Functions

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