高考數(shù)學(xué)復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 2.4 導(dǎo)數(shù)及其應(yīng)用（壓軸題）練習(xí) 文-人教版高三數(shù)學(xué)試題

2.4導(dǎo)數(shù)及其應(yīng)用(壓軸題)高考命題規(guī)律1.每年必考考題,一般在21題位置作為壓軸題呈現(xiàn).2.解答題,12分,高檔難度.3.全國高考有4種命題角度,分布如下表.2020年高考必備2015年2016年2017年2018年2019年Ⅰ卷Ⅱ卷Ⅰ卷Ⅱ卷Ⅲ卷Ⅰ卷Ⅱ卷Ⅲ卷Ⅰ卷Ⅱ卷Ⅲ卷Ⅰ卷Ⅱ卷Ⅲ卷命題角度1利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性2120命題角度2函數(shù)的單調(diào)性與極值、最值的綜合應(yīng)用21202121命題角度3利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)或方程的根21212120命題角度4導(dǎo)數(shù)與不等式21212121命題角度5恒成立與存在性問題命題角度1利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性高考真題體驗(yàn)·對方向1.(2019全國Ⅲ·20)已知函數(shù)f(x)=2x3-ax2+2.(1)討論f(x)的單調(diào)性;(2)當(dāng)0<a<3時,記f(x)在區(qū)間[0,1]的最大值為M,最小值為m,求M-m的取值范圍.解(1)f'(x)=6x2-2ax=2x(3x-a).令f'(x)=0,得x=0或x=a3若a>0,則當(dāng)x∈(-∞,0)∪a3,+∞時,f'(當(dāng)x∈0,a3時,f'(x)故f(x)在(-∞,0),a3,+若a=0,f(x)在(-∞,+∞)單調(diào)遞增;若a<0,則當(dāng)x∈-∞,a3∪(0,+∞)時,f'(當(dāng)x∈a3,0時,f'(x)故f(x)在-∞,a3,(0,+∞(2)當(dāng)0<a<3時,由(1)知,f(x)在0,a3單調(diào)遞減,在a3,1單調(diào)遞增,所以f(x)在[0,1]的最小值為fa3=-a327+2,最大值為f于是m=-a327+2,所以M-m=2當(dāng)0<a<2時,可知2-a+a3所以M-m的取值范圍是827當(dāng)2≤a<3時,a327單調(diào)遞增,所以M-m的取值范圍是綜上,M-m的取值范圍是8272.(2017全國Ⅱ·21)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,f(x)≤ax+1,求a的取值范圍.解(1)f'(x)=(1-2x-x2)ex.令f'(x)=0得x=-1-2,x=-1+2.當(dāng)x∈(-∞,-1-2)時,f'(x)<0;當(dāng)x∈(-1-2,-1+2)時,f'(x)>0;當(dāng)x∈(-1+2,+∞)時,f'(x)<0.所以f(x)在(-∞,-1-2),(-1+2,+∞)內(nèi)單調(diào)遞減,在(-1-2,-1+2)內(nèi)單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a≥1時,設(shè)函數(shù)h(x)=(1-x)ex,h'(x)=-xex<0(x>0),因此h(x)在[0,+∞)內(nèi)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,設(shè)函數(shù)g(x)=ex-x-1,g'(x)=ex-1>0(x>0),所以g(x)在[0,+∞)內(nèi)單調(diào)遞增,而g(0)=0,故ex≥x+1.當(dāng)0<x<1時,f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5-4a-12,則x0∈(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0當(dāng)a≤0時,取x0=5-12,則x0∈(0,1),f(x0)>(1-x0)(1+x0)2=1≥ax0綜上,a的取值范圍是[1,+∞).典題演練提能·刷高分1.已知函數(shù)f(x)=13x3+x2+ax+1(1)若曲線y=f(x)在點(diǎn)(0,1)處切線的斜率為-3,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若函數(shù)f(x)在區(qū)間[-2,a]上單調(diào)遞增,求a的取值范圍.解(1)因?yàn)閒(0)=1,所以曲線y=f(x)經(jīng)過點(diǎn)(0,1),又f'(x)=x2+2x+a,曲線y=f(x)在點(diǎn)(0,1)處的切線的斜率為-3,所以f'(0)=a=-3,所以f'(x)=x2+2x-3.當(dāng)x變化時,f'(x),f(x)的變化情況如下表:x(-∞,-3)-3(-3,1)1(1,+∞)f'(x)+0-0+f(x)增極大值減極小值增所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(-∞,-3),(1,+∞),單調(diào)遞減區(qū)間為(-3,1).(2)因?yàn)楹瘮?shù)f(x)在區(qū)間[-2,a]上單調(diào)遞增,所以f'(x)≥0.即對x∈[-2,a],只要f'(x)min≥0.因?yàn)楹瘮?shù)f'(x)=x2+2x+a的對稱軸為x=-1,當(dāng)-2≤a≤-1時,f'(x)在[-2,a]上的最小值為f'(a),由f'(a)=a2+3a≥0,得a≥0或a≤-3,所以此種情況不成立;當(dāng)a>-1時,f'(x)在[-2,a]上的最小值為f'(-1),由f'(-1)=1-2+a≥0得a≥1,綜上,實(shí)數(shù)a的取值范圍是[1,+∞).2.