Question

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 \(x\in\mathbb{R}\), \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) and \(0\le\left(\sin^{-1}x-\frac{\pi}{4}\right)^2\le\frac{9\pi^2}{16}\).

• A. Both statements I and II are true.
• B. Both statements I and II are false.
• C. Statement I is true and statement II is false.
• D. Statement I is false and statement II is true.

Hint:

Put \(t=\sin^{-1}x\in[-\pi/2,\pi/2]\), so \(\cos^{-1}x=\pi/2-t\), and find the range of \(t^3+(\pi/2-t)^3\); it is bounded above by \(7\pi^3/8\).

Answer 1

The Correct Option is D

Let \(t=\sin^{-1}x\), so \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\) and \(\cos^{-1}x=\frac{\pi}{2}-t\).
\(t^3+(\frac{\pi}{2}-t)^3=(\frac{\pi}{2})^3-3t(\frac{\pi}{2}-t)\cdot\frac{\pi}{2}\).
Now \(t(\frac{\pi}{2}-t)=\frac{\pi^2}{16}-(t-\frac{\pi}{4})^2\).
So the expression \(=\frac{\pi^3}{8}-\frac{3\pi}{2}\left[\frac{\pi^2}{16}-(t-\frac{\pi}{4})^2\right]=\frac{\pi^3}{32}+\frac{3\pi}{2}(t-\frac{\pi}{4})^2\).
For \(t\in[-\frac{\pi}{2},\frac{\pi}{2}]\): \(0\le(t-\frac{\pi}{4})^2\le(\frac{3\pi}{4})^2=\frac{9\pi^2}{16}\) (attained at \(t=-\frac{\pi}{2}\)).
Hence \(\frac{\pi^3}{32}\le a\pi^3\le\frac{\pi^3}{32}+\frac{3\pi}{2}\cdot\frac{9\pi^2}{16}=\frac{7\pi^3}{8}\), i.e. \(\frac{1}{32}\le a\le\frac{7}{8}\).
Statement I claims a solution for ALL \(a\ge\frac{1}{32}\); for \(a>\frac{7}{8}\) there is none, so Statement I is false.
Statement II: \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) is the standard identity, and the bound \(0\le(\sin^{-1}x-\frac{\pi}{4})^2\le\frac{9\pi^2}{16}\) was shown above, so Statement II is true.
Answer: (4) Statement I is false and statement II is true.