極值點偏移與拐點偏移問題

11(2)函數(shù)f(x)中存在x1，x2且x1≠x2，滿足f(x1)=f(x2)，求證：x1+x2>2x0(x0為函數(shù)f(x)x,-f'(x0)>0；x-f'(x0)>0.F(x)=f(x)-f(2x,-x).(2024·江蘇揚州·模擬預(yù)測)已知函數(shù)f(x(=ln(mx(-x(m>0(.(1)若f(x(≤0恒成立，求m的取值范圍；(2)若f(x(有兩個不同的零點x1,x2，證明x1+x2>2.22已知f(x(=(x-1(ex+ax2.(1)討論函數(shù)f(x(的單調(diào)性；題目3(2024·全國·模擬預(yù)測)已知函數(shù)f(x)=>0)，且f(x)有兩個相異零點x1,x2.x2>33 題目4(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=-x2+2lnx,g(x(=a(x2+2x(.(1)若曲線f(x(在點(1,-1(處的切線與曲線g(x(有且只有一個公共點，求實數(shù)a的值.(2)若方程g(x(-f(x(=1有兩個不相等的實數(shù)根x1,x2， (1)討論f(x)的單調(diào)性；2-x1<44 題目2(2024·湖南株洲·一模)已知函數(shù)f(x(=(x+a(ebx在(1,f(1((處的切線方程為y=e(x-1(，其中e為自2是方程f(x(=kx2-2(k>2)的兩個不相等的正實根，證明：|x1-x2|>ln(2024·北京朝陽·二模)已知函數(shù)f(x(=ax-ln(1-x((a∈R(.(1)求曲線y=f(x(在點(0,f(0((處的切線方程；(3)若f(x(有兩個不同的零點x1,x2，且|x2-x1|>e-1，求a的取值范圍.55(2024·河南·模擬預(yù)測)已知b>0，函數(shù)f(x(=(x+a(ln(x+b(的圖象在點(1,f(1((處的切線方程為xln2-y-ln2=0.(2)若方程f(x(=為自然對數(shù)的底數(shù))有兩個實數(shù)根x1,x2，且x1<x2，證明：x2-x1<1+(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=ax2-(lnx(2(a∈R(.(2)若f(x(有兩個極值點x1,x2.66②當x1>4x21x>16e3. 若f(x1(=f(x2(，證明：x1?x277 題目4(2024·安徽合肥·模擬預(yù)測)已知函數(shù)f(x)=a(1-2(1)討論f(x)的單調(diào)性；2(x1≠x2(為函數(shù)=kx2+-lnx的兩個零點，求證：(x1x2(4>12e4. 1(2023·全國·模擬預(yù)測)已知函數(shù)f(x(=(1)設(shè)函數(shù)g(x(=ekx-(k>0(，若f(x(≤g(x(恒成立，求k的最小值；88 題目2(23-24高二下·河南平頂山·階段練習(xí))已知函數(shù)f(x(=x2lnx-a(a∈R(. 3(2023·湖北武漢·三模)已知函數(shù)f(x(=ax+(a-1(lnx+，a∈R.(1)討論函數(shù)f(x(的單調(diào)性；(2)若關(guān)于x的方程f(x(=xex-lnx+有兩個不相等的實數(shù)根x1、x2，99已知f(x(=x2- (1)討論f(x(的單調(diào)性；(2)若方程f(x(=1有兩個不同的根x1,x2. 2(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=(1)求函數(shù)f(x(在[1,3[上的值域；(2024·四川涼山·二模)已知函數(shù)f(x(=x+asinx.(1)若函數(shù)f(x(在R上是增函數(shù)，求a的取值范圍；設(shè)g(x(=x-sinx-lnx，若g(x1(=g(x2((x1≠x2(，證明： 題目4(2024·黑龍江哈爾濱·模擬預(yù)測)已知函數(shù)f(x(=x2lnx-m有兩個不同的零點x1，x2，且t=x+x.(2024·四川·一模)已知函數(shù)f(x(=ax2+x-lnx-a.(2)若f(x(有2個零點x1,x2，證明：a(x1+x2(2+(x1+x2(>2. 2(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=x2(alnx-有兩個極值點x1,x2，且x1<x2.(2)證明：x1lnx1+x2lnx2>x1+x2. 題目3(2024·貴州·模擬預(yù)測)已知函數(shù)f(x)=xex+1.(1)求函數(shù)f(x(的單調(diào)區(qū)間； (2)若f(x(存在兩個極值點x1,x2(x2>x1>0(.2-x1+2> 題目1(23-24高三下·四川成都·期末)已知函數(shù)f(x(=ex-ax2.(2)設(shè)函數(shù)g(x(=f(x(-sinx，當a=時，若g(x1(+g(x2(= 的極值點，則(α,φ(α((為曲線y=φ(x(的拐點.已知函數(shù)f(x(=aex-xlnx有兩個極值點x1,x2，且Q(x0,f(x0((為曲線C：y=f(x(的拐點.(2024·全國·模擬預(yù)測)已知函數(shù)f(x)=2lnx+x2+x.(1)求曲線y=f(x)在點(1,f(1))處的切線方程.(2)若正實數(shù)x1,x2滿足f(x1)+f(x2)=4，求證：x1+x2≥2. 題目4(23-24高二下·貴州貴陽·階段練習(xí))設(shè)f/(x(是函數(shù)f(x(的導(dǎo)函數(shù)，若f/(x(可導(dǎo)，則稱函數(shù)f/(x(的導(dǎo)函數(shù)為f(x(的二階導(dǎo)函數(shù)，記為f″(x(.若f″(x(有變號零點x=x0，則稱點(x0,f(x0((為曲線y=f(x(的數(shù)y=f(x(的圖象的對稱中心.已知函數(shù)f(x(=x3+bx2-24x+d的圖象的對稱中心為(1,3(，求函數(shù)f(x((2)已知函數(shù)g(x(=emx-1+mx3-x2+若g(x1(+g(x2(=-2(x1≠x2(，求證：x1+x2< A.a>1B.x1+x2<C.x1?x2<1D.x2-x1>-1 2(2024·全國·模擬預(yù)測)若函數(shù)f(x(=alnx+x2-2x有兩個不同的極值點x1,x2，且t-f(x1(+x2<f(x2(-x1恒成立，則實數(shù)t的取值范圍為()A.(-∞,-5(B.(-∞,-5[C.(-∞,2-2ln2(D.