2025版新高考版高考總復(fù)習(xí)數(shù)學(xué) 函數(shù)的綜合運(yùn)用

2025版新高考版高考總復(fù)習(xí)數(shù)學(xué)4.3導(dǎo)數(shù)的綜合運(yùn)用考點(diǎn)1利用導(dǎo)數(shù)證明不等式1.(2023天津,20,16分)已知函數(shù)f(x)=1x+12ln((1)求曲線y=f(x)在x=2處切線的斜率;(2)當(dāng)x>0時(shí),證明:f(x)>1;(3)證明:56<lnn!-n+12lnn+n≤1(n∈解析(1)f'(x)=x+22x(x+1)?1x2ln(x+1),曲線f(x)在x(2)證明:當(dāng)x>0時(shí),f(x)>1?ln(x+1)-2xx+2>0,令g(x)=ln(x+1)-2xx+2,而g'(x)=x2(x+1)(x+2)2>0,故g(x)在(0,+∞)上單調(diào)遞增,(3)證法一:令中間待證式為某數(shù)列{an}的前n項(xiàng)和Sn,則Sn=lnn!-n+12lnn+n,則a1=S當(dāng)n≥2時(shí),an=Sn-Sn-1=1+n?由(2)知an<0(n≥2,n∈N*),故Sn≤S1=1,不等式右邊得證.欲證56<Sn,只需證:對(duì)任意的n≥2,n∈N*,k=2n(-ak)=k=2nf1k?1?1<16,令h(x)=ln(x+1)-x(x+2)2(x+1)(x>0),h'(x)=?x22(x+1)2,當(dāng)x>0時(shí),h'(x)<0,h(x)<h(0)=0,ln(x+1)當(dāng)n≥4時(shí),累加得k=4n(-ak)=k=4nf1k?1?1<1又-a2=f(1)-1=32ln2?1<32×0.694-1=0.041,-a3=52ln32?1<52×故k=2n(-ak)=(-a2)+(-a3)+k=4n(-ak)<0.041+0.0175+0.1=0.1585<證法二:只需證56<ln(n-1)!-n?12lnn+n≤1,n=1時(shí)顯然成立,先證ln(n-1)!-n?12lnn+當(dāng)n≥2時(shí),由(2)可得f1i?1>1(2≤i≤nn?12[lnn-ln(n-1n?1?12[ln(n-1)-ln(n-2……32[ln2-ln(2-1)]>1累加可得n?12lnn-ln(n-1)!>所以ln(n-1)!-n?12lnn+n≤1(n方便起見,以下x∈(0,1],令g(x)=ln(x+1)-2xg'(x)=x3(故g(x)單調(diào)遞減,所以g(x)<g(0)=0,所以f(x)<x+2又-16x2+x+2>x+1,所以f(x)<1+x則f1i?1<1+1i?12n?12[lnn-ln(n-1)]n?1?12[ln(n-1)-ln(n-2)]……32(ln2-ln1)<1+1累加可得n?12lnn-ln(n-1)!<n即56<ln(n-1)!-n?12綜上,56<lnn!-n+12lnn+2.(2021全國(guó)乙理,20,12分)設(shè)函數(shù)f(x)=ln(a-x),已知x=0是函數(shù)y=xf(x)的極值點(diǎn).(1)求a;(2)設(shè)函數(shù)g(x)=x+f(x)xf(x)解析(1)由題意得y=xf(x)=xln(a-x),x∈(-∞,a),∴y'=ln(a-x)+x·1a?x·(-1)=ln(a-x)-xa?x,x∵x=0是函數(shù)y=xf(x)的極值點(diǎn),∴l(xiāng)n(a-0)-0a?0=0,可得a當(dāng)a=1時(shí),y'=ln(1-x)-x1?x,x∈(-∞,令p(x)=ln(1-x)-x1?則p'(x)=1x易知當(dāng)x∈(-∞,1)時(shí),p'(x)<0恒成立.∴p(x)在(-∞,1)上為減函數(shù),又p(0)=0,∴當(dāng)x∈(-∞,0)時(shí),p(x)>0;當(dāng)x∈(0,1)時(shí),p(x)<0,∴函數(shù)y=xf(x)=xln(1-x)在(-∞,0)上為增函數(shù),在(0,1)上為減函數(shù).∴當(dāng)a=1時(shí),x=0是函數(shù)y=xf(x)的極值點(diǎn).∴a=1.(2)證明:由(1)知a=1,∴f(x)=ln(1-x),x∈(-∞,1),當(dāng)x∈(0,1)時(shí),f(x)=ln(1-x)<0,∴xf(x)<0,當(dāng)x∈(-∞,0)時(shí),f(x)=ln(1-x)>0,∴xf(x)<0,∴要證g(x)=x+f(x)xf(x)<1,只需證x+f只需證x+ln(1-x)>xln(1-x),只需證x+(1-x)ln(1-x)>0,令h(x)=x+(1-x)ln(1-x),則h'(x)=1-ln(1-x)-1=-ln(1-x),∴當(dāng)x∈(0,1)時(shí),h'(x)>0,h(x)單調(diào)遞增,當(dāng)x∈(-∞,0)時(shí),h'(x)<0,h(x)單調(diào)遞減,∴當(dāng)x∈(-∞,0)∪(0,1)時(shí),h(x)>h(0)=0,∴x+(1-x)ln(1-x)>0在(-∞,0)∪(0,1)上恒成立.∴g(x)<1.名師點(diǎn)撥:在論證較為復(fù)雜的不等式時(shí),可考慮數(shù)學(xué)證明中的分析法,將問題轉(zhuǎn)化,構(gòu)造函數(shù),通過求函數(shù)最值達(dá)到解決問題的目的.3.(2021新高考Ⅰ,22,12分)已知函數(shù)f(x)=x(1-lnx).(1)討論f(x)的單調(diào)性;(2)設(shè)a,b為兩個(gè)不相等的正數(shù),且blna-alnb=a-b,證明:2<1a+解題指導(dǎo):(1)首先確定函數(shù)f(x)的定義域,然后求其導(dǎo)數(shù)f'(x),再分別令f'(x)>0與f'(x)<0,解不等式,進(jìn)而得出函數(shù)f(x)的單調(diào)性;(2)先將已知條件進(jìn)行等價(jià)轉(zhuǎn)化,得到f1a=f1b,設(shè)x1=1a,x2=1b,將條件轉(zhuǎn)化為方程f(x)=k的兩個(gè)實(shí)根為x1,x2,然后結(jié)合函數(shù)的單調(diào)性分別證明x1+x2>2和x解析(1)函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=-lnx,令f'(x)>0,解得0<x<1,令f'(x)<0,解得x>1,所以函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.(2)證明:由blna-alnb=a-b得1a(1+lna)=1b(1+ln即1a令x1=1a,x2=1b,則x1,x2為f(x)=k的兩個(gè)實(shí)根,當(dāng)x→0+時(shí),f(x)→0+,當(dāng)x→+∞時(shí),f(x)→-∞,且f(1)=1,故k∈(0,不妨令x1∈(0,1),x2∈(1,e),則2-x1>1,e-x1>1,先證明x1+x2>2,即證x2>2-x1,即證f(x2)=f(x1)<f(2-x1).令h(x)=f(x)-f(2-x),x∈(0,1),則h'(x)=f'(x)+f'(2-x)=-lnx-ln(2-x)=-ln[x(2-x)].∵x∈(0,1),∴x(2-x)∈(0,1),∴h'(x)>0恒成立,∴h(x)為增函數(shù),∴h(x)<h(1)=0.∴f(x2)<f(2-x1),∴x2>2-x1,∴x1+x2>2.再證x1+x2<e,即證x2<e-x1,即證f(x2)=f(x1)>f(e-x1).令φ(x)=f(x)-f(e-x),x∈(0,1),則φ'(x)=-ln[x(e-x)],∵x→0+時(shí),φ'(x)→+∞,φ'(1)=-ln(e-1)<0,φ'(x)在(0,1)上單調(diào)遞減,∴在(0,1)上必存在唯一x0,使φ'(x0)=0,且當(dāng)x∈(0,x0)時(shí),φ'(x)>0,φ(x)單調(diào)遞增,當(dāng)x∈(x0,1)時(shí),φ'(x)<0,φ(x)單調(diào)遞減,又x→0+時(shí),f(x)→0+,且f(e)=0,∴x→0+時(shí),φ(x)→0+,又φ(1)=f(1)-f(e-1)>0,∴φ(x)>0恒成立,∴f(x2)>f(e-x1),∴x2<e-x1,∴x1+x2<e.綜上,2<1a+1方法總結(jié):利用導(dǎo)數(shù)證明不等式時(shí),首先要轉(zhuǎn)化為函數(shù)的單調(diào)性問題,然后結(jié)合函數(shù)的最值、函數(shù)的零點(diǎn)問題解決,注意構(gòu)造函數(shù)在證明過程中的應(yīng)用.4.