If 5 f x + 4 f 1 x = x 2 − 2 , ∀ x ≠ 0 and y = 9 x 2 f x , then y is strictly increasing in:

Given: $5f\left(x\right)+4f\left(\frac{1}{x}\right)=x^2-2 ...\left(i\right)$

Replacing $x$ by $\frac{1}{x}$.

$\Rightarrow 5f\left(\frac{1}{x}\right)+4f\left(x\right)={\left(\frac{1}{x}\right)}^2-2 ...\left(ii\right)$

Using, $\left(i\right)\times 5-\left(ii\right)\times 4$

$\Rightarrow 25f\left(x\right)+20f\left(\frac{1}{x}\right)-16f\left(x\right)-20f\left(\frac{1}{x}\right)=5x^2-10-\frac{4}{x^2}+8$

$\Rightarrow 9f\left(x\right)=5x^2-\frac{4}{x^2}-2$

Now, $y=9f\left(x\right)\times x^2$

$\Rightarrow y=\left(5x^2-2-\frac{4}{x^2}\right)\times x^2$

$\Rightarrow y=5x^4-2x^2-4$

$\Rightarrow y^{\text{'}}=20x^3-4x$

$\Rightarrow y^{\text{'}}=4x\left(5x^2-1\right)$

For function to be increasing $y^{\text{'}}>0$

$\Rightarrow x\in \left(-\sqrt{\frac{1}{5}},0\right)\cup \left(\sqrt{\frac{1}{5}},\infty \right)$