(-∞,2-2ln2[ 4(2024·遼寧·三模)已知函數(shù)f(x)=lnx+x2-ax存在兩個極值點，若對任意滿足f(x1)=f(x2)=(x1<x2<x3)，均有f(ex)<f(ex)<f(ex)，則實數(shù)a的取值范圍為() lnx-+2|-m(0<m<3)有兩個不同的零點x1，x2(x1<x2①<e2m②x1>③e<x2<-④x1x2>1A.1B.2C.3D.4 f(x(=ex-mx2有兩個極值點x1，x2(0<x1<x2)，函數(shù)g(x(=xlnx-x2-x有兩個極值點x3，x4(0<x3<x4)，設(shè)M=+，則()A.0<M<B.0<M<C.M>D.M>e 7(2023·四川成都·一模)已知函數(shù)f(x(=(lnx(2-xlnx+x2有三個零點x1、x2、x3且x1<x2<x3，則++的取值范圍是()A.B. lnx-+2|-m(0<m<3)有兩個不同的零點x1，x2(x1<x2①<e2m②x1>③x1x2>1A.0B.1C.2D.3 A.f(x(≤B.a的取值范圍是D.x1+x2>2 題目10(2024·山西太原·三模)已知x1是函數(shù)f(x(=x3+mx+n(m<0(的極值點，若f(x2(=f(x1((x1≠x2(，則下列結(jié)論正確的是()A.f(x(的對稱中心為(0,n(B.f(-x1(>f(x1(C.2x1+x2=0D.x1+x2>0 題目11(23-24高二上·湖北武漢·期中)已知函數(shù)f(x)=xlnx-x與y=a有兩個不同的交點，交點坐標分別為(x1,y1(，(x2,y2((x1<x2(，下列說法正確的有()A.f(x)在(0,1)上單調(diào)遞減，在(1,+∞)上單調(diào)遞增B.a的取值范圍為(-1,0]C.x2-x1>ae+eD.x2-x1<2a+e+ 題目15(2024·陜西安康·模擬預(yù)測)已知函數(shù)f(x(=x2e-x-a(x2-4x(在定義域(0,+∞(上僅有1個極大值(2)若f(x1(=f(x2(,1<x1<2<x2，證明：x1+x2>4. (2)若x1,x2是f(x)的兩個相異零點，求證：|x2-x1|<1-(3)當a>0時，若滿足f(x1(=f(x2((x1<x2(，求證：x1+x2<2lna.(2024·安徽合肥·模擬預(yù)測)已知函數(shù)f(x)=a(1-2lnx)+4x6(a∈R).(1)討論f(x)的單調(diào)性；2(x1≠x2(為函數(shù)=kx2+-lnx的兩個零點，求證：(x1x2(4>12e4. (1)求函數(shù)f(x)的單調(diào)區(qū)間；(2)若x1<x2，滿足f(x1(=f(x2(=0.11(2)函數(shù)f(x)中存在x1，x2且x1≠x2，滿足f(x1)=f(x2)，求證：x1+x2>2x0(x0為函數(shù)f(x)x,-f'(x0)>0；x-f'(x0)>0.F(x)=f(x)-f(2x,-x). 題目1(2024·江蘇揚州·模擬預(yù)測)已知函數(shù)f(x(=ln(mx(-x(m>0(.(1)若f(x(≤0恒成立，求m的取值范圍；(2)若f(x(有兩個不同的零點x1,x2，證明x1+x2>2.(2)構(gòu)造p(t(=f(1+t(-f(1-t(并證明t>0時f(1+t(>f(1-t(【解答過程】(1)首先由m>0可知f(x(的定義域是(0,+∞(，從而f(x(=ln(mx(-x=lnx-x+lnm.故f(x(=ln(mx(-x=-1=，從而當0<x<1時f(x(>0，當x>1時f(x(<0.故f(x(在(0,1(上遞增，在(1,+∞(上遞減，所以f(x(具有最大值f(1(=lnm-1.所以m的取值范圍是(0,e[.(2)不妨設(shè)x1<x2，由于f(x(在(0,1(上遞增，在(1,+∞(上遞減，故一定有0<x1<1<x2.22在-1<t<1的范圍內(nèi)定義函數(shù)p(t(=f(1+t(-f(1-t(.則p/(t(=f/(1+t(+f/(1-t(=+-=->0，所以p(t(單調(diào)遞增.這表明t>0時p(t(>p(0(=f(1(-f(1(=0，即f(1+t(>f(1-t(.又因為f(2-x1(=f(1+(1-x1((>f(1-(1-x1((=f(x1(=0=f(x2(，且2-x1和x2都大于1，故由f(x(在(1,+∞(上的單調(diào)性知2-x1<x2，即x1+x2>2. 2(2024·遼寧·三模)已知f(x(=(x-1(ex+ax2.(1)討論函數(shù)f(x(的單調(diào)性；(2)先用零點存在性定理證明結(jié)論，再構(gòu)造新函數(shù)討論f(x1)與f(-x2)大小關(guān)系，利用f(x(在(0,+∞)上單調(diào)所以f(x(在(0,+∞)上單調(diào)遞增，在(-∞,0)上單調(diào)遞減；當ln(-a(<0，即-1<a<0時，由f/(x(>0得x∈(-∞,ln(-a((∪(0,+∞(，f/(x(<0得x∈(ln(-a(,0(，所以f(x(在(-∞,ln(-a((和(0,+∞(上單調(diào)遞增，在(ln(-a(,0(上單調(diào)遞減；當ln(-a(=0，即a=-1時，f/(x(≥0恒成立，f(x(在R上單調(diào)遞增；當ln(-a(>0，即a<-1時，由f/(x(>0得x∈(-∞,0(∪(ln(-a(,+∞(，由f/(x(<0得x∈(0,ln(-a((，所以f(x(在(-∞,0(和(ln(-a(,+∞(上單調(diào)遞增，在(0,ln(-a((上單調(diào)遞減.綜上，當a≥0時，f(x(在(0,+∞(上單調(diào)遞增，在(-∞,0(上單調(diào)遞減；當-1<a<0時，f(x(在(-∞,ln(-a((和(0,+∞)上單調(diào)遞增，在(ln(-a(,0(上單調(diào)遞減；當a=-1時，f(x(在R上單調(diào)遞增；當a<-1時，f(x(在(-∞,0)和(ln(-a(,+∞(上單調(diào)遞增，在(0,ln(-a((上單調(diào)遞減.(2)由第(1)問中a>0時，f(x(在(0,+∞(上單調(diào)遞增，在(-∞,0(上單調(diào)遞減，使得f(x2(=0；則f(x(=(x-1(ex+ax2>(x-1)+ax2=ax2+x-1，33其中>0,<0，a取b=-1-、1+2a<0，af(b(=(b-1(eb+ab2>ab2+b-1=0，即f(b(>0，且f(0(=-1<0，使得f(x1(=0；所以當a>0時，函數(shù)f(x(有且僅有兩個零點；所以ax=(1-x2)ex2，所以f(-x2)=(-x2-1)e-x2++ax=(-x2-1)e-x2+(1-x2)ex2，令g(x(=(-x-1(e-x+(1-x(ex，g(x(=x(e-x-ex(，所以(-x2-1)e-x+(1-x2)ex<0，所以f(-x2)<0，所以f(x1)=0>f(-x2)，所以x1<-x2，所以x1+x2<0.+x2>.則f(x)min=f(1)=e-a.f(x2)=0，44于是g(a)>0，函數(shù)g(a)=ea-a2在(e,+∞(上單調(diào)遞增，g(a)>g(e)=ee-e2>0，(令函數(shù)h(x)=xlnx-lna.x+1，求導(dǎo)得h(x)=lnx+1-lna在(0,+∞(上單調(diào)遞增，且h=0，不妨設(shè)x1<x2要證x1+x2>2>-x1>，即證h(x2)>h-x1(.又h(x1(=h(x2(，則即證h(x1)>h-x1(，令函數(shù)F(x)=h(x)-h-x(，x∈(0,，則F(x)=h(x)+h-x(=lnx+1-lna+ln-x(+1-lna=lnx-x2(+ln，而x-x2=-(x-2+<，則F(x)<ln+ln=0，因此函數(shù)F(x)在(0,上單調(diào)遞減，即F(x)>F=0，則h(x1)>h-x1(，所以x1+x2>. 