问题
解答题
设函数f(x)=lnx-ax,(a∈R). (Ⅰ)判断函数f(x)的单调性; (Ⅱ)当lnx<ax对于x∈(0,+∞)上恒成立时,求a的取值范围; (Ⅲ)若k,n∈N*,且1≤k≤n,证明:
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答案
(Ⅰ)求导函数,可得f′(x)=
-a(x>0)1 x
当a≤0时,f′(x)>0,f(x)在(0,+∞)上是增函数;
当a>0时,由f′(x)>0可得0<x<
,由f′(x)>0可得x>1 a
,1 a
∴当a≤0时,函数f(x)的单调增区间是(0,+∞);当a>0时,函数f(x)的单调增区间是(0,
),单调减区间是(1 a
,+∞);1 a
(Ⅱ)lnx<ax对于x∈(0,+∞)上恒成立,等价于f(x)max<0
由上知,a≤0时,不成立;
a>0时,f(x)max=f(
)=ln1 a
-1<0,∴a>1 a
;1 e
(Ⅲ)证明:∵函数f(x)=lnx-ax,由(Ⅱ)知,a=1时,f(x)max=f(
)=ln1 a
-1=-11 a
∴lnx-x<-1
∴lnx<x-1
令x=1+
,则ln(1+k n
)<k n
,∴nln(1+k n
)<k,∴ln(1+k n
)n<kk n
∴(1+
)n<ek,∴k n
>1 (1+
)nk n 1 ek
∴
+1 (1+
)n1 n
+…+1 (1+
)n2 n
+…+1 (1+
)nk n
>1 (1+
)nn n
+1 e
+…+1 e2
+1 e2
=1 2n
+
(1-1 e
)1 en-1 1- 1 e 1 2n
当n→+∞时,
→
(1-1 e
)1 en-1 1- 1 e
.1 e-1
∴
+1 (1+
)n1 n
+…+1 (1+
)n2 n
+…+1 (1+
)nk n
>1 (1+
)nn n
.1 e-1