（新課標(biāo)）廣西2019高考數(shù)學(xué)二輪復(fù)習(xí) 專題對點(diǎn)練7 導(dǎo)數(shù)與不等式及參數(shù)范圍.docx

專題對點(diǎn)練7導(dǎo)數(shù)與不等式及參數(shù)范圍1.已知函數(shù)f(x)=12x2+(1-a)x-aln x.(1)討論f(x)的單調(diào)性;(2)設(shè)a<0,若對x1,x2(0,+),|f(x1)-f(x2)|4|x1-x2|,求a的取值范圍.2.設(shè)函數(shù)f(x)=(1-x2)ex.(1)求f(x)的單調(diào)區(qū)間;(2)當(dāng)x0時(shí),f(x)ax+1,求a的取值范圍.3.(2018北京,文19)設(shè)函數(shù)f(x)=ax2-(3a+1)x+3a+2ex.(1)若曲線y=f(x)在點(diǎn)(2,f(2)處的切線斜率為0,求a;(2)若f(x)在x=1處取得極小值,求a的取值范圍.4.已知函數(shù)f(x)=ln x+ax2+(2a+1)x.(1)討論f(x)的單調(diào)性;(2)當(dāng)a<0時(shí),證明f(x)-34a-2.專題對點(diǎn)練7答案1.解 (1)f(x)的定義域?yàn)?0,+),f(x)=x+1-a-ax=x2+(1-a)x-ax=(x+1)(x-a)x,若a0,則f(x)>0,此時(shí)f(x)在(0,+)內(nèi)單調(diào)遞增;若a>0,則由f(x)=0得x=a,當(dāng)0<x<a時(shí),f(x)<0,當(dāng)x>a時(shí),f(x)>0,此時(shí)f(x)在(0,a)內(nèi)單調(diào)遞減,在(a,+)內(nèi)單調(diào)遞增.(2)不妨設(shè)x1x2,而a<0,由(1)知,f(x)在(0,+)內(nèi)單調(diào)遞增,f(x1)f(x2),|f(x1)-f(x2)|4|x1-x2|4x1-f(x1)4x2-f(x2),令g(x)=4x-f(x),則g(x)在(0,+)內(nèi)單調(diào)遞減,g(x)=4-f(x)=4-x+1-a-ax=ax-x+3+a,g(x)=ax-x+3+a0對x(0,+)恒成立,ax2-3xx+1對x(0,+)恒成立,ax2-3xx+1min.又x2-3xx+1=x+1+4x+1-52(x+1)4x+1-5=-1,當(dāng)且僅當(dāng)x+1=4x+1,即x=1時(shí),等號成立.a-1,故a的取值范圍為(-,-1.2.解 (1)f(x)=(1-2x-x2)ex.令f(x)=0得x=-1-2,x=-1+2.當(dāng)x(-,-1-2)時(shí),f(x)<0;當(dāng)x(-1-2,-1+2)時(shí),f(x)>0;當(dāng)x(-1+2,+)時(shí),f(x)<0.所以f(x)在(-,-1-2),(-1+2,+)內(nèi)單調(diào)遞減,在(-1-2,-1+2)內(nèi)單調(diào)遞增.(2)f(x)=(1+x)(1-x)ex.當(dāng)a1時(shí),設(shè)函數(shù)h(x)=(1-x)ex,h(x)=-xex<0(x>0),因此h(x)在0,+)內(nèi)單調(diào)遞減,而h(0)=1,故h(x)1,所以f(x)=(x+1)h(x)x+1ax+1.當(dāng)0<a<1時(shí),設(shè)函數(shù)g(x)=ex-x-1,g(x)=ex-1>0(x>0),所以g(x)在0,+)內(nèi)單調(diào)遞增,而g(0)=0,故exx+1.當(dāng)0<x<1時(shí),f(x)>(1-x)(1+x)2,(1-x)(1+x)2-ax-1=x(1-a-x-x2),取x0=5-4a-12,則x0(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.當(dāng)a0時(shí),取x0=5-12,則x0(0,1),f(x0)>(1-x0)(1+x0)2=1ax0+1.綜上,a的取值范圍是1,+).3.解 (1)因?yàn)閒(x)=ax2-(3a+1)x+3a+2ex,所以f(x)=ax2-(a+1)x+1ex.所以f(2)=(2a-1)e2.由題設(shè)知f(2)=0,即(2a-1)e2=0,解得a=12.(2)(方法一)由(1)得f(x)=ax2-(a+1)x+1ex=(ax-1)(x-1)ex.若a>1,則當(dāng)x1a,1時(shí),f(x)<0;當(dāng)x(1,+)時(shí),f(x)>0.所以f(x)在x=1處取得極小值.若a1,則當(dāng)x(0,1)時(shí),ax-1x-1<0,所以f(x)>0.所以1不是f(x)的極小值點(diǎn).綜上可知,a的取值范圍是(1,+).(方法二)由(1)得f(x)=(ax-1)(x-1)ex.當(dāng)a=0時(shí),令f(x)=0,得x=1.f(x),f(x)隨x的變化情況如下表:x(-,1)1(1,+)f(x)+0-f(x)極大值f(x)在x=1處取得極大值,不合題意.當(dāng)a>0時(shí),令f(x)=0,得x1=1a,x2=1.當(dāng)x1=x2,即a=1時(shí),f(x)=(x-1)2ex0,f(x)在R上單調(diào)遞增,f(x)無極值,不合題意.當(dāng)x1>x2,即0<a<1時(shí),f(x),f(x)隨x的變化情況如下表:x(-,1)11,1a1a1a,+f(x)+0-0+f(x)極大值極小值f(x)在x=1處取得極大值,不合題意.當(dāng)x1<x2,即a>1時(shí),f(x),f(x)隨x的變化情況如下表:x-,1a1a1a,11(1,+)f(x)+0-0+f(x)極大值極小值f(x)在x=1處取得極小值,即a>1滿足題意.當(dāng)a<0時(shí),令f(x)=0,得x1=1a,x2=1.f(x),f(x)隨x的變化情況如下表:x-,1a1a1a,11(1,+)f(x)-0+0-f(x)極小值極大值f(x)在x=1處取得極大值,不合題意.綜上所述,a的取值范圍為(1,+).4.(1)解 f(x)的定義域?yàn)?0,+),f(x)=1x+2ax+2a+1=(x+1)(2ax+1)x.若a0,則當(dāng)x(0,+)時(shí),f(x)>0,故f(x)在(0,+)單調(diào)遞增.若a<0,則當(dāng)x0,-12a時(shí),f(x)>0;當(dāng)x-12a,+時(shí),f(x)<0.故f(x)在0,-12a單調(diào)遞增,在-12a,+單調(diào)遞減.(2)證明 由(1)知,當(dāng)a<0時(shí),f(x)在x=-12a取得最大值,最大值為f-12a=ln-12a-1-14a.所以f(x)-34a-2等價(jià)于ln-12a-1-14a-34a-2,即ln-12a+12a+10.設(shè)g(x)=ln x-x+1,則g(x)=1x-1.當(dāng)x(0,1)時(shí),g(x)>0;當(dāng)x(1,+)時(shí),g(x)<0.所以g(x)在(0,1)單調(diào)遞增,在(1,+)單調(diào)遞減.故當(dāng)x=1時(shí),g(x)取得最大值,最大值為g(1)=0.所以當(dāng)x>0時(shí),g(x)0.從而當(dāng)a<0時(shí),ln-12a+12a+10,即f(x)-34a-2.