A.(0,e]B.(0,e)C.(0,1(令f(x(=，則f(x(在(0,a(上單調(diào)遞增，2.已知函數(shù)f(x(=ex(x2+1((其中e為自然對數(shù)的底數(shù)(1)討論函數(shù)y=f(x(+(a-2(x.ex(a∈R(的單調(diào)性；2x-e2x>λ|f(x1(-f(x2(|恒成立，求實(shí)數(shù)λ的取值范圍.f(x(+(a-2(x.ex=[x2+(a-2(x+1[ex，x∈R，則hI(x(=(x2+ax+a-1(ex=(x+a-1((x+1(ex，令hI(x(=0，解得x=1-a或x=-1.I(x(≥0恒成立且hI(x(不恒為零，所以，函數(shù)h(x(的增區(qū)間為(-∞,1-a(、(-1,+∞(，減區(qū)間為(1-a,-1(；當(dāng)1-a>-1時(shí)，即a<2時(shí)，由hI(x(>0可得x<-1或x>1-a，由hI(x(<0可得-1<x<1-a.所以，函數(shù)h(x(的增區(qū)間為(-∞,-1(、(1-a,+∞(，減區(qū)間為(-1,1-a(.當(dāng)a>2時(shí)，函數(shù)h(x(的增區(qū)間為(-∞,1-a(、(-1,+∞(，減區(qū)間為(1-a,-1(；當(dāng)a<2時(shí)，函數(shù)h(x(的增區(qū)間為(-∞,-1(、(1-a,+∞(，減區(qū)間為(-1,1-a(.所以f(x(在(0,+∞(上單調(diào)遞增，且f(x(>f(0(=1>0.，所以f(x1(>f(x2(，則不等式e2x-e2x>λ|f(x1(-f(x2(|可化為e2x-e2x>λ[f(x1(-f(x2([，2x-λf(x1(>e2x-λf(x2(.令g(x(=e2x-λf(x(，則問題等價(jià)于函數(shù)g(x(在(0,+∞(上單調(diào)遞增，即gI(x(=2e2x-λfI(x(≥0在(0,+∞(上恒成立， 令p(xx∈(0,+∞(，則p/(xx(<0，函數(shù)p(x(在(0,1(上單調(diào)遞減；x(>0，函數(shù)p(x(在(1,+∞(上單調(diào)遞增；(1)?x∈D，m≤f(x(?m≤f(x(min；(2)?x∈D，m≥f(x(?m≥f(x(max；(3)?x∈D，m≤f(x(?m≤f(x(max；(4)?x∈D，m≥f(x(?m≥f(x(min.間(mln2,+∞(上單調(diào)遞增；(2),+∞(．將a=2代入f(x(的解析式，得f(x(=log2x+-，求導(dǎo)得f當(dāng)m≤0時(shí)，f/(x(>0，故f(x(在(0,+∞(上單調(diào)遞增；f/(x(>0，于是f(x(在區(qū)間(0,mln2(上單調(diào)遞(mln2,+∞(上單調(diào)遞增．所以g/(t(=-≤0在(0,+∞(上恒成立，可得2m≥t-tlnt在(0,+∞(上恒成立，設(shè)h(t(=t-tlnt，則h/(t(=-lnt，令h/(t(=0，得t=1．所以h(t(max=h(1(=1，得m≥.所以實(shí)數(shù)m的取值范圍為,+∞(． (5)寫出函數(shù)f(x(的單調(diào)區(qū)間，解決恒成立問題的策略：若f(x(在閉區(qū)間D上有最值，則f(x(>0在區(qū)間D上恒成立?f(x(min>0；f(x(<0在區(qū)間D上恒成立f(x(min<0．解決恒成立與存在性問題時(shí)，要注意理解4.已知函數(shù)f(x)=(a+1)lnx+ax2+1．)-f(x2)|≥4|x1-x2|．求導(dǎo)f/(x)=+4x，切線斜率k=f/(1)=7∴曲線y=f(x)在(1,f(1)(處的切線方程為y=7x-4．(2)∵a≤-2，f(x)的定義域?yàn)?0,+∞)，求導(dǎo)f/(x)=ax∴f(x)在(0,+∞)上單調(diào)遞減．