-17分111.如果函數(shù)f(x)在區(qū)間[a,b]上,f(x)≥0恒成立,則f(a)≥0或f(b)≥0.2.如果函數(shù)f(x)在區(qū)問[a,b]上,f(x)≥0恒成立,且f(a)=0(或f(b)=0),則f(a)≥0(或f(b)≤0(.3.如果函數(shù)f(x)在區(qū)問[a,b]上,f(x)≥0恒成立,且f(a)=0,f(a)=0(或f(b)=0,f(b)≤0(則fⅡ(a)≥0(或f22若xlnx-2mx(x-1(+ex-1-x≥0對(duì)?x≥1恒成立，則實(shí)數(shù)m的取值范圍是. 33 44(2020·全國(guó)·統(tǒng)考高考真題)已知函數(shù)f(x)=ex+ax2-x.(2024·全國(guó)甲卷·高考真題)已知函數(shù)f(x(=(1-ax(ln(1+x(-x.55(全國(guó)·高考真題)已知函數(shù)f(x)=2sinx-xcosx-x，f′(x)為f(x)的導(dǎo)數(shù).66(2024·浙江寧波·模擬預(yù)測(cè))已知函數(shù)f(x)=ex-ax-1.(1)討論f(x)的單調(diào)性;(2024·河南·模擬預(yù)測(cè))已知函數(shù)f(x(=alnx+x-1(a∈R(.(1)討論f(x(的單調(diào)性；f(x(≥-2lnx+(lnx(2，求a的取值范圍.77(2024·廣西·三模)已知函數(shù)f(x(=ex-x.(1)求函數(shù)f(x(的極值；(2)若對(duì)任意x>0,f(x(>ax2+1，求a的取值范圍.(2024·四川綿陽(yáng)·模擬預(yù)測(cè))已知函數(shù)f(x)=2sinx+ln(x+1)-ax.88 5(2024·云南昆明·一模)已知函數(shù)f(x(=(1)求曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程；(2)當(dāng)x≥1時(shí)，f(x(≤a(x-1(，求a的取值范圍. 6(2024·安徽池州·模擬預(yù)測(cè))設(shè)函數(shù)f(x(=(1)當(dāng)a=1時(shí)，求曲線y=f(x(在點(diǎn)(0,f(0((處的切線方程；(2)當(dāng)x≥0時(shí)，若f(x(≤a恒成立，求實(shí)數(shù)a的取值范圍.99 (全國(guó)·高考真題)已知函數(shù)f(x)=ex(ex(全國(guó)·高考真題)設(shè)函數(shù)f(x(=ex-e-x(1)求證：f(x)的導(dǎo)數(shù)f(x(≥2；(2)若對(duì)任意x≥0都有f(x)≥ax,求a的取值范圍.(全國(guó)·高考真題)設(shè)函數(shù)f(x)=x(ex-1(-ax2(Ⅱ)若當(dāng)x≥0時(shí)f(x)≥0，求a的取值范圍 -17分111.如果函數(shù)f(x)在區(qū)間[a,b]上,f(x)≥0恒成立,則f(a)≥0或f(b)≥0.2.如果函數(shù)f(x)在區(qū)問[a,b]上,f(x)≥0恒成立,且f(a)=0(或f(b)=0),則f(a)≥0(或f(b)≤0(.3.如果函數(shù)f(x)在區(qū)問[a,b]上,f(x)≥0恒成立,且f(a)=0,f(a)=0(或f(b)=0,f(b)≤0(則fⅡ(a)≥0(或f22若xlnx-2mx(x-1(+ex-1-x≥0對(duì)?x≥1恒成立，則實(shí)數(shù)m的取值范圍是.【方法一】解：因?yàn)閤lnx-2mx(x-1(+ex-1-x≥0對(duì)?x≥1恒成立，即lnx-2m(x-1(+-1≥0對(duì)?x≥1恒成立，記f(x(=lnx-2m(x-1(+-1，x∈[1,+∞(，所以fI(x(=-2m+，令g(x(=-2m+，則gI(x(=≥=>0所以fI(x(在[1,+∞(上是增函數(shù)，所以fI(x(≥fI(1(=1-2m當(dāng)1-2m≥0，即m≤時(shí)，f(x(在[1,+∞(上是增函數(shù)，所以f(x(≥f(1(=0符合題意；即當(dāng)x∈(1,x0(時(shí)fI(x(<0，f(x(單調(diào)遞減，此時(shí)f(x(<f(1(=0，所以m>不符合題意，綜上可得m≤,即m∈【方法二-端點(diǎn)效應(yīng)】因?yàn)閤lnx-2mx(x-1(+ex-1-x≥0對(duì)?x≥1恒成立，即lnx-2m(x-1(+-1≥0對(duì)?x≥1恒成立，記f(x(=lnx-2m(x-1(+-1，x∈[1,+∞(，因?yàn)閒(1(=0，f(x(≥0欲在x∈[1,+∞(恒成立，則f(x(要在x∈[1,+∞(單調(diào)遞增再證明充分性，當(dāng)m≤,能否有xlnx-2mx(x-1(+ex-1-x≥0對(duì)?x≥1恒成立(證明略)綜上可得m≤,即m∈33 