高考數(shù)學(xué)大二輪復(fù)習(xí) 大題專(zhuān)項(xiàng)練（五）函數(shù)與導(dǎo)數(shù) 理-人教版高三數(shù)學(xué)試題

大題專(zhuān)項(xiàng)練(五)函數(shù)與導(dǎo)數(shù)A組基礎(chǔ)通關(guān)1.(2017全國(guó)Ⅰ,理21)已知函數(shù)f(x)=ae2x+(a-2)ex-x.(1)討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求a的取值范圍.解(1)f(x)的定義域?yàn)?-∞,+∞),f'(x)=2ae2x+(a-2)ex-1=(aex-1)(2ex+1).(ⅰ)若a≤0,則f'(x)<0,所以f(x)在(-∞,+∞)單調(diào)遞減.(ⅱ)若a>0,則由f'(x)=0得x=-lna.當(dāng)x∈(-∞,-lna)時(shí),f'(x)<0;當(dāng)x∈(-lna,+∞)時(shí),f'(x)>0,所以f(x)在(-∞,-lna)單調(diào)遞減,在(-lna,+∞)單調(diào)遞增.(2)(ⅰ)若a≤0,由(1)知,f(x)至多有一個(gè)零點(diǎn).(ⅱ)若a>0,由(1)知,當(dāng)x=-lna時(shí),f(x)取得最小值,最小值為f(-lna)=1-1a+lna①當(dāng)a=1時(shí),由于f(-lna)=0,故f(x)只有一個(gè)零點(diǎn);②當(dāng)a∈(1,+∞)時(shí),由于1-1a+lna>即f(-lna)>0,故f(x)沒(méi)有零點(diǎn);③當(dāng)a∈(0,1)時(shí),1-1a+lna<0,即f(-lna)<0又f(-2)=ae-4+(a-2)e-2+2>-2e-2+2>0,故f(x)在(-∞,-lna)有一個(gè)零點(diǎn).設(shè)正整數(shù)n0滿足n0>ln3a則f(n0)=en0(aen0+a-2)-n0>en0-n0>由于ln3a-1>-lna,因此f(x)在(-lna,+綜上,a的取值范圍為(0,1).2.在某次水下科研考查活動(dòng)中,需要潛水員潛入水深為60米的水底進(jìn)行作業(yè),根據(jù)以往經(jīng)驗(yàn),潛水員下潛的平均速度為v(米/單位時(shí)間),每單位時(shí)間的用氧量為v103+1(升),在水底作業(yè)10個(gè)單位時(shí)間,每單位時(shí)間用氧量為0.9(升),返回水面的平均速度為v2(米/單位時(shí)間),每單位時(shí)間用氧量為1.5(升),記該潛水員在此次考查活動(dòng)中的總用氧量為y(升).(1)求y關(guān)于v的函數(shù)關(guān)系式;(2)若c≤v≤15(c>0),求當(dāng)下潛速度v取什么值時(shí),總用氧量最少.解(1)由題意,得下潛用時(shí)60v用氧量為v103+1×60v=3v水底作業(yè)時(shí)的用氧量為10×0.9=9(升);返回水面用時(shí)60v用氧量為120v×1.5=180∴總用氧量y=3v250+240(2)y'=3v令y'=0,得v=1032當(dāng)0<v<1032時(shí),y'<當(dāng)v>1032時(shí),y'>∴當(dāng)0<c<1032時(shí),函數(shù)在(c,103在(1032∴當(dāng)v=1032當(dāng)c≥1032時(shí),y在[c∴當(dāng)v=c時(shí)總用氧量最少.綜上,若0<c<1032則當(dāng)v=1032時(shí)總用氧量最少;若c≥103則當(dāng)v=c時(shí)總用氧量最少.3.