第四章第四節(jié)第1課時(shí)　利用導(dǎo)數(shù)研究恒(能)成立問題

第四節(jié)導(dǎo)數(shù)的綜合應(yīng)用第1課時(shí)利用導(dǎo)數(shù)研究恒(能)成立問題【命題分析】恒(能)成立問題與有解問題是高考數(shù)學(xué)的重要知識(shí),其中不等式恒成立問題經(jīng)常與導(dǎo)數(shù)及其幾何意義、函數(shù)、方程等知識(shí)相交匯,綜合考查學(xué)生分析問題、解決問題的能力,一般作為壓軸題出現(xiàn).【核心考點(diǎn)·分類突破】題型一分離參數(shù)求參數(shù)范圍[例1]已知f(x)=xlnx,g(x)=x3+ax2x+2.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若對任意x∈(0,+∞),2f(x)≤g'(x)+2恒成立,求實(shí)數(shù)a的取值范圍.【解析】(1)因?yàn)閒(x)=xlnx,所以f'(x)=lnx+1.令f'(x)<0,得lnx+1<0,解得0<x<1e,所以f(x)的單調(diào)遞減區(qū)間是(0,1e).令f'(x)>0,得lnx+1>0,解得x>1e,所以f(x)的單調(diào)遞增區(qū)間是(綜上,f(x)的單調(diào)遞減區(qū)間是(0,1e),單調(diào)遞增區(qū)間是(1e(2)因?yàn)間'(x)=3x2+2ax1,由題意得2xlnx≤3x2+2ax+1恒成立.因?yàn)閤>0,所以a≥lnx32x12x在(0,+∞)上恒成立.設(shè)h(x)=lnx32x12x(x>0),則h'(x)=1x32+12x2=(x-1)(3x+1當(dāng)x變化時(shí),h'(x),h(x)的變化情況如表:x(0,1)1(1,+∞)h'(x)+0h(x)↗極大值↘所以當(dāng)x=1時(shí),h(x)取得極大值,也是最大值,且h(x)max=h(1)=2,所以若a≥h(x)在(0,+∞)上恒成立,則a≥h(x)max=2,即a≥2,故實(shí)數(shù)a的取值范圍是[2,+∞).【解題技法】分離參數(shù)法解決恒(能)成立問題的策略(1)分離變量.構(gòu)造函數(shù),直接把問題轉(zhuǎn)化為函數(shù)的最值問題.(2)a≥f(x)恒成立?a≥f(x)max;a≤f(x)恒成立?a≤f(x)min;存在x,a≥f(x)成立?a≥f(x)min;存在x,a≤f(x)成立?a≤f(x)max.【對點(diǎn)訓(xùn)練】(2023·成都診斷節(jié)選)已知函數(shù)f(x)=sinx2ax,a∈R,若關(guān)于x的不等式f(x)≤cosx1在區(qū)間(π2,π)上恒成立,求a的取值范圍【解析】不等式f(x)≤cosx1在區(qū)間(π2,π)上恒成立,即2a≥sinx-cosx+1設(shè)g(x)=sinx-cosx+1x,則設(shè)h(x)=xcosx+xsinxsinx+cosx1,則h'(x)=xsinx+xcosx=x(cosxsinx).因?yàn)楫?dāng)x∈(π2,π)時(shí),sinx>0,cosx<0,所以當(dāng)x∈(π2,π)時(shí),h'(所以函數(shù)h(x)在(π2,π)上單調(diào)遞減因?yàn)閔(π2)=π22<0,所以當(dāng)x∈(π2,π)時(shí),h(x)<0,即g'(x)<0,所以函數(shù)g(x)在(π因?yàn)間(π2)=4π,所以當(dāng)x∈(π2,π)時(shí),有g(shù)(x)<4π,所以a≥2π,所以a【加練備選】(2020·全國Ⅰ卷節(jié)選)已知函數(shù)f(x)=ex+ax2x.當(dāng)x≥0時(shí),f(x)≥12x3+1恒成立,求a的取值范圍【解析】由f(x)≥12x3+1得ex+ax2x≥12x3+1,其中①當(dāng)x=0時(shí),不等式為1≥1,顯然成立,此時(shí)a∈R.