2.2函數(shù)的單調(diào)性和最值考點(diǎn)函數(shù)的單調(diào)性1.(2023新課標(biāo)Ⅰ,4,5分,易)設(shè)函數(shù)f(x)=2x(x-a)在區(qū)間(0,1)單調(diào)遞減,則a的取值范圍是()A.(-∞,-2]B.[-2,0)C.(0,2]D.[2,+∞)答案Df(x)=2x(x-a)=2(x?a2)2?a24,由復(fù)合函數(shù)的單調(diào)性知函數(shù)y=x?a22-a24在(0,1)上單調(diào)遞減,所以2.(2024新課標(biāo)Ⅰ,6,5分,中)已知函數(shù)f(x)=-x2-2ax-a,x<0,eA.(-∞,0] B.[-1,0] C.[-1,1] D.[0,+∞)6B當(dāng)x≥0時(shí),函數(shù)f(x)顯然是增函數(shù);當(dāng)x<0時(shí),f(x)=-x2-2ax-a=-(x+a)2+a2-a,而f(x)在R上單調(diào)遞增,所以-a≥0,-a≤e0+ln(0+1),則-1易錯(cuò)警示該題容易只考慮當(dāng)x≥0時(shí),函數(shù)f(x)是增函數(shù),及當(dāng)x<0時(shí),函數(shù)f(x)是增函數(shù),從而得到a≤0,而忽視了函數(shù)分界點(diǎn)處的函數(shù)值大小.3.(2023全國(guó)甲文,11,5分,中)已知函數(shù)f(x)=e?(x?1)2.記a=f22,b=f32,c=fA.b>c>aB.b>a>cC.c>b>aD.c>a>b答案A∵f(x)=e?(x?1)2是由y=eu和u=-(x∴f(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,又知f(2-x)=e?(2?x?1)2=e?(1?∴f(x)的圖象關(guān)于直線x=1對(duì)稱,∴f62=f2?又∵22<2-62<32<1,∴f22<f2?6即a<c<b,故選A.4.(2023北京,4,4分,易)下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞增的是()A.f(x)=-lnxB.f(x)=1C.f(x)=-1xD.f(x)=3|x答案C對(duì)于A,f(x)=-lnx在(0,+∞)上單調(diào)遞減,不符合題意;對(duì)于B,f(x)=12x在(0,+∞)上單調(diào)遞減,對(duì)于C,f(x)=-1x在(0,+∞)上單調(diào)遞增,符合題意對(duì)于D,f(x)=3|x-1|=3x?1,x≥1,13x?1,5.(2021全國(guó)甲文,4,5分)下列函數(shù)中是增函數(shù)的為()A.f(x)=-xB.f(x)=2C.f(x)=x2D.f(x)=3答案D解題指導(dǎo):排除法,利用基本初等函數(shù)的性質(zhì)逐一判斷四個(gè)選項(xiàng).解析對(duì)于f(x)=-x,由正比例函數(shù)的性質(zhì)可知,f(x)是減函數(shù),故A不符合題意;對(duì)于f(x)=23x,由指數(shù)函數(shù)的單調(diào)性可知,f(x)是減函數(shù),故B對(duì)于f(x)=x2,由二次函數(shù)的圖象可知,f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,故C不符合題意;對(duì)于f(x)=3x=x13,由冪函數(shù)的性質(zhì)可知,f(x)在(-∞,+∞)方法總結(jié):一次函數(shù)y=kx+b(k≠0)單調(diào)性的判斷:若k>0,則函數(shù)在R上單調(diào)遞增;若k<0,則函數(shù)在R上單調(diào)遞減.指數(shù)函數(shù)y=ax(a>0且a≠1)單調(diào)性的判斷:若a>1,則函數(shù)在R上單調(diào)遞增;若0<a<1,則函數(shù)在R上單調(diào)遞減.