高中總復(fù)習(xí)理科數(shù)學(xué)配人教版(老高考舊教材)課后習(xí)題Word題型練8　大題專項(xiàng)(六)　函數(shù)與導(dǎo)數(shù)綜合問題

1、 題型練8大題專項(xiàng)(六)函數(shù)與導(dǎo)數(shù)綜合問題題型練第72頁一、解答題1.設(shè)函數(shù)f(x)=ax2-(4a+1)x+4a+3ex.(1)若曲線y=f(x)在點(diǎn)(1,f(1)處的切線與x軸平行,求a;(2)若f(x)在x=2處取得極小值,求a的取值范圍.解：(1)因?yàn)閒(x)=ax2-(4a+1)x+4a+3ex,所以f(x)=2ax-(4a+1)ex+ax2-(4a+1)x+4a+3ex=ax2-(2a+1)x+2ex(xR).f(1)=(1-a)e.由題設(shè)知f(1)=0,即(1-a)e=0,解得a=1.此時(shí)f(1)=3e0,所以a的值為1.(2)由(1)得f(x)=ax2-(2a+1)x+2ex=

2、(ax-1)(x-2)ex.若a12,則當(dāng)x1a,2時(shí),f(x)0.所以f(x)在x=2處取得極小值.若a12,則當(dāng)x(0,2)時(shí),x-20,ax-112x-10.所以2不是f(x)的極小值點(diǎn).綜上可知,a的取值范圍是12,+.2.已知a3,函數(shù)F(x)=min2|x-1|,x2-2ax+4a-2,其中minp,q=p,pq,q,pq.(1)求使得等式F(x)=x2-2ax+4a-2成立的x的取值范圍.(2)求F(x)的最小值m(a);求F(x)在區(qū)間0,6上的最大值M(a).解：(1)由于a3,故當(dāng)x1時(shí),(x2-2ax+4a-2)-2|x-1|=x2+2(a-1)(2-x)0,當(dāng)x1時(shí),(

3、x2-2ax+4a-2)-2|x-1|=(x-2)(x-2a).所以,使得等式F(x)=x2-2ax+4a-2成立的x的取值范圍為2,2a.(2)設(shè)函數(shù)f(x)=2|x-1|,g(x)=x2-2ax+4a-2,則f(x)min=f(1)=0,g(x)min=g(a)=-a2+4a-2,所以,由F(x)的定義知m(a)=minf(1),g(a),即m(a)=0,3a2+2,-a2+4a-2,a2+2.當(dāng)0 x2時(shí),F(x)f(x)maxf(0),f(2)=2=F(2),當(dāng)2x6時(shí),F(x)g(x)maxg(2),g(6)=max2,34-8a=maxF(2),F(6).所以,M(a)=34-8a

4、,3a4,2,a4.3.已知函數(shù)f(x)=ex+ax2-x.(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí),f(x)12x3+1,求a的取值范圍.解：(1)當(dāng)a=1時(shí),f(x)=ex+x2-x,f(x)=ex+2x-1.故當(dāng)x(-,0)時(shí),f(x)0.所以f(x)在(-,0)單調(diào)遞減,在(0,+)單調(diào)遞增.(2)f(x)12x3+1等價(jià)于12x3-ax2+x+1e-x1.設(shè)函數(shù)g(x)=12x3-ax2+x+1e-x(x0),則g(x)=- 12x3-ax2+x+1-32x2+2ax-1e-x=-12xx2-(2a+3)x+4a+2e-x=-12x(x-2a-1)(x-2)e-x.若2

5、a+10,即a-12,則當(dāng)x(0,2)時(shí),g(x)0.所以g(x)在(0,2)單調(diào)遞增,而g(0)=1,故當(dāng)x(0,2)時(shí),g(x)1,不合題意.若02a+12,即-12a12,則當(dāng)x(0,2a+1)(2,+)時(shí),g(x)0.所以g(x)在(0,2a+1),(2,+)單調(diào)遞減,在(2a+1,2)單調(diào)遞增.因?yàn)間(0)=1,所以g(x)1當(dāng)且僅當(dāng)g(2)=(7-4a)e-21,即a7-e24.所以當(dāng)7-e24a12時(shí),g(x)1.若2a+12,即a12,則g(x)12x3+x+1e-x.由于07-e24,12,故由可得12x3+x+1e-x1.故當(dāng)a12時(shí),g(x)1.綜上,a的取值范圍是7-e

