廣西2020版高考數(shù)學(xué)一輪復(fù)習(xí) 高考大題專項(xiàng)練一 高考中的函數(shù)與導(dǎo)數(shù) 文.docx

高考大題專項(xiàng)練一高考中的函數(shù)與導(dǎo)數(shù)1.(2018北京,文19)設(shè)函數(shù)f(x)=ax2-(3a+1)x+3a+2ex.(1)若曲線y=f(x)在點(diǎn)(2,f(2)處的切線斜率為0,求a;(2)若f(x)在x=1處取得極小值,求a的取值范圍.解(1)因?yàn)閒(x)=ax2-(3a+1)x+3a+2ex,所以f(x)=ax2-(a+1)x+1ex,f(2)=(2a-1)e2.由題設(shè)知f(2)=0,即(2a-1)e2=0,解得a=12.(2)由(1)得f(x)=ax2-(a+1)x+1ex=(ax-1)(x-1)ex.若a>1,則當(dāng)x1a,1時(shí),f(x)<0;當(dāng)x(1,+)時(shí),f(x)>0.所以f(x)在x=1處取得極小值.若a1,則當(dāng)x(0,1)時(shí),ax-1x-1<0,所以f(x)>0.所以1不是f(x)的極小值點(diǎn).綜上可知,a的取值范圍是1,+.2.(2018全國(guó),文21)已知函數(shù)f(x)=ax2+x-1ex.(1)求曲線y=f(x)在點(diǎn)(0,-1)處的切線方程;(2)證明:當(dāng)a1時(shí),f(x)+e0.(1)解f(x)=-ax2+(2a-1)x+2ex,f(0)=2.因此曲線y=f(x)在(0,-1)處的切線方程是2x-y-1=0.(2)證明當(dāng)a1時(shí),f(x)+e(x2+x-1+ex+1)e-x.令g(x)=x2+x-1+ex+1,則g(x)=2x+1+ex+1.當(dāng)x<-1時(shí),g(x)<0,g(x)單調(diào)遞減;當(dāng)x>-1時(shí),g(x)>0,g(x)單調(diào)遞增;所以g(x)g(-1)=0.因此f(x)+e0.3.已知函數(shù)f(x)=ln x+12ax2-x-m(mZ).(1)若f(x)是增函數(shù),求a的取值范圍;(2)若a<0,且f(x)<0恒成立,求m的最小值.解(1)f(x)=1x+ax-1,依題設(shè)可得a1x-1x2max,而1x-1x2=-1x-122+1414,當(dāng)x=2時(shí),等號(hào)成立.所以a的取值范圍是14,+.(2)由(1)可知f(x)=1x+ax-1=ax2-x+1x,設(shè)g(x)=ax2-x+1,則g(0)=1>0,g(1)=a<0,g(x)=ax-12a2+1-14a在(0,+)內(nèi)單調(diào)遞減.因此g(x)在(0,1)內(nèi)有唯一的解x0,使得ax02=x0-1,而且當(dāng)0<x<x0時(shí),f(x)>0,當(dāng)x>x0時(shí),f(x)<0,所以f(x)f(x0)=lnx0+12ax02-x0-m=lnx0+12(x0-1)-x0-m=lnx0-12x0-12-m.設(shè)r(x)=lnx-12x-12-m,則r(x)=1x-12=2-x2x>0.所以r(x)在(0,1)內(nèi)單調(diào)遞增.所以r(x)<r(1)=-1-m,由已知可得-1-m0,所以m-1,即m的最小值為-1.4.函數(shù)f(x)=x22+ax+2ln x(aR)在x=2處取得極值.(1)求實(shí)數(shù)a的值及函數(shù)f(x)的單調(diào)區(qū)間;(2)若方程f(x)=m有三個(gè)實(shí)根,求m的取值范圍.解(1)由已知f(x)=x+a+2x,f(2)=2+a+22=0,故a=-3,所以f(x)=x-3+2x=x2-3x+2x=(x-2)(x-1)x,x>0,由f(x)>0,得0<x<1或x>2;由f(x)<0,得1<x<2.所以函數(shù)f(x)的單調(diào)遞增區(qū)間是(0,1),(2,+),單調(diào)遞減區(qū)間是(1,2).(2)由(1)可知極小值f(2)=2ln2-4,極大值為f(1)=-52.因?yàn)榉匠蘤(x)=m有三個(gè)實(shí)根,所以2ln2-4<m<-52.5.(2018全國(guó),文21)已知函數(shù)f(x)=aex-ln x-1.(1)設(shè)x=2是f(x)的極值點(diǎn),求a,并求f(x)的單調(diào)區(qū)間;(2)證明:當(dāng)a1e時(shí),f(x)0.(1)解f(x)的定義域?yàn)?0,+),f(x)=aex-1x.由題設(shè)知,f(2)=0,所以a=12e2.從而f(x)=12e2ex-lnx-1,f(x)=12e2ex-1x.當(dāng)0<x<2時(shí),f(x)<0;當(dāng)x>2時(shí),f(x)>0.所以f(x)在(0,2)上單調(diào)遞減,在(2,+)上單調(diào)遞增.(2)證明當(dāng)a1e時(shí),f(x)exe-lnx-1.設(shè)g(x)=exe-lnx-1,則g(x)=exe-1x.當(dāng)0<x<1時(shí),g(x)<0;當(dāng)x>1時(shí),g(x)>0.所以x=1是g(x)的最小值點(diǎn).故當(dāng)x>0時(shí),g(x)g(1)=0.因此,當(dāng)a1e時(shí),f(x)0.6.定義在實(shí)數(shù)集上的函數(shù)f(x)=x2+x,g(x)=13x3-2x+m.