（通用版）2020版高考數(shù)學(xué)大二輪復(fù)習(xí) 專題突破練8 應(yīng)用導(dǎo)數(shù)求參數(shù)的值或范圍 文

專題突破練8應(yīng)用導(dǎo)數(shù)求參數(shù)的值或范圍1.(2019北京順義統(tǒng)考二,文18)設(shè)函數(shù)f(x)=ax-ln x,aR.(1)若點(diǎn)(1,1)在曲線y=f(x)上,求在該點(diǎn)處曲線的切線方程;(2)若f(x)有極小值2,求a.2.(2019湖南三湘名校大聯(lián)考一,文21)已知函數(shù)f(x)=xln x.(1)略;(2)若當(dāng)x1e時(shí),f(x)ax2-x+a-1,求實(shí)數(shù)a的取值范圍.3.(2019山東濰坊二模,文21)已知函數(shù)f(x)=xex-aln x(無(wú)理數(shù)e=2.718).(1)若f(x)在(0,1)單調(diào)遞減,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a=-1時(shí),設(shè)g(x)=x(f(x)-xex)-x3+x2-b,若函數(shù)g(x)存在零點(diǎn),求實(shí)數(shù)b的最大值.4.已知函數(shù)f(x)=ax+ln x(aR).(1)略;(2)設(shè)g(x)=x2-2x+2,若對(duì)任意x1(0,+),均存在x20,1,使得f(x1)<g(x2),求a的取值范圍.5.(2019河北唐山三模,文21)已知函數(shù)f(x)=xln x-a(x2-x)+1,函數(shù)g(x)=f'(x).(1)若a=1,求f(x)的極大值;(2)當(dāng)0<x<1時(shí),g(x)有兩個(gè)零點(diǎn),求a的取值范圍.6.(2019四川第二次診斷,理21)已知f(x)=xln x.(1)求f(x)的極值;(2)若f(x)-axx=0有兩個(gè)不同解,求實(shí)數(shù)a的取值范圍.7.(2019山東德州一模,理21,文21)已知函數(shù)f(x)=e2x-3-(2x-3)2.(1)證明:當(dāng)x32時(shí),f(x)1;(2)設(shè)g(x)=14+lnx2,若存在實(shí)數(shù)x1,x2,使得f(x1)+(2x1-3)2=g(x2),求x2-x1的最小值.專題突破練8應(yīng)用導(dǎo)數(shù)求參數(shù)的值或范圍1.解(1)因?yàn)辄c(diǎn)(1,1)在曲線y=f(x)上,所以a=1,f(x)=x-lnx.又f'(x)=x2x-1x=x-22x,所以f'(1)=-12.在該點(diǎn)處曲線的切線方程為y-1=-12(x-1),即x+2y-3=0.(2)f(x)的定義域?yàn)?0,+),f'(x)=ax2x-1x=ax-22x.討論:當(dāng)a0時(shí),f'(x)<0,此時(shí)f(x)在(0,+)上單調(diào)遞減,所以不存在極小值.當(dāng)a>0時(shí),令f'(x)=0可得x=4a2,當(dāng)x發(fā)生變化時(shí),f'(x),f(x)的變化情況如下表:x0,4a24a24a2,+f'(x)-0+f(x)單調(diào)遞減極小值單調(diào)遞增所以f(x)在0,4a2上單調(diào)遞減,在4a2,+上單調(diào)遞增,所以f(x)極小值=f4a2=2-ln4a2,所以2-ln4a2=2,解得a=2(負(fù)值舍去).2.解(1)略.(2)由已知得axlnx+x+1x2+1,設(shè)h(x)=xlnx+x+1x2+1,則h'(x)=(1-x)(xlnx+lnx+2)(x2+1)2.y=xlnx+lnx+2是增函數(shù),且x1e,y-1e-1+2>0.當(dāng)x1e,1時(shí),h'(x)>0;當(dāng)x(1,+)時(shí),h'(x)<0,h(x)在x=1處取得最大值h(1)=1,a1.3.解(1)f'(x)=(x+1)ex-ax=(x2+x)ex-ax.由題意可得f'(x)0,x(0,1)恒成立.即(x2+x)ex-a0,也就是a(x2+x)ex在x(0,1)恒成立.設(shè)h(x)=(x2+x)ex,則h'(x)=(x2+3x+1)ex.當(dāng)x(0,1)時(shí),x2+3x+1>0,h'(x)>0在x(0,1)單調(diào)遞增.即h(x)<h(1)=2e.故a2e.(2)當(dāng)a=-1時(shí),f(x)=xex+lnx.g(x)=xlnx-x3+x2-b,由題意得問(wèn)題等價(jià)于方程b=xlnx-x3+x2,在(0,+)上有解.先證明lnxx-1.設(shè)u(x)=lnx-x+1,x(0,+),則u'(x)=1x-1=1-xx.可得當(dāng)x=1時(shí),函數(shù)u(x)取得極大值,u(x)u(1)=0.因此lnxx-1,所以b=xlnx-x3+x2x(x-1)-x3+x2=-x(x2-2x+1)0.當(dāng)x=1時(shí),取等號(hào).故實(shí)數(shù)b的最大值為0.4.解(1)略.(2)由已知,轉(zhuǎn)化為f(x)max<g(x)max.g(x)=(x-1)2+1,x0,1,g(x)max=g(0)=2.