專(zhuān)題能力訓(xùn)練21不等式選講(選修45)一、能力突破訓(xùn)練1.(2019廣東汕頭二模,23)已知函數(shù)f(x)=|2x+2|+|x-1|.(1)畫(huà)出y=f(x)的圖象;(2)若y=f(x)的最小值為m,當(dāng)正數(shù)a,b滿(mǎn)足2b+1a=m時(shí),求a+2b的最小值.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(mǎn)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.(2019山東青島質(zhì)檢,23)已知a>0,b>0,c>0,函數(shù)f(x)=|a-x|+|x+b|+c.(1)當(dāng)a=b=c=2時(shí),求不等式f(x)<8的解集;(2)若函數(shù)f(x)的最小值為1,證明:a2+b2+c213.7.已知函數(shù)f(x)=|x-3|-|x-a|.(1)當(dāng)a=2時(shí),解不等式f(x)-12;(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.(2019全國(guó),文23)設(shè)x,y,zR,且x+y+z=1.(1)求(x-1)2+(y+1)2+(z+1)2的最小值;(2)若(x-2)2+(y-1)2+(z-a)213成立,證明:a-3 或a-1.專(zhuān)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)f(x)=-3x-1,x<-1,x+3,-1x1,3x+1,x>1.畫(huà)出y=f(x)的圖象如圖所示.(2)由(1)知f(x)min=f(-1)=2,m=2.2b+1a=2.a+2b=12(a+2b)2b+1a=ab+ba+522ab·ba+52=92,當(dāng)且僅當(dāng)ab=ba,即a=b=32時(shí)等號(hào)成立.a+2b的最小值為92.2.(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+1a,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.(1)解當(dāng)a=b=c=2時(shí),f(x)=|x-2|+|x+2|+2=2-2x,x-2,6,-2<x<2,2x+2,x2,f(x)<8x-2,2-2x<8或-2<x<2,6<8或x2,2x+2<8.不等式的解集為x|-3<x<3.(2)證明a>0,b>0,c>0,f(x)=|a-x|+|x+b|+c|a-x+x+b|+c=|a+b|+c=a+b+c,當(dāng)且僅當(dāng)(a-x)(x+b)0時(shí)等號(hào)成立.f(x)的最小值為1,a+b+c=1,(a+b+c)2=a2+b2+c2+2ab+2ac+2bc=1.2aba2+b2,2bcb2+c2,2aca2+c2,當(dāng)且僅當(dāng)a=b=c時(shí)等號(hào)成立,1=a2+b2+c2+2ab+2ac+2bc3(a2+b2+c2).a2+b2+c213.7.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.(1)解由于(x-1)+(y+1)+(z+1)2=(x-1)2+(y+1)2+(z+1)2+2(x-1)(y+1)+(y+1)(z+1)+(z+1)(x-1)3(x-1)2+(y+1)2+(z+1)2,故由已知得(x-1)2+(y+1)2+(z+1)243,當(dāng)且僅當(dāng)x=53,y=-13,z=-13時(shí)等號(hào)成立.所以(x-1)2+(y+1)2+(z+1)2的最小值為43.(2)證明由于(x-2)+(y-1)+(z-a)2=(x-2)2+(y-1)2+(z-a)2+2(x-2)(y-1)+(y-1)(z-a)+(z-a)(x-2)3(x-2)2+(y-1)2+(z-a)2,故由已知得(x-2)2+(y-1)2+(z-a)2(2+a)23,當(dāng)且僅當(dāng)x=4-a3,y=1-a3,z=2a-23時(shí)等號(hào)成立.因此(x-2)2+(y-1)2+(z-a)2的最小值為(2+a)23.由題設(shè)知(2+a)2313,解得a-3或a-1.