2025高考數(shù)學(xué)二輪復(fù)習(xí)-專(zhuān)題1 函數(shù)與導(dǎo)數(shù) 第4講 利用導(dǎo)數(shù)研究不等式【課件】

第4講利用導(dǎo)數(shù)研究不等式2025基礎(chǔ)回扣?考教銜接以題梳點(diǎn)?核心突破目錄索引

基礎(chǔ)回扣?考教銜接1.(人A選必二5.3節(jié)習(xí)題改編)證明下列不等式:(1)ex>1+x,x≠0;(2)lnx<x<ex,x>0.證明

(1)設(shè)f(x)=ex-1-x,x∈R,則f'(x)=ex-1.令f'(x)=0,解得x=0.當(dāng)x>0時(shí),f'(x)>0,函數(shù)f(x)=ex-1-x在區(qū)間(0,+∞)上單調(diào)遞增;當(dāng)x<0時(shí),f'(x)<0,函數(shù)f(x)=ex-1-x在區(qū)間(-∞,0)上單調(diào)遞減.所以,當(dāng)x=0時(shí),f(x)取得最小值.所以,當(dāng)x≠0時(shí),f(x)>f(0)=0,即ex>1+x,x≠0.(2)設(shè)f(x)=ln

x-x,x>0,則f'(x)=-1.令f'(x)=0,解得x=1.當(dāng)0<x<1時(shí),f'(x)>0,f(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增;當(dāng)x>1時(shí),f'(x)<0,f(x)在區(qū)間(1,+∞)上單調(diào)遞減.所以,當(dāng)且僅當(dāng)x=1時(shí),f(x)取得最大值.所以,當(dāng)x>0時(shí),f(x)=ln

x-x≤f(1)=-1<0.因此,當(dāng)x>0時(shí),ln

x<x.由(1)可知,當(dāng)x>0時(shí),ex>x+1>x.綜上,ln

x<x<ex,x>0.2.(人A選必二復(fù)習(xí)第五章習(xí)題)已知函數(shù)f(x)=ex-ln(x+m).當(dāng)m≤2時(shí),求證:f(x)>0.證明

當(dāng)m≤2,x∈(-m,+∞)時(shí),ln(x+m)≤ln(x+2),則有f(x)=ex-ln(x+m)≥ex-ln(x+2).故只需證明當(dāng)m=2時(shí),f(x)>0.又f'(-1)<0,f'(0)>0,故f'(x)=0在區(qū)間(-2,+∞)上有唯一實(shí)根x0,且x0∈(-1,0).當(dāng)x∈(-2,x0)時(shí),f'(x)<0;當(dāng)x∈(x0,+∞)時(shí),f'(x)>0,從而當(dāng)x=x0時(shí),f(x)取得最小值.綜上,當(dāng)m≤2時(shí),f(x)>0.3.(人B選必三第六章習(xí)題)已知函數(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,求實(shí)數(shù)a的取值范圍.真題體驗(yàn)1.(2020·全國(guó)Ⅱ,文21(1))已知函數(shù)f(x)=2lnx+1.若f(x)≤2x+c,求c的取值范圍.解

設(shè)h(x)=f(x)-2x-c,則h(x)=2ln

x-2x+1-c,其定義域?yàn)?0,+∞),h'(x)=2.當(dāng)0<x<1時(shí),h'(x)>0;當(dāng)x>1時(shí),h'(x)<0.所以h(x)在區(qū)間(0,1)單調(diào)遞增,在區(qū)間(1,+∞)單調(diào)遞減.從而當(dāng)x=1時(shí),h(x)取得最大值,最大值為h(1)=-1-c.故當(dāng)且僅當(dāng)-1-c≤0,即c≥-1時(shí),f(x)≤2x+c.所以c的取值范圍為[-1,+∞).2.(2023·新高考Ⅰ,19)已知函數(shù)f(x)=a(ex+a)-x.(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)a>0時(shí),f(x)>2lna+.(1)解

f'(x)=aex-1,x∈R.①當(dāng)a≤0時(shí),f'(x)≤0對(duì)任意x∈R恒成立,所以f(x)在(-∞,+∞)內(nèi)單調(diào)遞減.②當(dāng)a>0時(shí),令f'(x)=0,得x=ln=-ln

