新高考數(shù)學(xué)一輪復(fù)習(xí)專題三導(dǎo)數(shù)及其應(yīng)用3-3導(dǎo)數(shù)的綜合應(yīng)用練習(xí)含答案

3.3導(dǎo)數(shù)的綜合應(yīng)用五年高考高考新風(fēng)向(創(chuàng)新考法)(2024新課標(biāo)Ⅰ,18,17分,難)已知函數(shù)f(x)=lnx2?x+ax+b(x-1)(1)若b=0,且f'(x)≥0,求a的最小值;(2)證明:曲線y=f(x)是中心對稱圖形;(3)若f(x)>-2當(dāng)且僅當(dāng)1<x<2,求b的取值范圍.解析(1)f(x)的定義域?yàn)?0,2).當(dāng)b=0時,f(x)=lnx2?x+所以f'(x)=2?xxx2?x'+a因?yàn)閒'(x)≥0,所以a≥2x(x依題意得a≥2x所以當(dāng)x=1時,a取到最小值-2.(2)證明:因?yàn)閒(x)=lnx2?x+ax+b(x-1)所以f(2-x)=ln2?xx+a(2-x)+b(1-x)所以f(2-x)+f(x)=2a,所以曲線y=f(x)關(guān)于(1,a)中心對稱,即曲線y=f(x)是中心對稱圖形.(3)由(2)知曲線y=f(x)關(guān)于(1,a)中心對稱,且f(x)在(0,2)內(nèi)連續(xù),故f(1)=a=-2,即f(x)=lnx2?x-2x+b(x-1)3,x∈(1,f'(x)=2x(2?x)+3b易知y=2x(2?x)∈(所以當(dāng)b≥-23時,f'(x)>0在(1,2)上恒成立,f(x)單調(diào)遞增所以f(x)>f(1)=-2成立.當(dāng)b<-23時,存在x0∈(1,2),使得2x0(2?即3b=2x在(1,x0)上,f'(x)<0,f(x)單調(diào)遞減,又f(1)=-2,所以在(1,x0)上有f(x)<-2,不符合題意.綜上,可得b≥-23考點(diǎn)1利用導(dǎo)數(shù)研究不等式1.(2023新課標(biāo)Ⅰ,19,12分,中)已知函數(shù)f(x)=a(ex+a)-x.(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)a>0時,f(x)>2lna+32解析(1)由已知得函數(shù)f(x)的定義域?yàn)镽,f'(x)=aex-1.①當(dāng)a≤0時,f'(x)<0,f(x)在R上單調(diào)遞減;②當(dāng)a>0時,令f'(x)=0,則x=ln1a當(dāng)x<ln1a時,f'(x)<0,f(x)單調(diào)遞減當(dāng)x>ln1a時,f'(x)>0,f(x)單調(diào)遞增綜上所述,當(dāng)a≤0時,f(x)在R上單調(diào)遞減;當(dāng)a>0時,f(x)在?∞,ln1a上單調(diào)遞減,(2)證明:由(1)知,當(dāng)a>0時,f(x)在?∞,ln1a上單調(diào)遞減,在ln1a,+∞上單調(diào)遞增,則f(x)min=fln1a要證明f(x)>2lna+32,只需證明1+a2+lna>2lna+3即證a2-lna-12>0令g(x)=x2-lnx-12(x>0),則g'(x)=2x-1x=當(dāng)0<x<22時,g'(x)<0,g(x)單調(diào)遞減當(dāng)x>22時,g'(x)>0,g(x)單調(diào)遞增∴g(x)min=g22=12-ln22-12=-ln2∴g(x)>0在(0,+∞)上恒成立,即a2-lna-12>0∴f(x)>2lna+322.(2021新高考Ⅰ,22,12分,難)已知函數(shù)f(x)=x(1-lnx).(1)討論f(x)的單調(diào)性;(2)設(shè)a,b為兩個不相等的正數(shù),且blna-alnb=a-b,證明:2<1a+1b解析(1)函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=-lnx,令f'(x)>0,解得0<x<1,令f'(x)<0,解得x>1,所以函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.(2)證明:由blna-alnb=a-b(兩邊同除以ab)得1a(1+lna)=1b(1+lnb),即1a令x1=1a,x2=1b,則x1,x2為f(x)=k的兩個實(shí)根,當(dāng)x→0+時,f(x)→0+,當(dāng)x→+∞時,f(x)→-∞,且f(1)=1,故k∈(0,不妨令x1∈(0,1),x2∈(1,e),則2-x1>1,e-x1>1,先證明x1+x2>2,即證x2>2-x1,即證f(x2)=f(x1)<f(2-x1).令h(x)=f(x)-f(2-x),x∈(0,1),則h'(x)=f'(x)+f'(2-x)=-lnx-ln(2-x)=-ln[x(2-x)].∵x∈(0,1),∴x(2-x)∈(0,1),∴h'(x)>0恒成立,∴h(x)為增函數(shù),∴h(x)<h(1)=0.