Let f : R R be a function such that f(x) + 3f(2 - x) = x, x R.

Question

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

Answer: 16

Accepted Answer

From $f(x) + 3f\left(\dfrac{\pi}{2}-x\right) = \sin x$ \ldots (1), replace $x \to \dfrac{\pi}{2}-x$:

$$f\left(\frac{\pi}{2}-x\right) + 3f(x) = \cos x \quad \text{\ldots (2)}$$

Multiply (1) by 3 and subtract (2): $8f\left(\dfrac{\pi}{2}-x\right) = 3\sin x - \cos x$, so replacing back,

$$8f(x) = 3\cos x - \sin x \implies f(x) = \frac{1}{8}(3\cos x - \sin x)$$

The amplitude of $3\cos x - \sin x$ is $\sqrt{3^2+1^2} = \sqrt{10}$, so $f_{\max} = \alpha = \dfrac{\sqrt{10}}{8}$.

For $g(x) = x^2$, $h(x) = \beta x^3$ ($\beta>0$), they meet at $x=0$ and $x = 1/\beta$. Area between them:

$$\Delta = \int_0^{1/\beta} (\beta x^3 - x^2)\,dx = \frac{1}{12\beta^3}$$

Setting $\Delta = \alpha^2 = \dfrac{10}{64} = \dfrac{5}{32}$:

$$\frac{1}{12\beta^3} = \frac{5}{32} \implies \beta^3 = \frac{8}{15} \implies 30\beta^3 = 30 \times \frac{8}{15} = 16$$

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