(2016課標(biāo)Ⅲ文,21,12分)設(shè)函數(shù)f(x)=lnx-x+1.(1)討論f(x)的單調(diào)性;(2)證明當(dāng)x∈(1,+∞)時(shí),1<x?(3)設(shè)c>1,證明當(dāng)x∈(0,1)時(shí),1+(c-1)x>cx.解析(1)由題設(shè)知,f(x)的定義域?yàn)?0,+∞),f'(x)=1x-1,令f'(x)=0,解得當(dāng)0<x<1時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x>1時(shí),f'(x)<0,f(x)單調(diào)遞減.(4分)(2)證明:由(1)知f(x)在x=1處取得最大值,最大值為f(1)=0.所以當(dāng)x≠1時(shí),lnx<x-1.故當(dāng)x∈(1,+∞)時(shí),lnx<x-1,ln1x<1x-1,即1<x?(3)證明:由題設(shè)c>1,設(shè)g(x)=1+(c-1)x-cx,則g'(x)=c-1-cxlnc,令g'(x)=0,解得x0=lnc當(dāng)x<x0時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x>x0時(shí),g'(x)<0,g(x)單調(diào)遞減.(9分)由(2)知1<c?1lnc<c,又g(0)=g(1)=0,故當(dāng)0<x<1時(shí),g(x)>0.所以當(dāng)x∈(0,1)時(shí),1+(c-1)x>cx.(12分)疑難突破在(3)中,首先要解方程g'(x)=0,為了判定g(x)的單調(diào)性,必須比較極值點(diǎn)x0與區(qū)間(0,1)的關(guān)系,注意到g(0)=g(1)=0是求解本題的突破點(diǎn).評(píng)析本題考查了導(dǎo)數(shù)的綜合運(yùn)用及不等式的證明.5.(2018浙江,22,15分)已知函數(shù)f(x)=x-lnx.(1)若f(x)在x=x1,x2(x1≠x2)處導(dǎo)數(shù)相等,證明:f(x1)+f(x2)>8-8ln2;(2)若a≤3-4ln2,證明:對(duì)于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).解析(1)函數(shù)f(x)的導(dǎo)函數(shù)f'(x)=12x-由f'(x1)=f'(x2)得12x1-1x1因?yàn)閤1≠x2,所以1x1+1x由基本不等式得12x1x2=x1因?yàn)閤1≠x2,所以x1x2>256.由題意得f(x1)+f(x2)=x1-lnx1+x2-lnx2=12x1設(shè)g(x)=12x-lnx,則g'(x)=14x所以x(0,16)16(16,+∞)g'(x)-0+g(x)↘2-4ln2↗所以g(x)在[256,+∞)上單調(diào)遞增,故g(x1x2)>g(256)=8-8ln2,即f(x1)+f(x2)>8-8ln2.(2)令m=e-(|a|+k),n=|a則f(m)-km-a>|a|+k-k-a≥0,f(n)-kn-a<n1n?an所以,存在x0∈(m,n)使f(x0)=kx0+a,所以,對(duì)于任意的a∈R及k∈(0,+∞),直線y=kx+a與曲線y=f(x)有公共點(diǎn).由f(x)=kx+a得k=x?設(shè)h(x)=x?則h'(x)=lnx?x其中g(shù)(x)=x2-ln由(1)可知g(x)≥g(16),又a≤3-4ln2,故-g(x)-1+a≤-g(16)-1+a=-3+4ln2+a≤0,所以h'(x)≤0,即函數(shù)h(x)在(0,+∞)上單調(diào)遞減,因此方程f(x)-kx-a=0至多1個(gè)實(shí)根.綜上,當(dāng)a≤3-4ln2時(shí),對(duì)于任意k>0,直線y=kx+a與曲線y=f(x)有唯一公共點(diǎn).一題多解(1)f'(x)=12x-1x,且f'(x1)=f'(x2)(x1≠設(shè)f'(x1)=t,則12x-1x=t的兩根為x1即2t(x)2-x+2=0有兩個(gè)不同的正根x1,x2.∴Δ=1?∴f(x1)+f(x2)=x1+x2-ln(x1x=12t+2lnt設(shè)g(t)=12t+2lnt則g'(t)=-12t2+2∴g(t)在0,116上為減函數(shù),∴g(t)>g116∴f(x1)+f(x2)>8-8ln2.(2)設(shè)h(x)=f(x)-kx-a=x-lnx-kx-a,只需證明:當(dāng)a≤3-4ln2時(shí),對(duì)于任意的k>0,函數(shù)h(x)在(0,+∞)上只有唯一的零點(diǎn).取m=e-|a|-k,則h(m)=e?|a≥e?|a|?又x>0時(shí),x-kx<12k-k·12即h(x)<14k-a-lnx,取n=則h(n)<14k-a-ln而-|a|-k≤-a-k<-a+14由于h(m)>0,h(n)<0,∴h(x)在(m,n)上至少有一個(gè)零點(diǎn),即h(x)在(0,+∞)上至少有一個(gè)零點(diǎn).∵h(yuǎn)'(x)=12x-1x-k≤12×14∴當(dāng)k≥116時(shí),h(x)在(0,+∞)上單調(diào)遞減即當(dāng)k≥116時(shí),h(x)在(0,+∞)上只有一個(gè)零點(diǎn)當(dāng)0<k<116時(shí),h'(x)=0有兩個(gè)不同的正根α,β(其中此時(shí)h(x)在(0,α)上為減函數(shù),在(α,β)上為增函數(shù),在(β,+∞)上為減函數(shù).∵h(yuǎn)'(x)=0,∴k=12α-則h(α)=α-lnα-kα-a=12α-lnh'(α)=14α-1α∴h(α)在(0,16)上為減函數(shù),在(16,+∞)上為增函數(shù),∴h(α)≥h(16)=3-ln16-a=3-4ln2-a≥0.又當(dāng)α=16時(shí),k=116,又0<k<116,∴α故h(α)>0,∴x∈(0,β]時(shí),h(x)>0.即h(x)在(0,β]上沒有零點(diǎn),但h(x)在(β,+∞)上有一個(gè)零點(diǎn).∴當(dāng)0<k<116時(shí),h(x)在(0,+∞)上也只有一個(gè)零點(diǎn)∴對(duì)于任意的k>0,直線y=kx+a與曲線y=f(x)有唯一的公共點(diǎn).6.(2022新高考Ⅱ,22,12分)已知函數(shù)f(x)=xeax-ex.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x>0時(shí),f(x)<-1,求a的取值范圍;(3)設(shè)n∈N*,證明:112+1+12解析(1)當(dāng)a=1時(shí),f(x)=xex-ex,則f'(x)=xex,當(dāng)x∈(-∞,0)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(0,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增.(2)當(dāng)x>0時(shí),f(x)<-1,即xeax-ex<-1在(0,+∞)上恒成立,令F(x)=xeax-ex+1(x>0),則F(x)<0在(0,+∞)上恒成立.易得F(0)=0,F'(x)=eax+axeax-ex,F'(0)=0,F″(x)=aeax+aeax+a2xeax-ex,F″(0)=2a-1.若F″(0)>0,則F'(x)必定存在一個(gè)單調(diào)遞增區(qū)間(0,x0),又F'(0)=0,∴F(x)也必定存在一個(gè)單調(diào)遞增區(qū)間(0,x'0).于是F(x)>F(0)=0在(0,x'0)上恒成立,與F(x)<0矛盾,∴F″(0)≤0,∴a≤12∵eax≤ex2在(0,+∞)上成立,∴F(x)≤xex2-ex+1在(0,故只需證xex2-ex+1<0在(0,+∞)令G(x)=xex2-ex+1(x則G'(x)=ex∵ex>x+1在(0,+∞)上成立,∴ex2>x2+1在(0∴G'(x)<0,故G(x)在(0,+∞)上單調(diào)遞減,∴G(x)<G(0)=0.∴xex2-ex+1<0在(0,+∞)故當(dāng)a≤12時(shí),xeax-ex<-1在(0,+∞)成立∴a的取值范圍為?∞(3)證明:構(gòu)造函數(shù)h(x)=x-1x-2lnx(x>1則h'(x)=1+1x易知h'(x)>0,∴h(x)在(1,+∞)上單調(diào)遞增,∴h(x)>h(1)=0,∴x-1x>2lnx令x=1+1n,則有∴1n∴112+1+12原式得證.7.