（天津?qū)Ｓ茫?020屆高考數(shù)學(xué)一輪復(fù)習(xí) 考點(diǎn)規(guī)范練3 基本不等式及其應(yīng)用（含解析）新人教A版

考點(diǎn)規(guī)范練3基本不等式及其應(yīng)用一、基礎(chǔ)鞏固1.下列不等式一定成立的是()A.lgx2+14>lg x(x>0)B.sin x+1sinx2(xk,kZ)C.x2+12|x|(xR)D.1x2+1>1(xR)2.已知a>0,b>0,a+b=2,則y=1a+4b的最小值是()A.72B.4C.92D.53.已知a>0,b>0,a+b=1a+1b,則1a+2b的最小值為()A.4B.22C.8D.164.已知不等式2x+m+8x-1>0對一切x(1,+)恒成立,則實(shí)數(shù)m的取值范圍是()A.m>-10B.m<-10C.m>-8D.m<-85.若正數(shù)x,y滿足4x2+9y2+3xy=30,則xy的最大值是()A.43B.53C.2D.546.若兩個(gè)正實(shí)數(shù)x,y滿足2x+1y=1,且x+2y>m2+2m恒成立,則實(shí)數(shù)m的取值范圍是()A.(-,-2)4,+)B.(-,-42,+)C.(-2,4)D.(-4,2)7.設(shè)x,yR,a>1,b>1,若ax=by=3,a+b=23,則1x+1y的最大值為()A.2B.32C.1D.128.已知x>1,則logx9+log27x的最小值是. 9.已知a>0,b>0,且2a+b=1,求證:2+1a1+2b16+83.二、能力提升10.已知不等式2x2-axy+y20對任意x1,2及y1,3恒成立,則實(shí)數(shù)a的取值范圍是()A.a22B.a22C.a113D.a9211.(2018天津,文13)已知a,bR,且a-3b+6=0,則2a+18b的最小值為. 12.已知實(shí)數(shù)x,y滿足x>y>0,且x+y=1,求4x+3y+1x-y的最小值.13.某工廠某種產(chǎn)品的年固定成本為250萬元,每生產(chǎn)x千件,需另投入成本為C(x)(單位:萬元),當(dāng)年產(chǎn)量不足80千件時(shí),C(x)=13x2+10x.當(dāng)年產(chǎn)量不少于80千件時(shí),C(x)=51x+10000x-1 450.每件商品售價(jià)為0.05萬元.通過市場分析,該廠生產(chǎn)的商品能全部售完.(1)寫出年利潤L(x)(單位:萬元)關(guān)于年產(chǎn)量x(單位:千件)的函數(shù)解析式;(2)當(dāng)年產(chǎn)量為多少千件時(shí),該廠在這一商品的生產(chǎn)中所獲利潤最大?三、高考預(yù)測14.已知正實(shí)數(shù)a,b,c滿足a2-ab+4b2-c=0,當(dāng)cab取最小值時(shí),a+b-c的最大值為()A.2B.34C.38D.14考點(diǎn)規(guī)范練3基本不等式及其應(yīng)用1.C解析因?yàn)閤>0,所以x2+142·x·12=x,所以lgx2+14lgx(x>0),故選項(xiàng)A不正確;當(dāng)xk,kZ時(shí),sinx的正負(fù)不定,故選項(xiàng)B不正確;由基本不等式可知選項(xiàng)C正確;當(dāng)x=0時(shí),有1x2+1=1,故選項(xiàng)D不正確.2.C解析由題意,得1a+4b=121a+4b(a+b)=125+ba+4ab125+2ba·4ab=92,當(dāng)且僅當(dāng)a+b=2,ba=4ab,a>0,b>0,即a=23,b=43時(shí)取等號(hào),故1a+4b的最小值是92.3.B解析由a>0,b>0,a+b=1a+1b=a+bab,得ab=1.則1a+2b21a·2b=22,當(dāng)且僅當(dāng)1a=2b,即a=22,b=2時(shí)等號(hào)成立.故選B.4.A解析原不等式可化為-m<2x+8x-1.令f(x)=2x+8x-1(x>1),則f(x)=2(x-1)+8x-1+222(x-1)·8x-1+2=10,即當(dāng)2(x-1)=8x-1時(shí),f(x)取最小值10.因此要使不等式恒成立,應(yīng)滿足-m<10,解得m>-10.5.C解析由x>0,y>0,得4x2+9y2+3xy2×(2x)×(3y)+3xy(當(dāng)且僅當(dāng)2x=3y時(shí)等號(hào)成立).