∴f(x1)-f(2-x1)<0,即f(x1)<f(2-x1),∴f(x2)<f(2-x1),∴x2>2-x1,∴x1+x2>2.再證x1+x2<e.設(shè)x2=tx1,則t>1,結(jié)合lna+1a=lnb+1b,1a=x可得x1(1-lnx1)=x2(1-lnx2),即1-lnx1=t(1-lnt-lnx1),故lnx1=t?1?要證x1+x2<e,即證(t+1)x1<e,即證ln(t+1)+lnx1<1,即證ln(t+1)+t?1?tlntt?1<1,即證(t-1)ln(t+1)令S(t)=(t-1)ln(t+1)-tlnt,t>1,則S'(t)=ln(t+1)+t?1t+1-1-lnt=ln1+因?yàn)閘n(x+1)≤x(x>-1,當(dāng)且僅當(dāng)x=0時等號成立),所以可得當(dāng)t>1時,ln1+1t≤1t<2t+1,故S'(故S(t)在(1,+∞)上為減函數(shù),故S(t)<S(1)=0,故(t-1)ln(t+1)-tlnt<0成立,即x1+x2<e成立.綜上所述,2<1a+1b3.(2020課標(biāo)Ⅱ理,21,12分,難)已知函數(shù)f(x)=sin2xsin2x.(1)討論f(x)在區(qū)間(0,π)的單調(diào)性;(2)證明:|f(x)|≤33(3)設(shè)n∈N*,證明:sin2xsin22xsin24x·…·sin22nx≤3n解析(1)f'(x)=cosx(sinxsin2x)+sinx(sinx·sin2x)'=2sinxcosxsin2x+2sin2xcos2x=2sinxsin3x.當(dāng)x∈0,π3∪2π3,π時,f'(x)>0;當(dāng)x∈π3,2π3時,f'(x)<0.所以f(x)在區(qū)間0,(2)證明:因?yàn)閒(0)=f(π)=0,由(1)知,f(x)在區(qū)間[0,π]的最大值為fπ3=338,最小值為f2π3=-338.而f(x)是周期為π的周期函數(shù),故|f(3)證明:由于(sin2xsin22x…sin22nx)=|sin3xsin32x…sin32nx|=|sinx||sin2xsin32x…sin32n-1xsin2nx||sin22nx|=|sinx||f(x)f(2x)…f(2n-1x)||sin22nx|≤|f(x)f(2x)…f(2n-1x)|,所以sin2xsin22x…sin22nx≤33824.(2022新高考Ⅱ,22,12分,難)已知函數(shù)f(x)=xeax-ex.(1)當(dāng)a=1時,討論f(x)的單調(diào)性;(2)當(dāng)x>0時,f(x)<-1,求a的取值范圍;(3)設(shè)n∈N*,證明:112+1+122+2+…+1解析(1)當(dāng)a=1時,f(x)=xex-ex,則f'(x)=xex,當(dāng)x∈(-∞,0)時,f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(0,+∞)時,f'(x)>0,f(x)單調(diào)遞增.(2)解法一:不等式f(x)<-1等價于xeax-ex+1<0,令g(x)=xeax-ex+1,x>0,則g'(x)=(ax+1)eax-ex,且g'(x)=eax(ax+1-ex-ax)≤eax[ax+1-(x-ax+1)]=(2a-1)xeax,①當(dāng)a≤12時,g'(x)≤(2a-1)xeax≤0,g(x)在(0,+∞)上單調(diào)遞減,所以g(x)<g(0)=0,滿足題意②當(dāng)a>12時,令h(x)=g'(x)=(ax+1)eax-ex則h'(x)=(a2x+2a)eax-ex,故h'(0)=2a-1>0,所以存在δ>0,使得當(dāng)x∈(0,δ)時,h'(x)>0,所以h(x)在(0,δ)上單調(diào)遞增,故當(dāng)x∈(0,δ)時,h(x)>h(0)=0,所以g(x)在(0,δ)上單調(diào)遞增,故g(x)>g(0)=0,不滿足題意.綜上可知,a的取值范圍是?∞解法二:函數(shù)f(x)的定義域?yàn)镽,f'(x)=(1+ax)eax-ex.(導(dǎo)函數(shù)中含有參數(shù),要根據(jù)參數(shù)對導(dǎo)函數(shù)取值符號的影響分段討論)對于x∈(0,+∞),當(dāng)a≥1時,f'(x)=(1+ax)eax-ex>eax-ex≥ex-ex=0,∴f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增.∵f(0)=-1,∴f(x)>-1,不滿足題意.當(dāng)a≤0時,f'(x)≤eax-ex<1-ex<0且等號不恒成立,∴f(x)在(0,+∞)上單調(diào)遞減.∵f(0)=-1,∴f(x)<-1,滿足題意.