(2011課標(biāo)文,21,12分)已知函數(shù)f(x)=alnxx+1+bx,曲線y=f(x)在點(diǎn)(1)求a,b的值;(2)證明:當(dāng)x>0,且x≠1時(shí),f(x)>lnx解析(1)f'(x)=ax+1x由于直線x+2y-3=0的斜率為-12,且過點(diǎn)故f(1)=1,f解得a=1,b=1.(2)證明:由(1)知f(x)=lnxx+1所以f(x)-lnxx?考慮函數(shù)h(x)=2lnx-x2則h'(x)=2x-=-(x所以當(dāng)x≠1時(shí),h'(x)<0.而h(1)=0,故當(dāng)x∈(0,1)時(shí),h(x)>0,可得11當(dāng)x∈(1,+∞)時(shí),h(x)<0,可得11從而當(dāng)x>0,且x≠1時(shí),f(x)-lnxx?1>0,評(píng)析本題考查函數(shù)的概念、性質(zhì)及導(dǎo)數(shù)等基礎(chǔ)知識(shí),含字母的代數(shù)式的運(yùn)算要求較高,對(duì)考生的整體處理、分類整合的數(shù)學(xué)能力與技巧的要求很高.屬難題.考點(diǎn)2利用導(dǎo)數(shù)研究不等式恒（能）成立問題1.(2015課標(biāo)Ⅰ理,12,5分)設(shè)函數(shù)f(x)=ex(2x-1)-ax+a,其中a<1,若存在唯一的整數(shù)x0使得f(x0)<0,則a的取值范圍是()A.?32e,1B.?答案D由f(x0)<0,即ex0(2x0-1)-a(x0-1)<0得ex0(2x當(dāng)x0=1時(shí),得e<0,顯然不成立,所以x0≠1.若x0>1,則a>ex令g(x)=ex(2x?1)當(dāng)x∈1,32時(shí),g'(x)<0,g(x)當(dāng)x∈32,+∞時(shí)要滿足題意,則x0=2,此時(shí)需滿足g(2)<a≤g(3),得3e2<a≤52e3,與a<1矛盾,所以x0因?yàn)閤0<1,所以a<ex易知,當(dāng)x∈(-∞,0)時(shí),g'(x)>0,g(x)為增函數(shù),當(dāng)x∈(0,1)時(shí),g'(x)<0,g(x)為減函數(shù),要滿足題意,則x0=0,此時(shí)需滿足g(-1)≤a<g(0),得32e≤a<1(滿足a<1).評(píng)析本題主要考查導(dǎo)數(shù)的應(yīng)用及分類討論思想,分離參變量是解決本題的關(guān)鍵,本題綜合性較強(qiáng),屬難題.2.(2023全國(guó)甲理,21,12分)已知函數(shù)f(x)=ax-sinxcos3x(1)當(dāng)a=8時(shí),討論f(x)的單調(diào)性;(2)若f(x)<sin2x,求a的取值范圍.解析(1)當(dāng)a=8時(shí),f(x)=8x-sinx∴f'(x)=8-cos4x?3cos2x(?sin2x)co令f'(x)>0,得cosx>22,又x∈0,π2,∴x令f'(x)<0,得cosx<22,又x∈0,π2,∴x∴f(x)在0,π4上單調(diào)遞增,在π(2)令g(x)=sin2x-f(x)=sin2x-ax+sinxcos3x∴g'(x)=2cos2x-a+co=4cos2x+3?2cos2xco令t=cos2x,∵x∈0,π2,∴t∈(0,令h(t)=4t+3?2tt2-∴h'(t)=4+2t?6t3=又∵t∈(0,1),∴h'(t)<0在(0,1)上恒成立,∴h(t)在(0,1)上單調(diào)遞減,又t=cos2x在0,π2∴g'(x)在0,π2且g'(0)=4+1-a-2=3-a,①當(dāng)a≤3時(shí),g'(x)>g'(0)=3-a≥0在0,π2∴g(x)在0,π2上為增函數(shù),又g(0)=0,∴g(x)>0在0,π2上恒成立,即f(x②當(dāng)a>3時(shí),由于g'(x)在0,π2上單調(diào)遞增,且g'(0)=3-a∴?x0∈0,π2,使得g'(x0)則g'(x)在(0,x0)上恒小于0,故g(x)在(0,x0)上單調(diào)遞減,∴當(dāng)x∈(0,x0)時(shí),g(x)<g(0)=0,不合題意,舍去.綜上,實(shí)數(shù)a的取值范圍為(-∞,3].3.(2015山東理,21,14分)設(shè)函數(shù)f(x)=ln(x+1)+a(x2-x),其中a∈R.(1)討論函數(shù)f(x)極值點(diǎn)的個(gè)數(shù),并說明理由;(2)若?x>0,f(x)≥0成立,求a的取值范圍.解析(1)由題意知函數(shù)f(x)的定義域?yàn)?-1,+∞),f'(x)=1x+1+a(2x-1)=令g(x)=2ax2+ax-a+1,x∈(-1,+∞).①當(dāng)a=0時(shí),g(x)=1,此時(shí)f'(x)>0,函數(shù)f(x)在(-1,+∞)單調(diào)遞增,無(wú)極值點(diǎn).②當(dāng)a>0時(shí),Δ=a2-8a(1-a)=a(9a-8).(i)當(dāng)0<a≤89時(shí),Δ≤0,g(x)≥f'(x)≥0,函數(shù)f(x)在(-1,+∞)單調(diào)遞增,無(wú)極值點(diǎn).(ii)當(dāng)a>89時(shí)設(shè)方程2ax2+ax-a+1=0的兩根為x1,x2(x1<x2),因?yàn)閤1+x2=-12,所以x1<-14,x2>-由g(-1)=1>0,可得-1<x1<-14所以當(dāng)x∈(-1,x1)時(shí),g(x)>0,f'(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈(x1,x2)時(shí),g(x)<0,f'(x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)x∈(x2,+∞)時(shí),g(x)>0,f'(x)>0,函數(shù)f(x)單調(diào)遞增.因此函數(shù)有兩個(gè)極值點(diǎn).③當(dāng)a<0時(shí),Δ>0,由g(-1)=1>0,可得x1<-1.當(dāng)x∈(-1,x2)時(shí),g(x)>0,f'(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈(x2,+∞)時(shí),g(x)<0,f'(x)<0,函數(shù)f(x)單調(diào)遞減.所以函數(shù)有一個(gè)極值點(diǎn).綜上所述,當(dāng)a<0時(shí),函數(shù)f(x)有一個(gè)極值點(diǎn);當(dāng)0≤a≤89時(shí),函數(shù)f(x)無(wú)極值點(diǎn)當(dāng)a>89時(shí),函數(shù)f(x)有兩個(gè)極值點(diǎn)(2)由(1)知,①當(dāng)0≤a≤89時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增因?yàn)閒(0)=0,所以x∈(0,+∞)時(shí),f(x)>0,符合題意.②當(dāng)89<a≤1時(shí),由g(0)≥0,得x2≤所以函數(shù)f(x)在(0,+∞)上單調(diào)遞增.又f(0)=0,所以x∈(0,+∞)時(shí),f(x)>0,符合題意.③當(dāng)a>1時(shí),由g(0)<0,可得x2>0.所以x∈(0,x2)時(shí),函數(shù)f(x)單調(diào)遞減.因?yàn)閒(0)=0,所以x∈(0,x2)時(shí),f(x)<0,不合題意.④當(dāng)a<0時(shí),設(shè)h(x)=x-ln(x+1).因?yàn)閤∈(0,+∞)時(shí),h'(x)=1-1x+1=所以h(x)在(0,+∞)上單調(diào)遞增.因此當(dāng)x∈(0,+∞)時(shí),h(x)>h(0)=0,即ln(x+1)<x.可得f(x)<x+a(x2-x)=ax2+(1-a)x,當(dāng)x>1-1a時(shí),ax2此時(shí)f(x)<0,不合題意.綜上所述,a的取值范圍是[0,1].4.(2014課標(biāo)Ⅱ理,21,12分)已知函數(shù)f(x)=ex-e-x-2x.(1)討論f(x)的單調(diào)性;(2)設(shè)g(x)=f(2x)-4bf(x),當(dāng)x>0時(shí),g(x)>0,求b的最大值;(3)已知1.4142<2<1.4143,估計(jì)ln2的近似值(精確到0.001).