高考數(shù)學(xué)一輪復(fù)習(xí)專題二不等式3基本不等式與不等式的綜合應(yīng)用專題檢測(cè)含解析新人教A版

PAGEPAGE4基本不等式與不等式的綜合應(yīng)用專題檢測(cè)1.(2020山東師大附中第一次月考,12)下列不等式一定成立的是()A.lgx2+14>lgx(x>0)B.sinx+1sinx≥2(xC.x2+1≥2|x|(x∈R)D.1x2+1>1(答案C本題主要考查應(yīng)用基本不等式求最值,考查的核心素養(yǎng)是邏輯推理.對(duì)于A,由于x2+14≥2x2·14=x,當(dāng)且僅當(dāng)x=12時(shí),取“=對(duì)于B,當(dāng)x∈(π,2π)時(shí),sinx<0,sinx+1sinx≤-2,故B不正確;對(duì)于C,x2+1-2|x|=(|x|-1)2≥0恒成立,故C對(duì)于D,當(dāng)x=0時(shí),1x2+1=1,故2.(2020西南四省八校9月聯(lián)考,12)若x>0,y>0,x+2y=1,則xy2x+yA.14B.15C.19答案Cxy2x+y=12y+1x,∵x>0,y>0,x+2y=1,∴1x+2y=1x+2y·1=1x+2y(x+2y)=5+2y∴12y+1x≤19,故xy3.(2020山東青島期初調(diào)研,8)函數(shù)f(x)=x2+x+2x+4x2(x>0)A.4+22B.42C.8D.2+2答案A∵x>0,∴f(x)=x2+x+2x+4x2=x2+4x2+x+2x≥2x2·4x2+2x·2x=4+22,當(dāng)且僅當(dāng)x2=4x4.(2018福建廈門(mén)外國(guó)語(yǔ)中學(xué)模擬,10)已知實(shí)數(shù)a>0,b>0,1a+1+1b+1=1,則a+2bA.32B.22C.3D.2答案B∵a>0,b>0,∴a+1>1,b+1>1,又∵1a+1+1b+1=1,∴a+2b=[(a+1)+2(b+1)]-3=[(a+1)+2(b+1)]·1a+1+1b+1-3=1+2(b+1)a+1+a+1b5.(2018河北大名一中月考)已知關(guān)于x的不等式x2-4ax+3a2<0(a<0)的解集為(x1,x2),則x1+x2+ax1x2A.63B.233C.4答案D由題意知x1,x2是方程x2-4ax+3a2=0的兩根.由根與系數(shù)的關(guān)系得x1x2=3a2,x1+x2=4a,∴x1+x2+ax1x2=4a+13a,∵a<0,∴-4a+13a≥24a·13a=433,即4a+13a故x1+x2+ax1x2的最大值為-6.(2019晉冀魯豫名校期末聯(lián)考,10)已知函數(shù)f(x)=x2ex,若a>0,b>0,p=fa2+b22,q=fa+b22,r=fA.q≤r≤pB.q≤p≤rC.r≤p≤qD.r≤q≤p答案D因?yàn)閍2+b22-a+b2所以a2+b22≥a+b22,又a+b2≥ab(a>0,b>0),所以a+b22≥ab.易得函數(shù)f(x)=x2ex在(0,+∞)上單調(diào)遞增,所以f7.(2020河南濮陽(yáng)第二次檢測(cè),9)已知a>2,b>2,則a2b-2+bA.2B.4C.6D.16答案D因?yàn)閍>2,b>2,所以a-2>0,b-2>0.令x=b-2,y=a-2,則x>0,y>0.原式=(y+2)2x+(x+2)2y≥2(y+2)2當(dāng)且僅當(dāng)x=y=2時(shí)取等號(hào).故選D.思路分析利用換元思想,設(shè)x=b-2,y=a-2,則x>0,y>0,將原式化為(y+2)2x8.(2019新疆昌吉教育共同體聯(lián)考,9)在1和17之間插入(n-2)個(gè)數(shù),使這n個(gè)數(shù)成等差數(shù)列,若這(n-2)個(gè)數(shù)中第一個(gè)為a,第(n-2)個(gè)為b,當(dāng)1a+25b取最小值時(shí),n的值為 (A.6B.7C.8D.9答案D由已知得a+b=18,則1a+25b=1a+25b×a+b18=1181+25+ba+25ab≥1189.