全國(guó)統(tǒng)考高考數(shù)學(xué)大一輪復(fù)習(xí)第3章導(dǎo)數(shù)及其應(yīng)用第3講導(dǎo)數(shù)的綜合應(yīng)用2備考試題文含解析

一輪復(fù)習(xí)精品資料(高中)PAGEPAGE1第三章導(dǎo)數(shù)及其應(yīng)用第二講導(dǎo)數(shù)的綜合應(yīng)用1.〖2021惠州市二調(diào)〗若函數(shù)f(x)=ex(x2-2x+1-a)-x恒有2個(gè)零點(diǎn),則a的取值范圍是()A.(-1e,+∞)C.(0,1e) D.(-∞,-12.〖2021陜西百校聯(lián)考〗已知銳角x1,x2滿(mǎn)足sinx1-cosx2<x1+x2-π2,則下列結(jié)論一定正確的是()A.sinx1<sin(x1+x2)B.tanx1>tanxC.sinx1+cosx1>sinx2+cosx2D.sinx1+sinx2>cosx1+cosx23.〖2021大同市調(diào)研測(cè)試〗已知函數(shù)f(x)=ax3-3x2+1,若f(x)存在唯一的零點(diǎn)x0,且x0<0,則a的取值范圍是()A.(2,+∞) B.(-∞,-2)C.(1,+∞) D.(-∞,-1)4.〖2020廣東七校第二次聯(lián)考〗設(shè)定義在R上的函數(shù)y=f(x)滿(mǎn)足?x∈R,f(x+2)=1f(x),且x∈(0,4〗時(shí),f'(x)>f(x)x,則6A.6f(2017)<3f(2018)<2f(2019) B.3f(2018)<6f(2017)<2f(2019) C.2f(2019)<3f(2018)<6f(2017) D.2f(2019)<6f(2017)<3f(2018)5.〖2020鄭州市三?！皆O(shè)函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),當(dāng)x>0時(shí),f'(x)lnx<-1xf(x).則使得(x2-4)f(x)>0成立的x的取值范圍是()A.(-2,0)∪(0,2) B.(-∞,-2)∪(2,+∞)C.(-2,0)∪(2,+∞) D.(-∞,-2)∪(0,2)6.已知函數(shù)f(x)=(a-12)x2+lnx,若函數(shù)f(x)在區(qū)間(1,+∞)上的圖象恒在直線(xiàn)y=2ax的下方,則實(shí)數(shù)a的取值范圍是7.〖2021晉南高中聯(lián)考〗已知函數(shù)f(x)=ex-ax,g(x)=1+xlnx.(1)討論函數(shù)f(x)的單調(diào)性;(2)若當(dāng)x>0時(shí),方程f(x)=g(x)有實(shí)數(shù)解,求實(shí)數(shù)a的取值范圍.8.〖2020貴陽(yáng)市高三模擬〗〖交匯題〗已知f(x)=ex,g(x)=x+1.(e為自然對(duì)數(shù)的底數(shù))(1)求證:f(x)≥g(x)恒成立.(2)設(shè)m是正整數(shù),對(duì)任意的正整數(shù)n,(1+13)(1+132)·…·(1+13n)<9.〖2021江西紅色七校第一次聯(lián)考〗若存在兩個(gè)正實(shí)數(shù)x,y使得等式x(2+lnx)=xlny-ay成立,則實(shí)數(shù)a的取值范圍是()A.(0,1e2) B.(-∞,C.(0,1e3) D.(-∞,10.〖2021洛陽(yáng)市統(tǒng)考〗已知函數(shù)f(x)=xex-2-tx-t有2個(gè)零點(diǎn)a,b,且在區(qū)間(a,b)上有且僅有2個(gè)正整數(shù),則實(shí)數(shù)t的取值范圍是A.〖23,e2) B.(23C.〖34e,23) D.(34e11.〖2021江西紅色七校聯(lián)考〗已知函數(shù)f(x)=ax2+bx-lnx.(1)當(dāng)a=-2時(shí),函數(shù)f(x)在(0,+∞)上是減函數(shù),求b的取值范圍;(2)若方程f(x)=0的兩個(gè)根分別為x1,x2(x1<x2),求證:f'(x1+12.〖2021濟(jì)南名校聯(lián)考〗已知f(x)=lnx+ax,g(x)=ex(1)若函數(shù)f(x)的圖象在x=e處的切線(xiàn)與直線(xiàn)2x-y+8=0垂直,求f(x)的極值;(2)當(dāng)x>0時(shí),g(x)≥f(x)恒成立,求實(shí)數(shù)a的取值范圍.13.