已知函數(shù)f(x)=(2-m)lnx+1x+2mx(1)當(dāng)f'(1)=0時,求實(shí)數(shù)m的值及曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)討論函數(shù)f(x)的單調(diào)性.解(1)函數(shù)y=f(x)的定義域?yàn)?0,+∞),f'(x)=2mx2+(2-m從而f(1)=-1,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=-1.(2)由f'(x)=(mx+1)(當(dāng)m≥0時,函數(shù)y=f(x)的減區(qū)間為0,12,增區(qū)間為12,+∞.當(dāng)m<0時,由f'(x)=(mx+1)(2x-1)當(dāng)m<-2時,y=f(x)的減區(qū)間為0,-1m和12,+∞,增區(qū)間為-1m,12;當(dāng)m=-2時,y=f(x)的減區(qū)間為(0,+∞),沒有增區(qū)間.當(dāng)-2<m<0時,y=f(x)的減區(qū)間為0,12和-1m,+∞,增區(qū)間為12,-1m.綜上可知:當(dāng)m≥0時,函數(shù)y=f(x)的減區(qū)間為0,12,增區(qū)間為12,+∞;當(dāng)m<-2時,y=f(x)的減區(qū)間為0,-1m和12,+∞,增區(qū)間為-1m,12;當(dāng)m=-2時,y=f(x)的減區(qū)間為(0,+∞),沒有增區(qū)間;當(dāng)-2<m<0時,y=f(x)的減區(qū)間為0,12和-1m,+∞,增區(qū)間為12,-1m.3.已知函數(shù)f(x)=lnx+ax-x+1-a(a∈R)(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若存在x>1,使f(x)+x<1-xx成立,求整數(shù)解(1)由題意可知,x>0,f'(x)=1x-ax2方程-x2+x-a=0對應(yīng)的Δ=1-4a,當(dāng)Δ=1-4a≤0,即a≥14時,當(dāng)x∈(0,+∞)時,f'(x∴f(x)在(0,+∞)上單調(diào)遞減;當(dāng)0<a<14時,方程-x2+x-a=0的兩根為1±1-4此時,f(x)在1-1-4a2,1+1-4a2在0,1-1-4a2,1+1-4a2,+∞上f'(x當(dāng)a≤0時,1-1-4a此時當(dāng)x∈0,1+1-4a2,f'(x)>0,f當(dāng)x∈1+1-4a2,+∞時,f'(x)<0,f(綜上:當(dāng)a≤0時,x∈0,1+1-4a2,f(x)單調(diào)遞增,當(dāng)x∈1+1-4a2,+當(dāng)0<a<14時,f(x)在1-1-4a2,1+1-4a2上單調(diào)遞增,在當(dāng)a≥14時,f(x)在(0,+∞)上單調(diào)遞減(2)原式等價于(x-1)a>xlnx+2x-1,即存在x>1,使a>xlnx設(shè)g(x)=xlnx+2x-1x-1(設(shè)h(x)=x-lnx-2,則h'(x)=1-1x=∴h(x)在(1,+∞)上單調(diào)遞增.又h(3)=3-ln3-2=1-ln3<0,h(4)=4-ln4-2=2-2ln2>0,根據(jù)零點(diǎn)存在性定理,可知h(x)在(1,+∞)上有唯一零點(diǎn),設(shè)該零點(diǎn)為x0,則x0∈(3,4),且h(x0)=x0-lnx0-2=0,即x0-2=lnx0,∴g(x)min=x0lnx0+2由題意可知a>x0+1.又x0∈(3,4),a∈Z,∴a的最小值為5.4.已知函數(shù)f(x)=3x-1ex+ax(x>0,a∈R).(1)若f(x)在(0,+∞)上單調(diào)遞減,求a的取值范圍;(2)當(dāng)a∈(-3,-e)時,判斷關(guān)于x的方程f(x)=2的解的個數(shù).解(1)∵f(x)=3x-1ex+ax(x>0),∴f'(x)=ex3x-1-3x2-ax2=-x由題意得f'(x)=-x2+3x-3x2·即a≥(-x2+3x-3)·ex在(0,+∞)恒成立,設(shè)g(x)=(-x2+3x-3)·ex,則g'(x)=ex(-x2+x),∴g(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,∴g(x)max=g(1)=-e,∴a≥-e.∴實(shí)數(shù)a的取值范圍為[-e,+∞).(2)由題意得f(x)=3x-1ex+ax=2(x>0),∴a=2x-(3-x)ex(x>0),令h(x)=2x-(3-x)ex,則h'(x)=2+(x-2)ex,令φ(x)=h'(x)=2+(x-2)ex(x>0),則φ'(x)=(x-1)ex,∴h'(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,∴h'(x)min=h'(1)=2-e<0.又h'(0)=0,h'(2)=2>0,∴存在x0∈(0,2),使得x∈(0,x0)時h'(x)<0,h(x)單調(diào)遞減;當(dāng)x∈(x0,+∞)時,h'(x)>0,h(x)單調(diào)遞增,又h(0)=-3,h(x0)<0,當(dāng)x→+∞時,h(x)→+∞,∴當(dāng)x>0,a∈(-3,-e)時,方程a=2x-(3-x)ex有一個解,即當(dāng)a∈(-3,-e)時,方程f(x)=2只有一個解.命題角度2函數(shù)的單調(diào)性與極值、最值的綜合應(yīng)用高考真題體驗(yàn)·對方向1.(2019天津·20)設(shè)函數(shù)f(x)=lnx-a(x-1)ex,其中a∈R.(1)若a≤0,討論f(x)的單調(diào)性;(2)若0<a<1e①證明f(x)恰有兩個零點(diǎn);②設(shè)x0為f(x)的極值點(diǎn),x1為f(x)的零點(diǎn),且x1>x0,證明3x0-x1>2.(1)解由已知,f(x)的定義域?yàn)?0,+∞),且f'(x)=1x-[aex+a(x-1)ex]=1因此當(dāng)a≤0時,1-ax2ex>0,從而f'(x)>0,所以f(x)在(0,+∞)內(nèi)單調(diào)遞增.(2)證明①由(1)知,f'(x)=1-ax2exx.