不妨假設(shè)x1≥x2，∴|f(x1(-f(x2(|≥4|x1-x2|等價(jià)于f(x2(-f(x1(≥4x1-4x2．即f(x2(+4x2≥f(x1(+4x1．令g(x)=f(x)+4x，則g/(x(=+2ax+4=．∵a≤-2，x>0，∴g/(x(≤=≤0(當(dāng)且僅當(dāng)x=時(shí)等號成立)．從而g(x)在(0,+∞)單調(diào)遞減，故g(x1)≤g(x2)，即f(x2(+4x2故對任意x1,x2∈(0,+∞(,|f(x1(-f(x2(|≥4|x1-x2|．5.已知函數(shù)f(x(=+lnx.(1)求出f(x(的極值點(diǎn)；，若f(x1(=f(x2(，則x1+x2>4. ，f(x(單調(diào)遞減；，f(x(單調(diào)遞增，因?yàn)閒(x1(=f(x2(，不妨設(shè)0<x1<2<x2，x2-4>0令H(t(=2t2-2-4tlnt(t>1)，H/(t(=4t-4lnt-4(t>1)，因?yàn)镠Ⅱ(t>0，所以H/(t(在(1,+∞(上是增函數(shù)，所以H/(t(>H/(1)=0，所以H(t(在(1,+∞(上是增函數(shù)，(1)討論函數(shù)y=f(x)的單調(diào)性；f(x)在(lna,+∞)上是單調(diào)增函數(shù)；(x)=ex-a．，f(x)在(-∞,lna)上是單調(diào)減函數(shù)；<lna<x2，x+xx2-x1，也就是證e<ex2x2-x1，x-xx-x又x2-x1>0，只要證e2-e-2>x2-x1(*)．x-xx-x設(shè)g(t)=(et-e-t)-2t(t>0)，g(t)=(et+e-t)-2>0，∴x1+x2<2lna．7.已知函數(shù)f(x(=lnx-x-1.(1)討論f(x(的單調(diào)性；則f(xf(x)>0，則f(x(在(0,上單調(diào)遞增，(2)根據(jù)題意，g(x(=xf(x(+a得到g(x(=lnx-ax， 令h(t(=lnt-,，令h/(t(=0t(>0在(0,1(上成立，8.已知函數(shù)g(x)=ex-ax2-ax，h(x)=ex-2x-lnx．其中e為自然對數(shù)的底數(shù)．(1)若f(x)=h(x)-g(x)，討論f(x)的單調(diào)性；【詳解】解：(1)f(x)=h(x)-g(x)=ex-2x-lnx-ex+ax2+ax=ax2+(a-2)x-lnx(x>0)，f/(x)=2ax+(a-2)-==(x>0)，x-x，即證(x1-x2(e2>ex-x-1. t令t=x1-x2(t<0(，即證te2-et+1>0，(t)=，:p(t)>p(0)=0:h/(t)<0，:h(t)在(-∞,0)上遞減，結(jié)果.9.已知函數(shù)f(x)=x-a+2lnx((a∈R)在定義域上有兩個(gè)極值點(diǎn)x1,x2，則f(x1.x2(的取值范圍是()A.(-∞,1)B.(-∞,0)C.(0,+∞)D.(1,+∞)令g(x)=x2-2ax+a，(Δ=(-2a)2-4a>0:f(x1.x2(=f(a)=a-a+2lna(=a-2alna-1，令T(a)=a-2alna-1(a>1)，:T'(a)=-1-2lna<0:T(a)<T(1)=0，故f(x1.x2(的取值范圍是(-∞,0)，10.已知函數(shù)f(x(=+(a-3(x+alnx(a∈R(，在定義域上有兩個(gè)極值點(diǎn)x1,x2,且x1<x2．(2)求證:f(x1(+f(x2(+5>0 x>x2故f(x1)+f(x2)=+(a-3)x1+alnx1++(a-3)x2+alnx2=-x1x2+(a-3)(x1+x2)=-a+(a-3)(3-a)+alna=alna-+2a-，由于h(e-3(=-e-3<0，h(1)=2>0，所以當(dāng)0<a<1時(shí)，g(a)≥g(t)=tlnt-+2t-=t(t-3)-+2t-=-t-，-3,1)，-t-=(t-1)2-5>-5，所以f(x1)+f(x2)+5>0．