解析(1)f'(x)=ex+e-x-2≥0,等號(hào)僅當(dāng)x=0時(shí)成立.所以f(x)在(-∞,+∞)上單調(diào)遞增.(2)g(x)=f(2x)-4bf(x)=e2x-e-2x-4b(ex-e-x)+(8b-4)x,g'(x)=2[e2x+e-2x-2b(ex+e-x)+(4b-2)]=2(ex+e-x-2)(ex+e-x-2b+2).(i)當(dāng)b≤2時(shí),g'(x)≥0,等號(hào)僅當(dāng)x=0時(shí)成立,所以g(x)在(-∞,+∞)上單調(diào)遞增.而g(0)=0,所以對(duì)任意x>0,g(x)>0.(ii)當(dāng)b>2時(shí),若x滿足2<ex+e-x<2b-2,即0<x<ln(b-1+b2?2b)時(shí),g'(x)<0.而g(0)=0,因此當(dāng)0<x≤ln(b-1+綜上,b的最大值為2.(3)由(2)知,g(ln2)=32-22b+2(2b-1)ln當(dāng)b=2時(shí),g(ln2)=32-42+6lnln2>82?3當(dāng)b=324+1時(shí),ln(b-1+b2g(ln2)=-32-22+(32+2)lnln2<18+228<0.693所以ln2的近似值為0.693.5.(2020江蘇,19,16分)已知關(guān)于x的函數(shù)y=f(x),y=g(x)與h(x)=kx+b(k,b∈R)在區(qū)間D上恒有f(x)≥h(x)≥g(x).(1)若f(x)=x2+2x,g(x)=-x2+2x,D=(-∞,+∞),求h(x)的表達(dá)式;(2)若f(x)=x2-x+1,g(x)=klnx,h(x)=kx-k,D=(0,+∞),求k的取值范圍;(3)若f(x)=x4-2x2,g(x)=4x2-8,h(x)=4(t3-t)x-3t4+2t2(0<|t|≤2),D=[m,n]?[-2,2],求證:n-m≤7.解析本題主要考查利用導(dǎo)數(shù)研究函數(shù)的性質(zhì),考查綜合運(yùn)用數(shù)學(xué)思想方法分析與解決問題以及邏輯推理能力.(1)由條件f(x)≥h(x)≥g(x),得x2+2x≥kx+b≥-x2+2x,取x=0,得0≥b≥0,所以b=0.由x2+2x≥kx,得x2+(2-k)x≥0,此式對(duì)一切x∈(-∞,+∞)恒成立,所以(2-k)2≤0,則k=2,此時(shí)2x≥-x2+2x恒成立,所以h(x)=2x.(2)h(x)-g(x)=k(x-1-lnx),x∈(0,+∞).令u(x)=x-1-lnx,則u'(x)=1-1x,令u'(x)=0,得x(0,1)1(1,+∞)u'(x)-0+u(x)↘極小值↗所以u(píng)(x)min=u(1)=0.則x-1≥lnx恒成立,所以當(dāng)且僅當(dāng)k≥0時(shí),h(x)≥g(x)恒成立.另一方面,f(x)≥h(x)恒成立,即x2-x+1≥kx-k恒成立,也即x2-(1+k)x+1+k≥0恒成立.因?yàn)閗≥0,對(duì)稱軸為x=1+k所以(1+k)2-4(1+k)≤0,解得-1≤k≤3.因此,k的取值范圍是0≤k≤3.(3)證明:①當(dāng)1≤t≤2時(shí),由g(x)≤h(x),得4x2-8≤4(t3-t)x-3t4+2t2,整理得x2-(t3-t)x+3t4令Δ=[-(t3-t)]2-(3t4-2t2-8),則Δ=t6-5t4+3t2+8.記φ(t)=t6-5t4+3t2+8(1≤t≤2),則φ'(t)=6t5-20t3+6t=2t(3t2-1)(t2-3)<0恒成立,所以φ(t)在[1,2]上是減函數(shù),則φ(2)≤φ(t)≤φ(1),即2≤φ(t)≤7.所以不等式(*)有解,設(shè)解的范圍為x1≤x≤x2,因此n-m≤x2-x1=Δ≤7.②當(dāng)0<t<1時(shí),f(-1)-h(-1)=3t4+4t3-2t2-4t-1.設(shè)v(t)=3t4+4t3-2t2-4t-1,v'(t)=12t3+12t2-4t-4=4(t+1)(3t2-1),令v'(t)=0,得t=33當(dāng)t∈0,33時(shí),v'(t)<0,v(t)當(dāng)t∈33,1時(shí),v'(t)>0,v(t)v(0)=-1,v(1)=0,則當(dāng)0<t<1時(shí),v(t)<0.(或證:v(t)=(t+1)2(3t+1)(t-1)<0)則f(-1)-h(-1)<0,因此-1?(m,n).因?yàn)閇m,n]?[-2,2],所以n-m≤2+1<7.③當(dāng)-2≤t<0時(shí),因?yàn)閒(x),g(x)均為偶函數(shù),因此n-m≤7也成立.綜上所述,n-m≤7.6.(2019課標(biāo)Ⅰ文,20,12分)已知函數(shù)f(x)=2sinx-xcosx-x,f'(x)為f(x)的導(dǎo)數(shù).(1)證明:f'(x)在區(qū)間(0,π)存在唯一零點(diǎn);(2)若x∈[0,π]時(shí),f(x)≥ax,求a的取值范圍.解析(1)設(shè)g(x)=f'(x),則g(x)=cosx+xsinx-1,g'(x)=xcosx.當(dāng)x∈0,π2時(shí),g'(x)>0;當(dāng)x∈π所以g(x)在0,π2在π2,又g(0)=0,gπ2>0,g(π)=-2,故g(x)在(0,π)存在唯一零點(diǎn)所以f'(x)在(0,π)存在唯一零點(diǎn).(2)由題設(shè)知f(π)≥aπ,f(π)=0,可得a≤0.由(1)知,f'(x)在(0,π)只有一個(gè)零點(diǎn),設(shè)為x0,且當(dāng)x∈(0,x0)時(shí),f'(x)>0;當(dāng)x∈(x0,π)時(shí),f'(x)<0,所以f(x)在(0,x0)單調(diào)遞增,在(x0,π)單調(diào)遞減.又f(0)=0,f(π)=0,所以,當(dāng)x∈[0,π]時(shí),f(x)≥0.又當(dāng)a≤0,x∈[0,π]時(shí),ax≤0,故f(x)≥ax.因此,a的取值范圍是(-∞,0].7.(2017課標(biāo)Ⅱ文,21,12分)設(shè)函數(shù)f(x)=(1-x2)ex.(1)討論f(x)的單調(diào)性;(2)當(dāng)x≥0時(shí),f(x)≤ax+1,求a的取值范圍.解析本題考查函數(shù)的單調(diào)性,恒成立問題.(1)f'(x)=(1-2x-x2)ex.令f'(x)=0,得x=-1-2或x=-1+2.當(dāng)x∈(-∞,-1-2)時(shí),f'(x)<0;當(dāng)x∈(-1-2,-1+2)時(shí),f'(x)>0;當(dāng)x∈(-1+2,+∞)時(shí),f'(x)<0.所以f(x)在(-∞,-1-2),(-1+2,+∞)單調(diào)遞減,在(-1-2,-1+2)單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a≥1時(shí),設(shè)函數(shù)h(x)=(1-x)ex,h'(x)=-xex<0(x>0),因此h(x)在[0,+∞)單調(diào)遞減,而h(0)=1,故h(x)≤1,所以f(x)=(x+1)h(x)≤x+1≤ax+1.當(dāng)0<a<1時(shí),設(shè)函數(shù)g(x)=ex-x-1,g'(x)=ex-1>0(x>0),所以g(x)在[0,+∞)單調(diào)遞增,而g(0)=0,故ex≥x+1.當(dāng)0<x<1時(shí),f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5?則x0∈(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.當(dāng)a≤0時(shí),取x0=5?則x0∈(0,1),f(x0)>(1-x0)(1+x0)2=1≥ax0+1.綜上,a的取值范圍是[1,+∞).解題思路(1)求f'(x),令f'(x)>0,求出f(x)的單調(diào)增區(qū)間,令f'(x)<0,求出f(x)的單調(diào)減區(qū)間.(2)對(duì)參數(shù)a的取值進(jìn)行分類討論,當(dāng)a≥1時(shí),構(gòu)造函數(shù)可知(1-x)·ex≤1,所以f(x)=(x+1)(1-x)·ex≤x+1≤ax+1成立;當(dāng)0<a<1時(shí),構(gòu)造函數(shù)可知ex≥x+1,通過舉反例x0=5?4a?12,f(x0)>ax0+1,從而說明命題不成立;當(dāng)a≤0時(shí),舉反例疑難突破(1)求單調(diào)區(qū)間的一般步驟:①求定義域;②求f'(x),令f'(x)>0,求出f(x)的增區(qū)間,令f'(x)<0,求出f(x)的減區(qū)間;③寫出結(jié)論,注意單調(diào)區(qū)間不能用“∪”連接.