②當(dāng)x>0時(shí),分離參數(shù)a,得a≥ex記g(x)=ex則g'(x)=(x令h(x)=ex12x2x1(x>0),則h'(x)=exx令H(x)=exx1,H'(x)=ex1>0,故h'(x)在(0,+∞)上單調(diào)遞增,因此h'(x)>h'(0)=0,故函數(shù)h(x)在(0,+∞)上單調(diào)遞增,所以h(x)>h(0)=0,即ex12x2x1>0恒成立故當(dāng)x∈(0,2)時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x∈(2,+∞)時(shí),g'(x)<0,g(x)單調(diào)遞減,因此,g(x)max=g(2)=7-綜上,實(shí)數(shù)a的取值范圍是[7-e題型二等價(jià)轉(zhuǎn)化求參數(shù)范圍[例2]已知函數(shù)f(x)=alnxexx+ax,a∈(1)當(dāng)a<0時(shí),討論f(x)的單調(diào)性;(2)設(shè)g(x)=f(x)+xf'(x),若關(guān)于x的不等式g(x)≤ex+x22+(a1)x在[1,2]上有解,求實(shí)數(shù)a【解析】(1)由題意知,f'(x)=axxex-exx當(dāng)a<0時(shí),axex<0恒成立,所以當(dāng)x>1時(shí),f'(x)<0,當(dāng)0<x<1時(shí),f'(x)>0,故函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.(2)因?yàn)間(x)=f(x)+xf'(x),所以g(x)=alnxex+2axa,由題意知,存在x∈[1,2],使得g(x)≤ex+x22+(a1)x成立,即存在x∈[1,2],使得alnx+(a+1)xx22令h(x)=alnx+(a+1)xx22a,x則h'(x)=-ax+a+1x=x∈[1,2].①當(dāng)a≤1時(shí),h'(x)≤0,所以函數(shù)h(x)在[1,2]上單調(diào)遞減,所以h(x)min=h(2)=aln2+a≤0成立,解得a≤0.②當(dāng)1<a<2時(shí),令h'(x)>0,解得1<x<a;令h'(x)<0,解得a<x<2,所以函數(shù)h(x)在[1,a]上單調(diào)遞增,在[a,2]上單調(diào)遞減.又因?yàn)閔(1)=12,所以h(2)=aln2+a≤0,解得a≤0,與1<a<2矛盾,故舍去③當(dāng)a≥2時(shí),h'(x)≥0,所以函數(shù)h(x)在[1,2]上單調(diào)遞增,所以h(x)min=h(1)=12>0,不符合題意綜上所述,實(shí)數(shù)a的取值范圍為(∞,0].【解題技法】等價(jià)轉(zhuǎn)化法解決恒成立問題的策略(1)遇到f(x)≥g(x)型的不等式恒成立問題時(shí),一般采用作差法,構(gòu)造“左減右”的函數(shù)h(x)=f(x)g(x)或“右減左”的函數(shù)u(x)=g(x)f(x),進(jìn)而只需滿足h(x)min≥0或u(x)max≤0.(2)將比較法的思想融入函數(shù)中,轉(zhuǎn)化為求解函數(shù)最值的問題,適用范圍較廣,但是往往需要對參數(shù)進(jìn)行分類討論.【對點(diǎn)訓(xùn)練】已知函數(shù)f(x)=axlnx.(1)討論函數(shù)f(x)的單調(diào)性;(2)若x=1為函數(shù)f(x)的極值點(diǎn),當(dāng)x∈[e,+∞)時(shí),不等式x[f(x)x+1]≤m(ex)恒成立,求實(shí)數(shù)m的取值范圍.【解析】(1)f'(x)=a1x=ax-1①當(dāng)a≤0時(shí),f'(x)<0恒成立,所以f(x)在(0,+∞)上單調(diào)遞減;②當(dāng)a>0時(shí),令f'(x)>0,得x>1a,令f'(x)<0,得0<x<1所以f(x)在(1a,+∞)上單調(diào)遞增,在(0,1a)綜上,當(dāng)a≤0時(shí),f(x)在(0,+∞)上單調(diào)遞減;當(dāng)a>0時(shí),f(x)在(1a,+∞)上單調(diào)遞增,在(0,1a)(2)因?yàn)閤=1為函數(shù)f(x)的極值點(diǎn),所以f'(1)=0,所以a=1,f(x)=xlnx,x[f(x)x+1]=x(1lnx),當(dāng)x∈[e,+∞)時(shí),不等式x[f(x)x+1]≤m(ex)?x(1lnx)≤m(ex),即x(1lnx)m(ex)≤0,令g(x)=x(1lnx)m(ex),g(e)=0,g'(x)=mlnx,x∈[e,+∞),若m≤1,g'(x)≤0在[e,+∞)上恒成立,則g(x)在[e,+∞)上單調(diào)遞減,所以g(x)≤g(e)=0,滿足題意.若m>1,由g'(x)>0,可得e≤x<em,則g(x)在[e,em)上單調(diào)遞增,所以在[e,em)上存在x0使得g(x0)>g(e)=0,與題意不符,綜上,實(shí)數(shù)m的取值范圍為(∞,1].