2024高考數(shù)學(xué)一輪復(fù)習(xí)第三章導(dǎo)數(shù)及其應(yīng)用3.2第1課時(shí)利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性學(xué)案文含解析新人教A版

3.2導(dǎo)數(shù)在探討函數(shù)中的應(yīng)用第1課時(shí)利用導(dǎo)數(shù)探討函數(shù)的單調(diào)性必備學(xué)問(wèn)預(yù)案自診學(xué)問(wèn)梳理函數(shù)的單調(diào)性與導(dǎo)數(shù)的關(guān)系(1)已知函數(shù)f(x)在某個(gè)區(qū)間內(nèi)可導(dǎo),①假如f'(x)>0,那么函數(shù)y=f(x)在這個(gè)區(qū)間上;

②假如f'(x)<0,那么函數(shù)y=f(x)在這個(gè)區(qū)間上;

③假如f'(x)=0,那么函數(shù)y=f(x)在這個(gè)區(qū)間上是常數(shù)函數(shù).(2)可導(dǎo)函數(shù)f(x)在[a,b]上單調(diào)遞增,則有f'(x)≥0在[a,b]上恒成立.(3)可導(dǎo)函數(shù)f(x)在[a,b]上單調(diào)遞減,則有f'(x)≤0在[a,b]上恒成立.(4)若函數(shù)y=f(x)在區(qū)間(a,b)上具有單調(diào)性,則f'(x)在該區(qū)間上不變號(hào).可導(dǎo)函數(shù)f(x)在(a,b)上單調(diào)遞增(減)的充要條件是?x∈(a,b),都有f'(x)≥0(f'(x)≤0)且f'(x)在(a,b)的任何子區(qū)間上都不恒為零.考點(diǎn)自診1.推斷下列結(jié)論是否正確,正確的畫(huà)“√”,錯(cuò)誤的畫(huà)“×”.(1)在(a,b)上f'(x)≤0,且f'(x)=0的根有有限個(gè),則f(x)在(a,b)上單調(diào)遞減.()(2)若函數(shù)f(x)在定義域上都有f'(x)<0,則函數(shù)f(x)在定義域上肯定單調(diào)遞減.()(3)已知函數(shù)f(x)在區(qū)間[a,b]上單調(diào)遞增,則f'(x)>0恒成立.()(4)在某區(qū)間上f'(x)>0(f'(x)<0)是函數(shù)f(x)在此區(qū)間上單調(diào)遞增(減)的充分不必要條件.()(5)函數(shù)f(x)=sinx-2x在(0,π)上單調(diào)遞減.()2.函數(shù)f(x)=lnx+2x的單調(diào)遞減區(qū)間是A.1e,+∞ B.(1,+∞)C.0,1e D.(0,1)3.函數(shù)f(x)=ex-ex,x∈R的單調(diào)遞增區(qū)間是()A.(0,+∞) B.(-∞,0)C.(-∞,1) D.(1,+∞)4.(2024天津河北區(qū)線(xiàn)上測(cè)試,6)已知函數(shù)f(x)=3x+2cosx,若a=f(32),b=f(2),c=f(log27),則a,b,c的大小關(guān)系是(A.a<b<c B.c<a<bC.b<a<c D.b<c<a5.已知函數(shù)f(x)=12x2+2ax-lnx,若f(x)在區(qū)間13,2上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為.

關(guān)鍵實(shí)力學(xué)案突破考點(diǎn)利用導(dǎo)數(shù)求函數(shù)的單調(diào)區(qū)間【例1】已知函數(shù)f(x)=lnx-2x2+3,則函數(shù)f(x)的單調(diào)遞增區(qū)間為.

