2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 專題突破練7 應(yīng)用導(dǎo)數(shù)求參數(shù)的值或參數(shù)的范圍 文1.(2018遼寧撫順3月模擬,文21節(jié)選)已知函數(shù)f(x)=ax-2ln x(aR).(1)略;(2)若f(x)+x3>0對(duì)任意x(1,+)恒成立,求a的取值范圍.2.(2018安徽蕪湖期末,文21節(jié)選)已知函數(shù)f(x)=x3-aln x(aR).(1)略;(2)若函數(shù)y=f(x)在區(qū)間(1,e上存在兩個(gè)不同零點(diǎn),求實(shí)數(shù)a的取值范圍.3.已知函數(shù)f(x)=xln x,g(x)=-x2+ax-2(e為自然對(duì)數(shù)的底數(shù),aR).(1)判斷曲線y=f(x)在點(diǎn)(1,f(1)處的切線與曲線y=g(x)的公共點(diǎn)個(gè)數(shù);(2)當(dāng)x時(shí),若函數(shù)y=f(x)-g(x)有兩個(gè)零點(diǎn),求a的取值范圍.4.(2018寧夏石嘴山一模,文21)已知函數(shù)f(x)=ex+ax-a(aR且a0).(1)若函數(shù)f(x)在x=0處取得極值,求實(shí)數(shù)a的值;并求此時(shí)f(x)在-2,1上的最大值;(2)若函數(shù)f(x)不存在零點(diǎn),求實(shí)數(shù)a的取值范圍.5.(2018江西南昌一模,文21節(jié)選)已知函數(shù)f(x)=ex-aln x-e(aR),其中e為自然對(duì)數(shù)的底數(shù).(1)略(2)若當(dāng)x1,+)時(shí),f(x)0恒成立,求a的取值范圍.6.(2018山西太原一模,文21)已知函數(shù)f(x)=ln x-ax2+(2-a)x,g(x)=-2.(1)求函數(shù)f(x)的極值;(2)若對(duì)任意給定的x0(0,e,方程f(x)=g(x0)在(0,e上總有兩個(gè)不相等的實(shí)數(shù)根,求實(shí)數(shù)a的取值范圍.參考答案專題突破練7應(yīng)用導(dǎo)數(shù)求參數(shù)的值或參數(shù)的范圍1.解 (1)略.(2)由題意f(x)+x3>0,即a>-x2+對(duì)任意x(1,+)恒成立,記p(x)=-x2+,定義域?yàn)?1,+),則p'(x)=-2x+,設(shè)q(x)=-2x3+2-2ln x,q'(x)=-6x2-,則當(dāng)x>1時(shí),q(x)單調(diào)遞減,所以當(dāng)x>1時(shí),q(x)<q(1)=0,故p'(x)<0在(1,+)上恒成立,所以函數(shù)p(x)=-x2+在(1,+)上單調(diào)遞減,所以當(dāng)x>1時(shí),p(x)<p(1)=-1,得a-1,所以a的取值范圍是-1,+).2.解 (1)略.(2)由f(x)=0,得a=在區(qū)間(1,e上有兩個(gè)不同實(shí)數(shù)解,即函數(shù)y=a的圖象與函數(shù)g(x)=的圖象有兩個(gè)不同的交點(diǎn),因?yàn)間'(x)=,令g'(x)=0得x=.所以當(dāng)x(1,)時(shí),g'(x)<0,函數(shù)在(1,)上單調(diào)遞減,當(dāng)x(,e時(shí),g'(x)>0,函數(shù)在(,e上單調(diào)遞增;則g(x)min=g()=3e,而g()=27>27,且g(e)=e3<27,要使函數(shù)y=a的圖象與函數(shù)g(x)=的圖象有兩個(gè)不同的交點(diǎn),a的取值范圍為(3e,e3.3.解 (1)f'(x)=ln x+1,所以切線斜率k=f'(1)=1.又f(1)=0,所以曲線在點(diǎn)(1,0)處的切線方程為y=x-1.由得x2+(1-a)x+1=0.由=(1-a)2-4=a2-2a-3=(a+1)(a-3),可知:當(dāng)>0,即a<-1或a>3時(shí),有兩個(gè)公共點(diǎn);當(dāng)=0,即a=-1或a=3時(shí),有一個(gè)公共點(diǎn);當(dāng)<0,即-1<a<3時(shí),沒有公共點(diǎn).