（黃岡名師）高考數(shù)學(xué)大題規(guī)范滿分練一理（含解析）新人教A版.docx

大題規(guī)范滿分練(一)函數(shù)與導(dǎo)數(shù)綜合問題1.(2018全國卷)已知函數(shù)f=-x+aln x.(1)討論f的單調(diào)性.(2)若f存在兩個極值點(diǎn)x1,x2,證明:2,令f(x)=0得,x=或x=.當(dāng)x,+時,f(x)0.所以f(x)在,上單調(diào)遞減,在上單調(diào)遞增.(2)由(1)知,f(x)存在兩個極值點(diǎn),當(dāng)且僅當(dāng)a2.由于f(x)的兩個極值點(diǎn)x1,x2滿足x2-ax+1=0,所以x1x2=1,不妨設(shè)x11.由于=-1+a=-2+a=-2+a,所以a-2等價于-x2+2ln x20.設(shè)函數(shù)g(x)=-x+2ln x,由(1)知,g(x)在(0,+)上單調(diào)遞減,又g(1)=0,從而當(dāng)x(1,+)時,g(x)0.所以-x2+2ln x20,即0.(1)求f(x)的單調(diào)區(qū)間和極值.(2)證明:若f(x)存在零點(diǎn),則f(x)在區(qū)間(1,上僅有一個零點(diǎn).【解析】(1)由f(x)=-kln x(k0),得x0且f(x)=x-=.由f(x)=0,解得x=(負(fù)值舍去).f(x)與f(x)在區(qū)間(0,+)上的情況如下:x(0,)(,+)f(x)-0+f(x)所以f(x)的單調(diào)遞減區(qū)間是(0,),單調(diào)遞增區(qū)間是(,+).f(x)在x=處取得極小值f()=,無極大值.(2)由(1)知,f(x)在區(qū)間(0,+)上的最小值為f()=.因為f(x)存在零點(diǎn),所以0,從而ke.當(dāng)k=e時,f(x)在區(qū)間(1,上單調(diào)遞減,且f()=0,所以x=是f(x)在區(qū)間(1,上的唯一零點(diǎn).當(dāng)ke時,f(x)在區(qū)間(0,上單調(diào)遞減,且f(1)=0,f()=0,所以f(x)在區(qū)間(1,上僅有一個零點(diǎn).綜上可知,若f(x)存在零點(diǎn),則f(x)在區(qū)間(1,上僅有一個零點(diǎn).3.已知函數(shù)f(x)=ax+ln x,其中a為常數(shù).(1)當(dāng)a=-1時,求f(x)的單調(diào)遞增區(qū)間;(2)當(dāng)0-0,當(dāng)a=-1時,f(x)=-x+ln x(x0),f(x)=(x0);當(dāng)0x0;當(dāng)x1時,f(x)0),令f(x)=0,解得x=-;由f(x)0,解得0x-;由f(x)0,解得-xe.從而f(x)的單調(diào)遞增區(qū)間為0,-,遞減區(qū)間為-,e,所以f(x)max=f-=-1+ln-=-3.解得a=-e2.(3)由(1)知當(dāng)a=-1時,f(x)max=f(1)=-1,所以|f(x)|1.令g(x)=+,則g(x)=.當(dāng)0x0;當(dāng)xe時,g(x)0.從而g(x)在(0,e)上單調(diào)遞增,在(e,+)上