課時規(guī)范練55不等式選講基礎(chǔ)鞏固組1.(2017山西呂梁二模,23)已知函數(shù)f(x)=|x-1|+|x-a|.(1)若a=-1,解不等式f(x)3;(2)如果xR,使得f(x)<2成立,求實(shí)數(shù)a的取值范圍.導(dǎo)學(xué)號241908462.(2017遼寧鞍山一模,文23)設(shè)函數(shù)f(x)=x-52+|x-a|,xR.(1)當(dāng)a=-12時,求不等式f(x)4的解集;(2)若關(guān)于x的不等式f(x)a在R上恒成立,求實(shí)數(shù)a的最大值.3.已知函數(shù)f(x)=|x+a|+|x-2|.(1)當(dāng)a=-3時,求不等式f(x)3的解集;(2)若f(x)|x-4|的解集包含1,2,求a的取值范圍.4.已知函數(shù)f(x)=|2x-a|+a.(1)當(dāng)a=2時,求不等式f(x)6的解集;(2)設(shè)函數(shù)g(x)=|2x-1|.當(dāng)xR時,f(x)+g(x)3,求a的取值范圍.5.(2017山西臨汾三模,23)已知函數(shù)f(x)=|x+2|-m,mR,且f(x)0的解集為-3,-1.(1)求m的值;(2)設(shè)a,b,c為正數(shù),且a+b+c=m,求3a+1+3b+1+3c+1的最大值.導(dǎo)學(xué)號24190847綜合提升組6.(2017遼寧沈陽一模,23)設(shè)不等式-2<|x-1|-|x+2|<0的解集為M,a,bM,(1)證明:13a+16b<14;(2)比較|1-4ab|與2|a-b|的大小,并說明理由.7.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,|a+b|<|1+ab|.8.設(shè)a,b,c均為正數(shù),且a+b+c=1.證明:(1)ab+bc+ac13;(2)a2b+b2c+c2a1.導(dǎo)學(xué)號24190848創(chuàng)新應(yīng)用組9.已知函數(shù)f(x)=|x+1|-2|x-a|,a>0.(1)當(dāng)a=1時,求不等式f(x)>1的解集;(2)若f(x)的圖象與x軸圍成的三角形面積大于6,求a的取值范圍.10.(2017河北邯鄲二模,文23)已知函數(shù)f(x)=|x+1|+|x-3|,g(x)=a-|x-2|.(1)若關(guān)于x的不等式f(x)<g(x)有解,求實(shí)數(shù)a的取值范圍;(2)若關(guān)于x的不等式f(x)<g(x)的解集為b,72,求a+b的值.答案：1.解 (1)若a=-1,f(x)3,即為|x-1|+|x+1|3,當(dāng)x-1時,1-x-x-13,即有x-32;當(dāng)-1<x<1時,1-x+x+1=23不成立;當(dāng)x1時,x-1+x+1=2x3,解得x32.綜上可得,f(x)3的解集為-,-3232,+;(2)xR,使得f(x)<2成立,即有2>f(x)min,由函數(shù)f(x)=|x-1|+|x-a|x-1-x+a|=|a-1|,當(dāng)(x-1)(x-a)0時,取得最小值|a-1|,則|a-1|<2,即-2<a-1<2,解得-1<a<3.則實(shí)數(shù)a的取值范圍為(-1,3).2.解 (1)f(x)=x-52+x+12=-2x+2,x<-12,3,-12x52,2x-2,x>52.由f(x)4得x<-12,-2x+24或x>52,2x-24.解得x-1或x3,所以不等式的解集為x|x1或x3.(2)由絕對值的性質(zhì)得f(x)=x-52+|x-a|x-52-(x-a)=a-52,所以f(x)的最小值為a-52,從而a-52a,解得a54,因此a的最大值為54.3.解 (1)當(dāng)a=-3時,f(x)=-2x+5,x2,1,2<x<3,2x-5,x3.當(dāng)x2時,由f(x)3得-2x+53,解得x1;當(dāng)2<x<3時,f(x)3無解;當(dāng)x3時,由f(x)3得2x-53,解得x4;所以f(x)3的解集為x|x1x|x4.(2)f(x)|x-4|x-4|-|x-2|x+a|.