課時(shí)規(guī)范練25利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)高考總復(fù)習(xí)優(yōu)化設(shè)計(jì)GAOKAOZONGFUXIYOUHUASHEJI202512341.(2024·浙江杭州模擬)設(shè)函數(shù)f(x)=(x-1)2ex-ax,若曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=-2x+b.(1)求實(shí)數(shù)a,b的值;(2)判斷函數(shù)f(x)零點(diǎn)的個(gè)數(shù).解

(1)由題意可得f'(x)=(x2-1)ex-a.因?yàn)榍€y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=-2x+b,123412342.(2024·河北唐山模擬)已知函數(shù)f(x)=ex(x+2),g(x)=x2+3x+2.(1)證明:當(dāng)x≥0時(shí),f(x)≥g(x);(2)若函數(shù)h(x)=f(x)-4ex-m有兩個(gè)零點(diǎn),求m的取值范圍.(1)證明

令k(x)=f(x)-g(x)=ex(x+2)-x2-3x-2,則k'(x)=ex(x+3)-x-3=(x+3)(ex-1),因?yàn)閤≥0,所以k'(x)=(x+3)(ex-1)≥0,故k(x)單調(diào)遞增,又k(0)=2-2=0,故f(x)-g(x)≥0,即f(x)≥g(x).1234(2)解

h(x)=ex(x+2)-4ex-m=ex(x-2)-m,令h(x)=0,得ex(x-2)=m.函數(shù)h(x)=f(x)-4ex-m有兩個(gè)零點(diǎn),即直線y=m與函數(shù)y=ex(x-2)的圖象有兩個(gè)不同的交點(diǎn).令w(x)=ex(x-2),則w'(x)=ex(x-1),當(dāng)x>1時(shí),w'(x)>0,當(dāng)x<1時(shí),w'(x)<0,所以w(x)在(-∞,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,因此w(x)在x=1處取得極小值,也是最小值,且w(1)=-e,又當(dāng)x<2時(shí),w(x)=ex(x-2)<0恒成立,當(dāng)x→+∞時(shí),w(x)→+∞,當(dāng)x→-∞時(shí),w(x)→0,要使函數(shù)h(x)=f(x)-4ex-m有兩個(gè)零點(diǎn),則m∈(-e,0).12343.(2024·四川成都模擬)已知函數(shù)f(x)=xe-x+asinx,e是自然對(duì)數(shù)的底數(shù),若x=0恰為f(x)的極值點(diǎn).(1)求實(shí)數(shù)a的值;(2)求f(x)在區(qū)間(-∞,)上零點(diǎn)的個(gè)數(shù).1234123412344.(2024·北京陳經(jīng)綸中學(xué)模擬)已知函數(shù)f(x)=xex-

(x+1)2(m≥0).(1)當(dāng)m=0時(shí),求函數(shù)f(x)的極小值;(2)若函數(shù)f(x)在區(qū)間(-∞,1)上有且只有一個(gè)零點(diǎn),求m的取值范圍.

1234(2)若m=0,f(x)=xex,令f(x)=0,解得x=0,當(dāng)x<0時(shí),f(x)<0;當(dāng)x>0時(shí),f(x)>0,所以函數(shù)f(x)在區(qū)間(-∞,1)上有且只有一個(gè)零點(diǎn),符合題意.若m>0,由f'(x)=(x+1)(ex-m),令f'(x)=0,解得x=-1或x=ln

m.-1<x<ln