題目4(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=-x2+2lnx,g(x(=a(x2+2x(.(1)若曲線f(x(在點(1,-1(處的切線與曲線g(x(有且只有一個公共點，求實數(shù)a的值.(2)若方程g(x(-f(x(=1有兩個不相等的實數(shù)根x1,x2，(2)①構(gòu)造函數(shù)，令h(x(=2lnx-(a+1(x2-2ax+1，則h(x(有兩個零點x1,x2，進行求解即可；②由①得-1<a<0，則>2，分析得出：需【解答過程】(1)函數(shù)f(x(的定義域為(0,+∞(,f(x(=-2x+,f(1(=-1,f(1(=0，所以曲線f(x(在點(1,-1(處的切線方程為y=-1.因為切線y=-1與曲線g(x(=a(x2+2x(有且只有一個公共點，所以g(-1(=-a=-1，故a=1.(2)①方程g(x(-f(x(=1有兩個不相等的實數(shù)根x1,x2，即方程2lnx-(a+1(x2-2ax+1=0有兩個不相等55令h(x(=2lnx-(a+1(x2-2ax+1，則h(x(有兩個零點x1,x2.hI(x(=-2(a+1(x-2a=-，因為x>0,x+1>0，所以hI(x(的正負取決于(a+1(x-1的正負.因為函數(shù)h(x(有兩個零點，所以h>0，即2ln-(a+12-2a?+1>0，化簡得2ln(a+1(+<0.令m(x(=2ln(x+1(+,x>-1，則mI(x(=+>0，則函數(shù)m(x(在(-1,+∞(上單調(diào)遞又m(0(=0，則不等式2ln(a+1(+<0的解集為{a|-1<a<0{.因為h(x(=2lnx-(a+1(x2-2ax+1<2lnx-2ax+1，所以h(e-2(<2lne-2-+1=-3-<0.設(shè)φ(x(=lnx-x+1,x>0，則φI(x(=-1=，即h(x(=2lnx-(a+1(x2-2ax+1<2(x-+1(x2-2ax+1=x[2(1-a(-(a+1(x[，取x0>>，則h(x0(<x0[2(1-a(-(a+1(x0[<0.所以h(x(共有兩個零點.②由①得-1<a<0，則>2，不妨設(shè)0<x1<<x2，要證明x1+x2>2，只需證明x1+x2>，即證明x1>-x2.66所以只需證明h(x1(>h-x2(．又h(x1(=h(x2(，所以需證h(x2(-h-x2(>0.記F(x(=h(x(-h-x(,x∈,+∞(，則F(x(=h(x(+h-x(=+-4(a+1(=-4(a+1(>-4(a+1(=0，則F(x(>0，所以F(x2(>0，即h(x2(-h-x2(>0，所以x1+x2>，故x1+x2>2. (1)討論f(x)的單調(diào)性；(2)設(shè)g(x)=-f(x)+x2-(a+1)x-2a+(a-1)lnx，若g(x)存在兩個不同的零點x1，x2，且x1<x2.(ii)證明：x2-x1<-.【解答過程】(1)由題f(x)的定義域為(0,+∞)，f(x)=2x-(2+a)+== (2x-a)(x-1)77當0<x<1或x>時，f(x)>0；當1<x<時，f(x)<0，(2)(i)由題意知g(x)=-lnx+x-2a，所以g(x)=-+1==，則g(x)min=g(1)=e+1-2a，(ii)下面找出兩個點m，n(0<m<1<n)，使得g(m)>0，g(n)>0,①g(2a)>0?-ln(2a)>0，設(shè)m(x)=-lnx(x>2)x-x2-x(x>0)，則h(x)=ex-x-1，故h″(x)=ex-1>0，故h(x)>h(2)=e2-4>0，即ex>x2+x(x>2)，因此m(x)=-lnx>x+1-lnx，設(shè)u(x)=x+1-lnx(x>2)，則u(x)=-=>0，因此m(x)=-lnx>0，又2a>e又f(1)<0，所以1<x2<2a.所以m(x)在(2,+∞)上單調(diào)遞增，88②g=(2a-1-ln+-2a，設(shè)t(x)=lnx-x+1，則t(x)=，所以t(x)≤t(1)=0，即lnx≤x-1，因此g≥(2a-12a-1)=(2a-1-1(>0，，即-1<-x1<-，于是x2-x1<2a-=-. 題目2(2024·湖南株洲·一模)已知函數(shù)f(x(=(x+a(ebx在(1,f(1((處的切線方程為y=e(x-1(，其中e為自2是方程f(x(=kx2-2(k>2)的兩個不相等的正實根，證明：|x1-x2|>ln.得f(x(的最小值；【解答過程】(1)f(x(=ebx+b(x+a(ebx=(bx+ab+1(ebx，由題意有f(1(=(b+ab+1(eb=e及f(1(=(1+a(eb=e(1-1(=0，故a=-1、b=1，則f(x(=(x-1(ex，f(x(=(x-1+1(ex=xex，f(x(<0，故f(x(在(-∞,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增，故f(x(有最小值f(0(=(0-1(e0=-1；(2)令g(x(=f(x(-kx2+2=(x-1(ex-kx2+2，x>0，k>2，則g(x(=xex-2kx=x(ex-2k(，99故g(x(在(0,ln2k(上單調(diào)遞減，在(ln2k,+∞(上單調(diào)遞增，故g(ln2k(=(ln2k-1(eln2k-k(ln2k(2+2=2k(ln2k-1(-k(ln2k(2+2=-k[(ln2k(2-2ln2k+2[+2=-k[(ln2k-1(2+1[+2，(ln2k(=-k[(ln2k-1(2+1[+2<-k+2<0，故g(x(有兩個零點，不妨設(shè)兩零點x1<x2又g(1(=(1-1(e1-k+2=-k+2<0，由1<2ln2<2lnk，故g(ln2k(<g(則x1<1<2ln2<x2-x2|>2ln2-1=ln. 題目3(2024·北京朝陽·二模)已知函數(shù)f(x(=ax-ln(1-x((a∈R(.(1)求曲線y=f(x(在點(0,f(0((處的切線方程；(3)若f(x(有兩個不同的零點x1,x2，且|x2-x1|>e-1，求a的取值范圍.單調(diào)性可得f(x)min=a+1+ln(-a(≥0，設(shè)φ(x(=x+1+ln(-x((x<0)，利用導(dǎo)數(shù)研究函數(shù)φ(x(的性質(zhì)即【解答過程】(1)由f(x)=ax-ln(1-x)，得f/(x(=a+-(x<1)，因為f(0)=0,f/(0)=a+1，所以曲線y=f(x)在點(0,f(0))處的切線方程為y=(a+1)x；①當a≥0時，f(-1)=-a-ln2<0，不符合題意.f/(x)<0，f(x)在區(qū)間上單調(diào)遞減，/(x)>0所以當x=1+時，f(x)取得最小值f(1+=a+1+ln(-a(；若f(x(≥0恒成立，則a+1+ln(-a(≥0，設(shè)φ(x(=x+1+ln(-x((x<0)，則φ/(x(=1+=，所以φ(x(≤φ(-1(=0，即a+1+ln(-a(≥0的解為a=-1.所以a=-1；若x2<02<1-e， >0，函數(shù)f(x(=(x+a(ln(x+b(的圖象在點(1,f(1((處的切線方程為xln2-y-ln2=0.