The complete solution set of the inequality ( | vedclass

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(A) Let $y = \sec^{-1}x$. The inequality becomes $(y - 4)(y - 1)(y - 2) \ge 0$.Since the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,th

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$[\sec 2, \sec 1]$

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(A) Let $y = \sec^{-1}x$. The inequality becomes $(y - 4)(y - 1)(y - 2) \ge 0$.Since the domain of $\sec^{-1}x$ is $(-\infty, -1] \cup [1, \infty)$,the range is $[0, \pi/2) \cup (\pi/2, \pi]$.However,note that $y = \sec^{-1}x$ implies $y \in [0, \pi]$ and $y 
eq \pi/2$.Since $y \le \pi \approx 3.14$,the term $(y - 4)$ is always negative because $y Dividing the inequality by $(y - 4)$,the sign of the inequality reverses: $(y - 1)(y - 2) \le 0$.This holds for $y \in [1, 2]$.Since $y = \sec^{-1}x$,we have $1 \le \sec^{-1}x \le 2$.Taking the secant of all parts,we get $\sec 1 \le x \le \sec 2$.Thus,the solution set is $[\sec 1, \sec 2]$.

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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