Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Function (on its domain) | Range | ||
| (A) | \(\tan^{-1}\left(x^{2}\right)\) | (I) | \(\left[-\displaystyle\frac{\pi}{2}, 0\right]\) |
| (B) | \(\cos^{-1}\left(|x|\right)\) | (II) | \(\left(0, \displaystyle\frac{\pi}{2}\right]\) |
| (C) | \(\sin^{-1}\left(-|x|\right)\) | (III) | \(\left[0, \displaystyle\frac{\pi}{2}\right)\) |
| (D) | \(\cot^{-1}\left(x^{2}\right)\) | (IV) | \(\left[0, \displaystyle\frac{\pi}{2}\right]\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
First find the set of values of the inside expression, then read the values of the inverse function from its graph. An end is included only if the inside expression actually reaches the matching value.
(A) \(\tan^{-1}\left(x^{2}\right)\)
This matches (III) \(\left[0, \frac{\pi}{2}\right)\): \(x^{2}\) takes all values in \([0, \infty)\), and \(\tan^{-1}\) is increasing, starting at \(\tan^{-1} 0 = 0\) and only approaching \(\frac{\pi}{2}\).
(B) \(\cos^{-1}\left(|x|\right)\)
This matches (IV) \(\left[0, \frac{\pi}{2}\right]\): on the domain \([-1, 1]\), \(|x|\) takes all values in \([0, 1]\), and \(\cos^{-1}\) falls from \(\cos^{-1} 0 = \frac{\pi}{2}\) to \(\cos^{-1} 1 = 0\), both reached.
(C) \(\sin^{-1}\left(-|x|\right)\)
This matches (I) \(\left[-\frac{\pi}{2}, 0\right]\): \(-|x|\) takes all values in \([-1, 0]\), and \(\sin^{-1}\) rises from \(\sin^{-1}(-1) = -\frac{\pi}{2}\) to \(\sin^{-1} 0 = 0\).
(D) \(\cot^{-1}\left(x^{2}\right)\)
This matches (II) \(\left(0, \frac{\pi}{2}\right]\): \(x^{2}\) takes all values in \([0, \infty)\), and \(\cot^{-1}\) is decreasing, starting at \(\cot^{-1} 0 = \frac{\pi}{2}\) and only approaching 0.