Question

If $2f\left(x^2\right)+3f\left(\frac{1}{x^2}\right)=x^2-1,x\in R_0$, then $f\left(x^2\right)$ is

• A. $\frac{1-x^4}{5x^2}$
• B. $\frac{1-x^2}{5x}$
• C. $\frac{5x^2}{1-x^4}$
• D. $\frac{3-2x^4-x^2}{5x^2}$

Answer 1

The correct option is D $\frac{3-2x^4-x^2}{5x^2}$

$2f\left(x^2\right)+3f\left(\frac{1}{x^2}\right)=x^2-1$ ........ (i)

Replacing $x$ by $\frac{1}{x}$, we get

$2f\left(\frac{1}{x^2}\right)+3f\left(x^2\right)=\frac{1}{x^2}-1$ ....... (ii)

Solving (i) and (ii) we get

$f\left(x^2\right)=\frac{3-2x^4-x^2}{5x^2}$