令g(x)由0<a<1e,可知g(x)在(0,+∞)內(nèi)單調(diào)遞減,又g(1)=1-ae>且gln1a=1-aln1a21a=1-ln1a2<0,故g(x)=0在(0,+∞)內(nèi)有唯一解,從而f'(x)=0在(0,+∞)內(nèi)有唯一解,不妨設(shè)為x0,則1<x0<ln1a當(dāng)x∈(0,x0)時,f'(x)=g(x所以f(x)在(0,x0)內(nèi)單調(diào)遞增;當(dāng)x∈(x0,+∞)時,f'(x)=g(x所以f(x)在(x0,+∞)內(nèi)單調(diào)遞減,因此x0是f(x)的唯一極值點(diǎn).令h(x)=lnx-x+1,則當(dāng)x>1時,h'(x)=1x-1<0,故h(x)在(1,+∞)內(nèi)單調(diào)遞減,從而當(dāng)x>1時,h(x)<h(1)=0,所以x<x-1從而fln1a=lnln1a-aln1a-1eln1a=lnln1a-ln1a+1=hln1a<0,又因?yàn)閒(x0)>f(1)=0,所以f(x)在(x0,+∞)內(nèi)有唯一零點(diǎn).又f(x)在(0,x0)內(nèi)有唯一零點(diǎn)1,從而,f(x)在(0,+∞)內(nèi)恰有兩個零點(diǎn).②由題意,f'(x0)=0,f(因?yàn)楫?dāng)x>1時,lnx<x-1,又x1>x0>1,故ex得lnex1-x0<lnx02,于是x1-x0<2lnx0<2(x0-1),整理得32.(2019全國Ⅱ·21)已知函數(shù)f(x)=(x-1)lnx-x-1.證明:(1)f(x)存在唯一的極值點(diǎn);(2)f(x)=0有且僅有兩個實(shí)根,且兩個實(shí)根互為倒數(shù).證明(1)f(x)的定義域?yàn)?0,+∞).f'(x)=x-1x+lnx-1=ln因?yàn)閥=lnx單調(diào)遞增,y=1x所以f'(x)單調(diào)遞增.又f'(1)=-1<0,f'(2)=ln2-12=ln4-12>0,故存在唯一x0∈(1,2),使得f'(又當(dāng)x<x0時,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x>x0時,f'(x)>0,f(x)單調(diào)遞增.因此,f(x)存在唯一的極值點(diǎn).(2)由(1)知f(x0)<f(1)=-2,又f(e2)=e2-3>0,所以f(x)=0在區(qū)間(x0,+∞)內(nèi)存在唯一根x=α.由α>x0>1得1α<1<x0又f1α=1α-1ln1α-1α-1=f故1α是f(x)=0在(0,x0)的唯一根綜上,f(x)=0有且僅有兩個實(shí)根,且兩個實(shí)根互為倒數(shù).3.(2017北京·20)已知函數(shù)f(x)=excosx-x.(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)求函數(shù)f(x)在區(qū)間0,π解(1)因?yàn)閒(x)=excosx-x,所以f'(x)=ex(cosx-sinx)-1,f'(0)=0.又因?yàn)閒(0)=1,所以曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=1.(2)設(shè)h(x)=ex(cosx-sinx)-1,則h'(x)=ex(cosx-sinx-sinx-cosx)=-2exsinx.當(dāng)x∈0,π2時,h'(x)<0,所以h(x)在區(qū)間所以對任意x∈0,π2有h(x)<h即f'(x)<0.所以函數(shù)f(x)在區(qū)間0,π因此f(x)在區(qū)間0,π2上的最大值為f(0)=1,最小值為fπ4.(2017全國Ⅰ·21)已知函數(shù)f(x)=ex(ex-a)-a2x.(1)討論f(x)的單調(diào)性;(2)若f(x)≥0,求a的取值范圍.解(1)函數(shù)f(x)的定義域?yàn)?-∞,+∞),f'(x)=2e2x-aex-a2=(2ex+a)(ex-a).①若a=0,則f(x)=e2x,在(-∞,+∞)單調(diào)遞增.②若a>0,則由f'(x)=0得x=lna.當(dāng)x∈(-∞,lna)時,f'(x)<0;當(dāng)x∈(lna,+∞)時,f'(x)>0.故f(x)在(-∞,lna)單調(diào)遞減,在(lna,+∞)單調(diào)遞增.③若a<0,則由f'(x)=0得x=ln-a當(dāng)x∈-∞,ln-a2時,當(dāng)x∈ln-a2,+∞時,f'故f(x)在-∞,ln-a2單調(diào)遞減,在ln-a(2)①若a=0,則f(x)=e2x,所以f(x)≥0.②若a>0,則由(1)得,當(dāng)x=lna時,f(x)取得最小值,最小值為f(lna)=-a2lna.從而當(dāng)且僅當(dāng)-a2lna≥0,即a≤1時,f(x)≥0.③若a<0,則由(1)得,當(dāng)x=ln-a2時,f(x)取得最小值,最小值為fln-a2=a234-ln-a2.從而當(dāng)且僅當(dāng)a234綜上,a的取值范圍是[-2e34典題演練提能·刷高分1.(2019湖北八校聯(lián)考一)已知函數(shù)f(x)=x3+32x2-4ax+1(a∈R)(1)若函數(shù)f(x)有兩個極值點(diǎn),且都小于0,求a的取值范圍;(2)若函數(shù)h(x)=a(a-1)lnx-x3+3x+f(x),求函數(shù)h(x)的單調(diào)區(qū)間.解(1)由f(x)有兩個極值點(diǎn)且都小于0,得f'(x)=3x2+3x-4a=0有兩個不相等的負(fù)實(shí)根,∴Δ解得-316<a<0故a的取值范圍為-316,0.