29 微專題26　零點問題 【正文】教師

微專題26零點問題[備選理由]例1考查與分段函數(shù)有關(guān)的零點問題,考查當(dāng)參數(shù)不同時判斷函數(shù)零點個數(shù)以及根據(jù)零點個數(shù)求參數(shù)的取值范圍,考查數(shù)形結(jié)合思想;例2考查能轉(zhuǎn)化為零點(方程的解)問題的圖象問題,需要根據(jù)不同的條件構(gòu)造不同的函數(shù),考查函數(shù)的性質(zhì)和導(dǎo)數(shù)的應(yīng)用,綜合性較強;例3考查指數(shù)復(fù)合函數(shù)有關(guān)的根據(jù)零點個數(shù)求參數(shù)的取值范圍以及零點偏移問題,注意分類討論思想在解決零點個數(shù)問題中的應(yīng)用及同構(gòu)、指對函數(shù)放縮方法的應(yīng)用;例4考查導(dǎo)數(shù)的幾何意義,考查根據(jù)零點個數(shù)求參數(shù)的取值范圍,考查數(shù)形結(jié)合思想.1[配例1使用](多選題)[2023·遼陽二模]已知函數(shù)f(x)=xlnx,x>0,-x2-2x+1,x≤0,函數(shù)g(x)=[f(x)]2-(a-1)f(A.若a<-1e,則g(x)恰有2B.若1≤a<2,則g(x)恰有4個零點C.若g(x)恰有3個零點,則a的取值范圍是[0,1)D.若g(x)恰有2個零點,則a的取值范圍是-∞,-1e∪(2,[解析]令g(x)=[f(x)]2-(a-1)f(x)-a=0,得[f(x)-a][f(x)+1]=0,解得f(x)=-1或f(x)=a.當(dāng)x>0時,f(x)=xlnx,f'(x)=lnx+1,由f'(x)>0,得x>1e,由f'(x)<0,得0<x<1e,則f(x)在0,1e上單調(diào)遞減,在1e,+∞上單調(diào)遞增,當(dāng)x=1e時,f(x)取得極小值,極小值為f1e=-1e.當(dāng)x≤0時,f(x)=-x2-2x+1,f(x)在(-∞,-1)上單調(diào)遞增,在(-1,0]上單調(diào)遞減,所以當(dāng)x=-1時,f(x)取得極大值,極大值為f(-1)=2,又f(0)=1,f(1)=0,故f(x)的圖象如圖所示.由圖可知,f(x)=-1有且僅有1個實根.當(dāng)a=-1時,g(x)恰有1個零點,故A中說法錯誤;當(dāng)1≤a<2時,f(x)=a有3個實根,則g(x)恰有4個零點,故B中說法正確;若g(x)恰有3個零點,則f(x)=a恰有2個實根,故a=2或0≤a<1或a=-1e,故C中說法錯誤;若g(x)恰有2個零點,則f(x)=a恰有1個實根,且a≠-1,故a<-1或-2[補充使用](多選題)[2023·廣州一模]已知函數(shù)f(x)=x2+2(x≥0),g(x)=ae-x(a>0),點P,Q分別在函數(shù)y=f(x)和y=g(x)的圖象上,O為坐標(biāo)原點,則下列說法正確的是 (BCD)A.若關(guān)于x的方程f(x)-g(x)=0在[0,1]上無解,則a>3eB.存在點P,Q關(guān)于直線y=x對稱C.若存在點P,Q關(guān)于y軸對稱,則0<a≤2D.若存在點P,Q滿足∠POQ=90°,則0<a≤2[解析]函數(shù)f(x)=x2+2(x≥0),g(x)=ae-x(a>0).對于A,關(guān)于x的方程f(x)-g(x)=0在[0,1]上有解?h(x)=x2+2-ae-x=0在[0,1]上有解,顯然函數(shù)h(x)在[0,1]上單調(diào)遞增,則只需h(0)=2-a≤0,h(1)=3-ae-1≥0,解得2≤a≤3e,故若關(guān)于x的方程f(x)-g(x)=0在[0,1]上無解,則0<a<2或a>3e,A錯誤.對于B,設(shè)點Q(t,ae-t),若Q關(guān)于直線y=x的對稱點(ae-t,t)在函數(shù)f(x)=x2+2(x≥0)的圖象上,則關(guān)于t的方程t=a2e-2t+2有解,即a2=(t-2)e2t有解,此時t>2.令函數(shù)φ(t)=(t-2)e2t,t>2,則φ'(t)=(2t-3)e2t>0,故函數(shù)φ(t)在(2,+∞)上單調(diào)遞增,故φ(t)>φ(2)=0,即函數(shù)φ(t)的值域為(0,+∞),又a2>0,故a2=(t-2)e2t在(2,+∞)上有解,所以存在點P,Q關(guān)于直線y=x對稱,B正確.對于C,設(shè)點P(u,u2+2),u≥0,若點P關(guān)于y軸的對稱點(-u,u2+2)在函數(shù)g(x)=ae-x(a>0)的圖象上,則關(guān)于u的方程aeu=u2+2有解,即a=u2+2eu有解,令F(u)=u2+2eu,u≥0,則F'(u)=-u2+2u-2eu=-(u-1)2+1eu<0,故函數(shù)F(u)在[0,+∞)上單調(diào)遞減,所以F(u)max=F(0)=2,又對任意u∈[0,+∞),恒有F(u)>0,所以F(u)的值域為(0,2],故0<a≤2,C正確.