(2)恒成立問題的三種常見解法:①分離參數(shù),化為最值問題求解,如a≥φ(x)max或a≤φ(x)min;②構(gòu)造函數(shù),分類討論,如f(x)≥g(x),即F(x)=f(x)-g(x),求F(x)min≥0;③轉(zhuǎn)變主元,選取適當(dāng)?shù)闹髟墒箚栴}簡(jiǎn)化.8.(2017課標(biāo)Ⅲ文,21,12分)已知函數(shù)f(x)=lnx+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時(shí),證明f(x)≤-34解析(1)f(x)的定義域?yàn)?0,+∞),f'(x)=1x+2ax+2a+1=(若a≥0,則當(dāng)x∈(0,+∞)時(shí),f'(x)>0,故f(x)在(0,+∞)單調(diào)遞增.若a<0,則當(dāng)x∈0,?12a時(shí),當(dāng)x∈?12a,+∞時(shí)故f(x)在0,?12a單調(diào)遞增,(2)由(1)知,當(dāng)a<0時(shí),f(x)在x=-12a取得最大值,最大值為f?12a所以f(x)≤-34a-2等價(jià)于ln?12a-1-14a≤-34a設(shè)g(x)=lnx-x+1,則g'(x)=1x當(dāng)x∈(0,1)時(shí),g'(x)>0;當(dāng)x∈(1,+∞)時(shí),g'(x)<0.所以g(x)在(0,1)單調(diào)遞增,在(1,+∞)單調(diào)遞減.故當(dāng)x=1時(shí),g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時(shí),g(x)≤0.從而當(dāng)a<0時(shí),ln?12a+12a+1≤0,即9.(2017天津理,20,14分)設(shè)a∈Z,已知定義在R上的函數(shù)f(x)=2x4+3x3-3x2-6x+a在區(qū)間(1,2)內(nèi)有一個(gè)零點(diǎn)x0,g(x)為f(x)的導(dǎo)函數(shù).(1)求g(x)的單調(diào)區(qū)間;(2)設(shè)m∈[1,x0)∪(x0,2],函數(shù)h(x)=g(x)(m-x0)-f(m),求證:h(m)h(x0)<0;(3)求證:存在大于0的常數(shù)A,使得對(duì)于任意的正整數(shù)p,q,且pq∈[1,x0)∪(x0,2],滿足pq?解析本小題主要考查導(dǎo)數(shù)的運(yùn)算、利用導(dǎo)數(shù)研究函數(shù)的性質(zhì)、證明不等式等基礎(chǔ)知識(shí)和方法.考查函數(shù)思想和化歸思想.考查抽象概括能力、綜合分析問題和解決問題的能力.(1)由f(x)=2x4+3x3-3x2-6x+a,可得g(x)=f'(x)=8x3+9x2-6x-6,進(jìn)而可得g'(x)=24x2+18x-6.令g'(x)=0,解得x=-1或x=14當(dāng)x變化時(shí),g'(x),g(x)的變化情況如下表:x(-∞,-1)?1g'(x)+-+g(x)↗↘↗所以,g(x)的單調(diào)遞增區(qū)間是(-∞,-1),14,+∞,(2)證明:由h(x)=g(x)(m-x0)-f(m),得h(m)=g(m)(m-x0)-f(m),h(x0)=g(x0)(m-x0)-f(m).令函數(shù)H1(x)=g(x)(x-x0)-f(x),則H1'(x)=g'(x)(x-x0).由(1)知,當(dāng)x∈[1,2]時(shí),g'(x)>0,故當(dāng)x∈[1,x0)時(shí),H1'(x)<0,H1(x)單調(diào)遞減;當(dāng)x∈(x0,2]時(shí),H1'(x)>0,H1(x)單調(diào)遞增.因此,當(dāng)x∈[1,x0)∪(x0,2]時(shí),H1(x)>H1(x0)=-f(x0)=0,可得H1(m)>0,即h(m)>0.令函數(shù)H2(x)=g(x0)(x-x0)-f(x),則H2'(x)=g(x0)-g(x).由(1)知g(x)在[1,2]上單調(diào)遞增,故當(dāng)x∈[1,x0)時(shí),H2'(x)>0,H2(x)單調(diào)遞增;當(dāng)x∈(x0,2]時(shí),H2'(x)<0,H2(x)單調(diào)遞減.因此,當(dāng)x∈[1,x0)∪(x0,2]時(shí),H2(x)<H2(x0)=0,可得H2(m)<0,即h(x0)<0.所以,h(m)h(x0)<0.(3)證明:對(duì)于任意的正整數(shù)p,q,且pq∈[1,x0)∪(x0,2],令m=pq,函數(shù)h(x)=g(x)(m-x0)-f(m).由(2)知,當(dāng)m∈[1,x0)時(shí),h(x)在區(qū)間(m,x0)內(nèi)有零點(diǎn);當(dāng)m∈(x0,2]時(shí),h(x)在區(qū)間(x0,m)內(nèi)有零點(diǎn).所以h(x)在(1,2)內(nèi)至少有一個(gè)零點(diǎn),不妨設(shè)為x1,則h(x1)=g(x1)pq?由(1)知g(x)在[1,2]上單調(diào)遞增,故0<g(1)<g(x1)<g(2),于是pq?x0=|2p因?yàn)楫?dāng)x∈[1,2]時(shí),g(x)>0,故f(x)在[1,2]上單調(diào)遞增,所以f(x)在區(qū)間[1,2]上除x0外沒有其他的零點(diǎn),而pq≠x0,故fpq≠0.又因?yàn)閜,q,a均為整數(shù),所以|2p4+3p3q-3p2q2-6pq3+aq4|是正整數(shù),從而|2p4+3p3q-3p2q2-6pq3+aq4|≥1.所以pq?x0≥1g(2)q4思路分析(1)求出函數(shù)f(x)的導(dǎo)函數(shù)g(x)=f'(x)=8x3+9x2-6x-6,求出使導(dǎo)函數(shù)為零的x的值,通過列表求出單調(diào)區(qū)間即可.(2)由h(x)推出h(m),h(x0),令函數(shù)H1(x)=g(x)(x-x0)-f(x),H2(x)=g(x0)(x-x0)-f(x),求出導(dǎo)函數(shù)H1'(x),H2'(x),由此可推出h(m)h(x0)<0.(3)對(duì)于任意的正整數(shù)p,q,令m=pq,函數(shù)h(x)=g(x)·(m-x0)-f(m),由(2)可推出h(x)在(1,2)內(nèi)至少有一個(gè)零點(diǎn),結(jié)合(1)可得pq?x0=fpqg(x1)≥fpqg(2)=|2p4+310.(2016四川文,21,14分)設(shè)函數(shù)f(x)=ax2-a-lnx,g(x)=1x-eex,其中a∈R(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)x>1時(shí),g(x)>0;(3)確定a的所有可能取值,使得f(x)>g(x)在區(qū)間(1,+∞)內(nèi)恒成立.解析(1)f'(x)=2ax-1x=2當(dāng)a≤0時(shí),f'(x)<0,f(x)在(0,+∞)內(nèi)單調(diào)遞減.當(dāng)a>0時(shí),由f'(x)=0有x=12當(dāng)x∈0,12a時(shí),f'(x)<0,當(dāng)x∈12a,+∞時(shí),f'(x)>0,(2)令s(x)=ex-1-x,則s'(x)=ex-1-1.當(dāng)x>1時(shí),s'(x)>0,所以ex-1>x,從而g(x)=1x-1(3)由(2),當(dāng)x>1時(shí),g(x)>0.當(dāng)a≤0,x>1時(shí),f(x)=a(x2-1)-lnx<0.故當(dāng)f(x)>g(x)在區(qū)間(1,+∞)內(nèi)恒成立時(shí),必有a>0.當(dāng)0<a<12時(shí),1由(1)有f12a<f(1)=0,而g所以此時(shí)f(x)>g(x)在區(qū)間(1,+∞)內(nèi)不恒成立.當(dāng)a≥12時(shí),令h(x)=f(x)-g(x)(x≥當(dāng)x>1時(shí),h'(x)=2ax-1x+1x2-e1-x>x-1x+1x2-因此,h(x)在區(qū)間(1,+∞)內(nèi)單調(diào)遞增.