全國通用版2019版高考數(shù)學(xué)總復(fù)習(xí)專題八鴨內(nèi)容8.2不等式選講精選刷題練理.doc

8.2不等式選講命題角度1含絕對值不等式的圖象與解法高考真題體驗對方向1.(2018全國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,f(x)ax+b,求a+b的最小值.解(1)f(x)=-3x,x-12,x+2,-12x1,3x,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,f(x)ax+b在0,+)成立,因此a+b的最小值為5.2.(2017全國23)已知函數(shù)f(x)=-x2+ax+4,g(x)=|x+1|+|x-1|.(1)當(dāng)a=1時,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.解(1)當(dāng)a=1時,不等式f(x)g(x)等價于x2-x+|x+1|+|x-1|-40.當(dāng)x1時,式化為x2+x-40,從而11的解集.解(1)f(x)=x-4,x-1,3x-2,-132,y=f(x)的圖象如圖所示.(2)由f(x)的表達(dá)式及圖象,當(dāng)f(x)=1時,可得x=1或x=3;當(dāng)f(x)=-1時,可得x=13或x=5,故f(x)1的解集為x|1x3;f(x)-1的解集為xx5.所以|f(x)|1的解集為xx13或1x5.新題演練提能刷高分1.(2018安徽淮南一模)設(shè)函數(shù)f(x)=|2x-4|+1.(1)畫出函數(shù)y=f(x)的圖象;(2)若不等式f(x)ax的解集非空,求a的取值范圍.解(1)由于f(x)=-2x+5,x2,2x-3,x2,則y=f(x)的圖象如圖所示:(2)由函數(shù)y=f(x)與函數(shù)y=ax的圖象可知,當(dāng)且僅當(dāng)a12或a-2時,函數(shù)y=f(x)與函數(shù)y=ax的圖象有交點(diǎn),故不等式f(x)ax的解集非空時,a的取值范圍是(-,-2)12,+.2.(2018河北邯鄲一模)已知函數(shù)f(x)=|x-4|+|x-1|-3.(1)求不等式f(x)2的解集;(2)若直線y=kx-2與函數(shù)f(x)的圖象有公共點(diǎn),求k的取值范圍.解(1)由f(x)2,得x1,2-2x2或1x4,02或x4,2x-82,解得0x5,故不等式f(x)2的解集為0,5.(2)f(x)=|x-4|+|x-1|-3=2-2x,x1,0,1x4,2x-8,x4,作出函數(shù)f(x)的圖象,如圖所示,直線y=kx-2過定點(diǎn)C(0,-2),當(dāng)此直線經(jīng)過點(diǎn)B(4,0)時,k=12;當(dāng)此直線與直線AD平行時,k=-2.故由圖可知,k(-,-2)12,+.3.(2018安徽蚌埠模擬)已知函數(shù)f(x)=|x+1|-2|x|.(1)求不等式f(x)-6的解集;(2)若f(x)的圖象與直線y=a圍成的圖形的面積不小于14,求實(shí)數(shù)a的取值范圍.解(1)f(x)=|x+1|-2|x|=x-1,x0.則不等式f(x)-6等價于x0,1-x-6,解得x-5或x7.故不等式f(x)-6的解集為x|x-5或x7.(2)作出函數(shù)f(x)的圖象,如圖.若f(x)的圖象與直線y=a圍成的圖形是三角形,則當(dāng)a=-2時,ABC的面積取得最大值1243=6,f(x)的圖象與直線y=a圍成圖形的面積不小于14,該圖形一定是四邊形,即a-2.ABC的面積是6,梯形ABED的面積不小于8.AB=4,D(1+a,a),E(1-a,a),DE=-2a,12(4-2a)(-2-a)14-6=8,a212.又a-2,則a-23,故實(shí)數(shù)a的取值范圍是(-,-23.4.(2018福建漳州期末調(diào)研)已知函數(shù)f(x)=|2x-1|+2|x+2|.(1)求函數(shù)f(x)的最小值;(2)解不等式f(x)8.解(1)因為f(x)=|2x-1|+2|x+2|(2x-1)-2(x+2)|=5,所以f(x)=-4x-3,x12,(2)當(dāng)x-2時,由-4x-3-114,即-114x-2;當(dāng)-2x12時,512時,由4x+38,解得x54,即12x54,所以原不等式的解集為-114,54.5.(2018江西九校聯(lián)考)已知函數(shù)f(x)=|2x|-|x+3|.(1)若對于任意的實(shí)數(shù)x,都有f(x)2m2-7m成立,求m的取值范圍;(2)若g(x)=ax,方程f(x)=g(x)有兩個不同的實(shí)數(shù)解,求a的取值范圍.解(1)由于f(x)=|2x|-|x+3|=3-x,x0,所以f(x)的最小值為f(0)=-3.又因為對任意的實(shí)數(shù)x,都有f(x)2m2-7m成立,只需2m2-7m-3,即2m2-7m+30,解得12m3,故m的取值范圍為12,3.