(2)由h(x)=a(a-1)lnx+32x2-(4a-3)x+1,x>則h'(x)=a(a-1)x+3x-(4a-3)=1x(3令(3x-a)[x-(a-1)]=0,得x=a3或x=a-1,令a3=a-1,得a=①當(dāng)a3≤0,a-1≤0,即a≤0時,在(0,+∞②當(dāng)0<a3,a-1≤0,即0<a≤1時,當(dāng)x>a3時,h'(x)>0,當(dāng)0③當(dāng)a3>a-1>0,即1<a<32,當(dāng)0<x<a-1或x>a3時,h'(x)>0,當(dāng)a-1<x<a3時,h'(④當(dāng)a3=a-1>0,即a=32時,h'(x)⑤當(dāng)a-1>a3>0,即a>32,當(dāng)0<x<a3或x>a-1時,h'(x)>0,當(dāng)a3<x<a-1時,h'(x綜上所述:當(dāng)a≤0或a=32時,h(x)在(0,+∞當(dāng)0<a≤1時,h(x)在a3,+∞上單調(diào)遞增,在0,a3上單調(diào)遞減;當(dāng)1<a<32時,h(x)在(0,a-1),a3,+∞上單調(diào)遞增,在a-1,a3上單調(diào)遞減;當(dāng)a>32時,h(x)在0,a3,(a-1,+∞)上單調(diào)遞增,在a3,a-1上單調(diào)遞減.2.已知函數(shù)f(x)=-x2+ax-aex(1)當(dāng)a=1時,求函數(shù)f(x)的極值;(2)設(shè)g(x)=f(x)+f'(x)x-1,若函數(shù)g(x)在(0,1)∪(1,+∞)內(nèi)有兩個極值點(diǎn)x1,x2,求證:g((1)解當(dāng)a=1時,f(x)=-x2+x∴f'(x)=(-2x+1)當(dāng)x∈(0,1),(2,+∞)時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(1,2)時,f'(x)<0,f(x)單調(diào)遞減.所以f(x)在(0,+∞)上有極大值f(1)=-1e,極小值f(2)=-3(2)證明由題意得g(x)=f(∴g'(x)=2x設(shè)h(x)=2x2-(2+a)x+2,∵函數(shù)g(x)在(0,1)∪(1,+∞)內(nèi)有兩個極值點(diǎn)x1,x2,∴方程h(x)=2x2-(2+a)x+2=0在(0,1)∪(1,+∞)上有兩個不相等的實(shí)根x1,x2,且1不能是方程的根,Δ=(2+a)2-16>0.∴x1+x2∴g(x1)g(x2)=(-=4=2(∵a>2,∴4e∴g(x1)g(x2)=4e3.(2019安徽江淮十校聯(lián)考一)已知函數(shù)f(x)=ax2+xlnx(a為常數(shù),a∈R,e為自然對數(shù)的底數(shù),e=2.71828…).(1)若函數(shù)f(x)≤0恒成立,求實(shí)數(shù)a的取值范圍;(2)若曲線y=f(x)在點(diǎn)(e,f(e))處的切線方程為y=(2e+2)x-e2-e,k∈Z且k<f(x)x-1解(1)函數(shù)f(x)≤0恒成立,即ax2+xlnx≤0恒成立,可得a≤-lnxx設(shè)g(x)=-lnxx,g'(x)=當(dāng)0<x<e時,g'(x)<0,g(x)遞減;當(dāng)x>e時,g'(x)>0,g(x)遞增,可得當(dāng)x=e時g(x)取得最小值,且g(e)=-lnee=-1所以a≤-1e(2)f(x)的導(dǎo)數(shù)為f'(x)=2ax+1+lnx,曲線y=f(x)在點(diǎn)(e,f(e))處的切線斜率為2ae+2=2e+2,可得a=1,即f(x)=x2+xlnx.又由k<f(x)x-1對任意x>1都成立,可得設(shè)h(x)=x2+xlnxx-1,x>設(shè)k(x)=x2-x-lnx-1,x>1,k'(x)=2x-1-1x=可得k(x)在(1,+∞)內(nèi)遞增,由k(1.8)=0.44-ln1.8.由1.8>e可得ln1.8>12即有k(1.8)<0,k(2)=1-ln2>0,則存在m∈(1.8,2),使得k(m)=0,則1<x<m,k(x)<0,h'(x)<0,即h(x)在(1,m)內(nèi)遞減;當(dāng)x>m時,k(x)>0,h'(x)>0,h(x)在x>m遞增,可得h(x)min=h(m)=m2又k(m)=m2-m-lnm-1=0,即有m2-1=m+lnm,可得h(x)min=m2+m在(1.8.2)遞增,可得h(x)min∈(5.04,6),由k<h(x)min,k∈Z,故k的最大值為5.4.(2019山東濰坊二模)已知函數(shù)f(x)=xex-alnx(無理數(shù)e=2.718…).(1)若f(x)在(0,1)單調(diào)遞減,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a=-1時,設(shè)g(x)=x(f(x)-xex)-x3+x2-b,若函數(shù)g(x)存在零點(diǎn),求實(shí)數(shù)b的最大值.解(1)f'(x)=(x+1)ex-ax由題意可得f'(x)≤0,x∈(0,1)恒成立.即(x2+x)ex-a≤0,也就是a≥(x2+x)ex在x∈(0,1)恒成立.設(shè)h(x)=(x2+x)ex,則h'(x)=(x2+3x+1)ex.當(dāng)x∈(0,1)時,x2+3x+1>0,h'(x)>0在x∈(0,1)單調(diào)遞增.即h(x)<h(1)=2e.故a≥2e.(2)當(dāng)a=-1時,f(x)=xex+lnx.g(x)=xlnx-x3+x2-b,由題意得問題等價于方程b=xlnx-x3+x2,在(0,+∞)上有解.先證明lnx≤x-1.設(shè)u(x)=lnx-x+1,x∈(0,+∞),則u'(x)=1x-1=1可得當(dāng)x=1時,函數(shù)u(x)取得極大值,∴u(x)≤u(1)=0.因此lnx≤x-1,所以b=xlnx-x3+x2≤x(x-1)-x3+x2=-x(x2-2x+1)≤0.當(dāng)x=1時,取等號.故實(shí)數(shù)b的最大值為0.5.(2019湘贛十四校聯(lián)考一)已知函數(shù)f(x)=lnx-mx-n(m,n∈R).(1)若n=1時,函數(shù)f(x)有極大值為-2,求m;(2)若對任意實(shí)數(shù)x>0,都有f(x)≤0,求m+n的最小值.解(1)當(dāng)n=1時,f(x)=lnx-mx-1.∵函數(shù)f(x)有極大值為-2,由f'(x)=1x-m=0知x=1m∴f1m=-lnm-1-1=-2,∴m=1.經(jīng)檢驗(yàn),m=1滿足題意.