第十一節(jié)導(dǎo)數(shù)切線放縮問(wèn)題總結(jié)講義-高二下學(xué)期數(shù)學(xué)人教A版選擇性（答案詳解）

第十一節(jié)專題:導(dǎo)數(shù)的切線放縮問(wèn)題重點(diǎn)題型專練【1】?e,+∞解析:方法1:切線法?即y=ex的圖像在又因?yàn)椤呒磞=ex與y只需要?a方法2:分離參數(shù)法原題等價(jià)于?x①x=0時(shí),f②x>0時(shí),a>?exx令gx=?exx∴gx在(0,1)上遞增,在[∴g【2】[解析:方法1:切線法?即?x即y=lnx在y又∵x>0時(shí),∴只需m?1≥?1方法2:分離參數(shù)法?x>0,ln令gx=lnx?x∴gx在(0,1)上遞增,在[1,+∞)【3】e,+∞解析:切線法:即y=ax與y=ln∵x>0時(shí),lnx≤【4】[解析:切線法:∵?x>令t=即?t>即?t>即y=lnt在y∵t>lnt≤t?1【5】詳見(jiàn)解析解析:方法1:單獨(dú)放縮lnx證明:當(dāng)x>0時(shí),lnx≤x?1(放縮),∴ex?lnx≥ex?x?1,只需證x>0時(shí),ex?x?1>2方法2:單獨(dú)放縮ex證明:當(dāng)x>0時(shí),ex>x+1(放縮),∵ex≥x+1∴ex?lnx>x+1?lnx,故只需證x>0時(shí),x方法3:同時(shí)放縮ex與lnx證明:當(dāng)x>0時(shí),ex【6】證明見(jiàn)解析:因?yàn)閙≤2,要證fx>0,只要滿足ex?lnx+2>0即可.設(shè)gx=ex?lnx+2,則只需證明gx=ex?lnx+2>0即可.事實(shí)上,當(dāng)x>0時(shí),lnx≤x?1,故解析:∵x>0時(shí),令?x=1?xlnx?x∵?令?′x>0得lnx<?2,∴0∴?xmax=?e?2【8】ABD解析:對(duì)于A,設(shè)fx=1?1當(dāng)x>1時(shí),f當(dāng)0<x<1時(shí),所以當(dāng)x=1時(shí),函數(shù)有最大值,即fx=1?1x?lnx對(duì)于B,設(shè)Fx=x?1當(dāng)x>1時(shí),F當(dāng)0<x<1時(shí),所以當(dāng)x=1時(shí),函數(shù)有最小值,即Fx=x?1?lnx≥對(duì)于C,設(shè)gx=ex?當(dāng)x>0時(shí),g′x>0,gx所以當(dāng)x=0時(shí),函數(shù)有最小值,即gx=ex?x對(duì)于D,設(shè)Gx因?yàn)?<x<π,所以所以當(dāng)0<x<π時(shí),有Gx<G0=0,即【9】BD解析:A選項(xiàng),令fx=lnf當(dāng)0<x<3時(shí),f′x又f0=0,∴B選項(xiàng),令fx則f′x所以fx=ln又f1=0,所以當(dāng)x∈0,1時(shí),fx>fC選項(xiàng),當(dāng)x=3時(shí),e3>1+3+32D選項(xiàng),令fx=cosx?1令?x=f′x=?sinf′x又f′0=0,所以當(dāng)x>0時(shí),f當(dāng)x<0時(shí),f′x<0f因此cosx≥1?12x【10】證明見(jiàn)解析解析:(1)令fx=sinx?x所以fx在0,π2單調(diào)遞減,故fx(2)令gx則g=由(1)知x>sin∴g則gx在0,π2單調(diào)遞增,故gx(3)令?x則?′由(2)知:3x<tanx+2sin則?=1所以?x在0,π2單調(diào)遞增,故?x【11】(1)單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為1,+∞(1)由題意,函數(shù)fx=lnx?x+1f當(dāng)x>1時(shí),f′x<0;當(dāng)0<x<1時(shí),f′x>0.