又因?yàn)閔(1)=0,所以當(dāng)x>1時(shí),h(x)=f(x)-g(x)>0,即f(x)>g(x)恒成立.綜上,a∈1211.(2015課標(biāo)Ⅱ理,21,12分)設(shè)函數(shù)f(x)=emx+x2-mx.(1)證明:f(x)在(-∞,0)單調(diào)遞減,在(0,+∞)單調(diào)遞增;(2)若對(duì)于任意x1,x2∈[-1,1],都有|f(x1)-f(x2)|≤e-1,求m的取值范圍.解析(1)f'(x)=m(emx-1)+2x.若m≥0,則當(dāng)x∈(-∞,0)時(shí),emx-1≤0,f'(x)<0;當(dāng)x∈(0,+∞)時(shí),emx-1≥0,f'(x)>0.若m<0,則當(dāng)x∈(-∞,0)時(shí),emx-1>0,f'(x)<0;當(dāng)x∈(0,+∞)時(shí),emx-1<0,f'(x)>0.所以,f(x)在(-∞,0)單調(diào)遞減,在(0,+∞)單調(diào)遞增.(2)由(1)知,對(duì)任意的m,f(x)在[-1,0]單調(diào)遞減,在[0,1]單調(diào)遞增,故f(x)在x=0處取得最小值.所以對(duì)于任意x1,x2∈[-1,1],|f(x1)-f(x2)|≤e-1的充要條件是f即em設(shè)函數(shù)g(t)=et-t-e+1,則g'(t)=et-1.當(dāng)t<0時(shí),g'(t)<0;當(dāng)t>0時(shí),g'(t)>0.故g(t)在(-∞,0)單調(diào)遞減,在(0,+∞)單調(diào)遞增.又g(1)=0,g(-1)=e-1+2-e<0,故當(dāng)t∈[-1,1]時(shí),g(t)≤0.當(dāng)m∈[-1,1]時(shí),g(m)≤0,g(-m)≤0,即①式成立;當(dāng)m>1時(shí),由g(t)的單調(diào)性,g(m)>0,即em-m>e-1;當(dāng)m<-1時(shí),g(-m)>0,即e-m+m>e-1.綜上,m的取值范圍是[-1,1].12.(2014課標(biāo)Ⅰ文,21,12分)設(shè)函數(shù)f(x)=alnx+1?a2x2-bx(a≠1),曲線y=f(x)在點(diǎn)(1,(1)求b;(2)若存在x0≥1,使得f(x0)<aa?1,求解析(1)f'(x)=ax由題設(shè)知f'(1)=0,解得b=1.(2)f(x)的定義域?yàn)?0,+∞),由(1)知,f(x)=alnx+1?a2x2-x,f'(x)=a(i)若a≤12,則a1?a≤1,故當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f(x)所以,存在x0≥1,使得f(x0)<aa?1的充要條件為f(1)<aa?1,即1?a2(ii)若12<a<1,則a1?a>1,故當(dāng)x∈1,a1?a時(shí),f'(x)<0;當(dāng)x∈a1?a,+∞時(shí)所以,存在x0≥1,使得f(x0)<aa?1的充要條件為fa而fa1?a=alna1?a+a2(iii)若a>1,則f(1)=1?a2-1=?綜上,a的取值范圍是(-2-1,2-1)∪(1,+∞).13.(2019浙江,22,15分)已知實(shí)數(shù)a≠0,設(shè)函數(shù)f(x)=alnx+1+x(1)當(dāng)a=-34時(shí),求函數(shù)f(x)的單調(diào)區(qū)間(2)對(duì)任意x∈1e2,+∞均有f(x)≤x2注:e=2.71828…為自然對(duì)數(shù)的底數(shù).解析本題主要考查函數(shù)的單調(diào)性,導(dǎo)數(shù)的運(yùn)算及其應(yīng)用,同時(shí)考查邏輯思維能力和綜合應(yīng)用能力.考查數(shù)學(xué)抽象、邏輯推理、數(shù)學(xué)運(yùn)算的核心素養(yǎng).(1)當(dāng)a=-34時(shí),f(x)=-34lnx+f'(x)=-34x+12所以,函數(shù)f(x)的單調(diào)遞減區(qū)間為(0,3),單調(diào)遞增區(qū)間為(3,+∞).(2)由f(1)≤12a,得0<a≤當(dāng)0<a≤24時(shí),f(x)≤x2a等價(jià)于xa2-2令t=1a,則t≥22.設(shè)g(t)=t2x-2t1+x-2lnx,t≥22,則g(t)=xt?1+(i)當(dāng)x∈17,+∞時(shí),1+1則g(t)≥g(22)=8x-421+x-2ln記p(x)=4x-221+x-lnx,x≥則p'(x)=2x-2x+1-=(x故x111(1,+∞)p'(x)-0+p(x)p1單調(diào)遞減極小值p(1)單調(diào)遞增所以,p(x)≥p(1)=0.因此,g(t)≥g(22)=2p(x)≥0.(ii)當(dāng)x∈1e2g(x)≥g1+1x=令q(x)=2xlnx+(x+1),x∈1e則q'(x)=lnx故q(x)在1e2所以q(x)≤q17由(i)得,q17=-277p1所以,q(x)<0.因此,g(t)≥g1+1x=-由(i)(ii)知對(duì)任意x∈1e2,+∞,t∈[22,+∞),g(t)≥0,即對(duì)任意均有f(x)≤x2綜上所述,所求a的取值范圍是0,2疑難突破(1)導(dǎo)函數(shù)f'(x)通分后,對(duì)分子的因式分解比較困難,可以選擇先求分子等于0時(shí)的根,從而確定根兩側(cè)導(dǎo)函數(shù)的正負(fù).(2)先對(duì)本題復(fù)雜不等式化簡(jiǎn)變形是解題的切入點(diǎn),由于a的范圍太大,借助恒成立的條件用特殊值縮小a的范圍是解題的關(guān)鍵,另外,對(duì)雙變量不等式,合理確定主元,是解決本題的思維轉(zhuǎn)折點(diǎn).考點(diǎn)3利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)問題1.(2014課標(biāo)Ⅰ,理11,文12,5分)已知函數(shù)f(x)=ax3-3x2+1,若f(x)存在唯一的零點(diǎn)x0,且x0>0,則a的取值范圍是()A.(2,+∞)B.(1,+∞)C.(-∞,-2)D.(-∞,-1)答案C(1)當(dāng)a=0時(shí),顯然f(x)有兩個(gè)零點(diǎn),不符合題意.(2)當(dāng)a≠0時(shí),f'(x)=3ax2-6x,令f'(x)=0,解得x1=0,x2=2a當(dāng)a>0時(shí),2a>0,所以函數(shù)f(x)=ax3-3x2+1在(-∞,0)與2a,+∞上為增函數(shù),在0,2a上為減函數(shù),因?yàn)閒(x)存在唯一零點(diǎn)x0,且x0>0,當(dāng)a<0時(shí),2a<0,所以函數(shù)f(x)=ax3-3x2+1在?∞,2a和(0,+∞)上為減函數(shù),在2a,0上為增函數(shù),因?yàn)閒(x)存在唯一零點(diǎn)x0,且x0>0,則f2a>0,即a·8a3-3·4a2+1>0,2.(2022全國(guó)乙文,20,12分)已知函數(shù)f(x)=ax-1x-(a+1)ln(1)當(dāng)a=0時(shí),求f(x)的最大值;(2)若f(x)恰有一個(gè)零點(diǎn),求a的取值范圍.解析(1)當(dāng)a=0時(shí),f(x)=-1x-lnx(x>0∴f'(x)=1x2?1令f'(x)=0,得x=1,x∈(0,1)時(shí),f'(x)>0,x∈(1,+∞)時(shí),f'(x)<0,∴f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.∴f(x)max=f(1)=-1.(2)解法一:f'(x)=a+1x(i)當(dāng)a≤0時(shí),ax-1≤0恒成立,∴0<x<1時(shí),f'(x)>0,f(x)單調(diào)遞增,x>1時(shí),f'(x)<0,f(x)單調(diào)遞減,∴f(x)max=f(1)=a-1<0.此時(shí)f(x)無(wú)零點(diǎn),不合題意.(ii)當(dāng)a>0時(shí),令f'(x)=0,解得x=1或x=1a①當(dāng)0<a<1時(shí),1<1a∴1<x<1a時(shí),f'(x)<0,f(x)單調(diào)遞減0<x<1或x>1a時(shí),f'(x)>0,f(x)單調(diào)遞增∴f(x)在(0,1),1a,+∞上單調(diào)遞增,在1,1a上單調(diào)遞減,f(x)的極大值為f(1x→+∞時(shí),f(x)>0,∴f(x)恰有1個(gè)零點(diǎn).