專題能力訓(xùn)練23不等式選講(選修45)一、能力突破訓(xùn)練1.設(shè)a>0,|x-1|<,|y-2|<,求證:|2x+y-4|<a.2.已知函數(shù)f(x)=|x-1|+|x+3|,xR.(1)解不等式f(x)5;(2)若不等式t2+3t>f(x)在xR上有解,求實(shí)數(shù)t的取值范圍.3.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.4.(2018全國,理23)設(shè)函數(shù)f(x)=|2x+1|+|x-1|.(1)畫出y=f(x)的圖象;(2)當(dāng)x0,+)時(shí),f(x)ax+b,求a+b的最小值.5.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|.6.設(shè)關(guān)于x的不等式|2x-a|+|x+3|2x+4的解集為A.(1)若a=1,求A;(2)若A=R,求a的取值范圍.7.已知函數(shù)f(x)=|2x-1|+|x-a|,aR.(1)當(dāng)a=3時(shí),解不等式f(x)4;(2)若f(x)=|x-1+a|,求x的取值范圍.二、思維提升訓(xùn)練8.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.9.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.10.設(shè)函數(shù)f(x)=|x-1|+|x-a|.(1)若a=-1,解不等式f(x)3;(2)如果xR,f(x)2,求a的取值范圍.專題能力訓(xùn)練23不等式選講(選修45)一、能力突破訓(xùn)練1.證明 因?yàn)閨x-1|<,|y-2|<,所以|2x+y-4|=|2(x-1)+(y-2)|2|x-1|+|y-2|<2a3+a3=a.2.解 (1)原不等式等價(jià)于x<-3,-2-2x5或-3x1,45或x>1,2x+25,得-72x<-3或-3x1或1<x32,因此不等式的解集為-72,32.(2)f(x)=|x-1|+|x+3|x-1-(x+3)|=4,要使t2+3t>f(x)在xR上有解,只需t2+3t大于f(x)的最小值,t2+3t>f(x)min=4t2+3t-4>0t<-4或t>1.3.(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+212.4.解 (1)f(x)=-3x,x<-12,x+2,-12x<1,3x,x1.y=f(x)的圖象如圖所示.(2)由(1)知,y=f(x)的圖象與y軸交點(diǎn)的縱坐標(biāo)為2,且各部分所在直線斜率的最大值為3,故當(dāng)且僅當(dāng)a3且b2時(shí),f(x)ax+b在0,+)成立,因此a+b的最小值為5.5.(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|.6.解 (1)當(dāng)x12時(shí),2x-1+x+32x+4,解得x2.當(dāng)-3<x<12時(shí),1-2x+x+32x+4,解得-3<x0.當(dāng)x-3時(shí),1-2x-x-32x+4,解得x-3.綜上,原不等式的解集A=x|x0或x2.(2)當(dāng)x-2時(shí),|2x-a|+|x+3|02x+4成立.當(dāng)x>-2時(shí),|2x-a|+|x+3|=|2x-a|+x+32x+4,即|2x-a|x+1,得xa+1或xa-13,所以a+1-2或a+1a-13,得a-2.綜上,a的取值范圍為a-2.7.解 (1)當(dāng)a=3時(shí),函數(shù)f(x)=|2x-1|+|x-3|=3x-4,x3,x+2,12<x<3,4-3x,x12,如圖,由于直線y=4和函數(shù)f(x)的圖象交于點(diǎn)(0,4),(2,4),故不等式f(x)4的解集為(0,2).(2)由f(x)=|x-1+a|,可得|2x-1|+|x-a|=|x-1+a|.由于|2x-1|+|x-a|(2x-1)-(x-a)|=|x-1+a|,當(dāng)且僅當(dāng)(2x-1)(x-a)0時(shí)取等號(hào),故有(2x-1)(x-a)0.當(dāng)a=12時(shí),可得x=12,故x的取值范圍為12;當(dāng)a>12時(shí),可得12xa,故x的取值范圍為12,a;當(dāng)a<12時(shí),可得ax12,故x的取值范圍為a,12.二、思維提升訓(xùn)練8.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0.9.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-,32.10.解 (1)當(dāng)a=-1時(shí),f(x)=|x-1|+|x+1|,f(x)=-2x,x<-1,2,-1x1,2x,x>1.作出函數(shù)f(x)=|x-1|+|x+1|的圖象.由圖象可知,不等式f(x)3的解集為xx-32或x32.(2)若a=1,則f(x)=2|x-1|,不滿足題設(shè)條件;若a<1,則f(x)=-2x+a+1,xa,1-a,a<x<1,2x-(a+1),x1,f(x)的最小值為1-a;若a>1,則f(x)=-2x+a+1,x1,a-1,1<x<a,2x-(a+1),xa,f(x)的最小值為a-1.故對(duì)于xR,f(x)2的充要條件是|a-1|2,a的取值范圍是(-,-13,+).