(2)函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=1x-m①當(dāng)m<0時,當(dāng)x∈(0,+∞)時,f'(x)>0,∴f(x)在(0,+∞)上單調(diào)遞增,令x=en,則f(en)=lnen-men-n=-men>0,舍去;②當(dāng)m=0時,當(dāng)x∈(0,+∞)時f'(x)>0,∴f(x)在(0,+∞)上單調(diào)遞增,令x=en+1,則f(en+1)=lnen+1-n=1>0,舍去;③當(dāng)m>0時,當(dāng)x∈0,1m時,f'(x)>0;當(dāng)x∈1m,+∞時,f'(x)<0;∴f(x)在0,1m上單調(diào)遞增,在1m,+∞上單調(diào)遞減,∴f(x)的最大值為f1m=-lnm-1-n≤0,即n≥-lnm-1,∴m+n≥m-lnm-1.設(shè)h(m)=m-lnm-1,令h'(m)=1-1m=0,則m=1當(dāng)m∈(0,1)時,h'(m)<0,∴h(m)在(0,1)上單調(diào)遞減,當(dāng)m∈(1,+∞)時,h'(m)>0,∴h(m)在(1,+∞)上單調(diào)遞增.∴h(m)的最小值為h(1)=0,綜上所述,當(dāng)m=1,n=-1時m+n的最小值為0.命題角度3利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)或方程的根高考真題體驗(yàn)·對方向1.(2019全國Ⅰ·20)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,f(x)≥ax,求a的取值范圍.(1)證明設(shè)g(x)=f'(x),則g(x)=cosx+xsinx-1,g'(x)=xcosx.當(dāng)x∈0,π2時,g'(x當(dāng)x∈π2,π時,g'(x所以g(x)在0,π2單調(diào)遞增,在又g(0)=0,gπ2>0,g(π)=-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)解由題設(shè)知f(π)≥aπ,f(π)=0,可得a≤0.由(1)知,f'(x)在(0,π)只有一個零點(diǎn),設(shè)為x0,且當(dāng)x∈(0,x0)時,f'(x)>0;當(dāng)x∈(x0,π)時,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,f(x)≥0.又當(dāng)a≤0,x∈[0,π]時,ax≤0,故f(x)≥ax.因此,a的取值范圍是(-∞,0].2.(2016全國Ⅰ·21)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)有兩個零點(diǎn),求a的取值范圍.解(1)f'(x)=(x-1)ex+2a(x-1)=(x-1)(ex+2a).(ⅰ)設(shè)a≥0,則當(dāng)x∈(-∞,1)時,f'(x)<0;當(dāng)x∈(1,+∞)時,f'(x)>0.所以f(x)在(-∞,1)單調(diào)遞減,在(1,+∞)單調(diào)遞增.(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-所以f(x)在(-∞,+∞)單調(diào)遞增.②若a>-e2,則ln(-2a)<故當(dāng)x∈(-∞,ln(-2a))∪(1,+∞)時,f'(x)>0;當(dāng)x∈(ln(-2a),1)時,f'(x)<0.所以f(x)在(-∞,ln(-2a)),(1,+∞)單調(diào)遞增,在(ln(-2a),1)單調(diào)遞減.③若a<-e2,則ln(-2a)>故當(dāng)x∈(-∞,1)∪(ln(-2a),+∞)時,f'(x)>0;當(dāng)x∈(1,ln(-2a))時,f'(x)<0,所以f(x)在(-∞,1),(ln(-2a),+∞)單調(diào)遞增,在(1,ln(-2a))單調(diào)遞減.(2)(ⅰ)設(shè)a>0,則由(1)知,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則f(b)>a2(b-2)+a(b-1)2=ab2所以f(x)有兩個零點(diǎn).(ⅱ)設(shè)a=0,則f(x)=(x-2)ex,所以f(x)只有一個零點(diǎn).(ⅲ)設(shè)a<0,若a≥-e2,則由(1)知,f(x)在(1,+∞)單調(diào)遞增又當(dāng)x≤1時f(x)<0,故f(x)不存在兩個零點(diǎn);若a<-e2,則由(1)知,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f(x)<0,故f(x)不存在兩個零點(diǎn).綜上,a的取值范圍為(0,+∞).典題演練提能·刷高分1.已知函數(shù)f(x)=x2a-2lnx(a∈R,a≠(1)討論函數(shù)f(x)的單調(diào)性;(2)若函數(shù)f(x)有最小值,記為g(a),關(guān)于a的方程g(a)+a-29a-1=m有三個不同的實(shí)數(shù)根,求實(shí)數(shù)m解(1)f'(x)=2xa-當(dāng)a<0時,f'(x)<0,知f(x)在(0,+∞)上是遞減的;當(dāng)a>0時,f'(x)=2(x+a)(x-a)ax,知f((2)由(1)知,a>0,f(x)min=f(a)=1-lna,即g(a)=1-lna,方程g(a)+a-29a-1=m,即m=a-lna-29令F(a)=a-lna-29a(a>0),則F'(a)=1-知F(a)在0,13F(a)極大=F13=-13+ln3,F(a)極小=F23=依題意得13-ln2+ln3<m<-13+2.已知函數(shù)f(x)=ex,g(x)=1x(1)設(shè)函數(shù)F(x)=f(x)+g(x),試討論函數(shù)F(x)零點(diǎn)的個數(shù);(2)若a=-2,x>0,求證:f(x)·g(x)>x+1(1)解函數(shù)F(x)的定義域?yàn)?-∞,a)∪(a,+∞).