Question

设函数f(x)=2的X次方+2的X次方分之a再 —1(a为常数)

Answer 1

(1) a=0时,f(x) = 2^x - 1

x>0时,g(x)=f(x) = 2^x - 1

x0

f(x) = t + a/t - 1 = 0

t² - t + a = 0

t = [1±√(1-4a) ] / 2

[1-√(1-4a) ] / 2 < 0 舍去

t= 1+√(1-4a) = 2^x

x = log₂ [ 1+√(1-4a) ]