已知函數(shù)f(x)=lnx,g(x)=x-m.(1)當(dāng)m=0時(shí),求函數(shù)y=f((2)設(shè)h(x)=f(x)-g(x),若x1<x2且h(x1)=h(x2)=0,求證:ln(em+x1-x2)+m>0.答案第三章導(dǎo)數(shù)及其應(yīng)用 第三講導(dǎo)數(shù)的綜合應(yīng)用1.A由f(x)=0,得x2-2x+1-a=xex.令g(x)=xex,則函數(shù)f(x)=ex(x2-2x+1-a)-x恒有2個(gè)零點(diǎn)等價(jià)于函數(shù)y=x2-2x+1-a與y=g(x)的圖象有2個(gè)交點(diǎn),g'(x)=1-xex,令g'(x)>0,得x<1,令g'(x)<0,得x>1,所以g(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以g(x)max=g(1)=1e.作出函數(shù)y=x2-2x+1-a=(x-1)2-a與y=g(x)的圖象,如圖D3-3-2所示,數(shù)形結(jié)合可得-圖D3-3-22.D解法一因?yàn)閟inx1-cosx2<x1+x2-π2,即sinx1-x1<sin(π2-x2)-(π2-x2),所以可構(gòu)造函數(shù)f(x)=sinx-x,x∈(0,π2),因?yàn)閒'(x)=cosx-1<0,所以f(x)在(0,π2)上單調(diào)遞減,因?yàn)閤1∈(0,π2),π2-x2∈(0,π2),所以0<π2-x2<x1<π2,所以sin(π2-x2)<sinx1,即cosx2<sinx1,同理cosx1解法二取x1=x2=π3,則sinx1-cosx2=3-12<π6=π3+π3-π2,此時(shí)sinx1=sin(x1+x2),tanx13.A由題意易知x=0不是函數(shù)f(x)的零點(diǎn),則f(x)=ax3-3x2+1=0?a=3x2-1x3(x≠0),令y=3x2-1x3-a(x≠0),因此f(x)的零點(diǎn)與y=3x2-1x3-a(x≠0)的零點(diǎn)相同.設(shè)g(x)=3x2-1x3(x≠0),則g'(x)=-3(x-1)(x+1)x4,則當(dāng)x∈(-∞,-1)∪(1,+∞)時(shí),g'(x)<0;當(dāng)x∈(-1,0)∪(0,1)時(shí),g'(x)>0,故g(x)在(-1,0),(0,1)上單調(diào)遞增,在(-∞,-1),(1,+∞)上單調(diào)遞減,又g(-1)=-2,g(1)=2,當(dāng)g(x)=0時(shí),x=±33,當(dāng)x→+∞時(shí),圖D3-3-34.A因?yàn)閒(x+2)=1f(x),所以f(x)是以4為周期的周期函數(shù),故6f(2017)=6f(1),3f(2018)=3f(2),2f(2019)=2f(3).令g(x)=f(x)x(x∈(0,4〗),則g'(x)=xf'(x)-f(x)x2,因?yàn)閤∈(0,4〗時(shí),f'(x)>f(x)x,所以g'(x)>0,所以5.D設(shè)函數(shù)g(x)=f(x)lnx,則g'(x)=f'(x)lnx+1xf(x).于是,當(dāng)x>0時(shí),由f'(x)lnx<-1xf(x)可得g'(x)<0,所以函數(shù)g(x)在(0,+∞)上單調(diào)遞減.從而,當(dāng)x>1時(shí),有g(shù)(x)<g(1)=0,即f(x)lnx<0,又lnx>0,所以此時(shí)f(x)<0;當(dāng)0<x<1時(shí),有g(shù)(x)>g(1)=0,即f(x)lnx>0,又lnx<0,所以此時(shí)f(x)<0.對(duì)于不等式f'(x)lnx<-1xf(x),取x=1可得f'(1)ln1<-11f(1),化簡(jiǎn)得f(1)<0,即當(dāng)x=1時(shí),f(x)<0.于是,由上述討論可知,當(dāng)x>0時(shí),f(x)<0,故由(x2-4)f(x)>0(x>0)得x2-4<0,解得0<x<2.當(dāng)x<0時(shí),由f(x)為奇函數(shù)及“當(dāng)x>0時(shí),f(x)<0”可得f(x)>0,故由(x2-4)f(x)>0(x<0)得x2-4>0,解得x<-2.