由f(x)=ax+lnx,得f'(x)=a+1x=ax+1x(x>0),當(dāng)a0時(shí),由于x>0,故ax+1>0,f'(x)>0,f(x)在(0,+)單調(diào)遞增,值域?yàn)镽,不合題意.當(dāng)a<0時(shí),f(x)在0,-1a上單調(diào)遞增,在-1a,+上單調(diào)遞減,故f(x)的極大值即為最大值,f-1a=-1+ln-1a=-1-ln(-a),所以2>-1-ln(-a),解得a<-1e3.5.解(1)f(x)=xlnx-x2+x+1,g(x)=f'(x)=lnx-2x+2,g'(x)=1x-2=1-2xx,當(dāng)x0,12時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x12,+時(shí),g'(x)<0,g(x)單調(diào)遞減.注意到g(1)=f'(1)=0,則當(dāng)x12,1時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x(1,+)時(shí),f'(x)<0,f(x)單調(diào)遞減;故當(dāng)x=1時(shí),f(x)取得極大值f(1)=1.(2)g(x)=f'(x)=lnx+1-2ax+a,g'(x)=1x-2a=1-2axx,若a0,則g'(x)>0,g(x)單調(diào)遞增,至多有一個(gè)零點(diǎn),不合題意.若a>0,則當(dāng)x0,12a時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)x12a,+時(shí),g'(x)<0,g(x)單調(diào)遞減;則g12ag12=ln12+1=lne2>0.不妨設(shè)g(x1)=g(x2),x1<x2,則0<x1<12a<x2<1.一方面,需要g(1)<0,得a>1.另一方面,由(1)得,當(dāng)x>1時(shí),lnx<x-1<x,則x<ex,進(jìn)而,有2a<e2a,則e-2a<12a,且g(e-2a)=-2ae-2a+1-a<0,故存在x1,使得0<e-2a<x1<12a.綜上,a的取值范圍是(1,+).6.解(1)f(x)的定義域是(0,+),f'(x)=lnx+1.令f'(x)>0,解得x>1e.令f'(x)<0,解得0<x<1e.所以f(x)在0,1e內(nèi)遞減,在1e,+內(nèi)遞增,故當(dāng)x=1e時(shí),f(x)極小值=f1e=-1e.(2)記t=xlnx,t-1e,則et=exlnx=(elnx)x=xx,故f(x)-axx=0,即t-aet=0,a=tet.令g(t)=tet,g'(t)=1-tet.令g'(t)>0,解得0<t<1.令g'(t)<0,解得t>1.故g(t)在(0,1)遞增,在(1,+)內(nèi)遞減,故g(t)max=g(1)=1e.由t=xlnx,t-1e,a=g(t)=tet的圖象和性質(zhì)有:0<a<1e,y=a和g(t)有兩個(gè)不同交點(diǎn)(t1,a),(t2,a),且0<t1<1<t2,t1=xlnx,t2=xlnx各有一解,即f(x)-axx=0有2個(gè)不同解.-e1-ee<a<0,y=a和g(t)=tet僅有1個(gè)交點(diǎn)(t3,a),且-1e<t3<0,t3=xlnx有2個(gè)不同的解,即f(x)-axx=0有兩個(gè)不同解.a取其他值時(shí),f(x)-axx=0最多1個(gè)解,綜上,a的范圍是(-e1-ee,0)0,1e.7.解(1)令t=2x-3,當(dāng)x32時(shí),f(x)1等價(jià)于當(dāng)t0時(shí),et-t2-10.設(shè)函數(shù)u(t)=et-t2-1,則u'(t)=et-2t.u'(t)'=et-2.當(dāng)t0,ln2)時(shí),u'(t)為減函數(shù),當(dāng)t(ln2,+)時(shí),u'(t)為增函數(shù).所以u(píng)'(t)u'(ln2)=2-2ln2>0.所以u(píng)(t)在0,+)內(nèi)為增函數(shù),所以u(píng)(t)u(0)=0.即當(dāng)x32時(shí),f(x)1.(2)設(shè)f(x1)+(2x1-3)2=g(x2)=m,則e2x1-3=14+lnx22=m.因?yàn)閤1R,所以e2x1-3>0,即m>0,所以2x1-3=lnm,lnx22=m-14,所以x1=lnm+32,x2=2em-14,x2-x1=2em-14-lnm+32(m>0).令h(x)=2ex-14-lnx+32(x>0),則h'(x)=2ex-14-12x,所以h'(x)'=2ex-14+12x2>0,所以h'(x)在(0,+)內(nèi)為增函數(shù),且h'14=0.當(dāng)x>14時(shí),h'(x)>0;當(dāng)0<x<14時(shí),h'(x)<0.所以,h(x)在0,14內(nèi)為減函數(shù),在14,+內(nèi)為增函數(shù).故當(dāng)x=14時(shí),h(x)min=h14=ln4+12,即x2-x1的最小值為ln4+12.14