當(dāng)0<a≤12時,f'(x)≤1+x2ex(當(dāng)導(dǎo)函數(shù)的正負(fù)不能直接判斷時,可考慮構(gòu)造新函數(shù),通過研究新函數(shù)的單調(diào)性判斷)令g(x)=1+x2-ex2,則g'(x)=12∴g(x)在(0,+∞)上單調(diào)遞減,∴g(x)<g(0)=0,∴f'(x)=ex2·g(x)<0,f(x)在(0,+∞)∵f(0)=-1,∴f(x)<-1,滿足題意.當(dāng)12<a<1時,f'(x)=eax[1+ax-e(1-a)x]令h(x)=1+ax-e(1-a)x,則h'(x)=a+(a-1)e(1-a)x.∵h(yuǎn)'(x)為減函數(shù),又h'(0)=2a-1>0,x→+∞,h'(x)<0,∴?x0∈(0,+∞),使得h'(x0)=0.(當(dāng)導(dǎo)函數(shù)的零點(diǎn)存在,但不易求出時,可引入虛擬零點(diǎn))∴當(dāng)x∈(0,x0)時,h'(x)>0,h(x)在(0,x0)上單調(diào)遞增,h(x)>h(0)=0,∴當(dāng)x∈(0,x0)時,f'(x)=eax·h(x)>0,f(x)在(0,x0)上單調(diào)遞增.∵f(0)=-1,∴f(x)>-1,不滿足題意.綜上,a的取值范圍是?∞(3)證法一:構(gòu)造函數(shù)h(x)=x-1x-2lnx(x>1則h'(x)=1+1x2-2x=x易知h'(x)>0,∴h(x)在(1,+∞)上單調(diào)遞增,∴h(x)>h(1)=0,∴x-1x>2lnx令x=1+1n,則有1+1n-∴1n2+∴112+1+122+2+…+1n2+n>ln21+ln證法二:由(2)可知,當(dāng)a=12時,f(x)=xe12x-ex<-1令x=ln1+1n(n∈N得ln1+1n·e12整理得1+1nln1+1n∴1n1+1n>ln1+1n∴k=1n1k2+k>k=1n即112+1+122+2+…+1考點(diǎn)2利用導(dǎo)數(shù)研究函數(shù)零點(diǎn)1.(2023全國乙文,8,5分,中)函數(shù)f(x)=x3+ax+2存在3個零點(diǎn),則a的取值范圍是(B)A.(-∞,-2)B.(-∞,-3)C.(-4,-1)D.(-3,0)2.(2022全國乙文,20,12分,中)已知函數(shù)f(x)=ax-1x-(a+1)ln(1)當(dāng)a=0時,求f(x)的最大值;(2)若f(x)恰有一個零點(diǎn),求a的取值范圍.解析(1)當(dāng)a=0時,f(x)=-1x-lnx(x>0∴f'(x)=1x2-1x(x>0),令f'(x)=0,得x∈(0,1)時,f'(x)>0,x∈(1,+∞)時,f'(x)<0,∴f(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.∴f(x)max=f(1)=-1.(2)f'(x)=a+1x2-a+1(i)當(dāng)a≤0時,ax-1≤0恒成立,∴0<x<1時,f'(x)>0,f(x)單調(diào)遞增,x>1時,f'(x)<0,f(x)單調(diào)遞減,∴f(x)max=f(1)=a-1<0.此時f(x)無零點(diǎn),不合題意.(ii)當(dāng)a>0時,令f'(x)=0,解得x=1或x=1a①當(dāng)0<a<1時,1<1a∴1<x<1a時,f'(x)<0,f(x)單調(diào)遞減0<x<1或x>1a時,f'(x)>0,f(x)單調(diào)遞增∴f(x)在(0,1),1a,+∞上單調(diào)遞增,在1,1a上單調(diào)遞減,f(x)的極大值為f(1x→+∞時,f(x)>0,∴f(x)恰有1個零點(diǎn).②當(dāng)a=1時,1=1a,f(x)在(0,+∞)上單調(diào)遞增,f(1)=0,符合題意③當(dāng)a>1時,1a<1,f(x)在0,1a,(1,+∞)上單調(diào)遞增,在f(x)的極小值為f(1)=a-1>0,x→0時,f(x)→-∞,∴f(x)恰有1個零點(diǎn).綜上所述,a>0.3.(2020課標(biāo)Ⅲ理,21,12分,難)設(shè)函數(shù)f(x)=x3+bx+c,曲線y=f(x)在點(diǎn)12,f1(1)求b;(2)若f(x)有一個絕對值不大于1的零點(diǎn),證明:f(x)所有零點(diǎn)的絕對值都不大于1.解析(1)f'(x)=3x2+b.依題意得f'12=0,即34+b=0.故b=-(2)證法一:由(1)知f(x)=x3-34x+c,f'(x)=3x2-3令f'(x)=0,解得x=-12或x=1f'(x)與f(x)隨x的變化情況為:x?-1?11f'(x)+0-0+f(x)↗c+1↘c-1↗因?yàn)閒(1)=f?12=c+14,所以當(dāng)c<-14時,f(x)因?yàn)閒(-1)=f12=c-14,所以當(dāng)c>14時,f(x)只有小于由題設(shè)可知-14≤c≤1當(dāng)c=-14時,f(x)只有兩個零點(diǎn)-12和當(dāng)c=14時,f(x)只有兩個零點(diǎn)-1和1當(dāng)-14<c<14時,f(x)有三個零點(diǎn)x1,x2,x3,且x1∈?1,?12,x2∈?12綜上,若f(x)有一個絕對值不大于1的零點(diǎn),則f(x)所有零點(diǎn)的絕對值都不大于1.