(2019安徽淮北模擬)已知函數(shù)f(x)=ax-1+(1)若函數(shù)f(x)在(e,+∞)內(nèi)有極值,求實(shí)數(shù)a的取值范圍;(2)在(1)的條件下,對(duì)任意t∈(1,+∞),s∈(0,1),求證:f(t)-f(s)>e+2-1e(1)解由定義域?yàn)?0,1)∪(1,+∞),f'(x)=1x設(shè)h(x)=x2-(a+2)x+1,要使y=f(x)在(e,+∞)上有極值,則x2-(a+2)x+1=0有兩個(gè)不同的實(shí)根x1,x2,∴Δ=(a+2)2-4>0,∴a>0或a<-4,①且至少有一根在區(qū)間(e,+∞)上,又∵x1·x2=1,∴只有一根在區(qū)間(e,+∞)上,不妨設(shè)x2>e,∴0<x1<1e<e<x2又h(0)=1,∴只需h1e<0,即1e2-(a+2)1e+1∴a>e+1e-2,聯(lián)立①②可得a>e+1e-2即實(shí)數(shù)a的取值范圍是e+1e-2,+∞.(2)證明由(1)知,當(dāng)x∈(1,x2)時(shí),f'(x)<0,f(x)單調(diào)遞減,當(dāng)x∈(x2,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,∴f(x)在(1,+∞)上有最小值f(x2),即?t∈(1,+∞),都有f(t)≥f(x2),又當(dāng)x∈(0,x1)時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)x∈(x1,1)時(shí),f'(x)<0,f(x)單調(diào)遞減,∴f(x)在(0,1)上有最大值f(x1),即對(duì)?s∈(0,1),都有f(s)≤f(x1),又∵x1+x2=2+a,x1x2=1,x1∈0,1e,x2∈(e,+∞),∴f(t)-f(s)≥f(x2)-f(x1)=lnx2+ax2-1-ln=lnx=lnx22+x2-1x2(設(shè)k(x)=lnx2+x-1x=2lnx+x-1x(則k'(x)=2x+1+1x2>∴k(x)在(e,+∞)上單調(diào)遞增,∴k(x)>k(e)=2+e-1e∴f(t)-f(s)>e+2-1e4.(2019河南商丘模擬)已知函數(shù)f(x)=(2x+1)ln(2x+1)-a(2x+1)2-x(a>0).(1)如圖,設(shè)直線x=-12,y=-x將坐標(biāo)平面分成Ⅰ,Ⅱ,Ⅲ,Ⅳ四個(gè)區(qū)域(不含邊界),若函數(shù)y=f(x)的圖象恰好位于其中一個(gè)區(qū)域內(nèi),判斷其所在的區(qū)域并求對(duì)應(yīng)的a(2)當(dāng)a>12時(shí),求證:?x1,x2∈(0,+∞)且x1≠x2,有f(x1)+f(x2)<2fx1+(1)解函數(shù)f(x)的定義域?yàn)?12,+∞,且當(dāng)x=0時(shí),f(0)=-a<0.又∵直線y=-x恰好通過(guò)原點(diǎn),∴函數(shù)y=f(x)的圖象應(yīng)位于區(qū)域Ⅳ內(nèi),于是可得f(x)<-x,即(2x+1)ln(2x+1)-a(2x+1)2-x<-x.∵2x+1>0,∴a>ln(令h(x)=ln(2x+1)則h'(x)=2-2ln(2x∴當(dāng)x∈-12,e-12時(shí),h'(x)>0,當(dāng)x∈e-12,+∞時(shí),h'(x)<0,h(x)單調(diào)遞減∴h(x)max=he-12=1∴a的取值范圍是1e,+∞.(2)證明∵f'(x)=2ln(2x+1)-4a(2x+1)+1,設(shè)u(x)=2ln(2x+1)-4a(2x+1)+1,則u'(x)=42x+1-8ax>-1∵當(dāng)x>0時(shí),42x+1<4,當(dāng)a>1∴u'(x)=42x+1-∴當(dāng)x>0時(shí),f'(x)為減函數(shù),不妨設(shè)x2>x1>0,令g(x)=f(x)+f(x1)-2fx+x12(x>x可得g(x1)=0,g'(x)=f'(x)-f'x+x1∵x>x+x12且f'(x∴g'(x)<0,∴當(dāng)x>x1時(shí),g(x)為減函數(shù),∴g(x2)<g(x1)=0,即f(x1)+f(x2)<2fx1+x5.