課時(shí)規(guī)范練19利用導(dǎo)數(shù)研究不等式恒(能)成立問題1.已知函數(shù)f(x)=ax-sinxcos3x,x∈(1)當(dāng)a=8時(shí),討論f(x)的單調(diào)性;(2)若f(x)<sin2x,求a的取值范圍.2.已知f(x)=ex-ax2-x-1.(1)當(dāng)a=e2時(shí),求f(x)的極值點(diǎn)個(gè)數(shù)(2)當(dāng)x∈[0,+∞)時(shí),f(x)≥0,求實(shí)數(shù)a的取值范圍.3.已知函數(shù)f(x)=(x+1)e-ax,a≠0,若對任意的x≥0,不等式f(x)≤12x+1恒成立,求實(shí)數(shù)a的取值范圍4.(2024廣東深圳模擬)已知f(x)=ax-lnx,a∈R.(1)討論f(x)的單調(diào)性和極值;(2)當(dāng)x∈(0,e]時(shí),f(x)≤3有解,求a的取值范圍.

課時(shí)規(guī)范練19利用導(dǎo)數(shù)研究不等式恒(能)成立問題1.解(1)a=8時(shí),f(x)=8x-sinxcos3x0f'(x)=(4co當(dāng)f'(x)>0時(shí),cosx>22,0<x<π當(dāng)f'(x)<0時(shí),cosx<22,∴f(x)在0,π4內(nèi)為增函數(shù),在(2)(方法1)令g(x)=f(x)-sin2x=ax-sinxcos3x-sin則g'(x)=a-3-2cos2x令t=cos2x,則t∈(0,1),則h(t)=a-3-2tth'(t)=-令φ(t)=-4t3-2t+6,則φ'(t)=-12t2-2<0,∴φ(t)單調(diào)遞減,φ(t)>φ(1)=0,∴h'(t)>0,h(t)單調(diào)遞增.又當(dāng)x∈0,π2時(shí),y=cos∴g'(x)在0,π∴g'(x)<g'(0)=a-3.當(dāng)a≤3時(shí),g'(0)≤0,g(x)單調(diào)遞減,g(x)<g(0)=0,滿足題意.當(dāng)a>3時(shí),g'(0)>0,且x→π2時(shí),g'(x)→-∴存在x0∈0,π2使得g'(x0∴當(dāng)x∈(0,x0)時(shí),g'(x)>0,g(x)單調(diào)遞增,∴此時(shí)存在x,使得g(x)>g(0)=0,不滿足題意.綜上,a≤3.(方法2)f(x)=ax-sinx(sin2x+cos令g(x)=sin2x-f(x),由f(x)<sin2x,得g(x)>0,即2tanxtan2x+1+tan∴g'(x)=4-2(tan2x+1)tan2x令t=tan2x+1(t>1),則h(t)=4-2tt+3t2-2t-a,h'(t令φ(t)=3t3-t2-2,則φ'(t)=9t2-2t,當(dāng)t∈(1,+∞)時(shí),φ'(t)>0,∴φ(t)在(1,+∞)上單調(diào)遞增,又φ(1)=0,∴φ(t)>0,∴h'(t)>0,∴h(t)在(1,+∞)上單調(diào)遞增,h(t)>h(1)=3-a,當(dāng)a≤3時(shí),3-a≥0,h(t)>0,∴g'(x)>0,g(x)在0,π2內(nèi)單調(diào)遞增,g(x)>0對x當(dāng)a>3時(shí),若g(x)>g(0)=0對x∈0,π2恒成立,則h(1)=又h(t)在(1,+∞)上單調(diào)遞增,且t→+∞時(shí),h(t)→+∞,∴存在t0∈(1,+∞)使得h(t0)=0,∴存在x0∈0,π2使得g'(x0與g(x)>g(0)=0對x∈0,綜上,a≤3.(方法3)由于?x∈0,π2,f(x)<sin2x,即a<2sinx當(dāng)x→0時(shí),sinxx→1,故猜想下證a≤3時(shí),f(x)<sin2x成立,由a≤3得ax≤3x,即ax-sinxcos3令g(x)=3x-sinxcos3x-sin2x只需證g(x)<g(0),g'(x)=3-cos2x+3sin2xcog(x)在0,π2內(nèi)單調(diào)遞減,g(x)<g∴3x-sinxcos3x-sin綜上,a≤3時(shí),f(x)<sin2x成立.2.解(1)當(dāng)a=e2時(shí),f(x)=ex-e2x2-x-1,則f'(x)=ex-ex-1.