Let $$f: (0, \infty) \rightarrow (0, \infty)$$ be a strictly decreasing function. Consider $$h(x) = \frac{f\left(\frac{x}{1 + x}\right)}{1 + f\left(\frac{x}{1 + x}\right)}$$ Which one of the following is always true?

$$\ h\left(x\right)=\ \ \frac{\ f\left(\ \frac{\ x}{1+x}\right)}{1+f\left(\ \frac{\ x}{1+x}\right)}=\ \frac{1+\ f\left(\ \frac{\ x}{1+x}\right)-1}{1+f\left(\ \frac{\ x}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ \frac{\ x}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ \frac{\ 1+x-1}{1+x}\right)}=1-\ \frac{\ 1}{1+f\left(\ 1-\frac{\ 1}{1+x}\right)}$$.

Now, Since f is a strictly decreasing function and its domain is positive, then, $$f\left(\ \frac{\ 1}{x+1}\right)$$ will be a strictly increasing function. Again, $$f\left(\ 1-\frac{\ 1}{x+1}\right)$$ will be a strictly decreasing function.

Let us assume $$1+f\left(\ 1-\frac{\ 1}{x+1}\right)$$= g(x).

So, g(x) is a strictly decreasing function.

h(x)= $$1-\ \frac{\ 1}{g\left(x\right)}$$. 

 $$\frac{\ 1}{g\left(x\right)}$$ becomes strictly increasing function, but $$1-\frac{\ 1}{g\left(x\right)}$$ will finally be strictly decreasing function. 

Therefore, h(x) becomes strictly decreasing function- Option A.