已知函數(shù)f(x)=lnx+ax,g(x)=e-x+bx,a,b∈R,e為自然對(duì)數(shù)的底數(shù)(1)若函數(shù)y=g(x)在R上存在零點(diǎn),求實(shí)數(shù)b的取值范圍;(2)若函數(shù)y=f(x)在x=1e處的切線方程為ex+y-2+b=0.求證:對(duì)任意的x∈(0,+∞),總有f(x)>g(x)(1)解易得g'(x)=-e-x+b=b-1e若b=0,則g(x)=1ex∈(0,+若b<0,則g(0)=1>0,g-1b=e1b-1<若b>0,令g'(x)=-e-x+b=0,得x=-lnb.∴g(x)在(-∞,-lnb)上單調(diào)遞減;在(-lnb,+∞)上單調(diào)遞增,則g(x)min=g(-lnb)=elnb-blnb=b-blnb≤0,∴b≥e.綜上所述,實(shí)數(shù)b的取值范圍是(-∞,0)∪[e,+∞).(2)證明易得f'(x)=1x則由題意,得f'1e=e-ae2=-e,解得a=2e.∴f(x)=lnx+2ex,從而f1e=即切點(diǎn)為1e,1.將切點(diǎn)坐標(biāo)代入ex+y-2+b=0中,解得b=0.∴g(x)=e-x.要證f(x)>g(x),即證lnx+2ex>e-x(x∈(0,+只需證xlnx+2e>xe-x(x∈(0,+∞))令u(x)=xlnx+2e,v(x)=xe-x,x∈(0,+∞)則由u'(x)=lnx+1=0,得x=1e∴u(x)在0,1e上單調(diào)遞減,在1e,+∞上單調(diào)遞增,∴u(x)min=u1e=1e.又由v'(x)=e-x-xe-x=e-x(1-x)=0,得x=1,∴v(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,∴v(x)max=v(1)=1e∴u(x)≥u(x)min=v(x)max≥v(x),顯然,上式的等號(hào)不能同時(shí)取到.故對(duì)任意的x∈(0,+∞),總有f(x)>g(x).6.(2019安徽馬鞍山模擬)已知函數(shù)g(x)=xlnx,h(x)=ax2-1(1)若g(x)<h(x)對(duì)x∈(1,+∞)恒成立,求a的取值范圍;(2)證明:不等式1+1n21+2n2…1+nn2<e34對(duì)于正整數(shù)n恒成立,其中e=2.(1)解當(dāng)x∈(1,+∞)時(shí),g(x)<h(x)等價(jià)于a>2x令F(x)=2xlnxF'(x)=2(x-記m(x)=x-1-xlnx(x>1),則m'(x)=-lnx<0,∴m(x)在(1,+∞)上單調(diào)遞減,∴m(x)<m(1)=0,∴F'(x)<0,即F(x)在(1,+∞)上單調(diào)遞減,F(x)<F(1)=1,故a∈[1,+∞).(2)證明由(1)知取a=1,當(dāng)x∈(1,+∞)時(shí),g(x)<h(x)恒成立,即xlnx<x2-12即ln(1+x)<(x+1)2-12由此,ln1+kn2<(kn2)

