Statement I: The equation (sin^{-1} x)^3 + (c | vedclass

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(D) For Statement $II$: We know that $\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Subtracting $\frac{\pi}{4}$ from all sides,we get $-\frac{3\pi}

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Statement $I$ is false and statement $II$ is true.

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(D) For Statement $II$: We know that $\sin^{-1} x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.Subtracting $\frac{\pi}{4}$ from all sides,we get $-\frac{3\pi}{4} \le \sin^{-1} x - \frac{\pi}{4} \le \frac{\pi}{4}$.Squaring the inequality,we get $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}$. Thus,Statement $II$ is true.For Statement $I$: Let $u = \sin^{-1} x$. Then $\cos^{-1} x = \frac{\pi}{2} - u$. The equation becomes $u^3 + (\frac{\pi}{2} - u)^3 = a\pi^3$.Expanding this,we get $u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = a\pi^3$.$\frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8} - a\pi^3 = 0$.Dividing by $\frac{3\pi}{2}$,we get $u^2 - \frac{\pi}{2}u + \frac{\pi^2}{12} - \frac{2a\pi^2}{3} = 0$.Completing the square: $(u - \frac{\pi}{4})^2 - \frac{\pi^2}{16} + \frac{\pi^2}{12} - \frac{2a\pi^2}{3} = 0$.$(u - \frac{\pi}{4})^2 = \frac{2a\pi^2}{3} - \frac{\pi^2}{48} = \frac{\pi^2}{48}(32a - 1)$.Since $0 \le (u - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}$,we have $0 \le \frac{\pi^2}{48}(32a - 1) \le \frac{9\pi^2}{16}$.$0 \le 32a - 1 \le 27$,which implies $\frac{1}{32} \le a \le \frac{7}{8}$.Since the equation has a solution only for $a \in [\frac{1}{32}, \frac{7}{8}]$,the statement that it has a solution for all $a \ge \frac{1}{32}$ is false.

Statement $I:$ The equation $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a\pi^3 = 0$ has a solution for all $a \ge \frac{1}{32}.$
Statement $II:$ For any $x \in [-1, 1],$ $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}.$

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Statement $I:$ The equation $(\sin^{-1} x)^3 + (\cos^{-1} x)^3 - a\pi^3 = 0$ has a solution for all $a \ge \frac{1}{32}.$Statement $II:$ For any $x \in [-1, 1],$ $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ and $0 \le (\sin^{-1} x - \frac{\pi}{4})^2 \le \frac{9\pi^2}{16}.$