(2)方程f(x)=g(x)有兩個不同的實(shí)數(shù)解,即函數(shù)y=f(x)與y=g(x)的圖象有兩個不同的交點(diǎn),作出這兩個函數(shù)圖象,由圖象可知,a的取值范圍是-1,1)-2.6.(2018湖北天門、仙桃、潛江聯(lián)考)已知函數(shù)f(x)=x-a,其中a1.(1)當(dāng)a=2時,求不等式f(x)4-|x-4|的解集;(2)已知關(guān)于x的不等式|f(2x+a)-2f(x)|2的解集為x|1x2,求a的值.解(1)當(dāng)a=2時,f(x)+|x-4|=-2x+6,x2,2,2x4,2x-6,x4.當(dāng)x2時,由f(x)4-|x-4|得-2x+64,解得x1.當(dāng)2x4時,f(x)4-|x-4|無解.當(dāng)x4時,由f(x)4-|x-4|得2x-64,解得x5.f(x)4-|x-4|的解集為x|x1或x5.(2)記h(x)=f(2x+a)-2f(x),則h(x)=-2a,x0,4x-2a,0x1的解集;(2)若x(0,1)時不等式f(x)x成立,求a的取值范圍.解(1)當(dāng)a=1時,f(x)=|x+1|-|x-1|,即f(x)=-2,x-1,2x,-1x1的解集為xx12.(2)當(dāng)x(0,1)時|x+1|-|ax-1|x成立等價于當(dāng)x(0,1)時|ax-1|0,|ax-1|1的解集為0x2a,所以2a1,故0a2.綜上,a的取值范圍為(0,2.2.(2018全國23)設(shè)函數(shù)f(x)=5-|x+a|-|x-2|.(1)當(dāng)a=1時,求不等式f(x)0的解集;(2)若f(x)1,求a的取值范圍.解(1)當(dāng)a=1時,f(x)=2x+4,x-1,2,-12.可得f(x)0的解集為x|-2x3.(2)f(x)1等價于|x+a|+|x-2|4.而|x+a|+|x-2|a+2|,且當(dāng)x=2時等號成立.故f(x)1等價于|a+2|4.由|a+2|4可得a-6或a2.所以a的取值范圍是(-,-62,+).3.(2017全國23)已知函數(shù)f(x)=|x+1|-|x-2|.(1)求不等式f(x)1的解集;(2)若不等式f(x)x2-x+m的解集非空,求m的取值范圍.解(1)f(x)=-3,x2.當(dāng)x2時,由f(x)1解得x2.所以f(x)1的解集為x|x1.(2)由f(x)x2-x+m得m|x+1|-|x-2|-x2+x.而|x+1|-|x-2|-x2+x|x|+1+|x|-2-x2+|x|=-|x|-322+5454,且當(dāng)x=32時,|x+1|-|x-2|-x2+x=54.故m的取值范圍為-,54.4.(2016全國24)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.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.(分類討論)當(dāng)a1時,等價于1-a+a3,無解.當(dāng)a1時,等價于a-1+a3,解得a2.所以a的取值范圍是2,+).新題演練提能刷高分1.(2018江西新課程質(zhì)量監(jiān)測)已知函數(shù)f(x)=|x+1|-|x-a|,其中a為實(shí)數(shù).(1)當(dāng)a=1時,解不等式f(x)1;(2)當(dāng)x0,+)時,不等式f(x)2恒成立,求a的取值范圍.解(1)當(dāng)a=1時,f(x)=|x+1|-|x-1|=-2,x1,故f(x)1x12,即不等式f(x)1的解集是12,+.(2)當(dāng)x0,+)時,f(x)2|x+1|-|x-a|2x+1-|x-a|x-1,當(dāng)x0,1)時,x-1x-1或a-xx-1,求得a1.綜上,a(-,1).2.(2018山東濟(jì)南一模)已知函數(shù)f(x)=|2x-2|-|x+2|.(1)求不等式f(x)6的解集;(2)當(dāng)xR時,f(x)-x+a恒成立,求實(shí)數(shù)a的取值范圍.解(1)當(dāng)x-2時,f(x)=-x+4,f(x)6-x+46x-2,故x-2;當(dāng)-2x1時,f(x)=-3x,f(x)6-3x6x-2,故x;當(dāng)x1時,f(x)=x-4,f(x)6x-46x10,故x10.綜上可知:f(x)6的解集為(-,-210,+).(2)由(1)知:f(x)=-x+4,x-2,-3x,-2x1,x-4,x1,當(dāng)x-2時,-x+4-x+a恒成立,a4,當(dāng)-2x1時,-3x-x+a恒成立,a-2,當(dāng)x1時,x-4-x+a恒成立,a-2.綜上,實(shí)數(shù)a的取值范圍為(-,-2.3.(2018山西一模)已知函數(shù)f(x)=|x-1|-a(aR).(1)若f(x)的最小值不小于3,求a的最大值;(2)若g(x)=f(x)+2|x+a|+a的最小值為3,求a的值.