易知f(0)=0,所以x=0不滿(mǎn)足(x2-4)f(x)>0〖解后反思〗一般地,若題設(shè)中出現(xiàn)了與導(dǎo)數(shù)有關(guān)的不等式,則很可能是根據(jù)導(dǎo)數(shù)的運(yùn)算法則提前計(jì)算后而精心設(shè)計(jì)的,所以應(yīng)多從這個(gè)角度考慮如何構(gòu)造函數(shù),以便順利解決問(wèn)題.6.〖-12,12〗由題意知,對(duì)于任意x∈(1,+∞),f(x)<2ax,即(a-12)x2+lnx-2ax<0在(1,+∞)上恒成立.設(shè)g(x)=(a-12)x2+lnx-2ax,x∈(1,+∞),則g(x)的最大值小于0,g'(x)=(①當(dāng)a≤12時(shí),g'(x)<0,∴g(x)在(1,+∞)上單調(diào)遞減,∴g(x)<-a-12≤0,即a≥-12,∴-②當(dāng)a≥1時(shí),g'(x)>0,∴g(x)在(1,+∞)上單調(diào)遞增,最大值可無(wú)窮大,不滿(mǎn)足題意.③當(dāng)12<a<1時(shí),易知g(x)在(1,12a-綜上,實(shí)數(shù)a的取值范圍是〖-12,17.(1)函數(shù)f(x)的定義域?yàn)镽,f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,則f(x)在(-∞,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=ex-a=0,得x=lna,則f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.綜上,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞增,當(dāng)a>0時(shí),f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.(2)由f(x)=g(x),得ax=ex-xlnx-1,因?yàn)閤>0,所以a=exx-ln令h(x)=exx-lnx-1x,x>0,則h'(x)=xex-ex-x+1x2=(ex-1)(x-1)x2.令h'所以h(x)min=h(1)=e-1.又h(x)=exx-lnx-1x=ex-所以當(dāng)x→0時(shí),h(x)→+∞.所以函數(shù)h(x)的值域?yàn)椤糴-1,+∞),因此實(shí)數(shù)a的取值范圍為〖e-1,+∞).8.(1)令h(x)=f(x)-g(x)=ex-x-1,則h'(x)=ex-1,當(dāng)x∈(-∞,0)時(shí),h'(x)<0,當(dāng)x∈(0,+∞)時(shí),h'(x)>0,故h(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以h(x)min=h(0)=0,即h(x)≥0恒成立,所以f(x)≥g(x)恒成立.(2)由(1)可知x=13n時(shí),0<1+(1+13)(1+132)·…·(1+13n)<e易知比e大的最小正整數(shù)是2,所以m的最小值為2.9.D因?yàn)閤,y均為正數(shù),所以等式x(2+lnx)=xlny-ay可化為2+lnx=lny-ayx,即lnyx-2=ayx,即a=lnyx-2yx.令yx則f'(t)=3-lntt2,令f'(t)=0,解得t=e3,當(dāng)t∈(0,e3)時(shí),f'(t)>0,f(t)在(0,e3)上單調(diào)遞增,當(dāng)t∈(e3,+∞)時(shí),f'(t)<0,f(t)在(e3,+∞)上單調(diào)遞減,所以f(t)max=f(e3)=1e3,且當(dāng)t→0時(shí),f(10.C由題意知函數(shù)f(x)=xex-2-tx-t有2個(gè)互異的零點(diǎn)a,b等價(jià)于函數(shù)g(x)=xex-2與h(x)=tx+t的圖象有2個(gè)不同的交點(diǎn).因?yàn)間(x)=xex-2,所以g'(x)=1-xex-2.令g'(x)>0,可得x<1;令g'(x)<0,可得x>1.所以函數(shù)g(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以g(x)max=g(1)=e.