(1)求函數(shù)f(x)的圖象在x=1處的切線方程;(2)若f(x)g(x)對(duì)任意的x-4,4恒成立,求實(shí)數(shù)m的取值范圍.解(1)f(x)=x2+x,當(dāng)x=1時(shí),f(1)=2,f(x)=2x+1,f(1)=3,所求切線方程為y-2=3(x-1),即3x-y-1=0.(2)令h(x)=g(x)-f(x)=13x3-x2-3x+m,則h(x)=(x-3)(x+1).當(dāng)-4<x<-1時(shí),h(x)>0;當(dāng)-1<x<3時(shí),h(x)<0;當(dāng)3<x<4時(shí),h(x)>0.要使f(x)g(x)恒成立,即h(x)max0,由上知h(x)的最大值在x=-1或x=4處取得,而h(-1)=m+53,h(4)=m-203,故m+530,即m-53,故實(shí)數(shù)m的取值范圍為-,-53.7.已知函數(shù)f(x)=12ax2-(2a+1)x+2ln x(aR).(1)求f(x)的單調(diào)區(qū)間;(2)設(shè)g(x)=x2-2x,若對(duì)任意x1(0,2,均存在x2(0,2,使得f(x1)<g(x2),求a的取值范圍.解f(x)=ax-(2a+1)+2x(x>0).(1)f(x)=(ax-1)(x-2)x(x>0).當(dāng)a0時(shí),x>0,ax-1<0,在區(qū)間(0,2)內(nèi),f(x)>0,在區(qū)間(2,+)內(nèi),f(x)<0,故f(x)的單調(diào)遞增區(qū)間是(0,2),單調(diào)遞減區(qū)間是(2,+).當(dāng)0<a<12時(shí),1a>2,在區(qū)間(0,2)和1a,+內(nèi),f(x)>0,在區(qū)間2,1a內(nèi),f(x)<0,故f(x)的單調(diào)遞增區(qū)間是(0,2)和1a,+,單調(diào)遞減區(qū)間是2,1a.當(dāng)a=12時(shí),f(x)=(x-2)22x,故f(x)的單調(diào)遞增區(qū)間是(0,+).當(dāng)a>12時(shí),0<1a<2,在區(qū)間0,1a和(2,+)內(nèi),f(x)>0,在區(qū)間1a,2內(nèi),f(x)<0,故f(x)的單調(diào)遞增區(qū)間是0,1a和(2,+),單調(diào)遞減區(qū)間是1a,2.(2)對(duì)任意x1(0,2,均存在x2(0,2,使得f(x1)<g(x2)在(0,2上有f(x)max<g(x)max.由題意可知g(x)max=0,由(1)可知,當(dāng)a12時(shí),f(x)在(0,2上單調(diào)遞增.故f(x)max=f(2)=2a-2(2a+1)+2ln2=-2a-2+2ln2,所以-2a-2+2ln2<0,解得a>ln2-1.故ln2-1<a12.當(dāng)a>12時(shí),f(x)在0,1a上單調(diào)遞增,在1a,2上單調(diào)遞減,故f(x)max=f1a=12a-(2a+1)1a+2ln1a=-12a-2-2lna<0因?yàn)楫?dāng)a>12時(shí),12a+2lna>12a+2lne-1=12a-2>-2.故a>12時(shí)滿足題意.綜上,a的取值范圍為(ln2-1,+).8.設(shè)a,bR,|a|1.已知函數(shù)f(x)=x3-6x2-3a(a-4)x+b,g(x)=exf(x).(1)求f(x)的單調(diào)區(qū)間;(2)已知函數(shù)y=g(x)和y=ex的圖象在公共點(diǎn)(x0,y0)處有相同的切線,求證:f(x)在x=x0處的導(dǎo)數(shù)等于0;若關(guān)于x的不等式g(x)ex在區(qū)間x0-1,x0+1上恒成立,求b的取值范圍.解(1)由f(x)=x3-6x2-3a(a-4)x+b,可得f(x)=3x2-12x-3a(a-4)=3(x-a)x-(4-a).令f(x)=0,解得x=a或x=4-a.由|a|1,得a<4-a.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x(-,a)(a,4-a)(4-a,+)f(x)+-+f(x)所以,f(x)的單調(diào)遞增區(qū)間為(-,a),(4-a,+),單調(diào)遞減區(qū)間為(a,4-a).(2)證明:因?yàn)間(x)=ex(f(x)+f(x),由題意知g(x0)=ex0,g(x0)=ex0,所以f(x0)ex0=ex0,ex0(f(x0)+f(x0)=ex0,解得f(x0)=1,f(x0)=0.所以,f(x)在x=x0處的導(dǎo)數(shù)等于0.因?yàn)間(x)ex,xx0-1,x0+1,由ex>0,可得f(x)1.又因?yàn)閒(x0)=1,f(x0)=0,故x0為f(x)的極大值點(diǎn),由(1)知x0=a.另一方面,由于|a|1,故a+1<4-a,由(1)知f(x)在(a-1,a)內(nèi)單調(diào)遞增,在(a,a+1)內(nèi)單調(diào)遞減,故當(dāng)x0=a時(shí),f(x)f(a)=1在a-1,a+1上恒成立,從而g(x)ex在x0-1,x0+1上恒成立.由f(a)=a3-6a2-3a(a-4)a+b=1,得b=2a3-6a2+1,-1a1.令t(x)=2x3-6x2+1,x-1,1,所以t(x)=6x2-12x,令t(x)=0,解得x=2(舍去)或x=0.因?yàn)閠(-1)=-7,t(1)=-3,t(0)=1,所以t(x)的值域?yàn)?7,1.故b的取值范圍是-7,1.