2. D Let $f(1)=a$, set $x=1$, then $f(1) f(f(1)+1)=1, f(a+1)=\frac{1}{a}$; set $x=a+1$, $f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1, \frac{1}{a} f\left(\frac{1}{a}+\frac{1}{a+1}\right)=1, f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1)$, by the fact that $f(x)$ is an increasing function, we know $\frac{1}{a}+\frac{1}{a+1}=1, a=\frac{1 \pm \sqrt{5}}{2}$. But when $f(1)=\frac{1+\sqrt{5}}{2}$, $1<a=f(1)<f(a+1)=\frac{1}{a}<1$, contradiction!
2. Given that the function $f(x)$ is defined and increasing on $(0,+\infty)$. It satisfies $f(x) \cdot f\left(f(x)+\frac{1}{x}\right)=1$, then $f(1)=$
A. 1
B. 0
C. $\frac{1+\sqrt{5}}{2}$
D. $\frac{1-\sqrt{5}}{2}$