題型三雙變量的恒(能)成立問題[例3]已知函數(shù)f(x)=a(x2-x-1)(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若?x1,x2∈[0,4],不等式|f(x1)f(x2)|<1恒成立,求實(shí)數(shù)a的取值范圍.【解析】(1)因?yàn)閒(x)=a(x2-x所以f'(x)=-ax(x-3)因?yàn)閍>0,所以令f'(x)>0,得0<x<3;令f'(x)<0,得x<0或x>3,所以f(x)的單調(diào)遞增區(qū)間為(0,3),單調(diào)遞減區(qū)間為(∞,0)和(3,+∞).(2)由(1)知,f(x)在(0,3)上單調(diào)遞增,在(3,4)上單調(diào)遞減,所以f(x)在[0,4]上的最大值是f(3)=5a又f(0)=a<0,f(4)=11ae4>0,所以f(0)<f(4),所以f(x)在[0,4]上的最小值為f(0)=a.若?x1,x2∈[0,4],不等式|f(x1)f(x2)|<1恒成立,則需f(x)maxf(x)min<1在[0,4]上恒成立,即f(3)f(0)<1,即5ae3+a<1,解得a又a>0,所以0<a<e3故實(shí)數(shù)a的取值范圍為(0,e35+【解題技法】雙變量的恒(能)成立問題中常見的轉(zhuǎn)化(1)?x1∈M,?x2∈N,f(x1)>g(x2)?f(x)min>g(x)min.(2)?x1∈M,?x2∈N,f(x1)>g(x2)?f(x)min>g(x)max.(3)?x1∈M,?x2∈N,f(x1)>g(x2)?f(x)max>g(x)min.(4)?x1∈M,?x2∈N,f(x1)>g(x2)?f(x)max>g(x)max.【對點(diǎn)訓(xùn)練】已知函數(shù)f(x)=x(a+1)lnxax(a∈R),g(x)=12x2+exxe(1)當(dāng)x∈[1,e]時(shí),求f(x)的最小值;(2)當(dāng)a<1時(shí),若存在x1∈[e,e2],使得對任意的x2∈[2,0],f(x1)<g(x2)恒成立,求a的取值范圍.【解析】(1)f(x)的定義域?yàn)?0,+∞),f'(x)=(x①若a≤1,當(dāng)x∈[1,e]時(shí),f'(x)≥0,f(x)在[1,e]上單調(diào)遞增,f(x)min=f(1)=1a.②若1<a<e,當(dāng)x∈[1,a]時(shí),f'(x)≤0,f(x)單調(diào)遞減;當(dāng)x∈[a,e]時(shí),f'(x)≥0,f(x)單調(diào)遞增,所以f(x)min=f(a)=a(a+1)lna1.③若a≥e,當(dāng)x∈[1,e]時(shí),f'(x)≤0,f(x)在[1,e]上單調(diào)遞減,f(x)min=f(e)=e(a+1)ae綜上,當(dāng)a≤1時(shí),f(x)min=1a;當(dāng)1<a<e時(shí),f(x)min=a(a+1)lna1;當(dāng)a≥e時(shí),f(x)min=e(a+1)ae(2)由題意知f(x)(x∈[e,e2])的最小值小于g(x)(x∈[2,0])的最小值.由(1)知f(x)在[e,e2]上單調(diào)遞增,所以f(x)min=f(e)=e(a+1)ae又g'(x)=(1ex)x,所以當(dāng)x∈[2,0]時(shí),g'(x)≤0,g(x)在[2,0]上單調(diào)遞減,則g(x)min=g(0)=1,所以e(a+1)ae<1,解得a>e所以a的取值范圍為(e2-【加練備選】(2023·石家莊質(zhì)檢)已知函數(shù)f(x)=ax2lnx與g(x)=x2bx.(1)若f(x)與g(x)在x=1處有相同的切線,求a,b,并證明f(x)≥g(x);(2)若對?x∈[1,e],都?b∈[1,e2]使f(x)≥g(x)恒成立,求a的取值范圍【解析】(1)f'(x)=2axlnx+ax,g'(x)=2xb,因?yàn)楹瘮?shù)f(x)與g(x)在x=1處有相同的切線,所以f(1解得a此時(shí)f(x)=x2lnx,g(x)=x2x.要證f(x)≥g(x),即證x2lnx≥x2x,即xlnx≥x1,令h(x)=xlnxx+1,則h'(x)=lnx,且h'(1)=0,當(dāng)x∈(0,1)時(shí),h'(x)<0,當(dāng)x∈(1,+∞)時(shí),h'(x)>0,所以h(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以h(x)≥h(1)=0,即f(x)≥g(x).(2)欲使f(x)≥g(x)恒成立,即ax2lnx≥x2bx恒成立,即axlnxx≥b恒成立,因?yàn)?b∈[1,e2]使f(x)≥g(x)恒成立所以axlnxx≥e2恒成立當(dāng)x=1時(shí)