If f(x) = 2 + |sin^{-1} x| and A = {x in R : | vedclass

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(B) For $f^{-1}(x)$ to exist,the function $f(x)$ must be a bijection (both one-to-one and onto). Given $f(x) = 2 + |\sin^{-1} x|$. The domain of $\sin

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$[-1, 1]$

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(B) For $f^{-1}(x)$ to exist,the function $f(x)$ must be a bijection (both one-to-one and onto). Given $f(x) = 2 + |\sin^{-1} x|$. The domain of $\sin^{-1} x$ is $[-1, 1]$. For $x \in [-1, 0]$,$f(x) = 2 - \sin^{-1} x$. This is a strictly decreasing function. For $x \in [0, 1]$,$f(x) = 2 + \sin^{-1} x$. This is a strictly increasing function. Since the function decreases on $[-1, 0]$ and increases on $[0, 1]$,it is not one-to-one on the interval $[-1, 1]$. Specifically,$f(x_1) = f(x_2)$ for some $x_1 \in [-1, 0)$ and $x_2 \in (0, 1]$. Therefore,$f(x)$ is not invertible on its entire domain $[-1, 1]$. However,if the question implies the set $A$ where the inverse exists locally or is defined,typically such functions are not invertible. Given the options provided,there is no valid interval where the function is globally invertible. If we assume the question implies the domain where the function is defined,it is $[-1, 1]$. But strictly speaking,$f^{-1}(x)$ does not exist for the whole domain.

If $f(x) = 2 + |\sin^{-1} x|$ and $A = \{x \in R : f^{-1}(x) \text{ exists} \}$,then $A = $

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