数学导数系数取值问题求解已知函数f(x)=(lnx)/(x+1)+1/x,如果x大于0x≠1时,f(x)大于lnx/(x-1)+k/x,求k的取值范围.要过程,分不是问题,谢了.

Question

数学导数系数取值问题求解
已知函数f(x)=(lnx)/(x+1)+1/x,如果x大于0 x ≠1时,f(x)大于lnx/(x-1)+k/x,求k的取值范围. 要过程,分不是问题,谢了.

Answer 1

令y=(lnx)/(x+1) + 1/x - lnx/(x-1) - k/x→当x＞0,且x≠1时,xlnx/(x-1)＜(1-k)/2.令z=xlnx/(x-1)→z'=[x - 1 - (x+1)lnx]/(x-1) 令w=x - 1 - (x+1)lnx→w'=x - 1/x - 2xlnx→w''=1/x - 2lnx - 1→w'''= -2(x+1)/x 易知：当x＞0时,w'''＜0→w''单调递减.当x=1时,w''=0→当0＜x＜1时,w''＞0→w'单调递增；当x＞1时,w''＜0→w'单调递减.当x=1时,w'=0→当x＞0,且x≠1时,w'＜0→w单调递减.当x=1时,w=0→当0＜x＜1时,w＞0→z单调递增；当x＞1时,w＜0→z单调递减.易得：lim(x→1) xlnx/(x-1)=1/2→当x＞0,且x≠1时,xlnx/(x-1)＜1/2 由xlnx/(x-1)＜(1-k)/2,xlnx/(x-1)＜1/2→k≤0 故：k的取值范围是k≤0