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六导数(B)1.(2018广西二模)已知函数f(x)=ln (x+a)-x(aR),直线l:y=-23x+ln 3-23是曲线y=f(x)的一条切线.(1)求a的值;(2)设函数g(x)=xex-2x-f(x-a)-a+2,证明:函数g(x)无零点.2.(2018咸阳一模)已知f(x)=ex-aln x(aR).(1)求函数f(x)在点(1,f(1)处的切线方程;(2)当a=-1时,若不等式f(x)e+m(x-1)对任意x(1,+)恒成立,求实数m的取值范围.3.(2018凯里市校级三模)已知函数f(x)=lnxmx(m0).(1)试讨论函数f(x)的单调性;(2)对a,b(e,+),且aba.4.(2018辽宁模拟)已知函数f(x)=-x+aln x(aR).(1)讨论f(x)的单调性;(2)设g(x)=x2-2x+2a,若对任意x1(0,+),均存在x20,1,使得f(x1)0,g(x)=(x+1)ex-1-1x=(x+1)ex-1x,可设ex-1x=0的根为m,即有em=1m,即有m=-ln m,当xm时,g(x)递增,0x0恒成立,则函数g(x)无零点.2.解:(1)由f(x)=ex-aln x,则f(x)=ex-ax,f(1)=e-a,切点为(1,e),所求切线方程为y-e=(e-a)(x-1),即(e-a)x-y+a=0.(2)由f(x)=ex-aln x,原不等式即为ex+ln x-e-m(x-1)0,记F(x)=ex+ln x-e-m(x-1),F(1)=0,依题意有F(x)0对任意x1,+)恒成立,求导得F(x)=ex+1x-m,F(1)=e+1-m,F(x)=ex-1x2,当x1时,F(x)0,则F(x)在(1,+)上单调递增,有F(x)F(1)=e+1-m,若me+1,则F(x)0,若F(x)在(1,+)上单调递增,且F(x)F(1)=0,适合题意;若me+1,则F(1)0,故存在x1(1,ln m)使F(x)=0,当1xx1时,F(x)0,得F(x)在(1,x1)上单调递减,得F(x)0时,对x(0,e),有f(x)0,故函数f(x)在(0,e)上单调递增;对x(e,+),有f(x)0,故函数f(x)在(e,+)上单调递减;当m0时,对x(0,e),有f(x)0,故函数f(x)在(e,+)上单调递增.(2)证明:对a,b(e,+),且af(b),所以lnaalnbb,所以bln aaln b,所以abba.4.解:(1)f(x)=-1+ax=-(x-a)x(x0),a0时,由于x0,故x-a0,f(x)0时,由f(x)=0,解得x=a,在区间(0,a)上,f(x)0,在区间(a,+)上,f(x)0时,函数f(x)在(0,a)上递增,在(a,+)上递减.(2)依题意,要满足对任意x1(0,+),均存在x20,1,使得f(x1)g(x2),转化为f(x)maxg(x)max,因为g(x)=x2-2x+2a,x0,1,所以g(x)max=2a,由(1)得a0时,f(x)在(0,+)上递减,值域是R,不合题意,a=0时,f(x)=-x0时,f(x)在(0,a)上递增,在(a,+)上递减,故f(x)的极大值即为最大值,f(a)=-a+aln a,故2a-a+aln a,解得0ae3.综上,a的范围是0,e3.
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