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1、(全國通用版)2022高考數(shù)學(xué)二輪復(fù)習(xí) 壓軸大題突破練(四)函數(shù)與導(dǎo)數(shù)(2)文1(2018成都模擬)已知f(x)ln xax1(aR)(1)討論函數(shù)的單調(diào)性;(2)證明:當(dāng)a2,且x1時,f(x)ex12恒成立(1)解 f(x)ln xax1,aR,f(x)a,當(dāng)a0時,f(x)的增區(qū)間為(0,),無減區(qū)間,當(dāng)a0時,增區(qū)間為,減區(qū)間為.(2)證明當(dāng)x1,)時,由(1)可知當(dāng)a2時,f(x)在1,)上單調(diào)遞減,f(x)f(1)1,再令G(x)ex12,在x1,)上,G(x)ex10,G(x)單調(diào)遞增,所以G(x)G(1)1,所以G(x)f(x)恒成立,當(dāng)x1時取等號,所以原不等式恒成立2(20
2、18合肥模擬)已知函數(shù)f(x)xln x,g(x)(x21)(為常數(shù))(1)若函數(shù)yf(x)與函數(shù)yg(x)在x1處有相同的切線,求實數(shù)的值;(2)當(dāng)x1時,f(x)g(x),求實數(shù)的取值范圍解(1)由題意得f(x)ln x1,g(x)2x,又f(1)g(1)0,且函數(shù)yf(x)與yg(x)在x1處有相同的切線,f(1)g(1),則21,即.(2)設(shè)h(x)xln x(x21),則h(x)0對x1,)恒成立h(x)1ln x2x,且h(1)0,h(1)0,即120,.另一方面,當(dāng)時,記(x)h(x),則(x)2.當(dāng)x1,)時,(x)0,(x)在1,)內(nèi)為減函數(shù),當(dāng)x1,)時,(x)(1)120
3、,即h(x)0,h(x)在1,)內(nèi)為減函數(shù),當(dāng)x1,)時,h(x)h(1)0恒成立,符合題意當(dāng)時,若0,則h(x)1ln x2x0對x1,)恒成立,h(x)在1,)內(nèi)為增函數(shù),當(dāng)x1,)時,h(x)h(1)0恒成立,不符合題意若00,則1x(1)120,即h(x)0,h(x)在內(nèi)為增函數(shù),當(dāng)x時,h(x)h(1)0,不符合題意,綜上所述,的取值范圍是.3(2018山東省名校聯(lián)盟模擬)已知f(x)xexa(x1)2.(1)若函數(shù)f(x)在x1處取得極值,求a的值;(2)當(dāng)x2時,f(x)0,求a的取值范圍解(1)f(x)(x1)ex2a(x1)(x1)(ex2a),若函數(shù)f(x)在x1處取得極值
4、,則f(1)0,所以a,經(jīng)檢驗,當(dāng)a時,函數(shù)f(x)在x1處取得極值(2)f(x)(x1)ex2a(x1)(x1)(ex2a),a0時,當(dāng)2x1時,f(x)1時,f(x)0,f(x)為增函數(shù);又f(1)0,當(dāng)x2時,f(x)0成立a1,即a時,當(dāng)2xln(2a)時,f(x)0;當(dāng)1xln(2a)時,f(x)0,則f(x)在(2,1),(ln(2a),)上為增函數(shù),在(1,ln(2a)上為減函數(shù),又f(1)0,f(x)在(1,ln(2a)上小于零,不符合題意,舍去當(dāng)ln(2a)1,即a時,當(dāng)2x1時,f(x)0,f(x)在(2,)上單調(diào)遞增,又f(1)0,當(dāng)x(2,1)時,f(x)0,不符合題意
5、,舍去;當(dāng)2ln(2a)1,即a時,當(dāng)2x1時,f(x)0,當(dāng)ln(2a)x1時,f(x)0,則f(x)在(2,ln(2a),(1,)上為增函數(shù),在(ln(2a),1)上為減函數(shù),又f(1)0,要使f(x)0恒成立,則f(2)0,則a,又a,a.當(dāng)ln(2a)2,即a1時,f(x)0,當(dāng)2x1時,f(x)0時,xexeln xx3x2.(1)解由題意可知,g(x) f(x)xaaex,則g(x)1aex,當(dāng)a0時,g(x)0,g(x)在(,)上單調(diào)遞增;當(dāng)a0時,當(dāng)x0,當(dāng)xln a時,g(x)0時,g(x)的單調(diào)遞增區(qū)間為(,ln a),單調(diào)遞減區(qū)間為(ln a,)(2)解由(1)可知,a0
6、,且g(x)在xln a處取得最大值,g(ln a)ln aaaealn a1,即aln a10,觀察可得當(dāng)a1時,方程成立,令h(a)aln a1(a0),h(a)1,當(dāng)a(0,1)時,h(a)0,h(a)在(0,1)上單調(diào)遞減,在(1,)上單調(diào)遞增,h(a)h(1)0,當(dāng)且僅當(dāng)a1時,aln a10,f(x)x2xex,由題意可知f(x)g(x)0,f(x)在0,)上單調(diào)遞減,f(x)在x0處取得最大值f(0)1.(3)證明由(2)知,若a1,當(dāng)x0時,f(x)1,即x2xex1,x3x2xexx,x3x2xexeln xeln xx,令F(x)eln xx,F(xiàn)(x)1,當(dāng)0x0;當(dāng)xe時
7、,F(xiàn)(x)0,F(xiàn)(x)在(0,e)上單調(diào)遞增,在(e,)上單調(diào)遞減,F(xiàn)(x)F(e)0,即eln xx0,x3x2xexeln x0時,xexeln xx3x2.5(2018四省名校大聯(lián)考)已知函數(shù)f(x)a(x1)2ex(aR)(1)當(dāng)a時,判斷函數(shù)f(x)的單調(diào)性;(2)若f(x)有兩個極值點x1,x2(x1x2)求實數(shù)a的取值范圍;證明:f(x1)0,得x0,由g(x)1ex0,g(x)即f(x)在區(qū)間(,0)上單調(diào)遞增,在區(qū)間(0,)上單調(diào)遞減f(x)maxf(0)0.對xR,f(x)0,f(x)在R上單調(diào)遞減(2)解f(x)有兩個極值點,關(guān)于x的方程f(x)2a(x1)ex0有兩個根
8、x1,x2,設(shè)(x)2a(x1)ex,則(x)2aex,當(dāng)a0時,(x)2aex0 時,由(x)0,得xln 2a,由(x)ln 2a,(x)即f(x)在區(qū)間(,ln 2a)上單調(diào)遞增,在區(qū)間(ln 2a,)上單調(diào)遞減且當(dāng)x時,f(x),當(dāng)x時,f(x),要使f(x)0有兩個不同的根,必有f(x)maxf(ln 2a)2a(ln 2a1)2a2aln 2a0,解得a,實數(shù)a的取值范圍是.證明f(1)0,1x10,又f(x1)2a(x11)ex10,a,f(x1)a(x11)2(x11)(x11)(1x10),令h(x)(x1)ex(1x0),則h(x)xex0,h(x)在區(qū)間(1,0)上單調(diào)遞減,f(0)f(x1),f(1),f(x1).