1. Given functional equation

We have $5f(x)+4f\left(\frac{1}{x}\right)=x^2-2, \quad x\ne 0.$

We need to study $y=9x^2f(x)$ and find where it is strictly increasing.

2. Form a second equation

Replace $x$ by $\frac{1}{x}$ in the given relation: $5f\left(\frac{1}{x}\right)+4f(x)=\frac{1}{x^2}-2.$

So we have the system: $5f(x)+4f\left(\frac{1}{x}\right)=x^2-2 \quad ...(1)$ $4f(x)+5f\left(\frac{1}{x}\right)=\frac{1}{x^2}-2 \quad ...(2)$

3. Solve for $f(x)$

Let $a=f(x), \qquad b=f\left(\frac{1}{x}\right).$ Then $5a+4b=x^2-2$ $4a+5b=\frac{1}{x^2}-2.$

Eliminate $b$:

Multiply the first by $5$ and the second by $4$: $25a+20b=5x^2-10$ $16a+20b=\frac{4}{x^2}-8$ Subtracting, $9a=5x^2-2-\frac{4}{x^2}.$ Hence, $f(x)=a=\frac{1}{9}\left(5x^2-2-\frac{4}{x^2}\right).$

4. Compute $y$

Given $y=9x^2f(x),$ substitute $f(x)$: $y=9x^2\cdot \frac{1}{9}\left(5x^2-2-\frac{4}{x^2}\right).$ So, $y=x^2\left(5x^2-2-\frac{4}{x^2}\right)=5x^4-2x^2-4.$

5. Differentiate

$y'=\frac{d}{dx}(5x^4-2x^2-4)=20x^3-4x=4x(5x^2-1).$

For $y$ to be strictly increasing, $y'>0 \iff 4x(5x^2-1)>0 \iff x(5x^2-1)>0.$

Critical points are $x=0, \quad x=\pm \frac{1}{\sqrt{5}}.$

6. Sign analysis

Check intervals:

• For $x< -\frac{1}{\sqrt{5}}$: $x<0, \quad 5x^2-1>0 \implies y'<0.$

• For $-\frac{1}{\sqrt{5}}<x<0$: $x<0, \quad 5x^2-1<0 \implies y'>0.$

• For $0<x<\frac{1}{\sqrt{5}}$: $x>0, \quad 5x^2-1<0 \implies y'<0.$

• For $x>\frac{1}{\sqrt{5}}$: $x>0, \quad 5x^2-1>0 \implies y'>0.$

Thus $y$ is strictly increasing on $\left(-\frac{1}{\sqrt{5}},0\right)\cup\left(\frac{1}{\sqrt{5}},\infty\right).$

7. Match with options

This corresponds to Option B.

8. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.