(2)y=f(x)-g(x)=x2-ax+2+xln x,由y=0,得a=x+ln x.令h(x)=x+ln x,則h'(x)=.當(dāng)x時(shí),由h'(x)=0,得x=1.所以h(x)在上單調(diào)遞減,在1,e上單調(diào)遞增,因此h(x)min=h(1)=3.由h+2e-1,h(e)=e+1,比較可知h>h(e),所以,結(jié)合函數(shù)圖象可得,當(dāng)3<ae+1時(shí),函數(shù)y=f(x)-g(x)有兩個(gè)零點(diǎn).4.解 (1)函數(shù)f(x)的定義域R,f'(x)=ex+a,f'(0)=e0+a=0,a=-1.在(-,0)上f'(x)<0,f(x)單調(diào)遞減,在(0,+)上f'(x)>0,f(x)單調(diào)遞增,所以當(dāng)x=0時(shí)f(x)取極小值.所以f(x)在-2,0)上單調(diào)遞增,在(0,1上單調(diào)遞減;又f(-2)=+3,f(1)=e,f(-2)>f(1).當(dāng)x=-2時(shí),f(x)在-2,1的最大值為+3.(2)f'(x)=ex+a由于ex>0,當(dāng)a>0時(shí),f'(x)>0,f(x)是增函數(shù),且當(dāng)x>1時(shí),f(x)=ex+a(x-1)>0,當(dāng)x<0時(shí),f(x)=ex+a(x-1)<1+a(x-1)<0,x<-+1,取x=-,則f<1+a=-a<0,所以函數(shù)f(x)存在零點(diǎn).當(dāng)a<0時(shí),f'(x)=ex+a=0,x=ln(-a).在(-,ln(-a)上f'(x)<0,f(x)單調(diào)遞減,在(ln(-a),+)上f'(x)>0,f(x)單調(diào)遞增,所以x=ln(-a)時(shí)f(x)取最小值.令f(x)min=f(ln(-a)<0,解得-e2<a<0.綜上所述:所求的實(shí)數(shù)a的取值范圍是-e2<a<0.5.解 (1)略.(2)由f(x)=ex-aln x-e(aR),得f'(x)=ex-,()當(dāng)a<0時(shí),f'(x)=ex->0,f(x)在x1,+)上遞增,f(x)min=f(1)=0(符合題意).()當(dāng)a>0時(shí),f'(x)=ex-=0,當(dāng)x1,+)時(shí),y=exe.當(dāng)a(0,e時(shí),因?yàn)閤1,+),所以y=e,f'(x)=ex-0,f(x)在1,+)上遞增,f(x)min=f(1)=0(符合題意).當(dāng)a(e,+)時(shí),存在x01,+),滿足f'(x)=ex-=0,f(x)在1,x0)上遞減,(x0,+)上遞增,故f(x0)<f(1)=0.不滿足x1,+)時(shí),f(x)0恒成立,綜上所述,a的取值范圍是(-,e.6.解 (1)函數(shù)的定義域是(0,+),f'(x)=-2ax+(2-a)=,當(dāng)a0時(shí),f'(x)>0,f(x)單調(diào)遞增,f(x)無極值;當(dāng)a>0時(shí),由f'(x)>0,得0<x<,故f(x)在遞增,在,+遞減,所以f(x)的極大值為f=ln-1.綜上所述,當(dāng)a0時(shí),f(x)無極值;當(dāng)a>0時(shí),f(x)的極大值為f=ln-1.(2)由g(x)=-2,得g'(x)=,當(dāng)x(-,1)時(shí),g'(x)>0,g(x)遞增,當(dāng)x(1,+)時(shí),g'(x)<0,g(x)遞減.當(dāng)x=1時(shí),g(x)max=g(1)=-2.x0(0,e,g(0)=-2,g(e)=-2<2,g(x).方程f(x)=g(x0)在(0,e上總有兩個(gè)不相等的實(shí)數(shù)根,f(e)=1-ae2+2e-ea2,a,由f=ln-1>-2,即ln a-<1,令h(a)=ln a-,可知h(a)遞增,且h(e)=1,h(a)<1=h(e),0<a<e.綜上所述,a<e.