證法二(分離參數(shù)法):設(shè)x0為f(x)的一個絕對值不大于1的零點(diǎn),根據(jù)題意,f(x0)=x03-34x0+c=0,且|x則c=-x03+34x0,且|x0|≤1,令c(x)=-x3+34x(則c'(x)=-3x2+34=-3x當(dāng)x∈?1,?12∪12,1時,c'(x)<0,c(當(dāng)x∈?12,12時,c'(x)>0,c又c(-1)=14,c(1)=-14,c?12=-14,所以-14≤c≤1設(shè)x1為f(x)的零點(diǎn),則必有f(x1)=x13-34x1+即-14≤c=-x13+34x即4x13?3x1|x1|≤1.所以f(x)所有零點(diǎn)的絕對值都不大于1.4.(2021新高考Ⅱ,22,12分,難)已知函數(shù)f(x)=(x-1)ex-ax2+b.(1)討論函數(shù)f(x)的單調(diào)性;(2)從下面兩個條件中選一個,證明:f(x)有一個零點(diǎn).①12<a≤e22,b②0<a<12,b≤2解析(1)∵f(x)=(x-1)ex-ax2+b,∴f'(x)=xex-2ax=x(ex-2a).①當(dāng)a≤0時,ex-2a>0對任意x∈R恒成立,當(dāng)x∈(-∞,0)時,f'(x)<0,當(dāng)x∈(0,+∞)時,f'(x)>0.因此y=f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.②當(dāng)a>0時,令ex-2a=0?x=ln(2a).(i)當(dāng)0<a<12時,ln(2a)<0y=f'(x)的大致圖象如圖1所示.圖1圖2因此當(dāng)x∈(-∞,ln(2a))∪(0,+∞)時,f'(x)>0,當(dāng)x∈(ln(2a),0)時,f'(x)<0,所以f(x)在(-∞,ln(2a))和(0,+∞)上單調(diào)遞增,在(ln(2a),0)上單調(diào)遞減.(ii)當(dāng)a=12時,ln(2a)=0,此時f'(x)≥0對任意x∈R恒成立,故f(x)在R上單調(diào)遞增(iii)當(dāng)a>12時,ln(2a)>0y=f'(x)的大致圖象如圖2所示.因此,當(dāng)x∈(-∞,0)∪(ln(2a),+∞)時,f'(x)>0,當(dāng)x∈(0,ln(2a))時,f'(x)<0,所以f(x)在(-∞,0)和(ln(2a),+∞)上單調(diào)遞增,在(0,ln(2a))上單調(diào)遞減.(2)選①.證明:由(1)知,f(x)在(-∞,0)上單調(diào)遞增,在(0,ln(2a))上單調(diào)遞減,在(ln(2a),+∞)上單調(diào)遞增,又f(0)=b-1>0,f?ba=?所以f(x)在(-∞,0]上有唯一零點(diǎn).當(dāng)x∈(0,+∞)時,f(x)≥f(ln(2a))=[ln(2a)-1]·2a-aln2(2a)+b=aln(2a)[2-ln(2a)]+b-2a>aln(2a)[2-ln(2a)].因?yàn)?2<a≤e22,所以0<ln(2a所以f(x)>0對任意x>0恒成立.綜上,f(x)在R上有一個零點(diǎn).選②.證明:由(1)知f(x)在(-∞,ln(2a))上單調(diào)遞增,在(ln(2a),0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增,f(0)=b-1<0,當(dāng)x→+∞時,f(x)→+∞,所以一定存在x0∈(0,+∞),使得f(x0)=0.結(jié)合單調(diào)性知f(x)在[0,+∞)上有唯一零點(diǎn).當(dāng)x∈(-∞,0)時,f(x)≤f(ln(2a))=aln(2a)·[2-ln(2a)]+b-2a<0,即f(x)<0對任意x<0恒成立.綜上,f(x)在R上有一個零點(diǎn).5.(2022新高考Ⅰ,22,12分,難)已知函數(shù)f(x)=ex-ax和g(x)=ax-lnx有相同的最小值.(1)求a;(2)證明:存在直線y=b,其與兩條曲線y=f(x)和y=g(x)共有三個不同的交點(diǎn),并且從左到右的三個交點(diǎn)的橫坐標(biāo)成等差數(shù)列.解析(1)f'(x)=ex-a,g'(x)=a-1x當(dāng)a≤0時,f'(x)>0恒成立,f(x)在R上無最小值,不符合題意.∴a>0.令f'(x)=0,得x=lna,令g'(x)=0,得x=1a易知f(x)min=f(lna)=a-alna,g(x)min=g1a=1+lna∴a-alna=1+lna,即lna=a?1a令h(x)=lnx-x?1x+1(則h'(x)=1x-2(x+1)∴h(x)在(0,+∞)上單調(diào)遞增,則h(x)最多有一個零點(diǎn).