高考解答題專(zhuān)項(xiàng)一　第3課時(shí)　利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)

1、 高考解答題專(zhuān)項(xiàng)一函數(shù)與導(dǎo)數(shù)中的綜合問(wèn)題第3課時(shí)利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)1.(2021北京海淀高三期末)已知函數(shù)f(x)=x2-3x+ln x.(1)求曲線(xiàn)y=f(x)在點(diǎn)(3,f(3)處的切線(xiàn)方程;(2)求函數(shù)f(x)的單調(diào)區(qū)間,并判斷函數(shù)f(x)的零點(diǎn)個(gè)數(shù).2.(2021全國(guó)甲,文20)設(shè)函數(shù)f(x)=a2x2+ax-3ln x+1,其中a0.(1)討論f(x)的單調(diào)性;(2)若y=f(x)的圖象與x軸沒(méi)有公共點(diǎn),求a的取值范圍.3.(2021山西長(zhǎng)治高三期末)已知函數(shù)f(x)=ex,g(x)=ln(f(x)+a)(a為常數(shù)),g(x)是R上的奇函數(shù).(1)證明:f(x)x+1(xR);(2

2、)討論關(guān)于x的方程:ln g(x)=g(x)(x2-2ex+m)(mR)的實(shí)數(shù)根的個(gè)數(shù).4.(2021浙江寧波高三期末)設(shè)函數(shù)f(x)=ax2-ln x,其中aR.(1)若a=12,求函數(shù)f(x)的單調(diào)區(qū)間;(2)若方程f(x)=x恰有兩個(gè)不等實(shí)數(shù)根,求實(shí)數(shù)a的取值范圍.5.(2021山東煙臺(tái)高三期中)已知函數(shù)f(x)=ax+2ex+1(aR).(1)若函數(shù)f(x)在區(qū)間(1,+)上單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)當(dāng)a0時(shí),討論函數(shù)g(x)=f(x)-a-3的零點(diǎn)個(gè)數(shù),并給予證明.6.(2021吉林長(zhǎng)春高三期末)已知函數(shù)f(x)=x-a(1+ln x).(1)討論函數(shù)f(x)的單調(diào)性;(2

3、)若函數(shù)f(x)有兩個(gè)零點(diǎn)x1,x2,求實(shí)數(shù)a的取值范圍.第3課時(shí)利用導(dǎo)數(shù)研究函數(shù)的零點(diǎn)1.解 (1)因?yàn)楹瘮?shù)的定義域?yàn)?0,+),f(3)=ln 3,所以切點(diǎn)為(3,ln 3).又因?yàn)閒(x)=2x-3+1x,所以f(3)=103,即切線(xiàn)斜率為k=103,所以切線(xiàn)方程是y=103(x-3)+ln 3,即10 x-3y+3ln 3-30=0.(2)由(1)知f(x)=2x2-3x+1x=(2x-1)(x-1)x,令f(x)=0,得x1=12,x2=1.x0,121212,11(1,+)f(x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減極小值單調(diào)遞增如表格,函數(shù)f(x)的單調(diào)遞增區(qū)間是0,12和(

4、1,+),單調(diào)遞減區(qū)間是12,1.又因?yàn)閒(x)的極大值f12=-54+ln120,所以當(dāng)0 x1時(shí)f(x)0恒成立.又因?yàn)閒(x)在區(qū)間(1,+)上單調(diào)遞增,f(1)0,所以存在x0(1,3),使得f(x0)=0,即函數(shù)f(x)有且只有一個(gè)零點(diǎn).2.解(1)f(x)=a2x2+ax-3ln x+1,x(0,+),f(x)=2a2x+a-3x=2a2x2+ax-3x=(ax-1)(2ax+3)x.a0,x0,2ax+3x0,當(dāng)x0,1a時(shí),f(x)0,函數(shù)f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增.(2)y=f(x)的圖象與x軸沒(méi)有公共點(diǎn),函數(shù)f(x)在(0,+)上沒(méi)有零點(diǎn),由(1)可