解題心得1.利用導(dǎo)數(shù)求單調(diào)區(qū)間(或證明單調(diào)性),只要在函數(shù)定義域內(nèi)解(或證明)不等式f'(x)>0或f'(x)<0.2.求f(x)的單調(diào)區(qū)間,需知f'(x)的正負(fù),若f'(x)不含參數(shù),但又不好推斷正負(fù),將f'(x)中正負(fù)不定的部分設(shè)為g(x),對(duì)g(x)再進(jìn)行一次或二次求導(dǎo),由g'(x)的正負(fù)及g(x)的零點(diǎn)推斷出g(x)的單調(diào)性及最值,進(jìn)而得出f'(x)的正負(fù).對(duì)點(diǎn)訓(xùn)練1(2024陜西西安中學(xué)六模,文21)設(shè)函數(shù)f(x)=x+lnx(1)求函數(shù)f(x)在x=1處的切線(xiàn)方程;(2)求函數(shù)f(x)的單調(diào)區(qū)間.考點(diǎn)利用導(dǎo)數(shù)探討含參數(shù)的函數(shù)的單調(diào)性【例2】設(shè)a>0,探討函數(shù)f(x)=lnx+a(1-a)x2-2(1-a)x的單調(diào)性.解題心得對(duì)于含參數(shù)的函數(shù)的單調(diào)性的探討,常見(jiàn)的分類(lèi)探討點(diǎn)按探討的先后依次有以下三個(gè):分類(lèi)探討點(diǎn)1:求導(dǎo)后,考慮f'(x)=0是否有實(shí)數(shù)根,從而引起分類(lèi)探討;分類(lèi)探討點(diǎn)2:求導(dǎo)后,f'(x)=0有實(shí)數(shù)根,但不清晰f'(x)=0的實(shí)數(shù)根是否落在定義域內(nèi),從而引起分類(lèi)探討;分類(lèi)探討點(diǎn)3:求導(dǎo)后,f'(x)=0有實(shí)數(shù)根,f'(x)=0的實(shí)數(shù)根也落在定義域內(nèi),但不清晰這些實(shí)數(shù)根的大小關(guān)系,從而引起分類(lèi)探討.對(duì)點(diǎn)訓(xùn)練2(2024全國(guó)2,文21)已知函數(shù)f(x)=2lnx+1.(1)若f(x)≤2x+c,求實(shí)數(shù)c的取值范圍;(2)設(shè)a>0,探討函數(shù)g(x)=f(x考點(diǎn)函數(shù)單調(diào)性的應(yīng)用(多考向探究)考向1比較大小或解不等式【例3】(1)(2024湖南長(zhǎng)郡中學(xué)四模,11)若0<x<1,則ln3+13,x2A.xB.xC.ln3+1D.ln3+1(2)(2024河北保定二模,文12)設(shè)函數(shù)f(x)是定義在R上的函數(shù),其導(dǎo)函數(shù)為f'(x),若f(x)+f'(x)>1,f(0)=2020,則不等式exf(x)>ex+2019的解集為()A.(-∞,0)B.(-∞,0)∪(2019,+∞)C.(2019,+∞)D.(0,+∞)解題心得利用導(dǎo)數(shù)比較大小或解不等式的常用技巧利用題目條件,構(gòu)造協(xié)助函數(shù),把比較大小或求解不等式的問(wèn)題轉(zhuǎn)化為先利用導(dǎo)數(shù)探討函數(shù)的單調(diào)性,再由單調(diào)性比較大小或解不等式的問(wèn)題.對(duì)點(diǎn)訓(xùn)練3(1)設(shè)f(x),g(x)在[a,b]上可導(dǎo),且f'(x)>g'(x),則當(dāng)a<x<b時(shí),有()A.f(x)>g(x) B.f(x)<g(x)C.f(x)+g(a)>g(x)+f(a) D.f(x)+g(b)>g(x)+f(b)(2)(2024山東煙臺(tái)模擬,16)設(shè)定義域?yàn)镽的函數(shù)f(x)滿(mǎn)意f'(x)>f(x),則不等式ex-1f(x)<f(2x-1)的解集為.

