Area Bounded by Bounded Curves and Maximum Value of Function

Given the functional equation for $f: \mathbb{R} \to \mathbb{R}$: $f(x) + 3f\left(\frac{\pi}{2} - x\right) = \sin x \quad \text{--- (1)}$

Replacing $x$ with $\frac{\pi}{2} - x$ in equation (1), we get: $f\left(\frac{\pi}{2} - x\right) + 3f(x) = \sin\left(\frac{\pi}{2} - x\right) = \cos x \quad \text{--- (2)}$

From equation (2), express $f\left(\frac{\pi}{2} - x\right)$ in terms of $f(x)$: $f\left(\frac{\pi}{2} - x\right) = \cos x - 3f(x)$

Substituting this into equation (1): $f(x) + 3\left(\cos x - 3f(x)\right) = \sin x$ $f(x) + 3\cos x - 9f(x) = \sin x$ $-8f(x) = \sin x - 3\cos x$ $f(x) = \frac{3\cos x - \sin x}{8}$

The maximum value of a function of the form $A\cos x + B\sin x$ is $\sqrt{A^2 + B^2}$. Here, $A = 3$ and $B = -1$, so the maximum value of $3\cos x - \sin x$ is $\sqrt{3^2 + (-1)^2} = \sqrt{10}$.

Thus, the maximum value of $f(x)$ on $\mathbb{R}$ is: $\alpha = \frac{\sqrt{10}}{8}$

Squaring $\alpha$, we obtain: $\alpha^2 = \left(\frac{\sqrt{10}}{8}\right)^2 = \frac{10}{64} = \frac{5}{32}$

Next, we calculate the area bounded by the curves $g(x) = x^2$ and $h(x) = \beta x^3$ with $\beta > 0$.

Finding the points of intersection by setting $g(x) = h(x)$: $x^2 = \beta x^3 \implies x^2(1 - \beta x) = 0$ So, the intersection points are $x = 0$ and $x = \frac{1}{\beta}$.

For $x \in \left[0, \frac{1}{\beta}\right]$, we have $x^2 \ge \beta x^3$. The bounded area is: $\text{Area} = \int_{0}^{\frac{1}{\beta}} (x^2 - \beta x^3) \, dx$ $\text{Area} = \left[ \frac{x^3}{3} - \frac{\beta x^4}{4} \right]_0^{\frac{1}{\beta}} = \frac{1}{3\beta^3} - \frac{\beta}{4\beta^4} = \frac{1}{12\beta^3}$

We are given that this area is equal to $\alpha^2$: $\frac{1}{12\beta^3} = \frac{5}{32}$

Solving for $\beta^3$: $\beta^3 = \frac{32}{12 \times 5} = \frac{8}{15}$

Finally, calculating the value of $30\beta^3$: $30\beta^3 = 30 \times \frac{8}{15} = 16$