6、24,+.4.已知函數(shù)f(x)=12ax-a+1-lnxx.(1)若函數(shù)f(x)為減函數(shù),求實(shí)數(shù)a的取值范圍;(2)若函數(shù)f(x)有兩個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍.解：(1)f(x)=12ax-a+1-lnxx,f(x)=12a-1-lnxx2.函數(shù)f(x)為減函數(shù),f(x)0,即12a1-lnxx2對(duì)x(0,+)恒成立.設(shè)m(x)=1-lnxx2,則m(x)=2lnx-3x3.m(x)在區(qū)間0,e32內(nèi)單調(diào)遞減,在區(qū)間(e32,+)內(nèi)單調(diào)遞增.m(x)min=m(e32)=-12e3.12a-12e3,即a-e-3,故實(shí)數(shù)a的取值范圍是(-,-e-3.(2)易知函數(shù)f(x)的定義域?yàn)?0

7、,+),f(x)=12ax2-(a-1)x-lnxx.設(shè)h(x)=12ax2-(a-1)x-ln x,則原命題等價(jià)于函數(shù)h(x)有兩個(gè)不同的零點(diǎn),求實(shí)數(shù)a的取值范圍,可知h(x)=ax-(a-1)-1x=ax2-(a-1)x-1x=(ax+1)(x-1)x,當(dāng)a0時(shí),函數(shù)h(x)在區(qū)間(0,1)內(nèi)單調(diào)遞減,在區(qū)間(1,+)內(nèi)單調(diào)遞增.若函數(shù)h(x)有兩個(gè)不同的零點(diǎn),則必有h(1)=-12a+12.此時(shí),在x(1,+)內(nèi),有h(2)=2a-2(a-1)-ln 2=2-ln 20;在x(0,1)內(nèi),h(x)=12a(x2-2x)+x-ln x,-1x2-2x-12a+x-ln x,h(e-12a)

8、-12a+e-12a-ln(e-12a)=e-12a0,h(x)在區(qū)間(0,1),(1,+)內(nèi)各有一個(gè)零點(diǎn),故a2符合題意;當(dāng)a=-1時(shí),可知函數(shù)h(x)在區(qū)間(0,+)內(nèi)單調(diào)遞減,函數(shù)h(x)至多有一個(gè)零點(diǎn),不符合題意;當(dāng)-1a0,函數(shù)h(x)至多有一個(gè)零點(diǎn),不符合題意;當(dāng)a0,函數(shù)h(x)至多有一個(gè)零點(diǎn),不符合題意.綜上所述,實(shí)數(shù)a的取值范圍是(2,+).5.設(shè)函數(shù)f(x)=ablnxx,g(x)=-12x+(a+b)(其中e為自然對(duì)數(shù)的底數(shù),a,bR,且a0),曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程為y=ae(x-1).(1)求b的值;(2)若對(duì)任意x1e,+,f(x)與g(x)

9、有且只有兩個(gè)交點(diǎn),求a的取值范圍.解：(1)由f(x)=ablnxx,得f(x)=ab(1-lnx)x2,由題意得f(1)=ab=ae.a0,b=e.(2)令h(x)=xf(x)-g(x)=12x2-(a+e)x+aeln x,則任意x1e,+,f(x)與g(x)有且只有兩個(gè)交點(diǎn),等價(jià)于函數(shù)h(x)在區(qū)間1e,+有且只有兩個(gè)零點(diǎn).由h(x)=12x2-(a+e)x+aeln x,得h(x)=(x-a)(x-e)x,當(dāng)a1e時(shí),由h(x)0得xe;由h(x)0得1exe.此時(shí)h(x)在區(qū)間1e,e內(nèi)單調(diào)遞減,在區(qū)間(e,+)內(nèi)單調(diào)遞增.因?yàn)閔(e)=12e2-(a+e)e+aeln e=-12e

10、20(或當(dāng)x+時(shí),h(x)0亦可),所以要使得h(x)在區(qū)間1e,+內(nèi)有且只有兩個(gè)零點(diǎn),則只需h1e=12e2-a+ee+aeln1e=(1-2e2)-2e(1+e2)a2e20,即a1-2e22e(1+e2).當(dāng)1ea0得1exe;由h(x)0得axe.此時(shí)h(x)在區(qū)間(a,e)內(nèi)單調(diào)遞減,在區(qū)間1e,a和(e,+)內(nèi)單調(diào)遞增.此時(shí)h(a)=-12a2-ae-aeln a-12a2-ae+aeln e=-12a2e時(shí),由h(x)0得1exa,由h(x)0得exa,此時(shí)h(x)在區(qū)間1e,e和(a,+)內(nèi)單調(diào)遞增,在區(qū)間(e,a)上單調(diào)遞減,且h(e)=-12e20;當(dāng)x3,23時(shí),f(x)0.所以f(x)在區(qū)間0,3,23,單調(diào)遞增,在區(qū)間3,23單調(diào)遞減.(2)證明因?yàn)閒(0)=f()=0,由(1)知,f(x)在區(qū)間0,的最大值為f3=338,最小值為f23=-338.而f(x)是周期為的周期函數(shù),故|f(x)|338.(3)證明由于(sin2xsi