Question

Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) + 3f\left( \frac{\pi}{2} - x \right) = \sin x, x \in \mathbb{R}$. Let the maximum value of $f$ on $\mathbb{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x) = x^2$ and $h(x) = \beta x^3, \beta>0$, is $\alpha^2$, then $30\beta^3$ is equal to :}

Answer 1

Correct Answer: 16

Step 1: Understanding the Concept:
We first determine the function $f(x)$ using functional equations, find its maximum value $\alpha$, and then solve the area-related integral to find $\beta$.
Step 2: Detailed Explanation:
Given: $f(x) + 3f(\frac{\pi}{2} - x) = \sin x \quad \dots(1)$
Replace $x$ with $\frac{\pi}{2} - x$:
$f(\frac{\pi}{2} - x) + 3f(x) = \sin(\frac{\pi}{2} - x) = \cos x \quad \dots(2)$
From (2), $f(\frac{\pi}{2} - x) = \cos x - 3f(x)$. Substitute into (1):
$f(x) + 3(\cos x - 3f(x)) = \sin x \implies -8f(x) = \sin x - 3 \cos x$
$f(x) = \frac{3 \cos x - \sin x}{8}$.
Maximum value $\alpha = \frac{\sqrt{3^2 + (-1)^2}}{8} = \frac{\sqrt{10}}{8}$.
Now, $\alpha^2 = \frac{10}{64} = \frac{5}{32}$.
Curves $y = x^2$ and $y = \beta x^3$ intersect at $x=0$ and $x = 1/\beta$.
Area $= \int_0^{1/\beta} (x^2 - \beta x^3) dx = [ \frac{x^3}{3} - \frac{\beta x^4}{4} ]_0^{1/\beta}$
Area $= \frac{1}{3\beta^3} - \frac{\beta}{4\beta^4} = \frac{1}{3\beta^3} - \frac{1}{4\beta^3} = \frac{1}{12\beta^3}$.
Equating Area to $\alpha^2$:
$\frac{1}{12\beta^3} = \frac{5}{32} \implies \beta^3 = \frac{32}{60} = \frac{8}{15}$.
We need $30\beta^3 = 30 \times \frac{8}{15} = 16$.
Step 3: Final Answer:
The value of $30\beta^3$ is 16.