2022年高考數(shù)學(xué)二輪復(fù)習(xí) 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 理1.(2018陜西咸陽一模,理21節(jié)選)已知f(x)=ex-aln x(aR).(1)略;(2)當(dāng)a=-1時,若不等式f(x)>e+m(x-1)對任意x(1,+)恒成立,求實數(shù)m的取值范圍.2.(2018山西太原一模,理21)f(x)=a(x-1),g(x)=(ax-1)ex,aR.(1)證明:存在唯一實數(shù)a,使得直線y=f(x)和曲線y=g(x)相切;(2)若不等式f(x)>g(x)有且只有兩個整數(shù)解,求a的范圍.3.已知函數(shù)f(x)=xln x,g(x)=-x2+ax-2(e為自然對數(shù)的底數(shù),aR).(1)判斷曲線y=f(x)在點(1,f(1)處的切線與曲線y=g(x)的公共點個數(shù);(2)當(dāng)x時,若函數(shù)y=f(x)-g(x)有兩個零點,求a的取值范圍.4.設(shè)函數(shù)f(x)=x2+ax+b,g(x)=ex(cx+d).若曲線y=f(x)和曲線y=g(x)都過點P(0,2),且在點P處有相同的切線y=4x+2.(1)求a,b,c,d的值;(2)若x-2時,f(x)kg(x),求k的取值范圍.5.(2018江西南昌一模,理21)已知函數(shù)f(x)=ln(ax)+bx在點(1,f(1)處的切線是y=0.(1)求函數(shù)f(x)的極值;(2)當(dāng)f(x)+x(m<0)恒成立時,求實數(shù)m的取值范圍(e為自然對數(shù)的底數(shù)).6.(2018山東濰坊一模,理21)函數(shù)f(x)=exsin x,g(x)=(x+1)cos x-ex.(1)求f(x)的單調(diào)區(qū)間;(2)對x1,x2,使f(x1)+g(x2)m成立,求實數(shù)m的取值范圍;(3)設(shè)h(x)=·f(x)-n·sin 2x在上有唯一零點,求正實數(shù)n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f(x)=ex-aln x,原不等式即為ex+ln x-e-m(x-1)>0,記F(x)=ex+ln x-e-m(x-1),F(1)=0,依題意有F(x)>0對任意x1,+)恒成立,求導(dǎo)得F'(x)=ex+-m,F'(1)=ex+1-m,F(x)=ex-,當(dāng)x>1時,F(x)>0,則F'(x)在(1,+)上單調(diào)遞增,有F'(x)>F'(1)=ex+1-m,若me+1,則F'(x)>0,若F(x)在(1,+)上單調(diào)遞增,且F(x)>F(1)=0,適合題意;若m>e+1,則F'(1)<0,又F'(ln m)=>0,故存在x1(1,ln m),使F'(x)=0,當(dāng)1<x<x1時,F'(x)<0,得F(x)在(1,x1)上單調(diào)遞減,F(x)<F(1)=0,舍去,綜上,實數(shù)m的取值范圍是me+1.2.解 (1)設(shè)切點為(x0,y0),則y0=a(x0-1)=(ax0-1),a(x0-x0+1)=,y=f(x)和y=g(x)相切,則a=g'(x0)=(a+ax0-1),a(x0-1)=,所以x0-x0+1=x0-1,即+x0-2=0.令h(x)=ex+x-2,h'(x)=ex+1>0,所以h(x)單遞遞增.又因為h(0)=-1<0,h(1)=e-1>0,所以,存在唯一實數(shù)x0,使得+x0-2=0,且x0(0,1).所以只存在唯一實數(shù)a,使成立,即存在唯一實數(shù)a使得y=f(x)和y=g(x)相切.(2)令f(x)>g(x),即a(x-1)>(ax-1)ex,所以a<1,令m(x)=x-,則m'(x)=,由(1)可知,m(x)在(-,x0)上單調(diào)遞減,在(x0,+)上單調(diào)遞增,且x0(0,1),故當(dāng)x0時,m(x)m(0)=1,當(dāng)x1時,m(x)m(1)=1,當(dāng)a<0時,要求整數(shù)解,m(x)在xZ時,m(x)1,am(x)<1有無窮多個整數(shù)解,舍去;當(dāng)0<a<1時,m(x)<,又>1,m(0)=m(1)=1,所以兩個整數(shù)解為0和1,即a,即a,當(dāng)a1時,m(x)<,因為1,m(x)在xZ內(nèi)大于或等于1,m(x)<無整數(shù)解,舍去,綜上,a3.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(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,有兩個公共點;當(dāng)=0,即a=-1或a=3時,有一個公共點;當(dāng)<0,即-1<a<3時,沒有公共點.