②當(dāng)a=1時(shí),1=1a,f(x)在(0,+∞)上單調(diào)遞增,f(1)=0,符合題意③當(dāng)a>1時(shí),1a<1,f(x)在0,1a,(1,+∞)上單調(diào)遞增,在f(x)的極小值為f(1)=a-1>0,x→0時(shí),f(x)→-∞,∴f(x)恰有1個(gè)零點(diǎn).綜上所述,a>0.解法二:f(x)=ax-1x-(a+1)lnx只有一個(gè)零點(diǎn)即a(x-lnx)=1x+lnx在(0,+∞)上只有一個(gè)解由lnx≤x-1,得x-lnx≥1,∴a=1x+lnxx?lnx,令g則g'(x)=?=x?1令h(x)=1-lnxx?則h'(x)=lnx?xx2,h'(x)<0在(0∴h(x)在(0,+∞)上單調(diào)遞減,且h(1)=0,∴在(0,1)上,h(x)>0,在(1,+∞)上,h(x)<0,故在(0,1)上,g'(x)<0,在(1,+∞)上,g'(x)<0,∴g(x)在(0,+∞)上單調(diào)遞減,x→0,g(x)→+∞,x→+∞,g(x)→0,∴g(x)>0,∴a>0時(shí),a(x-lnx)=1x+lnx恰有一解故a>0.3.(2022全國(guó)甲理,21,12分)已知函數(shù)f(x)=exx-lnx+x(1)若f(x)≥0,求a的取值范圍;(2)證明:若f(x)有兩個(gè)零點(diǎn)x1,x2,則x1x2<1.解析(1)∵f(x)=exx-lnx+x-a,∴函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=令f'(x)=0,得x=1,f(x),f'(x)的變化情況如下:x(0,1)1(1,+∞)f'(x)-0+f(x)↘↗∴當(dāng)x=1時(shí),f(x)min=e+1-a.∵f(x)≥0,∴e+1-a≥0,∴a≤e+1.(2)f(x)=exx-lnx+x-a=ex-lnx+x-lnx-設(shè)t=x-lnx,則y=et+t-a,y'=et+1.

∵f(x)有兩個(gè)零點(diǎn)x1,x2,∴由(1)知a>e+1.∵y'=et+1>0,∴y=et+t-a為增函數(shù),∴x1-lnx1=x2-lnx2.由t=x-lnx得t'=1-1x=x?1x,令t'=0t,t'的變化情況如下:x(0,1)1(1,+∞)t'-0+t↘1↗證法一:不妨設(shè)0<x1<1<x2,∴x2-x1=lnx2-lnx1.可以證明:x2?x1lnx2?lnx1>x證明如下:∵x2∴只需證x2設(shè)m=x2x1(從而只要證m?1m>lnm(m>1設(shè)g(m)=m?1m-lnm(m則g'(m)=2m∵g'(m)>0,∴g(m)在(1,+∞)上單調(diào)遞增,∴g(m)>g(1)=0,∴x2?故x1x2<1.證法二:不妨設(shè)0<x1<1<x2,F(x)=f(x)-f1x,0<x<1則F'(x)=f'(x)+1x2f'1x=x?1x2(ex+x設(shè)φ(x)=ex+x-1-xe1x,x∈(0,則φ'(x)=ex+1-e1∵0<x<1,∴φ'(x)>0,∴φ(x)在(0,1)上單調(diào)遞增,∴φ(x)<φ(1)=0,又x?1x2<0,∴F'(x∴F(x)在(0,1)上單調(diào)遞增.∴F(x)<F(1)=0,∴f(x)<f1x∴f(x2)=f(x1)<f1x由(1)可知,f(x)在(1,+∞)上單調(diào)遞增,∵x2>1,1x1>1,∴x2<1x1,∴x14.(2020課標(biāo)Ⅲ文,20,12分)已知函數(shù)f(x)=x3-kx+k2.(1)討論f(x)的單調(diào)性;(2)若f(x)有三個(gè)零點(diǎn),求k的取值范圍.解析(1)f'(x)=3x2-k.當(dāng)k=0時(shí),f(x)=x3,故f(x)在(-∞,+∞)單調(diào)遞增.當(dāng)k<0時(shí),f'(x)=3x2-k>0,故f(x)在(-∞,+∞)單調(diào)遞增.當(dāng)k>0時(shí),令f'(x)=0,得x=±3k3.當(dāng)x∈?∞,?3k3時(shí),f'(x)>0;當(dāng)x∈?3k3,3k3時(shí),f'(x)<0;當(dāng)x∈3k3,+(2)由(1)知,當(dāng)k≤0時(shí),f(x)在(-∞,+∞)單調(diào)遞增,f(x)不可能有三個(gè)零點(diǎn).當(dāng)k>0時(shí),x=-3k3為f(x)的極大值點(diǎn),x=3k3此時(shí),-k-1<-3k3<3k3<k+1且f(-k-1)<0,f(k+1)>0,根據(jù)f(x)的單調(diào)性,當(dāng)且僅當(dāng)f3k3<0,即k2-2k3k9<0時(shí),f(x)有三個(gè)零點(diǎn),解得k<45.(2020浙江,22,15分)已知1<a≤2,函數(shù)f(x)=ex-x-a,其中e=2.71828…是自然對(duì)數(shù)的底數(shù).(1)證明:函數(shù)y=f(x)在(0,+∞)上有唯一零點(diǎn);(2)記x0為函數(shù)y=f(x)在(0,+∞)上的零點(diǎn),證明:(i)a?1≤x0≤(ii)x0f(ex0)解析本題主要考查函數(shù)的單調(diào)性、零點(diǎn),導(dǎo)數(shù)的運(yùn)算及其應(yīng)用,同時(shí)考查數(shù)學(xué)抽象、邏輯推理與數(shù)學(xué)運(yùn)算等素養(yǎng).(1)因?yàn)閒(0)=1-a<0,f(2)=e2-2-a≥e2-4>0,所以y=f(x)在(0,+∞)上存在零點(diǎn).因?yàn)閒'(x)=ex-1,所以當(dāng)x>0時(shí),f'(x)>0,故函數(shù)f(x)在[0,+∞)上單調(diào)遞增,所以函數(shù)y=f(x)在(0,+∞)上有唯一零點(diǎn).(2)(i)令g(x)=ex-12x2-x-1(x≥0),g'(x)=ex-x-1=f(x)+a-1,由(1)知函數(shù)g'(x)在[0,+∞)上單調(diào)遞增,故當(dāng)x>0時(shí),g'(x)>g'(0)=0,所以函數(shù)g(x)在[0,+∞)單調(diào)遞增,故g(x)≥由g(2(a?1))≥0得f(2(a?1))=e2(因?yàn)閒(x)在[0,+∞)單調(diào)遞增,故2(a?1)≥令h(x)=ex-x2-x-1(0≤x≤1),h'(x)=ex-2x-1,令h1(x)=ex-2x-1(0≤x≤1),h'1(x)=ex-2,所以x0(0,ln2)ln2(ln2,1)1h'1(x)-1-0+e-2h1(x)0↘↗e-3故當(dāng)0<x<1時(shí),h1(x)<0,即h'(x)<0,所以h(x)在[0,1]單調(diào)遞減,因此當(dāng)0≤x≤1時(shí),h(x)≤h(0)=0.由h(a?1)≤0得f(a?1)=ea?1因?yàn)閒(x)在[0,+∞)單調(diào)遞增,故a?1≤x綜上,a?1≤x0≤(ii)令u(x)=ex-(e-1)x-1,u'(x)=ex-(e-1),所以當(dāng)x>1時(shí),u'(x)>0,故函數(shù)u(x)在區(qū)間[1,+∞)上單調(diào)遞增,因此u(x)≥u(1)=0.由ex0=x0x0f(ex0)=x0f(x0+a)=(ea-1)x02+a(ea-2)x0由x0≥a?1得x0f(ex6.(2019課標(biāo)Ⅱ理,20,12分)已知函數(shù)f(x)=lnx-x+1(1)討論f(x)的單調(diào)性,并證明f(x)有且僅有兩個(gè)零點(diǎn);(2)設(shè)x0是f(x)的一個(gè)零點(diǎn),證明曲線y=lnx在點(diǎn)A(x0,lnx0)處的切線也是曲線y=ex的切線.解析本題考查利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性,求函數(shù)零點(diǎn)以及導(dǎo)數(shù)的幾何意義.考查學(xué)生分析、解決問題的能力,考查邏輯推理能力和運(yùn)算求解能力,體現(xiàn)了邏輯推理和數(shù)學(xué)運(yùn)算的核心素養(yǎng).(1)f(x)的定義域?yàn)?0,1)∪(1,+∞).因?yàn)閒'(x)=1x+2(x?1)2因?yàn)閒(e)=1-e+1e?1<0,f(e2)=2-e2+1e2?1=e2?3e2?1>0,所以f(x)在(1,+∞)有唯一零點(diǎn)x1,即f(x1)=0.