If 5f(x) + 4f((1)/(x)) = x ^ 2 - 2, x ≠ 0 and y = 9 x ^ 2 f(x),...

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

Replacing $x$ by $\dfrac{1}{x}$.

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

Using, \left(i\right) \times 5 - \left(ii\right) \times 4$</p><p>$⇒ 25f\left(x\right) + 20f\left(\dfrac{1}{x}\right) - 16f\left(x\right) - 20f\left(\dfrac{1}{x}\right) = 5{x}^{2} - 10 - \dfrac{4}{{x}^{2}} + 8$</p><p>$⇒ 9f\left(x\right) = 5{x}^{2} - \dfrac{4}{{x}^{2}} - 2$</p><p>Now,$y = 9f\left(x\right) \times {x}^{2}$</p><p>$⇒ y = \left(5{x}^{2} - 2 - \dfrac{4}{{x}^{2}}\right) \times {x}^{2}$</p><p>$⇒ y = 5{x}^{4} - 2{x}^{2} - 4$</p><p>$⇒ {y}^{'} = 20{x}^{3} - 4x$</p><p>$⇒ {y}^{'} = 4x\left(5{x}^{2} - 1\right)$</p><p>For function to be increasing${y}^{'} > 0$</p><p>$⇒ x \in \left(- \sqrt{\dfrac{1}{5}}, 0\right) \cup \left(\sqrt{\dfrac{1}{5}}, \infty\right)$