備戰(zhàn)2025年高考二輪復(fù)習(xí)數(shù)學(xué)專題突破練4

專題突破練(分值:60分)學(xué)生用書P1411.(13分)(2024·陜西二模)證明下列兩個(gè)不等式:(1)-xln(-x)≥-1e(2)ex-x+xln(-x)<3.證明(1)令f(x)=-xln(-x),定義域?yàn)?-∞,0),f'(x)=-ln(-x)+(-x)·1-x×(-1)=-ln(-x)-令f'(x)>0,得-1e<x<0,此時(shí)函數(shù)f(x)單調(diào)遞增,令f'(x)<0,得x<-1e,此時(shí)函數(shù)f(x)所以f(x)≥f-1e=-1e(2)令h(x)=ex-x+xln(-x)(x<0),則h'(x)=ex+ln(-x).當(dāng)x<-1時(shí),h'(x)=ex+ln(-x)>0.當(dāng)-1<x<0時(shí),令φ(x)=h'(x),則φ'(x)=ex+1x.因?yàn)?x<-1,ex<1,所以φ'(x)<0,即h'(x)又h'(-1)=1e>0,h'-1e=e-1e+ln1e=e-1e-1<0,所以存在當(dāng)x∈(-∞,x0)時(shí),h'(x)>0,函數(shù)h(x)單調(diào)遞增,當(dāng)x∈(x0,0)時(shí),h'(x)<0,函數(shù)h(x)單調(diào)遞減,即可得h(x)max=h(x0)=ex0-x0+x0ln(-x0因?yàn)閔'(x0)=0,所以ex0=-ln(-x即x0ex0=-x0ln(-x0)=ex0lnex0,由(1)可知f(x0)=f(-ex0).因?yàn)閤0∈-1,-1e,所以-e所以x0=-ex0,同時(shí)ln(-x0)=x0,可得h(x0)=x02-2x0.因?yàn)閤0∈-1,-1e,所以h(x0)=x02-又因?yàn)閔(x)≤h(x0),所以h(x)<3.2.(15分)(2024·湖南婁底一模)已知函數(shù)f(x)=xex,其中e=2.71828…(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)證明:f(x)≤ex-1;(3)設(shè)g(x)=f(x)-e2x+2aex-4a2+1(a∈R),若存在實(shí)數(shù)x0使得g(x0)≥0,求a的最大值.(1)解f(x)的定義域?yàn)镽,f'(x)=1-xex,所以當(dāng)x<1時(shí),f'(x)>0;當(dāng)x>1時(shí),f'(x)<0.所以f(x)的增區(qū)間為(-∞,1),減區(qū)間為(2)證明要證f(x)≤ex-1,即證xex≤ex-1,令h(x)=xex-ex+1,即證h(x)≤0,h'(x)=1-x-e2xex,令m(x)=1-x-e2x,則m'(x)=-1-2e2x<0,所以m(x)在R上單調(diào)遞減,又m(0)=0,所以當(dāng)x<0時(shí),m(x)>0,h'(x)>0;當(dāng)x>0時(shí),m(x)<0,h'(x)<0.所以h(x)在(-∞,0)上單調(diào)遞增,在(0,+∞)上單調(diào)遞減,所以h(x)≤h(0)=0,所以xex(3)解當(dāng)0≤a≤12時(shí),g(0)=2a-4a2=2a(1-2a)≥0,即存在x0=0滿足題意當(dāng)a>12時(shí),由(2)知,g(x)=f(x)-e2x+2aex-4a2+1≤(ex-1)-e2x+2aex-4a2+1=-e2x+(2a+1)ex-4a2=-ex此時(shí)g(x)<0恒成立,不滿足題意.綜上,a的最大值為123.(15分)(2024·山東二模)已知函數(shù)f(x)=a2xex-x-lnx.(1)當(dāng)a=1e時(shí),求f(x)的單調(diào)區(qū)間(2)當(dāng)a>0時(shí),f(x)≥2-a,求a的取值范圍.解(1)當(dāng)a=1e時(shí),f(x)=xex-1-x-lnx,x>0,則f'(x)=(x+1)ex-1-1-1x=x+1x(設(shè)g(x)=xex-1-1,則g'(x)=(x+1)ex-1>0恒成立,又g(1)=e0-1=0,所以當(dāng)x∈(0,1)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以f(x)的減區(qū)間為(0,1),增區(qū)間為(1,+∞).(2)f'(x)=a2(x+1)ex-1-1x=x+1x(a2設(shè)h(x)=a2xex-1,則h'(x)=a2(x+1)ex>0,所以h(x)在(0,+∞)上單調(diào)遞增,又h(0)=-1<0,h1a2=e所以存在x0∈0,1a2,使得h(x即a2x0ex0-1當(dāng)x∈(0,x0)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(x0,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以當(dāng)x=x0時(shí),f(x)取得極小值,也是最小值,所以f(x)≥f(x0)=a2x0ex0-x0-lnx0=1-ln(x0ex0)=1+2lna,所以1+2lna≥2-a,即a+2lna-1≥0,令F(a)=a+2lna-1,易知F(a)單調(diào)遞增,且所以要滿足F(a)≥F(1),則a≥1.綜上,a≥1.4.(17分)(2023·全國甲,文20)已知函數(shù)f(x)=ax-sinxcos2x,x∈(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)+sinx<0,求a的取值范圍.解(1)(第一步:由導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系,先求導(dǎo))當(dāng)a=1時(shí),f(x)=x-sinxf'(x)=1-cos=co=co=(cos(第二步:利用導(dǎo)數(shù)符號(hào)判斷單調(diào)性)當(dāng)0<x<π2時(shí),∵cosx-1<∴cos2x+2cosx+2=(cosx+1)2+1>0,∴f'(x)<0.故當(dāng)a=1時(shí),f(x)在0,π(2)構(gòu)建g(x)=f(x)+sinx=ax-sinxcos2x+sinx(則g'(x)=a-1+sin2xcos3若g(x)=f(x)+sinx<0,且g(0)=f(0)+sin0=0,則g'(0)=a-1+1=a≤0,解得a≤0.當(dāng)a=0時(shí),∵sinx-sinxcos2x=又x∈0,π2,∴0<sinx<1,0<cos則1cos∴f(x)+