當(dāng)x<0時(shí),g(x)<0,當(dāng)x>0時(shí),g(x)>0,且g(0)=0,x?+∞時(shí),g(x)?0.由h(x)=tx+t=t(x+1),知函數(shù)h(x)的圖象為過(guò)定點(diǎn)(-1,0)的一條直線(xiàn),在同一平面直角坐標(biāo)系中,分別作出函數(shù)g(x)與h(x)的圖象如圖D3-3-4所示,若滿(mǎn)足h(x),圖D3-3-411.(1)當(dāng)a=-2時(shí),f(x)=-2x2+bx-lnx,∵f(x)在(0,+∞)上單調(diào)遞減,∴f'(x)=-4x+b-1x≤0對(duì)x∈(0,+∞)恒成立,即b≤4x+1x∴b≤(4x+1x)min∵x>0,∴4x+1x≥4,當(dāng)且僅當(dāng)x=1∴b≤4,即b的取值范圍為(-∞,4〗.(2)f(x)的定義域?yàn)?0,+∞),由題意可得f即lnx1=ax12+bx1,lnx2=ax22+bx2,兩式相減,得lnx1x2=a(x1+x2)(x1-x由f'(x)=2ax+b-1x知,f'(x1+x22)=a(x1+x2)+b-2x1+x2設(shè)t=x1x2∈(0,1),則lnx1令g(t)=lnt-2(t則g'(t)=1t∴g(t)在(0,1)上單調(diào)遞增,∴g(t)<g(1)=0,即lnx1又x1-x2<0,∴1x1-x2即f'(x1+12.(1)∵f(x)的定義域?yàn)?0,+∞),f'(x)=1-∴f'(e)=-ae2,由已知可得f'(e)×2=-1,即a由f'(x)=0,可得x=e1x(0,e1e(e1f'(x)+0-f(x)單調(diào)遞增極大值單調(diào)遞減∴f(x)的極大值為f(e1-e2(2)當(dāng)x>0時(shí),g(x)≥f(x),即ex+2x-1≥化簡(jiǎn)可得,a≤x(ex-1)-lnx+2.令F(x)=x(ex-1)-lnx+2(x>0),只需a≤F(x)min.F'(x)=(x+1)(ex-1x),令h(x)=ex-1則h'(x)=ex+1x2>0,∴h(∵h(yuǎn)(12)=e12∴存在唯一的x0∈(0,+∞),使得h(x0)=ex易知F(x)在區(qū)間(0,x0)上單調(diào)遞減,在區(qū)間(x0,+∞)上單調(diào)遞增,∴F(x)min=F(x0)=x0(ex0-1)-lnx0+2=x0ex0-(由ex0-1x兩邊取對(duì)數(shù)得x0+lnx0=0,∴F(x)min=F(x0)=3,∴a≤3,即實(shí)數(shù)a的取值范圍是(-∞,3〗.13.(1)當(dāng)m=0時(shí),y=f(x)g(當(dāng)x>e時(shí),y'<0;當(dāng)0<x<e時(shí),y'>0.∴函數(shù)y=lnxx∴ymax=y

x(2)由題可知x1,x2是函數(shù)h(x)=lnx-x+m的零點(diǎn).h'(x)=1x-1=當(dāng)x>1時(shí),h'(x)<0;當(dāng)0<x<1時(shí),h'(x)>0,∴函數(shù)h(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,故函數(shù)h(x)要有兩個(gè)零點(diǎn),必有h(1)=-1+m>0,即m>1.要證ln(em+x1-x2)+m>0,只需證ln(em+x1-x2)>lne-m,即x2-x1<em-e-m,只需證e-m<x1<1<x2<em,①由于m>1,e-m∈(0,1),h(e-m)=-m-e-m+m<0,h(1)=-1+m>0,∴函數(shù)h(x)在(e-m,1)上存在唯一零點(diǎn)x1,即e-m<x1<1.②由(1)知,lnxx≤1e,所以lnx≤x∴h(em)=ln(em)-em+m<eme-em+m∴函數(shù)h(x)在(1,em)上存在唯一零點(diǎn)x2,即1<x2<em.③由②③可知①成立,故ln(em+x1-x2)+m>0.第三章導(dǎo)數(shù)及其應(yīng)用第二講導(dǎo)數(shù)的綜合應(yīng)用1.〖2021惠州市二調(diào)〗若函數(shù)f(x)=ex(x2-2x+1-a)-x恒有2個(gè)零點(diǎn),則a的取值范圍是()A.(-1e,+∞)C.(0,1e) D.(-∞,-12.