11.已知函數(shù)f(x(=2lnx，g(x(=ax2-2x(a>0)．(2)設(shè)函數(shù)h(x(=f(x(+g(x(-(a+2(x+3．將切線方程y=2ex-4代入g(x(=ax2-2x(a>0)中，可得ax2-(2e+2(x+4=0.，(2)(i)h(x(=2lnx+ax2-(a+4(x+3，設(shè)x1<x2,x2(x(<0；即函數(shù)h(x(有兩個(gè)極值點(diǎn)x1，x2.(ii)h(x1(+h(x2(=2(lnx1+lnx2(+a(x+x(-(a+4((x1+x2(+6=2lnx1x2+a[(x1+x2(2-2x1x2[-(a+4((x1+x2(+6若h(x1(+h(x2(≤-a-3恒成立，則2ln+-+3≤0恒成立，(1)討論函數(shù)f(x)的單調(diào)性；(2)設(shè)函數(shù)g(x)=2f(x)-x2+2ax+，若g(x)存在兩個(gè)極值點(diǎn)x1,x2(x1<x2(，證明：g(x1(-g(x2(>2(a-2)(x1-x2(.(x)>0， lllll從而gl(x)=-2-=,x>0.2-16≤0ll0<x1<2,x2>2，所以g(x1(-g(x2(>2(a-2)(x1-x2(等價(jià)于2aa，令h(x)=-x+2lnx(x>2),hl(x)=--1+=<0，故g(x1(-g(x2(>2(a-2)(x1-x2(成立.13.已知函數(shù)f(x(=ex-x，g(x(=(x+k(ln(x+k(-x．+l(x(=ex-1，gl(x(=ln(x+k(，l(t(=gl(t(…①,得et-ln(t+1(-1=0，令φ(t(=et-ln(t+1(-1，則φl(t(=et-，因?yàn)?t(=etl(t(在(-1,+∞(單調(diào)遞增，t(>0，φ(t(單調(diào)遞增；t(<0，φ(t(單調(diào)遞減；(2)解法一：令h(x(=f(x(-bx+g(b(-f(0(-g(0((x>0)，則hl(x(=ex-(b+1(，所以當(dāng)x>ln(b+1(時(shí)，hl(x故h(x(≥h(ln(b+1((=f(ln(b+1((+g(b(-f(0(-g(0(-bln(b+1(=(b+k(ln(b+k(-(b+1(ln(b+1(-klnk．令t(x(=(x+k(ln(x+k(-(x+1(ln(x+1(-klnk(x>0)，則tl(x(=ln(x+k(-ln(x+1(．l(x(>0，t(x(在(0,+∞(單調(diào)遞增，(iii)若0<k<1時(shí)，tl(x(<0，t(x(在(0,+∞(單調(diào)遞減，x-x-1≥0，x-1≥lnx，x-xlnx-1≤0…(*)．令φ(x(=ex-x-1，x(=ex-1≥0，φ(x(單調(diào)遞增，x(=ex-1≤0，φ(x(單調(diào)遞減，又由上式得，當(dāng)x>0時(shí)，-1≥ln，1-x≥-xlnx，x-xlnx-1≤0.令h(x(=g(x(-ax+f(a(-f(0(-g(0((x>0)，則hl(x(=ln(x+k(-a，a-k時(shí)，hl(x(>0，h(x(單調(diào)故h(x(≥h(ea-k(=g(ea-k(-a(ea-k(+f(a(-f(0(-g(0(=(k-1(a+k-1-klnk．(ii)若0<a≤lnk時(shí)，hl(x所以h(x(>h(0(=f(a(-f(0(=ea-a-1．-1(a+k-1-klnk≥0，則當(dāng)a>lnk時(shí)，h(x(≥(k-1(a+k-1-klnk>(k-1(lnk+k-1-klnk=-lnk+k-1>0(由(*)知)； 也即證b-a>2.