備戰(zhàn)2025年高考二輪復(fù)習(xí)數(shù)學(xué)專(zhuān)題突破練5

專(zhuān)題突破練(分值:60分)學(xué)生用書(shū)P1431.(13分)(2024·浙江杭州模擬)設(shè)函數(shù)f(x)=(x-1)2ex-ax,若曲線(xiàn)y=f(x)在點(diǎn)(0,f(0))處的切線(xiàn)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€(xiàn)y=f(x)在點(diǎn)(0,f(0))處的切線(xiàn)方程為y=-2x+b,所以f'(0(2)由(1)知f(x)=(x-1)2ex-x,所以f(x)=ex(x-1)2-xex,令g(x)=(x-1)2-xex,則f(x)的零點(diǎn)個(gè)數(shù)就是g(x)的零點(diǎn)個(gè)數(shù).由于g'(x)=(x-1)2+1ex,所以當(dāng)x<1時(shí),g'(x)<0,g(x)單調(diào)遞減;當(dāng)x>1時(shí),g'(x)>0,g(x)單調(diào)遞增.又g(0)=1>0,g(1)=-1e<0,g(2)=1-2e2>0,所以g(0)g(1)<0,g(1)g(2)<0,由零點(diǎn)存在定理可得,?x1∈(0,1),使得g(x1)=0,?x22.(15分)已知函數(shù)f(x)=alnx-2x.(1)若a=2,求曲線(xiàn)y=f(x)在(1,f(1))處的切線(xiàn)方程;(2)若函數(shù)f(x)在(0,16]上有兩個(gè)零點(diǎn),求實(shí)數(shù)a的取值范圍.解(1)當(dāng)a=2時(shí),f(x)=2lnx-2x,該函數(shù)的定義域?yàn)?0,+∞),f'(x)=2x-1x,又f(1)=-2,f'(1)=1,因此,曲線(xiàn)y=f(x)在(1,f(1))處的切線(xiàn)方程為y+2=x-1,即x-y-(2)①當(dāng)a≤0時(shí),f'(x)=ax-則f(x)在(0,+∞)上單調(diào)遞減,不符合題意;②當(dāng)a>0時(shí),由f(x)=alnx-2x=0可得2a=lnxx,令g(x)=lnxx,其中x>0,則直線(xiàn)y=2a與曲線(xiàn)y=g(x)的圖象在(0,16]上有兩個(gè)交點(diǎn),g'(x)=xx-lnx2xx=2-lnx2xx,令g'(xx(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)=2e,且g(16)=ln2,作出g(x)的圖象如圖所示由圖可知,當(dāng)ln2≤2a即e<a≤2ln2時(shí),直線(xiàn)y=2a與曲線(xiàn)y=g(x)的圖象在(0,16]上有兩個(gè)交點(diǎn),即f(x)在(0,16]上有兩個(gè)零點(diǎn),因此,實(shí)數(shù)a的取值范圍是e,2ln23.(15分)(2024·湖北黃石三模)已知函數(shù)f(x)=x-lnx+m有兩個(gè)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(shí)m的取值范圍.解(1)令f(x)=0,即m=lnx-x,令g(x)=lnx-x,則g'(x)=1x-1=1-xx,當(dāng)0<x<1時(shí),g'(x)>0,當(dāng)x>1時(shí),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(shí),g(x)→-∞,當(dāng)x→+∞時(shí),g(x)→-∞,又y=g(x)的圖象與直線(xiàn)y=m有兩個(gè)交點(diǎn),所以,實(shí)數(shù)m的取值范圍是(-∞(2)由(1)可得m=lnx1-x1,m=lnx2-x2,又0<x1<1<x2≤2x1,所以lnx1-x1=lnx2-x2,即lnx2x1=x2-x1,令t=x2x1,t∈(1,2],則lnt=(t-1)x1,所以x1=lntt-1,x2=tlntt-1,記h(t)=lntt-1,t∈(1,2],則h'(t)=1-1t-lnt(t-1)2,令H(t)=1-1t-lnt,t∈(1,2],則H'(t)=1t2-1t=1-tt2,所以,當(dāng)t∈(1,2]時(shí),H'(t)<0,即H(t)單調(diào)遞減,由于H(1)=0,所以當(dāng)t∈(1,2]時(shí),H(t)<0,所以h'(t)=1-1t-lnt(t-1)2<0,所以函數(shù)h(t)在區(qū)間(1,2]上單調(diào)遞減,故x1=h(t)≥h(2)=ln2,即ln2≤x1<1,而m=g(x4.(17分)(2024·四川成都一模)已知函數(shù)f(x)=exsinx(e是自然對(duì)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(gè)數(shù).(參考數(shù)據(jù):eπ2≈4解(1)f(x)=exsinx,定義域?yàn)镽.f'(x)=ex(sinx+cosx)=2exsinx+令f'(x)<0,解得sinx+π解得2kπ+3π4<x<7π4+2kπ(k∈Z).∴f(x)的單調(diào)遞減區(qū)間為3π4(2)由已知g(x)=exsinx-ax,∴g'(x)=ex(sinx+cosx)-a.令h(x)=g'(x),則h'(x)=2excosx.∵x∈(0,π),∴當(dāng)x∈0,π2時(shí),h'(x當(dāng)x∈π2,π時(shí),h'(x∴h(x)在0,π2上單調(diào)遞增,在π2,π上單調(diào)遞減,即g'(x)在0,π2上單調(diào)遞增,在π2,π上單調(diào)遞減.∵g'(0)=①當(dāng)1-a≥0,即0<a≤1時(shí),g'(0)≥0,∴g'π2>0.∴?x0∈π2,π,使得g'(x∴當(dāng)x∈(0,x0)時(shí),g'(x)>0;當(dāng)x∈(x0,π)時(shí),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(shí)g(x)在(0,π)上僅有一個(gè)零點(diǎn).②當(dāng)1<a<3時(shí),g'(0)=1-a<0,∵g'(x)在0,π2上單調(diào)遞增,在π2,π上單調(diào)遞減∴?x1∈0,π2,x2∈π2,π,使得g'(x1)=0,g'且當(dāng)x∈(0,x1),x∈(x2,π)時(shí),g'(x)<0;當(dāng)x∈(x1,x2)時(shí),g'(x)>0,∴g(x)在(0,x1)和(x2,π)上單調(diào)遞減,在(x1,x2)上單調(diào)遞增.∵g(0)=0,∴g(x1)<0.∵gπ2=eπ2-π2a>eπ