Question

If $2f\left(x^2\right)+3f\left(\frac{1}{x^2}\right)=x^2-1$ for all $x\in R-\left\{0\right\},$ then $f\left(x^2\right)$ is equal to

• A. $\frac{1-x^2}{5x^2}$
• B. $\frac{(1-x^2)(3+2x^2)}{5x^2}$
• C. $\frac{(1-x^2)(3-2x^2)}{3x^2}$
• D. none of these

Answer 1

Answer: (B)

$\frac{(1-x^2)(3+2x^2)}{5x^2}$

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

Replace $x^2$ by $\frac{1}{x^2}$

$2f\left(\frac{1}{x^2}\right)+3f\left(x^2\right)=\frac{1}{x^2}-1=\frac{1-x^2}{x^2}$
Eliminating $f\left(\frac{1}{x^2}\right)$ from the above two,we have

$(9-4)f\left(x^2\right)=\frac{(1-x^2)(3+2x^2)}{x^2}$

$\therefore f\left(x^2\right)=\frac{(1-x^2)(3+2x^2)}{5x^2}$