2020版高考數(shù)學(xué)一輪復(fù)習(xí) 課時(shí)規(guī)范練64 不等式選講 理 北師大版.doc

課時(shí)規(guī)范練64不等式選講基礎(chǔ)鞏固組1.(2018河南最后一次模擬,23)已知函數(shù)f(x)=|2x+4|+|2x-a|.(1)當(dāng)a=6時(shí),求f(x)12的解集; (2)已知a>-2,g(x)=x2+2ax+,若對(duì)于x-1, ,都有f(x)g(x)成立,求a的取值范圍.2.(2018湖南長(zhǎng)沙模擬二,23)已知函數(shù)f(x)=|x-1|,關(guān)于x的不等式f(x)<3-|2x+1|的解集記為A.(1)求A;(2)已知a,bA,求證:f(ab)>f(a)-f(b).3.(2018安徽淮南二模,23)已知函數(shù)f(x)=|x-2|-|x+1|.(1)解不等式f(x)+x>0.(2)若關(guān)于x的不等式f(x)a2-2a的解集為R,求實(shí)數(shù)a的取值范圍.4.(2018河北衡水中學(xué)三輪檢測(cè),23)已知函數(shù)f(x)=|ax-1|-(a-2)x.(1)當(dāng)a=3時(shí),求不等式f(x)>0的解集;(2)若函數(shù)f(x)的圖像與x軸沒(méi)有交點(diǎn),求實(shí)數(shù)a的取值范圍.綜合提升組5.已知函數(shù)f(x)=|x-a|.(1)當(dāng)a=-2時(shí),解不等式f(x)16-|2x-1|;(2)若關(guān)于x的不等式f(x)1的解集為0,2,求證:f(x)+f(x+2)2.6.(2018河南南陽(yáng)模擬,23)已知函數(shù)f(x)=|x-2a+1|+|x+2|,g(x)=3x+1.(1)當(dāng)a=1時(shí),求不等式f(x)g(x)的解集;(2)x-2,a),f(x)g(x),求a的取值范圍.7.已知函數(shù)f(x)=|2x+1|,g(x)=|x+1|,不等式f(x)g(x)+1的解集為A.(1)求A;(2)證明:對(duì)于任意的a,bRA,都有g(shù)(ab)>g(a)-g(-b)成立.創(chuàng)新應(yīng)用組8.已知函數(shù)f(x)=|x-2|-|x|+m(mR).(1)若m=0,解不等式f(x)x-1;(2)若方程f(x)=-x有三個(gè)不同的解,求實(shí)數(shù)m的取值范圍.9.(2018安徽安慶熱身考,23)若關(guān)于x的不等式|3x+2|+|3x-1|-t0的解集為R,記實(shí)數(shù)t的最大值為a.(1)求a的值;(2)若正實(shí)數(shù)m,n滿足4m+5n=a,求y=1m+2n+43m+3n的最小值.參考答案課時(shí)規(guī)范練64不等式選講1.解 (1)當(dāng)a=6時(shí),f(x)=|2x+4|+|2x-6|,f(x)12等價(jià)于|x+2|+|x-3|6,因?yàn)閨x+2|+|x-3|=2x-1,x>3,5,-2x3,-2x+1,x<-2,所以x>3,2x-16或-2x3,56或x<-2,-2x+16,解得x72或x-52,所以解集為xx-52或x72.(2)當(dāng)a>-2時(shí),且x-1,a2時(shí),f(x)=2x+4-(2x-a)=4+a,所以f(x)g(x),即4+ag(x).又g(x)=x2+2ax+74的最大值必為g(-1),ga2之一,所以4+a114-2a,4+a54a2+74,即3a-54,54a2-a-940,解得-512a95,所以a的取值范圍為-512,95.2.解 (1)由f(x)<3-|2x+1|,得|x-1|+|2x+1|<3,即x-12,1-x-2x-1<3或-12<x<1,1-x+2x+1<3或x1,x-1+2x+1<3,解得-1<x-12或-12<x<1,所以,集合A=xR|-1<x<1.(2)證明 a,bA,-1<ab<1,f(ab)=|ab-1|=1-ab,f(a)=|a-1|=1-a,f(b)=|b-1|=1-b,f(ab)-f(a)-f(b)=1-ab-1+a+1-b=(1+a)(1-b)>0,f(ab)>f(a)-f(b).3.解 (1)不等式f(x)+x>0可化為|x-2|+x>|x+1|.當(dāng)x<-1時(shí),-(x-2)+x>-(x+1),解得x>-3,即-3<x<-1;當(dāng)-1x2時(shí),-(x-2)+x>x+1,解得x<1,即-1x<1;當(dāng)x>2時(shí),x-2+x>x+1,解得x>3,即x>3.綜上所述:不等式f(x)+x>0的解集為x|-3<x<1或x>3.(2)由不等式f(x)a2-2a可得|x-2|-|x+1|a2-2a,|x-2|-|x+1|x-2-x-1|=3,a2-2a3,即a2-2a-30.解得a3或a-1.故實(shí)數(shù)a的取值范圍是a3或a-1.4.解 (1)當(dāng)a=3時(shí),不等式可化為|3x-1|-x>0,即|3x-1|>x.3x-1<-x或3x-1>x,即x<14或x>12.即不等式f(x)>0的解集是xx<14或x>12.