(shù)滿足的不等式,往往能取得意想不到的效果對點訓(xùn)練1設(shè)函數(shù)f (x)ax2aln x,其中aR.(1)討論f (x)的單調(diào)性;(2)確定a的所有可能取值,使得f (x)e1x在區(qū)間(1,)內(nèi)恒成立(e2.718為自然對數(shù)的底數(shù))解:(1)由題意,f (x)2ax,x0,當(dāng)a0時,2ax210,f (x)0,f (x)在(0,)上單調(diào)遞減當(dāng)a0時,f (x),當(dāng)x時,f (x)0.故f (x)在上單調(diào)遞減,在上單調(diào)遞增綜上所述,當(dāng)a0時,f (x)在(0,)上單調(diào)遞減;當(dāng)a0時,f (x)在上單調(diào)遞減,在上單調(diào)遞增(2)原不等式等價于f (x)e1x0在(1,)上恒成立一方面,令g(x)f (x)e1xax2ln xe1xa,只需g(x)在(1,)上恒大于0即可又g(1)0,故g(x)在x1處必大于等于0.令F(x)g(x)2axe1x,由g(1)0,可得a.另一方面,當(dāng)a時,F(xiàn)(x)2ae1x1e1xe1x,因為x(1,),故x3x20.又e1x0,故F(x)在a時恒大于0.所以當(dāng)a時,F(xiàn)(x)在(1,)上單調(diào)遞增所以F(x)F(1)2a10,故g(x)也在(1,)上單調(diào)遞增所以g(x)g(1)0,即g(x)在(1,)上恒大于0.綜上所述,a.故實數(shù)a的取值范圍為.題型策略(二)已知函數(shù)f (x)xaln x,g(x)(aR)若在1,e上存在一點x0,使得f (x0)g(x0)成立,求a的取值范圍破題思路求什么想什么求a的取值范圍,想到建立關(guān)于a的不等式給什么用什么題目條件中,給出存在x01,e,使f (x0)g(x0)成立,想到利用f (x0)g(x0)建立關(guān)于a的不等式差什么找什么要建立關(guān)于a的不等式,可令h(x)f (x)g(x),轉(zhuǎn)化為h(x)的最值問題求解規(guī)范解答依題意,只需f (x0)g(x0)min0,x01,e即可令h(x)f (x)g(x)xaln x,x1,e,則h(x)1.令h(x)0,得xa1.若a11,即a0時,h(x)0,h(x)單調(diào)遞增,h(x)minh(1)a20,得a2;若1a1e,即0a2,x(0,e1)與h(x)0不符,故舍去若a1e,即ae1時,h(x)在1,e上單調(diào)遞減,則h(x)minh(e)ea,令h(e)e1成立綜上所述,a的取值范圍為(,2).題后悟通思路受阻分析本題構(gòu)造函數(shù)后,求解a的取值范圍時,需對a分類討論此處往往因不會分類討論或討論不全而導(dǎo)致解題失誤技法關(guān)鍵點撥不等式能成立問題的解題關(guān)鍵點對點訓(xùn)練2(2019屆高三河北“五個一名校聯(lián)盟”模擬)已知a為實數(shù),函數(shù)f (x)aln xx24x.(1)若x3是函數(shù)f (x)的一個極值點,求實數(shù)a的值;(2)設(shè)g(x)(a2)x,若存在x0,使得f (x0)g(x0)成立,求實數(shù)a的取值范圍解:(1)函數(shù)f (x)的定義域為(0,),f (x)2x4.x3是函數(shù)f (x)的一個極值點,f (3)0,解得a6.經(jīng)檢驗,當(dāng)a6時,x3是函數(shù)f (x)的一個極小值點,符合題意,故a6.(2)由f (x0)g(x0),得(x0ln x0)ax2x0,記F(x)xln x(x0),則F(x)(x0),當(dāng)0x1時,F(xiàn)(x)1時,F(xiàn)(x)0,F(xiàn)(x)單調(diào)遞增F(x)F(1)10,a.