故f(x)≥f(0),f(x)=.所以，f(x)=f(a-1)=lna-a+1=lna-(a-1)，【方法二-端點(diǎn)效應(yīng)】 4故f(x(min=2+a，而f(x(≥0成立，故a+24所以a的最小值為-2.，設(shè)P(m,n(為y=f(x(圖象上任意一點(diǎn)，P(m,n(關(guān)于(1,a(的對(duì)稱點(diǎn)為Q(2-m,2a-n(，而f(2-m(=ln+a(2-m(+b(2-m=-n+2a，所以Q(2-m,2a-n(也在y=f(x(圖象上，因?yàn)閒(x(>-2當(dāng)且僅當(dāng)1<x<2，故x=1為f(x(=所以f(1(=-2即a=-2，先考慮1<x<2時(shí)，f(x(>-2恒成立.此時(shí)f(x(>-2即為ln-+2(1-x(+b(x-1(3>0在(1,2(上恒成立，1-t21-t2，1-t21-t2，故gI(t(>0恒成立，故g(t(在(0,1(上為增函數(shù)，故g(t(>g(0(=0即f(x(>-2在(1,2(上恒成立.故gI(t(≥0恒成立，故g(t(在(0,1(上為增函數(shù)，故g(t(>g(0(=0即f(x(>-2在(1,2(上恒成立.綜上，f(x(>-2在(1,2(上恒成立時(shí)b≥-.5即f(x(>-2的解為(1,2(.5故a=-2.即≥0,即fI≥0,f是上的單調(diào)遞增函數(shù)，f(x)>f(1)=-2,符合題意；--b得當(dāng)1<x<時(shí)，fI(x)<0,此時(shí)，f(x)<f(1)=-2,不符合題意由題意得：0<x≤1,必有f(x)≤-2,所以f(1)=-2,解得a=-2,問題等價(jià)于研究當(dāng)1<x<2,f(x)>-2恒成立，僅需[f(x)+2]min>0,x2-x令h(x)=f(x)+2,hI(x)=1+x2-x3x(2-x)3x(2-x)3x(2-x),3x(2-x),66 (2)(-∞,3]=a-=a-2x=t,則t則f(x)=g(t)=a-=當(dāng)a=8,f(x)=g(t)==所以f(x)在(0,上單調(diào)遞增,在,上單調(diào)遞減【法一】設(shè)g(x)=f(x)-sin2xg(x)=f(x)-2cos2x=g(t)-2(2cos2x-1(=-2(2t-1)=a+2-4t+-設(shè)φ(t)=a+2-4t+-φ(t)=-4-+==->0∈(-∞,3],g(x)=φ(t)<a-3≤0即g(x)在(0,上單調(diào)遞減,所以g(x)<g(0)=0.所以當(dāng)a∈(-∞,3],f(x)<sin2x,符合題意.當(dāng)t→0,-=-3-2+→-∞,所以φ(t)→-∞.所以?t0g(x)>g(0)=0,不合題意.77綜上,a的取值范圍為(-∞,3].(2)f(x)<sin2x?ax-<sin2x?g(x)=ax-sin2x-<0g(x)=a-2cos2x-,∴g(x)>g(0)=0,不成立.∴g(x)單調(diào)遞減,∴g(x)≤g(0)=0.特上 【答案】(1)f(x(在(0,上單調(diào)遞減則f(x(=1-=1-3x-cos2x-2(1-cos2x(=cos3x+cos2x-23x，2x-2=t3+t2-2=t3-t2+2t2-2=t2(t-1(+2(t+1((t-1(=(t2+2t+2((t-1(，3x=t3>0，所以f(x(=<0在(0,上恒成立，所以f(x(在(0,上單調(diào)遞減.88構(gòu)建g(x(=f(x(+sinx=ax-+sinx(0<x<，則g(x(=a-+cosx(0<x<，若g(x(=f(x(+sinx<0，且g(0(=f(0(+sin0=0，所以f(x(+sinx=sinx-<0，滿足題意；所以f(x(+sinx=ax-+sinx<sinx-<0，滿足題意；所以a的取值范圍為(-∞,0[.