當(dāng)x1,2時,|x-4|-|x-2|x+a|4-x-(2-x)|x+a|-2-ax2-a.由條件得-2-a1且2-a2,即-3a0.故滿足條件的a的取值范圍為-3,0.4.解 (1)當(dāng)a=2時,f(x)=|2x-2|+2.解不等式|2x-2|+26得-1x3.因此f(x)6的解集為x|-1x3.(2)當(dāng)xR時,f(x)+g(x)=|2x-a|+a+|1-2x|2x-a+1-2x|+a=|1-a|+a,當(dāng)x=12時等號成立,所以當(dāng)xR時,f(x)+g(x)3等價于|1-a|+a3.當(dāng)a1時,等價于1-a+a3,無解.當(dāng)a>1時,等價于a-1+a3,解得a2.所以a的取值范圍是2,+).5.解(1)由題意,|x+2|mm0,-m-2xm-2,由f(x)0的解集為-3,-1,得-m-2=-3,m-2=-1,解得m=1.(2)由(1)可得a+b+c=1,由柯西不等式可得(3a+1+3b+1+3c+1)(12+12+12)(3a+1+3b+1+3c+1)2,3a+1+3b+1+3c+132,當(dāng)且僅當(dāng)3a+1=3b+1=3c+1,即a=b=c=13時等號成立,3a+1+3b+1+3c+1的最大值為32.6.(1)證明 記f(x)=|x-1|-|x+2|=3,x-2,-2x-1,-2<x<1,-3,x1,由-2<-2x-1<0解得-12<x<12,則M=-12,12.a,bM,|a|<12,|b|<12.13a+16b13|a|+16|b|<1312+1612=14.(2)解 由(1)得a2<14,b2<14.因?yàn)閨1-4ab|2-4|a-b|2=(1-8ab+16a2b2)-4(a2-2ab+b2)=(4a2-1)(4b2-1)>0,所以|1-4ab|2>4|a-b|2,故|1-4ab|>2|a-b|.7.解 (1)f(x)=-2x,x-12,1,-12<x<12,2x,x12,當(dāng)x-12時,由f(x)<2得-2x<2,解得x>-1;當(dāng)-12<x<12時,f(x)<2;當(dāng)x12時,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)由(1)知,當(dāng)a,bM時,-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|.8.證明 (1)由a2+b22ab,b2+c22bc,c2+a22ca,得a2+b2+c2ab+bc+ca.由題設(shè)得(a+b+c)2=1,即a2+b2+c2+2ab+2bc+2ca=1,所以3(ab+bc+ca)1,即ab+bc+ca13.(2)因?yàn)閍2b+b2a,b2c+c2b,c2a+a2c,故a2b+b2c+c2a+(a+b+c)2(a+b+c),即a2b+b2c+c2aa+b+c.所以a2b+b2c+c2a1.9.解 (1)當(dāng)a=1時,f(x)>1化為|x+1|-2|x-1|-1>0.當(dāng)x-1時,不等式化為x-4>0,無解;當(dāng)-1<x<1時,不等式化為3x-2>0,解得23<x<1;當(dāng)x1時,不等式化為-x+2>0,解得1x<2.所以f(x)>1的解集為x23<x<2.(2)由題設(shè)可得,f(x)=x-1-2a,x<-1,3x+1-2a,-1xa,-x+1+2a,x>a.所以函數(shù)f(x)的圖象與x軸圍成的三角形的三個頂點(diǎn)分別為A2a-13,0,B(2a+1,0),C(a,a+1),ABC的面積為23(a+1)2.由題設(shè)得23(a+1)2>6,故a>2.所以a的取值范圍為(2,+).10.解 (1)當(dāng)x=2時,g(x)=a-|x-2|取最大值為a,f(x)=|x+1|+|x-3|4,當(dāng)且僅當(dāng)-1x3,f(x)取最小值4,又關(guān)于x的不等式f(x)<g(x)有解,a>4,即實(shí)數(shù)a的取值范圍是(4,+).(2)當(dāng)x=72時,f(x)=5,則g72=-32+a=5,解得a=132,當(dāng)x<2時,g(x)=x+92,令g(x)=x+92=4,得x=-12(-1,3),b=-12,則a+b=6.