2025屆高考數(shù)學(xué)二輪總復(fù)習(xí)專題1函數(shù)與導(dǎo)數(shù)專題突破練5利用導(dǎo)數(shù)證明問題課件

專題突破練5利用導(dǎo)數(shù)證明問題12341.(17分)(2024廣東廣州一模)已知函數(shù)f(x)=cosx+xsinx,x∈(-π,π).(1)求f(x)的單調(diào)區(qū)間和極小值;(2)證明:當(dāng)x∈[0,π)時(shí),2f(x)≤ex+e-x.12341234(2)證明

當(dāng)x∈[0,π)時(shí),令F(x)=ex+e-x-2(cos

x+xsin

x),求導(dǎo)得F'(x)=ex-e-x-2xcos

x≥ex-e-x-2x,令φ(x)=ex-e-x-2x,求導(dǎo)得φ'(x)=ex+e-x-2≥函數(shù)φ(x)在[0,π)內(nèi)單調(diào)遞增,則φ(x)≥φ(0)=0,F'(x)≥0,F(x)在[0,π)內(nèi)單調(diào)遞增,因此F(x)≥F(0)=0,所以2f(x)≤ex+e-x.12341234設(shè)g(x)=-x2+ax-1,注意到g(0)=-1,①當(dāng)a≤0時(shí),g(x)<0恒成立,即f'(x)<0恒成立,此時(shí)函數(shù)f(x)在(0,+∞)內(nèi)單調(diào)遞減;②當(dāng)a>0時(shí),判別式Δ=a2-4,(ⅰ)當(dāng)0<a≤2時(shí),Δ≤0,即g(x)≤0,即f'(x)≤0恒成立,此時(shí)函數(shù)f(x)在(0,+∞)內(nèi)單調(diào)遞減;12341234123412341234(1)解

g(x)≥f(x)在[0,+∞)內(nèi)恒成立,理由如下:令h(x)=g(x)-f(x)=-1+cos

x,x∈[0,+∞),則h'(x)=x-sin

x,x∈[0,+∞),令q(x)=h'(x),則q'(x)=1-cos

x≥0在[0,+∞)內(nèi)恒成立,故q(x)=h'(x)在[0,+∞)內(nèi)單調(diào)遞增,其中h'(0)=0,故h'(x)≥0在[0,+∞)內(nèi)恒成立,故h(x)在[0,+∞)內(nèi)單調(diào)遞增,故h(x)≥h(0)=0,即g(x)≥f(x)恒成立.1234只需證(m-n)cos

n+sin

n-sin

m>0.令r(x)=(x-n)cos

n-sin

x+sin

n,0<x<n,則只需證明r(m)>0,r'(x)=cos

n-cos

x,令p(x)=cos

n-cos

x,則函數(shù)p(x)在(0,)內(nèi)單調(diào)遞增,所以當(dāng)0<x<n時(shí),p(x)<p(n)=0,所以r'(x)<0,所以r(x)在(0,n)內(nèi)單調(diào)遞減,所以r(x)>r(n)=0,故r(m)>r(n)=0,123412344.(17分)(2024福建廈門模擬)已知函數(shù)f(x)=aex+2x-1(其中常數(shù)e=2.71828…是自然對數(shù)的底數(shù)).(1)討論函數(shù)f(x)的單調(diào)性;(2)證明:對任意的a≥1,當(dāng)x>0時(shí),f(x)≥(x+ae)x.1234(1)解

由f(x)=aex+2x-1,得f'(x)=aex+2.①當(dāng)a≥0時(shí),f'(x)>0,函數(shù)f(x)在R上單調(diào)遞增;1234當(dāng)a≥1時(shí),aex-x-1≥ex-x-1.令h(x)=ex-x-1,則當(dāng)x>0時(shí),h'(x)=ex-1>0,h(x)單調(diào)遞增,因此h(x)>h(0)=0,于是當(dāng)0<x<1時(shí),g'(x)<0,g(x)單調(diào)遞減,當(dāng)x=1時(shí),g'(x)=0,當(dāng)x>1時(shí),g'