專題能力訓(xùn)練21不等式選講一、能力突破訓(xùn)練1.若a>0,b>0,且1a+1b=ab.(1)求a3+b3的最小值;(2)是否存在a,b,使得2a+3b=6?并說明理由.2.設(shè)函數(shù)f(x)=x+1a+|x-a|(a>0).(1)證明:f(x)2;(2)若f(3)<5,求a的取值范圍.3.已知關(guān)于x的不等式m-|x-2|1,其解集為0,4.(1)求m的值;(2)若a,b均為正實(shí)數(shù),且滿足a+b=m,求a2+b2的最小值.4.已知函數(shù)f(x)=x-12+x+12,M為不等式f(x)<2的解集.(1)求M;(2)證明:當(dāng)a,bM時(shí),|a+b|<|1+ab|.5.(2018全國(guó),文23)已知f(x)=|x+1|-|ax-1|.(1)當(dāng)a=1時(shí),求不等式f(x)>1的解集;(2)若x(0,1)時(shí)不等式f(x)>x成立,求a的取值范圍.二、思維提升訓(xùn)練6.已知函數(shù)f(x)=x,x1,1x,0<x<1,g(x)=af(x)-|x-2|,aR.(1)當(dāng)a=0時(shí),若g(x)|x-1|+b對(duì)任意x(0,+)恒成立,求實(shí)數(shù)b的取值范圍;(2)當(dāng)a=1時(shí),求函數(shù)y=g(x)的最小值.7.已知函數(shù)f(x)=|x-3|-|x-a|.(1)當(dāng)a=2時(shí),解不等式f(x)-;(2)若存在實(shí)數(shù)a,使得不等式f(x)a成立,求實(shí)數(shù)a的取值范圍.8.已知函數(shù)f(x)=-x2+ax+4,g(x) =|x+1|+|x-1|.(1)當(dāng)a=1時(shí),求不等式f(x)g(x)的解集;(2)若不等式f(x)g(x)的解集包含-1,1,求a的取值范圍.專題能力訓(xùn)練21不等式選講(選修45)一、能力突破訓(xùn)練1.解 (1)由ab=1a+1b2ab,得ab2,且當(dāng)a=b=2時(shí)等號(hào)成立.故a3+b32a3b342,且當(dāng)a=b=2時(shí)等號(hào)成立.所以a3+b3的最小值為42.(2)由(1)知,2a+3b26ab43.由于43>6,從而不存在a,b,使得2a+3b=6.2.(1)證明 由a>0,有f(x)=x+1a+|x-a|x+1a-(x-a)=1a+a2.故f(x)2.(2)解 f(3)=3+1a+|3-a|.當(dāng)a>3時(shí),f(3)=a+,由f(3)<5,得3<a<5+212.當(dāng)0<a3時(shí),f(3)=6-a+1a,由f(3)<5,得1+52<a3.綜上,a的取值范圍是1+52,5+212.3.解 (1)不等式m-|x-2|1可化為|x-2|m-1,1-mx-2m-1,即3-mxm+1.其解集為0,4,3-m=0,m+1=4,m=3.(2)由(1)知a+b=3.(方法一:利用基本不等式)(a+b)2=a2+b2+2ab(a2+b2)+(a2+b2)=2(a2+b2),a2+b292,當(dāng)且僅當(dāng)a=b=32時(shí)取等號(hào),a2+b2的最小值為92.(方法二:消元法求二次函數(shù)的最值)a+b=3,b=3-a,a2+b2=a2+(3-a)2=2a2-6a+9=2a-322+9292,a2+b2的最小值為92.4.(1)解 f(x)=-2x,x-12,1,-12<x<12,2x,x12.當(dāng)x-12時(shí),由f(x)<2得-2x<2,解得x>-1;當(dāng)-12<x<12時(shí),f(x)<2;當(dāng)x12時(shí),由f(x)<2得2x<2,解得x<1.所以f(x)<2的解集M=x|-1<x<1.(2)證明 由(1)知,當(dāng)a,bM時(shí),-1<a<1,-1<b<1,從而(a+b)2-(1+ab)2=a2+b2-a2b2-1=(a2-1)(1-b2)<0.因此|a+b|<|1+ab|.5.解 (1)當(dāng)a=1時(shí),f(x)=|x+1|-|x-1|,即f(x)=-2,x-1,2x,-1<x<1,2,x1.故不等式f(x)>1的解集為xx>12.(2)當(dāng)x(0,1)時(shí)|x+1|-|ax-1|>x成立等價(jià)于當(dāng)x(0,1)時(shí)|ax-1|<1成立.若a0,則當(dāng)x(0,1)時(shí)|ax-1|1;若a>0,|ax-1|<1的解集為0<x<2a,所以2a1,故0<a2.綜上,a的取值范圍為(0,2.二、思維提升訓(xùn)練6.解 (1)當(dāng)a=0時(shí),g(x)=-|x-2|(x>0),g(x)|x-1|+b-b|x-1|+|x-2|.|x-1|+|x-2|(x-1)-(x-2)|=1,當(dāng)且僅當(dāng)1x2時(shí)等號(hào)成立.故實(shí)數(shù)b的取值范圍是-1,+).(2)當(dāng)a=1時(shí),g(x)=1x+x-2,0<x<1,2x-2,1x2,2,x>2.當(dāng)0<x<1時(shí),g(x)=1x+x-2>2x1x-2=0;當(dāng)x1時(shí),g(x)0,當(dāng)且僅當(dāng)x=1時(shí)等號(hào)成立;故當(dāng)x=1時(shí),函數(shù)y=g(x)取得最小值0.7.解 (1)a=2,f(x)=|x-3|-|x-2|=1,x2,5-2x,2<x<3,-1,x3,f(x)-12等價(jià)于x2,1-12或5-2x-12,2<x<3或x3,-1-12.解得114x<3或x3,不等式的解集為xx114.(2)由不等式性質(zhì)可知f(x)=|x-3|-|x-a|(x-3)-(x-a)|=|a-3|,若存在實(shí)數(shù)x,使得不等式f(x)a成立,則|a-3|a,解得a32.實(shí)數(shù)a的取值范圍是-,32.8.解 (1)當(dāng)a=1時(shí),不等式 f(x)g(x)等價(jià)于x2-x+|x+1|+|x-1|-40.當(dāng)x<-1時(shí),式化為x2-3x-40,無解;當(dāng)-1x1時(shí),式化為x2-x-20,從而-1x1;當(dāng)x>1時(shí),式化為x2+x-40,從而1<x-1+172.所以f(x)g(x)的解集為x-1x-1+172.(2)當(dāng)x-1,1時(shí),g(x)=2.所以f(x)g(x)的解集包含-1,1,等價(jià)于當(dāng)x-1,1時(shí)f(x)2.又f(x)在-1,1的最小值必為f(-1)與f(1)之一,所以f(-1)2且f(1)2,得-1a1.所以a的取值范圍為-1,1.