冪函數(shù)y=xα單調(diào)性的判斷:若α>0,則函數(shù)在(0,+∞)上單調(diào)遞增;若α<0,則函數(shù)在(0,+∞)上單調(diào)遞減.6.(2021全國(guó)乙文,8,5分)下列函數(shù)中最小值為4的是()A.y=x2+2x+4B.y=|sinx|+4C.y=2x+22-xD.y=lnx+4答案C解題指導(dǎo):對(duì)于A,利用配方法或二次函數(shù)的單調(diào)性求最值,對(duì)于B,C,D,利用換元法轉(zhuǎn)化為對(duì)勾函數(shù)進(jìn)行判斷.解析對(duì)于A,y=x2+2x+4=(x+1)2+3≥3,所以它的最小值為3,所以A不符合題意;對(duì)于B,設(shè)|sinx|=t,則0<t≤1,y=|sinx|+4|sinx=t+4t,t∈(0,1],易知y=t+4t在(0,1]上單調(diào)遞減,故t=1時(shí),ymin=1+41=5,所以B不符合題意;對(duì)于C,令2x=t(t>0),則y=2x+22-x=t+4t,t>0,易知y=t+4t在(0,2)上單調(diào)遞減,在(2,+∞)上單調(diào)遞增,所以當(dāng)t=2時(shí),y取最小值,ymin=2+42=4,故C符合題意;對(duì)于D,令lnx=t,t∈R且t≠0,則y=lnx+4ln7.(2020新高考Ⅰ,8,5分)若定義在R的奇函數(shù)f(x)在(-∞,0)單調(diào)遞減,且f(2)=0,則滿足xf(x-1)≥0的x的取值范圍是()A.[-1,1]∪[3,+∞)B.[-3,-1]∪[0,1]C.[-1,0]∪[1,+∞)D.[-1,0]∪[1,3]答案D∵f(x)是定義在R上的奇函數(shù),∴f(x-1)的圖象關(guān)于點(diǎn)(1,0)中心對(duì)稱,又∵f(x)在(-∞,0)上單調(diào)遞減,∴f(x-1)在(-∞,1)上單調(diào)遞減,在(1,+∞)上也單調(diào)遞減,且過(guò)(-1,0)和(3,0),f(x-1)的大致圖象如圖:當(dāng)-1≤x≤0時(shí),f(x-1)≤0,∴xf(x-1)≥0;當(dāng)1≤x≤3時(shí),f(x-1)≥0,∴xf(x-1)≥0.綜上,滿足xf(x-1)≥0的x的取值范圍是[-1,0]∪[1,3].故選D.8.(2019北京文,3,5分)下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞增的是()A.y=x12C.y=log12答案A本題主要考查指數(shù)函數(shù)、對(duì)數(shù)函數(shù)、冪函數(shù)的單調(diào)性,考查數(shù)形結(jié)合的思想.考查的核心素養(yǎng)是直觀想象.A選項(xiàng),12>0,所以冪函數(shù)y=x12在B選項(xiàng),指數(shù)函數(shù)y=2-x=12x在(0,+∞)C選項(xiàng),因?yàn)?<12<1,所以對(duì)數(shù)函數(shù)y=log12x在D選項(xiàng),反比例函數(shù)y=1x在(0,+∞)上單調(diào)遞減解題關(guān)鍵熟練掌握基本初等函數(shù)的圖象和性質(zhì)是解決本題的關(guān)鍵.9.(2016北京文,4,5分)下列函數(shù)中,在區(qū)間(-1,1)上為減函數(shù)的是()A.y=11?xC.y=ln(x+1)D.y=2-x答案D選項(xiàng)A中,y=11?x=1?(x?1)的圖象是將y=-1x的圖象向右平移1個(gè)單位得到的,故y=11?x在(-1,1)上為增函數(shù),不符合題意;選項(xiàng)B中,y=cosx在(-1,0)上為增函數(shù),在(0,1)上為減函數(shù),不符合題意;選項(xiàng)C中,y=ln(x+1)的圖象是將y=lnx的圖象向左平移1個(gè)單位得到的,評(píng)析本題考查了基本函數(shù)的圖象和性質(zhì)以及圖象的變換,屬中檔題.10.