Question

我要解题过程

Answer 1

f(x) = a - 1/(2^x-1)
分母不为零，定义域x≠0
定义在(-∞,-1] U [1,+∞)上的奇函数
f(-x)=-f(x)
a - 1/(2^(-x)-1) = - a + 1/(2^x-1)
2a = 1/(2^(-x)-1) + 1/(2^x-1) = 2^x/(1-2^x) + 1/(2^x-1) = (2^x-1)/(1-2^x) = -1
a = -1/2

追答

等等，计算错了

追问

确定是这样？好像不是啊

追答

f(x) = a - 1/(2^x+1)
f(-x)=-f(x)
a - 1/(2^(-x)+1) = - a + 1/(2^x+1)
2a = 1/(2^(-x)+1) + 1/(2^x+1) = 2^x/(1+2^x) + 1/(2^x+1) = (2^x+1)/(1+2^x) = 1
a = 1/2