，故只需證(b-a((eb-a+1(>2(eb-a-1(①,整理得：(t-2(et+t+2>0②,令h(t(=(t-2(et+t+2，t>0，則h/(t(=(t-1(et+1，hⅡ(t(=tet>∴h/(t(在(0,+∞(上單調(diào)遞增，又h/(0(=0，，從而h(t(在(0,+∞(上單調(diào)遞增，即證(b-a((eb+ea(-2(eb-ea(>0，令u(x(=(x-a((ex+ea(-2(ex-ea(，x>a，則u/(x(=(ex+ea(+(x-a(ex-2ex=(x-a-1(ex+ea，uⅡ(x(=(x-a(ex>0，∵b>a，∴u(b(=(b-a((eb+ea(-2(eb-ea(>0，故>．(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程；又f/(x)=ex(ln(1+x)+，∴切線斜率k=f/(0)=1 令h=ln(3)解：原不等式等價(jià)于f(s+t)-f(s)>f(t)-f(0)，令m(x)=f(x+t)-f(x)，(x,t>0)，即證m(x)>m(0)，∵m(x)=f(x+t)-f(x)=ex+tln(1+x+t)-exln(1+x)，m/(x)=ex+tln(1+x+t)+-exln(1+x)-=g(x+t)-g(x)，由(2)知g(x)=f/(x)=ex(ln(1+x)+在[0,+∞(上單調(diào)遞增，∴g(x+t)>g(x)，(x)>0因?yàn)閒(x(有兩個(gè)極值點(diǎn)，所以f/(x(=0有兩個(gè)根，. 所以函數(shù)y=在(-∞,1(單調(diào)遞增，在(1,+∞(單調(diào)遞減，x1-lna①,x2=lnx2-lna②,①-②得x1-x2=lnx1-lnx2=ln，令g(t(=tlnt+2lnt-3(t-1(，則g(t(=lnt+-2，令h(t(=lnt+-2，則h(t(=-=<0，所以g(t(在(0,1(單調(diào)遞減，所以g(t(>g(1(=0，所以g(t(在(0,1(單調(diào)遞增，所以g(t(<g(1(=0，即tlnt+2lnt-3(t-1(<0，得tlnt+2lnt<3(t-1(17.已知f(x(=x-alnx-x2.(2)若函數(shù)y=f(x(存在兩個(gè)不同的極值點(diǎn)x1,x2，求f(x1(+f(x2(的取值范圍.又曲線y=f(x(在x=1處的切線與y=x垂直，(2)因?yàn)閒,(x(=1--ax=，x>0，所以f(x(的兩個(gè)極值點(diǎn)x1,x2是方程-ax2+x-a=0的兩正根，所以f(x1(+f(x2(=x1-alnx1-x+x2-alnx2-x=(x1+x2(-aln(x1x2(-[(x1+x2(2-2x1x2[所以g(a(在(0,單調(diào)遞減，+所以f(x1(+f(x2(的取值范圍是,+∞(.曲線y=f(x(在x=e處切線的斜率為k=f,(e(=3+lne=4，又“f(e(=2e+elne=3e,:切線方程為y-3e=4(x-e(，即曲線y=f(x(在x=e處的切線方程為4x-y-e=0；(2)若f(x(有兩個(gè)零點(diǎn)x1,x2，得-a故lnx，則lnx2=ln(tx1(=lnt+lnx1=lnt+令h(tt>3(，則h,(t，令g(t(=-2lnt+t-(t>3(，則g,(t:g(t(在(3,+∞(上單調(diào)遞增，:g(t(>g(3(=-2ln3+3-=(4-3ln3(=(lne4-ln33(>0， (t-1)2∴h/(t(=g(t(>0，則h(t(在(t-1)2故x1x2>.19.已知函數(shù)f(x(=lnx+x2-2ax,a∈R，(2)若函數(shù)f(x(有兩個(gè)極值點(diǎn)x1,x2(x1<x2(，求2f(x1(-f(x2(的最小值.(2)-【詳解】(1)因?yàn)閒(x(=lnx+x2所以f/(x)=+2x-2a=,此時(shí)f(x)在(0,+∞)上單調(diào)遞增，(x)>0；3<x<x4所以f(x)在(0,x3(,(x4,+∞(上單調(diào)遞增，在(x3,x4(上單調(diào)遞減；則2f(x1(-f(x2(=2(lnx1+x-2ax1(-(lnx2+x-2ax2(=2(lnx1+x-2x-1(-(lnx2+x-2x-1(=-2x+2lnx1-lnx2+x-1=x-22+2ln-lnx2-1=x-2-lnx-2ln2-1lnt-2ln2-1，tlnt-2ln2-1， 所以g(t(min=g所以2f(x1(-f(x2(的最小值為-.