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(A) Let $y = \sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,so $y \in [0, \pi/2) \cup (\pi/2, \pi]$.Given inequality: $(y

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(A) Let $y = \sec^{-1}x$. The domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,so $y \in [0, \pi/2) \cup (\pi/2, \pi]$.Given inequality: $(y - 4)(y - 1)(y - 2) \ge 0$.Since $y \in [0, \pi] \approx [0, 3.14]$,the term $(y - 4)$ is always negative.Dividing by $(y - 4)$,the inequality reverses: $(y - 1)(y - 2) \le 0$.This holds for $y \in [1, 2]$.Since $y = \sec^{-1}x$,we have $1 \le \sec^{-1}x \le 2$.Applying the secant function (which is increasing on $[0, \pi/2)$ and $[1, \pi]$),we get $\sec 1 \le x \le \sec 2$.However,we must respect the domain of $\sec^{-1}x$. Since $1 \le y \le 2$,$x$ must be in the range of $\sec y$. For $y \in [1, 2]$,$x \in [\sec 1, \sec 2]$ is valid.

The complete solution set of the inequality $(\sec^{-1}x - 4)(\sec^{-1}x - 1)(\sec^{-1}x - 2) \ge 0$ is

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