(2019遼寧沈陽(yáng)東北育才學(xué)校五模,9)已知函數(shù)f(x)=2x-12x+1+x+sinx,若正實(shí)數(shù)a,b滿足f(4a)+f(b-9)=0,則1A.1B.92C.9答案A因?yàn)閒(x)=2x-12x+1+x+sinx,所以f(-x)=2-x-12-x+1-x-sinx=-2x-12x+1+x+sinx=-f(x),所以函數(shù)f(x)為奇函數(shù),易知f(x)單調(diào)遞增,又正實(shí)數(shù)a,b滿足f(4a)+f(b-9)=0,所以4a+b-9=0,所以1a+1b=191a+1b(4a+b)=10.(2020黑龍江道里檢測(cè),10)設(shè)a,b,c,d均為大于零的實(shí)數(shù),且abcd=1,令m=a(b+c+d)+b(c+d)+cd,則a2+b2+m的最小值為 ()A.8B.4+23C.5+23D.43答案B∵a,b,c,d均大于零且abcd=1,m=a(b+c+d)+b(c+d)+cd,∴a2+b2+m=a2+b2+(a+b)(c+d)+ab+cd≥2ab+2ab·2cd+ab+cd=4+3ab+cd≥4+23abcd=4+23,當(dāng)且僅當(dāng)a=b,c=d,3ab=cd,即a=b=1314,c=d=314時(shí)取等號(hào),∴a2+b2+m的最小值為11.(多選題)(2020山東煙臺(tái)期中,11)下列結(jié)論正確的是 ()A.若a>b>0,c<d<0,則一定有bc>B.若x>y>0,且xy=1,則x+1y>y2x>log2(xC.設(shè){an}是等差數(shù)列,若a2>a1>0,則a2>aD.若x∈[0,+∞),則ln(1+x)≥x-18x答案AC對(duì)于A,∵c<d<0,∴-c>-d>0,∴-1d>-1又∵a>b>0,∴-ad>-bc>0,∴bc>ad,對(duì)于B,∵x>y>0,且xy=1,∴可取x=2,y=12,此時(shí)x+1y=4,y2x=124=18,log2(x+y)=log252>log22=1,故不滿足x+1y>y2x對(duì)于C,∵{an}是等差數(shù)列,∴a2=a1+a32.又∵a3-a2=a2-a1>0,∴a3>a2>a1>0,∴a1+a32>a1對(duì)于D,令f(x)=ln(1+x)-x+18x2,x≥0,則f'(x)=11+x-1+14x=1-(1+x)+14x(1+x)1+x=14x2-34x1+x=x2-3x4(1+x),x>0,令∵f(0)=ln1-0+0=0,∴當(dāng)x∈(0,3]時(shí),f(x)<0恒成立,故當(dāng)x∈[0,+∞)時(shí),ln(1+x)≥x-18x2不恒成立,故D不正確,故選AC12.(2019湖北黃岡元月調(diào)研,15)若關(guān)于x的不等式x+4x-a≥5在x∈(a,+∞)上恒成立,則實(shí)數(shù)a答案1解析關(guān)于x的不等式x+4x-a≥5在x∈(a,+∞)上恒成立,即x-a+4x-a≥5-a在x∈(a,+∞)上恒成立,由x>a可得x-a>0,則x-a+4x-a≥2(x-a)·4x-a=4,當(dāng)且僅當(dāng)x-a=2,即x=a+2時(shí)13.(2020上海復(fù)旦大學(xué)附中9月綜合練,8)已知a2+2a+2x≤4x2-x+1對(duì)于任意的x答案[-3,1]解析由已知a2+2a+2x≤4x2-x+1對(duì)于任意的x∈(1,+∞)恒成立可知,a2+2a+2≤4x-1+x對(duì)于任意的x∈(1,+∞)恒成立,令g(x)=4x-1+x,x>1,則g(x)=4x-1+x-1+1≥24x-1·(x-1)+1=5,當(dāng)且僅當(dāng)x=3時(shí)取“=”,∴14.(2019安徽黃山八校聯(lián)考,16)不等式(acos2x-3)sinx≥-3對(duì)任意x∈R恒成立,則實(shí)數(shù)a的取值范圍是.

答案-解析令g(x)=(acos2x-3)sinx,sinx=t,-1≤t≤1,則原函數(shù)化為g(t)=(-at2+a-3)t,即