Question

已知定义域为R的函数f(x)=2的x次方-1/a+2的x+1次方是奇函数.(1)求a的值；（2）求证：f(x)在R上是增

Answer 1

f(x)=(2^x-1)/[(a+2^(x+1)]是奇函数

f(-x)=-f(x)

[2^(-x)-1]/[(a+2^(-x+1)] = -(2^x-1)/[(a+2^(x+1)]

- [(a+2^(-x+1)] (2^x-1) = [2^(-x)-1] [(a+2^(x+1)]

-a*2^x -2 + a+2*2^(-x) = a*2^(-x)-a+2-2*2^x

(a-2)*2^x + (a-2)*2^(-x) +2(a-2) = 0

(a-2){2^x + 2^(-x) +2} = 0

a = 2

f(x)=(2^x-1)/[(2+2^(x+1)] = (2^x-1)/{2(2^x+1)} = (2^x+1-2)/{2(2^x+1)} = 1/2 - 1/(2^x+1)}

2^x+1单调增,1/(2^x+1)} 单调减,1/2 - 1/(2^x+1)} 单调增