(2015課標(biāo)Ⅱ文,12,5分)設(shè)函數(shù)f(x)=ln(1+|x|)-11+x2,則使得f(x)>f(2x-1)成立的A.13,1B.?C.?13,13答案A當(dāng)x>0時(shí),f(x)=ln(1+x)-11+x2,∴f'(x)=11+x+2x(1+x2)2∴|x|>|2x-1|,即3x2-4x+1<0,解得13<x<1,故選11.(2016浙江,7,5分)已知函數(shù)f(x)滿足:f(x)≥|x|且f(x)≥2x,x∈R.()A.若f(a)≤|b|,則a≤bB.若f(a)≤2b,則a≤bC.若f(a)≥|b|,則a≥bD.若f(a)≥2b,則a≥b答案B依題意得f(a)≥2a,若f(a)≤2b,則2a≤f(a)≤2b,∴2a≤2b,又y=2x是R上的增函數(shù),∴a≤b.故選B.12.(2023北京,15,5分,難)設(shè)a>0,函數(shù)f(x)=x+2,x①f(x)在區(qū)間(a-1,+∞)上單調(diào)遞減;②當(dāng)a≥1時(shí),f(x)存在最大值;③設(shè)M(x1,f(x1))(x1≤a),N(x2,f(x2))(x2>a),則|MN|>1;④設(shè)P(x3,f(x3))(x3<-a),Q(x4,f(x4))(x4≥-a).若|PQ|存在最小值,則a的取值范圍是0,1其中所有正確結(jié)論的序號(hào)是.

答案②③解析f(x)的大致圖象如圖所示,易知f(x)在(-∞,-a)上單調(diào)遞增,在[-a,0)上單調(diào)遞增,在[0,a]上單調(diào)遞減,在(a,+∞)上單調(diào)遞減.對(duì)于①,當(dāng)12<a<1時(shí),f(x)在(a-1,0)上單調(diào)遞增,故①錯(cuò)誤對(duì)于②,當(dāng)x<-a時(shí),f(x)<-a+2≤1,當(dāng)-a≤x≤a時(shí),0≤f(x)≤a,當(dāng)x>a時(shí),f(x)<-a-1≤-2.綜上,x=0時(shí),f(x)取得最大值a,故②正確.對(duì)于③,令M'(a,0),N'(a,-a-1),顯然|MN|>|M'N'|=a+1>1,故③正確.對(duì)于④,若|PQ|存在最小值,則點(diǎn)(0,0)到直線x+2=y的距離大于a,且直線y=-x與y=x+2的交點(diǎn)(-1,1)在射線y=x+2(x<-a)上,則21+1>a,且-1<-a,又a>0,所以0<a<1,故④錯(cuò)誤綜上,所有正確結(jié)論的序號(hào)是②③.13.(2016北京文,10,5分)函數(shù)f(x)=xx?1(x≥2)答案2解析解法一:∵f'(x)=?1(x?1)2,∴x≥2時(shí)∴f(x)在[2,+∞)上單調(diào)遞減,∴f(x)在[2,+∞)上的最大值為f(2)=2.解法二:∵f(x)=xx?1=x∴f(x)的圖象是將y=1x的圖象向右平移1個(gè)單位,再向上平移1個(gè)單位得到的.∵y=1x在[2,+∞)∴f(x)在[2,+∞)上單調(diào)遞減,故f(x)在[2,+∞)上的最大值為f(2)=2.解法三:由題意可得f(x)=1+1x∵x≥2,∴x-1≥1,∴0<1x?∴1<1+1x?1≤2,即1<故f(x)在[2,+∞)上的最大值為2.評(píng)析本題考查函數(shù)的最值,有多種解法,屬中檔題.14.(2015浙江文,12,6分)已知函數(shù)f(x)=x2,x≤1,x+6答案-12;26解析f(-2)=(-2)2=4,f(f(-2))=f(4)=4+64-6=-1當(dāng)x≤1時(shí),f(x)=x2≥0,當(dāng)x>1時(shí),f(x)=x+6x-6≥26當(dāng)且僅當(dāng)x=6時(shí),