Question

If $2f\left(x^2\right)+3f\left(\frac{1}{x^2}\right)=x^2-1(x\ne 0)$ then $f\left(x^2\right)$ is:

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

Answer 1

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

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

Replace $x^2$ by $\frac{1}{x^2}$, we get,

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

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

Multiply (1) by 2 and (2) by 3, we get,

$4f\left(x^2\right)+6f\left(\frac{1}{x^2}\right)=2x^2-2$ (3)

$9f\left(x^2\right)+6f\left(\frac{1}{x^2}\right)=\frac{3(1-x^2)}{x^2}$ (4)

Subtract (3) from (4), we get,

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

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

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

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

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