2019版高考數(shù)學(xué)二輪復(fù)習(xí) 專題二 函數(shù)與導(dǎo)數(shù) 專題對點(diǎn)練8 導(dǎo)數(shù)與函數(shù)的零點(diǎn)及參數(shù)范圍 文.doc

專題對點(diǎn)練8導(dǎo)數(shù)與函數(shù)的零點(diǎn)及參數(shù)范圍1.(2018全國,文21)已知函數(shù)f(x)= x3-a(x2+x+1).(1)若a=3,求f(x)的單調(diào)區(qū)間;(2)證明:f(x)只有一個(gè)零點(diǎn).2.已知函數(shù)f(x)=ax+x2-xln a-b(a,bR,a>1),e是自然對數(shù)的底數(shù).(1)當(dāng)a=e, b=4時(shí),求函數(shù)f(x)零點(diǎn)個(gè)數(shù);(2)若b=1,求f(x)在-1,1上的最大值.3.已知函數(shù)f(x)=xln x,g(x)=-x2+ax-2(e為自然對數(shù)的底數(shù),aR).(1)判斷曲線y=f(x)在點(diǎn)(1,f(1)處的切線與曲線y=g(x)的公共點(diǎn)個(gè)數(shù);(2)當(dāng)x1e,e時(shí),若函數(shù)y=f(x)-g(x)有兩個(gè)零點(diǎn),求a的取值范圍.4.(2018天津,文20)設(shè)函數(shù)f(x)=(x-t1)(x-t2)(x-t3),其中t1,t2,t3R,且t1,t2,t3是公差為d的等差數(shù)列.(1)若t2=0,d=1,求曲線y=f(x)在點(diǎn)(0,f(0)處的切線方程;(2)若d=3,求f(x)的極值;(3)若曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn),求d的取值范圍.專題對點(diǎn)練8答案1.解 (1)當(dāng)a=3時(shí),f(x)= x3-3x2-3x-3,f(x)=x2-6x-3.令f(x)=0,解得x=3-23或x=3+23.當(dāng)x(-,3-23)(3+23,+)時(shí),f(x)>0;當(dāng)x(3-23,3+23)時(shí),f(x)<0.故f(x)在(-,3-23),(3+23,+)單調(diào)遞增,在(3-23,3+23)單調(diào)遞減.(2)由于x2+x+1>0,所以f(x)=0等價(jià)于x3x2+x+1-3a=0.設(shè)g(x)=x3x2+x+1-3a,則g(x)=x2(x2+2x+3)(x2+x+1)20,僅當(dāng)x=0時(shí)g(x)=0,所以g(x)在(-,+)單調(diào)遞增,故g(x)至多有一個(gè)零點(diǎn),從而f(x)至多有一個(gè)零點(diǎn).又f(3a-1)=-6a2+2a-=-6a-162-16<0,f(3a+1)= >0,故f(x)有一個(gè)零點(diǎn).綜上,f(x)只有一個(gè)零點(diǎn).2.解 (1)由題意f(x)=ex+x2-x-4,f(x)=ex+2x-1,f(0)=0,當(dāng)x>0時(shí), ex>1,f(x)>0,故f(x)是(0,+)上的增函數(shù);當(dāng)x<0時(shí),ex<1,f(x)<0,故f(x)是(-,0)上的減函數(shù).f(1)=e-4<0,f(2)=e2-2>0,存在x1(1,2)是f(x)在(0,+)上的唯一零點(diǎn);f(-2)=1e2+2>0,f(-1)= -2<0,存在x2(-2,-1)是f(x)在(-,0)上的唯一零點(diǎn).f(x)的零點(diǎn)個(gè)數(shù)為2.(2)當(dāng)b=1時(shí),f(x)=ax+x2-xln a-1,f(x)=axln a+2x-ln a=2x+(ax-1)ln a,當(dāng)x>0時(shí),由a>1,可知ax-1>0,ln a>0,f(x)>0;當(dāng)x<0時(shí),由a>1,可知ax-1<0,ln a>0,f(x)<0;當(dāng)x=0時(shí),f(x)=0,f(x)是-1,0上的減函數(shù),0,1上的增函數(shù).當(dāng)x-1,1時(shí),f(x)min=f(0),f(x)max為f(-1)和f(1)中的較大者.