（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 專項(xiàng)強(qiáng)化練四 導(dǎo)函數(shù)不可求零點(diǎn)的導(dǎo)數(shù)綜合問題.docx

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專項(xiàng)強(qiáng)化練四導(dǎo)函數(shù)不可求零點(diǎn)的導(dǎo)數(shù)綜合問題1.已知函數(shù)f(x)=x2ex-ln x,求證:當(dāng)x0時(shí),不等式f(x)1.證明f (x)=x(x+2)ex-1x,x0.則f (x)=(x2+4x+2)ex+1x20,故f (x)在(0,+)上單調(diào)遞增.又f 14=916e14-40,根據(jù)零點(diǎn)存在性定理可知,存在x014,12,使得f (x0)=0.當(dāng)x(0,x0)時(shí), f (x)0,故f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=x02ex0-ln x0.由f (x0)=0,得x0(x0+2)ex0-1x0=0,即x0(x0+2)ex0=1x0,ex0=1x02(x0+2).故f(x0)=x02ex0-ln x0=1x0+2-ln x0,其中x014,12.令g(x)=1x+2-ln x,x14,12.則g(x)=-1(x+2)2-1xg12=25-ln121,即f(x0)1.綜上,有f(x)min1,則當(dāng)x0時(shí),不等式f(x)1.2.已知函數(shù)f(x)=e2x-aln x,求證:當(dāng)a0時(shí), f(x)2a+aln2a.證明f (x)=2e2x-ax,x0.f (x)有零點(diǎn),等價(jià)于方程2e2x-ax=0有實(shí)根,等價(jià)于方程2e2x=ax有實(shí)根,等價(jià)于函數(shù)y=2e2x與函數(shù)y=ax的圖象有交點(diǎn).顯然,當(dāng)a0時(shí),兩個(gè)函數(shù)圖象有一個(gè)交點(diǎn).因此,當(dāng)a0時(shí), f (x)只有一個(gè)零點(diǎn).當(dāng)a0時(shí), f (x)在(0,+)上單調(diào)遞增,且只有一個(gè)零點(diǎn),設(shè)此零點(diǎn)為x0,則f (x0)=0.當(dāng)x(0,x0)時(shí), f (x)0, f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=e2x0-aln x0.由f (x0)=0,得2e2x0-ax0=0,即e2x0=a2x0,即ln e2x0=ln a-ln 2x0,化簡得ln x0=ln a-ln 2-2x0.故f(x0)=a2x0-a(ln a-ln 2-2x0)=a2x0+2ax0+aln 2a2a+aln2a.故f(x)min2a+aln2a,即當(dāng)a0時(shí), f(x)2a+aln2a.3.已知函數(shù)f(x)=1+ln(x+1)x,當(dāng)x0時(shí), f(x)kx+1恒成立,求正整數(shù)k的最大值.解析由已知得k0上恒成立.令h(x)=(x+1)1+ln(x+1)x,x0,只需k0,得(x)在(0,+)上單調(diào)遞增.又(2)=1-ln 30,根據(jù)零點(diǎn)存在性定理可知,存在x0(2,3),使得(x0)=0.當(dāng)x(0,x0)時(shí),(x)0,h(x)0,h(x)0,h(x)在(x0,+)上單調(diào)遞增.故h(x)min=h(x0)=(x0+1)1+ln(x0+1)x0.由(x0)=0,得x0-1-ln(x0+1)=0,即x0=1+ln(x0+1).則h(x0)=x0+1(3,4).故正整數(shù)k的最大值為3.4.已知函數(shù)f(x)=aex+a+1x-2(a+1)0對任意的x(0,+)恒成立,其中a0.求a的取值范圍.解析f (x)=aex-a+1x2=ax2ex-(a+1)x2,令g(x)=ax2ex-(a+1),其中x0,a0.則g(x)=a(2x+x2)ex0,故g(x)在(0,+)上單調(diào)遞增.