设f(x)在[1,+∞)可导,[xf(x)]≤-kf(x)(x>1),在(1,+∞)的子区间上不恒等,又f(1)≤M,其中后k,M为常数,求证:f(x)<(x>1).

admin2016-10-26  26

问题 设f(x)在[1,+∞)可导,[xf(x)]≤-kf(x)(x>1),在(1,+∞)的子区间上不恒等,又f(1)≤M,其中后k,M为常数,求证:f(x)<(x>1).

选项

答案已知xf′(x)+(k+1)f(x)≤0(x>1),在(1,+∞)[*]子区间上不恒为零,要证f(x)xk+1<M(x>1).令F(x)=f(x)xk+1[*]F′(x)=xk+1f′(x)+(k+1)xkf(x)=xk[xf′(x)+(k+1)f(x)]≤0(x>1),在(1,+∞)[*]子区间上不恒为零,又F(x)在[1,+∞)连续[*]F(x)在[1,+∞)单调下降[*]F(x)<F(1)=f(1)≤M (x>1).

解析
转载请注明原文地址:https://kaotiyun.com/show/2Gu4777K
0

最新回复(0)