(2)當(dāng)a>0時(shí),f(x)=2x-1,x1a,2(1-a)x+1,x<1a,要使函數(shù)f(x)與x軸無(wú)交點(diǎn),只需2a-1>0,2(1-a)0,即1a<2.當(dāng)a=0時(shí),f(x)=2x+1,函數(shù)f(x)與x軸有交點(diǎn).當(dāng)a<0時(shí),f(x)=2x-1,x1a,2(1-a)x+1,x>1a,要使函數(shù)f(x)與x軸無(wú)交點(diǎn),只需2a-1<0,2(1-a)0,此時(shí)a無(wú)解.綜上可知,當(dāng)1a<2時(shí),函數(shù)f(x)與x軸無(wú)交點(diǎn).5.(1)解 當(dāng)a=-2時(shí),不等式為|x+2|+|2x-1|16,當(dāng)x-2時(shí),原不等式可化為-x-2-2x+116,解得x-173,當(dāng)-2<x12時(shí),原不等式可化為x+2-2x+116,解得x-13,不滿足,舍去;當(dāng)x>12時(shí),原不等式可化為x+2+2x-116,解得x5.綜上不等式的解集為xx-173或x5.(2)證明 f(x)1即|x-a|1,解得a-1xa+1,而f(x)1的解集是0,2,所以a-1=0,a+1=2,解得a=1,從而f(x)=|x-1|.于是證明f(x)+f(x+2)2,即證|x-1|+|x+1|2,因?yàn)閨x-1|+|x+1|=|1-x|+|x+1|1-x+x+1|=2,所以|x-1|+|x+1|2,所以原不等式得證.6.解 (1)當(dāng)a=1時(shí),f(x)=|x-1|+|x+2|,當(dāng)x-2時(shí),f(x)=-2x-1,由-2x-13x+1,知此時(shí)無(wú)解;當(dāng)-2<x<1時(shí),f(x)=3,由33x+1,解得23x<1;當(dāng)x1時(shí),f(x)=2x+1,由2x+13x+1,解得x1,綜上所述,不等式的解集為xx23.(2)當(dāng)x-2,a)時(shí),f(x)=|x-2a+1|+x+23x+1,即|x-2a+1|2x-1.當(dāng)-2<a12時(shí),2x-1<0,|x-2a+1|2x-1恒成立;當(dāng)a>12,x-2,12時(shí),2x-1<0,|x-2a+1|2x-1恒成立;x12,a時(shí),|x-2a+1|2(2x-1)2恒成立,即3x2+2(2a-3)x-4a(a-1)0恒成立,令g(x)=3x2+2(2a-3)x-4a(a-1),g(x)的最大值只可能是g12或g(a),g120,g(a)=3a2-2a0,得0a23.又a>12,所以12<a23.綜上所述,a的取值范圍是a-2<a23.7.(1)解 不等式f(x)g(x)+1,即|x+1|-|2x+1|+10.當(dāng)x<-1時(shí),不等式可化為-x-1+(2x+1)+10,解得x-1,x無(wú)解;當(dāng)-1x-12,不等式可化為x+1+(2x+1)+10,解得x-1,-1x-12;當(dāng)x>-12時(shí),不等式可化為x+1-(2x+1)+10,解得x1,-12<x1.不等式f(x)g(x)+1的解集A=x|-1x1.(2)證明 g(a)-g(-b)=|a+1|-|-b+1|a+1-(-b+1)|=|a+b|,要證g(ab)>g(a)-g(-b)成立,只需證|ab+1|>|a+b|,即證|ab+1|2>|a+b|2,也就是證明a2b2+2ab+1>a2+2ab+b2成立,即證a2b2-a2-b2+1>0,即證(a2-1)(b2-1)>0.A=x|-1x1,a,bRA,|a|>1,|b|>1,a2>1,b2>1,(a2-1)(b2-1)>0成立.從而對(duì)于任意的a,bRA,都有g(shù)(ab)>g(a)-g(-b)成立.8.解 (1)因?yàn)閙=0,所以f(x)=|x-2|-|x|,有x>2,x-2-xx-1或0x2,-x+2-xx-1或x<0,-x+2+xx-1.解得,x或0x1或x<0.所以不等式f(x)x-1的解集為(-,1.(2)因?yàn)閒(x)=|x-2|-|x|+m,所以方程f(x)=-x有三個(gè)不同的解等價(jià)于函數(shù)g(x)=|x-2|-|x|的圖像與直線y=-x-m有三個(gè)不同的交點(diǎn),作圖可知,當(dāng)直線y=-x-m經(jīng)過(guò)點(diǎn)A(0,2)時(shí),m=-2;當(dāng)直線y=-x-m經(jīng)過(guò)點(diǎn)B(2,-2)時(shí),m=0.所以實(shí)數(shù)m的取值范圍是(-2,0).9.解 (1)由題意得|3x+2|+|3x-1|t對(duì)xR恒成立,又|3x+2|+|3x-1|=|3x+2|+|1-3x|3,t3.a=3.(2)由(1)得4m+5n=3,且m,n>0,3y=1m+2n+43m+3n(4m+5n)=1m+2n+43m+3n(m+2n)+(3m+3n)=5+3m+3nm+2n+4(m+2n)3m+3n5+23m+3nm+2n4(m+2n)3m+3n=9.當(dāng)且僅當(dāng)3m+3nm+2n=4(m+2n)3m+3n且4m+5n=3,即m=n=13時(shí)等號(hào)成立.y3,即y=1m+2n+43m+3n的最小值為3.