(2)若方程f(x(=(e為自然對數(shù)的底數(shù))有(2)由題意得x1<0<x0<1<x2，構(gòu)造函數(shù)證明不等式(x-1(ln(x+1(≥(x-1(?ln2，當且僅當x=1時，等號成立；(x-1(ln(x+1(≥-x，當且僅當x=0時，等號成立；從而可分別得到x=1+>x2，x=-<【解答過程】(1)因為fI(x(=+ln(x+b(，所以fI(1(=+ln(1+b(=ln2，聯(lián)立方程組+ln(1+b(=ln2，解得a=-1,b=1.由題意知f(1(=0，所以f(1(=(1+聯(lián)立方程組+ln(1+b(=ln2，解得a=-1,b=1.(2)由(1)可知f(x(=(x-1(ln(x+1(,x>-1，f(0(=0,f(1(=0，fI(x(=1-+ln(x+1(，設(shè)fI(x(=u(x(，uI(x(=+>0，所以u(x(即fI(x(在(-1,+∞(上單調(diào)遞增.00設(shè)h(x(=(x-1(?ln2，令F(x(=f(x(-h(x(=(x-1(ln(x+1(-(x-1(?ln2，則F/(x(=+ln(x+1(-ln2=ln(x+1(-+1-ln2，因為f/(x(在(-1,+∞(上單調(diào)遞增，所以F/(x(在(-1,+∞(上單調(diào)遞增.所以F(x(在(-1,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增.故F(x(≥F(1(=0，即(x-1(ln(x+1(≥(x-1(?ln2，當且僅當x=1時，等號成立.因為方程f(x(=有兩個實數(shù)根x1,也就是f(x2(=f(x1(=>f(1(=f(0(=0，且注意到f(x(在(1,+∞(上單調(diào)遞增，所以x1<0<x0<1<x2，所以(x2-1(ln(x2+1(>(x2-1(ln2，即f(x2(>h(x2(.又h(x(在(-1,+∞(上單調(diào)遞增，所以h(x(=f(x2(>h(x2(，故x>x2①.易知f(x(的圖象在坐標原點處的切線方程為g(x(=-x，令T(x(=f(x(-g(x(=(x-1(ln(x+1(+x，則T/(x(=+ln(x+1(=2-+ln(x+1(，因為f/(x(在(-1,+∞(上單調(diào)遞增，所以T/(x(在(-1,+∞(上單調(diào)遞增.所以當-1<x<0時，T/(x(<0，當x>0時，T/(x(>0，所以T(x(在(-1,0(上單調(diào)遞減，在(0,+∞(上單調(diào)遞增.所以T(x(≥T(0(=0，(x-1(ln(x+1(≥-x，當且僅當x=0時，等號成立.因為x1<0-1(ln(x1+1(>-x1，即f(x1(>g(x1(.設(shè)g(x(=的根為x，則x=-，又g(x(在(-1,+∞(上單調(diào)遞減，所以g(x(=f(x1(>g(x1(，所以x<x1，從而-x>-x1②.2-x1<x-x=1++. 題目1(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=ax2-(lnx(2(a∈R(.(2)若f(x(有兩個極值點x1,x2.lnx-1，求得h(x(為(0,+∞(上的增函數(shù)，且h(1(=0，進而求得f(x(單調(diào)區(qū)間；t(m>1(，令s(m(=lnm-(m>1(，利用導(dǎo)數(shù)求得函數(shù)的單調(diào)性與最值，即可求解.設(shè)g(x(=2x-，則g(x(=，令h(x(=x2+lnx-1，可得h(x(=2x+>0恒成立，所以h(x(為(0,+∞(上的增函數(shù)，且h(1(=0，所以g(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，所以g(x)min=g(1(=2，所以f(x)min=2>0，所以f(x(>0，所以f(x(在(0,+∞(上單調(diào)遞增.(2)解：①因為函數(shù)f(x(=ax2-(lnx(2，可得f(x(=2ax-，設(shè)p(x(=，可得p(x(=,x>0，因為f(x(有兩個極值點x1,x2，則直線y=a與函數(shù)p(x(=的圖象有兩個不同的交點，所以p(x(在(0,e(上單調(diào)遞增，在(e,+∞(上單調(diào)遞減，所以p(x)max=p(e(=.②由函數(shù)f/(x(有兩個零點x1,x2，所以2ax=lnx,2ax=lnx，只需證明t1t2>e2，不妨令t1>t2，由2at1=lnt1,2at2=lnt2得2a=要證t1t2>e2，只需證明lnt1+lnt2>2，即證lnt1+lnt2=2a(t1+t2(=(t1+t2(?>2，即證lnt1-lnt2>，即證ln，令，則m>1，只需證明lnm>(m>1(，所以s(m(在(1,+∞(上單調(diào)遞增，所以s(m(>ln1-=0， ②當x1>4x21x>16e3.(x)=lnx-2ax-1，切線方程為y-(x0lnx0-ax-2x0)=(lnx0-2ax0-1)(x-x0)則-x0lnx0+ax+2x0=(lnx0-2ax0-1)(1-x0)，即有ax-2ax0-x0+lnx0-1=0，由過點(1,0)可作曲線y=f(x)兩條切線，得方程ax-2ax0-x0+lnx0-1=0有兩個不相等的實數(shù)根，令g(x)=ax2-2ax-x+lnx-1，則函數(shù)g(x)有2個零點，求導(dǎo)得gl(x)=2ax-2a-1+==， 又g=2-2a?-+ln-1=-ln2a--2<0，③若0<a<，由gl(x)>0，得0<x<1或x>，由gl(x)<0，得1<x<，當0<x<1時，g(x)=a(x-1)2-x-a-1+lnx<-a-1+lnx，當x>1時，令y=lnx-x，求導(dǎo)得yl=-1<0，函數(shù)y=lnx-x在(1,+∞)上單調(diào)遞減，則lnx-x<-1，g(x)=a(x-1)2-a-1+lnx-x<a(x-1)2-a-2，所以過點(1,0(可作曲線y=f(x(兩條切線時，a的取值范圍是(-∞,-2(.fl(x)=lnx-2ax-1，2max=，要證明x1x>16e3，只需證明lnx1+2lnx2>4ln2+3，令t=(x1>4x2)，則t>4，欲證明lnx1+2lnx2>4ln2+3，令h(t)=lnt-4ln2?(t>4)，求導(dǎo)得hI(t(=-4ln2?==，則φ(t)=t+4+-12ln2在t>4時單調(diào)遞增，故φ(t)>φ(4)=9-12ln2>0，所以x1x>16e3. (2)若f(x1(=f(x2(2<.(2)由f(x1(=f(x2(，可得lnx1+lnx2=?(lnx1-lnx2(，由x1?x2<?lnx1+lnx2<-2，然后換元變所以fI(1(=0+1+a=1,∴a=0，則f(x(=xlnx，定義域為(0,+f(x(=lnx+1，令f(x(=lnx+1=0，解得x=，令f(x(>0，解得x>，令f(x(<0，解得0<x<，∴f(x(在x=時取得極小值f=ln=-，無極大值.(2)由已知，令f(x1(=f(x2(=m，則x1lnx1=m，即lnx1=，x2lnx2=m，即lnx2=，兩式相減可得，lnx1+lnx2=m，兩式相加可得，lnx1-lnx2=m，+lnx2=?(lnx1-lnx2(，2<?lnx1+lnx2<-2，因此只需證明?(lnx1-lnx2(<-2即可，x1而?