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1、(全國(guó)通用版)2022高考數(shù)學(xué)二輪復(fù)習(xí) 壓軸大題突破練(三)函數(shù)與導(dǎo)數(shù)(1)文1(2018咸陽模擬)已知函數(shù)f(x)a(x1)ln xx1(aR)(1)當(dāng)a2時(shí),求函數(shù)f(x)在點(diǎn)(1,f(1)處的切線方程;(2)當(dāng)a時(shí),求證:對(duì)任意的x1,f(x)0恒成立(1)解由f(x)2(x1)ln xx1,得f(x)2ln x1,切點(diǎn)為(1,0),斜率為f(1)3,所求切線方程為y3(x1),即3xy30.(2)證明當(dāng)a時(shí),f(x)(x1)ln xx1(x1),欲證:f(x)0,注意到f(1)0,只要f(x)f(1)即可,f(x)a1(x1),令g(x)ln x1(x1),則g(x)0(x1),知g(
2、x)在1,)上單調(diào)遞增,有g(shù)(x)g(1)2,所以f(x)2a10,可知f(x)在1,)上單調(diào)遞增,所以f(x)f(1)0,綜上,當(dāng)a時(shí),對(duì)任意的x1,f(x)0恒成立2(2018濰坊模擬)已知函數(shù)f(x)ln xx2ax(aR),g(x)exx2.(1)討論函數(shù)f(x)極值點(diǎn)的個(gè)數(shù);(2)若對(duì)x0,不等式f(x)g(x)恒成立,求實(shí)數(shù)a的取值范圍解(1)f(x)xa(x0),令f(x)0,即x2ax10,a24,當(dāng)a240,即2a2時(shí),x2ax10恒成立,即f(x)0,此時(shí)f(x)在(0,)上單調(diào)遞增,無極值點(diǎn),當(dāng)a240,即a2時(shí),若a2,設(shè)方程x2ax10的兩根為x1,x2,且x10,x
3、20,此時(shí)x(0,x1),f(x)0,f(x)單調(diào)遞增,x(x1,x2),f(x)0,f(x)單調(diào)遞增,故x1,x2分別為f(x)的極大值點(diǎn)和極小值點(diǎn),因此a2,設(shè)方程x2ax10的兩根為x1,x2,且x1x2,由根與系數(shù)的關(guān)系得故x10,x20,此時(shí)f(x)無極值點(diǎn),綜上,當(dāng)2a2時(shí),f(x)無極值點(diǎn),當(dāng)a0恒成立設(shè)h(x),h(x),當(dāng)x(0,1)時(shí),ex(x1)ln xx210,即h(x)0,即h(x)0,h(x)單調(diào)遞增,因此x1為h(x)的極小值點(diǎn),即h(x)h(1)e1,故ae1.3(2018亳州模擬)已知函數(shù)f(x)在x1處取得極值(1)求a的值,并討論函數(shù)f(x)的單調(diào)性;(2
4、)當(dāng)x1,)時(shí),f(x)恒成立,求實(shí)數(shù)m的取值范圍解(1)由題意知f(x),又f(1)1a0,即a1, f(x)(x0),令f(x)0,得0x1;令f(x)1,函數(shù)f(x)在(0,1)上單調(diào)遞增,在(1,)上單調(diào)遞減(2)依題意知,當(dāng)x1,)時(shí),f(x)恒成立,即m恒成立,令g(x)(x1),只需g(x)minm即可,又g(x),令h(x)xln x,h(x)10(x1),h(x)在1,)上單調(diào)遞增, h(x)h(1)10, g(x)0,g(x)在1,)上單調(diào)遞增,g(x)ming(1)2,故m2.4(2018福建省百校模擬)已知函數(shù)f(x)x1aex.(1)討論f(x)的單調(diào)性;(2)當(dāng)a1
5、時(shí),設(shè)1x10且f(x1)f(x2)5,證明:x12x24.(1)解f(x)1aex,當(dāng)a0時(shí),f(x)0,則f(x)在R上單調(diào)遞增當(dāng)a0,得xln,則f(x)的單調(diào)遞增區(qū)間為,令f(x)ln,則f(x)的單調(diào)遞減區(qū)間為.(2)證明方法一設(shè)g(x)f(x)2xex3x1,則g(x)ex3,由g(x)ln 3;由g(x)0得xln 3,故g(x)maxg(ln 3)3ln 340,從而得g(x)f(x)2x0,f(x1)f(x2)5,f(x2)2x25f(x1)2x24.方法二f(x1)f(x2)5,x1x23,x12x23x23,設(shè)g(x)ex3x,則g(x)ex3,由g(x)0得x0得xln
6、 3,故g(x)ming(ln 3)33ln 3.1x10,x12x2e133ln 333ln 3,3ln 3ln 274.5(2018江南十校模擬)已知函數(shù)f(x),g(x)mx.(1)求函數(shù)f(x)的單調(diào)區(qū)間;(2)當(dāng)a0時(shí),f(x)g(x)恒成立,求實(shí)數(shù)m的取值范圍;(3)當(dāng)a1時(shí),求證:當(dāng)x1時(shí),(x1)f(x)2.(1)解f(x)的定義域?yàn)?0,),且f(x).由f(x)0得1ln xa0,即ln x1a,解得0x0得0x2,等價(jià)于.令p(x),則p(x),令(x)xln x,則(x)1,x1,(x)0,(x)在(1,)上單調(diào)遞增,(x)(1)10,p(x)0,p(x)在(1,)上單調(diào)遞增,p(x)p(1)2,令h(x),則h(x),x1,1ex0,h(x)1時(shí),h(x)h(x),即(x1)f(x)2,x1.