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(參考版)
2025-04-05 06:02本頁面
【正文】 (k)0,函數(shù)h(k)單調(diào)遞增,當(dāng)kln2時(shí),h39。(x)0,解得1xek.∴g(x)在(1,ek)上單調(diào)遞減,在(ek,+∞)上單調(diào)遞增.∴g(x)min=g(ek)=kek(k+1)ek+2k=2kek0,令h(k)=2kek,k0,則h39。當(dāng)k0時(shí),令g39。(x)=lnxk,當(dāng)k≤0時(shí),g39。(x)=1xax1,∴f39。(x)0.故f(x)在∞,a3,(0,+∞)上單調(diào)遞增,在a3,0上單調(diào)遞減.(2)滿足題設(shè)條件的a,b存在.①當(dāng)a≤0時(shí),由(1)知,f(x)在[0,1]上單調(diào)遞增,所以f(x)在區(qū)間[0,1]上的最小值為f(0)=b,最大值為f(1)=2a+,b滿足題設(shè)條件當(dāng)且僅當(dāng)b=1,2a+b=1,即a=0,b=1.②當(dāng)a≥3時(shí),由(1)知,f(x)在[0,1]上單調(diào)遞減,所以f(x)在區(qū)間[0,1]上的最大值為f(0)=b,最小值為f(1)=2a+,b滿足題設(shè)條件當(dāng)且僅當(dāng)2a+b=1,b=1,即a=4,b=1.③當(dāng)0a3時(shí),由(1)知,f(x)在[0,1]上的最小值為fa3=a327+b,最大值為b或2a+b.若a327+b=1,b=1,則a=332,與0a3矛盾.若a327+b=1,2a+b=1,則a=33或a=33或a=0,與0a3矛盾.綜上,當(dāng)且僅當(dāng)a=0,b=1或a=4,b=1時(shí),f(x)在[0,1]上的最小值為1,最大值為1.對(duì)點(diǎn)訓(xùn)練4解(1)∵函數(shù)f(x)在[1,+∞)上單調(diào)遞增,∴f39。(x)0。當(dāng)x∈0,a3時(shí),f39。(x)=0,得x=0或x=a3.若a0,則當(dāng)x∈(∞,0)∪a3,+∞時(shí),f39。x2+2a(x+2)=2a0.故f(x)在(lna,+∞)上存在唯一零點(diǎn).從而f(x)在(∞,+∞)上有兩個(gè)零點(diǎn).綜上,a的取值范圍是1e,+∞.例4解(1)f39。(x)0.所以f(x)在(∞,lna)上單調(diào)遞減,在(lna,+∞)上單調(diào)遞增,故當(dāng)x=lna時(shí),f(x)取得最小值,最小值為f(lna)=a(1+lna).①若0a≤1e,則f(lna)≥0,f(x)在(∞,+∞)上至多存在1個(gè)零點(diǎn),不合題意.②若a1e,則f(lna)0.由于f(2)=e20,所以f(x)在(∞,lna)上存在唯一零點(diǎn).由(1)知,當(dāng)x2時(shí),exx20,所以當(dāng)x4且x2ln(2a)時(shí),f(x)=ex2(x)0。(x)0,所以f(x)在(∞,+∞)上單調(diào)遞增,故f(x)至多存在1個(gè)零點(diǎn),不合題意.當(dāng)a0時(shí),由f39。(x)0.所以f(x)在(∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.(2)f39。(x)0。(x)0,所以g(x)在(∞,0)上單調(diào)遞增,在(0,+∞)上單調(diào)遞減.g(0)=1,當(dāng)x→∞時(shí),g(x)→∞,當(dāng)x→+∞時(shí),g(x)→0+.函數(shù)g(x)的簡圖如圖所示.若f(x)有兩個(gè)零點(diǎn),則y=a與g(x)有兩個(gè)交點(diǎn),所以a的取值范圍是(0,1).對(duì)點(diǎn)訓(xùn)練3解(1)當(dāng)a=1時(shí),f(x)=exx2,則f39。(x)=ex+10,所以h(x)在R上單調(diào)遞增,而h(0)=0,所以當(dāng)x0時(shí),h(x)0,當(dāng)x0時(shí),h(x)0,于是當(dāng)x0時(shí),g39。因?yàn)閒ln3a1=a3a12+(a2)3a1ln3a1=3a1ln3a10,且ln3a1ln1a,所以f(x)在ln1a,+∞上有1個(gè)零點(diǎn).綜上所述,a的取值范圍為(0,1).(方法2)ae2x+(a2)exx=0?ae2x+aex=2ex+x?a=2ex+xe2x+ex.令g(x)=2ex+xe2x+ex,則g39。(x)0得xln1a,所以f(x)在∞,ln1a上單調(diào)遞減,在ln1a,+∞上單調(diào)遞增.(2)(方法1)①當(dāng)a≤0時(shí),由(1)可知,f(x)在R上單調(diào)遞減,不可能有兩個(gè)零點(diǎn).②當(dāng)a0時(shí),f(x)min=fln1a=11a+lna,令g(a)=f(x)min,則g39。(x)0,所以f(x)在R上單調(diào)遞減.②當(dāng)a0時(shí),由f39。(x)0.∴g(x)在(0,x1)和(x2,π)上單調(diào)遞減,在(x1,x2)上單調(diào)遞增.∵g(0)=0,∴g(x1)0.∵gπ2=eπ2π0,∴g(x2)(π)=2π0,由零點(diǎn)存在定理得,g(x)在(x1,x2)和(x2,π)上各有一個(gè)零點(diǎn),∴函數(shù)g(x)在(0,π)上有兩個(gè)零點(diǎn).例3解(1)f39。(x)0。(x1)=0,g39。(π)=eπ20,∴g39。(0)=120,g39。當(dāng)x∈π2,π時(shí),g″(x)0.∴g39。(x)=ex(sinx+cosx)2,∴g″(x)=2
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