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：本題解決的關(guān)鍵是，利用韋達(dá)定理將雙變量的2f(x1(-f(x2(轉(zhuǎn)化為關(guān)于單變量x的函 x1ex2-x2ex1<m(x2-x1(?<令函數(shù)f(x)=，則上述不等式等價(jià)于f(x2)<f(x1)(當(dāng)x1<x2時(shí))，即f(x)=在[1,2[上單調(diào)遞減.x2x2.f/(x)=ex.x-(ex+m(=ex(x2x2.2>0(x,2[(，故f/(x)≤0等價(jià)于：ex(x-1(-m≤0?m≥ex(x-1(.令k(x)=ex(x-1(，求導(dǎo)：k/(x)=ex(x-1(+ex=xex.x>0且x∈[1,2[時(shí)x>0，故k/(x)>0，即k(x)=ex(x-1(在[1,2[上單調(diào)遞增.k(2)=e2(2-1(=e2.由m≥ex(x-1(恒成立，且ex(x-1(的最大值為e2，得到m≥e221.已知函數(shù)f(x(=(x2-ax-a(ex.(1)討論函數(shù)f(x(的單調(diào)性；∈[-4,0[，都有|f(x2(-f(x1(|<4e-2+mea恒成立，求m的取值范圍.(2)m>【詳解】(1)函數(shù)f(x(=(x2-ax-a(ex的定義域?yàn)镽，f/(x(=[x2+(2-a(x-2a[ex=(x-a((x+2(ex.令f/(x(=0，解得x=a或x=-2.當(dāng)a<-2，令f/(x(>0得x<a或x>-2；令f/(x(<0得a<x<-2，∴函數(shù)f(x(在(-∞,a(和(-2,+∞(上單調(diào)遞增，在(a,-2(上單調(diào)遞減；當(dāng)a=-2，f/(x(=(x+2(2ex≥0恒成立，∴函數(shù)f(x(在(-∞,+∞(上單調(diào)遞增；當(dāng)a>-2，令f/(x(>0得x<-2或x>a；令f/(x(<0得-2<x<a，∴函數(shù)f(x(在(-∞,-2(和(a,+∞(上單調(diào)遞增，在(-2,a(上單調(diào)遞減.綜上，當(dāng)a<-2，函數(shù)f(x(在(-∞,a(和(-2,+∞(上單調(diào)遞增，在(a,-2(上單調(diào)遞減；當(dāng)a=-2，函數(shù)f(x(在(-∞,+∞(上單調(diào)遞增； 當(dāng)a>-2，函數(shù)f(x(在(-∞,-2(和(a,+∞(上單調(diào)遞增，在(-2,a(上單調(diào)遞減.f(x(在(-4,-2(上單調(diào)遞增，在(-2,0(上單調(diào)遞減.∴f(x(max=f(-2(=(4+a(e-2.又f(-4(=(16+3a(e-4>0，f(0(=-a<0，∴f(x(min=f(0(=-a.∈[-4,0[，|f(x2(-f(x1(|max=|f(-2(-f(0(|=(4+a(e-2+a.∈[-4,0[，都有|f(x2(-f(x1(|<4e-2+mea恒成立，(e-2+1((1-a(a.∴函數(shù)g(a)=在(0,1(上單調(diào)遞增，在(1,2(上單調(diào)遞減，max=g22.已知f(x(=ex-ax，g(x(=lnx-ax，若對任意x1∈(0,+∞(，都存在x2∈(0,+∞(，使得f(x1(g(x2(=x1x2【答案】(-∞,0[則F/(x)=，故F(x)的值域?yàn)閇e-a,+∞(；故G(x)的值域?yàn)?a；若a≥e，則e-a≤0，故F(x)值域[e-a,+∞(?[0,+∞(，且恒有G(x)≤-a<0.