又g(0)=-(a+1)0,當(dāng)x+時(shí),g(x)+,故存在x0(0,+),使得g(x0)=0.當(dāng)x(0,x0)時(shí),g(x)0, f (x)0, f (x)0, f(x)在(x0,+)上單調(diào)遞增.故f(x)min=f(x0)=aex0+a+1x0-2(a+1).由g(x0)=0,得ax02ex0-(a+1)=0,即aex0=a+1x02.則f(x0)=aex0+a+1x0-2(a+1)=a+1x02+a+1x0-2(a+1),令a+1x02+a+1x0-2(a+1)0,由x00,a0,得0x01.因?yàn)間(x)=ax2ex-(a+1)在(0,+)上單調(diào)遞增,00.解析(1)f (x) =ex-1x+m.由x=0是f(x)的極值點(diǎn),得f (0)=0,所以m=1.于是f(x)=ex-ln(x+1),定義域?yàn)?-1,+),f (x)=ex-1x+1.函數(shù)f (x)=ex-1x+1在(-1,+)上單調(diào)遞增,且f (0)=0,因此當(dāng)x(-1,0)時(shí), f (x)0.所以f(x)在(-1,0)上單調(diào)遞減,在(0,+)上單調(diào)遞增.(2)證明:當(dāng)m2,x(-m,+)時(shí),ln(x+m)ln(x+2),故只需證明當(dāng)m=2時(shí), f(x)0.當(dāng)m=2時(shí),函數(shù)f (x)=ex-1x+2在(-2,+)上單調(diào)遞增.又f (-1)0,故f (x)=0在(-2,+)上有唯一實(shí)根x0,且x0(-1,0).當(dāng)x(-2,x0)時(shí), f (x)0,從而當(dāng)x=x0時(shí), f(x)取得極小值,也是最小值.f (x)、 f(x)的大致圖象如圖.由f (x0)=0得ex0=1x0+2,即ln(x0+2)=-x0,故f(x)f(x0)=1x0+2+x0=(x0+1)2x0+20.所以當(dāng)m2時(shí), f(x)0.6.已知函數(shù)f(x)=x2-ax+ln x(aR).(1)當(dāng)a=1時(shí),求曲線f(x)在點(diǎn)P(1,0)處的切線方程;(2)若函數(shù)f(x)有兩個(gè)極值點(diǎn)x1,x2,求f(x1+x2)的取值范圍.解析(1)當(dāng)a=1時(shí), f(x)=x2-x+ln x,則f (x)=2x-1+1x.所以f (1)=2.因此曲線f(x)在點(diǎn)P(1,0)處的切線方程為2x-y-2=0.(2)由題意得f (x)=2x-a+1x,令f (x)=0,則2x-a+1x=0,即2x2-ax+1=0(x0).由題意知2x2-ax+1=0有兩個(gè)不等的實(shí)根,為x1,x2.由根與系數(shù)的關(guān)系得=a2-80,x1+x2=a20,x1x2=120,解得a22.故f(x1+x2)=(x1+x2)2-a(x1+x2)+ln(x1+x2)=-a24+lna2.設(shè)g(a)=-a24+lna2(a22),則g(a)=-a2+1a=2-a22a0.故g(a)在(22,+)上單調(diào)遞減,所以g(a)g(22)=-2+ln 2.因此f(x1+x2)的取值范圍是(-,-2+ln 2).7.已知函數(shù)f(x)=ex-56x2-2x,g(x)=-13x2-x+1.(1)設(shè)h(x)=f(x)-g(x),求h(x)在-1,0上的最大值;(2)當(dāng)-1x1時(shí),求證:e-176f(x)0,得xln53,由m(x)=ex-530,得xln53.m(x)在-1,ln53上單調(diào)遞減,在ln53,1上單調(diào)遞增,又mln53=53-53ln53-20,m(1)=e-53-20,所以m(x)在-1,ln53上必有一零點(diǎn)x0,在(-1,x0)上f(x)單調(diào)遞增,在(x0,1)上 f(x)單調(diào)遞減.所以f(x)min=minf(-1), f(1)=e-176;f(x)max=f(x0)g(x0),又因?yàn)間(x)在(-1,x0)上單調(diào)遞減,所以f(x)maxg(x0)g(-1)=53.綜上所述,當(dāng)-1x1時(shí),e-176f(x)53.