當(dāng)x∈(a,+∞)時,ex>0,1x-所以F(x)=ex+1x-a>0,即F(x)在區(qū)間(a,+當(dāng)x∈(-∞,a)時,F(x)=ex+1x-a=ex(x-a)+1x-只要討論h(x)的零點(diǎn)即可.h'(x)=ex(x-a+1),h'(a-1)=0,當(dāng)x∈(-∞,a-1)時,h'(x)<0,h(x)是減函數(shù);當(dāng)x∈(a-1,a)時,h'(x)>0,h(x)是增函數(shù).所以h(x)在區(qū)間(-∞,a)的最小值為h(a-1)=1-ea-1.顯然,當(dāng)a=1時,h(a-1)=0,所以x=a-1是F(x)的唯一的零點(diǎn);當(dāng)a<1時,h(a-1)=1-ea-1>0,所以F(x)沒有零點(diǎn);當(dāng)a>1時,h(a-1)=1-ea-1<0,所以F(x)有兩個零點(diǎn).(2)證明若a=-2,x>0,要證f(x)·g(x)>x+1+x2-82x+4,即要證ex∵x+1<設(shè)M(x)=ex-(x+2)x2+1-12x2+4=ex-x2-2x+2,M'(x)=ex-2x-2,令φ(x)=ex-2x-2,φ'(x)=ex-2,φ(x)在(-∞,ln2)上單調(diào)遞減,在(ln2,+∞)上單調(diào)遞增.∵φ(1)φ(2)<0,φ(-1)φ(0)<0,∴M'(x)在(0,+∞)上只有一個零點(diǎn)x0(1<x0<2),ex0-2x0-2=0,∴M(x)在(0,x0)上單調(diào)遞減,在(x0,+∞∴M(x)≥ex0-x02-2x0+2=∴ex>(x+2)x2+1+12x2-4.∵x2+1>x∴ex>(x+2)x+1+12∴f(x)·g(x)>x+1+3.(2019河北唐山三模)已知函數(shù)f(x)=xlnx-a(x2-x)+1,函數(shù)g(x)=f'(x).(1)若a=1,求f(x)的極大值;(2)當(dāng)0<x<1時,g(x)有兩個零點(diǎn),求a的取值范圍.解(1)f(x)=xlnx-x2+x+1,g(x)=f'(x)=lnx-2x+2,g'(x)=1x-2=1當(dāng)x∈0,12時,g'(x)>0,g(x)單調(diào)遞增;當(dāng)x∈12,+∞時,g'(x)<0,g(x)單調(diào)遞減.注意到g(1)=f'(1)=0,則當(dāng)x∈12,1時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(1,+∞)時,f'(x)<0,f(x)單調(diào)遞減;故當(dāng)x=1時,f(x)取得極大值f(1)=1.(2)g(x)=f'(x)=lnx+1-2ax+a,g'(x)=1x-2a=1①若a≤0,則g'(x)>0,g(x)單調(diào)遞增,至多有一個零點(diǎn),不合題意.②若a>0,則當(dāng)x∈0,12a時,g'(x)>0,g(x)單調(diào)遞增;當(dāng)x∈12a,+∞時,g'(x)<0,g(x)單調(diào)遞減;則g12a≥g12=ln12+1=lne2>0不妨設(shè)g(x1)=g(x2),x1<x2,則0<x1<12a<x2<一方面,需要g(1)<0,得a>1.另一方面,由(1)得,當(dāng)x>1時,lnx<x-1<x,則x<ex,進(jìn)而,有2a<e2a,則e-2a<12且g(e-2a)=-2ae-2a+1-a<0,故存在x1,使得0<e-2a<x1<12a.綜上,a的取值范圍是(1,+∞4.(2019山西運(yùn)城二模)已知函數(shù)f(x)=xex-a(lnx+x),a∈R.(1)當(dāng)a=e時,求f(x)的單調(diào)區(qū)間;(2)若f(x)有兩個零點(diǎn),求實(shí)數(shù)a的取值范圍.解(1)f(x)定義域?yàn)?0,+∞),當(dāng)a=e時,f'(x)=(1+∴當(dāng)0<x<1時,f'(x)<0,當(dāng)x>1時,f'(x)>0,∴f(x)在(0,1)上為減函數(shù);在(1,+∞)上為增函數(shù).(2)記t=lnx+x,則t=lnx+x在(0,+∞)上單調(diào)遞增,且t∈R.∴f(x)=xex-a(lnx+x)=et-at=g(t).∴f(x)在(0,+∞)上有兩個零點(diǎn)等價于g(t)=et-at在t∈R上有兩個零點(diǎn).①當(dāng)a=0時,g(t)=et在R上單調(diào)遞增,且g(t)>0,故g(t)無零點(diǎn);②當(dāng)a<0時,g'(t)=et-a>0恒成立,∴g(t)在R上單調(diào)遞增,又g(0)=1>0,g1a=e1a-1<0,故g(t)在R③當(dāng)a>0時,由g'(t)=et-a=0可知g(t)在t=lna時有唯一的一個極小值g(lna)=a(1-lna),若0<a<e,g(t)極小值=a(1-lna)>0,g(t)無零點(diǎn);若a=e,g(t)極小值=0,g(t)只有一個零點(diǎn);若a>e,g(t)極小值=a(1-lna)<0,而g(0)=1>0,由y=lnxx在(e,+∞)上為減函數(shù),可知當(dāng)a>e時,ea>ae>a從而g(a)=ea-a2>0,∴g(t)在(0,lna)和(lna,+∞)上各有一個零點(diǎn).綜上可知,當(dāng)a>e時,f(x)有兩個點(diǎn),故所求a的取值范圍是(e,+∞).命題角度4導(dǎo)數(shù)與不等式高考真題體驗(yàn)·對方向1.(2017全國Ⅲ·21)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,證明f(x)≤-34a-(1)解f(x)的定義域?yàn)?0,+∞),f'(x)=1x+2ax+2a+1=(若a≥0,則當(dāng)x∈(0,+∞)時,f'(x)>0,故f(x)在(0,+∞)單調(diào)遞增.