但當(dāng)F(x1)≥0時(shí)，恒有F(x1)G(x2)≤0，故答案為：(-∞,0[.2∈B，使得M(x1)=N(x2)?{M(x)|x∈A{?{N(x)|x∈B{，即M(x)在定義域A上的值域是函數(shù)N(x)在定義域B上的值域的子集.23.已知函數(shù)f(x(=-mx+lnx+1，g(x(=cosx+xsinx-1．(1)求g(x(在區(qū)間[0,π[的最大值和最小值；(2)討論函數(shù)f(x(的單調(diào)區(qū)間與極值；,π[，使得不等式f(x1(-g(x2(>1成立，試求實(shí)數(shù)m的所以g(x)max=g=-1，因?yàn)間(0(=0,g(π(=-2，所以g(x(min=-2，g(x(在區(qū)間[0,π[的最大值和最小值分別是-1,-2；(2)由f(x(=-mx+lnx+1?f/(x(=-m+，函數(shù)f(x(的定義域?yàn)槿w正實(shí)數(shù)集，當(dāng)x>時(shí)，f/(x(<0,f(x),+∞(上在單調(diào)遞減，該函數(shù)有極大值f=-lnm，無極小值， ,π[，使得不等式f(x1(-g(x2(>1成立，等價(jià)于f(x(在[1，2[上的最小值與g(x(在[0,π[上的最小值的差大于1，當(dāng)m≥1時(shí)，則有0<≤1，由(2)可知f(x(在[1，2[上的最小值為f(x(min=f(2(=-2m+1+ln2，由(1)可知g(x(在[0,π[上的最小值為-2，所以有-2m+1+ln2-(-2(>1?m<1ln2?1≤mln2；-f(1(=ln2-m，所以當(dāng)<m<ln2時(shí)，f(x(min=f(1(=-m+1，此時(shí)有-m+1-(-2(>1?m<2?m<ln2；當(dāng)ln2≤m<1時(shí)，f(x(min=f(2(=-2m+1+ln2，則有-2m+1+ln2-(-2(>1?m<1+ln2?l 24.已知函數(shù)f(x)=，g(x)=ax+b，設(shè)F(x)=f(x)-g(x).(2)求f(x)在[2,m](m>2)上的最小值；【答案】(1)-1-b；【詳解】(1)依題意，函數(shù)F(x)=f(x)-g(x)=-ax-b，其定義域?yàn)?0,+∞)，當(dāng)a=1時(shí)，F(xiàn)(x)=-x-b，求導(dǎo)得F/(x)=-1，(x)<0，所以F(x)的最大值為F(1)=-1-b.f/(x)<0，所以當(dāng)2<m≤4時(shí)，f(x)min=f(2)=；當(dāng)m>4時(shí)，f(x)min=f(m)=.兩式相減得lnx2-lnx1=(x2-x1)[a(x2+x1)+b]?=a(x2+xx1x1x2+x1x1tx1+x1t+1，令t=x2>12-2(x2-x1)=lntx1-2(txx1x1x2+x1x1tx1+x1t+1，所以(x1+x2)g(x1+x2)>2.25.已知函數(shù)f(x)=ex-ax．(1)討論f(x)的單調(diào)性；(lna,+∞(單調(diào)遞增(2)[0,e[【詳解】(1)由題知函數(shù)f(x)的定義域?yàn)镽，f/(x)=(x)=ex-a>0，函數(shù)f(x)在R上單調(diào)遞增；(x)=ex-a在R上單調(diào)遞增，且f/(lna)=elna-a=a-a=0，(x)=ex-a>0，函數(shù)f(x)單調(diào)遞增，當(dāng)a>0時(shí)，函數(shù)f(x)在(-∞,lna(上單調(diào)遞減，在(lna,+∞(單調(diào)遞增.