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1、第27練導(dǎo)數(shù)與函數(shù)的單調(diào)性、極值、最值壓軸大題突破練明晰考情1.命題角度:討論函數(shù)的單調(diào)性、極值、最值以及利用導(dǎo)數(shù)求參數(shù)范圍是高考的熱點(diǎn).2.題目難度:偏難題.考點(diǎn)一利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性方法技巧(1)函數(shù)單調(diào)性的判定方法:在某個(gè)區(qū)間(a,b)內(nèi),如果f(x)0,那么函數(shù)yf(x)在此區(qū)間內(nèi)單調(diào)遞增;如果f(x)0,討論f(x)在(0,2)上的單調(diào)性.解因?yàn)閒(x)1(x0,k0).當(dāng)0kk0,且2,所以當(dāng)x(0,k)時(shí),f(x)0,所以函數(shù)f(x)在(0,k)上是減函數(shù),在(k,2)上是增函數(shù);當(dāng)k2時(shí),k2,f(x)2時(shí),0,所以當(dāng)x時(shí),f(x)0,所以函數(shù)f(x)在上是減函數(shù),在上是增
2、函數(shù).綜上可知,當(dāng)0k2時(shí),f(x)在上是減函數(shù),在上是增函數(shù).2.已知函數(shù)f(x)aln(x1)axx2,討論f(x)在定義域上的單調(diào)性.解f(x)a2x,令f(x)0,得x0或x,又f(x)的定義域?yàn)?1,),當(dāng)1,即當(dāng)a0時(shí),若x(1,0),f(x)0,則f(x)單調(diào)遞增;若x(0,),f(x)0,則f(x)單調(diào)遞減.當(dāng)10,即2a0時(shí),若x,f(x)0,則f(x)單調(diào)遞減;若x,f(x)0,則f(x)單調(diào)遞增;若x(0,),f(x)0,則f(x)單調(diào)遞減.當(dāng)0,即a2時(shí),f(x)0,f(x)在(1,)上單調(diào)遞減.當(dāng)0,即a2時(shí),若x(1,0),f(x)0,則f(x)單調(diào)遞減;若x,f(
3、x)0,則f(x)單調(diào)遞增;若x,f(x)0,則f(x)單調(diào)遞減.綜上,當(dāng)a0時(shí),f(x)在(1,0)上單調(diào)遞增,在(0,)上單調(diào)遞減;當(dāng)2a0時(shí),f(x)在上單調(diào)遞減,在上單調(diào)遞增,在(0,)上單調(diào)遞減;當(dāng)a2時(shí),f(x)在(1,)上單調(diào)遞減;當(dāng)a2時(shí),f(x)在(1,0)上單調(diào)遞減,在上單調(diào)遞增,在上單調(diào)遞減.3.設(shè)函數(shù)f(x)(aR).(1)若f(x)在x0處取得極值,確定a的值,并求此時(shí)曲線yf(x)在點(diǎn)(1,f(1)處的切線方程;(2)若f(x)在3,)上為減函數(shù),求a的取值范圍.解(1)對f(x)求導(dǎo),得f(x),因?yàn)閒(x)在x0處取得極值,所以f(0)0,即a0.當(dāng)a0時(shí),f(
4、x),f(x),故f(1),f(1),從而f(x)在點(diǎn)(1,f(1)處的切線方程為y(x1),化簡得3xey0.(2)由(1)知,f(x).令g(x)3x2(6a)xa,由g(x)0,解得x1,x2.當(dāng)xx1時(shí),g(x)0,即f(x)0,故f(x)為減函數(shù);當(dāng)x1xx2時(shí),g(x)0,即f(x)0,故f(x)為增函數(shù);當(dāng)xx2時(shí),g(x)0,即f(x)0,故f(x)為減函數(shù).由f(x)在3,)上為減函數(shù)知,x23,解得a,故a的取值范圍為.4.已知函數(shù)f(x)x22aln x(a2)x.(1)當(dāng)a1時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;(2)是否存在實(shí)數(shù)a,使函數(shù)g(x)f(x)ax在(0,)上單調(diào)遞
5、增?若存在,求出a的取值范圍;若不存在,請說明理由.解(1)當(dāng)a1時(shí),f(x)x22ln x3x(x0),則f(x)x3.當(dāng)0x2時(shí),f(x)0,f(x)單調(diào)遞增;當(dāng)1x2時(shí),f(x)0時(shí),求函數(shù)f(x)的單調(diào)遞增區(qū)間;(2)當(dāng)a0,x0,0,令f(x)0,即x10,得x1,f(x)的單調(diào)遞增區(qū)間為(1,).(2)由(1)可得f(x),a1,即a0時(shí),f(x)在(0,1)上是減函數(shù),f(x)在上的最小值為f(1)1a;當(dāng)1,即1a時(shí),當(dāng)x時(shí),f(x)0;當(dāng)x時(shí),f(x)0,因此f(x)在上是減函數(shù),在上是增函數(shù),f(x)的最小值為f1ln(2a);當(dāng),即a0,均有x(2ln aln x)a恒成