又h(1)=ln1-1?11+1=0∴方程①有且僅有一解,為a=1,即為所求.(2)證明:由(1)知,f(x)=ex-x,g(x)=x-lnx,當(dāng)x<0時,f(x)單調(diào)遞減,當(dāng)x>0時,f(x)單調(diào)遞增;當(dāng)0<x<1時,g(x)單調(diào)遞減,當(dāng)x>1時,g(x)單調(diào)遞增.不妨設(shè)直線y=b與y=f(x)的圖象的兩交點(diǎn)的橫坐標(biāo)分別為x1,x2,與y=g(x)的圖象的兩交點(diǎn)的橫坐標(biāo)分別為x2,x3,且x1<x2<x3.則ex1-x1=ex2-x2=x2-lnx2=x3∴ex1-x1=x2-lnx2=elnx易知x1∈(-∞,0),x2∈(0,1),則lnx2∈(-∞,0),又f(x)在(-∞,0)上單調(diào)遞減,∴x1=lnx2,同理x2=lnx3,x3=ex又ex2-x2=x2-lnx2,∴l(xiāng)nx2+ex2∴l(xiāng)nx2+ex2=x1+x3=2x2.∴x1,x2,x3∴存在直線y=b,其與兩條曲線y=f(x)和y=g(x)共有三個不同的交點(diǎn),并且從左到右的三個交點(diǎn)的橫坐標(biāo)成等差數(shù)列.6.(2023新課標(biāo)Ⅱ,22,12分,難)(1)證明:當(dāng)0<x<1時,x-x2<sinx<x;(2)已知函數(shù)f(x)=cosax-ln(1-x2),若x=0是f(x)的極大值點(diǎn),求a的取值范圍.解析(1)證明:令g(x)=x-x2-sinx,0<x<1,則g'(x)=1-2x-cosx,令G(x)=g'(x),得G'(x)=-2+sinx<0在區(qū)間(0,1)上恒成立,所以g'(x)在區(qū)間(0,1)上單調(diào)遞減,因?yàn)間'(0)=0,所以g'(x)<0在區(qū)間(0,1)上恒成立,所以g(x)在區(qū)間(0,1)上單調(diào)遞減,所以g(x)<g(0)=0,即當(dāng)0<x<1時,x-x2<sinx.令h(x)=sinx-x,0<x<1,則h'(x)=cosx-1<0在區(qū)間(0,1)上恒成立,所以h(x)在區(qū)間(0,1)上單調(diào)遞減,所以h(x)<h(0)=0,即當(dāng)0<x<1時,sinx<x.綜上,當(dāng)0<x<1時,x-x2<sinx<x.(2)函數(shù)f(x)的定義域?yàn)?-1,1).當(dāng)a=0時,f(x)=1-ln(1-x2),f(x)在(-1,0)上單調(diào)遞減,在(0,1)上單調(diào)遞增,x=0不是f(x)的極大值點(diǎn),所以a≠0.當(dāng)a>0時,f'(x)=-asinax+2x1?x2,x∈(-1(i)當(dāng)0<a≤2時,取m=min1a,1,x∈(0,則ax∈(0,1),由(1)可得f'(x)=-asinax+2x1?x2>-a2x+因?yàn)閍2x2>0,2-a2≥0,1-x2>0,所以f'(x)>0,所以f(x)在(0,m)上單調(diào)遞增,不合題意.(ii)當(dāng)a>2時,取x∈0,1a?(0,1),則ax∈(0,由(1)可得f'(x)=-asinax+2x1?x2<-a(ax-a2x=x1?x2(-a3x3+a2x2+a3x+2-設(shè)h(x)=-a3x3+a2x2+a3x+2-a2,x∈0,1則h'(x)=-3a3x2+2a2x+a3,因?yàn)閔'(0)=a3>0,h'1a=a3-a>0,且h'(x)的圖象是開口向下的拋物線,所以?x∈0,1a,均有h'(x)>0,所以h(x)在因?yàn)閔(0)=2-a2<0,h1a=2>0,所以h(x)在0,1當(dāng)x∈(0,n)時,h(x)<0,又因?yàn)閤>0,1-x2>0.則f'(x)<x1?x2(-a3x3+a2x2+a3x+2-a2即當(dāng)x∈(0,n)?(0,1)時,f'(x)<0,則f(x)在(0,n)上單調(diào)遞減.又因?yàn)閒(x)是偶函數(shù),所以f(x)在(-n,0)上單調(diào)遞增,所以x=0是f(x)的極大值點(diǎn).綜合(i)(ii)知a>2.當(dāng)a<0時,由于將f(x)中的a換為-a所得解析式不變,所以a<-2符合要求.故a的取值范圍為(-∞,-2)∪(2,+∞).

