高考二輪復(fù)習(xí)理科數(shù)學(xué)課件考點(diǎn)突破練21利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)問題

考點(diǎn)突破練21利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)問題1.(2023江西南昌一模)已知函數(shù)f(x)=(x-a)2+bex(a,b∈R).(1)若a=0時(shí),函數(shù)y=f(x)有2個(gè)極值點(diǎn),求b的取值范圍;解

(1)a=0時(shí),f(x)=x2+bex,f'(x)=2x+bex,則2x+bex=0有兩實(shí)根,(2)當(dāng)a>2時(shí),①求證:f(x)有唯一的極值點(diǎn)x1;∴g(x)在(0,+∞)上單調(diào)遞減.∵g(e-a)=1+ea>0,g(1)=2-a<0,∴g(x)在(0,+∞)上有唯一的零點(diǎn),∴f'(x)=0在(0,+∞)上有唯一解,不妨設(shè)為x1,x1∈(e-a,1).f'(x)與f(x)的情況如下,x(0,x1)x1(x1,+∞)f'(x)+0-f(x)單調(diào)遞增極大值單調(diào)遞減∴f(x)有唯一的極值點(diǎn)x1.3.已知函數(shù)f(x)=ex+cosx-ex,f'(x)是f(x)的導(dǎo)函數(shù).(1)證明:函數(shù)f(x)只有一個(gè)極值點(diǎn);(2)若關(guān)于x的方程f(x)=t(t∈R)在(0,π)上有兩個(gè)不相等的實(shí)數(shù)根x1,x2,證明:

4.(2023江西上饒一模)已知f(x)=ex-ax,g(x)=ex(1-sinx).(1)討論f(x)的單調(diào)性;(2)若a∈(0,3),h(x)=f(x)-g(x),試討論h(x)在(0,π)內(nèi)的零點(diǎn)個(gè)數(shù).(參考數(shù)據(jù)解

(1)由已知可知f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,f(x)在R上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=0,則x=ln

a,當(dāng)x∈(-∞,ln

a)時(shí),f'(x)<0,f(x)在(-∞,ln

a)上單調(diào)遞減,當(dāng)x∈(ln

a,+∞)時(shí),f'(x)>0,f(x)在(ln

a,+∞)上單調(diào)遞增.綜上,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞增,當(dāng)a>0時(shí),f(x)在(-∞,ln

a)上單調(diào)遞減,在(ln

a,+∞)上單調(diào)遞增.由零點(diǎn)存在性定理可得,h(x)在(x1,x2)和(x2,π)上各有一個(gè)零點(diǎn),即h(x)在(0,π)上有2個(gè)零點(diǎn),綜上所述,當(dāng)0<a≤1時(shí),h(x)在(0,π)上僅有一個(gè)零點(diǎn),當(dāng)1<a<3時(shí),h(x)在(0,π)上有2個(gè)零點(diǎn).5.(2023江西南昌二模)已知函數(shù)f(x)=a(x2-1)-lnx(x>0).當(dāng)0<x<1時(shí),f'(x)<0,則f(x)在(0,1)上單調(diào)遞減;當(dāng)x>1時(shí),f'(x)>0,則f(x)在(1,+∞)上單調(diào)遞增;∴f(x)的極小值為f(1)=0,無極大值.6.已知函數(shù)f(x)=ex-a(x+cosx),其中a>0,且滿足對(duì)?x∈[0,+∞)時(shí),f(x)≥0恒成立.(1)求實(shí)數(shù)a的取值范圍;解

(1)一方面,當(dāng)x=0時(shí),f(0)=1-a≥0,所以0<a≤1.另一方面,f'(x)=ex-a(1-sin

x)=ex+asin

x-a.令φ(x)=f'(x),則φ'(x)=ex+acos

x.當(dāng)x∈[0,+∞)時(shí),φ'(x)>0,所以φ(x)即f'(x)在[0,+∞)上單調(diào)遞增.又因?yàn)閒'(0)