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(A) Given $f(x) + 3f(\frac{\pi}{2} - x) = \sin x$  $(1)$. Replacing $x$ with $\frac{\pi}{2} - x$, we get $f(\frac{\pi}{2} - x) + 3f(x) = \cos x$  $(2)

Vedclass · Accepted Answer

$16$

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(A) Given $f(x) + 3f(\frac{\pi}{2} - x) = \sin x$  $(1)$. Replacing $x$ with $\frac{\pi}{2} - x$, we get $f(\frac{\pi}{2} - x) + 3f(x) = \cos x$  $(2)$. Multiply $(2)$ by $3$: $3f(\frac{\pi}{2} - x) + 9f(x) = 3\cos x$  $(3)$. Subtract $(1)$ from $(3)$: $8f(x) = 3\cos x - \sin x$. Thus, $f(x) = \frac{3}{8}\cos x - \frac{1}{8}\sin x$. The maximum value of $f(x) = A\cos x + B\sin x$ is $\sqrt{A^2 + B^2}$. So, $\alpha = \sqrt{(\frac{3}{8})^2 + (-\frac{1}{8})^2} = \sqrt{\frac{9+1}{64}} = \frac{\sqrt{10}}{8}$. Then $\alpha^2 = \frac{10}{64} = \frac{5}{32}$. The curves $g(x) = x^2$ and $h(x) = \beta x^3$ intersect at $x^2 = \beta x^3$, which gives $x = 0$ and $x = \frac{1}{\beta}$. The area is $\int_0^{1/\beta} (x^2 - \beta x^3) dx = [\frac{x^3}{3} - \frac{\beta x^4}{4}]_0^{1/\beta} = \frac{1}{3\beta^3} - \frac{1}{4\beta^3} = \frac{1}{12\beta^3}$. Given $\frac{1}{12\beta^3} = \alpha^2 = \frac{5}{32}$. Therefore, $12\beta^3 = \frac{32}{5} = 6.4$, so $\beta^3 = \frac{6.4}{12} = \frac{8}{15}$. Finally, $30\beta^3 = 30 	imes \frac{8}{15} = 16$.

Let $f: R \to R$ be a function such that $f(x) + 3f(\frac{\pi}{2} - x) = \sin x, x \in R$. Let the maximum value of $f$ on $R$ 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 ————

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