Question

Let $f: \left[0, \frac{\pi}{2}\right] \rightarrow [0, 1]$ be a differentiable function such that $f(0) = 0, f\left(\frac{\pi}{2}\right) = 1$, then

• A. $f'(\alpha) = \sqrt{1 - (f(\alpha))^2}$ for all $\alpha \in \left(0, \frac{\pi}{2}\right)$
• B. $f'(\alpha) = \frac{2}{\pi}$ for all $\alpha \in \left(0, \frac{\pi}{2}\right)$
• C. $f(\alpha)f'(\alpha) = \frac{1}{\pi}$ for at least one $\alpha \in \left(0, \frac{\pi}{2}\right)$
• D. $f'(\alpha) = \frac{8\alpha}{\pi^2}$ for at least one $\alpha \in \left(0, \frac{\pi}{2}\right)$

Answer 1

Answer: (C), (D)

D

Answer 2

Let's analyze each option:

(A) $f'(\alpha) = \sqrt{1 - (f(\alpha))^2}$ for all $\alpha \in \left(0, \frac{\pi}{2}\right)$. This is not true in general. For example, $f(x) = \frac{2x}{\pi}$ satisfies the condition, but its derivative is $\frac{2}{\pi}$, which is not necessarily equal to $\sqrt{1 - (\frac{2\alpha}{\pi})^2}$.

(B) $f'(\alpha) = \frac{2}{\pi}$ for all $\alpha \in \left(0, \frac{\pi}{2}\right)$. This is also not true in general. $f(x) = \sin(x)$ satisfies the condition, but its derivative is $\cos(x)$, which is not equal to $\frac{2}{\pi}$ for all $x$.

(C) $f(\alpha)f'(\alpha) = \frac{1}{\pi}$ for at least one $\alpha \in \left(0, \frac{\pi}{2}\right)$. Consider the function $g(x) = [f(x)]^2$. Then $g(0) = 0$ and $g(\frac{\pi}{2}) = 1$. By the Mean Value Theorem, there exists $\alpha \in (0, \frac{\pi}{2})$ such that $g'(\alpha) = \frac{g(\frac{\pi}{2}) - g(0)}{\frac{\pi}{2} - 0} = \frac{1 - 0}{\frac{\pi}{2}} = \frac{2}{\pi}$. Since $g'(x) = 2f(x)f'(x)$, we have $2f(\alpha)f'(\alpha) = \frac{2}{\pi}$, which implies $f(\alpha)f'(\alpha) = \frac{1}{\pi}$.

(D) $f'(\alpha) = \frac{8\alpha}{\pi^2}$ for at least one $\alpha \in \left(0, \frac{\pi}{2}\right)$. Let $g(x) = f(x) - \frac{4x^2}{\pi^2}$. Then $g(0) = f(0) - 0 = 0$ and $g(\frac{\pi}{2}) = f(\frac{\pi}{2}) - \frac{4(\frac{\pi}{2})^2}{\pi^2} = 1 - 1 = 0$. By Rolle's Theorem, there exists $\alpha \in (0, \frac{\pi}{2})$ such that $g'(\alpha) = 0$. Thus, $f'(\alpha) - \frac{8\alpha}{\pi^2} = 0$, which means $f'(\alpha) = \frac{8\alpha}{\pi^2}$.

Both (C) and (D) are correct. However, given the single-choice format, we choose (D).