三年模擬練思維1.(2024江蘇南通第二次適應(yīng)性調(diào)研,16)已知函數(shù)f(x)=lnx-ax,g(x)=2ax,a≠0(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)若a>0且f(x)≤g(x)恒成立,求a的最小值.解析(1)f'(x)=1x-a=1?axx(a≠0),當(dāng)a<0時,f'(x)>0恒成立,從而f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>0時,由f'(x)>0得0<x<1a;由f'(x)<0得x>1從而f(x)在0,1a上單調(diào)遞增,在1綜上,當(dāng)a<0時,f(x)的單調(diào)遞增區(qū)間為(0,+∞),沒有單調(diào)遞減區(qū)間;當(dāng)a>0時,f(x)的單調(diào)遞增區(qū)間為0,1a,單調(diào)遞減區(qū)間為(2)令h(x)=f(x)-g(x)=lnx-ax-2ax,要使f(x)≤g(x)恒成立,只要使h(x)≤0恒成立,只要使h(x)max≤0h'(x)=1x-a+2ax由于a>0,x>0,所以ax+1>0恒成立,當(dāng)0<x<2a時,h'(x)>0,當(dāng)x>2a時,h'(x)所以h(x)max=h2a=ln2a-3≤0,解得a≥所以a的最小值為2e2.(2024東北三省四市一模,16)已知函數(shù)f(x)=xex-aex,a(1)當(dāng)a=0時,求曲線f(x)在x=1處的切線方程;(2)當(dāng)a=1時,求f(x)的單調(diào)區(qū)間和極值;(3)若對任意x∈R,有f(x)≤ex-1恒成立,求a的取值范圍.解析(1)當(dāng)a=0時,f(x)=xex,則f'(x)=1?xex,f'(1)=0,f(所以切線方程為y=1e.(3分)(2)當(dāng)a=1時,f(x)=xe-x-ex,f'(x)=(1-x)e-x-ex=1?x?e2x令g(x)=1-x-e2x,g'(x)=-1-2e2x<0,故g(x)在R上單調(diào)遞減,而g(0)=0,因此0是g(x)在R上的唯一零點(diǎn),即0是f'(x)在R上的唯一零點(diǎn).(6分)當(dāng)x變化時,f'(x),f(x)的變化情況如表:x(-∞,0)0(0,+∞)f'(x)+0-f(x)↗極大值↘f(x)的單調(diào)遞增區(qū)間為(0,+∞);單調(diào)遞減區(qū)間為(-∞,0).(8分)f(x)的極大值為f(0)=-1,無極小值.(9分)(3)由題意知xe-x-aex≤ex-1,即a≥xe?x?ex?1e設(shè)m(x)=xe2x-1e,則m'(x)=e2x?2x令m(x)=0,解得x=12當(dāng)x∈?∞,12時,m'(x)>0,m(x)單調(diào)遞增,當(dāng)x∈12,+∞時,m'(x)<0所以m(x)max=m12=12e-1e=-12e,(所以a≥-12e.(15分)3.(2024廣東廣州一模,17)已知函數(shù)f(x)=cosx+xsinx,x∈(-π,π).(1)求f(x)的單調(diào)區(qū)間和極小值;(2)證明:當(dāng)x∈[0,π)時,2f(x)≤ex+e-x.解析(1)f'(x)=-sinx+sinx+xcosx=xcosx,由f'(x)=0得x=0或π2或-π當(dāng)-π<x<-π2時,f'(x)>0,f(x)單調(diào)遞增當(dāng)-π2<x<0時,f'(x)<0,f(x)單調(diào)遞減當(dāng)0<x<π2時,f'(x)>0,f(x)單調(diào)遞增當(dāng)π2<x<π時,f'(x)<0,f(x)單調(diào)遞減∴f(x)的單調(diào)遞增區(qū)間為?π,?π2,0,π2;單調(diào)遞減區(qū)間為?π2,0,π2,π,f(x(2)證明:令F(x)=ex+e-x-2(cosx+xsinx),x∈[0,π),則F(0)=0,F'(x)=ex-e-x-2xcosx.因?yàn)閤∈[0,π),所以cosx≤1,所以2xcosx≤2x,從而F'(x)=ex-e-x-2xcosx≥ex-e-x-2x,當(dāng)且僅當(dāng)x=0時取等號.令h(x)=ex-e-x-2x,x∈[0,π),顯然h(0)=0.h'(x)=ex+e-x-2≥2ex·e?x-2=0,所以h(x)在[0,π)上單調(diào)遞增,所以h(x)≥h(0)=0,從而F'(x)≥h(x)≥0,所以F(x)在[0,π)上單調(diào)遞增,所以F(x)≥F(0)=0,故ex+e-x≥2f(x).4.(2024湖南九校聯(lián)盟聯(lián)考,17)已知函數(shù)f(x)=x3+ax2+bx+c(a,b,c∈R),其圖象的對稱中心為(1,-2).(1)求a-b-c的值;(2)判斷函數(shù)f(x)的零點(diǎn)個數(shù).解析(1)∵函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,-2)中心對稱,∴y=f(x+1)+2為奇函數(shù).從而有f(x+1)+2+f(-x+1)+2=0,即f(x+1)+f(-x+1)=-4,又f(x+1)=(x+1)3+a(x+1)2+b(x+1)+c=x3+(a+3)x2+(2a+b+3)x+a+b+c+1,f(1-x)=(1-x)3+a(1-x)2+b(1-x)+c=-x3+(a+3)x2-(2a+b+3)x+a+b+c+1,∴2a+6=0,∴a-b-c=-3.