(lnx1-lnx2(<-2?-?ln<-2，x1不妨設(shè)x1<x2,t=，則由0<x1<x2可知0<t<1，-?ln<-2??lnt<-2?(t+1(lnt-2t+2<0，令g(t(=(t+1(lnt-2t+2,0<t<1，g(t(=lnt+-1，令h(t(=lnt+-1，則h(t(=-=<0，t(在(0,1(上遞增，g(t(<g(1(=0，即(t+1(lnt-2t+2<0， 2(為函數(shù)g(x)=kx2+-lnx的兩個零點，求證：(x1x2(4>12e4.lnx1lnx2當f(x)>0時，x>6，當f(x)<0時，6,+∞6,+∞(2)設(shè)0<x1<x2，則g(x1(=kx+-lnx1=0，g(x2(=kx+-lnx2=0，所以k=-=-，lnx1lnx2lnx1lnx211x-xx-x1xlnx1lnx211x-xx-x1xxxx(x1x2(4，x-x(x1x2(4，x2(4>12e4，只需證->-，只需證->-1，只需證lnx1+x<lnx2+xx-xt4xt4xt4.=4>0=23=e，又0<x1<x2，所以h(x1(<h(x2(，所以(x1x2(4>12e4成立. (1)設(shè)函數(shù)g(x(=ekx-(k>0(，若f(x(≤g(x(恒成立，求k的最小值；e2lnx+x-2lnx-x-1≥0成立，進而分類討論0<k<1與k>1兩種情況，從而得解；(2)利用導(dǎo)數(shù)研究函數(shù)f(x(的性質(zhì)可得0<m<1，由題意可得m(x+x)<2x1x2(1-lnm)，原不等式變形為1+lnx1x2<x1x2-x1x2lnm，利用分析法，構(gòu)造函數(shù)h(x)=lnx-x+1(x>0)證明lnx+1≤x，即1+lnx1x2≤x1x2，結(jié)合x1x2<x1x2-x1x2lnm即可證明.等價于1+2lnx≤x2ekx-?1+2lnx≤e2lnx+kx-?e2lnx+kx--(1+2lnx)≥0恒成立.設(shè)u(x)=ex-x-1，則u(x)x-x-1≥0，又v=2ln+=-2+<0，v(1(=1>0，0所以u(v(x)(≥u(v(x0((=0，即e2lnx+x-2lnx-x-1≥0，且e2lnx+x-2lnx0-x0-1=0；當0<k<1時，令m(k(=e2lnx+kx--(1+2lnx)，易得關(guān)于k的函數(shù)y=e2lnx+kx與y=-在(0,1(上單調(diào)遞增，則m(k(<m(1(=e2lnx+x-2lnx-x-1，2lnx+kx--(1+2lnx)<0，不滿足題意；當k>1時，易得m(k(>m(1(=e2lnx+x-2lnx-x-1≥0，即e2lnx+kx--(1+2lnx)≥0恒成立；f(x)=，則f(x)=(x>0)，令f(x)>0?0<x<1，令f(x)<0?x>1，max=f(1)=1，當x>1時，易得f(x)=>0恒成立，當x=時，f(x)=<0，又函數(shù)f(x)=m有兩個不同的實根x1,x2，即f(x(與y=m的圖像有兩個交點，作出f(x(與y=m的部分圖像如圖：所以0<m<1，且m=,m=，得mx=1+2lnx1,mx=1+2lnx2，有m(x+x)=2+2lnx1x2.要證+<，即證m(x+x)<2x1x2(1-lnm)，即證2+2lnx1x2<2x1x2(1-lnm)，即證1+lnx1x2<x1x2-x1x2lnm，由x1x2>0,lnm<0，得x1x2<x1x2-x1x2lnm.設(shè)h(x)=lnx-x+1(x>0)，則h(x)=-1(x>0)，令h(x)>0?0<x<1，令h(x)<0?x>1，所以h(x)max=h(1)=0，則h(x)≤0，即lnx+1≤x，所以1+lnx1x2≤x1x2，則1+lnx1x2≤x1x2<x1x2-x1x2lnm，即1+lnx1x2<x1x2-x1x2lnm，即證. +<-.(2)由題意，將命題成立轉(zhuǎn)化為+<-成立>0，令F(t(=lnt+，利用導(dǎo)數(shù)證明F(t(>0即可.則g(x(=2xlnx+x=x(2lnx+1(，1且g(e-=-，又f(x(恰有兩個零點，即x2lnx=a(a∈R(有兩個根，故函數(shù)g(x(與y=a有兩個交點，(2)因為f(x(的兩個零點分別為x1,x2(x1<x2)，所以xlnx1-a=0，xlnx2-a=0，所以lnx1=，lnx2=，故lnx1-lnx2=--xx令F(t(=lnt+，又F(1(=0，F(xiàn)I(t(=+×=，所以4t4+(4-3e(t2+1=4m2+(4-3e(m+1，而y=4m2+(4-3e(m+1，其對稱軸為m=<1，所以y=4m2+(4-3e(m+1在(1,+∞(上單調(diào)遞增，所以4t4+(4-3e(t2+1>4+4-3e+1>0，于是FI(t(>0在(1,+∞(上恒成立，因此F(t(在(1,+∞(上單調(diào)遞增，所以F(t(>F(1(=0，所以原命題得證. 3(2023·湖北武漢·三模)已知函數(shù)f(x(=ax+(a-1(lnx+，a∈R.(1)討論函數(shù)f(x(的單調(diào)性；(2)若關(guān)于x的方程f(x(=xex-lnx+有兩個不相等的實數(shù)根x1、x2，函數(shù)f(x(的增區(qū)間和減區(qū)間；(2)(i)將方程變形為ex+lnx=a(x+lnx(，令t(x(=x+ln【解答過程】(1)解：因為f(x(=ax+(a-1(lnx+，所以f(x(=a+-==，其中x>0.②當a>0時，由f(x(>0得x>，由f(x(<0可得0<x<.所以函數(shù)f(x(的增區(qū)間為,+∞(，減區(qū)間為(0,.(2)解：(i)方程f(x(=xex-lnx+可化為xex=ax+alnx，即ex+lnx=a(x+lnx(.令t(x(=x+lnx，因為函數(shù)t(x(在(0,+∞(上單調(diào)遞增，易知函數(shù)t(x(=x+lnx的值域為R，t=at(*)有兩個不等的實根.所以，函數(shù)g(t(在(-∞,0(和(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增.作出函數(shù)g(t(和y=a的圖象如下圖所示：ex2>t1<1<t2.令(t>1(，只需證lnp>.所以h(p(在(1,+∞(上單調(diào)遞增，故h(p(>h(1(=0，即h(p(>0在(1,+∞(上恒成立.所以原不等式得證.已知f(x(=x2-(2)①由題意0<x1<1<x2，0<<1，作差得f(x1(-f((=f(x2(-f(x2+x2--2lnx2(，由此即可構(gòu)造函數(shù)g(x(=x--2lnx(x>1(比較大小，進而結(jié)合函數(shù)f(x(單 -h<0，所以構(gòu)造函數(shù)H(x(=h(x(-h=-x(-lnx+1(即可得證.