If $2 \mathrm{f}(x)+3 \mathrm{f}\left(\frac{1}{x}\right)=x^2+1, x \neq 0$ and $\mathrm{y}=5 x^2 \mathrm{f}(x)$, then y is strictly increasing in

The function $y$ is strictly increasing when its derivative $\frac{dy}{dx} > 0$.

From the equation $2f(x) + 3f(\frac{1}{x}) = x^2 + 1$, replacing $x$ with $\frac{1}{x}$ yields $2f(\frac{1}{x}) + 3f(x) = \frac{1}{x^2} + 1$. Solving this system gives $f(x) = \frac{1}{5}(\frac{3}{x^2} + 1 - 2x^2)$.

Substituting into $y = 5x^2 f(x)$ yields $y = 3 + x^2 - 2x^4$. Differentiating gives $\frac{dy}{dx} = 2x - 8x^3 = 2x(1 - 4x^2) = 2x(1 - 2x)(1 + 2x)$.

The critical points are $x = -\frac{1}{2}$, $x = 0$, and $x = \frac{1}{2}$. Testing intervals shows $\frac{dy}{dx} > 0$ on $(-\infty, -\frac{1}{2})$ and $(0, \frac{1}{2})$.

Among the options, only A lies within an interval where $y$ is strictly increasing.