因?yàn)閟inx-sinx=sinxcos2x-sinx=sinx(故sinx-<0在(0,上恒成立，所以f(x(+sinx=ax-+sinx<sinx-<0，滿足題意；當(dāng)a>0時(shí)，因?yàn)閒(x(+sinx=ax-+sinx=ax-，令g(x(=ax-0<x<，則g(x(=a-，注意到g(0(=a-=a>0，<0，此時(shí)g(x(在(0,x1(上有g(shù)(x(>0，所以g(x(在(0,x1(上單調(diào)遞增，99 【答案】(1)當(dāng)x∈(-∞,0(時(shí)，f'(x(<0,f(x(單調(diào)遞減，當(dāng)x∈(0,+∞(時(shí)，f'(x(>0,f(x(單調(diào)遞增.(2),+∞(-x，fI(x(=ex+2x-1，當(dāng)x∈(-∞,0(時(shí)，fI(x(<0,f(x(單調(diào)遞減，fI(x(>0,f(x(單調(diào)遞增.x-3-x-令h(x(=ex-x2-x-1(x≥0(，由h(x(≥0可得：ex-x2-x-1≥0恒成立，只需證當(dāng)a≥時(shí)，f(x)≥x3+1恒成立.當(dāng)a≥時(shí)，f(x)=ex+ax2-x≥ex+?x2-x.只需證明ex+x2-x≥x3+1(x≥0)⑤式成立.(e2-7(x2+4x+(e2-7(x2+4x+2x3+4令h(x)=(x≥0)，則hI(x)===，綜上a≥.當(dāng)x≥0時(shí)，f(x)≥x3+1恒成立?ex≥x3+1-ax2+x?x3-ax2+x+1(e-x≤1，記g(x(=x3-ax2+x+1(e-x(x≥0)，gI(x(x3-ax2+x+1-x2+2ax-1(e-x=-x[x2-(2a+3(x+4a+2[e-x=-x(x-2a-1((x-2(e-x，所以若滿足g(x(≤1，只需g(2(≤1，即g(2(=(7-4a)e-2≤1?a≥,所以當(dāng)?≤a<時(shí)，③當(dāng)2a+1≥2即a≥時(shí)，g(x(=x3-ax2+x+1(e-x≤x3+x+1(e-x，又由②可知≤a<+x+1(e-x≤1恒成立， 2(2024·全國(guó)甲卷·高考真題)已知函數(shù)f(x(=(1-ax(ln(1+x(-x.(2)a≤-(2)求出函數(shù)的二階導(dǎo)數(shù)，就a≤-、-<a<0、a≥0分類討論后可得參數(shù)的取值范圍.(1)當(dāng)a=-2時(shí)，f(x)=(1+2x)ln(1+x)-x，故f(x)=2ln(1+x)+-1=2ln(1+x)-+1，因?yàn)閥=2ln(1+x),y=-+1在(-1,+∞(上為增函數(shù)，故f(x)在(-1,+∞(上為增函數(shù)，而f(0)=0，故當(dāng)-1<x<0時(shí)，f(x)<0，當(dāng)x>0時(shí)，f(x)>0，故f(x(在x=0處取極小值且極小值為f(0(=0，無(wú)極大值.(2)f(x(=-aln(1+x(+-1=-aln(1+x(-,x>0，設(shè)s(x(=-aln(1+x(-,x>0，則s(x(=-=-=-，當(dāng)a≤-時(shí)，s(x(>0，故s(x(在(0,+∞(上為增函數(shù)，故s(x(>s(0(=0，即f(x(>0，所以f(x(在[0,+∞(上為增函數(shù)，故f(x(≥f(0(=0.當(dāng)-<a<0時(shí)，當(dāng)0<x<-時(shí)，s(x(<0，故s(x(在(0,-上為減函數(shù)，故在(0,-上s(x(<s(0(，即在(0,-上f(x(<0即f(x(為減函數(shù)，同理可得在(0,+∞(上f(x(<f(0(=0恒成立綜上，a≤-. 3(全國(guó)·高考真題)已知函數(shù)f(x)=2sinx-xcosx-x，f′(x)為f(x)的導(dǎo)數(shù).(2)a∈(-∞,0[.