又0<1x綜上,f(x)有且僅有兩個(gè)零點(diǎn).(2)因?yàn)?x0=e?lnx0,故點(diǎn)B由題設(shè)知f(x0)=0,即lnx0=x0故直線AB的斜率k=1x0?lnx曲線y=ex在點(diǎn)B?lnx0,1x0處切線的斜率是1x0,曲線y=lnx在點(diǎn)A(x0,lnx0)處切線的斜率也是1x0,所以曲線y=lnx在點(diǎn)A(x解后反思(1)先判斷函數(shù)的單調(diào)性,然后結(jié)合零點(diǎn)存在性定理證明函數(shù)f(x)有且僅有兩個(gè)零點(diǎn).(2)要證明曲線y=lnx在點(diǎn)A(x0,lnx0)處的切線也是曲線y=ex的切線,首先求得這條切線的斜率k=1x0,所以必須在曲線y=ex上找一點(diǎn)B(x1,ex1),使ex1=1x0,從而求得B點(diǎn)的坐標(biāo)為?lnx0,1x0,然后證明曲線y=lnx在點(diǎn)7.(2018課標(biāo)Ⅱ文,21,12分)已知函數(shù)f(x)=13x3-a(x2(1)若a=3,求f(x)的單調(diào)區(qū)間;(2)證明:f(x)只有一個(gè)零點(diǎn).解析(1)當(dāng)a=3時(shí),f(x)=13x3-3x2-3x-3,f'(x)=x2令f'(x)=0,解得x=3-23或x=3+23.當(dāng)x∈(-∞,3-23)∪(3+23,+∞)時(shí),f'(x)>0;當(dāng)x∈(3-23,3+23)時(shí),f'(x)<0.故f(x)在(-∞,3-23),(3+23,+∞)單調(diào)遞增,在(3-23,3+23)單調(diào)遞減.(2)由于x2+x+1>0,所以f(x)=0等價(jià)于x3設(shè)g(x)=x3x2+x+1-3a,則g'(x)=x2(x2+2x+3)(x2+x+1)又f(3a-1)=-6a2+2a-13=-6a?162-16<0,f(3a+1)=綜上,f(x)只有一個(gè)零點(diǎn).方法總結(jié)利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)的方法:方法一:(1)利用導(dǎo)數(shù)求出函數(shù)f(x)的單調(diào)區(qū)間和極值;(2)根據(jù)函數(shù)f(x)的性質(zhì)作出圖象;(3)判斷函數(shù)零點(diǎn)的個(gè)數(shù).方法二:(1)利用導(dǎo)數(shù)求出函數(shù)f(x)的單調(diào)區(qū)間和極值;(2)分類討論,判斷函數(shù)零點(diǎn)的個(gè)數(shù).8.(2016課標(biāo)Ⅰ理,21,12分)已知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個(gè)零點(diǎn).(1)求a的取值范圍;(2)設(shè)x1,x2是f(x)的兩個(gè)零點(diǎn),證明:x1+x2<2.解析(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(2分)(i)設(shè)a=0,則f(x)=(x-2)ex,f(x)只有一個(gè)零點(diǎn).(3分)(ii)設(shè)a>0,則當(dāng)x∈(-∞,1)時(shí),f'(x)<0;當(dāng)x∈(1,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,1)單調(diào)遞減,在(1,+∞)單調(diào)遞增.又f(1)=-e,f(2)=a,取b滿足b<0且b<lna2,f(b)>a2(b-2)+a(b-1)2=ab故f(x)存在兩個(gè)零點(diǎn).(4分)(iii)設(shè)a<0,由f'(x)=0得x=1或x=ln(-2a).若a≥-e2,則ln(-2a)≤1,故當(dāng)x∈(1,+∞)時(shí),f'(x)>0,因此f(x)在(1,+∞)單調(diào)遞增.又當(dāng)x≤1時(shí)f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn).(6分若a<-e2,則ln(-2a)>1,故當(dāng)x∈(1,ln(-2a))時(shí),f'(x)<0;當(dāng)x∈(ln(-2a),+∞)時(shí),f因此f(x)在(1,ln(-2a))單調(diào)遞減,在(ln(-2a),+∞)單調(diào)遞增.又當(dāng)x≤1時(shí)f(x)<0,所以f(x)不存在兩個(gè)零點(diǎn).綜上,a的取值范圍為(0,+∞).(8分)(2)不妨設(shè)x1<x2.由(1)知,x1∈(-∞,1),x2∈(1,+∞),2-x2∈(-∞,1),f(x)在(-∞,1)單調(diào)遞減,所以x1+x2<2等價(jià)于f(x1)>f(2-x2),即f(2-x2)<0.由于f(2-x2)=-x2e2?x2+a(x2-1)2,而f(x2)=(x2-2)ex2+a(x2-1)2=0,所以f(2-x2)=-x2e2設(shè)g(x)=-xe2-x-(x-2)ex,則g'(x)=(x-1)(e2-x-ex).所以當(dāng)x>1時(shí),g'(x)<0,而g(1)=0,故當(dāng)x>1時(shí),g(x)<0.從而g(x2)=f(2-x2)<0,故x1+x2<2.(12分)9.(2016課標(biāo)Ⅰ文,21,12分)已知函數(shù)f(x)=(x-2)ex+a(x-1)2.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.解析(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(i)設(shè)a≥0,則當(dāng)x∈(-∞,1)時(shí),f'(x)<0;當(dāng)x∈(1,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.(2分)(ii)設(shè)a<0,由f'(x)=0得x=1或x=ln(-2a).①若a=-e2,則f'(x)=(x-1)(ex-e),所以f(x)在(-∞,+∞)上單調(diào)遞增②若a>-e2,則ln(-2a)<1,故當(dāng)x∈(-∞,ln(-2a))∪(1,+∞)時(shí),f'(x)>0;當(dāng)x∈(ln(-2a),1)時(shí),f'(x)<0.所以f(x)在(-∞,ln(-2a)),(1,+∞)上單調(diào)遞增,在(ln(-2a),1)上單調(diào)遞減.(4分③若a<-e2,則ln(-2a)>1,故當(dāng)x∈(-∞,1)∪(ln(-2a),+∞)時(shí),f'(x)>0;當(dāng)x∈(1,ln(-2a))時(shí),f'(x)<0.所以f(x)在(-∞,1),(ln(-2a),+∞)上單調(diào)遞增,在(1,ln(-2a))上單調(diào)遞減.(6分(2)(i)設(shè)a>0,則由(1)知,f(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.又f(1)=-e,f(2)=a,取b滿足b<0且b<lna2則f(b)>a2(b-2)+a(b-1)2=ab所以f(x)有兩個(gè)零點(diǎn).(8分)(ii)設(shè)a=0,則f(x)=(x-2)ex,所以f(x)只有一個(gè)零點(diǎn).(9分)(iii)設(shè)a<0,若a≥-e2,則由(1)知,f(x)在(1,+∞)上單調(diào)遞增,又當(dāng)x≤1時(shí)f(x)<0,故f(x)不存在兩個(gè)零點(diǎn);(10分若a<-e2,則由(1)知,f(x)在(1,ln(-2a))上單調(diào)遞減,在(ln(-2a),+∞)上單調(diào)遞增,又當(dāng)x≤1時(shí)f(x)<0,故f(x)不存在兩個(gè)零點(diǎn).(11分綜上,a的取值范圍為(0,+∞).(12分)疑難突破(1)分類討論時(shí)臨界點(diǎn)的選取是關(guān)鍵,易忽略a=-e2的情形.(2)在討論a>0時(shí)函數(shù)零點(diǎn)的個(gè)數(shù)時(shí),注意利用不等式的放縮評(píng)析本題考查函數(shù)的單調(diào)性、零點(diǎn)等知識(shí)點(diǎn),解題時(shí)要認(rèn)真審題、仔細(xì)解答,注意分類討論和等價(jià)轉(zhuǎn)化.10.(2015課標(biāo)Ⅰ理,21,12分)