高考數(shù)學(xué)藝體生文化課總復(fù)習(xí)第六章導(dǎo)數(shù)第4節(jié)導(dǎo)數(shù)綜合解答題點(diǎn)金課件

第六章

導(dǎo)數(shù)第4節(jié)導(dǎo)數(shù)綜合解答題1.(2020新課標(biāo)Ⅰ卷,文)已知函數(shù)f(x)=ex-a(x+2).當(dāng)a=1時(shí),討論f(x)的單調(diào)性.【解析】當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.2.(2020新課標(biāo)Ⅰ卷,理)已知函數(shù)f(x)=ex+ax2-x.當(dāng)a=1時(shí),討論f(x)的單調(diào)性.【解析】當(dāng)a=1時(shí),f(x)=ex+x2-x,則f'(x)=ex+2x-1.故當(dāng)x∈(-∞,0)時(shí),f'(x)<0;當(dāng)x∈(0,+∞)時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.3.(2018深圳模擬)設(shè)函數(shù)f(x)=ex-1-alnx,其中e為自然對(duì)數(shù)的底數(shù).若a=1,求f(x)的單調(diào)區(qū)間.【解析】若a=1,則f(x)=ex-1-lnx(x>0),∴f'(x)=(x>0).令t(x)=xex-1-1(x>0),則t'(x)=(x+1)ex-1(x>0),當(dāng)x>0時(shí),t'(x)>0,即t(x)單調(diào)遞增,又t(1)=0,∴當(dāng)x∈(0,1)時(shí),t(x)<0,f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),t(x)>0,f'(x)>0,f(x)單調(diào)遞增.∴f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞).

4.(2020新課標(biāo)Ⅱ卷,理)已知函數(shù)f(x)=sin2xsin2x.討論f(x)在區(qū)間(0,π)的單調(diào)性.5.(2017廣州模擬)已知函數(shù)f(x)=lnx+(a>0).若函數(shù)f(x)有零點(diǎn),求實(shí)數(shù)a的取值范圍.6.(2018惠州模擬)已知函數(shù)f(x)=4lnx-mx2+1(m∈R).討論函數(shù)f(x)的單調(diào)性.7.(2015新課標(biāo)Ⅱ卷)設(shè)函數(shù)f(x)=emx+x2-mx.證明:f(x)在(-∞,0)單調(diào)遞減,在(0,+∞)單調(diào)遞增.【證明】

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)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.8.(2020新課標(biāo)Ⅱ卷,文)已知函數(shù)f(x)=2lnx+1.若f(x)≤2x+c,求c的取值范圍.【解析】設(shè)h(x)=f(x)-2x-c,則h(x)=2lnx-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)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(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,+∞).9.已知函數(shù)f(x)=lnx,h(x)=ax(a∈R).函數(shù)f(x)與h(x)的圖象無(wú)公共點(diǎn),試求實(shí)數(shù)a的取值范圍.x(0,e)e(e,+∞)t'(x)+0-t(x)單調(diào)遞增極大值單調(diào)遞減10.已知函數(shù)f(x)=lnx-ax2+x,a∈R.討論函數(shù)f(x)的單調(diào)性.11.(2016新課標(biāo)Ⅰ卷,理)已知函數(shù)f(x)=(x-2)ex+a(x-1)2有兩個(gè)零點(diǎn).求a的取值范圍.12.已知函數(shù)f(x)=lnx-(a+1)x,其中a∈R.試討論函數(shù)f(x)的單調(diào)性及最值.13.(2017新課標(biāo)Ⅱ卷)已知函數(shù)f(x)=ax2-ax-xlnx,且f(x)≥0.求a.14.(2017新課標(biāo)Ⅲ卷)已知函數(shù)f(x)=lnx+ax2+(2a+1)x.討論f(x)的單調(diào)性.15.(2018新課標(biāo)Ⅱ卷)已知函數(shù)f(x)=x3-a(x2+x+1).(1)若a=3,求f(x)的單調(diào)區(qū)間;15.(2018新課標(biāo)Ⅱ卷)已知函數(shù)f(x)=x3-a(x2+x+1).(2)證明:f(x)只有一個(gè)零點(diǎn).16.(2018新課標(biāo)Ⅰ卷)已知函數(shù)f(x)=aex-lnx-1.(1)設(shè)x=2是f(x)的極值點(diǎn).求a,并求f(x)的單調(diào)區(qū)間;16.(2018新課標(biāo)Ⅰ卷)已知函數(shù)f(x)=aex-lnx-1.(2)證明