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1、專題跟蹤檢測(四) “導(dǎo)數(shù)與不等式”考法面面觀1(2019屆高三唐山模擬)已知f(x)x2a2ln x,a0.(1)求函數(shù)f(x)的最小值;(2)當(dāng)x2a時(shí),證明:a.解:(1)函數(shù)f(x)的定義域?yàn)?0,),f(x)x.當(dāng)x(0,a)時(shí),f(x)0,f(x)單調(diào)遞增所以當(dāng)xa時(shí),f(x)取得極小值,也是最小值,且f(a)a2a2ln a.(2)證明:由(1)知,f(x)在(2a,)上單調(diào)遞增,則所證不等式等價(jià)于f(x)f(2a)a(x2a)0.設(shè)g(x)f(x)f(2a)a(x2a),則當(dāng)x2a時(shí),g(x)f(x)axa0,所以g(x)在(2a,)上單調(diào)遞增,當(dāng)x2a時(shí),g(x)g(2a)0
2、,即f(x)f(2a)a(x2a)0,故a.2已知函數(shù)f(x)xex2xaln x,曲線yf(x)在點(diǎn)P(1,f(1)處的切線與直線x2y10垂直(1)求實(shí)數(shù)a的值;(2)求證:f(x)x22.解:(1)因?yàn)閒(x)(x1)ex2,所以曲線yf(x)在點(diǎn)P(1,f(1)處的切線斜率kf(1)2e2a.而直線x2y10的斜率為,由題意可得(2e2a)1,解得a2e.(2)證明:由(1)知,f(x)xex2x2eln x.不等式f(x)x22可化為xex2x2eln xx220.設(shè)g(x)xex2x2eln xx22,則g(x)(x1)ex22x.記h(x)(x1)ex22x(x0),則h(x)(
3、x2)ex2,因?yàn)閤0,所以x22,ex1,故(x2)ex2,又0,所以h(x)(x2)ex20,所以函數(shù)h(x)在(0,)上單調(diào)遞增又h(1)2e22e20,所以當(dāng)x(0,1)時(shí),h(x)0,即g(x)0,即g(x)0,函數(shù)g(x)單調(diào)遞增所以g(x)g(1)e22eln 112e1,顯然e10,所以g(x)0,即xex2x2eln xx22,也就是f(x)x22.3(2018武漢模擬)設(shè)函數(shù)f(x)(1xx2)ex(e2.718 28是自然對(duì)數(shù)的底數(shù))(1)討論f(x)的單調(diào)性;(2)當(dāng)x0時(shí),f(x)ax12x2恒成立,求實(shí)數(shù)a的取值范圍解:(1)f(x)(2xx2)ex(x2)(x1)
4、ex.當(dāng)x1時(shí),f(x)0;當(dāng)2x0.所以f(x)在(,2),(1,)上單調(diào)遞減,在(2,1)上單調(diào)遞增(2)設(shè)F(x)f(x)(ax12x2),F(xiàn)(0)0,F(xiàn)(x)(2xx2)ex4xa,F(xiàn)(0)2a,當(dāng)a2時(shí),F(xiàn)(x)(2xx2)ex4xa(x2)(x1)ex4x2(x2)(x1)exx2(x2)(x1)ex1,設(shè)h(x)(x1)ex1,h(x)xex0,所以h(x)在0,)上單調(diào)遞增,h(x)(x1)ex1h(0)0,即F(x)0在0,)上恒成立,F(xiàn)(x)在0,)上單調(diào)遞減,F(xiàn)(x)F(0)0,所以f(x)ax12x2在0,)上恒成立當(dāng)a0,而函數(shù)F(x)的圖象在(0,)上連續(xù)且x,F(xiàn)(