〖2021陜西百校聯(lián)考〗已知銳角x1,x2滿(mǎn)足sinx1-cosx2<x1+x2-π2,則下列結(jié)論一定正確的是()A.sinx1<sin(x1+x2)B.tanx1>tanxC.sinx1+cosx1>sinx2+cosx2D.sinx1+sinx2>cosx1+cosx23.〖2021大同市調(diào)研測(cè)試〗已知函數(shù)f(x)=ax3-3x2+1,若f(x)存在唯一的零點(diǎn)x0,且x0<0,則a的取值范圍是()A.(2,+∞) B.(-∞,-2)C.(1,+∞) D.(-∞,-1)4.〖2020廣東七校第二次聯(lián)考〗設(shè)定義在R上的函數(shù)y=f(x)滿(mǎn)足?x∈R,f(x+2)=1f(x),且x∈(0,4〗時(shí),f'(x)>f(x)x,則6A.6f(2017)<3f(2018)<2f(2019) B.3f(2018)<6f(2017)<2f(2019) C.2f(2019)<3f(2018)<6f(2017) D.2f(2019)<6f(2017)<3f(2018)5.〖2020鄭州市三模〗設(shè)函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),當(dāng)x>0時(shí),f'(x)lnx<-1xf(x).則使得(x2-4)f(x)>0成立的x的取值范圍是()A.(-2,0)∪(0,2) B.(-∞,-2)∪(2,+∞)C.(-2,0)∪(2,+∞) D.(-∞,-2)∪(0,2)6.已知函數(shù)f(x)=(a-12)x2+lnx,若函數(shù)f(x)在區(qū)間(1,+∞)上的圖象恒在直線(xiàn)y=2ax的下方,則實(shí)數(shù)a的取值范圍是7.〖2021晉南高中聯(lián)考〗已知函數(shù)f(x)=ex-ax,g(x)=1+xlnx.(1)討論函數(shù)f(x)的單調(diào)性;(2)若當(dāng)x>0時(shí),方程f(x)=g(x)有實(shí)數(shù)解,求實(shí)數(shù)a的取值范圍.8.〖2020貴陽(yáng)市高三模擬〗〖交匯題〗已知f(x)=ex,g(x)=x+1.(e為自然對(duì)數(shù)的底數(shù))(1)求證:f(x)≥g(x)恒成立.(2)設(shè)m是正整數(shù),對(duì)任意的正整數(shù)n,(1+13)(1+132)·…·(1+13n)<9.〖2021江西紅色七校第一次聯(lián)考〗若存在兩個(gè)正實(shí)數(shù)x,y使得等式x(2+lnx)=xlny-ay成立,則實(shí)數(shù)a的取值范圍是()A.(0,1e2) B.(-∞,C.(0,1e3) D.(-∞,10.〖2021洛陽(yáng)市統(tǒng)考〗已知函數(shù)f(x)=xex-2-tx-t有2個(gè)零點(diǎn)a,b,且在區(qū)間(a,b)上有且僅有2個(gè)正整數(shù),則實(shí)數(shù)t的取值范圍是A.〖23,e2) B.(23C.〖34e,23) D.(34e11.〖2021江西紅色七校聯(lián)考〗已知函數(shù)f(x)=ax2+bx-lnx.(1)當(dāng)a=-2時(shí),函數(shù)f(x)在(0,+∞)上是減函數(shù),求b的取值范圍;(2)若方程f(x)=0的兩個(gè)根分別為x1,x2(x1<x2),求證:f'(x1+12.〖2021濟(jì)南名校聯(lián)考〗已知f(x)=lnx+ax,g(x)=ex(1)若函數(shù)f(x)的圖象在x=e處的切線(xiàn)與直線(xiàn)2x-y+8=0垂直,求f(x)的極值;(2)當(dāng)x>0時(shí),g(x)≥f(x)恒成立,求實(shí)數(shù)a的取值范圍.13.已知函數(shù)f(x)=lnx,g(x)=x-m.(1)當(dāng)m=0時(shí),求函數(shù)y=f((2)設(shè)h(x)=f(x)-g(x),若x1<x2且h(x1)=h(x2)=0,求證:ln(em+x1-x2)+m>0.答案第三章導(dǎo)數(shù)及其應(yīng)用 第三講導(dǎo)數(shù)的綜合應(yīng)用1.A由f(x)=0,得x2-2x+1-a=xex.