2+2(kn2于是ln1+1n21+2n2…1+nn2=ln1+1n2+ln1+2n2+…+ln1+nn2<121n2+2n2+=14=1=143-n3=143-n(n-1故1+1n21+2n2…1+nn2<e3B組能力提升7.(2019廣東深中、華附、省實(shí)、廣雅四校聯(lián)考)已知函數(shù)f(x)=x-1-a6ex+1,其中e=2.718…為自然對(duì)數(shù)的底數(shù),常數(shù)a>0.(1)求函數(shù)f(x)在區(qū)間(0,+∞)上的零點(diǎn)個(gè)數(shù);(2)函數(shù)F(x)的導(dǎo)數(shù)F'(x)=(ex-a)f(x),是否存在無(wú)數(shù)個(gè)a∈(1,4),使得lna為函數(shù)F(x)的極大值點(diǎn)?請(qǐng)說(shuō)明理由.解(1)f'(x)=x-a6ex,當(dāng)0<x<a6時(shí),f'(x)<0,f(x當(dāng)x>a6時(shí),f'(x)>0,f(x所以當(dāng)x∈(0,+∞)時(shí),f(x)min=fa6,因?yàn)閒a6<f(0)=-a6<0,f1+a6=1>0,所以存在x0∈a6,1+a6,使f(x0)=0,且當(dāng)0<x<x0時(shí),f(x)<0,當(dāng)x>x0時(shí),f(x)>0.故函數(shù)f(x)在(0,+∞)上有1個(gè)零點(diǎn),即x0.(2)當(dāng)a>1時(shí),lna>0.因?yàn)楫?dāng)x∈(0,lna)時(shí),ex-a<0;當(dāng)x∈(lna,+∞)時(shí),ex-a>0.由(1)知,當(dāng)x∈(0,x0)時(shí),f(x)<0;當(dāng)x∈(x0,+∞)時(shí),f(x)>0.下面證:當(dāng)a∈(1,e)時(shí),lna<x0,即證f(lna)<0.f(lna)=lna-1-a6a+1=alna-a-a26+記g(x)=xlnx-x-x26+1,xg'(x)=lnx-x3,x∈令h(x)=g'(x),則h'(x)=3-x所以g'(x)在(1,e)上單調(diào)遞增,由g'(1)=-13<0,g'(e)=1-e3所以存在唯一零點(diǎn)t0∈(1,e),使得g'(t0)=0,且x∈(1,t0)時(shí),g'(x)<0,g(x)單調(diào)遞減,x∈(t0,e)時(shí),g'(x)>0,g(x)單調(diào)遞增.所以當(dāng)x∈(1,e)時(shí),g(x)<max{g(1),g(e)}.由g(1)=-16<0,g(e)=6-得當(dāng)x∈(1,e)時(shí),g(x)<0.故f(lna)<0,0<lna<x0.當(dāng)0<x<lna時(shí),ex-a<0,f(x)<0,F'(x)=(ex-a)f(x)>0,F(x)單調(diào)遞增;當(dāng)lna<x<x0時(shí),ex-a>0,f(x)<0,F'(x)=(ex-a)f(x)<0,F(x)單調(diào)遞減.所以存在a∈(1,e)?(1,4),使得lna為F(x)的極大值點(diǎn).8.已知函數(shù)f(x)=ex(1+alnx),其中a>0,設(shè)f'(x)為f(x)的導(dǎo)函數(shù).(1)設(shè)g(x)=e-xf'(x),若g(x)≥2恒成立,設(shè)a的取值范圍;(2)設(shè)函數(shù)f(x)的零點(diǎn)為x0,函數(shù)f'(x)的極小值點(diǎn)為x1,當(dāng)a>2時(shí),求證:x0>x1.解(1)由題設(shè)知,f'(x)=ex1+ax+alnx(x>0),g(x)=e-xf'(x)=1+ax+alnxg'(x)=a(x-1當(dāng)x∈(0,1)時(shí),g'(x)<0,g(x)在區(qū)間(0,1)上單調(diào)遞減,當(dāng)x∈(1,+∞)時(shí),g'(x)>0,g(x)在區(qū)間(1,+∞)上單調(diào)遞增,故g(x)在x=1處取得最小值,且g(1)=1+a.由于g(x)≥2恒成立,所以1+a≥2,得a≥1,即a的取值范圍為[1,+∞).(2)設(shè)h(x)=f'(x)=ex1+ax+alnx,則h'(x)=ex1+2ax-ax2設(shè)H(x)=1+2ax-ax2則H'(x)=-2ax故H(x)在(0,+∞)