(2)由(1)可知,f(x)=x3-3x2-cx+c,f'(x)=3x2-6x-c,令f'(x)=0,則Δ=36+12c,①當(dāng)Δ=36+12c≤0,即c≤-3時,f'(x)≥0,∴f(x)在R上單調(diào)遞增,∵f(1)=-2<0,f(3)=27-3×9-3c+c=-2c>0,∴函數(shù)f(x)有且僅有一個零點(diǎn);②當(dāng)Δ=36+12c>0,即c>-3時,設(shè)f'(x)=0的兩根為x1,x2,則x1+x2=2>0,x1x2=-c3(i)當(dāng)c=0時,f(x)=x3-3x2,令f(x)=x3-3x2=0,解得x=0或x=3,∴f(x)有兩個零點(diǎn);(ii)當(dāng)-3<c<0時,x1+x2=2>0,x1x2=-c3>0∴f'(x)=0有兩個正根,不妨令0<x1<x2,則3x12-6x1-c∴函數(shù)f(x)在(-∞,x1)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增,∵f(x1)=x13-3x12-(x1-1)(3x12-6x1)=-2x1(x12-3x1+3)<0,∴函數(shù)f(x)有且僅有一個零點(diǎn);(iii)當(dāng)c>0時,x1+x2=2,x1x2=-c3<0∴f'(x)=0有一個正根和一個負(fù)根,不妨令x1<0<x2,∴函數(shù)f(x)在(-∞,x1)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增,∵f(x1)>f(0)=c>0,f(x2)<f(1)=-2<0,∴函數(shù)f(x)有且僅有三個零點(diǎn).綜上,當(dāng)c>0時,函數(shù)f(x)有三個零點(diǎn);當(dāng)c=0時,函數(shù)f(x)有兩個零點(diǎn);當(dāng)c<0時,函數(shù)f(x)有一個零點(diǎn).5.(2024浙江杭州二模,16)已知函數(shù)f(x)=aln(x+2)-12x2(a∈R)(1)討論函數(shù)f(x)的單調(diào)性.(2)若函數(shù)f(x)有兩個極值點(diǎn),(i)求實(shí)數(shù)a的取值范圍;(ii)證明:函數(shù)f(x)有且只有一個零點(diǎn).解析(1)函數(shù)f(x)的定義域?yàn)?-2,+∞),f'(x)=ax+2-x=當(dāng)a≤-1時,f(x)在(-2,+∞)上單調(diào)遞減;當(dāng)-1<a<0時,令f'(x)=0,即-(x+1)2+a+1=0,解得x1=-a+1-1,x2=a+1-1.因?yàn)?1<a<0,所以0<a+1<1,則-2<-a所以當(dāng)x∈(-2,-a+1-1)時,f'(x)<0當(dāng)x∈(-a+1-1,a+1-1)時,f'(x)當(dāng)x∈(a+1-1,+∞)時,f'(x)<0所以f(x)在(-2,-a+1-1)上單調(diào)遞減,在(-a+1-1,a+1-1)上單調(diào)遞增,在(a+1-1,當(dāng)a≥0時,-a+1-1≤-2,所以x∈(-2,a+1-1)時,f'(x)>0,當(dāng)x∈(a+1-1,+∞)時,f'(x)<0,所以f(x)在(-2,a+1-1)上單調(diào)遞增,在(a+1-1(2)(i)由(1)知-1<a<0.(ii)證明:由(1)知f(x)在x=a+1-1處取得極大值,在x=-a+1-1因?yàn)?1<a<0,所以0<a+1<1,則1<a+1+1<2因?yàn)閒(x)極大值=f(a+1-1)=aln(a+1+1)-12(a+1-1)2<0,f(-a+1-1)<f(所以f(x)在(-a+1-1,+∞)上沒有零點(diǎn),又-1<a<0,則4a<-4,則0<e4a<e-4,-2<e4a-2<e-4-2,則0<(e4a-2)2<4,所以f(e4a-2)=4-12(e4a-2)2>0,所以f綜上,函數(shù)f(x)有且只有一個零點(diǎn).6.(2024山東青島二模,18)已知函數(shù)f(x)=ex-ax2-x,f'(x)為f(x)的導(dǎo)數(shù).(1)討論f'(x)的單調(diào)性;(2)若x=0是f(x)的極大值點(diǎn),求a的取值范圍;(3)若θ∈0,π2,證明:esinθ-1+ecosθ-1+ln(sinθcosθ)解析(1)由題知f'(x)=ex-2ax-1,令g(x)=f'(x)=ex-2ax-1,則g'(x)=ex-2a,當(dāng)a≤0時,g'(x)>0,f'(x)在區(qū)間(-∞,+∞)上單調(diào)遞增,當(dāng)a>0時,令g'(x)=0,解得x=ln(2a),當(dāng)x∈(-∞,ln(2a))時,g'(x)<0,當(dāng)x∈(ln(2a),+∞)時,g'(x)>0,所以f'(x)在區(qū)間(-∞,ln(2a))上單調(diào)遞減,在區(qū)間(ln(2a),+∞)上單調(diào)遞增.