適用于新高考新教材廣西專(zhuān)版2025屆高考數(shù)學(xué)一輪總復(fù)習(xí)第四章一元函數(shù)的導(dǎo)數(shù)及其應(yīng)用課時(shí)規(guī)范練18利用導(dǎo)數(shù)證明不等式

課時(shí)規(guī)范練18利用導(dǎo)數(shù)證明不等式1.(2023新高考Ⅰ,19)已知函數(shù)f(x)=a(ex+a)-x.(1)討論f(x)的單調(diào)性;(2)證明:當(dāng)a>0時(shí),f(x)>2lna+322.已知函數(shù)f(x)=(x-1)ex,g(x)=ax2+xlnx-12(1)判斷是否存在實(shí)數(shù)a,使得g(x)在x=1處取得極值?若存在,求出實(shí)數(shù)a;若不存在,請(qǐng)說(shuō)明理由;(2)若a≤12,當(dāng)x≥1時(shí),求證:f(x)≥g(x)3.(2023廣西玉林三模)設(shè)函數(shù)f(x)=(x+a)lnx+b,曲線(xiàn)y=f(x)在點(diǎn)(1,f(1))處的切線(xiàn)方程為x+y-2=0.(1)求y=f(x)的解析式;(2)證明:f(x)-4.已知函數(shù)f(x)=(x+1)lnx.(1)求曲線(xiàn)y=f(x)在點(diǎn)(1,f(1))處的切線(xiàn)方程;(2)證明:ln21+ln76+…+ln(n2-2)

課時(shí)規(guī)范練18利用導(dǎo)數(shù)證明不等式1.(1)解f'(x)=aex-1,x∈R.①當(dāng)a≤0時(shí),f'(x)≤0對(duì)任意x∈R恒成立,所以f(x)在(-∞,+∞)內(nèi)單調(diào)遞減.②當(dāng)a>0時(shí),令f'(x)=0,得x=ln1a=-lna隨x的變化,f'(x),f(x)的變化如下表:x(-∞,-lna)-lna(-lna,+∞)f'(x)-0+f(x)↘極小值↗所以函數(shù)f(x)的單調(diào)遞增區(qū)間是(-lna,+∞),單調(diào)遞減區(qū)間是(-∞,-lna).綜上,當(dāng)a≤0時(shí),f(x)的單調(diào)遞減區(qū)間是(-∞,+∞),無(wú)單調(diào)遞增區(qū)間;當(dāng)a>0時(shí),f(x)的單調(diào)遞增區(qū)間是(-lna,+∞),單調(diào)遞減區(qū)間是(-∞,-lna).(2)證明當(dāng)a>0時(shí),要證f(x)>2lna+32恒成立,即證f(x)min>2lna+32當(dāng)a>0時(shí),由(1)知,f(x)的極小值同時(shí)也是最小值,是f(-lna),下面證明f(-lna)>2lna+32f(-lna)=a(e-lna+a)-(-lna)=1+a2+lna.令g(a)=f(-lna)-2lna-32=a2-lna-12,a∈(0,+∞),則g'(a)=2a-令g'(a)=0,得a=22隨a的變化,g'(a),g(a)的變化如下表:a0,22222,+∞g'(a)-0+g(a)↘極小值↗所以在a=22時(shí),g(a)取最小值g(a)min=g22=12-ln22-12=-ln12=ln2因此f(-lna)>2lna+32成立因此當(dāng)a>0時(shí),f(x)>2lna+322.(1)解假設(shè)存在這樣的實(shí)數(shù)a,則有g(shù)'(1)=2a+1=0,解得a=-12當(dāng)a=-12時(shí),g(x)=-x22+xlnx-12,則g'(x)=-x+令t(x)=-x+lnx+1,x>0,則t'(x)=-1+1x令t'(x)=0,得x=1.當(dāng)0<x<1時(shí),t'(x)>0;當(dāng)x>1時(shí),t'(x)<0,所以t(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減,所以t(x)max=t(1)=0,即g'(x)≤0,所以函數(shù)g(x)在區(qū)間(0,+∞)上為減函數(shù),因此不存在這樣的實(shí)數(shù)a.(2)證明因?yàn)閤2≥0,a≤12,所以ax2≤12x要證f(x)≥g(x),即證(x-1)ex≥12x2+xlnx-1即證(x-1)ex-12x2-xlnx+12≥令m(x)=(x-1)ex-12x2-xlnx+12,x≥1,則m'(x)=xex-x-lnx-1=ex+lnx-(x+lnx)-令t=x+lnx,t∈[1,+∞),h(t)=et-t-1,則h'(t)=et-1.因?yàn)楫?dāng)t≥1,h'(t)>0,所以函數(shù)h(t)在區(qū)間[1,+∞)上單調(diào)遞增,所以h(t)min=h(1)=e-2>0,所以m'(x)>0,從而得函數(shù)m(x)在區(qū)間[1,+∞)上單調(diào)遞增,所以m(x)min=m(1)=0,所以m(x)≥0,即f(x)≥g(x)得證.3.(1)解∵f'(x)=lnx+1+ax∴f'(1)=1+a=-1,∴a=-2.又點(diǎn)(1,f(1))在切線(xiàn)x+y-2=0上,∴1+b-2=0,∴b=1,∴y=f(x)的解析式為f(x)=(x-2)lnx+1.(2)證明令g(x)=x-ex(x>0),則g'(x)=1-ex,∴當(dāng)x>0時(shí),g'(x)<0,則g(x)在(0,+∞)上單調(diào)遞減,∴g(x)<g(0)=-1<0,∴要證f(x)-1x-ex<1,即證f(即證f(x)-1>-1,即證f(x)>0.由(1)知f'(x)=lnx+1-2x,令h(x)=lnx+1-2x(則h'(x)=1x+∴h(x)在(0,+∞)上單調(diào)遞增,即f'(x)在(0,+∞)上單調(diào)遞增.又f'(1)=-1<0,f'(2)=ln2>0,∴存在x0∈(1,2),使得f'(x0)=0.當(dāng)0<x<x0時(shí),f'(x)<0,此時(shí)f(x)單調(diào)遞減,當(dāng)x>x0時(shí),f'(x)>0,此時(shí)f(x)單調(diào)遞增,故f(x)在x0處取到極小值,也是最小值.由f'(x0)=0,得lnx0=2x0∴f(x)≥f(x0)=(x0-2)lnx0+1=(x0-2)2x0-1+1=5-x0+4x0.令r(x)=x+4x(1<x<2),則對(duì)勾函數(shù)r(x)在區(qū)間(1,2)內(nèi)單調(diào)遞減∴r(x)<r(1)=5,于是f(x)≥5-x0+4x0>5-5=0.綜上,對(duì)任意x∈(0,+∞),f(x)-4.(1)解函數(shù)f(x)的定義域?yàn)?0,+∞),f'(x)=lnx+x+1x,所以曲線(xiàn)y=f(x)在點(diǎn)(1,f(1))處的切線(xiàn)斜率k=f'(1)=2,又因?yàn)閒(1)=0,所以該切線(xiàn)方程為y=2(x-(2)證明設(shè)F(x)=(x+1)lnx-2x+2(x>1),則F'(x)=lnx+1x-1令g(x)=F'(x),x>1,則g'(x)=1x-1x2=x-1x2,所以g(x)在區(qū)間(1,+∞)上單調(diào)遞增,又因?yàn)間(1)=0,所以g(x)=F'(x)>0,即F(x)在區(qū)間(1,+∞)上單調(diào)遞增,所以F(x)>F(1)=0,故當(dāng)x>1時(shí),(x+1)