Question

Domain of $ \sin^{-1}(2x^2 - 3) $ is:

• A. $(-1, 0) \cup (1, \sqrt{2})$
• B. $(-\sqrt{2}, -1) \cup (0, 1)$
• C. $[-\sqrt{2}, -1] \cup [1, \sqrt{2}]$
• D. $(-\sqrt{2}, -1) \cup (1, \sqrt{2})$

Hint:

To determine the domain of inverse trigonometric functions, ensure that the input expression lies within the domain of the original function. For $ \sin^{-1}(y) $, the valid input range is $-1 \leq y \leq 1$.

Answer 1

The Correct Option is C

The domain of the expression $ \sin^{-1}(2x^2 - 3) $ is determined by the requirement that the argument of the inverse sine function must be within the interval $ [-1, 1] $. Therefore, we establish the inequality:\[-1 \leq 2x^2 - 3 \leq 1\]Adding 3 to all parts of the inequality yields:\[2 \leq 2x^2 \leq 4\]Dividing by 2, we get:\[1 \leq x^2 \leq 2\]Taking the square root of all parts and considering both positive and negative values of $x$, we obtain:\[1 \leq |x| \leq \sqrt{2}\]This inequality implies that $x$ must be in the intervals $ [-\sqrt{2}, -1] $ or $ [1, \sqrt{2}] $. The domain of the function is thus the union of these two intervals:\[[-\sqrt{2}, -1] \cup [1, \sqrt{2}]\]