Match the LIST-I with LIST-II
| LIST-I Functions | LIST-II Principal value | ||
| A. | \( \sin^{-1}\left(\displaystyle\frac{1}{2}\right) \) | I. | \( \displaystyle\frac{\pi}{4} \) |
| B. | \( \sin^{-1}\left(-\displaystyle\frac{1}{\sqrt{2}}\right) \) | II. | \( \displaystyle\frac{\pi}{6} \) |
| C. | \( \tan^{-1}(1) \) | III. | \( -\displaystyle\frac{\pi}{4} \) |
| D. | \( \cos^{-1}\left(-\displaystyle\frac{1}{2}\right) \) | IV. | \( \displaystyle\frac{2\pi}{3} \) |
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-I, D-IV
The principal value branches are
\( \displaystyle \sin^{-1}x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] \),
\( \displaystyle \tan^{-1}x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right) \),
\( \displaystyle \cos^{-1}x\in[0,\pi] \).
(A) \( \sin^{-1}\left(\frac{1}{2}\right) \)
Since \( \sin\frac{\pi}{6}=\frac{1}{2} \), this is \( \frac{\pi}{6} \), which matches (II).
(B) \( \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
Using \( \sin^{-1}(-x)=-\sin^{-1}x \), this is \( -\frac{\pi}{4} \), which matches (III).
(C) \( \tan^{-1}(1) \)
Since \( \tan\frac{\pi}{4}=1 \), this is \( \frac{\pi}{4} \), which matches (I).
(D) \( \cos^{-1}\left(-\frac{1}{2}\right) \)
Using \( \cos^{-1}(-x)=\pi-\cos^{-1}x \), this is
\( \displaystyle \pi-\frac{\pi}{3}=\frac{2\pi}{3} \)
which matches (IV).
So the matching is (A)-(II), (B)-(III), (C)-(I), (D)-(IV).