5、得函數(shù)f(x)在0,1a上單調(diào)遞減,在1a,+上單調(diào)遞增,f1a=3-3ln1a=3+3ln a0,ln a-1,a1e,即實(shí)數(shù)a的取值范圍是1e,+.3.(1)證明 設(shè)F(x)=f(x)-x-1,則F(x)=ex-1,令F(x)=0,得x=0.所以當(dāng)x(-,0)時(shí),F(x)0,F(x)單調(diào)遞增;所以F(x)在x=0處取到最小值,即F(x)F(0)=0,故f(x)x+1.(2)解 因?yàn)間(x)是R上的奇函數(shù),所以有g(shù)(0)=0,則a=0,g(x)=x.所以原方程為ln x=x(x2-2ex+m),即lnxx=x2-2ex+m.設(shè)h(x)=lnxx,則由h(x)=1-lnxx2=0,得x=e,當(dāng)

6、x(0,e)時(shí),h(x)0,h(x)單調(diào)遞增;當(dāng)x(e,+)時(shí),h(x)e2+1e時(shí),方程無(wú)解;當(dāng)m=e2+1e時(shí),方程有且只有一個(gè)實(shí)數(shù)根x=e;當(dāng)me2+1e時(shí),方程有兩個(gè)實(shí)數(shù)根.4.解 (1)當(dāng)a=12,f(x)=12x2-ln x,則f(x)=x-1x=x2-1x=(x-1)(x+1)x.f(x)的定義域?yàn)?0,+),當(dāng)x(0,1)時(shí),f(x)0,即f(x)的單調(diào)遞增區(qū)間為(1,+).(2)若f(x)=x恰有兩個(gè)不等實(shí)數(shù)根,則ax2-ln x=x,即a=lnx+xx2恰有兩個(gè)不等實(shí)數(shù)根.令g(x)=lnx+xx2,則g(x)=(1x+1)x2-2x(lnx+x)x4=1-x-2lnxx3

7、,令h(x)=1-x-2ln x,因?yàn)閔(x)=-1-2x0,當(dāng)x(1,+)時(shí),h(x)0,g(x)單調(diào)遞增;當(dāng)x(1,+)時(shí),g(x)0,g(x)單調(diào)遞減,故g(x)max=g(1)=1,且g(x)在(1,+)上恒大于0.要使方程f(x)=x有兩個(gè)不等實(shí)數(shù)根,即直線(xiàn)y=a與函數(shù)g(x)的圖象有兩個(gè)不同的交點(diǎn),故實(shí)數(shù)a的取值范圍是(0,1).5.解 (1)f(x)=a-2ex.由題意得f(x)0,即a2ex在區(qū)間(1,+)上恒成立.當(dāng)x(1,+)時(shí),2ex0,2e,所以a2e,故實(shí)數(shù)a的取值范圍為2e,+.(2)由已知得g(x)=ax+2ex-a-2,則g(x)=a-2ex=aex-2ex.當(dāng)

8、a0時(shí),g(x)0,g(1)=2e-20時(shí),令g(x)0,得x0,得xln2a,函數(shù)g(x)單調(diào)遞增,而gln2a=aln2a-2a0.由于xln x,所以a+2a2aln2a,所以g(x)在ln2a,a+2a上存在一個(gè)零點(diǎn).又因?yàn)間ln2a2+a+2=aa-lna2+a+22,且ln2a2+a+20在(0,+)上恒成立,故h(a)在(0,+)上單調(diào)遞增.而h(0)=0,所以h(a)0在(0,+)上恒成立,所以gln2a2+a+20,所以g(x)在ln2a2+a+2,ln2a上存在一個(gè)零點(diǎn).綜上所述,當(dāng)a0時(shí),函數(shù)g(x)有兩個(gè)零點(diǎn).6.解 f(x)=x-a(1+ln x)的定義域?yàn)?0,+),f(x)=1-ax.(1)若a0,則f(x)=1-ax0,f(x)在(0,+)上單調(diào)遞增;若a0,令f(x)=1-ax=0,得x=a,當(dāng)0 xa時(shí),f(x)a時(shí),f(x)0,所以f(x)在(0,a)上單調(diào)遞減,在(a,+)上單調(diào)遞增.(2)由(1)可得:當(dāng)a0時(shí),f(x)在(0,+)上單調(diào)遞增,至多有1個(gè)零點(diǎn),不合題意;當(dāng)a0時(shí),f(x)min=f(a)=-aln a,(i)當(dāng)0a0,無(wú)零點(diǎn),不合題意;(ii)當(dāng)a=1時(shí),f(x)min=f(a)=-aln a=0,有1個(gè)零點(diǎn),不合題意;