导数30问lnx已知函数f(x)=，若f(x)=a有两个不同的零点x，x，且x0)∴g(x)=-a(x>0,a∈R)x①若a≤0,则g(x)>0恒成立,g(x)在(0,+∞)上单调递增,此时g(x)不可能有两个零点;1②若a>0,则当x∈(0,)时,g(x)>0,g(x)单调递增,a1当x∈(,+∞)时,g(x)<0,g(x)单调递减;a11当x=时,g(x)取得最大值,且g(x)=f()=-lna-1amaxa11①当-lna-1≤0即a≥时,g(x)至多只有一个零点x=,不合题意;ea1②当-lna-1>0即00e∵∀x∈(0,+∞),lnx12a1【解析】f(x)=a有两个不同的零点x,x,则有a>0且0,由对均不等式易证:x1+x2>x1-x2x1+x2a3.求证:x1+x2>2e112【解析】∵0e,由(1)得x+x>>2e,∴x+x>2eea12a1224.求证:x+x>12a【解析】lnx1-ax1=lnx2-ax2，x2x2a(x2-x1)=(lnx2-lnx1)=ln=2ln>x1x14x2-1x14x2-x14(x2-x1)x=a(x2-x1)(x2+x1)>2x2+x1x2+x1x1+12整理得x+x>12a115.求证:+>2ax1x2x21x2x1【解析】a(x2-x1)=(lnx2-lnx1)=ln<(-)x12x1x211整理得+>2ax1x226.求证:e2,lnx1fx1=ax=alnx-ax=0∵⇔1⇔11lnxfx2=a2=alnx2-ax2=0x2lnx1=ax12∴即证a(x1+x2)>2,即证x1+x2>,由(2)知成立lnx2=ax2a17.求证:xx<(反向对均不等式).12a21x2-x1x2-x11【解析】x1x2<2,>x1x2,>x1x2,∴x1x2<得证alnx2-lnx1a(x2-x1)a38.求证:-e0①,ax2-12e-x1x1+e2令x2=e得<,整理得-ax1-(ae-3)⋅x1-e>0②,1-ax1222①+②得:a(x2-x1)+(ae-3)x2-(ae-3)x1>03-ae整理得a(x+x)(x-x)>(ae-3)(x-x),∴x+x>12211212ax1x12(e-1)法二:证明:左边:∵00②,3-ae①-②得x+x>.右边同法一12a-2lna9.求证：x+x<12alnx=ax-2lna11lnx1+lnx2-2lna【解析】要证x1+x2<,∵要证<，alnx2=ax2aa1即证lnxx<1na-2,即证xx<,由(5)得证.1212a21-lna10.求证:x+x>12a1-lna1lna【解析】x+x>=-12aaa11x2-x1x1+x21x2-ax2+a∵，<,先令x1=,<，lnx2-lnx12aax2+lna2111a-x1x1+a2lna2再令x2=，<，ax2+(lna-1)⋅x2++>0①，a-lna-ax12aalna2-ax2-(lna-1)⋅x-->0②,11aa1-lna①+②得:x+x>.12a11.求证:lnx1+lnx2>1-lna.lnx1=ax11-lna【解析】∵∴ax1+ax2>1-lna，x1+x2>由(7)得证.lnx2=axae12.求证:xx>12aee【解析】xx>,即证lnxx>ln.即lnx+lnx>1-lna,由(8)得证.12a12a1211213.求证：+>x1x2e11211【解析】法一：要证+>,令=t1，=t2，x1x2ex1x2由题意得lnx1=ax1，lnx2=ax2，1111代入则有ln=a，ln=at1t1t2t2aa即0=lnt1+，0=lnt2+，t1t2aa两式作差有lnt2-Int1-+=0，t1t2lnt2-lnt1a1=<，a2∴t1+t2>2t1t2>，即+>eeex1x2e1111t1法二：另解:令t=,则t1=,t2=,有0;12e-ax111t1-t2由题意lnt1=,lnt2=,lnt1-lnt2=-a(-)=a,t1t2t1t2t1+t2t1t1-t2t1+t2即ln=a;lnt1+lnt2=-a;t2t1+t2t1⋅t2t2ln(tt)t+tt+1t1212即21，∴t=-ln(t1t2)=-t⋅lnln1t1-t21-1t2t2t1t1-1t2(t-1)t1又121，ln>t(0<<1)∴ln(t1t2)<-2∴t1t2<2t21+1t2et22又t+t>2tt即t+t>,得证.121212e14.