高中數(shù)學(xué)選修2-2配人教A版-課后習(xí)題word-模塊復(fù)習(xí)課第2課時　利用導(dǎo)數(shù)研究不等式問題和函數(shù)零點(diǎn)問題

1、第2課時利用導(dǎo)數(shù)研究不等式問題和函數(shù)零點(diǎn)問題課后篇鞏固提升1.已知函數(shù)f(x)=ln x+ax-x2(aR).(1)若f(x)0恒成立,求實(shí)數(shù)a的取值范圍;(2)證明:ln(n+1)0),g(x)=1-1-lnxx2=x2+lnx-1x2,令h(x)=x2+ln x-1(x0),h(x)=2x+1x0,h(x)在(0,+)內(nèi)單調(diào)遞增,而h(1)=0,x(0,1)時,h(x)0,即x(0,1)時,g(x)0,g(x)在(0,1)內(nèi)單調(diào)遞減,在(1,+)內(nèi)單調(diào)遞增,g(x)g(1)=1,a1.故實(shí)數(shù)a的取值范圍是(-,1.(2)證明由(1)得a=1時,f(x)=ln x+x-x20,故ln xx2

2、-x,即ln(x+1)x(x+1)(當(dāng)且僅當(dāng)x=0時,等號成立),對任意正整數(shù)n,取x=1n0,得ln1n+11n+1n2,lnn+1nln 2+ln32+ln43+lnn+1n=ln(n+1),即ln(n+1)212+322+n+1n2.2.已知函數(shù)f(x)=aln x+x,g(x)=xex-a.(1)若x=1是f(x)的極值點(diǎn),求f(x)的單調(diào)區(qū)間;(2)若a=1,證明f(x)g(x).解(1)由已知可得,函數(shù)f(x)的定義域為(0,+),且f(x)=ax+1.因為x=1是f(x)的極值點(diǎn),所以f(1)=a+1=0,解得a=-1,此時f(x)=-1x+1=x-1x.故當(dāng)0 x1時,f(x)

3、1時,f(x)0.所以f(x)的單調(diào)遞增區(qū)間為(1,+),單調(diào)遞減區(qū)間為(0,1).(2)若a=1,則f(x)=ln x+x,g(x)=xex-1.設(shè)h(x)=f(x)-g(x)=ln x+x-xex+1,x(0,+),則h(x)=1x+1-(x+1)ex=(x+1)1x-ex.令t(x)=1x-ex,x(0,+),則t(x)=-1x2-ex0,t(1)=1-e0,所以x012,1,使得t(x0)=1x0ex0=0,即1x0=ex0,則ln1x0=ln ex0,即-ln x0=x0.因此,當(dāng)0 x0,即h(x)0,則h(x)單調(diào)遞增;當(dāng)xx0時,t(x)0,即h(x)e.(1)解定義域為(0,

4、+),f(x)=1x-2cx=1-2cx2x,當(dāng)c0時,f(x)0,f(x)在(0,+)內(nèi)單調(diào)遞增,x0時,f(x)-,x+時,f(x)+,f(x)有且只有1個零點(diǎn);當(dāng)c0時,由f(x)=0,得x=2c2c,當(dāng)0 x0,f(x)單調(diào)遞增,當(dāng)x2c2c時,f(x)0,解得c12e,當(dāng)c12e時,f(x)有2個零點(diǎn),當(dāng)c=12e時,f(x)有1個零點(diǎn),當(dāng)0c12e時,f(x)沒有零點(diǎn).綜上,當(dāng)c0或c=12e時,f(x)有1個零點(diǎn);當(dāng)0c12e時,f(x)有2個零點(diǎn).(2)證明設(shè)x1x2,ln x1-cx12=0,ln x2-cx22=0,ln x1+ln x2=cx12+cx22,ln x1-l

5、n x2=cx12-cx22,則c=ln x1-ln x2x12-x22.欲證明x1x2e,即證ln x1+ln x21,因為ln x1+ln x2=c(x12+x22),即證c1x12+x22,原命題等價于證明ln x1-ln x2x12-x221x12+x22,即證lnx1x2x12-x22x12+x22(x1x20),令x1x2=t,則t1,設(shè)g(t)=ln t-t2-1t2+1=ln t+2t2+1-1(t1),g(t)=(t2-1)2t(t2+1)20,g(t)在(1,+)內(nèi)單調(diào)遞增,又g(1)=0,g(t)g(1)=0,ln tt2-1t2+1,x1x2e.4.已知函數(shù)f(x)=x

6、-1-aln x.(1)若f(x)0,求a的值;(2)設(shè)m為整數(shù),且對于任意正整數(shù)n,1+121+1221+12nm,求m的最小值.解(1)f(x)的定義域為(0,+).若a0,因為f12=-12+aln 20,由f(x)=1-ax=x-ax知,當(dāng)x(0,a)時,f(x)0.所以f(x)在(0,a)單調(diào)遞減,在(a,+)單調(diào)遞增.故x=a是f(x)在(0,+)的唯一最小值點(diǎn).由于f(1)=0,所以當(dāng)且僅當(dāng)a=1時,f(x)0.故a=1.(2)由(1)知當(dāng)x(1,+)時,x-1-ln x0.令x=1+12n得ln1+12n12n.從而ln1+12+ln1+122+ln1+12n12+122+12

7、n=1-12n1.故1+121+1221+12n2,所以m的最小值為3.5.設(shè)aZ,已知定義在R上的函數(shù)f(x)=2x4+3x3-3x2-6x+a在區(qū)間(1,2)內(nèi)有一個零點(diǎn)x0,g(x)為f(x)的導(dǎo)函數(shù).(1)求g(x)的單調(diào)區(qū)間;(2)設(shè)m1,x0)(x0,2,函數(shù)h(x)=g(x)(m-x0)-f(m),求證:h(m)h(x0)0,故當(dāng)x1,x0)時,H1(x)0,H1(x)單調(diào)遞增.因此,當(dāng)x1,x0)(x0,2時,H1(x)H1(x0)=-f(x0)=0,可得H1(m)0,即h(m)0.令函數(shù)H2(x)=g(x0)(x-x0)-f(x),則H2(x)=g(x0)-g(x).由(1)

8、知g(x)在1,2上單調(diào)遞增,故當(dāng)x1,x0)時,H2(x)0,H2(x)單調(diào)遞增;當(dāng)x(x0,2時,H2(x)0,H2(x)單調(diào)遞減.因此,當(dāng)x1,x0)(x0,2時,H2(x)H2(x0)=0,可得H2(m)0,即h(x0)0.所以,h(m)h(x0)0.(3)證明對于任意的正整數(shù)p,q,且pq1,x0)(x0,2,令m=pq,函數(shù)h(x)=g(x)(m-x0)-f(m).由(2)知,當(dāng)m1,x0)時,h(x)在區(qū)間(m,x0)內(nèi)有零點(diǎn);當(dāng)m(x0,2時,h(x)在區(qū)間(x0,m)內(nèi)有零點(diǎn).所以h(x)在(1,2)內(nèi)至少有一個零點(diǎn),不妨設(shè)為x1,則h(x1)=g(x1)pq-x0-fpq=0.由(1)知g(x)在1,2上單調(diào)遞增,故0g(1)g(x1)0,故f(x)在1,2上單調(diào)遞增,所以f(x)在區(qū)間1,2上除x0外沒有其他