Inverse Trigonometric Functions

$A = \{ x \in \mathbb{R} \mid \sin^{-1}(\sqrt{x^2+x+1}) \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \}$ and $B = \{ y \in \mathbb{R} \mid y = \sin^{-1}(\sqrt{x^2+x+1}), x \in A \}$ Since, domain of $\sin^{-1}(x)$ is $[-1,1]$ So, domain of $\sin^{-1}\sqrt{x^{2}+x+1}$ is real, so so $\sqrt{x^2+x+1}\;\in[-1,1]$ $\Rightarrow \sqrt{x^2+x+1}\;\in[0,1]$ ( ∵ square root is always non-negative) $\Rightarrow x^2+x+1 \le 1$ $\Rightarrow x^2+x \le 0$ $\Rightarrow x(x+1) \le 0$ So, $x \in[-1,0]$ for set $A$ Thus, $A=[-1,0]$ Now, for set $B$, since $x \in A=[-1,0]$, the range values for $\sqrt{x^2+x+1}$ for $x \in[-1,0]$ we will find. $\text{At}\; x=-1, \sqrt{(-1)^2-1+1}=\sqrt{1-1+1}=1$ At $x=0, \sqrt{0^2+0+1}=\sqrt{0+1}=1$ At $x=-\frac{1}{2}, \sqrt{\left(\frac{-1}{2}\right)^2-\frac{1}{2}+1}$ $= \sqrt{\frac{1}{4}-\frac{1}{2}+1} = \sqrt{\frac{3}{4}}$So, $\sqrt{x^2+x+1} \in \left[\sqrt{\frac{3}{4}}, 1\right] = \left[\frac{\sqrt{3}}{2}, 1\right]$Thus, $y = \sin^{-1}(\sqrt{x^2 + x + 1})$$\in \left[\sin^{-1}\left(\frac{\sqrt{3}}{2}\right), \sin^{-1}(1)\right]$$= \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$Hence, $B = \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$$\Rightarrow \text{Now,}\; A \cap B = [-1,0] \cap \left[\frac{\pi}{3}, \frac{\pi}{2}\right] = \emptyset$So, $A \cap B \neq \emptyset$ is false.$\Rightarrow A \cap B^C = [-1,0] \cap \left[\frac{\pi}{3}, \frac{\pi}{2}\right]^C$ and the intersection with any set outside this cannot be $[0,1]$,So, $A \cap B^C = [0,1]$ is false.$\Rightarrow A^C$ is everything except $[-1,0]$.Since $B = \left(\frac{\pi}{3}, \frac{\pi}{2}\right) \subset \mathbb{R} \setminus (-1,0)$$\Rightarrow A^C \cap B$ is true.$\Rightarrow A \cup B = \mathbb{R} \setminus \left\{[-1,0] \cup \left[\frac{\pi}{3}, \frac{\pi}{2}\right]\right\}$So, union of $A$ and $B$ is the complement of their own union.That's impossible.Thus, $A^C \cap B = \left[\frac{\pi}{3}, \frac{\pi}{2}\right]$, option (c) is correct.