對于D,設(shè)P(x1,x12+2),Q(x2,ae-x2),若∠POQ=90°,則OP·OQ=x1x2+ae-x2(x12+2)=0,顯然x1x2≠0,且x1>0,x2<0,故a=x1x12+2·-x2e-x2.令G(x)=xex,x>0,則G'(x)=1-xex,當(dāng)0<x<1時,G'(x)>0,函數(shù)G(x)單調(diào)遞增,當(dāng)x>1時,G'(x)<0,函數(shù)G(x)單調(diào)遞減,因此G(x)max=G3[配例3使用][2023·山東日照三模]已知函數(shù)f(x)=ax2ex-1-lnx-ln(1)求a的取值范圍;(2)設(shè)函數(shù)f(x)的三個零點由小到大依次是x1,x2,x3,證明:aex1x解:(1)由題意知a>0,f(x)的定義域為(0,+∞),f'(x)=ae(i)當(dāng)x≥2時,f'(x)<0,f(x)單調(diào)遞減.(ii)當(dāng)x∈(0,2)時,令g(x)=2x2-x3ex-1,0<x<當(dāng)x∈(0,1)時,g'(x)>0,當(dāng)x∈(1,2)時,g'(x)<0,所以g(x)在(0,1)上單調(diào)遞增,在(1,2)上單調(diào)遞減,所以g(x)≤g(1)=1,又g(0)=0,g(2)=0,所以0<g(x)≤1.當(dāng)a∈(0,1]時,f'(x)≤0在(0,2)上恒成立,則f(x)在(0,2)上單調(diào)遞減,又f(x)在[2,+∞)上單調(diào)遞減,故f(x)至多有一個零點,與題設(shè)矛盾;當(dāng)a>1時,f'(x)=ag(x)-1ax(0<x<2),記f'(x)的兩個零點為m,n,且0<m<1<n<2,則f(x)在(0,m)上單調(diào)遞減,在(m,n)上單調(diào)遞增,在(n,+∞)上單調(diào)遞減,所以f(n)>f(1)>0.令p(x)=xe1-x-1(0<x<1),則p'(x)=(1-x)e1-x>0(0<x<1),所以p(x)在(0,1)上單調(diào)遞增,因為0<1a<1,所以p1a=1ae1-1a-1<p(1)=e1-1-1=0,所以f1a所以f(m)<0,又當(dāng)x→0時,f(x)→+∞,當(dāng)x→+∞時,f(x)→-∞,故函數(shù)f(x)有三個零點.綜上所述,a>1.(2)證明:f(x)=0等價于aexex=ln(a令t(x)=lnxx,則t'(x)=所以t(x)在(0,e)上單調(diào)遞增,在(e,+∞)上單調(diào)遞減,由(1)可得0<x1<1a<x2<1<x3,則aex1<e,ex1<e,aex3>e,所以t(aex1)=t(ex1),t(aex3)=t(ex3),所以aex1=ex1,a則x1,x3滿足x1-ln要證aex1x3>e,只需證x1x令q(x)=x-lnx,則q'(x)=x-令q'(x)<0,得0<x<1,令q'(x)>0,得x>1,所以函數(shù)q(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增.下面證明x1+x3>1+k,即證x3>1+k-x1,又x1<1<x3,k>1,即證q(x3)>q(1+k-x1),即證k>1+k-x1-ln(1+k-x1),即證0>1-x1-ln(1+x1-lnx1-x1)=1-x1-ln(1-lnx1),即證e1-x1-1+lnx1<0,x令c(x)=e1-x-1+lnx,x∈(0,1),則c'(x)=-xe1-x+1x,令d(x)=-xe1-x+1,x∈(0,1),則d'(x)=(x-1)e1-x<0,所以d(x)單調(diào)遞減,故d所以c'(x)=-xe1-x+1x>0,則c(x)在(0,1)上單調(diào)遞增,故c(x)<c(1)=0,所以e1-x1-1所以x1+x3>1+k,所以x1x3-1≥ln(x1x3)=x1+x3-2k>1+k-2k=1-k,所以x1x3>2-k,得證.4[補充使用]已知函數(shù)f(x)=alnx+bx2e1-x,a,b∈R.(1)若曲線y=f(x)在點(2,f(2))處的切線方程是y=x+ln2,求a和b的值;(2)若a=e,且f(x)的導(dǎo)函數(shù)f'(x)恰有兩個零點,求b的取值范圍.解:(1)因為f'(x)=ax+bx(2-x)e1-x,所以f'(2)=a因為曲線y=f(x)在點(2,f(2))處的切線方程是y=x+ln2,所以f'(2)=1,f(2)=2+ln2,即a2=1,aln2+4be=2+解得a=2,b=(2(2)當(dāng)a=e時,由f'(x)=ex+bx(2-x)e1-x=0,得exx=bx(x-2),顯然x≠故