2019版高考數(shù)學(xué)二輪復(fù)習(xí)專(zhuān)題二函數(shù)與導(dǎo)數(shù)專(zhuān)題對(duì)點(diǎn)練7導(dǎo)數(shù)與不等式及參數(shù)范圍文

1、專(zhuān)題對(duì)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)=x2+(1-a)x-alnx.談?wù)揻(x)的單一性;設(shè)a0,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.求f(x)的單一區(qū)間;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(xiàn)y=f(x)在點(diǎn)(2,f(2)處的切線(xiàn)斜率為0,求a;若f(x)在x=1處獲取極小值,求a的取值范圍.4.已知函數(shù)f(x)=lnx+ax2+(2a+1)x.談?wù)揻(x)的單一性;(2)當(dāng)a0,此時(shí)f

2、(x)在(0,+)內(nèi)單一遞加;若a0,則由f(x)=0得x=a,當(dāng)0 xa時(shí),f(x)a時(shí),f(x)0,此時(shí)f(x)在(0,a)內(nèi)單一遞減,在(a,+)內(nèi)單一遞加.(2)不如設(shè)xx,而a0,由(1)知,f(x)在(0,+)內(nèi)單一遞加,f(x)f(x),(1)(12?41(1)42(12|fx-fx2)412x-fx2),|x-x|x-fx令g(x)=4x-f(x),則g(x)在(0,+)內(nèi)單一遞減,g(x)=4-f(x)=4-x+3+a,g()=-x+30對(duì)?x(0,+)恒建立,x+aa對(duì)?x(0,+)恒建立,a.又=x+1+-52-5=-1,當(dāng)且僅當(dāng)1,即1時(shí),等號(hào)建立.x+=x=a-1,故

3、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+,+)時(shí),f(x)0.因此f(x)在(-,-1-),(-1+,+)內(nèi)單一遞減,在(-1-,-1+)內(nèi)單一遞加.(2)f(x)=(1+x)(1-x)ex.xx0),當(dāng)a1時(shí),設(shè)函數(shù)h(x)=(1-x)e,h(x)=-xe因此h(x)在0,+)內(nèi)單一遞減,而(0)1,故(x)1,h=h因此f(x)=(x+1)h(x)x+1ax+1.當(dāng)0a0(x0),因此g(x)在0,+)內(nèi)單一遞加,而g(0)=0,x故ex+1.當(dāng)01時(shí),f(x)(1-x)(

4、1)2,(1-x)(1)21(12),取0=,則x0 x+x+x-ax-=x-a-x-xx(0,1),(1-x0)(1+x0)2-ax0-1=0,故f(x0)ax0+1.當(dāng)a0時(shí)0000020,取x=,則x(0,1),f(x)(1-x)(1+x)=1ax+1.綜上,a的取值范圍是1,+).3.解(1)因?yàn)閒(x)=ax2-(3a+1)x+3a+2ex,因此f(x)=ax2-(a+1)x+1ex.因此f(2)=(2a-1)e2.a-1)e2=0,解得a=.由題設(shè)知f(2)=0,即(2x(1)ex(2)(方法一)由(1)得f()2(1)1e1)(x-x=ax-a+x+=ax-.若1,則當(dāng)x時(shí),f(

5、)0;ax0.因此f(x)在x=1處獲取極小值.若a1,則當(dāng)x(0,1)時(shí),ax-1x-10.因此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的變化情況以下表:x(-,1)1(1,+)f(x)+0-f(x)極大值f(x)在x=1處獲取極大值,不合題意.當(dāng)a0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上單一遞加,f(x)無(wú)極值,不合題意.當(dāng)x1x2,即0a1時(shí),f(x),f(x)隨x的變化情況以下表:x(-,1)1f(x)+0-0+f(x)極大值極小值f(x)在x=1處獲取極大值,不合題意.當(dāng)12,即1時(shí),f(x),f(x)隨x的變化情況以下表:xx1(1,+)f(x)+0-0+f(x)極大值極小值f(x)在x=1處獲取極小值,即a1滿(mǎn)足題意.當(dāng)a0,故f(x)在(0,+)單一遞加.若a0;當(dāng)x時(shí),f