所以fx在(0,1)上單調(diào)遞增,在1,+∞上單調(diào)遞減,即fx的單調(diào)遞增區(qū)間為(0,1),單調(diào)遞減區(qū)間為1,+∞;(2)證明:由(1)得fx在(0,1)上單調(diào)遞增,在1,+∞上單調(diào)遞減,所以fx即x+12>0【12】詳見(jiàn)解析解析:∵點(diǎn)1,f∴f∵點(diǎn)(1,e)在曲線上∴∴f(2)證明:(2)由(1)得fx∵x>0時(shí),即可證x+1?lnxx>1,即證x>令gx=x2∴gx【13】(1)fx的單調(diào)遞增區(qū)間是1,+∞,單調(diào)遞減區(qū)間是(0,1)(1)當(dāng)a=1時(shí),fx=2xf由f′x>0可得x>1;由所以fx的單調(diào)遞增區(qū)間是1,+∞(2)設(shè)gx=lnx?x由g′x>0可得0<x<1;所以gx在(0,1)上單調(diào)遞增,在1,+∞故gx≤g1=0,即當(dāng)且僅當(dāng)x=1要證fx≥0,即2x+a因?yàn)閍≥4,x>0,所以當(dāng)且僅當(dāng)x=2因?yàn)棰佗谌〉玫忍?hào)的條件不同,所以當(dāng)a≥4時(shí),【14】(1)m≥?1解析:(1)令Fx則F所以Fx在1,+∞上單調(diào)遞減,在(0,1)上單調(diào)遞增,Fx在x=1若fx≤gx恒成立,則?1(2)證明:由(1)可知lnx≤x?1要證ex+2?ex設(shè)?x=ex?設(shè)mx則m′x=ex?2,易得?′x又?′0=3?e>0,?x0∈0,ln2所以當(dāng)x∈0,x0∪1,+∞時(shí),?′故?x在0,x0上單調(diào)遞增,在x0,又?0=?1=因此,當(dāng)x>0時(shí),故當(dāng)x>0時(shí),【15】(1)y=?x；(2)a解析:(1)a=2時(shí),fx=e所以k=而f0所以切線方程為y=?(2)因?yàn)閒x=ex?axf當(dāng)a≤1時(shí),在[0,+∞)上,f′x∴fx在[0,+∞)上單調(diào)遞增,又x∈[0當(dāng)a>1時(shí),令f′x=ex?時(shí),f′x<0,∴fx在0,ln綜上所示,實(shí)數(shù)a的取值范圍是a≤(3)由(2)得,a=1時(shí),fx≥0對(duì)于任意的又∵x≥0時(shí),∴為證exlnx+1令gx則g′當(dāng)x>0時(shí)g′x=lnx又∵g0=0+1ln0+1所以當(dāng)x≥0時(shí),e【16】(1)a=1解析:(1)設(shè)Fx=ex?當(dāng)a≤0時(shí),Fx在R上單調(diào)遞增,且F0=0,當(dāng)x<0當(dāng)a>0時(shí),令F′x≥0,則x≥lna;令F′x<0,則x<lna故Fx若Fx≥0,則只需Fx最小值則?′a=1?ln1,+∞上單調(diào)遞減,故?因此,只有當(dāng)a=1時(shí)滿足題意,即(2)由(1)知,?t∈R,et令t=x?lnx,代入可得ex由(1)知,ex≥x+1>x,即因此exx>x?ln【17】(1)函數(shù)fx的極小值為解析:(1)∵f∴f′x=ex?2?當(dāng)x>2時(shí),f′x>0,函數(shù)f當(dāng)x<2時(shí),f′x<0,函數(shù)f∴當(dāng)x=2時(shí),函數(shù)fx取極小值,極小值為1,函數(shù)(2)設(shè)gx=x?1當(dāng)x>1時(shí),g′x>0,函數(shù)g當(dāng)0<x<1時(shí),g′x∴gx≥g要證明fx>xlnx只需證明4ex?2?只需證明ex設(shè)?x=ex?ex,則?′x=當(dāng)x>1時(shí),?′x>0,函數(shù)?當(dāng)0<x<1時(shí),?′x∴?∴當(dāng)x>0時(shí)e∴f【18】(1)當(dāng)0<t≤1?ee時(shí),fxmin=et?lnt;當(dāng)(2)見(jiàn)詳解解析:(1)由fx=ex?lnx令f′x>0,得ex?1>0,即x>1e,因此函數(shù)fx①當(dāng)0<t+1≤1e,即0<t≤1?e②當(dāng)t≥1e時(shí),函數(shù)fx在f③當(dāng)t<1e<t+1,即1?ee1e,t+1綜上所述,當(dāng)0<t≤1?ee時(shí),fxmin=et?lnt;當(dāng)(2)證明:設(shè)?x=ex?ex,x∈0,+∞,則?