If $f\left(x\right)$  is an even function and satisfies the relation $x^2.f\left(x\right)-2.f\left(\frac{1}{x}\right)=g\left(x\right)$ where $g\left(x\right)$  is an odd function is, then the value of f (5)

$x^{2}f\left(x\right)-2.f\left(\frac{1}{x}\right)=g\left(x\right) \mathrm{...}\left(i\right)$  
replacing ‘x’ by $\frac{1}{x}$  then $\frac{1}{x^2}.f\left(\frac{1}{x}\right)-2f\left(x\right)=g\left(\frac{1}{x}\right)$  (or)
 $2f\left(\frac{1}{x}\right)-4x^{2}f\left(x\right)=2x^2.g\left(\frac{1}{x}\right) \mathrm{........}\left(2\right)$
Solve (i), (ii)
$\begin{array}{l}\Rightarrow f\left(x\right)=-\left(\frac{g\left(x\right)+2x^2.g\left(\frac{1}{x}\right)}{3x^2}\right)\\ f(-x)=-\left(\frac{g(-x)+2x^2.g(-\frac{1}{x})}{3x^2}\right)\\ f\left(x\right)=-f\left(x\right)\end{array}$ 
($\therefore g\left(x\right)$ is an odd and f(x) is an even function)
 $f\left(x\right)=0\Rightarrow f\left(5\right)=0$