Q2: Domain of $\sec^{-1}(2x)$

$\sec\theta = \frac{1}{\cos\theta}$ and $|\cos\theta| \le 1$⁠, so $|\sec\theta| \ge 1$⁠. A secant never lies strictly between $-1$ and $1$⁠, so $\sec^{-1}t$ is defined only for

that is, for $|t| \ge 1$⁠.

$|t| \ge 1$ means $t$ is at least 1 away from 0, on either side: $t \le -1$ or $t \ge 1$⁠.

Dividing by 2, a positive number, keeps both inequality signs the same way.

The end points are included: at $x = \pm\frac{1}{2}$⁠, $2x = \pm 1$⁠, and $\sec^{-1}1 = 0$⁠, $\sec^{-1}(-1) = \pi$ both exist. So the brackets at $\pm\frac{1}{2}$ are square.

Watch out: (C) leaves out the end points, and (B) is exactly the gap where $\sec^{-1}(2x)$ is not defined.