記G(x),x,則G(x).x,22ln x2(1ln x)0,x2ln x20,當(dāng)x時,G(x)0,G(x)單調(diào)遞增G(x)minG(1)1,aG(x)min1,故實數(shù)a的取值范圍為1,)題型策略(三)已知函數(shù)f (x)ln xmx,g(x)x(a0)(1)求函數(shù)f (x)的單調(diào)區(qū)間;(2)若m,對x1,x22,2e2都有g(shù)(x1)f (x2)成立,求實數(shù)a的取值范圍破題思路第(1)問求什么想什么求f (x)的單調(diào)區(qū)間,想到解不等式f (x)0或f (x)0,所以f (x)m,當(dāng)m0時,f (x)0,f (x)在(0,)上單調(diào)遞增當(dāng)m0時,由f (x)0得x;由得0x.所以f (x)在上單調(diào)遞增,在上單調(diào)遞減綜上所述,當(dāng)m0時,f (x)的單調(diào)遞增區(qū)間為(0,),無單調(diào)遞減區(qū)間;當(dāng)m0時,f (x)的單調(diào)遞增區(qū)間為,單調(diào)遞減區(qū)間為.(2)若m,則f (x)ln xx.對x1,x22,2e2都有g(shù)(x1)f (x2)成立,等價于對x2,2e2都有g(shù)(x)minf (x)max,由(1)知在2,2e2上f (x)的最大值為f (e2),又g(x)10(a0),x2,2e2,所以函數(shù)g(x)在2,2e2上是增函數(shù),所以g(x)ming(2)2.由2,得a3,又a0,所以a(0,3,所以實數(shù)a的取值范圍為(0,3題后悟通(一)思路受阻分析本題(2)中不會或不能準確地將已知條件“x1,x22,2e2都有g(shù)(x1)f (x2)成立”進行轉(zhuǎn)化,而導(dǎo)致無法求解此題(二)技法關(guān)鍵點撥1最值定位法解雙參不等式恒成立問題的思路策略(1)用最值定位法解雙參不等式恒成立問題是指通過不等式兩端的最值進行定位,轉(zhuǎn)化為不等式兩端函數(shù)的最值之間的不等式,列出參數(shù)所滿足的不等式,從而求解參數(shù)的取值范圍(2)有關(guān)兩個函數(shù)在各自指定范圍內(nèi)的不等式恒成立問題,這里兩個函數(shù)在指定范圍內(nèi)的自變量是沒有關(guān)聯(lián)的,這類不等式的恒成立問題就應(yīng)該通過最值進行定位,對于任意的x1a,b,x2m,n,不等式f (x1)g(x2)恒成立,等價于f (x)min(xa,b)g(x)max(xm,n),列出參數(shù)所滿足的不等式,便可求出參數(shù)的取值范圍2常見的雙變量不等式恒成立問題的類型(1)對于任意的x1a,b,總存在x2m,n,使得f (x1)g(x2)f (x1)maxg(x2)max.(2)對于任意的x1a,b,總存在x2m,n,使得f (x1)g(x2)f (x1)ming(x2)min.(3)若存在x1a,b,對任意的x2m,n,使得f (x1)g(x2)f (x1)ming(x2)min.(4)若存在x1a,b,對任意的x2m,n,使得f (x1)g(x2)f (x1)maxg(x2)max.(5)對于任意的x1a,b,x2m,n,使得f (x1)g(x2)f (x1)maxg(x2)min.(6)對于任意的x1a,b,x2m,n,使得f (x1)g(x2)f (x1)ming(x2)max.對點訓(xùn)練3已知函數(shù)f (x)x(a1)ln x(aR),g(x)x2exxex.(1)當(dāng)x1,e時,求f (x)的最小值;(2)當(dāng)a1時,若存在x1e,e2,使得對任意的x22,0,f (x1)g(x2)恒成立,求a的取值范圍解:(1)f (x)的定義域為(0,),f (x)1.當(dāng)a1時,x1,e,f (x)0,f (x)為增函數(shù),所以f (x)minf (1)1a.當(dāng)1ae時,x1,a時,f (x)0,f (x)為減函數(shù);xa,e時,f (x)0,f (x)為增函數(shù)所以f (x)minf (a)a(a1)ln a1.當(dāng)ae時,x1,e,f (x)0,f (x)在1,e上為減函數(shù),所以f (x)minf (e)e(a1).