求证：2lnx1+lnx2>ex2-x11【解析】易知=，要证2lnx1+lnx2>e，lnx2-lnx1aax2-x1即证2lnx1+lnx2>e⋅，lnx2-lnx1x2-x1x2-x1即2ax1+ax2>ae，即证2x1+x2>e，lnx2-lnx1lnx2-lnx1xe(t-1)令2=t∈(1,+∞)，即只需证lnt>.x1t+2243x-1(x-1)令g(x)=lnx-2(x>1)，g(x)=22>0，x+4x+1xx+4x+1g(x)在(1,+∞)单调递增，∴g(x)>g(1)=0，2e(t-1)3t-1e(t-1)lnt->-=t+2t2+4t+1t+2t-122(3-e)t+(9-4e)t+6-e>0t+4t+1(t+2)∴原式得证.1115.求证：+>2aelnx1lnx2111111【解析】要证+>2ae，证+>2a2e，由(4)得+>2a，lnx1lnx2x1x2x1x211又ae∈(0,1)故+>2a2e，原式得证.x1x23216.求证：x+1x+1<-+112a2a3232【解析】要证x+1x+1<-+1即证x+x+xx<-，12a2a1212a2a12lna2lna22由(7)知xx<①，由(9)知x+x<-，∴-<-12a212aaa2a111由lnx>1-(x≠1)⇔-lnx<-1+(x≠1)⇔-lna<-1+(a≠1)xxa2lna2232得-<-②，由①+②得x+x+xx<-，原式得证.aa2a1212a2a2e17.求证：x2x+x2x>1221a22e【解析】由(2)知x+x>；由(12)知xx>，12a12a222e则有xx+xx=xxx+x>12211212a2x18.求证：1,∴x-x<+1a2a12aax1lnx2法二：由题可得e>x1,满足.1-1-ae19.求证:x<﹔1a【解析】法一：因为x1,x2是lnx=ax的两个根,且11-ax⇔ax2-2x+e>01a111⇔x1lnx1-2x1+e>0,构造函数h(x)=xlnx-2x+e,x∈(1,e),则h(x)=lnx-1<0,h(x)在(1,e)上递减,∴h(x)=h(e)=0,原不等式成立.法二:因为xlnx≥x-1当且仅当x=1时等号成立.xxxxln≥-1,当且仅当=1时等号成立，eeee即x(lnx-1)≥x-e当且仅当x=e时等号成立,即xlnx≥2x-e当且仅当x=e时等号成立,∵10,原不等式成立.1+1-ae20.求证:x>;2a1+1-ae【解析】要证x>，即证ax-1>1-ae.2a2lnx即证ax2-2x+e>0∴只需证ax2=xlnx>2x-e,f(x)=222222x1-lnx∵f(x)=,∴f(x)在(0,e)上单调递增,在(e,+∞)上单调递减.x211∴f(x)=f(e)=,0e,∴x2lnx2>2x2-e得证21-ae21.求证:x-x>;21alnx1-lnx【解析】法一：f(x)=∴f(x)=xx2∴f(x)在(0,e)上单调递增,在(e,+∞)上单调递减.11∴f(x)=f(e)=,∴01-,xxxxxe由ax=lnx得ax-1=lnx-1=ln>1-,ex2∵x1>0,∴上式化简为ax1-2x1+e>0,又∵a>0,Δ=4-4ae>0,2±4-4ae1+1-ae1-1-ae∴x=∴x=或x=,12a1a2a11-1-ae∵0∴12>.12a2ax+x11-1-eax+x-2x1-ea∴12-x>-