【解答過程】(1)fI(x(=2x所以當0<x<1時，g(x(<g(1(=0，fI(x(<0，f(x(單調(diào)遞減，所以f(x(min=f(1(=1-a，若f(x(≥0，則f(x(min=f(1(=1-a≥0，解得a≤1，(1)可知f(x(min=f(1(=1-a<0?a>1，且0<x1<1<x2，0<<1所以f(x1(-f=x2+x2--2lnx2(，設(shè)g(x(=x--2lnx(x>1(，gI(x(=1+-=>0,(x>1(，所以函數(shù)g(x(單調(diào)遞增，所以g(x2(>g(1(=0，所以f(x1(-f>0，即f(x1(>f，又函數(shù)f(x(在(0,1(上面單調(diào)遞減，所以0<x1<<1，所以1<；②注意到x--a=x--a，只需x2lnx1-x1lnx2<x1-x2，即只需<，令h(x(=，則hI(x(=-，又>x2>1，所以h<h(x2(，所以要證h(x1(<h(x2(，只需h(x1(<h，即h(x1(-h<0，不妨設(shè)H(x(=h(x(-h=-x(-lnx+1(，則H(x(=-+lnx-1+1=lnx?,當0<x<1時，H(x(>0，H(x(單調(diào)遞增，當x>1時，H(x(>0，H(x(單調(diào)遞增，因為0<x1<1，所以H(x1(=h(x1(-h<H(1(=0，即h(x1(<h，又因為h<h(x2(，所以h(x1(=<=h(x2(，綜上所述，命題<x1+x2得證. (1,2(時，p(x(=g(x(-g(2-x((0<x<1(，利用導(dǎo)數(shù)說明函數(shù)的單調(diào)性，即可證明x1+x2>2，再由基本不若a>0，則當0<x<1時，f(x(>0,f(x(單調(diào)遞增，當x>1時，f(x(<0,f(x(單調(diào)遞減；若a<0，則當0<x<1時，f(x(<0,f(x(單調(diào)遞減，當x>1時，f(x(>0,f(x(單調(diào)遞增.(ii)不妨設(shè)x1<x2，則0<x1<1<x2，且=.+x>x≥4>2，即x+x>2；-x2∈(0設(shè)p(x(=g(x(-g(2-x(=+---,0<x<1,則p(x(=-->--=->0,則p(x(<p(1(=0，即g(x(<g(2-x(，所以g(2-x1(>g(x1(=g(x2(,-x1>1,x2>1,g(x(在區(qū)間(1,+∞(內(nèi)單調(diào)遞減，所以2-x1<x2，即x1+x2>2，，所以x+x>2x1x2，故2x+2x>x+x+2x1x2=(x1+x2(2>4，所以x+x>2，得證.設(shè)h(x(=g(x(-g=-x(1-lnx(，x∈(0,+∞(，則h(x(=+lnx=lnx?≥0，所以h(x1(=g(x1(-g<0，即g(x1(<g.又g(x2(=g(x1(，所以g(x2(<g，，所以x+x>2x1x2>2，得證.x1+x22≥x1-x2x1+x22≥,所以fI(x(=x(2ex(.,所以fI(x(=x(2ex(.fI(x(<0,f(x(單調(diào)遞減，fI(x(>0,f(x(f(x)max=f(2(又f(1(=<f(3(=3，所以當x=2f(x)min=f(1(=e1e1e所以f(x(在[1,3[上的值域是由(1)可知0<x1<2<x2，且a>2.x1-x2lnx1-lnx2x1-x2lnx1-lnx2要證a(x+x(>2e2，即證ex+ex>2e2.22lnx1-lnx2即證x1+x2≥2lnx1-lnx2<0，t-1(t+1>lnt.t-1(t+1>lnt.故a(x+x(>2e2.(1)若函數(shù)f(x(在R上是增函數(shù)，求a的取值范圍；(2)設(shè)g(x(=x-sinx-lnx，若g(x1(=g(x2((x1≠x2(，證明：、x1x2<2.(2)由(1)的信息可得x2-sinx2>x1-sinx1(x2>x1)，利用分析法推理所以a的取值范圍為[-1,1].(2)函數(shù)g(x)=x-sinx-lnx的定義域為(0,+∞)，由g(x1)=g(x2)，得lnx2-lnx1=(x2-x1)-(sinx2-sinx1)，由(1)知，函數(shù)f(x)=x-sinx在(0,+∞)上是增函數(shù)，不妨令x2>x1>02-sinx2>x1-sinx1，即x2-x1>sinx2-sinx1，亦即(x2-x1)>(sinx2-sinx1)，則(x2-x1)-(sinx2-sinx1)>(x2-x1)，于是lnx2-lnx1>(x2-x1)，則<2，-1x2x1x2x1x2-x1x1x2-1x2x1x2x1x2-x1x1x2x1x2>求導(dǎo)得φI(t)=-1-=-<0，則函數(shù)φ(t)在(1,+∞)上單調(diào)遞(2024·黑龍江哈爾濱·模擬預(yù)測)已知函數(shù)f(x(=x2lnx-m有兩個不同的零點x1，x2，且t=x+x.【解答過程】(1)f(x(=x2lnx-m=x2lnx2-m=(x2lnx2-2m(，u=x2，考慮函數(shù)g(u(=ulnu-2m，f(x(有兩個零點x1,x2，u1=x,u2=x，則g(u(有兩個零點u1,u2，g(u(=lnu+1,g(u(在(0,上單調(diào)遞減，,+∞(單調(diào)遞增，所以g(u(min=g=--2m<0，g(u(→-∞,所以-<m<0；即lnu1+ln(1+λ(<0①,∵u1lnu1=u2lnu2=λu1lnλu1，∴l(xiāng)nu1=λ(lnλ+lnu1(,∴l(xiāng)nu1=-，構(gòu)造φ(λ(=(λ>1(，φ(λ(=，所以y=1--lnλ在λ>1時單調(diào)遞減，所以y<1--ln1=0，φ(λ(在(1,+∞(單調(diào)遞減，∴φ(λ(>φ(λ+1(得證.(3)先證u1+u2>,0<u1<<u2，構(gòu)造F(u(=g(u(-g-u(，0<u<，F(xiàn)(u(=g(u(+g-u(=lnu-u(+2≤0，∴F(u(在(0,單調(diào)遞減，∴F(u1(>F=0，即g(u1(>g-u1(，∴g(u2(>g-u1(,∵g(u(在,+∞(單調(diào)遞增，∴u2>-u1,∴u1+u2>.由0<u1<<u2得0<eu1<1,eu2>1，所以ln(eu1(<，整理得：eu-2meu1-3u1-2m>0③,所以y=lnx->0，即lnx>，可得：eu-2meu2-3u2-2m<由③-④得：e(u-u(>(2me+3((u1-u2(， 題目1(2024·四川·一模)已知函數(shù)f(x(=ax2+x-lnx-a.(2)若f(x(有2個零點x1,x2，證明：a(x1+x2(2+(x1+x2(>2.-2(t-1(<0成立.【解答過程】(1)當a=1，函數(shù)f(x(=x2+x-lnx-1(x>0(，則f(x(=2x+1-==，可知當0<x<時，f(x(<0，f(x(單調(diào)遞減；當x>時，f(x(>0，f(x(單調(diào)遞增，則當x=時，f(x(取得極小值f=ln2-，也即為最小值，所以f(x(的最小值為ln2-；,x2是f(x(=ax2+x-lnx-a的兩個零點，則ax+x1-lnx1-a=0，ax+x2-lnx2-a=0，+x2((x1-x2(+(x1-x2(-(lnx1-lnx2(=0，整理得a(x1+x2(=-1，欲證明a(x1+x2(2+(x1+x2(>2，只需證明不等式-1((x1+x2(+(x1+x2(>2，t+1(>2，即證明(t+1(lnt-2(t-1(<0(0<t<1(即可，令h(t(=(t+1(lnt-2(t-1((0<t<1(，則h(t(=lnt+-1，又令u(t(=h(t(=lnt+-1(0<t<1(，則u(t(=-=<0，故當0<t<1時，h(t(單調(diào)遞增，則h+x2(2+(x1+x2(>2得證. 2(2024·全國·模擬預(yù)測)已知函數(shù)f(x(=x2(alnx-有兩個極值點x1,x2，且x1<x2.