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1、課后限時(shí)集訓(xùn)19利用導(dǎo)數(shù)解決函數(shù)的零點(diǎn)問(wèn)題建議用時(shí):45分鐘1(2019全國(guó)卷)已知函數(shù)f(x)(x1)ln xx1.證明:(1)f(x)存在唯一的極值點(diǎn);(2)f(x)0有且僅有兩個(gè)實(shí)根,且兩個(gè)實(shí)根互為倒數(shù)解(1)證明:f(x)的定義域?yàn)?0,)f(x)ln x1ln x.因?yàn)閥ln x在(0,)上單調(diào)遞增,y在(0,)上單調(diào)遞減,所以f(x)在(0,)上單調(diào)遞增又f(1)10,f(2)ln 20,故存在唯一x0(1,2),使得f(x0)0.又當(dāng)xx0時(shí),f(x)x0時(shí),f(x)0,f(x)單調(diào)遞增,因此,f(x)存在唯一的極值點(diǎn)(2)證明:由(1)知f(x0)0,所以,f(x)0在(x0,
2、)內(nèi)存在唯一根x.由x01得1x0.又fln 10,故是f(x)0在(0,x0)的唯一根綜上,f(x)0有且僅有兩個(gè)實(shí)根,且兩個(gè)實(shí)根互為倒數(shù)2已知函數(shù)f(x)x3x2axb.(1)當(dāng)a1時(shí),求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)若函數(shù)f(x)的圖像與直線yax恰有兩個(gè)不同的交點(diǎn),求實(shí)數(shù)b的值解(1)當(dāng)a1時(shí),f(x)x3x2xb,則f(x)3x22x1,由f(x)0,得x1或x,所以函數(shù)f(x)的單調(diào)遞增區(qū)間為(,1)和.(2)函數(shù)f(x)的圖像與直線yax恰有兩個(gè)不同的交點(diǎn),等價(jià)于f(x)ax0有兩個(gè)不等的實(shí)根令g(x)f(x)axx3x2b,則g(x)3x22x.由g(x)0,得x或x0;由
3、g(x)0,得x0.所以函數(shù)g(x)在和(0,)上單調(diào)遞增,在上單調(diào)遞減所以當(dāng)x時(shí),函數(shù)g(x)取得極大值gb;當(dāng)x0時(shí),函數(shù)g(x)取得極小值為g(0)b.要滿足題意,則需gb0或g(0)b0,所以b或b0.3(2019武漢調(diào)研)已知函數(shù)f(x)exax1(aR)(e2.718 28是自然對(duì)數(shù)的底數(shù))(1)求f(x)的單調(diào)區(qū)間;(2)討論g(x)f(x)在區(qū)間0,1上零點(diǎn)的個(gè)數(shù)解(1)f(x)exax1,f(x)exa,當(dāng)a0時(shí),f(x)0恒成立,f(x)的單調(diào)遞增區(qū)間為(,),無(wú)單調(diào)遞減區(qū)間;當(dāng)a0時(shí),令f(x)0,得xln a,令f(x)0,得xln a,f(x)的單調(diào)遞減區(qū)間為(,ln
4、 a),單調(diào)遞增區(qū)間為(ln a,)(2)令g(x)0,得f(x)0或x,先考慮f(x)在區(qū)間0,1上的零點(diǎn)個(gè)數(shù),當(dāng)a1時(shí),f(x)在(0,)上單調(diào)遞增且f(0)0,f(x)在0,1上有一個(gè)零點(diǎn)當(dāng)ae時(shí),f(x)在(,1)上單調(diào)遞減,f(x)在0,1上有一個(gè)零點(diǎn)當(dāng)1ae時(shí),f(x)在(0,ln a)上單調(diào)遞減,在(ln a,1)上單調(diào)遞增而f(1)ea1,當(dāng)ea10,即1ae1時(shí),f(x)在0,1上有兩個(gè)零點(diǎn);當(dāng)ea10,即e1ae時(shí),f(x)在0,1上有一個(gè)零點(diǎn)再考慮x時(shí),由f0,得a2(1)綜上所述,當(dāng)a1或ae1或a2(1)時(shí),g(x)在0,1上有兩個(gè)零點(diǎn);當(dāng)1ae1且a2(1)時(shí),g(x)在0,1上有三個(gè)零點(diǎn).- 3 -