當(dāng)a>0時(shí)，函數(shù)f(x)在(-∞,lna(上單調(diào)遞減，在(lna,+∞(單調(diào)遞增，所以f(x)min=f(lna(=a-alna=a(1-lna(≥0，解得0<a≤e，f/(x)=ex-2=0?x=ln2，所以f(x)在(-∞,ln2(上單調(diào)遞減，在(ln2,+∞(單調(diào)遞增，f(x)min=f(ln2(=2-2ln2，∵f(x)=t(t∈R)有兩個(gè)實(shí)根x1和x2，∴f(x1)=ex-2x1=t,f(x2)=ex-2x2=t，不妨取x1<ln2<x2，要證|x1-x2|<3t-1，即證x2-x1<3t-1=2t-(1-t(，現(xiàn)證明x2<2t2<2(ex-2x2(?2ex-5x2>0，令u(x(=2ex-5x,x>ln2，u/(x(=2ex-5=0?x=ln，再證明x1≥1-t，即x1≥1-ex+2x1?ex-x1-1≥0，令v(x(=ex-x-1,x<ln2，v/(x(=ex-1=0?x=0， 所以v(x(在(-∞,0(單調(diào)遞減，在(0,ln2(單調(diào)遞增，即v(x(≥v(0(=e0-0-1=0，所以ex-x1-1≥0，即x1≥1-t，-x1≤t-1，所以x2-x1<3t-1，即得證|x1-x2|<3t-1.26.已知函數(shù)f(x(=2x-ex．(1)求函數(shù)f(x(在x=1處的切線方程；(2)求函數(shù)f(x(的極值；(3)若關(guān)于x的方程f(x(=a(a∈R(有兩個(gè)根x1和x2，求證：|x1【答案】(1)(2-e(x-y=0又f(1(=2-e，故f(x(在x=1處的切線方程為y-(2-e(=(2-e((x-1(，即(2-e(x-y=0．(2)函數(shù)f(x(=2x-ex的定義域?yàn)镽，又f/(x(=2-ex，故f(x(在(-∞,ln2(上單調(diào)遞增，在(ln2,+∞(上單調(diào)遞減，所以f(x(的極大值為f(ln2(=2ln2-eln2=2ln2-2f(x1(=f(x2(=a，構(gòu)造g(x(=f(x(-(2-e(x=ex-ex，x∈(ln2,+∞(，g(x(在(ln2,1(上單調(diào)遞增，在(1,+∞(上單調(diào)遞減，g(x(≤g(1(=0，=f(x2(-(2-e(x2≤0，所以x2≤構(gòu)造h(x(=f(x(-x+1=x+1-ex，x∈(-∞,ln2(，hh(x(在(-∞,0(上單調(diào)遞增，在(0,ln2(上單調(diào)遞減，h(x(≤h(0(=0，∈(-∞,ln2(，則h(x1(=f(x1(-x1+1≤0，所以|x1-x2|=x2-x1≤,又等號不同時(shí)成立，所以|x1-xa-1．(2)構(gòu)造新的函數(shù)h(x(；(1)討論f(x)的單調(diào)性；,+∞)． n.解法一：設(shè)g(x)=f(12-x)-f(x)=6ln(12-x)-(12-x)-(6lnx-x(=6ln(12-x)-6lnx+2x-12,x∈所以g(x)在(0,6)上單調(diào)遞減，所以g(x)>g(6)=f(12-6)-f(6)=0，則f(12-x1(>f(x1(=f(x2(，又12-x1,x2∈(6,+∞)，且f(x)在(6,+∞)上單調(diào)遞減，2-x1的最小值g(x2(=m>02x=m2-1=m，所以x1=lnm，x2=m+1，x2-x1=m+1-l 令h(m(=m+1-lnm(m>0(，所以h(m(=1-，所以當(dāng)m=時(shí)，h(m(有最小值+ln2， 即x2-x1的最小值為+ln2.30.已知函數(shù)f(x)=ex-(x>0),g(x)=lnx．(1)求函數(shù)h(x)=f(x)+g(x)在[1,e]上的最值及其零點(diǎn)個(gè)數(shù)；(2)若對于任意的0<x1<x2，均有f(x1(-ag(x1(<f(x2(-ag(x2(，求a的取值范圍．