2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 專題對(duì)點(diǎn)練7 導(dǎo)數(shù)與不等式及參數(shù)范圍 文

2022年高考數(shù)學(xué)二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 專題對(duì)點(diǎn)練7 導(dǎo)數(shù)與不等式及參數(shù)范圍 文1.已知函數(shù)f(x)= x2+(1-a)x-aln x.(1)討論f(x)的單調(diào)性;(2)設(shè)a<0,若對(duì)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)-2.專題對(duì)點(diǎn)練7答案1.解 (1)f(x)的定義域?yàn)?0,+),f'(x)=x+1-a-,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+3+a,g'(x)= -x+3+a0對(duì)x(0,+)恒成立,a對(duì)x(0,+)恒成立,a.又=x+1+-52-5=-1,當(dāng)且僅當(dāng)x+1=,即x=1時(shí),等號(hào)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-,x=-1+.當(dāng)x(-,-1-)時(shí),f'(x)<0;當(dāng)x(-1-,-1+)時(shí),f'(x)>0;當(dāng)x(-1+,+)時(shí),f'(x)<0.所以f(x)在(-,-1-),(-1+,+)內(nèi)單調(diào)遞減,在(-1-,-1+)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=,則x0(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)>ax0+1.當(dāng)a0時(shí),取x0=,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=.(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)x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=,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)1f'(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:x1(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=,x2=1.f'(x),f(x)隨x的變化情況如下表:x1(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)= +2ax+2a+1=.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)x時(shí),f'(x)>0;當(dāng)x時(shí),f'(x) <0.故f(x)在單調(diào)遞增,在單調(diào)遞減.(2)證明 由(1)知,當(dāng)a<0時(shí),f(x)在x=-取得最大值,最大值為f=ln-1-.所以f(x)-2等價(jià)于ln-1-2,即ln+10.設(shè)g(x)=ln x-x+1,則g'(x)= -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+10,即f(x)-2.