If x ≤ -1, what is the set of all values that sec^-1 x takes?

Question 59

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If $x \le -1$ , what is the set of all values that $\sec^{-1} x$ takes?

\left(\frac{\pi}{2}, \pi\right]

Correct answerYour answer

\left[-\frac{\pi}{2}, 0\right)

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\left[\frac{\pi}{2}, \pi\right)

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\left[0, \frac{\pi}{2}\right)

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\sec^{-1} x = \cos^{-1} \frac{1}{x}$ , and for $x \le -1$ , $\frac{1}{x}$ lies in $[-1, 0)$ .
$\cos^{-1}$ of a number in $[-1, 0)$ lies in $\left(\frac{\pi}{2}, \pi\right]$ ; $\sec^{-1}(-1) = \pi$ is included, $\frac{\pi}{2}$ is not.

Worth noting

$\left[-\frac{\pi}{2}, 0\right)$ gives $\sec^{-1}$ a negative angle; its branch is $[0, \pi]$ , like that of $\cos^{-1}$ .

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