考向2已知函數(shù)單調(diào)性求參數(shù)取值范圍【例4】已知函數(shù)f(x)=lnx,g(x)=12ax2+2x.若函數(shù)h(x)=f(x)-g(x)在[1,4]上單調(diào)遞減,則實(shí)數(shù)a的取值范圍為.解題心得利用函數(shù)單調(diào)性求參數(shù)取值范圍的兩類(lèi)熱點(diǎn)問(wèn)題的處理方法(1)函數(shù)f(x)在區(qū)間D上存在單調(diào)遞增(減)區(qū)間.方法一:轉(zhuǎn)化為“f'(x)>0(<0)在區(qū)間D上有解”;方法二:轉(zhuǎn)化為“存在區(qū)間D的一個(gè)子區(qū)間使f'(x)>0(或f'(x)<0)成立”.(2)函數(shù)f(x)在區(qū)間D上單調(diào)遞增(減).方法一:轉(zhuǎn)化為“f'(x)≥0(≤0)在區(qū)間D上恒成立”;方法二:轉(zhuǎn)化為“區(qū)間D是函數(shù)f(x)的單調(diào)遞增(減)區(qū)間的子集”.變式發(fā)散1將例4中的“在[1,4]上單調(diào)遞減”改為“存在單調(diào)遞減區(qū)間”,其他不變.變式發(fā)散2例4中的已知條件不變,把后面的問(wèn)題改為:探討函數(shù)h(x)=f(x)-g(x)的單調(diào)性.對(duì)點(diǎn)訓(xùn)練4(1)(2024湖南湘潭三模,文12)已知函數(shù)f(x)=12ax2+3x-3ex是減函數(shù),則正數(shù)a=(A.9 B.e2 C.3 D.e(2)若函數(shù)f(x)=x-13sin2x+asinx在區(qū)間(-∞,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是(A.[-1,1] B.-1,13C.-13,13 D.-1,-1閱讀下列四個(gè)在抽象函數(shù)中構(gòu)造協(xié)助函數(shù),利用協(xié)助函數(shù)解決問(wèn)題的案例,思索如何構(gòu)造協(xié)助函數(shù).你能不能從詳細(xì)的實(shí)例中抽象出構(gòu)造協(xié)助函數(shù)的數(shù)學(xué)結(jié)論?【例1】已知f(x)是定義在R上的奇函數(shù),f(-1)=0,當(dāng)x<0時(shí),f(x)+xf'(x)<0,則使得f(x)>0成立的x的取值范圍是()A.(-∞,-1)∪(0,1) B.(-1,0)∪(1,+∞)C.(-∞,-1)∪(-1,0) D.(0,1)∪(1,+∞)答案B解析構(gòu)造函數(shù)F(x)=xf(x).當(dāng)x<0時(shí),F'(x)=f(x)+xf'(x)<0,F(x)單調(diào)遞減.又因?yàn)閒(-1)=0,所以F(-1)=0.所以當(dāng)-1<x<0時(shí),F(x)<0,所以當(dāng)-1<x<0時(shí),f(x)>0.因?yàn)閒(x)為奇函數(shù),所以F(x)=xf(x)為偶函數(shù),所以當(dāng)x>1時(shí),F(x)>0,所以當(dāng)x>1時(shí),f(x)>0.綜上可知,f(x)>0的解集為(-1,0)∪(1,+∞).故選B.【例2】已知函數(shù)f(x)滿(mǎn)意f(x)+2f'(x)>0,則下列不等式成立的是()A.f(1)>f(B.f(2)<fC.f(1)>ef(2) D.f(0)>e2f(4)答案A解析設(shè)F(x)=2e12xf(x),則F'(x)=e12xf(x)+2e12xf'(x)=e12x因?yàn)閒(x)+2f'(x)>0,所以F'(x)>0,所以F(x)在定義域上單調(diào)遞增.所以F(1)>F(0),即2ef(1)>2f(0),所以f(1)>f(0)e【例3】已知f(x)為定義在R上的可導(dǎo)函數(shù),且f(x)<f'(x)對(duì)于x∈R恒成立,e為自然對(duì)數(shù)的底數(shù),則()A.f(1)>ef(0)B.f(1)=ef(0)C.f(1)<ef(0)D.f(1)與ef(0)的大小不確定答案A解析設(shè)F(x)=f(x)ex,則F'(因?yàn)閒(x)<f'(x),所以F'(x)>0,所以F(x)是R上的增函數(shù).