2025高考數(shù)學(xué)二輪復(fù)習(xí)-專題突破練5 利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)【課件】

專題突破練5利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)202512341.(13分)(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)的個數(shù).123412342.(15分)已知函數(shù)f(x)=alnx-2.(1)若a=2,求曲線y=f(x)在(1,f(1))處的切線方程;(2)若函數(shù)f(x)在(0,16]上有兩個零點(diǎn),求實(shí)數(shù)a的取值范圍.12341234x(0,e2)e2(e2,16]g'(x)+0-g(x)單調(diào)遞增極大值單調(diào)遞減所以函數(shù)g(x)在區(qū)間(0,16]上的極大值為g(e2)=,且g(16)=ln

2,作出g(x)的圖象如圖所示.12343.(15分)(2024·湖北黃石三模)已知函數(shù)f(x)=x-lnx+m有兩個零點(diǎn)x1,x2.(1)求實(shí)數(shù)m的取值范圍;(2)如果x1<x2≤2x1,求此時m的取值范圍.解

(1)令f(x)=0,即m=ln

x-x,令g(x)=ln

x-x,則g'(x)=,當(dāng)0<x<1時,g'(x)>0,當(dāng)x>1時,g'(x)<0,所以g(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減.又g(1)=-1,且x→0時,g(x)→-∞,當(dāng)x→+∞時,g(x)→-∞,又y=g(x)的圖象與直線y=m有兩個交點(diǎn),所以,實(shí)數(shù)m的取值范圍是(-∞,-1).123412344.(17分)(2024·四川成都一模)已知函數(shù)f(x)=exsinx(e是自然對數(shù)的底數(shù)).(1)求f(x)的單調(diào)遞減區(qū)間;(2)記g(x)=f(x)-ax,若0<a<3,試討論g(x)在(0,π)上的零點(diǎn)個數(shù).12341234(2)由已知g(x)=exsin

x-ax,∴g'(x)=ex(sin

x+cos

x)-a.令h(x)=g'(x),則h'(x)=2excos

x.1234①當(dāng)1-a≥0,即0<a≤1時,g'(0)≥0,∴當(dāng)x∈(0,x0)時,g'(x)>0;當(dāng)x∈(x0,π)時,g'(x)<0,∴g(x)在(0,x0)上單調(diào)遞增,在(x0,π)上單調(diào)遞減.∵g(0)=0,∴g(x0)>0.又g(π)=-aπ<0,∴由零點(diǎn)存在定理可得,此時g(x)在(0,π)上僅有一個零點(diǎn).1234又g(π)=-aπ<0,由零點(diǎn)存在定理可得,g(x)在(x1,x2)和(x2,π)內(nèi)各有一個零點(diǎn),即此時g(x)在(0,π)內(nèi)有兩個零點(diǎn).綜上,當(dāng)0<a