在念念不忘中遗忘
幼苗
共回答了11个问题采纳率:90.9% 举报
(1)
f(x) = [1-2^x]/ [2^(x+1) + a]
f(1) = -1/(4+a)
f(-1) = (1/2)/(1+a) = 1/2(1+a)
f(1) = -f(-1)
1/(4+a) = 1/2(1+a)
2(1+a) = 4+a
a = 2 #
f(x) = [1-2^x]/ [2^(x+1) + 2]
= [1-2^x]/ 2[1+ 2^x]
f(t^2-6t-1)+f(2t^2-k) < 0
put t=0
f(-1) + f(-k) < 0
(1/2)/ 2(3/2) + (1-2^-k)/ 2(1+2^-k) 4
2^(-k+1) > 2^2
-k+1 > 2
k < -1
1年前
3