Question

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

• A. $0$
• B. $\frac{50}{75}$
• C. $\frac{49}{75}$
• D. $-1$

Answer 1

The correct option is A $0$

$x^{2}f\left(x\right)-2f\left(\frac{1}{x}\right)=g\left(x\right)\dots \left(i\right)$
Replacing $x$ by $-x$
$x^{2}f(-x)-2f(-\frac{1}{x})=g(-x)\Rightarrow x^{2}f\left(x\right)-2f\left(\frac{1}{x}\right)=-g\left(x\right)\dots \left(ii\right)(\because f(x)$ is an even function and $g\left(x\right)$ is an odd function$)$
Now adding equations $\left(i\right)$ and $\left(ii\right)$, we have
$2x^{2}f\left(x\right)-4f\left(\frac{1}{x}\right)=0\Rightarrow x^{2}f\left(x\right)-2f\left(\frac{1}{x}\right)=0\dots \left(iii\right)$
Replacing $x$ by $\frac{1}{x}$
$\Rightarrow {\left(\frac{1}{x}\right)}^{2}f\left(\frac{1}{x}\right)-2f\left(x\right)=0\Rightarrow f\left(\frac{1}{x}\right)-2x^{2}f\left(x\right)=0\dots \left(iv\right)$
From equations $\left(iii\right)$ and $\left(iv\right)$, we have
$x^{2}f\left(x\right)=0\therefore f\left(5\right)=0$