若a<0,則當(dāng)x∈0,-12a時,f'(當(dāng)x∈-12a,+∞時,f'故f(x)在0,-12a(2)證明由(1)知,當(dāng)a<0時,f(x)在x=-12a取得最大值,最大值為f-12a=ln-所以f(x)≤-34a-2等價于ln-12a-1-1即ln-12a設(shè)g(x)=lnx-x+1,則g'(x)=1x-1當(dāng)x∈(0,1)時,g'(x)>0;當(dāng)x∈(1,+∞)時,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,g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時,g(x)≤0.從而當(dāng)a<0時,ln-12即f(x)≤-34a-2.(2016全國Ⅲ·21)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,1<x-1(3)設(shè)c>1,證明當(dāng)x∈(0,1)時,1+(c-1)x>cx.(1)解由題設(shè),f(x)的定義域?yàn)?0,+∞),f'(x)=1x-令f'(x)=0解得x=1.當(dāng)0<x<1時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x>1時,f'(x)<0,f(x)單調(diào)遞減.(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,lnx<x-1.故當(dāng)x∈(1,+∞)時,lnx<x-1,ln1x<即1<x-1(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,g'(x)>0,g(x)單調(diào)遞增;當(dāng)x>x0時,g'(x)<0,g(x)單調(diào)遞減.由(2)知1<c-1lnc<c,故0<x又g(0)=g(1)=0,故當(dāng)0<x<1時,g(x)>0.所以當(dāng)x∈(0,1)時,1+(c-1)x>cx.典題演練提能·刷高分1.(2019江西九江一模)已知函數(shù)f(x)=x2-(2a-1)x-alnx(a∈R).(1)試討論函數(shù)f(x)的單調(diào)性;(2)若函數(shù)f(x)存在最小值f(x)min,求證:f(x)min<34解(1)f'(x)=(2①當(dāng)a≤0時,f'(x)>0在(0,+∞)恒成立,故f(x)在(0,+∞)遞增,②當(dāng)a>0時,由f'(x)>0,解得x>a;由f'(x)<0,解得0<x<a,故f(x)在(0,a)內(nèi)遞減,在(a,+∞)內(nèi)遞增.(2)由(1)知要使f(x)存在最小值,則a>0且f(x)min=f(a)=a-a2-alna.令g(x)=x-x2-xlnx(x>0),則g'(x)=-2x-lnx在(0,+∞)內(nèi)遞減.又g'1e=1-2e>0,g'12=ln2-1<0,∴存在x0∈1e,12使得g'(x0)故g(x)在(0,x0)內(nèi)遞增,在(x0,+∞)內(nèi)遞減.∵g'(x0)=0,∴-2x0-lnx0=0,∴l(xiāng)nx0=-2x0,故g(x)max=g(x0)=x0+122-14,又x0∈1e,12,∴g(x)max=x0+122-14<12+122-14=34,故2.(2019河南八市重點(diǎn)高中高三五模)已知函數(shù)f(x)=ex-ax2,且曲線y=f(x)在點(diǎn)x=1處的切線與直線x+(e-2)y=0垂直.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)求證:x>0時,ex-ex-1≥x(lnx-1).(1)解由f(x)=ex-ax2,得f'(x)=ex-2ax.因?yàn)榍€y=f(x)在點(diǎn)x=1處的切線與直線x+(e-2)y=0垂直,所以f'(1)=e-2a=e-2,所以a=1,即f(x)=ex-x2,f'(x)=ex-2x.令g(x)=ex-2x,則g'(x)=ex-2.所以x∈(-∞,ln2)時,g'(x)<0,g(x)單調(diào)遞減;x∈(ln2,+∞)時,g'(x)>0,g(x)單調(diào)遞增.所以g(x)min=g(ln2)=2-2ln2>0.所以f'(x)>0,f(x)單調(diào)遞增.即f(x)的單調(diào)增區(qū)間為(-∞,+∞),無減區(qū)間.(2)證明由(1)知f(x)=ex-x2,f(1)=e-1,所以y=f(x)在x=1處的切線方程為y-(e-1)=(e-2)(x-1),即y=(e-2)x+1.令h(x)=ex-x2-(e-2)x-1,則h'(x)=ex-2x-(e-2)=ex-e-2(x-1),且h'(1)=0,h″(x)=ex-2.x∈(-∞,ln2)時,h″(x)<0,h'(x)單調(diào)遞減;x∈(ln2,+∞)時,h″(x)>0,h'(x)單調(diào)遞增.因?yàn)閔'(1)=0,所以h'(x)min=h'(ln2)=4-e-2ln2<0.因?yàn)閔'(0)=3-e>0,所以存在x0∈(0,1),使x∈(0,x0)時,h'(x)>0,h(x)單調(diào)遞增;x∈(x0,1)時,h'(x)<0,h(x)單調(diào)遞減;x∈(1,+∞)時,h'(x)>0,h(x)單調(diào)遞增.又h(0)=h(1)=0,所以x>0時,h(x)≥0,即ex-x2-(e-2)x-1≥0,所以ex-(e-2)x-1≥x2.令φ(x)=lnx-x,則φ'(x)=1x-1=1所以x∈(0,1)時,φ'(x)>0,φ(x)單調(diào)遞增;x∈(1,+∞)時,φ'(x)<0,φ(x)單調(diào)遞減,所以φ(x)≤φ(1)=-1,即lnx+1≤x.因?yàn)閤>0,所以x(lnx+1)≤x2.所以x>0時,ex-(e-2)x-1≥x(lnx+1),即x>0時,ex-ex-1≥x(lnx-1).3.(2019陜西咸陽一模)設(shè)函數(shù)f(x)=x+1-mex,m∈R.(1)當(dāng)m=1時,求f(x)的單調(diào)區(qū)間;(2)求證:當(dāng)x∈(0,+∞)時,lnex(1)解當(dāng)m=1時,f(x)=x+1-ex,f'(x)=1-ex.令f'(x)=0,則x=0.當(dāng)x<0時,f'(x)>0;當(dāng)x>0時,f'(x)<0,故函數(shù)f(x)的單調(diào)遞增區(qū)間是(-∞,0);單調(diào)遞減區(qū)間是(0,+∞).