新高考數(shù)學(xué)一輪復(fù)習(xí)專題三導(dǎo)數(shù)及其應(yīng)用微專題一隱零點(diǎn)問題練習(xí)含答案

微專題一隱零點(diǎn)問題1.(2024廣東茂名二模,16)已知函數(shù)f(x)=exsinx-ax.(1)若曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為x+y=0,求實(shí)數(shù)a的值;(2)若a=32,求函數(shù)f(x)在區(qū)間0,π解析(1)因?yàn)閒(x)=exsinx-ax,所以f'(x)=ex(sinx+cosx)-a,所以f'(0)=1-a.因?yàn)榍€y=f(x)在點(diǎn)(0,f(0))處的切線方程為x+y=0,所以f'(0)=-1,所以1-a=-1,所以a=2.(2)當(dāng)a=32時(shí),令h(x)=f'(x)=ex(sinx+cosx)-3h'(x)=ex(sinx+cosx+cosx-sinx)=2excosx,當(dāng)x∈0,π2時(shí),h'(x)≥0,h(x)又h(0)=1-32=-12<0,hπ2=eπ2-所以存在唯一的x0∈0,π2,使得h(x0)當(dāng)x∈[0,x0)時(shí),h(x)<0,f(x)單調(diào)遞減;當(dāng)x∈x0,π2時(shí),h(x)>0,f(又f(0)=0,fπ2=eπ2-3π4>e-10所以f(x)max=fπ2=eπ22.(2024遼寧撫順一模,19)已知函數(shù)f(x)=xex+1-ax2-2ax.(1)當(dāng)a=12時(shí),判斷f(x)的單調(diào)性(2)若x∈(0,+∞)時(shí),f'(x-1)≥1+lnx-x恒成立,求實(shí)數(shù)a的取值范圍.解析(1)函數(shù)f(x)的定義域?yàn)镽,f'(x)=ex+1+xex+1-2ax-2a=(x+1)(ex+1-2a),當(dāng)a=12時(shí),f'(x)=(x+1)(ex+1-1)當(dāng)x+1≥0時(shí),有ex+1-1≥0,得f'(x)≥0,函數(shù)f(x)在區(qū)間[-1,+∞)上單調(diào)遞增;當(dāng)x+1<0時(shí),有ex+1-1<0,得f'(x)>0,函數(shù)f(x)在區(qū)間(-∞,-1)上單調(diào)遞增.綜上,當(dāng)a=12時(shí),f(x)在R上單調(diào)遞增(2)當(dāng)x∈(0,+∞)時(shí),f'(x-1)≥1+lnx-x恒成立,即2a≤xex因此只需2a≤xe令g(x)=xex?1?lnx+xx,x>0,則令h(x)=x2ex+lnx,x>0,則h'(x)=(x2+2x)ex+1x>0所以h(x)在區(qū)間(0,+∞)上單調(diào)遞增.因?yàn)閔(1)=e>0,h12=14e12-ln2<0,所以h(x)在(0,+∞)內(nèi)存在唯一零點(diǎn)x0,且所以當(dāng)0<x<x0時(shí),h(x)<0,即g'(x)<0,函數(shù)g(x)在區(qū)間(0,x0)上單調(diào)遞減,當(dāng)x>x0時(shí),h(x)>0,即g'(x)>0,函數(shù)g(x)在區(qū)間(x0,+∞)上單調(diào)遞增.又因?yàn)閔(x0)=x02ex0即x0ex0=-lnx0x0=-1x0lnx0=設(shè)m(x)=xex,當(dāng)x>0時(shí),m'(x)=(x+1)ex>0,所以m(x)在(0,+∞)上單調(diào)遞增,因?yàn)閤0∈12,1,所以x0>0,-lnx0由(*)知m(x0)=m(-lnx0),所以x0=-lnx0,即ex0=所以g(x)min=g(x0)=x0ex0所以2a≤2,即a≤1,故實(shí)數(shù)a的取值范圍為(-∞,1].3.(2024江西贛州一模,19)已知函數(shù)f(x)=ex-1-lnx.(1)求f(x)的單調(diào)區(qū)間;(2)已知m>0.若函數(shù)g(x)=f(x)-m(x-1)有唯一的零點(diǎn)x0.證明:1<x0<2.解析(1)易知函數(shù)的定義域?yàn)?0,+∞).∵f(x)=ex-1-lnx,∴f'(x)=ex-1-1x令l(x)=ex-1-1x,x>0,則l'(x)=ex-1+1x∴當(dāng)x>0時(shí),l(x)即f'(x)為增函數(shù),又∵f'(1)=0,∴當(dāng)x∈(0,1)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(1,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,∴f(x)的單調(diào)遞減區(qū)間為(0,1),單調(diào)遞增區(qū)間為(1,+∞).(2)證明:∵g(x)=f(x)-m(x-1)=ex-1-lnx-mx+m(m>0),∴g'(x)=ex-1-1x-m(x>0,m>0由(1)可知g'(x)在(0,+∞)上單調(diào)遞增,且g'(1)=-m<0,又g'(1+m)=em-11+m-m>em-(m+1)∴存在唯一的t∈(1,1+m)?(1,+∞)使得g'(t)=0,∴當(dāng)x∈(0,t)時(shí)g'(t)<0,g(x)單調(diào)遞減;當(dāng)x∈(t,+∞)時(shí)g'(t)>0,g(x)單調(diào)遞增,∴g(x)min=g(t)=et-1-lnt-mt+m.若g(x)=ex-1-lnx-mx+m=0有唯一的實(shí)數(shù)解x0,則x0=t>1,∴g'消去m可得(2-t)et-1-lnt+1-1t=0(t>1令h(t)=(2-t)et-1-lnt+1-1t(t>1則h'(t)=(1-t)et-1-1t+1t2=(1-t)∴h(t)在t∈(1,+∞)上為減函數(shù),且h(1)=1>0,h(2)=12-ln2<0∴當(dāng)h(t)=0時(shí)t∈(1,2),即1<x0<2.4.(2024四川南充二模)設(shè)函數(shù)f(x)=x?2x+1ex,g(x)(1)若函數(shù)f(x)在區(qū)間(a-1,a+2)上是單調(diào)函數(shù),求a的取值范圍;(2)設(shè)0≤m<e2,證明:函數(shù)g(x)在區(qū)間(0,+∞)上存在最小值A(chǔ),且e2<A≤解析(1)f(x)=x?2x+1ex的定義域?yàn)?-∞,-1)∪(-1,+∞),f'(x)=x+1?x+2(x+1)2ex+所以函數(shù)f(x)在(-∞,-1),(-1,+∞)上單調(diào)遞增.由函數(shù)f(x)在區(qū)間(a-1,a+2)上是單調(diào)函數(shù),得a+2≤-1或a-1≥-1,即a≤-3或a≥0.故a的取值范圍是(-∞,-3]∪[0,+∞).(2)證明:由g(x)=2ex?2mx得g'(x)=2xex因?yàn)閤>0,所以x+1x設(shè)φ(x)=x?2x+1ex+m,x>0,由(1)得φ(x)=x?2x+1ex+m在(又0≤m<e2,則φ(1)=-e2+m<0,φ(2)=m所以?x0∈(1,2],使得φ(x0)=x0?2x0+1ex0+且當(dāng)x<x0時(shí),g