2025屆高考數(shù)學二輪總復習專題突破練5利用導數(shù)證明問題

專題突破練5利用導數(shù)證明問題(分值:68分)1.(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)的單調區(qū)間和極小值;(2)證明:當x∈[0,π)時,2f(x)≤ex+e-x.(1)解函數(shù)f(x)=cosx+xsinx,x∈(-π,π),求導得f'(x)=-sinx+sinx+xcosx=xcosx,當-π<x<-π2時,f'(x)>0,f(x)單調遞增;當-π2<x<0時,f'(x)<0,f(x)當0<x<π2時,f'(x)>0,f(x)單調遞增;當π2<x<π時,f'(x)<0,f(x)所以f(x)的單調遞增區(qū)間為-π,-π2,0,π2,單調遞減區(qū)間為-π2,0,π2,π,f(x)的極小值為f(0)=1.(2)證明當x∈[0,π)時,令F(x)=ex+e-x-2(cosx+xsinx),求導得F'(x)=ex-e-x-2xcosx≥ex-e-x-2x,令φ(x)=ex-e-x-2x,求導得φ'(x)=ex+e-x-2≥2ex·e-函數(shù)φ(x)在[0,π)內單調遞增,則φ(x)≥φ(0)=0,F'(x)≥0,F(x)在[0,π)內單調遞增,因此F(x)≥F(0)=0,所以2f(x)≤ex+e-x.2.(17分)(2024陜西西安一模)已知函數(shù)f(x)=alnx+1x-x+m(a,m∈R)(1)討論f(x)的單調性;(2)若f(x)存在兩個極值點x1,x2,證明:f(x1(1)解函數(shù)f(x)的定義域為(0,+∞),f'(x)=ax-1x2設g(x)=-x2+ax-1,注意到g(0)=-1,①當a≤0時,g(x)<0恒成立,即f'(x)<0恒成立,此時函數(shù)f(x)在(0,+∞)內單調遞減;②當a>0時,判別式Δ=a2-4,(ⅰ)當0<a≤2時,Δ≤0,即g(x)≤0,即f'(x)≤0恒成立,此時函數(shù)f(x)在(0,+∞)內單調遞減;(ⅱ)當a>2時,令f'(x)>0,得a-a2令f'(x)<0,得0<x<a-a2-所以當a>2時,f(x)在區(qū)間(a-a2-42,a+a2-4綜上所述,當a≤2時,f(x)在(0,+∞)內單調遞減,當a>2時,在區(qū)間(0,a-a2-42),(在區(qū)間(a-a2(2)證明由(1)知a>2,0<x1<1<x2,x1x2=1,則f(x1)-f(x2)=alnx1+1x1-x1-alnx2-1x2+x2=(x2-x1)(1+1x1x2)+a(lnx1-lnx2)=2(x2-x1)+a(lnx1-lnx2),則f(x1即證明lnx1-lnx2>x1-x2,則lnx1-ln1x1>x1-即lnx1+lnx1>x1-1x1,即證2lnx1>x1-1x1設h(x)=2lnx-x+1x,0<x<1,其中h(1)=求導得h'(x)=2x-1-1x2=-x2則h(x)在(0,1)上單調遞減,所以h(x)>h(1),即2lnx-x+1x>故2lnx>x-1x,則f(x13.(17分)(2024河南鄭州一模)已知函數(shù)f(x)=-cosx,g(x)=x22-1,x∈[0,+∞(1)判斷g(x)≥f(x)是否對?x∈[0,+∞)恒成立,并給出理由;(2)證明:①當0<m<n<π2時,sinm-sin②當ai=12i(i∈N*),ki=f'(ai+1)-f'(ai)a(1)解g(x)≥f(x)在[0,+∞)內恒成立,理由如下:令h(x)=g(x)-f(x)=x22-1+cosx,x∈[0,則h'(x)=x-sinx,x∈[0,+∞),令q(x)=h'(x),則q'(x)=1-cosx≥0在[0,+∞)內恒成立,故q(x)=h'(x)在[0,+∞)內單調遞增,其中h'(0)=0,故h'(x)≥0在[0,+∞)內恒成立,故h(x)在[0,+∞)內單調遞增,故h(x)≥h(0)=0,即g(x)≥f(x)恒成立.(2)證明①0<m<n<π2時,因為y=sinx單調遞增,所以sinm<sinn又m-n<0,cosn>0,故要證sinm-sinn只需證(m-n)cosn+sinn-sinm>0.令r(x)=(x-n)cosn-sinx+sinn,0<x<n,則只需證明r(m)>0,r'(x)=cosn-cosx,令p(x)=cosn-cosx,則函數(shù)p(x)在0,π2內單調遞增,所以當0<x<n時,p(x)<p(n)=0,所以r'(x)<0,所以r(x)在(0,n)內單調遞減,所以r(x)>r(n)=0,故r(m)>r(n)=0,所以當0<m<n<π2時,sinm-②由(1)知,cosx>1-x22,x>0,f'(x)=sinx,由于0<所以ki=f'(ai+1)-f'(a所以∑i=1n-1ki>(1-123)+(1-125)+…+(1-122n-1)=n-1-(4.(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)的單調性;(2)證明:對任意的a≥1,當x>0時,f(x)≥(x+ae)x.(1)解由f(x)=aex+2x-1,得f'(x)=aex+2.①當a≥0時,f'(x)>0,函數(shù)f(x)在R上單調遞增;②當a<0時,由f'(x)>0,解得x<ln-2a,由f'(x)<0,解得x>ln-2a,故f(x)在-∞,ln-2a內單調遞增,在ln-2a,+∞內單調遞減.綜上所述,當a≥0時,函數(shù)f(x)在R上單調遞增;當a<0時,f(x)在-∞,ln-2a內單調遞增,在ln-2a,+∞內單調遞減.(2)證明f(x)≥(x+ae)x,其中a≥1,x>0?exx-xa-1ax+2a-e≥0.令g(x)=當a≥1時,aex-x-1≥ex-x-1.令h(x)=ex-x-1,則當x>0時,h