20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：高考大題專項(xiàng)(一)-突破2-利用導(dǎo)數(shù)研究與函數(shù)零點(diǎn)有關(guān)的問題(編輯修改稿)
2025-04-05 06:02 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 g(b)2ga+b2中以b為主元構(gòu)造函數(shù),設(shè)F(x)=g(a)+g(x)2ga+x2,則F39。(x)=g39。(x)2ga+x239。=lnxlna+x2.當(dāng)0xa時(shí),F39。(x)0,因此F(x)在(0,a)上單調(diào)遞減,當(dāng)xa時(shí),F39。(x)0,因此F(x)在(a,+∞)上為增函數(shù),從而當(dāng)x=a時(shí),F(x)有極小值F(a).因?yàn)镕(a)=0,ba,所以F(b)0,即g(a)+g(b)2ga+b20.設(shè)G(x)=F(x)(xa)ln2,則G39。(x)=lnxlnx+a2ln2=lnxln(x+a),當(dāng)x0時(shí),G39。(x)0.因此G(x)在(0,+∞)上為減函數(shù).因?yàn)镚(a)=0,ba,所以G(b)0,即g(a)+g(b)2ga+b2(ba)ln2.注:本題以b為主元構(gòu)造函數(shù),當(dāng)然也可以以a為主元構(gòu)造函數(shù).突破2 利用導(dǎo)數(shù)研究與函數(shù)零點(diǎn)有關(guān)的問題關(guān)鍵能力學(xué)案突破例1解(1)略.(2)函數(shù)f(x)的定義域?yàn)镽,f39。(x)=ex+(x1)exkx=xexkx=x(exk),①當(dāng)0k≤1時(shí),令f39。(x)0,解得xlnk或x0.∴f(x)在(∞,lnk)和(0,+∞)上單調(diào)遞增,在[lnk,0]上單調(diào)遞減.由f(0)=1,當(dāng)x∈(∞,0)時(shí),f(x)≤f(x)max=f(lnk)=(lnk1)kk2ln2k=k2[(lnk1)2+1]0,此時(shí)f(x)無零點(diǎn),當(dāng)x∈[0,+∞)時(shí),f(2)=e22k≥e220.又f(x)在[0,+∞)上單調(diào)遞增,∴f(x)在[0,+∞)上有唯一的零點(diǎn),∴函數(shù)f(x)在定義域(∞,+∞)上有唯一的零點(diǎn).②當(dāng)k1時(shí),令f39。(x)0,解得x0或xlnk,∴f(x)在(∞,0)和(lnk,+∞)上單調(diào)遞增,在[0,lnk]上單調(diào)遞減.當(dāng)x∈(∞,lnk)時(shí),f(x)≤f(x)max=f(0)=10,此時(shí)f(x)無零點(diǎn).當(dāng)x∈[lnk,+∞)時(shí),f(lnk)f(0)=10,f(k+1)=kek+1k(k+1)22=kek+1(k+1)22.令g(t)=et12t2,t=k+12,則g39。(t)=ett,g″(t)=et1,∵t2,g″(t)0,g39。(t)在(2,+∞)上單調(diào)遞增,g39。(t)g39。(2)=e220,∴g(t)在(2,+∞)上單調(diào)遞增,得g(t)g(2)=e220,即f(k+1)0.∴f(x)在[lnk,+∞]上有唯一的零點(diǎn),故函數(shù)f(x)在定義域(∞,+∞)上有唯一的零點(diǎn).綜合①②可知,當(dāng)k0時(shí),函數(shù)f(x)在定義域(∞,+∞)上有且只有一個(gè)零點(diǎn).對(duì)點(diǎn)訓(xùn)練1解(1)F(x)=lnxx12x,即F(x)=lnx+12x12(x0),則F39。(x)=1x12x2=2x12x2,令F39。(x)=0,解得x=12.當(dāng)x∈0,12,F39。(x)0,F(x)在0,12上單調(diào)遞減。當(dāng)x∈12,+∞,F39。(x)0,F(x)在12,+∞∈(0,+∞)時(shí),F(x)min=F12=12ln2.因?yàn)?2ln2=lne12ln20,所以F(x)min0.又因?yàn)镕1e2=2+e2212=e2520,F(e)=1+12e12=12e+120,所以F1e2F120,F(e)F120,所以F(x)分別在區(qū)間1e2,12,12,e上各存在一個(gè)零點(diǎn),函數(shù)F(x)存在兩個(gè)零點(diǎn).(2)假設(shè)f(x)≥g(x)對(duì)任意x∈[1,+∞)恒成立,即lnxmxm2x≥0對(duì)任意x∈[1,+∞)恒成立.令h(x)=lnxmxm2x(x≥1),則h39。(x)=1xm2x2=2xm2x2.①當(dāng)m≤2,即2xm≥0時(shí),則h39。(x)≥0且h39。(x)不恒為0,所以函數(shù)h(x)=lnxmxm2x在區(qū)間[1,+∞)上單調(diào)遞增.又因?yàn)閔(1)=ln1m1m21=0,所以h(x)≥0對(duì)任意x∈[1,+∞)≤2不符合題意.②當(dāng)m2時(shí),令h39。(x)=2xm2x20,得1≤xm2。令h39。(x)=2xm2x20,得x(x)=lnxmxm2x在區(qū)間1,m2上單調(diào)遞減,在區(qū)間m2,+∞上單調(diào)遞增,所以hm2h(1)=0,即當(dāng)m2時(shí),存在x0≥1,使h(x0)0,即f(x0)g(x0).故m2符合題意.綜上可知,實(shí)數(shù)m的取值范圍是(2,+∞).例2證明(1)設(shè)g(x)=f39。(x),則g(x)=cosx11+x,g39。(x)=sinx+1(1+x)2.當(dāng)x∈1,π2時(shí),g39。(x)單調(diào)遞減,而g39。(0
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