令m(x)=ex-ex-1,則m'(x)=ex-e.令m'(x)=0,得x=當(dāng)x<1時(shí),m'(x)<0,m(x)在(-∞,1)上單調(diào)遞減;當(dāng)x>1時(shí),m'(x)>0,m(x)在(1,+∞)上單調(diào)遞增.因?yàn)閒'(0)=0,f'(1)=-1,f'(2)=e2-2e-1>0,所以存在x0∈(1,2),使f'(x0)=0,所以,當(dāng)x∈(-∞,0)時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(0,x0)時(shí),f'(x)<0,f(x)單調(diào)遞減;當(dāng)x∈(x0,+∞)時(shí),f'(x)>0,f(x)單調(diào)遞增,所以0和x0是f(x)的極值點(diǎn),所以f(x)有兩個(gè)極值點(diǎn).(2)f(x)=ex-ax2-x-1,f'(x)=ex-2ax-1,設(shè)h(x)=f'(x)=ex-2ax-1(x≥0),則h'(x)=ex-2a單調(diào)遞增,又因?yàn)閔'(0)=1-2a,所以當(dāng)a≤12時(shí),h'(x)≥0,h(x)在[0,+∞)上單調(diào)遞增,所以h(x)≥h(0)=0,即f'(x)≥0,則f(x)在[0,+∞)上單調(diào)遞增,所以f(x)≥f(0)=0,當(dāng)a>12時(shí),令h'(x)=0,解得x=ln2a當(dāng)x∈[0,ln2a)時(shí),h'(x)<0,h(x)在區(qū)間[0,ln2a)內(nèi)單調(diào)遞減,f'(x)=h(x)≤h(0)=0,f(x)在區(qū)間(0,ln2a)內(nèi)單調(diào)遞減,所以當(dāng)x∈(0,ln2a)時(shí),f(x)<f(0)=0,不符合題意.所以實(shí)數(shù)a的取值范圍是-∞,12.3.解對任意的x≥0,不等式f(x)≤12x+1恒成立,即對任意的x≥0,(x+1)e-ax≤12當(dāng)x≥0時(shí),(x+1)e-ax>0,12x+1>0,將不等式兩邊取對數(shù)得ln(x+1)-ax≤ln12x+1,即ln(x+1)-ln12x+1-ax≤0.因此只需證明當(dāng)x≥0時(shí),不等式ln(x+1)-ln12x+1-ax≤0恒成立即可.令g(x)=ln(x+1)-ln12x+1-ax,x≥0,則g'(x)=1x+1-1x若a≥12,因?yàn)楫?dāng)x≥0時(shí),所以g'(x)=1(x+1)(x+2)-a≤0,所以g(x)在[0,+∞)上單調(diào)遞減,因此g(x若a<0,因?yàn)楫?dāng)x≥0時(shí),1(x+1)(x+2)>0,所以g'(x)=1(x+1)(x+2)-a>0,所以g(x)在[0,+∞)若0<a<12,令g'(x)=1(x+1)(x+2)-a=0,得x=4+a4a-32x=-4+a4a-32舍去,當(dāng)x∈0,4+a4a-32時(shí),g'(x)=1(x+1綜上,當(dāng)不等式f(x)≤12x+1恒成立時(shí),實(shí)數(shù)a的取值范圍是12,+∞4.解(1)f'(x)=a-1x=ax當(dāng)a≤0時(shí),f'(x)<0恒成立,函數(shù)f(x)在區(qū)間(0,+∞)上單調(diào)遞減,無極值;當(dāng)a>0時(shí),令f'(x)=0,得x=1a,令f'(x)<0,得0<x<1a,函數(shù)f(x)在區(qū)間0,1a內(nèi)單調(diào)遞減令f'(x)>0,得x>1a,函數(shù)f(x)在區(qū)間1a,+∞上單調(diào)遞增,當(dāng)x=1a時(shí),函數(shù)f(x)取得極小值f1a=1+lna.綜上可知,當(dāng)a≤0時(shí),函數(shù)f(x)的單調(diào)遞減區(qū)間是(0,+∞),無增區(qū)間,無極值;當(dāng)a>0時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為1a,+∞,單調(diào)遞減區(qū)間為0,1a,極小值為1+lna,無極大值.(2)由題意可知,ax-lnx≤3,在x∈(0,e]時(shí)有解,則a≤3x