(2)證明：x1lnx1+x2lnx2>x1+x2.證即證lnt->0，構(gòu)造函數(shù)利用導(dǎo)數(shù)證明不等式.【解答過程】(1)函數(shù)f(x(=x2(alnx-x-的定義域是(0,+∞(，f(x(=2x(alnx-x-+x2-=2x(alnx-x(，因為函數(shù)f(x(有兩個不相等的極值點x1,x2，所以方程alnx-x=0在(0,+∞(上有兩個不相等的實數(shù)根x1,x2，所以a≠0，即直線y=與函數(shù)y=的圖象有兩個交點.當x∈(0,e(時，g(x(>0,g(x(單調(diào)遞增；當x∈(e,+∞(時，g(x(<0,g(x(單調(diào)遞減.，即a(lnt+lnx1(=tx1，整理得lnx1=(*(，要證x1lnx1+x2lnx2>x1+x2，只需證x1lnx1+tx1(lnt+lnx1(>x1(t+1(，t(lnt+lnx1(>t+1，即證(t+1(lnx1+tlnt>t+1，即證lnt->0，設(shè)h(t(=lnt-,(t>1(，因為h(t(=-=>0，所以h(t(在(1,+∞(上單調(diào)遞增，所以h(t(>ln1-=0，所以原不等式得證. 題目3(2024·貴州·模擬預(yù)測)已知函數(shù)f(x)=xex+1.(1)求函數(shù)f(x(的單調(diào)區(qū)間；要證x1+x2+ln(x1x2)>2，ex>e21t2>e2，只需證lnt1+lnt2>2.不妨設(shè)t1>t2>0，則只需證lnt1+lnt2=ln>2，即證ln>=.所以h(s)>h(1)=0，故x1+x2+ln(x1x2)>2. (2)若f(x(存在兩個極值點x1,x2(x2>x1>0(.2-x1+2>；f(x(>-+-2.<x1<1<x2，分析可知只需證明x2>-1，即證f(x(在(1,+∞(上的單調(diào)性，從而得到f(x(≥f(x2(=【解答過程】(1)函數(shù)f(x(=+a(lnx+-2的定義域為(0,+∞(，則fI(x(=++a-=，所以g(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，所以fI(x(≥0在(0,+∞(上恒成立，所以f(x(在(0,+∞(上單調(diào)遞增；(2)(ⅰ)由(1)可知g(x(在(0,+∞(上的最小值為g(1(=a-1，若f(x(存在兩個極值點x1,x2(x2>x1>0(，則g(x(=0有兩個不相等的實數(shù)根x1,x2(x2>x1>0(，且當0<x<x1時g(x(>0，即fI(x(>0，則f(x(單調(diào)遞增，當x1<x<x2時g(x(<0，即fI(x(<0，則f(x(單調(diào)遞減，當x>x2時g(x(>0，即fI(x(>0，則f(x(單調(diào)遞增，所以x1為f(x(的極大值點，x2為f(x(的極小值點， 1+lnx1x11+lnx2x 1+lnx1x11+lnx2x2(a-(x1==0=0La- 1+La-只需證x2>-1，令p(x(=lnx-(x>1(，則pI(x(=-=>0，所以p(x(在(1,+∞(上單調(diào)遞增，所以p(x(>p(1(=0，即lnx2->0成立，所以x2-x1+2>；>1，a=，且當1<x<x2時f(x(<0，當x>x2時f(x(>0，所以f(x(≥f(x2(=+a(lnx2+-2=+lnx2+-2=2lnx2+(lnx2(2+2-2=x2，令H(x(=lnx+(x>1(，則H(x(=-=>0，所以H(x(在(1,+∞(上單調(diào)遞增，所以H(x(>H(1(=1，即lnx>1->0(x>1(，所以>=-+-2，所以f(x(>-+-2. x-x2-sinx，通過研究函數(shù)h(x(=ex-x的最值情況，可得g(x(在R上單調(diào)遞增，又由，可設(shè)x1<0<x2.則x1+x2<0?-x2>x1?g(-x2(>g(x1(?g(-x2(>2-g(x2(，故證明當x>0時，g(x(+g(-x(-2>0即可.則在x=0處的切線方程為x-y+1=0；x-x2-sinx，則g(x(=ex-x-cosx.令h(x(=ex-x，則h(x(=ex-1.令h(x(>0?x>0，得h(x(在(0,+∞(上單調(diào)遞h(x(<0?x<0，得h(x(在(-∞,0(上單調(diào)遞減.則h(x(≥h(0(=1.=h(x(-cosx≥g(0(=0，當且僅當x=0時取等號.則不妨設(shè)x1<0<x2.令p(x(=g(x(+g(-x(-2=ex+e-x-x2-2，其中x∈(0,+∞(.則p(x(=ex-e-x-2x.令n(x(=ex-e-x-2x，x∈(0,+∞(.則n(x(=ex+e-x-2>2、ex?e-x-2=0，得n(x(在(0,+∞(上單調(diào)遞增，則p(x(=n(x(>n(0(=0，得p(x(在(0,+∞(上單調(diào)遞增，有p(x(>p(0(=0，即x>0時，g(x(+g(-x(-2>0.g(x2(+g(-x2(-2>0?g(-x2(>2-g(x2(，又g(x1(+g(x2(=2，則g(-x2(>g(x1(，又注意到g(x(在R上單調(diào)遞增，則-x2>x1?x1+x2<0. 已知函數(shù)f(x(=aex-xlnx有兩個極值點x1,x2，且Q(x0,f(x0((為曲線C：y=f(x(的拐點.令g(x(=(x>0(?g(x(=，易知y=-lnx-1(x>0(單調(diào)遞減，且x=1時y=0，x∈(1,+∞(時g(x(<0，即此時g(x(單調(diào)遞減，所以g(x(≤g(1(=，(2)令f(x(=aex-lnx-1=h(x(?h(x(=aex-，所以C在Q處的切線方程為：y=f(x0((x-x0(+f(x0(，令m(x(=aex-xlnx-f(x0((x-x0(-f(x0(，即證y=m(x(只有一個零點，易知m(x(=f(x(-f(x0(，即f(x(≥f(x0(?m(x(≥0，所以y=m(x(單調(diào)遞增，顯然m(x0(=0，所以y=m(x(只有一個零點，結(jié)合(2)知0<x1<x0<x2，根據(jù)f(x(的單調(diào)性知f(x0(<0=f(x1(=f(x2(，所以f(x0(<0<，證畢. 2滿足f(x1)+f(x2)=4，求證：x1+x2≥2.(2)首先將f(x1)+f(x2)=4化簡為(x1+x2)2+(x1+x2)=4+2(x1x2-lnx1x2)，令x1x2=t，g(t)=t-lnt，tfI(x)=+2x+1，k=fI(1)=5.(2)f(x1)+f(x2)=2lnx1+x+x1+2lnx2+x+x2=42lnx1+x+x1+2lnx2+x+x2=4.(x1+x2)2+(x1+x2)=4+2(x1x2-lnx1x2)令x1x2=t，g(t)=t-lnt，t>0，(t)=1-=，3)(x1+x2-2)≥0，得到x1+x2-2≥0 題目4(23-24高二下·貴州貴陽·階段練習(xí))設(shè)fI(x(是函數(shù)f(x(的導(dǎo)函數(shù)，若fI(x(可導(dǎo)，則稱函數(shù)fI(x(的導(dǎo)函數(shù)為f(x(的二階導(dǎo)函數(shù)，記為f″(x(.