所以F(1)>F(0),即f(1)e>f(0),所以f(1)>ef(0)【例4】定義在區(qū)間0,π2上的函數(shù)f(x),f'(x)是其導(dǎo)函數(shù),恒有f(x)>f'(x)tanx成立,則A.3fB.f(1)>2fπ6C.2fπ6D.3fπ答案A解析因?yàn)閤∈0,π2,所以sinx>0,cos由f(x)>f'(x)tanx,得f(x)cosx-f'(x)sinx>0.設(shè)F(x)=f(x)sinx,則F'(x所以F(x)在區(qū)間0,π所以Fπ4>Fπ即3fπ4>數(shù)學(xué)抽象的思維過(guò)程細(xì)致視察和思索例1和例2的解法,它們有一個(gè)共同特點(diǎn):采納導(dǎo)數(shù)的積運(yùn)算法則,即[f(x)·g(x)]'=f'(x)g(x)+f(x)g'(x).例3和例4的解法,它們也有一個(gè)共同點(diǎn):采納導(dǎo)數(shù)的商運(yùn)算法則,即f(x)g(x)'=f'(x)g(x)-f(x)g'(x)[g(x)]2(g(x)≠0).由此可見(jiàn),對(duì)于含有f(x)和f'(x)的不等式,將不等式的右邊化為0,若左邊是u(x)f(x)+v(x)f'(x)的形式,其中u(x)和v在例1中,f(x)+xf'(x)<0,依據(jù)導(dǎo)數(shù)的積運(yùn)算法則(箭頭指向方向?yàn)楹瘮?shù)的導(dǎo)函數(shù),后面不再說(shuō)明),可以看出f(x)的導(dǎo)數(shù)為f'(x),x的導(dǎo)數(shù)為1,從而構(gòu)造出函數(shù)F(x)=xf(x).在例2中,f(x)+2f'(x)>0,依據(jù)導(dǎo)數(shù)的積運(yùn)算法則,可以看出f(x)的導(dǎo)數(shù)為f'(x),2的導(dǎo)數(shù)為1明顯不成立,則不等式兩邊肯定約去了一個(gè)不為0的變量,則猜想到y(tǒng)=ex,但這里還要考慮系數(shù)1和2,進(jìn)一步猜想到復(fù)合函數(shù)y=e12x,給上述不等式兩邊同乘從而構(gòu)造出函數(shù)F(x)=2e12xf(在例3中,由f(x)<f'(x),得f(x)-f'(x)<0,依據(jù)導(dǎo)數(shù)的商運(yùn)算法則,可以看出f(x)的導(dǎo)數(shù)為f'(x),ex的導(dǎo)數(shù)為ex,從而構(gòu)造出函數(shù)F(x)=f(在例4中,由f(x)>f'(x)tanx,得f(x)cosx-f'(x)sinx>0,且sinx>0,依據(jù)導(dǎo)數(shù)的商運(yùn)算法則,可以看出f(x)的導(dǎo)數(shù)為f'(x),sinx的導(dǎo)數(shù)為cosx,從而構(gòu)造出函數(shù)F(x)=f(數(shù)學(xué)抽象的結(jié)論依據(jù)題設(shè)條件,并借助初等函數(shù)的導(dǎo)數(shù)公式和導(dǎo)數(shù)的基本運(yùn)算法則,相應(yīng)地構(gòu)造函數(shù)如下.(1)對(duì)于不等式f'(x)>k(k≠0),構(gòu)造函數(shù)g(x)=f(x)-kx+b.(2)對(duì)于不等式xf'(x)+f(x)>0,構(gòu)造函數(shù)g(x)=xf(x).(3)對(duì)于不等式xf'(x)-f(x)>0,構(gòu)造函數(shù)g(x)=f(x)x((4)對(duì)于不等式xf'(x)+nf(x)>0,構(gòu)造函數(shù)g(x)=xnf(x).(5)對(duì)于不等式xf'(x)-nf(x)>0,構(gòu)造函數(shù)g(x)=f(x)xn(6)對(duì)于不等式f'(x)+f(x)>0,構(gòu)造函數(shù)g(x)=exf(x).(7)對(duì)于不等式f'(x)-f(x)>0,構(gòu)造函數(shù)g(x)=f((8)對(duì)于不等式f'(x)+kf(x)>0,構(gòu)造函數(shù)g(x)=ekxf(x).(9)對(duì)于不等式f'(x)+2xf(x)>0,構(gòu)造函數(shù)g(x)=ex2f(x(10)對(duì)于不等式f'(x)+lnaf(x)>0(a>0),構(gòu)造函數(shù)g(x)=axf(x).(11)對(duì)于不等式f(x)+f'(x)tanx>0,構(gòu)造函數(shù)g(x)=f(x)sinx.