Match List-I with List-II
| List-I | List-II | ||
| (A) | Domain of \( \sin^{-1}(2x-3) \) | (I) | [-1, 0] |
| (B) | Domain of \( \sin^{-1}(1+2x) \) | (II) | [-1, 1] |
| (C) | Domain of \( \cos^{-1}(2x-1) \) | (III) | [1, 2] |
| (D) | Domain of \( \sin^{-1}(x^{2}) \) | (IV) | [0, 1] |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
\( \sin^{-1}t \) and \( \cos^{-1}t \) are defined only for \( -1\le t\le 1 \), so each argument must lie between −1 and 1.
(A) Domain of \( \sin^{-1}(2x-3) \)
This matches (III) [1, 2].
\( \displaystyle -1\le 2x-3\le 1 \)
\( \displaystyle 2\le 2x\le 4 \)
\( \displaystyle 1\le x\le 2 \)
(B) Domain of \( \sin^{-1}(1+2x) \)
This matches (I) [-1, 0].
\( \displaystyle -1\le 1+2x\le 1 \)
\( \displaystyle -2\le 2x\le 0 \)
\( \displaystyle -1\le x\le 0 \)
(C) Domain of \( \cos^{-1}(2x-1) \)
This matches (IV) [0, 1].
\( \displaystyle -1\le 2x-1\le 1 \)
\( \displaystyle 0\le 2x\le 2 \)
\( \displaystyle 0\le x\le 1 \)
(D) Domain of \( \sin^{-1}(x^{2}) \)
This matches (II) [-1, 1].
\( \displaystyle 0\le x^2\le 1 \)
\( \displaystyle -1\le x\le 1 \)
Here negative x is allowed as well, since \( x^2 \) is never negative.
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).