If 0 le x le 1 and (sin^{-1} x)^3 + (cos^{-1} | vedclass

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Question

(A) Let $u = \sin^{-1} x$. Since $0 \le x \le 1$, we have $0 \le u \le \frac{\pi}{2}$.Then $\cos^{-1} x = \frac{\pi}{2} - u$.The expression becomes $f

Vedclass · Accepted Answer

$a \ge \frac{1}{32}$

Vedclass · Answer

(A) Let $u = \sin^{-1} x$. Since $0 \le x \le 1$, we have $0 \le u \le \frac{\pi}{2}$.Then $\cos^{-1} x = \frac{\pi}{2} - u$.The expression becomes $f(u) = u^3 + (\frac{\pi}{2} - u)^3$ for $u \in [0, \frac{\pi}{2}]$.$f(u) = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$.To find the range, find the critical points: $f'(u) = 3\pi u - \frac{3\pi^2}{4} = 0 \implies u = \frac{\pi}{4}$.At $u = 0$, $f(0) = \frac{\pi^3}{8}$.At $u = \frac{\pi}{4}$, $f(\frac{\pi}{4}) = (\frac{\pi}{4})^3 + (\frac{\pi}{4})^3 = 2(\frac{\pi^3}{64}) = \frac{\pi^3}{32}$.At $u = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = (\frac{\pi}{2})^3 + 0 = \frac{\pi^3}{8}$.Thus, the minimum value is $\frac{\pi^3}{32}$ and the maximum value is $\frac{\pi^3}{8}$.Given $a\pi^3 = f(u)$, we have $\frac{\pi^3}{32} \le a\pi^3 \le \frac{\pi^3}{8}$, so $\frac{1}{32} \le a \le \frac{1}{8}$.This implies $a \ge \frac{1}{32}$.

If $0 \le x \le 1$ and $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a\pi^3$, then find the range of $a$.

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