Xavier Aptitude Test 2021 Paper

$\frac{\mathit{f}\left(\mathit{x}\right)}{\mathit{g}\left(\mathit{x}\right)}=\frac{\left({\mathit{x}}^2+1\right)}{\left({\mathit{x}}^2-1\right)}.\frac{\left(\mathit{x}-1\right)}{\left(\mathit{x}+1\right)}=\frac{\left({\mathit{x}}^2+1\right)}{{\left(\mathit{x}+1\right)}^2}$
This function is definitely greater than $0$
Let $\mathit{y}=\frac{\left({\mathit{x}}^2+1\right)}{{\left(\mathit{x}=1\right)}^2}$
$\Rightarrow {\mathit{x}}^2\left(\mathit{y}-1\right)+2\mathit{y}\mathit{x}+\left(\mathit{y}-1\right)=0$ which is quadratic in $\mathit{x}$
Discriminant should be greater than $0$
$4{\mathit{y}}^2-4{\left(\mathit{y}-1\right)}^2\ge 0$
$\Rightarrow \mathit{y}\ge \frac{1}{2}$
When $\mathit{x}=1,\frac{\mathit{f}\left(\mathit{x}\right)}{\mathit{g}\left(\mathit{x}\right)}=\frac{1}{3}$
Hence either the value should be greater than $\frac{1}{2}$ or should be equal to $\frac{1}{3}$