導(dǎo)數(shù)與不等式專題能力訓(xùn)練8利用解及參數(shù)的取值范圍

專題能力訓(xùn)練 利用導(dǎo)數(shù)解不等式及參數(shù)的取值范2.已知函數(shù)f(x)=ex-e-x-的切線的斜率為3. ?? .n>m>1(m,n∈N)時(shí),證明√>.√?? √y=2x+1沒有公共點(diǎn),k的取值范圍2若a=1,對(duì)任意的x1>x2>0,不等式m[g(x1)-g(x2)]>x1f(x1)-x2f(x2)恒成立.求m(m∈Z,m≤1)值解3

(2)a<0時(shí),x∈(01)∪(,1),f(x)=f( 2????2+(2-??)??- (????+1)(2??-1.解:(1)f'(x)=2ax+(2-a)-1= = 當(dāng)-1<,a<-2時(shí),f(x)的單調(diào)遞增區(qū)間為(11),單調(diào)遞減區(qū)間為(01,+ ?? 當(dāng)-1=1,a=-2時(shí),f(x)在(0,+∞)上單調(diào)遞減 當(dāng)-1>1,0>a>-2時(shí),f(x)的單調(diào)遞增區(qū)間為(11),單調(diào)遞減區(qū)間為(011,

即 g(x)=ln??+??,g'(x)=-??(??- ∵x∈(1,e),∴g'(x)<0,即在區(qū)間(1,e)內(nèi)g(x)單調(diào)遞減∴a<g(1)=1.故實(shí)數(shù)a的取值范圍為解:(1)f'(x)=ex+e-x-2≥0,當(dāng)且僅當(dāng)x=0時(shí)等號(hào)成立,故f(x)在(-∞,+∞)上單調(diào)遞增g(x)=f(2x)-4bf(x)=e2x-e-2x-4b(ex-e-x)+(8b-g'(x)=2[e2x+e-2x-2b(ex+e-x)+(4b-2)]=2(ex+e-x-2)(ex+e-x-①當(dāng)b≤2時(shí),g'(x)≥0,當(dāng)且僅當(dāng)x=0時(shí)等號(hào)成立,則g(x)在(-∞,+∞)上單調(diào)遞增②b>2時(shí),x滿足2<ex+e-x<2b-2,0<x<ln(b-1+√??2-2??)時(shí)因?yàn)間(0)=0,所以當(dāng)0<x≤ln(b-1+√??2-2??)時(shí),g(x)<0,不符合題意2 ln2<18+√2<0.693f(x)≤kx2對(duì)任意x>0成立,k≥1+ln??對(duì)任意x>0成立1·- 令 ∴g(x)在(0,1)內(nèi)是增函數(shù)∴g(x)在(1,+∞)上是減函數(shù)

??

故g(x)在x=1處取得最大值g(1)=1,∴k≥1即為所求令h(x)=??ln??,則h'(x)=??-1-??- (??-由(2)知,x≥1+ln∴h(x)是(1,+∞)上的增函數(shù)∵n>m>1,∴h(n)>h(m),即??ln??>??- ??-∴mnlnn-nlnn>mnlnm-mlnmnlnn+mlnm>mnlnm+nln整理,得

m∴√m√??√

解:(1)函數(shù)f(x)的定義域?yàn)?0,+∞),且f'(x)=??-∵當(dāng)0<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí)∴f(x)在區(qū)間(0,1)上為減函數(shù),在(1,+∞)上為增函數(shù)2 ∵g(x)在其定義域內(nèi)為減函數(shù)∴ax2+x+a≤0?a(x2+1)≤-x?a≤-???a≤[-?? ??2+1

又??=1≤1,-??≥-

當(dāng)且僅當(dāng)x=1時(shí)取等號(hào).∴a≤-2+ 有實(shí)數(shù)解①k=2時(shí),方程(*)可化為1=0,其在R上沒有實(shí)數(shù)解②k≠2時(shí),方程(*)1??-令g(x)=xex,則有x--0+↘-↗ 11)時(shí),方程(*)無實(shí)數(shù)解,k的取值范圍是(2-??- 綜合①②,可知k的取值范圍是(2-2化簡(jiǎn),a(x-lnx)≥1x2-21111??- ??-因而 .設(shè) 1 ??-1 (??-1)(??-ln??)-(1-1)(??- (??-1)(??+1-則 ?? (??- (??-2∴y'>0在x∈[1,e]時(shí)成立 2(2)a=1時(shí),f(x)=ln由m[g(x1)-g(x2)]>x1f(x1)-x2f(x2)恒成立,得mg(x1)-x1f(x1)>mg(x2)-x2f(x2)恒成立t(x)=??x2-xln2由題意知x1>x2>0,則當(dāng)x∈(0,+∞)時(shí)函數(shù)t(x)單調(diào)遞增因此,記h(x)=ln??+1,得h'(x)=- ∵函數(shù)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減∴函數(shù)h(x)在x=1處取得極大值,并且這個(gè)極大值就是函數(shù)h(x)的最大值由此可得h(x)max=h(1)=1,故m≥1,結(jié)合已知條件m∈Z,m≤1,可得g(x)=f'(x)=2(x-a)-2lnx-2(1+ 41 4

2??=2(??-)+2(??-

0<a<1時(shí),g(x)在區(qū)間(01-√1-4??1+√1-4??,+∞)上單調(diào)遞增 在區(qū)間(1-√1-4??1+√1-4??)上單調(diào)遞減 當(dāng)a≥1時(shí),g(x)在區(qū)間(0,+∞)上單調(diào)遞增 -(2)證明:f'(x)=2(x-a)-2lnx-2(1+)=0,a=??1 ??-1-

1+??- ??-1-)lnx+x- )x-)lnx+x- )x- - e- 故存在x0∈(1,e),使得0a=??0-1-ln??0,u(x)=x-1-ln01+??-00=??(1)<??(??0=a<??(e)

e-

01+e- 1+e-a=a0時(shí),故當(dāng)x∈(1,x0)時(shí),f'(x)<0,從而當(dāng)x∈(x0,+∞)時(shí),f'(x)>0,從而7.解:(1)f'(x)=x2+2x+a,方程x2+2x+a=0的判別式為Δ=4-①當(dāng)a≥1時(shí),Δ≤0,則f'(x)≥0,此時(shí)f(x)在R上是增函數(shù)②a<1時(shí),方程x2+2x+a=0兩根分別為x1=-1-√1-??,x2=-1+√1- 1 1(2)f(x)-f()

??3+??2+ax+1-1×[()]?()-a·1-3 1=1[??3-()]+[??2-()]+a(??- 0=1(??-1)(??2+??0+1)+ - 0

0 (??0+)+a(??0-)=(??0-)·(0 +x0++??) (??0- x∈(01)∪(1,1),f(x)=f(1),4??2+14x+7+12a=0在(01)∪(1,1)內(nèi)有解 ∵a<0,∴Δ=142-16(7+12a)=4(21-4??2+14x7+12a=0x-7-√21-48??,x'=-7+√21- ∵x>0,∴x=x'=-

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