數(shù)h(x(=f(x(-ax，通過(guò)二次求導(dǎo)可判斷出hI(x(min=hI(π(=-2-a，hI(x(max=hI=-a；分別在a≤-2，-2<a≤0，0<a<和a≥的情況下根據(jù)導(dǎo)函數(shù)的符號(hào)判斷h(x(單調(diào)性，從而確定【詳解】(1)fI(x(=2cosx-cosx+xsinx-1=cosx+xsinx-1令g(x(=cosx+xsinx-1，則gI(x(=-sinx+sinx+xcosx=xcosxI(x(<0fI(x(無(wú)零點(diǎn)=0又g(x(在,π(上單調(diào)遞減∴x=x0為g(x(，即fI(x(在,π(上的唯一零點(diǎn)綜上所述：fI(x(在區(qū)間(0,π(存在唯一零點(diǎn)令h(x(=f(x(-ax=2sinx-xcosx-(a+1(x且h(0(=-a，h=-a，h(π(=-2-a∴h(x(min=h(π(=-2-a，h(x(max=h=-a①當(dāng)a≤-2時(shí)，h(x(min=h(π(=-2-a≥0，即h(x(≥0在[0,π[上恒成立∴h(x(在[0,π[上單調(diào)遞增∴h(x(≥h(0(=0，即f(x(-ax≥0，此時(shí)f(x(≥ax恒成立∴?x1=0∴h(x(在[0,x1(上單調(diào)遞增，在(x1,π[上單調(diào)遞減∴h(x(≥0在[0,π[上恒成立，即f(x(≥ax恒成立(-a>0∴?x2=0∴h(x(在[0,x2(上單調(diào)遞減，在(x2,上單調(diào)遞增f(x(≥ax不恒成立∴h(x(在(0,上單調(diào)遞減∴h(x(<h(0(=0可知f(x(≥ax不恒成立(2024·浙江寧波·模擬預(yù)測(cè))已知函數(shù)f(x)=ex-ax-1.(1)討論f(x)的單調(diào)性;【詳解】(1)f(x(=ex-a，所以當(dāng)x∈(lna,+∞(時(shí)，f(x(>0，f(x(單調(diào)遞增；當(dāng)x∈(-∞,lna(時(shí)，f(x(<0，f(x(單調(diào)遞減；綜上，當(dāng)a≤0時(shí)，f(x(在R上單調(diào)遞增；當(dāng)a>0時(shí)，f(x(在(lna,+∞(上單調(diào)遞增，在(-∞,lna(上單調(diào)遞再令h(x(=(x-1(ex+1，則h(x(則h(x(在(0,+∞(單調(diào)遞增，所以h(x(>h(0(=0，所以g(x(在(0,+∞(上單調(diào)遞增，所以a≤1 2(2024·河南·模擬預(yù)測(cè))已知函數(shù)f(x(=alnx+x-1(1)討論f(x(的單調(diào)性；f(x(≥-2lnx+(lnx(2，求a的取值范圍.(2)[-e,+∞(當(dāng)a≥0時(shí)，f(x(>0，所以f(x(在(0,+∞(上單調(diào)遞增；所以f(x(在(0,-a(上單調(diào)遞減，在(-a,+∞(上單調(diào)遞增.設(shè)h(x(=x-lnx-1(x>1(，由(1)可知，h(x(在(1,+∞(上單調(diào)遞增，所以h(x(>h(1(=0，則g(x(max=g(e(=-e，所以a≥-e， (2)對(duì)任意x>0,f(x(>ax2+1，即ex-x-ax2-1>0，設(shè)g(x(=ex-x-ax2-1,x>0,g(x(=ex-1-ax,x>0， 4(2024·四川綿陽(yáng)·模擬預(yù)測(cè))已知函數(shù)f(x)=2sinx+ln(x+1)-ax.(2)a≥3則g(x(=-2sinx-，≤0，g(x(在(0,單調(diào)遞減，即f(x(在(0,單調(diào)遞減，且f(0(=1>0，f=-2<0，∴?x0∴f(x(在(0,x0(單調(diào)遞增，(x0,單調(diào)遞減；∴f(x(在(0,有1個(gè)零點(diǎn)；f(x(≤0.∴此時(shí)f(x(在(0,+∞(單調(diào)遞減，∴此時(shí)f(x(≤f(0(=0.當(dāng)a≤3時(shí)，f(0(=3-a>0，必存在x1∈(0,+∞(，使f(x(在(0,x1(單調(diào)遞增，那么?x∈(0,x1( (1)求曲線y=f(x(在點(diǎn)(1,f(1((處的切線方程；(2)當(dāng)x≥1時(shí)，f(x(≤a(x-1(，求a的取值范圍.