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 :

Question

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

Accepted Answer

$5 f(x)+4 f(1 / x)=x^2-2$ ........(1)

Replace $x$ by $1 / x$

$$
5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2
$$ ..........(2)

Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)

$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}$

$\begin{aligned} & y=9 x^2 f(x) \\ & y=5 x^4-2 x^2-4 \\ & y^{\prime}=20 x^3-4 x \\ & \text { Put } y^{\prime}>0 \\ & 20 x^3-4 x>0 \\ & 5 x^3-x>0 \\ & x\left(5 x^2-1\right)>0\end{aligned}$

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

Solutions · 1