(12)對(duì)于不等式f'(x)-f(x)tanx>0,構(gòu)造函數(shù)g(x)=f(x)cosx.(13)對(duì)于不等式f'(x)f(x)>0(f(x)>0),構(gòu)造函數(shù)g(x)(14)對(duì)于不等式f'(x)lnx+f(x)x>0(x>0),構(gòu)造函數(shù)g(x)=f(3.2導(dǎo)數(shù)在探討函數(shù)中的應(yīng)用第1課時(shí)利用導(dǎo)數(shù)探討函數(shù)的單調(diào)性必備學(xué)問(wèn)·預(yù)案自診學(xué)問(wèn)梳理(1)①單調(diào)遞增②單調(diào)遞減考點(diǎn)自診1.(1)√(2)×(3)×(4)√(5)√2.Af'(x)=-1+lnxx2,由x>0及f'(x)<0得x>1e3.D由題意知,f'(x)=ex-e,令f'(x)>0,解得x>1.故選D.4.D由題意,得f'(x)=3-2sinx,所以f'(x)在R上恒為正,所以f(x)是R上的增函數(shù).又因?yàn)?=log24<log27<3<32,所以b<c<a,故選D5.43,+∞由題意知f'(x)=x+2a-1x≥0在13,2上恒成立,即2a≥-x+1x在13,2上恒成立.令g(x)=-x+1x,則g'(x)=-1-1x2,g'(x)<0在13,2上恒成立,所以g(x)=-x+1x在13,2上單調(diào)遞減.故g(x)≤g13=83.所以2a≥83,即a≥43.故實(shí)數(shù)a的取值范圍為43,+關(guān)鍵實(shí)力·學(xué)案突破例10,12依題意,f'(x)=1x-4x=1-4x2x=(1+2x)(1-2x)x,x∈(0,+∞).令f'(x)>對(duì)點(diǎn)訓(xùn)練1解(1)f'(x)=1+1-lnxx則切線(xiàn)斜率k=f'(1)=2.又因?yàn)閒(1)=1,所以切線(xiàn)方程為y-1=2(x-1),即2x-y-1=0.(2)f'(x)=1+1-lnxx令g(x)=x2+1-lnx(x>0),則g'(x)=2x2-1當(dāng)0<x<22時(shí),g'(x)<當(dāng)x>22時(shí),g'(x)>0即g(x)在區(qū)間0,22上單調(diào)遞減,在22,+∞上單調(diào)遞增.故g(x)min=g22=32-ln22>所以f'(x)>0在(0,+∞)上恒成立.所以,函數(shù)f(x)的單調(diào)遞增區(qū)間為(0,+∞),無(wú)單調(diào)遞減區(qū)間.例2解f(x)的定義域是(0,+∞).f'(x)=1x+2a(1-a)x-2(1-a)=2令g(x)=2a(1-a)x2-2(1-a)x+1,為確定函數(shù)g(x)的函數(shù)類(lèi)型,對(duì)a進(jìn)行分類(lèi)探討.(1)當(dāng)a=1時(shí),g(x)是常數(shù)函數(shù),此時(shí)g(x)=1>0,f'(x)=1x>0,于是f(x)在(0,+∞)上單調(diào)遞增(2)當(dāng)a≠1時(shí),g(x)是二次函數(shù),首先探討f'(x)=0是否有實(shí)數(shù)根,方程g(x)=0對(duì)應(yīng)的Δ=4(a-1)(3a-1).①當(dāng)Δ<0,即13<a<1時(shí),g(x)=0無(wú)實(shí)數(shù)根,g(x)的圖象在x軸上方,即f'(x)>0,f(x)在(0,+∞)上單調(diào)遞增②當(dāng)Δ=0,即a=13時(shí),g(x)=0有兩個(gè)相等的實(shí)根x1=x2=32,于是f'(x)≥0,所以f(x)在(0,+∞)③當(dāng)Δ>0,即0<a<13或a>1時(shí),g(x)=0有兩個(gè)不相等的實(shí)根分別為x1=12a-(a因?yàn)閤1+x2=1a,x1x2=1所以當(dāng)0<a<13時(shí),有x1+x2>0且x1x2>0,所以x1>0,x2>0由x1與x2的表達(dá)式知x1<x2,由f'(x)>0,可得0<x<x1或x>x2,所以f(x)在(0,x1)和(x2,+∞)上單調(diào)遞增;由f'(x)<0,可得x1<x<x2,所以f(x)在(x1,x2)上單調(diào)遞減.