故切線方程為y-0=(x-1)，即y=x-.f(x(≤a(x-1(等價(jià)于lnx≤a(x2-1(， 可得fI(x(==，所以fI(0(=-2，f(0(=1，所以曲線y=f(x(在點(diǎn)(0,f(0((處的切線方程為y-1=-2(x-0(，即y=-2x+1.x+x-1>e0+0-1=0.所以命題等價(jià)于a≥對(duì)x>0恒成立，x+2x2-2x-x2ex-x2=2xex+x2-2x-x2ex=-2x(1-ex(+x2(1-ex((ex+x-1(2(ex+x-1(2(ex+x-1(2=(x2-2x((1-ex(=x(x-2((1-ex((ex+x-1(2(ex+x-1(2，所以h(x)max=h(2(=.. (x-x+1,fI(x)=ex-1,fI(1)=e-1,f(1)=e，x-x2(≥x-1恒成立.令g(x)=ex-x2，x-2x,φI(x)=ex-2(ln2)=2-2ln2>0.x-x>0，. (2)(-∞,1[【詳解】(1)由f(x(=ex-ax-2，得f/(x(=ex-a，x(>0，則f(x(單調(diào)遞增，f(x(不存，f(x(單調(diào)遞減；，f(x(單調(diào)遞增，所以x=lna是f(x(的極小值點(diǎn)，(2)由f(x(>x-sinx-cosx在x∈(0,+∞(時(shí)恒成立，x+cosx+sinx-(a+1(x-2>0在x∈(x-sinx+cosx-(a+1(，令m(x(=gI(x(=ex-sinx+cosx-(a+1(，則mI(x(=ex-cosx-sinx，令n(x(=mI(x(=ex-cosx-sinx，則nI(x(=ex+sinx-cosx，所以x∈(0,+∞(時(shí)，nI(x(>0，則n(x(即mI(x(單調(diào)遞增，所以mI(x(>mI(0(=0，則m(x(即gI(x(單調(diào)遞增，所以gI(x(>gI(0(=1-a，所以g(x(>g(0(=0，所以f(x(>x-sinx-cosx在x∈(0,+∞(時(shí)恒成立，-a<0，I[ln(a+3([=a+3-sin[ln(a+3([+cos[ln(a+3([-(a+1(故在區(qū)間(0,ln(a+3((上函數(shù)gI(x(存在零點(diǎn)x0，即gI(x0(=0，故函數(shù)g(x(在區(qū)間(0,x0(上單調(diào)遞減，進(jìn)行分類討論. 【答案】(1)f(x(在(-∞,+∞(上單調(diào)遞增.(2)將問題轉(zhuǎn)化為?x∈[0,+∞(,ex-x2-2ax-cosx≥0恒成立，構(gòu)造函數(shù)F(x(=ex-x2-2ax-cosx,x∈令m(x(=ex-2x，所以mI(x(=ex-2，所以m(x(在(-∞,ln2(上單調(diào)遞減，在(ln2,+∞(上單調(diào)遞增，所以m(x(≥m(ln2(=2-2ln2>0，即fI(x(>0，從而函數(shù)f(x(在(-∞,+∞(上單調(diào)遞增.(2)因?yàn)閒(x(=ex-ax2+2，所以fI(x(=ex-2ax，又g(x(=x2+cosx，?x∈[0,+∞(,fI(x(≥g(x(恒成立等價(jià)于?x∈[0,+∞(,ex-x2-2ax-cosx≥0恒成立.記F(x(=ex-x2-2ax-cosx,x∈[0,+∞(，所以FI(x(=ex-2x-2a+sinx.令h(x(=ex-2x-2a+sinx,x∈[0,+∞(，所以hI(x(=ex+cosx-2.設(shè)r(x(=ex+cosx-2,x∈[0,+∞(，從而rI(x(=ex-sinx>0，故有r(x(≥r(0(=0，則h(x(在[0,+∞(上單調(diào)遞增，即FI(x(在[0,+∞(上單調(diào)遞增，故有FI(x(≥FI(0(=1-2a.I(x(≥FI(0(=1-2a≥0，此時(shí)F(x(單調(diào)遞增，從而F(x(≥F(0(=0，滿足題意.I(0(=1-2a<0，且FI(x(在(0,+∞(上單調(diào)遞增，x→+∞,FI(x(→+∞,∈(0,+∞(滿足FI(x0(=0，,FI(x(<0，則F(x(在(0,x0(上單調(diào)遞減，時(shí)，F(xiàn)(x(≤F(0(=0，不滿足題意. (2)(-∞,-1].(2)函數(shù)f(x)=+x-1，f(x)≥a+1?a(lnx-x)≥2x-x2，x∈[1,+∞)，所以a≤-1，即實(shí)數(shù)a的取值范圍