綜上,當(dāng)a1時,f (x)min1a;當(dāng)1ae時,f (x)mina(a1)ln a1;當(dāng)ae時,f (x)mine(a1).(2)由題意知f (x)(xe,e2)的最小值小于g(x)(x2,0)的最小值由(1)知,當(dāng)a1時,f (x)在e,e2上單調(diào)遞增,所以f (x)minf (e)e(a1).由題意知g(x)(1ex)x.當(dāng)x2,0時,g(x)0,g(x)為減函數(shù),g(x)ming(0)1,所以e(a1),所以a的取值范圍為.專題跟蹤檢測(對應(yīng)配套卷P171)1(2019屆高三唐山模擬)已知f (x)x2a2ln x,a0.(1)求函數(shù)f (x)的最小值;(2)當(dāng)x2a時,證明:a.解:(1)函數(shù)f (x)的定義域為(0,),f (x)x.當(dāng)x(0,a)時,f (x)0,f (x)單調(diào)遞增所以當(dāng)xa時,f (x)取得極小值,也是最小值,且f (a)a2a2ln a.(2)證明:由(1)知,f (x)在(2a,)上單調(diào)遞增,則所證不等式等價于f (x)f (2a)a(x2a)0.設(shè)g(x)f (x)f (2a)a(x2a),則當(dāng)x2a時,g(x)f (x)axa0,所以g(x)在(2a,)上單調(diào)遞增,當(dāng)x2a時,g(x)g(2a)0,即f (x)f (2a)a(x2a)0,故a.2已知函數(shù)f (x)xex2xaln x,曲線yf (x)在點P(1,f (1)處的切線與直線x2y10垂直(1)求實數(shù)a的值;(2)求證:f (x)x22.解:(1)因為f (x)(x1)ex2,所以曲線yf (x)在點P(1,f (1)處的切線斜率kf (1)2e2a.而直線x2y10的斜率為,由題意可得(2e2a)1,解得a2e.(2)證明:由(1)知,f (x)xex2x2eln x.不等式f (x)x22可化為xex2x2eln xx220.設(shè)g(x)xex2x2eln xx22,則g(x)(x1)ex22x.記h(x)(x1)ex22x(x0),則h(x)(x2)ex2,因為x0,所以x22,ex1,故(x2)ex2,又0,所以h(x)(x2)ex20,所以函數(shù)h(x)在(0,)上單調(diào)遞增又h(1)2e22e20,所以當(dāng)x(0,1)時,h(x)0,即g(x)0,即g(x)0,函數(shù)g(x)單調(diào)遞增所以g(x)g(1)e22eln 112e1,顯然e10,所以g(x)0,即xex2x2eln xx22,也就是f (x)x22.3(2018武漢模擬)設(shè)函數(shù)f (x)(1xx2)ex(e2.718 28是自然對數(shù)的底數(shù))(1)討論f (x)的單調(diào)性;(2)當(dāng)x0時,f (x)ax12x2恒成立,求實數(shù)a的取值范圍解:(1)f (x)(2xx2)ex(x2)(x1)ex.當(dāng)x1時,f (x)0;當(dāng)2x0.所以f (x)在(,2),(1,)上單調(diào)遞減,在(2,1)上單調(diào)遞增(2)設(shè)F(x)f (x)(ax12x2),F(xiàn)(0)0,F(xiàn)(x)(2xx2)ex4xa,F(xiàn)(0)2a,當(dāng)a2時,F(xiàn)(x)(2xx2)ex4xa(x2)(x1)ex4x2(x2)(x1)exx2(x2)(x1)ex1,設(shè)h(x)(x1)ex1,h(x)xex0,所以h(x)在0,)上單調(diào)遞增,h(x)(x1)ex1h(0)0,即F(x)0在0,)上恒成立,F(xiàn)(x)在0,)上單調(diào)遞減,F(xiàn)(x)F(0)0,所以f (x)ax12x2在0,)上恒成立當(dāng)a0,而函數(shù)F(x)的圖象在(0,)上連續(xù)且x,F(xiàn)(x)逐漸趨近負無窮,必存在正實數(shù)x0使得F(x0)0且在(0,x0)上F(x)0,所以F(x)在(0,x0)上單調(diào)遞增,此時F(x)F(0)0,f (x)ax12x2有解,不滿足題意綜上,a的取值范圍是2,)4(2018南昌模擬)設(shè)函數(shù)f (x)2ln xmx21.