令g(x)=xex,則函數(shù)f(x)=ex(x2-2x+1-a)-x恒有2個(gè)零點(diǎn)等價(jià)于函數(shù)y=x2-2x+1-a與y=g(x)的圖象有2個(gè)交點(diǎn),g'(x)=1-xex,令g'(x)>0,得x<1,令g'(x)<0,得x>1,所以g(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以g(x)max=g(1)=1e.作出函數(shù)y=x2-2x+1-a=(x-1)2-a與y=g(x)的圖象,如圖D3-3-2所示,數(shù)形結(jié)合可得-圖D3-3-22.D解法一因?yàn)閟inx1-cosx2<x1+x2-π2,即sinx1-x1<sin(π2-x2)-(π2-x2),所以可構(gòu)造函數(shù)f(x)=sinx-x,x∈(0,π2),因?yàn)閒'(x)=cosx-1<0,所以f(x)在(0,π2)上單調(diào)遞減,因?yàn)閤1∈(0,π2),π2-x2∈(0,π2),所以0<π2-x2<x1<π2,所以sin(π2-x2)<sinx1,即cosx2<sinx1,同理cosx1解法二取x1=x2=π3,則sinx1-cosx2=3-12<π6=π3+π3-π2,此時(shí)sinx1=sin(x1+x2),tanx13.A由題意易知x=0不是函數(shù)f(x)的零點(diǎn),則f(x)=ax3-3x2+1=0?a=3x2-1x3(x≠0),令y=3x2-1x3-a(x≠0),因此f(x)的零點(diǎn)與y=3x2-1x3-a(x≠0)的零點(diǎn)相同.設(shè)g(x)=3x2-1x3(x≠0),則g'(x)=-3(x-1)(x+1)x4,則當(dāng)x∈(-∞,-1)∪(1,+∞)時(shí),g'(x)<0;當(dāng)x∈(-1,0)∪(0,1)時(shí),g'(x)>0,故g(x)在(-1,0),(0,1)上單調(diào)遞增,在(-∞,-1),(1,+∞)上單調(diào)遞減,又g(-1)=-2,g(1)=2,當(dāng)g(x)=0時(shí),x=±33,當(dāng)x→+∞時(shí),圖D3-3-34.A因?yàn)閒(x+2)=1f(x),所以f(x)是以4為周期的周期函數(shù),故6f(2017)=6f(1),3f(2018)=3f(2),2f(2019)=2f(3).令g(x)=f(x)x(x∈(0,4〗),則g'(x)=xf'(x)-f(x)x2,因?yàn)閤∈(0,4〗時(shí),f'(x)>f(x)x,所以g'(x)>0,所以5.D設(shè)函數(shù)g(x)=f(x)lnx,則g'(x)=f'(x)lnx+1xf(x).于是,當(dāng)x>0時(shí),由f'(x)lnx<-1xf(x)可得g'(x)<0,所以函數(shù)g(x)在(0,+∞)上單調(diào)遞減.從而,當(dāng)x>1時(shí),有g(shù)(x)<g(1)=0,即f(x)lnx<0,又lnx>0,所以此時(shí)f(x)<0;當(dāng)0<x<1時(shí),有g(shù)(x)>g(1)=0,即f(x)lnx>0,又lnx<0,所以此時(shí)f(x)<0.對(duì)于不等式f'(x)lnx<-1xf(x),取x=1可得f'(1)ln1<-11f(1),化簡(jiǎn)得f(1)<0,即當(dāng)x=1時(shí),f(x)<0.于是,由上述討論可知,當(dāng)x>0時(shí),f(x)<0,故由(x2-4)f(x)>0(x>0)得x2-4<0,解得0<x<2.當(dāng)x<0時(shí),由f(x)為奇函數(shù)及“當(dāng)x>0時(shí),f(x)<0”可得f(x)>0,故由(x2-4)f(x)>0(x<0)得x2-4>0,解得x<-2.易知f(0)=0,所以x=0不滿(mǎn)足(x2-4)f(x)>0〖解后反思〗一般地,若題設(shè)中出現(xiàn)了與導(dǎo)數(shù)有關(guān)的不等式,則很可能是根據(jù)導(dǎo)數(shù)的運(yùn)算法則提前計(jì)算后而精心設(shè)計(jì)的,所以應(yīng)多從這個(gè)角度考慮如何構(gòu)造函數(shù),以便順利解決問(wèn)題.6.〖-12,12〗由題意知,對(duì)于任意x∈(1,+∞),f(x)<2ax,即(a-12)x2+lnx-2ax<0在(1,+∞)上恒成立.