5、x)逐漸趨近負(fù)無窮,必存在正實(shí)數(shù)x0使得F(x0)0且在(0,x0)上F(x)0,所以F(x)在(0,x0)上單調(diào)遞增,此時(shí)F(x)F(0)0,f(x)ax12x2有解,不滿足題意綜上,a的取值范圍是2,)4(2018南昌模擬)設(shè)函數(shù)f(x)2ln xmx21.(1)討論函數(shù)f(x)的單調(diào)性;(2)當(dāng)f(x)有極值時(shí),若存在x0,使得f(x0)m1成立,求實(shí)數(shù)m的取值范圍解:(1)函數(shù)f(x)的定義域?yàn)?0,),f(x)2mx,當(dāng)m0時(shí),f(x)0,f(x)在(0,)上單調(diào)遞增;當(dāng)m0時(shí),令f(x)0,得0x,令f(x),f(x)在上單調(diào)遞增,在上單調(diào)遞減(2)由(1)知,當(dāng)f(x)有極值時(shí),
6、m0,且f(x)在上單調(diào)遞增,在上單調(diào)遞減f(x)maxf2lnm1ln m,若存在x0,使得f(x0)m1成立,則f(x)maxm1.即ln mm1,ln mm10),g(x)10,g(x)在(0,)上單調(diào)遞增,且g(1)0,0m0時(shí),對(duì)任意的x,恒有f(x)e1成立,求實(shí)數(shù)b的取值范圍解:(1)函數(shù)f(x)的定義域?yàn)?0,)當(dāng)b2時(shí),f(x)aln xx2,所以f(x)2x.當(dāng)a0時(shí),f(x)0,所以函數(shù)f(x)在(0,)上單調(diào)遞增當(dāng)a0時(shí),令f(x)0,解得x (負(fù)值舍去),當(dāng)0x 時(shí),f(x)時(shí),f(x)0,所以函數(shù)f(x)在上單調(diào)遞增綜上所述,當(dāng)b2,a0時(shí),函數(shù)f(x)在(0,)上
7、單調(diào)遞增;當(dāng)b2,a0時(shí),f(x)bln xxb,f(x)bxb1.令f(x)0,得0x0,得x1.所以函數(shù)f(x)在上單調(diào)遞減,在(1,e上單調(diào)遞增,f(x)max為fbeb與f(e)beb中的較大者f(e)febeb2b.令g(m)emem2m(m0),則當(dāng)m0時(shí),g(m)emem2220,所以g(m)在(0,)上單調(diào)遞增,故g(m)g(0)0,所以f(e)f,從而f(x)maxf(e)beb所以bebe1,即ebbe10.設(shè)(t)ette1(t0),則(t)et10,所以(t)在(0,)上單調(diào)遞增又(1)0,所以ebbe10的解集為(0,1所以b的取值范圍為(0,16(2018開封模擬)
8、已知函數(shù)f(x)axx2xln a(a0,a1)(1)當(dāng)ae(e是自然對(duì)數(shù)的底數(shù))時(shí),求函數(shù)f(x)的單調(diào)區(qū)間;(2)若存在x1,x21,1,使得|f(x1)f(x2)|e1,求實(shí)數(shù)a的取值范圍解:(1)f(x)axln a2xln a2x(ax1)ln a.當(dāng)ae時(shí),f(x)2xex1,其在R上是增函數(shù),又f(0)0,f(x)0的解集為(0,),f(x)1時(shí),ln a0,y(ax1)ln a在R上是增函數(shù),當(dāng)0a1時(shí),ln a1或0a0),g(a)120,g(a)a2ln a在(0,)上是增函數(shù)而g(1)0,故當(dāng)a1時(shí),g(a)0,即f(1)f(1);當(dāng)0a1時(shí),g(a)0,即f(1)1時(shí),f(x)maxf(x)minf(1)f(0)e1,即aln ae1,函數(shù)yaln a在(1,)上是增函數(shù),解得ae;當(dāng)0a1時(shí),f(x)maxf(x)minf(1)f(0)e1,即ln ae1,函數(shù)yln a在(0,1)上是減函數(shù),解得0a.綜上可知,實(shí)數(shù)a的取值范圍為e,)