If 5 f(x)+4 fleft(frac{1}{x}right)=x^2-2, for | vedclass

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(B) Given the equation: $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2$ ....$(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$5 f\left(\frac{1}{x}\rig

Vedclass · Accepted Answer

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

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(B) Given the equation: $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2$ ....$(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$:$5 f\left(\frac{1}{x}\right)+4 f(x)=\frac{1}{x^2}-2$ ....$(2)$Multiply $(1)$ by $5$ and $(2)$ by $4$:$25 f(x)+20 f\left(\frac{1}{x}\right)=5x^2-10$$16 f(x)+20 f\left(\frac{1}{x}\right)=\frac{4}{x^2}-8$Subtracting the two equations:$9 f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 = 5x^2 - 2 - \frac{4}{x^2}$Given $y = 9x^2 f(x)$,substitute $9 f(x)$:$y = x^2 \left(5x^2 - 2 - \frac{4}{x^2}\right) = 5x^4 - 2x^2 - 4$To find where $y$ is strictly increasing,calculate the derivative:$\frac{dy}{dx} = 20x^3 - 4x = 4x(5x^2 - 1)$For strictly increasing,$\frac{dy}{dx} > 0$:$4x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$The critical points are $x = -\frac{1}{\sqrt{5}}, 0, \frac{1}{\sqrt{5}}$.Testing intervals,we find $\frac{dy}{dx} > 0$ for $x \in \left(-\frac{1}{\sqrt{5}}, 0\right) \cup \left(\frac{1}{\sqrt{5}}, \infty\right)$.

If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x)$,then $y$ is strictly increasing in :

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