當(dāng)a>1時(shí),有x1+x2>0且x1x2<0,此時(shí)x2<0<x1,由f'(x)>0,可得0<x<x1,所以f(x)在(0,x1)上單調(diào)遞增;由f'(x)<0可得x>x1,所以f(x)在(x1,+∞)上單調(diào)遞減.綜上所述,當(dāng)0<a<13時(shí),f(x)在(0,x1)和(x2,+∞)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減當(dāng)13≤a≤1時(shí),f(x)在(0,+∞)上單調(diào)遞增當(dāng)a>1時(shí),f(x)在(0,x1)上單調(diào)遞增,在(x1,+∞)上單調(diào)遞減.其中x1=12a-(a-對(duì)點(diǎn)訓(xùn)練2解設(shè)h(x)=f(x)-2x-c,則h(x)=2lnx-2x+1-c,其定義域?yàn)?0,+∞),h'(x)=2x-2(1)當(dāng)0<x<1時(shí),h'(x)>0;當(dāng)x>1時(shí),h'(x)<0.所以h(x)在區(qū)間(0,1)上單調(diào)遞增,在區(qū)間(1,+∞)上單調(diào)遞減.從而當(dāng)x=1時(shí),h(x)取得最大值,最大值為h(1)=-1-c.故當(dāng)且僅當(dāng)-1-c≤0,即c≥-1時(shí),f(x)≤2x+c.所以c的取值范圍為[-1,+∞).(2)g(x)=f(x)-f(a)x-a=2(lng'(x)=2x取c=-1得h(x)=2lnx-2x+2,h(1)=0,則由(1)知,當(dāng)x≠1時(shí),h(x)<0,即1-x+lnx<0.故當(dāng)x∈(0,a)∪(a,+∞)時(shí),1-ax+lnax<0,從而g'(x)<0.所以g(x)在區(qū)間(0,a),(a,+∞)例3(1)B(2)D(1)設(shè)f(x)=x+1ex,則f'(x)=-xex.令f'(x)>0,得x<0;令f'(x)<0,得x>0.則f(x)在(-∞,0)上單調(diào)遞增,在(0,+∞)上單調(diào)遞減.若0<x<1,則0<x2<x<1<ln3,因此f(x2)>f(x)>f(ln3),(2)設(shè)g(x)=exf(x)-ex,則g'(x)=exf(x)+exf'(x)-ex=ex[f(x)+f'(x)-1].∵f(x)+f'(x)>1,ex>0,∴g'(x)=ex[f(x)+f'(x)-1]>0,∴g(x)是R上的增函數(shù).又g(0)=f(0)-1=2024,∴g(x)>2024的解集為(0,+∞),即不等式exf(x)>ex+2024的解集為(0,+∞).故選D.對(duì)點(diǎn)訓(xùn)練3(1)C(2)(1,+∞)(1)∵f'(x)>g'(x),∴[f(x)-g(x)]'>0.∴f(x)-g(x)在[a,b]上單調(diào)遞增.∴f(a)-g(a)<f(x)-g(x),即f(x)+g(a)>g(x)+f(a).(2)設(shè)F(x)=f(x)ex,則F'(∵f'(x)>f(x),∴F'(x)>0,即函數(shù)F(x)在定義域上單調(diào)遞增.∵ex-1f(x)<f(2x-1),∴f(x)ex<f(2x-1)e2x-1,即F(x)<F(2x-1),∴x<2x-1,即x>1,∴例4-716,+∞由h(x)在[1,4]上單調(diào)遞減得,當(dāng)x∈[1,4]時(shí),h'(x)=1x-ax-2≤0恒成立,即a≥1x2設(shè)G(x)=1x2-2x=1x-所以a≥G(x)max.因?yàn)閤∈[1,4],所以1x∈14,1,所以G(x)max=-716(此時(shí)x=所以a≥-716變式發(fā)散1(-1,+∞)h(x)=lnx-12ax2-2x,x∈(0,+∞所以h'(x)=1x-ax-2由h(x)在(0,+∞)上存在單調(diào)遞減區(qū)間得,當(dāng)x∈(0,+∞)時(shí),1x-ax-2<0有解,即a>1x2-2x有解.設(shè)G(x)=1x2-而G(x)=1x-12-1,所以G(x)min=-1.所以a>-1.變式發(fā)散2解h(x)=lnx-12ax2-2x,x∈(0,+∞所以h'(x)=1x-ax-2=-當(dāng)a=0時(shí),h'(x)=-2x+1x,則h(x)在0,12上單調(diào)遞增,在12,+∞當(dāng)a≤-1時(shí),Δ≤0,則h'(x)≥