解(1)因為f(x)min=f(1)=-a,所以-a3,解得a-3,即amax=-3;(2)g(x)=f(x)+2|x+a|+a=|x-1|+2|x+a|,當(dāng)a=-1時,g(x)=3|x-1|0,03,所以a=-1不符合題意,當(dāng)a-1時,g(x)=(x-1)+2(x+a),x-a,(x-1)-2(x+a),1x-a,-(x-1)-2(x+a),x1,即g(x)=3x-1+2a,x-a,-x-1-2a,1x-a,-3x+1-2a,x-1時,同法可知g(x)min=g(-a)=a+1=3,解得a=2.綜上,a=2或-4.4.(2018山東濰坊一模)設(shè)函數(shù)f(x)=|ax+1|+|x-a|(a0),g(x)=x2+x.(1)當(dāng)a=1時,求不等式g(x)f(x)的解集;(2)已知f(x)32,求a的取值范圍.解(1)當(dāng)a=1時,不等式g(x)f(x),即x2+x|x+1|+|x-1|,當(dāng)x1時,x2+x2x,x2-x0,x1或x0,此時,x1,不等式的解集為x|x-3或x1.(2)f(x)=|ax+1|+|x-a|=-(a+1)x+a-1,xa.若01,則f(x)min=f-1a=a+1a232,a1.綜上所述,a22.5.(2018山西太原二模)已知函數(shù)f(x)=|x+m|+|2x-1|(m0).(1)當(dāng)m=1時,解不等式f(x)3;(2)當(dāng)xm,2m2時,不等式12f(x)|x+1|恒成立,求實(shí)數(shù)m的取值范圍.解(1)當(dāng)m=1時,f(x)=|x+1|+|2x-1|=-3x,x12,由f(x)3解得x-1或x1.(2)由題意得m0,m12,當(dāng)xm,2m2時,不等式12f(x)|x+1|等價于x+m+2x-12(x+1),x3-m,2m23-m,解得120,b0,a3+b3=2.證明:(1)(a+b)(a5+b5)4;(2)a+b2.證明(1)(a+b)(a5+b5)=a6+ab5+a5b+b6=(a3+b3)2-2a3b3+ab(a4+b4)=4+ab(a2-b2)24.(2)因為(a+b)3=a3+3a2b+3ab2+b3=2+3ab(a+b)2+3(a+b)24(a+b)=2+3(a+b)34,所以(a+b)38,因此a+b2.2.(2016全國24)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|.(1)證明f(x)=-2x,x-12,1,-12x12,2x,x12.當(dāng)x-12時,由f(x)2得-2x-1;當(dāng)-12x12時,f(x)2;當(dāng)x12時,由f(x)2得2x2,解得x1.所以f(x)2的解集M=x|-1x1.(2)解由(1)知,當(dāng)a,bM時,-1a1,-1b1,從而(a+b)2-(1+ab)2=a2+b2-a2b2-1=(a2-1)(1-b2)0.因此|a+b|1+ab|.新題演練提能刷高分1.(2018湖南、江西十四校第一次聯(lián)考)已知函數(shù)f(x)=|x-1|-|x+2|.(1)若不等式f(x)|m-1|有解,求實(shí)數(shù)m的最大值M;(2)在(1)的條件下,若正實(shí)數(shù)a,b滿足3a2+b2=M,證明:3a+b4.(1)解若不等式f(x)|m-1|有解,只需f(x)的最大值f(x)max|m-1|即可.因為|x-1|-|x+2|(x-1)-(x+2)|=3,所以|m-1|3,解得-2m4,所以實(shí)數(shù)m的最大值M=4.(2)證明根據(jù)(1)知正實(shí)數(shù)a,b滿足3a2+b2=4,由柯西不等式可知(3a2+b2)(3+1)(3a+b)2,所以,(3a+b)216,因為a,b均為正實(shí)數(shù),所以3a+b4(當(dāng)且僅當(dāng)a=b=1時取“=”).2.(2018安徽江南十校3月聯(lián)考)已知函數(shù)f(x)=|x+2|+|2x+a|,aR.(1)當(dāng)a=1時,解不等式f(x)2;(2)求證:f(x)|a-2|-12|a|.(1)解當(dāng)a=1時,f(x)=|x+2|+|2x+1|2x-2,-3x-32或-2x-12,-x+12或x-12,3x+32x-2或-2x-1或x-13x-1或x-13,所以不等式的解集為xx-1或x-13.(2)證明f(x)=|x+2|+|2x+a|=|x+2|+x+a2+x+a22-a2+x+a22-a2=a2-2=(a-2)-12a|a-2|-12a=|a-2|-12|a|.3.(2018吉林長春質(zhì)量監(jiān)測二)已知函數(shù)f(x)=|2x-3|+|3x-6|.(1)求f(x)2的解集;(2)若f(x)的最小值為T,正數(shù)a,b滿足a+b=12,求證:a+bT.(1)解f(x)=|2x-3|+|3x-6|=3-2x+6-3xx2)=-5x+9x2).由圖象可知:f(x)2的解集為75,115.(2)證明由圖象可知f(x