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(A) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$ where $x \in [-1, 1]$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Let $u = \sin^{-1}

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(A) Let $f(x) = (\sin^{-1} x)^3 + (\cos^{-1} x)^3$ where $x \in [-1, 1]$.We know that $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Let $u = \sin^{-1} x$,where $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Then $f(u) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$.This is a quadratic in $u$. The vertex is at $u = -\frac{b}{2a} = \frac{3\pi^2/4}{3\pi} = \frac{\pi}{4}$.Since $u \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $u = \frac{\pi}{4}$:$f(\frac{\pi}{4}) = \frac{3\pi}{2}(\frac{\pi^2}{16}) - \frac{3\pi^2}{4}(\frac{\pi}{4}) + \frac{\pi^3}{8} = \frac{3\pi^3}{32} - \frac{3\pi^3}{16} + \frac{\pi^3}{8} = \frac{3\pi^3 - 6\pi^3 + 4\pi^3}{32} = \frac{\pi^3}{32}$.The maximum value occurs at the boundary $u = -\frac{\pi}{2}$:$f(-\frac{\pi}{2}) = (-\frac{\pi}{2})^3 + (\pi)^3 = -\frac{\pi^3}{8} + \pi^3 = \frac{7\pi^3}{8}$.Thus,the range of $f(x)$ is $[\frac{\pi^3}{32}, \frac{7\pi^3}{8}]$.Given $a Since the minimum value of the expression is $\frac{\pi^3}{32}$,there are no values of $x$ such that $f(x) = a\pi^3$ when $a Therefore,the number of solutions is $0$.

If $a < \frac{1}{32},$ then the number of solutions of $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 = a\pi^3$ is

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