(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,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,函數(shù)y=f(x)-g(x)有兩個零點.4.解 (1)由已知得f(0)=2,g(0)=2,f'(0)=4,g'(0)=4.而f'(x)=2x+a,g'(x)=ex(cx+d+c),故b=2,d=2,a=4,d+c=4.從而a=4,b=2,c=2,d=2.(2)由(1)知,f(x)=x2+4x+2,g(x)=2ex(x+1).設(shè)函數(shù)F(x)=kg(x)-f(x)=2kex(x+1)-x2-4x-2,則F'(x)=2kex(x+2)-2x-4=2(x+2)(kex-1).由題設(shè)可得F(0)0,即k1.令F'(x)=0得x1=-ln k,x2=-2.若1k<e2,則-2<x10.從而當(dāng)x(-2,x1)時,F'(x)<0;當(dāng)x(x1,+)時,F'(x)>0.即F(x)在(-2,x1)單調(diào)遞減,在(x1,+)單調(diào)遞增.故F(x)在-2,+)的最小值為F(x1).而F(x1)=2x1+2-4x1-2=-x1(x1+2)0.故當(dāng)x-2時,F(x)0,即f(x)kg(x)恒成立.若k=e2,則F'(x)=2e2(x+2)(ex-e-2).從而當(dāng)x>-2時,F'(x)>0,即F(x)在(-2,+)單調(diào)遞增.而F(-2)=0,故當(dāng)x-2時,F(x)0,即f(x)kg(x)恒成立.若k>e2,則F(-2)=-2ke-2+2=-2e-2(k-e2)<0.從而當(dāng)x-2時,f(x)kg(x)不可能恒成立.綜上,k的取值范圍是1,e2.5.解 (1)f(x)=ln(ax)+bx,f'(x)=+b=+b,點(1,f(1)處的切線是y=0,f'(x)=1+b=0,且f(1)=ln a+b=0,a=e,b=-1,即f(x)=ln x-x+1(x>0),f'(x)=-1=,f(x)在(0,1)上遞增,在(1,+)上遞減.所以f(x)的極大值為f(1)=ln e-1=0,無極小值.(2)由(1)知f(x)=ln x-x+1,當(dāng)f(x)+x(m<0)恒成立時,即ln x-x+1+x(m<0)在x(0,+)恒成立,同除以x得-2+設(shè)g(x)=,h(x)=-2,則g'(x)=,h'(x)=-,又m<0,當(dāng)0<x<1時,g'(x)<0,h'(x)>0;當(dāng)x>1時,g'(x)>0,h'(x)<0.g(x)在(0,1)上單調(diào)遞減,在(1,+)上單調(diào)遞增,g(x)min=g(1)=;h(x)在(0,1)上單調(diào)遞增,在(1,+)上單調(diào)遞減,h(x)max=h(1)=-1.g(x),h(x)均在x=1處取得最值,所以要使g(x)h(x)恒成立,只需g(x)minh(x)max,即-1,解得m1-e.又m<0,實數(shù)m的取值范圍是1-e,0).6.解 (1)f'(x)=exsin x+excos x=ex(sin x+cos x)=sinex,當(dāng)2kx+2k,即x時,f'(x)0,f(x)單調(diào)遞增;當(dāng)+2kx+2+2k,即x時,f'(x)<0,f(x)單調(diào)遞減;綜上,f(x)的單調(diào)遞增區(qū)間為+2k,kZ;f(x)的單調(diào)遞減區(qū)間為+2k,+2k,kZ.(2)f(x1)+g(x2)m,即f(x1)m-g(x2),設(shè)t(x)=m-g(x),則原問題等價于f(x)mint(x)min,x,一方面由(1)可知,當(dāng)x時,f'(x)0,故f(x)在單調(diào)遞增,f(x)min=f(0)=0.另方面:t(x)=m-(x+1)cos x+ex,t'(x)=-cos x+(x+1)sin x+ex,由于-cos x-1,0,ex,-cos x+ex>0.又(x+1)sin x0,當(dāng)x時,t'(x)>0,t(x)在為增函數(shù),t(x)min=t(0)=m-1+,所以m-1+0,m1-(3)h(x)=2xex-nsin 2x,x0,h'(x)=2(ex+xex)-2ncos 2x=2(x+1)ex-2ncos 2x.若0<n1,則h'(x)>0,h(x)單調(diào)遞增,h(x)>h(0)=0無零點,若n>1,設(shè)k(x)=2(x+1)ex-2ncos 2x,則k'(x)=2ex(x+2)+4nsin 2x>0,故k(x)單調(diào)遞增,k(0)=2-2n<0,k=2>0,存在x0,使k(x0)=0,因此當(dāng)x(0,x0)時,k(x)<0,即h'(x)<0,h(x)單調(diào)遞減;當(dāng)x時,k(x)>0,即h'(x)>0,h(x)單調(diào)遞增.故當(dāng)x(0,x0)時,h(x)<h(0)=0無零點,當(dāng)x時,h(x0)<0,h=>0,存在唯一零點,綜上,當(dāng)n>1時,有唯一零點.