若f″(x(有變號零點x=x0，則稱點(x0,f(x0((為曲線y=f(x(的數(shù)y=f(x(的圖象的對稱中心.已知函數(shù)f(x(=x3+bx2-24x+d的圖象的對稱中心為(1,3(，求函數(shù)f(x((2)已知函數(shù)g(x(=emx-1+mx3-x2+x--1(m>0).(ii)若g(x1(+g(x2(=-2(x1≠x2(，求證：x1+x2<.(ⅱ)由(i)可得函數(shù)g(x(在R上單調(diào)遞增，將要證的不等式轉(zhuǎn)化為g(x1(+g-x1(>-2，構(gòu)造函數(shù)h(x(=又函數(shù)f(x(=x3+bx2-24x+d的圖象的對稱中心為(1,3∴f(x(=x3-3x2-24x+29，∴fI(x(=3x2-6x-24=3(x-4((x+2(，∵函數(shù)f/(x(在(-∞,-2(上為正，在(-2,4(上為負，在(4,+∞(上為正，∴函數(shù)f(x(在(-∞,-2(上單調(diào)遞增，在(-2,4(上單調(diào)遞減，在(4,+∞(上單調(diào)遞增.(2)(i)∵g(x(=emx-1+mx3-x2+x--1，/(x(=emx-1+mx2-2x+，∴g″(x(=emx-1+mx-2.″(x(=emx-1+mx-2在R上單調(diào)遞增，又g=e0+m×-+×--1=-1，∴曲線y=g(x(的拐點是,-1(.″(x(<0,g/(x(單調(diào)遞減；″(x(>0,g/(x(單調(diào)遞增；∴函數(shù)g(x(在R上單調(diào)遞增，不妨設(shè)x1<<x2.要證x1+x2<2<-x1，即證g(x2(<g-x1(，又g(x1(+g(x2(=-2，即證-2-g(x1(<g-x1(，即證g(x1(+g-x1(>-2.令h(x(=g(x(+g-x(，則h/(x(=g/(x(-g/-x(，∴h″(x(=g″(x(+g″-x(=(emx-1+mx-2(+e+m-x(-2e=emx-1+e1-mx-2=emx-1+-2≥0，e∴函數(shù)h/(x(=g/(x(-g/-x(在R上單調(diào)遞增，又h/=g/g/-=0，∴函數(shù)h(x(=g(x(+g-x(在上單調(diào)遞減，在,+∞(上單調(diào)遞增.∴h(x1(=g(x1(+g-x1(>h=g+g-=-2得證，即x1+x2<成立. A.a>1B.x1+x2<C.x1?x2<1D.x2-x1>則直線y=a與函數(shù)g(x(的圖象有兩個交點，g(x(=-，由圖可知，<x1<1<x2，因為f(x(=-a=，由f(x(>0可得0<x<，由f(x(<0可得x>，(，則必有0<x1<<x2，所以，0<x1<，則-x1>，令h(x(=f-x(-f(x(=ln-x(-a-x(-lnx+ax，其中0<x<，所以，h(x1(>h=0，即f-x1(-f(x1(>0，即f(x1(<f-x1(，又f(x2(=f(x1(=0，可得f(x2(<f-x1(，因為函數(shù)f(x(的單調(diào)遞減區(qū)間為,+∞(，則x2>-x1，即x1+x2>，故B錯誤；+x2=>，2>1由圖可知<x1<1<x2，則-x1>-1，又因為x2>，所以，x2-x1>-1，故D正確. 2(2024·全國·模擬預(yù)測)若函數(shù)f(x(=alnx+x2-2x有兩個不同的極值點x1,x2，且t-f(x1(+x2<f(x2(-x1恒成立，則實數(shù)t的取值范圍為()A.(-∞,-5(B.(-∞,-5[C.(-∞,2-2ln2(D.(-∞,2-2ln2[【解題思路】首先對f(x(求導(dǎo)，得fI(x(=(x>0(，根據(jù)題意得到方程x2-2x+a=0有兩個不相等f(x1(+f(x2(-x1-x2關(guān)于參數(shù)a的表達式，從而構(gòu)造函數(shù)，利用導(dǎo)數(shù)知識進行求解.【解答過程】依題意得fI(x(=+x-2=(x>0(，若函數(shù)f(x(有兩個不同的極值點x1,x2，則方程x2-2x+a=0有兩個不相等的正實數(shù)根x1,x2，(Δ=4-4a>0x1+x2=2>0，解得0<a<1，1x2=a>0因為t-f(x1(+x2<f(x2(-x1，可得t<f(x1(+f(x2(-x1-x2=alnx1+x-2x1+alnx2+x-2x2-x1-x2=aln(x1x2(+(x+x(-3(x1+x2(=aln(x1x2(+(x1+x2(2-x1x2-3(x1+x2(=alna+×22-a-3×2=alna-a-4.設(shè)h(a(=alna-a-4(0<a<1(，則hI(a(=lna<0，則h(a(單調(diào)遞減，h(a(>h(1(=-5，可知t≤-5. 【解答過程】函數(shù)f(x)=(x2+2((x3-3ax2+b(有且僅有兩個不同的零點x1，x2，(x(=x3-3ax2+b，即g(x(有且僅有兩個不同的零點x1，x2，gI(x(=3x2-6ax=3x(x-2a(=0得x=0或x=2a，所以g(x(在(-∞,0(,(2a,+∞(上單調(diào)遞增，在(0,2a(上單調(diào)遞減，同理若a<0，g(x(在(-∞,2a(,(0,+∞(上單調(diào)遞增，在(2a,0(上單調(diào)遞減，3則g(x(有一根是確定的為2a，又因為g(x(=x3-3ax2+b=(x-2a(2(x+a(，所以g(x(的另一根為-a，-1<-1<-1. 4(2024·遼寧·三模)已知函數(shù)f(x)=lnx+x2-ax存在兩個極值點，若對任意滿足f(x1)=f(x2)=f(x3)的x1,x2,x3(x1<x2<x3)，均有f(ex)<f(ex)<f(ex)，則實數(shù)a的取值范圍為()【解答過程】函數(shù)f(x)=lnx+21ex2-ax由函數(shù)f(x)存在兩個極值點，得fI(x)=0，即x2-eax+e=0有兩個不等的正根t1,t2(t1<t2)，(Δ=e2a2-4e>0Lt1t2=et1+t2=ea>0x>t2I(x)Lt1t2=e于是函數(shù)f(x)在(0,t1),(t2,+∞)上單調(diào)遞增，在(t1,t2)上單調(diào)遞減，令f(x1)=f(x2)=f(x3)=m，(x1<x2<x3)，則直線y=m與f(x)的圖象有3個公共點，此時0<x1<t1<x2<t2<x3，顯然ex>et，令y=ex-x,x>0，求導(dǎo)得yI=ex-1>0，由x1<x2<x3x<ex<ex，因為對任意滿足f(x1)=f(x2)=f(x3)的x1,x2,x3(x1<x2<x3)，均有f(ex)<f(ex)<f(ex)，t>t2， 5(2023·四川南充·一模)已知函數(shù)f(x)=|lnx-+2|-m(0<m<3)有兩個不同的零點x1，x2(x1<x2)A.1B.2C.3D.4函數(shù)f(x(=|lnx-+2|-m(0<m<3(有兩個不同零點x1,x2(x1<x2(，轉(zhuǎn)化為|lnx-+2|=m有兩個交lnx-+2【解答過程】由函數(shù)f(x(=|lnx-+2|-m(0<m<3(有兩個不同零點x1,x2(x1<x2(，lnx