If $2 f\left(x^2\right)+3 f\left(\frac{1}{x^2}\right)=x^2-1, \forall x \in R-\{0\}$, then $f\left(x^8\right)$ is equal to

• Correct answer
  A

  $\frac{\left(1-x^8\right)\left(2 x^8+3\right)}{5 x^8}$
• B

  $\frac{\left(1+x^8\right)\left(2 x^8-3\right)}{5 x^8}$
• C

  $\frac{\left(1-x^8\right)\left(2 x^8-3\right)}{5 x^8}$
• D

  None of these

Solution

Replacing $x$ by $\frac{1}{x}$, we get
On multiplying Eq. (i) by 2, Eq. (ii) by 3 and then subtracting Eq. (i) from Eq. (ii), we get $\begin{array}{rlrl} & & 5 f\left(x^2\right) & =3\left(\frac{1}{x^2}-1\right)-2\left(x^2-1\right) \\ \Rightarrow & \quad 5 f\left(x^2\right) & =\frac{3}{x^2}-2 x^2-1 \\ \Rightarrow & \quad f\left(x^2\right) & =\frac{1}{5}\left(\frac{3}{x^2}-2 x^2-1\right) \\ \Rightarrow & \quad f\left(x^2\right) & =\frac{1}{5 x^2}\left(3-2 x^4-x^2\right) \\ \Rightarrow & \quad f\left(x^2\right) & =\frac{\left(2 x^2+3\right)\left(1-x^2\right)}{5 x^2} \\ \therefore & & f\left(x^8\right) & =\frac{\left(1-x^8\right)\left(2 x^8+3\right)}{5 x^8} \end{array}$

Hence, the correct answer is option: 'A'.