帮忙看下这道题，第2问。

Answer 1

容易求出a = -1

x在3, 4]上时, f(x) = -(1/ln2)[ln(x + 1)- ln(x- 1)]
f'(x) = -(1/ln2)[1/(x + 1) - 1/(x - 1)] = (2/ln2)/[(x + 1)(x - 1)]
f'(x)在此区间内为正, f(x)为增函数，其最小值为n = f(3) = -(1/ln2)(ln4 - ln2) = -1

令g(x) = (1/2)^x + m = 2^(-x) + m
g'(x) = -(ln2) [2^(-x)]
在同一区间内g'(x) < 0, 为减函数, g(x)在区间内的最大值N = g(3) = 1/8 + m
要使不等式成立, 须n > N, 即-1 > 1/8 + m, m < -9/8