而f(1)-f(-1)=a-2ln a,設(shè)g(x)=x-2ln x(x>0).g(x)=1+1x2-2x=1x-120(當(dāng)且僅當(dāng)x=1時(shí)等號成立),g(x)在(0,+)上單調(diào)遞增,而g(1)=0,當(dāng)x>1時(shí),g(x)>0,即當(dāng)a>1時(shí),a-2ln a>0,f(1)>f(-1).f(x)max=f(1)=a+1-ln a-1=a-ln a.3.解 (1) f(x)=ln x+1,所以切線斜率k=f(1)=1.又f(1)=0,所以曲線在點(diǎn)(1,0)處的切線方程為y=x-1.由y=-x2+ax-2,y=x-1,得x2+(1-a)x+1=0.由=(1-a)2-4=a2-2a-3=(a+1)(a-3),可知:當(dāng)>0,即a<-1或a>3時(shí),有兩個(gè)公共點(diǎn);當(dāng)=0,即a=-1或a=3時(shí),有一個(gè)公共點(diǎn);當(dāng)<0,即-1<a<3時(shí),沒有公共點(diǎn).(2)y=f(x)-g(x)=x2-ax+2+xln x,由y=0,得a=x+ln x.令h(x)=x+ln x,則h(x)=(x-1)(x+2)x2.當(dāng)x1e,e時(shí),由h(x)=0,得x=1.所以h(x)在1e,1上單調(diào)遞減,在1,e上單調(diào)遞增,因此h(x)min=h(1)=3.由h1e=1e+2e-1,h(e)=e+1,比較可知h1e>h(e),所以,結(jié)合函數(shù)圖象可得,當(dāng)3<ae+2e+1時(shí),函數(shù)y=f(x)-g(x)有兩個(gè)零點(diǎn).4.解 (1)由已知,可得f(x)=x(x-1)(x+1)=x3-x,故f(x)=3x2-1.因此f(0)=0,f(0)=-1.又因?yàn)榍€y=f(x)在點(diǎn)(0,f(0)處的切線方程為y-f(0)=f(0)(x-0),故所求切線方程為x+y=0.(2)由已知可得f(x)=(x-t2+3)(x-t2)(x-t2-3)=(x-t2)3-9(x-t2)=x3-3t2x2+(3t22-9)x-t23+9t2.故f(x)=3x2-6t2x+3t22-9.令f(x)=0,解得x=t2-3,或x=t2+3.當(dāng)x變化時(shí),f(x),f(x)的變化情況如下表:x(-,t2-3)t2-3(t2-3,t2+3)t2+3(t2+3,+)f(x)+0-0+f(x)極大值極小值所以函數(shù)f(x)的極大值為f(t2-3)=(-3)3-9(-3)=63;函數(shù)f(x)的極小值為f(t2+3)=(3)3-93=-63. (3)曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn)等價(jià)于關(guān)于x的方程(x-t2+d)(x-t2)(x-t2-d)+(x-t2)+63=0有三個(gè)互異的實(shí)數(shù)解.令u=x-t2,可得u3+(1-d2)u+63=0.設(shè)函數(shù)g(x)=x3+(1-d2)x+63,則曲線y=f(x)與直線y=-(x-t2)-63有三個(gè)互異的公共點(diǎn)等價(jià)于函數(shù)y=g(x)有三個(gè)零點(diǎn).g(x)=3x2+(1-d2).當(dāng)d21時(shí),g(x)0,這時(shí)g(x)在R上單調(diào)遞增,不合題意.當(dāng)d2>1時(shí),令g(x)=0,解得x1=-d2-13,x2=d2-13.易得,g(x)在(-,x1)上單調(diào)遞增,在x1,x2上單調(diào)遞減,在(x2,+)上單調(diào)遞增.g(x)的極大值g(x1)=g-d2-13=23(d2-1)329+63>0.g(x)的極小值g(x2)=gd2-13=-23(d2-1)329+63.若g(x2)0,由g(x)的單調(diào)性可知函數(shù)y=g(x)至多有兩個(gè)零點(diǎn),不合題意.若g(x2)<0,即(d2-1)32>27,也就是|d|>10,此時(shí)|d|>x2,g(|d|)=|d|+63>0,且-2|d|<x1,g(-2|d|)=-6|d|3-2|d|+63<-6210+63<0,從而由g(x)的單調(diào)性,可知函數(shù)y=g(x)在區(qū)間(-2|d|,x1),(x1,x2),(x2,|d|)內(nèi)各有一個(gè)零點(diǎn),符合題意.所以,d的取值范圍是(-,-10)(10,+).