Let $\mathrm{f}(\mathrm{x})=\frac{x^2+1}{x^2-1}$, if $\mathrm{x} \neq 1,-1$, and 1 if $\mathrm{x}=1,-1$. Let $\mathrm{g}(\mathrm{x})=\frac{x+1}{x-1}$ if $\mathrm{x} \neq 1$ and 3 if $x=1$.
What is the minimum possible value of $\frac{f(x)}{g(x)}$ ?

A. 1

B. -1

C. 1/4

D. 1/3

D. 1/3

$\frac{f(x)}{g(x)}=\frac{(x^2+1)(x-1)}{(x^2-1)(x+1)} =\frac{(x^2+1)}{(x+1)^2}$

i.e x cannot be -1

we observe when x=1(which is possible here), we get f(x)=1 and g(x)=3

Hence, $\frac{f(x)}{g(x)} =\frac{1}{3}$