(2)證明由(1)知,當(dāng)m=1時,f(x)max=f(0)=0,∴當(dāng)x∈(0,+∞)時,x+1-ex<0,即ex>x+1.當(dāng)x∈(0,+∞)時,要證lnex只需證ex-1>xex令F(x)=ex-1-xex2=ex-x(e)xF'(x)=ex-(e)x-x(e)xlne=(e)x(e)x-1-x2=ex2ex2-1由ex>x+1可得,ex2>1+則x∈(0,+∞)時,F'(x)>0恒成立,即F(x)在(0,+∞)內(nèi)單調(diào)遞增,∴F(x)>F(0)=0.即ex-1>xex2,∴l(xiāng)n4.(2019山東泰安二模)已知函數(shù)f(x)=(x-m)lnx(m≤0).(1)若函數(shù)f(x)存在極小值點(diǎn),求m的取值范圍;(2)當(dāng)m=0時,證明:f(x)<ex-1.(1)解函數(shù)的定義域?yàn)?0,+∞),f'(x)=x-mx+lnx=1-mx①當(dāng)m=0時,f'(x)=0得x=1e,當(dāng)x∈0,1e時,f'(x)<0;當(dāng)x∈1e,+∞時,f'(x)>0,∴x=1e是函數(shù)f(x)的極小值點(diǎn),滿足題意②當(dāng)m<0時,令g(x)=f'(x),g'(x)=mx令g'(x)=0,解得x=-m.當(dāng)x∈(0,-m)時,g'(x)<0,當(dāng)x∈(-m,+∞)時,g'(x)>0.∴g(x)min=g(-m)=2+ln(-m),若g(-m)≥0,即m≤-e-2,則f'(x)=g(x)≥0恒成立,∴f(x)在(0,+∞)上單調(diào)遞增,無極值點(diǎn),不滿足題意.若g(-m)=2+ln(-m)<0,即-e-2<m<0,g(1-m)=1-m1-m+ln(1-m∴g(-m)·g(1-m)<0.又g(x)在(-m,+∞)上單調(diào)遞增,∴g(x)在(-m,+∞)上恰有一個零點(diǎn)x1.當(dāng)x∈(-m,x1)時,f'(x)=g(x)<0,當(dāng)x∈(x1,+∞)時,f'(x)=g(x)>0,∴x1是f(x)的極小值點(diǎn),滿足題意,綜上,-e-2<m≤0.(2)證明當(dāng)m=0時,f(x)=xlnx.①當(dāng)x∈(0,1]時,ex-1>0,xlnx≤0,∴f(x)<ex-1;②當(dāng)x∈(1,+∞)時,令h(x)=ex-xlnx-1,h'(x)=ex-lnx-1,令φ(x)=h'(x),則φ'(x)=ex-1x∵φ'(x)在(1,+∞)上是增函數(shù),∴φ'(x)>φ'(1)=e-1>0,∴φ(x)在(1,+∞)上單調(diào)遞增,h'(x)=φ(x)>φ(1)=e-1>0,∴h(x)在(1,+∞)上單調(diào)遞增,∴h(x)>h(1)=e-1>0,∴當(dāng)x>1時,xlnx<ex-1成立,綜上,f(x)<ex-1.命題角度5恒成立與存在性問題高考真題體驗(yàn)·對方向(2019浙江·22)已知實(shí)數(shù)a≠0,設(shè)函數(shù)f(x)=alnx+1+x,x>0(1)當(dāng)a=-34時,求函數(shù)f(x(2)對任意x∈1e2,+∞均有f(x)≤x2a,求a注:e=2.71828…為自然對數(shù)的底數(shù).解(1)當(dāng)a=-34時,f(x)=-34lnx+1+x,f'(x)=-3=(1+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,f(x)≤x2a等價于xa2令t=1a,則t≥22設(shè)g(t)=t2x-2t1+x-2lnx,t≥22g(t)=xt-1+1x2-1+xx-2ln①當(dāng)x∈17,+∞時,1+1x≤22g(t)≥g(22)=8x-421+x-2ln記p(x)=4x-221+x-lnx,x≥p'(x)=2=2=(x故x117,11(1,+∞)p'(x)-0+p(x)p17單調(diào)遞減極小值p(1)單調(diào)遞增所以,p(x)≥(1)=0.因此,g(t)≥g(22)=2p(x)≥0.②當(dāng)x∈1e2,g(t)≥g1+1x=-2令q(x)=2xlnx+(x+1),x∈1e2,q'(x)=lnx+2x+故q(x)在1e2,所以q(x)≤q17.由①得,q17=-277p17<-277p(1)=所以,q(x)<0.因此,g(t)≥g1+1x=-q(x)由①②知對任意x∈1e2,+∞,t∈[22,+∞),g(t)≥0,即對任意x∈1e2,+∞,均有f(x)≤x2綜上所述,所求a的取值范圍是0,24.典題演練提能·刷高分1.(2019河南八市重點(diǎn)高中高三五模)已知函數(shù)f(x)=x(lnx+a)+b,曲線y=f(x)在點(diǎn)(1,f(1))處的切線為2x-y-1=0.(1)求a,b的值;(2)若對任意的x∈(1,+∞),f(x)≥m(x-1)恒成立,求正整數(shù)m的最大值.解(1)由f(x)=x(lnx+a)+b,得f'(x)=lnx+a+1,由切線方程可知:f(1)=2-1=1,∴f解得a(2)由(1)知f(x)=x(lnx+1),則x∈(1,+∞)時,f(x)≥m(x-1)恒成立等價于x∈(1,+∞)時,m≤x(ln令g(x)=x(lnx則g'(x)=x-令h(x)=x-lnx-2,則h'(x)=1-1x=x-1x,∴當(dāng)x∈(1,+∞)時,h'(x)∵h(yuǎn)(3)=1-ln3<0,h(4)=2-2ln2>0,∴?x0∈(3,4),使得h(x0)=0.當(dāng)x∈(1,x0)時,g'(x)<0;x∈(x0,+∞)時,g'(x)>0,∴g(x)min=g(x0)=x0∵h(yuǎn)(x0)=x0-lnx0-2=0,∴l(xiāng)nx0=x0-2.∴g(x)min=g(x0)=x0(x0-∴m≤x0∈(3,4),即正整數(shù)m的最大值為3.2.已知函數(shù)f(x)=ex-1,g(x)=lnx+a.(1)設(shè)F(x)=xf(x),求F(x)的最小值;(2)證明:當(dāng)a<1時,總存在兩條直線與曲線y=f(x)與y=g(x)都相切