a.隨x的變化,f'(x),f(x)的變化如下表:x(-∞,-ln

a)-ln

a(-ln

a,+∞)f'(x)-0+f(x)單調(diào)遞減極小值單調(diào)遞增所以函數(shù)f(x)的單調(diào)遞增區(qū)間是(-ln

a,+∞),單調(diào)遞減區(qū)間是(-∞,-ln

a).綜上,當(dāng)a≤0時(shí),f(x)的單調(diào)遞減區(qū)間是(-∞,+∞),無(wú)單調(diào)遞增區(qū)間;當(dāng)a>0時(shí),f(x)的單調(diào)遞增區(qū)間是(-ln

a,+∞),單調(diào)遞減區(qū)間是(-∞,-ln

a).以題梳點(diǎn)?核心突破考點(diǎn)一利用導(dǎo)數(shù)證明不等式例1(2024·全國(guó)甲,文20)已知函數(shù)f(x)=a(x-1)-lnx+1.(1)求f(x)的單調(diào)區(qū)間;(2)若a≤2時(shí),證明:當(dāng)x>1時(shí),f(x)<ex-1.(2)證明

∵a≤2,∴當(dāng)x>1時(shí),ex-1-f(x)=ex-1-a(x-1)+ln

x-1≥ex-1-2x+ln

x+1,x>0.h'(x)在區(qū)間(1,+∞)上單調(diào)遞增,則h'(x)>h'(1)=0,即h(x)在區(qū)間(1,+∞)上單調(diào)遞增,故g'(x)>g'(1)=0,即g(x)在區(qū)間(1,+∞)上單調(diào)遞增,∴g(x)>g(1)=0,即當(dāng)x>1時(shí),f(x)<ex-1.[對(duì)點(diǎn)訓(xùn)練1](2021·全國(guó)乙,理20)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),證明:g(x)<1.(體現(xiàn)定義域分類(lèi)討論變形)所以當(dāng)x<0時(shí),y'>0,y=xf(x)單調(diào)遞增,當(dāng)x>0時(shí),y'<0,y=xf(x)單調(diào)遞減,所以x=0是函數(shù)y=xf(x)的極值點(diǎn),符合題意.所以a=1.因?yàn)閤>0,ln(1-x)<0,所以xln(1-x)<0,即證x+ln(1-x)>xln(1-x),即證x+(1-x)ln(1-x)>0.同理,當(dāng)x∈(-∞,0)時(shí),因?yàn)閤<0,ln(1-x)>0,所以xln(1-x)<0,即證x+ln(1-x)>xln(1-x),即證x+(1-x)ln(1-x)>0.令h(x)=x+(1-x)ln(1-x),x<1且x≠0,再令t=1-x,則t∈(0,1)∪(1,+∞),x=1-t,令φ(t)=1-t+tln

t,φ'(t)=-1+ln

t+1=ln

t,當(dāng)t∈(0,1)時(shí),φ'(t)<0,φ(t)單調(diào)遞減,故φ(t)>φ(1)=0;當(dāng)t∈(1,+∞)時(shí),φ'(t)>0,φ(t)單調(diào)遞增,故φ(t)>φ(1)=0.x+(1-x)ln(1-x)>0即證.考點(diǎn)二不等式恒(能)成立問(wèn)題當(dāng)m≤0時(shí),f'(x)>0恒成立,此時(shí)f(x)單調(diào)遞增,f(x)無(wú)極值;當(dāng)m>0時(shí),令f'(x)=0,得x=m.故當(dāng)x∈(0,m)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(m,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,此時(shí)f(x)在x=m處取到極小值ln

m-m+1,無(wú)極大值.[對(duì)點(diǎn)訓(xùn)練2](1)(能成立問(wèn)題)(2024·湖北模擬預(yù)測(cè)改編)若?x∈(0,+∞),使lnx≤成立,求a的最小值.解

由題意a≥x(ln

x+1)在(0,+∞)上有解,故當(dāng)x∈(0,+∞)時(shí),a≥[x(ln

x+1)]min.令h(x)=x(ln

x+1),x>0,h'(x)=ln

x+2,(2)(2024·山東濟(jì)南三模)已知函數(shù)f(x)=ax+2x-2,其中a>0且a≠1.①若f(x)是偶函數(shù),求a的值;②