則12xy+3xy30,即xy2,故xy的最大值為2.6.D解析因?yàn)閤>0,y>0,2x+1y=1,所以x+2y=(x+2y)2x+1y=2+4yx+xy+28,當(dāng)且僅當(dāng)4yx=xy,即x=2y時(shí)等號(hào)成立.由x+2y>m2+2m恒成立,可知m2+2m<8,即m2+2m-8<0,解得-4<m<2.7.C解析由ax=by=3,得1x+1y=1loga3+1logb3=lga+lgblg3=lg(ab)lg3.又a>1,b>1,所以aba+b22=3,當(dāng)且僅當(dāng)a=b時(shí)等號(hào)成立.所以lg(ab)lg3,從而1x+1ylg3lg3=1,當(dāng)且僅當(dāng)a=b=3時(shí)等號(hào)成立.8.263解析x>1,logx9+log27x=2lg3lgx+lgx3lg3223=263,當(dāng)且僅當(dāng)x=36時(shí)等號(hào)成立.logx9+log27x的最小值為263.9.證明2+1a1+2b=2+2a+ba1+2(2a+b)b=4+ba3+4ab=12+16ab+3ba+4=16+16ab+3ba.因?yàn)閍>0,b>0,所以16ab+3ba216ab·3ba=83,當(dāng)且僅當(dāng)16ab=3ba,即3b=4a時(shí)取等號(hào).所以2+1a1+2b=16+16ab+3ba16+83.10.A解析因?yàn)?x2-axy+y20,且y0,所以2xy2-a·xy+10.令t=xy,則不等式變?yōu)?t2-at+10.由x1,2,y1,3,可知t13,2,即2t2-at+10在t13,2時(shí)恒成立.由2t2-at+10可得a2t2+1t,即a2t+1t.又2t+1t22t·1t=22,當(dāng)且僅當(dāng)2t=1t,即t=22時(shí)等號(hào)成立,所以2t+1t取得最小值22,所以有a22,故選A.11.14解析a-3b+6=0,a-3b=-6.a,bR,2a>0,18b>0.2a+18b22a-3b=22-6=14,當(dāng)且僅當(dāng)2a=18b,即a=-3,b=1時(shí)取等號(hào).12.解x>y>0,x+y=1,4x+3y+1x-y=2(x+3y+x-y)x+3y+x+3y+x-y2(x-y)=2+2(x-y)x+3y+12+x+3y2(x-y)=2(x-y)x+3y+x+3y2(x-y)+522+52=92,當(dāng)且僅當(dāng)2(x-y)x+3y=x+3y2(x-y),即x=56,y=16時(shí)等號(hào)成立.4x+3y+1x-y的最小值是92.13.解(1)因?yàn)槊考唐肥蹆r(jià)為0.05萬元,所以x千件商品的銷售額為(0.05×1000x)萬元.依題意得,當(dāng)0<x<80時(shí),L(x)=(0.05×1000x)-13x2-10x-250=-13x2+40x-250;當(dāng)x80時(shí),L(x)=(0.05×1000x)-51x-10000x+1450-250=1200-x+10000x,則L(x)=-13x2+40x-250,0<x<80,1200-x+10000x,x80.(2)當(dāng)0<x<80時(shí),L(x)=-13(x-60)2+950,此時(shí),當(dāng)x=60時(shí),L(x)取得最大值L(60)=950.當(dāng)x80時(shí),L(x)=1200-x+10000x1200-2x·10000x=1200-200=1000,當(dāng)且僅當(dāng)x=10000x,即x=100時(shí),L(x)取得最大值1000.因?yàn)?50<1000,所以當(dāng)年產(chǎn)量為100千件時(shí),該廠在這一商品的生產(chǎn)中所獲利潤最大,最大利潤為1000萬元.14.C解析由正實(shí)數(shù)a,b,c滿足a2-ab+4b2-c=0,可得c=a2-ab+4b2.故cab=a2-ab+4b2ab=ab+4ba-12ab·4ba-1=3,當(dāng)且僅當(dāng)a=2b時(shí)取等號(hào).則當(dāng)a=2b時(shí),cab取得最小值,且c=6b2.因此,a+b-c=2b+b-6b2=-6b2+3b=-6b-142+38,當(dāng)b=14時(shí),a+b-c有最大值為38.故選C.6