高考二輪復(fù)習(xí)文科數(shù)學(xué)課件（老高考舊教材）增分2利用導(dǎo)數(shù)證明不等式

增分2利用導(dǎo)數(shù)證明不等式考點(diǎn)一不含參數(shù)的一元不等式的證明例1設(shè)函數(shù)f(x)=aexlnx+

,曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y=e(x-1)+2.(1)求a,b的值;(2)證明:f(x)>1.先證ex≥ex,x∈R,設(shè)d(x)=ex-ex,則d'(x)=ex-e,令d'(x)=0,則x=1.當(dāng)x∈(-∞,1)時(shí),d'(x)<0,故d(x)在(-∞,1)上單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),d'(x)>0,故d(x)在(1,+∞)上單調(diào)遞增.∴[d(x)]min=d(1)=0,故d(x)≥0,即ex≥ex,x∈R(當(dāng)且僅當(dāng)x=1時(shí)取等號(hào)).規(guī)律方法利用導(dǎo)數(shù)證明不等式f(x)>g(x)的基本方法

對(duì)點(diǎn)訓(xùn)練1(2023山東濰坊一模)已知函數(shù)f(x)=ex-1lnx,g(x)=x2-x.(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)x∈(0,2)時(shí),f(x)≤g(x).對(duì)點(diǎn)訓(xùn)練2已知函數(shù)f(x)=ex-x-1.(1)求函數(shù)f(x)的單調(diào)區(qū)間和極值;(2)當(dāng)x≥0時(shí),求證:f(x)+x+1≥

x2+cosx.(1)解

函數(shù)f(x)的定義域?yàn)镽,由f(x)=ex-x-1,得f'(x)=ex-1,令f'(x)=ex-1>0,得x>0,f(x)在(0,+∞)上單調(diào)遞增;令f'(x)=ex-1<0,得x<0,f(x)在(-∞,0)上單調(diào)遞減;即函數(shù)f(x)的極小值為f(0)=0,f(x)沒(méi)有極大值.考點(diǎn)二含參數(shù)的一元不等式的證明解題技巧含參數(shù)的一元函數(shù)不等式的證明的一般思路是:首先利用參數(shù)的范圍,通過(guò)放縮法把參數(shù)消去,將問(wèn)題轉(zhuǎn)化為不含參數(shù)的一元不等式的證明,再利用證明不含參數(shù)的一元不等式的方法證明.對(duì)點(diǎn)訓(xùn)練3已知函數(shù)f(x)=3lnx+1+.(1)求函數(shù)f(x)的圖象在點(diǎn)(1,f(1))處的切線方程;(2)若0<a≤1,g(x)=f(x)-2ax,證明:當(dāng)x∈(0,1)時(shí),g(x)≥2-3ln2.對(duì)點(diǎn)訓(xùn)練4(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;(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)↘極小值↗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).隨a的變化,g'(a),g(a)的變化如下表:考點(diǎn)三二元不等式的證明解題技巧

對(duì)點(diǎn)訓(xùn)練5(2023浙江寧波二模改編)已知函數(shù)f(x)=lnx-ax2.(1)討論函數(shù)f(x)的單調(diào)性;對(duì)點(diǎn)訓(xùn)練6(2021新高考Ⅰ,22)已知函數(shù)f(x)=x(1-lnx).(1)討論f(x)的單調(diào)性;(1)解

由條件知,函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=-ln

x.當(dāng)x∈(0,1)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.即在區(qū)間(0,1)內(nèi),函數(shù)f(x)單調(diào)遞增;在區(qū)間(1,+∞)內(nèi),函數(shù)f(x)單調(diào)遞減.令f'(x)=0,得x=1.且f(e)=0.結(jié)合(1)中的f(x)的單調(diào)性,易知,0<x1<1<x2<e.待證結(jié)論?2<x1+x2<e.下面證明x1+x2>2.令g(x)=f(x)-f(2-x),x∈(0,1),則g'(x)=-ln(x(2-x))>0,所以g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,所以0=g(1)>g(x1)=f(x1)-f(2-x1),即f(2-x1)>f(x1)=f(x2).又f(x)在區(qū)間(1,+∞)上單調(diào)遞減,所以2-x1<x2,即x1+x2>2.再證明x1+x2<e.(方法一)當(dāng)x2≤e-1時(shí),結(jié)論顯然成立;當(dāng)x2∈(e-1,e)時(shí),x1<e-x2?f(x1)<f(e-x2)?f(x2)<f(e-x2),x2∈(e-1,e),令h(x)=f(x)-f(e-x),x∈(e-1,e),h'(x)=-ln(x(e-x)),則h(x)在區(qū)間(e-1,e)內(nèi)先單調(diào)遞減后單調(diào)遞增,故h(x)<0.對(duì)x∈(e-1,e),則h(x2)<0,即f(x2)<f(e-x2).故f(x1)<f(e-x2),結(jié)合當(dāng)x∈(0,1)時(shí),f(x)單調(diào)遞增,有x1<e-x2,即x1+x2<e.(方法二)f(x)在點(diǎn)(e,0)處的切線φ(x)=e-x,令F(x)=f(x)-φ(x)=2x-xln

x-e,x∈(0,e),F'(x)=1-ln

x>0,所以F(x)在區(qū)間(0,e)內(nèi)單調(diào)遞增,即F(x)<F(e)=0,所以當(dāng)x∈(0,e)時(shí),f(x)<φ(x).令t=f