設(shè)g(x)=(a-12)x2+lnx-2ax,x∈(1,+∞),則g(x)的最大值小于0,g'(x)=(①當(dāng)a≤12時(shí),g'(x)<0,∴g(x)在(1,+∞)上單調(diào)遞減,∴g(x)<-a-12≤0,即a≥-12,∴-②當(dāng)a≥1時(shí),g'(x)>0,∴g(x)在(1,+∞)上單調(diào)遞增,最大值可無(wú)窮大,不滿(mǎn)足題意.③當(dāng)12<a<1時(shí),易知g(x)在(1,12a-綜上,實(shí)數(shù)a的取值范圍是〖-12,17.(1)函數(shù)f(x)的定義域?yàn)镽,f'(x)=ex-a,當(dāng)a≤0時(shí),f'(x)>0,則f(x)在(-∞,+∞)上單調(diào)遞增;當(dāng)a>0時(shí),令f'(x)=ex-a=0,得x=lna,則f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.綜上,當(dāng)a≤0時(shí),f(x)在R上單調(diào)遞增,當(dāng)a>0時(shí),f(x)在(-∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增.(2)由f(x)=g(x),得ax=ex-xlnx-1,因?yàn)閤>0,所以a=exx-ln令h(x)=exx-lnx-1x,x>0,則h'(x)=xex-ex-x+1x2=(ex-1)(x-1)x2.令h'所以h(x)min=h(1)=e-1.又h(x)=exx-lnx-1x=ex-所以當(dāng)x→0時(shí),h(x)→+∞.所以函數(shù)h(x)的值域?yàn)椤糴-1,+∞),因此實(shí)數(shù)a的取值范圍為〖e-1,+∞).8.(1)令h(x)=f(x)-g(x)=ex-x-1,則h'(x)=ex-1,當(dāng)x∈(-∞,0)時(shí),h'(x)<0,當(dāng)x∈(0,+∞)時(shí),h'(x)>0,故h(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,所以h(x)min=h(0)=0,即h(x)≥0恒成立,所以f(x)≥g(x)恒成立.(2)由(1)可知x=13n時(shí),0<1+(1+13)(1+132)·…·(1+13n)<e易知比e大的最小正整數(shù)是2,所以m的最小值為2.9.D因?yàn)閤,y均為正數(shù),所以等式x(2+lnx)=xlny-ay可化為2+lnx=lny-ayx,即lnyx-2=ayx,即a=lnyx-2yx.令yx則f'(t)=3-lntt2,令f'(t)=0,解得t=e3,當(dāng)t∈(0,e3)時(shí),f'(t)>0,f(t)在(0,e3)上單調(diào)遞增,當(dāng)t∈(e3,+∞)時(shí),f'(t)<0,f(t)在(e3,+∞)上單調(diào)遞減,所以f(t)max=f(e3)=1e3,且當(dāng)t→0時(shí),f(10.C由題意知函數(shù)f(x)=xex-2-tx-t有2個(gè)互異的零點(diǎn)a,b等價(jià)于函數(shù)g(x)=xex-2與h(x)=tx+t的圖象有2個(gè)不同的交點(diǎn).因?yàn)間(x)=xex-2,所以g'(x)=1-xex-2.令g'(x)>0,可得x<1;令g'(x)<0,可得x>1.所以函數(shù)g(x)在(-∞,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,所以g(x)max=g(1)=e.當(dāng)x<0時(shí),g(x)<0,當(dāng)x>0時(shí),g(x)>0,且g(0)=0,x?+∞時(shí),g(x)?0.由h(x)=tx+t=t(x+1),知函數(shù)h(x)的圖象為過(guò)定點(diǎn)(-1,0)的一條直線(xiàn),在同一平面直角坐標(biāo)系中,分別作出函數(shù)g(x)與h(x)的圖象如圖D3-3-4所示,若滿(mǎn)足h(x),圖D3-3-411.(1)當(dāng)a=-2時(shí),f(x)=-2x2+bx-lnx,∵f(x)在(0,+∞)上單調(diào)遞減,∴f'(x)=-4x+b-1x≤0對(duì)x∈(0,+∞)恒成立,即b≤4x+1x∴b≤(4x+1x)min∵x>0,∴4x+1x≥4,當(dāng)且僅當(dāng)x