6、立,求正數(shù)a的取值范圍.解(1)f(x),x(0,).當(dāng)a0時(shí),f(x)0,f(x)在(0,)上為增函數(shù),無極值;當(dāng)a0,x(0,a)時(shí),f(x)0,f(x)在(a,)上為增函數(shù),所以f(x)在(0,)上有極小值,無極大值,f(x)的極小值為f(a)ln a1.(2)若對任意x0,均有x(2ln aln x)a恒成立,即對任意x0,均有2ln aln x恒成立,由(1)可知f(x)的最小值為ln a1,問題轉(zhuǎn)化為2ln aln a1,即ln a1,故0ae,故正數(shù)a的取值范圍是(0,e.典例(12分)設(shè)函數(shù)f(x)a2x2ln x(aR).(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)如果函數(shù)f(x)
7、的圖象不在x軸的下方,求實(shí)數(shù)a的取值范圍.審題路線圖(1)(2)規(guī)范解答評分標(biāo)準(zhǔn)解(1)f(x)a2x(x0).1分當(dāng)a0時(shí),f(x)0,故f(x)在(0,)上單調(diào)遞減.當(dāng)a0時(shí),f(x),由f(x)0,得x;由f(x)0,得0x.3分所以f(x)的單調(diào)遞減區(qū)間為,單調(diào)遞增區(qū)間為.當(dāng)a0時(shí),f(x),由f(x)0,得x;由f(x)0,得0x0時(shí),h(x)0;當(dāng)x0時(shí),h(x)0.當(dāng)a0時(shí),g(x)(xa)(xsin x),當(dāng)x(,a)時(shí),xa0,g(x)單調(diào)遞增;當(dāng)x(a,0)時(shí),xa0,g(x)0,g(x)0,g(x)單調(diào)遞增.所以當(dāng)xa時(shí),g(x)取到極大值,極大值是g(a)a3sin a
8、;當(dāng)x0時(shí),g(x)取到極小值,極小值是g(0)a.當(dāng)a0時(shí),g(x)x(xsin x),當(dāng)x(,)時(shí),g(x)0,g(x)單調(diào)遞增;所以g(x)在(,)上單調(diào)遞增,g(x)無極大值也無極小值.當(dāng)a0時(shí),g(x)(xa)(xsin x),當(dāng)x(,0)時(shí),xa0,g(x)單調(diào)遞增;當(dāng)x(0,a)時(shí),xa0,g(x)0,g(x)0,g(x)單調(diào)遞增.所以當(dāng)x0時(shí),g(x)取到極大值,極大值是g(0)a;當(dāng)xa時(shí),g(x)取到極小值,極小值是g(a)a3sin a.綜上所述,當(dāng)a0時(shí),函數(shù)g(x)在(,0)和(a,)上單調(diào)遞增,在(0,a)上單調(diào)遞減,函數(shù)既有極大值,又有極小值,極大值是g(0)a,
9、極小值是g(a)a3sin a.4.已知函數(shù)f(x)axln xx2.(1)若a1,求函數(shù)f(x)的極值;(2)若a1,x1(1,2),x2(1,2),使得f(x1)xmx2mx(m0),求實(shí)數(shù)m的取值范圍.解(1)依題意知,當(dāng)a1時(shí),f(x)xln xx2,f(x)12x,因?yàn)閤(0,),故當(dāng)x(0,1)時(shí),f(x)0,當(dāng)x(1,)時(shí),f(x)0,故當(dāng)x1時(shí),f(x)有極小值,極小值為f(1)0,無極大值.(2)當(dāng)a1時(shí),f(x)xln xx2.因?yàn)閤1(1,2),x2(1,2),使得f(x1)xmx2mx(m0),故ln x1x1mxmx2.設(shè)h(x)ln xx,g(x)mx3mx,當(dāng)x(1,2)時(shí),h(x)10,即函數(shù)h(x)在(1,2)上單調(diào)遞減,故h(x)的值域?yàn)锳(ln 22,1).又g(x)mx2mm(x1)(x1).當(dāng)m0時(shí),g(x)在(1,2)上單調(diào)遞減,此時(shí)g(x)的值域?yàn)锽,因?yàn)锳B,又01,故ln 22,即mln 23;當(dāng)m0時(shí),g(x)在(1,2)上單調(diào)遞增,此時(shí)g(x)的值域?yàn)锽,因?yàn)锳B,又01,故ln 22,故m(ln 22)3ln 2.綜上所述,實(shí)數(shù)m的取值范圍為.