【詳解】(1)易知h(x)=f(x)+g(x)=ex-+lnx(x>0(，所以h(x(在[1,e]上單調(diào)遞增，則h(x(≤h(e(=ee-1,h(x(≥h(1(=-e，(2)設(shè)m(x(=f(x(-ag(x(，則對于任意的0<x1<x2，均有m(x1(<m(x2(，即m(x(在(0,+∞(上單調(diào)遞增，所以m'(x(=f'(x(-ag'(x(=ex+-≥0在(0,+∞(上恒成立，即a≤xex+在(0,+∞(上恒成立，令n(x(=xex+(x>0(，則n'(x(=(x+1(ex-，令u(x(=n'(x(，則u'(x(=(x+2(ex+>0，即n'(x(在(0,+∞(上單調(diào)遞增，又n'(1(=0，則n(x(在(0,1(上單調(diào)遞減，在(1,+∞(上單調(diào)遞增，31.已知函數(shù)f(x(=x2-1+aln(1+x(，(3)若f(x(存在兩個(gè)不同的極值點(diǎn)x1,。x2，x1<x2，且f(x1(<mx2，求實(shí)數(shù)m的取值范f'(x(=0的解不連續(xù)，則a≥[-2x(1+x([max=|-2(x+2+max=，令h(x(=,x∈(-1,+∞(，則h'(x(=,x∈(-1,+∞(，所以當(dāng)x∈(-1,0(時(shí)，h'(x(>0，則(3)因?yàn)閒(x(存在兩個(gè)不同的極值點(diǎn)x1,。x2,。x1<x2， 所以Δ=4-8a>0,x1+x2=-1,x1x2=，且-1<x1<-，-<x2<0，又由f(x1(<mx2可得m<，1+x1(，令φ(x(=1-x+2xln(1+x，x(=-1+2ln(1+x(+ln(1+xln(1+x(，則(xln(1+x(<0，所以φ(x(在區(qū)間(-1,-上單調(diào)遞減，32.設(shè)函數(shù)f(x)=x(lnx)2-(a+2)xlnx+(a+3)x,a>0．(2)討論f(x)的單調(diào)性；因此曲線y=f(x)在點(diǎn)(e,f(e))處的切線方程為y-2e=(2-a)(x-e)，(2)f/(x)=(lnx-2+1-．(x)≥0,f(x)在(0,+∞)單調(diào)遞增．lnx1>0,lnx2>0.-<等價(jià)于(lnx1(2-(a+2)lnx1-(lnx2lnx其中(lnx1(2-(lnx2(2=(lnx1-lnx2((lnx1+lnx2(=a(lnx1-lnx2(．.lnx2=1且x1<x2得x2>e，記h(x)=(lnx)2-+1,(x>e)，則h/(x)=-=，33.已知函數(shù)f(x(=ex-mx,g(x(=ex(-sinx+cosx+a(≥f(x2(成立，求a的取值范圍.(2)[0,+∞(【詳解】(1)f(x)=ex-mx，f/(x)=ex-m，f(x)在x=0處取得極值，則f/(0)=0,m=1.∴f/(x(=ex-1，當(dāng)f/(x)<0,x<0，f/(x)>0,x>0所以f(x)的減區(qū)間為(-∞,0(，增區(qū)間為(0,+∞(符合題意.(2)由(1)知，函數(shù)f(x(min=f(0(=1?x1≥f(x2(成立等價(jià)于不等式ex(-sinx+cosx+a(≥1在x∈0,時(shí)有解即不等式a≥sinx-cosx+e-x在x∈0,]時(shí)有解...設(shè)F(x(=sinx-cosx+e-x，F(xiàn)/(x(=sinx+cosx-e-x,2[而e-x≤1所以F/(x(≥0恒成立即F(x)在0，|上是增函數(shù)，則F(x)min=F(0)=034.已知函數(shù)f(x)=lnx-ax+1(a∈R).且函數(shù)f(x)有兩個(gè)零點(diǎn)，【詳解】(1)函數(shù)f(x)的定義域?yàn)?0,+∞)，對函數(shù)f(x)求導(dǎo)得f/(x)=-a， 35.已知函數(shù)f(x)=ae-x-x2,g(x)=xex-asinx，其中a∈R1-x2>2【詳解】(1)令f(x)=0,即ae-x-x2=0,即a=x2ex,令h(x)=x2ex-a,h/(x)=(x2+2x)ex,且h(0(=-a<0,h(a)=aea-a=a(ea-1(,ex有唯一解,所以f(x)在(0,+∞)上存在唯一的零點(diǎn).2)=