′x=ex?e要證xfx<gx因?yàn)閑x≥ex,所以故只需證ex2?xlnx≤ex令φx=xlnx,則φ′x=lnx+在1e,+∞上單調(diào)遞增,因此φx≥φ因此當(dāng)x>0時(shí),【19】(1)答案見(jiàn)解析(2)證明見(jiàn)解析解析:(1)解:gx∴g當(dāng)x∈0,1∴當(dāng)1?2a≥0?∴gx∴g當(dāng)1?2agx在0,ln2a單調(diào)遞減,在∴g當(dāng)e?2a≤0?a≥e2∴g綜上所述:a≤12時(shí),12<a<ea≥e2時(shí),(2)解:由(1)知,當(dāng)a≤12時(shí),∴b=1且x>0即ex要證不等式ex?1lnx只需證明x2+2x2>設(shè)Fx則F′所以當(dāng)x>0時(shí),F故Fx在0,+∞上單調(diào)遞增,又∴Fx∴原不等式成立.【20】(1)y=e解析:(1)因?yàn)閒x=ex?x2,所以∴fx在x=1(2)f′x=ex?2x,f′′所以f′x≥f′ln2=2?所以fxmax=f1=e?1,x∈0,1.故可猜測(cè):當(dāng)x>0,x≠1時(shí),f下證:當(dāng)x>0時(shí),設(shè)gx=g′x=ex?ln2,+∞g′所以,存在x0∈0,ln2所以,當(dāng)x∈0,x0∪1,+∞時(shí),g′x>0;當(dāng)x∈x0,1時(shí),g′又g0=g1=0,∴gx=ex?x2?e?2x?1增,在1,+∞上單調(diào)遞減,所以?xmax=?1立,即x≥lnx+1,即ex+【21】(1)單調(diào)遞增區(qū)間為1,+∞(2)[(3)證明見(jiàn)解析解析:(1)當(dāng)a=1時(shí),顯然x>1時(shí),f′x>0,此時(shí)f0<x<1時(shí),f′x即fx的單調(diào)遞增區(qū)間為1,+∞(2)由題意知f′若a≤0?f′x即fx≤若1>a>0,則1a>1,易知x∈1,則f1a若a≥1,則1a≤1,x∈1,+∞所以fx≥綜上,a(3)由(1)、(2)可知lnx所以lnx即lnn所以ln2累加得lnn+【22】(1)答案見(jiàn)解析(2)最小值為e,證明見(jiàn)解析解析:(1)fx的定義域?yàn)?①當(dāng)m≤0時(shí),f′x<0恒成立,∴②當(dāng)m>0時(shí),當(dāng)x>1m時(shí),f′x>0∴fx在0,1m綜上可得:當(dāng)m≤0時(shí),fx在當(dāng)m>0時(shí),fx在0,1m(2)當(dāng)m=1時(shí),由(1)可知fx在(0,1]故fx所以當(dāng)m=1時(shí),函數(shù)fx的最小值為因?yàn)閤?lnx+e?1≥e當(dāng)x≥2時(shí),0<lnx<即1ln令x=n,則所以1ln故當(dāng)n≥2時(shí),即k【23】(1)(?∞,1]解析:(1)對(duì)任意的x≥1,fx設(shè)gx=lnx?a當(dāng)a≤1時(shí),g′x≥0對(duì)任意的x≥1恒成立,此時(shí),函數(shù)gx在[1,+∞)上單調(diào)遞增,∴gx≥g1=0滿足條件;當(dāng)a>1時(shí),令g′x=0,得x=a,當(dāng)1≤x<a時(shí),g′x<0;當(dāng)x>a時(shí),g′x>0,∴gx在區(qū)間1,a上單調(diào)遞減,在區(qū)間[a,+∞)上單調(diào)遞增,∴gxmin=ga<g1=0,與已知矛盾.綜上所述,a的取值范圍是(?∞,1];(2)由(1)可知,當(dāng)a=1,x>1時(shí)fx>0,即xlnx?x+1>0對(duì)x>1恒成立,故有l(wèi)nx>x?1x.令x=nn?1>1n≥2,得lnnn?1>1n,所以