Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function (on its domain) | Range | ||
| (A) | \(2\sin^{-1} x\) | (I) | \(\left(-\displaystyle\frac{\pi}{2}, \displaystyle\frac{\pi}{2}\right)\) |
| (B) | \(\cos^{-1} x - \displaystyle\frac{\pi}{2}\) | (II) | \([-\pi, \pi]\) |
| (C) | \(2\tan^{-1} x\) | (III) | \(\left[-\displaystyle\frac{\pi}{2}, \displaystyle\frac{\pi}{2}\right]\) |
| (D) | \(\cot^{-1} x - \displaystyle\frac{\pi}{2}\) | (IV) | \((-\pi, \pi)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Each range comes from the range of the principal value branch. Multiplying by 2 doubles both ends, and subtracting \(\frac{\pi}{2}\) moves both ends down by \(\frac{\pi}{2}\). A closed end stays closed and an open end stays open.
(A) \(2\sin^{-1} x\)
This matches (II) \([-\pi, \pi]\): \(\sin^{-1} x\) takes all values in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), and doubling gives \([-\pi, \pi]\), with the ends taken at \(x = \mp 1\).
(B) \(\cos^{-1} x - \frac{\pi}{2}\)
This matches (III) \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\): \(\cos^{-1} x\) takes all values in \([0, \pi]\), and \(0 - \frac{\pi}{2} = -\frac{\pi}{2}\), \(\pi - \frac{\pi}{2} = \frac{\pi}{2}\).
(C) \(2\tan^{-1} x\)
This matches (IV) \((-\pi, \pi)\): \(\tan^{-1} x\) takes all values in the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), and doubling gives \((-\pi, \pi)\).
(D) \(\cot^{-1} x - \frac{\pi}{2}\)
This matches (I) \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\): \(\cot^{-1} x\) takes all values in the open interval \((0, \pi)\), and moving down by \(\frac{\pi}{2}\) gives \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).