綜上,當(dāng)a≤0時,f'(x)在區(qū)間(-∞,+∞)上單調(diào)遞增;當(dāng)a>0時,f'(x)在區(qū)間(-∞,ln(2a))上單調(diào)遞減,在區(qū)間(ln(2a),+∞)上單調(diào)遞增.(2)由題知f'(0)=0,當(dāng)a≤0時,由(1)知,當(dāng)x∈(-∞,0)時,f'(x)<0,f(x)在(-∞,0)上單調(diào)遞減,當(dāng)x∈(0,+∞)時,f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增,所以x=0是函數(shù)f(x)的極小值點(diǎn),不符合題意;當(dāng)0<a<12時,ln(2a)<0由(1)知,當(dāng)x∈(ln(2a),0)時,f'(x)<0,f(x)在(ln(2a),0)上單調(diào)遞減,當(dāng)x∈(0,+∞)時,f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增,所以x=0是函數(shù)f(x)的極小值點(diǎn),不符合題意;當(dāng)a=12時,ln(2a)=0,則當(dāng)x∈(-∞,+∞)時,f'(x)≥0,f(x)在(-∞,+∞)上單調(diào)遞增,所以f(x)無極值點(diǎn),不符合題意當(dāng)a>12時,ln(2a)>0當(dāng)x∈(-∞,0)時,f'(x)>0,f(x)在(-∞,0)上單調(diào)遞增,當(dāng)x∈(0,ln(2a))時,f'(x)<0,f(x)在(0,ln(2a))上單調(diào)遞減,所以x=0是函數(shù)f(x)的極大值點(diǎn),符合題意.綜上所述,a的取值范圍是12(3)證明:要證esinθ-1+ecosθ-1+ln(sinθcosθ)<1,只要證esinθ-1+ecosθ-1+lnsinθ+lncosθ<sin2θ+cos2θ,只要證esinθ-1+lnsinθ<sin2θ,ecosθ-1+lncosθ<cos2θ,因?yàn)棣取?,π2,則sinθ∈(0,1),cosθ∈(0,所以只要證對任意0<x<1,有ex-1+lnx<x2,只要證對任意-1<x<0,有ex+ln(1+x)<(1+x)2(※),因?yàn)橛?2)知:當(dāng)a=1時,若x<0,則f(x)<f(0)=1,所以ex-x2-x<1,即ex<x2+x+1①,令h(x)=ln(1+x)-x(-1<x<0),則h'(x)=11+x-1=?x1+x,所以當(dāng)-1<x<0時,h'所以h(x)在(-1,0)上單調(diào)遞增;則h(x)<h(0)=0,即ln(1+x)<x(-1<x<0)②,由①+②得ex+ln(1+x)<x2+2x+1=(x+1)2,所以(※)成立,所以esinθ-1+ecosθ-1+ln(sinθcosθ)<1成立.7.(2024山東泰安一模,18)已知函數(shù)f(x)=aeax(a≠0).(1)若a>0,曲線y=f(x)在點(diǎn)(0,f(0))處的切線與直線x+y-2=0垂直,證明:f(x)>ln(x+2);(2)若對任意的x1,x2且x1<x2,函數(shù)g(x)=f(x)-eax1?eax2x1?x2,證明:函數(shù)g解析(1)證明:∵f'(x)=a2eax,∴f'(0)=a2=1,由a>0解得a=1,∴f(x)=ex.設(shè)μ(x)=ex-x-1,則μ'(x)=ex-1,∴當(dāng)x∈(-∞,0)時,μ'(x)<0,μ(x)在(-∞,0)上單調(diào)遞減,當(dāng)x∈(0,+∞)時,μ'(x)>0,μ(x)在(0,+∞)上單調(diào)遞增,∴μ(x)≥μ(0)=0,即ex≥x+1,x=0時等號成立.設(shè)m(x)=lnx-x+1,則m'(x)=1x-1=1?∴當(dāng)x∈(0,1)時,m'(x)>0,m(x)在(0,1)上單調(diào)遞增,當(dāng)x∈(1,+∞)時,m'(x)<0,m(x)在(1,+∞)上單調(diào)遞減,∴m(x)≤m(1)=0,即lnx≤x-1,∴l(xiāng)n(x+2)≤x+1,當(dāng)x=-1時等號成立.∴ex>x+1>ln(x+2)(取等的條件不能同時成立),即f(x)>ln(x+2).(2)證明:∵g(x)=aeax-eax1?∴g'(x)=a2eax>0,(注意:x1,x2是常數(shù),故分式求導(dǎo)后為0)∴g(x)在(x1,x2)上單調(diào)遞增,又∵g(x1)=aeax=a=-eax1x2?x1[a(x1=-eax1x2?x1[ea(xg(x2)=eax2x2?x1[ea(x由(1)知當(dāng)x≠0時,ex>x+1,即ex-x-1>0,∴ea(x2?x1)-a(x2-x1)-1>0,ea(x又eax1x2∴g(x1)<0,g(x2)>0,又∵g(x)在(x1,x2)上單調(diào)遞增,∴函數(shù)g(x)在(x1,x2)上存在唯一零點(diǎn).8.(2024安徽皖北協(xié)作區(qū)聯(lián)考,19