(1)討論函數(shù)f (x)的單調(diào)性;(2)當(dāng)f (x)有極值時,若存在x0,使得f (x0)m1成立,求實數(shù)m的取值范圍解:(1)函數(shù)f (x)的定義域為(0,),f (x)2mx,當(dāng)m0時,f (x)0,f (x)在(0,)上單調(diào)遞增;當(dāng)m0時,令f (x)0,得0x,令f (x),f (x)在上單調(diào)遞增,在上單調(diào)遞減(2)由(1)知,當(dāng)f (x)有極值時,m0,且f (x)在上單調(diào)遞增,在上單調(diào)遞減f (x)maxf 2lnm1ln m,若存在x0,使得f (x0)m1成立,則f (x)maxm1.即ln mm1,ln mm10),g(x)10,g(x)在(0,)上單調(diào)遞增,且g(1)0,0m0時,對任意的x,恒有f (x)e1成立,求實數(shù)b的取值范圍解:(1)函數(shù)f (x)的定義域為(0,)當(dāng)b2時,f (x)aln xx2,所以f (x)2x.當(dāng)a0時,f (x)0,所以函數(shù)f (x)在(0,)上單調(diào)遞增當(dāng)a0時,令f (x)0,解得x (負值舍去),當(dāng)0x 時,f (x)時,f (x)0,所以函數(shù)f (x)在上單調(diào)遞增綜上所述,當(dāng)b2,a0時,函數(shù)f (x)在(0,)上單調(diào)遞增;當(dāng)b2,a0時,f (x)bln xxb,f (x)bxb1.令f (x)0,得0x0,得x1.所以函數(shù)f (x)在上單調(diào)遞減,在(1,e上單調(diào)遞增,f (x)max為f beb與f (e)beb中的較大者f (e)f ebeb2b.令g(m)emem2m(m0),則當(dāng)m0時,g(m)emem2220,所以g(m)在(0,)上單調(diào)遞增,故g(m)g(0)0,所以f (e)f ,從而f (x)maxf (e)beb所以bebe1,即ebbe10.設(shè)(t)ette1(t0),則(t)et10,所以(t)在(0,)上單調(diào)遞增又(1)0,所以ebbe10的解集為(0,1所以b的取值范圍為(0,16(2018開封模擬)已知函數(shù)f (x)axx2xln a(a0,a1)(1)當(dāng)ae(e是自然對數(shù)的底數(shù))時,求函數(shù)f (x)的單調(diào)區(qū)間;(2)若存在x1,x21,1,使得|f (x1)f (x2)|e1,求實數(shù)a的取值范圍解:(1)f (x)axln a2xln a2x(ax1)ln a.當(dāng)ae時,f (x)2xex1,其在R上是增函數(shù),又f (0)0,f (x)0的解集為(0,),f (x)1時,ln a0,y(ax1)ln a在R上是增函數(shù),當(dāng)0a1時,ln a1或0a0),g(a)120,g(a)a2ln a在(0,)上是增函數(shù)而g(1)0,故當(dāng)a1時,g(a)0,即f (1)f (1);當(dāng)0a1時,g(a)0,即f (1)1時,f (x)maxf (x)minf (1)f (0)e1,即aln ae1,函數(shù)yaln a在(1,)上是增函數(shù),解得ae;當(dāng)0a1時,f (x)maxf (x)minf (1)f (0)e1,即ln ae1,函數(shù)yln a在(0,1)上是減函數(shù),解得0a.綜上可知,實數(shù)a的取值范圍為e,)- 1.請仔細閱讀文檔,確保文檔完整性,對于不預(yù)覽、不比對內(nèi)容而直接下載帶來的問題本站不予受理。
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